**Perturbed Differential Equations with Singular Points**

Keldibay Alymkulov and Dilmurat Adbillajanovich Tursunov

Additional information is available at the end of the chapter

http://dx.doi.org/10.5772/67856

Dedicated to academician of National Academy Sciences Kyrgyz Republic and Corresponding member of RAS Imanaliev Murzabek

#### Abstract

tion methods are widely used on the basis of the so-called Green's function approach. In modern theory of strongly interacting systems, the perturbation expansions are combined with ideas of scaling and renormalization, and thus these expansions are in the basis of the so-called renormalization group. The latter is a powerful tool of investigation of the effect of

Once introduced and highly developed in physics, perturbation methods of study are also spread in chemistry—mainly in quantum chemistry, in physical chemistry, in chemical physics, and in biophysics. In the last three–four decades, new interdisciplinary research fields appeared, for example, sociophysics and econophysics, where perturbation theories together with numerical analysis and computer simulations will undoubtedly be very im‐

The book contains seven chapters, written by noted experts and young researchers who present their recent studies of both pure mathematical problems of perturbation theories and application of perturbation methods to the study of important topics in physics, for ex‐ ample, renormalization group theory and applications to basic models in theoretical physics (Y. Takashi), the quantum gravity and its detection and measurement (F. Bulnes), atom-pho‐ ton interactions (E. G. Thrapsaniotis), treatment of spectra and radiation characteristics by relativistic perturbation theory (A. V. Glushkov et al.), and Green's function approach and some applications (Jing Huang). The pure mathematical issues are related to the problem of generalization of the boundary layer function method for bisingularly perturbed differential equations (K. Alymkulov and D. A. Torsunov) and to the development of new homotopy

> **Dimo I. Uzunov** Professor of Physics

> > Sofia, Bulgaria

Bulgarian Academy of Sciences

strong interactions in field theories.

asymptotic methods and their applications (Baojian Hong).

portant.

VIII Preface

Here, we generalize the boundary layer functions method (or composite asymptotic expansion) for bisingular perturbed differential equations (BPDE that is perturbed differential equations with singular point). We will construct a uniform valid asymptotic solution of the singularly perturbed first-order equation with a turning point, for BPDE of the Airy type and for BPDE of the second-order with a regularly singular point, and for the boundary value problem of Cole equation with a weak singularity.A uniform valid expansion of solution of Lighthill model equation by the method of uniformization and the explicit solution—this one by the generalization method of the boundary layer function—is constructed. Furthermore, we construct a uniformly convergent solution of the Lagerstrom model equation by the method of fictitious parameter.

Keywords: turning point, singularly perturbed, bisingularly perturbed, Cauchy problem, Dirichlet problem, Lagerstrom model equation, Lighthill model equation, Cole equation, generalization boundary layer functions

## 1. Preliminary

## 1.1. Symbols O, o, ~. Asymptotic expansions of functions

Let a function fðxÞ and ϕðxÞ be defined in a neighborhood of x ¼ 0. Definition 1. If lim<sup>x</sup>!<sup>0</sup> fðxÞ <sup>ϕ</sup>ðx<sup>Þ</sup> <sup>¼</sup> <sup>M</sup>, then write <sup>f</sup>ðxÞ ¼ <sup>O</sup>ðϕðxÞÞ, x ! 0, and <sup>M</sup> is constant. If lim<sup>x</sup>!<sup>0</sup> fðxÞ <sup>ϕ</sup>ðx<sup>Þ</sup> <sup>¼</sup> 0, then write <sup>f</sup>ðxÞ ¼ <sup>o</sup>ðϕðxÞÞ, x ! 0. If lim<sup>x</sup>!<sup>0</sup> fðxÞ <sup>ϕ</sup>ðx<sup>Þ</sup> <sup>¼</sup> 1, then write <sup>f</sup>ðx<sup>Þ</sup> <sup>e</sup>ϕðxÞ, x ! 0.

Definition 2. The sequence {δnðεÞ}, where δnðεÞ defined in some neighborhood of zero, is called the asymptotic sequence in ε ! 0, if

> © 2017 The Author(s). Licensee InTech. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

$$\lim\_{\varepsilon \to 0} \frac{\delta\_{n+1}(\varepsilon)}{\delta\_n(\varepsilon)} = 0, \quad \forall n = 1, 2, \dots$$

For example.

$$\{\varepsilon^n\}\_{\prime} \quad \left\{ \left(1/\ln(1/\varepsilon)\right)^n \right\}\_{\prime} \quad \left\{ \left(\varepsilon\ln(1/\varepsilon)\right)^n \right\}.$$

Note 1. Everywhere below ε denotes a small parameter.

Definition 3. We say that fðxÞ function can be expanded in an asymptotic series by the asymptotic sequence {ϕnðxÞ}, x ! 0, if there exists a sequence of numbers {f <sup>n</sup>} and has the relation

$$f(\mathbf{x}) = \sum\_{k=0}^{n} f\_k \varphi\_k(\mathbf{x}) + \mathcal{O}(\varphi\_{n+1}(\mathbf{x})), \quad \mathbf{x} \to \mathbf{0},$$

and write

$$f(\mathbf{x}) \stackrel{\sim}{\sum\_{k=0}^{\infty}} f\_k \varphi\_k(\mathbf{x}), \quad \mathbf{x} \to \mathbf{0}.$$

#### 1.2. The asymptotic expansion of infinitely differentiable functions

Theorem (Taylor (1715) and Maclaurin (1742)). If the function <sup>f</sup>ðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup> in some neighborhood of <sup>x</sup> <sup>¼</sup> 0, then it can be expanded in an asymptotic series for the asymptotic sequence {xn}, i.e.,

$$f(\mathbf{x}) - \sum\_{n=1}^{\infty} f\_n \mathbf{x}^n \text{, where } f\_n = f^{(n)}(0) / n! \dots$$

Thus, the concept of an asymptotic expansion was given for the first time by Taylor and Maclaurin,although an explicit definition was given by Poincaré in 1886.

#### 1.3. The asymptotic expansion of the solution of the ordinary differential equation

Consider the Cauchy problem for a normal ordinary differential equation

$$y'(\mathbf{x}) = f(\mathbf{x}, y, \varepsilon), \quad y(0) = 0. \tag{1}$$

The function fðx, y, εÞ is infinitely differentiable on the variables x, y, ε in some neighborhood Oð0, 0, 0Þ. It is correct next.

Theorem 1. The solution y ¼ yðx, εÞ of problem (1) exists and unique in some neighborhood point <sup>O</sup>ð0, <sup>0</sup>, <sup>0</sup><sup>Þ</sup> and <sup>y</sup>ðx, <sup>ε</sup>Þ∈C<sup>∞</sup>, for small x, <sup>ε</sup>.

Corollary. The solution of problem (1) can be expanded in an asymptotic series by the small parameter ε, i.e.,

$$y(\mathbf{x}, \varepsilon) = \sum\_{k=1}^{\infty} \varepsilon^k y\_k(\mathbf{x}). \tag{2}$$

Here and below, the equality is understood in an asymptotic sense.

Note 2. Theorem 1 for the case when fðx, y, εÞ is analytical was given in [1] by Duboshin.

Note 3. This theorem 1 is not true if fðx, y, εÞ is not smooth at ε. For example, the solution of a singularly perturbed equation

$$
\varepsilon y'(\mathbf{x}) = -y(\mathbf{x}), \quad y(0) = a
$$

function <sup>y</sup>ðxÞ ¼ ae�x=<sup>ε</sup> and is not expanded in an asymptotic series in powers of <sup>ε</sup>, because here fðx, y, ε޼�yðxÞ=ε and f have a pole of the first order with respect to ε.

Note 4. The series 2 is a uniform asymptotic expansion of the function yðxÞ in a neighborhood of x ¼ 0.

For example. Series

lim ε!0

�

<sup>ε</sup><sup>n</sup> f g,

<sup>f</sup>ðxÞ ¼ <sup>X</sup><sup>n</sup>

k¼0

<sup>f</sup>ðxÞ<sup>e</sup> X∞ k¼0

1.2. The asymptotic expansion of infinitely differentiable functions

Maclaurin,although an explicit definition was given by Poincaré in 1886.

Consider the Cauchy problem for a normal ordinary differential equation

y0

Here and below, the equality is understood in an asymptotic sense.

ðnÞ ð0Þ=n!.

Note 1. Everywhere below ε denotes a small parameter.

For example.

2 Recent Studies in Perturbation Theory

and write

<sup>f</sup>ðx<sup>Þ</sup> <sup>e</sup>

X∞ n¼1

<sup>f</sup> <sup>n</sup>xn, where <sup>f</sup> <sup>n</sup> <sup>¼</sup> <sup>f</sup>

Oð0, 0, 0Þ. It is correct next.

parameter ε, i.e.,

point <sup>O</sup>ð0, <sup>0</sup>, <sup>0</sup><sup>Þ</sup> and <sup>y</sup>ðx, <sup>ε</sup>Þ∈C<sup>∞</sup>, for small x, <sup>ε</sup>.

δ<sup>n</sup>þ<sup>1</sup>ðεÞ

1=lnð1=εÞ n o �<sup>n</sup>

<sup>δ</sup>nðε<sup>Þ</sup> <sup>¼</sup> <sup>0</sup>, <sup>∀</sup><sup>n</sup> <sup>¼</sup> <sup>1</sup>, <sup>2</sup>, …

, �

<sup>f</sup> <sup>k</sup>ϕkðxÞ þ <sup>O</sup>ðϕ<sup>n</sup>þ<sup>1</sup>ðxÞÞ, x ! <sup>0</sup>,

f <sup>k</sup>ϕkðxÞ, x ! 0:

Definition 3. We say that fðxÞ function can be expanded in an asymptotic series by the asymptotic sequence {ϕnðxÞ}, x ! 0, if there exists a sequence of numbers {f <sup>n</sup>} and has the relation

Theorem (Taylor (1715) and Maclaurin (1742)). If the function <sup>f</sup>ðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup> in some neighborhood of <sup>x</sup> <sup>¼</sup> 0, then it can be expanded in an asymptotic series for the asymptotic sequence {xn}, i.e.,

Thus, the concept of an asymptotic expansion was given for the first time by Taylor and

The function fðx, y, εÞ is infinitely differentiable on the variables x, y, ε in some neighborhood

Theorem 1. The solution y ¼ yðx, εÞ of problem (1) exists and unique in some neighborhood

Corollary. The solution of problem (1) can be expanded in an asymptotic series by the small

k¼1 εk

<sup>y</sup>ðx, <sup>ε</sup>Þ ¼ <sup>X</sup><sup>∞</sup>

1.3. The asymptotic expansion of the solution of the ordinary differential equation

εlnð1=εÞ n o �<sup>n</sup>

:

ðxÞ ¼ fðx, y, εÞ, yð0Þ ¼ 0: ð1Þ

ykðxÞ: ð2Þ

$$y(\mathbf{x}, \varepsilon) = \mathbf{1} + \varepsilon \mathbf{x}^{-1} \ + (\varepsilon \mathbf{x}^{-1})^2 + \dots + (\varepsilon \mathbf{x}^{-1})^n + \dots$$

It is not uniform valid asymptotic series on the interval [0, 1], but it is a uniform valid asymptotic expansion of the segment <sup>½</sup>εα, <sup>1</sup>�, where 0 <sup>&</sup>lt; <sup>α</sup> <sup>&</sup>lt; 1.

#### 1.4. Singularly perturbed ordinary differential equations

We divide such equations into three types:

(I) Singular perturbations of ordinary differential equations such as the Prandtl-Tikhonov [2–56], i.e., perturbed equations that contain a small parameter at the highest derivative, i.e., equations of the form

$$y'(\mathbf{x}) = f(\mathbf{x}, y, \varepsilon), \quad y(0) = 0, \qquad \varepsilon z'(\mathbf{x}) = \mathbf{g}(\mathbf{x}, y, \varepsilon), \quad z(0) = 0,$$

where f, g are infinitely differentiable in the variables x, y, ε in the neighborhood of Oð0, 0, 0Þ. It is obvious that unperturbed equation (ε ¼ 0)

$$y\_0'(\mathbf{x}) = f(\mathbf{x}, y\_\prime \, 0), \quad \mathbf{0} = \mathbf{g}(\mathbf{x}, y\_\prime \, 0)$$

is a first order.

Definition 4. Singularly perturbed equation will be called bisingulary perturbed if the corresponding unperturbed differential equation has a singular point, or this one is an unbounded solution in the considering domain.

For example


$$
\varepsilon y''(\mathfrak{x}) + (1 - y^2(\mathfrak{x}))y'(\mathfrak{x}) + y(\mathfrak{x}) = 0.
$$

It is a bisingularly perturbed ordinary differential equation with singular points, if yðx޼�1.


For example, a Lighthill model equation

$$p(\mathbf{x} + \varepsilon y(\mathbf{x}))y'(\mathbf{x}) + p(\mathbf{x})y(\mathbf{x}) = r(\mathbf{x}), \quad y(1) = a$$

where <sup>x</sup> <sup>∈</sup>½0, <sup>1</sup>�, <sup>p</sup>ðxÞ, rðxÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�. For unperturbed equation

$$x y\_0'(\mathbf{x}) + p(\mathbf{x}) y\_0(\mathbf{x}) = r(\mathbf{x}),$$

point x ¼ 0 is a regular singular point.

(III) A singularly perturbed equation with a small parameter is considered on an infinite interval. For example, the Lagerstrom equation [70–81]

$$\begin{cases} y''(\mathbf{x}) + n\mathbf{x}^{-1}y'(\mathbf{x}) + y(\mathbf{x})y'(\mathbf{x}) = \beta(y'(\mathbf{x}))^2, \\ y(\boldsymbol{\varepsilon}) = 0, y(\boldsymbol{\varepsilon}) = 1. \end{cases}$$

where 0 < β is a given number and n is the dimension space.

Remark. The division into such classes is conditional, because singularly perturbed equation of Van der Pol in the neighborhood of points y ¼ �1 leads to an equation of Lighthill type [2, 3].

## 1.5. Methods of construction of asymptotic expansions of solutions of singularly perturbed differential equations


It should be noted that, for the first time, the uniform valid asymptotic expansion of the solution of Eq. (5) is constructed by Vasil'eva (1960) [50] after Wasow [69] and Sibuya in 1963 [68] by the method of matching.

This method is constructive and understandable for the applied scientists.


3. εy<sup>0</sup>

4 Recent Studies in Perturbation Theory

ðxÞ � xyðxÞ ¼ 1, x ∈½0, 1� is a bisingularly perturbed equation, because the

.

unperturbed equation has an unbounded solution <sup>y</sup>0ðx޼�x�<sup>1</sup>

a singular point in the considering domain.

ðx þ εyðxÞÞy<sup>0</sup>

where <sup>x</sup> <sup>∈</sup>½0, <sup>1</sup>�, <sup>p</sup>ðxÞ, rðxÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�. For unperturbed equation

xy<sup>0</sup>

interval. For example, the Lagerstrom equation [70–81]

yðεÞ ¼ 0, yð∞Þ ¼ 1:

ever, this method is relatively complex for applied scientists.

Imanaliev [24], O'Malley (1971) [38], and Hoppenstedt (1971) [42].

where 0 < β is a given number and n is the dimension space.

<sup>y</sup>00ðxÞ þ nx�<sup>1</sup>y<sup>0</sup>

For example, a Lighthill model equation

point x ¼ 0 is a regular singular point.

Lighthill type [2, 3].

1963 [68] by the method of matching.

differential equations

4. εy00ðxÞ � xyðxÞ ¼ 1, x ∈½0, 1� is a bisingularly perturbed equation also. (II) Singularly perturbed differential equations such as the Lighthill's type [57–69], in

<sup>0</sup>ðxÞ þ pðxÞy0ðxÞ ¼ rðx),

(III) A singularly perturbed equation with a small parameter is considered on an infinite

ðxÞ þ yðxÞy<sup>0</sup>

1.5. Methods of construction of asymptotic expansions of solutions of singularly perturbed

1. The method of matching of outer and inner expansions [13, 19, 28, 29, 37, 49] is the most common method for constructing asymptotic expansions of solutions of singularly perturbed differential equations. Justification for this method is given by Il'in [22]. How-

2. The boundary layer function method (or composite asymptotic expansion)dates back to the work of many mathematicians. For the first time, this method for a singularly perturbed differential equations in partial derivatives is developed by Vishik and Lyusternik [52] and for nonlinear integral-differential equations (thus for the ordinary differential equations)

It should be noted that, for the first time, the uniform valid asymptotic expansion of the solution of Eq. (5) is constructed by Vasil'eva (1960) [50] after Wasow [69] and Sibuya in

Remark. The division into such classes is conditional, because singularly perturbed equation of Van der Pol in the neighborhood of points y ¼ �1 leads to an equation of

which the order of the corresponding unperturbed equation is not reduced, but has

ðxÞ þ pðxÞyðxÞ ¼ rðxÞ, yð1Þ ¼ a

ðxÞ ¼ βðy<sup>0</sup>

ðxÞÞ<sup>2</sup> , 6. The averaging method is applicable to the construction of solutions of a singularly perturbed equation on a large but finite interval.

Here, we consider a bisingularly perturbed differential equations and types of equations of Lighthill and Lagerstrom.

Here, we generalize the boundary layer function method for bisingular perturbed equations. We will construct a uniform asymptotic solution of the Lighthill model equation by the method of uniformization and construct the explicit solution of this one by the generalized method of the boundary layer functions.

Furthermore, we construct a uniformly convergent solution of the Lagerstrom model equation by the method of fictitious parameter.

## 2. Bisingularly perturbed ordinary differential equations

#### 2.1. Singularly perturbed of the first-order equation with a turning point

Consider the Cauchy problem [5]

$$
\varepsilon y'(\mathbf{x}) + \mathbf{x}y(\mathbf{x}) = f(\mathbf{x}), \quad 0 < \mathbf{x} \le 1, \quad y(0) = a,\tag{3}
$$

where <sup>f</sup>ðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�, <sup>f</sup>(x)=X<sup>∞</sup> k¼0 <sup>f</sup> <sup>k</sup>xk , f <sup>k</sup> ¼ f ðkÞ ð0Þ=k!, f <sup>0</sup> 6¼ 0; a is the constant

Explicit solution of the problem (3) has the form: <sup>y</sup>ðxÞ ¼ ae�x2=2<sup>ε</sup> <sup>þ</sup> <sup>1</sup> ε ðx 0 e <sup>ð</sup>s2�x2Þ=2<sup>ε</sup> fðsÞds:

The corresponding unperturbed equation (ε ¼ 0)

$$-\mathfrak{x}\tilde{y}(\mathfrak{x}) + f(\mathfrak{x}) = 0,$$

has a solution ~yðxÞ ¼ fðxÞ=x, which is unbounded at x ¼ 0.

If you seek a solution to problem (1) in the form

$$\mathbf{y}(\mathbf{x}) = y\_0(\mathbf{x}) + \varepsilon y\_1(\mathbf{x}) + \varepsilon^2 y\_2(\mathbf{x}) + \dots \tag{4}$$

then

$$y\_0(\mathbf{x}) = \frac{f(\mathbf{x})}{\mathbf{x}} \sim f\_0 \mathbf{x}^{-1}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_1(\mathbf{x}) = \mathbf{x}^{-1} y\_0'(\mathbf{x}) \sim f\_0 \mathbf{x}^{-3}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_2(\mathbf{x}) = \mathbf{x}^{-1} y\_1'(\mathbf{x}) \sim 3f\_0 \mathbf{x}^{-5}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_3(\mathbf{x}) = \mathbf{x}^{-1} y\_2'(\mathbf{x}) \sim 3 \cdot 5f\_0 \mathbf{x}^{-5}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_n(\mathbf{x}) = \mathbf{x}^{-1} y\_{n-1}'(\mathbf{x}) \sim 3 \cdot 5 \cdot \dots \cdot (2n-1) f\_0 \mathbf{x}^{-(2n+1)}, \quad \mathbf{x} \to \mathbf{0},$$

and a series of Eq. (4) is asymptotic in the segment <sup>ð</sup> ffiffiffi <sup>ε</sup> <sup>p</sup> , <sup>1</sup>�, and the point <sup>x</sup>0= ffiffiffi <sup>ε</sup> <sup>p</sup> <sup>¼</sup> <sup>μ</sup> is singular point of the asymptotic series of Eq. (4). Therefore, the solution of problem (3) we will seek in the form

$$y(\mathbf{x}) = \mu^{-1}\pi\_{-1}(t) + Y\_0(\mathbf{x}) + \pi\_0(t) + \mu\left(Y\_1(\mathbf{x}) + \pi\_1(t)\right) + \mu^2\left(Y\_2(\mathbf{x}) + \pi\_2(t)\right) + \dots, \quad \mu \to 0,\tag{5}$$

where YkðxÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup> <sup>½</sup>0, <sup>1</sup>�, <sup>π</sup>kðtÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup> <sup>½</sup>0, <sup>μ</sup>�<sup>1</sup>�, x <sup>¼</sup> <sup>μ</sup><sup>t</sup> and boundary layer functions <sup>π</sup>kðt<sup>Þ</sup> decreasing by power law as t ! ∞, that is,πkðtÞ ¼ Oðt �<sup>m</sup>Þ, t ! <sup>∞</sup>, m <sup>∈</sup> <sup>N</sup>.

Substituting Eq. (5) into Eq. (3), we obtain

$$\begin{cases} \pi\_{-1}'(t) + \mu^2 Y\_0'(\mathbf{x}) + \mu \pi\_0'(t) + \mu^3 Y\_1'(\mathbf{x}) + \mu^2 \pi\_1'(t) + \mu^4 Y\_2'(\mathbf{x}) + \mu^3 \pi\_2'(t) + \mu^5 Y\_3'(\mathbf{x}) + \mu^4 \pi\_3'(t) + \dots \\ \pi\_1 + x Y\_0(\mathbf{x}) + \mu \pi Y\_1(\mathbf{x}) + \mu^2 x Y\_2(\mathbf{x}) + \mu^3 x Y\_3(\mathbf{x}) + \dots + t\pi\_{-1}(t) + \mu t \pi\_0(t) + \mu^2 t \pi\_1(t) + \mu^3 t \pi\_2(t) \\ \pi\_1 + \mu^4 t \pi\_3(t) + \dots = f(\mathbf{x}). \end{cases} (6)$$

The initial conditions for the functions π<sup>k</sup>�<sup>1</sup>ðtÞ, k ¼ 0, 1, … we take in the next form

$$
\pi\_{-1}(0) = 0,\ \pi\_0(0) = a - Y\_0(0),\ \pi\_k(0) = -Y\_k(0),\ \ k = 1,2,\ldots
$$

From Eq. (6), we have

$$
\mu^0: \quad \pi\_{-1}'(t) + t\pi\_{-1}(t) + xY\_0(\mathbf{x}) = f(\mathbf{x}), \tag{7.-1}
$$

$$
\mu^1: \quad \pi\_0'(t) + t\pi\_0(t) + \mathbf{x}Y\_1(\mathbf{x}) = \mathbf{0}, \tag{7.0}
$$

$$\mu^{k+1}: \quad \pi'\_k(t) + t\pi\_k(t) + \mathbf{x}Y\_{k+1}(\mathbf{x}) + Y'\_{k-1}(\mathbf{x}) = \mathbf{0}, \quad k = 1, 2, \ldots \tag{7.18}$$

To Y0ðxÞ function has been smooth, and we define it from the equation

$$\mathbf{x}\mathbf{Y}\_0(\mathbf{x}) = f(\mathbf{x}) - f\_0 \Rightarrow \mathbf{Y}\_0(\mathbf{x}) = (f(\mathbf{x}) - f\_0) / \mathbf{x}\_0$$

and then from Eq. (7.�1), we have obtained the equation

Perturbed Differential Equations with Singular Points http://dx.doi.org/10.5772/67856 7

$$
\pi\_{-1}'(t) + t\pi\_{-1}(t) = f\_{0,1}
$$

Therefore

<sup>y</sup>0ðxÞ ¼ <sup>f</sup>ðx<sup>Þ</sup>

y0

y0

y0

�

<sup>1</sup>ðxÞ þ <sup>μ</sup><sup>2</sup>π<sup>0</sup>

The initial conditions for the functions π<sup>k</sup>�<sup>1</sup>ðtÞ, k ¼ 0, 1, … we take in the next form

<sup>k</sup>ðtÞ þ tπkðtÞ þ xYkþ<sup>1</sup>ðxÞ þ Y<sup>0</sup>

<sup>y</sup>1ðxÞ ¼ <sup>x</sup>�<sup>1</sup>

<sup>y</sup>2ðxÞ ¼ <sup>x</sup>�<sup>1</sup>

<sup>y</sup>3ðxÞ ¼ <sup>x</sup>�<sup>1</sup>

y0

and a series of Eq. (4) is asymptotic in the segment <sup>ð</sup> ffiffiffi

π�<sup>1</sup>ðtÞ þ Y0ðxÞ þ π0ðtÞ þ μ

<sup>½</sup>0, <sup>1</sup>�, <sup>π</sup>kðtÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup>

<sup>0</sup>ðtÞ þ <sup>μ</sup><sup>3</sup>Y<sup>0</sup>

μ<sup>0</sup> : π<sup>0</sup>

and then from Eq. (7.�1), we have obtained the equation

μ<sup>1</sup> : π<sup>0</sup>

To Y0ðxÞ function has been smooth, and we define it from the equation

decreasing by power law as t ! ∞, that is,πkðtÞ ¼ Oðt

Substituting Eq. (5) into Eq. (3), we obtain

μ<sup>k</sup>þ<sup>1</sup> : π<sup>0</sup>

<sup>0</sup>ðxÞ þ μπ<sup>0</sup>

ynðxÞ ¼ <sup>x</sup>�<sup>1</sup>

the form

π0

<sup>y</sup>ðxÞ ¼ <sup>μ</sup>�<sup>1</sup>

6 Recent Studies in Perturbation Theory

where YkðxÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup>

�<sup>1</sup>ðtÞ þ <sup>μ</sup><sup>2</sup>Y<sup>0</sup>

<sup>þ</sup> <sup>μ</sup><sup>4</sup>tπ3ðtÞ þ … <sup>¼</sup> <sup>f</sup>ðxÞ:

From Eq. (6), we have

<sup>x</sup> <sup>e</sup> <sup>f</sup> <sup>0</sup>x�<sup>1</sup>

<sup>0</sup>ðx<sup>Þ</sup> <sup>e</sup> <sup>f</sup> <sup>0</sup>x�<sup>3</sup>

<sup>1</sup>ðx<sup>Þ</sup> <sup>e</sup> <sup>3</sup><sup>f</sup> <sup>0</sup>x�<sup>5</sup>

<sup>2</sup>ðx<sup>Þ</sup> <sup>e</sup> <sup>3</sup> � <sup>5</sup><sup>f</sup> <sup>0</sup>x�<sup>5</sup>

<sup>n</sup>�<sup>1</sup>ðx<sup>Þ</sup> <sup>e</sup> <sup>3</sup> � <sup>5</sup> � … � ð2<sup>n</sup> � <sup>1</sup>Þ<sup>f</sup> <sup>0</sup>x�ð2nþ1<sup>Þ</sup>

point of the asymptotic series of Eq. (4). Therefore, the solution of problem (3) we will seek in

Y1ðxÞ þ π1ðtÞ

<sup>1</sup>ðtÞ þ <sup>μ</sup><sup>4</sup>Y<sup>0</sup>

<sup>þ</sup> xY0ðxÞ þ <sup>μ</sup>xY1ðxÞ þ <sup>μ</sup><sup>2</sup>xY2ðxÞ þ <sup>μ</sup><sup>3</sup>xY3ðxÞ þ … <sup>þ</sup> <sup>t</sup>π�<sup>1</sup>ðtÞ þ <sup>μ</sup>tπ0ðtÞ þ <sup>μ</sup><sup>2</sup>tπ1ðtÞ þ <sup>μ</sup><sup>3</sup>tπ2ðt<sup>Þ</sup>

π�<sup>1</sup>ð0Þ ¼ 0, π0ð0Þ ¼ a � Y0ð0Þ, πkð0޼�Ykð0Þ, k ¼ 1, 2, …

xY0ðxÞ ¼ fðxÞ � f <sup>0</sup> ) Y0ðxÞ¼ðfðxÞ � f <sup>0</sup>Þ=x,

� <sup>þ</sup> <sup>μ</sup><sup>2</sup> �

, x ! 0,

, x ! 0,

, x ! 0,

, x ! 0,

<sup>ε</sup> <sup>p</sup> , <sup>1</sup>�, and the point <sup>x</sup>0= ffiffiffi

Y2ðxÞ þ π2ðtÞ

<sup>½</sup>0, <sup>μ</sup>�<sup>1</sup>�, x <sup>¼</sup> <sup>μ</sup><sup>t</sup> and boundary layer functions <sup>π</sup>kðt<sup>Þ</sup>

<sup>2</sup>ðtÞ þ <sup>μ</sup><sup>5</sup>Y<sup>0</sup>

�<sup>m</sup>Þ, t ! <sup>∞</sup>, m <sup>∈</sup> <sup>N</sup>.

�<sup>1</sup>ðtÞ þ <sup>t</sup>π�<sup>1</sup>ðtÞ þ xY0ðxÞ ¼ <sup>f</sup>ðx), <sup>ð</sup>7:-1<sup>Þ</sup>

<sup>0</sup>ðtÞ þ tπ0ðtÞ þ xY1ðxÞ ¼ 0, ð7:0Þ

<sup>k</sup>�<sup>1</sup>ðxÞ ¼ <sup>0</sup>, k <sup>¼</sup> <sup>1</sup>, <sup>2</sup>, … <sup>ð</sup>7:k<sup>Þ</sup>

<sup>2</sup>ðxÞ þ <sup>μ</sup><sup>3</sup>π<sup>0</sup>

�

, x ! 0,

<sup>ε</sup> <sup>p</sup> <sup>¼</sup> <sup>μ</sup> is singular

þ …, μ ! 0,

<sup>3</sup>ðxÞ þ <sup>μ</sup><sup>4</sup>π<sup>0</sup>

ð5Þ

<sup>3</sup>ðtÞ þ …

ð6Þ

$$\pi\_{-1}(t) = f\_0 e^{-t^2/2} \int\_0^t e^{s^2/2} ds \in \mathbb{C}^\approx[0, \mu^{-1}]\_\prime$$

Obviously, this function bounded and is infinitely differentiable on the segment <sup>½</sup>0, <sup>μ</sup>�<sup>1</sup>�, and

$$
\pi\_{-1}(t) = -\frac{f\_0}{t}(1 + \frac{1}{t^2} + \frac{3}{t^4} + \dots), \quad t \to \infty.
$$

This asymptotic expression can be obtained by integration by parts the integral expression for π�<sup>1</sup>ðtÞ.

Eq. (7.0) define Y1ðxÞ and π0ðtÞ. Let Y1ðxÞ � 0, then

$$
\pi\_0'(t) + t\pi\_0(t) = 0, \ \pi\_0(0) = a - f\_{1,1}
$$

Hence, we find

$$
\pi\_0(t) = (a - f\_1)e^{-t^2/2a}
$$

From Eq. (7c) for k ¼ 1, we have

$$
\pi\_1'(t) + t\pi\_1(t) + \mathfrak{x}Y\_2(\mathfrak{x}) + Y\_0'(\mathfrak{x}) = \mathbf{0}.
$$

Let xY2ðxÞ ¼ Y<sup>0</sup> <sup>0</sup>ð0Þ � Y<sup>0</sup> <sup>0</sup>ðxÞ, then π<sup>0</sup> <sup>1</sup>ðtÞ þ tπ1ðt޼�Y<sup>0</sup> <sup>0</sup>ð0Þ.

From these, we get

$$Y\_2(\mathbf{x}) = (Y\_0'(0) - Y\_0'(\mathbf{x}))/\mathbf{x}, \quad \pi\_1(t) = -f\_2 e^{-t^2/2} \int\_0^t e^{s^2/2} ds \in \mathbb{C}^\circ[0, \mu^{-1}],$$

and

$$
\pi\_{-1}(t) = \frac{f\_2}{t} (1 + \frac{1}{t^2} + \frac{3}{t^4} + \dots), \quad t \to \infty.
$$

From Eq. (7c) for k ¼ 2, we have

$$
\pi\_2'(t) + t\pi\_2(t) + \mathbf{x}\mathbf{Y}\_3(\mathbf{x}) + \mathbf{Y}\_1'(\mathbf{x}) = \mathbf{0} \text{ or } \pi\_2'(t) + t\pi\_2(t) + \mathbf{x}\mathbf{Y}\_3(\mathbf{x}) = \mathbf{0}.
$$

Let Y3ðxÞ � 0, then

$$
\pi' \,\_2(t) + t\pi\_2(t) = 0, \,\, \pi\_2(0) = -Y\_2(0) = \mathcal{Y}\_3 \,.
$$

From this, we get

$$
\pi\_2(t) = 2f\_3 e^{-t^2/2}.
$$

Analogously continuing this process, we determine the others of the functions YkðxÞ, πkðtÞ.

In order to show that the constructed series of [Eq. (5)] is asymptotic series, we consider remainder term RmðxÞ ¼ yðxÞ � ymðxÞ,

$$\text{where } y\_m(\mathbf{x}) = \frac{1}{\mu}\pi\_{-1}(t) + Y\_0(\mathbf{x}) + \pi\_0(t) + \mu \left(Y\_1(\mathbf{x}) + \pi\_1(t)\right) + \dots + \mu^m \left(Y\_m(\mathbf{x}) + \pi\_m(t)\right).$$

For the remainder term RmðxÞ, we obtain a problem:

$$
\varepsilon \mathcal{R}'\_{\mathfrak{m}}(\mathbf{x}) + \mathbf{x} \mathcal{R}\_{\mathfrak{m}}(\mathbf{x}) = -\mu^{\mathfrak{m}+2} Y'\_{\mathfrak{m}}(\mathbf{x}), \quad 0 < \mathbf{x} \le \mathbf{1}, \ \mathcal{R}\_{\mathfrak{m}}(\mathbf{0}) = \mathbf{0}. \tag{8}
$$

We note that if m is odd, then Y<sup>0</sup> <sup>m</sup>ðxÞ � 0.

The problem (8) has a unique solution

$$R\_m(\mathbf{x}) = -\mu^m e^{-\mathbf{x}^2/2\varepsilon} \int\_0^\varepsilon Y'\_m(s) e^{s^2/2\varepsilon} ds,$$

and from this, we have RmðxÞ ¼ <sup>O</sup>ðμ<sup>m</sup>Þ, <sup>μ</sup> ! <sup>0</sup>, x <sup>∈</sup>½0, <sup>1</sup>�:

#### 2.2. Bisingularly perturbed in a homogenous differential equation of the Airy type

Consider the boundary value problem for the second-order ordinary in a homogenous differential equation with a turning point

$$
\varepsilon y''(\mathbf{x}) - \varepsilon y(\mathbf{x}) = f(\mathbf{x}), \quad \mathbf{x} \in (0, 1), \tag{9}
$$

$$y(0) = 0, \quad y(1) = 0. \tag{10}$$

where <sup>f</sup>ðxÞ ¼ <sup>X</sup><sup>∞</sup> k¼0 <sup>f</sup> <sup>k</sup>xk , x ! 0, f <sup>k</sup> ¼ f ðkÞ ð0Þ=k!, f <sup>0</sup> 6¼ 0.

Note 5. It is the general case of this one was considered in Ref. [8, 45–47].

Without loss of generality, we consider the homogeneous boundary conditions, since <sup>y</sup>ð0Þ ¼ a, yð1Þ ¼ b, a<sup>2</sup> <sup>þ</sup> <sup>b</sup><sup>2</sup> 6¼ <sup>0</sup>, using transformation

$$y(\mathbf{x}) = a + (b - a)\mathbf{x} + z(\mathbf{x})\_{\prime}$$

can lead to conditions (10).

If the asymptotic solution of the problems (9)–(10) we seek in the form

$$y(\mathbf{x}) = y\_0(\mathbf{x}) + \varepsilon y\_1(\mathbf{x}) + \varepsilon^2 y\_2(\mathbf{x}) + \dots,\tag{11}$$

then we have

π<sup>0</sup> <sup>2</sup>ðtÞ þ tπ2ðtÞ ¼ 0, π2ð0޼�Y2ð0Þ ¼ 2f <sup>3</sup>:

�t <sup>2</sup>=2 :

Y1ðxÞ þ π1ðtÞ

�

<sup>þ</sup> … <sup>þ</sup> <sup>μ</sup><sup>m</sup>

εy00ðxÞ � xyðxÞ ¼ fðxÞ, x∈ð0, 1Þ, ð9Þ

yð0Þ ¼ 0, yð1Þ ¼ 0: ð10Þ

�

<sup>m</sup>ðxÞ, 0 < x ≤ 1, Rmð0Þ ¼ 0: ð8Þ

YmðxÞ þ πmðtÞ

� .

π2ðtÞ ¼ 2f <sup>3</sup>e

Analogously continuing this process, we determine the others of the functions YkðxÞ, πkðtÞ. In order to show that the constructed series of [Eq. (5)] is asymptotic series, we consider

�

Y0

�x2=2<sup>ε</sup> ð x

Consider the boundary value problem for the second-order ordinary in a homogenous

ð0Þ=k!, f <sup>0</sup> 6¼ 0.

Without loss of generality, we consider the homogeneous boundary conditions, since

yðxÞ ¼ a þ ðb � aÞx þ zðxÞ,

2.2. Bisingularly perturbed in a homogenous differential equation of the Airy type

ðkÞ

Note 5. It is the general case of this one was considered in Ref. [8, 45–47].

0 Y0 <sup>m</sup>ðsÞe s2=2ε ds;

From this, we get

8 Recent Studies in Perturbation Theory

where ymðxÞ ¼ <sup>1</sup>

where <sup>f</sup>ðxÞ ¼ <sup>X</sup><sup>∞</sup>

k¼0

can lead to conditions (10).

<sup>f</sup> <sup>k</sup>xk

remainder term RmðxÞ ¼ yðxÞ � ymðxÞ,

εR<sup>0</sup>

The problem (8) has a unique solution

differential equation with a turning point

We note that if m is odd, then Y<sup>0</sup>

<sup>μ</sup> π�<sup>1</sup>ðtÞ þ Y0ðxÞ þ π0ðtÞ þ μ

and from this, we have RmðxÞ ¼ <sup>O</sup>ðμ<sup>m</sup>Þ, <sup>μ</sup> ! <sup>0</sup>, x <sup>∈</sup>½0, <sup>1</sup>�:

, x ! 0, f <sup>k</sup> ¼ f

<sup>y</sup>ð0Þ ¼ a, yð1Þ ¼ b, a<sup>2</sup> <sup>þ</sup> <sup>b</sup><sup>2</sup> 6¼ <sup>0</sup>, using transformation

<sup>m</sup>ðxÞ þ xRmðx޼�μ<sup>m</sup>þ<sup>2</sup>

<sup>m</sup>ðxÞ � 0.

Rmðx޼�μme

For the remainder term RmðxÞ, we obtain a problem:

$$y\_0(\mathbf{x}) = -\frac{f(\mathbf{x})}{\mathbf{x}} \sim f\_0 \mathbf{x}^{-1}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_1(\mathbf{x}) = \mathbf{x}^{-1} y\_0''(\mathbf{x}) \sim \mathbf{1} \cdot 2f\_0 \mathbf{x}^{-4}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_2(\mathbf{x}) = \mathbf{x}^{-1} y\_1''(\mathbf{x}) \sim \mathbf{1} \cdot \mathbf{2} \cdot \mathbf{4} \cdot \mathbf{5} f\_0 \mathbf{x}^{-7}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_3(\mathbf{x}) = \mathbf{x}^{-1} y\_2''(\mathbf{x}) \sim \mathbf{1} \cdot \mathbf{2} \cdot \mathbf{4} \cdot \mathbf{5} \cdot \mathbf{7} \cdot 8 f\_0 \mathbf{x}^{-10}, \quad \mathbf{x} \to \mathbf{0},$$

$$y\_n(\mathbf{x}) = \mathbf{x}^{-1} y\_{n-1}''(\mathbf{x}) \sim \mathbf{1} \cdot \mathbf{2} \cdot \mathbf{4} \cdot \mathbf{5} \cdot \mathbf{7} \cdot 8 \cdot \dots \cdot (3n-2) \cdot (3n-1) f\_0 \mathbf{x}^{-(3n+1)}, \; 0 < n, \quad \mathbf{x} \to \mathbf{0},$$

and the series (11) is asymptotic in the segment <sup>ð</sup> ffiffi <sup>ε</sup> <sup>p</sup><sup>3</sup> , 1�. The point <sup>x</sup>0= ffiffi <sup>ε</sup> <sup>p</sup><sup>3</sup> <sup>¼</sup> <sup>μ</sup> is singular point of asymptotic series (11).

The solution of problems (9) and (10) will be sought in the form

$$y(\mathbf{x}) = \mu^{-1}\pi\_{-1}(t) + \sum\_{k=0}^{\bullet} \mu^k \left( Y\_k(\mathbf{x}) + \pi\_k(t) \right) + \sum\_{k=0}^{\bullet} \lambda^k w\_k(\eta), \tag{12}$$

where <sup>t</sup> <sup>¼</sup> <sup>x</sup>=μ, <sup>μ</sup> <sup>¼</sup> ffiffi <sup>ε</sup> <sup>p</sup><sup>3</sup> , <sup>η</sup> ¼ ð<sup>1</sup> � <sup>x</sup>Þ=λ, <sup>λ</sup> <sup>¼</sup> ffiffi <sup>ε</sup> <sup>p</sup> . Here, YkðxÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�, <sup>π</sup>kðt<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>=μ� is boundary layer function in a neighborhood of t ¼ 0 and decreases by the power law as t ! ∞, and the function wkðtÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>=λ� is boundary function in a neighborhood of <sup>η</sup> <sup>¼</sup> 0 and decreases exponentially as η ! ∞.

Substituting Eq. (12) in Eq. (9), we get

$$\sum\_{k=0}^{\bullet} \mu^k(\pi^\*\_{k-1}(t) - t\pi\_{k-1}(t)) + \sum\_{k=0}^{\bullet} \mu^{k+3} Y\_k(\mathbf{x}) - \mathbf{x} \sum\_{k=0}^{\bullet} \mu^k Y\_k(\mathbf{x}) = f(\mathbf{x}) \tag{13}$$

$$\sum\_{k=0}^{\infty} \lambda^k \left( w\_{\; \lambda}^{\prime \prime}(\eta) - (1 - \lambda \eta) w\_{\; \lambda}(\eta) \right) = 0. \tag{14}$$

From Eq. (13), we have

$$
\mu^0: \quad \pi^\*\_{-1}(t) - t\pi\_{-1}(t) - \mathbf{x}\mathbf{Y}\_0(\mathbf{x}) = f(\mathbf{x}).\tag{15.-1}
$$

$$
\mu^1: \quad \pi^"\mathfrak{o}(t) - t\pi\_0(t) - \mathfrak{x}Y\_1(\mathfrak{x}) = \mathbf{0},
\tag{15.0}
$$

$$
\mu^2: \quad \pi^"{.}(t) - t\pi\_1(t) - \text{x}\,\text{Y}\_2(\text{x}) = 0,\tag{15.1}
$$

$$
\mu^3: \quad \pi^"{}\_2(t) - t\pi\_2(t) + \mathcal{Y}^"{}\_0(\mathbf{x}) - \mathbf{x} \mathcal{Y}\_3(\mathbf{x}) = \mathbf{0},\tag{15.2}
$$

$$\mu^k: \quad \pi^\*\_{k-1}(t) - t\pi\_{k-1}(t) + Y^\*\_{k-3}(\mathbf{x}) - \mathbf{x}Y\_k(\mathbf{x}) = \mathbf{0}, \quad k > \mathbf{3},\tag{15.1}$$

Boundary conditions for functions π<sup>k</sup>�<sup>1</sup>ðtÞ, k ¼ 0, 1, … we take next form

$$
\pi\_{-1}(0) = 0,\ \pi\_k(0) = -Y\_k(0),\ \lim\_{\mu \to 0} \pi\_{k-1}(1/\mu) = 0,\ \ k = 0,1,2,\ldots
$$

To Y0ðxÞ function has been smooth; therefore, we define it from the equation

$$-\mathbf{x}\,\mathbf{Y}\_0(\mathbf{x}) = f(\mathbf{x}) - f\_0 \Rightarrow \mathbf{Y}\_0(\mathbf{x}) = -(f(\mathbf{x}) - f\_0)/\mathbf{x}\_1$$

then from Eq. (15.1), we have the equation

$$
\pi\_{-1}''(t) - t\pi\_{-1}(t) = f\_0,
$$

Let us prove an auxiliary lemma.

Lemma 1. Next boundary value problem

$$z''(t) - tz(t) = b, \quad 0 < t < 1/\mu, \text{ here } b \text{ is the constant}, \tag{16}$$

$$z(0) = z^0, \quad z(1/\mu) \to 0, \quad \mu \to 0 \tag{17}$$

will have the unique solution and this one have next form

$$z(t) = z^0 \frac{Ai(t)}{Ai(0)} - \tau b \Big( Ai(t) \Big|\_{0}^{t} Bi(s)ds + Bi(t) \Big|\_{t}^{1/\mu} Ai(s)ds - Ai(t)\sqrt{3} \Big|\_{0}^{1/\mu} Ai(s)ds\Big),$$

and <sup>z</sup>ðtÞ∈C<sup>∞</sup>½0, <sup>μ</sup>�<sup>1</sup>�.

Proof. We verify the boundary conditions:

$$z(0) = z^0 - \pi b \left( Bi(0) \int\_0^{1/\mu} Ai(s) ds - Ai(0) \sqrt{3} \int\_0^{1/\mu} Ai(s) ds \right),$$

as Bið0Þ ¼ Aið0<sup>Þ</sup> ffiffiffi 3 <sup>p</sup> , so <sup>z</sup>ð0Þ ¼ <sup>z</sup>0.

$$z(1/\mu) = z^0 \frac{Ai(1/\mu)}{Ai(0)} - \pi b(1 - \sqrt{3})Ai(1/\mu) \int\_0^{1/\mu} Bi(s)ds,$$

as Aiðt<sup>Þ</sup> <sup>e</sup> <sup>t</sup> �1=<sup>4</sup>e�<sup>2</sup> 3 t 3=2 , Biðt<sup>Þ</sup> <sup>e</sup> <sup>t</sup> �1=<sup>4</sup>e 2 3 t 3=2 , t ! ∞, so zð1=μÞ ¼ OðμÞ, μ ! 0.

Now we show that z(t) satisfies Eq. (16). For this, we compute derivatives:

$$z'(t) = z^0 \frac{Ai'(t)}{Ai(0)} - \pi b \left( A i'(t) \int\_0^t Bi(s)ds + Bi'(t) \int\_t^{1/\mu} Ai(s)ds - A i'(t) \sqrt{3} \int\_0^{1/\mu} Ai(s)ds \right)$$

$$z''(t) = z^0 \frac{Ai''(t)}{Ai(0)} - \pi b \left( A i'(t) \int\_0^t Bi(s)ds + Bi'(t) \int\_t^{1/\mu} Ai(s)ds - \frac{1}{\pi} - A i'(t) \sqrt{3} \int\_0^{1/\mu} Ai(s)ds \right)$$

Substituting the expressions for z00ðtÞ and zðtÞ in Eq. (17), and given that Ai00ðtÞ � tAiðtÞ � 0 and Bi00ðtÞ � tBiðtÞ � 0, we get: b � b.

The uniqueness of zðtÞ the solution is proved by contradiction. Let uðtÞalso be a solution of problems (16) and (17), zðtÞ 6¼ uðtÞ. Considering the function rðtÞ ¼ zðtÞ � uðtÞ, for the function rðtÞ, we obtain the problem

$$r''(t) - tr(t) = 0, \quad 0 < t < 1/\mu, \quad r(0) = 0, \quad r(1/\mu) \to 0, \quad \mu \to 0.$$

The general solution of the homogeneous equation is

rðtÞ ¼ c1AiðtÞ þ c2BiðtÞ; c1, <sup>2</sup> is the constant.

Boundary conditions for functions π<sup>k</sup>�<sup>1</sup>ðtÞ, k ¼ 0, 1, … we take next form

To Y0ðxÞ function has been smooth; therefore, we define it from the equation

π<sup>00</sup>

<sup>z</sup>ð0Þ ¼ <sup>z</sup><sup>0</sup>

� Bið0Þ

<sup>z</sup>ð1=μÞ ¼ <sup>z</sup><sup>0</sup> Aið1=μ<sup>Þ</sup>

�1=<sup>4</sup>e 2 3 t 3=2

> ðt 0

Now we show that z(t) satisfies Eq. (16). For this, we compute derivatives:

BiðsÞds þ BiðtÞ

ð<sup>1</sup>=<sup>μ</sup> 0

Aið0<sup>Þ</sup> � <sup>π</sup>bð<sup>1</sup> � ffiffiffi

BiðsÞds þ Bi<sup>0</sup>

BiðsÞds þ Bi00ðtÞ

will have the unique solution and this one have next form

� πb � AiðtÞ ðt 0

<sup>z</sup>ð0Þ ¼ <sup>z</sup><sup>0</sup> � <sup>π</sup><sup>b</sup>

<sup>p</sup> , so <sup>z</sup>ð0Þ ¼ <sup>z</sup>0.

, Biðt<sup>Þ</sup> <sup>e</sup> <sup>t</sup>

� πb � Ai<sup>0</sup> ðtÞ ðt 0

Aið0<sup>Þ</sup> � <sup>π</sup>b Ai00ðt<sup>Þ</sup>

ðtÞ Aið0Þ μ!0

�xY0ðxÞ ¼ fðxÞ � f <sup>0</sup> ) Y0ðxÞ ¼ �ðfðxÞ � f <sup>0</sup>Þ=x,

�<sup>1</sup>ðtÞ � <sup>t</sup>π�<sup>1</sup>ðtÞ ¼ <sup>f</sup> <sup>0</sup>:

z00ðtÞ � tzðtÞ ¼ b, 0 < t < 1=μ, here b is the constant, ð16Þ

ð<sup>1</sup>=<sup>μ</sup> t

AiðsÞds � Aið0<sup>Þ</sup> ffiffiffi

3 <sup>p</sup> <sup>Þ</sup>Aið1=μ<sup>Þ</sup>

ðtÞ ð<sup>1</sup>=<sup>μ</sup> t

> ð<sup>1</sup>=<sup>μ</sup> t

, t ! ∞, so zð1=μÞ ¼ OðμÞ, μ ! 0.

, zð1=μÞ ! 0, μ ! 0 ð17Þ

3 p ð<sup>1</sup>=<sup>μ</sup> 0

AiðsÞds � ,

BiðsÞds,

<sup>ð</sup>t<sup>Þ</sup> ffiffiffi 3 p ð<sup>1</sup>=<sup>μ</sup> 0

<sup>π</sup> � Ai00ðt<sup>Þ</sup> ffiffiffi

3 p ð<sup>1</sup>=<sup>μ</sup> 0

AiðsÞds �

AiðsÞds

AiðsÞds � ,

AiðsÞds � Aiðt<sup>Þ</sup> ffiffiffi

3 p ð<sup>1</sup>=<sup>μ</sup> 0

> ð<sup>1</sup>=<sup>μ</sup> 0

AiðsÞds � Ai<sup>0</sup>

AiðsÞds � <sup>1</sup>

!

π<sup>k</sup>�<sup>1</sup>ð1=μÞ ¼ 0, k ¼ 0, 1, 2, …

π�<sup>1</sup>ð0Þ ¼ 0, πkð0޼�Ykð0Þ, lim

then from Eq. (15.1), we have the equation

Lemma 1. Next boundary value problem

Let us prove an auxiliary lemma.

10 Recent Studies in Perturbation Theory

<sup>z</sup>ðtÞ ¼ <sup>z</sup><sup>0</sup> Aiðt<sup>Þ</sup>

and <sup>z</sup>ðtÞ∈C<sup>∞</sup>½0, <sup>μ</sup>�<sup>1</sup>�.

as Bið0Þ ¼ Aið0<sup>Þ</sup> ffiffiffi

�1=<sup>4</sup>e�<sup>2</sup> 3 t 3=2

<sup>ð</sup>tÞ ¼ <sup>z</sup><sup>0</sup> Ai<sup>0</sup>

<sup>z</sup>00ðtÞ ¼ <sup>z</sup><sup>0</sup> Ai00ðt<sup>Þ</sup>

as Aiðt<sup>Þ</sup> <sup>e</sup> <sup>t</sup>

z0

Aið0Þ

Proof. We verify the boundary conditions:

3

Considering the boundary condition rð1=μÞ ! 0, μ ! 0, we have c<sup>2</sup> ¼ 0; rðtÞ ¼ c1AiðtÞ. And the second condition rð0Þ ¼ 0, c<sup>1</sup> ¼ 0 follows. This implies that rðtÞ � 0.

Therefore, <sup>z</sup>ðtÞ � <sup>u</sup>ðtÞ. It is obvious that <sup>z</sup>ðt<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>μ</sup>�<sup>1</sup>�. Lemma 1 is proved.

This Lemma 1 implies the existence and uniqueness of <sup>π</sup>�<sup>1</sup>ðtÞ∈C<sup>∞</sup>½0, <sup>μ</sup>�<sup>1</sup>� solution of the problem:

$$
\pi\_{-1}^{\prime\prime}(t) - t\pi\_{-1}(t) = f\_{0\prime} \quad 0 < t < 1/\mu, \quad \pi\_{-1}(0) = 0, \quad \pi\_{-1}(1/\mu) \to 0, \quad \mu \to 0.
$$

This function bounded and is infinitely differentiable on the segment <sup>½</sup>0, <sup>μ</sup>�<sup>1</sup>�, and as <sup>t</sup> ! <sup>∞</sup>:

$$
\pi\_{-1}(t) = -\frac{f\_0}{t} \left( 1 + \frac{1 \cdot 2}{t^3} + \frac{1 \cdot 2 \cdot 4 \cdot 5}{t^6} + \dots \right).
$$

This asymptotic expression can be obtained by integration by parts the integral expression for π�<sup>1</sup>ðtÞ.

From Eq. (15.0), we define Y1ðxÞ and π0ðtÞ. Let Y1ðxÞ � 0, then

$$
\pi\_{0}^{\prime\prime}(t) - t\pi\_{0}(t) = 0, \quad \pi\_{0}(0) = f\_{1\prime} \quad \pi\_{0}(1/\mu) \to 0, \quad \mu \to 0,
$$

And by Lemma 1, we have

$$
\pi\_0(t) = f\_1 Ai(t) / Ai(0).
$$

Analogously, from Eq. (15.1), we define Y2ðxÞ and π1ðtÞ. Let Y2ðxÞ � 0, then

$$
\pi\_{1}^{\prime\prime}(t) - t\pi\_{1}(t) = 0, \quad \pi\_{1}(0) = 0, \quad \pi\_{0}(1/\mu) \to 0, \quad \mu \to 0 \dots
$$

In view of Lemma 1, we have π1ðtÞ � 0.

To Y3ðxÞ function has been smooth; as above, we define it from the equation

$$\text{tr}\,Y\_3(\mathbf{x}) = \boldsymbol{Y}\_{\;0}^{\;\prime}(\mathbf{x}) - \boldsymbol{Y}\_{\;0}^{\;\prime}(\mathbf{0}) \Rightarrow \boldsymbol{Y}\_3(\mathbf{x}) = (\boldsymbol{Y}\_{\;0}^{\;\prime}(\mathbf{x}) - \boldsymbol{Y}\_{\;0}^{\;\prime}(\mathbf{0})) / \text{x}, \quad (\boldsymbol{Y}\_{\;0}^{\;\prime}(\mathbf{0}) = -2\boldsymbol{f}\_3),$$

then Eq. (15.2) to π2ðtÞhase the problem

$$
\pi\_{\mathfrak{n}}^{''}(t) - t\pi\_{\mathfrak{2}}(t) = \mathfrak{2}f\_{\mathfrak{3}'} \quad \pi\_{\mathfrak{2}}(0) = 0, \quad \pi\_{\mathfrak{2}}(1/\mu) \to 0, \quad \mu \to 0.
$$

By Lemma 1, we can write an explicit solution to this problem, and this solution bounded and is infinitely differentiable on the segment <sup>½</sup>0, <sup>μ</sup>�<sup>1</sup>�, and as <sup>t</sup> ! <sup>∞</sup>:

$$
\pi\_2(t) = -\frac{2f\_3}{t} \left( 1 + \frac{1 \cdot 2}{t^3} + \frac{1 \cdot 2 \cdot 4 \cdot 5}{t^6} + \dots \right).
$$

Analogously continuing this process, we determine the rest of the functions YkðxÞ, πkðtÞ.

Now we will define functions wkðηÞ from the equality (14) by using the boundary conditions yð1Þ ¼ 0 We state problems

$$Lw\_0 \equiv w\_{\
u}^"(\eta) - w\_0(\eta) = 0, \ w\_0(0) = Y\_0(1), \ \lim\_{\eta \to \infty} w\_0(\eta) = 0\tag{18.0}$$

$$Lw\_k = -\eta w\_{k-1}(\eta), \ w\_{2i}(0) = Y\_{3i}(1), \ w\_{2i-1}(0) = 0, \ \lim\_{\eta \to \infty} w\_k(\eta) = 0, \ k, i \in \mathcal{N}.\tag{18.k}$$

One can easily make sure that all these problems (18.0) and (18.k) have unique solutions such that wkðηÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>∞</sup>Þ, wkðηÞ ¼ <sup>O</sup>ðe�<sup>η</sup><sup>Þ</sup> with <sup>η</sup> ! <sup>∞</sup>.

Thus, all functions YkðxÞ, wkðηÞ, and πkðtÞ in equality (12) are defined, i.e., a formally asymptotic expansion is constructed. Let us justify the constructed expansion. Let

$$y\_m(\mathbf{x}) = \mu^{-1}\pi\_{-1}(t) + \sum\_{k=0}^{3m} \mu^k \left( Y\_k(\mathbf{x}) + \pi\_k(t) \right) + \sum\_{k=0}^{2m} \lambda^k w\_k(\eta), \quad r\_m(\mathbf{x}) = y(\mathbf{x}) - y\_m(\mathbf{x}).$$

Then for the remainder term, we state the following problem:

$$
\varepsilon r\_m^\*(\mathbf{x}) - \mathbf{x}r\_m(\mathbf{x}) = O(\varepsilon^{m+1/2}), \quad \varepsilon \to 0, \quad \mathbf{x} \in (0, 1). \tag{19}
$$

$$r\_m(0) = O(e^{-1/\sqrt{\varepsilon}}), \quad r\_m(1) = O(\varepsilon^{m+1}), \quad \varepsilon \to 0. \tag{20}$$

Let rmðxÞ¼ð<sup>2</sup> � <sup>x</sup><sup>2</sup>ÞRmðxÞ=2, and then problems (19) and (20) take the form

$$\begin{aligned} \varepsilon \boldsymbol{R}\_{\;m}^{\prime}(\mathbf{x}) - \frac{4\mathbf{x}\varepsilon}{2 - \mathbf{x}^{2}} \boldsymbol{R}\_{\;m}^{\prime}(\mathbf{x}) - \left(\frac{2\varepsilon}{2 - \mathbf{x}^{2}} + \mathbf{x}\right) \boldsymbol{R}\_{\;m}(\mathbf{x}) &= O(\varepsilon^{m + 1/2}), \quad \varepsilon \to 0, \\\\ \boldsymbol{R}\_{\;m}(0) &= O(\varepsilon^{-1/\sqrt{\varepsilon}}), \quad \boldsymbol{R}\_{\;m}(1) = O(\varepsilon^{m + 1}), \quad \varepsilon \to 0. \end{aligned}$$

According to the maximum principle [23, p. 117, 82], we have RmðxÞ ¼ <sup>O</sup>ðε<sup>m</sup>�1=<sup>2</sup>Þ, <sup>ε</sup> ! <sup>0</sup>, x<sup>∈</sup> <sup>½</sup>0, <sup>1</sup>�. Hence, we get rmðxÞ ¼ <sup>O</sup>ðε<sup>m</sup>�1=<sup>2</sup>Þ, <sup>ε</sup> ! <sup>0</sup>, x <sup>∈</sup>½0, <sup>1</sup>�.

Thus, we have proved.

xY3ðxÞ ¼ <sup>Y</sup>″

12 Recent Studies in Perturbation Theory

yð1Þ ¼ 0 We state problems

ymðxÞ ¼ <sup>μ</sup>�<sup>1</sup>

εR″

Lw<sup>0</sup> � <sup>w</sup>″

that wkðηÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>∞</sup>Þ, wkðηÞ ¼ <sup>O</sup>ðe�<sup>η</sup><sup>Þ</sup> with <sup>η</sup> ! <sup>∞</sup>.

<sup>π</sup>�<sup>1</sup>ðtÞ þ<sup>X</sup>

εr ″

<sup>m</sup>ðxÞ � <sup>4</sup>x<sup>ε</sup>

3m

k¼0 μk �

rmð0Þ ¼ Oðe

<sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>R</sup><sup>0</sup>

Rmð0Þ ¼ Oðe

Then for the remainder term, we state the following problem:

then Eq. (15.2) to π2ðtÞhase the problem

π″

<sup>0</sup>ðxÞ � <sup>Y</sup>″

is infinitely differentiable on the segment <sup>½</sup>0, <sup>μ</sup>�<sup>1</sup>�, and as <sup>t</sup> ! <sup>∞</sup>:

<sup>π</sup>2ðt޼� <sup>2</sup><sup>f</sup> <sup>3</sup>

Lwk ¼ �ηwk�<sup>1</sup>ðηÞ, w2ið0Þ ¼ Y3ið1Þ, w2i�<sup>1</sup>ð0Þ ¼ 0, lim

totic expansion is constructed. Let us justify the constructed expansion. Let

<sup>m</sup>ðxÞ � xrmðxÞ ¼ <sup>O</sup>ðε<sup>m</sup>þ1=<sup>2</sup>

�1<sup>=</sup> ffiffi ε p

Let rmðxÞ¼ð<sup>2</sup> � <sup>x</sup><sup>2</sup>ÞRmðxÞ=2, and then problems (19) and (20) take the form

<sup>m</sup>ðxÞ � <sup>2</sup><sup>ε</sup>

�1<sup>=</sup> ffiffi ε p

YkðxÞ þ πkðtÞ

t

1 þ 1 � 2 t <sup>3</sup> þ

Analogously continuing this process, we determine the rest of the functions YkðxÞ, πkðtÞ.

Now we will define functions wkðηÞ from the equality (14) by using the boundary conditions

<sup>0</sup>ðηÞ � w0ðηÞ ¼ 0, w0ð0Þ ¼ Y0ð1Þ, lim

One can easily make sure that all these problems (18.0) and (18.k) have unique solutions such

Thus, all functions YkðxÞ, wkðηÞ, and πkðtÞ in equality (12) are defined, i.e., a formally asymp-

� þ<sup>X</sup> 2m

<sup>Þ</sup>, rmð1Þ ¼ <sup>O</sup>ðε<sup>m</sup>þ<sup>1</sup>

<sup>Þ</sup>, Rmð1Þ ¼ <sup>O</sup>ðε<sup>m</sup>þ<sup>1</sup>

<sup>2</sup> � <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>x</sup> � �

k¼0 λk

<sup>0</sup>ð0Þ ) <sup>Y</sup>3ðxÞ¼ðY″

<sup>0</sup>ðxÞ � <sup>Y</sup>″

1 � 2 � 4 � 5 t

� �

<sup>6</sup> þ …

η!∞

η!∞

RmðxÞ ¼ <sup>O</sup>ðε<sup>m</sup>þ1=<sup>2</sup>

Þ, ε ! 0:

:

<sup>2</sup>ðtÞ � tπ2ðtÞ ¼ 2f <sup>3</sup>, π2ð0Þ ¼ 0, π2ð1=μÞ ! 0, μ ! 0:

By Lemma 1, we can write an explicit solution to this problem, and this solution bounded and

<sup>0</sup>ð0ÞÞ=x, <sup>ð</sup>Y″

<sup>0</sup>ð0޼�2f <sup>3</sup>Þ;

w0ðηÞ ¼ 0 ð18:0Þ

wkðηÞ ¼ 0, k, i∈ N: ð18:kÞ

wkðηÞ, rmðxÞ ¼ yðxÞ � ymðxÞ:

Þ, ε ! 0, x∈ ð0, 1Þ: ð19Þ

Þ, ε ! 0: ð20Þ

Þ, ε ! 0,

Theorem 2. Let fð0Þ 6¼ 0, then the solution to problem (9) and (10) will have next form

$$y(\mathbf{x}) = \frac{1}{\sqrt[3]{\varepsilon}} \pi\_{-1} \left(\frac{\mathbf{x}}{\sqrt[3]{\varepsilon}}\right) + \sum\_{k=0}^{\infty} \sqrt[3]{\varepsilon^k} \left(y\_k(\mathbf{x}) + \pi\_k \left(\frac{\mathbf{x}}{\sqrt[3]{\varepsilon}}\right)\right) + \sum\_{k=0}^{\infty} \sqrt{\varepsilon^k} w\_k \left(\frac{1-\mathbf{x}}{\sqrt{\varepsilon}}\right).$$

Example. Consider the problem

$$
\varepsilon y''(\mathbf{x}) - \varepsilon y(\mathbf{x}) = \mathbf{1} + \mathbf{x}, \quad \mathbf{x} \in (0, 1), \quad y(0) = 0, \quad y(1) = 0.
$$

The asymptotic solution this problem we can represent in the form <sup>y</sup>ðxÞ ¼ <sup>μ</sup>�<sup>1</sup>π�<sup>1</sup>ðtÞþ X 3 k¼0 μk <sup>ð</sup>YkðxÞþπkðtÞÞ þ <sup>w</sup>0ðηÞ þ <sup>λ</sup>w1ðηÞ þ <sup>λ</sup><sup>2</sup> w2ðηÞ þ RðxÞ.

We have got Y0ðxÞ ¼ �ð1 þ x � 1Þ=x ¼ �1, Y1,2,3ðxÞ � 0,

$$
\pi\_{-1}(t) = -\pi \left( Ai(t) \int\_0^t Bi(s)ds + Bi(t) \int\_t^{1/\mu} Ai(s)ds - Ai(t)\sqrt{3} \right)^{1/\mu} Ai(s)ds
$$

$$
\pi\_0(t) = Ai(t)/Ai(0), \quad \pi\_{1,2,3}(t) \equiv 0,\\
\pi\_0(\eta) = 2e^{-\eta}, \text{ } \pi\_k(\eta) = O(e^{-\eta}), \ k = 1, 2.
$$

$$
\varepsilon \mathbb{R}''(\mathbf{x}) - \varepsilon \mathbb{R}(\mathbf{x}) = O(\varepsilon^{3/2}),\\
0 < \mathbf{x} < 1,\\
\mathcal{R}(0) = O(e^{-1/\sqrt{\varepsilon}}),\\
\mathcal{R}(1) = O(\varepsilon^2), \ \varepsilon \to 0.
$$

We have

$$y(\mathbf{x}) = \varepsilon^{-1/3}\pi\_{-1}(t) - 1 + 2e^{-(1-\mathbf{x})/\sqrt{\varepsilon}} + \pi\_0(t) + \sqrt{\varepsilon}w\_1(\eta) + \varepsilon w\_2(\eta) + O(\sqrt{\varepsilon}), \varepsilon \to 0.$$

## 2.3. Bisingularly perturbed equation of the second order with a regularly singular point Consider the boundary value problem [6, 7]

$$L\_{\varepsilon}y \equiv \varepsilon y' + xy' - q(\mathbf{x})y = f(\mathbf{x}), \quad \mathbf{x} \in [0, 1]. \tag{21}$$

$$y(0) = 0, \quad y(1) = 0,\tag{22}$$

where <sup>q</sup>ðxÞ, fðxÞ∈C<sup>∞</sup>½0, <sup>1</sup>�.

Here, for simplicity, we consider the case qð0Þ ¼ 1, qðxÞ ≥ 1.

The solution of the unperturbed problem

$$My \equiv xy' - q(\mathbf{x})y = f(\mathbf{x})\_{\mathbf{x}}$$

represented as

$$y\_0(\mathbf{x}) \equiv \exp(\mathbf{x}) \int\_1^\mathbf{x} r(\mathbf{s}) \mathbf{s}^{-2} d\mathbf{s},\tag{23}$$

where

$$p(\mathbf{x}) = p^{-1}(\mathbf{x}) f(\mathbf{x}), \quad p(\mathbf{x}) = \exp\left\{ \int\_1^\mathbf{x} \left( q(\mathbf{x}) - 1 \right) \mathbf{s}^{-1} d\mathbf{s} \right\}.$$

Extracting in Eq. (23), the main part of the integral in the sense of Hadamard [34], it can be represented as

$$y\_0(\mathbf{x}) = a(\mathbf{x}) + r\_1 \mathbf{x} p(\mathbf{x}) \ln \mathbf{x},\tag{24}$$

where

$$\begin{aligned} a(\mathbf{x}) &= \mathbf{x}p(\mathbf{x}) \int\_{1}^{\mathbf{x}} \left( r(\mathbf{s}) - r\_0 - r\_1 \mathbf{s} \right) \mathbf{s}^{-2} d\mathbf{s} + r\_0 p(\mathbf{x}) [\mathbf{x} - 1], \\ r\_0 &= r(\mathbf{0}), \ r\_1 = r'(\mathbf{0}) = p(\mathbf{0})^{-1} [f'(\mathbf{0}) - q'(\mathbf{0}) f(\mathbf{0})]. \end{aligned} \tag{25}$$

Function <sup>a</sup>ðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�.

Theorem 3. Suppose that the conditions referred to the above with respect to qðxÞ and fðxÞ. Then the asymptotic behavior of the solution of the problems (21) and (22) can be written as:

$$\sum\_{k=0}^{\infty} \mu^k \left( z\_k(\mathbf{x}) + \pi\_k(t) \right), \quad \varepsilon = \mu^2, \ \mathbf{x} = \mu t,\tag{26}$$

where zkðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�, <sup>π</sup>kðtÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>μ</sup>�<sup>1</sup>�.

Function z0ðxÞis a solution of equation

$$M\mathbf{z}\_0 = f(\mathbf{x}) - c\_0 \mathbf{x} p(\mathbf{x}),$$

wherec<sup>0</sup> ¼ pð0Þ �1 ½f 0 ð0Þ � q<sup>0</sup> ð0Þfð0Þ�.

The coefficients zkðxÞof the series (26) will be determined as the solution of equations

$$M\mathbf{z}\_k = -\mathbf{z}\_{k-1}^\prime(\mathbf{x}) - c\_k \mathbf{x} p(\mathbf{x})\_k$$

where ck ¼ pð0Þ �1 ½�z‴ <sup>k</sup>�<sup>1</sup>ð0Þ þ <sup>z</sup>″ <sup>k</sup>�<sup>1</sup>ð0Þq<sup>0</sup> ð0Þ�, with boundary conditions zkð1Þ ¼ 0, k ≥ 1. Functions πkðtÞ is the solution of the equations

$$L\pi\_k \equiv \pi\_k'(t) + t\pi\_k'(t) - q(\mu t)\pi\_k(t) - c\_k\mu tp(\mu t)$$

with boundary conditions <sup>π</sup>kð0޼�zkð0Þ, <sup>π</sup>kðμ�<sup>1</sup>Þ ¼ 0.

Next, we use the following lemma.

Lemma 2. The problem

My � xy<sup>0</sup> � qðxÞy ¼ fðxÞ,

ðx 1 rðsÞs �2

Extracting in Eq. (23), the main part of the integral in the sense of Hadamard [34], it can be

rðsÞ � r<sup>0</sup> � r1s

ð0Þ ¼ pð0Þ

zkðxÞ þ πkðtÞ

� s �2

�1 ½f 0 ð0Þ � q<sup>0</sup>

Theorem 3. Suppose that the conditions referred to the above with respect to qðxÞ and fðxÞ. Then the asymptotic behavior of the solution of the problems (21) and (22) can be written as:

�

Mz<sup>0</sup> ¼ fðxÞ � c0xpðxÞ,

The coefficients zkðxÞof the series (26) will be determined as the solution of equations

Mzk ¼ �z″

<sup>k</sup>�<sup>1</sup>ð0Þq<sup>0</sup>

, <sup>ε</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup>

<sup>k</sup>�<sup>1</sup>ðxÞ � ckxpðxÞ,

ð0Þ�, with boundary conditions zkð1Þ ¼ 0, k ≥ 1.

ðx 1 �

qðxÞ � 1

� �

� s �1 ds

y0ðxÞ ¼ aðxÞ þ r1xpðxÞlnx, ð24Þ

ds þ r0pðxÞ½x � 1�,

ð0Þfð0Þ�:

ds, ð23Þ

:

, x ¼ μt, ð26Þ

ð25Þ

y0ðxÞ � xpðxÞ

ðxÞfðxÞ, pðxÞ ¼ exp

<sup>r</sup>ðxÞ ¼ <sup>p</sup>�<sup>1</sup>

aðxÞ ¼ xpðxÞ

r<sup>0</sup> ¼ rð0Þ, r<sup>1</sup> ¼ r<sup>0</sup>

X∞ k¼0 μk �

ð0Þfð0Þ�.

<sup>k</sup>�<sup>1</sup>ð0Þ þ <sup>z</sup>″

Functions πkðtÞ is the solution of the equations

where zkðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�, <sup>π</sup>kðtÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>μ</sup>�<sup>1</sup>�.

Function z0ðxÞis a solution of equation

�1 ½f 0 ð0Þ � q<sup>0</sup>

�1 ½�z‴ ðx 1 �

represented as

14 Recent Studies in Perturbation Theory

represented as

Function <sup>a</sup>ðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�.

wherec<sup>0</sup> ¼ pð0Þ

where ck ¼ pð0Þ

where

where

$$My = f(\mathbf{x}) - r\_1 x p(\mathbf{x})$$

It has a unique solution <sup>y</sup>ðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�.

The proof of Lemma 2 follows from Eqs. (24) and (25).

Lemma 3. A boundary value problem

$$L\_0 \upsilon \equiv \upsilon'' + t\upsilon' - \upsilon(t) = 0, \quad \upsilon(0) = a, \quad \upsilon(1/\mu) = 0, \dots$$

has solution vðtÞ ¼ aXðtÞ, where

$$X(t) = t \int\_{t}^{\mu^{-1}} s^{-2} \exp\left(\frac{-s^2}{2}\right) ds, \ 0 \le X(t) \le 1, \quad X(0) = 1.$$

The proof of Lemma 3 is obvious.

Lemma 4. In order to solve the boundary value problem

$$L\_0 W = -\mu t \,\, \mathcal{W}(0) = \mathcal{W}(\mu^{-1}) = 0,$$

we have the estimate

$$0 \le \mathcal{W}(\mu, t) \le e^{-1} \ln \mu^{-1}.$$

Proof. This follows from the fact that the solution of this problem existsuniquely by the maximum principle [23, 82] and will be represented in the form

$$\mathcal{W}(\mu, t) = \mu t \int\_{t}^{\mu^{-1}} y^{-2} \exp\left(-\frac{y^2}{2}\right) \int\_{0}^{y} s^2 \exp\left(\frac{s^2}{2}\right) ds d\mathbf{y} \dots$$

Lemma 5. The estimate

$$|\pi\_k(\mu, t)| < B\_{k\nu}$$

where 0 < Bkis constant.

Proof. Consider the function

$$\mathcal{V}\_{\pm}(\mu, t) = \gamma\_1 \mathcal{W}(\mu, t) + \gamma\_2 \mathcal{X}(t) \pm \pi\_k(\mu, t),$$

where γ<sup>1</sup> and γ<sup>2</sup> are positive constants such that

$$\mathcal{V}\_1 > \max\_{\left[0,1\right]} \left| p(\mathfrak{x}) \right| \, \_\prime \mathcal{V}\_2 > \left| z\_k(0) \right|.$$

It is obvious that

$$V\_{\pm}(\mu,0) > 0, \ V\_{\pm}(\mu,\mu^{-1}) > 0, \ L\_0 V\_{\pm} \equiv V\_{\pm}(t) + tV\_{\pm}(t) - V\_{\pm}(t) < 0$$

From the maximum principle, it follows that jπkðμ, tÞj < γ1Wðμ, tÞ þ γ2XðtÞ: Now the proof of the lemma 5 follows from estimates of Wðμ, tÞ and XðtÞ. If we introduce the notation

$$Y\_n(\mathbf{x}, \varepsilon) = \sum\_{k=0}^n \varepsilon^k \left( z\_k(\mathbf{x}) + \pi\_k(\boldsymbol{\mu}, t) \right),$$

where zkðxÞ, πkðμ, tÞare constructed above functions, then

$$L\_{\varepsilon}Y\_{n}(\mathfrak{x},\mathfrak{e}) = f(\mathfrak{x}) + \varepsilon^{n+1}z''\_{n}\mathfrak{e}$$

Let yðx, εÞbe the solution of the problems (21) and (22). Then

$$\left| L\_{\varepsilon} \left( Y\_n(\mathbf{x}, \varepsilon) - y(\mathbf{x}, \varepsilon) \right) \right| < B\_n \varepsilon^{n+1}, \quad Y\_n(0, \varepsilon) - y(0, \varepsilon) = Y\_n(1, \varepsilon) - y(1, \varepsilon) = 0.$$

Therefore, <sup>j</sup>Ynðx, <sup>ε</sup>Þ � <sup>y</sup>ðx, <sup>ε</sup>Þj <sup>&</sup>lt; Bnε<sup>n</sup>þ<sup>1</sup>:

#### 2.4. The bisingular problem of Cole equation with a weak singularity

The following problem is considered [9, 13, 28, 29],

$$
\epsilon y''(\mathbf{x}) + \sqrt{\mathbf{x}} y'(\mathbf{x}) - y(\mathbf{x}) = \mathbf{0}, \ \mathbf{0} < \mathbf{x} < 1,\tag{27}
$$

$$y(0) = a, \ y(1) = b \tag{28}$$

where x∈ ½0, 1�; a, b are the given constants.

The unperturbed equation ffiffiffi x p y<sup>0</sup> ðxÞ � yðxÞ ¼ 0, 0 < x < 1, has the general solution

$$y\_0(x) = c e^{2\sqrt{x}}, \ c - \text{const.}$$

This is a nonsmooth function in ½0, 1�.

We seek asymptotic representation of the solution of the problems (27) and (28) in the form:

$$y(\mathbf{x}) = \sum\_{k=0}^{n} \varepsilon^{k} y\_{k}(\mathbf{x}) + \sum\_{k=0}^{3(n+1)} \mu^{k} \pi\_{k}(t) + \mathcal{R}(\mathbf{x}, \varepsilon), \tag{29}$$

where <sup>t</sup> <sup>¼</sup> <sup>x</sup>=μ<sup>2</sup>, <sup>ε</sup> <sup>¼</sup> <sup>μ</sup><sup>3</sup>, ykðxÞ∈C½0, <sup>1</sup>�, <sup>π</sup>kðt<sup>Þ</sup> <sup>∈</sup>C½0, <sup>1</sup>=μ<sup>2</sup>�, Rðx, <sup>ε</sup><sup>Þ</sup> is the reminder term.

Substituting Eq. (29) into Eq. (27), we have

$$\begin{aligned} \sum\_{k=0}^{n} \varepsilon^{k} (\varepsilon \mathbf{y}\_{\cdot k}^{\star}(\mathbf{x}) + \sqrt{\mathbf{x}} \mathbf{y}\_{\cdot k}^{\prime}(\mathbf{x}) - y\_{k}(\mathbf{x})) + \frac{1}{\mu} \Big( \pi^{\star} \boldsymbol{o}(t) + \sqrt{\ell} \pi \mathbf{r}\_{\cdot}^{\prime} \boldsymbol{o}(t) \Big) \\ + \sum\_{k=1}^{3(n+1)} \mu^{k-1} \Big( \pi^{\star} \boldsymbol{k}(t) + \sqrt{\ell} \pi \mathbf{r}\_{\cdot}^{\prime}(t) - \pi\_{k-1}(t) \Big) - \mu^{3(n+1)} \pi\_{3(n+1)}(t) + \varepsilon \mathsf{R}^{\boldsymbol{\eta}}(\mathbf{x}, \varepsilon) + \sqrt{\mathsf{x}} \mathsf{R}^{\boldsymbol{\prime}}(\mathbf{x}, \varepsilon) \Big) \\ - \mathsf{R}(\mathbf{x}, \varepsilon) - h(\mathbf{x}, \varepsilon) + h(\mathbf{x}, \varepsilon) = 0 \end{aligned} \tag{30}$$

By the method of generalized boundary layer function, we put the term <sup>h</sup>ðx, <sup>ε</sup>Þ ¼ <sup>X</sup><sup>n</sup>�<sup>1</sup> k¼0 εk hkðxÞ into the equation. We choose functions hkðxÞ so that ykðxÞ∈C½0, 1�.

Taking into account the boundary condition (28), from Eq. (30), we obtain

$$
\sqrt{\mathbf{x}} y\_0'(\mathbf{x}) - y\_0(\mathbf{x}) = \mathbf{0}, \quad \mathbf{0} < \mathbf{x} < \mathbf{1}, \quad y\_0(\mathbf{1}) = b. \tag{31}
$$

$$
\sqrt{\mathbf{x}} \boldsymbol{y}\_k'(\mathbf{x}) - \boldsymbol{y}\_k(\mathbf{x}) = \boldsymbol{h}\_{k-1}(\mathbf{x}) - \boldsymbol{y}\_{k-1}^\*(\mathbf{x}), \quad 0 < \mathbf{x} < 1, \ k \in \mathbb{N}, \quad \boldsymbol{y}\_k(1) = 0. \tag{32}
$$

The solution of the problems (31) and (32) exists. It is unique and has the form

$$y\_0(\mathbf{x}) = b e^{2(\sqrt{\mathbf{x}}-1)} , \\ y\_k(\mathbf{x}) = e^{2\sqrt{\mathbf{x}}} \Big|\_{1}^{\mathbf{x}} \frac{h\_{k-1}(\mathbf{s}) - y\_{\mathbf{s}-1}^{\mathbf{r}}(\mathbf{s})}{\sqrt{\mathbf{s}}} e^{-2\sqrt{\mathbf{s}}} d\mathbf{s} \quad k \in \mathcal{N}.$$

We choose indefinite functions hk(x) as follows: y<sup>00</sup> <sup>k</sup>�<sup>1</sup>ðxÞ � hk�<sup>1</sup>ðx<sup>Þ</sup> <sup>∈</sup>C½0, <sup>1</sup>�. We can represent

$$y\_0(\mathbf{x}) = b e^{-2} \left( 1 + 2\sqrt{\mathbf{x}} + \frac{\left( 2\sqrt{\mathbf{x}} \right)^2}{2!} + \frac{\left( 2\sqrt{\mathbf{x}} \right)^3}{3!} + \frac{\left( 2\sqrt{\mathbf{x}} \right)^4}{4!} + \dots + \frac{\left( 2\sqrt{\mathbf{x}} \right)^n}{n!} + \dots \right),$$

Let <sup>h</sup>1ðxÞ ¼ be�<sup>2</sup> <sup>2</sup> ffiffiffi <sup>x</sup> <sup>p</sup> <sup>þ</sup> <sup>ð</sup><sup>2</sup> ffiffi x p Þ 3 3! � �″ ¼ �be�<sup>2</sup> <sup>1</sup> <sup>2</sup> ffiffiffi <sup>x</sup><sup>3</sup> <sup>p</sup> � <sup>1</sup>ffiffi x p � �.

Then

V�ðμ, tÞ ¼ γ1Wðμ, tÞ þ γ2XðtÞ � πkðμ, tÞ,

<sup>Þ</sup> <sup>&</sup>gt; <sup>0</sup>, L0V� � <sup>V</sup>″

jpðxÞj, γ<sup>2</sup> > jzkð0Þj:

�ðtÞ þ tV<sup>0</sup>

� ,

, Ynð0, εÞ � yð0, εÞ ¼ Ynð1, εÞ � yð1, εÞ ¼ 0:

ðxÞ � yðxÞ ¼ 0, 0 < x < 1, ð27Þ

yð0Þ ¼ a, yð1Þ ¼ b ð28Þ

zkðxÞ þ πkðμ, tÞ

z″ n: �ðtÞ � V�ðtÞ < 0

γ<sup>1</sup> > max <sup>½</sup><sup>0</sup>, <sup>1</sup>�

From the maximum principle, it follows that jπkðμ, tÞj < γ1Wðμ, tÞ þ γ2XðtÞ:

Now the proof of the lemma 5 follows from estimates of Wðμ, tÞ and XðtÞ.

Ynðx, <sup>ε</sup>Þ ¼ <sup>X</sup><sup>n</sup>

where zkðxÞ, πkðμ, tÞare constructed above functions, then

Let yðx, εÞbe the solution of the problems (21) and (22). Then

�

<sup>j</sup> <sup>&</sup>lt; Bnε<sup>n</sup>þ<sup>1</sup>

2.4. The bisingular problem of Cole equation with a weak singularity

<sup>x</sup> <sup>p</sup> <sup>y</sup><sup>0</sup>

ðxÞ � yðxÞ ¼ 0, 0 < x < 1,

x p

, c � const:

<sup>y</sup>0ðxÞ ¼ ce<sup>2</sup> ffiffi

<sup>ε</sup>y00ðxÞ þ ffiffiffi

x p y<sup>0</sup>

Ynðx, εÞ � yðx, εÞ

The following problem is considered [9, 13, 28, 29],

where x∈ ½0, 1�; a, b are the given constants.

The unperturbed equation ffiffiffi

This is a nonsmooth function in ½0, 1�.

has the general solution

Therefore, <sup>j</sup>Ynðx, <sup>ε</sup>Þ � <sup>y</sup>ðx, <sup>ε</sup>Þj <sup>&</sup>lt; Bnε<sup>n</sup>þ<sup>1</sup>:

k¼0 εk �

<sup>L</sup>εYnðx, <sup>ε</sup>Þ ¼ <sup>f</sup>ðxÞ þ <sup>ε</sup><sup>n</sup>þ<sup>1</sup>

where γ<sup>1</sup> and γ<sup>2</sup> are positive constants such that

<sup>V</sup>�ðμ, <sup>0</sup><sup>Þ</sup> <sup>&</sup>gt; <sup>0</sup>, V�ðμ, <sup>μ</sup>�<sup>1</sup>

It is obvious that

16 Recent Studies in Perturbation Theory

If we introduce the notation

jLε �

$$y\_0''(\mathbf{x}) - h\_0(\mathbf{x}) \in \mathbb{C}[0, 1], \mu^3 h\_1(t\mu^2) = -c\_1 \left(\frac{1}{2\sqrt{t^3}} - \frac{\mu^2}{\sqrt{t}}\right), \ c\_1 = b e^{-2}\mu,$$

$$y\_1(\mathbf{x}) = c\_1 e^{2\sqrt{\mathbf{x}}} \int\_{-1}^{\mathbf{x}} \left(-\frac{1}{2s^2} + \frac{1}{s} + \frac{1}{2s^2} e^{2\sqrt{s}} - \frac{1}{\sqrt{s^3}} e^{2\sqrt{s}}\right) e^{-2\sqrt{s}} ds.$$

We can rewrite y1(x) in the form:

$$y\_1(\mathbf{x}) = y\_{1,0} + y\_{1,1}(2\sqrt{\mathbf{x}}) + y\_{1,2}(2\sqrt{\mathbf{x}})^2 + y\_{1,3}(2\sqrt{\mathbf{x}})^3 + \dots$$

where <sup>y</sup>1, <sup>0</sup> <sup>¼</sup> <sup>3</sup> <sup>2</sup> <sup>þ</sup> <sup>1</sup> 2e<sup>2</sup> � �c1, y1, <sup>1</sup> <sup>¼</sup> <sup>1</sup> <sup>6</sup> <sup>þ</sup> <sup>1</sup> 2e<sup>2</sup> � �c1, y1, <sup>2</sup> <sup>¼</sup> �<sup>1</sup> <sup>6</sup> <sup>þ</sup> <sup>1</sup> 4e<sup>2</sup> � �c1, y1, <sup>3</sup> <sup>¼</sup> �<sup>1</sup> <sup>10</sup> <sup>þ</sup> <sup>1</sup> 12e<sup>2</sup> � �c1:

Analogously, we have obtained

$$h\_1(\mathbf{x}) = \left( y\_{1,1}(2\sqrt{\mathbf{x}}) + y\_{1,3}(2\sqrt{\mathbf{x}})^3 \right)^\* = -\frac{y\_{1,1}}{2\sqrt{\mathbf{x}^3}} + \frac{6y\_{1,3}}{\sqrt{\mathbf{x}}} \dots$$

Then

$$y\_{"2}^{\prime\prime}(\mathbf{x}) - h\_2(\mathbf{x}) \in \mathbb{C}[0, 1], \mu^6 h\_2(t\mu^2) = -\frac{\mu^3 y\_{1,1}}{2\sqrt{t^3}} + \frac{\mu^5 y\_{1,3}}{\sqrt{t}}.$$

Continuing this process, we have

$$h\_{k-1}(\mathbf{x}) = -\frac{y\_{k-1,1}}{2\sqrt{\mathbf{x}^3}} + \frac{6y\_{k-1,3}}{\sqrt{\mathbf{x}}}, k = 4, \dots, n\_1$$

where yk�1, <sup>1</sup>, yk�1,<sup>3</sup> are corresponding coefficients of the expansion of yk�1,1ðx<sup>Þ</sup> in powers of (2 ffiffiffi x p ).

From Eq. (30), we have the following equations for the boundary functions πkðtÞ:

$$L\pi\_0 \equiv \pi\_{\ 0}^\*(t) + \sqrt{t}\pi\_{\ 0}^\prime(t) = 0, \quad 0 < t < \bar{\mu}, \quad \pi\_0(0) = a - y\_0(0), \ \pi\_0(\bar{\mu}) = 0, \ \bar{\mu} = 1/\mu^2,\tag{33}$$

$$L\pi\_{3k+1}(t) = \pi\_{3k}(t) + \frac{y\_{k,1}}{2\sqrt{t^3}}, \quad 0 < t < \tilde{\mu}, \quad \pi\_{3k+1}(0) = 0, \quad \pi\_{3k+1}(\tilde{\mu}) = 0, \quad k = 0, 1, \ldots, n \quad (34)$$

$$L\pi\_{3k+2}(t) = \pi\_{3k+1}(t), \quad 0 < t < \tilde{\mu}, \quad \pi\_{3k+2}(0) = 0, \quad \pi\_{3k+2}(\tilde{\mu}) = 0, \quad k = 0, 1, \ldots, n \tag{35}$$

$$L\pi\_{3k+3} = \pi\_{3k+2}(t) - \frac{y\_{k,3}}{\sqrt{t}}, \quad 0 < t < \tilde{\mu}, \quad \pi\_{3k}(0) = -y\_k(0), \quad \pi\_{3k}(\tilde{\mu}) = 0, \quad k = 0, 1, \ldots, n-1 \tag{36}$$

$$L\pi\_{3(n+1)}(t) = \pi\_{3n+2}(t) - \frac{y\_{n,3}}{\sqrt{t}}, \quad 0 < t < \ddot{\mu}, \quad \pi\_{3n}(0) = 0, \quad \pi\_{3n}(\ddot{\mu}) = 0 \tag{37}$$

The solution of problem (33) is represented in the form

$$\pi\_0(t) = (a - be^{-2})A \int\_t^{\tilde{\mu}} e^{-\frac{2}{\tilde{\sigma}}s^{3/2}} ds, \quad A = \left( \int\_0^{\tilde{\mu}} e^{-\frac{2}{\tilde{\sigma}}s^{3/2}} ds \right)^{-1}.$$

We note that π0ðtÞ will exponentially decrease as t ! μ~.

Lemma 6. The general solution of this equation LzðtÞ ¼ 0 will have zðtÞ ¼ c1YðtÞ þ c2XðtÞ; here c1, c<sup>2</sup> are constants, and

$$Y(t) = 1 - X(t),\ \ X(t) = \alpha \int\_{-t}^{\tilde{\mu}} e^{-\frac{2}{3}\tilde{\mathfrak{s}}^{3/2}} ds \ (\quad \alpha \int\_{-0}^{\tilde{\mu}} e^{-\frac{2}{3}\tilde{\mathfrak{s}}^{3/2}} ds = 1).$$

Two linearly independent solutions and YðtÞ ¼ OðtÞ, t ! 0, 0 < XðtÞ ≤ 1,

$$X(t) \ = t^{-\frac{1}{2}} e^{\frac{2}{3}t^{3/2}} \left( 1 - \frac{1}{2} t^{-\frac{3}{2}} + \dots + \frac{(-1)^{\mathfrak{t}}}{2^{\mathfrak{t}}} \prod\_{k=1}^{\mathfrak{n}} 1 \cdot 4 \cdot \dots \cdot (3k - 2) t^{-\frac{3\mathfrak{n}}{2}} + \dots \right), \quad t \to \tilde{\mu} \tag{38}$$

Lemma 7. The boundary problem LzðtÞ ¼ 0, zð0Þ ¼ zðμ~Þ ¼ 0 will have only trivial solution.

The proofs of Lemmas 6 and 7 are evident.

Theorem 4. The problem

<sup>y</sup>1ðxÞ ¼ <sup>y</sup>1, <sup>0</sup> <sup>þ</sup> <sup>y</sup>1,1ð<sup>2</sup> ffiffiffi

<sup>6</sup> <sup>þ</sup> <sup>1</sup> 2e<sup>2</sup> � �

<sup>y</sup>1,1ð<sup>2</sup> ffiffiffi

<sup>2</sup>ðxÞ � <sup>h</sup>2ðxÞ<sup>∈</sup> <sup>C</sup>½0, <sup>1</sup>�, <sup>μ</sup><sup>6</sup>

hk�<sup>1</sup>ðx޼� yk�1, <sup>1</sup>

2 ffiffiffiffiffi <sup>x</sup><sup>3</sup> <sup>p</sup> <sup>þ</sup>

From Eq. (30), we have the following equations for the boundary functions πkðtÞ:

ffiffi

ÞA ð μ~

t e �2 3s3=<sup>2</sup>

<sup>c</sup>1, y1, <sup>1</sup> <sup>¼</sup> <sup>1</sup>

�

h1ðxÞ ¼

y00

where <sup>y</sup>1, <sup>0</sup> <sup>¼</sup> <sup>3</sup>

Then

(2 ffiffiffi x p ).

<sup>L</sup>π<sup>0</sup> � <sup>π</sup>″

<sup>2</sup> <sup>þ</sup> <sup>1</sup> 2e<sup>2</sup> � �

Analogously, we have obtained

18 Recent Studies in Perturbation Theory

Continuing this process, we have

<sup>0</sup>ðtÞ þ ffiffi t <sup>p</sup> <sup>π</sup><sup>0</sup>

> 2 ffiffiffiffi t

> > ffiffi

<sup>L</sup>π<sup>3</sup>ðnþ<sup>1</sup>ÞðtÞ ¼ <sup>π</sup>3nþ<sup>2</sup>ðtÞ � yn,<sup>3</sup>

The solution of problem (33) is represented in the form

<sup>π</sup>0ðtÞ¼ð<sup>a</sup> � be�<sup>2</sup>

We note that π0ðtÞ will exponentially decrease as t ! μ~.

<sup>L</sup>π3kþ<sup>1</sup>ðtÞ ¼ <sup>π</sup>3kðtÞ þ yk, <sup>1</sup>

<sup>L</sup>π3kþ<sup>3</sup> <sup>¼</sup> <sup>π</sup>3kþ<sup>2</sup>ðtÞ � yk,<sup>3</sup>

c1, c<sup>2</sup> are constants, and

<sup>x</sup> <sup>p</sup> Þ þ <sup>y</sup>1, <sup>2</sup>ð<sup>2</sup> ffiffiffi

<sup>c</sup>1, y1, <sup>2</sup> <sup>¼</sup> �<sup>1</sup>

x p Þ 3 �″

<sup>h</sup>2ðtμ<sup>2</sup>

<sup>6</sup>yk�1, <sup>3</sup> ffiffiffi

where yk�1, <sup>1</sup>, yk�1,<sup>3</sup> are corresponding coefficients of the expansion of yk�1,1ðx<sup>Þ</sup> in powers of

<sup>x</sup> <sup>p</sup> Þ þ <sup>y</sup>1, <sup>3</sup>ð<sup>2</sup> ffiffiffi

x p Þ

<sup>6</sup> <sup>þ</sup> <sup>1</sup> 4e<sup>2</sup> � �

<sup>2</sup> <sup>þ</sup> <sup>y</sup>1,3ð<sup>2</sup> ffiffiffi

¼ � <sup>y</sup>1, <sup>1</sup> 2 ffiffiffiffiffi <sup>x</sup><sup>3</sup> <sup>p</sup> <sup>þ</sup>

޼� <sup>μ</sup><sup>3</sup>y1,<sup>1</sup> 2 ffiffiffiffi t <sup>3</sup> <sup>p</sup> <sup>þ</sup> <sup>μ</sup><sup>5</sup>y1,<sup>3</sup> ffiffi

<sup>x</sup> <sup>p</sup> , k <sup>¼</sup> <sup>4</sup>, …, n,

<sup>0</sup>ðtÞ ¼ <sup>0</sup>, <sup>0</sup> <sup>&</sup>lt; <sup>t</sup> <sup>&</sup>lt; <sup>μ</sup>~, <sup>π</sup>0ð0Þ ¼ <sup>a</sup> � <sup>y</sup>0ð0Þ, <sup>π</sup>0ðμ~Þ ¼ <sup>0</sup>, <sup>μ</sup><sup>~</sup> <sup>¼</sup> <sup>1</sup>=μ<sup>2</sup>

Lπ3kþ<sup>2</sup>ðtÞ ¼ π3kþ<sup>1</sup>ðtÞ, 0 < t < μ~, π3kþ<sup>2</sup>ð0Þ ¼ 0, π3kþ<sup>2</sup>ðμ~Þ ¼ 0, k ¼ 0, 1, …, n ð35Þ

ds, A ¼

Lemma 6. The general solution of this equation LzðtÞ ¼ 0 will have zðtÞ ¼ c1YðtÞ þ c2XðtÞ; here

<sup>3</sup> <sup>p</sup> , <sup>0</sup> <sup>&</sup>lt; <sup>t</sup> <sup>&</sup>lt; <sup>μ</sup>~, <sup>π</sup>3kþ<sup>1</sup>ð0Þ ¼ <sup>0</sup>, <sup>π</sup>3kþ<sup>1</sup>ðμ~Þ ¼ <sup>0</sup>, k <sup>¼</sup> <sup>0</sup>, <sup>1</sup>, …, n <sup>ð</sup>34<sup>Þ</sup>

<sup>t</sup> <sup>p</sup> , <sup>0</sup> <sup>&</sup>lt; <sup>t</sup> <sup>&</sup>lt; <sup>μ</sup>~, <sup>π</sup>3kð0޼�ykð0Þ, <sup>π</sup>3kðμ~Þ ¼ <sup>0</sup>, k <sup>¼</sup> <sup>0</sup>, <sup>1</sup>, …, n–<sup>1</sup> <sup>ð</sup>36<sup>Þ</sup>

ð μ~

0 B@

0 e �2 3s3=<sup>2</sup> ds

<sup>t</sup> <sup>p</sup> , <sup>0</sup> <sup>&</sup>lt; <sup>t</sup> <sup>&</sup>lt; <sup>μ</sup>~, <sup>π</sup>3nð0Þ ¼ <sup>0</sup>, <sup>π</sup>3nðμ~Þ ¼ <sup>0</sup> <sup>ð</sup>37<sup>Þ</sup>

1 CA

�1

:

x p Þ

<sup>c</sup>1, y1, <sup>3</sup> <sup>¼</sup> �<sup>1</sup>

<sup>3</sup> <sup>þ</sup> …,

6y1,<sup>3</sup> ffiffiffi <sup>x</sup> <sup>p</sup> :

<sup>t</sup> <sup>p</sup> :

<sup>10</sup> <sup>þ</sup> <sup>1</sup> 12e<sup>2</sup> � �

c1:

, ð33Þ

$$Lz(t) = f(t), \ z(0) = 0, \ z(\tilde{\mu}) = 0.$$

will have the unique solution and this one has the next form

$$z(t) = \int\_{0}^{\bar{\mu}} G(t, s) e^{\frac{2}{3}t^{3/2}} f(s) ds,$$

and <sup>G</sup>ðt, sÞ ¼ �YðtÞXðsÞ, <sup>0</sup> <sup>≤</sup> <sup>t</sup> <sup>≤</sup> s, �YðsÞXðtÞ, s ≤ t ≤ μ~, �

is the function of Green andfðtÞ∈ Cð0, μ~� .

Theorem 4 implies the existence and uniqueness of the solution of problem (34)–(37): jπkðtÞj < l ¼ const, t∈ ½0, μ~�.

Lemma 8. Asymptotical expansions of functions πkðtÞ, t ! μ~ (k ¼ 1, 2, …) will have the next forms

$$\pi\_1(t) = -\frac{y\_{0,1}}{2t} \left( 1 + \frac{4}{5\sqrt{t^3}} + \frac{7}{4t^3} + \frac{42}{11\sqrt{t^9}} + \frac{39}{2t^7} + \dots \right),$$

$$\pi\_2(t) = \frac{y\_{0,1}}{\sqrt{t}} \left( 1 + \frac{23}{40\sqrt{t^3}} + \frac{173}{2t^3} + \dots \right), \\ \pi\_3(t) = -\frac{23y\_{0,1}}{60\sqrt{t^3}} + O\left(\frac{1}{t^3}\right),$$

$$\pi\_{3k+1}(t) = t^{-1} \sum\_{j=0}^{\infty} l\_{3k+1,j} t^{-\frac{2j}{2}}, \\ \pi\_{3k+2}(t) = t^{-1/2} \sum\_{j=0}^{\infty} l\_{3k+2,j} t^{-\frac{2j}{2}}, \\ \pi\_{3k}(t) = \sum\_{j=1}^{\infty} l\_{3k,j} t^{-\frac{2j}{2}}.$$

Proof for Lemma 8.

Firs proof. We can prove this lemma by applying formulas (38) and Theorem 4.

Second proof. We can receive these representations from Eqs. (34)–(37) directly.

Now we will prove the boundedness of the reminder function Rðx, εÞ. This function will satisfy the next equation:

$$
\varepsilon \mathbb{R}''(\mathbf{x}, \varepsilon) + \sqrt{\pi} \mathbb{R}'(\mathbf{x}, \varepsilon) - \mathbb{R}(\mathbf{x}, \varepsilon) = \mu^{3(n+1)} \pi\_{3(n+1)}(t) + \varepsilon^{n+1} (h\_n(\mathbf{x}) - \boldsymbol{y}^\*\_n(\mathbf{x})),
$$

$$
\mathbb{R}(\mathbf{0}, \varepsilon) = \mathbf{0}, \ \mathbb{R}(\mathbf{1}, \varepsilon) = \mathbf{0}.
$$

Applying to this problem theorem [23, p.117, 82], we obtained

$$|\mathcal{R}(\mathbf{x},\varepsilon)| \le \varepsilon^{n+1} \mathbb{C} \max\_{\substack{0 \le \mathbf{x} \le 1 \\ 0 \le \mathbf{x} \le 1 \\ 0 \le t \le \tilde{\mu}}} |\pi\_{3(n+1)}(t) + h\_n(\mathbf{x}) - \boldsymbol{y}^\*\_{\boldsymbol{n}}(\mathbf{x})|.$$

Therefore, we have <sup>R</sup>ðx, <sup>ε</sup>Þ ¼ <sup>O</sup>ðε<sup>n</sup>þ<sup>1</sup>Þ, <sup>ε</sup> ! <sup>0</sup>, x<sup>∈</sup> <sup>½</sup>0, <sup>1</sup>�.

We prove next.

Theorem 5. The asymptotical expansion of the solution of the problems (27) and (28) and will have the next form

$$y(\mathbf{x}) = \sum\_{k=0}^{n} \varepsilon^k y\_k(\mathbf{x}) + \sum\_{k=0}^{3(n+1)} \mu^k \pi\_k(t) + \mathcal{O}(\varepsilon^{n+1}), \quad \varepsilon \to 0.$$

## 3. Singularly perturbed differential equations Lighthill type

#### 3.1. The idea of the method of Poincare

Consider the equation

$$M y(\mathbf{x}) := y''(\mathbf{x}) + y(\mathbf{x}) - \varepsilon y^3(\mathbf{x}) = 0. \tag{39}$$

Unperturbed equation has solutions y0ðxÞ ¼ a<sup>1</sup> cos x þ b<sup>1</sup> sin x (where a1, b<sup>1</sup> are arbitrary constants) with period 2π. We are looking for the periodic solution of the equation yðx, εÞ with a period of ωðεÞ ¼ ωð0Þ ¼ 2π.

Note that the operator <sup>M</sup> transforms Fourier series <sup>X</sup><sup>∞</sup> k¼1 ak cos kx and <sup>X</sup><sup>∞</sup> k¼1 ak sin kx in itself.

Poincare's method reduces the existence of periodic solutions of differential equations to the existence of the solution of an algebraic equation.

We will seek a periodic solution of Eq. (39) with the initial condition

$$y(0) = 1, \ y'(0) = 0.$$

If we seek the solution in the form

$$y(\mathbf{x}) = y\_0(\mathbf{x}) + \varepsilon y\_1(\mathbf{x}) + \varepsilon^2 y\_2(\mathbf{x}) + \dots$$

with the initial conditions

Perturbed Differential Equations with Singular Points http://dx.doi.org/10.5772/67856 21

$$y\_0(0) = 1, \ y\_0'(1) = 0, \ y\_k(0) = y\_k'(1) = 0, \ k = 1, 2, \dots$$

then for ysðxÞ, s ¼ 0, 1, … we have next equations

$$L y\_0 := y\_{0}^{\prime\prime}(\mathbf{x}) + y\_{0}(\mathbf{x}) = 0 \Rightarrow y\_{0}(\mathbf{x}) = \cos \mathbf{x}$$

$$L y\_1 = \cos^3 \mathbf{x} = \frac{3}{4} \cos \mathbf{x} + \frac{1}{4} \cos 3 \mathbf{x} \Rightarrow y\_1(\mathbf{x}) = \frac{3}{8} \mathbf{x} \sin \mathbf{x} - \frac{1}{32} \cos 3 \mathbf{x} + \frac{1}{32} \cos 3 \mathbf{x}$$

Thus, <sup>y</sup>ðxÞ ¼ cos <sup>x</sup> <sup>þ</sup> <sup>ε</sup> <sup>8</sup> <sup>3</sup><sup>x</sup> sin <sup>x</sup> � <sup>1</sup> <sup>4</sup> cos 3<sup>x</sup> <sup>þ</sup> <sup>1</sup> <sup>4</sup> cos <sup>x</sup> � � <sup>þ</sup> …it is not a uniform expansion of the <sup>y</sup> (x) on the segment ½�∞, ∞�, since the term εx sin x is present here.

If these secular terms do not appear in Eq. (39), it is necessary to make the substitution

$$\mathbf{x} = t(1 + \varepsilon \alpha\_1 + \varepsilon^2 \alpha\_2 + \dots)$$

where the constant α<sup>k</sup> should be selected so as not to have secular terms in t.

Thus, the solution of Eq. (39) must be sought in the form

$$\begin{aligned} y(t) &= y\_0(t) + \varepsilon y\_1(t) + \varepsilon^2 y\_2(t) + \dots \\ \mathbf{x} &= t(1 + \varepsilon a\_1 + \varepsilon^2 a\_2 + \dots) \end{aligned} \tag{40}$$

Then Eq. (39) has the form

$$z''(t) + (1 + \alpha\_1 \varepsilon + \alpha\_2 \varepsilon^2 + \dots)z(t) = \varepsilon(1 + \alpha\_1 \varepsilon + \alpha\_2 \varepsilon^2 + \dots)z^3(t)$$

where yðwðεÞtÞ ¼ zðtÞ.

We will seek the 2π periodic solution of this equation in the form

$$z(t) = z\_0(t) + \varepsilon z\_1(t) + \varepsilon^2 z\_2(t) + \dots$$

Then

<sup>ε</sup>R00ðx, <sup>ε</sup>Þ þ ffiffiffi

20 Recent Studies in Perturbation Theory

We prove next.

have the next form

Consider the equation

period of ωðεÞ ¼ ωð0Þ ¼ 2π.

If we seek the solution in the form

with the initial conditions

<sup>x</sup> <sup>p</sup> <sup>R</sup><sup>0</sup>

Applying to this problem theorem [23, p.117, 82], we obtained

<sup>j</sup>Rðx, <sup>ε</sup>Þj <sup>≤</sup> <sup>ε</sup><sup>n</sup>þ<sup>1</sup>

Therefore, we have <sup>R</sup>ðx, <sup>ε</sup>Þ ¼ <sup>O</sup>ðε<sup>n</sup>þ<sup>1</sup>Þ, <sup>ε</sup> ! <sup>0</sup>, x<sup>∈</sup> <sup>½</sup>0, <sup>1</sup>�.

<sup>y</sup>ðxÞ ¼ <sup>X</sup><sup>n</sup>

3.1. The idea of the method of Poincare

k¼0 εk ykðxÞ þ

Note that the operator <sup>M</sup> transforms Fourier series <sup>X</sup><sup>∞</sup>

We will seek a periodic solution of Eq. (39) with the initial condition

existence of the solution of an algebraic equation.

<sup>ð</sup>x, <sup>ε</sup>Þ � <sup>R</sup>ðx, <sup>ε</sup>Þ ¼ <sup>μ</sup><sup>3</sup>ðnþ1<sup>Þ</sup>

C max 0 ≤ x ≤ 1 0 ≤ t ≤ μ~

Rð0, εÞ ¼ 0, Rð1, εÞ ¼ 0:

Theorem 5. The asymptotical expansion of the solution of the problems (27) and (28) and will

3ð X<sup>n</sup>þ1<sup>Þ</sup> k¼0 μk

Myðx<sup>Þ</sup> :<sup>¼</sup> <sup>y</sup>00ðxÞ þ <sup>y</sup>ðxÞ � <sup>ε</sup>y<sup>3</sup>

Unperturbed equation has solutions y0ðxÞ ¼ a<sup>1</sup> cos x þ b<sup>1</sup> sin x (where a1, b<sup>1</sup> are arbitrary constants) with period 2π. We are looking for the periodic solution of the equation yðx, εÞ with a

Poincare's method reduces the existence of periodic solutions of differential equations to the

<sup>y</sup>ðxÞ ¼ <sup>y</sup>0ðxÞ þ <sup>ε</sup>y1ðxÞ þ <sup>ε</sup><sup>2</sup>y2ðxÞ þ …

yð0Þ ¼ 1, y<sup>0</sup>

3. Singularly perturbed differential equations Lighthill type

<sup>π</sup><sup>3</sup>ðnþ<sup>1</sup>ÞðtÞ þ <sup>ε</sup><sup>n</sup>þ<sup>1</sup>

<sup>j</sup>π<sup>3</sup>ðnþ<sup>1</sup>ÞðtÞ þ hnðxÞ � <sup>y</sup>″

<sup>π</sup>kðtÞ þ <sup>O</sup>ðε<sup>n</sup>þ<sup>1</sup>

k¼1

ð0Þ ¼ 0:

<sup>ð</sup>hnðxÞ � <sup>y</sup>″

<sup>n</sup>ðxÞj:

Þ, ε ! 0:

ak cos kx and <sup>X</sup><sup>∞</sup>

ðxÞ ¼ 0: ð39Þ

k¼1

ak sin kx in itself.

<sup>n</sup>ðxÞÞ,

$$Lz\_0 := z\_0^{\prime\prime}(t) + z\_0(t) = 0 \Rightarrow z\_0(t) = \cos t.$$

$$Lz\_1(t) = \alpha\_1 \cos t + \frac{3}{4} \cos t + \frac{1}{4} \cos 3t.$$

The function Z1(t) will have the periodical solution we take <sup>α</sup><sup>1</sup> ¼ �3=4. Then <sup>z</sup>1ðt޼� <sup>1</sup> <sup>32</sup> cos 3t. Similarly, from equations

$$\alpha z\_n(t) = -\alpha\_n \cos t + \operatorname{g}(\alpha\_1, \alpha\_2, \dots, \alpha\_{n-1}) \cos t + \sum\_{m=1}^{2n+1} \beta\_n \cos mt$$

α<sup>n</sup> and etc. are uniquely determined.

Theorem 6. Equation (39) has a unique 2π=ω periodic solution, and it can be represented in the form (40).

#### 3.2. The idea of the Lighthill method

Lighthill in 1949 [67] reported an important generalization of the method of Poincare.

He considered the model equation [67, 82]:

$$(\mathbf{x} + \varepsilon y(\mathbf{x}))y'(\mathbf{x}) + q(\mathbf{x})y(\mathbf{x}) = r(\mathbf{x}), \quad y(\mathbf{1}) = a \tag{41}$$

where <sup>x</sup><sup>∈</sup> <sup>½</sup>0, <sup>1</sup>�qðxÞ, rðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�.

Lighthill proposed to seek the solution of Eq. (41) in the form

$$\begin{aligned} y(\xi) &= y\_0(\xi) + \varepsilon y\_1(\xi) + \varepsilon^2 y\_2(\xi) + \dots \\ \mathbf{x} &= \xi + \varepsilon \mathbf{x}\_1(\xi) + \varepsilon^2 \mathbf{x}\_2(\xi) + \dots \end{aligned} \tag{42}$$

It is obvious that Eq. (42) has generalized the Poincare ideas (see, the transformation Eq. (40)). At first, we consider the example

$$y'(\mathbf{x} + \varepsilon y(\mathbf{x}))y'(\mathbf{x}) + y(\mathbf{x}) = \mathbf{0}, \quad y(\mathbf{1}) = b. \tag{43}$$

It has exact solution

$$y(\mathbf{x}) = (\sqrt{\mathbf{x}^2 + 2b\varepsilon + \varepsilon^2 b^2} - \mathbf{x})/\varepsilon. \tag{44}$$

It is obvious that for b > 0, the solution (43) exists on the interval ½0, 1� and

$$y(0) = \sqrt{2b + \varepsilon b^2} / \sqrt{\varepsilon}.$$

The solution of Eq. (43) is obtained by the method of small parameter that can be obtained from Eq. (44). For this purpose, we write Eq. (44) in the form

$$y(\mathbf{x}) = \frac{\mathbf{x}}{\varepsilon} \left( -1 + \sqrt{1 + 2b\frac{\varepsilon}{\mathbf{x}} + b^2 \left(\frac{\varepsilon}{\mathbf{x}}\right)^2} \right)$$

and considering x<sup>2</sup> > 2εb, this expression can be expanded in powers of ε, and then we have

$$y(\mathbf{x}) = \frac{b}{\mathbf{x}} + \frac{b^2}{2\mathbf{x}} \frac{\varepsilon}{\mathbf{x}^2} (\mathbf{x}^2 - 1) + \dots + O\left(\frac{1}{\mathbf{x}} \left(\frac{\varepsilon}{\mathbf{x}^2}\right)^n\right) + \dots \tag{45}$$

The series (45) is uniformly convergent asymptotic series only on the segment <sup>½</sup>εα, <sup>1</sup>�, <sup>0</sup> <sup>&</sup>lt; <sup>α</sup> <sup>&</sup>lt; <sup>1</sup>=2.

First, we write Eq. (43) in the form

Perturbed Differential Equations with Singular Points http://dx.doi.org/10.5772/67856 23

$$(\mathbf{x} + \varepsilon y(\xi))y'(\xi) + y(\xi)\mathbf{x}'(\xi) = \mathbf{0} \tag{46}$$

Substituting Eq. (42) into Eq. (46):

Theorem 6. Equation (39) has a unique 2π=ω periodic solution, and it can be represented in the

<sup>y</sup>ðξÞ ¼ <sup>y</sup>0ðξÞ þ <sup>ε</sup>y1ðξÞ þ <sup>ε</sup><sup>2</sup>y2ðξÞ þ …

It is obvious that Eq. (42) has generalized the Poincare ideas (see, the transformation Eq. (40)).

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>2</sup>b<sup>ε</sup> <sup>þ</sup> <sup>ε</sup><sup>2</sup>b<sup>2</sup> <sup>p</sup>

> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup><sup>b</sup> <sup>þ</sup> <sup>ε</sup>b<sup>2</sup> <sup>p</sup>

> > 1 þ 2b ε x

! r

The solution of Eq. (43) is obtained by the method of small parameter that can be obtained

and considering x<sup>2</sup> > 2εb, this expression can be expanded in powers of ε, and then we have

<sup>x</sup><sup>2</sup> <sup>ð</sup>x<sup>2</sup> � <sup>1</sup>Þ þ … <sup>þ</sup> <sup>O</sup> <sup>1</sup>

The series (45) is uniformly convergent asymptotic series only on the segment

= ffiffiffi <sup>ε</sup> <sup>p</sup> :

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

<sup>þ</sup> <sup>b</sup><sup>2</sup> <sup>ε</sup> x � �<sup>2</sup>

> x ε x2 � �<sup>n</sup> � �

ðxÞ þ qðxÞyðxÞ ¼ rðxÞ, yð1Þ ¼ a ð41Þ

ðxÞ þ yðxÞ ¼ 0, yð1Þ ¼ b: ð43Þ

� xÞ=ε: ð44Þ

þ … ð45Þ

<sup>x</sup> <sup>¼</sup> <sup>ξ</sup> <sup>þ</sup> <sup>ε</sup>x1ðξÞ þ <sup>ε</sup><sup>2</sup>x2ðξÞ þ … <sup>ð</sup>42<sup>Þ</sup>

Lighthill in 1949 [67] reported an important generalization of the method of Poincare.

form (40).

3.2. The idea of the Lighthill method

22 Recent Studies in Perturbation Theory

where <sup>x</sup><sup>∈</sup> <sup>½</sup>0, <sup>1</sup>�qðxÞ, rðx<sup>Þ</sup> <sup>∈</sup>C<sup>∞</sup>½0, <sup>1</sup>�.

At first, we consider the example

It has exact solution

<sup>½</sup>εα, <sup>1</sup>�, <sup>0</sup> <sup>&</sup>lt; <sup>α</sup> <sup>&</sup>lt; <sup>1</sup>=2.

First, we write Eq. (43) in the form

He considered the model equation [67, 82]:

ðx þ εyðxÞÞy<sup>0</sup>

Lighthill proposed to seek the solution of Eq. (41) in the form

ðx þ εyðxÞÞy<sup>0</sup>

yðxÞ¼ð

from Eq. (44). For this purpose, we write Eq. (44) in the form

<sup>y</sup>ðxÞ ¼ <sup>b</sup> x þ b2 2x ε

<sup>y</sup>ðxÞ ¼ <sup>x</sup>

It is obvious that for b > 0, the solution (43) exists on the interval ½0, 1� and

yð0Þ ¼

<sup>ε</sup> �<sup>1</sup> <sup>þ</sup>

$$\begin{aligned} &(\xi + \varepsilon(y\_0(\xi) + \mathbf{x}\_1(\xi)) + \dots + \varepsilon^n(y\_{n-1}(\xi) + \mathbf{x}\_n(\xi)) + \dots)(y\_0'(\xi) + \varepsilon y\_1'(\xi) + \dots + \\ &+ \varepsilon^n y\_n'(\xi) + \dots) + (y\_0(\xi) + \varepsilon y\_1'(\xi) + \dots \varepsilon^n y\_n'(\xi) + \dots)(1 + \varepsilon \mathbf{x}\_1'(\xi) + \dots + \varepsilon^n \mathbf{x}\_n'(\xi) + \dots) = 0, \end{aligned}$$

and equating coefficients of the same powers ε,we have

$$
\xi y\_0'(\xi) + y\_0(\xi) = 0 \tag{47}
$$

$$\left(\xi y'\_n(\xi) + y\_n(\xi) + \sum\_{i=0}^{n-1} \left( (y\_i(\xi) + \mathbf{x}\_{i+1}(\xi)) y'\_{n-1-i}(\xi) + y\_i(\xi) \mathbf{x}'\_{n-i}(\xi) \right) \right) = 0, \quad y\_n(1) = 0, \ n = 1, 2, \dots \tag{48}$$

From Eq. (47), we have

$$y\_0(\xi) = b\xi^{-1}.$$

Using Eq. (47), Eq. (48) for n ¼ 1 can be written as

$$
\xi y\_1'(\xi) + y\_1(\xi) = (\xi x\_1'(\xi) - x\_1(\xi) + y\_0(\xi)) \\
y\_0'(\xi) = 0, \quad y\_1(1) = 0. \tag{49}
$$

If we put x1ðξÞ ¼ 0 in Eq. (49), we obtain

$$
\xi y\_1'(\xi) + y\_1(\xi) = -b^2 \xi^{-3}, \quad y\_1(1) = 0.
$$

Hence, solving this equation, we have

$$y\_1(\xi) = b^2 (2\xi)^{-1} - b^2 (2\xi^3)^{-1}.$$

Since differentiation increased singularity of nonsmooth function, we select x1ðξÞ so that the expression in the right side of Eq. (49) is equal to zero, i.e.,

$$
\xi \mathfrak{x}\_1'(\xi) - \mathfrak{x}\_1(\xi) + y\_0(\xi) = 0, \quad \mathfrak{x}\_1(1) = 0.
$$

Hence, we have

$$\mathbf{x}\_1(\xi) = \mathbf{2}^{-1} b \xi - (\mathbf{2}\xi)^{-1} b.$$

Then Eq. (49) takes the form

$$
\xi y\_1'(\xi) + y\_1(\xi) = 0, \quad y\_1(1) = 0.
$$

Hence, we obtain y1ðξÞ ¼ 0.

Now Eq. (48) for n ¼ 2 takes the form

$$
\xi y\_2'(\xi) + y\_2(\xi) = (\xi x\_2'(\xi) - x\_2(\xi))y\_0'(\xi) = 0, \quad y\_2(1) = 0.
$$

Let x2ðξÞ ¼ 0, and then y1ðξÞ ¼ 0. Further also choose xiðξÞ ¼ yi ðξÞ ¼ 0 ði ¼ 3, 4, …Þ, as they also satisfy the initial conditions. Thus, we have found that

$$y(\xi) = b\xi^{-1} \tag{50}$$

$$\mathfrak{x}(\xi) = \xi + \frac{b}{2} \left( \xi - \frac{1}{\xi} \right) \varepsilon. \tag{51}$$

Putting in Eq. (51) x ¼ 0, we have

$$
\eta = \sqrt{b\varepsilon/(2+b\varepsilon)}.\tag{52}
$$

For b > 0, the point x ¼ 0 is achieved. Moreover, the except in variable ξ from Eq. (50) and to Eq. (51) setting ξ, we obtain the exact solution (44).

Now we will present the main idea of the Lighthill method to Eq. (41) under conditions:qðxÞ, rðxÞ∈C<sup>∞</sup>½0, <sup>1</sup>� and <sup>q</sup><sup>0</sup> <sup>¼</sup> <sup>q</sup>ð0<sup>Þ</sup> <sup>&</sup>gt; 0. We will write it in the form of

$$(\mathbf{x}(\xi) + \varepsilon y(\xi))y'(\xi) = [r(\mathbf{x}(\xi)) - q(\mathbf{x}(\xi))y(\xi)] \mathbf{x}'(\xi), \quad y(1) = y^0. \tag{53}$$

It is obvious that we have one equation for two unknown functions, yðξÞ, x(ξ). Now we substitute the series (42) to Eq. (53):

$$\begin{aligned} &\left(\xi + \sum\_{k=0}^{\bullet} \varepsilon^{k} (\mathsf{y}\_{k}(\xi) + \mathsf{x}\_{k}(\xi))\right) \sum\_{k=0}^{\bullet} \varepsilon^{k} \mathsf{y}\_{k}^{\prime}(\xi) = \\ &= \left(\sum\_{j=0}^{\bullet} r\_{j}(\xi) \left(\sum\_{k=0}^{\bullet} \mathsf{x}\_{k}(\xi) \varepsilon^{k}\right)^{j} - \sum\_{j=0}^{\bullet} q\_{j}(\xi) \left(\sum\_{k=0}^{\bullet} \mathsf{x}\_{k}(\xi) \varepsilon^{k}\right)^{j}\right) \left(1 + \sum\_{k=0}^{\bullet} \mathsf{x}\_{k}^{\prime}(\xi) \varepsilon^{k}\right). \end{aligned}$$

where qj ¼ qj <sup>ð</sup>ξÞ ¼ <sup>1</sup> <sup>j</sup>! <sup>q</sup>ðj<sup>Þ</sup> <sup>ð</sup>ξÞ, rj <sup>¼</sup> rjðξÞ ¼ <sup>1</sup> j!rðj<sup>Þ</sup> ðξÞ.

Hence, equating the coefficients of equal powers has ε

$$L\mu\_0 \equiv \xi y\_0'(\xi) + \eta(\xi)y\_0(\xi) = r(\xi), \quad y\_0(1) = y^0,\tag{54}$$

$$L y\_1 = [\xi y\_0' \mathbf{x}\_1' - y\_0' \mathbf{x}\_1 - y\_0 y\_0'] + (r\_1 - q\_1 y\_0) \mathbf{x}\_1, \quad y\_1(1) = 0,\tag{55}$$

$$\begin{array}{ll} \mathrm{L}y\_2 = [\xi y\_0' \mathbf{x}\_2' - (y\_0 + \mathbf{x}\_1) y\_1' - (y\_1 + \mathbf{x}\_2) y\_0' + ((r\_1 - q\_1 y\_0) \mathbf{x}\_1 - q y\_1) \mathbf{x}\_1'] + \\\ + \{r\_1 \mathbf{x}\_2 + r\_2 \mathbf{x}\_1^2 - q\_1 \mathbf{x}\_1 y\_1 - (q\_1 \mathbf{x}\_2 + q\_2 \mathbf{x}\_1^2) y\_0\} & y\_2(1) = \mathbf{0}, \end{array} \tag{56}$$

…

$$\begin{array}{l} \text{Ly}\_{n} = [\mathbf{y}\_{0}^{\prime}\mathbf{x}\_{n}^{\prime} - y\_{0}^{\prime}\mathbf{x}\_{n} + f\_{n}(\mathbf{y}\_{0^{\prime}}, \dots, \mathbf{y}\_{n-1^{\prime}}, \mathbf{x}\_{1^{\prime}}, \dots, \mathbf{x}\_{n-1^{\prime}}, y\_{0^{\prime}}^{\prime}, \dots, y\_{n-1^{\prime}}^{\prime}, \mathbf{x}\_{1^{\prime}}^{\prime}, \dots, \mathbf{x}\_{n-1}^{\prime})] + \\ + \{\mathbf{g}\_{n}(\mathbf{y}\_{0^{\prime}}, \dots, \mathbf{y}\_{n-1^{\prime}}, \mathbf{x}\_{1^{\prime}}, \dots, \mathbf{x}\_{n-1})\}\_{\nu} \ \mathbf{y}\_{n}(1) = \mathbf{0}; \dots \end{array} \tag{57}$$

where q ¼ q0, r ¼ r0,

f <sup>n</sup> ¼ �ðy<sup>0</sup> þ x1Þy<sup>0</sup> <sup>n</sup>�<sup>1</sup> � ðy<sup>1</sup> <sup>þ</sup> <sup>x</sup>2Þy<sup>0</sup> <sup>n</sup>�<sup>2</sup> � … � ðyn�<sup>2</sup> <sup>þ</sup> xn�<sup>1</sup>Þy<sup>0</sup> <sup>1</sup> � yn�<sup>1</sup>y<sup>0</sup> <sup>0</sup>þ þðr1x<sup>1</sup> � qy<sup>1</sup> � q1x1y0Þx<sup>0</sup> <sup>n</sup>�<sup>1</sup> þ ðr1x<sup>2</sup> <sup>þ</sup> <sup>r</sup>2x<sup>2</sup> <sup>1</sup> � qy<sup>2</sup> � <sup>q</sup>1x1y<sup>1</sup> � ðq1x<sup>2</sup> <sup>þ</sup> <sup>q</sup>2x<sup>2</sup> 1Þy0Þx<sup>0</sup> <sup>n</sup>�<sup>2</sup> þ … þðr1xn�<sup>1</sup> <sup>þ</sup> <sup>2</sup>r2x1xn�<sup>2</sup> <sup>þ</sup> <sup>2</sup>r2x2xn�<sup>3</sup> <sup>þ</sup> … <sup>þ</sup> rn�<sup>1</sup>xn�<sup>1</sup> <sup>1</sup> � q1y1xn�<sup>2</sup> � ðq1xn�<sup>1</sup> þ 2q2x1xn�<sup>2</sup> þ … <sup>þ</sup>qn�<sup>1</sup>xn�<sup>1</sup> <sup>1</sup> Þy<sup>0</sup> 0Þx<sup>0</sup> 1, gn <sup>¼</sup> <sup>r</sup>1xn <sup>þ</sup> <sup>2</sup>r2x1xn�<sup>1</sup> <sup>þ</sup> … <sup>þ</sup> rnxn <sup>1</sup> � <sup>q</sup>1x1yn�<sup>1</sup> � ðq1x<sup>2</sup> <sup>þ</sup> <sup>q</sup>2x<sup>2</sup> <sup>1</sup>Þyn�<sup>2</sup> � … �ðq1xn <sup>þ</sup> <sup>2</sup>q2x1xn�<sup>1</sup> <sup>þ</sup> … <sup>þ</sup> qnxn <sup>1</sup> Þy0:

In these equations, the coefficient rðξÞ � qðξÞy0ðξÞ of the derivative x<sup>0</sup> <sup>n</sup>ðξÞ ðn ¼ 1, 2, …Þwas replaced by Eq. (54) on ξy<sup>0</sup> <sup>0</sup>ðξÞ.

From Eq. (57) for n = 1,2,…, it follows that if we want to define functions xnðξÞ ðn ¼ 1, 2, …Þ from this differential equations, then we must assume that

$$
\xi y\_0'(\xi) = r(\xi) - q(\xi)y\_0(\xi) \neq 0, \ \xi \in (0, 1]. \tag{58}
$$

And this condition cannot be avoided by applying the Lighthill method to Eq. (41). Condition (58) first appeared in [69], justifying Lighthill method, then in the works Habets [66] and Sibuya, Takahashi [68]. Comstock [65] on the example shows that the condition (58) is not necessary for the existence of solutions on the interval ½0, 1�. Further assume that the condition (58) holds. Note that the right-handside of Eq. (57) is linear with respect to xnðξÞ, and f <sup>n</sup> function depends from y<sup>0</sup> <sup>0</sup>, …, y<sup>0</sup> <sup>n</sup>�<sup>1</sup>, x<sup>0</sup> <sup>1</sup>, …, x<sup>0</sup> <sup>n</sup>�<sup>1</sup> only.

The solution of Eq. (54) can be written as

$$y\_0(\xi) = \xi^{-\mathfrak{q}\_0} g(\xi) (y^0 + \int\_1^{\xi} s^{\mathfrak{q}\_0 - 1} r(s) g^{-1}(s) ds) := \xi^{-\mathfrak{q}\_0} w(\xi), \tag{59}$$

where <sup>g</sup>ðξÞ ¼ exp �ð<sup>ξ</sup> 1 � q<sup>0</sup> � qðsÞ � s �1 ds� .

Let

Now Eq. (48) for n ¼ 2 takes the form

24 Recent Studies in Perturbation Theory

Putting in Eq. (51) x ¼ 0, we have

the series (42) to Eq. (53):

� <sup>ξ</sup> <sup>þ</sup>X<sup>∞</sup> k¼0 εk

¼ �X<sup>∞</sup> j¼0 rjðξÞ

<sup>ð</sup>ξÞ ¼ <sup>1</sup> <sup>j</sup>! <sup>q</sup>ðj<sup>Þ</sup>

Ly<sup>2</sup> ¼ ½ξy<sup>0</sup>

Lyn ¼ ½y<sup>0</sup>

where q ¼ q0, r ¼ r0,

<sup>þ</sup>{r1x<sup>2</sup> <sup>þ</sup> <sup>r</sup>2x<sup>2</sup>

0x0 <sup>n</sup> � y<sup>0</sup>

where qj ¼ qj

…

ξy<sup>0</sup>

<sup>2</sup>ðξÞ þ y2ðξÞ¼ðξx<sup>0</sup>

also satisfy the initial conditions. Thus, we have found that

Eq. (51) setting ξ, we obtain the exact solution (44).

ðxðξÞ þ εyðξÞÞy<sup>0</sup>

ðykðξÞ þ xkðξÞÞ

xkðξÞε<sup>k</sup> �j �X<sup>∞</sup> j¼0 qj ðξÞ �X<sup>∞</sup> k¼0

<sup>ð</sup>ξÞ, rj <sup>¼</sup> rjðξÞ ¼ <sup>1</sup>

�X<sup>∞</sup> k¼0

Hence, equating the coefficients of equal powers has ε

Ly<sup>1</sup> ¼ ½ξy<sup>0</sup>

0x0

Lu<sup>0</sup> � ξy<sup>0</sup>

0x0 <sup>1</sup> � y<sup>0</sup>

<sup>2</sup> � ðy<sup>0</sup> þ x1Þy<sup>0</sup>

Let x2ðξÞ ¼ 0, and then y1ðξÞ ¼ 0. Further also choose xiðξÞ ¼ yi

<sup>2</sup>ðξÞ � x2ðξÞÞy<sup>0</sup>

b <sup>2</sup> <sup>ξ</sup> � <sup>1</sup> ξ � �

<sup>η</sup> <sup>¼</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

For b > 0, the point x ¼ 0 is achieved. Moreover, the except in variable ξ from Eq. (50) and to

Now we will present the main idea of the Lighthill method to Eq. (41) under

ðξÞ¼½rðxðξÞÞ � qðxðξÞÞyðξÞ�x<sup>0</sup>

It is obvious that we have one equation for two unknown functions, yðξÞ, x(ξ). Now we substitute

xkðξÞε<sup>k</sup>

<sup>0</sup>ðξÞ þ <sup>q</sup>ðξÞy0ðξÞ ¼ <sup>r</sup>ðξÞ, y0ð1Þ ¼ <sup>y</sup><sup>0</sup>

<sup>þ</sup>{gnðy0, …, yn�<sup>1</sup>, <sup>x</sup>1, …, xn�<sup>1</sup>Þ}, ynð1Þ ¼ 0; … <sup>ð</sup>57<sup>Þ</sup>

�<sup>j</sup>��

xðξÞ ¼ ξ þ

conditions:qðxÞ, rðxÞ∈C<sup>∞</sup>½0, <sup>1</sup>� and <sup>q</sup><sup>0</sup> <sup>¼</sup> <sup>q</sup>ð0<sup>Þ</sup> <sup>&</sup>gt; 0. We will write it in the form of

�X<sup>∞</sup> k¼0 εk y0 <sup>k</sup>ðξÞ ¼

> j!rðj<sup>Þ</sup> ðξÞ.

<sup>0</sup>x<sup>1</sup> � y0y<sup>0</sup>

<sup>1</sup> � <sup>q</sup>1x1y<sup>1</sup> � ðq1x<sup>2</sup> <sup>þ</sup> <sup>q</sup>2x<sup>2</sup>

<sup>1</sup> � ðy<sup>1</sup> þ x2Þy<sup>0</sup>

<sup>0</sup>xn <sup>þ</sup> <sup>f</sup> <sup>n</sup>ðy0, …, yn�<sup>1</sup>, x1,…, xn�<sup>1</sup>, y<sup>0</sup>

<sup>0</sup>ðξÞ ¼ 0, y2ð1Þ ¼ 0:

<sup>y</sup>ðξÞ ¼ <sup>b</sup>ξ�<sup>1</sup> <sup>ð</sup>50<sup>Þ</sup>

<sup>b</sup>ε=ð<sup>2</sup> <sup>þ</sup> <sup>b</sup>ε<sup>Þ</sup> <sup>p</sup> : <sup>ð</sup>52<sup>Þ</sup>

<sup>ð</sup>ξÞ, yð1Þ ¼ <sup>y</sup><sup>0</sup>

<sup>1</sup> <sup>þ</sup>X<sup>∞</sup> k¼0 x0 kðξÞε<sup>k</sup> � ,

<sup>0</sup>�þðr<sup>1</sup> � q1y0Þx1, y1ð1Þ ¼ 0, ð55Þ

<sup>n</sup>�<sup>1</sup>, x<sup>0</sup>

<sup>1</sup>Þy0}, <sup>y</sup>2ð1Þ ¼ 0, <sup>ð</sup>56<sup>Þ</sup>

<sup>1</sup>, …, x<sup>0</sup>

<sup>0</sup> þ ððr<sup>1</sup> � q1y0Þx<sup>1</sup> � qy1Þx<sup>0</sup>

<sup>0</sup>, …, y<sup>0</sup>

ðξÞ ¼ 0 ði ¼ 3, 4, …Þ, as they

: ð53Þ

, ð54Þ

<sup>1</sup>�þ

<sup>n</sup>�<sup>1</sup>Þ�þ

ε: ð51Þ

$$w\_0 = y^0 - \int\_0^1 s^{\eta\_0 - 1} r(s) g^{-1}(s) ds \neq 0 \Leftrightarrow w\_0 = w(0) \neq 0.$$

Hence, we have

$$y\_0(\xi) \sim \xi^{-q\_0} w\_{0\prime} \quad \xi \to 0. \tag{60}$$

Since the differentiation of y0ðξÞ increased of its singularity at the point ξ ¼ 0, it is better to choose such that the first brace in Eq. (55) is equal to zero, i.e.,

$$
\xi \mathfrak{x}\_1' = \mathfrak{x}\_1 + y\_{0'} \quad \mathfrak{x}(1) = 0.
$$

Hence, using Eq. (60), we obtain

$$\mathbf{x}\_1(\xi) = \xi + \xi \int\_1^{\xi} s^{-2} y\_0(s) ds \sim -\frac{\varpi\_0}{1 + q\_0} \xi^{-q\_0}.\tag{61}$$

Then Eq. (55) takes the form

$$Ly\_1 = (r\_1 - q\_1 y\_0) x\_1 \dots \tilde{a}\_1 \xi^{-2q\_0} y\_1$$

where ~a1=const. Hence, we have

$$y\_1(\xi) \prec a\_1 \ \xi^{-2q\_0}(a\_1 = \text{const}), \xi \to 0. \tag{62}$$

Now equating to zero the expression in the first brace in the right-hand side of Eq. (56), we have

$$\mathbf{x}'\_2\mathbf{x}'\_2 - \mathbf{x}\_2 = y\_1 + ((y\_0 + \mathbf{x}\_1)y'\_1 - ((r\_1 - q\_1y\_0)\mathbf{x}\_1 - qy\_1)\mathbf{x}'\_1)(y'\_0)^{-1} \dots \tilde{b}\_2 \mathbf{\xi}'^{-2q\_0}, \ \tilde{b}\_2 = \text{const.} $$

From this, we get

$$\propto\_2(\xi)^{\sim}b\_2\xi^{-2q\_0},\ b\_2 = \text{const},\ \xi \to 0. \tag{63}$$

Now Eq. (56) takes the form

$$L y\_2 = g\_2(y\_{0'} y\_{1'}, x\_1, x\_2) \sim \tilde{a}\_2 \xi^{-3\eta\_0}, \ \tilde{a}\_2 = const, \ \xi \to 0$$

Solving this equation, we have

$$y\_2(\xi) - a\_2 \xi^{-3q\_0}, \ a\_2 = \text{const}, \ \xi \to 0 \tag{64}$$

Next, the method of induction, it is easy to show that

$$\infty\_j(\xi) \sim b\_j \xi^{-jq\_0}, \quad y\_j(\xi) \sim a\_j \xi^{-(j+1)q\_0}, \quad j = 1, 2, \dots \tag{65}$$

Thus, the series (42) has the asymptotic

$$y(\xi) \sim \xi^{-q\_0} (w\_0 + a\_1 \varepsilon \xi^{-q\_0} + \dots + a\_n (\varepsilon \xi^{-q\_0})^n + \dots), \quad \xi \to 0,\tag{66}$$

$$\propto -\xi - \frac{w\_0}{1 + q\_0} \xi^{-q\_0} \varepsilon + b\_2 (\varepsilon \xi^{-q\_0})^2 + \dots + b\_n (\varepsilon \xi^{-q\_0})^n + \dots \tag{67}$$

From Eq. (67), it follows that the point x ¼ 0 corresponds to the root of the equation

$$
\sigma + \varepsilon \mathbf{x}\_1(\eta) + \varepsilon^2 \mathbf{x}\_2(\eta) + \dots = \mathbf{0} \tag{68}
$$

Moreover, this equation should have a positive root and if the solution of Eq. (41) exists on the interval ð0, 1�. Solving Eq. (68), we obtain

$$
\eta \sim (\mathfrak{w}\_0 \varepsilon/\mathbf{1} + q\_0)^{1/(1+q\_0)}, \quad \varepsilon \to 0. \tag{69}
$$

And, under the conditionw<sup>0</sup> > 0, η<sup>0</sup> will be positive. It is obvious that on the interval ½ξ0, 1� series (42) or (66) and (67) remains asymptotic. Substituting Eq. (69) into Eq. (66), we have

$$y(0) \sim w\_0 \left(\frac{w\_0 \varepsilon}{1 + q\_0}\right)^{-q\_0/(1 + q\_0)}, \quad \varepsilon \to 0.$$

If w<sup>0</sup> < 0 the point x ¼ 0 does not have the positive root of Eq. (68), so that the solution of Eq. (41) goes to infinity, before reaching the point x ¼ 0.

We have the

x1ðξÞ ¼ ξ þ ξ

Then Eq. (55) takes the form

26 Recent Studies in Perturbation Theory

where ~a1=const. Hence, we have

<sup>2</sup> � x<sup>2</sup> ¼ y<sup>1</sup> þ ððy<sup>0</sup> þ x1Þy<sup>0</sup>

have

ξx<sup>0</sup>

From this, we get

Now Eq. (56) takes the form

Solving this equation, we have

Thus, the series (42) has the asymptotic

x

interval ð0, 1�. Solving Eq. (68), we obtain

Next, the method of induction, it is easy to show that

<sup>e</sup> <sup>ξ</sup> � <sup>w</sup><sup>0</sup> 1 þ q<sup>0</sup>

xjðξ<sup>Þ</sup> <sup>e</sup> bjξ�jq<sup>0</sup> , yj

<sup>y</sup>ðξ<sup>Þ</sup> <sup>e</sup> <sup>ξ</sup>�q<sup>0</sup> <sup>ð</sup>w<sup>0</sup> <sup>þ</sup> <sup>a</sup>1εξ�q<sup>0</sup> <sup>þ</sup> … <sup>þ</sup> anðεξ�q<sup>0</sup> <sup>Þ</sup>

<sup>ξ</sup>�q<sup>0</sup> <sup>ε</sup> <sup>þ</sup> <sup>b</sup>2ðεξ�q<sup>0</sup> <sup>Þ</sup>

From Eq. (67), it follows that the point x ¼ 0 corresponds to the root of the equation

Moreover, this equation should have a positive root and if the solution of Eq. (41) exists on the

<sup>η</sup> <sup>þ</sup> <sup>ε</sup>x1ðηÞ þ <sup>ε</sup><sup>2</sup>

ðξ 1 s �2 <sup>y</sup>0ðsÞds <sup>e</sup> � <sup>w</sup><sup>0</sup>

Ly<sup>1</sup> ¼ ðr<sup>1</sup> � <sup>q</sup>1y0Þx<sup>1</sup> <sup>e</sup>~a1ξ�2q<sup>0</sup> ,

Now equating to zero the expression in the first brace in the right-hand side of Eq. (56), we

<sup>1</sup> � ððr<sup>1</sup> � q1y0Þx<sup>1</sup> � qy1Þx<sup>0</sup>

Ly<sup>2</sup> <sup>¼</sup> <sup>g</sup>2ðy0, y1, x1, x2<sup>Þ</sup> <sup>e</sup>~a2ξ�3q<sup>0</sup> , <sup>~</sup>a<sup>2</sup> <sup>¼</sup> const, <sup>ξ</sup> ! <sup>0</sup>

1 þ q<sup>0</sup>

<sup>y</sup>1ðξ<sup>Þ</sup> <sup>e</sup> <sup>a</sup><sup>1</sup> <sup>ξ</sup>�2q<sup>0</sup> <sup>ð</sup>a<sup>1</sup> <sup>¼</sup> constÞ, <sup>ξ</sup> ! <sup>0</sup>: <sup>ð</sup>62<sup>Þ</sup>

1Þðy<sup>0</sup> 0Þ �1 e

<sup>x</sup>2ðξÞeb2ξ�2q<sup>0</sup> , b<sup>2</sup> <sup>¼</sup> const, <sup>ξ</sup> ! <sup>0</sup>: <sup>ð</sup>63<sup>Þ</sup>

<sup>y</sup>2ðξ<sup>Þ</sup> <sup>e</sup> <sup>a</sup>2ξ�3q<sup>0</sup> , a<sup>2</sup> <sup>¼</sup> const, <sup>ξ</sup> ! <sup>0</sup> <sup>ð</sup>64<sup>Þ</sup>

<sup>2</sup> <sup>þ</sup> … <sup>þ</sup> bnðεξ�q<sup>0</sup> <sup>Þ</sup>

<sup>ð</sup>ξ<sup>Þ</sup> <sup>e</sup> ajξ�ðjþ1Þq<sup>0</sup> , j <sup>¼</sup> <sup>1</sup>, <sup>2</sup>,…: <sup>ð</sup>65<sup>Þ</sup>

x2ðηÞ þ … ¼ 0 ð68Þ

<sup>n</sup> <sup>þ</sup> …Þ, <sup>ξ</sup> ! <sup>0</sup>, <sup>ð</sup>66<sup>Þ</sup>

<sup>n</sup> <sup>þ</sup> … <sup>ð</sup>67<sup>Þ</sup>

<sup>ξ</sup>�q<sup>0</sup> : <sup>ð</sup>61<sup>Þ</sup>

<sup>~</sup>b2ξ�2q<sup>0</sup> , <sup>~</sup>b<sup>2</sup> <sup>¼</sup> const:

Theorem 7. Suppose that the conditions (1) <sup>q</sup>ðxÞ, rðxÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�; (2) <sup>q</sup><sup>0</sup> <sup>&</sup>gt; 0; (3) <sup>w</sup><sup>0</sup> <sup>&</sup>gt; 0; (4) ξy<sup>0</sup> <sup>0</sup> 6¼ 0, ξ∈ ½0, 1�. Then the solution of problem (41) exists on the interval ½0, 1�, and it can be represented in the asymptotic series (42), (66) and (67).

Theorem 7 proved by Wasow [69], Sibuya and Takahashi [68] in the case where qðxÞ, rðxÞ are analytic functions on <sup>½</sup>0, <sup>1</sup>�; proved by Habets [66] in the case <sup>q</sup>ðxÞ, rðxÞ∈C<sup>2</sup> ½0, 1�. Moreover, instead of the condition (3) Wasow impose a stronger condition: a >> 1.

In the proof of Theorem 7, we will not stop because it is held by Majorant method.

From the foregoing, it follows that Wasow condition y<sup>0</sup> <sup>0</sup>ðξÞ 6¼ 0, ξ∈ð0, 1� is essential in the Lighthill method.

Comment 2. Prytula and later Martin [65] proposed the following variant of the Lighthill method. At first direct expansion determined using by the method of small parameter

$$y(\mathbf{x}) = y\_0(\mathbf{x}) + \varepsilon y\_1(\mathbf{x}) + \varepsilon^2 y\_2(\mathbf{x}) + \dots \tag{70}$$

and further at second they will make transformation

$$\mathbf{x} = \xi + \varepsilon \mathbf{x}\_1(\xi) + \varepsilon^2 \mathbf{x}\_2(\xi) + \dots \tag{71}$$

Here unknowns xjðξÞ are determined from the condition that function yj ðξÞ was less singular function yj�<sup>1</sup>ðξÞ. We show that using the method Prytula or Martin, also cannot avoid Wasow conditions. Really, substituting Eq. (71) into Eq. (70) and expanding in a Taylor series in powers of ε, we have

$$y(\xi) = y\_0(\xi) + \varepsilon \{ y\_1(\xi) + y\_0'(\xi) \mathbf{x}\_1(\xi) \} + O(\varepsilon^2).$$

Hence, to obtain a uniform representation of the solution to the second order by ε, we must to put to zero the expression in the curly brackets, i.e., x1ðξ޼�y1ðξÞ=y<sup>0</sup> <sup>0</sup>ðξÞ. Therefore, <sup>y</sup>ðξÞ ¼ <sup>y</sup>0ðξÞ þ <sup>O</sup>ðε<sup>2</sup>Þ. Hence, it is clear that we must make the condition of Wasow: <sup>y</sup><sup>0</sup> <sup>0</sup>ðξÞ 6¼ 0 in the method of Prytula or Martin also.

### 3.3. Uniformization method for a Lighthill model equation

We will consider the problem (41) again [3, 58–60], i.e.,

$$(\mathbf{x} + \varepsilon y(\mathbf{x}))y'(\mathbf{x}) = r(\mathbf{x}) - q(\mathbf{x})y(\mathbf{x}), \quad y(1) = a,\tag{72}$$

Theorem 8. Suppose that the problem (72) has a parametric representation of the solution y ¼ yðξÞ, x ¼ xðξÞ, where ξ∈½η, 1�, η ¼ ηðεÞ > 0, then the problem (72) is equivalent to the problem

$$\begin{cases} \xi y'(\xi) = r(\mathbf{x}(\xi)) - q(\mathbf{x}(\xi))y(\xi), & y(1) = y^0, \\ \xi \mathbf{x}'(\xi) = \mathbf{x}(\xi) + \varepsilon y(\xi), & \mathbf{x}(1) = 1, \quad \xi \in [\eta, 1]. \end{cases} \tag{73}$$

where η ¼ ηðεÞ is the root equation xðηÞ ¼ 0 and if the root η ¼ ηðεÞ > 0 and xðξÞ þ εyðξÞ 6¼ 0 on the interval ½η, 1�.

Proof. Sufficiency. Let the solution of the problem (72) exists and xðξÞ, yðξÞ are a parametric representation of the solution of the problem (72). Then introducing the variable-parameter ξ, we obtain the problem (73).

Necessity. Let it fulfill the conditions of Theorem 8. Then dividing the first equation by second one, we get Eq. (72). Theorem 8 is proved.

Equation (73) on the proposal of the Temple [43], we will call uniformizing equation for the problem (72).

We have the following

Theorem 9. Suppose that the first three conditions of Theorem 8. i.e.,(1) <sup>q</sup>ðxÞ, rðxÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�; (2) q<sup>0</sup> > 0; (3) w<sup>0</sup> > 0. Then the solution of problem (72) is represented in the form of an asymptotic series (42) and its solution can be obtained from uniformizing equation (73).

The proof of this theorem is completely analogous to the proof of Theorem 8, even more easily.

Only it remains to show that under the conditions of Theorem 9 we can get an explicit solution y ¼ yðx, εÞ. Really, since

$$
\widetilde{\mathfrak{X}}\,\,\xi - \frac{w\_0}{1+q\_0} \,\xi^{-q\_0} \varepsilon \,\,\,\xi \to 0.
$$

Let

$$F(\mathbf{x}, \boldsymbol{\xi}, \varepsilon) = \mathbf{x} - \boldsymbol{\xi} + \frac{\varpi\_0}{1 + q\_0} \boldsymbol{\xi}^{-q\_0} \varepsilon + O\left((\varepsilon \boldsymbol{\xi}^{-q\_0})^2\right), \quad \boldsymbol{\xi} \to \mathbf{0},\\\eta = \sqrt[q\_0]{\frac{\varpi\_0}{1 + q\_0} \varepsilon}, \; \varepsilon \to 0.$$

then

$$\frac{\partial F(\mathbf{x}, \xi, \varepsilon)}{\partial \xi}|\_{\xi = \eta(\varepsilon)} = -1 - q\_0 + \mathcal{O}\left(\varepsilon^{1/(1+q\_0)}\right) \neq 0, \ \xi \in [\eta, 1].$$

Therefore, by the implicit function theorem, we can express ξ : ξ ¼ ϕðx, εÞ.

Then when we put it in first equality (42), we obtain an explicit solution y ¼ yðx, εÞ. Comment 3. Explicit asymptotic solution that this problem obtained in Section 3.4. Example 43. Uniformized equation is

$$\begin{cases} \xi y'(\xi) = -y(\xi), \quad y(1) = b, \\ \xi x'(\xi) = x(\xi) + \varepsilon y(\xi), \ x(1) = 1, \quad \xi \in [\eta, 1]. \end{cases}$$

It is easy to integrate this system, and we obtain

$$y(\xi) = b\xi^{-1}, \mathfrak{x}(\xi) = \left(1 + 2^{-1}b\varepsilon\right)\xi - \left(2\xi\right)^{-1}b\varepsilon\zeta$$

Hence, excluding variable ξ, we have an exact solution (44).

Example 2 [37, 43])

$$y(\mathbf{x} + \varepsilon y(\mathbf{x}))y'(\mathbf{x}) + (2 + \mathbf{x})y(\mathbf{x}) = \mathbf{0}, \ y(1) = e^{-1}.$$

Uniformized equation is

$$\begin{cases} \xi x'(\xi) = x + \varepsilon y(\xi), & x(1) = 1, \\ \xi y'(\xi) = -(2 + x(\xi)) y(\xi), & y(1) = e^{-1}, \quad \xi \in [\eta, 1]. \end{cases} \tag{74}$$

Let

3.3. Uniformization method for a Lighthill model equation

ðx þ εyðxÞÞy<sup>0</sup>

Theorem 8. Suppose that the problem (72) has a parametric representation of the solution y ¼ yðξÞ, x ¼ xðξÞ, where ξ∈½η, 1�, η ¼ ηðεÞ > 0, then the problem (72) is equivalent to the

<sup>ð</sup>ξÞ ¼ <sup>r</sup>ðxðξÞÞ � <sup>q</sup>ðxðξÞÞyðξÞ, yð1Þ ¼ <sup>y</sup><sup>0</sup>,

where η ¼ ηðεÞ is the root equation xðηÞ ¼ 0 and if the root η ¼ ηðεÞ > 0 and xðξÞ þ εyðξÞ 6¼ 0

Proof. Sufficiency. Let the solution of the problem (72) exists and xðξÞ, yðξÞ are a parametric representation of the solution of the problem (72). Then introducing the variable-parameter ξ,

Necessity. Let it fulfill the conditions of Theorem 8. Then dividing the first equation by second

Equation (73) on the proposal of the Temple [43], we will call uniformizing equation for the

Theorem 9. Suppose that the first three conditions of Theorem 8. i.e.,(1) <sup>q</sup>ðxÞ, rðxÞ<sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�; (2) q<sup>0</sup> > 0; (3) w<sup>0</sup> > 0. Then the solution of problem (72) is represented in the form of an asymp-

The proof of this theorem is completely analogous to the proof of Theorem 8, even more easily. Only it remains to show that under the conditions of Theorem 9 we can get an explicit solution

<sup>ξ</sup>�q<sup>0</sup> <sup>ε</sup>, <sup>ξ</sup> ! <sup>0</sup>:

� ε<sup>1</sup>=ð1þq0<sup>Þ</sup> �

, ξ ! 0, η ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi w<sup>0</sup> 1 þ q<sup>0</sup> ε <sup>q</sup>0þ<sup>1</sup> r

6¼ 0, ξ∈ ½η, 1�:

, ε ! 0:

totic series (42) and its solution can be obtained from uniformizing equation (73).

<sup>x</sup><sup>e</sup> <sup>ξ</sup> � <sup>w</sup><sup>0</sup> 1 þ q<sup>0</sup>

<sup>ξ</sup>�q<sup>0</sup> <sup>ε</sup> <sup>þ</sup> <sup>O</sup>

<sup>∂</sup><sup>ξ</sup> <sup>j</sup><sup>ξ</sup>¼ηðε<sup>Þ</sup> ¼ �<sup>1</sup> � <sup>q</sup><sup>0</sup> <sup>þ</sup> <sup>O</sup>

Therefore, by the implicit function theorem, we can express ξ : ξ ¼ ϕðx, εÞ.

� <sup>ð</sup>εξ�q<sup>0</sup> <sup>Þ</sup> 2 �

w0 1 þ q<sup>0</sup>

∂Fðx, ξ, εÞ

ðξÞ ¼ xðξÞ þ εyðξÞ, xð1Þ ¼ 1, ξ∈ ½η, 1�,

ðxÞ ¼ rðxÞ � qðxÞyðxÞ, yð1Þ ¼ a, ð72Þ

ð73Þ

We will consider the problem (41) again [3, 58–60], i.e.,

ξy<sup>0</sup>

�

ξx<sup>0</sup>

problem

on the interval ½η, 1�.

28 Recent Studies in Perturbation Theory

problem (72).

We have the following

y ¼ yðx, εÞ. Really, since

Fðx, ξ, εÞ ¼ x � ξ þ

Let

then

we obtain the problem (73).

one, we get Eq. (72). Theorem 8 is proved.

$$\begin{cases} \mathbf{x}(\xi) = \mathbf{x}\_0(\xi) + \varepsilon \mathbf{x}\_1(\xi) + O(\varepsilon^2), \\ y(\xi) = y\_0(\xi) + \varepsilon y\_1(\xi) + O(\varepsilon^2). \end{cases} \tag{75}$$

Substituting Eq. (75) into Eq. (74), we have

$$\mathbf{x}\_0(\xi) = \xi,\ \mathbf{x}\_1(\xi) = \xi \int\_1^{\xi} e^{-s} s^{-4} ds,\ \ y\_0(\xi) = e^{-\xi} \xi^{-2},\ \ y\_1(\xi) = -e^{-\xi} \xi^{-2} \int\_1^{\xi} e^{-s} s^{-4} ds.$$

Hence if ξ ! 0, we obtain

$$\mathbf{x}\_0(\xi) = \xi, \quad \mathbf{x}\_1(\xi) = -\frac{1}{3}\xi^{-2} + \dots, \quad y\_0(\xi) = \xi^{-2} + \dots, \quad y\_1(\xi) = -\frac{1}{6}\xi^{-4} + \dots$$

From the equation <sup>x</sup>ðηÞ ¼ 0, we find <sup>η</sup> : <sup>η</sup><sup>e</sup> ffiffiffiffiffiffiffi ε=3 p<sup>3</sup> .

We prove that xðξÞ þ εyðξÞ 6¼ 0 on the interval ½η, 1�.

Really,

$$
\varepsilon x(\xi) + \varepsilon y(\xi)^{\sim} \xi + \varepsilon \xi^{-2} \neq 0, \ \xi \in [\eta, 1].
$$

#### 3.4. It is construction explicit form of the solution of the model Lighthill equation

We will consider the problem [57], i.e., (41) again

$$(\mathbf{x} + \varepsilon y(\mathbf{x}))y'(\mathbf{x}) + q(\mathbf{x})y(\mathbf{x}) = r(\mathbf{x}), \ y(\mathbf{1}) = b \tag{76}$$

where b is given constant, x ∈½0, 1�, y<sup>0</sup> ðxÞ ¼ dy=dx . Given functions are subjected to the conditions U: <sup>q</sup>ðxÞ, rðxÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup> ½0, 1�.

Here, we consider the case q<sup>0</sup> ¼ �1; this is done to provide a detailed illustration of the idea of the application of the method. We search for the solution of problem (76) in the form

$$y(\mathbf{x}) = \mu^{-1}\pi\_{-1}(t) + \sum\_{k=0}^{\infty} \left(\pi\_k(t) + \mu\_k(\mathbf{x})\right)\mu^k,\tag{77}$$

where <sup>t</sup> <sup>¼</sup> <sup>x</sup>=μ, <sup>ε</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup>, ukðxÞ∈Cð∞<sup>Þ</sup> <sup>½</sup>0, <sup>1</sup>� and <sup>π</sup>kðtÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup> ½0, μ0�, μ<sup>0</sup> ¼ 1=μ:

Note that πkðtÞ ¼ πkðt, μÞ , i.e., πkðtÞ depends also on μ, but this dependence is not indicated. The initial conditions for the functions πjðtÞ are taken as

$$
\pi\_{-1}(1/\mu) = b\mu, \quad b = \mu^0 - \sum\_{k=0}^{\infty} \mu^k \mu\_k(1), \quad \pi\_k(\mu\_0) = 0, \quad k = 0, 1, \dots \tag{78}
$$

Substituting Eq. (77) into Eq. (76), we obtain to determine the functions πkðtÞ, k ¼ �1, 0, 1, …, unðxÞ, n ¼ 0, 1, …,

we have the following equations:

$$\left(t + \pi\_{-1}(t)\right)\pi\_{-1}'(t) = q(\mu \, t)\pi\_{-1}(t), \quad \pi\_{-1}(\mu\_0) = b\mu,\tag{79.-1}$$

$$\mathrm{Lu}\_{0}(\mathbf{x}) := \mathbf{x}u\_{0}'(\mathbf{x}) - q(\mathbf{x})u\_{0}(\mathbf{x}) = r(\mathbf{x}), \quad u\_{0}(\mathbf{x}) \in \mathbb{C}^{(\approx)}[0, 1] \tag{80.0}$$

$$D\pi\_0(t) := \left(t + \pi\_{-1}(t)\right)\pi\_0'(t) + \left(\pi\_{-1}'(t) - q(\mu \, t)\right)\pi\_0(t) = -\mu\_0(t\mu)\pi\_{-1}'(t), \quad \pi\_0(\mu\_0) = 0 \quad (79.0)$$

$$L u\_1(\mathbf{x}) = 0, \quad u\_1(\mathbf{x}) \in \mathbb{C}^{(\circ)}[0, 1]. \tag{80.1}$$

$$D\pi\_1(t) = -\mu\_0(t\mu)\pi\_0'(t) + \pi\_0(t)\pi\_1'(t) - \mu\_1(t\mu)\pi\_{-1}'(t), \quad \pi\_1(\mu\_0) = 0\tag{79.1}$$

$$L\mu\_2(\mathbf{x}) := -\mu\_0(\mathbf{x})\mu\_0'(\mathbf{x}), \quad \mu\_2(\mathbf{x}) \in \mathbb{C}^{(\circ)}[0,1] \tag{80.2}$$

$$D\pi\_2(t) := -u\_0(t\mu)\pi\_{-1}'(t) - \pi\_0(t)\pi\_1'(t) - u\_1(t\mu)\pi\_0'(t) - \pi\_1(t)\pi\_0'(t) - u\_2(t\mu)\pi\_{-1}'(t), \ \pi\_2(\mu\_0) = 0\tag{79.2}$$

$$L u\_3(\mathbf{x}) := -\mu\_0(\mathbf{x}) u\_1'(\mathbf{x}) - u\_0'(\mathbf{x}) u\_1(\mathbf{x}), \quad u\_3(\mathbf{x}) \in \mathbb{C}^{(\approx)}[0, 1]. \tag{80.3}$$

$$D\pi\_3(t) = \sum\_{\substack{i+j=2\\i\ge 0, j\ge -2}} u\_i(\mu|t)\pi\_i^j(t) + \sum\_{\substack{i+j=2\\i,j\ge 0}} \pi\_i(t)\pi\_i^j(t), \quad \pi\_3(\mu\_0) = 0,\tag{79.3}$$

We solve these problems successively. We write problem (79.�1) as

$$t z'(t) - q(\mu t) z(t) = -z(t) z'(t), \quad z(\mu\_0) = b\mu\_r$$

where

ðx þ εyðxÞÞy<sup>0</sup>

<sup>y</sup>ðxÞ ¼ <sup>μ</sup>�<sup>1</sup>

The initial conditions for the functions πjðtÞ are taken as

<sup>π</sup>�<sup>1</sup>ð1=μÞ ¼ <sup>b</sup>μ, <sup>b</sup> <sup>¼</sup> <sup>u</sup><sup>0</sup> �X<sup>∞</sup>

Here, we consider the case q<sup>0</sup> ¼ �1; this is done to provide a detailed illustration of the idea of

k¼0

<sup>½</sup>0, <sup>1</sup>� and <sup>π</sup>kðtÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup>

Note that πkðtÞ ¼ πkðt, μÞ , i.e., πkðtÞ depends also on μ, but this dependence is not indicated.

Substituting Eq. (77) into Eq. (76), we obtain to determine the functions πkðtÞ, k ¼ �1, 0, 1, …,

<sup>0</sup>ðxÞ � <sup>q</sup>ðxÞu0ðxÞ ¼ <sup>r</sup>ðxÞ, u0ðxÞ∈Cð∞<sup>Þ</sup>

�

<sup>0</sup>ðtÞ � u1ðtμÞπ<sup>0</sup>

<sup>0</sup>ðtÞ � π1ðtÞπ<sup>0</sup>

<sup>0</sup>ðxÞu1ðxÞ, u3ðx<sup>Þ</sup> <sup>∈</sup>Cð∞<sup>Þ</sup>

<sup>π</sup>iðtÞπ<sup>j</sup>

<sup>0</sup>ðxÞ, u2ðx<sup>Þ</sup> <sup>∈</sup>Cð∞<sup>Þ</sup>

i þ j ¼ 2 i, j ≥ 0

�

πkðtÞ þ ukðxÞ

the application of the method. We search for the solution of problem (76) in the form

<sup>π</sup>�<sup>1</sup>ðtÞ þX<sup>∞</sup>

k¼0 μk

�<sup>1</sup>ðtÞ � qðμ tÞ

Lu1ðxÞ ¼ <sup>0</sup>, u1ðxÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup>

<sup>0</sup>ðtÞ þ π0ðtÞπ<sup>0</sup>

<sup>1</sup>ðtÞ � u1ðtμÞπ<sup>0</sup>

<sup>1</sup>ðxÞ � u<sup>0</sup>

0ðtÞ þ <sup>X</sup>

uið<sup>μ</sup> <sup>t</sup>Þπ<sup>j</sup>

Lu2ðxÞ :¼ �u0ðxÞu<sup>0</sup>

We solve these problems successively. We write problem (79.�1) as

where b is given constant, x ∈½0, 1�, y<sup>0</sup>

where <sup>t</sup> <sup>¼</sup> <sup>x</sup>=μ, <sup>ε</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup>, ukðxÞ∈Cð∞<sup>Þ</sup>

we have the following equations:

t þ π�<sup>1</sup>ðtÞ

�

t þ π�<sup>1</sup>ðtÞ

Lu0ðxÞ :¼ xu<sup>0</sup>

� π0 <sup>0</sup>ðtÞ þ � π0

Dπ1ðt޼�u0ðtμÞπ<sup>0</sup>

<sup>D</sup>π3ðtÞ ¼ <sup>X</sup>

�<sup>1</sup>ðtÞ � π0ðtÞπ<sup>0</sup>

Lu3ðxÞ :¼ �u0ðxÞu<sup>0</sup>

i þ j ¼ 2 i ≥ 0, j ≥ � 2

� π0

½0, 1�.

tions U: <sup>q</sup>ðxÞ, rðxÞ<sup>∈</sup> <sup>C</sup>ð∞<sup>Þ</sup>

30 Recent Studies in Perturbation Theory

unðxÞ, n ¼ 0, 1, …,

Dπ0ðtÞ :¼

�

Dπ2ðtÞ :¼ �u0ðtμÞπ<sup>0</sup>

ðxÞ þ qðxÞyðxÞ ¼ rðxÞ, yð1Þ ¼ b ð76Þ

ðxÞ ¼ dy=dx . Given functions are subjected to the condi-

� μk

½0, μ0�, μ<sup>0</sup> ¼ 1=μ:

�<sup>1</sup>ðtÞ ¼ qðμ tÞπ�<sup>1</sup>ðtÞ, π�<sup>1</sup>ðμ0Þ ¼ bμ, ð79:-1Þ

π0ðt޼�u0ðtμÞπ<sup>0</sup>

ukð1Þ, πkðμ0Þ ¼ 0, k ¼ 0, 1, … ð78Þ

, ð77Þ

½0, 1� ð80:0Þ

�<sup>1</sup>ðtÞ, π0ðμ0Þ ¼ 0 ð79:0Þ

½0, 1�, ð80:1Þ

<sup>0</sup>ðtÞ � u2ðtμÞπ<sup>0</sup>

0ðtÞ, π3ðμ0Þ ¼ 0,

�<sup>1</sup>ðtÞ, π1ðμ0Þ ¼ 0 ð79:1Þ

½0, 1� ð80:2Þ

�<sup>1</sup>ðtÞ, π2ðμ0Þ ¼ 0

½0, 1�, ð80:3Þ

ð79:2Þ

ð79:3Þ

$$z = \pi\_{-1}(t), \quad \mu\_0 = \mu^{-1}.$$

The fundamental solution of the homogeneous equation corresponding to this equation is of the form

$$z^0(t) = \exp\left\{\int\_{\mu\_0}^t q(\mu s) \frac{ds}{s}\right\} = \exp\left\{\int\_{\mu\_0}^t \left(q(\mu s) + 1\right) \frac{ds}{s} - \int\_{\mu\_0}^t \frac{ds}{s}\right\} = \frac{p(t, \mu)}{\mu t}.$$

where

$$p(t,\mu) = \exp\left\{ \int\_{\mu\_0}^t \left( q(\mu s) + 1 \right) \frac{ds}{s} \right\}.$$

Using the expression for <sup>z</sup><sup>0</sup>ðtÞ, the solution of the inhomogeneous equation for <sup>z</sup>ðt<sup>Þ</sup> can be written as

$$z(t) = \frac{p(t,\mu)}{\mu t} [z(\mu\_0) + \mu \int\_{\mu\_0}^t p^{-1}(s,\mu) z(s) z'(s) ds] \ ,$$

$$\text{Or } tz(t) = p(t,\mu)b - p(t,\mu) \int\_{\mu\_0}^t p^{-1}(s,\mu)z(s)z'(s)ds.$$

After integrating by parts, we reduce the last expression to the following equation:

$$tz(t) = p(t,\mu)b - \frac{z^2(t)}{2} + p(t,\mu)\frac{b^2\mu^2}{2} + \frac{p(t,\mu)}{2}\int\_{\mu\_0}^t \frac{1+q(\mu s)}{s}p^{-1}(s,\mu)z^2(s)ds$$

or

$$z^2(t) + 2tz(t) - p(t, \mu)b\_0 = p(t, \mu) \int\_{\mu\_0}^t \phi(s, \mu) p^{-1}(s, \mu) z^2(s) ds := p(t, \mu)T(t, z^2) \tag{81}$$

where <sup>ϕ</sup>ðs, <sup>μ</sup>Þ¼ð<sup>1</sup> <sup>þ</sup> <sup>q</sup>ðμsÞÞ=s, b<sup>0</sup> <sup>¼</sup> <sup>2</sup><sup>b</sup> <sup>þ</sup> <sup>b</sup><sup>2</sup> μ2.

Let b<sup>0</sup> > 0. Let us introduce the notation z0ðt޼�t þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t <sup>2</sup> <sup>þ</sup> <sup>b</sup>0pðt, <sup>μ</sup><sup>Þ</sup> q . This function satisfies the inequality 0 <sup>&</sup>lt; <sup>z</sup>0ðt<sup>Þ</sup> <sup>≤</sup> Mt�<sup>1</sup> <sup>ð</sup><sup>t</sup> <sup>&</sup>gt; <sup>0</sup><sup>Þ</sup> and is a strictly decreasing bounded function on the closed interval ½0, μ0�. Here and elsewhere, all constants independent of the small parameter μ are denoted by M. Let S<sup>μ</sup> be the set of functions zðtÞ satisfying the condition

$$||z - z\_0|| \le M\mu\_\prime \text{ where } ||z|| = \max\_{0 \le t \le \mu\_0} |z(t)|\nu\_t$$

Theorem 10. If b<sup>0</sup> > 0, then there exists a unique constraint of the solution of problem (79.-1) from the set Sμ.

Proof. Equation (81) is equivalent to the equation z ¼ F½t, z�, where

$$F[t, z] = -t + \sqrt{t^2 + bp(t, \mu) + p(t, \mu)T(t, z^2)}.$$

Suppose that <sup>k</sup>ϕðt, <sup>μ</sup>Þk <sup>≤</sup> <sup>M</sup>μ, <sup>0</sup> <sup>&</sup>lt; <sup>m</sup> <sup>≤</sup> <sup>p</sup>ðt, <sup>μ</sup><sup>Þ</sup> <sup>≤</sup> M, <sup>k</sup>p�<sup>1</sup>ðtÞk <sup>≤</sup> <sup>M</sup>: First, let us estimate <sup>T</sup>ðt, z<sup>2</sup><sup>Þ</sup> on the set Sμ. We have

$$\begin{aligned} |T(t, z^2)| &\leq \int\_{t}^{\mu\_0} |\varphi(s, \mu)| |p^{-1}(s, \mu)| |z(s)|^2 ds \leq M\mu \int\_{t}^{\mu\_0} |z(s)|^2 ds \leq M\mu \int\_{0}^{\mu\_0} |z(s)|^2 ds \leq M\mu\\ &\leq M\mu \int\_{0}^{1} |z(s)|^2 ds + M\mu \int\_{1}^{\mu\_0} |z(s)|^2 ds \leq M\mu. \end{aligned}$$

Here, we have used the triangle inequality

$$|z(t)| \le |z(t) - z\_0(t)| + |z\_0(t)|.$$

as well as the inequality

$$|z\_0(t)| \le Mt^{-1} \quad (t>0).$$

The Fréchet derivative of the operator Fðt, zÞ with respect to z at the point z0ðtÞ is a linear operator:

$$F\_z(t, z\_0)h = -p(t, \mu) \int\_t^{\mu\_0} \rho(s, \mu) p^{-1}(s, \mu) z\_0(s) h(s) \frac{ds}{\sqrt{t^2 + p(t, \mu) \left(b + T(t, z^2)\right)}}$$

where hðtÞ is a continuous function on the closed interval ½0, μ0�. Note that, in view of <sup>T</sup>ðt, z<sup>2</sup> <sup>0</sup>Þ ¼ OðμÞ, the denominator of this expression is strictly positive on the closed interval ½0, μ0�. For F<sup>0</sup> <sup>z</sup>ðt, z0Þ, we can obtain the estimate <sup>k</sup>Fzðt, z0Þk <sup>≤</sup> <sup>M</sup>μlnμ�<sup>1</sup> in the same way as the estimate for <sup>T</sup>ðt, z<sup>2</sup>Þ. Hence, in turn, it follows from the Lagrange inequality that the operator is a contraction operator in the set Sμ. Therefore, by the fixed-point principle, Eq. (81) has a unique solution from the class Sμ. The theorem is proved.

Corollary. The following inequalities hold:

1. zðtÞ ¼ π�<sup>1</sup>ðtÞ ≥ M > 0 for all t∈ ½0, μ0�;

$$\mathbf{2}. \quad \pi\_{-1}(t) \le Mt^{-1} \ \ (t>0).$$

The other function πjðtÞ, ujðxÞ, j ¼ 0, 1, 2, … is determined from the inhomogeneous linear equations; therefore, the following lemmas are needed.

Lemma 9. For any function fðx<sup>Þ</sup> <sup>∈</sup>Cð∞<sup>Þ</sup> ½0, 1�, the equation Lξ ¼ fðxÞ has a unique bounded solution <sup>ξ</sup>ðxÞ∈Cð∞<sup>Þ</sup> ½0, 1� expressible as

Perturbed Differential Equations with Singular Points http://dx.doi.org/10.5772/67856 33

$$\mathcal{L}(\mathbf{x}) = Q(\mathbf{x}) \int\_0^\mathbf{x} Q^{-1}(\mathbf{s}) f(\mathbf{s}) \frac{d\mathbf{s}}{\mathbf{x}'} \, Q(\mathbf{x}) = \exp\left\{ \int\_1^\mathbf{x} \left( q(\mathbf{s}) + 1 \right) \frac{d\mathbf{s}}{\mathbf{s}} \right\}.$$

Proof. The proof follows from the fact that the general solution of the equation under consideration is expressed as

$$\xi(\mathbf{x}) = Q(\mathbf{x})\mathbf{x}^{-1}[\xi(1) + \int\_{-1}^{\mathbf{x}} Q^{-1}(s)f(s)ds].$$

If we choose

Theorem 10. If b<sup>0</sup> > 0, then there exists a unique constraint of the solution of problem (79.-1) from the

Suppose that <sup>k</sup>ϕðt, <sup>μ</sup>Þk <sup>≤</sup> <sup>M</sup>μ, <sup>0</sup> <sup>&</sup>lt; <sup>m</sup> <sup>≤</sup> <sup>p</sup>ðt, <sup>μ</sup><sup>Þ</sup> <sup>≤</sup> M, <sup>k</sup>p�<sup>1</sup>ðtÞk <sup>≤</sup> <sup>M</sup>: First, let us estimate <sup>T</sup>ðt, z<sup>2</sup><sup>Þ</sup>

ds ≤ Mμ:

jzðtÞj ≤ jzðtÞ � z0ðtÞj þ jz0ðtÞj,

<sup>j</sup>z0ðtÞj <sup>≤</sup> Mt�<sup>1</sup> <sup>ð</sup><sup>t</sup> <sup>&</sup>gt; <sup>0</sup>Þ:

The Fréchet derivative of the operator Fðt, zÞ with respect to z at the point z0ðtÞ is a linear

where hðtÞ is a continuous function on the closed interval ½0, μ0�. Note that, in view of

estimate for <sup>T</sup>ðt, z<sup>2</sup>Þ. Hence, in turn, it follows from the Lagrange inequality that the operator is a contraction operator in the set Sμ. Therefore, by the fixed-point principle, Eq. (81) has a

The other function πjðtÞ, ujðxÞ, j ¼ 0, 1, 2, … is determined from the inhomogeneous linear

<sup>0</sup>Þ ¼ OðμÞ, the denominator of this expression is strictly positive on the closed interval

<sup>z</sup>ðt, z0Þ, we can obtain the estimate <sup>k</sup>Fzðt, z0Þk <sup>≤</sup> <sup>M</sup>μlnμ�<sup>1</sup> in the same way as the

ds ≤ Mμ

t

<sup>ð</sup>s, <sup>μ</sup>ÞjjzðsÞj<sup>2</sup>

jzðsÞj<sup>2</sup>

q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

<sup>2</sup> <sup>þ</sup> bpðt, <sup>μ</sup>Þ þ <sup>p</sup>ðt, <sup>μ</sup>ÞTðt, z<sup>2</sup>Þ:

ð<sup>μ</sup><sup>0</sup> t

jzðsÞj<sup>2</sup>

<sup>ð</sup>s, <sup>μ</sup>Þz0ðsÞhðs<sup>Þ</sup> ds ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

<sup>2</sup> <sup>þ</sup> <sup>p</sup>ðt, <sup>μ</sup><sup>Þ</sup>

½0, 1�, the equation Lξ ¼ fðxÞ has a unique bounded solution

�

<sup>r</sup> �,

b þ Tðt, z<sup>2</sup>Þ

t

ds ≤ Mμ

ð<sup>μ</sup><sup>0</sup> 0

jzðsÞj<sup>2</sup> ds ≤

Proof. Equation (81) is equivalent to the equation z ¼ F½t, z�, where

F½t, z�¼�t þ

<sup>j</sup>ϕðs, <sup>μ</sup>Þjjp�<sup>1</sup>

ð<sup>μ</sup><sup>0</sup> 1

ð<sup>μ</sup><sup>0</sup> t

unique solution from the class Sμ. The theorem is proved.

equations; therefore, the following lemmas are needed.

<sup>ϕ</sup>ðs, <sup>μ</sup>Þp�<sup>1</sup>

ds þ Mμ

set Sμ.

on the set Sμ. We have

32 Recent Studies in Perturbation Theory

<sup>j</sup>Tðt, z<sup>2</sup>Þj <sup>≤</sup>

≤ Mμ ð1 0 jzðsÞj<sup>2</sup>

as well as the inequality

F0

operator:

<sup>T</sup>ðt, z<sup>2</sup>

½0, μ0�. For F<sup>0</sup>

<sup>ξ</sup>ðxÞ∈Cð∞<sup>Þ</sup>

ð<sup>μ</sup><sup>0</sup> t

Here, we have used the triangle inequality

<sup>z</sup>ðt, z0Þh ¼ �pðt, μÞ

Corollary. The following inequalities hold:

Lemma 9. For any function fðx<sup>Þ</sup> <sup>∈</sup>Cð∞<sup>Þ</sup>

½0, 1� expressible as

2. <sup>π</sup>�<sup>1</sup>ðt<sup>Þ</sup> <sup>≤</sup> Mt�<sup>1</sup> <sup>ð</sup><sup>t</sup> <sup>&</sup>gt; <sup>0</sup>Þ.

1. zðtÞ ¼ π�<sup>1</sup>ðtÞ ≥ M > 0 for all t∈ ½0, μ0�;

$$\xi(1) = \int\_0^1 Q^{-1}(s) f(s) ds.$$

then we obtain the required result.

This lemma implies that all the functions ukðxÞ, k ¼ 0, 1, … are uniquely determined and belong to the class <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�.

Lemma 10. The problem

$$\left(t + \pi\_{-1}(t)\right)\eta'(t) + \left(\pi'\_{-1}(t) - q(\mu t)\right)\eta(t) = k(t), \quad \eta(\mu\_0) = 0,\tag{82}$$

where the function kðt<sup>Þ</sup> belongs to C<sup>∞</sup>½0, <sup>1</sup>� is continuous and bounded, and if <sup>j</sup>kðtÞj <sup>≤</sup> Mt�<sup>2</sup>, t ! <sup>∞</sup>, has a unique uniformly bounded solutionηðtÞ ¼ ηðt, μÞ on the closed interval t ∈½0, μ0�for a small μ.

Proof. The fundamental solution of the homogeneous equation (82) is of the form

$$\mathcal{O}(t) = \frac{(1+\mu^2 b)g(t,\mu)}{\mu\left(t+\pi\_{-1}(t)\right)}, \quad \mathcal{g}(t,\mu) = \exp\left\{-\int\_{t}^{\mu\_0} \left(1+q(\mu s)\right) \frac{ds}{s+\pi\_{-1}(s)}\right\}.$$

Obviously, <sup>k</sup>gðt, <sup>μ</sup>Þk <sup>≤</sup> <sup>M</sup> and <sup>g</sup>�<sup>1</sup>ðt, <sup>μ</sup><sup>Þ</sup> <sup>≤</sup> <sup>M</sup> for <sup>t</sup> <sup>∈</sup>½0, <sup>μ</sup>0�and <sup>μ</sup>aresmall. The solution of problem (82) can be expressed as

$$\eta(t) = \frac{g(t,\mu)}{t + \pi\_{-1}(t)} \int\_{\mu\_0}^{t} g^{-1}(s,\mu)k(s)ds. \tag{83}$$

The estimate of the integral term in Eq. (83) shows that it is bounded by the constant M. Hence, it also follows that <sup>j</sup>ηðtÞj <sup>≤</sup> Mt�<sup>1</sup> <sup>ð</sup><sup>t</sup> <sup>&</sup>gt; <sup>0</sup>Þ. The solution of problem (79.0) is defined by the integral Eq. (83), where

$$k(t) = -\mu\_0(t\mu)\pi\_{-1}(t) = -\mu\_0(t\mu)q(\mu t)\frac{\pi\_{-1}(t)}{t + \pi\_{-1}(t)},$$

satisfies the assumptions of the lemma. Therefore, the function π0ðtÞ is bounded on ½0, μ0�. The boundedness of the other functions πkðtÞ, k ¼ 1, 2,… is proved in a similar way, because the right-hand sides of the equations defining these functions satisfy the assumptions of Lemma 10. The estimate of the asymptotic behavior of the series (77) is also carried out using Lemma 10.

Let us introduce the notation

$$y(\mathbf{x}) = \mu^{-1}\pi\_{-1}(t) + \sum\_{k=0}^{n} \mu^{k} \left(\pi\_{k}(t) + u\_{k}(\mathbf{x})\right) + \mu^{n+1} R\_{n+1}(\mathbf{x}, \boldsymbol{\mu}).\tag{84}$$

The following statement holds.

Theorem 11. Let b<sup>0</sup> > 0 (for this, it suffices that the condition b<sup>0</sup> :¼ b � y0ð1Þ > 0holds). Then the solution of problem (76) exists on the closed interval ½0, 1�and its asymptotics can be expressed as Eq. (84) andjRnþ<sup>1</sup>ðx, μÞj ≤ M for all x ∈½0, 1�.

Example. Consider the equation

$$y\left(\mathbf{x} + \varepsilon \, y(\mathbf{x})\right)y'(\mathbf{x}) + y(\mathbf{x}) = 1, \quad y(1) = b\_{\prime\prime}$$

This equation is integrated exactly

$$y(\mathbf{x}) = \varepsilon^{-1} \left[ -\mathbf{x} + \sqrt{\mathbf{x}^2 + 2b\_0 \varepsilon + \varepsilon^2 \left( y^{(0)} \right)^2 + 2\varepsilon \mathbf{x}} \right] \mathbf{x}$$

where b<sup>0</sup> ¼ b � 1. If b<sup>0</sup> > 0, then the solution of problem (1) exists on the closed interval ½0, 1�, which is confirmed by Theorem 11. The equation for π�<sup>1</sup>ðtÞ is of the form

$$\left(t + \pi\_{-1}(t)\right)\pi\_{-1}'(t) + \pi\_{-1}(t) = 0, \quad \pi\_{-1}(\mu\_0) = b\mu.$$

The solution of this problem can be expressed as

$$
\pi\_{-1}(t) = -t + \sqrt{t^2 + 2b + b^2 \mu^2}.
$$

The equation for <sup>u</sup>0(x) has the solution <sup>y</sup>0ðxÞ ¼ <sup>1</sup><sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�. Further,

$$
\pi\_0(t) = \frac{-\pi\_{-1}(t) + b\mu}{t + \pi\_{-1}(t)}, \quad u\_k(\mathbf{x}) = 0, \quad k = 1, 2, \dots, n
$$

where b ¼ b0. The asymptotics of the solutions of problem (76) can be expressed as <sup>y</sup>ðxÞ ¼ <sup>μ</sup>�<sup>1</sup>π�<sup>1</sup>ðx=μÞ þ <sup>1</sup> <sup>þ</sup> <sup>π</sup>0ðx=μÞ þ <sup>o</sup>ðμ<sup>Þ</sup> for all <sup>x</sup> <sup>∈</sup>½0, <sup>1</sup>�, <sup>μ</sup> ! <sup>0</sup>:

## 4. Lagerstrom model problem

The problem [32]

$$
\sigma''(r) + \frac{k}{r}\sigma'(r) + \upsilon(r)\upsilon'(r) = \beta[\upsilon'(r)]^2, \ \upsilon(\varepsilon) = 0, \ \upsilon(\circ) = 1,\tag{85}
$$

where 0 < β is constant, k∈ N.

right-hand sides of the equations defining these functions satisfy the assumptions of Lemma 10. The estimate of the asymptotic behavior of the series (77) is also carried out using Lemma

Theorem 11. Let b<sup>0</sup> > 0 (for this, it suffices that the condition b<sup>0</sup> :¼ b � y0ð1Þ > 0holds). Then the solution of problem (76) exists on the closed interval ½0, 1�and its asymptotics can be expressed as Eq.

πkðtÞ þ ukðxÞ

ðxÞ þ yðxÞ ¼ 1, yð1Þ ¼ b,

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

�<sup>1</sup>ðtÞ þ π�<sup>1</sup>ðtÞ ¼ 0, π�<sup>1</sup>ðμ0Þ ¼ bμ:

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

μ2

:

<sup>2</sup> <sup>þ</sup> <sup>2</sup><sup>b</sup> <sup>þ</sup> <sup>b</sup><sup>2</sup>

<sup>t</sup> <sup>þ</sup> <sup>π</sup>�<sup>1</sup>ðt<sup>Þ</sup> , ukðxÞ ¼ <sup>0</sup>, k <sup>¼</sup> <sup>1</sup>, <sup>2</sup>, …,

� yð0<sup>Þ</sup> �2

þ 2εx

,

x<sup>2</sup> þ 2b0ε þ ε<sup>2</sup>

where b<sup>0</sup> ¼ b � 1. If b<sup>0</sup> > 0, then the solution of problem (1) exists on the closed interval ½0, 1�,

t

q

" # r

� <sup>þ</sup> <sup>μ</sup><sup>n</sup>þ<sup>1</sup>

Rnþ<sup>1</sup>ðx, μÞ: ð84Þ

10.

Let us introduce the notation

34 Recent Studies in Perturbation Theory

The following statement holds.

Example. Consider the equation

This equation is integrated exactly

(84) andjRnþ<sup>1</sup>ðx, μÞj ≤ M for all x ∈½0, 1�.

<sup>y</sup>ðxÞ ¼ <sup>μ</sup>�<sup>1</sup>

�

�

The solution of this problem can be expressed as

4. Lagerstrom model problem

The problem [32]

x þ ε yðxÞ

<sup>y</sup>ðxÞ ¼ <sup>ε</sup>�<sup>1</sup> �<sup>x</sup> <sup>þ</sup>

t þ π�<sup>1</sup>ðtÞ

<sup>π</sup>�<sup>1</sup>ðtÞ þX<sup>n</sup>

k¼0 μk �

� y0

which is confirmed by Theorem 11. The equation for π�<sup>1</sup>ðtÞ is of the form

π�<sup>1</sup>ðt޼�t þ

where b ¼ b0. The asymptotics of the solutions of problem (76) can be expressed as

� π0

The equation for <sup>u</sup>0(x) has the solution <sup>y</sup>0ðxÞ ¼ <sup>1</sup><sup>∈</sup> <sup>C</sup><sup>∞</sup>½0, <sup>1</sup>�. Further,

<sup>π</sup>0ðtÞ ¼ �π�<sup>1</sup>ðtÞ þ <sup>b</sup><sup>μ</sup>

<sup>y</sup>ðxÞ ¼ <sup>μ</sup>�<sup>1</sup>π�<sup>1</sup>ðx=μÞ þ <sup>1</sup> <sup>þ</sup> <sup>π</sup>0ðx=μÞ þ <sup>o</sup>ðμ<sup>Þ</sup> for all <sup>x</sup> <sup>∈</sup>½0, <sup>1</sup>�, <sup>μ</sup> ! <sup>0</sup>:

It has been proposed as a model for Lagerstrom Navier-Stokes equations at low Reynolds numbers. It can be interpreted as a problem of distribution of a stationary temperature vðrÞ.

The first two terms in Eq. (1) is ðk þ 1Þ dimensional Laplacian depending only on the radius, and the other two members—some nonlinear heat loss.

It turns out that not only the asymptotic solution but also convergent solutions of Eq. (1) can be easily constructed by a fictitious parameter [70]. The basic idea of this method is as follows. The initial problem is entered fictitious parameter λ∈ ½0, 1� with the following properties: 1. λ ¼ 0, the solution of the equation satisfies all initial and boundary conditions;

2. The solution of the problem can be expanded in integral powers of the parameter λ for all λ ∈½0, 1�.

It is convenient in Eq. (85) to make setting r ¼ εx, v ¼ 1 � u, then

$$
\mu''(\mathbf{x}) + (k\mathbf{x}^{-1} + \varepsilon)u'(\mathbf{x}) - \lambda\varepsilon u(\mathbf{x})u'(\mathbf{x}) = \left[u'(\mathbf{x})\right]^2, \ u(\mathbf{1}) = 1, \ u(\circ \circ) = 0. \tag{86}
$$

We have the following

Theorem 12. For small ε > 0, the solution of problem (86) can be represented in the form of absolutely and uniformly convergent series

$$u(\mathbf{x}) = u\_0(\mathbf{x}, \varepsilon) + v\_k(\varepsilon)u\_1(\mathbf{x}, \varepsilon) + \dots + v\_k^n(\varepsilon)u\_n(\mathbf{x}, \varepsilon) + \dots, \varepsilon$$

for the sufficiently small parameter ε, where

$$v\_1(\varepsilon) \sim \left(\ln \frac{1}{\varepsilon}\right)^{-1}, \quad v\_2 \sim \varepsilon \ln \frac{1}{\varepsilon}, \quad v\_k \sim \frac{k-1}{k-2} \; \varepsilon(j > 2) \; ; \; u\_k(\mathbf{x}, \varepsilon) = O(1), \forall \mathbf{x} \in [1, \infty).$$

Note that the function unðx, εÞ also depends on k, but for simplicity, this dependence is not specified.

Proof. We introduce Eq. (86) parameter λ, i.e., consider the problem

$$
\mu''(\mathbf{x}) + (\mathbf{k}\mathbf{x}^{-1} + \varepsilon)\mu'(\mathbf{x}) - \beta[\mu'(\mathbf{x})]^2 = \lambda\varepsilon u(\mathbf{x})\mu'(\mathbf{x}), \ \mathfrak{u}(\mathbf{1}) = \mathbf{1}, \ \mathfrak{u}(\mathbf{\omega}) = \mathbf{0} \tag{87}
$$

Here, we will prove this Theorem 12 in the case β ¼ 0 only for simplicity.

Setting λ ¼ 0 in Eq. (87), we have

$$
\stackrel{\circ}{u}\stackrel{\circ}{\ 0} + (\text{x}^{-1}k + \varepsilon)u'\_0 = 0, \quad u\_0(1) = 1, \quad u\_0(\ast \omega) = 0. \tag{88}
$$

It has a unique solution

$$\mu\_0 = X(\mathbf{x}, \varepsilon) := 1 - X\_1(\mathbf{X}, \varepsilon), \quad X\_1 = \mathbb{C}\_0 \int\_1^{\mathbf{x}} s^{-k} e^{-\varepsilon s} ds, \quad \mathbb{C}\_0^{-1} = \int\_1^{\infty} s^{-k} e^{-\varepsilon s} ds.$$

Therefore, Eq. (88) with zero boundary conditions is the Green's function

$$K(\mathbf{x}, \mathbf{s}, \varepsilon) = \begin{cases} \mathbb{C}\_0^{-1} X\_1(\mathbf{x}, \varepsilon) X(\mathbf{s}, \varepsilon), & 1 \le \mathbf{x} \le \mathbf{s}, \\\mathbb{C}\_0^{-1} X\_1(\mathbf{s}, \varepsilon) X(\mathbf{x}, \varepsilon), & \mathbf{s} < \mathbf{x} < \mathbf{s}. \end{cases}$$

Hence, the problem (87) is reduced to the system of integral equations

$$\begin{aligned} u(\mathbf{x}) &= X(\mathbf{x}, \varepsilon) + \lambda \varepsilon \int\_{1}^{\infty} G(\mathbf{x}, \mathbf{s}, \varepsilon) u(\mathbf{s}) u'(\mathbf{s}) d\mathbf{s}, \\ u'(\mathbf{x}) &= X'(\mathbf{x}, \varepsilon) + \lambda \varepsilon \int\_{1}^{\infty} G\_{\mathbf{x}}(\mathbf{x}, \mathbf{s}, \varepsilon) u(\mathbf{s}) u'(\mathbf{s}) d\mathbf{s}, \end{aligned} \tag{89}$$

where

$$G(\mathbf{x}, \mathbf{s}, \varepsilon) = \begin{cases} X\_1(\mathbf{x}, \varepsilon)X(\mathbf{s}, \varepsilon) / X'(\mathbf{s}, \varepsilon), & 1 \le \mathbf{x} \le \mathbf{s}, \\ X\_1(\mathbf{s}, \varepsilon)X(\mathbf{x}, \varepsilon) / X'(\mathbf{s}, \varepsilon), & \mathbf{s} < \mathbf{x} < \infty. \end{cases}$$

In Eq. (89), we make the substitution u ¼ Xðx, εÞϕðxÞ, u<sup>0</sup> ¼ X<sup>0</sup> ðx, εÞψðxÞ, and then we have

$$\begin{aligned} \varphi(\mathbf{x}) &= 1 + \lambda \varepsilon \int\_{1}^{\infty} Q\_1(\mathbf{x}, \mathbf{s}, \varepsilon) \varphi(\mathbf{s}) \psi(\mathbf{s}) d\mathbf{s} := 1 + \lambda \varepsilon Q\_1(q\psi), \\ \psi(\mathbf{x}) &= 1 + \lambda \varepsilon \int\_{1}^{\infty} Q\_2(\mathbf{x}, \mathbf{s}, \varepsilon) \varphi(\mathbf{s}) \psi(\mathbf{s}) d\mathbf{s} := 1 + \lambda \varepsilon Q\_2(\lambda \psi), \end{aligned} \tag{90}$$

where

$$\begin{array}{l} Q\_1 = X^{-1}(\mathfrak{x}, \varepsilon) G(\mathfrak{x}, \mathfrak{s}, \varepsilon) X(\mathfrak{s}, \varepsilon) X'(\mathfrak{s}, \varepsilon), \\ Q\_2 = X\_{\mathfrak{x}}^{-1}(\mathfrak{x}, \varepsilon) G\_{\mathfrak{x}}(\mathfrak{x}, \mathfrak{s}, \varepsilon) X(\mathfrak{s}, \varepsilon) X'(\mathfrak{s}, \varepsilon). \end{array}$$

To prove the theorem, we need next

Lemma 11. The following estimate holds

$$\int\_{1}^{\infty} |Q\_{j}(\mathbf{x}, \mathbf{s}, \varepsilon)| ds \le \int\_{1}^{\infty} X(\mathbf{s}, \varepsilon) ds \qquad (j = 1, 2) \tag{91}$$

Given that, we have 0 ≤ X1ðx, εÞ ≤ 1, jX<sup>0</sup> ðx, εÞj ¼ X<sup>0</sup> ðx, εÞ, X<sup>0</sup> ðx, εÞ ≤ 0, x∈ ½1, ∞Þ, we have

$$\begin{split} &\int\_{1}^{\infty} |Q\_{1}(\mathbf{x},\mathbf{s},\varepsilon)| ds \leq \int\_{1}^{\varepsilon} \frac{X\_{1}(\mathbf{s},\varepsilon)}{X\_{1}'(\mathbf{s},\varepsilon)} |X'(\mathbf{s},\varepsilon)| X(\mathbf{s},\varepsilon) ds + \\ &+ \int\_{\varepsilon}^{\infty} X^{-1}(\mathbf{x},\varepsilon) \frac{X^{2}(\mathbf{s},\varepsilon) |X'(\mathbf{s},\varepsilon)|}{X\_{1}'(\mathbf{s},\varepsilon)} ds \leq \int\_{1}^{\infty} X(\mathbf{s},\varepsilon) ds + \int\_{\varepsilon}^{\infty} X(\mathbf{s},\varepsilon) ds = \int\_{1}^{\infty} X(\mathbf{s},\varepsilon) ds. \end{split}$$

Inequality Eq. (91) for j ¼ 2 is proved similarly.

Further, by integrating by parts, we have

$$\int\_{1}^{\infty} X(s,\varepsilon)ds = -1 + \mathbb{C}\_{0} \int\_{1}^{\infty} s^{-k+1} e^{-\varepsilon s} ds \leq \int\_{1}^{\infty} s^{-k+1} e^{-\varepsilon s} ds / \int\_{1}^{\infty} s^{-k} e^{-\varepsilon s} ds := \frac{\upsilon\_{k}(\varepsilon)}{\varepsilon}.$$

Consequently,

u<sup>0</sup> ¼ Xðx, εÞ :¼ 1 � X1ðX, εÞ, X<sup>1</sup> ¼ C<sup>0</sup>

36 Recent Studies in Perturbation Theory

Therefore, Eq. (88) with zero boundary conditions is the Green's function

(

C�<sup>1</sup>

ðx, εÞ þ λε

<sup>G</sup>ðx, s, <sup>ε</sup>Þ ¼ <sup>X</sup>1ðx, <sup>ε</sup>ÞXðs, <sup>ε</sup>Þ=X<sup>0</sup>

�

ð∞ 1

ð∞ 1

<sup>Q</sup><sup>1</sup> <sup>¼</sup> <sup>X</sup>�<sup>1</sup>

<sup>Q</sup><sup>2</sup> <sup>¼</sup> <sup>X</sup>�<sup>1</sup>

jQjðx, s, εÞjds ≤

X1ðs, εÞ X0 <sup>1</sup>ðs, εÞ

ðs, εÞj

<sup>1</sup>ðs, <sup>ε</sup><sup>Þ</sup> ds <sup>≤</sup>

jX0

ð∞ 1

ð∞ 1

X1ðs, εÞXðx, εÞ=X<sup>0</sup>

<sup>K</sup>ðx, s, <sup>ε</sup>Þ ¼ <sup>C</sup>�<sup>1</sup>

Hence, the problem (87) is reduced to the system of integral equations

uðxÞ ¼ Xðx, εÞ þ λε

ðxÞ ¼ X<sup>0</sup>

In Eq. (89), we make the substitution u ¼ Xðx, εÞϕðxÞ, u<sup>0</sup> ¼ X<sup>0</sup>

ϕðxÞ ¼ 1 þ λε

ψðxÞ ¼ 1 þ λε

ð∞ 1

> ðx 1

ðs, εÞjX<sup>0</sup>

X0

To prove the theorem, we need next

Lemma 11. The following estimate holds

Given that, we have 0 ≤ X1ðx, εÞ ≤ 1, jX<sup>0</sup>

jQ1ðx, s, εÞjds ≤

Inequality Eq. (91) for j ¼ 2 is proved similarly.

ð∞ 1

þ ð∞ x X�<sup>1</sup> ðx, εÞ X2

u0

where

where

ðx 1 s �k e �εs

<sup>0</sup> X1ðx, εÞXðs, εÞ, 1 ≤ x ≤ s,

<sup>0</sup> X1ðs, εÞXðx, εÞ, s < x < ∞:

Gðx, s, εÞuðsÞu<sup>0</sup>

Gxðx, s, εÞuðsÞu<sup>0</sup>

Q1ðx, s, εÞϕðsÞψðsÞds :¼ 1 þ λεQ1ðϕψÞ,

Q2ðx, s, εÞϕðsÞψðsÞds :¼ 1 þ λεQ2ðλψÞ,

ðx, εÞ, X<sup>0</sup>

ð∞ x

Xðs, εÞds ¼

ðx, εÞGðx, s, εÞXðs, εÞX<sup>0</sup>

<sup>x</sup> ðx, εÞGxðx, s, εÞXðs, εÞX<sup>0</sup>

ð∞ 1

ðx, εÞj ¼ X<sup>0</sup>

ðx 1

ðs, εÞjXðs, εÞdsþ

Xðs, εÞds þ

ds, C�<sup>1</sup> <sup>0</sup> ¼

ðsÞds,

ðs, εÞ, 1 ≤ x ≤ s,

ðs, εÞ, s < x < ∞:

ðs, εÞ,

ðs, εÞ:

Xðs, εÞds ðj ¼ 1, 2Þ ð91Þ

ðx, εÞ ≤ 0, x∈ ½1, ∞Þ, we have

ð∞ 1

Xðs, εÞds:

ðsÞds,

ðx, εÞψðxÞ, and then we have

ð89Þ

ð90Þ

ð∞ 1 s �k e �εs ds:

$$
\varepsilon \int\_{1}^{\infty} X(\varkappa, \varepsilon) ds \le \upsilon\_{k}(\varepsilon). \tag{92}
$$

It is from integral expressing of vkðεÞ we can obtain the asymptotic behavior such as indicated in the theorem.

With the solution of Eq. (90), we can expand in series

$$\begin{array}{l} \varphi(\mathbf{x}) = 1 + \varphi\_1(\mathbf{x}, \varepsilon)\lambda + \varphi\_2(\mathbf{x}, \varepsilon)\lambda^2 + \dots \\ \Psi(\mathbf{x}) = 1 + \Psi\_1(\mathbf{x}, \varepsilon)\lambda + \Psi\_2(\mathbf{x}, \varepsilon)\lambda^2 + \dots \end{array}$$

The coefficients of this series are uniquely determined from the equations ϕ<sup>0</sup> ¼ Ψ<sup>0</sup> ¼ 1, ϕ<sup>1</sup> ¼ εQ1ð1Þ, Ψ<sup>1</sup> ¼ Q2ð1Þ, <sup>ϕ</sup><sup>n</sup> <sup>¼</sup> <sup>ε</sup>Q1ðϕ<sup>n</sup>�<sup>1</sup>Þ þ <sup>ε</sup>Q1ðΨ<sup>n</sup>�<sup>1</sup>Þ þ <sup>ε</sup>Q1ðϕ1Ψ<sup>n</sup>�<sup>2</sup>Þ þ … <sup>þ</sup> <sup>ε</sup>Q1ðϕ<sup>n</sup>�<sup>2</sup>Ψ1Þ, <sup>Ψ</sup><sup>n</sup> <sup>¼</sup> <sup>ε</sup>Q2ðϕ<sup>n</sup>�<sup>1</sup>Þ þ <sup>ε</sup>Q2ðΨ<sup>n</sup>�<sup>1</sup>Þ þ <sup>ε</sup>Q2ðϕ1Ψ<sup>n</sup>�<sup>2</sup>Þ þ … <sup>þ</sup> <sup>ε</sup>Q2ðϕ<sup>n</sup>�<sup>2</sup>Ψ1Þ, <sup>ð</sup><sup>n</sup> <sup>¼</sup> <sup>2</sup>, <sup>3</sup>, …Þ:

Let z ¼ sup 1 ≤ x<∞ fjϕðxÞj, jΨðxÞjg, then by using Eq. (92) we have a Majorant equation: <sup>z</sup> <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>λ</sup>vkðεÞz2. The solution of this equation can be expanded in powers <sup>λ</sup> the under condition 8vkðεÞ ≤ 1 for all λ∈ ½0, 1�.

If we call unðx, <sup>ε</sup>Þ ¼ <sup>X</sup>ðx, <sup>ε</sup>Þϕn<sup>ð</sup>x, <sup>ε</sup><sup>Þ</sup> vn <sup>k</sup> <sup>ð</sup>ε<sup>Þ</sup> , we get the proof of the theorem.

## Author details

Keldibay Alymkulov\* and Dilmurat Adbillajanovich Tursunov

\*Address all correspondence to: keldibay@mail.ru

Institute of the Fundamental and Applied Researches, Osh State University, Osh, Kyrgyzstan

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[80] Teruhiko, K. An asymptotic approach on Lagerstrom mathematical model for viscous flow at Reynolds numbers. Bull. Univ. Osaka Prefect. Ser. A. 1988, Vol. 36, No. 2, pp. 83-97. [81] Tsien, H.S. The Poincare-Lighthill-Kuo Method. Advances in Applied Mechanics. Aca-

[82] Protter, M.H., Weinberger, H.F. Maximum Principles in Differential Equations. Prentice

viscous flow numbers. SIAM. J.Appl. Math. 1990, Vol. 50, No. 1, pp.48-63.

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Differ. Equ. May 20, 2004, Vol. 199, No. 2, pp. 290-325.

J. Appl. Math. 1975, Vol. 29, No. 1, pp. 110-120.

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Appl. 1975, Vol. 49, No. 2, pp. 286-294.

Hall, Englewood Cliffs, New Jersey, 1967.

207.

42 Recent Studies in Perturbation Theory

No. 4, pp. 531-565.

Dekker, New York, 1991.

Additional information is available at the end of the chapter

http://dx.doi.org/10.5772/67876

## Abstract

As we all know, perturbation theory is closely related to methods used in the numerical analysis fields. In this chapter, we focus on introducing two homotopy asymptotic methods and their applications. In order to search for analytical approximate solutions of two types of typical nonlinear partial differential equations by using the famous homotopy analysis method (HAM) and the homotopy perturbation method (HPM), we consider these two systems including the generalized perturbed Kortewerg-de Vries-Burgers equation and the generalized perturbed nonlinear Schrödinger equation (GPNLS). The approximate solution with arbitrary degree of accuracy for these two equations is researched, and the efficiency, accuracy and convergence of the approximate solution are also discussed.

Keywords: homotopy analysis method, homotopy perturbation method, generalized KdV-Burgers equation, generalized perturbed nonlinear Schrödinger equation, approximate solutions, Fourier transformation

## 1. Introduction

In the past decades, due to the numerous applications of nonlinear partial differential equations (NPDEs) in the areas of nonlinear science [1, 2], many important phenomena can be described successfully using the NPDEs models, such as engineering and physics, dielectric polarization, fluid dynamics, optical fibers and quantitative finance and so on [3–5]. Searching for analytical exact solutions of these NPDEs plays an important and a significant role in all aspects of this subject. Many authors presented various powerful methods to deal with this problem, such as inverse scattering transformation method, Hirota bilinear method, homogeneous balance method, Bäcklund transformation, Darboux transformation, the generalized Jacobi elliptic function expansion method, the mapping deformation method and so on [6–10]. But once people noticed the complexity of nonlinear terms of NPDEs, they could not find the exact analytic solutions for many of them, especially with disturbed terms. Researchers had to

© 2017 The Author(s). Licensee InTech. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

develop some approximate and numerical methods for nonlinear theory; a great deal of efforts has been proposed for these problems, such as the multiple-scale method, the variational iteration method, the indirect matching method, the renormalization method, the Adomian decomposition method (ADM), the generalized differential transform method and so forth [11–13], among them the perturbation method [14], including the regular perturbation method, the singular perturbation method and the homotopy perturbation method (HPM) and so on.

Perturbation theory is widely used in numerical analysis as we all know. The earliest perturbation theory was built to deal with the unsolvable mathematical problems in the calculation of the motions of planets in the solar system [15]. The gradually increasing accuracy of astronomical observations led to incremental demands in the accuracy of solutions to Newton's gravitational equations, which extended and generalized the methods of perturbation theory. In the nineteenth century, Charles-Eugène Delaunay discovered the problem of small denominators which appeared in the nth term of the perturbative expansion when he was studying the perturbative expansion for the Earth-Moon-Sun system [16]. These welldeveloped perturbation methods were adopted and adapted to solve new problems arising during the development of Quantum Mechanics in the twentieth century. In the middle of the twentieth century, Richard Feynman realized that the perturbative expansion could be given a dramatic and beautiful graphical representation in terms of what are now called Feynman diagrams [17]. In the late twentieth century, because the broad questions about perturbation theory were found in the quantum physics community, including the difficulty of the nth term of the perturbative expansion and the demonstration of the convergent about the perturbative expansion, people had to pay more attention to the area of non-perturbative analysis, and much of the theoretical work goes under the name of quantum groups and non-commutative geometry [18]. As we all know, the solutions of the famous Korteweg-de Vries (KdV) equation cannot be reached by perturbation theory, even if the perturbations were carried out. Now, we can divide the perturbation theory to regular and singular perturbation theory; singular perturbation theory concerns those problems which depend on a parameter (here called ε) and whose solutions at a limiting value have a non-uniform behavior when the parameter tends to a pre-specified value. For regular perturbation problems, the solutions converge to the solutions of the limit problem as the parameter tends to the limit value. Both of these two methods are frequently used in physics and engineering today. There is no guarantee that perturbative methods lead to a convergent solution. In fact, the asymptotic series of the solution is the norm. In order to obtain the perturbative solution, we involve two distinct steps in general. The first is to assume that there is a convergent power asymptotic series about the parameter ε expressing the solution; then, the coefficients of the nth power of ε exist and can be computed via finite computation. The second step is to prove that the formal asymptotic series converges for ε small enough or to at least find a summation rule for the formal asymptotic series, thus providing a real solution to the problem.

The homotopy analysis method (HAM) was firstly proposed in 1992 by Liao [19], which yields a rapid convergence in most of the situations [20]. It also showed a high accuracy to solutions of the nonlinear differential systems. After this, many types of nonlinear problems were solved with HAM by others, such as nonlinear Schrödinger equation, fractional KdV- Burgers-Kuramoto equation, a generalized Hirota-Satsuma coupled KdV equation, discrete KdV equation and so on [21–24]. With this basic idea of HAM (as ℏ ¼ �1 and Hðx, tÞ ¼ 1), Jihuan He proposed the homotopy perturbation method(HPM) [25] which has been widely used to handle the nonlinear problems arising in the engineering and mathematical physics [26, 27].

In this chapter, we extend the applications of HAM and HPM with the aid of Fourier transformation to solve the generalized perturbed KdV-Burgers equation with power-law nonlinearity and a class of disturbed nonlinear Schrödinger equations in nonlinear optics. Many useful results are researched.

## 1.1. The homotopy analysis method (HAM)

Let us consider the following nonlinear equation

$$N[\mu(\mathbf{x}, t)] = 0,\tag{1}$$

where N is a nonlinear operator, uðx, tÞ is an unknown function and xand t denote spatial and temporal independent variables, respectively.

With the basic idea of the traditional homotopy method, we construct the following zero-order deformation equation

$$
\mu(1-q)L[\phi(\mathbf{x},t;q) - \mu\_0(\mathbf{x},t)] = q\hbar H(\mathbf{x},t)\mathcal{N}[\phi(\mathbf{x},t;q)]\tag{2}
$$

where ℏ 6¼ 0 is a non-zero auxiliary parameter, q∈½0, 1� is the embedding parameter, Hðx, tÞ is an auxiliary function, L is an auxiliary linear operator, u~0ðx, tÞ is an initial guess of uðx, tÞ and φðx, t; qÞ is an unknown function. Obviously, when q ¼ 0 and q ¼ 1, it holds

$$
\phi(\mathbf{x},t;\mathbf{0}) = u\_0(\mathbf{x},t), \\
\phi(\mathbf{x},t;\mathbf{1}) = u(\mathbf{x},t). \tag{3}
$$

Thus, as q increases from 0 to 1, the solution φðx, t; qÞ varies from the initial guess u0ðx, tÞ to the solution uðx, tÞ. Expanding φðx, t; qÞ in Taylor series with respect to q, we have

$$\begin{aligned} \phi(\mathbf{x}, t; q) &= u\_0 + \sum\_{m=1}^{\infty} u\_m q^m \\ \mathbf{x} &= u\_0 + q\mathbf{u}\_1 + q^2 \mathbf{u}\_2 + \cdots; \mathbf{u}\_0 = \tilde{u}\_0(\mathbf{x}, t), \mathbf{u}\_m = \mathbf{u}\_m(\mathbf{x}, t). \end{aligned} \tag{4}$$

where

develop some approximate and numerical methods for nonlinear theory; a great deal of efforts has been proposed for these problems, such as the multiple-scale method, the variational iteration method, the indirect matching method, the renormalization method, the Adomian decomposition method (ADM), the generalized differential transform method and so forth [11–13], among them the perturbation method [14], including the regular perturbation method, the singular perturbation method and the homotopy perturbation method (HPM) and so on.

44 Recent Studies in Perturbation Theory

Perturbation theory is widely used in numerical analysis as we all know. The earliest perturbation theory was built to deal with the unsolvable mathematical problems in the calculation of the motions of planets in the solar system [15]. The gradually increasing accuracy of astronomical observations led to incremental demands in the accuracy of solutions to Newton's gravitational equations, which extended and generalized the methods of perturbation theory. In the nineteenth century, Charles-Eugène Delaunay discovered the problem of small denominators which appeared in the nth term of the perturbative expansion when he was studying the perturbative expansion for the Earth-Moon-Sun system [16]. These welldeveloped perturbation methods were adopted and adapted to solve new problems arising during the development of Quantum Mechanics in the twentieth century. In the middle of the twentieth century, Richard Feynman realized that the perturbative expansion could be given a dramatic and beautiful graphical representation in terms of what are now called Feynman diagrams [17]. In the late twentieth century, because the broad questions about perturbation theory were found in the quantum physics community, including the difficulty of the nth term of the perturbative expansion and the demonstration of the convergent about the perturbative expansion, people had to pay more attention to the area of non-perturbative analysis, and much of the theoretical work goes under the name of quantum groups and non-commutative geometry [18]. As we all know, the solutions of the famous Korteweg-de Vries (KdV) equation cannot be reached by perturbation theory, even if the perturbations were carried out. Now, we can divide the perturbation theory to regular and singular perturbation theory; singular perturbation theory concerns those problems which depend on a parameter (here called ε) and whose solutions at a limiting value have a non-uniform behavior when the parameter tends to a pre-specified value. For regular perturbation problems, the solutions converge to the solutions of the limit problem as the parameter tends to the limit value. Both of these two methods are frequently used in physics and engineering today. There is no guarantee that perturbative methods lead to a convergent solution. In fact, the asymptotic series of the solution is the norm. In order to obtain the perturbative solution, we involve two distinct steps in general. The first is to assume that there is a convergent power asymptotic series about the parameter ε expressing the solution; then, the coefficients of the nth power of ε exist and can be computed via finite computation. The second step is to prove that the formal asymptotic series converges for ε small enough or to at least find a summation rule for the formal asymptotic

The homotopy analysis method (HAM) was firstly proposed in 1992 by Liao [19], which yields a rapid convergence in most of the situations [20]. It also showed a high accuracy to solutions of the nonlinear differential systems. After this, many types of nonlinear problems were solved with HAM by others, such as nonlinear Schrödinger equation, fractional KdV-

series, thus providing a real solution to the problem.

$$\left.u\_{m}(\mathbf{x},t)\right| = \frac{1}{m!} \frac{\partial^{m}}{\partial q^{m}} \phi(\mathbf{x},t;q)\Big|\_{q=0} = \mathbf{0} \tag{5}$$

If the auxiliary linear operator, the initial guess, the auxiliary parameter and the auxiliary function are so properly chosen such that they are smooth enough, the Taylor's series (4) with respect to q converges at q ¼ 1, and we have

$$\mu = \phi(\mathbf{x}, t; 1) = \sum\_{m=0}^{\infty} \mu\_{m\nu} \tag{6}$$

which must be one of the solutions of the original nonlinear equation, as proved by Liao. As ℏ ¼ �1 and Hðx, tÞ ¼ 1, Eq. (2) becomes

$$
\dot{\phi}(1-q)L[\phi(\mathbf{x},t;q) - u\_0(\mathbf{x},t)] + q\mathcal{N}[\phi(\mathbf{x},t;q)] = 0. \tag{7}
$$

Eq. (7) is used mostly in the HPM, whereas the solution is obtained directly, without using Taylor's series. As Hðx, tÞ ¼ 1, Eq. (2) becomes

$$(1 - q)L[\phi(\mathbf{x}, t; q) - \mu\_0(\mathbf{x}, t)] = q\hbar \mathcal{N}[\phi(\mathbf{x}, t; q)]\_\prime \tag{8}$$

which is used in the HAM when it is not introduced in the set of base functions. According to definition (5), the governing equation can be deduced from Eq. (2). Define the vector

$$
\overrightarrow{\boldsymbol{\mu}}\_{m}(\mathbf{x}, \mathbf{t}) = \{ \boldsymbol{\mu}\_{0\prime} \boldsymbol{\mu}\_{1\prime} \boldsymbol{\mu}\_{2\prime} \cdots \boldsymbol{\mu}\_{m} \}.\tag{9}
$$

Differentiating Eq. (2) m times with respect to the embedding parameter q and then setting q ¼ 0 and finally dividing them by m!, we have the so-called mth-order deformation equation

$$L[\boldsymbol{\mu}\_m(\mathbf{x}, t) - \chi\_m \boldsymbol{\mu}\_{m-1}(\mathbf{x}, t)] = \hbar H(\mathbf{x}, t) R\_{m-1}(\overrightarrow{\boldsymbol{\mu}}\_{m-1}, \mathbf{x}, t), \tag{10}$$

where

$$R\_{m-1}(\overrightarrow{\boldsymbol{\mu}}\_{m-1}, \mathbf{x}, t) = \frac{1}{(m-1)!} \frac{\partial^{m-1}}{\partial q^{m-1}} N[\phi(\mathbf{x}, t; q)] \Big|\_{q} = \mathbf{0} \tag{11}$$

And

$$\chi\_m = \begin{cases} 0, \ge 1 \\ 1, \ge 2 \end{cases} \tag{12}$$

It should be emphasized that umðx, tÞ for m ≥ 1 is governed by the linear Eq. (10) with the linear boundary conditions that come from the original problem, which can be easily solved by symbolic computation software such as Mathematica and Matlab.

#### 1.2. The homotopy perturbation method

To illustrate the basic concept of the homotopy perturbation method, consider the following nonlinear system of differential equations with boundary conditions

$$\begin{cases} A(u) = f(r), r \in \Omega, \\ B(u\_r \frac{\partial u}{\partial n}) = 0, r \in \Gamma = \partial \Omega \end{cases} \tag{13}$$

where B is a boundary operator and Γ is the boundary of the domain Ω, fðrÞ is a known analytical function. The differential operator A can be divided into two parts, L and N, in general, where L is a linear and N is a nonlinear operator. Eq. (13) can be rewritten as follows:

$$L(\mu) + N(\mu) = f(r). \tag{14}$$

We construct the following homotopy mapping Hðφ, qÞ:Ω � ½0, 1� ! R, which satisfies

$$H(\phi, q) = (1 - q)[L(\upsilon) - L(\tilde{\mu}\_0)] + q[A(\upsilon) - f(r)] = 0, q \in [0, 1], r \in \Omega,\tag{15}$$

where u~0is an initial approximation of Eq. (13), and is the embedding parameter; we have the following power series presentation for φ,

$$\phi = \sum\_{i=0}^{\bullet} \mu\_i(\mathbf{x}, t) \boldsymbol{\eta}^i = \mu\_0 + \eta \boldsymbol{u}\_1 + \eta^2 \boldsymbol{u}\_2 + \cdots. \tag{16}$$

The approximate solution can be obtained by setting q ¼ 1, that is

$$\mu = \lim\_{q \to 1} \phi = \mu\_0 + \mu\_1 + \mu\_2 + \cdots. \tag{17}$$

If we let u0ðx, tÞ ¼ u~0ðx, tÞ,notice the analytic properties of f , L, u~<sup>0</sup> and mapping (15), we know that the series of (17) is convergence in most cases when q∈½0, 1� [28]. We obtain the solution of Eq. (13).

To study the convergence of the method, let us state the following theorem.

Theorem (Sufficient Condition of Convergence).

Suppose that X and Y are Banach spaces and N : X ! Y is a contract nonlinear mapping that is

$$\forall \mu, \mu\* \in \mathcal{X}: \|N(\mu) - N(\mu\*)\| \le \gamma \|\mu - \mu\*\|, 0 < \gamma < 1. \tag{18}$$

Then, according to Banach's fixed point theorem, N has a unique fixed point u, that is NðuÞ ¼ u. Assume that the sequence generated by homotopy perturbation method can be written as

$$\mathcal{U}\mathcal{U}\_n = \mathcal{N}(\mathcal{U}\_{n-1}), \mathcal{U}\_n = \sum\_{i=0}^n \mu\_i, u\_i \in \mathcal{X}, n = 1, 2, 3, \cdots,\tag{19}$$

and suppose that

<sup>u</sup> <sup>¼</sup> <sup>φ</sup>ðx, t; <sup>1</sup>Þ ¼ <sup>X</sup><sup>∞</sup>

which must be one of the solutions of the original nonlinear equation, as proved by Liao. As

Eq. (7) is used mostly in the HPM, whereas the solution is obtained directly, without using

which is used in the HAM when it is not introduced in the set of base functions. According to

Differentiating Eq. (2) m times with respect to the embedding parameter q and then setting q ¼ 0 and finally dividing them by m!, we have the so-called mth-order deformation equation

definition (5), the governing equation can be deduced from Eq. (2). Define the vector

L½umðx, tÞ � χmum�<sup>1</sup>ðx, tÞ� ¼ ℏHðx, tÞRm�<sup>1</sup>ðu

ðm � 1Þ!

<sup>χ</sup><sup>m</sup> <sup>¼</sup> <sup>0</sup>, x <sup>≤</sup> <sup>1</sup> 1, x ≥ 2 :

It should be emphasized that umðx, tÞ for m ≥ 1 is governed by the linear Eq. (10) with the linear boundary conditions that come from the original problem, which can be easily solved by

To illustrate the basic concept of the homotopy perturbation method, consider the following

AðuÞ ¼ fðrÞ, r ∈ Ω, ð13:1Þ

Þ ¼ 0, r∈ Γ ¼ ∂Ω ð13:2Þ

,

�

∂<sup>m</sup>�<sup>1</sup>

u !

<sup>m</sup>�<sup>1</sup>, x, tÞ ¼ <sup>1</sup>

symbolic computation software such as Mathematica and Matlab.

nonlinear system of differential equations with boundary conditions

8 < :

Bðu, ∂u ∂n

ℏ ¼ �1 and Hðx, tÞ ¼ 1, Eq. (2) becomes

46 Recent Studies in Perturbation Theory

where

And

Taylor's series. As Hðx, tÞ ¼ 1, Eq. (2) becomes

Rm�<sup>1</sup>ðu !

1.2. The homotopy perturbation method

m¼0

ð1 � qÞL½φðx, t; qÞ � u0ðx, tÞ� þ qN½φðx, t; qÞ� ¼ 0: ð7Þ

ð1 � qÞL½φðx, t; qÞ � u0ðx, tÞ� ¼ qℏN½φðx, t; qÞ�, ð8Þ

<sup>m</sup>ðx, tÞ ¼ {u0, u1, u2, ⋯, um}: ð9Þ

!

<sup>m</sup>�<sup>1</sup>, x, tÞ, ð10Þ

ð12Þ

ð13Þ

<sup>∂</sup>qm�<sup>1</sup> <sup>N</sup>½φðx, t; qÞ�j <sup>q</sup> <sup>¼</sup> <sup>0</sup> : <sup>ð</sup>11<sup>Þ</sup>

um, ð6Þ

$$\mathcal{U}\_0 = \mu\_0 \in \mathcal{B}\_r(\mu), \mathcal{B}\_r(\mu) = \{ \mu \* \in \mathcal{X} | |\| \mu \* - \mu \| < \gamma \}\tag{20}$$

$$\text{then, we have (i) } \mathcal{U}\_n \in B\_r(\mu) \text{ (ii)} \lim\_{n \to \infty} \mathcal{U}\_n = \mu. \tag{21}$$

Proof. (i) By inductive approach, for n ¼ 1, we have

kU<sup>1</sup> � uk¼kNðU0Þ � NðuÞk ≤ γkU<sup>0</sup> � uk and then <sup>k</sup>Un � <sup>u</sup>k¼kNðUn�<sup>1</sup>Þ � <sup>N</sup>ðuÞk <sup>≤</sup> <sup>γ</sup><sup>n</sup>kU<sup>0</sup> � <sup>u</sup><sup>k</sup> <sup>≤</sup> <sup>γ</sup>nr ) Un <sup>∈</sup>Brðu<sup>Þ</sup>

(ii) Because of 0 <sup>&</sup>lt; <sup>γ</sup> <sup>&</sup>lt; 1, we have lim<sup>n</sup>!<sup>∞</sup> <sup>k</sup>Un � <sup>u</sup>k ¼ 0 that is lim<sup>n</sup>!<sup>∞</sup> Un ¼ u.

## 2. Application to the generalized perturbed KdV-Burgers equation

Consider the following generalized perturbed KdV-Burgers equation

$$
\mu\_t + \alpha u^p u\_x + \beta u^{2p} u\_x + \gamma u\_{xx} + \delta u\_{xxx} = f(t, x, u). \tag{22}
$$

where α, β, γ, δ, p are arbitrary constants, and f ¼ fðt, x, uÞ is a disturbed term, which is a sufficiently smooth function in a corresponding domain.

This equation with p ≥ 1 is a model for long-wave propagation in nonlinear media with dispersion and dissipation. Eq. (22) arises in a variety of physical contexts which include a number of equations, and many valuable results about Eq. (22) have been studied by many authors in [29–31]. In fact, if one takes different value of α, β, γ, δ, p and f , Eq.(22) represents a large number of equations, such as KdV equation, MKdV equation, CKdV equation, Burgers equation, KdV-Burgers equation and the equations as the following forms.

Fitzhugh-Nagumo equation [32]:

$$
\mu\_t - \mu\_{xx} = f = \mu(\mu - \alpha)(1 - \mu), \tag{23}
$$

Burgers-Huxley equation [33]

$$
\lambda u\_t + \alpha u^\delta u\_x - \lambda u\_{xx} = f = \beta u (1 - u^\delta)(\eta u^\delta - \gamma) \tag{24}
$$

Burgers-Fisher equation [34]

$$
\mu\_t + \alpha u^\delta u\_x - u\_{xx} = f = \beta u(1 - u^\delta) \tag{25}
$$

It's significant for us to handle Eq. (22).

#### 2.1. The generalized KdV-Burgers equation

If we let f ¼ 0 in Eq. (22), we can obtain the famous generalized KdV-Burgers equation with nonlinear terms of any order [35, 36].

$$
\mu\_t + \alpha \mu^p u\_x + \beta u^{2p} u\_x + \gamma u\_{xx} + \delta u\_{xxx} = 0. \tag{26}
$$

Eq. (26) is solved on the infinite line �∞ < x < ∞ together with the initial condition uðx, 0Þ ¼ fðxÞ, � ∞ < x < ∞ by using the HAM. We first introduce the traveling wave transform

$$
\xi = \mathfrak{x} + \mathfrak{c}t + \xi\_0. \tag{27}
$$

where c are constants to be determined later and ξ<sup>0</sup> ∈ C are arbitrary constants. Secondly, we make the following transformation:

$$
\mu(\xi) = v^{1/p}(\xi). \tag{28}
$$

Eq. (26) is reduced to the following form:

kU<sup>1</sup> � uk¼kNðU0Þ � NðuÞk ≤ γkU<sup>0</sup> � uk and then

(ii) Because of 0 <sup>&</sup>lt; <sup>γ</sup> <sup>&</sup>lt; 1, we have lim<sup>n</sup>!<sup>∞</sup>

48 Recent Studies in Perturbation Theory

Fitzhugh-Nagumo equation [32]:

Burgers-Huxley equation [33]

Burgers-Fisher equation [34]

It's significant for us to handle Eq. (22).

nonlinear terms of any order [35, 36].

2.1. The generalized KdV-Burgers equation

<sup>k</sup>Un � <sup>u</sup>k¼kNðUn�<sup>1</sup>Þ � <sup>N</sup>ðuÞk <sup>≤</sup> <sup>γ</sup><sup>n</sup>kU<sup>0</sup> � <sup>u</sup><sup>k</sup> <sup>≤</sup> <sup>γ</sup>nr ) Un <sup>∈</sup>Brðu<sup>Þ</sup>

where α, β, γ, δ, p are arbitrary constants, and f ¼ fðt, x, uÞ is a disturbed term, which is a

This equation with p ≥ 1 is a model for long-wave propagation in nonlinear media with dispersion and dissipation. Eq. (22) arises in a variety of physical contexts which include a number of equations, and many valuable results about Eq. (22) have been studied by many authors in [29–31]. In fact, if one takes different value of α, β, γ, δ, p and f , Eq.(22) represents a large number of equations, such as KdV equation, MKdV equation, CKdV equation, Burgers equa-

ux � <sup>λ</sup>uxx <sup>¼</sup> <sup>f</sup> <sup>¼</sup> <sup>β</sup>uð<sup>1</sup> � <sup>u</sup><sup>δ</sup>

If we let f ¼ 0 in Eq. (22), we can obtain the famous generalized KdV-Burgers equation with

Eq. (26) is solved on the infinite line �∞ < x < ∞ together with the initial condition uðx, 0Þ ¼

ux <sup>þ</sup> <sup>β</sup>u<sup>2</sup><sup>p</sup>

fðxÞ, � ∞ < x < ∞ by using the HAM. We first introduce the traveling wave transform

ux � uxx <sup>¼</sup> <sup>f</sup> <sup>¼</sup> <sup>β</sup>uð<sup>1</sup> � <sup>u</sup><sup>δ</sup>

2. Application to the generalized perturbed KdV-Burgers equation

Consider the following generalized perturbed KdV-Burgers equation

tion, KdV-Burgers equation and the equations as the following forms.

ut <sup>þ</sup> <sup>α</sup>u<sup>δ</sup>

ut <sup>þ</sup> <sup>α</sup>u<sup>p</sup>

ux <sup>þ</sup> <sup>β</sup>u<sup>2</sup><sup>p</sup>

ut <sup>þ</sup> <sup>α</sup>up

sufficiently smooth function in a corresponding domain.

ut <sup>þ</sup> <sup>α</sup>u<sup>δ</sup>

<sup>k</sup>Un � <sup>u</sup>k ¼ 0 that is lim<sup>n</sup>!<sup>∞</sup>

Un ¼ u.

ux þ γuxx þ δuxxx ¼ fðt, x, uÞ: ð22Þ

ut � uxx ¼ f ¼ uðu � αÞð1 � uÞ, ð23Þ

Þðηu<sup>δ</sup> � <sup>γ</sup>Þ ð24<sup>Þ</sup>

Þ ð25Þ

ux þ γuxx þ δuxxx ¼ 0: ð26Þ

$$\begin{aligned} p(p+1)(2p+1)\delta v(\xi)v''(\xi) + (p+1)(2p+1)\delta (1-p)v^2(\xi) \\ + p(p+1)(2p+1)\gamma v(\xi)v'(\xi) + cp^2(p+1)(2p+1)v^2(\xi) \\ + p^2(2p+1)\alpha v^3(\xi) + p^2(p+1)\beta v^4(\xi) = 0 \end{aligned} \tag{29}$$

where the derivatives are performed with respect to the coordinate ξ. We can conclude that Eq. (26) has the following solution, by using the deformation mapping method:

$$\tilde{u}\_0 = \left\{-\frac{c(1+p)}{2\alpha} + \frac{d(1+p)\gamma}{p\alpha} \sqrt{\frac{c^2 p^2}{4d^2 \gamma^2}} \tanh(d\sqrt{\frac{c^2 p^2}{4d^2 \gamma^2}} (\mathbf{x} + ct + \xi\_0))\right\}^{\frac{1}{p}}.\tag{30}$$

#### 2.2. The approximate solutions by using HAM

To solve Eq. (22) by means of HAM, we choose the initial approximation

$$
\mu\_0(\mathbf{x}, t) = \tilde{u}\_0(\mathbf{x}, t)\Big|\_{t=0} = \mathbf{g}(\mathbf{x}), \tag{31}
$$

where u~0ðx, tÞ is an arbitrary exact solution of Eq. (23).

According to Eq. (1), we define the nonlinear operator

$$N[\phi] = \phi\_t + a\phi^p\phi\_x + \beta\phi^{2p}\phi\_x + \gamma\phi\_{xx} + \delta\phi\_{xxx} - f(\phi), \phi = \phi(\mathbf{x}, t; q). \tag{32}$$

It is reasonable to express the solution uðx, tÞ by set of base functions gnðxÞt n, n ≥ 0, under the rule of solution expression; it is straightforward to choose Hðx, tÞ ¼ 1 and the linear operator

$$L[\phi(\mathbf{x}, t; q)] = \frac{\partial \phi(\mathbf{x}, t; q)}{\partial t} \tag{33}$$

with the property

$$L[c(\mathbf{x})] = \mathbf{0}.\tag{34}$$

From Eqs. (10, 11 and 32), we have

$$\begin{split} R\_{m-1}(\overrightarrow{\boldsymbol{\mu}}\_{m-1}, \mathbf{x}, t) &= \boldsymbol{u}\_{m-1, t} + \gamma \boldsymbol{u}\_{m-1, \mathbf{x}\mathbf{x}} + \delta \boldsymbol{u}\_{m-1, \mathbf{x}\mathbf{x}} + a \boldsymbol{D}\_{m-1}(\phi^p \phi\_x) \\ &+ \beta \boldsymbol{D}\_{m-1}(\phi^{2p} \phi\_x) - F(\boldsymbol{u}\_0, \boldsymbol{u}\_1, \cdots, \boldsymbol{u}\_{m-1}), \end{split} \tag{35}$$

where

$$D\_{m-1}(\phi^n \phi\_x) = \sum\_{k\_1=0}^n \sum\_{k\_2=0}^{k\_1} \sum\_{k\_3=0}^{k\_2} \cdots \sum\_{k\_{m-1}=0}^{k\_{m-2}} \sum\_{i=0}^{m-1} \mathbb{C}\_n^{k\_1} \mathbb{C}\_{k\_1}^{k\_2} \mathbb{C}\_{k\_2}^{k\_3} \cdots \mathbb{C}\_{k\_{m-2}}^{k\_{m-1}} u\_0^{n-k\_1} u\_1^{k\_1-k\_2} \cdots u\_{m-1}^{k\_{m-1}} u\_{i\xi} \tag{36}$$

and n ≥ k<sup>1</sup> ≥ k<sup>2</sup> ≥ ⋯ ≥ km�<sup>1</sup> ≥ 0 ∈ N, with

$$\begin{aligned} \sum\_{j=1}^{m-1} k\_j + i &= m - 1, i = 0, \cdots, m - 1 \\\\ F(u\_0, u\_1, \cdots, u\_{m-1}) &= \frac{1}{(n-1)!} \frac{\partial^{(m-1)}}{\partial q^{m-1}} f(\mathbf{x}, t, u) \bigg|\_{q=0} \end{aligned} \tag{37}$$

Now, the solution of the mth-order deformation in Eq. (10) with initial condition umðx, tÞ ¼ 0 for m ≥ 1 becomes

$$
\mu\_m = \chi\_m \mu\_{m-1} + L^{-1} [\hbar R\_{m-1}(\overrightarrow{\mu}\_{m-1}, \ge, t)],\tag{38}
$$

Thus, from Eqs. (31, 35 and 38), we can successively obtain

$$
\mu\_0 = \tilde{\mu}\_0(\mathbf{x}, \mathbf{0}) = \mathbf{g}(\mathbf{x}), \tag{39}
$$

$$
\mu\_1 = -\hbar t[\tilde{u}\_{0t} + f(\mu\_0)],\\\tilde{u}\_{0t} = \frac{\partial}{\partial t}\tilde{u}\_0(\mathbf{x}, t)|\_{t=0\prime} \tag{40}
$$

$$u\_2 = (1+\hbar)u\_1 + \hbar(au\_0^p u\_{1,x} + \beta u\_0^{2p} u\_{1,x} + \gamma u\_{1,xx} + \delta u\_{1,xxx} - f\_u(u\_0)u\_1)t \tag{41}$$

$$u\_m = (1+\hbar)u\_{m-1} + \hbar[\gamma u\_{1,xx} + \delta u\_{1,xxx} + aD\_{m-1}(\phi^p \phi\_x) + \beta D\_{m-1}(\phi^{2p} \phi\_x) - F(u\_0, u\_1, \dots, u\_{m-1})]t \tag{42}$$

⋮

⋮

We obtain the mth-order approximate solution and exact solution of Eq. (22) as follows

$$
\mu\_{m,appr} = \sum\_{k=0}^{m} \mu\_{k\prime} \mu\_{exact} = \phi(\mathbf{x}, t; 1) = \lim\_{m \to \infty} \sum\_{k=0}^{m} \mu\_k \tag{43}
$$

if we choose

$$\tilde{u}\_0(\mathbf{x},0) = \left\{ -\frac{c(1+p)}{2\alpha} + \frac{d(1+p)\gamma}{p\alpha} \sqrt{\frac{c^2 p^2}{4d^2 \gamma^2}} \tanh(d\sqrt{\frac{c^2 p^2}{4d^2 \gamma^2}} \mathbf{x}) \right\}^{\frac{1}{p}}.\tag{44}$$

From Eqs. (39–44), we can obtain the corresponding approximate solution of Eq. (22).

#### 2.3. Example

where

for m ≥ 1 becomes

if we choose

Dm�<sup>1</sup>ðφ<sup>n</sup>φxÞ ¼ <sup>X</sup><sup>n</sup>

and n ≥ k<sup>1</sup> ≥ k<sup>2</sup> ≥ ⋯ ≥ km�<sup>1</sup> ≥ 0 ∈ N, with

50 Recent Studies in Perturbation Theory

k1¼0

Xm�1 j¼1

X k1

X k2

⋯ X km�<sup>2</sup>

km�1¼0

kj þ i ¼ m � 1, i ¼ 0, ⋯, m � 1

Xm�1 i¼0 Ck<sup>1</sup> <sup>n</sup> Ck<sup>2</sup> k1 Ck3 k2 ⋯Ckm�<sup>1</sup> km�<sup>2</sup> un�k<sup>1</sup> <sup>0</sup> uk1�k<sup>2</sup>

ðn � 1Þ!

Now, the solution of the mth-order deformation in Eq. (10) with initial condition umðx, tÞ ¼ 0

∂ðm�1<sup>Þ</sup>

½ℏRm�<sup>1</sup>ðu !

∂t

<sup>φ</sup>xÞ þ <sup>β</sup>Dm�<sup>1</sup>ðφ<sup>2</sup><sup>p</sup>

<sup>∂</sup>qm�<sup>1</sup> <sup>f</sup>ðx, t, u<sup>Þ</sup>

q ¼ 0 :

� � � � �

u<sup>0</sup> ¼ u~0ðx, 0Þ ¼ gðxÞ, ð39Þ

Xm k¼0

> ffiffiffiffiffiffiffiffiffiffiffiffi c<sup>2</sup>p<sup>2</sup> 4d<sup>2</sup> γ2

xÞ

p

s

<sup>0</sup> u1, <sup>x</sup> þ γu1, xx þ δu1, xxx � f <sup>u</sup>ðu0Þu1Þt ð41Þ

<sup>1</sup> <sup>⋯</sup>ukm�<sup>1</sup>

<sup>m</sup>�<sup>1</sup>, x, tÞ�, ð38Þ

u~0ðx, tÞj<sup>t</sup>¼<sup>0</sup>, ð40Þ

φxÞ � Fðu0, u1, ⋯, um�<sup>1</sup>Þ�t

uk ð43Þ

: ð44Þ

ð42Þ

<sup>m</sup>�<sup>1</sup>ui<sup>ξ</sup> <sup>ð</sup>36<sup>Þ</sup>

ð37Þ

k3¼0

<sup>F</sup>ðu0, u1, <sup>⋯</sup>, um�<sup>1</sup>Þ ¼ <sup>1</sup>

um <sup>¼</sup> <sup>χ</sup>mum�<sup>1</sup> <sup>þ</sup> <sup>L</sup>�<sup>1</sup>

<sup>u</sup><sup>1</sup> ¼ �ℏt½u~0<sup>t</sup> <sup>þ</sup> <sup>f</sup>ðu0Þ�, <sup>u</sup>~0<sup>t</sup> <sup>¼</sup> <sup>∂</sup>

<sup>0</sup>u1, <sup>x</sup> þ βu

2p

⋮

⋮

uk, uexact <sup>¼</sup> <sup>φ</sup>ðx, t; <sup>1</sup>Þ ¼ lim<sup>m</sup>!<sup>∞</sup>

ffiffiffiffiffiffiffiffiffiffiffiffi c<sup>2</sup>p<sup>2</sup> 4d<sup>2</sup> γ2

tanhðd

s

( )<sup>1</sup>

We obtain the mth-order approximate solution and exact solution of Eq. (22) as follows

Thus, from Eqs. (31, 35 and 38), we can successively obtain

<sup>u</sup><sup>2</sup> ¼ ð<sup>1</sup> <sup>þ</sup> <sup>ℏ</sup>Þu<sup>1</sup> <sup>þ</sup> <sup>ℏ</sup>ðαup

um ¼ ð<sup>1</sup> <sup>þ</sup> <sup>ℏ</sup>Þum�<sup>1</sup> <sup>þ</sup> <sup>ℏ</sup>½γu1, xx <sup>þ</sup> <sup>δ</sup>u1, xxx <sup>þ</sup> <sup>α</sup>Dm�<sup>1</sup>ðφ<sup>p</sup>

um, appr <sup>¼</sup> <sup>X</sup><sup>m</sup>

<sup>u</sup>~0ðx, <sup>0</sup>Þ¼ � <sup>c</sup>ð<sup>1</sup> <sup>þ</sup> <sup>p</sup><sup>Þ</sup>

k¼0

2α

þ

From Eqs. (39–44), we can obtain the corresponding approximate solution of Eq. (22).

dð1 þ pÞγ pα

k2¼0

In the following, three examples are presented to illustrate the effectiveness of the HAM. We first plot the so-called ℏ curves of u 00 apprð0, 0Þ and u 000 apprð0, 0Þ to discover the valid region of ℏ, which corresponds to the line segment nearly parallel to the horizontal axis. The simulate comparison between the initial exact solution, exact solution and the fourth order of approximation solution is given.

Now, we consider the small perturbation term <sup>f</sup> <sup>¼</sup> <sup>ε</sup>~<sup>f</sup> in Eq. (22).

Example 1. Consider the CKdV equation with small disturbed term

$$
\mu\_t + 6\mu u\_x - 6\mu^2 u\_x + u\_{xxx} = \varepsilon \mu^2,\\
0 < \varepsilon \ll 1 \tag{45}
$$

with the initial exact solution

$$
\tilde{u}\_0(\mathbf{x}, t) = \frac{1}{2} - \frac{1}{2} \tanh[\frac{1}{2}(\mathbf{x} - t)]. \tag{46}
$$

From Section 2.2, we have

$$
\mu\_0 = \frac{1}{2} - \frac{1}{2}\tanh\left(\frac{1}{2}\mathbf{x}\right),
\tilde{\mu}\_{0t} = \frac{1}{4}\operatorname{sech}^2\left(\frac{1}{2}\mathbf{x}\right),\tag{47}
$$

$$u\_1 = -\hbar \left\{ \frac{1}{4} \text{sech}^2 \left(\frac{1}{2}\mathbf{x}\right) + \varepsilon \left[\frac{1}{2} - \frac{1}{2} \tanh\left(\frac{1}{2}\mathbf{x}\right)\right]^2 \right\} t \tag{48}$$

u<sup>2</sup> ¼ �ð1 þ ℏÞℏt 1 4 sech2 <sup>1</sup> 2 x � � þ ε 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � <sup>2</sup> ( ) � <sup>ℏ</sup><sup>2</sup> t 2 ( 6 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � 1 4 sech<sup>2</sup> <sup>1</sup> 2 x � � þ ε 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � <sup>2</sup> ( )x <sup>þ</sup> <sup>6</sup>ℏ<sup>2</sup> t <sup>2</sup> 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � <sup>2</sup> <sup>1</sup> 4 sech2 <sup>1</sup> 2 x � � þ ε 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � <sup>2</sup> ( )x � <sup>ℏ</sup><sup>2</sup> t <sup>2</sup> 1 4 sech<sup>2</sup> <sup>1</sup> 2 x � � þ ε 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � <sup>2</sup> ( )xxx <sup>þ</sup> <sup>2</sup>εℏ<sup>2</sup> t <sup>2</sup> 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � 1 4 sech<sup>2</sup> <sup>1</sup> 2 x � � þ ε 1 2 � 1 2 tanh <sup>1</sup> 2 x � � � � <sup>2</sup> ( ) ¼ ℏt <sup>32</sup> cosh <sup>x</sup> 2 � � � sinh <sup>x</sup> 2 h i � � sec <sup>h</sup><sup>5</sup> <sup>x</sup> 2 � �( ℏð Þ� 5t � 3 � 3ε 3 � 3ε <sup>þ</sup> <sup>2</sup>ℏtεð Þþ <sup>1</sup> <sup>þ</sup> <sup>ε</sup> 2coshð Þ<sup>x</sup> <sup>2</sup><sup>ε</sup> � <sup>2</sup> � <sup>2</sup>ℏð Þþ <sup>1</sup> <sup>þ</sup> <sup>ε</sup> <sup>ℏ</sup><sup>t</sup> <sup>2</sup>ε<sup>2</sup> <sup>þ</sup> <sup>7</sup><sup>ε</sup> � <sup>3</sup> � � � � <sup>þ</sup> <sup>ℏ</sup> <sup>t</sup> � <sup>ε</sup> � <sup>1</sup> <sup>þ</sup> <sup>2</sup>tε<sup>2</sup> � � � <sup>ε</sup> � <sup>1</sup> � �cosh 2ð Þ� <sup>x</sup> 2sinh <sup>x</sup> 2 � �½<sup>1</sup> � <sup>ε</sup> <sup>þ</sup> <sup>ℏ</sup> � <sup>ε</sup><sup>ℏ</sup> <sup>þ</sup> <sup>ℏ</sup><sup>t</sup> <sup>2</sup> � <sup>3</sup><sup>ε</sup> <sup>þ</sup> <sup>2</sup>ε<sup>2</sup> � � <sup>þ</sup> ð Þ <sup>1</sup> � <sup>ε</sup> cosh<sup>x</sup> <sup>þ</sup> <sup>ℏ</sup> <sup>1</sup> � <sup>t</sup> � <sup>ε</sup> <sup>þ</sup> <sup>2</sup>tε<sup>2</sup> � �coshxÞ�) ð49Þ

$$\begin{split} u\_{qpr} &= \frac{1}{2} - \frac{1}{2}\tanh\left(\frac{1}{2}\chi\right) - \hbar\left\{\frac{1}{4}\operatorname{sech}^2\left(\frac{1}{2}\chi\right) + -\varepsilon\left[\frac{1}{2} - \frac{1}{2}\tanh\left(\frac{1}{2}\chi\right)\right]^2\right\}t \\ &+ \frac{\hbar t}{32}\left[\cosh\left(\frac{\chi}{2}\right) - \sinh\left(\frac{\chi}{2}\right)\operatorname{sech}^5\left(\frac{\chi}{2}\right)\right]\hbar(5t - 3 - 3\varepsilon - 3 - 3\varepsilon + \\ 2\hbar t\varepsilon(1 + \varepsilon) + 2\cosh(\chi)\left[2\varepsilon - 2 - 2\hbar(1 + \varepsilon) + \hbar t\big(2\varepsilon^2 + 7\varepsilon - 3\}\right] \\ &+ \left[\hbar\left(t - \varepsilon - 1 + 2t\varepsilon^2\right) - \varepsilon - 1\right]\cosh(2\chi) - 2\sinh\left(\frac{\chi}{2}\right)\left[1 - \varepsilon + \hbar - \varepsilon\hbar\right.\end{split} \tag{50}$$

$$+ \hbar t\left(2 - 3\varepsilon + 2\varepsilon^2\right) + (1 - \varepsilon)\cosh\chi + \hbar\left(1 - t - \varepsilon + 2t\varepsilon^2\right)\cosh\chi\right] \bigg| + \cdots$$

Example 2. Consider the KdV-Burgers equation with small disturbed term

$$
\mu\_t + \theta \mu \mu\_x + \mu\_{xx} - \mu\_{xxx} = \varepsilon \sin \mu \tag{51}
$$

with the initial exact solution

uappr <sup>¼</sup> <sup>1</sup>

52 Recent Studies in Perturbation Theory

The ℏ curves of u

Figure 1. (a) The ℏ curves of u

00

and the fourth order of approximation solution.

apprð0, 0Þ and u

000

in Figure 1(b).

þ ℏt <sup>32</sup> cosh <sup>x</sup>

00

apprð0, 0Þ and u

tanh <sup>1</sup> 2 x � �

> 2 � �

h i � �

000

� <sup>ℏ</sup> <sup>1</sup> 4

� sinh <sup>x</sup> 2 sech2 <sup>1</sup> 2 x � �

> sec <sup>h</sup><sup>5</sup> <sup>x</sup> 2 � �(

<sup>2</sup>ℏtεð Þþ <sup>1</sup> <sup>þ</sup> <sup>ε</sup> 2coshð Þ<sup>x</sup> <sup>2</sup><sup>ε</sup> � <sup>2</sup> � <sup>2</sup>ℏð Þþ <sup>1</sup> <sup>þ</sup> <sup>ε</sup> <sup>ℏ</sup><sup>t</sup> <sup>2</sup>ε<sup>2</sup> <sup>þ</sup> <sup>7</sup><sup>ε</sup> � <sup>3</sup> � � � �

<sup>þ</sup> <sup>ℏ</sup><sup>t</sup> <sup>2</sup> � <sup>3</sup><sup>ε</sup> <sup>þ</sup> <sup>2</sup>ε<sup>2</sup> � � <sup>þ</sup> ð Þ <sup>1</sup> � <sup>ε</sup> cosh<sup>x</sup> <sup>þ</sup> <sup>ℏ</sup> <sup>1</sup> � <sup>t</sup> � <sup>ε</sup> <sup>þ</sup> <sup>2</sup>tε<sup>2</sup> � �cosh<sup>x</sup>

ison between the initial exact solution and the fourth order of approximation solution is shown

<sup>þ</sup> <sup>ℏ</sup> <sup>t</sup> � <sup>ε</sup> � <sup>1</sup> <sup>þ</sup> <sup>2</sup>tε<sup>2</sup> � � � <sup>ε</sup> � <sup>1</sup> � �cosh 2ð Þ� <sup>x</sup> 2sinh <sup>x</sup>

þ �ε 1 2 � 1 2

� � � � <sup>2</sup> ( )

tanh <sup>1</sup> 2 x

ℏð Þ� 5t � 3 � 3ε 3 � 3εþ

2 � �

apprð0, 0Þ in Eq. (45) are shown in Figure 1(a), and the compar-

apprð0, 0Þat the fourth order of approximation. (b) The initial exact solution

t

ð50Þ

½1 � ε þ ℏ � εℏ

�� ) þ ⋯

$$
\tilde{u}\_0(\mathbf{x}, t) = \frac{1}{50} \left\{ 1 - \coth[-\frac{1}{10}(\mathbf{x} - \frac{6}{25}t)] \right\}^2 \tag{52}
$$

From Section 2.2, we have

$$
\mu\_0 = \frac{1}{50} [1 - \coth(-\frac{1}{10}x)]^2,\\
\tilde{u}\_{0t} = \frac{3}{3125} \text{csch}^2(\frac{1}{10}x)[1 + \coth(\frac{1}{10}x)] \tag{53}
$$

$$u\_1 = -\hbar\varepsilon\sin\left\{\frac{1}{50}[1-\coth(\frac{-1}{10}x)]^2\right\}t - \frac{3\hbar t}{3125}\text{csch}^2(\frac{1}{10}x)[1+\coth(\frac{1}{10}x)]\tag{54}$$

$$u\_2 = (1+\hbar)u\_1 + \hbar t(\delta u\_0 u\_{1,\ge} + u\_{1,\ge x} - u\_{1,\ge x} - \varepsilon u\_1 \cos u\_0) \tag{55}$$

$$\begin{split} u\_{\text{appr}} &= \frac{1}{50} \left[ 1 - \coth\left( -\frac{1}{10} \mathbf{x} \right) \right]^2 - \hbar \varepsilon \sin\left\{ \frac{1}{50} \left[ 1 - \coth\left( -\frac{1}{10} \mathbf{x} \right) \right]^2 \right\} t \\ &- \frac{3}{3125} \hbar t \text{csch}^2 \left( \frac{1}{10} \mathbf{x} \right) \left[ 1 + \coth\left( \frac{1}{10} \mathbf{x} \right) \right] + u\_2 + \cdots \end{split} \tag{56}$$

The ℏ curves of u 00 apprð0, 0Þ and u 000 apprð0, 0Þ in Eq. (51) are shown in Figure 2(a); the comparison between the initial exact solution and the fourth order of approximation solution is shown in Figure 2(b).

Figure 2. (a) The ℏ curves of u 00 apprð10ln2, 0Þ and u 000 apprð10ln2, 0Þ at the fourth order of approximation. (b) The initial exact solution and the fourth order of approximation solution.

Example 3. Consider the Burgers-Fisher equation

$$
\mu\_t + \mu^2 \mu\_x - \mu\_{xx} = \varepsilon \mathfrak{u} (1 - \mathfrak{u}^2) \tag{57}
$$

with the exact solution and the initial exact solution

$$
\mu\_{1\_{\text{carr}}} = \sqrt{\frac{1}{2} - \frac{1}{2}\tanh\left[\frac{1}{3}x - \frac{1 + 9\varepsilon}{9}t + \xi\_0\right]}\tag{58}
$$

$$
\mu\_{2\_{\text{root}}} = \sqrt{\frac{1}{2} - \frac{1}{2}\coth\left[\frac{1}{3}x - \frac{1 + 9\varepsilon}{9}t + \xi\_0\right]}\tag{59}
$$

$$
\tilde{u}\_0(\mathbf{x}, t) = \sqrt{\frac{1}{2} - \frac{1}{2}\tanh\left[\frac{1}{3}\mathbf{x} - \frac{1}{9}t + \xi\_0\right]}\tag{60}
$$

From Section 2.2, we have

$$
\mu\_0 = \sqrt{\frac{1}{2} - \frac{1}{2}\tanh\left(\frac{1}{3}x\right)}, \qquad \tilde{u}\_{0t} = \mathrm{sech}^2\left(\frac{1}{3}x\right) / 18\sqrt{2 - 2\tanh\left(\frac{1}{3}x\right)}\tag{61}
$$

$$u\_1 = -\frac{\hbar t \text{sech}^2(\frac{1}{3}\mathbf{x})}{18\sqrt{2 - 2\tanh(\frac{1}{3}\mathbf{x})}} - \hbar t \varepsilon \sqrt{\frac{1}{2} - \frac{1}{2}\tanh\left(\frac{1}{3}\mathbf{x}\right)} \left(\frac{1}{2} + \frac{1}{2}\tanh\left(\frac{1}{3}\mathbf{x}\right)\right) \tag{62}$$

$$u\_2 = (1+\hbar)u\_1 + \hbar(\alpha u\_0 u\_{1,x} - u\_{1,xx} - \varepsilon u\_1 + \mathfrak{Z}\varepsilon u\_0^2 u\_1) \tag{63}$$

Figure 3. (a) The ℏ curves of u 00 apprð0, 0Þ and u 000 apprð0, 0Þ at the fourth order of approximation. (b) The exact solution, initial exact solution and the fourth order of approximation solution.

#### Homotopy Asymptotic Method and Its Application http://dx.doi.org/10.5772/67876 55

$$\mu\_{appr} = \sqrt{\frac{1}{2} - \frac{1}{2}\tanh\left(\frac{1}{3}x\right)} - \frac{\hbar t \text{sech}^2\left(\frac{1}{3}x\right)}{18\sqrt{2 - 2\tanh\left(\frac{1}{3}x\right)}}\tag{64}$$

$$-\hbar t \varepsilon \sqrt{\frac{1}{2} - \frac{1}{2}\tanh\left(\frac{1}{3}x\right)}\left(\frac{1}{2} + \frac{1}{2}\tanh\left(\frac{1}{3}x\right)\right) + u\_2 + \cdots$$

The ℏ curves of u 00 apprð0, 0Þ and u 000 apprð0, 0Þ in Eq. (57) are shown in Figure 3(a), the comparison between the initial exact solution and the fourth order of approximation solution is shown in Figure 3(b).

## 3. Application to the generalized perturbed NLS equation

In this section, we will use the HPM and Fourier's transformation to search for the solution of the generalized perturbed nonlinear Schrödinger equation (GPNLS)

$$i\frac{\partial u}{\partial z} + \frac{1}{2}\beta(z)\frac{\partial^2 u}{\partial t^2} + \delta(z)u|u|^2 - i\alpha(z)u = \beta(z)f(u, z, t). \tag{65}$$

If we let t ! x, z ! t,Eq. (65) turns to the following form

$$i\hbar\frac{\partial u}{\partial t} + \frac{1}{2}\beta(t)\frac{\partial^2 u}{\partial x^2} + \delta(t)u|u|^2 - i\alpha(t)u = \beta(t)f(u,t,x). \tag{66}$$

where disturbed term f is a sufficiently smooth function in a corresponding domain. αðtÞ represents the heat-insulating amplification or loss. βðtÞ and δðtÞ are the slowly increasing dispersion coefficient and nonlinear coefficient, respectively. The transmission of soliton in the real communication system of optical soliton is described by Eq. (66) with f ¼ 0 [37–39].

$$i\frac{\partial u}{\partial t} + \frac{1}{2}\beta(t)\frac{\partial^2 u}{\partial x^2} + \delta(t)u|u|^2 - i\alpha(t)u = 0. \tag{67}$$

We make the transformation

Example 3. Consider the Burgers-Fisher equation

with the exact solution and the initial exact solution

From Section 2.2, we have

54 Recent Studies in Perturbation Theory

Figure 3. (a) The ℏ curves of u

00

exact solution and the fourth order of approximation solution.

apprð0, 0Þ and u

000

u<sup>0</sup> ¼

<sup>u</sup><sup>1</sup> ¼ � <sup>ℏ</sup>tsech<sup>2</sup> <sup>1</sup>

18

u<sup>1</sup>exact ¼

u<sup>2</sup>exact ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

s � �

tanh <sup>1</sup> 3 x

<sup>3</sup> <sup>x</sup> � �

3 x <sup>q</sup> � � � <sup>ℏ</sup>t<sup>ε</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup> � 2tanh <sup>1</sup>

u~0ðx, tÞ ¼

ut <sup>þ</sup> <sup>u</sup><sup>2</sup>

ux � uxx <sup>¼</sup> <sup>ε</sup>uð<sup>1</sup> � <sup>u</sup><sup>2</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

s � �

tanh <sup>1</sup> 3 <sup>x</sup> � <sup>1</sup> 9 t þ ξ<sup>0</sup>

s � �

<sup>x</sup> � <sup>1</sup> <sup>þ</sup> <sup>9</sup><sup>ε</sup>

<sup>x</sup> � <sup>1</sup> <sup>þ</sup> <sup>9</sup><sup>ε</sup>

3 x � �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

<sup>s</sup> � � <sup>1</sup>

tanh <sup>1</sup> 3 x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

s � �

<sup>9</sup> <sup>t</sup> <sup>þ</sup> <sup>ξ</sup><sup>0</sup>

<sup>9</sup> <sup>t</sup> <sup>þ</sup> <sup>ξ</sup><sup>0</sup>

=18

apprð0, 0Þ at the fourth order of approximation. (b) The exact solution, initial

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup> � 2tanh <sup>1</sup>

> tanh <sup>1</sup> 3 x

� � � �

s � �

3 x

<sup>0</sup>u1Þ ð63Þ

tanh <sup>1</sup> 3

coth <sup>1</sup> 3

, <sup>u</sup>~0<sup>t</sup> <sup>¼</sup> sech2 <sup>1</sup>

<sup>u</sup><sup>2</sup> ¼ ð<sup>1</sup> <sup>þ</sup> <sup>ℏ</sup>Þu<sup>1</sup> <sup>þ</sup> <sup>ℏ</sup>tðαu0u1, <sup>x</sup> � <sup>u</sup>1, xx � <sup>ε</sup>u<sup>1</sup> <sup>þ</sup> <sup>3</sup>εu<sup>2</sup>

Þ ð57Þ

ð58Þ

ð59Þ

ð60Þ

ð61Þ

ð62Þ

$$
\mu = A(t)\varphi(\xi)\mathbf{c}^{i\eta}, \xi = k\_1\mathbf{x} + c\_1(t), \eta = k\_2\mathbf{x} + c\_2(t) \tag{68}
$$

With the following consistency conditions,

$$A(t) = c\varepsilon \ell\_0^{\int\_0^t a(\tau)d\tau}, c\_1(t) = -k\_1 k\_2 \int\_0^t \beta(\tau)d\tau, c\_2(t) = \frac{1}{2} (a\_2 k\_1^2 - k\_2^2) \int\_0^t \beta(\tau)d\tau,\\ \delta(t) = \frac{-a\_4 k\_1^2}{c^2} \beta(t) e^{-2 \int\_0^t a(\tau)d\tau} \tag{69}$$

where k1, k2, a2, a4, c are arbitrary non-zero constants.

If we let <sup>f</sup>ðu, t, xÞ ¼ <sup>1</sup> 2 k2 <sup>1</sup>fðϕÞe<sup>i</sup><sup>η</sup>, substituting Eq. (68) into Eq. (67), we have

$$
\rho \overset{\circ}{\xi}\_{\xi\xi} - a\_2 \varphi - 2a\_4 \varphi^3 = f(\varphi). \tag{70}
$$

By using the general mapping deformation method [10, 40], we can obtain the following solutions of the corresponding undisturbed Eq. (70) when f ¼ 0.

$$
\tilde{\varphi}\_0 = c n[k\_1 \mathbf{x} - k\_1 k\_2 \int\_0^t \beta(\tau) d\tau]. \tag{71}
$$

In order to obtain the solution of Eq. (70), we introduce the following homotopic mapping Hðϕ, pÞ: R � I ! R,

$$H(\boldsymbol{\varphi}, p) = \mathcal{L}\boldsymbol{\varphi} - \mathcal{L}\tilde{\boldsymbol{\varphi}}\_0 + \eta \Big(\mathcal{L}\tilde{\boldsymbol{\varphi}}\_0 - 2a\_4 \boldsymbol{\varphi}^3 - f(\boldsymbol{\varphi})\Big). \tag{72}$$

where R ¼ ð�∞, þ ∞Þ, I ¼ ½0, 1�,ϕ~ <sup>0</sup> is an initial approximate solution to Eq. (70), and the linear operator L is expressed as

$$L(\mu) = \boldsymbol{\wp}\_{\xi\xi}^{''} - \boldsymbol{a}\_2 \boldsymbol{\wp}. \tag{73}$$

Obviously, from mapping Eq. (72), Hðϕ, 1Þ ¼ 0 is the same as Eq. (70). Thus, the solution of Eq. (70) is the same as the solution of Hðϕ, qÞ as q ! 1.

#### 3.1. Approximate solution

In order to obtain the solution of Eq. (70), set

$$\varphi = \sum\_{i=0}^{\infty} \varphi\_i(\xi) q^i = \varphi\_0 + q\varphi\_1 + q^2 \varphi\_2 + \dotsb \tag{74}$$

If we let ϕ<sup>0</sup> ¼ ϕ~ <sup>0</sup>,notice the analytical properties of f ,ϕ~ <sup>0</sup>, and mapping Eq. (72), we can deduce that the series of Eq. (74) are uniform convergence when q∈½0, 1�. Substituting expression (74) into Hðu, qÞ ¼ 0 and expanding nonlinear terms into the power series in powers of q, we compare the coefficients of the same power of q on both sides of the equation and we have

$$\mathfrak{q}^0: \mathcal{L}\mathfrak{q}\_0 = \mathcal{L}\tilde{\mathfrak{q}}\_{0'} \tag{75}$$

$$\mathfrak{q}^1: L\mathfrak{q}\_1 = f(\mathfrak{q}\_0),\tag{76}$$

$$\mathfrak{q}^2: \mathcal{L}\wp\_2 = \mathfrak{6}a\_4\wp\_0^2\wp\_1 + f\_{\mathfrak{q}}(\wp\_0)\wp\_{1'} \tag{77}$$

$$\begin{aligned} \cdots\\ q^n : \operatorname{Lep}\_n &= F(\varphi\_0, \varphi\_1, \cdots, \varphi\_{n-1}) + 2a\_4 \sum\_{k\_1=0}^3 \sum\_{k\_2=0}^{k\_1} \sum\_{k\_3=0}^{k\_2} \cdots\\ \sum\_{k\_{n-1}=0}^{k\_{n-2}} \operatorname{C}\_3^{k\_1} \operatorname{C}\_{k\_1}^{k\_2} \cdots \operatorname{C}\_{k\_{n-2}}^{k\_{n-1}} \varphi\_0^{3-k\_1} \varphi\_1^{k\_1-k\_2} \varphi\_2^{k\_2-k\_3} \cdots \varphi\_{n-2}^{k\_{n-2}-k\_{n-1}} \varphi\_{n-1}^{k\_{n-1}}\\ &\cdots \end{aligned} \tag{78}$$

$$\text{where} \quad 3 \ge k\_1 \ge k\_2 \ge \cdots \ge k\_{n-1} \ge 0 \in N\_\prime \quad \sum\_{j=1}^{n-1} k\_j = n - 1, n \in N^+ \quad \text{and} \quad F(\boldsymbol{\varphi}\_0, \boldsymbol{\varphi}\_1, \dots, \boldsymbol{\varphi}\_{n-1}) = \frac{1}{(n-1)!} \frac{\partial^{(n-1)}}{\partial \boldsymbol{\eta}^{n-1}}$$

<sup>f</sup>ðϕ0,ϕ1, <sup>⋯</sup>,ϕ<sup>n</sup>�<sup>1</sup>Þj <sup>p</sup> <sup>¼</sup> <sup>0</sup> .

If we let <sup>f</sup>ðu, t, xÞ ¼ <sup>1</sup>

56 Recent Studies in Perturbation Theory

Hðϕ, pÞ: R � I ! R,

operator L is expressed as

3.1. Approximate solution

2 k2

<sup>1</sup>fðϕÞe<sup>i</sup><sup>η</sup>, substituting Eq. (68) into Eq. (67), we have

By using the general mapping deformation method [10, 40], we can obtain the following

In order to obtain the solution of Eq. (70), we introduce the following homotopic mapping

�

where R ¼ ð�∞, þ ∞Þ, I ¼ ½0, 1�,ϕ~ <sup>0</sup> is an initial approximate solution to Eq. (70), and the linear

Obviously, from mapping Eq. (72), Hðϕ, 1Þ ¼ 0 is the same as Eq. (70). Thus, the solution of

<sup>ð</sup>ξÞq<sup>i</sup> <sup>¼</sup> <sup>ϕ</sup><sup>0</sup> <sup>þ</sup> <sup>q</sup>ϕ<sup>1</sup> <sup>þ</sup> <sup>q</sup><sup>2</sup>

If we let ϕ<sup>0</sup> ¼ ϕ~ <sup>0</sup>,notice the analytical properties of f ,ϕ~ <sup>0</sup>, and mapping Eq. (72), we can deduce that the series of Eq. (74) are uniform convergence when q∈½0, 1�. Substituting expression (74) into Hðu, qÞ ¼ 0 and expanding nonlinear terms into the power series in powers of q, we compare the coefficients of the same power of q on both sides of the equation and we have

⋯

⋯

X 3

X k1

X k2

k3¼0 ⋯

<sup>n</sup>�<sup>2</sup> <sup>ϕ</sup>kn�<sup>1</sup> n�1

k2¼0

<sup>2</sup> <sup>⋯</sup>ϕkn�2�kn�<sup>1</sup>

k1¼0

<sup>1</sup> <sup>ϕ</sup><sup>k</sup>2�k<sup>3</sup>

<sup>q</sup><sup>2</sup> : <sup>L</sup>ϕ<sup>2</sup> <sup>¼</sup> <sup>6</sup>a4ϕ<sup>2</sup>

<sup>q</sup><sup>n</sup> : <sup>L</sup>ϕ<sup>n</sup> <sup>¼</sup> <sup>F</sup>ðϕ0,ϕ1, <sup>⋯</sup>,ϕ<sup>n</sup>�<sup>1</sup>Þ þ <sup>2</sup>a<sup>4</sup>

<sup>L</sup>ðuÞ ¼ <sup>ϕ</sup><sup>00</sup>

ðt 0

<sup>L</sup>ϕ<sup>~</sup> <sup>0</sup> � <sup>2</sup>a4ϕ<sup>3</sup> � <sup>f</sup>ðϕ<sup>Þ</sup>

ϕ~ <sup>0</sup> ¼ cn½k1x � k1k<sup>2</sup>

ξξ � <sup>a</sup>2<sup>ϕ</sup> � <sup>2</sup>a4ϕ<sup>3</sup> <sup>¼</sup> <sup>f</sup>ðϕÞ: <sup>ð</sup>70<sup>Þ</sup>

βðτÞdτ�: ð71Þ

: ð72Þ

�

ξξ � a2ϕ: ð73Þ

<sup>q</sup><sup>0</sup> : <sup>L</sup>ϕ<sup>0</sup> <sup>¼</sup> <sup>L</sup>ϕ<sup>~</sup> <sup>0</sup>, <sup>ð</sup>75<sup>Þ</sup>

<sup>q</sup><sup>1</sup> : <sup>L</sup>ϕ<sup>1</sup> <sup>¼</sup> <sup>f</sup>ðϕ0Þ, <sup>ð</sup>76<sup>Þ</sup>

<sup>0</sup>ϕ<sup>1</sup> þ f <sup>ϕ</sup>ðϕ0Þϕ1, ð77Þ

: ð78Þ

ϕ<sup>2</sup> þ ⋯ ð74Þ

ϕ00

Hðϕ, pÞ ¼ Lϕ � Lϕ~ <sup>0</sup> þ q

Eq. (70) is the same as the solution of Hðϕ, qÞ as q ! 1.

<sup>ϕ</sup> <sup>¼</sup> <sup>X</sup><sup>∞</sup> i¼0 ϕi

In order to obtain the solution of Eq. (70), set

X kn�<sup>2</sup>

Ck1 <sup>3</sup> <sup>C</sup><sup>k</sup><sup>2</sup> k1 Ck3 k2 ⋯Ckn�<sup>1</sup> kn�<sup>2</sup> ϕ<sup>3</sup>�k<sup>1</sup> <sup>0</sup> <sup>ϕ</sup><sup>k</sup>1�k<sup>2</sup>

kn�1¼0

solutions of the corresponding undisturbed Eq. (70) when f ¼ 0.

From Eq. (75) we have ϕ0ðξÞ ¼ ϕ~ <sup>0</sup>ðξÞ. If we select ϕ1j <sup>ξ</sup>¼<sup>0</sup> <sup>¼</sup> 0, by using Fourier transformation and from Eq. (76), we have

$$\varphi\_1 = \frac{1}{\sqrt{a\_2}} \int\_0^\zeta f(\varphi\_0) (e^{\sqrt{a\_2}(\xi - \tau)} - e^{-\sqrt{a\_2}(\xi - \tau)}) d\tau, \quad a\_2 \neq 0, \quad f(\varphi\_0) = f(\varphi\_0(\tau)). \tag{79}$$

If we select ϕ2j <sup>ξ</sup>¼<sup>0</sup> <sup>¼</sup> 0, from Eq. (77) we have

$$\varphi\_2 = \frac{1}{\sqrt{a\_2}} \int\_0^\xi [6a\_4 \varphi\_0^2 \varphi\_1 + f\_\varphi(\varphi\_0)\varphi\_1](e^{\sqrt{a\_2}(\xi-\tau)} - e^{-\sqrt{a\_2}(\xi-\tau)})d\tau. \tag{80}$$

where a<sup>2</sup> 6¼ 0,ϕ<sup>0</sup> ¼ ϕ0ðτÞ,ϕ<sup>1</sup> ¼ ϕ1ðτÞ.

We obtain the first- and second-order approximate solutions u1homðx, tÞ and u2homðx, tÞ of the Eq. (70) as follows:

$$\varphi\_{1\text{hom}}(\mathbf{x},t) = \tilde{\varphi}\_0 + \frac{1}{\sqrt{2m^2 - 1}} \int\_0^\xi f(\rho\_0) (e^{\sqrt{2m^2 - 1}(\xi - \tau)} - e^{-\sqrt{2m^2 - 1}(\xi - \tau)}) d\tau \tag{81}$$

$$u\_{1\text{hom}}(\mathbf{x},t) = \text{cc}\_0^{\int\_0^t a(\tau)d\tau + i[k\_2x + \frac{1}{2}]} \mathbb{J}\_0^t(2m^2 - 1)\mathbf{k}\_1^2 - k\_2^2)\mathbb{K}(\tau)d\tau] \tag{82}$$

$$\begin{split} q\_{2\text{hom}}(\mathbf{x},t) &= \ddot{\varphi}\_{0} + \frac{1}{\sqrt{2m^{2}-1}} \int\_{0}^{\xi} f(\varphi\_{0}) (e^{\sqrt{2m^{2}-1}(\xi-\tau)} - e^{-\sqrt{2m^{2}-1}(\xi-\tau)}) d\tau \\ &+ \frac{1}{\sqrt{2m^{2}-1}} \int\_{0}^{\xi} [-6m^{2}\varphi\_{0}^{2}\varphi\_{1} + f\_{\varphi}(\varphi\_{0})\varphi\_{1}] (e^{\sqrt{2m^{2}-1}(\xi-\tau)} - e^{-\sqrt{2m^{2}-1}(\xi-\tau)}) d\tau \\ &\mu\_{2\text{hom}}(\mathbf{x},t) = c\varepsilon e^{\int\_{0}^{t} a(\tau)d\tau + i[k\_{2}\mathbf{x} + \frac{1}{2}\int\_{0}^{t} ((2m^{2}-1)k\_{1}^{2} - k\_{2}^{2})\mathcal{S}(\tau)d\tau} \boldsymbol{q}\_{2\text{hom}}(\mathbf{x},t) \end{split} \tag{84}$$

With the same process, we can also obtain the N-order approximate solution

ϕnhomðx, tÞ ¼ ϕ~ <sup>0</sup> þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m<sup>2</sup> � <sup>1</sup> <sup>p</sup> ðξ 0 <sup>f</sup>ðϕ0Þð<sup>e</sup> ffiffiffi a2 <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> � <sup>e</sup> � ffiffiffi a2 <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> Þdτ þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m<sup>2</sup> � <sup>1</sup> <sup>p</sup> ðξ 0 ½�6m<sup>2</sup> ϕ2 <sup>0</sup>ϕ<sup>1</sup> þ f <sup>ϕ</sup>ðϕ0Þϕ1�ðe ffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> � <sup>e</sup> � ffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> Þdτ þ ⋯ þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m<sup>2</sup> � <sup>1</sup> <sup>p</sup> ðξ 0 ðe ffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> � <sup>e</sup> � ffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> Þ½Fðϕ0,ϕ1, <sup>⋯</sup>,ϕ<sup>n</sup>�<sup>1</sup>Þ � <sup>2</sup>m<sup>2</sup> X 3 k1¼0 X k1 k2¼0 X k2 k3¼0 ⋯ X kn�<sup>2</sup> kn�1¼0 Ck1 <sup>3</sup> Ck<sup>2</sup> k1 Ck3 k2 ⋯Ckn�<sup>1</sup> kn�<sup>2</sup> ϕ<sup>3</sup>�k<sup>1</sup> <sup>0</sup> <sup>ϕ</sup><sup>k</sup>1�k<sup>2</sup> <sup>1</sup> <sup>ϕ</sup><sup>k</sup>2�k<sup>3</sup> <sup>2</sup> <sup>⋯</sup>ϕkn�2�kn�<sup>1</sup> <sup>n</sup>�<sup>2</sup> <sup>ϕ</sup>kn�<sup>1</sup> <sup>n</sup>�<sup>1</sup>�d<sup>τ</sup> ð85Þ unhomðx, tÞ ¼ ce Ðt <sup>0</sup> <sup>α</sup>ðτÞdτþi½k2xþ<sup>1</sup> 2 Ðt 0 ðð2m2�1Þk<sup>2</sup> 1�k<sup>2</sup> <sup>2</sup>ÞβðτÞdτ� ϕnhomðx, tÞ ð86Þ

$$\begin{aligned} \text{where } &3 \ge k\_1 \ge k\_2 \ge \cdots \ge k\_{n-1} \ge 0 \in N, \sum\_{j=1}^{n-1} k\_j = n-1, n \in \mathbb{N}^+ \text{ and} \\\\ F(\boldsymbol{\varphi}\_{0'} \boldsymbol{\varphi}\_{1'} \cdots \boldsymbol{\varphi}\_{n-1}) &= \frac{1}{(n-1)!} \frac{\partial^{(n-1)}}{\partial \boldsymbol{\eta}^{n-1}} f(\boldsymbol{\varphi}\_{0'} \boldsymbol{\varphi}\_{1'} \cdots \boldsymbol{\varphi}\_{n-1}) \bigg|\_{\boldsymbol{\theta}} \quad \tag{87} \end{aligned} \tag{87}$$

ðn � 1Þ!

#### 3.2. Comparison of accuracy

In order to explain the accuracy of the expressions of the approximate solution represented by Eq. (86), we consider the small perturbation term

$$i\frac{\partial u}{\partial t} + \frac{1}{2}\beta(t)\frac{\partial^2 u}{\partial x^2} + \delta(t)u|u|^2 - i\alpha(t)u = \frac{1}{2}\varepsilon k\_1^2 \beta(t)e^{i\eta}\sin^n\phi\tag{88}$$

�

where n∈ Nþ,ϕ ¼ e � Ðt <sup>0</sup> <sup>α</sup>ðτÞdτ�iðk2xþ<sup>1</sup> <sup>2</sup>ða2k 2 1�k<sup>2</sup> 2Þ Ðt <sup>0</sup> <sup>β</sup>ðτÞdτ<sup>Þ</sup> u=c,0 < ε ≪ 1.

From the discussion of Section 3.1, we obtain the second-order approximate Jacobi-like elliptic function solution of Eq. (88) as follows

$$\begin{split} \varphi\_{2\text{hom}}(\mathbf{x},t) &= c\mathfrak{n}[k\_1\mathbf{x} - k\_1k\_2] \int\_0^t \beta(\tau)d\tau] + \frac{\varepsilon}{\sqrt{2m^2 - 1}} \Big|\_0^\xi \sin^n(\varphi\_0)(e^{\sqrt{2m^2 - 1}(\xi - \tau)} \\ &- e^{-\sqrt{2m^2 - 1}(\xi - \tau)})d\tau + \frac{1}{\sqrt{2m^2 - 1}} \int\_0^\xi [-\epsilon m^2 \varphi\_0^2 \varphi\_1 + \varepsilon n \sin^{n-1}(\varphi\_0) \\ &\qquad \cos\left(\varphi\_0\right)\varrho\_1\big](e^{\sqrt{2m^2 - 1}(\xi - \tau)} - e^{-\sqrt{2m^2 - 1}(\xi - \tau)})d\tau \\ \mathfrak{u}\_{2\text{hom}}(\mathbf{x},t) &= c\varrho\_0^\ell \int\_0^t a(\tau)d\tau + i[k\_2\mathbf{x} + \frac{1}{2}\int\_0^t ((2m^2 - 1)k\_1^2 - k\_2^2)\beta(\tau)d\tau]} \varrho\_{2\text{hom}}(\mathbf{x},t). \end{split} \tag{90}$$

Set <sup>ϕ</sup>exaðx, tÞ ¼ <sup>X</sup><sup>∞</sup> i¼0 ϕi ðx, tÞ to be an exact solution of Eq. (88), notice that

$$\begin{split} L(\boldsymbol{p}\_{\rm can} - \boldsymbol{p}\_{2\rm hom}) &= f(\boldsymbol{\varrho}) + 2a\_{4}\boldsymbol{\varrho}\_{\rm ex}^{-3} - [2a\_{4}\boldsymbol{\varrho}\_{0}^{-3} + f(\boldsymbol{\varrho}\_{0}) + 6a\_{4}\boldsymbol{\varrho}\_{0}^{2}\boldsymbol{\varrho}\_{1} \\ &+ f\_{\boldsymbol{\varrho}}(\boldsymbol{\varrho}\_{0})\boldsymbol{\varrho}\_{1}] = \boldsymbol{\varepsilon}\sin^{\boldsymbol{\mathfrak{u}}}\left(\sum\_{i=0}^{\boldsymbol{\mathfrak{u}}}\boldsymbol{\varrho}\_{i}\right) + 2a\_{4}\left(\sum\_{i=0}^{\boldsymbol{\mathfrak{u}}}\boldsymbol{\varrho}\_{i}\right)^{3} - [2a\_{4}\boldsymbol{\mathfrak{q}}\_{0}^{3} + \boldsymbol{\varepsilon}\sin^{\boldsymbol{\mathfrak{u}}}(\boldsymbol{\varrho}\_{0}), \ (91) \\ &+ 6a\_{4}\boldsymbol{\mathfrak{q}}\_{0}^{2}\boldsymbol{\mathfrak{q}}\_{1} + \boldsymbol{\varepsilon}\boldsymbol{\mathfrak{u}}\sin^{\boldsymbol{\mathfrak{u}}-1}(\boldsymbol{\varrho}\_{0})\cos(\boldsymbol{\varrho}\_{0})\boldsymbol{\varrho}\_{1}] = O(\boldsymbol{\varepsilon}^{2}) \end{split} (91)$$

where 0 < ε ≪ 1, selecting arbitrary constants such that ϕexað0Þ ¼ ϕ2homð0Þ, from the fixed point theorem [41], we have <sup>ϕ</sup>exa � <sup>ϕ</sup>2hom <sup>¼</sup> <sup>O</sup>ðε<sup>2</sup>Þ, then

$$\begin{split} |u|\_{\rm ext} - u\_{2\rm hom}| &= |A(t)e^{i\eta}|\varphi\_{\rm ext} - \varphi\_{2\rm hom}| \\ &= \left| \frac{e^2 A n \sin^{n-1}(\varphi\_0) \cos(\varphi\_0)}{\sqrt{2m^2 - 1}} \right|\_0^\zeta \sin^n(\varphi\_0) (e^{\sqrt{a}(\zeta - \tau)} - e^{-\sqrt{a}(\zeta - \tau)}) d\tau \bigg|\_{\zeta = 0} \end{split} \tag{92}$$

Figure 4. A comparison between the curves of solutions ju1homðξÞj (solid line) and ju0ðξÞj (dashed line) with ε ¼ 0:01.

Figure 5. A comparison between the curves of solutions ju1homðξÞj (solid line) and ju0ðξÞj (dashed line) with ε ¼ 0:001.

Therefore, from the above result, we know that the approximate solution,u2hom, obtained by asymptotic method and possesses better accuracy.

Set AðtÞ ¼ 1, k<sup>1</sup> ¼ k<sup>2</sup> ¼ 1, βðtÞ ¼ 1, m ! 1, n ¼ 1, ξ∈ ½0, 3� and ε ¼ 0:01, 0:001 for Eq. (90), and then, we will have the curves of solutions ju1homðξÞj and ju0ðξÞj and be able to compare them; see Figures 4 and 5. From Figures 4 and 5, it is easy to see that as 0 < ε ≪ 1 is a small parameter, and the solutions ju1homðξÞj and ju0ðξÞj are very close to each other. This behavior is coincident with that of the approximate solution of the weakly disturbed evolution in Eq. (88).

## 4. Conclusions

where 3 ≥ k<sup>1</sup> ≥ k<sup>2</sup> ≥ ⋯ ≥ kn�<sup>1</sup> ≥ 0 ∈ N,

58 Recent Studies in Perturbation Theory

3.2. Comparison of accuracy

where n∈ Nþ,ϕ ¼ e

Set <sup>ϕ</sup>exaðx, tÞ ¼ <sup>X</sup><sup>∞</sup>

i¼0 ϕi Xn�1 j¼1

<sup>F</sup>ðϕ0,ϕ1, <sup>⋯</sup>,ϕ<sup>n</sup>�<sup>1</sup>Þ ¼ <sup>1</sup>

Eq. (86), we consider the small perturbation term

<sup>0</sup> <sup>α</sup>ðτÞdτ�iðk2xþ<sup>1</sup>

� <sup>e</sup>� ffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup>

cos ðϕ0Þϕ1�ðe

Ðt

<sup>þ</sup> <sup>f</sup> <sup>ϕ</sup>ðϕ0Þϕ1� ¼ <sup>ε</sup> sin <sup>n</sup> <sup>X</sup><sup>∞</sup>

u2homðx, tÞ ¼ ce

<sup>L</sup>ðϕexa � <sup>ϕ</sup>2homÞ ¼ <sup>f</sup>ðϕÞ þ <sup>2</sup>a4ϕexa<sup>3</sup> � ½2a4ϕ<sup>0</sup>

<sup>þ</sup> <sup>6</sup>a4ϕ<sup>2</sup>

<sup>j</sup>uexa � <sup>u</sup>2homj¼jAðtÞe<sup>i</sup><sup>η</sup>½ϕexa � <sup>ϕ</sup>2hom�j

� � � � �

point theorem [41], we have <sup>ϕ</sup>exa � <sup>ϕ</sup>2hom <sup>¼</sup> <sup>O</sup>ðε<sup>2</sup>Þ, then

<sup>¼</sup> <sup>ε</sup><sup>2</sup>An sin <sup>n</sup>�<sup>1</sup>ðϕ0<sup>Þ</sup> cos <sup>ð</sup>ϕ0<sup>Þ</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m<sup>2</sup> � <sup>1</sup> <sup>p</sup>

i ∂u ∂t þ 1 2 βðtÞ ∂<sup>2</sup>u

� Ðt

function solution of Eq. (88) as follows

ϕ2homðx, tÞ ¼ cn½k1x � k1k<sup>2</sup>

kj ¼ n � 1, n∈ N<sup>þ</sup> and

∂ðn�1<sup>Þ</sup>

In order to explain the accuracy of the expressions of the approximate solution represented by

From the discussion of Section 3.1, we obtain the second-order approximate Jacobi-like elliptic

<sup>β</sup>ðτÞdτ� þ <sup>ε</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m<sup>2</sup> � <sup>1</sup> <sup>p</sup>

<sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> � <sup>e</sup>� ffiffiffiffiffiffiffiffiffiffi

2 Ðt 0 ðð2m2�1Þk<sup>2</sup>

ðx, tÞ to be an exact solution of Eq. (88), notice that

i¼0 ϕi !

<sup>0</sup>ϕ<sup>1</sup> <sup>þ</sup> <sup>ε</sup><sup>n</sup> sin <sup>n</sup>�<sup>1</sup>ðϕ0<sup>Þ</sup> cos <sup>ð</sup>ϕ0Þϕ1� ¼ <sup>O</sup>ðε<sup>2</sup><sup>Þ</sup>

where 0 < ε ≪ 1, selecting arbitrary constants such that ϕexað0Þ ¼ ϕ2homð0Þ, from the fixed

ðξ 0

<sup>2</sup> � <sup>i</sup>αðtÞ<sup>u</sup> <sup>¼</sup> <sup>1</sup>

2 εk 2 <sup>1</sup>βðtÞe

> ðξ 0

sin <sup>n</sup>ðϕ0Þð<sup>e</sup>

Þdτ

<sup>0</sup>ϕ<sup>1</sup>

� ½2a4ϕ<sup>0</sup>

� ffiffiffi a2 <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup>

X∞ i¼0 ϕi !<sup>3</sup>

ffiffiffiffiffiffiffiffiffiffi <sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup>

<sup>0</sup>ϕ<sup>1</sup> <sup>þ</sup> <sup>ε</sup><sup>n</sup> sin <sup>n</sup>�<sup>1</sup>

u=c,0 < ε ≪ 1.

<sup>2</sup>m<sup>2</sup> � <sup>1</sup> <sup>p</sup>

<sup>3</sup> <sup>þ</sup> <sup>f</sup>ðϕ0Þ þ <sup>6</sup>a4ϕ<sup>2</sup>

þ 2a<sup>4</sup>

sin <sup>n</sup>ðϕ0Þð<sup>e</sup> ffiffiffi

a2 <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup> � <sup>e</sup>

ðξ 0 ½�6m<sup>2</sup> ϕ2

<sup>2</sup>m2�<sup>1</sup> <sup>p</sup> <sup>ð</sup>ξ�τ<sup>Þ</sup>

1�k<sup>2</sup> <sup>2</sup>ÞβðτÞdτ�

<sup>∂</sup>pn�<sup>1</sup> <sup>f</sup>ðϕ0,ϕ1, <sup>⋯</sup>,ϕ<sup>n</sup>�<sup>1</sup><sup>Þ</sup> <sup>p</sup> <sup>¼</sup> <sup>0</sup>

� � � � �

<sup>i</sup><sup>η</sup> sin <sup>n</sup>ϕ, <sup>ð</sup>88<sup>Þ</sup>

ðϕ0Þ

ϕ2homðx, tÞ: ð90Þ

<sup>3</sup> <sup>þ</sup> <sup>ε</sup> sin <sup>n</sup>ðϕ0<sup>Þ</sup>

Þdτ

� � � � �

<sup>¼</sup> <sup>O</sup>ðε<sup>2</sup>Þ:

ð92Þ

ð87Þ

ð89Þ

, ð91Þ

ðn � 1Þ!

<sup>∂</sup>x<sup>2</sup> <sup>þ</sup> <sup>δ</sup>ðtÞuju<sup>j</sup>

<sup>2</sup>ða2k 2 1�k<sup>2</sup> 2Þ Ðt <sup>0</sup> <sup>β</sup>ðτÞdτ<sup>Þ</sup>

> ðt 0

> > Þdτ þ

ffiffiffiffiffiffiffiffiffiffi

<sup>0</sup> <sup>α</sup>ðτÞdτþi½k2xþ<sup>1</sup>

We research the generalized perturbed KdV-Burgers equation and GPNLS equation by using the HAM and HPM; these two powerful straightforward methods are much more simple and efficient than some other asymptotic methods such as perturbation method and Adomian decomposition method and so on. The Jacobi elliptic function and solitary wave approximate solution with arbitrary degree of accuracy for the disturbed equation are researched, which shows that these two methods have wide applications in science and engineering and also can be used in the soliton equation with complex variables, but it is still worth to research whether or not these two methods can be used in the system with high dimension and high order.

## Acknowledgements

The work is supported by the Scientific Research Foundation of Nanjing Institute of Technology (Grant No. ZKJ201513,2016YB22).

## Author details

Baojian Hong

Address all correspondence to: hbj@njit.edu.cn

Department of Mathematical and Physical Science, Nanjing Institute of Technology, Nanjing, China

## References


shows that these two methods have wide applications in science and engineering and also can be used in the soliton equation with complex variables, but it is still worth to research whether or not these two methods can be used in the system with high dimension and high order.

The work is supported by the Scientific Research Foundation of Nanjing Institute of Technology

Department of Mathematical and Physical Science, Nanjing Institute of Technology, Nanjing,

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60 Recent Studies in Perturbation Theory

Author details

Baojian Hong

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(Grant No. ZKJ201513,2016YB22).

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New York: Cambridge University Press; 1991.

Address all correspondence to: hbj@njit.edu.cn


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[24] S. Abbasbandy. The application of homotopy analysis method to solve a generalized Hirota–Satsuma coupled KdV equation. Physics Letters A, 361(6), 2007, 478–483.

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[26] D.D. Ganji. The application of He's homotopy perturbation method to nonlinear equa-

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[28] Jafar Biazar, Hossein Aminikhah. Study of convergence of homotopy perturbation method for systems of partial differential equations. Computers and Mathematics with Applica-

[29] B. Li, Y. Chen, H.Q. Zhang. Explicit exact solutions for compound KdV-type and compound KdV Burgers-type equations with nonlinear terms of any ord. Chaos, Solitons &

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[31] B.F. Feng, Takuji Kawahara. Stationary travelling-wave solutions of an unstable KdV-

[32] S. Abbasbandy. Soliton solutions for the Fitzhugh-Nagumo equation with the homotopy

[33] A. Molabahrami, F. Khani. The homotopy analysis method to solve the Burgers-Huxley

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[37] V. Serkin, A. Hasegawa. Novel soliton solutions of the nonlinear Schrödinger equation

[38] R.Y. Hao, L. Li, Z. Li, et al. A new approach to exact soliton solutions and soliton interaction for the nonlinear Schrödinger equation with variable coefficients. Optics

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62 Recent Studies in Perturbation Theory

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**Chapter 3**

## **Green Function**

## Jing Huang

Additional information is available at the end of the chapter

http://dx.doi.org/10.5772/68028

## Abstract

Both the scalar Green function and the dyadic Green function of an electromagnetic field and the transform from the scalar to dyadic Green function are introduced. The Green function of a transmission line and the propagators are also presented in this chapter.

Keywords: Green function, boundary condition, scatter, propagator, convergence

## 1. Introduction

In 1828, Green introduced a function, which he called a potential, for calculating the distribution of a charge on a surface bounding a region in Rn in the presence of external electromagnetic forces. The Green function has been an interesting topic in modern physics and engineering, especially for the electromagnetic theory in various source distributions (charge, current, and magnetic current), various construct conductors, and dielectric. Even though most problems can be solved without the use of Green functions, the symbolic simplicity with which they could be used to express relationships makes the formulations of many problems simpler and more compact. Moreover, it is easier to conceptualize many problems; especially the dyadic Green function is generalized to layered media of planar, cylindrical, and spherical configurations.

## 2. Definition of Green function

## 2.1. Mathematics definition

For the linear operator, there are: Lx^ <sup>¼</sup> <sup>f</sup>ðtÞ, <sup>t</sup> > 0;

$$\mathbf{x}(t)|\_{t=0} = y\_0 \mathbf{\dot{\cdot}} \cdots \mathbf{x}^{(n)}(t)|\_{t=0} = y\_n \tag{1}$$

Rewriting Eq. (1) as:

$$
\hat{L}\mathbf{x} = \int f(t')\delta(t - t')dt'\tag{2}
$$

Defining the Green function as:

$$
\hat{L}G(t, t') = \delta(t - t') \tag{3}
$$

So, the solution of Eq. (1) is:

$$\mathbf{x}(t) = \begin{cases} f(t')G(t, t')dt' \\ \end{cases} \tag{4}$$

We give several types of Green functions [1]

$$\begin{aligned} \hat{L} &= -(\frac{d^2}{dt^2} + 2\gamma \frac{d}{dt} + \omega\_0^2) & \mathbf{G}(t, t') &= \frac{1}{2\pi} \int\_{-\infty}^{+\infty} \frac{\exp\left[-i(t - t')k\right]}{k^2 + 2i\gamma k - \omega\_0^2} dk \\ \hat{L} &= -[f\_0(t)\frac{d^2}{dt^2} + f\_1(t)\frac{d}{dt} + f\_2(t)] & \mathbf{G}(t, t') &= -\frac{\mathbf{V}\_1(t)\mathbf{V}\_2(t) - \mathbf{V}\_2(t)\mathbf{V}\_1(t')}{f\_0(t')\|\mathbf{V}\_1(t')\mathbf{V}\_2(t) - \mathbf{V}\_1(t)\mathbf{V}\_2(t)} \\ \hat{L} &= -\frac{d}{dt}[(1 - t^2)\frac{d}{dt}] & \mathbf{G}(t, t') &= \frac{1}{2} + \sum\_{n=1}^{\infty} \frac{1}{n(n+1)} \cdot \frac{2n+1}{2} \mathbf{P}\_n(t)\mathbf{P}\_n(t') \end{aligned}$$

## 3. The scalar Green function

#### 3.1. The scalar Green function of an electromagnetic field

The Green function of a wave equation is the solution of the wave equation for a point source [2]. And when the solution to the wave equation due to a point source is known, the solution due to a general source can be obtained by the principle of linear superposition (see Figure 1).

This is merely a result of the linearity of the wave equation, and that a general source is just a linear superposition of point sources. For example, to obtain the solution to the scalar wave equation in V in Figure 1

$$(\nabla^2 + k^2)\varphi(\mathbf{r}) = \mathbf{s}(\mathbf{r})\tag{5}$$

we first seek the Green function in the same V, which is the solution to the following equation:

$$(\nabla^2 + k^2)\mathbf{g}(\mathbf{r}, \mathbf{r}') = \delta(\mathbf{r} - \mathbf{r}') \tag{6}$$

Given g (r, r<sup>0</sup> ), φ(r) can be found easily from the principle of linear superposition, since g (r, r<sup>0</sup> ) is the solution to Eq. (5) with a point source on the right-hand side. To see this more clearly, note that an arbitrary source s(r) is just

Figure 1. The radiation of a source s(r) in a volume V.

<sup>x</sup>ðtÞj<sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>y</sup>0; <sup>⋯</sup>xðn<sup>Þ</sup>

LG^ <sup>ð</sup>t, t<sup>0</sup>

xðtÞ ¼ ð fðt 0 ÞGðt, t<sup>0</sup>

<sup>0</sup><sup>Þ</sup> <sup>G</sup>ðt, t<sup>0</sup>

dt þ f <sup>2</sup>ðtÞÞ Gðt, t<sup>0</sup>

dt� <sup>G</sup>ðt, t<sup>0</sup>

Þ ¼ δðt � t

The Green function of a wave equation is the solution of the wave equation for a point source [2]. And when the solution to the wave equation due to a point source is known, the solution due to a general source can be obtained by the principle of linear superposition (see Figure 1). This is merely a result of the linearity of the wave equation, and that a general source is just a linear superposition of point sources. For example, to obtain the solution to the scalar wave

we first seek the Green function in the same V, which is the solution to the following equation:

the solution to Eq. (5) with a point source on the right-hand side. To see this more clearly, note

Þ ¼ δðr�r

), φ(r) can be found easily from the principle of linear superposition, since g (r, r<sup>0</sup>

0

<sup>ð</sup>∇<sup>2</sup> <sup>þ</sup> <sup>k</sup><sup>2</sup>

<sup>ð</sup>∇<sup>2</sup> <sup>þ</sup> <sup>k</sup> 2 Þgðr,r 0 0

Lx ^ <sup>¼</sup> ð fðt 0 Þδðt � t 0

Rewriting Eq. (1) as:

66 Recent Studies in Perturbation Theory

Defining the Green function as:

So, the solution of Eq. (1) is:

dt <sup>þ</sup> <sup>ω</sup><sup>2</sup>

dt<sup>2</sup> <sup>þ</sup> <sup>f</sup> <sup>1</sup>ðt<sup>Þ</sup> <sup>d</sup>

<sup>L</sup>^ ¼ �ð <sup>d</sup><sup>2</sup>

<sup>L</sup>^ ¼ � <sup>d</sup>

<sup>L</sup>^ ¼ �½<sup>f</sup> <sup>0</sup>ðt<sup>Þ</sup> <sup>d</sup><sup>2</sup>

dt<sup>2</sup> <sup>þ</sup> <sup>2</sup><sup>γ</sup> <sup>d</sup>

dt ½ð1 � t 2Þ d

We give several types of Green functions [1]

3. The scalar Green function

equation in V in Figure 1

that an arbitrary source s(r) is just

Given g (r, r<sup>0</sup>

3.1. The scalar Green function of an electromagnetic field

<sup>ð</sup>tÞj<sup>t</sup>¼<sup>0</sup> <sup>¼</sup> yn <sup>ð</sup>1<sup>Þ</sup>

Þdt<sup>0</sup> ð2Þ

Þ ð3Þ

Þdt<sup>0</sup> ð4Þ

k

޼� <sup>Ψ</sup><sup>1</sup> <sup>ð</sup>tÞΨ<sup>2</sup> <sup>ð</sup><sup>t</sup>

f <sup>0</sup> ðt 0 Þ½Ψ<sup>1</sup> ðt 0 <sup>Þ</sup>Ψ\_ <sup>2</sup> <sup>ð</sup><sup>t</sup> 0 Þ�Ψ\_ <sup>1</sup> <sup>ð</sup><sup>t</sup> 0 ÞΨ<sup>2</sup> ðt 0 Þ�

exp ½�iðt � t

0 Þ�Ψ<sup>2</sup> ðtÞΨ<sup>1</sup> ðt 0 Þ

1 nðn þ 1Þ � 2n þ 1 2

<sup>2</sup> <sup>þ</sup> <sup>2</sup>iγ<sup>k</sup> � <sup>ω</sup><sup>2</sup> 0 dk

0 Þk�

> PnðtÞPnðt 0 Þ

Þ ¼ <sup>1</sup> 2π ðþ<sup>∞</sup> �∞

Þ ¼ <sup>1</sup> <sup>2</sup> <sup>þ</sup>X<sup>∞</sup> n¼1

ÞϕðrÞ ¼ sðrÞ ð5Þ

Þ ð6Þ

) is

$$s(\mathbf{r}) = \int d\mathbf{r}' s(\mathbf{r}') \delta(\mathbf{r} - \mathbf{r}') \tag{7}$$

which is actually a linear superposition of point sources in mathematical terms. Consequently, the solution to Eq. (5) is just

$$\varphi(\mathbf{r}) = -\int\_{V} d\mathbf{r}' g(\mathbf{r}, \mathbf{r}') s(\mathbf{r}') \tag{8}$$

which is an integral linear superposition of the solution of Eq. (6). Moreover, it can be seen that g(r, r<sup>0</sup> ) � g(r<sup>0</sup> , r,) from reciprocity irrespective of the shape of V.

To find the solution of Eq. (6) for an unbounded, homogeneous medium, one solves it in spherical coordinates with the origin at r'. By so doing, Eq. (6) becomes

$$(\nabla^2 + k^2)\mathbf{g}(\mathbf{r}) = \delta(\mathbf{x})\delta(y)\delta(z) \tag{9}$$

But due to the spherical symmetry of a point source, g(r) must also be spherically symmetric. Then, for r 6¼ 0, adopt the proper coordinate origin (the vector r is replaced by the scalar r), the homogeneous, spherically symmetric solution to Eq. (9) is given by

$$g(r) = c\_1 \frac{e^{ikr}}{r} + c\_2 \frac{e^{-ikr}}{r} \tag{10}$$

Since sources are absent at infinity, physical grounds then imply that only an outgoing solution can exist; hence,

$$g(r) = c \frac{e^{ikr}}{r} \tag{11}$$

The constant c is found by matching the singularities at the origin on both sides of Eq. (9). To do this, we substitute Eq. (11) into Eq. (9) and integrate Eq. (9) over a small volume about the origin to yield

$$\int\_{\Delta V} dV \nabla \cdot \nabla \frac{c e^{ikr}}{r} + \int\_{\Delta V} dV k^2 \frac{c e^{ikr}}{r} = -1 \tag{12}$$

Note that the second integral vanishes when ΔV ! 0 because dV = 4πr2dr. Moreover, the first integral in Eq. (12) can be converted into a surface integral using Gauss theorem to obtain

$$\lim\_{r \to 0} 4\pi r^2 \frac{d}{dr} c \frac{e^{ikr}}{r} = -1 \tag{13}$$

or c = 1/(4π).

The solution to Eq. (6) must depend only on r – r<sup>0</sup> . Therefore, in general,

$$g(\mathbf{r}, \mathbf{r}') = g(\mathbf{r} - \mathbf{r}') = \frac{e^{ik(\mathbf{r} - \mathbf{r}')}}{4\pi(\mathbf{r} - \mathbf{r}')} \tag{14}$$

implying that g(r, r') is translationally invariant for unbounded, homogeneous media. Consequently, the solution to Eq. (5), from Eq. (9), is then

$$\varphi(\mathbf{r}) = -\int\_{V} d\mathbf{r}' \frac{e^{ik(\mathbf{r}-\mathbf{r}')}}{4\pi(\mathbf{r}-\mathbf{r}')} s(\mathbf{r}') \tag{15}$$

Once ϕ(r) and n^ � ∇ϕðrÞ are known on S, then ϕ(r<sup>0</sup> ) away from S could be found

$$\oint \boldsymbol{\varrho}(\mathbf{r}') = \oint\_{\mathcal{S}} d\mathbf{S} \hat{\boldsymbol{n}} \cdot \left[ \boldsymbol{g}(\mathbf{r}, \mathbf{r}') \nabla \boldsymbol{\varrho}(\mathbf{r}) - \boldsymbol{\varrho}(\mathbf{r}) \nabla \boldsymbol{g}(\mathbf{r}, \mathbf{r}') \right] \tag{16}$$

#### 3.2. The scalar Green functions of one-dimensional transmission lines

We consider a transmission line excited by a distributed current source, K(x), as sketched in Figure 2. The line may be finite or infinite, and it may be terminated at either end with impedance or by another line [3]. For a harmonically oscillating current source K(x), the voltage and the current on the line satisfy the following pair of equations:

Figure 2. Transmission line excited by a distributed current source, K(x).

$$\frac{d V(\mathbf{x})}{d\mathbf{x}} = \dot{\mathbf{u}} \omega \mathbf{L} I(\mathbf{x}) \tag{17}$$

$$\frac{dI(\mathbf{x})}{d\mathbf{x}} = i\omega \mathbf{C} V(\mathbf{x}) + K(\mathbf{x}) \tag{18}$$

L and C denote, respectively, the distributed inductance and capacitance of the line.

By eliminating I(x) between Eq. (17) and Eq. (18), there is

The constant c is found by matching the singularities at the origin on both sides of Eq. (9). To do this, we substitute Eq. (11) into Eq. (9) and integrate Eq. (9) over a small volume about the

ΔV

Note that the second integral vanishes when ΔV ! 0 because dV = 4πr2dr. Moreover, the first integral in Eq. (12) can be converted into a surface integral using Gauss theorem to obtain

dVk<sup>2</sup> ceikr

. Therefore, in general,

Þ

<sup>r</sup> ¼ �<sup>1</sup> <sup>ð</sup>12<sup>Þ</sup>

<sup>r</sup> ¼ �<sup>1</sup> <sup>ð</sup>13<sup>Þ</sup>

<sup>Þ</sup> <sup>ð</sup>14<sup>Þ</sup>

Þ ð15Þ

Þ� ð16Þ

ð

dV<sup>∇</sup> � <sup>∇</sup> ceik<sup>r</sup> r þ ð

> lim r!0 4πr <sup>2</sup> d dr c eikr

gðr,r 0

ϕðr޼�

Þ ¼ gðr � r

ð

V dr

0

We consider a transmission line excited by a distributed current source, K(x), as sketched in Figure 2. The line may be finite or infinite, and it may be terminated at either end with impedance or by another line [3]. For a harmonically oscillating current source K(x), the

dSn^ � ½gðr,r

3.2. The scalar Green functions of one-dimensional transmission lines

voltage and the current on the line satisfy the following pair of equations:

0

implying that g(r, r') is translationally invariant for unbounded, homogeneous media. Conse-

<sup>0</sup> <sup>e</sup>ikðr�r<sup>0</sup> Þ

4πðr � r<sup>0</sup>

Þ ¼ <sup>e</sup>ikðr�r<sup>0</sup>

4πðr � r<sup>0</sup>

Þ sðr 0

Þ∇ϕðrÞ � ϕðrÞ∇gðr,r

) away from S could be found

0

ΔV

The solution to Eq. (6) must depend only on r – r<sup>0</sup>

quently, the solution to Eq. (5), from Eq. (9), is then

Once ϕ(r) and n^ � ∇ϕðrÞ are known on S, then ϕ(r<sup>0</sup>

ϕðr 0 Þ ¼ ∮ S

Figure 2. Transmission line excited by a distributed current source, K(x).

origin to yield

68 Recent Studies in Perturbation Theory

or c = 1/(4π).

$$\frac{d^2V(\mathbf{x})}{d\mathbf{x}^2} + k^2V(\mathbf{x}) = \mathbf{i}\omega LK(\mathbf{x})\tag{19}$$

where <sup>k</sup> <sup>¼</sup> <sup>ω</sup> ffiffiffiffiffiffi LC <sup>p</sup> denotes the propagation constant of the line. Eq. (19) has been designated as an inhomogeneous one-dimensional scalar wave equation.

The Green function pertaining to a one-dimensional scalar wave equation of the form of Eq. (19), denoted by g(x, x<sup>0</sup> ), is a solution of the Eq. (9). The solution for g(x, x<sup>0</sup> ) is not completely determined unless there are two boundary conditions which the function must satisfy at the extremities of the spatial domain in which the function is defined. The boundary conditions which must be satisfied by g(x, x<sup>0</sup> ) are the same as those dictated by the original function which we intend to determine, namely, V(x) in the present case. For this reason, the Green functions are classified according to the boundary conditions, which they must obey. Some of the typical ones (for the transmission line) are illustrated in Figure 3.

In general, the subscript 0 designates infinite domain so that we have outgoing waves at x ! �∞, often called the radiation condition. Subscript 1 means that one of the boundary conditions satisfies the so-called Dirichlet condition, while the other satisfies the radiation condition. When one of the boundary conditions satisfies the so-called Neumann condition, we use subscript 2. Subscript 3 is reserved for the mixed type. Actually, we should have used a double subscript for two distinct boundary conditions. For example, case (b) of Figure 3 should be denoted by g01, indicating that one radiation condition and one Dirichlet condition are involved. With such an understanding, the simplified notation should be acceptable.

In case (d), a superscript becomes necessary because we have two sets of line voltage and current (V1, I1) and (V2, I2) in this problem, and the Green function also has different forms in the two regions. The first superscript denotes the region where this function is defined, and the second superscript denotes the region where the source is located.

Let the domain of x corresponds to (x1, x2). The function g(x, x<sup>0</sup> ) in Eq. (9) can represent any of the three types, g0, g1, and g2, illustrated in Figures 3a–c, respectively. The treatment of case (d) is slightly different, and it will be formulated later.

(a) By multiplying Eq. (19) by g(x, x<sup>0</sup> ) and Eq. (9) by V(x) and taking the difference of the two resultant equations, we obtain

$$\int\_{x\_1}^{x\_2} [V(\mathbf{x}) \frac{d^2 \mathbf{g}\_0(\mathbf{x}, \mathbf{x}')}{d\mathbf{x}^2} - \mathbf{g}\_0(\mathbf{x}, \mathbf{x}') \frac{d^2 V(\mathbf{x}, \mathbf{x}')}{d\mathbf{x}^2} \Big| d\mathbf{x} = -\int\_{x\_1}^{x\_2} V(\mathbf{x}) \delta(\mathbf{x} - \mathbf{x}') d\mathbf{x} - i\omega \mathcal{L} \Big|\_{x\_1}^{x\_2} K(\mathbf{x}) \mathbf{g}\_0(\mathbf{x}, \mathbf{x}') d\mathbf{x} \tag{20}$$

Figure 3. Classification of Green functions according to the boundary conditions.

The first term at the right-hand side of the above equation is simply V(xl), and the term at the left-hand side can be simplified by integration by parts, which gives

$$V(\mathbf{x'}) = -i\omega L \int\_{\mathbf{x}\_1}^{\mathbf{x}\_2} \mathbf{g}\_0(\mathbf{x}, \mathbf{x'}) K(\mathbf{x}) d\mathbf{x} \tag{21}$$

If we use the unprimed variable x to denote the position of a field point, as usually is the case, Eq. (21) can be changed to [4]

$$\begin{split} V(\mathbf{x}) &= -i\omega \mathcal{L} \int\_{\mathbf{x}\_1}^{\mathbf{x}\_2} \mathbf{g}(\mathbf{x}', \mathbf{x}) \mathcal{K}(\mathbf{x}') d\mathbf{x}' \\ &= -i\omega \mathcal{L} \int\_{\mathbf{x}\_1}^{\mathbf{x}\_2} \mathbf{g}\_0(\mathbf{x}, \mathbf{x}') \mathcal{K}(\mathbf{x}') d\mathbf{x}' \end{split} \tag{22}$$

The last identity is due to the symmetrical property of the Green function. The shifting of the primed and unprimed variables is often practiced in our work. For this reason, it is important to point out that g(x<sup>0</sup> , x), by definition, satisfies the Eq. (9).

The general solutions for Eq. (9) in the two regions (see Figure 3a) are

$$g\_0(\mathbf{x}, \mathbf{x}') = \begin{cases} \mathrm{i}/(2k)e^{ik(\mathbf{x}-\mathbf{x}')}, \mathbf{x} \ge \mathbf{x}'\\ \mathrm{i}/(2k)e^{-ik(\mathbf{x}-\mathbf{x}')}, \mathbf{x} \le \mathbf{x}' \end{cases} \tag{23}$$

The choice of the above functions is done with the proper satisfaction of boundary conditions at infinity. At x = x', the function must be continuous, and its derivative is discontinuous.

$$\text{They are: } \left[ \mathcal{g}\_0(\mathfrak{x}, \mathfrak{x}') \right]\_{\mathfrak{x}' - 0}^{\mathfrak{x}' + 0} = 0, \text{ and } \left[ \frac{d\mathcal{g}\_0(\mathfrak{x}, \mathfrak{x}')}{d\mathfrak{x}} \right]\_{\mathfrak{x}' - 0}^{\mathfrak{x}' + 0} = -1.$$

The physical interpretation of these two conditions is that the voltage at x' is continuous, but the difference of the line currents at x' must be equal to the source current.

(b) The choice of this type of function is done with the proper satisfaction of boundary conditions. At x = x', the function must be continuous, its derivative is discontinuous, and a Dirichlet condition is satisfied at x = 0.

$$\mathcal{g}\_1(\mathbf{x}, \mathbf{x}') = \begin{cases} \iota/(2\kappa) \left[ e^{\imath \kappa(\mathbf{x} - \mathbf{x}')} - e^{\imath \kappa(\mathbf{x} + \mathbf{x}')} \right], \mathbf{x} \ge \mathbf{x}' \\\iota/(2\kappa) \left[ e^{-\imath \kappa(\mathbf{x} - \mathbf{x}')} - e^{\imath \kappa(\mathbf{x} + \mathbf{x}')} \right], \mathbf{0} \le \mathbf{x} \le \mathbf{x}' \end{cases} \tag{24}$$

In view of Eq. (24), it can be interpreted as consisting of an incident and a scattered wave; that is

$$\mathbf{g}\_1(\mathbf{x}, \mathbf{x}') = \mathbf{g}\_0(\mathbf{x}, \mathbf{x}') + \mathbf{g}\_{1s}(\mathbf{x}, \mathbf{x}') \tag{25}$$

where g1sðx, x<sup>0</sup> Þ ¼ �<sup>i</sup> <sup>2</sup><sup>k</sup> eikðxþx<sup>0</sup> Þ .

The first term at the right-hand side of the above equation is simply V(xl), and the term at the

ð<sup>x</sup><sup>2</sup> x1

If we use the unprimed variable x to denote the position of a field point, as usually is the case,

g0ðx, x<sup>0</sup>

ÞKðxÞdx ð21Þ

left-hand side can be simplified by integration by parts, which gives

Figure 3. Classification of Green functions according to the boundary conditions.

Vðx<sup>0</sup>

Eq. (21) can be changed to [4]

70 Recent Studies in Perturbation Theory

޼�iωL

Such a notion is not only physically useful, but mathematically it offers a shortcut to finding a composite Green function. It is called as the shortcut method or the method of scattering superposition.

(c) Similarly, the method of scattering superposition suggests that we can start with

$$\mathbf{g\_2(x,x')} = \mathbf{g\_0(x,x')} + A e^{i\mathbf{k}x} \tag{26}$$

To satisfy the Neumann condition at x = 0, we require

#### 72 Recent Studies in Perturbation Theory

$$
\left[\frac{d\mathbf{g}\_0(\mathbf{x}, \mathbf{x}')}{d\mathbf{x}} + ikAe^{i\mathbf{x}}\right]\_{\mathbf{x}=\mathbf{0}} = \mathbf{0} \tag{27}
$$

Hence

$$A = \frac{\dot{\mathbf{i}}}{2k} e^{i\mathbf{k}\mathbf{x}'} \tag{28}$$

$$\mathcal{g}\_2(\mathbf{x}, \mathbf{x}') = \mathbf{i}/(2k) \begin{cases} e^{i\mathbf{k}(\mathbf{x} - \mathbf{x}')} + e^{i\mathbf{k}(\mathbf{x} + \mathbf{x}')}, & \mathbf{x} \ge \mathbf{x}' \\ e^{-i\mathbf{k}(\mathbf{x} - \mathbf{x}')} + e^{i\mathbf{k}(\mathbf{x} + \mathbf{x}')}, & 0 \le \mathbf{x} \le \mathbf{x}' \end{cases} \tag{29}$$

(d) In this case, we have two differential equations to start with

$$\frac{d^2V\_1(\mathbf{x})}{d\mathbf{x}^2} + k\_1^2V\_1(\mathbf{x}) = \mathrm{i}\omega L\_1K\_1(\mathbf{x}), \mathbf{x} \ge \mathbf{0} \tag{30}$$

$$\frac{d^2V\_2(\mathbf{x})}{d\mathbf{x}^2} + k\_2^2V\_2(\mathbf{x}) = 0, \mathbf{x} \le 0\tag{31}$$

It is assumed that the current source is located in region 1 (see Figure 3d). We introduce two Green functions of the third kind, denoted by g(11) (x, x') and g(21) (x, x'). g(21), the first number of the superscript corresponds to the region where the function is defined. The second number corresponds to the region where the source is located; then

$$\frac{d^2g^{(11)}(\mathbf{x},\mathbf{x}')}{d\mathbf{x}^2} + k\_1^2 g^{(11)}(\mathbf{x},\mathbf{x}') = -\delta(\mathbf{x} - \mathbf{x}'), \mathbf{x} \ge 0\tag{32}$$

$$\frac{d^2g^{(21)}(\mathbf{x},\mathbf{x}')}{d\mathbf{x}^2} + k\_2^2 g^{(21)}(\mathbf{x},\mathbf{x}') = \mathbf{0}, \mathbf{x} \le \mathbf{0} \tag{33}$$

At the junction corresponding to x = 0, g(11) and g(21) satisfy the boundary condition that

$$\mathbf{g}^{(11)}(\mathbf{x}, \mathbf{x}')\_{\mathbf{x}=\mathbf{0}} = \mathbf{g}^{(21)}(\mathbf{x}, \mathbf{x}')\_{\mathbf{x}=\mathbf{0}} \tag{34}$$

$$\frac{1}{L\_1} \frac{d g^{(11)}(\mathbf{x}, \mathbf{x}')}{d\mathbf{x}}\_{\mathbf{x}=0} = \frac{1}{L\_2} \frac{d g^{(21)}(\mathbf{x}, \mathbf{x}')}{d\mathbf{x}}\_{\mathbf{x}=0} \tag{35}$$

The last condition corresponds to the physical requirement that the current at the junction must be continuous. Again, by means of the method of scattering superposition, there are

$$\begin{split} \mathbf{g}^{(11)}(\mathbf{x}, \mathbf{x}') &= \mathbf{g}\_0(\mathbf{x}, \mathbf{x}') + \mathbf{g}\_s^{(11)}(\mathbf{x}, \mathbf{x}') \\ &= \frac{i}{2k\_1} \begin{cases} e^{i\mathbf{k}\_1(\mathbf{x} - \mathbf{x}')} + \mathbf{R} \mathbf{e}^{i\mathbf{k}\_1(\mathbf{x} - \mathbf{x}')}, & \mathbf{x} \ge \mathbf{x}' \\ e^{-i\mathbf{k}\_1(\mathbf{x} - \mathbf{x}')} + \mathbf{R} \mathbf{e}^{i\mathbf{k}\_1(\mathbf{x} + \mathbf{x}')}, & \mathbf{0} \le \mathbf{x} \le \mathbf{x}' \end{split} \end{split} \tag{36}$$

$$\log^{(21)}(\mathbf{x}, \mathbf{x}') = \frac{\mathbf{i}}{2k\_1} T e^{-i(k\_2 \mathbf{x} - k\_1 \mathbf{x}')}, \mathbf{x} \ge \mathbf{0} \tag{37}$$

The characteristic impedance of the lines, respectively, is

Green Function http://dx.doi.org/10.5772/68028 73

$$z\_1 = \left(\frac{L\_1}{\overline{C}\_1}\right)^{1/2}, z\_2 = \left(\frac{L\_2}{\overline{C}\_2}\right)^{1/2} \tag{38}$$

By the boundary condition, there are

dg0ðx,x<sup>0</sup> Þ dx <sup>þ</sup> ikAeikx � �

g2ðx, x<sup>0</sup>

(d) In this case, we have two differential equations to start with d2 V1ðxÞ dx<sup>2</sup> <sup>þ</sup> <sup>k</sup>

> d2 V2ðxÞ dx<sup>2</sup> <sup>þ</sup> <sup>k</sup>

number corresponds to the region where the source is located; then

gð11<sup>Þ</sup> ðx,x<sup>0</sup>

dgð11<sup>Þ</sup>

gð11<sup>Þ</sup> ðx, x<sup>0</sup>

gð21<sup>Þ</sup> ðx, x<sup>0</sup>

The characteristic impedance of the lines, respectively, is

ðx,x<sup>0</sup> Þ dx <sup>x</sup>¼<sup>0</sup>

d2 gð21<sup>Þ</sup> ðx, x<sup>0</sup> Þ dx<sup>2</sup> <sup>þ</sup> <sup>k</sup>

1 L1

d2 gð11<sup>Þ</sup> ðx, x<sup>0</sup> Þ dx<sup>2</sup> <sup>þ</sup> <sup>k</sup> <sup>A</sup> <sup>¼</sup> <sup>i</sup> 2k e ikx0

e�ikðx�x<sup>0</sup>

2

It is assumed that the current source is located in region 1 (see Figure 3d). We introduce two Green functions of the third kind, denoted by g(11) (x, x') and g(21) (x, x'). g(21), the first number of the superscript corresponds to the region where the function is defined. The second

ðx, x<sup>0</sup>

2 2gð21<sup>Þ</sup>

At the junction corresponding to x = 0, g(11) and g(21) satisfy the boundary condition that

<sup>Þ</sup><sup>x</sup>¼<sup>0</sup> <sup>¼</sup> <sup>g</sup>ð21<sup>Þ</sup>

¼ 1 L2

The last condition corresponds to the physical requirement that the current at the junction must be continuous. Again, by means of the method of scattering superposition, there are

Þ ¼ g0ðx, x<sup>0</sup>

¼ i 2k<sup>1</sup>

Þ ¼ <sup>i</sup> 2k<sup>1</sup> ޼�δðx � x<sup>0</sup>

ðx, x<sup>0</sup>

ðx,x<sup>0</sup>

ðx,x<sup>0</sup> Þ dx <sup>x</sup>¼<sup>0</sup>

Þ þ <sup>g</sup>ð11<sup>Þ</sup> <sup>s</sup> <sup>ð</sup>x, x<sup>0</sup>

Þ

eik1ðx�x<sup>0</sup>

e�ik1ðx�x<sup>0</sup>

Te�iðk2x�k1x<sup>0</sup>

Þ

Þ , x ≥ x<sup>0</sup>

> Þ , 0 ≤ x ≤ x<sup>0</sup>

( <sup>ð</sup>36<sup>Þ</sup>

, x ≥ 0 ð37Þ

<sup>Þ</sup> <sup>þ</sup> Reik1ðx�x<sup>0</sup>

<sup>Þ</sup> <sup>þ</sup> Reik1ðxþx<sup>0</sup>

dgð21<sup>Þ</sup>

Þ ¼ <sup>i</sup>=ð2k<sup>Þ</sup> <sup>e</sup>ikðx�x<sup>0</sup>

2

2 1gð11<sup>Þ</sup>

(

Hence

72 Recent Studies in Perturbation Theory

x¼0

<sup>Þ</sup> <sup>þ</sup> <sup>e</sup>ikðxþx<sup>0</sup>

<sup>Þ</sup> <sup>þ</sup> <sup>e</sup>ikðxþx<sup>0</sup>

Þ , x ≥ x<sup>0</sup>

Þ , 0 ≤ x ≤ x<sup>0</sup>

<sup>1</sup>V1ðxÞ ¼ iωL1K1ðxÞ, x ≥ 0 ð30Þ

<sup>2</sup>V2ðxÞ ¼ 0, x ≤ 0 ð31Þ

Þ, x ≥ 0 ð32Þ

Þ ¼ 0, x ≤ 0 ð33Þ

<sup>Þ</sup><sup>x</sup>¼<sup>0</sup> <sup>ð</sup>34<sup>Þ</sup>

¼ 0 ð27Þ

ð28Þ

ð29Þ

ð35Þ

$$R = \frac{z\_2 - z\_1}{z\_2 + z\_1}, T = \frac{2z\_2}{z\_2 + z\_1} \tag{39}$$

Example: Green function solution of nonlinear Schrodinger equation in the time domain [5]. The nonlinear Schrodinger equation including nonresonant and resonant nonlinear items is:

$$\begin{split} \frac{\partial A}{\partial \boldsymbol{z}} + \frac{i}{2} \beta\_2 \frac{\partial^2 A}{\partial t^2} - \frac{1}{6} \beta\_3 \frac{\partial^3 A}{\partial t^3} &= -\frac{a}{2} A + i \frac{3k\_0}{8n A\_{\rm eff}} \chi\_{\rm NR}^{(3)} |A|^2 A \\ + \frac{i k\_0 g(\omega\_0) [1 - \dot{\boldsymbol{\mu}}(\omega\_0)]}{2 n A\_{\rm eff}} A \int\_{-\nu\varepsilon}^{t} \chi\_{\rm R}^{(3)} (t - \tau) |A(\tau)|^2 d\tau \end{split} \tag{40}$$

Where A is the field, β<sup>2</sup> and β<sup>3</sup> are the second and third order dispersion, respectively. A(z) is the fiber absorption profile. k<sup>0</sup> ¼ ω0=c, ω<sup>0</sup> is the center frequency. Aeff is the effective core area. n is the refractive index.

$$f(\omega\_1 + \omega\_2 + \omega\_3) = \frac{2(\omega\_1 + \omega\_2 + \omega\_3)(1 - |\Gamma|)}{-2(\omega\_1 + \omega\_2 + \omega\_3)^2 - 2|\Gamma| + |\Gamma|^2} \tag{41}$$

$$\mathbf{g}(\omega\_1 + \omega\_2 + \omega\_3) = \left[ -\mathbf{2}(\omega\_1 + \omega\_2 + \omega\_3)^2 - 2|\Gamma| + |\Gamma|^2 \right] \tag{42}$$

where g(ω<sup>1</sup> + ω<sup>2</sup> + ω3) is the Raman gain and f(ω<sup>1</sup> + ω<sup>2</sup> + ω3) is the Raman nongain coefficient. Г is the attenuation coefficient.

The original nonlinear part is divided into the nonresonant and resonant susceptibility items χ<sup>ð</sup>3<sup>Þ</sup> NR and <sup>χ</sup><sup>ð</sup>3<sup>Þ</sup> <sup>R</sup> . The solution has the form:

$$A(z,t) = \varphi(t)e^{-i\mathbb{E}z} \tag{43}$$

Then, there is:

$$\frac{1}{2}\beta\_2\frac{\partial^2\phi}{\partial t^2} + \frac{i}{6}\beta\_3\frac{\partial^3\phi}{\partial t^3} - \frac{3k\_0}{8nA\_{\text{eff}}}\chi\_{\text{NR}}^{(3)}|\phi|^2\phi - \frac{k\_0g(\omega\_s)[1-\dot{\mathcal{H}}(\omega\_s)]}{2nA\_{\text{eff}}}\phi\Big|\_{-\alpha}^{+\alpha}\chi\_N^{(3)}(t-\tau)|\phi(\tau)|d\tau = E\phi \quad (44)$$

Let:

$$
\hat{H}\_0(t) = \frac{1}{2}\beta\_2 \frac{\partial^2}{\partial t^2} + \frac{i}{6}\beta\_3 \frac{\partial^3}{\partial t^3} \tag{45}
$$

$$\hat{V}(t) = \frac{-3k\_0}{8nA\_{\text{eff}}} \chi\_{\text{NR}}^{(3)} |\phi| - \frac{k\_0 g(\omega\_s)[1 - \text{if}(\omega\_s)]}{2nA\_{\text{eff}}} \int\_{-\infty}^{+\infty} \chi\_{\text{R}}^{(3)}(t - \tau) |\phi(\tau)|^2 d\tau \tag{46}$$

and taking the operator <sup>V</sup>^ <sup>ð</sup>tÞas a perturbation item, the eigenequation � X k n¼2 i n n! βn ∂nφ <sup>∂</sup>T<sup>n</sup> ¼ Eφ is

$$\frac{1}{2}\beta\_2\frac{\partial^2\phi}{\partial T^2} + \frac{i}{6}\beta\_3\frac{\partial^3\phi}{\partial T^3} = E\phi\tag{47}$$

Assuming E = 1, we get the corresponding characteristic equation:

$$-\frac{1}{2}\beta\_2 r^2 + \frac{\beta\_3}{6}r^3 = E\tag{48}$$

Its characteristic roots are r1,r2,r3. The solution can be represented as:

$$
\phi = c\_1 \phi\_1 + c\_2 \phi\_2 + c\_3 \phi\_3 \tag{49}
$$

where ϕ<sup>m</sup> ¼ exp ðirmtÞ, m ¼ 1, 2, 3, and c1,c2,c<sup>3</sup> are determined by the initial pulse. The Green function of Eq. (47) is:

$$\delta(E - \hat{H}\_0(t))\mathcal{G}\_0(t, t') = \delta(t - t') \tag{50}$$

Constructing the Green function as:

$$G\_0(t, t') = \begin{cases} a\_1 \phi\_1 + a\_2 \phi\_2 + a\_3 \phi\_3, t > t' \\ b\_1 \phi\_1 + b\_2 \phi\_2 + b\_3 \phi\_3, t < t' \end{cases} \tag{51}$$

At the point t = t 0 , there are:

$$a\_1\phi\_1(t') + a\_2\phi\_2(t') + a\_3\phi\_3(t') = b\_1\phi\_1(t') + b\_2\phi\_2(t') + b\_3\phi\_3(t')\tag{52}$$

$$a\_1 \phi'\_1(t') + a\_2 \phi'\_2(t') + a\_3 \phi'\_3(t') = b\_1 \phi'\_1(t') + b\_2 \phi'\_2(t') + b\_3 \phi'\_3(t')\tag{53}$$

$$a\_1 \boldsymbol{\phi}\_1^\*(t') + a\_2 \boldsymbol{\phi}\_2^\*(t') + a\_3 \boldsymbol{\phi}\_3^\*(t') - b\_1 \boldsymbol{\phi}\_1^\*(t') - b\_2 \boldsymbol{\phi}\_2^\*(t') - b\_3 \boldsymbol{\phi}\_3^\*(t') = -6i/\beta\_3\tag{54}$$

It is reasonable to let b<sup>1</sup> = b<sup>2</sup> = b<sup>3</sup> = 0, then:

$$a\_1 = \frac{\phi\_2 \dot{\phi}\_3 - \dot{\phi}\_2 \phi\_3}{W(t')},\\ a\_2 = \frac{\phi\_3 \dot{\phi}\_1 - \dot{\phi}\_3 \phi\_1}{W(t')},\\ a\_3 = \frac{\phi\_1 \dot{\phi}\_2 - \dot{\phi}\_1 \phi\_2}{W(t')}\tag{55}$$

$$\mathcal{W}(\mathbf{f}') = \begin{vmatrix} \phi\_1 & \phi\_2 & \phi\_3 \\ \phi\_1^{(1)} & \phi\_2^{(1)} & \phi\_3^{(1)} \\ \phi\_1^{(2)} & \phi\_2^{(2)} & \phi\_3^{(2)} \end{vmatrix} \tag{56}$$

Finally, the solution of Eq. (44) can be written with the eigenfunction and Green function:

$$\begin{split} \phi(t) &= \phi(t) + \int \mathsf{G}\_{0}(t, t') V(t') \phi(t') dt' \\ &= \varrho(t) + \int \mathsf{G}\_{0}(t, t', \underline{E}) V(t') \varrho(t') dt' + \int dt' \mathsf{G}\_{0}(t, t', \underline{E}) V(t') \left[ \mathsf{G}\_{0}(t', t', \underline{E}) V(t') \phi(t') dt' \\ &= \varrho(t) + \int \mathsf{G}\_{0}(t, t', \underline{E}) V(t') \varrho(t') dt' + \int dt' \mathsf{G}\_{0}(t, t', \underline{E}) V(t') \left[ \mathsf{G}\_{0}(t', t', \underline{E}) V(t') \phi(t') dt' + \cdots \right. \right. \\ &\left. + \underbrace{\int dt' \mathsf{G}\_{0}(t, t') V(t') \left[ \mathsf{G}\_{0}(t', t') V(t') dt' \cdots \right] \left[ \mathsf{G}\_{0}(t', t'^{1+}) V(t'^{1+}) \phi(t'^{1+}) dt'^{1+1} \right. \right. \end{split} (57)$$

The accuracy can be estimated by the last term of Eq. (57).

## 4. The dyadic Green function

and taking the operator <sup>V</sup>^ <sup>ð</sup>tÞas a perturbation item, the eigenequation �

1 2 β2 ∂<sup>2</sup>φ <sup>∂</sup>T<sup>2</sup> <sup>þ</sup>

> � 1 2 β2r <sup>2</sup> <sup>þ</sup> <sup>β</sup><sup>3</sup> 6 r

Assuming E = 1, we get the corresponding characteristic equation:

Its characteristic roots are r1,r2,r3. The solution can be represented as:

G0ðt, t<sup>0</sup>

Þ þ a2φ2ðt

2ðt 0 Þ þ <sup>a</sup>3φ″

Þ ¼

0

2ðt 0 Þ þ a3φ<sup>0</sup>

<sup>3</sup> � <sup>φ</sup>\_ <sup>2</sup>φ<sup>3</sup>

Wðt 0 (

Þ þ a3φ3ðt

3ðt 0 Þ � <sup>b</sup>1φ″

<sup>Þ</sup> , a<sup>2</sup> <sup>¼</sup> <sup>φ</sup>3φ\_

Wðt 0 Þ ¼ 0

3ðt 0 Þ ¼ b1φ<sup>0</sup>

Þ ¼ b1φ1ðt

1ðt 0 Þ � <sup>b</sup>2φ″

<sup>1</sup> � <sup>φ</sup>\_ <sup>3</sup>φ<sup>1</sup>

φ<sup>1</sup> φ<sup>2</sup> φ<sup>3</sup>

<sup>2</sup> <sup>φ</sup><sup>ð</sup>1<sup>Þ</sup> 3

<sup>2</sup> <sup>φ</sup><sup>ð</sup>2<sup>Þ</sup> 3

Wðt 0

φ<sup>ð</sup>1<sup>Þ</sup> <sup>1</sup> <sup>φ</sup><sup>ð</sup>1<sup>Þ</sup>

� � � � � � � �

φ<sup>ð</sup>2<sup>Þ</sup> <sup>1</sup> <sup>φ</sup><sup>ð</sup>2<sup>Þ</sup>

Finally, the solution of Eq. (44) can be written with the eigenfunction and Green function:

function of Eq. (47) is:

74 Recent Studies in Perturbation Theory

At the point t = t

Constructing the Green function as:

0

a1φ<sup>0</sup> 1ðt 0 Þ þ a2φ<sup>0</sup>

a1φ″ 1ðt 0 Þ þ <sup>a</sup>2φ″

, there are:

a1φ1ðt 0

It is reasonable to let b<sup>1</sup> = b<sup>2</sup> = b<sup>3</sup> = 0, then:

<sup>a</sup><sup>1</sup> <sup>¼</sup> <sup>φ</sup>2φ\_

i 6 β3 ∂<sup>3</sup>φ

where ϕ<sup>m</sup> ¼ exp ðirmtÞ, m ¼ 1, 2, 3, and c1,c2,c<sup>3</sup> are determined by the initial pulse. The Green

Þ ¼ δðt � t

a1φ<sup>1</sup> þ a2φ<sup>2</sup> þ a3φ3, t > t

b1φ<sup>1</sup> þ b2φ<sup>2</sup> þ b3φ3, t < t

0

1ðt 0 Þ þ b2φ<sup>0</sup>

Þ þ b2φ2ðt

2ðt 0 Þ � <sup>b</sup>3φ″

<sup>Þ</sup> , a<sup>3</sup> <sup>¼</sup> <sup>φ</sup>1φ\_

� � � � � � � � 0

<sup>ð</sup><sup>E</sup> � <sup>H</sup>^ <sup>0</sup>ðtÞÞG0ðt, t<sup>0</sup>

X k

i n n! βn ∂nφ <sup>∂</sup>T<sup>n</sup> ¼ Eφ is

n¼2

<sup>∂</sup>T<sup>3</sup> <sup>¼</sup> <sup>E</sup><sup>φ</sup> <sup>ð</sup>47<sup>Þ</sup>

<sup>3</sup> <sup>¼</sup> <sup>E</sup> <sup>ð</sup>48<sup>Þ</sup>

Þ ð50Þ

ð51Þ

ð56Þ

φ ¼ c1φ<sup>1</sup> þ c2φ<sup>2</sup> þ c3φ<sup>3</sup> ð49Þ

0

0

0

2ðt 0 Þ þ b3φ<sup>0</sup>

Þ þ b3φ3ðt

3ðt 0

<sup>2</sup> � <sup>φ</sup>\_ <sup>1</sup>φ<sup>2</sup>

Wðt 0 0

3ðt 0

Þ ð52Þ

Þ ð53Þ

޼�6i=β<sup>3</sup> ð54Þ

<sup>Þ</sup> <sup>ð</sup>55<sup>Þ</sup>

## 4.1. The dyadic Green function for the electromagnetic field in a homogeneous isotropic medium

The Green function for the scalar wave equation could be used to find the dyadic Green function for the vector wave equation in a homogeneous, isotropic medium [3]. First, notice that the vector wave equation in a homogeneous, isotropic medium is

$$\nabla \times \nabla \times \mathbf{E}(\mathbf{r}) - k^2 \mathbf{E}(\mathbf{r}) = i\omega\mu \mathbf{J}(\mathbf{r}) \tag{58}$$

Then, by using the fact that <sup>∇</sup> · <sup>∇</sup> · <sup>E</sup>ðr޼�∇<sup>2</sup><sup>E</sup> <sup>þ</sup> ∇∇ � <sup>E</sup> and that <sup>∇</sup> � <sup>E</sup> <sup>¼</sup> <sup>ρ</sup>=<sup>ε</sup> <sup>¼</sup> <sup>∇</sup> � <sup>J</sup>=iωε, which follows from the continuity equation, we can rewrite Eq. (58) as

$$\nabla^2 \mathbf{E}(\mathbf{r}) - k^2 \mathbf{E}(\mathbf{r}) = -i\omega\mu \left[\hat{\mathbf{I}} + \frac{\nabla \nabla}{k^2}\right] \cdot \mathbf{J}(\mathbf{r})\tag{59}$$

where ^I is an identity operator. In Cartesian coordinates, there are actually three scalar wave equations embedded in the above vector equation, each of which can be solved easily in the manner of Eq. (4). Consequently,

$$\mathbf{E}(\mathbf{r}) = -i\omega\mu \int\_{V} d\mathbf{r}' \mathcal{g}(\mathbf{r}'-\mathbf{r}) \left[\hat{\mathbf{I}} + \frac{\nabla'\nabla'}{k^2}\right] \cdot \mathbf{J}(\mathbf{r}) \tag{60}$$

where g(r<sup>0</sup> �r)is the unbounded medium scalar Green function. Moreover, by using the vector identities ∇gf ¼ f ∇g þ g∇f and ∇ � gF ¼ g∇ � F þ ð∇gÞ � F, it can be shown that

$$\int\_{V} d\mathbf{r}' \mathcal{g}(\mathbf{r}'-\mathbf{r}) \nabla' f(\mathbf{r}') = -\int\_{V} d\mathbf{r}' \nabla' \mathcal{g}(\mathbf{r}'-\mathbf{r}) f(\mathbf{r}') \tag{61}$$

and

$$\int\_{V} d\mathbf{r}' [\nabla' \mathcal{g}(\mathbf{r}'-\mathbf{r})] \nabla' \cdot \mathbf{J}(\mathbf{r}') = -\int\_{V} d\mathbf{r}' \mathbf{J}(\mathbf{r}') \cdot \nabla' \nabla' \mathcal{g}(\mathbf{r}'-\mathbf{r}) \tag{62}$$

Hence, Eq. (60) can be rewritten as

$$\mathbf{E}(\mathbf{r}) = i\omega\mu \int\_{V} d\mathbf{r}' \mathbf{J}(\mathbf{r}') \cdot \left[\hat{\mathbf{I}} + \frac{\nabla'\nabla'}{k^2}\right] \mathbf{g}(\mathbf{r}'-\mathbf{r})\tag{63}$$

It can also be derived using scalar and vector potentials.

Alternatively, Eq. (63) can be written as

$$\mathbf{E}(\mathbf{r}) = i\omega\mu \int\_{V} d\mathbf{r}' \mathbf{J}(\mathbf{r}') \cdot \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}', \mathbf{r}) \tag{64}$$

where

$$
\hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}) = \left[\overline{\mathbf{I}} + \frac{\nabla^{\prime}\nabla^{\prime}}{k^{2}}\right] \mathbf{g}(\mathbf{r}^{\prime} - \mathbf{r})\tag{65}
$$

is a dyad known as the dyadic Green function for the electric field in an unbounded, homogeneous medium. (A dyad is a 3 · 3 matrix that transforms a vector to a vector. It is also a second rank tensor). Even though Eq. (64) is established for an unbounded, homogeneous medium, such a general relationship also exists in a bounded, homogeneous medium. It could easily be shown from reciprocity that

$$
\begin{aligned}
\left< \mathbf{J}\_1(\mathbf{r}), \hat{\mathbf{G}}\_\varepsilon(\mathbf{r}, \mathbf{r}'), \mathbf{J}\_2(\mathbf{r}') \right> &= \left< \mathbf{J}\_2(\mathbf{r}), \hat{\mathbf{G}}\_\varepsilon(\mathbf{r}, \mathbf{r}'), \mathbf{J}\_1(\mathbf{r}') \right> \\ &= \left< \mathbf{J}\_1(\mathbf{r}), \hat{\mathbf{G}}\_\varepsilon^\dagger(\mathbf{r}, \mathbf{r}'), \mathbf{J}\_2(\mathbf{r}') \right>
\end{aligned} \tag{66}
$$

where

$$
\left\langle \mathbf{J}\_i(\mathbf{r}), \hat{\mathbf{G}}\_e(\mathbf{r}, \mathbf{r}'), \mathbf{J}\_j(\mathbf{r}') \right\rangle = \iint\_{VV} dr' dr \mathbf{J}\_i(\mathbf{r}') \cdot \hat{\mathbf{G}}\_e(\mathbf{r}', \mathbf{r}) \cdot \mathbf{J}\_j(\mathbf{r}) \tag{66a}
$$

is the relation between Ji and the electric field produced by Jj. Notice that the above equation implies [6]

$$
\hat{\mathbf{G}}\_{\varepsilon}^{\dagger}(\mathbf{r}', \mathbf{r}) = \hat{\mathbf{G}}\_{\varepsilon}(\mathbf{r}, \mathbf{r}') \tag{66b}
$$

Then, by taking transpose of Eq. (66b), Eq. (64) becomes

$$\mathbf{E}(\mathbf{r}) = i\omega\mu \int\_{V} d\mathbf{r}' \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') \cdot \mathbf{J}(\mathbf{r}') \tag{67}$$

Alternatively, the dyadic Green function for an unbounded, homogeneous medium can also be written as

Green Function http://dx.doi.org/10.5772/68028 77

$$\hat{\mathbf{G}}\_c(\mathbf{r}, \mathbf{r}') = \frac{1}{k^2} \left[ \nabla \times \nabla \times \hat{\mathbf{I}} \mathbf{g}(\mathbf{r} - \mathbf{r}') - \hat{\mathbf{I}} \delta(\mathbf{r} - \mathbf{r}') \right] \tag{68}$$

By substituting Eq. (67) back into Eq. (58) and writing

$$\mathbf{J(r)} = \int d\mathbf{r'} \hat{\mathbf{I}} \delta(\mathbf{r} - \mathbf{r'}) \cdot \mathbf{J(r')}\tag{69}$$

we can show quite easily that

ð

<sup>0</sup> � rÞ�∇<sup>0</sup> � Jðr

ð

V dr 0 Jðr 0 Þ � ^<sup>I</sup> <sup>þ</sup>

EðrÞ ¼ iωμ

<sup>G</sup>^ <sup>e</sup>ðrÞ ¼ <sup>I</sup> <sup>þ</sup>

Þ,J2ðr<sup>0</sup> Þ

> ¼ ð

> > ð

V dr 0 <sup>G</sup>^ <sup>e</sup>ðr,<sup>r</sup> 0 Þ � Jðr 0

Alternatively, the dyadic Green function for an unbounded, homogeneous medium can also be

V

ð

V dr<sup>0</sup> drJ<sup>i</sup> ðr 0 Þ � <sup>G</sup>^ <sup>e</sup>ð<sup>r</sup> 0 ,rÞ � J<sup>j</sup>

is the relation between Ji and the electric field produced by Jj. Notice that the above equation

,rÞ ¼ <sup>G</sup>^ <sup>e</sup>ðr,<sup>r</sup>

0

ð

V dr 0 Jðr 0 Þ � <sup>G</sup>^ <sup>e</sup>ð<sup>r</sup> 0

∇0 ∇0 k2 � �

is a dyad known as the dyadic Green function for the electric field in an unbounded, homogeneous medium. (A dyad is a 3 · 3 matrix that transforms a vector to a vector. It is also a second rank tensor). Even though Eq. (64) is established for an unbounded, homogeneous medium, such a general relationship also exists in a bounded, homogeneous medium. It could easily be

gðr

<sup>¼</sup> <sup>J</sup>2ðrÞ, <sup>G</sup>^ <sup>e</sup>ðr,r<sup>0</sup>

<sup>¼</sup> <sup>J</sup>1ðrÞ, <sup>G</sup>^ <sup>t</sup>

D E

eðr,r<sup>0</sup>

Þ,J1ðr<sup>0</sup> Þ

Þ,J2ðr<sup>0</sup> Þ D E <sup>ð</sup>66<sup>Þ</sup>

EðrÞ ¼ iωμ

It can also be derived using scalar and vector potentials.

<sup>J</sup>1ðrÞ, <sup>G</sup>^ <sup>e</sup>ðr,r<sup>0</sup>

<sup>ð</sup>rÞ, <sup>G</sup>^ <sup>e</sup>ðr,<sup>r</sup>

Then, by taking transpose of Eq. (66b), Eq. (64) becomes

Ji

D E

0 Þ,J<sup>j</sup> ðr 0 Þ

> G^ t eðr 0

EðrÞ ¼ iωμ

D E

0 ޼� ð

V dr 0 Jðr 0 Þ � ∇<sup>0</sup> ∇0 gðr

∇0 ∇0 k 2 � �

gðr

<sup>0</sup> � rÞ ð62Þ

<sup>0</sup> � rÞ ð63Þ

,rÞ ð64Þ

<sup>0</sup> � rÞ ð65Þ

ðrÞ ð66aÞ

Þ ð66bÞ

Þ ð67Þ

V dr 0 ½∇0 gðr

Hence, Eq. (60) can be rewritten as

76 Recent Studies in Perturbation Theory

Alternatively, Eq. (63) can be written as

shown from reciprocity that

where

where

implies [6]

written as

$$\nabla \times \nabla \times \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') - k^2 \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') = \hat{\mathbf{I}} \delta(\mathbf{r} - \mathbf{r}') \tag{70}$$

Equation (64) or (67), due to the ∇∇ operator inside the integration operating on g(r<sup>0</sup> �r), has a singularity of 1/|r<sup>0</sup> �r|<sup>3</sup> when <sup>r</sup><sup>0</sup> ! <sup>r</sup>. Consequently, it has to be redefined in this case for it does not converge uniformly, specifically, when r is also in the source region occupied by J(r). Hence, at this point, the evaluation of Eq. (67) in a source region is undefined.

And as the vector analog of Eq. (16)

$$\mathbf{E}(\mathbf{r}') = \oint\_{S} d\mathbf{S} \left[ \mathbf{n} \times \mathbf{E}(\mathbf{r}) \cdot \nabla \times \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') + i\omega\mu \mathbf{n} \times \mathbf{H}(\mathbf{r}) \cdot \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') \right] \tag{71}$$

#### 4.2. The boundary condition

The dyadic Green function is introduced mainly to formulate various canonical electromagnetic problems in a systematic manner to avoid treatments of many special cases which can be treated as one general problem [3, 7, 8]. Some typical problems are illustrated in Figure 4 where (a) shows a current source in the presence of a conducting sphere located in air, (b) shows a conducting cylinder with an aperture which is excited by some source inside the cylinder, (c) shows a rectangular waveguide with a current source placed inside the guide, and (d) shows two semi-infinite isotropic media in contact, such as air and "flat" earth with a current source placed in one of the regions.

Unless specified otherwise, we assume that for problems involving only one medium such as (a), (b), and (c) the medium is air, then the wave number k is equal to ωðμ0ε0Þ <sup>1</sup>=<sup>2</sup> <sup>¼</sup> <sup>2</sup>π=λ. The electromagnetic fields in these cases are solutions of the wave Eq. (62) and

$$
\nabla \times \nabla \times \mathbf{H}(\mathbf{r}) - k^2 \mathbf{H}(\mathbf{r}) = \nabla \times \mathbf{J}(\mathbf{r}) \tag{72}
$$

The fields must satisfy the boundary conditions required by these problems.

In general, using the notations G^ <sup>e</sup> and G^ <sup>m</sup> to denote, respectively, the electric and the magnetic dyadic Green functions; they are solutions of the dyadic differential equations

$$\nabla \times \nabla \times \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') - k^2 \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') = \hat{\mathbf{I}} \delta(\mathbf{r} - \mathbf{r}') \tag{73}$$

$$\nabla \times \nabla \times \hat{\mathbf{G}}\_m(\mathbf{r}, \mathbf{r}') - k^2 \hat{\mathbf{G}}\_m(\mathbf{r}, \mathbf{r}') = \nabla \times \left[\hat{\mathbf{I}} \delta(\mathbf{r} - \mathbf{r}')\right] \tag{74}$$

is the same as Eq. (70), and there is

Figure 4. Some typical boundary value problems.

$$
\hat{\mathbf{G}}\_{\mathfrak{m}} = \nabla \times \hat{\mathbf{G}}\_{\mathfrak{e}} \tag{75}
$$

(a) and (b): Electric dyadic Green function (the first kind, using the subscript 1 denotes G^ <sup>e</sup>1, G^ <sup>m</sup>1, and the subscript "0" represents the free-space condition that the environment does not have any scattering object) is required to satisfy the dyadic Dirichlet condition on Sd, namely,

$$\mathbf{n} \times \hat{\mathbf{G}}\_{\varepsilon 1} = 0, \mathbf{n} \times \hat{\mathbf{G}}\_{\mathbf{m} 1} = 0 \tag{76}$$

So, for (a)

$$\mathbf{E}(\mathbf{r}') = \int d\mathbf{r} \mathbf{J}(\mathbf{r}) \cdot \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') \tag{77}$$

and for (b)

$$\mathbf{E}(\mathbf{r}') = \oint\_{\mathcal{S}\_A} d\mathbf{S} \mathbf{n} \times \mathbf{E}(\mathbf{r}) \cdot \nabla \times \hat{\mathbf{G}}\_{\varepsilon}(\mathbf{r}, \mathbf{r}') \tag{78}$$

(c) the electric dyadic Green function is required to satisfy the dyadic boundary condition on Sd, namely,

$$\mathbf{n} \times \nabla \times \hat{\mathbf{G}}\_{\ell 2} = \mathbf{0} \; \mathbf{n} \times \nabla \times \hat{\mathbf{G}}\_{m2} = \mathbf{0} \tag{79}$$

$$\mathbf{H}(\mathbf{r}') = \int d\mathbf{r} \mathbf{J}(\mathbf{r}) \cdot \nabla \times \hat{\mathbf{G}}\_{\epsilon}(\mathbf{r}, \mathbf{r}') \tag{80}$$

(d) For problems involving two isotropic media such as the configuration shown in Figure 4d, there are two sets of fields [9]. The wave numbers in these two regions are denoted by k<sup>1</sup> ¼ ωðμ1ε1Þ <sup>1</sup>=<sup>2</sup> and <sup>k</sup><sup>2</sup> <sup>¼</sup> <sup>ω</sup>ðμ2ε2<sup>Þ</sup> 1=2 . There are four functions for the dyadic Green function of the electric type and another four functions for the magnetic type, denoted, respectively, by <sup>G</sup>^ <sup>11</sup> <sup>e</sup> <sup>G</sup>^ <sup>12</sup> <sup>e</sup> <sup>G</sup>^ <sup>21</sup> <sup>e</sup> and <sup>G</sup>^ <sup>22</sup> <sup>e</sup> , and <sup>G</sup>^ <sup>11</sup> <sup>m</sup> <sup>G</sup>^ <sup>12</sup> <sup>m</sup> <sup>G</sup>^ <sup>21</sup> <sup>m</sup> and <sup>G</sup>^ <sup>22</sup> <sup>m</sup> . The superscript notation in <sup>G</sup>^ <sup>11</sup> <sup>e</sup> means that both the field point and the source point are located in region 1. For <sup>G</sup>^ <sup>21</sup> <sup>e</sup> , it means that the field point is located in region 1 and the source point is located in region 2. A current source is located in region 1 only, and the two sets of wave equations are

$$\nabla \times \nabla \times \mathbf{E}\_1(\mathbf{r}) - k^2 \mathbf{E}\_1(\mathbf{r}) = i\omega\mu\_1 \mathbf{J}\_1(\mathbf{r}) \tag{81}$$

$$
\nabla \times \nabla \times \mathbf{H}\_1(\mathbf{r}) - k^2 \mathbf{H}\_1(\mathbf{r}) = \nabla \times \mathbf{J}\_1(\mathbf{r}) \tag{82}
$$

and

$$\nabla \times \nabla \times \mathbf{E}\_2(\mathbf{r}) - k^2 \mathbf{E}\_2(\mathbf{r}) = 0 \tag{83}$$

$$\nabla \times \nabla \times \mathbf{H}\_2(\mathbf{r}) - k^2 \mathbf{H}\_2(\mathbf{r}) = 0 \tag{84}$$

There are

<sup>G</sup>^ <sup>m</sup> <sup>¼</sup> <sup>∇</sup> · <sup>G</sup>^ <sup>e</sup> <sup>ð</sup>75<sup>Þ</sup>

<sup>n</sup> · <sup>G</sup>^ <sup>e</sup><sup>1</sup> <sup>¼</sup> <sup>0</sup>, <sup>n</sup> · <sup>G</sup>^ m1 <sup>¼</sup> <sup>0</sup> <sup>ð</sup>76<sup>Þ</sup>

Þ ð77Þ

Þ ð78Þ

Þ ð80Þ

(a) and (b): Electric dyadic Green function (the first kind, using the subscript 1 denotes G^ <sup>e</sup>1, G^ <sup>m</sup>1, and the subscript "0" represents the free-space condition that the environment does not have any

<sup>d</sup>rJðrÞ � <sup>G</sup>^ <sup>e</sup>ðr,<sup>r</sup>

dS<sup>n</sup> · <sup>E</sup>ðrÞ � <sup>∇</sup> · <sup>G</sup>^ <sup>e</sup>ðr,<sup>r</sup>

<sup>d</sup>rJðrÞ � <sup>∇</sup> · <sup>G</sup>^ <sup>e</sup>ðr,<sup>r</sup>

(c) the electric dyadic Green function is required to satisfy the dyadic boundary condition on Sd,

0

0

<sup>n</sup> · <sup>∇</sup> · <sup>G</sup>^ <sup>e</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup> <sup>n</sup> · <sup>∇</sup> · <sup>G</sup>^ <sup>m</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup> <sup>ð</sup>79<sup>Þ</sup>

0

scattering object) is required to satisfy the dyadic Dirichlet condition on Sd, namely,

Eðr 0 Þ ¼ ð

Eðr 0 Þ ¼ ∮ SA

> Hðr 0 Þ ¼ ð

So, for (a)

Figure 4. Some typical boundary value problems.

78 Recent Studies in Perturbation Theory

and for (b)

namely,

$$\nabla \times \nabla \times \hat{\mathbf{G}}\_{\varepsilon}^{11}(\mathbf{r}, \mathbf{r}') - k\_1^2 \hat{\mathbf{G}}\_{\varepsilon}^{11}(\mathbf{r}, \mathbf{r}') = \hat{\mathbf{I}} \delta(\mathbf{r} - \mathbf{r}') \tag{85}$$

$$\nabla \times \nabla \times \hat{\mathbf{G}}\_{\varepsilon}^{21}(\mathbf{r}, \mathbf{r}') - k\_2^2 \hat{\mathbf{G}}\_{\varepsilon}^{21}(\mathbf{r}, \mathbf{r}') = \mathbf{0} \tag{86}$$

At the interface, the electromagnetic field and the corresponding dyadic Green function satisfy the following boundary conditions

$$\mathbf{n} \times \left[ \hat{\mathbf{G}}\_{\varepsilon}^{11} - \hat{\mathbf{G}}\_{\varepsilon}^{21} \right] = \mathbf{0} \tag{87}$$

$$\mathbf{n} \times \left[ \nabla \times \hat{\mathbf{G}}\_{\epsilon}^{11} / \mu\_1 - \nabla \times \hat{\mathbf{G}}\_{\epsilon}^{21} / \mu\_2 \right] = \mathbf{0} \tag{88}$$

The electric fields are

$$\mathbf{E}\_1(\mathbf{r}') = i\omega\mu\_1 \int d\mathbf{r} \mathbf{J}(\mathbf{r}) \cdot \hat{\mathbf{G}}\_e^{11}(\mathbf{r}, \mathbf{r}') \tag{89}$$

$$\mathbf{E}\_2(\mathbf{r}') = i\omega\mu\_2 \int d\mathbf{r} \mathbf{J}(\mathbf{r}) \cdot \hat{\mathbf{G}}\_e^{21}(\mathbf{r}, \mathbf{r}') \tag{90}$$

## 5. Vector wave functions, L, M, and N

The vector wave functions are the building blocks of the eigenfunction expansions of various kinds of dyadic Green functions. These functions were first introduced by Hansen [10–12] in formulating certain electromagnetic problems.Three kinds of vector wave functions, denoted by L, M, and N, are solutions of the homogeneous vector Helmholtz equation. To derive the eigenfunction expansion of the magnetic dyadic Green functions that are solenoidal and satisfy with the vector wave equation, the L functions are not needed. If we try to find eigenfunction expansion of the electric dyadic Green functions then the L functions are also needed.

A vector wave function, by definition, is an eigenfunction or a characteristic function, which is a solution of the homogeneous vector wave equation <sup>∇</sup> · <sup>∇</sup> · <sup>F</sup> � <sup>κ</sup><sup>2</sup><sup>F</sup> <sup>¼</sup> 0.

There are two independent sets of vector wave functions, which can be constructed using the characteristic function pertaining to a scalar wave equation as the generating function. One kind of vector wave function, called the Cartesian or rectilinear vector wave function, is formed if we let

$$\mathbf{F} = \nabla \times (\Psi\_1 \mathbf{c}) \tag{91}$$

where ψ<sup>1</sup> denotes a characteristic function, which satisfies the scalar wave equation

$$
\nabla^2 \Psi + \kappa^2 \Psi = 0 \tag{92}
$$

And c denotes a constant vector, such as x, y, or z. For convenience, we shall designate c as the piloting vector and Ψ as the generating function. Another kind, designated as the spherical vector wavefunction, will be introduced later, whereby the piloting vector is identified as the spherical radial vector R.

Actually, substituting Eq. (91) into Eq. (92), it is

$$\nabla \times \left[ \mathbf{c} (\nabla^2 \Psi\_1 + \kappa^2 \Psi\_1) \right] = \mathbf{0} \tag{93}$$

The set of functions so obtained

$$\mathbf{M}\_1 = \nabla \times (\boldsymbol{\Psi}\_1 \mathbf{c}) \tag{94}$$

$$\mathbf{N}\_2 = \frac{1}{\kappa} \nabla \times \nabla \times (\boldsymbol{\Psi}\_2 \mathbf{c}) \tag{95}$$

$$\mathbf{L}\_3 = \nabla(\Psi\_3) \tag{96}$$

Ψ2,Ψ<sup>3</sup> denote the characteristic functions which also satisfy (92) but may be different from the function used to define M1.

In the following, the expressions for the dyadic Green functions of a rectangular waveguide will be derived asserting to the vector wave functions. The method and the general procedure would apply equally well to other bodies (cylindrical waveguide, circular cylinder in free space, and inhomogeneous media and moving medium).

Figure 5 shows the orientation of the guide with respect to the rectangular coordinate system, and we will choose the unit vector z to represent the piloting vector c.

The scalar wave function

$$\Psi = (A\cos k\_x \mathbf{x} + B\sin k\_x \mathbf{x})(\mathbf{C}\cos k\_y \mathbf{y} + D\sin k\_y \mathbf{y})e^{\hat{\theta}kz} \tag{97}$$

where k 2 <sup>x</sup> þ k 2 <sup>y</sup> <sup>þ</sup> <sup>h</sup><sup>2</sup> <sup>¼</sup> <sup>κ</sup>2.

eigenfunction expansion of the magnetic dyadic Green functions that are solenoidal and satisfy with the vector wave equation, the L functions are not needed. If we try to find eigenfunction

A vector wave function, by definition, is an eigenfunction or a characteristic function, which is

There are two independent sets of vector wave functions, which can be constructed using the characteristic function pertaining to a scalar wave equation as the generating function. One kind of vector wave function, called the Cartesian or rectilinear vector wave function, is formed if we let

And c denotes a constant vector, such as x, y, or z. For convenience, we shall designate c as the piloting vector and Ψ as the generating function. Another kind, designated as the spherical vector wavefunction, will be introduced later, whereby the piloting vector is identified as the

<sup>Ψ</sup><sup>1</sup> <sup>þ</sup> <sup>κ</sup><sup>2</sup>

Ψ2,Ψ<sup>3</sup> denote the characteristic functions which also satisfy (92) but may be different from the

In the following, the expressions for the dyadic Green functions of a rectangular waveguide will be derived asserting to the vector wave functions. The method and the general procedure would apply equally well to other bodies (cylindrical waveguide, circular cylinder in free

Figure 5 shows the orientation of the guide with respect to the rectangular coordinate system,

Ψ ¼ ðA cos kxx þ B sin kxxÞðC cos kyy þ D sin kyyÞe

F ¼ ∇ · ðΨ1cÞ ð91Þ

Ψ ¼ 0 ð92Þ

Ψ1Þ� ¼ 0 ð93Þ

ihz <sup>ð</sup>97<sup>Þ</sup>

M<sup>1</sup> ¼ ∇ · ðΨ1cÞ ð94Þ

∇ · ∇ · ðΨ2cÞ ð95Þ

L<sup>3</sup> ¼ ∇ðΨ3Þ ð96Þ

expansion of the electric dyadic Green functions then the L functions are also needed.

where ψ<sup>1</sup> denotes a characteristic function, which satisfies the scalar wave equation

∇2 <sup>Ψ</sup> <sup>þ</sup> <sup>κ</sup><sup>2</sup>

<sup>∇</sup> · <sup>½</sup>cð∇<sup>2</sup>

<sup>N</sup><sup>2</sup> <sup>¼</sup> <sup>1</sup> κ

spherical radial vector R.

80 Recent Studies in Perturbation Theory

The set of functions so obtained

function used to define M1.

The scalar wave function

<sup>y</sup> <sup>þ</sup> <sup>h</sup><sup>2</sup> <sup>¼</sup> <sup>κ</sup>2.

where k 2 <sup>x</sup> þ k 2

Actually, substituting Eq. (91) into Eq. (92), it is

space, and inhomogeneous media and moving medium).

and we will choose the unit vector z to represent the piloting vector c.

a solution of the homogeneous vector wave equation <sup>∇</sup> · <sup>∇</sup> · <sup>F</sup> � <sup>κ</sup><sup>2</sup><sup>F</sup> <sup>¼</sup> 0.

the constants kx and ky should have the following characteristic values

$$k\_x = \frac{m\pi}{a}, m = 0, 1, \dots \tag{98}$$

$$k\_y = \frac{n\pi}{b}, n = 0, 1, \dots \tag{99}$$

The complete expression and the notation for the set of functions M, which satisfy the vector Dirichlet condition are

$$\begin{split} \mathbf{M}\_{emn}(h) &= \nabla \times [\Psi\_{emn}\mathbf{z}] \\ &= (-k\_{\mathbf{y}}\mathbf{C}\_{\mathbf{x}}\mathbf{S}\_{\mathbf{y}}\mathbf{x} + k\_{\mathbf{x}}\mathbf{C}\_{\mathbf{y}}\mathbf{S}\_{\mathbf{x}}\mathbf{y})e^{il\mathbf{z}} \end{split} \tag{100}$$

where Sx ¼ sin kxx, Cx ¼ cos kxx, Sy ¼ sin kyy, Cy ¼ cos kyy. The subscript "e" attached to Memn is an abbreviation for the word "even," and "o" for "odd."

In a similar manner

$$\mathbf{N}\_{omn} = \frac{1}{\kappa} (ilk\_x \mathbf{C}\_x \mathbf{S}\_y \mathbf{x} + ilk\_y \mathbf{C}\_y \mathbf{S}\_x \mathbf{y} + (k\_x^2 + k\_y^2) \mathbf{S}\_x \mathbf{S}\_y \mathbf{z}) e^{ilz} \tag{101}$$

It is obvious that Memn represents the electric field of the TEmn mode, while Nomn represents that of the TMmn mode.

In summary, the vector wave functions, which can be used to represent the electromagnetic field inside a rectangular waveguide, are of the form

$$\mathbf{M}\_{e(o)mn} = \nabla \times [\Psi\_{e(o)mn}\mathbf{z}] \tag{102}$$

$$\mathbf{N}\_{\epsilon(o)mn} = \frac{1}{\kappa} \nabla \times \nabla \times \left[ \Psi\_{\epsilon(o)mn} \mathbf{z} \right] \tag{103}$$

Then

$$\hat{\mathbf{G}}\_{m2}(\mathbf{R}, \mathbf{R}') = \int\_{-\infty}^{+\infty} dh \sum\_{m,n} \frac{(2-\delta\_0)\kappa}{\pi ab(k\_x^2 + k\_y^2)} \cdot \left[ a(h)\mathbf{N}\_{emn}(h)\mathbf{M}'\_{emn}(-h) + b(h)\mathbf{M}\_{um}(h)\mathbf{N}'\_{omn}(-h) \right] \tag{104}$$

where <sup>a</sup>ðhÞ ¼ <sup>b</sup>ðhÞ ¼ <sup>1</sup> κ2�k 2, h ¼ �ðk <sup>2</sup> � <sup>k</sup> 2 <sup>x</sup> � k 2 yÞ 1=2 and <sup>δ</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup>, m <sup>¼</sup> <sup>0</sup>orn <sup>¼</sup> <sup>0</sup> 0, m 6¼ 0, n 6¼ 0 � .

M', N', m', n', h' denote another set of values, which may be distinct or the same as M, N, m, n, h.

## 6. Retarded and advanced Green functions

Green function is also utilized to solve the Schrödinger equation in quantum mechanics. Being completely equivalent to the Landauer scattering approach, the GF technique has the advantage that it calculates relevant transport quantities (e.g., transmission function) using effective numerical techniques. Besides, the Green function formalism is well adopted for atomic and molecular discrete-level systems and can be easily extended to include inelastic and manybody effects [13, 14].

(A) The definitions of propagators

The time-dependent Schrödinger equation is:

$$i\hbar \frac{\partial |\Psi(t)\rangle}{\partial t} = \hat{H} |\Psi(t)\rangle \tag{105}$$

The solution of this equation at time t can be written in terms of the solution at time t 0 :

$$|\Psi(t)\rangle = \hat{\mathcal{U}}(t, t')|\Psi(t')\rangle\tag{106}$$

where <sup>U</sup>^ <sup>ð</sup>t, t<sup>0</sup> Þ is called the time-evolution operator.

For the case of a time-independent Hermitian Hamiltonian H^ , so that the eigenstates <sup>j</sup>ΨnðtÞ〉 <sup>¼</sup> <sup>e</sup>�iEnt=<sup>ħ</sup>jΨn〉 with energies En are found from the stationary Schrödinger equation

$$
\hat{H}|\Psi\_n\rangle = E\_n|\Psi\_n\rangle\tag{107}
$$

The eigenfunctions jΨn〉are orthogonal and normalized, for discrete energy levels 1:

$$
\langle \Psi\_m | \Psi\_n \rangle = \delta\_{mn} \tag{108}
$$

and form a complete set of states (^I is the unity operator)

$$\sum\_{n} \langle \Psi\_{n} | \Psi\_{n} \rangle = 1 \tag{109}$$

The time-evolution operator for a time-independent Hamiltonian can be written as

Green Function http://dx.doi.org/10.5772/68028 83

$$
\hat{\mathcal{U}}(t - t') = e^{-i(t - t')\hat{H}/\hbar} \tag{110}
$$

This formal solution is difficult to use directly in most cases, but one can obtain the useful eigenstate representation from it. From the identity <sup>U</sup>^ <sup>¼</sup> <sup>U</sup>^ ^<sup>I</sup> and (107), (109), (110) it follows that

$$
\hat{\mathcal{U}}(t - t') = \sum\_{n} \epsilon^{i/\hbar E\_n(t - t')} |\Psi\_n\rangle\langle\Psi\_n| \tag{111}
$$

which demonstrates the superposition principle. The wave function at time t is

$$
\langle |\Psi(t)\rangle = \hat{\mathcal{U}}(t, t') |\Psi(t')\rangle = \sum\_{n} e^{-i/\hbar E\_n(t - t')} \langle \Psi\_n | \Psi(t') | \Psi\_n \rangle \tag{112}
$$

where 〈ΨnjΨðt 0 Þ〉 are the coefficients of the expansion of the initial function jΨðt 0 Þ〉on the basis of eigenstates.

It is equivalent and more convenient to introduce two Green operators, also called propagators, retarded <sup>G</sup>^ <sup>R</sup> ðt, t<sup>0</sup> <sup>Þ</sup> and advanced <sup>G</sup>^ <sup>A</sup> ðt, t<sup>0</sup> Þ:

$$\hat{\mathbf{G}}^{\hat{R}}(t, t') = -\frac{\dot{\mathbf{i}}}{\hbar} \Theta(t - t') \hat{\mathbf{U}}(t, t') = -\frac{\dot{\mathbf{i}}}{\hbar} \Theta(t - t') e^{-i(t - t') \hat{H}/\hbar} \tag{113}$$

$$
\hat{\mathbf{G}}^{A}(t, t') = \frac{\dot{\mathbf{i}}}{\hbar} \boldsymbol{\Theta}(t' - t) \hat{\mathbf{U}}(t, t') = \frac{\dot{\mathbf{i}}}{\hbar} \boldsymbol{\Theta}(t' - t) \mathbf{e}^{-i(t - t') \hat{H} / \hbar} \tag{114}
$$

so that at t > t <sup>0</sup> one has

<sup>G</sup>^ <sup>m</sup>2ðR, <sup>R</sup><sup>0</sup> Þ ¼

h.

where <sup>a</sup>ðhÞ ¼ <sup>b</sup>ðhÞ ¼ <sup>1</sup>

82 Recent Studies in Perturbation Theory

body effects [13, 14].

where <sup>U</sup>^ <sup>ð</sup>t, t<sup>0</sup>

(A) The definitions of propagators

The time-dependent Schrödinger equation is:

ðþ<sup>∞</sup> �∞

dhX m, <sup>n</sup>

κ2�k

ð2 � δ0Þκ πabðk 2 <sup>x</sup> þ k 2 yÞ

> <sup>2</sup> � <sup>k</sup> 2 <sup>x</sup> � k 2 yÞ 1=2

> > iħ

Þ is called the time-evolution operator.

and form a complete set of states (^I is the unity operator)

∂jΨðtÞ〉

The solution of this equation at time t can be written in terms of the solution at time t

<sup>j</sup>ΨðtÞ〉 <sup>¼</sup> <sup>U</sup>^ <sup>ð</sup>t, t<sup>0</sup>

The eigenfunctions jΨn〉are orthogonal and normalized, for discrete energy levels 1:

X n

The time-evolution operator for a time-independent Hamiltonian can be written as

For the case of a time-independent Hermitian Hamiltonian H^ , so that the eigenstates <sup>j</sup>ΨnðtÞ〉 <sup>¼</sup> <sup>e</sup>�iEnt=<sup>ħ</sup>jΨn〉 with energies En are found from the stationary Schrödinger equation

ÞjΨðt 0

2, h ¼ �ðk

6. Retarded and advanced Green functions

� ½aðhÞNemnðhÞM<sup>0</sup>

M', N', m', n', h' denote another set of values, which may be distinct or the same as M, N, m, n,

Green function is also utilized to solve the Schrödinger equation in quantum mechanics. Being completely equivalent to the Landauer scattering approach, the GF technique has the advantage that it calculates relevant transport quantities (e.g., transmission function) using effective numerical techniques. Besides, the Green function formalism is well adopted for atomic and molecular discrete-level systems and can be easily extended to include inelastic and many-

emnð�hÞ þ bðhÞMomnðhÞN<sup>0</sup>

0, m 6¼ 0, n 6¼ 0

<sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>H</sup>^ <sup>j</sup>ΨðtÞ〉 <sup>ð</sup>105<sup>Þ</sup>

<sup>H</sup>^ <sup>j</sup>Ψn〉 <sup>¼</sup> EnjΨn〉 <sup>ð</sup>107<sup>Þ</sup>

〈Ψ <sup>m</sup>jΨn〉 ¼ δmn ð108Þ

〈ΨnjΨn〉 ¼ 1 ð109Þ

Þ〉 ð106Þ

.

and <sup>δ</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup>, m <sup>¼</sup> <sup>0</sup>orn <sup>¼</sup> <sup>0</sup>

�

omnð�hÞ� ð104Þ

0 :

$$|\Psi(t)\rangle = i\hbar \hat{G}^{\,R}(t - t')|\Psi(t')\rangle\tag{115}$$

while at t < t <sup>0</sup> it follows

$$|\Psi(t)\rangle = i\hbar \hat{G}^A(t - t')|\Psi(t')\rangle\tag{116}$$

The operators <sup>G</sup>^ <sup>R</sup> ðt, t<sup>0</sup> Þ at t > t 0 and <sup>G</sup>^ <sup>A</sup> ðt, t<sup>0</sup> Þ at t < t <sup>0</sup> are the solutions of the equation

$$\left[i\hbar\frac{\partial}{\partial t} - \hat{H}\right]\hat{G}^{R(A)}(t, t') = \hat{I}\delta(t - t')\tag{117}$$

with the boundary conditions <sup>G</sup>^ <sup>R</sup> ðt, t<sup>0</sup> Þ ¼ 0 at t < t 0 , <sup>G</sup>^ <sup>A</sup> ðt, t<sup>0</sup> Þ ¼ 0 at t > t 0 . Indeed, at t > t <sup>0</sup> Eq. (118) satisfies the Schrödinger equation Eq. (105) due to Eq. (117). And integrating Eq. (117) from t ¼ t <sup>0</sup> � η to t ¼ t <sup>0</sup> þ η where η is an infinitesimally small positive number η ¼ 0þ, one gets

$$
\hat{\mathbf{G}}^{\mathbb{R}}(t+\eta, t') = \frac{1}{i\hbar}\hat{I} \tag{118}
$$

giving correct boundary condition at t = t 0 . Thus, if the retarded Green operator <sup>G</sup>^ <sup>R</sup> ðt, t<sup>0</sup> Þ is known, the time-dependent wave function at any initial condition is found (and makes many other useful things, as we will see below).

For a time-independent Hamiltonian, the Green function is a function of the time difference τ ¼ t�t 0 , and one can consider the Fourier transform

$$
\hat{G}^{R(A)}(E) = \int\_{-\r\nu}^{+\r\nu} \hat{G}^{R(A)}(\tau) e^{iE\tau/\hbar} d\tau \tag{119}
$$

This transform, however, can not be performed in all cases, because <sup>G</sup>^ <sup>R</sup>ðA<sup>Þ</sup> ðEÞ includes oscillating terms eiEτ=<sup>ħ</sup>. To avoid this problem we define the retarded Fourier transform

$$\hat{\mathbf{G}}^{\boldsymbol{R}}(E) = \lim\_{\eta \to 0+} \int\_{-\infty}^{+\infty} \hat{\mathbf{G}}^{\boldsymbol{R}}(\tau) e^{i(\mathbb{E} + i\eta)\tau/\hbar} d\tau \tag{120}$$

and the advanced one

$$\hat{\mathbf{G}}^{\mathcal{A}}(E) = \lim\_{\eta \to 0+} \int\_{-\infty}^{+\infty} \hat{\mathbf{G}}^{\mathcal{A}}(\tau) e^{i(E - i\eta)\tau/\hbar} d\tau \tag{121}$$

where the limit η ! 0 is assumed in the end of calculation. With this addition, the integrals are convergent. This definition is equivalent to the definition of a retarded (advanced) function as a function of complex energy variable at the upper (lower) part of the complex plain.

Applying this transform to Eq. (117), the retarded Green operator is

$$
\hat{\boldsymbol{G}}^{\boldsymbol{R}}(E) = \left[ (E + i\eta)\hat{\boldsymbol{I}} - \hat{\boldsymbol{H}} \right]^{-1} \tag{122}
$$

The advanced operator <sup>G</sup>^ <sup>A</sup> ðEÞ is related to the retarded one through

$$
\hat{\mathbf{G}}^{A}(E) = \hat{\mathbf{G}}^{R+}(E) \tag{123}
$$

Using the completeness propertyX n jΨn〉〈Ψnj ¼ 1, there is

$$\hat{G}^{\vec{R}}(E) = \sum\_{n} \frac{|\Psi\_{n}\rangle\langle\Psi\_{n}|}{(E+i\eta)\hat{I}-\hat{H}}\tag{124}$$

and

$$\hat{\boldsymbol{G}}^{\mathbb{R}}(E) = \sum\_{n} \frac{|\Psi\_{n}\rangle\langle\Psi\_{n}|}{E - E\_{n} + i\eta} \tag{125}$$

Apply the ordinary inverse Fourier transform to <sup>G</sup>^ <sup>R</sup> ðEÞ, the retarded function becomes

$$\hat{\mathbf{G}}^{R}(\tau) = \int\_{-\infty}^{+\infty} \hat{\mathbf{G}}^{R}(E) e^{-iE\tau/\hbar} \frac{dE}{2\pi\hbar} = -\frac{i}{\hbar} \Theta(\tau) \sum\_{n} e^{-iE\_{n}\tau/\hbar} |\Psi\_{n}\rangle\langle\Psi\_{n}| \tag{126}$$

Indeed, a simple pole in the complex E plain is at E ¼ En � iη, the residue in this point determines the integral at τ > 0 when the integration contour is closed through the lower half-plane, while at τ < 0 the integration should be closed through the upper half-plane and the integral is zero.

The formalism of retarded Green functions is quite general and can be applied to quantum systems in an arbitrary representation. For example, in the coordinate system Eq. (124) is

$$\hat{\mathbf{G}}^{R}(\mathbf{r},\mathbf{r}',E) = \frac{\sum\_{n} \langle \mathbf{r} | \Psi\_{n} \rangle \langle \Psi\_{n} | \mathbf{r}' \rangle}{E - E\_{n} + i\eta} = \sum\_{n} \frac{\Psi\_{n}(\mathbf{r}) \Psi\_{n}^{\*}(\mathbf{r})}{E - E\_{n} + i\eta} \tag{127}$$

(B) Path integral representation of the propagator

In the path integral representation, each path is assigned an amplitude e i Ð dtL, L is the Lagrangian function. The propagator is the sum of all the amplitudes associated with the paths connecting xa and xb (Figure 6). Such a summation is an infinite-dimensional integral.

The propagator satisfies

known, the time-dependent wave function at any initial condition is found (and makes many

For a time-independent Hamiltonian, the Green function is a function of the time difference

<sup>G</sup>^ <sup>R</sup>ðA<sup>Þ</sup> ðτÞe iEτ=ħ

iðEþiηÞτ=ħ

iðE�iηÞτ=ħ

dτ ð119Þ

dτ ð120Þ

dτ ð121Þ

�<sup>1</sup> <sup>ð</sup>122<sup>Þ</sup>

ðEÞ ð123Þ

<sup>ð</sup><sup>E</sup> <sup>þ</sup> <sup>i</sup>ηÞ^<sup>I</sup> � <sup>H</sup>^ <sup>ð</sup>124<sup>Þ</sup>

<sup>E</sup>�En <sup>þ</sup> <sup>i</sup><sup>η</sup> <sup>ð</sup>125<sup>Þ</sup>

jΨn〉〈Ψnj ð126Þ

ðEÞ, the retarded function becomes

ðEÞ includes oscillat-

ðþ<sup>∞</sup> �∞

> ðþ<sup>∞</sup> �∞ G^ R ðτÞe

> ðþ<sup>∞</sup> �∞ G^ A ðτÞe

where the limit η ! 0 is assumed in the end of calculation. With this addition, the integrals are convergent. This definition is equivalent to the definition of a retarded (advanced) function as

<sup>ð</sup>EÞ ¼ ½ð<sup>E</sup> <sup>þ</sup> <sup>i</sup>ηÞ^<sup>I</sup> � <sup>H</sup>^ �

ðEÞ is related to the retarded one through

<sup>ð</sup>EÞ ¼ <sup>G</sup>^ <sup>R</sup><sup>þ</sup>

jΨn〉〈Ψnj ¼ 1, there is

jΨn〉〈Ψnj

jΨn〉〈Ψnj

ħ θðτÞ X n e �iEnτ=ħ

a function of complex energy variable at the upper (lower) part of the complex plain.

G^ A

<sup>ð</sup>EÞ ¼ <sup>X</sup> n

> <sup>ð</sup>EÞ ¼ <sup>X</sup> n

�iEτ=<sup>ħ</sup> dE

<sup>2</sup>π<sup>ħ</sup> ¼ � <sup>i</sup>

n

G^ R

Apply the ordinary inverse Fourier transform to <sup>G</sup>^ <sup>R</sup>

ðþ<sup>∞</sup> �∞ G^ R ðEÞe

G^ R

other useful things, as we will see below).

84 Recent Studies in Perturbation Theory

, and one can consider the Fourier transform

G^ R

G^ A

Applying this transform to Eq. (117), the retarded Green operator is

G^ R

<sup>G</sup>^ <sup>R</sup>ðA<sup>Þ</sup>

This transform, however, can not be performed in all cases, because <sup>G</sup>^ <sup>R</sup>ðA<sup>Þ</sup>

ðEÞ ¼ lim η!0þ

ðEÞ ¼ lim η!0þ

ðEÞ ¼

ing terms eiEτ=<sup>ħ</sup>. To avoid this problem we define the retarded Fourier transform

τ ¼ t�t 0

and the advanced one

The advanced operator <sup>G</sup>^ <sup>A</sup>

and

Using the completeness propertyX

G^ R ðτÞ ¼

$$\mathrm{iG}(\mathbf{x}\_{\flat}, t\_{\flat\prime}, \mathbf{x}\_{\tt\prime}, t\_{\tt\imath}) = \int d\mathbf{x} \mathrm{iG}(\mathbf{x}\_{\flat\prime}, t\_{\flat\prime}, \mathbf{x}, t) \mathrm{iG}(\mathbf{x}, t, \mathbf{x}\_{\tt\imath}, t\_{\tt\imath}) \tag{128}$$

Let us divide the time interval [ta, tb] into N equal segments, each of length Δt ¼ ðtb � taÞ=N.

$$\begin{split} \mathrm{iG}(\mathbf{x}\_{b\prime}, t\_{b\prime}, \mathbf{x}\_{a\prime}, t\_{a}) &= \int d\mathbf{x}\_{1} \cdots d\mathbf{x}\_{N} \prod\_{j=1}^{N} \mathrm{iG}(\mathbf{x}\_{j\prime}, t\_{j\prime}, \mathbf{x}\_{j-1\prime}, t\_{j-1}) \\ &= A^{N} \Big[ \prod\_{j} d\mathbf{x}\_{j} \exp\left[ i \sum \Delta t L(t\_{j\prime} \frac{\mathbf{x}\_{j} + \mathbf{x}\_{j-1}}{2}, \frac{\mathbf{x}\_{j} - \mathbf{x}\_{j-1}}{2}) \right] \\ &= \int [D(\mathbf{x}) e^{i \int dt L(t, \mathbf{x}, \dot{\mathbf{x}})} \end{split} \tag{129}$$

Figure 6. The total amplitude is the sum of all amplitudes associated with thee paths connecting xa and xb.

where ln½iGðxj, tj, xj�<sup>1</sup>, tj�<sup>1</sup>Þ� ¼ <sup>i</sup>ΔtLðtj, xjþxj�<sup>1</sup> <sup>2</sup> , xj�xj�<sup>1</sup> <sup>2</sup> Þ.

Example: LC circuit-based metamaterials

In this section, we will use the relationship of current and voltage in the LC circuit to build the propagator of the LC circuit field coupled to an atom.

Figure 7 shows the LC-circuit.The following are valid:

$$I = -\frac{d\eta}{dt}\tag{130}$$

$$V = \frac{q}{\mathbb{C}} = L\frac{dI}{dt} \tag{131}$$

Thus:

$$C\frac{d^2\mathbf{x}}{dt^2} = -\frac{\mathbf{x}}{L} \tag{132}$$

where x ¼ LI, I is the current, V is the voltage, q is the charge quantity, L and C are the inductance and capacitance, respectively. Eq. (132) is equal to a harmonic, and the Lagrangian operator is:

$$L\_0(\mathbf{x}, \dot{\mathbf{x}}) = \frac{1}{2g} (\dot{\varepsilon}^2 - \Omega\_{L\mathbb{C}}^2 \varepsilon^2) \tag{133}$$

The Lagrangian operator describing the bipole is:

$$L\_0(\mathbf{x}, \dot{\mathbf{x}}) = \frac{m}{2} \dot{\mathbf{x}}^2 - \frac{m\Omega\_0^2}{2} \mathbf{x}^2 \tag{134}$$

where x is the coordinate of the bipole, ε is the LC field, m is the mass of an electron, and e is the unit of charge. <sup>g</sup> <sup>¼</sup> <sup>1</sup> c , and <sup>Ω</sup>LC <sup>¼</sup> <sup>1</sup>ffiffiffiffi LC <sup>p</sup> . Defining their action items as:

$$S\_{\rm LC} = \int dt \left[ \frac{1}{2g} \left( \dot{\varepsilon}^2 - \mathcal{Q}\_{\rm LC}^2 \varepsilon^2 \right) \right] \tag{135}$$

And

$$S\_0 = \int dt \left[\frac{m}{2} (\dot{\mathbf{x}}^2 - \Omega\_0^2 \mathbf{x}^2)\right] \tag{136}$$

Taking the coupling effect (exε) into account, the Green function of the coupled system is:

$$G(\mathbf{x}, \mathbf{c}) = \int D\mathbf{x} D\varepsilon e^{i\mathbf{S}\_{\mathrm{LC}} + i\mathbf{S}\_{\mathrm{U}} + i\int^{d[\mathrm{var}]}} \tag{137}$$

Where x represents the series coordinates x1,x2,…,and so on and ɛ represents ɛ1,ɛ2,…., and so on.

Figure 7. The coupled system, including an LC field and a bipole.

## 7. The recent applications of the Green function method

## 7.1. Convergence

where ln½iGðxj, tj, xj�<sup>1</sup>, tj�<sup>1</sup>Þ� ¼ <sup>i</sup>ΔtLðtj, xjþxj�<sup>1</sup>

86 Recent Studies in Perturbation Theory

Example: LC circuit-based metamaterials

Thus:

operator is:

the unit of charge. <sup>g</sup> <sup>¼</sup> <sup>1</sup>

And

on.

propagator of the LC circuit field coupled to an atom. Figure 7 shows the LC-circuit.The following are valid:

The Lagrangian operator describing the bipole is:

c

, and <sup>Ω</sup>LC <sup>¼</sup> <sup>1</sup>ffiffiffiffi

<sup>2</sup> , xj�xj�<sup>1</sup> <sup>2</sup> Þ.

In this section, we will use the relationship of current and voltage in the LC circuit to build the

<sup>I</sup> ¼ � dq

<sup>C</sup> <sup>¼</sup> <sup>L</sup> dI

<sup>V</sup> <sup>¼</sup> <sup>q</sup>

C d2 x dt<sup>2</sup> ¼ � <sup>x</sup>

<sup>L</sup>0ðx, <sup>x</sup>\_Þ ¼ <sup>1</sup>

<sup>L</sup>0ðx, <sup>x</sup>\_Þ ¼ <sup>m</sup>

LC

SLC ¼ ð dt <sup>1</sup> 2g ðε\_ <sup>2</sup> � <sup>Ω</sup><sup>2</sup> LCε<sup>2</sup> Þ

S<sup>0</sup> ¼ ð dt <sup>m</sup> 2 ðx\_

Gðx, εÞ ¼

where x ¼ LI, I is the current, V is the voltage, q is the charge quantity, L and C are the inductance and capacitance, respectively. Eq. (132) is equal to a harmonic, and the Lagrangian

> 2g ðε\_ <sup>2</sup> � <sup>Ω</sup><sup>2</sup> LCε<sup>2</sup>

2 x\_

where x is the coordinate of the bipole, ε is the LC field, m is the mass of an electron, and e is

Taking the coupling effect (exε) into account, the Green function of the coupled system is:

Where x represents the series coordinates x1,x2,…,and so on and ɛ represents ɛ1,ɛ2,…., and so

ð DxDεe

<sup>2</sup> � <sup>m</sup>Ω<sup>2</sup> 0

<sup>p</sup> . Defining their action items as:

� �

<sup>2</sup> � <sup>Ω</sup><sup>2</sup> 0x2 Þ

iSLCþiS0þi

Ð dt½exε�

h i

dt <sup>ð</sup>130<sup>Þ</sup>

dt <sup>ð</sup>131<sup>Þ</sup>

<sup>L</sup> <sup>ð</sup>132<sup>Þ</sup>

Þ ð133Þ

<sup>2</sup> <sup>x</sup><sup>2</sup> <sup>ð</sup>134<sup>Þ</sup>

ð135Þ

ð136Þ

ð137Þ

In the Green function, the high oscillation of Bessel/Hankel functions in the integrands results in quite time-consuming integrations along the Sommerfeld integration paths (SIP) which ensures that the integrands can satisfy the radiation condition in the direction normal to the interface of a medium. To facilitate the evaluation, the method of moments (MoM) [15], the steepest descent path (SDP) method, and the discrete complex image method (DCIM) [16, 17] are very important methods.

The technique for locating the modes is quite necessary for accurately calculating the spatial Green functions of a layered medium. The path tracking algorithm can obtain all the modes for the configuration shown in Figure 8, even when region 2 is very thick [18]. Like the method in Ref. [19], it does not involve a contour integration and could be extended to more complicated configurations.

The discrete complex image method (DCIM) has been shown to deteriorate sharply for distances between source and observation points larger than a few wavelengths [20]. So, the total least squares algorithm (TLSA) is applied to the determination of the proper and improper poles of spectral domain multilayered Green's functions that are closer to the branch point and to the determination of the residues at these poles [21].

The complex-plane kρ for the determination of proper and improper poles is shown in Figure 9. Since half the ellipse is in the proper sheet of the kρ-plane and half the ellipse is in the improper sheet, the poles will not only correctly capture the information of the proper poles but will also capture the information of those improper poles that are closer to the branch point kρ = k0.

For the 2-D dielectric photonic crystals as shown in Figure 10, the integral equation is written in terms of the unknown equivalent current sources flowing on the surfaces of the periodic

Figure 8. A general configuration with a three-layered medium: region 1 is free space, region 2 is a substrate with thickness h and relative permittivity ɛr1, and region 3 is a half space with relative permittivity ɛr2.

Figure 9. Elliptic path chosen in the complex kρ-plane when applying the total least squares algorithm. The upper half ellipse (solid line) is located in the proper Riemman sheet, and the lower half ellipse (dashed line) is located in the improper sheet.

2-D cylinders. The method of moments is then employed to solve for the unknown current distributions. The required Green function of the problem is represented in terms of a finite summation of complex images. It is shown that when the field-point is far from the periodic sources, it is just sufficient to consider the contribution of the propagating poles in the structure [22]. This will result in a summation of plane waves that has an even smaller size compared with the conventional complex images Green function. This provides an analyzed method for the dielectric periodic structures.

Figure 10. Typical (a) waveguide and (b) directional coupler in a rectangular lattice.

Others, since the Gaussian function is an eigenfunction of the Hankel transform operator, for the microstrip structures, the spectral Green's function can be expanded into a Gaussian series [23]. By introducing the mixed-form thin-stratified medium fast-multiple algorithm (MF-TSM-FMA), which includes the multipole expansion and the plane wave expansion in one multilevel tree, the different scales of interaction can be separated by the multilevel nature of the the fast multipole algorithm [24].

The vector wave functions, L, M, and N, are the solutions of the homogeneous vector Helmholtz equation. They can also be used for the analyses of the radiation in multilayer and this method avoids the finite integration in some cases.

#### 7.2. Multilayer structure

2-D cylinders. The method of moments is then employed to solve for the unknown current distributions. The required Green function of the problem is represented in terms of a finite summation of complex images. It is shown that when the field-point is far from the periodic sources, it is just sufficient to consider the contribution of the propagating poles in the structure [22]. This will result in a summation of plane waves that has an even smaller size compared with the conventional complex images Green function. This provides an analyzed

Figure 9. Elliptic path chosen in the complex kρ-plane when applying the total least squares algorithm. The upper half ellipse (solid line) is located in the proper Riemman sheet, and the lower half ellipse (dashed line) is located in the

Figure 8. A general configuration with a three-layered medium: region 1 is free space, region 2 is a substrate with

thickness h and relative permittivity ɛr1, and region 3 is a half space with relative permittivity ɛr2.

method for the dielectric periodic structures.

improper sheet.

88 Recent Studies in Perturbation Theory

The volume integral equation (VIE) can analyze electromagnetic radiation and scattering problems in inhomogeneous objects. By introducing an "impulse response" Green function, and invoking Green theorem, the Helmholtz equation can be cast into an equivalent volume integral equation including the source current or charges distribution. But the number of unknowns is typically large and the equation should be reformulated if there are in contrast both permittivity and permeability. At present, it is utilized to analyse the general scatterers in layered medium [25, 26].

When the inhomogeneity is one dimension, the Green function can be determined analytically in the spectral (Fourier) domain, and the spatial domain counterpart can be obtained by simply inverse Fourier transforming it.

Surface integral equation (SIE) method is another powerful method to handle electromagnetic problems. Similarly, by introducing the Green function, the Helmholtz equation can be cast into an equivalent surface integral equation, where the unknowns are pushed to the boundary of the scatterers [27].

Despite the convergence problem, the locations of the source and observation point may cause the change of Green function form, for example, for a source location either inside or outside the medium, the algebraic form of the Green functions changes as the receiver moves vertically in the direction of stratification from one layer to another [28].

First, we introduce the full-wave computational model [29]. A multilayer structure involving infinitely 1-D periodic chains of parallel circular cylinders in any given layer can be constructed as shown in Figure 11. Each layer consists of a homogeneous slab within which the circular cylinders are embedded. This is the typical aeronautic situation with fiberreinforced four-layer pile (with fibers orientated at 0, 45, 45, and 90), but any other arrangement is manageable likewise.

In the multilayered photonic crystals, the Rayleig's method and mode-matching are combined to produce scattering matrices. An S-matrix-based recursive matrix is developed for modeling electromagnetic scattering. Field expansions and the relationship between expansion coefficients are given.

There is a mix treatment for the inhomogeneous and homogeneous multilayered structure [30]. As shown in Figure 12, a substrate is divided into two regions. The top region is laterally inhomogeneous and for the finite-difference method (FDM) or the finite element method (FEM), the volume integral equation, is used. The bottom region is layerwise homogeneous,

Figure 11. (a) Sketch of a standard (0, 45, 45, 90) degree, four-layer fiber-reinforced composite laminate as in aeronautics. (b) General two-layer pile of interest exhibiting two different cylinder orientations and associated coordinate systems with geometrical parameters as indicated. (c) Cell defined in the lth layer of multilayered photonic crystals.

When the inhomogeneity is one dimension, the Green function can be determined analytically in the spectral (Fourier) domain, and the spatial domain counterpart can be obtained by

Surface integral equation (SIE) method is another powerful method to handle electromagnetic problems. Similarly, by introducing the Green function, the Helmholtz equation can be cast into an equivalent surface integral equation, where the unknowns are pushed to the boundary

Despite the convergence problem, the locations of the source and observation point may cause the change of Green function form, for example, for a source location either inside or outside the medium, the algebraic form of the Green functions changes as the receiver moves vertically

First, we introduce the full-wave computational model [29]. A multilayer structure involving infinitely 1-D periodic chains of parallel circular cylinders in any given layer can be constructed as shown in Figure 11. Each layer consists of a homogeneous slab within which the circular cylinders are embedded. This is the typical aeronautic situation with fiberreinforced four-layer pile (with fibers orientated at 0, 45, 45, and 90), but any other

In the multilayered photonic crystals, the Rayleig's method and mode-matching are combined to produce scattering matrices. An S-matrix-based recursive matrix is developed for modeling electromagnetic scattering. Field expansions and the relationship between expansion coeffi-

There is a mix treatment for the inhomogeneous and homogeneous multilayered structure [30]. As shown in Figure 12, a substrate is divided into two regions. The top region is laterally inhomogeneous and for the finite-difference method (FDM) or the finite element method (FEM), the volume integral equation, is used. The bottom region is layerwise homogeneous,

Figure 11. (a) Sketch of a standard (0, 45, 45, 90) degree, four-layer fiber-reinforced composite laminate as in aeronautics. (b) General two-layer pile of interest exhibiting two different cylinder orientations and associated coordinate systems

with geometrical parameters as indicated. (c) Cell defined in the lth layer of multilayered photonic crystals.

simply inverse Fourier transforming it.

arrangement is manageable likewise.

in the direction of stratification from one layer to another [28].

of the scatterers [27].

90 Recent Studies in Perturbation Theory

cients are given.

Figure 12. Substrate is divided into homogeneous and inhomogeneous regions in combined BEM/FEM and BEM/FDM methods.

and the boundary-element methods (BEM) are used. The two regions are connected such as a BEM panel is associated with an FEM node on the interface.

A Green function was derived for a layerwise uniform substrate and was then used in a layerwise nonuniform substrate with additional boundary conditions applied to the interface. Given that the lateral inhomogeneity is local, volume meshing is used only for the local inhomogeneous regions, BEM meshing is applied to the surfaces of these local regions.

For a field (observation) point in the jth layer and a source point in the kth layer, the Green function has the form:

$$\mathbf{G}\_{\vec{\mathbb{K}}}^{u,l} = \mathbf{G}\_{\vec{\mathbb{K}},0}^{u,l} + \sum\_{n=0 \atop m+n \neq 0}^{\circ} \sum\_{n=0}^{\circ} \frac{c\_{mn} \mathbf{q}\_{jk}^{u,l}}{ab \varepsilon\_{k\vec{\mathbb{K}}\searrow m}} \times \cos\frac{m\pi \mathbf{x}\_{f}}{a} \cos\frac{n\pi \mathbf{y}\_{f}}{b} \cos\frac{m\pi \mathbf{x}\_{s}}{a} \cos\frac{n\pi \mathbf{y}\_{s}}{b} \tag{138}$$

where the superscripts u and l indicate the upper and lower solutions, respectively, depending on whether the field point (or observation point) is above or below the source point. a and b are the substrate dimensions in the x- and y- directions, respectively, and more details can be found in Refs. [31, 32].

The electromagnetic field in a multilayer structure can be efficiently simplified by the assumption that the multilayer is grounded by a perfect electric conducto (PEC) plane [33, 34]. When the source and the field points are assumed to be inside the dielectric slab, in a layered medium as shown in Figure 13, by applying the boundary conditions, the 1-D Green functions is

$$(\mathbf{G\_x(x, x\_0; \lambda\_{x1}, \lambda\_{x2})} = (\mathbf{G\_x^{\rm PMC} + G\_x^{\rm REC})/2})/2 \tag{139}$$

where PMC represents the perfect magnetic conductor. The simplified Green function form can be deduced to the cae of (b).

Figure 13. (a) Geometry of an infinite dielectric slab of thickness d grounded by a PEC plane at x = d. (b) Geometry of a finite dielectric slab of thickness 2d and height 2L surrounded by regions □ and □.

The three-dimensional (3-D) Green function for a continuous, linearly stratified planar media, backed by a PEC ground plane, can also be expressed in terms of a single contour integral involving one-dimensional (1-D) green function. The constructure is shown in Figure 14.

The general formulation for a single electric current element has been worked out in detail in Ref. [35] which is based on the appropriate information from Ref. [36].

Figure 14. Representation of the continuous, linearly stratified media by discrete slabs of finite thickness and constant permittivity, ɛp and permeability μp for the pth layer of thickness hp. The thicknesses, permittivities and permeabilities are different for each layer.

## Author details

## Jing Huang

Address all correspondence to: jhuang@scut.edu.cn

Physics Department, South China University of Technology, Guangzhou, China

## References

The three-dimensional (3-D) Green function for a continuous, linearly stratified planar media, backed by a PEC ground plane, can also be expressed in terms of a single contour integral involving one-dimensional (1-D) green function. The constructure is shown in Figure 14.

Figure 13. (a) Geometry of an infinite dielectric slab of thickness d grounded by a PEC plane at x = d. (b) Geometry of a

The general formulation for a single electric current element has been worked out in detail in

Figure 14. Representation of the continuous, linearly stratified media by discrete slabs of finite thickness and constant permittivity, ɛp and permeability μp for the pth layer of thickness hp. The thicknesses, permittivities and permeabilities are

Ref. [35] which is based on the appropriate information from Ref. [36].

different for each layer.

finite dielectric slab of thickness 2d and height 2L surrounded by regions □ and □.

92 Recent Studies in Perturbation Theory


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