**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 limx!0 fðxÞ ϕðxÞ ¼ M, then write fðxÞ ¼ OðϕðxÞÞ, x ! 0, and M is constant. If limx!0 fðxÞ ϕðxÞ ¼ 0, then write fðxÞ ¼ oðϕðxÞÞ, x ! 0. If limx!0 fðxÞ ϕðxÞ ¼ 1, then write fðxÞ eϕðxÞ, 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 n} 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 fðxÞ ∈C∞ in some neighborhood of x ¼ 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 Oð0, 0, 0Þ and yðx, εÞ∈C∞, for small x, ε.

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 yðxÞ ¼ ae�x=ε and is not expanded in an asymptotic series in powers of ε, 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

�

εn f g,

fðxÞ ¼ Xn

k¼0

fðxÞe 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

fðxÞ e

X∞ n¼1

f nxn, where f n ¼ f

Oð0, 0, 0Þ. It is correct next.

parameter ε, i.e.,

point Oð0, 0, 0Þ and yðx, εÞ∈C∞, for small x, ε.

δnþ1ðεÞ

1=lnð1=εÞ n o �n

δnðεÞ ¼ 0, ∀n ¼ 1, 2, …

, �

f kϕkðxÞ þ Oðϕnþ1ðxÞÞ, x ! 0,

f kϕkðxÞ, 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 n} and has the relation

Theorem (Taylor (1715) and Maclaurin (1742)). If the function fðxÞ ∈C∞ in some neighborhood of x ¼ 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

yðx, εÞ ¼ X∞

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

εlnð1=εÞ n o �n

:

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 ½εα, 1�, where 0 < α < 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 x ∈½0, 1�, pðxÞ, rðxÞ∈ C∞½0, 1�. 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. εy0

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 y0ðx޼�x�1

a singular point in the considering domain.

ðx þ εyðxÞÞy0

where x ∈½0, 1�, pðxÞ, rðxÞ∈ C∞½0, 1�. For unperturbed equation

xy0

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.

y00ðxÞ þ nx�1y0

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

0ðxÞ þ pðxÞy0ðxÞ ¼ rðx),

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

ðxÞ þ yðxÞy0

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 βðy0

ðxÞÞ2 , 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 fðxÞ ∈C∞½0, 1�, f(x)=X∞ k¼0 f kxk , f k ¼ f ðkÞ ð0Þ=k!, f 0 6¼ 0; a is the constant

Explicit solution of the problem (3) has the form: yðxÞ ¼ ae�x2=2ε þ 1 ε ðx 0 e ðs2�x2Þ=2ε 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 ð ffiffiffi ε p , 1�, and the point x0= ffiffiffi ε p ¼ μ 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∈ Cð∞Þ ½0, 1�, πkðtÞ∈ Cð∞Þ ½0, μ�1�, x ¼ μt and boundary layer functions πkðtÞ decreasing by power law as t ! ∞, that is,πkðtÞ ¼ Oðt �mÞ, t ! ∞, m ∈ N.

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 πk�1ðtÞ, 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

y0ðxÞ ¼ fðxÞ

y0

y0

y0

�

1ðxÞ þ μ2π0

The initial conditions for the functions πk�1ðtÞ, k ¼ 0, 1, … we take in the next form

kðtÞ þ tπkðtÞ þ xYkþ1ðxÞ þ Y0

y1ðxÞ ¼ x�1

y2ðxÞ ¼ x�1

y3ðxÞ ¼ x�1

y0

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

π�1ðtÞ þ Y0ðxÞ þ π0ðtÞ þ μ

½0, 1�, πkðtÞ∈ Cð∞Þ

0ðtÞ þ μ3Y0

μ0 : π0

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

μ1 : π0

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

μkþ1 : π0

0ðxÞ þ μπ0

ynðxÞ ¼ x�1

the form

π0

yðxÞ ¼ μ�1

6 Recent Studies in Perturbation Theory

where YkðxÞ∈ Cð∞Þ

�1ðtÞ þ μ2Y0

þ μ4tπ3ðtÞ þ … ¼ fðxÞ:

From Eq. (6), we have

x e f 0x�1

0ðxÞ e f 0x�3

1ðxÞ e 3f 0x�5

2ðxÞ e 3 � 5f 0x�5

n�1ðxÞ e 3 � 5 � … � ð2n � 1Þf 0x�ð2nþ1Þ

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

Y1ðxÞ þ π1ðtÞ

1ðtÞ þ μ4Y0

þ xY0ðxÞ þ μxY1ðxÞ þ μ2xY2ðxÞ þ μ3xY3ðxÞ þ … þ tπ�1ðtÞ þ μtπ0ðtÞ þ μ2tπ1ðtÞ þ μ3tπ2ðtÞ

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

xY0ðxÞ ¼ fðxÞ � f 0 ) Y0ðxÞ¼ðfðxÞ � f 0Þ=x,

� þ μ2 �

, x ! 0,

, x ! 0,

, x ! 0,

, x ! 0,

ε p , 1�, and the point x0= ffiffiffi

Y2ðxÞ þ π2ðtÞ

½0, μ�1�, x ¼ μt and boundary layer functions πkðtÞ

2ðtÞ þ μ5Y0

�mÞ, t ! ∞, m ∈ N.

�1ðtÞ þ tπ�1ðtÞ þ xY0ðxÞ ¼ fðx), ð7:-1Þ

0ðtÞ þ tπ0ðtÞ þ xY1ðxÞ ¼ 0, ð7:0Þ

k�1ðxÞ ¼ 0, k ¼ 1, 2, … ð7:kÞ

2ðxÞ þ μ3π0

�

, x ! 0,

ε p ¼ μ is singular

þ …, μ ! 0,

3ðxÞ þ μ4π0

ð5Þ

3ðtÞ 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 ½0, μ�1�, 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�1ðtÞ.

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 Y0 0ð0Þ � Y0 0ðxÞ, then π0 1ðtÞ þ tπ1ðt޼�Y0 0ð0Þ.

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 Y0 mðxÞ � 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 OðμmÞ, μ ! 0, x ∈½0, 1�:

#### 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 fðxÞ ¼ X∞ k¼0 f kxk , x ! 0, f k ¼ f ðkÞ ð0Þ=k!, f 0 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 yð0Þ ¼ a, yð1Þ ¼ b, a2 þ b2 6¼ 0, 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

π0 2ðtÞ þ tπ2ðtÞ ¼ 0, π2ð0޼�Y2ð0Þ ¼ 2f 3:

�t 2=2 :

Y1ðxÞ þ π1ðtÞ

�

þ … þ μm

εy00ðxÞ � xyðxÞ ¼ fðxÞ, x∈ð0, 1Þ, ð9Þ

yð0Þ ¼ 0, yð1Þ ¼ 0: ð10Þ

�

mðxÞ, 0 < x ≤ 1, Rmð0Þ ¼ 0: ð8Þ

YmðxÞ þ πmðtÞ

� .

π2ðtÞ ¼ 2f 3e

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=2ε ð x

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

ð0Þ=k!, f 0 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 mðsÞe s2=2ε ds;

From this, we get

8 Recent Studies in Perturbation Theory

where ymðxÞ ¼ 1

where fðxÞ ¼ X∞

k¼0

can lead to conditions (10).

f kxk

remainder term RmðxÞ ¼ yðxÞ � ymðxÞ,

εR0

The problem (8) has a unique solution

differential equation with a turning point

We note that if m is odd, then Y0

μ π�1ðtÞ þ Y0ðxÞ þ π0ðtÞ þ μ

and from this, we have RmðxÞ ¼ OðμmÞ, μ ! 0, x ∈½0, 1�:

, x ! 0, f k ¼ f

yð0Þ ¼ a, yð1Þ ¼ b, a2 þ b2 6¼ 0, using transformation

mðxÞ þ xRmðx޼�μmþ2

mðxÞ � 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 ð ffiffi ε p3 , 1�. The point x0= ffiffi ε p3 ¼ μ 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 t ¼ x=μ, μ ¼ ffiffi ε p3 , η ¼ ð1 � xÞ=λ, λ ¼ ffiffi ε p . Here, YkðxÞ∈ C∞½0, 1�, πkðtÞ ∈C∞½0, 1=μ� is boundary layer function in a neighborhood of t ¼ 0 and decreases by the power law as t ! ∞, and the function wkðtÞ∈ C∞½0, 1=λ� is boundary function in a neighborhood of η ¼ 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 πk�1ðtÞ, 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 zðtÞ∈C∞½0, μ�1�.

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 Aið0Þ ffiffiffi 3 p , so zð0Þ ¼ z0.

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

as AiðtÞ e t �1=4e�2 3 t 3=2 , BiðtÞ e t �1=4e 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, 2 is the constant.

Boundary conditions for functions πk�1ðtÞ, k ¼ 0, 1, … we take next form

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

π00

zð0Þ ¼ z0

� Bið0Þ

zð1=μÞ ¼ z0 Aið1=μÞ

�1=4e 2 3 t 3=2

> ðt 0

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

BiðsÞds þ BiðtÞ

ð1=μ 0

Aið0Þ � πbð1 � ffiffiffi

BiðsÞds þ Bi0

BiðsÞds þ Bi00ðtÞ

will have the unique solution and this one have next form

� πb � AiðtÞ ðt 0

zð0Þ ¼ z0 � πb

p , so zð0Þ ¼ z0.

, BiðtÞ e t

� πb � Ai0 ðtÞ ðt 0

Aið0Þ � πb Ai00ðtÞ

ðtÞ Aið0Þ μ!0

�xY0ðxÞ ¼ fðxÞ � f 0 ) Y0ðxÞ ¼ �ðfðxÞ � f 0Þ=x,

�1ðtÞ � tπ�1ðtÞ ¼ f 0:

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

ð1=μ t

AiðsÞds � Aið0Þ ffiffiffi

3 p ÞAið1=μÞ

ðtÞ ð1=μ t

> ð1=μ t

, t ! ∞, so zð1=μÞ ¼ OðμÞ, μ ! 0.

, zð1=μÞ ! 0, μ ! 0 ð17Þ

3 p ð1=μ 0

AiðsÞds � ,

BiðsÞds,

ðtÞ ffiffiffi 3 p ð1=μ 0

π � Ai00ðtÞ ffiffiffi

3 p ð1=μ 0

AiðsÞds �

AiðsÞds

AiðsÞds � ,

AiðsÞds � AiðtÞ ffiffiffi

3 p ð1=μ 0

> ð1=μ 0

AiðsÞds � Ai0

AiðsÞds � 1

!

πk�1ð1=μÞ ¼ 0, k ¼ 0, 1, 2, …

π�1ð0Þ 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

zðtÞ ¼ z0 AiðtÞ

and zðtÞ∈C∞½0, μ�1�.

as Bið0Þ ¼ Aið0Þ ffiffiffi

�1=4e�2 3 t 3=2

ðtÞ ¼ z0 Ai0

z00ðtÞ ¼ z0 Ai00ðtÞ

as AiðtÞ e t

z0

Aið0Þ

Proof. We verify the boundary conditions:

3

Considering the boundary condition rð1=μÞ ! 0, μ ! 0, we have c2 ¼ 0; rðtÞ ¼ c1AiðtÞ. And the second condition rð0Þ ¼ 0, c1 ¼ 0 follows. This implies that rðtÞ � 0.

Therefore, zðtÞ � uðtÞ. It is obvious that zðtÞ ∈C∞½0, μ�1�. Lemma 1 is proved.

This Lemma 1 implies the existence and uniqueness of π�1ðtÞ∈C∞½0, μ�1� 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 ½0, μ�1�, and as t ! ∞:

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

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

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 ½0, μ�1�, and as t ! ∞:

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

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∈ C∞½0, ∞Þ, wkðηÞ ¼ Oðe�ηÞ with η ! ∞.

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 rmðxÞ¼ð2 � x2ÞRmðxÞ=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 Oðεm�1=2Þ, ε ! 0, x∈ ½0, 1�. Hence, we get rmðxÞ ¼ Oðεm�1=2Þ, ε ! 0, x ∈½0, 1�.

Thus, we have proved.

xY3ðxÞ ¼ Y″

12 Recent Studies in Perturbation Theory

yð1Þ ¼ 0 We state problems

ymðxÞ ¼ μ�1

εR″

Lw0 � w″

that wkðηÞ∈ C∞½0, ∞Þ, wkðηÞ ¼ Oðe�ηÞ with η ! ∞.

π�1ðtÞ þX

εr ″

mðxÞ � 4xε

3m

k¼0 μk �

rmð0Þ ¼ Oðe

2 � x2 R0

Rmð0Þ ¼ Oðe

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

then Eq. (15.2) to π2ðtÞhase the problem

π″

0ðxÞ � Y″

is infinitely differentiable on the segment ½0, μ�1�, and as t ! ∞:

π2ðt޼� 2f 3

Lwk ¼ �ηwk�1ðηÞ, w2ið0Þ ¼ Y3ið1Þ, w2i�1ð0Þ ¼ 0, lim

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

mðxÞ � xrmðxÞ ¼ Oðεmþ1=2

�1= ffiffi ε p

Let rmðxÞ¼ð2 � x2ÞRmðxÞ=2, and then problems (19) and (20) take the form

mðxÞ � 2ε

�1= ffiffi ε p

YkðxÞ þ πkðtÞ

t

1 þ 1 � 2 t 3 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

0ðηÞ � 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-

� þX 2m

Þ, rmð1Þ ¼ Oðεmþ1

Þ, Rmð1Þ ¼ Oðεmþ1

2 � x2 þ x � �

k¼0 λk

0ð0Þ ) Y3ðxÞ¼ðY″

0ðxÞ � Y″

1 � 2 � 4 � 5 t

� �

6 þ …

η!∞

η!∞

RmðxÞ ¼ Oðεmþ1=2

Þ, ε ! 0:

:

2ðtÞ � tπ2ðtÞ ¼ 2f 3, 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

0ð0ÞÞ=x, ðY″

0ð0޼�2f 3Þ;

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 yðxÞ ¼ μ�1π�1ðtÞþ X 3 k¼0 μk ðYkðxÞþπkðtÞÞ þ w0ðηÞ þ λw1ðηÞ þ λ2 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 qðxÞ, fðxÞ∈C∞½0, 1�.

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 aðxÞ ∈C∞½0, 1�.

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 zkðxÞ ∈C∞½0, 1�, πkðtÞ∈ C∞½0, μ�1�.

Function z0ðxÞis a solution of equation

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

wherec0 ¼ pð0Þ �1 ½f 0 ð0Þ � q0 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‴ k�1ð0Þ þ z″ k�1ð0Þq0 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 πkð0޼�zkð0Þ, πkðμ�1Þ ¼ 0.

Next, we use the following lemma.

Lemma 2. The problem

My � xy0 � 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 � r0 � r1s

ð0Þ ¼ pð0Þ

zkðxÞ þ πkðtÞ

� s �2

�1 ½f 0 ð0Þ � q0

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:

�

Mz0 ¼ fðxÞ � c0xpðxÞ,

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

Mzk ¼ �z″

k�1ð0Þq0

, ε ¼ μ2

k�1ðxÞ � 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

rðxÞ ¼ p�1

aðxÞ ¼ xpðxÞ

r0 ¼ rð0Þ, r1 ¼ r0

X∞ k¼0 μk �

ð0Þfð0Þ�.

k�1ð0Þ þ z″

Functions πkðtÞ is the solution of the equations

where zkðxÞ ∈C∞½0, 1�, πkðtÞ∈ C∞½0, μ�1�.

Function z0ðxÞis a solution of equation

�1 ½f 0 ð0Þ � q0

�1 ½�z‴ ðx 1 �

represented as

14 Recent Studies in Perturbation Theory

represented as

Function aðxÞ ∈C∞½0, 1�.

wherec0 ¼ pð0Þ

where ck ¼ pð0Þ

where

where

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

It has a unique solution yðxÞ ∈C∞½0, 1�.

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 γ1 and γ2 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, jYnðx, εÞ � yðx, εÞj < Bnεnþ1:

#### 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 y0 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 t ¼ x=μ2, ε ¼ μ3, ykðxÞ∈C½0, 1�, πkðtÞ ∈C½0, 1=μ2�, Rðx, εÞ 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 hðx, εÞ ¼ Xn�1 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: y00 k�1ðxÞ � hk�1ðxÞ ∈C½0, 1�.

Then

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

Þ > 0, L0V� � V″

jpðxÞj, γ2 > jzkð0Þj:

�ðtÞ þ tV0

� ,

, 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

γ1 > max ½0, 1�

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, εÞ ¼ Xn

where zkðxÞ, πkðμ, tÞare constructed above functions, then

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

�

j < Bnεnþ1

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

x p y0

ðxÞ � yðxÞ ¼ 0, 0 < x < 1,

x p

, c � const:

y0ðxÞ ¼ ce2 ffiffi

εy00ðxÞ þ ffiffiffi

x p y0

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, jYnðx, εÞ � yðx, εÞj < Bnεnþ1:

k¼0 εk �

LεYnðx, εÞ ¼ fðxÞ þ εnþ1

where γ1 and γ2 are positive constants such that

V�ðμ, 0Þ > 0, V�ðμ, μ�1

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 y1, 0 ¼ 3 2 þ 1 2e2 � �c1, y1, 1 ¼ 1 6 þ 1 2e2 � �c1, y1, 2 ¼ �1 6 þ 1 4e2 � �c1, y1, 3 ¼ �1 10 þ 1 12e2 � �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, 1, yk�1,3 are corresponding coefficients of the expansion of yk�1,1ðxÞ 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, c2 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

y1ðxÞ ¼ y1, 0 þ y1,1ð2 ffiffiffi

6 þ 1 2e2 � �

y1,1ð2 ffiffiffi

2ðxÞ � h2ðxÞ∈ C½0, 1�, μ6

hk�1ðx޼� yk�1, 1

2 ffiffiffiffiffi x3 p þ

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

ffiffi

ÞA ð μ~

t e �2 3s3=2

c1, y1, 1 ¼ 1

�

h1ðxÞ ¼

y00

where y1, 0 ¼ 3

Then

(2 ffiffiffi x p ).

Lπ0 � π″

2 þ 1 2e2 � �

Analogously, we have obtained

18 Recent Studies in Perturbation Theory

Continuing this process, we have

0ðtÞ þ ffiffi t p π0

> 2 ffiffiffiffi t

> > ffiffi

Lπ3ðnþ1ÞðtÞ ¼ π3nþ2ðtÞ � yn,3

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

π0ðtÞ¼ða � be�2

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

Lπ3kþ1ðtÞ ¼ π3kðtÞ þ yk, 1

Lπ3kþ3 ¼ π3kþ2ðtÞ � yk,3

c1, c2 are constants, and

x p Þ þ y1, 2ð2 ffiffiffi

c1, y1, 2 ¼ �1

x p Þ 3 �″

h2ðtμ2

6yk�1, 3 ffiffiffi

where yk�1, 1, yk�1,3 are corresponding coefficients of the expansion of yk�1,1ðxÞ in powers of

x p Þ þ y1, 3ð2 ffiffiffi

x p Þ

6 þ 1 4e2 � �

2 þ y1,3ð2 ffiffiffi

¼ � y1, 1 2 ffiffiffiffiffi x3 p þ

޼� μ3y1,1 2 ffiffiffiffi t 3 p þ μ5y1,3 ffiffi

x p , k ¼ 4, …, n,

0ðtÞ ¼ 0, 0 < t < μ~, π0ð0Þ ¼ a � y0ð0Þ, π0ðμ~Þ ¼ 0, μ~ ¼ 1=μ2

Lπ3kþ2ðtÞ ¼ π3kþ1ðtÞ, 0 < t < μ~, π3kþ2ð0Þ ¼ 0, π3kþ2ðμ~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

3 p , 0 < t < μ~, π3kþ1ð0Þ ¼ 0, π3kþ1ðμ~Þ ¼ 0, k ¼ 0, 1, …, n ð34Þ

t p , 0 < t < μ~, π3kð0޼�ykð0Þ, π3kðμ~Þ ¼ 0, k ¼ 0, 1, …, n–1 ð36Þ

ð μ~

0 B@

0 e �2 3s3=2 ds

t p , 0 < t < μ~, π3nð0Þ ¼ 0, π3nðμ~Þ ¼ 0 ð37Þ

1 CA

�1

:

x p Þ

c1, y1, 3 ¼ �1

3 þ …,

6y1,3 ffiffiffi x p :

t p :

10 þ 1 12e2 � �

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 Gðt, sÞ ¼ �YðtÞXðsÞ, 0 ≤ t ≤ 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 Rðx, εÞ ¼ Oðεnþ1Þ, ε ! 0, x∈ ½0, 1�.

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 a1 cos x þ b1 sin x (where a1, b1 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 M transforms Fourier series X∞ k¼1 ak cos kx and X∞ 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, yðxÞ ¼ cos x þ ε 8 3x sin x � 1 4 cos 3x þ 1 4 cos x � � þ …it is not a uniform expansion of the y (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 αk 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

εR00ðx, εÞ 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

x p R0

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

jRðx, εÞj ≤ εnþ1

Therefore, we have Rðx, εÞ ¼ Oðεnþ1Þ, ε ! 0, x∈ ½0, 1�.

yðxÞ ¼ Xn

3.1. The idea of the method of Poincare

k¼0 εk ykðxÞ þ

Note that the operator M transforms Fourier series X∞

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

existence of the solution of an algebraic equation.

ðx, εÞ � Rðx, εÞ ¼ μ3ðnþ1Þ

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 Xnþ1Þ k¼0 μk

MyðxÞ :¼ y00ðxÞ þ yðxÞ � εy3

Unperturbed equation has solutions y0ðxÞ ¼ a1 cos x þ b1 sin x (where a1, b1 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

yðxÞ ¼ y0ðxÞ þ εy1ðxÞ þ ε2y2ðxÞ þ …

yð0Þ ¼ 1, y0

3. Singularly perturbed differential equations Lighthill type

π3ðnþ1ÞðtÞ þ εnþ1

jπ3ðnþ1ÞðtÞ þ hnðxÞ � y″

πkðtÞ þ Oðεnþ1

k¼1

ð0Þ ¼ 0:

ðhnðxÞ � y″

nðxÞj:

Þ, ε ! 0:

ak cos kx and X∞

ðxÞ ¼ 0: ð39Þ

k¼1

ak sin kx in itself.

nðxÞÞ,

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

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

The function Z1(t) will have the periodical solution we take α1 ¼ �3=4. Then z1ðt޼� 1 32 cos 3t. Similarly, from equations

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

αn 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 x∈ ½0, 1�qðxÞ, rðxÞ ∈C∞½0, 1�.

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 x2 > 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 ½εα, 1�, 0 < α < 1=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

yðξÞ ¼ y0ðξÞ þ εy1ðξÞ þ ε2y2ðξÞ þ …

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

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 2bε þ ε2b2 p

> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2b þ εb2 p

> > 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 x2 > 2εb, this expression can be expanded in powers of ε, and then we have

x2 ðx2 � 1Þ þ … þ O 1

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

= ffiffiffi ε p :

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

þ b2 ε x � �2

> x ε x2 � �n � �

ðxÞ þ qðxÞyðxÞ ¼ rðxÞ, yð1Þ ¼ a ð41Þ

ðxÞ þ yðxÞ ¼ 0, yð1Þ ¼ b: ð43Þ

� xÞ=ε: ð44Þ

þ … ð45Þ

x ¼ ξ þ εx1ðξÞ þ ε2x2ðξÞ þ … ð42Þ

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 x∈ ½0, 1�qðxÞ, rðxÞ ∈C∞½0, 1�.

At first, we consider the example

It has exact solution

½εα, 1�, 0 < α < 1=2.

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

He considered the model equation [67, 82]:

ðx þ εyðxÞÞy0

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

ðx þ εyðxÞÞy0

yðxÞ¼ð

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

yðxÞ ¼ b x þ b2 2x ε

yðxÞ ¼ x

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

yð0Þ ¼

ε �1 þ

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

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∞½0, 1� and q0 ¼ qð0Þ > 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 ðξÞ ¼ 1 j! qðjÞ ðξÞ, rj ¼ rjðξÞ ¼ 1 j!rðjÞ 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 n ¼ �ðy0 þ x1Þy0 n�1 � ðy1 þ x2Þy0 n�2 � … � ðyn�2 þ xn�1Þy0 1 � yn�1y0 0þ þðr1x1 � qy1 � q1x1y0Þx0 n�1 þ ðr1x2 þ r2x2 1 � qy2 � q1x1y1 � ðq1x2 þ q2x2 1Þy0Þx0 n�2 þ … þðr1xn�1 þ 2r2x1xn�2 þ 2r2x2xn�3 þ … þ rn�1xn�1 1 � q1y1xn�2 � ðq1xn�1 þ 2q2x1xn�2 þ … þqn�1xn�1 1 Þy0 0Þx0 1, gn ¼ r1xn þ 2r2x1xn�1 þ … þ rnxn 1 � q1x1yn�1 � ðq1x2 þ q2x2 1Þyn�2 � … �ðq1xn þ 2q2x1xn�1 þ … þ qnxn 1 Þy0:

In these equations, the coefficient rðξÞ � qðξÞy0ðξÞ of the derivative x0 nðξÞ ðn ¼ 1, 2, …Þwas replaced by Eq. (54) on ξy0 0ðξÞ.

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 n function depends from y0 0, …, y0 n�1, x0 1, …, x0 n�1 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 gðξÞ ¼ exp �ðξ 1 � q0 � 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):

� ξ þX∞ k¼0 εk

¼ �X∞ j¼0 rjðξÞ

ðξÞ ¼ 1 j! qðjÞ

Ly2 ¼ ½ξy0

Lyn ¼ ½y0

where q ¼ q0, r ¼ r0,

þ{r1x2 þ r2x2

0x0 n � y0

where qj ¼ qj

…

ξy0

2ðξÞ þ y2ðξÞ¼ðξx0

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

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

ðxðξÞ þ εyðξÞÞy0

ðykðξÞ þ xkðξÞÞ

xkðξÞεk �j �X∞ j¼0 qj ðξÞ �X∞ k¼0

ðξÞ, rj ¼ rjðξÞ ¼ 1

�X∞ k¼0

Hence, equating the coefficients of equal powers has ε

Ly1 ¼ ½ξy0

0x0

Lu0 � ξy0

0x0 1 � y0

2 � ðy0 þ x1Þy0

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

2ðξÞ � x2ðξÞÞy0

b 2 ξ � 1 ξ � �

η ¼ 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�x0

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

xkðξÞεk

0ðξÞ þ qðξÞy0ðξÞ ¼ rðξÞ, y0ð1Þ ¼ y0

þ{gnðy0, …, yn�1, x1, …, xn�1Þ}, ynð1Þ ¼ 0; … ð57Þ

�j��

xðξÞ ¼ ξ þ

conditions:qðxÞ, rðxÞ∈C∞½0, 1� and q0 ¼ qð0Þ > 0. We will write it in the form of

�X∞ k¼0 εk y0 kðξÞ ¼

> j!rðjÞ ðξÞ.

0x1 � y0y0

1 � q1x1y1 � ðq1x2 þ q2x2

1 � ðy1 þ x2Þy0

0xn þ f nðy0, …, yn�1, x1,…, xn�1, y0

0ðξÞ ¼ 0, y2ð1Þ ¼ 0:

yðξÞ ¼ bξ�1 ð50Þ

bε=ð2 þ bεÞ p : ð52Þ

ðξÞ, yð1Þ ¼ y0

1 þX∞ k¼0 x0 kðξÞεk � ,

0�þðr1 � q1y0Þx1, y1ð1Þ ¼ 0, ð55Þ

n�1, x0

1Þy0}, y2ð1Þ ¼ 0, ð56Þ

1, …, x0

0 þ ððr1 � q1y0Þx1 � qy1Þx0

0, …, y0

ðξÞ ¼ 0 ði ¼ 3, 4, …Þ, as they

: ð53Þ

, ð54Þ

1�þ

n�1Þ�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 conditionw0 > 0, η0 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 w0 < 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

2 � x2 ¼ y1 þ ððy0 þ x1Þy0

have

ξx0

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

e ξ � w0 1 þ q0

xjðξÞ e bjξ�jq0 , yj

yðξÞ e ξ�q0 ðw0 þ a1εξ�q0 þ … þ anðεξ�q0 Þ

ξ�q0 ε þ b2ðεξ�q0 Þ

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

η þ εx1ðηÞ þ ε2

ðξ 1 s �2 y0ðsÞds e � w0

Ly1 ¼ ðr1 � q1y0Þx1 e~a1ξ�2q0 ,

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

1 � ððr1 � q1y0Þx1 � qy1Þx0

Ly2 ¼ g2ðy0, y1, x1, x2Þ e~a2ξ�3q0 , ~a2 ¼ const, ξ ! 0

1 þ q0

y1ðξÞ e a1 ξ�2q0 ða1 ¼ constÞ, ξ ! 0: ð62Þ

1Þðy0 0Þ �1 e

x2ðξÞeb2ξ�2q0 , b2 ¼ const, ξ ! 0: ð63Þ

y2ðξÞ e a2ξ�3q0 , a2 ¼ const, ξ ! 0 ð64Þ

2 þ … þ bnðεξ�q0 Þ

ðξÞ e ajξ�ðjþ1Þq0 , j ¼ 1, 2,…: ð65Þ

x2ðηÞ þ … ¼ 0 ð68Þ

n þ …Þ, ξ ! 0, ð66Þ

n þ … ð67Þ

ξ�q0 : ð61Þ

~b2ξ�2q0 , ~b2 ¼ const:

Theorem 7. Suppose that the conditions (1) qðxÞ, rðxÞ∈ C∞½0, 1�; (2) q0 > 0; (3) w0 > 0; (4) ξy0 0 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 ½0, 1�; proved by Habets [66] in the case qðxÞ, rðxÞ∈C2 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 y0 0ðξÞ 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�1ðξÞ. 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=y0 0ðξÞ. Therefore, yðξÞ ¼ y0ðξÞ þ Oðε2Þ. Hence, it is clear that we must make the condition of Wasow: y0 0ðξÞ 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) qðxÞ, rðxÞ∈ C∞½0, 1�; (2) q0 > 0; (3) w0 > 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 εyðxÞÞy0

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

ðξÞ ¼ rðxðξÞÞ � qðxðξÞÞyðξÞ, yð1Þ ¼ y0,

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) qðxÞ, rðxÞ∈ C∞½0, 1�; (2) q0 > 0; (3) w0 > 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

ξ�q0 ε, ξ ! 0:

� ε1=ð1þq0Þ �

, ξ ! 0, η ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi w0 1 þ q0 ε q0þ1 r

6¼ 0, ξ∈ ½η, 1�:

, ε ! 0:

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

xe ξ � w0 1 þ q0

ξ�q0 ε þ O

∂ξ jξ¼ηðεÞ ¼ �1 � q0 þ O

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

� ðεξ�q0 Þ 2 �

w0 1 þ q0

∂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.,

ξy0

�

ξx0

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 xðηÞ ¼ 0, we find η : ηe ffiffiffiffiffiffiffi ε=3 p3 .

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�, y0 ðxÞ ¼ dy=dx . Given functions are subjected to the conditions U: qðxÞ, rðxÞ∈ Cð∞Þ ½0, 1�.

Here, we consider the case q0 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 t ¼ x=μ, ε ¼ μ2, ukðxÞ∈Cð∞Þ ½0, 1� and πkðtÞ∈ Cð∞Þ ½0, μ0�, μ0 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 εyðxÞÞy0

yðxÞ ¼ μ�1

The initial conditions for the functions πjðtÞ are taken as

π�1ð1=μÞ ¼ bμ, b ¼ u0 �X∞

Here, we consider the case q0 ¼ �1; this is done to provide a detailed illustration of the idea of

k¼0

½0, 1� and πkðtÞ∈ Cð∞Þ

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, …,

0ðxÞ � qðxÞu0ðxÞ ¼ rðxÞ, u0ðxÞ∈Cð∞Þ

�

0ðtÞ � u1ðtμÞπ0

0ðtÞ � π1ðtÞπ0

0ðxÞu1ðxÞ, u3ðxÞ ∈Cð∞Þ

πiðtÞπj

0ðxÞ, u2ðxÞ ∈Cð∞Þ

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

π�1ðtÞ þX∞

k¼0 μk

�1ðtÞ � qðμ tÞ

Lu1ðxÞ ¼ 0, u1ðxÞ∈ Cð∞Þ

0ðtÞ þ π0ðtÞπ0

1ðtÞ � u1ðtμÞπ0

1ðxÞ � u0

0ðtÞ þ X

uiðμ tÞπj

Lu2ðxÞ :¼ �u0ðxÞu0

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

where b is given constant, x ∈½0, 1�, y0

where t ¼ x=μ, ε ¼ μ2, ukðxÞ∈Cð∞Þ

we have the following equations:

t þ π�1ðtÞ

�

t þ π�1ðtÞ

Lu0ðxÞ :¼ xu0

� π0 0ðtÞ þ � π0

Dπ1ðt޼�u0ðtμÞπ0

Dπ3ðtÞ ¼ X

�1ðtÞ � π0ðtÞπ0

Lu3ðxÞ :¼ �u0ðxÞu0

i þ j ¼ 2 i ≥ 0, j ≥ � 2

� π0

½0, 1�.

tions U: qðxÞ, rðxÞ∈ Cð∞Þ

30 Recent Studies in Perturbation Theory

unðxÞ, n ¼ 0, 1, …,

Dπ0ðtÞ :¼

�

Dπ2ðtÞ :¼ �u0ðtμÞπ0

ðxÞ þ qðxÞyðxÞ ¼ rðxÞ, yð1Þ ¼ b ð76Þ

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

� μk

½0, μ0�, μ0 ¼ 1=μ:

�1ðtÞ ¼ qðμ tÞπ�1ðtÞ, π�1ðμ0Þ ¼ bμ, ð79:-1Þ

π0ðt޼�u0ðtμÞπ0

ukð1Þ, πkðμ0Þ ¼ 0, k ¼ 0, 1, … ð78Þ

, ð77Þ

½0, 1� ð80:0Þ

�1ðtÞ, π0ðμ0Þ ¼ 0 ð79:0Þ

½0, 1�, ð80:1Þ

0ðtÞ � u2ðtμÞπ0

0ðtÞ, π3ðμ0Þ ¼ 0,

�1ðtÞ, π1ðμ0Þ ¼ 0 ð79:1Þ

½0, 1� ð80:2Þ

�1ðtÞ, 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 z0ðtÞ, the solution of the inhomogeneous equation for zðtÞ 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 ϕðs, μÞ¼ð1 þ qðμsÞÞ=s, b0 ¼ 2b þ b2 μ2.

Let b0 > 0. Let us introduce the notation z0ðt޼�t þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t 2 þ b0pðt, μÞ q . This function satisfies the inequality 0 < z0ðtÞ ≤ Mt�1 ðt > 0Þ 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 Sμ 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 b0 > 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 kϕðt, μÞk ≤ Mμ, 0 < m ≤ pðt, μÞ ≤ M, kp�1ðtÞk ≤ M: First, let us estimate Tðt, z2Þ 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 Tðt, z2 0Þ ¼ OðμÞ, the denominator of this expression is strictly positive on the closed interval ½0, μ0�. For F0 zðt, z0Þ, we can obtain the estimate kFzðt, z0Þk ≤ Mμlnμ�1 in the same way as the estimate for Tðt, z2Þ. 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�1ðtÞ ≥ 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 fðxÞ ∈Cð∞Þ ½0, 1�, the equation Lξ ¼ fðxÞ has a unique bounded solution ξðxÞ∈Cð∞Þ 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 b0 > 0, then there exists a unique constraint of the solution of problem (79.-1) from the

Suppose that kϕðt, μÞk ≤ Mμ, 0 < m ≤ pðt, μÞ ≤ M, kp�1ðtÞk ≤ M: First, let us estimate Tðt, z2Þ

ds ≤ Mμ:

jzðtÞj ≤ jzðtÞ � z0ðtÞj þ jz0ðtÞj,

jz0ðtÞj ≤ Mt�1 ðt > 0Þ:

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 Tðt, z2Þ. 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

0Þ ¼ OðμÞ, the denominator of this expression is strictly positive on the closed interval

zðt, z0Þ, we can obtain the estimate kFzðt, z0Þk ≤ Mμlnμ�1 in the same way as the

ds ≤ Mμ

t

ðs, μÞjjzðsÞj2

jzðsÞj2

q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 þ bpðt, μÞ þ pðt, μÞTðt, z2Þ:

ðμ0 t

jzðsÞj2

ðs, μÞz0ðsÞhðsÞ ds ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 þ pðt, μÞ

½0, 1�, the equation Lξ ¼ fðxÞ has a unique bounded solution

�

r �,

b þ Tðt, z2Þ

t

ds ≤ Mμ

ðμ0 0

jzðsÞj2 ds ≤

Proof. Equation (81) is equivalent to the equation z ¼ F½t, z�, where

F½t, z�¼�t þ

jϕðs, μÞjjp�1

ðμ0 1

ðμ0 t

unique solution from the class Sμ. The theorem is proved.

equations; therefore, the following lemmas are needed.

ϕðs, μÞp�1

ds þ Mμ

set Sμ.

on the set Sμ. We have

32 Recent Studies in Perturbation Theory

jTðt, z2Þj ≤

≤ Mμ ð1 0 jzðsÞj2

as well as the inequality

F0

operator:

Tðt, z2

½0, μ0�. For F0

ξðxÞ∈Cð∞Þ

ðμ0 t

Here, we have used the triangle inequality

zðt, z0Þh ¼ �pðt, μÞ

Corollary. The following inequalities hold:

Lemma 9. For any function fðxÞ ∈Cð∞Þ

½0, 1� expressible as

2. π�1ðtÞ ≤ Mt�1 ðt > 0Þ.

1. zðtÞ ¼ π�1ðtÞ ≥ 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 C∞½0, 1�.

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 kðtÞ belongs to C∞½0, 1� is continuous and bounded, and if jkðtÞj ≤ Mt�2, t ! ∞, 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, kgðt, μÞk ≤ M and g�1ðt, μÞ ≤ M for t ∈½0, μ0�and μaresmall. 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 jηðtÞj ≤ Mt�1 ðt > 0Þ. 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 b0 > 0 (for this, it suffices that the condition b0 :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) andjRnþ1ðx, 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 b0 ¼ b � 1. If b0 > 0, then the solution of problem (1) exists on the closed interval ½0, 1�, which is confirmed by Theorem 11. The equation for π�1ðtÞ 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 u0(x) has the solution y0ðxÞ ¼ 1∈ C∞½0, 1�. 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 yðxÞ ¼ μ�1π�1ðx=μÞ þ 1 þ π0ðx=μÞ þ oðμÞ for all x ∈½0, 1�, μ ! 0:

## 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 b0 > 0 (for this, it suffices that the condition b0 :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

�1ðtÞ þ π�1ðtÞ ¼ 0, π�1ðμ0Þ ¼ bμ:

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

μ2

:

2 þ 2b þ b2

t þ π�1ðtÞ , ukðxÞ ¼ 0, k ¼ 1, 2, …,

� yð0Þ �2

þ 2εx

,

x2 þ 2b0ε þ ε2

where b0 ¼ b � 1. If b0 > 0, then the solution of problem (1) exists on the closed interval ½0, 1�,

t

q

" # r

� þ μnþ1

Rnþ1ðx, 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) andjRnþ1ðx, μÞj ≤ M for all x ∈½0, 1�.

yðxÞ ¼ μ�1

�

�

The solution of this problem can be expressed as

4. Lagerstrom model problem

The problem [32]

x þ ε yðxÞ

yðxÞ ¼ ε�1 �x þ

t þ π�1ðtÞ

π�1ðtÞ þXn

k¼0 μk �

� y0

which is confirmed by Theorem 11. The equation for π�1ðtÞ is of the form

π�1ðt޼�t þ

where b ¼ b0. The asymptotics of the solutions of problem (76) can be expressed as

� π0

The equation for u0(x) has the solution y0ðxÞ ¼ 1∈ C∞½0, 1�. Further,

π0ðtÞ ¼ �π�1ðtÞ þ bμ

yðxÞ ¼ μ�1π�1ðx=μÞ þ 1 þ π0ðx=μÞ þ oðμÞ for all x ∈½0, 1�, μ ! 0:

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

$$\mu\_0 = X(\mathbf{x}, \varepsilon) := 1 - X\_1(\mathbf{X}, \varepsilon), \quad X\_1 = \mathbb{C}\_0 \int\_1^{\mathbf{x}} s^{-k} e^{-\varepsilon s} ds, \quad \mathbb{C}\_0^{-1} = \int\_1^{\infty} s^{-k} e^{-\varepsilon s} ds.$$