**2.1 Perturbation expansion**

It is possible to solve (6) exactly since it is of variables separable form. Here, we solve by an iterative process, known as perturbation expansion for the solution.

Let *v*ð Þ*<sup>i</sup>* denotes the *i*th iterate, which is obtained from the equation

$$\frac{dv^{(i)}}{dt} = -\mathbf{1} - \varepsilon v^{(i-1)}, \qquad v^{(i)}(\mathbf{0}) = \mathbf{1},\tag{7}$$

The justification for this iterative scheme is that the term *εv* involves the small multiplying coefficient *ε*, and so the term itself may be expected to be small. Thus, the term *εv*ð Þ*<sup>i</sup>* which should appear on the RHS of (7) to make it exact, may be replaced by *εv*ð Þ *<sup>i</sup>*�<sup>1</sup> with an error which is expected to be small.

The first iterate is obtained by neglecting the perturbation, thus

$$\frac{dv^{(0)}}{dt} = -\mathbf{1}, \quad v^{(0)} = \mathbf{1}.$$

This is known as the unperturbed problem, and direct integration yields

$$v^{(0)} = \mathbf{1} - t.$$

The next iterate *v*ð Þ<sup>1</sup> , satisfies

$$\frac{dv^{(1)}}{dt} = -\mathbf{1} - \varepsilon(\mathbf{1} - t), \quad v^{(1)} = \mathbf{1}.$$

and integration yields

$$v^{(1)} = \mathbf{1} - t(\mathbf{1} + \boldsymbol{\varepsilon}) + \frac{\mathbf{1}}{2}\boldsymbol{\varepsilon}^2$$

Similarly, *v*ð Þ<sup>2</sup> satisfies

$$\frac{dv^{(2)}}{dt} = -\mathbf{1} - \varepsilon \left[ \mathbf{1} - t(\mathbf{1} + \varepsilon) + \frac{\mathbf{1}}{2} \varepsilon t^2 \right], \quad v^{(2)} = \mathbf{1}.$$

Direct integration yields the solution

$$v^{(2)} = \mathbf{1} - t(\mathbf{1} + \varepsilon) + \varepsilon(\mathbf{1} + \varepsilon)\frac{t^2}{2} - \frac{1}{6}\varepsilon^2 t^3.$$

Rearranging the terms in these iterates in ascending powers of *ε*, we obtain

$$\begin{aligned} v^{(0)} &= \mathbf{1} - t, \\ v^{(1)} &= \mathbf{1} - t + \varepsilon \left(\frac{t^2}{2} - t\right), \\ v^{(2)} &= \mathbf{1} - t + \varepsilon \left(\frac{t^2}{2} - t\right) + \varepsilon^2 \left(\frac{t^2}{2} - \frac{t^3}{6}\right). \end{aligned} \tag{8}$$

Clearly as the iteration proceeds the expressions are refined by terms which involve increasing powers of *ε*. These terms become progressively smaller since *ε* is

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a small parameter. This is an example of a perturbation expansion. It will often be the case that perturbation expansions involve ascending integer powers of the small parameter, i.e., *ε*0, *ε*1, *ε*2, ⋯ � �. Such a sequence is called an *asymptotic sequence*. Although this is the most common sequence which we shall meet, it is by no means unique. Examples of other asymptotic sequences are *ε*1*<sup>=</sup>*2, *ε*, *ε*3*<sup>=</sup>*2, *ε*2, ⋯ � � and *ε*0, *ε*2, *ε*4, ⋯ � �. In each case the essential feature is that subsequent terms tend to zero faster than previous terms as *ε* ! 0.

An alternative procedure to that of developing the expansion by iteration is to assume the form of the expansion at the outset. Thus, if we assume that the perturbation expansion involves the standard asymptotic sequence *ε*0, *ε*1, *ε*2, ⋯ � �, then the solution *v*, which depends on the variable *t*, and the parameter *ε*, is expressed in the form

$$\boldsymbol{v}(t;\boldsymbol{\varepsilon}) = \boldsymbol{\varepsilon}^{0}\boldsymbol{v}\_{0}(t) + \boldsymbol{\varepsilon}^{1}\boldsymbol{v}\_{1}(t) + \boldsymbol{\varepsilon}^{2}\boldsymbol{v}\_{2}(t) + \cdots \tag{9}$$

The coefficients *v*0ð Þ*t* , *v*1ð Þ*t* , ⋯ of powers of *ε* are functions of *t* only. Substituting expansion (9) in the governing Eq. (6) yields the following

$$\begin{cases} \frac{dv\_0}{dt} + \varepsilon \frac{dv\_1}{dt} + \varepsilon^2 \frac{dv\_2}{dt} + \cdots = -\mathbf{1} - \varepsilon v\_0 - \varepsilon^2 v\_1 - \cdots \\\\ v\_0(\mathbf{0}) + \varepsilon v\_1(\mathbf{0}) + \varepsilon^2 v\_2(\mathbf{0}) + \cdots = \mathbf{1} \end{cases} \tag{10}$$

Thus, the coefficients of powers of *ε* can be equated on the left– and right–hand sides of (10):

$$\begin{cases} \varepsilon^0: & \frac{dv\_0}{dt} = -\mathbf{1}, \quad v\_0(\mathbf{0}) = \mathbf{1}, \\\\ \varepsilon^1: & \frac{dv\_1}{dt} = -v\_0, \quad v\_1(\mathbf{0}) = \mathbf{0} \\\\ \varepsilon^2: & \frac{dv\_2}{dt} = -v\_1, \quad v\_2(\mathbf{0}) = \mathbf{0}, \quad \text{etc.} \end{cases} \tag{11}$$

The proof of validity of this fundamental procedure can be developed by first setting *ε* ¼ 0 in (10) which yields the first equation of (11). This result allows the first member of the left– and right–hand side of Eq. (10) to be removed. Then, after dividing the remaining terms by *ε* we obtain the equation

$$\frac{dv\_1}{dt} + \varepsilon \frac{dv\_2}{dt} + \dots = -v\_0 - \varepsilon v\_1 - \dotsb$$

This is valid for all nonzero values of *ε* so that on taking the limit as *ε* ! 0 we obtain the second equation of (11). Repeating the procedure, we obtain the other equations.

Integrating the equations in (11), we obtain

$$v\_0 = \mathbf{1} - t, \quad v\_1 = t^2/2 - t, \quad v\_2 = t^2/2 / - t^3/6.$$

Using these values in (9), we obtain that

$$v(t; \varepsilon) = \mathbf{1} - t + \varepsilon(t^2 - t) + \varepsilon^2(t^2/2 - t^3/6) + \dotsb \tag{12}$$

This is the same as the expansion (8) which is generated by iteration.

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

The IVP (6) can be solved exactly as

$$\nu(t) = [(\mathbf{1} + \varepsilon)e^{-nt} - \mathbf{1}]e^{-1}.$$

The perturbation expansion can be obtained from (12) by replacing the exponential function by its Maclaurin expansion, i.e.,

$$\nu(t) = \frac{1}{\varepsilon} \left[ 1 - \varepsilon t + \frac{\varepsilon^2 t^2}{2} - \frac{\varepsilon^3 t^3}{6} + \dots + \varepsilon - \varepsilon^2 t + \frac{\varepsilon^3 t^2}{2} + \dots - 1 \right] \tag{13}$$

$$\mathbf{u} = (\mathbf{1} - t) + \varepsilon \left(\frac{t^2}{2} - t\right) + \varepsilon^2 \left(\frac{t^2}{2} - \frac{t^3}{6}\right) + \dotsb \tag{14}$$

This is the same as the expansion (12). Thus, the perturbation expansion approach is justified in this case. One can refer the books [1, 2].
