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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 120192, 16 pages
doi:10.1155/2012/120192
Research Article

Asymptotic Solutions of Singular Perturbed Problems with an Instable Spectrum of the Limiting Operator

Burkhan T. Kalimbetov, Marat A. Temirbekov, and Zhanibek O. Khabibullayev

Department of Mathematics, A. Yasawi International Kazakh-Turkish University, Turkestan 161200, Kazakhstan

Received 2 March 2012; Accepted 3 April 2012

Academic Editor: Allaberen Ashyralyev

Copyright © 2012 Burkhan T. Kalimbetov et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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., [1–19]).

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 form 𝜀 ̇ 𝑦 = 𝐴 ( 𝑡 ) 𝑦 + ℎ ( 𝑡 ) , 𝑦 ( 0 , 𝜀 ) = 𝑦 0 [ ] , , 𝑡 ∈ 0 , 𝑇 ( 1 . 1 ) where 𝑦 = { 𝑦 1 , … , 𝑦 𝑛 } , 𝐴 ( 𝑡 ) is a matrix of order ( 𝑛 × 𝑛 ) , ℎ ( 𝑡 ) = { ℎ 1 , … , ℎ 𝑛 } is a known function, 𝑦 0 ∈ 𝐶 𝑛 is a constant vector, and 𝜀 > 0 is a small parameter, in the case of violation of stability of a spectrum { 𝜆 𝑗 ( 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 𝜆 𝑖 ( 𝑡 ) ≠ 0 , 𝜆 𝑖 ( 𝑡 ) ≠ 𝜆 𝑗 ( 𝑡 ) , 𝑖 ≠ 𝑗 , 𝑖 , 𝑗 = 1 , 𝑛 f o r a l l 𝑡 ∈ [ 0 , 𝑇 ] ) is that the limiting system 0 = 𝐴 ( 𝑡 ) 𝑦 + ℎ ( 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 { 𝜆 𝑗 ( 𝑡 ) } of the limiting operator 𝐴 ( 𝑡 ) . As it was shown in [3–7], they were induced by instability points 𝑡 𝑗 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 ] , 𝐶 𝑛 ) , ℎ ( 𝑡 ) ∈ 𝐶 ∞ [ 0 , 𝑇 ] ; for any 𝑡 ∈ [ 0 , 𝑇 ] , the spectrum { 𝜆 𝑗 ( 𝑡 ) } of the operator 𝐴 ( 𝑡 ) satisfies the conditions:(ii) 𝜆 𝑖 ( 𝑡 ) = − ( 𝑡 − 𝑡 𝑖 ) 𝑠 𝑖 𝑘 𝑖 ( 𝑡 ) , 𝑘 𝑖 ( 𝑡 ) ≠ 0 , 𝑡 𝑖 ∈ [ 0 , 𝑇 ] , 𝑖 = 1 , 𝑚 , 𝑚 < 𝑛 (here 𝑠 𝑖 - are even natural numbers),(iii) 𝜆 𝑖 ( 𝑡 ) ≠ 0 , 𝑗 = 𝑚 + 1 , 𝑛 ,(iv) 𝜆 𝑖 ( 𝑡 ) ≠ 𝜆 𝑗 ( 𝑡 ) , 𝑖 ≠ 𝑗 , 𝑖 , 𝑗 = 1 , 𝑛 ,(v) R e 𝜆 𝑗 ( 𝑡 ) ⩽ 0 , 𝑗 = 1 , 𝑛 .

2. Regularization of the Problem

We introduce basic regularized variables by the spectrum of the limiting operator 𝜏 𝑗 = 𝜀 − 1 𝑡 0 𝜆 𝑗 𝜑 ( 𝑠 ) 𝑑 𝑠 ≡ 𝑗 ( 𝑡 ) 𝜀 , 𝑗 = 1 , 𝑛 . ( 2 . 1 ) Instable points 𝑡 𝑖 ∈ [ 0 , 𝑇 ] of the spectrum { 𝜆 𝑗 ( 𝑡 ) } induce additional regularized variables described by the formulas 𝜎 𝑖 𝑞 𝑖 = 𝑒 𝜑 𝑖 ( 𝑡 ) / 𝜀 𝑡 0 𝑒 − 𝜑 𝑖 ( 𝑠 ) / 𝜀 𝑠 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 𝑑 𝑠 ≡ 𝜓 𝑖 𝑞 𝑖 ( 𝑡 , 𝜀 ) , 𝑖 = 1 , 𝑚 , 𝑞 𝑖 = 0 , 𝑠 𝑖 − 1 . ( 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 that ̃ 𝑦 ( 𝑡 , 𝜏 , 𝜎 , 𝜀 ) ∣ 𝜏 = 𝜑 , 𝜎 = 𝜓 ≡ 𝑦 ( 𝑡 , 𝜀 ) . ( 2 . 3 )

For ̃ 𝑦 ( 𝑡 , 𝜏 , 𝜎 , 𝜀 ) , it is natural to set the following problem: 𝐿 𝜀 ̃ 𝑦 ( 𝑡 , 𝜏 , 𝜎 , 𝜀 ) ≡ 𝜀 𝜕 ̃ 𝑦 + 𝜕 𝑡 𝑛 𝑗 = 1 𝜆 𝑗 ( 𝑡 ) 𝜕 ̃ 𝑦 𝜕 𝜏 𝑗 + 𝑚 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝜆 𝑖 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 + 𝜀 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝜕 ̃ 𝑦 𝜕 𝜎 𝑖 𝑞 𝑖 − 𝐴 ( 𝑡 ) ̃ 𝑦 = ℎ ( 𝑡 ) , ̃ 𝑦 ( 0 , 0 , 0 , 𝜀 ) = 𝑦 0 . ( 2 . 4 ) We determine the solution of problem (2.4) in the form of a series ̃ 𝑦 ( 𝑡 , 𝜏 , 𝜎 , 𝜀 ) = ∞ 𝑘 = − 1 𝜀 𝑘 𝑦 𝑘 ( 𝑡 , 𝜏 , 𝜎 ) , ( 2 . 5 ) with coefficients 𝑦 𝑘 ( 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 𝑦 𝑘 ( 𝑡 , 𝜏 , 𝜎 ) : 𝐿 𝑦 − 1 ( 𝑡 , 𝜏 , 𝜎 ) ≡ 𝑛 𝑗 = 1 𝜆 𝑗 ( 𝑡 ) 𝜕 𝑦 − 1 𝜕 𝜏 𝑗 + 𝑚 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝜆 𝑖 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 𝜕 𝑦 − 1 𝜕 𝜎 𝑖 𝑞 𝑖 − 𝐴 ( 𝑡 ) 𝑦 − 1 = 0 , 𝑦 − 1 ( 0 , 0 , 0 ) = 0 , ( 𝜀 − 1 ) 𝐿 𝑦 0 ( 𝑡 , 𝜏 , 𝜎 ) = − 𝜕 𝑦 − 1 − 𝜕 𝑡 𝑚 ∑ 𝑠 𝑖 = 1 𝑖 − 1 ∑ 𝑞 𝑖 = 0 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝜕 𝑦 − 1 𝜕 𝜎 𝑖 𝑞 𝑖 + ℎ ( 𝑡 ) , 𝑦 0 ( 0 , 0 , 0 ) = 𝑦 0 , ⋮ ( 𝜀 0 ) 𝐿 𝑦 𝑘 + 1 ( 𝑡 , 𝜏 , 𝜎 ) = − 𝜕 𝑦 𝑘 − 𝜕 𝑡 𝑚 ∑ 𝑠 𝑖 = 1 𝑖 − 1 ∑ 𝑞 𝑖 = 0 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝜕 𝑦 𝑘 𝜕 𝜎 𝑖 𝑞 𝑖 , 𝑘 ⩾ 1 , 𝑦 𝑘 + 1 ⋮ ( 0 , 0 , 0 ) = 0 , ( 𝜀 𝑘 + 1 )

3. Resolvability of Iterative Problems

We solve each of the iterative problems ( 𝜀 𝑘 ) in the following space of functions: ⎧ ⎪ ⎨ ⎪ ⎩ 𝑈 = 𝑦 ( 𝑡 , 𝜏 , 𝜎 ) ∶ 𝑦 = 𝑛 𝑛 𝑘 = 1 𝑗 = 1 𝑦 𝑘 𝑗 ( 𝑡 ) 𝑐 𝑘 ( 𝑡 ) 𝑒 𝜏 𝑗 + 𝑛 𝑚 𝑘 = 1 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝑦 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑐 𝑘 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 + 𝑛 𝑘 = 1 𝑦 𝑘 ( 𝑡 ) 𝑐 𝑘 ( 𝑡 ) , 𝑦 𝑘 𝑗 ( 𝑡 ) , 𝑦 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) , 𝑦 𝑘 ( 𝑡 ) ∈ 𝐶 ∞ [ ] 0 , 𝑇 , 𝐶 1 , ( 3 . 1 ) where 𝑐 𝑘 ( 𝑡 ) are eigenvectors of the operator 𝐴 ( 𝑡 ) corresponding eigenvalues 𝜆 𝑘 ( 𝑡 ) , 𝑘 = 1 , 𝑛 . We represent 𝑈 in the form of 𝑈 ( 1 ) ⊕ 𝑈 ( 0 ) where 𝑈 ( 0 ) = 𝑦 ( 0 ) ( 𝑡 ) ∶ 𝑦 ( 0 ) = 𝑛 𝑗 = 1 𝑦 𝑗 ( 0 ) ( 𝑡 ) 𝑐 𝑗 ( 𝑡 ) , 𝑦 𝑗 ( 0 ) ( 𝑡 ) ∈ 𝐶 ∞ [ ] 0 , 𝑇 , 𝐶 1 , 𝑈 ( 1 ) = 𝑈 𝑈 ( 0 ) . ( 3 . 2 ) It is easy to note that each of the systems ( 𝜀 𝑘 + 1 ) can be written in the form 𝐿 𝑦 ( 𝑡 , 𝜏 , 𝜎 ) = ℎ ( 𝑡 , 𝜏 , 𝜎 ) , ( 3 . 3 ) where ℎ ( 𝑡 , 𝜏 , 𝜎 ) are the corresponding right hand side. Using representations of space 𝑈 , we can write system (3.3) in the equivalent form 𝐿 𝑦 ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) = ℎ ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) , ( 3 . 4 ) − 𝐴 ( 𝑡 ) 𝑦 ( 0 ) ( 𝑡 ) = ℎ ( 0 ) ( 𝑡 ) , ( 3 . 5 ) where 𝑦 ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) , ℎ ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) ∈ 𝑈 ( 1 ) , 𝑦 ( 0 ) ( 𝑡 ) , ℎ ( 0 ) ( 𝑡 ) ∈ 𝑈 ( 0 ) .

We have the following result.

Theorem 3.1. Let ℎ ( 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 ℎ ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) , 𝜈 𝑗 [ ] ( 𝑡 , 𝜏 , 𝜎 ) ≡ 0 ∀ 𝑡 ∈ 0 , 𝑇 , 𝑗 = ℎ 1 , 𝑛 , ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) , 𝜈 𝑖 𝑞 𝑖 ( 𝑡 , 𝜏 , 𝜎 ) ≡ 0 , 𝑖 = 1 , 𝑚 , 𝑞 𝑖 = 0 , 𝑠 𝑖 − 1 , ( 3 . 6 ) where 𝜈 𝑗 ( 𝑡 , 𝜏 , 𝜎 ) , 𝜈 𝑖 𝑞 𝑖 ( 𝑡 , 𝜏 , 𝜎 ) are basic elements of the kernel of the operator 𝐿 ∗ ≡ 𝑛 𝑗 = 1 𝜆 𝑗 𝜕 ( 𝑡 ) 𝜕 𝜏 𝑗 + 𝑚 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝜆 𝑖 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 𝜕 𝜕 𝜎 𝑖 𝑞 𝑖 − 𝐴 ∗ ( 𝑡 ) . ( 3 . 7 )

Proof. Let ℎ ( 1 ) ∑ ( 𝑡 , 𝜏 , 𝜎 ) = 𝑛 𝑘 = 1 ∑ 𝑛 𝑗 = 1 ℎ 𝑘 𝑗 ( 𝑡 ) 𝑐 𝑗 ( 𝑡 ) 𝑒 𝜏 𝑘 + ∑ 𝑛 𝑘 = 1 ∑ 𝑚 𝑖 = 1 ∑ 𝑠 𝑖 𝑞 − 1 𝑖 = 0 ℎ 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑐 𝑘 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 .
Determine solutions of system (3.4) in the form 𝑦 ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) = 𝑛 𝑛 𝑘 = 1 𝑗 = 1 𝑦 𝑘 𝑗 ( 𝑡 ) 𝑐 𝑘 ( 𝑡 ) 𝑒 𝜏 𝑗 + 𝑛 𝑚 𝑘 = 1 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝑦 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑐 𝑘 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 . ( 3 . 8 ) Substituting (3.8) in (3.4) and equating separately coefficients at 𝑒 𝜏 𝑗 and 𝜎 𝑖 𝑞 𝑖 , we obtain the equations 𝜆 𝑘 ( 𝑡 ) − 𝜆 𝑗 𝑦 ( 𝑡 ) 𝑘 𝑗 ( 𝑡 ) = ℎ 𝑘 𝑗 ( 𝑡 ) , 𝑘 , 𝑗 = 𝜆 1 , 𝑛 , 𝑖 ( 𝑡 ) − 𝜆 𝑘 𝑦 ( 𝑡 ) 𝑖 𝑞 𝑖 𝑘 ( 𝑡 ) = ℎ 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) , 𝑖 = 1 , 𝑚 , 𝑞 𝑖 = 0 , 𝑠 𝑖 − 1 , 𝑘 = 1 , 𝑛 . ( 3 . 9 )
One can see from this that obtained equations are solvable if and only if ℎ 𝑘 𝑘 ( 𝑡 ) ≡ 0 , 𝑘 = 1 , 𝑛 , ℎ 𝑖 𝑖 𝑞 𝑖 ( 𝑡 ) ≡ 0 , 𝑖 = 1 , 𝑚 , 𝑞 𝑖 = 0 , 𝑠 𝑖 − 1 , ( 3 . 1 0 ) 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 ) = 𝑛 𝑛 𝑘 = 1 𝑗 = 1 , 𝑗 ≠ 𝑘 ℎ 𝑘 𝑗 ( 𝑡 ) 𝜆 𝑘 ( 𝑡 ) − 𝜆 𝑗 𝑐 ( 𝑡 ) 𝑗 ( 𝑡 ) 𝑒 𝜏 𝑗 + 𝑛 𝑘 = 1 𝛼 𝑘 ( 𝑡 ) 𝑐 𝑘 ( 𝑡 ) 𝑒 𝜏 𝑘 + 𝑚 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑐 𝑖 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 + 𝑛 𝑚 𝑘 = 1 𝑠 𝑖 = 1 , 𝑖 ≠ 𝑘 𝑖 − 1 𝑞 𝑖 = 0 ℎ 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) 𝜆 𝑖 ( 𝑡 ) − 𝜆 𝑘 𝑐 ( 𝑡 ) 𝑘 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 , ( 3 . 1 1 ) where 𝛼 𝑘 ( 𝑡 ) , 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) ∈ 𝐶 ∞ ( [ 0 , 𝑇 ] , 𝐶 1 ) are arbitrary functions.
Consider now system (3.5). As d e t 𝐴 ( 𝑡 ) ≡ 0 in points 𝑡 = 𝑡 𝑖 , 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 ) = 𝑛 𝑗 = 1 𝑧 𝑗 ( 𝑡 ) 𝑐 𝑗 ( 𝑡 ) , 𝑧 𝑗 ( 𝑡 ) ∈ 𝐶 ∞ [ ] 0 , 𝑇 , 𝐶 1 , 𝑗 = 1 , 𝑛 , ( 3 . 1 2 ) having the properties 𝐷 𝑙 𝑖 𝑧 ( 0 ) ( 𝑡 ) , 𝑑 𝑖 ( 𝑡 ) 𝑡 = 𝑡 𝑖 = 𝐷 𝑙 𝑖 𝑧 𝑖 𝑡 𝑖 = 0 , ∀ 𝑙 𝑖 = 0 , 𝑠 𝑖 − 1 , 𝑖 = 1 , 𝑚 , ( 3 . 1 3 ) where 𝑑 𝑖 ( 𝑡 ) are eigenvectors of the operator 𝐴 ∗ ( 𝑡 ) with regard to eigenvalues 𝜆 𝑖 ( 𝑡 ) , 𝑖 = 1 , 𝑚 . Let ℎ ( 0 ) ∑ ( 𝑡 ) = 𝑛 𝑗 = 1 ℎ 𝑗 ( 𝑡 ) 𝑐 𝑗 ( 𝑡 ) ∈ 𝑉 ( 0 ) , that is, 𝐷 𝑙 𝑖 ℎ 𝑖 𝑡 𝑖 = 0 ∀ 𝑙 𝑖 = 0 , 𝑠 𝑖 − 1 , 𝑖 = 1 , 𝑚 . ( 3 . 1 4 ) Determine a solution of system (3.5) in the 𝑦 ( 0 ) ( 𝑡 ) = 𝑛 𝑗 = 1 𝑦 𝑗 ( 𝑡 ) 𝑐 𝑗 ( 𝑡 ) . ( 3 . 1 5 ) Substituting this function in (3.5), we obtain − 𝑛 𝑗 = 1 𝑦 𝑗 ( 𝑡 ) 𝜆 𝑗 ( 𝑡 ) 𝑐 𝑗 ( 𝑡 ) = 𝑛 𝑗 = 1 ℎ 𝑗 ( 𝑡 ) 𝑐 𝑗 ( 𝑡 ) . ( 3 . 1 6 ) Since { 𝑐 𝑗 ( 𝑡 ) } is a basis in 𝐶 𝑛 , we get − 𝜆 𝑖 ( 𝑡 ) 𝑦 𝑖 ( 𝑡 ) = ℎ 𝑖 ( 𝑡 ) , 𝑖 = 1 , 𝑚 , ( 3 . 1 7 ) − 𝜆 𝑗 ( 𝑡 ) 𝑦 𝑗 ( 𝑡 ) = ℎ 𝑗 ( 𝑡 ) , 𝑗 = 𝑚 + 1 , 𝑛 . ( 3 . 1 8 ) It is easy to see that (3.18) has the unique solution 𝑦 𝑗 ( 𝑡 ) = − ℎ 𝑗 ( 𝑡 ) 𝜆 𝑗 ( 𝑡 ) , 𝑗 = 𝑚 + 1 , 𝑛 . ( 3 . 1 9 ) By virtue of conditions (3.14), the function ℎ 𝑖 ( 𝑡 ) can be represented in the form ℎ 𝑖 ( 𝑡 ) = 𝑡 − 𝑡 𝑖 𝑠 𝑖 ℎ 𝑖 ( 𝑡 ) , 𝑖 = 1 , 𝑚 , ( 3 . 2 0 ) where ℎ 𝑖 ( 𝑡 ) ∈ 𝐶 ∞ ( [ 0 , 𝑇 ] , 𝐶 1 is the certain scalar function, − ( 𝑡 − 𝑡 𝑖 ) 𝑠 𝑖 𝑘 𝑖 ( 𝑡 ) 𝑦 𝑖 ( 𝑡 ) = ( 𝑡 − 𝑡 𝑖 ) 𝑠 𝑖 ℎ 𝑖 ( 𝑡 ) , and we see that 𝑦 𝑖 ⎧ ⎪ ⎨ ⎪ ⎩ − ℎ ( 𝑡 ) = 𝑖 ( 𝑡 ) 𝑘 𝑖 ( 𝑡 ) , 𝑡 ≠ 𝑡 𝑖 , 𝛾 𝑖 , 𝑡 = 𝑡 𝑖 , ( 3 . 2 1 ) where 𝛾 𝑖 are arbitrary constants, 𝑖 = 1 , 𝑚 . However, the solution of system (3.5) should belong to the space 𝑈 ( 0 ) , and it means that 𝑦 𝑖 ( 𝑡 ) ∈ 𝐶 ∞ ( [ 0 , 𝑇 ] , 𝐶 1 ) . Therefore, constants in (3.21) 𝛾 𝑖 ℎ = ( 𝑖 ( 𝑡 ) / 𝑘 𝑖 ( 𝑡 ) ) ∣ 𝑡 = 𝑡 𝑖 and functions are determined uniquely in the form 𝑦 𝑖 − ℎ ( 𝑡 ) = 𝑖 ( 𝑡 ) 𝑘 𝑖 [ ] ( 𝑡 ) , ∀ 𝑡 ∈ 0 , 𝑇 , 𝑖 = 1 , 𝑚 . ( 3 . 2 2 ) Thus, under conditions (3.14), system (3.5) has the solution 𝑦 ( 0 ) ( 𝑡 ) in 𝑈 ( 0 ) of 𝑦 ( 0 ) ( 𝑡 ) = − 𝑚 𝑖 = 1 ℎ 𝑖 ( 𝑡 ) 𝑘 𝑖 𝑐 ( 𝑡 ) 𝑖 ( 𝑡 ) − 𝑛 𝑗 = 𝑚 + 1 ℎ 𝑖 ( 𝑡 ) 𝜆 𝑖 𝑐 ( 𝑡 ) 𝑖 ( 𝑡 ) , ( 3 . 2 3 ) where ℎ 𝑖 ℎ ( 𝑡 ) = 𝑖 ( 𝑡 ) / ( 𝑡 − 𝑡 𝑖 ) 𝑠 𝑖 (in points 𝑡 = 𝑡 𝑖 , 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 ℎ ( 0 ) ( 𝑡 ) ∈ 𝑉 ( 0 ) , system (3.5) has the unique solution 𝑦 ( 0 ) ( 𝑡 ) in space 𝑈 ( 0 ) .

For uniquely determination of functions 𝛼 𝑗 ( 𝑡 ) , 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) , consider system (3.4) with additional conditions: 𝑦 ( 1 ) ( 0 , 0 , 0 ) = 𝑦 ∗ , − ( 3 . 2 4 ) 𝜕 𝑦 ( 1 ) 𝜕 𝑡 , 𝜈 𝑗 [ ] ( 𝑡 , 𝜏 , 𝜎 ) ≡ 0 ∀ 𝑡 ∈ 0 , 𝑇 , 𝑗 = − 1 , 𝑛 , ( 3 . 2 5 ) 𝜕 𝑦 ( 1 ) 𝜕 𝑡 , 𝜈 𝑖 𝑞 𝑖 ( 𝑡 , 𝜏 , 𝜎 ) ≡ 0 , 𝑖 = 1 , 𝑚 , 𝑞 𝑖 = 0 , 𝑠 𝑖 − 1 , ( 3 . 2 6 ) where 𝑦 ∗ ∈ 𝐶 𝑛 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 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑐 𝑖 ( 𝑡 ) 𝜎 𝑖 𝑞 𝑖 ( 𝑖 = 1 , 𝑚 , 𝑞 𝑖 = 0 , 𝑠 𝑖 − 1 ) . Functions 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) in the last summand are determined by the formula 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) = 𝛾 0 𝑖 𝑞 𝑖 ⋅ 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) + 𝑓 𝑖 𝑞 𝑖 ( 𝑡 ) , ( 3 . 2 7 ) where 𝑃 𝑖 𝑞 𝑖 ( 𝑡 ) , 𝑓 𝑖 𝑞 𝑖 ( 𝑡 ) are known functions, and 𝛾 0 𝑖 𝑞 𝑖 arbitrary constants.

Proof. Denote in (3.11) that 𝑔 𝑘 𝑗 ℎ ( 𝑡 ) = 𝑘 𝑗 ( 𝑡 ) 𝜆 𝑗 ( 𝑡 ) − 𝜆 𝑘 ( 𝑡 ) , 𝑔 𝑘 𝑖 𝑞 𝑖 ℎ ( 𝑡 ) = 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) 𝜆 𝑖 ( 𝑡 ) − 𝜆 𝑘 . ( 𝑡 ) ( 3 . 2 8 ) Using (3.11) and condition (3.24), we obtain the equality 𝑛 𝑛 𝑘 = 1 𝑗 = 1 𝑔 𝑘 𝑗 ( 0 ) 𝑐 𝑗 ( 0 ) + 𝑛 𝑘 = 1 𝛼 𝑘 ( 0 ) 𝑐 𝑘 ( 0 ) = 𝑦 ∗ . ( 3 . 2 9 ) Multiplying this equality scalarly by 𝑑 𝑠 ( 0 ) , we get 𝛼 𝑠 𝑦 ( 0 ) = ∗ , 𝑑 𝑠 − ( 0 ) 𝑛 𝑘 = 1 , 𝑘 ≠ 𝑠 𝑔 𝑘 𝑠 ( 0 ) ≡ 𝛼 0 𝑠 , 𝑠 = 1 , 𝑛 . ( 3 . 3 0 ) By (3.11) and conditions (3.25), we have − ̇ 𝛼 𝑠 ( 𝑡 ) − ̇ 𝑐 𝑠 ( 𝑡 ) , 𝑑 𝑠 𝛼 ( 𝑡 ) 𝑠 ( 𝑡 ) − 𝑛 𝑗 = 1 , 𝑗 ≠ 𝑠 𝑔 𝑠 𝑗 ( 𝑡 ) ̇ 𝑐 𝑗 ( 𝑡 ) , 𝑑 𝑠 ( 𝑡 ) = 0 , 𝑠 = 1 , 𝑛 . ( 3 . 3 1 ) Considering these equations with initial conditions (3.30), we can uniquely obtain functions 𝛼 𝑠 ( 𝑡 ) , 𝑠 = 1 , 𝑛 .
Now, using (3.11) and conditions (3.26), we get − ̇ 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) − ̇ 𝑐 𝑖 ( 𝑡 ) , 𝑑 𝑖 𝛾 ( 𝑡 ) 𝑖 𝑞 𝑖 ( 𝑡 ) − 𝑛 𝑘 = 1 , 𝑘 ≠ 𝑖 𝑔 𝑘 𝑖 𝑞 𝑖 ( 𝑡 ) ̇ 𝑐 𝑘 ( 𝑡 ) , 𝑑 𝑖 ( 𝑡 ) = 0 , 𝑖 = 1 , 𝑚 , 𝑞 𝑖 = 0 , 𝑠 𝑖 − 1 . ( 3 . 3 2 ) This implies that 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) have the form (3.27) where 𝑃 𝑖 𝑞 𝑖 ( 𝑡 ) = − 𝑡 𝑡 𝑖 ̇ 𝑐 𝑖 ( 𝑠 ) , 𝑑 𝑖 ( 𝑓 𝑠 ) 𝑑 𝑠 , 𝑖 𝑞 𝑖 ( 𝑡 ) = 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) 𝑡 𝑡 𝑖 𝑒 − 𝑃 𝑖 𝑖 𝑞 𝑛 ( 𝑠 ) 𝑘 = 1 , 𝑘 ≠ 𝑖 𝑔 𝑘 𝑖 𝑞 𝑖 ( 𝑠 ) ̇ 𝑐 𝑘 ( 𝑠 ) , 𝑑 𝑖 ( 𝑠 ) 𝑑 𝑠 . ( 3 . 3 3 ) Theorem 3.4 is proved.

Remark 3.5. If conditions (3.6) hold for ℎ ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) ∈ 𝑈 ( 1 ) and ℎ ( 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 . 3 4 ) 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 𝛼 𝑘 ( 𝑡 ) ∈ 𝐶 ∞ ( [ 0 , 𝑇 ] , 𝐶 1 ) are found uniquely in (3.11), and functions 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) are determined up to arbitrary constants 𝛾 0 𝑖 𝑞 𝑖 in the form of (3.27).

Let us give the following result.

Theorem 3.6. Let ℎ ( 0 ) ( 𝑡 ) ∈ 𝑈 ( 0 ) , ℎ ( 1 ) ( 𝑡 , 𝜏 , 𝜎 ) ∈ 𝑈 ( 1 ) , and conditions (i)–(iv), (3.6), (3.24)–(3.26) hold. Then, there exist unique numbers 𝛾 0 𝑖 𝑞 𝑖 involved in (3.27), such that the function (3.34) satisfies the condition 𝑃 𝑦 ≡ − 𝜕 𝑦 ( 0 ) − 𝜕 𝑡 𝑚 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝜕 𝑦 ( 1 ) 𝜕 𝜎 𝑖 𝑞 𝑖 + 𝐻 ( 0 ) ( 𝑡 ) ∈ 𝑉 ( 0 ) , ( 3 . 3 5 ) where 𝐻 ( 0 ) ( 𝑡 ) ∈ 𝑉 ( 0 ) is a fixed vector function.

Proof. To determine functions uniquely, calculate 𝑃 𝑦 ≡ − 𝑚 𝑖 = 1 ℎ 𝑖 ( 𝑡 ) 𝑘 𝑖 𝑐 ( 𝑡 ) 𝑖 ( 𝑡 ) − 𝑛 𝑗 = 𝑚 + 1 ℎ 𝑗 ( 𝑡 ) 𝜆 𝑗 𝑐 ( 𝑡 ) 𝑗 ( 𝑡 ) − 𝑚 𝑠 𝑖 = 1 𝑖 − 1 𝑞 𝑖 = 0 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑐 𝑖 + ( 𝑡 ) 𝑛 𝑚 𝑘 = 1 𝑠 𝑖 = 1 𝑖 ≠ 𝑘 𝑖 − 1 𝑞 𝑖 = 0 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! ⋅ ℎ 𝑖 𝑞 𝑖 ( 𝑡 ) 𝜆 𝑖 ( 𝑡 ) − 𝜆 𝑘 𝑐 ( 𝑡 ) 𝑘 ( 𝑡 ) + 𝐻 ( 0 ) ( 𝑡 ) , 𝑃 𝑦 , 𝑑 𝑖 ℎ ( 𝑡 ) ≡ − 𝑖 ( 𝑡 ) 𝑘 𝑖 ( 𝑡 ) − 𝑚 𝑖 = 1 ℎ 𝑖 ( 𝑡 ) 𝑘 𝑖 ( 𝑡 ) ̇ 𝑐 𝑖 ( 𝑡 ) , 𝑑 𝑖 − ( 𝑡 ) 𝑛 𝑗 = 𝑚 + 1 ℎ 𝑗 ( 𝑡 ) 𝑘 𝑗 ( 𝑡 ) ̇ 𝑐 𝑗 ( 𝑡 ) , 𝑑 𝑖 − ( 𝑡 ) 𝑠 𝑖 − 1 𝑞 𝑖 𝑚 = 0 𝑖 = 1 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝛾 𝑖 𝑞 𝑖 𝐻 ( 𝑡 ) + ( 0 ) ( 𝑡 ) , 𝑑 𝑖 ( 𝑡 ) , 𝑖 = 1 , 𝑚 . ( 3 . 3 6 ) Denote by 𝑟 𝑖 ( 𝑡 ) the known function 𝑟 𝑖 ℎ ( 𝑡 ) ≡ − 𝑖 ( 𝑡 ) 𝑘 𝑖 ( 𝑡 ) − 𝑚 𝑖 = 1 ℎ 𝑖 ( 𝑡 ) 𝑘 𝑖 ( 𝑡 ) ̇ 𝑐 𝑖 ( 𝑡 ) , 𝑑 𝑖 − ( 𝑡 ) 𝑛 𝑗 = 𝑚 + 1 ℎ 𝑗 ( 𝑡 ) 𝑘 𝑗 ( 𝑡 ) ̇ 𝑐 𝑗 ( 𝑡 ) , 𝑑 𝑖 + 𝐻 ( 𝑡 ) ( 0 ) ( 𝑡 ) , 𝑑 𝑖 , ( 𝑡 ) ( 3 . 3 7 ) and write the conditions (3.13) for ( 𝑃 𝑦 , 𝑑 𝑖 ( 𝑡 ) ) . Taking into account expression (3.27) for 𝛾 𝑖 𝑞 𝑖 ( 𝑡 ) , we get 𝑠 𝑖 − 1 𝑞 𝑖 = 0 𝛾 0 𝑖 𝑞 𝑖 𝐷 𝑙 𝑖 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑥 ) 𝑡 = 𝑡 𝑖 + 𝑠 𝑖 − 1 𝑞 𝑖 = 0 𝐷 𝑙 𝑖 𝑓 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑡 = 𝑡 𝑖 = 𝐷 𝑙 𝑖 𝑟 𝑖 ( 𝑡 ) 𝑡 = 𝑡 𝑖 , 𝑖 = 1 , 𝑚 , 𝑙 𝑖 = 0 , 𝑠 𝑖 − 1 . ( 3 . 3 8 ) Using the Leibnitz formula, we obtain that 𝐷 𝑙 𝑖 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) 𝑡 = 𝑡 𝑖 = ⎡ ⎢ ⎢ ⎣ 𝑙 𝑖 𝜈 = 0 𝐶 𝜈 𝑙 𝑖 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! ( 𝜈 ) 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) ( 𝑙 𝑖 − 𝜈 ) ⎤ ⎥ ⎥ ⎦ 𝑡 = 𝑡 𝑖 = ⎡ ⎢ ⎢ ⎣ 𝑞 𝑖 𝜈 = 0 𝐶 𝜈 𝑙 𝑖 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! ( 𝜈 ) 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) ( 𝑙 𝑖 − 𝜈 ) ⎤ ⎥ ⎥ ⎦ 𝑡 = 𝑡 𝑖 = 𝐶 𝑞 𝑖 𝑙 𝑖 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) ( 𝑙 𝑖 − 𝑞 𝑖 ) 𝑡 = 𝑡 𝑖 , ( 3 . 3 9 ) for 𝑙 𝑖 ≥ 𝑞 𝑖 , 𝐷 𝑙 𝑖 𝑡 − 𝑡 𝑖 𝑞 𝑖 𝑞 𝑖 ! 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) 𝑡 = 𝑡 𝑖 = 0 , ( 3 . 4 0 ) for 0 ≤ 𝑙 𝑖 ≤ 𝑞 𝑖 .
Therefore, previous equalities are written in the form of 𝑠 𝑖 − 1 𝑞 𝑖 = 0 𝛾 0 𝑖 𝑞 𝑖 𝐶 𝑞 𝑖 𝑙 𝑖 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) ( 𝑙 𝑖 − 𝑞 𝑖 ) 𝑡 = 𝑡 𝑖 = 𝑟 0 𝑖 𝑙 𝑖 𝑖 = 1 , 𝑚 , 𝑙 𝑖 = 0 , 𝑠 𝑖 − 1 , ( 3 . 4 1 ) where 𝑟 0 𝑖 𝑙 𝑖 = − 𝑠 𝑖 − 1 𝑞 𝑖 = 0 𝐷 𝑙 𝑖 𝑓 𝑖 𝑞 𝑖 ( 𝑡 ) 𝑡 = 𝑡 𝑖 − 𝐷 𝑙 𝑖 𝑟 𝑖 ( 𝑡 ) 𝑡 = 𝑡 𝑖 , f o r 𝑙 𝑖 = 0 , w e g e t 𝛾 0 𝑖 0 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 𝑖 ) = 𝑟 0 𝑖 0 ; f o r 𝑙 𝑖 = 1 , w e g e t 𝛾 0 𝑖 0 𝑐 0 1 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) 𝑡 = 𝑡 𝑖 + 𝛾 0 𝑖 1 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 𝑖 ) = 𝑟 0 𝑖 1 ; ⋮ f o r 𝑙 𝑖 = 𝑠 𝑖 − 1 , w e g e t 𝛾 0 𝑖 0 𝑐 0 𝑠 𝑖 − 1 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 ) 𝑠 𝑖 − 1 𝑡 = 𝑡 𝑖 + ⋯ + 𝛾 0 𝑖 𝑠 𝑖 − 1 𝑒 𝑃 𝑖 𝑖 𝑞 ( 𝑡 𝑖 ) = 𝑟 0 𝑖 𝑠 𝑖 − 1 . ( 3 . 4 2 )
We obtain from here sequentially the numbers 𝛾 0 𝑖 0 , … , 𝛾 0 𝑖 𝑠 𝑖 − 1 . 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 ℎ ( 0 ) ( 𝑡 ) ∈ 𝑈 ( 0 ) , ℎ ( 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 ( 𝜀 𝑘 ) and ( 𝜀 𝑘 + 1 ) are connected uniquely by conditions (3.23)–(3.25), (3.30); therefore, by virtue of Theorems 3.1–3.6, they are solvable uniquely in the space 𝑈 .

4. Asymptotical Character of Formal Solutions

Let 𝑦 − 1 ( 𝑡 , 𝜏 , 𝜎 ) , … , 𝑦 𝑘 ( 𝑡 , 𝜏 , 𝜎 ) be solutions of formal problems ( 𝜀 − 1 ), … , ( 𝜀 𝑘 ) in the space 𝑈 , respectively. Compose the partial sum for series (2.4): 𝑆 𝑛 ( 𝑡 , 𝜏 , 𝜎 ) = 𝑛 𝑘 = − 1 𝜀 𝑘 𝑦 𝑘 ( 𝑡 , 𝜏 , 𝜎 ) , ( 4 . 1 ) and take its restriction 𝑦 𝜀 𝑛 ( 𝑡 ) = 𝑆 𝑛 ( 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 ( 0 ≤ 𝜀 ≤ 𝜀 0 ) , the estimates ‖ ‖ 𝑦 ( 𝑡 , 𝜏 ) − 𝑦 𝜀 𝑛 ‖ ‖ ( 𝑡 ) 𝐶 [ 0 , 𝑇 ] ≤ 𝐶 𝑛 𝜀 𝑛 + 1 , 𝑛 = − 1 , 0 , 1 , … , ( 4 . 2 ) hold. Here, 𝑦 ( 𝑡 , 𝜀 ) is the exact solution of problem (1.1), and 𝑦 𝜀 𝑛 ( 𝑡 ) is the states above restriction of the 𝑛 th partial sum of series (2.4).

Proof. The restriction 𝑦 𝜀 𝑛 ( 𝑡 ) of series (2.4) satisfies the initial condition 𝑦 𝜀 𝑛 ( 0 ) = 𝑦 0 and system (1.1) up to terms containing 𝜀 𝑛 + 1 , that is, 𝜀 𝑑 𝑦 𝜀 𝑛 ( 𝑡 ) 𝑑 𝑡 = 𝐴 ( 𝑡 ) 𝑦 𝜀 𝑛 ( 𝑡 ) + 𝜀 𝑛 + 1 𝑅 𝑛 ( 𝑡 , 𝜀 ) + ℎ ( 𝑡 ) , ( 4 . 3 ) where 𝑅 𝑛 ( 𝑡 , 𝑠 ) is a known function satisfying the estimate ( ‖ 𝑅 𝑡 , 𝜀 ) ‖ 𝐶 [ 0 , 𝑇 ] ≤ 𝑅 𝑛 , 𝑅 𝑛 - - - c o n s t . ( 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 ) 𝑌 − 1 ( 𝑡 , 𝜀 ) of the system 𝜀 ̇ 𝑌 = 𝐴 ( 𝑡 ) 𝑌 , the estimate ‖ 𝑌 ( 𝑡 , 𝑠 , 𝜀 ) ‖ ≤ c o n s t ∀ ( 𝑡 , 𝜀 ) ∈ 𝑄 ≡ { 0 ≤ 𝑠 ≤ 𝑡 ≤ 𝑇 } , ∀ 𝜀 > 0 ∈ 0 , 𝜀 0 , ( 4 . 5 ) is valid. Here, 𝜀 0 > 0 − is sufficiently small. Now, write the equation 𝜀 𝑑 Δ ( 𝑡 , 𝜀 ) 𝑑 𝑡 = 𝐴 ( 𝑡 ) Δ ( 𝑡 , 𝜀 ) − 𝜀 𝑛 + 1 𝑅 𝑛 ( 𝑡 , 𝜀 ) , Δ ( 0 , 𝜀 ) = 0 , ( 4 . 6 ) for the remainder term Δ ( 𝑡 , 𝜀 ) ≡ 𝑦 ( 𝑡 , 𝜀 ) − 𝑦 𝜀 𝑛 ( 𝑡 ) . We obtain from this equation that Δ ( 𝑡 , 𝜀 ) = − 𝜀 𝑛 𝑡 0 𝑌 ( 𝑥 , 𝑠 , 𝜀 ) 𝑅 𝑛 ( 𝑠 , 𝜀 ) 𝑑 𝑠 , ( 4 . 7 ) whence we get the estimate ( ‖ Δ 𝑡 , 𝜀 ) ‖ 𝐶 [ 0 , 𝑇 ] ≤ − 𝜀 𝑛 𝑅 𝑛 , ( 4 . 8 ) where 𝑅 𝑛 = m a x ( 𝑡 , 𝑠 ) ∈ 𝑄 ‖ 𝑌 ( 𝑡 , 𝑠 , 𝜀 ) ‖ ⋅ ‖ 𝑅 𝑛 ( 𝑡 , 𝑠 ) ‖ ⋅ 𝑇 . So, we obtain the estimate ‖ ‖ 𝑦 ( 𝑡 , 𝜀 ) − 𝑦 𝜀 𝑛 ( ‖ ‖ 𝑡 ) 𝐶 [ 0 , 𝑇 ] ≤ 𝜀 𝑛 𝑅 𝑛 , 𝑛 = − 1 , 0 , 1 , … . ( 4 . 9 ) Taking instead of 𝑦 𝜀 𝑛 ( 𝑡 ) the partial sum 𝑦 𝜀 , 𝑛 + 1 ( 𝑡 ) ≡ 𝑦 𝜀 𝑛 ( 𝑡 ) + 𝜀 𝑛 + 1 𝑦 𝑛 + 1 𝑡 , 𝜑 ( 𝑡 ) 𝜀 , 𝜓 ( 𝑡 , 𝜀 ) , ( 4 . 1 0 ) we get ‖ ‖ ‖ 𝑦 ( 𝑡 , 𝜀 ) − 𝑦 𝜀 𝑛 ( 𝑡 ) − 𝜀 𝑛 + 1 𝑦 𝑛 + 1 𝑡 , 𝜑 ( 𝑡 ) 𝜀 ‖ ‖ ‖ , 𝜓 ( 𝑡 , 𝜀 ) ≤ 𝜀 𝑛 + 1 𝑅 𝑛 + 1 , ( 4 . 1 1 ) 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 problem 𝜀 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ ̇ 𝑦 ̇ 𝑧 − 5 𝑡 2 + 4 2 𝑡 2 − 2 − 1 0 𝑡 2 + 1 0 4 𝑡 2 ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ 𝑦 𝑧 ⎞ ⎟ ⎟ ⎠ + ⎛ ⎜ ⎜ ⎝ 𝑡 − 5 2 ℎ 1 0 ⎞ ⎟ ⎟ ⎠ ( 𝑡 ) , 𝑦 ( 0 , 𝜀 ) = 𝑦 0 , 𝑧 ( 0 , 𝜀 ) = 𝑧 0 , ( 5 . 1 ) where ℎ 1 ( 𝑡 ) ∈ 𝐶 ∞ [ 0 , 2 ] , 𝜀 > 0 is a small parameter. Eigenvalues of the matrix 𝐴 ( 𝑡 ) = − 5 𝑡 2 + 4 2 𝑡 2 − 2 − 1 0 𝑡 2 + 1 0 4 𝑡 2 − 5 are 𝜆 1 ( 𝑡 ) = − 𝑡 2 , 𝜆 2 ( 𝑡 ) = − 1 . Eigenvectors of matrices 𝐴 ( 𝑡 ) and 𝐴 ∗ ( 𝑡 ) , are, respectively, 𝜑 1 = ⎛ ⎜ ⎜ ⎝ 1 2 ⎞ ⎟ ⎟ ⎠ , 𝜑 2 = ⎛ ⎜ ⎜ ⎝ 2 5 ⎞ ⎟ ⎟ ⎠ , 𝜓 1 = ⎛ ⎜ ⎜ ⎝ 5 ⎞ ⎟ ⎟ ⎠ − 2 , 𝜓 2 = ⎛ ⎜ ⎜ ⎝ 1 ⎞ ⎟ ⎟ ⎠ − 2 . ( 5 . 2 ) We get ( ℎ ( 𝑡 ) , 𝜓 1 ( 𝑡 ) ) ≡ 5 𝑡 2 ℎ 1 ( 𝑡 ) . Therefore, ℎ ( 0 ) , 𝜓 1 𝑑 ( 0 ) = 0 , 𝑑 𝑡 ℎ ( 0 ) , 𝜓 1 ( 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 𝐾 𝑗 𝑖 ( 𝑡 ) . Since 𝜓 ( 𝑡 ) ≡ 𝜆 1 ( 𝑡 ) = − 𝑡 2 , there will be two such polynomials: 𝐾 0 0 ( 𝑡 ) and 𝐾 0 1 ( 𝑡 ) .

Take the arbitrary numbers 𝑎 0 0 ( 𝑡 ) and 𝑎 0 1 ( 𝑡 ) , and set the interpolation conditions for the polynomial 𝑟 ( 𝑡 ) , 𝑟 ( 𝑡 ) = 𝑎 0 0 , ̇ 𝑟 ( 1 ) = 𝑎 0 1 . ( 5 . 4 ) Expand 𝑟 ( 𝑡 ) onto partial fractions 𝑟 ( 𝑡 ) = 𝐴 𝜓 ( 𝑡 ) 𝑡 2 + 𝐵 𝑡 , ( 5 . 5 ) from where 𝑟 ( 𝑡 ) ≡ 𝐴 + 𝐵 𝑡 . ( 5 . 6 ) Use the interpolation polynomial (5.4). We get 𝐴 = 𝑎 0 0 , 𝐵 = 𝑎 0 1 . Hence, (5.6) takes the form 𝑟 ( 𝑡 ) ≡ 𝑎 0 0 + 𝑡 𝑎 0 1 . ( 5 . 7 ) Since numbers 𝑎 0 0 and 𝑎 0 1 are arbitrary, basic Lagrange-Silvestre polynomials will be coefficients standing before them, that is, 𝐾 0 0 ( 𝑡 ) ≡ 1 , 𝐾 0 1 ( 𝑡 ) ≡ 𝑡 . ( 5 . 8 )

Introduce the regularizing variables 𝜎 0 0 = 𝑒 ∫ ( 1 / 𝜀 ) 𝑡 0 𝜆 1 𝑑 𝑠 𝑡 0 𝑒 ∫ − ( 1 / 𝜀 ) 𝑠 0 𝜆 1 𝑑 𝑥 ⋅ 𝐾 0 0 ( 𝑠 ) 𝑑 𝑠 = 𝑒 − 𝑡 3 / 3 𝜀 𝑡 0 𝑒 𝑠 3 / 3 𝜀 𝑑 𝑠 ≡ 𝑝 0 0 ( 𝜎 𝑡 ) , 0 1 = 𝑒 ∫ ( 1 / 𝜀 ) 𝑡 0 𝜆 1 𝑑 𝑠 𝑡 0 𝑒 ∫ − ( 1 / 𝜀 ) 𝑠 0 𝜆 1 𝑑 𝑥 ⋅ 𝐾 0 1 ( 𝑠 ) 𝑑 𝑠 = 𝑒 − 𝑡 3 / 3 𝜀 𝑡 0 𝑒 𝑠 3 / 3 𝜀 ⋅ 𝑠 𝑑 𝑠 ≡ 𝑝 0 1 𝜏 ( 𝑡 ) , 1 = 1 𝜀 𝑡 0 𝜆 1 𝑡 𝑑 𝑠 = − 3 3 𝜀 ≡ 𝑞 1 ( 𝑡 ) , 𝜏 2 = 1 𝜀 𝑡 0 𝜆 2 𝑡 𝑑 𝑠 = − 𝜀 ≡ 𝑞 2 ( 𝑡 ) . ( 5 . 9 )

Construct the extended problem corresponding to problem (5.1): 𝜀 𝜕 𝑤 𝜕 𝑡 + 𝜆 1 ( 𝑡 ) 𝜕 𝑤 𝜕 𝜏 1 + 𝜆 2 ( 𝑡 ) 𝜕 𝑤 𝜕 𝜏 2 + 𝜆 1 ( 𝑡 ) 𝜎 0 0 𝜕 𝑤 𝜕 𝜎 0 0 + 𝜆 1 ( 𝑡 ) 𝜎 0 1 𝜕 𝑤 𝜕 𝜎 0 1 + 𝜀 𝜕 𝑤 𝜕 𝜎 0 0 + 𝜀 𝑡 𝜕 𝑤 𝜕 𝜎 0 1 − 𝐴 ( 𝑡 ) 𝑤 = ℎ ( 𝑡 ) , 𝑤 ( 0 , 0 , 0 , 𝜀 ) = 𝑤 0 , ( 5 . 1 0 ) where 𝜏 ≡ ( 𝜏 1 , 𝜏 2 ) , 𝜎 = ( 𝜎 0 0 , 𝜎 0 1 ) , 𝑤 = 𝑤 ( 𝑡 , 𝜏 , 𝜎 , 𝜀 ) .

Determining solutions of problem (5.10) in the form of a series 𝑤 ( 𝑡 , 𝜏 , 𝜎 , 𝜀 ) = ∞ 𝑘 = 0 𝜀 𝑘 𝑤 𝑘 ( 𝑡 , 𝜏 , 𝜎 ) , ( 5 . 1 1 ) we obtain the following iteration problems: 𝐿 𝑤 0 ≡ 𝜆 1 ( 𝑡 ) 𝜕 𝑤 0 + 𝜕 𝜏 𝜕 𝑤 0 𝜕 𝜎 0 0 + 𝑡 ⋅ 𝜎 0 1 𝜕 𝑤 0 𝜕 𝜎 0 1 + 𝜆 2 ( 𝑡 ) 𝜕 𝑤 0 𝜕 𝜏 2 − 𝐴 ( 𝑡 ) 𝑤 0 = ℎ ( 𝑡 ) , 𝑤 0 ( 0 , 0 , 0 ) = 𝑤 0 , ( 5 . 1 2 ) 𝐿 𝑤 1 = − 𝜕 𝑤 0 − 𝜕 𝑡 𝜕 𝑤 0 𝜕 𝜎 0 0 − 𝑡 𝜕 𝑤 0 𝜕 𝜎 0 1 , 𝑤 1 ⋮ ( 0 , 0 , 0 ) = 0 , ( 5 . 1 3 )

We determine solutions of iteration problems (5.12), (5.13), and so on in the space 𝑈 of functions in the form of 𝑤 ( 𝑡 , 𝜏 , 𝜎 ) = 𝑤 1 ( 𝑡 ) 𝑒 𝜏 1 + 𝑤 2 ( 𝑡 ) 𝑒 𝜏 2 + 𝑤 0 0 ( 𝑡 ) 𝜎 0 0 + 𝑤 0 1 ( 𝑡 ) 𝜎 0 1 + 𝑤 0 𝑤 ( 𝑡 ) , 0 ( 𝑡 ) , 𝑤 1 ( 𝑡 ) , 𝑤 2 ( 𝑡 ) , 𝑤 0 0 ( 𝑡 ) , 𝑤 0 1 ( 𝑡 ) ∈ 𝐶 ∞ [ ] 0 , 2 , 𝐶 2 . ( 5 . 1 4 )

Directly calculating, we obtain the solution of system (5.12) in the form of 𝑤 0 ( 𝑡 , 𝜏 , 𝜎 ) = 𝛼 1 ( 𝑡 ) 𝜑 1 𝑒 𝜏 1 + 𝛼 2 ( 𝑡 ) 𝜑 2 𝑒 𝜏 2 + 𝛾 0 0 ( 𝑡 ) 𝜑 1 𝜎 0 0 + 𝛾 0 1 ( 𝑡 ) 𝜑 1 𝜎 0 1 + 5 ℎ 1 ( 𝑡 ) 𝜑 1 − 2 𝑡 2 ℎ 1 ( 𝑡 ) 𝜑 2 , ( 5 . 1 5 ) where 𝛼 𝑗 ( 𝑡 ) , 𝛾 𝑗 𝑖 ( 𝑡 ) ∈ 𝐶 ∞ [ 0 , 2 ] are for now arbitrary functions.

To calculate the functions 𝛼 𝑗 ( 𝑡 ) and 𝛾 𝑖 𝑗 ( 𝑡 ) , we pass to the following iteration problem (5.13). Taking into account (5.15), it will be written in the form of 𝐿 𝑤 1 = − ̇ 𝛼 1 ( 𝑡 ) 𝜑 1 𝑒 𝜏 1 − ̇ 𝛼 2 ( 𝑡 ) 𝜑 2 𝑒 𝜏 2 − ̇ 𝛾 0 0 ( 𝑡 ) 𝜑 1 𝜎 0 0 − ̇ 𝛾 0 1 ( 𝑡 ) 𝜑 1 𝜎 0 1 ̇ ℎ − 5 1 ( 𝑡 ) 𝜑 1 − 2 𝑡 2 ℎ 1 ( 𝑡 ) 𝜑 2 − 𝛾 0 0 ( 𝑡 ) 𝜑 1 − 𝑡 𝛾 0 1 ( 𝑡 ) 𝜑 1 . ( 5 . 1 6 )

For solvability of problem (5.13) in the space 𝑈 , it is necessary and sufficient to fulfill the conditions − ̇ 𝛼 1 ( 𝑡 ) = 0 , − ̇ 𝛼 2 ( 𝑡 ) = 0 , − ̇ 𝛾 0 0 ( 𝑡 ) = 0 , − ̇ 𝛾 0 1 ̇ ℎ ( 𝑡 ) = 0 , − 5 1 ( 0 ) − 𝛾 0 0 ̈ ℎ ( 0 ) = 0 , − 5 1 ( 0 ) − ̇ 𝛾 0 0 ( 0 ) − 𝛾 0 1 ( 0 ) = 0 . ( 5 . 1 7 )

Using solution (5.15) and the initial condition 𝑤 0 ( 0 , 0 , 0 ) = 𝑤 0 , we obtain the equation 𝛼 1 ( 0 ) 𝜑 1 + 𝛼 2 ( 0 ) 𝜑 2 + 5 ℎ 1 ( 0 ) 𝜑 1 = 𝑤 0 . ( 5 . 1 8 )

Multiplying it (scalar) on 𝜓 1 and 𝜓 2 , we obtain the values 𝛼 1 𝑤 ( 0 ) = 0 , 𝜓 1 − 5 ℎ 1 ( 0 ) ≡ 5 𝑦 0 − 2 𝑧 0 − 5 ℎ 1 𝛼 ( 0 ) , 2 𝑤 ( 0 ) = 0 , 𝜓 2 = 𝑧 0 − 2 𝑦 0 . ( 5 . 1 9 )

Using equalities (5.17), and also the initial data (5.19), we obtain uniquely the functions 𝛼 𝑗 ( 𝑡 ) and 𝛾 𝑗 𝑖 ( 𝑡 ) : 𝛼 1 ( 𝑡 ) = 5 𝑦 0 − 2 𝑧 0 − 5 ℎ 1 ( 0 ) , 𝛼 2 ( 𝑡 ) = 𝑧 0 − 2 𝑦 0 . 𝛾 0 0 ̇ ℎ ( 𝑡 ) = − 5 1 ( 0 ) , 𝛾 0 1 ̈ ℎ ( 𝑡 ) = − 5 1 ( 0 ) . ( 5 . 2 0 )

Substituting these functions into (5.15), we obtain uniquely the solution of problem (5.12) in the space 𝑈 , 𝑤 0 ( 𝑡 , 𝜏 , 𝜎 ) = 5 𝑦 0 − 2 𝑧 0 − 5 ℎ 1 𝜑 ( 0 ) 1 𝑒 𝜏 1 + 𝑧 0 − 2 𝑦 0 𝜑 2 𝑒 𝜏 2 ̇ ℎ − 5 1 ( 0 ) 𝜑 1 𝜎 0 0 ̈ ℎ − 5 1 ( 0 ) 𝜑 1 𝜎 0 1 + 5 ℎ 1 ( 𝑡 ) 𝜑 1 − 2 𝑡 2 ℎ 1 ( 𝑡 ) 𝜑 2 . ( 5 . 2 1 )

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): 𝑤 0 𝜀 ( 𝑡 ) = 5 𝑦 0 − 2 𝑧 0 − 5 ℎ 1 𝜑 ( 0 ) 1 𝑒 − 𝑡 3 / 3 𝜀 + 𝑧 0 − 2 𝑦 0 𝜑 2 𝑒 − 𝑡 / 𝜀 ̇ ℎ − 5 1 ( 0 ) 𝜑 1 𝑒 − 𝑡 3 / 3 𝜀 𝑡 0 𝑒 𝑠 3 / 3 𝜀 ̈ ℎ 𝑑 𝑠 − 5 1 ( 0 ) 𝜑 1 𝑒 − 𝑡 3 / 3 𝜀 𝑡 0 𝑒 𝑠 3 / 3 𝜀 𝑠 𝑑 𝑠 + 5 ℎ 1 ( 𝑡 ) 𝜑 1 − 2 𝑡 2 ℎ 1 ( 𝑡 ) 𝜑 2 . ( 5 . 2 2 )

The zero-order asymptotical solution is obtained: it satisfies the estimate ‖ ‖ 𝑤 ( 𝑡 , 𝜀 ) − 𝑤 0 𝜀 ‖ ‖ ( 𝑡 ) 𝐶 [ 0 , 2 ] ≤ 𝐶 1 ⋅ 𝜀 , ( 5 . 2 3 ) where 𝑤 ( 𝑡 , 𝜀 ) is an exact solution of problem (1.1), and 𝐶 1 > 0 is a constant independent of 𝜀 at sufficiently small 𝜀 ( 0 < 𝜀 ≤ 𝜀 0 ) .

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