Asymptotic Solution of a Singularly Perturbed Integro-Differential Equation with Exponential Inhomogeneity

Abstract

The integro-differential Cauchy problem with exponential inhomogeneity and with a spectral value that turns zero at an isolated point of the segment of the independent variable is considered. The problem belongs to the class of singularly perturbed equations with an unstable spectrum and has not been considered before in the presence of an integral operator. A particular difficulty is its investigation in the neighborhood of the zero spectral value of inhomogeneity. Here, it is not possible to apply the well-known procedure of Lomov’s regularization method, so the authors have chosen the method of constructing the asymptotic solution of the initial problem based on the use of the regularized asymptotic solution of the fundamental solution of the corresponding homogeneous equation whose construction from the positions of the regularization method has not been considered so far. In the case of an unstable spectrum, it is necessary to take into account its point features. In this case, inhomogeneity plays an essential role. It significantly affects the type of singularities in the solution of the initial problem. The fundamental solution allows us to construct asymptotics regardless of the nature of the inhomogeneity (it can be both slowly changing and rapidly changing, for example, rapidly oscillating). The approach developed in the paper is universal with respect to arbitrary inhomogeneity. The first part of the study develops an algorithm for the regularization method to construct the asymptotic (of any order on the parameter) of the fundamental solution of the corresponding homogeneous integro-differential equation. The second part is devoted to constructing the asymptotics of the solution of the original problem. The main asymptotic term is constructed in detail, and the possibility of constructing its higher terms is pointed out. In the case of a stable spectrum, we can construct regularized asymptotics without using a fundamental solution.

1. Introduction

Consider the integro-differential problem

$$\begin{array}{c}\epsilon \frac{dy}{dt}=a\left(t\right)y+\underset{0}{\overset{t}{\int }}K\left(t,s\right)y\left(s,\epsilon \right)ds+h\left(t\right)e^{\frac{1}{\epsilon }\beta \left(t\right)},y\left(0,\epsilon \right)=y^0,t\in \left[0,1\right]\hfill \end{array}$$

(1)

where $\beta \left(t\right)=\underset{0}{\overset{t}{\int }}\mu \left(\theta \right)d\theta$ is some scalar function and $\mu \left(t\right)$ is the so-called spectral value of the inhomogeneity (see, for example, [1]). As was indicated in [2,3], the structure of the asymptotics of the solution of the problem (1) essentially depends on the behavior of the functions $a\left(t\right)$ and $\mu \left(t\right)$ on the segment $\left[0,1\right]$ (here, the segment $\left[0,1\right]$ is taken for the sake of simplicity; any segment $\left[0,T\right]$ can be taken instead). This becomes clear if we pass from problem (1) to the system

$$\begin{array}{c}\epsilon \frac{dy}{dt}=a\left(t\right)y+\underset{0}{\overset{t}{\int }}K\left(t,s\right)y\left(s,\epsilon \right)ds+h\left(t\right)z,y\left(0,\epsilon \right)=y^0,\hfill \\ \\ \epsilon \frac{dz}{dt}=\mu \left(t\right)z,z\left(0,\epsilon \right)=1.\hfill \end{array}$$

(2)

The matrix $\left(\begin{array}{cc}a\left(t\right)& 1\\ 0& \mu \left(t\right)\end{array}\right)$ of the differential part of this system can have both a stable spectrum and an unstable spectrum (see [1], pp. 39–40, 186–187). In each of these cases, the asymptotics has its structure. A particularly complicated structure arises in the case of spectrum instability, characterized by the presence of so-called point features. For example, if it has zero $t=0$ of the order r on the segment $\left[0,1\right]$$\left(\mu \left(t\right)=-l\left(t\right)t^r,l\left(t\right)>0\right)$, then in the asymptotics of the solution of the corresponding differential problem

$$\begin{array}{c}\epsilon \frac{dy}{dt}=a\left(t\right)y+h\left(t\right)z,y\left(0,\epsilon \right)=y^0,\hfill \\ \\ \epsilon \frac{dz}{dt}=\mu \left(t\right)z,z\left(0,\epsilon \right)=1.\hfill \end{array}$$

singularities appear, which are described by special functions of the type

$$e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}\mu \left(\theta \right)}\underset{0}{\overset{t}{\int }}{e}^{-\frac{1}{\epsilon }\underset{0}{\overset{s}{\int }}\mu \left(\theta \right)}\frac{s^j}{j!}ds,j=\overline{0,r-1}$$

absentees in the case $\mu \left(t\right)\ne 0.$ A description of singularities in the presence of an integral term in (2) becomes very difficult in this case. There are results for singularly perturbed integro-differential systems in which the exponential factor of the kernel of the integral operator is unstable, while the spectrum of the corresponding differential part is stable (see, for example, [4,5,6,7]). Problems with the spectrum instability of the limit operator were considered mainly for ordinary differential systems [8,9,10]. In them, the role of instability is played by the zeros of the coefficient $a\left(t\right)$ or of the matrices $A\left(t\right)$. Generalizing the algorithm of the regularization method [2,3], to singularly perturbed integral and integro-differential problems with rapidly oscillating coefficients [11,12,13,14,15,16], we studied the influence of rapidly oscillating inhomogeneities on the asymptotics of the solution [17,18,19,20,21,22]. An investigation of the instability of the general case of the spectrum $\left\{a\left(t\right),\mu \left(t\right)\right\}$ in the integro-differential problem has not yet been carried out. In our case, the instability is the exponential factor of inhomogeneity. The problem has not been considered in this formulation before. The study of the general case instability of the spectrum $\left\{a\left(t\right),\mu \left(t\right)\right\}$ in the integro-differential problem has not yet been carried out. We begin its study with problem (1), in which the spectral value of the inhomogeneity $\mu \left(t\right)={\beta }^{\prime }\left(t\right)$ is unstable: $\mu \left(t\right)=-l\left(t\right)t^r,l\left(t\right)>0,r\in {\mathbb{Z}}_{+}.$

The idea of constructing an asymptotic solution of the problem (1) under the indicated assumptions is based on the use of regularized asymptotics of the fundamental solution $V\left(t,s,\epsilon \right)$ of the corresponding homogeneous integro-differential system and the next statement.

Lemma 1.

Let $V(t,s,\epsilon )$ be the fundamental solution of a homogeneous integro-differential Volterra equation (s is a parameter):

$$\begin{array}{c}\epsilon \frac{dV\left(t,s,\epsilon \right)}{dt}=a\left(t\right)V\left(t,s,\epsilon \right)+\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)V\left(x,s,\epsilon \right)dx,\hfill \\ \\ V\left(s,s,\epsilon \right)=1,0\le s\le t\le 1.\hfill \end{array}$$

(3)

Then, the solution of the integro-differential problem

$$\epsilon \frac{dp}{dt}-a\left(t\right)p-\underset{0}{\overset{t}{\int }}{K}_0\left(t,s\right)p\left(s,\epsilon \right)ds=q\left(t\right),p\left(0,\epsilon \right)=b$$

(4)

is given by the formula

$$p\left(t,\epsilon \right)=V\left(t,0,\epsilon \right)b+\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}V\left(t,x,\epsilon \right)q\left(x\right)dx.$$

(5)

Proof. 

Although the justification for this fact is given in [1] (pp. 268–269), we will give it again since in [1] it is given very concisely. Substitute (5) into (4); this will result in:

$$\epsilon \left[\frac{dV\left(t,0,\epsilon \right)}{dt}b+\frac{1}{\epsilon }V\left(t,t,\epsilon \right)q\left(t\right)+\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}\frac{\partial V\left(t,x,\epsilon \right)}{\partial t}q\left(x\right)dx\right]$$

$$-a\left(t\right)\left[V\left(t,0,\epsilon \right)b+\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}V\left(t,x,\epsilon \right)q\left(x\right)dx\right]$$

$$-\underset{0}{\overset{t}{\int }}{K}_0\left(t,s\right)\left[V\left(s,0,\epsilon \right)b+\frac{1}{\epsilon }\underset{0}{\overset{s}{\int }}V\left(s,x,\epsilon \right)q\left(x\right)dx\right]ds$$

$$=\left[\underline{\underline{\epsilon \frac{dV\left(t,0,\epsilon \right)}{dt}-a\left(t\right)V\left(t,0,\epsilon \right)-\underset{0}{\overset{t}{\int }}{K}_0\left(t,s\right)V\left(t,0,\epsilon \right)ds}}\right]b+q\left(t\right)$$

$$+\underset{0}{\overset{t}{\int }}\frac{dV\left(t,x,\epsilon \right)}{dt}q\left(x\right)dx-\frac{a\left(t\right)}{\epsilon }\underset{0}{\overset{t}{\int }}V\left(t,x,\epsilon \right)q\left(x\right)$$

$$-\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}{K}_0\left(t,s\right)\left(\underset{0}{\overset{s}{\int }}V\left(s,x,\epsilon \right)q\left(x\right)dx\right)ds=0+q\left(t\right)$$

$$+\underset{0}{\overset{t}{\int }}\left[\underline{\underline{\frac{dV\left(t,x,\epsilon \right)}{dt}-\frac{a\left(t\right)}{\epsilon }V\left(t,x,\epsilon \right)-\frac{1}{\epsilon }\underset{s}{\overset{t}{\int }}{K}_0\left(t,s\right)V\left(s,x,\epsilon \right)ds}}\right]q\left(x\right)dx$$

$$=0+q\left(t\right)+0=q\left(t\right).$$

Here, we have changed the order of integration in the iterated integral:

$$\underset{0}{\overset{t}{\int }}{K}_0\left(t,s\right)\left(\underset{0}{\overset{s}{\int }}V\left(s,x,\epsilon \right)q\left(x\right)dx\right)ds=\underset{0}{\overset{t}{\int }}\left[K_0\left(t,s\right)\underset{s}{\overset{t}{\int }}V\left(s,x,\epsilon \right)ds\right]q\left(x\right)dx$$

and used equality (3) twice in the underlined expressions, replacing them with zeros. The lemma is proved. □

As mentioned earlier, to construct an asymptotic solution of the problem (1), we will use the first-order asymptotic of the fundamental solution $V\left(t,s,\epsilon \right)$. In this case, the main term of the asymptotic solution of the original problem (1) will be obtained. To construct the higher terms of the asymptotics, we also need the higher terms of the asymptotics of the fundamental solution. Therefore, in the following sections of our work, we construct an algorithm for constructing a complete asymptotic solution of the function $V\left(t,s,\epsilon \right).$

2. Regularization of the Problem

Problem (3) will be considered under the following conditions:

(i)

the function $a\left(t\right)\in C^{\infty }\left[0,1\right],$ the kernel $K_0\left(t,s\right)\in C^{\infty }\left(0\le s\le t\le 1\right);$

(ii)

$\mathrm{Re}a\left(t\right)\le 0,a\left(t\right)\ne 0\forall t\in \left[0,1\right].$

Following the regularization method of S.A. Lomov [2,3], we introduce an additional variable

$$\eta =\frac{1}{\epsilon }\underset{s}{\overset{t}{\int }}a\left(\theta \right)d\theta \equiv \frac{\phi \left(t,s\right)}{\epsilon }$$

(6)

and instead of problem (3), consider the “extended” problem

$$\begin{array}{c}\epsilon \frac{\partial \tilde{V}\left(t,s,\eta ,\epsilon \right)}{\partial t}+a\left(t\right)\frac{\partial \tilde{V}\left(t,s,\eta ,\epsilon \right)}{\partial \eta }=a\left(t\right)\tilde{V}\left(t,s,\eta ,\epsilon \right)\hfill \\ \\ +\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)\tilde{V}\left(x,s,\frac{\phi \left(x,s\right)}{\epsilon },\epsilon \right)dx,\tilde{V}\left(s,s,0,\epsilon \right)=1\hfill \end{array}$$

(7)

for the function $\tilde{V}\left(t,s,\eta ,\epsilon \right)$ such that the function $V\left(t,s,\epsilon \right)=\tilde{V}\left(t,s,\frac{\phi \left(t,s\right)}{\epsilon },\epsilon \right)$ is an exact solution of the problem (3). However, the problem (7) cannot be considered fully regularized since it has not been regularized integral operator

$$\begin{array}{c}J\tilde{V}\left(t,s,\eta ,\epsilon \right)\equiv J\left(\tilde{V}\left(t,s,\eta ,\epsilon \right){|}_{t=x,\eta =\frac{\phi \left(x,s\right)}{\epsilon }}\right)=\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)\tilde{V}\left(x,s,\frac{\phi \left(x,s\right)}{\epsilon },\epsilon \right)dx.\hfill \\ \hfill \end{array}$$

(8)

As was shown in [2], for its regularization it is necessary to introduce an invariant space $M_{\epsilon },$ with respect to the integral operator J (see [2], p. 62). It is done like this. Let us first introduce the space U of functions of the form

$$\begin{array}{c}v\left(t,s,\eta \right):v\left(t,s,\eta \right)=v_1\left(t,s\right)e^{\eta }+v_0\left(t,s\right),v_0\left(t,s\right),v_1\left(t,s\right)\in C^{\infty }\left(0\le s\le t\le 1,\mathbb{C}\right).\hfill \end{array}$$

(9)

As $M_{\epsilon }$ we take the class ${U|}_{\eta =\frac{\phi \left(t,s\right)}{\epsilon }}.$ It is necessary to show that the image

$$Jv\left(t,s,\eta \right)\equiv J\left(v\left(t,s,\eta \right){|}_{t=x,\eta =\phi \left(x,s\right)/\epsilon }\right)=\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v\left(x,\frac{\phi \left(x,s\right)}{\epsilon }\right)ds$$

(10)

is represented by a series of ${\displaystyle \sum _{k=0}^{\infty }}{\epsilon }^k\left(z^{\left(k\right)}(t,s)e^{\eta }+z_0^{\left(k\right)}(t,s)\right){|}_{\eta =\frac{\phi \left(t,s\right)}{\epsilon }},$ asymptotically converging to $Jv$ at $\epsilon \to +0$ (uniformly in $\left(t,s\right)\in \left\{0\le s\le t\le 1\right\}).$ Substituting the element $v\left(t,s,\eta \right)=v_1\left(t,s\right)e^{\eta }++v_0\left(t,s\right)$ of the space (9) into (10), we have

$$Jv\left(t,s,\eta \right)=\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_0\left(x,s\right)dx+\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_1\left(x,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }dx.$$

(11)

The second integral standing here can be expanded into an asymptotic series. Applying the operation integration by parts, we have

$$\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_1\left(x,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }dx=\epsilon \underset{s}{\overset{t}{\int }}\frac{K_0\left(t,x\right)}{a\left(x\right)}{v}_1\left(x,s\right)d_x{e}^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }$$

$$=\epsilon \left[\frac{K_0\left(t,x\right)}{a\left(x\right)}{v}_1\left(x,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }{|}_{x=s}^{x=t}\right]-\epsilon \underset{s}{\overset{t}{\int }}\frac{\partial }{\partial x}\left(\frac{K_0\left(t,x\right)}{a\left(x\right)}{v}_1\left(x,s\right)\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }dx$$

$$=\epsilon \left[\frac{K_0\left(t,t\right)}{a\left(t\right)}{v}_1\left(t,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{t}{\int }}a\left(\theta \right)d\theta }-\frac{K_0\left(t,s\right)}{a\left(s\right)}{v}_1\left(s,s\right)\right]$$

$$-\epsilon \underset{s}{\overset{t}{\int }}\frac{\partial }{\partial x}\left(\frac{K_0\left(t,x\right)}{a\left(x\right)}{v}_1\left(x,s\right)\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }dx.$$

Continuing this process further, we obtain the series

$$\begin{array}{c}\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_1\left(x,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }dx\hfill \\ \\ ={\displaystyle \sum _{m=0}^{\infty }}{\left(-1\right)}^m{\epsilon }^{m+1}[{\left(I^m\left(K_0\left(t,x\right)v_1\left(x,s\right)\right)\right)}_{x=t}{e}^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }-{\left(I^m\left(K_0\left(t,x\right)v_1\left(x,s\right)\right)\right)}_{x=s}]\hfill \end{array}$$

(12)

where the operators

$$I^0=\frac{1}{a\left(x\right)},I^1=\frac{1}{a\left(x\right)}\frac{\partial }{\partial x}{I}^0,I^m=\frac{1}{a\left(x\right)}\frac{\partial }{\partial x}{I}^{m-1},m\ge 2$$

(13)

are introduced. It is easy to show (see, for example, [4], pp. 291–293 ), that the series on the right-hand side of the equality (12) converges asymptotically to the integral $\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_1\left(x,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{x}{\int }}a\left(\theta \right)d\theta }dx,$ when $\epsilon \to +0$ (uniformly in $\left(t,s\right)\in \left\{0\le s\le t\le 1\right\}$). Thereby, it is shown that the class $M_{\epsilon }{=U|}_{\tau =\frac{\psi \left(t\right)}{\epsilon }}$ is asymptotically invariant with respect to the operator $J.$

The class $M_{\epsilon }$, which is invariant with respect to the operator J, allows us to regularize this operator, using its image on an element of the class $M_{\epsilon }.$ It is done like this. We introduce operators

$$\begin{array}{c}{R}_0\left(v\left(t,s,\eta \right)\right)=\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_0\left(x,s\right)dx,\hfill \\ \\ R_{m+1}\left(v\left(t,s,\eta \right)\right)\hfill \\ \\ ={\left(-1\right)}^m[{\left(I^m\left(K_0\left(t,x\right)v_1\left(x,s\right)\right)\right)}_{x=t}{e}^{\eta }-{\left(I^m\left(K_0\left(t,x\right)v_1\left(x,s\right)\right)\right)}_{x=s}],m\ge 0\hfill \end{array}$$

(14)

where operator $I^m$ has the form (13). These operators are called order operators because they are coefficients at the corresponding degree ${\epsilon }^{m+1}$ in $Jv\left(t,\tau ,\eta \right).$ Using operators $R_m,$ the operator $Jv\left(t,\tau ,\eta \right)$ can be written shorter:

$$Jv\left(t,s,\eta \right)={\left(R_{0}v\left(t,s,\eta \right)+\sum _{m=0}^{\infty }{\epsilon }^{m+1}{R}_{m+1}v\left(t,s,\eta \right)\right)}_{\eta =\frac{\phi \left(t,s\right))}{\epsilon }}.$$

Now, $\tilde{v}(t,s,\eta ,\epsilon )$ is an arbitrary continuous function in

$$(t,s,\eta )\in \left\{0\le s\le t\le 1\right\}\times \mathrm{\Pi }(\mathrm{\Pi }=\{\eta :\mathrm{Re}\eta \le 0\})$$

with asymptotic expansion

$$\tilde{v}(t,s,\eta ,\epsilon )=\sum _{k=0}^{\infty }{\epsilon }^k{v}_k(t,s,\eta ),v_k(t,s,\eta )\in U$$

(15)

converging at $\epsilon \to +0$ (uniformly in $(t,s,\eta )\in \left\{0\le s\le t\le 1\right\}\times \mathrm{\Pi }).$ Then, the image expands into an asymptotic series

$$J\tilde{v}(t,s,\eta ,\epsilon )=\sum _{k=0}^{\infty }{\epsilon }^{k}J{v}_k(t,s,\eta )=\sum _{r=0}^{\infty }{\epsilon }^r\sum _{k=0}^r{R}_{r-k}{v}_k{(t,s,\eta )|}_{\eta =\frac{\phi \left(t,s\right))}{\epsilon }}.$$

(16)

Equality (16) is the basis for introducing the following concept.

Definition 1.

A formal extension of an operator J is the operator $\tilde{J},$ acting for each function $\tilde{v}(t,s,\eta ,\epsilon )\in C(\left\{0\le s\le t\le 1\right\}\times \mathrm{\Pi })$ of the form (9) according to the law

$$\tilde{J}\tilde{v}(t,s,\eta ,\epsilon )\equiv \tilde{J}\left(\sum _{k=0}^{\infty }{\epsilon }^k{v}_k(t,s,\eta )\right)\stackrel{def}{=}\sum _{r=0}^{\infty }{\epsilon }^r\sum _{k=0}^r{R}_{r-k}{v}_k(t,s,\eta ).$$

(17)

From (16), the asymptotic equality $\tilde{J}\tilde{v}{(t,s,\eta ,\epsilon )|}_{\eta =\frac{\psi \left(t\right)}{\epsilon }}=J\tilde{v}(t,s,\eta ,\epsilon ){|}_{\eta =\frac{\psi \left(t\right)}{\epsilon }}$$(\epsilon \to +0)$ follows, which shows that the operator $\tilde{J}$, which we have introduced, is indeed an extension of the operator $J.$ Although the operator $\tilde{J}$ is formally defined, its usefulness is obvious since in practice usually the N-th approximation of the asymptotic solution (15) is constructed, in which only N-th partial sums of the series (15) are used, which have not a formal, but a true meaning.

Now, we can write the problem that is completely regularized with respect to the original problem (3):

$$\begin{array}{c}\epsilon \frac{\partial \tilde{V}\left(t,s,\eta ,\epsilon \right)}{\partial t}+a\left(t\right)\frac{\partial \tilde{V}\left(t,s,\eta ,\epsilon \right)}{\partial \eta }=a\left(t\right)\tilde{V}\left(t,s,\eta ,\epsilon \right)+\tilde{J}\tilde{V}\left(t,s,\eta ,\epsilon \right),\tilde{V}\left(s,s,0,\epsilon \right)=1\hfill \end{array}$$

(18)

where $\tilde{J}\tilde{V}\left(t,s,\eta ,\epsilon \right)$ has the form (17).

3. Solvability of Iterative Problems

Defining the solution of the problem (18) in the form of series (15), we obtain the following iterative problems:

$$L{v}_0(t,s,\eta )\equiv a\left(t\right)\frac{\partial v_0}{\partial \eta }-a\left(t\right)v_0-R_0{v}_0=0,v_0(s,s,0)=1;$$

(19a)

$$L{v}_1(t,s,\eta )=-\frac{\partial v_0}{\partial t}+R_1{v}_0,v_1(s,s,0)=0;$$

(19b)

$$L{v}_2(t,s,\eta )=-\frac{\partial v_1}{\partial t}+R_1{v}_1+R_2{v}_0,v_2(s,s,0)=0;$$

(19c)

                             ⋯

$$L{v}_k(t,s,\eta )=-\frac{\partial v_{k-1}}{\partial t}+R_1{v}_{k-1}+R_2{v}_{k-2}+\dots +R_k{v}_0,v_k(s,s,0)=0,$$

(19d)

$$L{v}_{k+1}(t,s,\eta )=-\frac{\partial v_k}{\partial t}+R_1{v}_k+R_2{v}_{k-1}+\dots +R_{k+1}{v}_0,v_{k+1}(s,s,0)=0,k\ge 2.$$

(19e)

Let us turn to the study of their solvability in the space $U.$

Each of the problems (19a–e) can be represented as

$$Lv(t,s,\eta )\equiv a\left(t\right)\frac{\partial v}{\partial \eta }-a\left(t\right)v-R_{0}v=H(t,s,\eta ),v(s,s,0)=b$$

(20)

where $H\left(t,s,\eta \right)=H_1\left(t,s\right)e^{\eta }+H_0\left(t,s\right)\in U$ is the known function of space $U,$ b is equal to zero or one, and $R_0$ is the operator, acting on each function (9) of the space U according to the law:

$$R_{0}v\equiv R_0\left(v_1\left(t,s\right)e^{\eta }+v_0\left(t,s\right)\right)\stackrel{def}{=}\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_0\left(x,s\right)dx.$$

In the space U, we introduce the scalar product (for each $\left(t,s\right)\in \left\{0\le s\le t\le 1\right\}$)

$$\begin{array}{c}〈v,w〉\equiv 〈v_1\left(t,s\right)e^{\eta }+v_0\left(t,s\right),w_1\left(t,s\right)e^{\eta }+w_0\left(t,s\right)〉\stackrel{def}{=}{v}_1\left(t,s\right)·{\overline{w}}_1\left(t,s\right)+v_0\left(t,s\right)·{\overline{w}}_0\left(t,s\right).\hfill \end{array}$$

We prove the following assertion (see, [5,8,9]).

Theorem 1.

Let conditions (i) and (ii) be satisfied and the right side $H\left(t,s,\eta \right)=H_1\left(t,s\right)e^{\eta }+H_0\left(t,s\right)$ of the Equation (20) belongs to the space $U.$ Then, for the solvability of the Equation (20) in the space U, it is necessary and sufficient that the identity

$$〈H\left(t,s,\eta \right),e^{\eta }〉\equiv 0\iff H_1\left(t,s\right)\equiv 0\left(\forall \left(t,s\right)\in \left\{0\le s\le t\le 1\right\}\right)$$

(21)

holds.

Proof. 

Let us substitute the element (9) of the space U into Equation (20); we obtain the equality

$$a\left(t\right)v_1\left(t,s\right)e^{\eta }-a\left(t\right)v_1\left(t,s\right)e^{\eta }-a\left(t\right)v_0\left(t,s\right)$$

$$-\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_0\left(x\right)ds=H_1\left(t,s\right)e^{\eta }+H_0\left(t,s\right).$$

Equating here separately the free terms and the coefficients at the exponent, we obtain the equations

$$\begin{array}{c}0·v_1\left(t,s\right)=H_1\left(t,s\right),\hfill \\ \\ -a\left(t\right)v_0\left(t,s\right)-\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_0\left(x,s\right)ds=H_0\left(t,s\right).\hfill \end{array}$$

(22)

The second Equation (22) is a Volterra-type integral equation; it is uniquely solvable in the space $C^{\infty }\left(0\le s\le t\le 1,\mathbb{C}\right)$. For the solvability of the first Equation (22) in this space, it is necessary and sufficient, that $H_1\left(t,s\right)\equiv 0,$ i.e., that condition (21) be satisfied. The theorem has proved. □

Remark 1.

When conditions (i) and (ii), as well as condition (21), hold, Equation (20) has the following solution in the space U:

$$v\left(t,s,\eta \right)=\alpha \left(t,s\right)e^{\eta }+v_0\left(t,s\right)$$

(23)

where $\alpha \left(t,s\right)\in C^{\infty }\left(0\le s\le t\le 1,\mathbb{C}\right)$ is an arbitrary function, $v_0\left(t,s\right)$ is the solution of the integral Equation (22).

We will not formulate a theorem on the unique solvability of the general iterative problem (20). Note that the application of Theorem 1 to two successive iterative problems $\left(19\mathrm{d}\right)$ and $\left(19\mathrm{e}\right)$ leads to a unique calculation of the solution of the iterative problem $\left(19\mathrm{d}\right)$ in the space $U.$ Let us show this is in the example of the problems (19a–e) (see, for example [16,20,22]).

The solution of equation $\left(19\mathrm{a}\right)$ has the form of the sum (23). Since there is no inhomogeneity in $\left(19\mathrm{a}\right)$, the integral Equation (22) will be homogeneous and, therefore, $v_0\left(t,s\right)\equiv 0,$ and the solution (23) itself takes form $v_0\left(t,s,\eta \right)=\alpha \left(t,s\right)e^{\eta }.$ Subordinating it to the initial condition $v_0(s,s)=1,$ we find that $\alpha \left(s,s\right)=1.$ Thus, the solution of the problem $\left(19\mathrm{a}\right)$ will be written in the form $v_0(t,s,\eta )=\alpha (t,s)e^{\eta },$ where the function $\alpha \left(t,s\right)$ is found only at the point $\left(t,s\right)=\left(s,s\right).$ For its final calculation, let us move on to the Equation $\left(19\mathrm{b}\right)$:

$$\begin{array}{c}L{v}_1(t,s,\eta )=-\frac{\partial v_0}{\partial t}+R_1{v}_0,v_1(s,s,0)=0\hfill \\ \\ \iff L{v}_1(t,s,\eta )=-\frac{\partial \alpha \left(t,s\right)}{\partial t}{e}^{\eta }+\left[\frac{K_0\left(t,t\right)}{a\left(t\right)}\alpha \left(t,s\right)e^{\eta }-\frac{K_0\left(t,s\right)}{a\left(s\right)}\right].\hfill \end{array}$$

(24)

By Theorem 1, this equation is solvable in the space U if and only if condition

$$-\frac{\partial \alpha \left(t,s\right)}{\partial t}+\frac{K_0\left(t,t\right)}{a\left(t\right)}\alpha \left(t,s\right)\equiv 0$$

holds. Attaching the initial condition $\alpha (s,s)=1$ to it, we uniquely find the function

$$\alpha (t,s)=exp\left\{\underset{s}{\overset{t}{\int }}\frac{K_0\left(x,x\right)}{a\left(x\right)}dx\right\}$$

(24a)

which means that we uniquely construct the solution $v_0\left(t,s,\eta \right)=\alpha \left(t,s\right)e^{\eta }$ of the problem $\left(19\mathrm{a}\right)$ in the space $U.$ In this case, the problem $\left(19\mathrm{b}\right)$ will be already heterogeneous:

$$\begin{array}{c}L{v}_1(t,s,\eta )=-\frac{\partial \alpha \left(t,s\right)}{\partial t}{e}^{\eta }+\left[\frac{K_0\left(t,t\right)}{a\left(t\right)}\alpha \left(t,s\right)e^{\eta }-\frac{K_0\left(t,s\right)}{a\left(s\right)}\right]\equiv -\frac{K_0\left(t,s\right)}{a\left(s\right)},v_1(s,s,0)=0.\hfill \end{array}$$

Since the right-hand side of this equation has no coefficient at the exponent $e^{\eta }$, condition (21) is satisfied, and we can calculate the solution to this equation (see Formula (23)):

$$v_1(t,s,\eta )={\alpha }_1\left(t,s\right)e^{\eta }+v_0\left(t,s\right)$$

(25)

where ${\alpha }_1\left(t,s\right)\in C^{\infty }\left(0\le s\le t\le 1,\mathbb{C}\right)$ is an arbitrary function, and $v_0\left(t,s\right)$ is the solution of the Volterra-type integral equation:

$$\begin{array}{c}-a\left(t\right)v_0\left(t,s\right)-\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_0\left(x,s\right)ds=-\frac{K_0\left(t,s\right)}{a\left(s\right)}\hfill \\ \\ \iff v_0\left(t,s\right)=\underset{s}{\overset{t}{\int }}\frac{K_0\left(t,x\right)}{-a\left(t\right)}{v}_0\left(x,s\right)ds+\frac{K_0\left(t,s\right)}{a\left(t\right)a\left(s\right)}.\hfill \end{array}$$

(26)

It exists, is unique, and belongs to the class $C^{\infty }\left(0\le s\le t\le 1,\mathbb{C}\right)$. Subordinating (25) to the initial condition $v_1(s,s,0)=0,$ we obtain the equation

$${\alpha }_1\left(s,s\right)+\frac{K_0\left(s,s\right)}{a^2\left(s\right)}=0\iff {\alpha }_1\left(s,s\right)=-\frac{K_0\left(s,s\right)}{a^2\left(s\right)}.$$

(27)

For the final calculation of the function (25), it is necessary to proceed to the following problem $\left(19\mathrm{c}\right)$:

$$L{v}_2(t,s,\eta )=-\frac{\partial v_1}{\partial t}+R_1{v}_1+R_2{v}_0,v_2(s,s,0)=0.$$

Given the type of operators $R_1$ and $R_2$:

$$R_1\left(v_1(t,s)e^{\eta }+v_0(t,s)\right)=\frac{K_0\left(t,t\right)}{a\left(t\right)}{v}_1\left(t,t\right)e^{\eta }-\frac{K_0\left(t,s\right)}{a\left(s\right)}{v}_1\left(s,s\right),$$

$$R_2\left(v_1(t,s)e^{\eta }+v_0(t,s)\right)=-{\left(I^1\left(K_0\left(t,x\right)v_1(x,s)\right)\right)}_{x=t}{e}^{\eta }+{\left(I^1\left(K_0\left(t,x\right)v_1(x,s)\right)\right)}_{x=s}$$

we select the coefficient at the exponent $e^{\eta }$ on the right side of the Equation $\left(19\mathrm{c}\right)$. It will be as follows:

$$-\frac{\partial }{\partial t}{\alpha }_1\left(t,s\right)+\frac{K_0\left(t,t\right)}{a\left(t\right)}{\alpha }_1\left(t,s\right)-{\left(I^1\left(K_0\left(t,x\right)\alpha (x,s)\right)\right)}_{x=t}=0$$

(27a)

where $\alpha \left(t,s\right)=exp\left\{\underset{s}{\overset{t}{\int }}\frac{K_0\left(x,x\right)}{a\left(x\right)}dx\right\}$ is a known function. Adding the initial condition (27) to this equation, we obtain the function ${\alpha }_1\left(t,s\right)$, and, therefore, we construct uniquely the solution (25) of the problem $\left(19\mathrm{c}\right)$. Similar solutions are calculated in the space U of the following iterative problems $\left(15\right)$ for $k\ge 2.$

4. Asymptotic Convergence of Formal Solutions to the Exact Solution

Applying Theorem 1 to iterative problems $\left(19\mathrm{d}\right)$, we uniquely calculate their solutions $v_k(t,s,\eta )$ in the space $U.$ Let us denote by $S_N(t,s,\eta ,\epsilon )$ the N-th partial sum of the series (15) and through $v_{\epsilon N}(t,s)=S_N(t,s,\frac{\phi \left(t,s\right)}{\epsilon },\epsilon )$ is the restriction of this sum at $\eta =\frac{\phi \left(t,s\right)}{\epsilon }.$ It is easy to prove the following assertion (see [2,4,14,16,20,21]).

Lemma 2.

Let conditions (i) and (ii) be satisfied. Then, the function $v_{\epsilon N}(t,s)$ is a formal asymptotic solution of the problem (3) of order $N$, i.e., $v_{\epsilon N}(t,s)$ satisfies the problem

$$\begin{array}{c}\epsilon \frac{d{v}_{\epsilon N}}{dt}=a\left(t\right)v_{\epsilon N}+\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)v_{\epsilon N}\left(x,s\right)dx+{\epsilon }^{N+1}{F}_N(t,s,\epsilon ),\hfill \\ \\ v_{\epsilon N}\left(s,s\right)=1,\left(s,t\right)\in \left\{0\le s\le t\le 1\right\}\hfill \end{array}$$

(28)

where $\left|\right|F_N{(t,s,\epsilon )\left|\right|}_{C\left(0\le s\le t\le 1\right)}\le \overline{F}(\overline{F}>0$ is a constant independent on ε for sufficiently small $\epsilon \in (0,{\epsilon }_0]$).

To estimate the difference $V\left(t,s,\epsilon \right)-v_{\epsilon N}\left(t,s\right)$ between the exact and approximate solutions of the problem (3), we must consider the integro-differential problem

$$\epsilon \frac{dz}{dt}=a\left(t\right)z+\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)z\left(x,s,\epsilon \right)dx+H(t,s,\epsilon ),z(s,s,\epsilon )=0$$

(29)

where $H(t,s,\epsilon )\in C\left(0\le s\le t\le 1,\mathbb{C}\right)$ is a known function. This problem has a unique solution $z=z\left(t,s,\epsilon \right)\in C^1\left(0\le s\le t\le 1,\mathbb{C}\right)$ for each $\epsilon >0$, and it is necessary to estimate the norm of the solution in terms of the right-hand side $H\left(t,s,\epsilon \right).$ Let us introduce another unknown function

$$u(t,s,\epsilon )=\underset{s}{\overset{t}{\int }}{K}_0\left(t,x\right)z\left(x,s,\epsilon \right)dx.$$

Differentiating it with respect to $t,$ we obtain the integro-differential problem:

$$\begin{array}{c}\frac{du(t,s,\epsilon )}{dt}=K_0\left(t,t\right)z\left(t,s,\epsilon \right)+\underset{s}{\overset{t}{\int }}\frac{\partial K_0\left(t,x\right)}{\partial t}z\left(x,s,\epsilon \right)ds,u\left(s,s,\epsilon \right)=0.\hfill \end{array}$$

(30)

As a result, for the vector function $\omega =\left\{z,u\right\}$ we obtain the integro-differential problem

$$\begin{array}{c}\epsilon \frac{d\omega (t,s,\epsilon )}{dt}=\left(\begin{array}{cc}a\left(t\right)& 1\\ 0& 0\end{array}\right)\omega (t,s,\epsilon )+\epsilon \left(\begin{array}{c}0\\ K_0\left(t,t\right)z\left(t,s,\epsilon \right)\end{array}\right)\hfill \\ \\ +\epsilon \underset{s}{\overset{t}{\int }}\left(\begin{array}{c}0\\ \\ \frac{\partial K_0\left(t,x\right)}{\partial t}z\left(x,s,\epsilon \right)dx\end{array}\right)+\left(\begin{array}{c}H\left(t,s,\epsilon \right)\\ 0\end{array}\right),\omega \left(s,s,\epsilon \right)=0.\hfill \end{array}$$

(31)

Denote by $Y\left(t,x,\epsilon \right)$ the Cauchy matrix of the differential problem

$$\epsilon \frac{dY(t,s,\epsilon )}{dt}=\left(\begin{array}{cc}a\left(t\right)& 1\\ 0& 0\end{array}\right)Y(t,s,\epsilon ),Y(s,s,\epsilon )=I,0\le s\le t\le 1.$$

The matrix $A\left(t\right)=\left(\begin{array}{cc}a\left(t\right)& 1\\ 0& 0\end{array}\right)$ is a matrix of simple structure with spectrum $\left\{a\left(t\right),0\right\}.$ Indeed, we have an obvious equality

$${\left(\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill 0& \hfill -a\end{array}\right)}^{-1}.\left(\begin{array}{cc}a& 1\\ 0& 0\end{array}\right).\left(\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill 0& \hfill -a\end{array}\right)=\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)$$

from which follows the simplicity of structure of the matrix $A\left(t\right).$ Since the spectrum $\left\{a\left(t\right),0\right\}$ of the matrix $A\left(t\right)$ lies in the half-plane $\mathrm{Re}\lambda \le 0,$ the matrix $Y\left(t,x,\epsilon \right)$ is uniformly bounded (see, for example, [4,13,18,19], pp. 119–120), i.e.,

$$∥Y\left(t,s,\epsilon \right)∥\le c_0=\mathrm{const}\left(\forall \left(t,s,\epsilon \right)\in \left\{0\le s\le t\le 1\right\}\times \left\{\epsilon >0\right\}\right).$$

Let us use this matrix to invert the problem (31); we obtain the integral system

$$\omega (t,s,\epsilon )=\underset{s}{\overset{t}{\int }}Y\left(t,x,\epsilon \right)\left(\left(\begin{array}{c}0\\ K_0\left(x,x\right)z\left(x,s,\epsilon \right)\end{array}\right)+\left(\begin{array}{c}0\\ \frac{\partial K_0\left(x,\xi \right)}{\partial t}z\left(\xi ,s,\epsilon \right)d\xi \end{array}\right)\right)dx$$

$$+\frac{1}{\epsilon }\underset{s}{\overset{t}{\int }}Y\left(t,x,\epsilon \right)\left(\begin{array}{c}H\left(x,s,\epsilon \right)\\ 0\end{array}\right)dx.$$

Passing here to norms, we obtain an integral inequality with respect to the norm $∥\omega (t,s,\epsilon )∥.$ Taking into account uniform boundedness of the matrix $Y\left(t,x,\epsilon \right),$ as well as the continuity of the functions $K_0\left(t,s\right)$ and $\frac{\partial K_0\left(x,\xi \right)}{\partial t}$ (and hence their boundedness) and applying the Gronwall–Bellman inequality [23], we obtain for sufficiently small $\epsilon \in \left(0,{\epsilon }_0\right]$ inequality

$${∥\omega (t,s,\epsilon )∥}_{C\left(0\le s\le t\le 1\right)}\le \frac{\nu }{\epsilon }{∥\left(\begin{array}{c}H\left(x,s,\epsilon \right)\\ 0\end{array}\right)∥}_{C\left(0\le s\le t\le 1\right)}$$

$(\nu =\mathrm{const}),$ from which follows the estimate

$${∥z(t,s,\epsilon )∥}_{C\left(0\le s\le t\le 1\right)}\le \frac{c_1}{\epsilon }{∥H\left(x,s,\epsilon \right)∥}_{C\left(0\le s\le t\le 1\right)}.$$

(32)

The following assertion is proved.

Lemma 3.

Let conditions (i) and (ii) be satisfied. Then, the solution $z(t,s,\epsilon )$ of the problem (29) exists; is unique for sufficiently small $\epsilon \in \left(0,{\epsilon }_0\right]$; and satisfies estimate (32), where the constant $c_1>0$ does not depend on ε for sufficiently small $\epsilon \in \left(0,{\epsilon }_0\right].$

Let us apply this lemma to prove the following assertion.

Theorem 2.

Let conditions (i) and (ii) be satisfied. Then, the problem (3) is uniquely solvable in the class ${\mathrm{C}}^1([0,1],\mathbb{C})$ and its solution satisfies the estimate

$$\left|\right|V(t,s,\epsilon )-v_{\epsilon N}(t,s){\left|\right|}_{C\left(0\le s\le t\le 1\right)}\le c_N{\epsilon }^{N+1},N=0,1,2,\dots$$

where the constant $c_N>0$ does not depend on ε for sufficiently small $\epsilon \in (0,{\epsilon }_0]$.

Proof. 

The problem (3) is uniquely solvable since it reduces to the problem (29) by replacing $V-1=z.$ By Lemma 3, for the difference ${\Delta }_N(t,s,\epsilon )=V(t,s,\epsilon )-v_{\epsilon N}(t,s)$ we obtain the problem

$$\begin{array}{c}\epsilon \frac{{\Delta }_N}{dt}=a\left(t\right){\Delta }_N(t,s,\epsilon )+\underset{s}{\overset{t}{\int }}K(t,x){\Delta }_N(x,s,\epsilon )dx-{\epsilon }^{N+1}{F}_N(t,s,\epsilon ),{\Delta }_N(s,s,\epsilon )=0.\hfill \end{array}$$

It has the form of the problem (29) with inhomogeneity $H(t,s,\epsilon )\equiv -{\epsilon }^{N+1}{F}_N(t,s,\epsilon ).$ By Lemma 3, we have the estimate

$$\left|\right|{\Delta }_N{(t,s,\epsilon )\left|\right|}_{C\left(0\le s\le t\le 1\right)}\equiv \left|\right|V(t,s,\epsilon )-v_{\epsilon N}(t,s){\left|\right|}_{C\left(0\le s\le t\le 1\right)}$$

$$\le \frac{c_2}{\epsilon }{\epsilon }^{N+1}\left|\right|F_N(t,s,\epsilon ){\left|\right|}_{C\left(0\le s\le t\le 1\right)}\le {\overline{c}}_0{\overline{F}}_N{\epsilon }^N\equiv {\overline{c}}_{N-1}{\epsilon }^N$$

and hence for ${\Delta }_{N+1}(t,s,\epsilon )=V(t,s,\epsilon )-v_{\epsilon ,N+1}(t,s)$ there will be an estimate

$$\left|\right|{\Delta }_{N+1}{(t,s,\epsilon )\left|\right|}_{C\left(0\le s\le t\le 1\right)}\left|\right|$$

$$\equiv (V(t,s,\epsilon )-v_{\epsilon N}(t,s))-{\epsilon }^{N+1}{v}_{N+1}(t,s,\frac{\phi \left(t,s\right)}{\epsilon }){\left|\right|}_{C\left(0\le s\le t\le 1\right)}\le {\overline{c}}_N{\epsilon }^{N+1}.$$

Consequently, the following relation follows from the resulting inequality:

$${\overline{c}}_N{\epsilon }^{N+1}\ge \left|\right|V(t,s,\epsilon )-v_{\epsilon N}{(t,s)\left|\right|}_{C\left(0\le s\le t\le 1\right)}-{\epsilon }^{N+1}\left|\right|v_{N+1}(t,s,\frac{\phi \left(t,s\right)}{\epsilon }){\left|\right|}_{C\left(0\le s\le t\le 1\right)}$$

or

$$\left|\right|V(t,s,\epsilon )-v_{\epsilon N}(t,s){\left|\right|}_{C\left(0\le s\le t\le 1\right)}\le c_N{\epsilon }^{N+1}$$

where $c_N={\overline{c}}_N+{\overline{v}}_N>0,\left|\right|v_{N+1}(t,s,\frac{\phi \left(t,s\right)}{\epsilon }){\left|\right|}_{C\left(0\le s\le t\le 1\right)}\le {\overline{v}}_N$ and the constant $c_N$ does not depend on $\epsilon \in (0,{\epsilon }_0],$ where ${\epsilon }_0>0$ is sufficiently small. The theorem has proved. □

Since in the next section we will use the first order asymptotics fundamental solution, we write it explicitly:

$$\begin{array}{c}{v}_{\epsilon 1}(t,s)=\left(v_0\left(t,s,\eta \right)+\epsilon v_0\left(t,s,\eta \right)\right){|}_{\eta =\frac{\phi \left(t,s\right)}{\epsilon }}\hfill \\ \\ ={\left[\alpha \left(t,s\right)e^{\eta }+\epsilon \left({\alpha }_1\left(t,s\right)e^{\eta }+v_0\left(t,s\right)\right)\right]}_{\eta =\frac{\phi \left(t,s\right)}{\epsilon }}\hfill \end{array}$$

(33)

where $\alpha \left(t,s\right)$ is the function $\left(24\mathrm{a}\right)$ ($K_0\left(t,s\right)\equiv K\left(t,s\right)$), $v_0\left(t,s\right)$ is the function (26), and the function ${\alpha }_1\left(t,s\right)$ satisfies the Equation $\left(27a\right)$ with the initial condition (27).

5. Construction of an Asymptotic Solution to the Original Problem

We will study the original problem (1) under the following assumptions:

(iii)

$a\left(t\right),\mu \left(t\right),h\left(t\right)\in C^{\infty }\left(\left[0,1\right],\mathbb{C}\right),K\left(t,s\right)\in C^{\infty }\left(0\le s\le t\le 1,\mathbb{C}\right);$

(iv)

$\mu \left(t\right)=-l\left(t\right)t^r,l\left(t\right)>0,\mathrm{Re}a\left(t\right)\le 0,a\left(t\right)\ne 0,a\left(t\right)\ne \mu \left(t\right)\left(\forall t\in \left[0,1\right]\right).$

By Lemma 1, the problem (1) has the following solution:

$$y\left(t,\epsilon \right)=V\left(t,0,\epsilon \right)y^0+\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}V\left(t,x,\epsilon \right)h\left(x\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{x}{\int }}\mu \left(\theta \right)d\theta }dx$$

(34)

where $V\left(t,s,\epsilon \right)$ is the solution of the problem (3) at $K_0\left(t,s\right)\equiv K\left(t,s\right).$ Let us use the first-order asymptotics (33) of the solution of the problem (3):

$$v_{\epsilon 1}(t,s)=\alpha \left(t,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{t}{\int }}a\left(\theta \right)d\theta }+\epsilon \left({\alpha }_1\left(t,s\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{t}{\int }}a\left(\theta \right)d\theta }+v_0\left(t,s\right)\right).$$

(35)

Substituting (35) into (34), we arrive at the equality

$$y\left(t,\epsilon \right)=\left[\alpha \left(t,0\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }+\epsilon \left({\alpha }_1\left(t,0\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }+v_0\left(t,0\right)\right)+O\left({\epsilon }^2\right)\right]y^0$$

$$+\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}\left[\alpha \left(t,x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }+\epsilon \left({\alpha }_1\left(t,x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }+v_0\left(t,x\right)\right)+O\left({\epsilon }^2\right)\right]h\left(x\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{x}{\int }}\mu \left(\theta \right)d\theta }dx,$$

or

$$y\left(t,\epsilon \right)=\left[\alpha \left(t,0\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }dx+\epsilon \left({\alpha }_1\left(t,0\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }+v_0\left(t,0\right)\right)\right]y^0$$

$$+\begin{array}{|c|}\hline \frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}\alpha \left(t,x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }dx+\underset{0}{\overset{t}{\int }}{\alpha }_1\left(t,x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }h\left(x\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{x}{\int }}\mu \left(\theta \right)d\theta }dx\\ \hline\end{array}$$

(36)

$$+\underset{0}{\overset{t}{\int }}{v}_0\left(t,x\right)h\left(x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx+O\left(\epsilon \right).$$

The integrals enclosed in a box can be integrated by parts; such an operation cannot be applied to the integral $\underset{0}{\overset{t}{\int }}{v}_0\left(t,x\right)h\left(x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx$ since the spectral value $\mu \left(t\right)$ turns to zero at the point $t=0$. Let us first deal with the integrals in the box. We have:

$$\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}\alpha \left(t,x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }dx=\underset{0}{\overset{t}{\int }}\frac{\alpha \left(t,x\right)}{-a\left(x\right)}d{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }=\left(\frac{\alpha \left(t,x\right)}{-a\left(x\right)}{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }\right){|}_{x=0}^{x=t}$$

$$-\underset{0}{\overset{t}{\int }}\frac{\partial }{\partial x}\left(\frac{\alpha \left(t,x\right)}{-a\left(x\right)}\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }dx=\left[\frac{\alpha \left(t,t\right)}{-a\left(t\right)}-\frac{\alpha \left(t,0\right)}{-a\left(0\right)}{e}^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\right]$$

(37)

$$-\underset{0}{\overset{t}{\int }}\frac{\partial }{\partial x}\left(\frac{\alpha \left(t,x\right)}{-a\left(x\right)}\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}a\left(\theta \right)d\theta }dx=\left[\frac{\alpha \left(t,t\right)}{-a\left(t\right)}-\frac{\alpha \left(t,0\right)}{-a\left(0\right)}{e}^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\right]+O\left(\epsilon \right).$$

For the second integral in the box, we introduce the notation $m\left(t,x\right)\equiv {\alpha }_1\left(t,x\right)h\left(x\right)$ and again apply the integration operation in parts:

$$e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\underset{0}{\overset{t}{\int }}m\left(t,x\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{x}{\int }}\left(\mu \left(\theta \right)-a\left(\theta \right)d\theta \right)}dx=\epsilon e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\underset{0}{\overset{t}{\int }}\frac{m\left(t,x\right)}{\mu \left(x\right)-a\left(x\right)}d{e}^{\frac{1}{\epsilon }\underset{0}{\overset{x}{\int }}\left(\mu \left(\theta \right)-a\left(\theta \right)d\theta \right)}$$

$$=\epsilon e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\left[\frac{m\left(t,x\right)}{\mu \left(x\right)-a\left(x\right)}{e}^{\frac{1}{\epsilon }\underset{0}{\overset{x}{\int }}\left(\mu \left(\theta \right)-a\left(\theta \right)d\theta \right)}{|}_{x=0}^{x=t}\right]$$

$$-\epsilon e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\underset{0}{\overset{t}{\int }}\frac{\partial }{\partial x}\left(\frac{m\left(t,x\right)}{\mu \left(x\right)-a\left(x\right)}\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{x}{\int }}\left(\mu \left(\theta \right)-a\left(\theta \right)d\theta \right)}$$

$$=\epsilon \left[\frac{m\left(t,t\right)}{\mu \left(t\right)-a\left(t\right)}{e}^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }-\frac{m\left(t,0\right)}{\mu \left(0\right)-a\left(0\right)}{e}^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\right]+O\left({\epsilon }^2\right).$$

Since we are going to construct the main term of the asymptotic of the solution of the original problem (1), then this term can be neglected. The most difficult is the construction of the asymptotics of the integral $\underset{0}{\overset{t}{\int }}{v}_0\left(t,x\right)h\left(x\right)e^{\frac{1}{\epsilon }\underset{s}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx$ since the classical operation of integration by parts does not work here. For this, as is known, it is necessary to apply the modified operation of integration by parts (see, for example, [4], Ch. 6). Let us denote for convenience $n\left(t,x\right)\equiv v_0\left(t,x\right)h\left(x\right).$ Let $\left\{K_{0j}\left(t\right),j=\overline{0,r-1}\right\}\equiv \left\{\frac{t^j}{j!},j=\overline{0,r-1}\right\}$ be the basic system of Lagrange–Sylvester polynomials for the spectral value $\mu \left(t\right)=-l\left(t\right)t^r,l\left(t\right)>0$ ([24], pp. 87–92). Then,

$$n\left(t,s\right)-\sum _{j=0}^{r-1}\frac{{\partial }^j}{\partial s^j}n\left(t,0\right)\frac{s^j}{j!}=l_0\left(t,s\right)\frac{s^r}{r!}$$

where $l_0\left(t,s\right)\in C^{\infty }\left(\left[0,1\right],{\mathbb{C}}^1\right)$ is some function. Let us do it the following operations:

$$\underset{0}{\overset{t}{\int }}n\left(t,x\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx=\underset{0}{\overset{t}{\int }}\left(n\left(t,x\right)-\sum _{j=0}^{r-1}\frac{{\partial }^{j}n\left(t,0\right)}{\partial x^j}\frac{x^j}{j!}\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx$$

$$+\underset{0}{\overset{t}{\int }}\sum _{j=0}^{r-1}\frac{{\partial }^{j}n\left(t,0\right)}{\partial x^j}\frac{x^j}{j!}{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx=\underset{0}{\overset{t}{\int }}\left(l_0\left(t,x\right)\frac{x^r}{r!}\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx$$

$$+\sum _{j=0}^{r-1}\frac{{\partial }^{j}n\left(t,0\right)}{\partial x^j}\underset{0}{\overset{t}{\int }}\frac{x^j}{j!}{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx.$$

It is already possible to apply the classical operation of integration by parts to the integral $P=\underset{0}{\overset{t}{\int }}\left(l_0\left(t,x\right)\frac{x^r}{r!}\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx$:

$$P=\epsilon \underset{0}{\overset{t}{\int }}\frac{\left(l_0\left(t,x\right)\frac{x^r}{r!}\right)}{-\mu \left(x\right)}d{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }=\epsilon \underset{0}{\overset{t}{\int }}\frac{\left(l_0\left(t,x\right)\frac{x^r}{r!}\right)}{-l\left(x\right)x^r}d{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }$$

$$=\epsilon \underset{0}{\overset{t}{\int }}\left(\frac{l_0\left(t,x\right)}{-l\left(x\right)r!}\right)d{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }=\epsilon \left(\frac{l_0\left(t,x\right)}{-l\left(x\right)r!}\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }{|}_{x=0}^{x=t}$$

$$-\epsilon \underset{0}{\overset{t}{\int }}\frac{\partial }{\partial x}\left(\frac{l_0\left(t,x\right)}{-l\left(x\right)r!}\right)e^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx.$$

Leaving in (36) $\epsilon$-independent terms, as well as the term

$$\sum _{j=0}^{r-1}\frac{{\partial }^{j}n\left(t,0\right)}{\partial x^j}\underset{0}{\overset{t}{\int }}\frac{x^j}{j!}{e}^{\frac{1}{\epsilon }{\int }_x^t\mu \left(\theta \right)d\theta }dx$$

we obtain the following equality:

$$y\left(t,\epsilon \right)=\left[\alpha \left(t,0\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }dx\right]y^0+\left[\frac{\alpha \left(t,t\right)}{-a\left(t\right)}-\frac{\alpha \left(t,0\right)}{-a\left(0\right)}{e}^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\right]$$

$$+\sum _{j=0}^{r-1}\frac{{\partial }^{j}n\left(t,0\right)}{\partial x^j}\underset{0}{\overset{t}{\int }}\frac{x^j}{j!}{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx+O\left(\epsilon \right)(\epsilon \to +0).$$

Here, the second term contains the so-called surge functions $\underset{0}{\overset{t}{\int }}\frac{x^j}{j!}{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx,$ which in the neighborhood of the point $t=0$ have a maximum proportional to the value

$$\sqrt[r+1]{{(r+1)}^{j+1}{\epsilon }^{j+1}},j=\overline{0,r-1}$$

(see [25], pp. 207–212). They do not belong to the class $O\left(\epsilon \right)$ and therefore should be included in the main term of the asymptotics of the solution of problem (1):

$$\begin{array}{cc}& y_{\epsilon 0}\left(t\right)=\left[\alpha \left(t,0\right)e^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }dx\right]y^0+\left[\frac{\alpha \left(t,t\right)}{-a\left(t\right)}-\frac{\alpha \left(t,0\right)}{-a\left(0\right)}{e}^{\frac{1}{\epsilon }\underset{0}{\overset{t}{\int }}a\left(\theta \right)d\theta }\right]\hfill \\ \\ & +\sum _{j=0}^{r-1}\frac{{\partial }^{j}n\left(t,0\right)}{\partial x^j}\underset{0}{\overset{t}{\int }}\frac{x^j}{j!}{e}^{\frac{1}{\epsilon }\underset{x}{\overset{t}{\int }}\mu \left(\theta \right)d\theta }dx\hfill \end{array}$$

(38)

where $\alpha \left(t,s\right)$ is the function $\left(24\mathrm{a}\right)$ $(K_0(t,s)\equiv K(t,s)).$

Remark 2.

We have constructed the main term (38) of the asymptotic solution of the problem (1) using the first-order asymptotics (35) for the fundamental solution $V\left(t,s,\epsilon \right).$ It is clear that in order to obtain the N-th order asymptotic for the solution $y\left(t,\epsilon \right),$ one should use the $\left(N+1\right)$-th order asymptotic for the function $V\left(t,s,\epsilon \right).$

6. Conclusions

In this paper, we consider the case of one exponential inhomogeneity. Similarly, the case of several inhomogeneities with unstable spectral values can be considered. However, in this case, different exponential inhomogeneities may have different points of instability belonging to the considered time interval. Therefore, it will not be easy to single out essentially special singularities in the solution of the original problem and to carry out further regularization of the problem. Using our method of constructing asymptotics with the help of a fundamental solution, this difficulty can also be overcome. The specific implementation of this case is the subject of a separate article. It will be more difficult to study the case of continual irreversibility of the spectral value, i.e., the case when the spectral value vanishes on some subset S of positive measure. We intend to study it in our future works.