Abstract

The regularization method is applied for the construction of algorithm for an asymptotical solution for linear singular perturbed systems with the irreversible limit operator. The main idea of this method is based on the analysis of dual singular points of investigated equations and passage in the space of the larger dimension, what reduces to study of systems of first-order partial differential equations with incomplete initial data.

1. Introduction

The investigation of singular perturbed systems for ordinary and partial differential equations occurring in systems with slow and fast variables, chemical kinetics, the mathematical theory of boundary layer, control with application of geoinformational technologies, quantum mechanics, and plasma physics (the Samarsky-Ionkin problem) has been studied by many researchers (see, e.g., [119]).

In this work, the algorithm for construction of an asymptotical solution for linear singular perturbed systems with the irreversible limit operator is given—the regularization method [1]. The main idea of this method is based on the analysis of dual singular points of investigated equations and passage in the space of the larger dimension, what reduces to the study of systems of first-order partial differential equations with incomplete (more exactly, point) initial data.

In this paper, we consider linear singular perturbed systems in the form𝜀̇𝑦=𝐴(𝑡)𝑦+(𝑡),𝑦(0,𝜀)=𝑦0[],,𝑡0,𝑇(1.1) where 𝑦={𝑦1,,𝑦𝑛},𝐴(𝑡) is a matrix of order (𝑛×𝑛),(𝑡)={1,,𝑛} is a known function, 𝑦0𝐶𝑛 is a constant vector, and 𝜀>0 is a small parameter, in the case of violation of stability of a spectrum {𝜆𝑗(𝑡)} of the limiting operator 𝐴(𝑡).

Difference of such type problems from similar problems with a stable spectrum (i.e., in the case of 𝜆𝑖(𝑡)0,𝜆𝑖(𝑡)𝜆𝑗(𝑡),𝑖𝑗,𝑖,𝑗=1,𝑛forall𝑡[0,𝑇]) is that the limiting system 0=𝐴(𝑡)𝑦+(𝑡)at violation of stability of the spectrum can have either no solutions or uncountable set of them. In the last case, presence of discontinuous on the segment [0,𝑇] solutions 𝑦(𝑡)of the limiting system is not excluded. Under conditions, one can prove (see, e.g., [1, 6]) that the exact solution 𝑦(𝑡,𝜀) of problem (1.1) tends (at 𝜀+0) to a smooth solution of the limiting system. However, there is a problematic problem about construction of an asymptotic solution of problem (1.1). When the spectrum is instable, essentially special singularities are arising in the solution of system (1.1). These singularities are not selected by the spectrum {𝜆𝑗(𝑡)} of the limiting operator 𝐴(𝑡). As it was shown in [37], they were induced by instability points 𝑡𝑗 of the spectrum.

In the present work, the algorithm of regularization method [1] is generalized on singular perturbed systems of the form (1.1), the limiting operator of which has some instable points of the spectrum. In order to construct the spectrum, we use the new algorithm requiring more constructive theory of solvability of iterative problems. These problems arose in application of the algorithm.

We will consider the problem (1.1) at the following conditions. Assume that(i)𝐴(𝑡)𝐶([0,𝑇],𝐶𝑛),(𝑡)𝐶[0,𝑇]; for any 𝑡[0,𝑇], the spectrum {𝜆𝑗(𝑡)} of the operator 𝐴(𝑡) satisfies the conditions:(ii)𝜆𝑖(𝑡)=(𝑡𝑡𝑖)𝑠𝑖𝑘𝑖(𝑡),𝑘𝑖(𝑡)0,𝑡𝑖[0,𝑇],𝑖=1,𝑚,𝑚<𝑛 (here 𝑠𝑖- are even natural numbers),(iii)𝜆𝑖(𝑡)0,𝑗=𝑚+1,𝑛,(iv)𝜆𝑖(𝑡)𝜆𝑗(𝑡),𝑖𝑗,𝑖,𝑗=1,𝑛,(v)Re𝜆𝑗(𝑡)0,𝑗=1,𝑛.

2. Regularization of the Problem

We introduce basic regularized variables by the spectrum of the limiting operator𝜏𝑗=𝜀1𝑡0𝜆𝑗𝜑(𝑠)𝑑𝑠𝑗(𝑡)𝜀,𝑗=1,𝑛.(2.1) Instable points 𝑡𝑖[0,𝑇] of the spectrum {𝜆𝑗(𝑡)} induce additional regularized variables described by the formulas𝜎𝑖𝑞𝑖=𝑒𝜑𝑖(𝑡)/𝜀𝑡0𝑒𝜑𝑖(𝑠)/𝜀𝑠𝑡𝑖𝑞𝑖𝑞𝑖𝑑𝑠𝜓𝑖𝑞𝑖(𝑡,𝜀),𝑖=1,𝑚,𝑞𝑖=0,𝑠𝑖1.(2.2) We consider a vector function ̃𝑦(𝑡,𝜏,𝜎,𝜀) instead of the solution 𝑦(𝑡,𝜀) to be found for problem (1.1). This vector function is such that̃𝑦(𝑡,𝜏,𝜎,𝜀)𝜏=𝜑,𝜎=𝜓𝑦(𝑡,𝜀).(2.3)

For ̃𝑦(𝑡,𝜏,𝜎,𝜀), it is natural to set the following problem:𝐿𝜀̃𝑦(𝑡,𝜏,𝜎,𝜀)𝜀𝜕̃𝑦+𝜕𝑡𝑛𝑗=1𝜆𝑗(𝑡)𝜕̃𝑦𝜕𝜏𝑗+𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝜆𝑖(𝑡)𝜎𝑖𝑞𝑖+𝜀𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝜕̃𝑦𝜕𝜎𝑖𝑞𝑖𝐴(𝑡)̃𝑦=(𝑡),̃𝑦(0,0,0,𝜀)=𝑦0.(2.4) We determine the solution of problem (2.4) in the form of a series̃𝑦(𝑡,𝜏,𝜎,𝜀)=𝑘=1𝜀𝑘𝑦𝑘(𝑡,𝜏,𝜎),(2.5) with coefficients 𝑦𝑘(𝑡,𝜏,𝜎)𝐶[0,𝑇].

If we substitute (2.5) in (2.4) and equate coefficients at identical degrees of 𝜀, we obtain the systems for coefficients 𝑦𝑘(𝑡,𝜏,𝜎):𝐿𝑦1(𝑡,𝜏,𝜎)𝑛𝑗=1𝜆𝑗(𝑡)𝜕𝑦1𝜕𝜏𝑗+𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝜆𝑖(𝑡)𝜎𝑖𝑞𝑖𝜕𝑦1𝜕𝜎𝑖𝑞𝑖𝐴(𝑡)𝑦1=0,𝑦1(0,0,0)=0,(𝜀1)𝐿𝑦0(𝑡,𝜏,𝜎)=𝜕𝑦1𝜕𝑡𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝜕𝑦1𝜕𝜎𝑖𝑞𝑖+(𝑡),𝑦0(0,0,0)=𝑦0,(𝜀0)𝐿𝑦𝑘+1(𝑡,𝜏,𝜎)=𝜕𝑦𝑘𝜕𝑡𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝜕𝑦𝑘𝜕𝜎𝑖𝑞𝑖,𝑘1,𝑦𝑘+1(0,0,0)=0,(𝜀𝑘+1)

3. Resolvability of Iterative Problems

We solve each of the iterative problems (𝜀𝑘) in the following space of functions:𝑈=𝑦(𝑡,𝜏,𝜎)𝑦=𝑛𝑛𝑘=1𝑗=1𝑦𝑘𝑗(𝑡)𝑐𝑘(𝑡)𝑒𝜏𝑗+𝑛𝑚𝑘=1𝑠𝑖=1𝑖1𝑞𝑖=0𝑦𝑘𝑖𝑞𝑖(𝑡)𝑐𝑘(𝑡)𝜎𝑖𝑞𝑖+𝑛𝑘=1𝑦𝑘(𝑡)𝑐𝑘(𝑡),𝑦𝑘𝑗(𝑡),𝑦𝑘𝑖𝑞𝑖(𝑡),𝑦𝑘(𝑡)𝐶[]0,𝑇,𝐶1,(3.1) where 𝑐𝑘(𝑡) are eigenvectors of the operator 𝐴(𝑡) corresponding eigenvalues 𝜆𝑘(𝑡),𝑘=1,𝑛. We represent 𝑈 in the form of 𝑈(1)𝑈(0) where𝑈(0)=𝑦(0)(𝑡)𝑦(0)=𝑛𝑗=1𝑦𝑗(0)(𝑡)𝑐𝑗(𝑡),𝑦𝑗(0)(𝑡)𝐶[]0,𝑇,𝐶1,𝑈(1)=𝑈𝑈(0).(3.2) It is easy to note that each of the systems (𝜀𝑘+1) can be written in the form𝐿𝑦(𝑡,𝜏,𝜎)=(𝑡,𝜏,𝜎),(3.3) where (𝑡,𝜏,𝜎) are the corresponding right hand side. Using representations of space 𝑈, we can write system (3.3) in the equivalent form𝐿𝑦(1)(𝑡,𝜏,𝜎)=(1)(𝑡,𝜏,𝜎),(3.4)𝐴(𝑡)𝑦(0)(𝑡)=(0)(𝑡),(3.5) where 𝑦(1)(𝑡,𝜏,𝜎),(1)(𝑡,𝜏,𝜎)𝑈(1),𝑦(0)(𝑡),(0)(𝑡)𝑈(0).

We have the following result.

Theorem 3.1. Let (1)(𝑡,𝜏,𝜎)𝑈(1) and satisfy conditions (i)–(iv). Then, system (3.4) is solvable in the 𝑈(1) if and only if (1)(𝑡,𝜏,𝜎),𝜈𝑗[](𝑡,𝜏,𝜎)0𝑡0,𝑇,𝑗=1,𝑛,(1)(𝑡,𝜏,𝜎),𝜈𝑖𝑞𝑖(𝑡,𝜏,𝜎)0,𝑖=1,𝑚,𝑞𝑖=0,𝑠𝑖1,(3.6) where 𝜈𝑗(𝑡,𝜏,𝜎),𝜈𝑖𝑞𝑖(𝑡,𝜏,𝜎) are basic elements of the kernel of the operator 𝐿𝑛𝑗=1𝜆𝑗𝜕(𝑡)𝜕𝜏𝑗+𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝜆𝑖(𝑡)𝜎𝑖𝑞𝑖𝜕𝜕𝜎𝑖𝑞𝑖𝐴(𝑡).(3.7)

Proof. Let (1)(𝑡,𝜏,𝜎)=𝑛𝑘=1𝑛𝑗=1𝑘𝑗(𝑡)𝑐𝑗(𝑡)𝑒𝜏𝑘+𝑛𝑘=1𝑚𝑖=1𝑠𝑖𝑞1𝑖=0𝑘𝑖𝑞𝑖(𝑡)𝑐𝑘(𝑡)𝜎𝑖𝑞𝑖.
Determine solutions of system (3.4) in the form 𝑦(1)(𝑡,𝜏,𝜎)=𝑛𝑛𝑘=1𝑗=1𝑦𝑘𝑗(𝑡)𝑐𝑘(𝑡)𝑒𝜏𝑗+𝑛𝑚𝑘=1𝑠𝑖=1𝑖1𝑞𝑖=0𝑦𝑘𝑖𝑞𝑖(𝑡)𝑐𝑘(𝑡)𝜎𝑖𝑞𝑖.(3.8) Substituting (3.8) in (3.4) and equating separately coefficients at 𝑒𝜏𝑗 and 𝜎𝑖𝑞𝑖, we obtain the equations 𝜆𝑘(𝑡)𝜆𝑗𝑦(𝑡)𝑘𝑗(𝑡)=𝑘𝑗(𝑡),𝑘,𝑗=𝜆1,𝑛,𝑖(𝑡)𝜆𝑘𝑦(𝑡)𝑖𝑞𝑖𝑘(𝑡)=𝑘𝑖𝑞𝑖(𝑡),𝑖=1,𝑚,𝑞𝑖=0,𝑠𝑖1,𝑘=1,𝑛.(3.9)
One can see from this that obtained equations are solvable if and only if 𝑘𝑘(𝑡)0,𝑘=1,𝑛,𝑖𝑖𝑞𝑖(𝑡)0,𝑖=1,𝑚,𝑞𝑖=0,𝑠𝑖1,(3.10) and these conditions coincide with conditions (3.6). Theorem 3.1 is proved.

Remark 3.2. Equations (1.1) imply that under conditions (3.6), system (3.4) has a solution in 𝑈(1) representable in the form 𝑦(1)(𝑡,𝜏,𝜎)=𝑛𝑛𝑘=1𝑗=1,𝑗𝑘𝑘𝑗(𝑡)𝜆𝑘(𝑡)𝜆𝑗𝑐(𝑡)𝑗(𝑡)𝑒𝜏𝑗+𝑛𝑘=1𝛼𝑘(𝑡)𝑐𝑘(𝑡)𝑒𝜏𝑘+𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝛾𝑖𝑞𝑖(𝑡)𝑐𝑖(𝑡)𝜎𝑖𝑞𝑖+𝑛𝑚𝑘=1𝑠𝑖=1,𝑖𝑘𝑖1𝑞𝑖=0𝑘𝑖𝑞𝑖(𝑡)𝜆𝑖(𝑡)𝜆𝑘𝑐(𝑡)𝑘(𝑡)𝜎𝑖𝑞𝑖,(3.11) where 𝛼𝑘(𝑡),𝛾𝑖𝑞𝑖(𝑡)𝐶([0,𝑇],𝐶1) are arbitrary functions.
Consider now system (3.5). As det𝐴(𝑡)0 in points 𝑡=𝑡𝑖, 𝑖=1,𝑚, this system does not always have a solution in 𝑈(0). Introduce the space 𝑉(0)𝑈(0) consisting of vector functions 𝑧(0)(𝑡)=𝑛𝑗=1𝑧𝑗(𝑡)𝑐𝑗(𝑡),𝑧𝑗(𝑡)𝐶[]0,𝑇,𝐶1,𝑗=1,𝑛,(3.12) having the properties 𝐷𝑙𝑖𝑧(0)(𝑡),𝑑𝑖(𝑡)𝑡=𝑡𝑖=𝐷𝑙𝑖𝑧𝑖𝑡𝑖=0,𝑙𝑖=0,𝑠𝑖1,𝑖=1,𝑚,(3.13) where 𝑑𝑖(𝑡) are eigenvectors of the operator 𝐴(𝑡) with regard to eigenvalues 𝜆𝑖(𝑡),𝑖=1,𝑚. Let (0)(𝑡)=𝑛𝑗=1𝑗(𝑡)𝑐𝑗(𝑡)𝑉(0), that is, 𝐷𝑙𝑖𝑖𝑡𝑖=0𝑙𝑖=0,𝑠𝑖1,𝑖=1,𝑚.(3.14) Determine a solution of system (3.5) in the 𝑦(0)(𝑡)=𝑛𝑗=1𝑦𝑗(𝑡)𝑐𝑗(𝑡).(3.15) Substituting this function in (3.5), we obtain 𝑛𝑗=1𝑦𝑗(𝑡)𝜆𝑗(𝑡)𝑐𝑗(𝑡)=𝑛𝑗=1𝑗(𝑡)𝑐𝑗(𝑡).(3.16) Since {𝑐𝑗(𝑡)} is a basis in 𝐶𝑛, we get 𝜆𝑖(𝑡)𝑦𝑖(𝑡)=𝑖(𝑡),𝑖=1,𝑚,(3.17)𝜆𝑗(𝑡)𝑦𝑗(𝑡)=𝑗(𝑡),𝑗=𝑚+1,𝑛.(3.18) It is easy to see that (3.18) has the unique solution 𝑦𝑗(𝑡)=𝑗(𝑡)𝜆𝑗(𝑡),𝑗=𝑚+1,𝑛.(3.19) By virtue of conditions (3.14), the function 𝑖(𝑡) can be represented in the form 𝑖(𝑡)=𝑡𝑡𝑖𝑠𝑖𝑖(𝑡),𝑖=1,𝑚,(3.20) where 𝑖(𝑡)𝐶([0,𝑇],𝐶1 is the certain scalar function, (𝑡𝑡𝑖)𝑠𝑖𝑘𝑖(𝑡)𝑦𝑖(𝑡)=(𝑡𝑡𝑖)𝑠𝑖𝑖(𝑡), and we see that 𝑦𝑖(𝑡)=𝑖(𝑡)𝑘𝑖(𝑡),𝑡𝑡𝑖,𝛾𝑖,𝑡=𝑡𝑖,(3.21) where 𝛾𝑖 are arbitrary constants, 𝑖=1,𝑚. However, the solution of system (3.5) should belong to the space 𝑈(0), and it means that 𝑦𝑖(𝑡)𝐶([0,𝑇],𝐶1). Therefore, constants in (3.21) 𝛾𝑖=(𝑖(𝑡)/𝑘𝑖(𝑡))𝑡=𝑡𝑖 and functions are determined uniquely in the form 𝑦𝑖(𝑡)=𝑖(𝑡)𝑘𝑖[](𝑡),𝑡0,𝑇,𝑖=1,𝑚.(3.22) Thus, under conditions (3.14), system (3.5) has the solution 𝑦(0)(𝑡) in 𝑈(0) of 𝑦(0)(𝑡)=𝑚𝑖=1𝑖(𝑡)𝑘𝑖𝑐(𝑡)𝑖(𝑡)𝑛𝑗=𝑚+1𝑖(𝑡)𝜆𝑖𝑐(𝑡)𝑖(𝑡),(3.23) where 𝑖(𝑡)=𝑖(𝑡)/(𝑡𝑡𝑖)𝑠𝑖 (in points 𝑡=𝑡𝑖,𝑖=1,𝑚, this equality is understood in the limiting sense). We summarize received outcome in the form of the following assertion.

Theorem 3.3. Let the operator 𝐴(𝑡) satisfy condition (i), and let its spectrum satisfy conditions (ii)–(iv). Then, for any vector function (0)(𝑡)𝑉(0), system (3.5) has the unique solution 𝑦(0)(𝑡) in space 𝑈(0).

For uniquely determination of functions 𝛼𝑗(𝑡),𝛾𝑖𝑞𝑖(𝑡), consider system (3.4) with additional conditions:𝑦(1)(0,0,0)=𝑦,(3.24)𝜕𝑦(1)𝜕𝑡,𝜈𝑗[](𝑡,𝜏,𝜎)0𝑡0,𝑇,𝑗=1,𝑛,(3.25)𝜕𝑦(1)𝜕𝑡,𝜈𝑖𝑞𝑖(𝑡,𝜏,𝜎)0,𝑖=1,𝑚,𝑞𝑖=0,𝑠𝑖1,(3.26) where 𝑦𝐶𝑛 is a constant vector.

We have the following result.

Theorem 3.4. Let conditions of Theorem 3.1 hold. Then, the system (3.4) with additional conditions (3.24)-(3.25) has solutions of the form (3.11) in which all summands are uniquely determinate except for 𝛾𝑖𝑞𝑖(𝑡)𝑐𝑖(𝑡)𝜎𝑖𝑞𝑖(𝑖=1,𝑚,𝑞𝑖=0,𝑠𝑖1). Functions 𝛾𝑖𝑞𝑖(𝑡) in the last summand are determined by the formula 𝛾𝑖𝑞𝑖(𝑡)=𝛾0𝑖𝑞𝑖𝑒𝑃𝑖𝑖𝑞(𝑡)+𝑓𝑖𝑞𝑖(𝑡),(3.27) where 𝑃𝑖𝑞𝑖(𝑡),𝑓𝑖𝑞𝑖(𝑡) are known functions, and 𝛾0𝑖𝑞𝑖arbitrary constants.

Proof. Denote in (3.11) that 𝑔𝑘𝑗(𝑡)=𝑘𝑗(𝑡)𝜆𝑗(𝑡)𝜆𝑘(𝑡),𝑔𝑘𝑖𝑞𝑖(𝑡)=𝑘𝑖𝑞𝑖(𝑡)𝜆𝑖(𝑡)𝜆𝑘.(𝑡)(3.28) Using (3.11) and condition (3.24), we obtain the equality 𝑛𝑛𝑘=1𝑗=1𝑔𝑘𝑗(0)𝑐𝑗(0)+𝑛𝑘=1𝛼𝑘(0)𝑐𝑘(0)=𝑦.(3.29) Multiplying this equality scalarly by 𝑑𝑠(0), we get 𝛼𝑠𝑦(0)=,𝑑𝑠(0)𝑛𝑘=1,𝑘𝑠𝑔𝑘𝑠(0)𝛼0𝑠,𝑠=1,𝑛.(3.30) By (3.11) and conditions (3.25), we have ̇𝛼𝑠(𝑡)̇𝑐𝑠(𝑡),𝑑𝑠𝛼(𝑡)𝑠(𝑡)𝑛𝑗=1,𝑗𝑠𝑔𝑠𝑗(𝑡)̇𝑐𝑗(𝑡),𝑑𝑠(𝑡)=0,𝑠=1,𝑛.(3.31) Considering these equations with initial conditions (3.30), we can uniquely obtain functions 𝛼𝑠(𝑡),𝑠=1,𝑛.
Now, using (3.11) and conditions (3.26), we get ̇𝛾𝑖𝑞𝑖(𝑡)̇𝑐𝑖(𝑡),𝑑𝑖𝛾(𝑡)𝑖𝑞𝑖(𝑡)𝑛𝑘=1,𝑘𝑖𝑔𝑘𝑖𝑞𝑖(𝑡)̇𝑐𝑘(𝑡),𝑑𝑖(𝑡)=0,𝑖=1,𝑚,𝑞𝑖=0,𝑠𝑖1.(3.32) This implies that 𝛾𝑖𝑞𝑖(𝑡) have the form (3.27) where 𝑃𝑖𝑞𝑖(𝑡)=𝑡𝑡𝑖̇𝑐𝑖(𝑠),𝑑𝑖(𝑓𝑠)𝑑𝑠,𝑖𝑞𝑖(𝑡)=𝑒𝑃𝑖𝑖𝑞(𝑡)𝑡𝑡𝑖𝑒𝑃𝑖𝑖𝑞𝑛(𝑠)𝑘=1,𝑘𝑖𝑔𝑘𝑖𝑞𝑖(𝑠)̇𝑐𝑘(𝑠),𝑑𝑖(𝑠)𝑑𝑠.(3.33) Theorem 3.4 is proved.

Remark 3.5. If conditions (3.6) hold for (1)(𝑡,𝜏,𝜎)𝑈(1)and (0)(𝑡)𝑈(0), then system (3.3) has a solution in the space 𝑈, representable in the form of 𝑦(𝑡,𝜏,𝜎)=𝑦(1)(𝑡,𝜏,𝜎)+𝑦(0)(𝑡),(3.34) where 𝑦(1)(𝑡,𝜏,𝜎)is a function in the form of (3.11), and 𝑦(0)(𝑡) is a function in the form of (3.23); moreover, functions 𝛼𝑘(𝑡)𝐶([0,𝑇],𝐶1) are found uniquely in (3.11), and functions 𝛾𝑖𝑞𝑖(𝑡) are determined up to arbitrary constants 𝛾0𝑖𝑞𝑖 in the form of (3.27).

Let us give the following result.

Theorem 3.6. Let (0)(𝑡)𝑈(0),(1)(𝑡,𝜏,𝜎)𝑈(1),and conditions (i)–(iv), (3.6), (3.24)–(3.26) hold. Then, there exist unique numbers 𝛾0𝑖𝑞𝑖 involved in (3.27), such that the function (3.34) satisfies the condition 𝑃𝑦𝜕𝑦(0)𝜕𝑡𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝜕𝑦(1)𝜕𝜎𝑖𝑞𝑖+𝐻(0)(𝑡)𝑉(0),(3.35) where 𝐻(0)(𝑡)𝑉(0)is a fixed vector function.

Proof. To determine functions uniquely, calculate 𝑃𝑦𝑚𝑖=1𝑖(𝑡)𝑘𝑖𝑐(𝑡)𝑖(𝑡)𝑛𝑗=𝑚+1𝑗(𝑡)𝜆𝑗𝑐(𝑡)𝑗(𝑡)𝑚𝑠𝑖=1𝑖1𝑞𝑖=0𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝛾𝑖𝑞𝑖(𝑡)𝑐𝑖+(𝑡)𝑛𝑚𝑘=1𝑠𝑖=1𝑖𝑘𝑖1𝑞𝑖=0𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝑖𝑞𝑖(𝑡)𝜆𝑖(𝑡)𝜆𝑘𝑐(𝑡)𝑘(𝑡)+𝐻(0)(𝑡),𝑃𝑦,𝑑𝑖(𝑡)𝑖(𝑡)𝑘𝑖(𝑡)𝑚𝑖=1𝑖(𝑡)𝑘𝑖(𝑡)̇𝑐𝑖(𝑡),𝑑𝑖(𝑡)𝑛𝑗=𝑚+1𝑗(𝑡)𝑘𝑗(𝑡)̇𝑐𝑗(𝑡),𝑑𝑖(𝑡)𝑠𝑖1𝑞𝑖𝑚=0𝑖=1𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝛾𝑖𝑞𝑖𝐻(𝑡)+(0)(𝑡),𝑑𝑖(𝑡),𝑖=1,𝑚.(3.36) Denote by 𝑟𝑖(𝑡)the known function 𝑟𝑖(𝑡)𝑖(𝑡)𝑘𝑖(𝑡)𝑚𝑖=1𝑖(𝑡)𝑘𝑖(𝑡)̇𝑐𝑖(𝑡),𝑑𝑖(𝑡)𝑛𝑗=𝑚+1𝑗(𝑡)𝑘𝑗(𝑡)̇𝑐𝑗(𝑡),𝑑𝑖+𝐻(𝑡)(0)(𝑡),𝑑𝑖,(𝑡)(3.37) and write the conditions (3.13) for (𝑃𝑦,𝑑𝑖(𝑡)).Taking into account expression (3.27) for 𝛾𝑖𝑞𝑖(𝑡), we get 𝑠𝑖1𝑞𝑖=0𝛾0𝑖𝑞𝑖𝐷𝑙𝑖𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝑒𝑃𝑖𝑖𝑞(𝑥)𝑡=𝑡𝑖+𝑠𝑖1𝑞𝑖=0𝐷𝑙𝑖𝑓𝑖𝑞𝑖(𝑡)𝑡=𝑡𝑖=𝐷𝑙𝑖𝑟𝑖(𝑡)𝑡=𝑡𝑖,𝑖=1,𝑚,𝑙𝑖=0,𝑠𝑖1.(3.38) Using the Leibnitz formula, we obtain that 𝐷𝑙𝑖𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝑒𝑃𝑖𝑖𝑞(𝑡)𝑡=𝑡𝑖=𝑙𝑖𝜈=0𝐶𝜈𝑙𝑖𝑡𝑡𝑖𝑞𝑖𝑞𝑖!(𝜈)𝑒𝑃𝑖𝑖𝑞(𝑡)(𝑙𝑖𝜈)𝑡=𝑡𝑖=𝑞𝑖𝜈=0𝐶𝜈𝑙𝑖𝑡𝑡𝑖𝑞𝑖𝑞𝑖!(𝜈)𝑒𝑃𝑖𝑖𝑞(𝑡)(𝑙𝑖𝜈)𝑡=𝑡𝑖=𝐶𝑞𝑖𝑙𝑖𝑒𝑃𝑖𝑖𝑞(𝑡)(𝑙𝑖𝑞𝑖)𝑡=𝑡𝑖,(3.39) for 𝑙𝑖𝑞𝑖, 𝐷𝑙𝑖𝑡𝑡𝑖𝑞𝑖𝑞𝑖!𝑒𝑃𝑖𝑖𝑞(𝑡)𝑡=𝑡𝑖=0,(3.40) for 0𝑙𝑖𝑞𝑖.
Therefore, previous equalities are written in the form of 𝑠𝑖1𝑞𝑖=0𝛾0𝑖𝑞𝑖𝐶𝑞𝑖𝑙𝑖𝑒𝑃𝑖𝑖𝑞(𝑡)(𝑙𝑖𝑞𝑖)𝑡=𝑡𝑖=𝑟0𝑖𝑙𝑖𝑖=1,𝑚,𝑙𝑖=0,𝑠𝑖1,(3.41) where 𝑟0𝑖𝑙𝑖=𝑠𝑖1𝑞𝑖=0𝐷𝑙𝑖𝑓𝑖𝑞𝑖(𝑡)𝑡=𝑡𝑖𝐷𝑙𝑖𝑟𝑖(𝑡)𝑡=𝑡𝑖,for𝑙𝑖=0,weget𝛾0𝑖0𝑒𝑃𝑖𝑖𝑞(𝑡𝑖)=𝑟0𝑖0;for𝑙𝑖=1,weget𝛾0𝑖0𝑐01𝑒𝑃𝑖𝑖𝑞(𝑡)𝑡=𝑡𝑖+𝛾0𝑖1𝑒𝑃𝑖𝑖𝑞(𝑡𝑖)=𝑟0𝑖1;for𝑙𝑖=𝑠𝑖1,weget𝛾0𝑖0𝑐0𝑠𝑖1𝑒𝑃𝑖𝑖𝑞(𝑡)𝑠𝑖1𝑡=𝑡𝑖++𝛾0𝑖𝑠𝑖1𝑒𝑃𝑖𝑖𝑞(𝑡𝑖)=𝑟0𝑖𝑠𝑖1.(3.42)
We obtain from here sequentially the numbers 𝛾0𝑖0,,𝛾0𝑖𝑠𝑖1.Theorem 3.6 is proved.

Thus, if conditions (3.24)–(3.26), (3.35) hold, all summands of solution (3.11) are defined uniquely.

So, if (0)(𝑡)𝑈(0),(1)(𝑡,𝜏,𝜎)𝑈(1),and conditions (3.6), (3.24)–(3.26), and (3.35) are valid, then the systems (3.4), (3.5) (and (3.3) together with them) are solvable uniquely in the class 𝑈=𝑈(1)𝑈(0).Two sequential problems (𝜀𝑘) and (𝜀𝑘+1) are connected uniquely by conditions (3.23)–(3.25), (3.30); therefore, by virtue of Theorems 3.13.6, they are solvable uniquely in the space 𝑈.

4. Asymptotical Character of Formal Solutions

Let 𝑦1(𝑡,𝜏,𝜎),,𝑦𝑘(𝑡,𝜏,𝜎) be solutions of formal problems (𝜀1), ,(𝜀𝑘) in the space 𝑈,respectively. Compose the partial sum for series (2.4):𝑆𝑛(𝑡,𝜏,𝜎)=𝑛𝑘=1𝜀𝑘𝑦𝑘(𝑡,𝜏,𝜎),(4.1) and take its restriction 𝑦𝜀𝑛(𝑡)=𝑆𝑛(𝑡,𝜑(𝑡)/𝜀,𝜓(𝑡,𝜀)).

We have the following result.

Theorem 4.1. Let conditions (i)–(v) hold. Then, for sufficiently small 𝜀(0𝜀𝜀0), the estimates 𝑦(𝑡,𝜏)𝑦𝜀𝑛(𝑡)𝐶[0,𝑇]𝐶𝑛𝜀𝑛+1,𝑛=1,0,1,,(4.2) hold. Here, 𝑦(𝑡,𝜀) is the exact solution of problem (1.1), and 𝑦𝜀𝑛(𝑡) is the states above restriction of the 𝑛th partial sum of series (2.4).

Proof. The restriction 𝑦𝜀𝑛(𝑡) of series (2.4) satisfies the initial condition 𝑦𝜀𝑛(0)=𝑦0 and system (1.1) up to terms containing 𝜀𝑛+1, that is, 𝜀𝑑𝑦𝜀𝑛(𝑡)𝑑𝑡=𝐴(𝑡)𝑦𝜀𝑛(𝑡)+𝜀𝑛+1𝑅𝑛(𝑡,𝜀)+(𝑡),(4.3) where 𝑅𝑛(𝑡,𝑠) is a known function satisfying the estimate (𝑅𝑡,𝜀)𝐶[0,𝑇]𝑅𝑛,𝑅𝑛---const.(4.4) Under conditions of Theorem 4.1 on the spectrum of the operator 𝐴(𝑡) for the fundamental matrix 𝑌(𝑡,𝑠,𝜀)𝑌(𝑡,𝜀)𝑌1(𝑡,𝜀) of the system 𝜀̇𝑌=𝐴(𝑡)𝑌, the estimate 𝑌(𝑡,𝑠,𝜀)const(𝑡,𝜀)𝑄{0𝑠𝑡𝑇},𝜀>00,𝜀0,(4.5) is valid. Here, 𝜀0>0is sufficiently small. Now, write the equation 𝜀𝑑Δ(𝑡,𝜀)𝑑𝑡=𝐴(𝑡)Δ(𝑡,𝜀)𝜀𝑛+1𝑅𝑛(𝑡,𝜀),Δ(0,𝜀)=0,(4.6) for the remainder term Δ(𝑡,𝜀)𝑦(𝑡,𝜀)𝑦𝜀𝑛(𝑡).We obtain from this equation that Δ(𝑡,𝜀)=𝜀𝑛𝑡0𝑌(𝑥,𝑠,𝜀)𝑅𝑛(𝑠,𝜀)𝑑𝑠,(4.7) whence we get the estimate (Δ𝑡,𝜀)𝐶[0,𝑇]𝜀𝑛𝑅𝑛,(4.8) where 𝑅𝑛=max(𝑡,𝑠)𝑄𝑌(𝑡,𝑠,𝜀)𝑅𝑛(𝑡,𝑠)𝑇. So, we obtain the estimate 𝑦(𝑡,𝜀)𝑦𝜀𝑛(𝑡)𝐶[0,𝑇]𝜀𝑛𝑅𝑛,𝑛=1,0,1,.(4.9) Taking instead of 𝑦𝜀𝑛(𝑡) the partial sum 𝑦𝜀,𝑛+1(𝑡)𝑦𝜀𝑛(𝑡)+𝜀𝑛+1𝑦𝑛+1𝑡,𝜑(𝑡)𝜀,𝜓(𝑡,𝜀),(4.10) we get 𝑦(𝑡,𝜀)𝑦𝜀𝑛(𝑡)𝜀𝑛+1𝑦𝑛+1𝑡,𝜑(𝑡)𝜀,𝜓(𝑡,𝜀)𝜀𝑛+1𝑅𝑛+1,(4.11) which implies the estimates (4.2). Theorem 4.1 is proved.

5. Example

Let it be required to construct the asymptotical solution for the Cauchy problem𝜀=̇𝑦̇𝑧5𝑡2+42𝑡2210𝑡2+104𝑡2𝑦𝑧+𝑡5210(𝑡),𝑦(0,𝜀)=𝑦0,𝑧(0,𝜀)=𝑧0,(5.1) where 1(𝑡)𝐶[0,2],𝜀>0is a small parameter. Eigenvalues of the matrix 𝐴(𝑡)=5𝑡2+42𝑡2210𝑡2+104𝑡25are 𝜆1(𝑡)=𝑡2,𝜆2(𝑡)=1.Eigenvectors of matrices 𝐴(𝑡)and 𝐴(𝑡), are, respectively,𝜑1=12,𝜑2=25,𝜓1=52,𝜓2=12.(5.2) We get ((𝑡),𝜓1(𝑡))5𝑡21(𝑡). Therefore,(0),𝜓1𝑑(0)=0,𝑑𝑡(0),𝜓1(0)=0.(5.3)

Hence, we can apply to problem (5.1) the above developed algorithm of the regularization method.

At first, obtain the basic Lagrange-Silvestre polynomials 𝐾𝑗𝑖(𝑡).Since 𝜓(𝑡)𝜆1(𝑡)=𝑡2,there will be two such polynomials: 𝐾00(𝑡)and 𝐾01(𝑡).

Take the arbitrary numbers 𝑎00(𝑡)and 𝑎01(𝑡),and set the interpolation conditions for the polynomial 𝑟(𝑡),𝑟(𝑡)=𝑎00,̇𝑟(1)=𝑎01.(5.4) Expand 𝑟(𝑡)onto partial fractions𝑟(𝑡)=𝐴𝜓(𝑡)𝑡2+𝐵𝑡,(5.5) from where𝑟(𝑡)𝐴+𝐵𝑡.(5.6) Use the interpolation polynomial (5.4). We get 𝐴=𝑎00,𝐵=𝑎01. Hence, (5.6) takes the form𝑟(𝑡)𝑎00+𝑡𝑎01.(5.7) Since numbers 𝑎00and 𝑎01 are arbitrary, basic Lagrange-Silvestre polynomials will be coefficients standing before them, that is,𝐾00(𝑡)1,𝐾01(𝑡)𝑡.(5.8)

Introduce the regularizing variables𝜎00=𝑒(1/𝜀)𝑡0𝜆1𝑑𝑠𝑡0𝑒(1/𝜀)𝑠0𝜆1𝑑𝑥𝐾00(𝑠)𝑑𝑠=𝑒𝑡3/3𝜀𝑡0𝑒𝑠3/3𝜀𝑑𝑠𝑝00(𝜎𝑡),01=𝑒(1/𝜀)𝑡0𝜆1𝑑𝑠𝑡0𝑒(1/𝜀)𝑠0𝜆1𝑑𝑥𝐾01(𝑠)𝑑𝑠=𝑒𝑡3/3𝜀𝑡0𝑒𝑠3/3𝜀𝑠𝑑𝑠𝑝01𝜏(𝑡),1=1𝜀𝑡0𝜆1𝑡𝑑𝑠=33𝜀𝑞1(𝑡),𝜏2=1𝜀𝑡0𝜆2𝑡𝑑𝑠=𝜀𝑞2(𝑡).(5.9)

Construct the extended problem corresponding to problem (5.1):𝜀𝜕𝑤𝜕𝑡+𝜆1(𝑡)𝜕𝑤𝜕𝜏1+𝜆2(𝑡)𝜕𝑤𝜕𝜏2+𝜆1(𝑡)𝜎00𝜕𝑤𝜕𝜎00+𝜆1(𝑡)𝜎01𝜕𝑤𝜕𝜎01+𝜀𝜕𝑤𝜕𝜎00+𝜀𝑡𝜕𝑤𝜕𝜎01𝐴(𝑡)𝑤=(𝑡),𝑤(0,0,0,𝜀)=𝑤0,(5.10) where 𝜏(𝜏1,𝜏2),𝜎=(𝜎00,𝜎01),𝑤=𝑤(𝑡,𝜏,𝜎,𝜀).

Determining solutions of problem (5.10) in the form of a series𝑤(𝑡,𝜏,𝜎,𝜀)=𝑘=0𝜀𝑘𝑤𝑘(𝑡,𝜏,𝜎),(5.11) we obtain the following iteration problems: 𝐿𝑤0𝜆1(𝑡)𝜕𝑤0+𝜕𝜏𝜕𝑤0𝜕𝜎00+𝑡𝜎01𝜕𝑤0𝜕𝜎01+𝜆2(𝑡)𝜕𝑤0𝜕𝜏2𝐴(𝑡)𝑤0=(𝑡),𝑤0(0,0,0)=𝑤0,(5.12)𝐿𝑤1=𝜕𝑤0𝜕𝑡𝜕𝑤0𝜕𝜎00𝑡𝜕𝑤0𝜕𝜎01,𝑤1(0,0,0)=0,(5.13)

We determine solutions of iteration problems (5.12), (5.13), and so on in the space 𝑈 of functions in the form of𝑤(𝑡,𝜏,𝜎)=𝑤1(𝑡)𝑒𝜏1+𝑤2(𝑡)𝑒𝜏2+𝑤00(𝑡)𝜎00+𝑤01(𝑡)𝜎01+𝑤0𝑤(𝑡),0(𝑡),𝑤1(𝑡),𝑤2(𝑡),𝑤00(𝑡),𝑤01(𝑡)𝐶[]0,2,𝐶2.(5.14)

Directly calculating, we obtain the solution of system (5.12) in the form of𝑤0(𝑡,𝜏,𝜎)=𝛼1(𝑡)𝜑1𝑒𝜏1+𝛼2(𝑡)𝜑2𝑒𝜏2+𝛾00(𝑡)𝜑1𝜎00+𝛾01(𝑡)𝜑1𝜎01+51(𝑡)𝜑12𝑡21(𝑡)𝜑2,(5.15) where 𝛼𝑗(𝑡),𝛾𝑗𝑖(𝑡)𝐶[0,2]are for now arbitrary functions.

To calculate the functions 𝛼𝑗(𝑡)and 𝛾𝑖𝑗(𝑡),we pass to the following iteration problem (5.13). Taking into account (5.15), it will be written in the form of𝐿𝑤1=̇𝛼1(𝑡)𝜑1𝑒𝜏1̇𝛼2(𝑡)𝜑2𝑒𝜏2̇𝛾00(𝑡)𝜑1𝜎00̇𝛾01(𝑡)𝜑1𝜎01̇51(𝑡)𝜑12𝑡21(𝑡)𝜑2𝛾00(𝑡)𝜑1𝑡𝛾01(𝑡)𝜑1.(5.16)

For solvability of problem (5.13) in the space 𝑈, it is necessary and sufficient to fulfill the conditionṡ𝛼1(𝑡)=0,̇𝛼2(𝑡)=0,̇𝛾00(𝑡)=0,̇𝛾01̇(𝑡)=0,51(0)𝛾00̈(0)=0,51(0)̇𝛾00(0)𝛾01(0)=0.(5.17)

Using solution (5.15) and the initial condition 𝑤0(0,0,0)=𝑤0,we obtain the equation𝛼1(0)𝜑1+𝛼2(0)𝜑2+51(0)𝜑1=𝑤0.(5.18)

Multiplying it (scalar) on 𝜓1and 𝜓2, we obtain the values𝛼1𝑤(0)=0,𝜓151(0)5𝑦02𝑧051𝛼(0),2𝑤(0)=0,𝜓2=𝑧02𝑦0.(5.19)

Using equalities (5.17), and also the initial data (5.19), we obtain uniquely the functions 𝛼𝑗(𝑡)and 𝛾𝑗𝑖(𝑡):𝛼1(𝑡)=5𝑦02𝑧051(0),𝛼2(𝑡)=𝑧02𝑦0.𝛾00̇(𝑡)=51(0),𝛾01̈(𝑡)=51(0).(5.20)

Substituting these functions into (5.15), we obtain uniquely the solution of problem (5.12) in the space 𝑈,𝑤0(𝑡,𝜏,𝜎)=5𝑦02𝑧051𝜑(0)1𝑒𝜏1+𝑧02𝑦0𝜑2𝑒𝜏2̇51(0)𝜑1𝜎00̈51(0)𝜑1𝜎01+51(𝑡)𝜑12𝑡21(𝑡)𝜑2.(5.21)

Producing here restriction on the functions 𝜏=𝑞(𝑡),𝜎=𝑝(𝑡),we obtain the principal term of the asymptotics for the solution of problem (5.1):𝑤0𝜀(𝑡)=5𝑦02𝑧051𝜑(0)1𝑒𝑡3/3𝜀+𝑧02𝑦0𝜑2𝑒𝑡/𝜀̇51(0)𝜑1𝑒𝑡3/3𝜀𝑡0𝑒𝑠3/3𝜀̈𝑑𝑠51(0)𝜑1𝑒𝑡3/3𝜀𝑡0𝑒𝑠3/3𝜀𝑠𝑑𝑠+51(𝑡)𝜑12𝑡21(𝑡)𝜑2.(5.22)

The zero-order asymptotical solution is obtained: it satisfies the estimate𝑤(𝑡,𝜀)𝑤0𝜀(𝑡)𝐶[0,2]𝐶1𝜀,(5.23) where 𝑤(𝑡,𝜀)is an exact solution of problem (1.1), and 𝐶1>0is a constant independent of 𝜀at sufficiently small 𝜀(0<𝜀𝜀0).