**4.3 Boundary layer thickness and the principle of least degeneracy**

The boundary layers which we have met so far have all had thickness *O*ð Þ*ε* . By this we mean that a variation of *O*ð Þ*ε* in the independent variable will encompass the region of rapid change in the dependent variable. The associated stretched independent variable *s*, appropriate for the boundary layer is related to *x* by a linear transformation involving division by *ε*.

There are practical situations where the boundary layer thickness will be of *<sup>O</sup> <sup>ε</sup><sup>p</sup>* ð Þ. This means that if the boundary layer is located at *<sup>x</sup>* <sup>¼</sup> *<sup>x</sup>*0, then the appropriate stretching transformation is *<sup>s</sup>* <sup>¼</sup> ð Þ *<sup>x</sup>* � *<sup>x</sup>*<sup>0</sup> *<sup>=</sup>ε<sup>p</sup>*. More generally, the choice of the function *δ ε*ð Þ to use in the stretching transformation *s* ¼ ð Þ *x* � *x*<sup>0</sup> *=δ ε*ð Þ is determined by the need to represent the region of rapid change correctly. We must ensure that the boundary layer solution contains rapidly varying functions. The form of the governing equation in the boundary layer region must have sufficient structure to allow such solutions.

Consider the example

$$\begin{cases} \varepsilon \frac{d^2 u}{d\mathfrak{x}^2} + \frac{du}{d\mathfrak{x}} + u = \mathfrak{x}, & (0, \mathfrak{1}) \\\\ u(\mathfrak{0}) = \mathfrak{1}, \quad u(\mathfrak{1}) = \mathfrak{2}. \end{cases} \tag{56}$$

Since the signs of the first and second derivatives are the same, and the boundary layer will occur at *x* ¼ 0. We are not going to assume at the outset that the boundary layer thickness is *O*ð Þ*ε* . Our intension is to deduce that the appropriate stretching variable is *s* ¼ *x=ε*.

The one-term outer expansion *u*<sup>0</sup> satisfies *du*<sup>0</sup> *dx* þ *u*<sup>0</sup> ¼ *x*, *u*0ð Þ¼ 1 2 The solution is

$$
\mu\_0(\mathbf{x}) = 2\mathbf{e}^{1-\mathbf{x}} + \mathbf{x} - \mathbf{1}.\tag{57}
$$

To determine the inner expansion we first wrongly assume that the boundary layer thickness is *<sup>O</sup> <sup>ε</sup>*<sup>1</sup>*=*<sup>2</sup> � �. The stretching transformation *<sup>s</sup>* <sup>¼</sup> *<sup>x</sup>=ε*<sup>1</sup>*=*<sup>2</sup> changes the original DE (56) into the following one:

$$\frac{d^2v}{ds^2} + \frac{1}{\varepsilon^{1/2}}\frac{dv}{ds} + v = \varepsilon^{1/2}s \tag{58}$$

### *A Collection of Papers on Chaos Theory and Its Applications*

If the appropriate stretching transformation has been used for the boundary layer then *dv=ds* and *d*<sup>2</sup> *<sup>v</sup>=ds*<sup>2</sup> will be of *<sup>O</sup>*ð Þ<sup>1</sup> within it. The leading order expansion *v*<sup>0</sup> will satisfy the dominant part of (58), i.e., the component of *O ε*�1*=*<sup>2</sup>

$$\frac{dv\_0}{ds} = 0, \quad v\_0(0) = 1.\tag{59}$$

The solution is *v*0ðÞ¼ *s* 1. This of course does not have the rapidly varying behavior which we anticipate in the boundary layer. Prandtl's matching condition cannot be satisfied since

$$\lim\_{\mathbf{x}\to\mathbf{0}}\left(2\mathbf{e}^{1-\mathbf{x}}+\mathbf{x}-\mathbf{1}\right)=2\mathbf{e}-\mathbf{1}\neq\lim\_{\mathbf{s}\to\mathbf{s}}\nu\_{0}(\mathbf{s})=\mathbf{1}.$$

Thus, we reject the assumption of a boundary layer of thickness *O ε*1*=*<sup>2</sup> .

Next, suppose that the boundary layer thickness is *<sup>O</sup> <sup>ε</sup>*<sup>2</sup> ð Þ and again we will discover that this is incorrect because the corresponding inner expansion cannot be matched to the outer expansion. Proceeding with the analysis we introduce the stretching transformation *<sup>s</sup>* <sup>¼</sup> *<sup>x</sup>=ε*<sup>2</sup> which leads to the equation

$$\frac{1}{\varepsilon^3} \frac{d^2 v}{ds^2} + \frac{1}{\varepsilon^2} \frac{dv}{ds} + v = \varepsilon^2 \mathfrak{s}.$$

Again we argue that if the appropriate stretching has been used then all derivatives are of *<sup>O</sup>*ð Þ<sup>1</sup> so that the governing equation for the leading term in *<sup>O</sup> <sup>ε</sup>*�<sup>3</sup> ð Þ, namely

$$\frac{d^2 v\_0}{ds^2} = 0, \quad v\_0(\mathbf{0}) = \mathbf{1}.\tag{60}$$

The solution is *v*0ðÞ¼ *s* 1 þ *As*, where the constant *A* is to be determined from matching. This solution is rapidly varying but the rapidity does not decay at the edge of the boundary layer (i.e., as *s* ! ∞). Indeed, we cannot match *v*<sup>0</sup> to the outer expansion because the term *As* becomes arbitrarily large as *s* ! ∞.

The correct choice of stretching transformation is *s* ¼ *x=ε* showing that the boundary layer thickness is *O*ð Þ*ε* . The boundary layer equation becomes

$$\frac{1}{\varepsilon} \frac{d^2 v}{ds^2} + \frac{1}{\varepsilon} \frac{dv}{ds} + v = \varepsilon \varepsilon.$$

The dominant equation satisfied by *v*<sup>0</sup> is *O*ð Þ 1*=ε* , namely

$$\frac{d^2v\_0}{ds^2} + \frac{dv\_0}{ds} = 0, \quad v\_0(0) = 1.\tag{61}$$

The solution is *<sup>v</sup>*0ðÞ¼ *<sup>s</sup>* <sup>1</sup> � *<sup>A</sup>* <sup>þ</sup> *Ae*�*<sup>s</sup>* . The last member provides the necessary rapid decay away from the point *x* ¼ *s* ¼ 0. Prandtl's matching condition requirest

$$\lim\_{x \to 0} \left( 2e^{1-x} + x - 1 \right) = \lim\_{s \to \infty} (1 - A + Ae^{-s}),$$

which leads to *A* ¼ 2 � 2*e*, and

$$v\_0(\mathbf{x}) = 2e - \mathbf{1} + \mathbf{2}(\mathbf{1} - e)e^{-\mathbf{x}/\varepsilon}.$$

The one-term composite expansion is

$$
\mu^{\rm comp} = \left( 2\varepsilon^{1-\chi} + \varkappa - \mathbf{1} \right) + \left( 2\varepsilon - \mathbf{1} \right) + \mathbf{2} \left( \mathbf{1} - \varepsilon \right) e^{-\mathbf{x}/\varepsilon} - \left( 2\varepsilon - \mathbf{1} \right). \tag{62}
$$

The leading order boundary layer equation associated with the stretching transformation *<sup>s</sup>* <sup>¼</sup> *<sup>x</sup>=ε*, (61) involves more terms than (59), associated with *<sup>s</sup>* <sup>¼</sup> *<sup>x</sup>=ε*1*<sup>=</sup>*2, and (60) associated with *<sup>s</sup>* <sup>¼</sup> *<sup>x</sup>=ε*2. The extra term in (61) allows sufficient structure in the solution to produce the required boundary layer behavior. An aid for choosing the boundary layer thickness is to seek a stretching transformation which retains the largest number of terms in the dominant equation governing *v*0. This referred to as the *principle of least degeneracy* by Van Dyke.

The composite expansion (62) can be verified by comparing with the exact solution of (56). The general solution of (56) is

$$
\mu^{\rm ex} = \mathbf{C}\_1 e^{m\_1 \chi} + \mathbf{C}\_2 e^{m\_2 \chi} + (\mathbf{x} - \mathbf{1}),
$$

where

$$m\_1 = \frac{-1 + \sqrt{1 - 4\varepsilon}}{2\varepsilon}, \quad m\_2 = \frac{-1 - \sqrt{1 - 4\varepsilon}}{2\varepsilon}.$$

We expand ffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup> � <sup>4</sup>*<sup>ε</sup>* <sup>p</sup> using the binomial series, ffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup> � <sup>4</sup>*<sup>ε</sup>* <sup>p</sup> <sup>¼</sup> <sup>1</sup> � <sup>2</sup>*<sup>ε</sup>* <sup>þ</sup> *<sup>O</sup> <sup>ε</sup>*<sup>2</sup> ð Þ, then

$$
\mathfrak{m}\_1 = -\mathbf{1} + \mathcal{O}(\varepsilon), \quad \text{and} \quad \mathfrak{m}\_2 = -\frac{\mathbf{1}}{\varepsilon} + \mathbf{1} + \mathcal{O}(\varepsilon),
$$

so that

$$
\mu^{\rm ex} = \mathbf{C}\_1 \mathbf{e}^{-\mathbf{x}} + \mathbf{C}\_2 \mathbf{e}^{-\mathbf{x}/\boldsymbol{\varepsilon}} \cdot \mathbf{e}^{\mathbf{x}} + (\mathbf{x} - \mathbf{1}) + \mathcal{O}(\boldsymbol{\varepsilon}).\tag{63}
$$

Using the boundary conditions and by neglecting the transcendentally small term *e*�1*=<sup>ε</sup>* , we have *C*<sup>1</sup> ¼ 2*e*,*C*<sup>2</sup> ¼ 2 1ð Þ � *e* . Then, (63) becomes

$$u^{\rm ex} = 2e^{1-\chi} + 2(1-e)e^{-\chi/\varepsilon} \cdot e^{\chi} + (\varkappa - \mathbf{1}) + O(\varepsilon). \tag{64}$$

There is an apparent discrepancy between (64) and the composite expansion (62) in the coefficient of the *e*�*x=<sup>ε</sup>* term. There is an extra term only contributes in the boundary layer where *<sup>x</sup>* <sup>¼</sup> *<sup>O</sup>*ð Þ*<sup>ε</sup>* so that the coefficient *ex* may to leading order, be replaced by unity. Thus, the leading order composite expansion and the leading order term in the exact solution are in complete agreement.

#### **4.4 Boundary layer of thickness of** *O* ffiffi *ε* p ð Þ

Consider the following two-point BVP:

$$\begin{cases} \varepsilon \frac{d^2 u}{d\mathbf{x}^2} + \mathfrak{x}^2 \frac{du}{d\mathbf{x}} - u = \mathbf{0}, \quad (\mathbf{0}, \mathbf{1}) \\\\ u(\mathbf{0}) = \mathbf{1}, \quad u(\mathbf{1}) = \mathbf{2}. \end{cases} \tag{65}$$

We seek a one-term composite expansion for the above BVP. We will tentatively assume that a boundary layer occurs at *x* ¼ 0 although the vanishing of the coefficient of the first derivative suggests the possibility of nonstandard behavior.

The one term outer expansion satisfies

$$
\propto^2 \frac{du\_0}{d\mathbf{x}} - u\_0 = \mathbf{0}, \quad u\_0(\mathbf{1}) = \mathbf{2}.
$$

Its exact solution is *<sup>u</sup>*0ð Þ¼ *<sup>x</sup>* <sup>2</sup>*e*ð Þ <sup>1</sup>�1*=<sup>x</sup>* .

Let us assume that the boundary layer thickness is of *<sup>O</sup> <sup>ε</sup><sup>p</sup>* ð Þ, where *<sup>p</sup>* is to be determined from the principle of least degeneracy. The stretched variable is *<sup>s</sup>* <sup>¼</sup> *<sup>x</sup>=εp*, and (65) becomes

$$
\epsilon^{1-2p} \frac{d^2 v}{ds^2} + \epsilon^p s^2 \frac{dv}{ds} - v = \mathbf{0}.
$$

The second-term is always dominated by the third, so the principle of degeneracy requires the first term to be of the same order as the third term (i.e., *O*ð Þ1 ). Thus, *p* ¼ 1*=*2, and the one-term inner expansion satisfies

$$\frac{d^2v\_0}{ds^2} + \frac{dv\_0}{ds} = \mathbf{0}, \quad v\_0(\mathbf{0}) = \mathbf{1}.$$

The solution of the above problem is *<sup>v</sup>*0ðÞ¼ *<sup>s</sup> Ae<sup>s</sup>* <sup>þ</sup> ð Þ <sup>1</sup> � *<sup>A</sup> <sup>e</sup>*�*<sup>s</sup>* . Prandtl's matching condition requires

$$\lim\_{x \to 0} 2e^{(1-1/x)} = \lim\_{t \to \infty} \left[ Ae^t + (1-A)e^{-t} \right]$$

which yields *A* ¼ 0. This example is rather special in that *A* will be zero for all boundary conditions.

The on-term composite expansion is

$$
\mu\_0^{\text{comp}} = \mathbf{2} \mathbf{c}^{(1-\mathbf{x})} + \mathbf{c}^{-\mathbf{x}/\sqrt{\mathbf{c}}}.
$$

We conclude this example with the observation that a choice for the value of the index *p* other than *p* ¼ 1*=*2 leads to boundary layer equations with insufficient structure to generate the required rapidly decaying behavior.

Thus, if *p* >1*=*2, the dominant equation becomes

$$\frac{d^2v\_0}{ds^2} = \mathbf{0}, \quad v\_0(\mathbf{0}) = \mathbf{1},$$

which gives *v*0ðÞ¼ *s* 1 þ *As*. It is obvious that Prandtl's matching condition cannot be used to determine *A*. Whereas, if *p*<1*=*2 the dominant equation degenerates to *v*0ðÞ¼ *s* 0 which does not satisfy the boundary condition at *s* ¼ 0.

#### **4.5 Interior layer**

Consider the BVP:

$$\begin{cases} \varepsilon \frac{d^2 u}{d\varkappa^2} + \varkappa \frac{du}{d\varkappa} + \varkappa u = 0, \quad (-1, 1) \\ u(-1) = \varepsilon, \quad u(1) = 2\varepsilon^{-1}. \end{cases} \tag{66}$$

The coefficient of the first derivative (convective term) is positive in 0, 1 ð Þ which indicates the occurrence of a boundary layer at the left hand limit *x* ¼ 0. *Perturbation Expansion to the Solution of Differential Equations DOI: http://dx.doi.org/10.5772/intechopen.94173*

While the corresponding coefficient is negative in the range �1< *x*<0 indicates a boundary layer located at the right-hand limit which is again is *x* ¼ 0. Thus, we are led to expr = ect two outer expansions for positive and negative *x* respectively and an inner expansion in the interior layer located at *x* ¼ 0. We denote the leading term in the outer expansion for positive *x* by *u*<sup>þ</sup> <sup>0</sup> , it satisfies

$$\frac{d\boldsymbol{u}\_0^+}{d\boldsymbol{x}} + \boldsymbol{u}\_0^+ = \mathbf{0}, \quad \boldsymbol{u}\_0^+(\mathbf{1}) = 2\boldsymbol{e}^{-1} \tag{67}$$

with the solution *u*<sup>þ</sup> <sup>0</sup> ð Þ¼ *<sup>x</sup>* <sup>2</sup>*e*�*x*.

The outer expansion for negative *x*, *u*� <sup>0</sup> satisfies

$$\frac{d\boldsymbol{u}\_0^-}{d\boldsymbol{x}} + \boldsymbol{u}\_0^- = \mathbf{0}, \quad \boldsymbol{u}\_0^+( -\mathbf{1}) = \boldsymbol{e} \tag{68}$$

with the solution *u*� <sup>0</sup> ð Þ¼ *<sup>x</sup> <sup>e</sup>*�*<sup>x</sup>*.

We suppose the boundary layer at *<sup>x</sup>* <sup>¼</sup> 0 has thickness *<sup>O</sup> <sup>ε</sup><sup>p</sup>* ð Þ and determine the index *<sup>p</sup>* using the principle of least degeneracy. Let *<sup>s</sup>* <sup>¼</sup> *<sup>x</sup>=ε<sup>p</sup>* so that the DE becomes

$$
\epsilon^{1-2p} \frac{d^2 v}{ds^2} + \mathfrak{s} \frac{dv}{ds} + \mathfrak{e}^p \mathfrak{s} v = \mathbf{0}.
$$

The third term is dominated by the second term. The first term has the same order as the second term if *p* ¼ 1*=*2. For this choice of *p* the leading term of the inner expansion *v*<sup>0</sup> satisfies

$$\frac{d^2v\_0}{ds^2} + s\frac{dv\_0}{ds} = 0.$$

Its solution can be given by

$$v\_0(\mathfrak{s}) = B \operatorname{erf} \left( \mathfrak{s} / \sqrt{2} \right) + v\_0(0),$$

Prandtl;s matching condition applied to the region *x*> 0 is

$$\lim\_{\mathfrak{s}\to+\infty} \upsilon\_0(\mathfrak{s}) = \lim\_{\mathfrak{x}\to \mathfrak{0}^+} \mathfrak{u}\_0^+(\mathfrak{x}),$$

and corresponding for *x*<0, we have

$$\lim\_{s \to -\infty} \nu\_0(s) = \lim\_{x \to 0^-} u\_0^-(x).$$

Using the limiting values *erf* ð Þ¼� �∞ 1 yields *v*0ð Þ¼ 0 1*:*5 and *B* ¼ 0*:*5. The leading order terms over the whole region are

$$\begin{aligned} u\_0^+(\mathbf{x}) &= 2e^{-\mathbf{x}}, & \mathbf{x} &> O(\sqrt{\varepsilon})\\ v\_0 &= \mathbf{0}.5\,\text{erf}\left(\mathbf{x}/\sqrt{2\varepsilon}\right) + \mathbf{1}.5, & \mathbf{x} &= O(\sqrt{\varepsilon})\\ u\_0^-(\mathbf{x}) &= e^{-\mathbf{x}}, & \mathbf{x} &< -O(\sqrt{\varepsilon}) \end{aligned}$$

A composite expansion cannot be formed in the standard way when there is more than one outer solution. However, the behavior of *v*<sup>0</sup> for ∣*x*∣> *O* ffiffi *ε* p ð Þ is as follows:

$$\begin{aligned} \upsilon\_0 \left[ \mathbf{x} > O(\sqrt{\varepsilon}) \right] &= \mathbf{0.5} + \mathbf{1.5} + \mathbf{T.S.T} \\ \upsilon\_0 \left[ \mathbf{x} < -O(\sqrt{\varepsilon}) \right] &= -\mathbf{0.5} + \mathbf{1.5} + \mathbf{T.S.T} \end{aligned}$$

Utilizing this enables a uniformly valid one-term composite expansion to be constructed which yields the correct coefficient of *e*�*<sup>x</sup>* outside the boundary layer and the correct leading order behavior within the boundary layer. It is

$$
\mu\_0^{\text{comp}} = \left[ \mathbf{0.5} \,\text{erf}\left(\mathbf{x} / \sqrt{2\varepsilon}\right) + \mathbf{1.5} \right] e^{-\mathbf{x}}.
$$

#### **4.6 Nonlinear differential equation**

Consider the following semilinear

$$\begin{cases} \varepsilon \frac{d^2 u}{dx^2} + \frac{du}{d\infty} + u^2 = 0, \quad (0, 1) \\ u(0) = 2, \quad u(1) = 1/2. \end{cases} \tag{69}$$

The coefficient of the first and second order derivatives have the same sign, so the boundary layer will occur at the left boundary *x* ¼ 0. The one-term outer expansion satisfies

$$\frac{d\mu\_0}{d\mathbf{x}} + \mu\_0^2 = \mathbf{0}, \quad \mu\_0(\mathbf{1}) = \mathbf{1}/2,$$

and the solution is *u*0ð Þ¼ *x* 1*=*ð Þ 1 þ *x* . The stretching transformation for the inner region will be *s* ¼ *x=ε* and therefore, the inner expansion satisfies

$$\frac{d^2v}{ds^2} + \frac{dv}{ds} + \epsilon v^2 = \mathbf{0}, \quad v(\mathbf{0}) = \mathbf{2}.$$

The one-term inner expansion *v*<sup>0</sup> satisfies the dominant part of this equation, i.e.,

$$\frac{d^2v\_0}{ds^2} + \frac{dv\_0}{ds} = \mathbf{0}, \quad v\_0(\mathbf{0}) = \mathbf{2}, \dots$$

which gives *<sup>v</sup>*0ðÞ¼ *<sup>s</sup> <sup>A</sup>* <sup>þ</sup> ð Þ <sup>2</sup> � *<sup>A</sup> <sup>e</sup>*�*<sup>s</sup>* . Prandtl's matching condition yields *A* ¼ 1, and the composite one-term uniformly valid expansion is

$$
\mu\_0^{\text{comp}} = \frac{1}{1+x} + e^{-x/x}.
$$

Next, consider the quasilinear problem

$$\begin{cases} \varepsilon \frac{d^2 u}{d\kappa^2} + 2u \frac{du}{d\kappa} - 4u = 0, \quad (0, 1) \\ u(0) = 0, \quad u(1) = 4. \end{cases} \tag{70}$$

The nonlinearity is associated with the first derivative term. The location of the boundary layer depends on the relative sign of the first and second derivative coefficients. If we assume that the dependent variable is nonnegative throughout the interval 0< *x*<1, then the boundary layer will occur at *x* ¼ 0. The one-term outer expansion satisfies

*Perturbation Expansion to the Solution of Differential Equations DOI: http://dx.doi.org/10.5772/intechopen.94173*

$$2\mu\_0 \frac{d\mu\_0}{d\mathbf{x}} - 4\mu\_0 = \mathbf{0}, \quad \mu\_0(\mathbf{1}) = 4\mathbf{s}$$

with the solution *u*0ð Þ¼ *x* 2*x* þ 2.

Assuming that the boundary layer thickness is *O*ð Þ*ε* , therefore, the dominantorder equation for the one-term inner expansion becomes

$$\frac{d^2 v\_0}{ds^2} + 2v\_0 \frac{dv\_0}{ds} = \mathbf{0}, \quad v\_0(\mathbf{0}) = \mathbf{0}.$$

Its solution is *v*0ðÞ¼ *s a* tanh ð Þ *as* . Prandtl's matching condition yields *a* ¼ 2. Thus, *v*0ðÞ¼ *s* 2 tanh 2ð Þ*s* , and the uniformly valid one-term composite expansion is

$$
\mu\_0^{\text{comp}} = 2\mathfrak{x} + 2 + 2\tanh\left(2\mathfrak{s}\right) - 2\mathfrak{s}
$$

Application of perturbation techniques to partial differential equations, and other types of problems can be seen in the books [5, 6].
