About this Journal Submit a Manuscript Table of Contents 
Abstract and Applied Analysis
Volume 2012 (2012), Article ID 846582, 29 pages
doi:10.1155/2012/846582
Research Article

A Note on the Second Order of Accuracy Stable Difference Schemes for the Nonlocal Boundary Value Hyperbolic Problem

Allaberen Ashyralyev1 and Ozgur Yildirim2

1Department of Mathematics, Fatih University, Buyukcekmece 34500, Istanbul, Turkey
2Department of Mathematics, Yildiz Technical University, Esenler 34210, Istanbul, Turkey

Received 23 March 2012; Accepted 11 June 2012

Academic Editor: Sergey Piskarev

Copyright © 2012 Allaberen Ashyralyev and Ozgur Yildirim. 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 second order of accuracy absolutely stable difference schemes are presented for the nonlocal boundary value hyperbolic problem for the differential equations in a Hilbert space H with the self-adjoint positive definite operator A. The stability estimates for the solutions of these difference schemes are established. In practice, one-dimensional hyperbolic equation with nonlocal boundary conditions and multidimensional hyperbolic equation with Dirichlet conditions are considered. The stability estimates for the solutions of these difference schemes for the nonlocal boundary value hyperbolic problem are established. Finally, a numerical method proposed and numerical experiments, analysis of the errors, and related execution times are presented in order to verify theoretical statements.

1. Introduction

Hyperbolic partial differential equations play an important role in many branches of science and engineering and can be used to describe a wide variety of phenomena such as acoustics, electromagnetics, hydrodynamics, elasticity, fluid mechanics, and other areas of physics (see [1–5] and the references given therein).

While applying mathematical modelling to several phenomena of physics, biology, and ecology, there often arise problems with nonclassical boundary conditions, which the values of unknown function on the boundary are connected with inside of the given domain. Such type of boundary conditions are called nonlocal boundary conditions. Over the last decades, boundary value problems with nonlocal boundary conditions have become a rapidly growing area of research (see, e.g., [6–16] and the references given therein).

In the present work, we consider the nonlocal boundary value problem 𝑑 2 𝑢 ( 𝑡 ) 𝑑 𝑡 2 𝑢 + 𝐴 𝑢 ( 𝑡 ) = 𝑓 ( 𝑡 ) ( 0 ≤ 𝑡 ≤ 1 ) , ( 0 ) = 𝑛 𝑗 = 1 𝛼 𝑗 𝑢 𝜆 𝑗 + 𝜑 , 𝑢 𝑡 ( 0 ) = 𝑛 𝑗 = 1 𝛽 𝑗 𝑢 𝑡 𝜆 𝑗 + 𝜓 , 0 < 𝜆 1 < 𝜆 2 < ⋯ < 𝜆 𝑛 ≤ 1 , ( 1 . 1 ) where 𝐴 is a self-adjoint positive definite operator in a Hilbert space 𝐻 .

A function 𝑢 ( 𝑡 ) is called a solution of the problem (1.1), if the following conditions are satisfied:(i) 𝑢 ( 𝑡 ) is twice continuously differentiable on the segment [ 0 , 1 ] . The derivatives at the endpoints of the segment are understood as the appropriate unilateral derivatives.(ii) The element 𝑢 ( 𝑡 ) belongs to 𝐷 ( 𝐴 ) , independent of 𝑡 , and dense in 𝐻 for all 𝑡 ∈ [ 0 , 1 ] and the function 𝐴 𝑢 ( 𝑡 ) is continuous on the segment [ 0 , 1 ] .(iii) 𝑢 ( 𝑡 ) satisfies the equation and nonlocal boundary conditions (1.1).

In the paper of [8], the following theorem on the stability estimates for the solution of the nonlocal boundary value problem (1.1) was proved.

Theorem 1.1. Suppose that 𝜑 ∈ 𝐷 ( 𝐴 ) , 𝜓 ∈ 𝐷 ( 𝐴 1 / 2 ) , and 𝑓 ( 𝑡 ) is a continuously differentiable function on [ 0 , 1 ] and the assumption 𝑛 𝑘 = 1 | | 𝛼 𝑘 + 𝛽 𝑘 | | + 𝑛 𝑚 = 1 | | 𝛼 𝑚 | | 𝑛 𝑘 = 1 𝑘 ≠ 𝑚 | | 𝛽 𝑘 | | < | | | | | 1 + 𝑛 𝑘 = 1 𝛼 𝑘 𝛽 𝑘 | | | | | ( 1 . 2 ) holds. Then, there is a unique solution of problem (1.1) and the stability inequalities m a x 0 ≤ 𝑡 ≤ 1 ‖ 𝑢 ( 𝑡 ) ‖ 𝐻 ≤ 𝑀 ‖ 𝜑 ‖ 𝐻 + ‖ ‖ 𝐴 − 1 / 2 𝜓 ‖ ‖ 𝐻 + m a x 0 ≤ 𝑡 ≤ 1 ‖ ‖ 𝐴 − 1 / 2 ‖ ‖ 𝑓 ( 𝑡 ) 𝐻 , m a x 0 ≤ 𝑡 ≤ 1 ‖ ‖ 𝐴 1 / 2 ‖ ‖ 𝑢 ( 𝑡 ) 𝐻 ‖ ‖ 𝐴 ≤ 𝑀 1 / 2 𝜑 ‖ ‖ 𝐻 + ‖ 𝜓 ‖ 𝐻 + m a x 0 ≤ 𝑡 ≤ 1 ‖ 𝑓 ( 𝑡 ) ‖ 𝐻 , m a x 0 ≤ 𝑡 ≤ 1 ‖ ‖ ‖ 𝑑 2 𝑢 ( 𝑡 ) 𝑑 𝑡 2 ‖ ‖ ‖ 𝐻 + m a x 0 ≤ 𝑡 ≤ 1 ‖ 𝐴 𝑢 ( 𝑡 ) ‖ 𝐻 ≤ 𝑀 ‖ 𝐴 𝜑 ‖ 𝐻 + ‖ ‖ 𝐴 1 / 2 𝜓 ‖ ‖ 𝐻 + ‖ 𝑓 ( 0 ) ‖ 𝐻 + 1 0 ‖ ‖ 𝑓 ‖ ‖ ( 𝑡 ) 𝐻 𝑑 𝑡 ( 1 . 3 ) hold, where 𝑀 does not depend on 𝜑 ,   𝜓 , and 𝑓 ( 𝑡 ) ,   𝑡 ∈ [ 0 , 1 ] .

Moreover, the first order of accuracy difference scheme 𝜏 − 2 𝑢 𝑘 + 1 − 2 𝑢 𝑘 + 𝑢 𝑘 − 1 + 𝐴 𝑢 𝑘 + 1 = 𝑓 𝑘 , 𝑓 𝑘 𝑡 = 𝑓 𝑘 , 𝑡 𝑘 𝑢 = 𝑘 𝜏 , 1 ≤ 𝑘 ≤ 𝑁 − 1 , 𝑁 𝜏 = 1 , 0 = 𝑛 𝑟 = 1 𝛼 𝑟 𝑢 [ 𝜆 𝑟 / 𝜏 ] 𝜏 + 𝜑 , − 1 𝑢 1 − 𝑢 0 = 𝑛 𝑟 = 1 𝛽 𝑟 𝑢 [ 𝜆 𝑟 / 𝜏 ] + 1 − 𝑢 [ 𝜆 𝑟 / 𝜏 ] 1 𝜏 + 𝜓 , ( 1 . 4 ) for the approximate solution of problem (1.1) was presented. The stability estimates for the solution of this difference scheme, under the assumption 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | + 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | + 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | < 1 , ( 1 . 5 ) were established.

In the development of numerical techniques for solving PDEs, the stability has been an important research topic (see [6–31]). A large cycle of works on difference schemes for hyperbolic partial differential equations, in which stability was established under the assumption that the magnitude of the grid steps 𝜏 and ℎ with respect to the time and space variables, are connected. In abstract terms, this particularly means that 𝜏 ‖ 𝐴 ℎ ‖ → 0 when 𝜏 → 0 .

We are interested in studying the high order of accuracy difference schemes for hyperbolic PDEs, in which stability is established without any assumption with respect to the grid steps 𝜏 and ℎ . Particularly, a convenient model for analyzing the stability is provided by a proper unconditionally absolutely stable difference scheme with an unbounded operator.

In the present paper, the second order of accuracy unconditionally stable difference schemes for approximately solving boundary value problem (1.1) is presented. The stability estimates for the solutions of these difference schemes and their first and second order difference derivatives are established. This operator approach permits one to obtain the stability estimates for the solutions of difference schemes of nonlocal boundary value problems, for one-dimensional hyperbolic equation with nonlocal boundary conditions in space variable and multidimensional hyperbolic equation with Dirichlet condition in space variables.

Some results of this paper without proof were presented in [7].

Note that nonlocal boundary value problems for parabolic equations, elliptic equations, and equations of mixed types have been studied extensively by many scientists (see, e.g., [11–16, 20–24, 32–38] and the references therein).

2. The Second Order of Accuracy Difference Scheme Generated by 𝐴 2

Throughout this paper for simplicity 𝜆 1 > 2 𝜏 and 𝜆 𝑛 < 1 will be considered. Let us associate boundary value problem (1.1) with the second order of accuracy difference scheme 𝜏 − 2 𝑢 𝑘 + 1 − 2 𝑢 𝑘 + 𝑢 𝑘 − 1 + 𝐴 𝑢 𝑘 + 𝜏 2 4 𝐴 2 𝑢 𝑘 + 1 = 𝑓 𝑘 , 𝑓 𝑘 𝑡 = 𝑓 𝑘 , 𝑡 𝑘 𝑢 = 𝑘 𝜏 , 1 ≤ 𝑘 ≤ 𝑁 − 1 , 𝑁 𝜏 = 1 , 0 = 𝑛 𝑚 = 1 𝛼 𝑚 𝑢 [ 𝜆 𝑚 / 𝜏 ] + 𝜏 − 1 𝑢 [ 𝜆 𝑚 / 𝜏 ] − 𝑢 [ 𝜆 𝑚 / 𝜏 ] − 1 × 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 + 𝜑 , 𝐼 + 2 𝐴 2 𝜏 − 1 𝑢 1 − 𝑢 0 − 𝜏 2 𝑓 0 − 𝐴 𝑢 0 = 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 − 1 𝑢 [ 𝜆 𝑘 / 𝜏 ] − 𝑢 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 × 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 𝐴 𝑢 [ 𝜆 𝑘 / 𝜏 ] 𝑓 + 𝜓 , 0 = 𝑓 ( 0 ) . ( 2 . 1 ) A study of discretization, over time only, of the nonlocal boundary value problem also permits one to include general difference schemes in applications, if the differential operator in space variables 𝐴 is replaced by the difference operator 𝐴 ℎ that act in the Hilbert space and are uniformly self-adjoint positive definite in ℎ for 0 < ℎ ≤ ℎ 0 .

In general, we have not been able to obtain the stability estimates for the solution of difference scheme (2.1) under assumption (1.5). Note that the stability of solution of difference scheme (2.1) will be obtained under the strong assumption 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | | | | | 𝜆 1 + 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | + 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | | | | | 𝜆 1 + 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | + 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | | | | | 𝜆 1 + 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | + 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | | | | | 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | | | | | 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | < 1 . ( 2 . 2 ) Now, let us give some lemmas that will be needed in the sequel.

Lemma 2.1. The following estimates hold: ‖ 𝑅 ‖ 𝐻 ↦ 𝐻 ‖ ‖ 𝑅 ‖ ‖ ≤ 1 , 𝐻 ↦ 𝐻 ‖ ‖ 𝑅 ≤ 1 , − 1 𝑅 ‖ ‖ 𝐻 ↦ 𝐻 ‖ ‖ 𝑅 ≤ 1 , − 1 𝑅 ‖ ‖ 𝐻 ↦ 𝐻 ‖ ‖ ≤ 1 , 𝜏 𝐴 1 / 2 𝑅 ‖ ‖ 𝐻 ↦ 𝐻 ‖ ‖ ≤ 1 , 𝜏 𝐴 1 / 2 𝑅 ‖ ‖ 𝐻 ↦ 𝐻 ≤ 1 . ( 2 . 3 ) Here, 𝐻 is the Hilbert space, 𝑅 = ( 𝐼 + 𝑖 𝜏 𝐴 1 / 2 − ( 𝜏 2 / 2 ) 𝐴 ) − 1 , and 𝑅 = ( 𝐼 − 𝑖 𝜏 𝐴 1 / 2 − ( 𝜏 2 / 2 ) 𝐴 ) − 1 .

Lemma 2.2. Suppose that assumption (2.2) holds. Denote 𝐵 𝜏 = 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 ⎡ ⎢ ⎢ ⎣ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 𝐴 1 / 2 2 ⎤ ⎥ ⎥ ⎦ + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 𝐼 + 2 𝐴 2 − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 − 1 2 𝑛 𝑛 𝑚 = 1 𝑘 = 1 𝛼 𝑚 𝛽 𝑘 𝜏 𝐼 + 2 𝐴 2 − 1 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝑖 𝐴 1 / 2 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎡ ⎢ ⎢ ⎣ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 + 𝑖 𝜏 𝐴 1 / 2 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 − 𝑖 𝜏 𝐴 1 / 2 2 ⎤ ⎥ ⎥ ⎦ + 𝑛 𝑘 = 1 𝛽 𝑘 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝑖 𝐴 1 / 2 𝜏 𝐼 + 2 𝐴 2 − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 + 1 4 𝑛 𝑛 𝑚 = 1 𝑘 = 1 𝛼 𝑚 𝛽 𝑘 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 2 × 𝑖 𝐴 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 2 𝑛 𝑛 𝑚 = 1 𝑘 = 1 𝛼 𝑚 𝛽 𝑘 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝑖 𝐴 1 / 2 𝜏 𝐼 + 2 𝐴 2 − 1 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 − 1 2 𝑛 𝑛 𝑚 = 1 𝑘 = 1 𝛼 𝑚 𝛽 𝑘 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝐴 𝜏 𝐼 + 2 𝐴 2 − 1 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 . ( 2 . 4 ) Then, the operator 𝐼 − 𝐵 𝜏 has an inverse 𝑇 𝜏 = 𝐼 − 𝐵 𝜏 − 1 , ( 2 . 5 ) and the following estimate holds: ‖ ‖ 𝑇 𝜏 ‖ ‖ 𝐻 ↦ 𝐻 ≤ 𝑀 . ( 2 . 6 )

Proof. The proof of estimate (2.6) is based on the estimate ‖ ‖ 𝐼 − 𝐵 𝜏 ‖ ‖ 𝐻 ↦ 𝐻 ≥ 1 − 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | | | | | 𝜆 1 + 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | − 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | | | | | 𝜆 1 + 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | − 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | | | | | 𝜆 1 + 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | − 𝑛 𝑘 = 1 | | 𝛼 𝑘 | | | | | | 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | 𝑛 𝑘 = 1 | | 𝛽 𝑘 | | | | | | 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 | | | | . ( 2 . 7 ) Estimate (2.7) follows from the triangle inequality and estimate (2.3). Lemma 2.2 is proved.
Now, we will obtain the formula for the solution of problem (2.1). It is easy to show that (see, e.g., [18]) there is unique solution of the problem 𝜏 − 2 𝑢 𝑘 + 1 − 2 𝑢 𝑘 + 𝑢 𝑘 − 1 + 𝐴 𝑢 𝑘 + 𝜏 2 4 𝐴 2 𝑢 𝑘 + 1 = 𝑓 𝑘 , 𝑓 𝑘 𝑡 = 𝑓 𝑘 , 𝑡 𝑘 𝜏 = 𝑘 𝜏 , 1 ≤ 𝑘 ≤ 𝑁 − 1 , 𝑁 𝜏 = 1 , 𝐼 + 2 𝐴 2 𝜏 − 1 𝑢 1 − 𝑢 0 − 𝜏 2 𝑓 0 − 𝐴 𝑢 0 = 𝜔 , 𝑓 0 = 𝑓 ( 0 ) , 𝑢 0 = 𝜇 , ( 2 . 8 ) and the following formula holds: 𝑢 0 = 𝜇 , 𝑢 1 = 𝜏 𝐼 + 2 𝐴 2 − 1 𝜏 𝜇 + 𝜏 𝜔 + 2 2 𝑓 0 , 𝑢 𝑘 = 𝜏 𝐼 + 2 𝐴 2 − 1 ⎛ ⎜ ⎜ ⎝ 𝑅 𝑘 − 1 𝐼 − 𝑖 𝜏 𝐴 1 / 2 + 𝑅 𝑘 − 1 𝐼 + 𝑖 𝜏 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ 𝜇 + 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 𝑅 𝑘 − 1 𝑅 − 1 − 𝑅 𝑘 − 1 𝑅 − 1 2 𝜏 𝜔 + 2 𝑓 0 − 𝑘 − 1 𝑠 = 1 𝜏 𝐴 2 𝑖 − 1 / 2 𝑅 𝑘 − 𝑠 − 𝑅 𝑘 − 𝑠 𝑓 𝑠 , 2 ≤ 𝑘 ≤ 𝑁 . ( 2 . 9 ) Applying formula (2.9) and the nonlocal boundary conditions in problem (2.1), we get 𝜇 = 𝑇 𝜏 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 2 𝑓 0 × 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 2 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ − ⎛ ⎜ ⎜ ⎝ 𝑛 𝑚 = 1 𝛼 𝑚 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑠 = 1 𝜏 𝐴 2 𝑖 − 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − s − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − s 𝑓 𝑠 − 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 × [ 𝜆 𝑚 / 𝜏 ] − 2 𝑠 = 1 𝜏 𝐴 2 𝑖 1 / 2 𝑖 𝐴 − 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − s 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − s 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 𝑓 𝑠 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝑅 𝑅 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 1 × − 𝜑 𝐼 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝐴 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 × 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 − 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎦ + 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 2 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎦ × 𝜏 2 𝑓 0 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ − 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝜏 𝐴 × 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 × 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 + 𝑛 𝑘 = 1 𝛽 𝑘 [ 𝜆 𝑘 / 𝜏 ] − 2 𝑠 = 1 𝜏 2 𝑖 × 𝐴 − 1 / 2 𝑖 𝐴 1 / 2 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 𝑓 𝑠 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 𝑅 𝑅 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝐴 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑠 = 1 𝜏 𝐴 2 𝑖 − 1 / 2 × 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝑓 𝑠 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 , + 𝜓 𝜔 = 𝑇 𝜏 𝐼 − 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 𝐴 1 / 2 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ − 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 + 𝑖 𝜏 𝐴 1 / 2 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 − 𝑖 𝜏 𝐴 1 / 2 2 ⎤ ⎥ ⎥ ⎦ × 𝜏 2 𝑓 0 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ − 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝜏 𝐴 × 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 × 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 + 𝑛 𝑘 = 1 𝛽 𝑘 [ 𝜆 𝑘 / 𝜏 ] − 2 𝑠 = 1 𝜏 2 𝑖 × 𝐴 − 1 / 2 𝑖 𝐴 1 / 2 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 𝑓 𝑠 + 𝑛 𝑘 = 1 𝛽 𝑘 𝑅 𝑅 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝐴 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑠 = 1 𝜏 2 𝑖 × 𝐴 − 1 / 2 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝑓 𝑠 + 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 − ⎡ ⎢ ⎢ ⎣ + 𝜓 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 𝐴 𝜏 𝐼 + 2 𝐴 2 − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 𝐴 1 / 2 2 − 1 2 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 𝐼 + 2 𝐴 2 − 1 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 𝐼 + 𝑖 𝜏 𝐴 1 / 2 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 𝐼 − 𝑖 𝜏 𝐴 1 / 2 × 𝜏 2 𝑓 0 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 2 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ − ⎛ ⎜ ⎜ ⎝ 𝑛 𝑚 = 1 𝛼 𝑚 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑠 = 1 𝜏 𝐴 2 𝑖 − 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 𝑓 𝑠 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 × [ 𝜆 𝑚 / 𝜏 ] − 2 𝑠 = 1 𝜏 𝐴 2 𝑖 − 1 / 2 𝑖 𝐴 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 𝑓 𝑠 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝑅 𝑅 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 1 . − 𝜑 ( 2 . 1 0 ) Thus, formulas (2.9) and (2.10) give a solution of problem (2.1).

Theorem 2.3. Suppose that assumption (2.2) holds and 𝜑 ∈ 𝐷 ( 𝐴 ) ,   𝜓 ∈ 𝐷 ( 𝐴 1 / 2 ) . Then, for the solution of difference scheme (2.1) the stability inequalities m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝑢 𝑘 ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑘 = 0 ‖ ‖ 𝐴 − 1 / 2 𝑓 𝑘 ‖ ‖ 𝐻 ‖ ‖ 𝐴 𝜏 + − 1 / 2 𝜓 ‖ ‖ 𝐻 + ‖ 𝜑 ‖ 𝐻 , ( 2 . 1 1 ) m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝐴 1 / 2 𝑢 𝑘 ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑘 = 0 ‖ ‖ 𝑓 𝑘 ‖ ‖ 𝐻 ‖ ‖ 𝐴 𝜏 + 1 / 2 𝜑 ‖ ‖ 𝐻 + ‖ 𝜓 ‖ 𝐻 , ( 2 . 1 2 ) m a x 1 ≤ 𝑘 ≤ 𝑁 − 1 ‖ ‖ 𝜏 − 2 𝑢 𝑘 + 1 − 2 𝑢 𝑘 + 𝑢 𝑘 − 1 ‖ ‖ 𝐻 + m a x 0 ≤ 𝑘 ≤ 𝑁 − 1 ‖ ‖ ‖ 𝐴 𝑢 𝑘 + 𝜏 2 4 𝐴 2 𝑢 𝑘 + 1 ‖ ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑘 = 1 ‖ ‖ 𝑓 𝑘 − 𝑓 𝑘 − 1 ‖ ‖ 𝐻 + ‖ ‖ 𝑓 0 ‖ ‖ 𝐻 + ‖ ‖ 𝐴 1 / 2 𝜓 ‖ ‖ 𝐻 + ‖ 𝐴 𝜑 ‖ 𝐻 ( 2 . 1 3 ) hold, where 𝑀 does not depend on 𝜏 ,   𝜑 ,   𝜓 , and 𝑓 𝑘 ,   0 ≤ 𝑘 ≤ 𝑁 − 1 .

Proof. By [18], the following estimates m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝑢 𝑘 ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑘 = 0 ‖ ‖ 𝐴 − 1 / 2 𝑓 𝑘 ‖ ‖ 𝐻 ‖ ‖ 𝐴 𝜏 + − 1 / 2 𝜔 ‖ ‖ 𝐻 + ‖ 𝜇 ‖ 𝐻 , ( 2 . 1 4 ) m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝐴 1 / 2 𝑢 𝑘 ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑘 = 0 ‖ ‖ 𝑓 𝑘 ‖ ‖ 𝐻 ‖ ‖ 𝐴 𝜏 + 1 / 2 𝜇 ‖ ‖ 𝐻 + ‖ 𝜔 ‖ 𝐻 , ( 2 . 1 5 ) m a x 1 ≤ 𝑘 ≤ 𝑁 − 1 ‖ ‖ 𝜏 − 2 𝑢 𝑘 + 1 − 2 𝑢 𝑘 + 𝑢 𝑘 − 1 ‖ ‖ 𝐻 + m a x 0 ≤ 𝑘 ≤ 𝑁 − 1 ‖ ‖ ‖ 𝐴 𝑢 𝑘 + 𝜏 2 4 𝐴 2 𝑢 𝑘 + 1 ‖ ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑘 = 1 ‖ ‖ 𝑓 𝑘 − 𝑓 𝑘 − 1 ‖ ‖ 𝐻 + ‖ ‖ 𝑓 0 ‖ ‖ 𝐻 + ‖ ‖ 𝐴 1 / 2 𝜔 ‖ ‖ 𝐻 + ‖ 𝐴 𝜇 ‖ 𝐻 ( 2 . 1 6 ) hold for the solution of (2.8). Using formulas of 𝜇 , 𝜔 , and (2.3) and (2.6) the following estimates obtained ‖ 𝜇 ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑠 = 0 ‖ ‖ 𝐴 − 1 / 2 𝑓 𝑠 ‖ ‖ 𝐻 ‖ ‖ 𝐴 𝜏 + − 1 / 2 𝜓 ‖ ‖ 𝐻 + ‖ 𝜑 ‖ 𝐻 , ‖ ‖ 𝐴 ( 2 . 1 7 ) − 1 / 2 𝜔 ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑠 = 0 ‖ ‖ 𝐴 − 1 / 2 𝑓 𝑠 ‖ ‖ 𝐻 ‖ ‖ 𝐴 𝜏 + − 1 / 2 𝜓 ‖ ‖ 𝐻 + ‖ 𝜑 ‖ 𝐻 . ( 2 . 1 8 ) Estimate (2.11) follows from (2.14), (2.17), and (2.18). In a similar manner, we obtain m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝐴 1 / 2 𝑢 𝑘 ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑘 = 0 ‖ ‖ 𝑓 𝑘 ‖ ‖ 𝐻 ‖ ‖ 𝐴 𝜏 + 1 / 2 𝜑 ‖ ‖ 𝐻 + ‖ 𝜓 ‖ 𝐻 . ( 2 . 1 9 ) Now, we obtain the estimates for ‖ 𝐴 𝜇 ‖ 𝐻 and ‖ 𝐴 1 / 2 𝜔 ‖ 𝐻 . First, applying 𝐴 to the formula of 𝜇 and using Abel’s formula, we can write 𝐴 𝜇 = 𝑇 𝜏 ⎧ ⎪ ⎨ ⎪ ⎩ 𝜏 2 𝑓 0 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 𝐴 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 + 2 𝑖 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝐴 𝜏 𝐼 + 2 𝐴 2 − 1 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ − ⎛ ⎜ ⎜ ⎝ 1 2 𝑛 𝑚 = 1 𝛼 𝑚 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑠 = 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − s 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − s 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 × 𝑓 𝑠 − 𝑓 𝑠 − 1 + 𝑛 𝑚 = 1 𝛼 𝑚 1 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 𝑓 1 − 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 4 − 1 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 1 + 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 × ⎡ ⎢ ⎢ ⎣ [ 𝜆 𝑚 / 𝜏 ] − 2 𝑠 = 2 1 𝐴 2 𝑖 1 / 2 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 𝑓 𝑠 − 𝑓 𝑠 − 1 + 1 𝐴 2 𝑖 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑓 1 + 𝑖 𝐴 1 / 2 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 2 ⎤ ⎥ ⎥ ⎦ + 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝐴 𝑅 𝑅 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 1 × − 𝐴 𝜑 𝐼 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝐴 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 − 1 × 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 − 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎤ ⎥ ⎥ ⎦ + 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 + 2 𝑖 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 × 𝐴 1 / 2 𝜏 𝐼 + 2 𝐴 2 − 1 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎤ ⎥ ⎥ ⎦ × ⎡ ⎢ ⎢ ⎣ 𝜏 2 𝑓 0 ⎛ ⎜ ⎜ ⎝ 𝑛 𝑘 = 1 𝛽 𝑘 𝐴 1 / 2 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ + 𝑛 𝑘 = 1 𝛽 𝑘 𝑖 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝜏 𝐴 × 𝐼 + 2 𝐴 2 − 1 × 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 − 𝑛 𝑘 = 1 𝛽 𝑘 [ 𝜆 𝑘 / 𝜏 ] − 2 𝑠 = 2 1 𝑅 2 𝑖 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝑓 𝑠 − 𝑓 𝑠 − 1 − 𝑛 𝑘 = 1 𝛽 𝑘 1 𝑅 2 𝑖 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑓 1 − 𝑛 𝑘 = 1 𝛽 𝑘 𝑖 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 2 + 𝑛 𝑘 = 1 𝛽 𝑘 𝐴 1 / 2 𝜏 𝑅 𝑅 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 × ⎡ ⎢ ⎢ ⎣ 1 2 𝐴 − 1 / 2 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑠 = 2 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 × 𝑓 𝑠 − 𝑓 𝑠 − 1 + 1 2 𝐴 − 1 / 2 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 𝑓 1 − 𝐴 − 1 / 2 𝜏 𝐼 + 2 𝐴 4 − 1 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝐴 1 / 2 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝐴 1 / 2 𝜓 ⎤ ⎥ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭ . ( 2 . 2 0 ) Second, applying 𝐴 1 / 2 to the formula of 𝜔 and using Abel’s formula, we can write 𝐴 1 / 2 𝜔 = 𝑇 𝜏 𝐼 − 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 𝐴 1 / 2 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ − 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 𝑖 𝐴 1 / 2 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 + 𝑖 𝜏 𝐴 1 / 2 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 − 𝑖 𝜏 𝐴 1 / 2 2 × 1 2 𝑓 0 𝑛 𝑘 = 1 𝛽 𝑘 𝐴 1 / 2 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝜏 𝐼 + 2 𝐴 2 − 1 × 𝐴 1 / 2 𝑖 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑅 − 1 2 − 𝑛 𝑘 = 1 𝛽 𝑘 [ 𝜆 𝑘 / 𝜏 ] − 2 𝑠 = 2 1 𝑅 2 𝑖 [ 𝜆 𝑘 / 𝜏 ] − s + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − s × 𝑓 𝑠 − 𝑓 𝑠 − 1 − 𝑛 𝑘 = 1 𝛽 𝑘 1 𝑅 2 𝑖 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑓 1 − 𝑛 𝑘 = 1 𝛽 𝑘 𝑖 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 2 + 𝑛 𝑘 = 1 𝛽 𝑘 𝐴 1 / 2 𝜏 𝑅 𝑅 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝑛 𝑘 = 1 𝛽 𝑘 𝜏 2 + 𝜆 𝑘 − 𝜆 𝑘 𝜏 𝜏 𝐴 − 1 / 2 × ⎡ ⎢ ⎢ ⎣ 1 2 [ 𝜆 𝑘 / 𝜏 ] − 1 𝑠 = 2 𝑅 [ 𝜆 𝑘 / 𝜏 ] − s 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 𝑠 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 × 𝑓 𝑠 − 𝑓 𝑠 − 1 + 1 2 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 𝑓 1 − 𝜏 𝐼 + 2 𝐴 4 − 1 𝑓 [ 𝜆 𝑘 / 𝜏 ] − 1 + 𝐴 1 / 2 𝜓 − ⎡ ⎢ ⎢ ⎣ 𝑛 𝑘 = 1 𝛽 𝑘 𝐴 1 / 2 𝜏 2 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ − 𝑛 𝑘 = 1 𝛽 𝑘 𝐴 − 1 / 2 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 + 𝑖 𝜏 𝐴 1 / 2 2 − 𝑅 [ 𝜆 𝑘 / 𝜏 ] − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 𝐼 − 𝑖 𝜏 𝐴 1 / 2 2 × 1 2 𝑓 0 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝐼 + 2 𝐴 2 − 1 × 𝜏 𝐴 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 − 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 + 2 𝑖 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝐴 𝜏 𝐼 + 2 𝐴 2 − 1 × ⎛ ⎜ ⎜ ⎝ 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 − ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑅 − 1 𝐼 + ( 𝑖 𝜏 / 2 ) 𝐴 1 / 2 2 ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ − 1 2 𝑛 𝑚 = 1 𝛼 𝑚 𝐴 − 1 / 2 × [ 𝜆 𝑚 / 𝜏 ] − 1 𝑠 = 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 × 𝑓 𝑠 − 𝑓 𝑠 − 1 + 1 2 𝑛 𝑚 = 1 𝛼 𝑚 𝐴 − 1 / 2 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 + 𝑖 𝜏 2 𝐴 1 / 2 − 1 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝐼 − 𝑖 𝜏 2 𝐴 1 / 2 − 1 𝑓 1 − 𝑛 𝑚 = 1 𝛼 𝑚 𝐴 − 1 / 2 𝜏 𝐼 + 2 𝐴 4 − 1 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 1 − 𝑛 𝑚 = 1 𝛼 𝑚 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 × ⎡ ⎢ ⎢ ⎣ [ 𝜆 𝑚 / 𝜏 ] − 2 𝑠 = 2 1 𝐴 2 𝑖 1 / 2 × 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 𝑠 𝑓 𝑠 − 𝑓 𝑠 − 1 + 1 𝐴 2 𝑖 1 / 2 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 + 𝑅 [ 𝜆 𝑚 / 𝜏 ] − 1 𝑓 1 + 𝑖 𝐴 1 / 2 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 2 ⎤ ⎥ ⎥ ⎦ + 𝑛 𝑚 = 1 𝛼 𝑚 𝜏 𝜆 𝑚 − 𝜆 𝑚 𝜏 𝜏 𝐴 𝑅 𝑅 𝑓 [ 𝜆 𝑚 / 𝜏 ] − 1 . − 𝐴 𝜑 ( 2 . 2 1 ) The following estimates ‖ 𝐴 𝜇 ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑠 = 1 ‖ ‖ 𝑓 𝑠 − 𝑓 𝑠 − 1 ‖ ‖ 𝐻 + ‖ ‖ 𝑓 0 ‖ ‖ 𝐻 + ‖ ‖ 𝐴 1 / 2 𝜓 ‖ ‖ 𝐻 + ‖ 𝐴 𝜑 ‖ 𝐻 , ‖ ‖ 𝐴 1 / 2 𝜔 ‖ ‖ 𝐻 ≤ 𝑀 𝑁 − 1 𝑠 = 1 ‖ ‖ 𝑓 𝑠 − 𝑓 𝑠 − 1 ‖ ‖ 𝐻 + ‖ ‖ 𝑓 0 ‖ ‖ 𝐻 + ‖ ‖ 𝐴 1 / 2 𝜓 ‖ ‖ 𝐻 + ‖ 𝐴 𝜑 ‖ 𝐻 ( 2 . 2 2 ) are obtained by using formulas (2.20), (2.21), (2.3), and (2.6).
Estimate (2.13) follows from (2.16), and (2.22). Theorem 2.3 is proved.
Now, let us consider the applications of Theorem 2.3. First, the nonlocal the mixed boundary value problem for hyperbolic equation 𝑢 𝑡 𝑡 − 𝑎 ( 𝑥 ) 𝑢 𝑥 𝑥 + 𝛿 𝑢 = 𝑓 ( 𝑡 , 𝑥 ) , 0 < 𝑡 < 1 , 0 < 𝑥 < 1 , 𝑢 ( 0 , 𝑥 ) = 𝑛 𝑚 = 1 𝛼 𝑚 𝑢 𝜆 𝑚 𝑢 , 𝑥 + 𝜑 ( 𝑥 ) , 0 ≤ 𝑥 ≤ 1 , 𝑡 ( 0 , 𝑥 ) = 𝑛 𝑘 = 1 𝛽 𝑘 𝑢 𝑡 𝜆 𝑘 , 𝑥 + 𝜓 ( 𝑥 ) , 0 ≤ 𝑥 ≤ 1 , 𝑢 ( 𝑡 , 0 ) = 𝑢 ( 𝑡 , 1 ) , 𝑢 𝑥 ( 𝑡 , 0 ) = 𝑢 𝑥 ( 𝑡 , 1 ) , 0 ≤ 𝑡 ≤ 1 , ( 2 . 2 3 ) under assumption (2.2), is considered. Here 𝑎 𝑟 ( 𝑥 ) ,   ( 𝑥 ∈ ( 0 , 1 ) ) ,   𝜑 ( 𝑥 ) , 𝜓 ( 𝑥 ) ( 𝑥 ∈ [ 0 , 1 ] ) and 𝑓 ( 𝑡 , 𝑥 ) ( 𝑡 ∈ ( 0 , 1 ) , 𝑥 ∈ ( 0 , 1 ) ) are given smooth functions and 𝑎 𝑟 ( 𝑥 ) ≥ 𝑎 > 0 ,   𝛿 > 0 . The discretization of problem (2.23) is carried out in two steps.
In the first step, the grid space is defined as follows: [ ] 0 , 1 ℎ = 𝑥 ∶ 𝑥 𝑟 = 𝑟 ℎ , 0 ≤ 𝑟 ≤ 𝐾 , 𝐾 ℎ = 1 . ( 2 . 2 4 ) We introduce the Hilbert space 𝐿 2 ℎ = 𝐿 2 ( [ 0 , 1 ] ℎ ) ,   𝑊 1 2 ℎ = 𝑊 1 2 ℎ ( [ 0 , 1 ] ℎ ) and 𝑊 2 2 ℎ = 𝑊 2 2 ℎ ( [ 0 , 1 ] ℎ ) of the grid functions 𝜑 ℎ ( 𝑥 ) = { 𝜑 𝑟 } 1 𝐾 − 1 defined on [ 0 , 1 ] ℎ , equipped with the norms ‖ ‖ 𝜑 ℎ ‖ ‖ 𝐿 2 ℎ = 𝐾 − 1 𝑟 = 1 | | 𝜑 ℎ | | ( 𝑥 ) 2 ℎ 1 / 2 , ‖ ‖ 𝜑 ℎ ‖ ‖ 𝑊 1 2 ℎ = ‖ ‖ 𝜑 ℎ ‖ ‖ 𝐿 2 ℎ + 𝐾 − 1 𝑟 = 1 | | | 𝜑 ℎ 𝑥 , 𝑗 | | | 2 ℎ 1 / 2 , ‖ ‖ 𝜑 ℎ ‖ ‖ 𝑊 2 2 ℎ = ‖ ‖ 𝜑 ℎ ‖ ‖ 𝐿 2 ℎ + 𝐾 − 1 𝑟 = 1 | | | 𝜑 ℎ 𝑥 , 𝑗 | | | 2 ℎ 1 / 2 + 𝐾 − 1 𝑟 = 1 | | | 𝜑 ℎ 𝑥 𝑥 , 𝑗 | | | 2 ℎ 1 / 2 , ( 2 . 2 5 ) respectively. To the differential operator 𝐴 generated by problem (2.23), we assign the difference operator 𝐴 𝑥 ℎ by the formula 𝐴 𝑥 ℎ 𝜑 ℎ ( 𝑥 ) = − ( 𝑎 ( 𝑥 ) 𝜑 𝑥 ) 𝑥 , 𝑟 + 𝛿 𝜑 𝑟 1 𝐾 − 1 , ( 2 . 2 6 ) acting in the space of grid functions 𝜑 ℎ ( 𝑥 ) = { 𝜑 𝑟 } 𝐾 0 satisfying the conditions 𝜑 0 = 𝜑 𝐾 ,   𝜑 1 − 𝜑 0 = 𝜑 𝐾 − 𝜑 𝐾 − 1 . With the help of 𝐴 𝑥 ℎ , we arrive at the nonlocal boundary value problem 𝑑 2 𝑣 ℎ ( 𝑡 , 𝑥 ) 𝑑 𝑡 2 + 𝐴 𝑥 ℎ 𝑣 ℎ ( 𝑡 , 𝑥 ) = 𝑓 ℎ [ ] ( 𝑡 , 𝑥 ) , 0 ≤ 𝑡 ≤ 1 , 𝑥 ∈ 0 , 1 ℎ , 𝑣 ℎ ( 0 , 𝑥 ) = 𝑛 𝑗 = 1 𝛼 𝑗 𝑣 ℎ 𝜆 𝑗 , 𝑥 + 𝜑 ℎ [ ] ( 𝑥 ) , 𝑥 ∈ 0 , 1 ℎ , 𝑣 ℎ 𝑡 ( 0 , 𝑥 ) = 𝑛 𝑗 = 1 𝛽 𝑗 𝑣 ℎ 𝑡 𝜆 𝑗 , 𝑥 + 𝜓 ℎ [ ] ( 𝑥 ) , 𝑥 ∈ 0 , 1 ℎ , ( 2 . 2 7 ) for an infinite system of ordinary differential equations.
In the second step, we replace problem (2.27) by difference scheme (2.28) 𝑢 ℎ 𝑘 + 1 ( 𝑥 ) − 2 𝑢 ℎ 𝑘 ( 𝑥 ) + 𝑢 ℎ 𝑘 − 1 ( 𝑥 ) 𝜏 2 + 𝐴 𝑥 ℎ 𝑢 ℎ 𝑘 𝜏 ( 𝑥 ) + 2 4 𝐴 𝑥 ℎ 2 𝑢 ℎ 𝑘 + 1 ( 𝑥 ) = 𝑓 ℎ 𝑘 [ ] ( 𝑥 ) , 𝑥 ∈ 0 , 1 ℎ , 𝑓 ℎ 𝑘 ( 𝑥 ) = 𝑓 ℎ 𝑡 𝑘 , 𝑥 , 𝑡 𝑘 𝑢 = 𝑘 𝜏 , 1 ≤ 𝑘 ≤ 𝑁 − 1 , 𝑁 𝜏 = 1 , ℎ 0 ( 𝑥 ) = 𝑛 𝑗 = 1 𝛼 𝑗 𝑢 ℎ 𝜆 𝑗 / 𝜏 ( 𝑥 ) + 𝜏 − 1 𝑢 ℎ [ 𝜆 𝑗 / 𝜏 ] ( 𝑥 ) − 𝑢 ℎ [ 𝜆 𝑗 / 𝜏 ] − 1 𝜆 ( 𝑥 ) 𝑗 − 𝜆 𝑗 𝜏 𝜏 + 𝜑 ℎ [ ] ( 𝑥 ) , 𝑥 ∈ 0 , 1 ℎ , 𝜏 𝐼 + 2 𝐴 𝑥 ℎ 2 𝑢 ℎ 1 ( 𝑥 ) − 𝑢 ℎ 0 ( 𝑥 ) 𝜏 − 𝜏 2 𝑓 ℎ 0 ( 𝑥 ) − 𝐴 𝑥 ℎ 𝑢 ℎ 0 = ( 𝑥 ) 𝑛 𝑗 = 1 𝛽 𝑗 ⎧ ⎪ ⎨ ⎪ ⎩ 𝑢 ℎ 𝜆 𝑗 / 𝜏 ( 𝑥 ) − 𝑢 ℎ 𝜆 𝑗 / 𝜏 − 1 ( 𝑥 ) 𝜏 + 𝜏 2 + 𝜆 𝑗 − 𝜆 𝑗 𝜏 𝜏 × 𝑓 ℎ [ 𝜆 𝑗 / 𝜏 ] ( 𝑥 ) − 𝐴 𝑥 ℎ 𝑢 ℎ [ 𝜆 𝑗 / 𝜏 ] ⎫ ⎪ ⎬ ⎪ ⎭ ( 𝑥 ) + 𝜓 ℎ 𝑓 ( 𝑥 ) , ℎ 0 ( 𝑥 ) = 𝑓 ℎ [ ] ( 0 , 𝑥 ) , 𝑥 ∈ 0 , 1 ℎ . ( 2 . 2 8 )

Theorem 2.4. Let 𝜏 and ℎ be sufficiently small positive numbers. Suppose that assumption (2.2) holds. Then, the solution of difference scheme (2.28) satisfies the following stability estimates: m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝑢 ℎ 𝑘 ‖ ‖ 𝐿 2 ℎ + m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝑢 ℎ 𝑘 𝑥 ‖ ‖ 𝐿 2 ℎ ≤ 𝑀 1 m a x 0 ≤ 𝑘 ≤ 𝑁 − 1 ‖ ‖ 𝑓 ℎ 𝑘 ‖ ‖ 𝐿 2 ℎ + ‖ ‖ 𝜓 ℎ ‖ ‖ 𝐿 2 ℎ + ‖ ‖ 𝜑 ℎ 𝑥 ‖ ‖ 𝐿 2 ℎ , m a x 1 ≤ 𝑘 ≤ 𝑁 − 1 ‖ ‖ 𝜏 − 2 𝑢 ℎ 𝑘 + 1 − 2 𝑢 ℎ 𝑘 + 𝑢 ℎ 𝑘 − 1 ‖ ‖ 𝐿 2 ℎ + m a x 0 ≤ 𝑘 ≤ 𝑁 ‖ ‖ 𝑢 ℎ 𝑘 𝑥 𝑥 ‖ ‖ 𝐿 2 ℎ ≤ 𝑀 1 ‖ ‖ 𝑓 ℎ 0 ‖ ‖ 𝐿 2 ℎ + m a x 1 ≤ 𝑘 ≤ 𝑁 − 1 ‖ ‖ 𝜏 − 1 𝑓 ℎ 𝑘 − 𝑓 ℎ 𝑘 − 1 ‖ ‖ 𝐿 2 ℎ + ‖ ‖ 𝜓 ℎ 𝑥 ‖ ‖ 𝐿 2 ℎ + ‖ ‖ 𝜑 ℎ 𝑥 𝑥 ‖ ‖ 𝐿 2 ℎ . ( 2 . 2 9 ) Here, 𝑀 1 does not depend on 𝜏 ,   ℎ ,   𝜑 ℎ ( 𝑥 ) ,   𝜓 ℎ ( 𝑥 ) and 𝑓 ℎ 𝑘 ,   0 ≤ 𝑘 < 𝑁 .

The proof of Theorem 2.4 is based on abstract Theorem 2.3 and symmetry properties of the operator 𝐴 𝑥 ℎ defined by (2.26).

Second, for the 𝑚 -dimensional hyperbolic equation under assumption (2.2) is considered. Let Ω be the unit open cube in the 𝑚 -dimensional Euclidean space ℝ 𝑚 { 𝑥 = ( 𝑥 1 , … , 𝑥 𝑚 ) ∶ 0 < 𝑥 𝑗 < 1 , 1 ≤ 𝑗 ≤ 𝑚 } with boundary 𝑆 , Ω = Ω ∪ 𝑆 . In [ 0 , 1 ] × Ω , the mixed boundary value problem for the multidimensional hyperbolic equation 𝜕 2 𝑢 ( 𝑡 , 𝑥 ) 𝜕 𝑡 2 − 𝑚 𝑟 = 1 𝑎 𝑟 ( 𝑥 ) 𝑢 𝑥 𝑟 𝑥 𝑟 𝑥 = 𝑓 ( 𝑡 , 𝑥 ) , 𝑥 = 1 , … , 𝑥 𝑚 ∈ Ω , 0 < 𝑡 < 1 , 𝑢 ( 0 , 𝑥 ) = 𝑛 𝑗 = 1 𝛼 𝑗 𝑢 𝜆 𝑗 , 𝑥 + 𝜑 ( 𝑥 ) , 𝑥 ∈ 𝑢 Ω ; 𝑡 ( 0 , 𝑥 ) = 𝑛 𝑘 = 1 𝛽 𝑘 𝑢 𝑡 𝜆 𝑘 , 𝑥 + 𝜓 ( 𝑥 ) , 𝑥 ∈ Ω ; 𝑢 ( 𝑡 , 𝑥 ) = 0 , 𝑥 ∈ 𝑆 ( 2 . 3 0 ) is considered.

Here, 𝑎 𝑟 ( 𝑥 ) ,   ( 𝑥 ∈ Ω ) ,   𝜑 ( 𝑥 ) ,   𝜓 ( 𝑥 )    ( 𝑥 ∈ Ω ) and 𝑓 ( 𝑡 , 𝑥 )