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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 276080, 26 pages
doi:10.1155/2012/276080
Research Article

A Note on the Inverse Problem for a Fractional Parabolic Equation

Department of Mathematics, Fatih University, 34500 Buyukcekmece, Istanbul, Turkey

Received 15 May 2012; Accepted 8 July 2012

Academic Editor: Ravshan Ashurov

Copyright © 2012 Abdullah Said Erdogan and Hulya Uygun. 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

For a fractional inverse problem with an unknown time-dependent source term, stability estimates are obtained by using operator theory approach. For the approximate solutions of the problem, the stable difference schemes which have first and second orders of accuracy are presented. The algorithm is tested in a one-dimensional fractional inverse problem.

1. Introduction

Inverse problems arise in many fields of science and engineering such as ion transport problems, chromatography, and heat determination problems with an unknown internal energy source. Different typed of inverse problems have been investigated, and the main results obtained in this field of research were given by many researchers (see [110]). More than three centuries the theory of fractional derivatives developed mainly as a pure theoretical field of mathematics. Fractional integrals and derivatives appear in the theory of control of dynamical systems, when the controlled system or/and the controller is described by a fractional differential equation (see [11]). Recently, many application areas such as bioengineering applications, image and signal processing are also related to fractional calculus. Methods of solutions of problems and theory of fractional calculus have been studied by many researchers [1128]. Among them finite difference method is used for solving several fractional differential equations (see [20, 22, 23, 27] and the references therein).

1.1. Statement of the Problem

Many scientists and researchers are trying to enhance mathematical models of real-life cases for investigating and understanding the behavior of them. Therefore, some phenomena have been modeled and investigated as fractional inverse problems (see [2933] and the references therein). In this paper, we consider the fractional parabolic inverse problem with the Dirichlet condition 𝜕 𝑢 ( 𝑡 , 𝑥 ) 𝜕 𝜕 𝑡 𝑎 2 𝑢 ( 𝑡 , 𝑥 ) 𝜕 𝑥 2 𝐷 𝑡 1 / 2 𝑢 𝑢 ( 𝑡 , 𝑥 ) + 𝜎 𝑢 ( 𝑡 , 𝑥 ) = 𝑝 ( 𝑡 ) 𝑞 ( 𝑥 ) + 𝑓 ( 𝑡 , 𝑥 ) , 0 < 𝑥 < 𝜋 , 0 < 𝑡 𝑇 , 𝑢 ( 𝑡 , 0 ) = 𝑢 ( 𝑡 , 𝜋 ) = 0 , 0 𝑡 𝑇 , 𝑢 ( 0 , 𝑥 ) = 𝜑 ( 𝑥 ) , 0 𝑥 𝜋 , 𝑡 , 𝑥 = 𝜌 ( 𝑡 ) , 0 < 𝑥 < 𝜋 , ( 1 . 1 ) where 𝑢 ( 𝑡 , 𝑥 ) and 𝑝 ( 𝑡 ) are unknown functions, 𝑎 ( 𝑥 ) 𝑎 > 0 , and 𝜎 > 0 is a sufficiently large number. Here, 𝐷 𝑡 1 / 2 = 𝐷 1 / 2 0 + is the standard Riemann-Liouville’s derivative of order 1 / 2 .

Theorems on the stability of problem (1.1) are analyzed by assuming that 𝑞 ( 𝑥 ) is a sufficiently smooth function, 𝑞 ( 0 ) = 𝑞 ( 𝜋 ) = 0 and 𝑞 ( 𝑥 ) 0 .

2. Main Results

In this section, stability estimates for the solution of (1.1) are investigated. For the mathematical substantiation, we introduce the Banach space 𝐶 𝛼 [ 0 , 𝜋 ] , 𝛼 ( 0 , 1 ) , of all continuous functions 𝜙 ( 𝑥 ) defined on [ 0 , 𝜋 ] with 𝜙 ( 0 ) = 𝜙 ( 𝜋 ) = 0 satisfying a Hölder condition for which the following norm is finite 𝜙 𝐶 𝛼 [ 0 , 𝜋 ] = 𝜙 𝐶 [ 0 , 𝜋 ] + s u p 0 < 𝑥 < 𝑥 + < 𝜋 | | | | 𝜙 ( 𝑥 + ) 𝜙 ( 𝑥 ) 𝛼 , ( 2 . 1 ) where 𝐶 [ 0 , 𝜋 ] is the space of all continuous function 𝜙 ( 𝑥 ) defined on [ 0 , 𝜋 ] with the norm 𝜙 𝐶 [ 0 , 𝜋 ] = m a x 0 𝑥 𝜋 | | | | . 𝜙 ( 𝑥 ) ( 2 . 2 ) With the help of a positive operator 𝐴 , we introduce the fractional spaces 𝐸 𝛼 , 0 < 𝛼 < 1 , consisting of all 𝑣 in a Banach space 𝐸 for which the following norm is finite: 𝑣 𝐸 𝛼 = 𝑣 𝐸 + s u p 𝜆 > 0 𝜆 1 𝛼 𝐴 e x p { 𝜆 𝐴 } 𝑣 𝐸 . ( 2 . 3 ) Throughout the paper, positive constants will be indicated by 𝑀 𝑖 ( 𝛼 , 𝛽 , ) . Here variables are used to focus on the fact that the constant depends only on 𝛼 , 𝛽 , and the subindex 𝑖 is used to indicate a different constant.

Theorem 2.1. Let 𝜑 𝐶 2 𝛼 + 2 [ 0 , 𝜋 ] , 𝐹 𝐶 ( [ 0 , 𝑇 ] , 𝐶 2 𝛼 [ 0 , 𝜋 ] ) , and 𝜌 𝐶 [ 0 , 𝑇 ] . Then for the solution of problem (1.1), the following coercive stability estimates 𝑢 𝑡 𝐶 ( [ 0 , 𝑇 ] , 𝐶 2 𝛼 [ 0 , 𝜋 ] ) + 𝑢 𝐶 ( [ 0 , 𝑇 ] , 𝐶 2 𝛼 + 2 [ 0 , 𝜋 ] ) 𝑥 𝑀 𝜌 , 𝑞 𝐶 [ 0 , 𝑇 ] + 𝑀 𝑎 , 𝛿 , 𝜎 , 𝛼 , 𝑥 × , 𝑞 , 𝑇 𝜑 𝐶 2 𝛼 + 2 [ 0 , 𝜋 ] + 𝐹 𝐶 ( [ 0 , 𝑇 ] , 𝐶 2 𝛼 [ 0 , 𝜋 ] ) + 𝜌 𝐶 [ 0 , 𝑇 ] , 𝑝 𝐶 [ 0 , 𝑇 ] 𝑥 𝑀 𝜌 , 𝑞 𝐶 [ 0 , 𝑇 ] + 𝑀 𝑎 , 𝛿 , 𝜎 , 𝛼 , 𝑥 × , 𝑞 , 𝑇 𝜑 𝐶 2 𝛼 + 2 [ 0 , 𝜋 ] + 𝐹 𝐶 ( [ 0 , 𝑇 ] , 𝐶 2 𝛼 [ 0 , 𝜋 ] ) + 𝜌 𝐶 [ 0 , 𝑇 ] ( 2 . 4 ) hold.

Proof. Let us search for the solution of inverse problem (1.1) in the following form (see [8]): 𝑢 ( 𝑡 , 𝑥 ) = 𝜂 ( 𝑡 ) 𝑞 ( 𝑥 ) + 𝑤 ( 𝑡 , 𝑥 ) , ( 2 . 5 ) where 𝜂 ( 𝑡 ) = 𝑡 0 𝑝 ( 𝑠 ) 𝑑 𝑠 . ( 2 . 6 ) Using the overdetermined condition, we get 𝜌 𝜂 ( 𝑡 ) = ( 𝑡 ) 𝑤 𝑡 , 𝑥 𝑞 ( 𝑥 ) , 𝜌 ( 2 . 7 ) 𝑝 ( 𝑡 ) = ( 𝑡 ) 𝑤 𝑡 𝑡 , 𝑥 𝑞 ( 𝑥 ) . ( 2 . 8 ) Using identity (2.8) and the triangle inequality, it follows that | | | | = | | | | 𝜌 𝑝 ( 𝑡 ) ( 𝑡 ) 𝑤 𝑡 𝑡 , 𝑥 𝑞 ( 𝑥 ) | | | | 𝑥 𝑀 | | 𝜌 , 𝑞 | | + | | 𝑤 ( 𝑡 ) 𝑡 𝑡 , 𝑥 | | 𝑥 𝑀 , 𝑞 m a x 0 𝑡 𝑇 | | 𝜌 | | ( 𝑡 ) + m a x 0 𝑡 𝑇 m a x 0 𝑥 𝜋 | | 𝑤 𝑡 | | 𝑥 ( 𝑡 , 𝑥 ) 𝑀 , 𝑞 m a x 0 𝑡 𝑇 | | 𝜌 | | ( 𝑡 ) + m a x 0 𝑡 𝑇 𝑤 𝑡 ( 𝑡 , ) 𝐶 2 𝛼 [ 0 , 𝜋 ] ( 2 . 9 ) for any 𝑡 , 𝑡 [ 0 , 𝑇 ] . Here, 𝑤 ( 𝑡 , 𝑥 ) is the solution of the following problem: 𝜕 𝑤 ( 𝑡 , 𝑥 ) 𝜕 𝜕 𝑡 𝑎 2 𝑤 ( 𝑡 , 𝑥 ) 𝜕 𝑥 2 𝜌 𝑎 ( 𝑡 ) 𝑤 𝑡 , 𝑥 𝑞 ( 𝑥 ) 𝑑 2 𝑞 ( 𝑥 ) 𝑑 𝑥 2 𝐷 𝑡 1 / 2 𝐷 𝑤 ( 𝑡 , 𝑥 ) 𝑡 1 / 2 𝜌 ( 𝑡 ) 𝐷 𝑡 1 / 2 𝑤 𝑡 , 𝑥 𝑞 ( 𝑥 ) 𝑞 ( 𝑥 ) + 𝜎 𝜌 ( 𝑡 ) 𝑤 𝑡 , 𝑥 𝑞 ( 𝑥 ) 𝑞 ( 𝑥 ) + 𝜎 𝑤 ( 𝑡 , 𝑥 ) = 𝑓 ( 𝑡 , 𝑥 ) , 0 < 𝑥 < 𝜋 , 0 < 𝑡 𝑇 , 𝑤 ( 𝑡 , 0 ) = 𝑤 ( 𝑡 , 𝜋 ) = 0 , 0 𝑡 𝑇 , 𝑤 ( 0 , 𝑥 ) = 𝜑 ( 𝑥 ) , 0 𝑥 𝜋 . ( 2 . 1 0 ) For simplicity, we assign 𝐹 ( 𝑡 , 𝑥 ) = 𝑎 𝜌 ( 𝑡 ) 𝑞 ( 𝑥 ) 𝑑 2 𝑞 ( 𝑥 ) 𝑑 𝑥 2 𝜎 𝜌 ( 𝑡 ) 𝑞 ( 𝑥 ) 𝐷 𝑞 ( 𝑥 ) + 𝑡 1 / 2 𝜌 ( 𝑡 ) 𝑞 ( 𝑥 ) 𝑞 ( 𝑥 ) + 𝑓 ( 𝑡 , 𝑥 ) , 𝐺 ( 𝑡 , 𝑥 ) = 𝑄 1 𝑞 , 𝜌 , 𝑥 , 𝑥 𝑤 , 𝑡 𝑡 , 𝑥 + 𝑄 2 𝑞 , 𝑥 , 𝑥 𝐷 𝑡 1 / 2 𝑤 𝑡 , 𝑥 + 𝐷 𝑡 1 / 2 𝑤 ( 𝑡 , 𝑥 ) , ( 2 . 1 1 ) where 𝑄 1 𝑞 , 𝜌 , 𝑥 , 𝑥 = 1 , 𝑡 𝑞 ( 𝑥 ) 𝑑 𝑎 2 𝑞 ( 𝑥 ) 𝑑 𝑥 2 , 𝑄 + 𝜎 𝜌 ( 𝑡 ) 2 𝑞 , 𝑥 , 𝑥 𝑞 = ( 𝑥 ) 𝑞 ( 𝑥 ) . ( 2 . 1 2 ) Note that functions 𝐹 ( 𝑡 , 𝑥 ) , 𝑄 1 ( 𝑞 , 𝜌 , 𝑥 , 𝑥 , 𝑡 ) and 𝑄 2 ( 𝑞 , 𝑥 , 𝑥 ) only contain given functions. Then, we can rewrite problem (2.10) as 𝜕 𝑤 ( 𝑡 , 𝑥 ) 𝜕 𝜕 𝑡 𝑎 2 𝑤 ( 𝑡 , 𝑥 ) 𝜕 𝑥 2 + 𝜎 𝑤 ( 𝑡 , 𝑥 ) = 𝐹 ( 𝑡 , 𝑥 ) + 𝐺 ( 𝑡 , 𝑥 ) , 0 < 𝑥 < 𝜋 , 0 < 𝑡 𝑇 , 𝑤 ( 𝑡 , 0 ) = 𝑤 ( 𝑡 , 𝜋 ) = 0 , 0 𝑡 𝑇 , 𝑤 ( 0 , 𝑥 ) = 𝜑 ( 𝑥 ) , 0 𝑥 𝜋 . ( 2 . 1 3 ) So, the end of proof of Theorem 2.1 is based on estimate (2.9) and the following theorem.

Theorem 2.2. For the solution of problem (2.10), the following coercive stability estimate 𝑤 𝑡 𝐶 2 𝛼 [ 0 , 𝜋 ] 𝑀 𝑎 , 𝛿 , 𝜎 , 𝛼 , 𝑥 × , 𝑞 , 𝑇 𝜑 𝐶 2 𝛼 + 2 [ 0 , 𝜋 ] + 𝐹 𝐶 ( [ 0 , 𝑇 ] , 𝐶 2 𝛼 [ 0 , 𝜋 ] ) + 𝜌 𝐶 [ 0 , 𝑇 ] ( 2 . 1 4 ) holds.

Proof. In a Banach space 𝐸 = 𝐶 [ 0 , 𝜋 ] , with the help of the positive operator 𝐴 defined by 𝜕 𝐴 𝑢 = 𝑎 ( 𝑥 ) 2 𝑢 ( 𝑡 , 𝑥 ) 𝜕 𝑥 2 + 𝜎 𝑢 , ( 2 . 1 5 ) with 𝐷 𝑢 ( 𝐴 ) = ( 𝑥 ) 𝑢 , 𝑢 , 𝑢 [ ] 𝐶 0 , 𝜋 , 𝑢 ( 0 ) = 𝑢 ( 𝜋 ) = 0 , ( 2 . 1 6 ) where 𝜎 is a positive constant, problem (2.10) can be written in the abstract form as an initial-value problem 𝑤 𝑡 𝑤 + 𝐴 𝑤 = 𝐹 ( 𝑡 ) + 𝐺 ( 𝑡 ) , 0 < 𝑡 𝑇 , ( 0 ) = 𝜑 . ( 2 . 1 7 ) By the Cauchy formula, the solution can be written as 𝑤 ( 𝑡 ) = 𝑒 𝑡 𝐴 𝜑 𝑡 0 𝑒 ( 𝑡 𝑠 ) 𝐴 ( 𝐹 ( 𝑠 ) + 𝐺 ( 𝑠 ) ) 𝑑 𝑠 . ( 2 . 1 8 ) Applying the formula 𝐷 𝑡 1 / 2 𝑢 ( 𝑡 ) = 𝑡 0 𝑢 ( 𝜁 ) 𝑑 𝜉 𝜋 ( 𝑡 𝜉 ) 1 / 2 , ( 2 . 1 9 ) we get the following presentation of the solution of abstract problem (2.17): 𝐷 1 / 2 𝑤 ( 𝑡 ) = 𝑡 0 𝐴 𝑒 𝜉 𝐴 𝜑 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝑡 0 𝐹 ( 𝜉 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝑡 0 𝐺 ( 𝜉 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 + 𝑡 0 𝜉 0 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝐹 ( 𝑠 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 + 𝑑 𝑠 𝑑 𝜉 𝑡 0 𝜉 0 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝐺 ( 𝑠 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝑠 𝑑 𝜉 . ( 2 . 2 0 ) Changing the order of integration, we obtain that 𝐷 1 / 2 𝑤 ( 𝑡 ) = 𝑡 0 𝐴 𝑒 𝜉 𝐴 𝜑 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝑡 0 𝐹 ( 𝜉 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 + 𝑑 𝜉 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐹 ( 𝑠 ) 𝑑 𝑠 𝑡 0 𝐺 ( 𝜉 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 + 𝑑 𝜉 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐺 ( 𝑠 ) 𝑑 𝑠 = 5 𝑘 = 1 𝐽 𝑘 , ( 2 . 2 1 ) where 𝐽 1 ( 𝑡 ) = 𝑡 0 𝐴 𝑒 𝜉 𝐴 𝜑 𝜋 ( 𝑡 𝑝 ) 1 / 2 𝐽 𝑑 𝜉 , 2 ( 𝑡 ) = 𝑡 0 𝐹 ( 𝜉 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝐽 𝑑 𝜉 , 3 ( 𝑡 ) = 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝐽 𝑑 𝜉 𝐹 ( 𝑠 ) 𝑑 𝑠 , 4 ( 𝑡 ) = 𝑡 0 𝐺 ( 𝜉 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝐽 𝑑 𝜉 , 5 ( 𝑡 ) = 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐺 ( 𝑠 ) 𝑑 𝑠 . ( 2 . 2 2 ) Now, we estimate 𝐽 𝑘 ( 𝑡 ) , 𝑘 = 1 , 2 , 3 , 4 , 5 separately. It is known that [13] 𝐴 𝛼 𝑒 𝑡 𝐴 𝐸 𝐸 𝑀 , 0 𝛼 1 . ( 2 . 2 3 ) Since operators 𝐴 and e x p ( 𝑡 𝐴 ) commute, 𝐴 𝑒 𝑡 𝐴 𝜑 𝐸 𝛼 𝑒 𝑡 𝐴 𝐸 𝛼 𝐸 𝛼 𝐴 𝜑 𝐸 𝛼 𝑒 𝑡 𝐴 𝐸 𝐸 𝐴 𝜑 𝐸 𝛼 . ( 2 . 2 4 ) Applying the definition of norm of the spaces 𝐸 𝛼 and (2.23) and (2.24), we get 𝐽 1 ( 𝑡 ) 𝐸 𝛼 = 𝑡 0 𝐴 𝑒 𝜉 𝐴 𝜑 𝜋 ( 𝑡 𝑝 ) 1 / 2 𝑑 𝜉 𝐸 𝛼 𝑀 1 𝐴 𝜑 𝐸 𝛼 ( 2 . 2 5 ) for any 𝑡 , 𝑡 [ 0 , 𝑇 ] . Estimation of 𝐽 2 ( 𝑡 ) is as follows: 𝐽 2 ( 𝑡 ) 𝐸 𝛼 = 𝑡 0 𝐹 ( 𝜉 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐸 𝛼 𝐹 ( 𝑡 ) 𝐶 ( 𝐸 𝛼 ) 𝑡 0 1 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝑀 2 𝐹 𝐶 ( 𝐸 𝛼 ) . ( 2 . 2 6 ) Let us estimate 𝐽 3 ( 𝑡 ) : 𝐽 3 ( 𝑡 ) 𝐸 𝛼 = 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐹 ( 𝑠 ) 𝑑 𝑠 𝐸 𝛼 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝑑 𝑠 𝐸 𝛼 𝐸 𝛼 𝐹 𝐶 ( 𝐸 𝛼 ) . ( 2 . 2 7 ) It is proven that (see [28]) 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐸 𝐸 𝑀 . 𝑡 𝑠 ( 2 . 2 8 ) Using the definition of norm of the spaces 𝐸 𝛼 , we can obtain that 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝑑 𝑠 𝐸 𝛼 𝐸 𝛼 = 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝑡 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐸 𝐸 𝑑 𝑠 + s u p 𝜆 > 0 𝑡 0 𝑡 𝑠 𝜆 1 𝛼 𝐴 𝑒 𝜆 𝐴 𝐴 𝑒 ( 𝑡 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐸 𝐸 𝑑 𝑠 . ( 2 . 2 9 ) Using estimates (2.23) and (2.28), we get 𝐽 3 ( 𝑡 ) 𝐸 𝛼 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝑑 𝑠 𝐸 𝛼 𝐸 𝛼 𝐹 𝐶 ( 𝐸 𝛼 ) 𝑀 3 𝐹 𝐶 ( 𝐸 𝛼 ) . ( 2 . 3 0 ) Expanding 𝐺 ( 𝑠 ) , estimation of 𝐽 4 ( 𝑡 ) is as follows: 𝐽 4 ( 𝑡 ) 𝐸 𝛼 𝑡 0 𝑄 1 𝑞 , 𝜌 , 𝑥 , 𝑥 𝑤 , 𝑡 𝜉 , 𝑥 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝐸 𝛼 + 𝑑 𝜉 𝑡 0 𝑄 2 𝑞 , 𝑥 , 𝑥 𝐷 𝑡 1 / 2 𝑤 𝜉 , 𝑥 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝐸 𝛼 𝑑 𝜉 + 𝑡 0 𝐷 𝑡 1 / 2 𝑤 ( 𝜉 , 𝑥 ) 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝐸 𝛼 𝑑 𝜉 . ( 2 . 3 1 ) It is known that (see [34]) 𝑤 𝐸 𝛼 𝐷 𝑀 𝑡 1 / 2 𝑤 𝐸 𝛼 . ( 2 . 3 2 ) Since 𝑄 1 ( 𝑞 , 𝜌 , 𝑥 , 𝑥 , 𝑡 ) and 𝑄 2 ( 𝑞 , 𝑥 , 𝑥 ) are known functions, it is easy to see that 𝐽 4 ( 𝑡 ) 𝐸 𝛼 𝑀 4 𝑞 , 𝜌 , 𝑥 , 𝑥 , 𝑇 𝑡 0 1 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝐷 𝑡 1 / 2 𝑤 ( 𝜉 ) 𝐸 𝛼 𝑑 𝜉 . ( 2 . 3 3 ) Estimation of 𝐽 5 ( 𝑡 ) can be given similar to the estimation of 𝐽 4 ( 𝑡 ) . By (2.23) and (2.32), 𝐽 5 ( 𝑡 ) 𝐸 𝛼 𝑡 0 𝑡 𝑠 𝐴 𝑒 ( 𝜉 𝑠 ) 𝐴 𝜋 ( 𝑡 𝜉 ) 1 / 2 𝑑 𝜉 𝐺 ( 𝑠 ) 𝑑 𝑠 𝑀 5 𝑞 , 𝜌 , 𝑥 , 𝑥 , 𝑇 𝑡 0 𝐷 𝑡 1 / 2 𝑤 ( 𝑠 ) 𝐸 𝛼 𝑑 𝑠 . ( 2 . 3 4 ) Finally combining estimates (2.25), (2.26), (2.30), (2.33), and (2.34), we get 𝐷 𝑡 1 / 2 𝑤 𝐸 𝛼 𝑀 1 𝐴 𝜑 𝐸 𝛼 + 𝑀 2 + 𝑀 3 𝐹 𝐶 ( 𝐸 𝛼 ) + 𝑡 0 𝑀 4 𝜋 ( 𝑡 𝜉 ) 1 / 2 + 𝑀 5 𝐷 𝑡 1 / 2 𝑤 ( 𝑠 ) 𝐸 𝛼 𝑑 𝑠 . ( 2 . 3 5 ) Using the Gronwall’s inequality, we can write 𝐷 𝑡 1 / 2 𝑤 𝐸 𝛼 𝑒 𝑀 6 𝑀 1 𝐴 𝜑 𝐸 𝛼 + 𝑀 7 𝐹 𝐶 ( 𝐸 𝛼 ) . ( 2 . 3 6 ) From the last estimate, we can obtain the estimate for 𝑤 𝑡 ( 𝑡 ) by using problem (2.17) and well-posedness of the Cauchy problem in 𝐶 ( 𝐸 𝛼 ) (see [35]). So the following theorem finishes the proof of Theorem 2.2.

Theorem 2.3 (see, [36]). For 0 < 𝛼 < 1 / 2 , the norms of the spaces 𝐸 𝛼 ( 𝐶 [ 0 , 𝜋 ] , 𝐴 ) and 𝐶 2 𝛼 [ 0 , 𝜋 ] are equivalent.

3. Numerical Results

We have not been able to obtain a sharp estimate for the constants figuring in the stability inequalities. So we will provide the following results of numerical experiments of the following problem: 𝜕 𝑢 ( 𝑡 , 𝑥 ) = 𝜕 𝜕 𝑡 2 𝑢 ( 𝑡 , 𝑥 ) 𝜕 𝑥 2 𝑢 ( 𝑡 , 𝑥 ) + 𝐷 𝑡 1 / 2 1 𝑢 ( 𝑡 , 𝑥 ) + 𝑝 ( 𝑡 ) s i n 𝑥 + 𝑓 ( 𝑡 , 𝑥 ) , 𝑓 ( 𝑡 , 𝑥 ) = 3 𝑡 𝜋 𝑡 1 / 2 + 2 𝜋 𝑡 1 / 2 ] , [ ] , [ ] , 𝑢 𝜋 s i n 𝑥 , 𝑥 ( 0 , 𝜋 ) , 𝑡 ( 0 , 1 𝑢 ( 0 , 𝑥 ) = s i n 𝑥 , 𝑥 0 , 𝜋 𝑢 ( 𝑡 , 0 ) = 𝑢 ( 𝑡 , 𝜋 ) = 0 , 𝑡 0 , 1 𝑡 , 2 = 1 𝑡 . ( 3 . 1 ) The exact solution of the given problem is 𝑢 ( 𝑡 , 𝑥 ) = ( 1 𝑡 ) s i n 𝑥 and for the control parameter 𝑝 ( 𝑡 ) is 1 + 𝑡 .

3.1. The First Order of Accuracy Difference Scheme

For the approximate solution of the problem (3.1), the Rothe difference scheme 𝑢 𝑘 𝑛 𝑢 𝑛 𝑘 1 𝜏 = 𝑢 𝑘 𝑛 + 1 2 𝑢 𝑘 𝑛 + 𝑢 𝑘 𝑛 1 2 𝑢 𝑘 𝑛 + 𝐷 𝜏 1 / 2 𝑢 𝑘 𝑛 + 𝑝 𝑘 𝑞 𝑛 𝑡 + 𝑓 𝑘 , 𝑥 𝑛 , 𝑓 𝑡 𝑘 , 𝑥 𝑛 = 3 𝑡 𝑘 1 𝜋 𝑡 𝑘 1 / 2 + 2 𝜋 𝑡 𝑘 1 / 2 s i n 𝑥 𝑛 , 𝑝 𝑘 𝑡 = 𝑝 𝑘 , 𝑞 𝑛 = s i n 𝑥 𝑛 , 𝑥 𝑛 = 𝑛 , 𝑡 𝑘 𝑢 = 𝑘 𝜏 , 1 𝑘 𝑁 , 1 𝑛 𝑀 1 , 𝑀 = 𝜋 , 𝑁 𝜏 = 1 , 0 𝑛 = s i n 𝑥 𝑛 𝑢 , 0 𝑛 𝑀 , 𝑘 0 = 𝑢 𝑘 𝑀 𝑢 = 0 , 0 𝑘 𝑁 , 𝑘 𝑠 𝑡 = 𝜌 𝑘 𝑡 , 𝜌 𝑘 = 1 𝑡 𝑘 𝜋 , 0 𝑘 𝑁 , 𝑠 = , 2 ( 3 . 2 ) where 𝑥 denotes greatest integer less than 𝑥 is constructed. Throughout the paper, let us denote 𝜌 𝑡 𝑘 = 1 𝑡 𝑘 , 𝑞 𝑛 = s i n 𝑥 𝑛 , 𝑡 𝑘 = 𝑡 𝑘 , 𝑥 = 𝑘 𝜏 , 0 𝑘 𝑁 , 𝑁 𝜏 = 1 𝑛 = 𝑥 𝑛 , 𝑓 𝑡 = 𝑛 , 0 𝑛 𝑀 1 , 𝑀 = 𝜋 𝑘 , 𝑥 𝑛 = 3 𝑡 𝑘 1 𝜋 𝑡 𝑘 1 / 2 + 2 𝜋 𝑡 𝑘 1 / 2 s i n 𝑥 𝑛 , 𝐹 𝑡 𝑘 , 𝑥 𝑛 = 𝜌 𝑡 𝑘 𝑥 s i n 𝑠 𝑥 s i n 𝑛 + 1 𝑥 2 s i n 𝑛 𝑥 + s i n 𝑛 1 2 𝑥 s i n 𝑛 1 𝜋 𝑘 𝑚 = 1 Γ ( 𝑘 𝑚 + ( 1 / 2 ) ) 𝜏 ( 𝑘 𝑚 ) ! 1 / 2 𝑥 s i n 𝑠 𝑥 s i n 𝑛 𝑡 + 𝑓 𝑘 , 𝑥 𝑛 . ( 3 . 3 ) We search the solution of (3.2) in the following form: 𝑢 𝑘 𝑛 = 𝜂 𝑘 𝑞 𝑛 + 𝑤 𝑘 𝑛 , ( 3 . 4 ) where 𝜂 𝑘 = 𝑘 𝑖 = 1 𝑝 𝑖 𝜏 , 1 𝑘 𝑁 , 𝜂 0 = 0 . ( 3 . 5 ) Moreover for the interior grid point 𝑢 𝑘 𝑠 , we have that 𝑢 𝑘 𝑠 = 𝜂 𝑘 𝑞 𝑠 + 𝑤 𝑘 𝑠 𝑡 = 𝜌 𝑘 . ( 3 . 6 ) From (3.4), (3.5), and the condition 𝑢 𝑘 𝑠 = 𝜌 ( 𝑡 𝑘 ) , it follows that 𝜂 𝑘 = 𝜌 𝑡 𝑘 𝑤 𝑘 𝑠 𝑞 𝑠 , 𝑝 ( 3 . 7 ) 𝑘 = 𝜂 𝑘 𝜂 𝑘 1 𝜏 𝑢 , 1 𝑘 𝑁 , ( 3 . 8 ) 𝑘 𝑛 = 𝜌 𝑡 𝑘 𝑤 𝑘 𝑠 𝑞 𝑠 𝑞 𝑛 + 𝑤 𝑘 𝑛 , 0 𝑘 𝑁 , 0 𝑛 𝑀 , ( 3 . 9 ) where 𝑤 𝑘 𝑛 , 0 𝑘 𝑁 , 0 𝑛 𝑀 is the solution of the difference scheme 𝑤 𝑘 𝑛 𝑤 𝑛 𝑘 1 𝜏 = 𝑤 𝑘 𝑛 + 1 2 𝑤 𝑘 𝑛 + 𝑤 𝑘 𝑛 1 2 𝑤 𝑘 𝑛 𝑤 𝑘 𝑠 𝑥 s i n 𝑠 𝑥 s i n 𝑛 + 1 𝑥 2 s i n 𝑛 𝑥 + s i n 𝑛 1 2 𝑥 s i n 𝑛 1 𝜋 𝑘 𝑚 = 1 Γ ( 𝑘 𝑚 + ( 1 / 2 ) ) 𝑤 ( 𝑘 𝑚 ) ! 𝑚 𝑠 𝑤 𝑠 𝑚 1 𝑥 s i n 𝑠 𝜏 1 / 2 𝑥 s i n 𝑛 𝑤 𝑚 𝑛 𝑤 𝑛 𝑚 1 𝜏 1 / 2 𝑡 + 𝐹 𝑘 , 𝑥 𝑛 𝑤 , 1 𝑘 𝑁 , 1 𝑛 𝑀 1 , 𝑘 0 = 𝑤 𝑘 𝑀 𝑤 = 0 , 0 𝑘 𝑁 , 0 𝑛 𝑥 = s i n 𝑛 , 0 𝑛 𝑀 . ( 3 . 1 0 ) First, applying the first order of accuracy difference scheme (3.10), we obtain ( 𝑀 + 1 ) × ( 𝑀 + 1 ) system of linear equations and we write them in the matrix form 𝐴 𝑤 𝑘 + 𝑘 1 𝑗 = 0 𝐵 𝑗 𝑤 𝑗 = 𝐷 𝜑 𝑘 , 1 𝑘 𝑁 , 𝑤 0 = 𝑥 s i n 𝑛 𝑀 𝑛 = 0 , ( 3 . 1 1 ) where 𝐴 = 1 0 0 0 . 0 . 0 0 0 𝑥 𝑦 𝑥 0 . 𝑧 1 . 0 0 0 0 𝑥 𝑦 𝑥 . 𝑧 2 . 0 0 0 . . . . . . . . . . 0 0 0 0 . 𝑧 𝑀 1 . 𝑥 𝑦 𝑥 0 0 0 0 . 0 . 0 0 1 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) , 𝐵 0 = 0 0 0 . 0 . 0 0 0 𝑎 0 . 𝑓 1 . 0 0 0 0 𝑎 . 𝑓 2 . 0 0 . . . . . . . . 0 0 0 . 𝑓 𝑠 + 𝑎 . 0 0 . . . . . . . . 0 0 0 . 𝑓 𝑀 1 . 0 𝑎 0 0 0 . 0 . 0 0 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) , 𝐵 𝑗 = 0 0 0 . 0 . 0 0 0 𝑐 0 . 𝑑 1 . 0 0 0 0 𝑐 . 𝑑 2 . 0 0 . . . . . . . . 0 0 0 . 𝑑 𝑠 + 𝑐 . 0 0 . . . . . . . . 0 0 0 . 𝑑 𝑀 1 . 0 𝑐 0 0 0 . 0 . 0 0 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) , ( 3 . 1 2 ) for any 𝑗 = 1 , 2 , , 𝑘 2 , and 𝐵 𝑘 1 = 0 0 0 . 0 . 0 0 0 𝑣 0 . 𝑐 1 . 0 0 0 0 𝑣 . 𝑐 2 . 0 0 . . . . . . . . 0 0 0 . 𝑐 𝑠 + 𝑣 . 0 0 . . . . . . . . 0 0 0 . 𝑐 𝑀 1 . 0 𝑣 0 0 0 . 0 . 0 0 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) . ( 3 . 1 3 ) Here, for any 𝑛 = 1 , 2 , , 𝑀 1 , 1 𝑥 = 2 1 , 𝑦 = 𝜏 + 2 2 1 + 1 + 𝜏 , 𝑧 𝑛 = 𝑥 s i n 𝑛 + 1 𝑥 2 s i n 𝑛 𝑥 + s i n 𝑛 1 𝑥 s i n 𝑠 2 𝑥 s i n 𝑛 𝑥 s i n 𝑠 𝑥 s i n 𝑛 𝑥 s i n 𝑠 𝜏 Γ , i n ( 𝑠 + 1 ) t h c o l u m n , 𝑎 = ( 𝑘 1 / 2 ) 𝜏 𝜋 ( 𝑘 1 ) ! , 𝑓 𝑛 𝑥 = s i n 𝑛 Γ ( 𝑘 1 / 2 ) 𝑥 𝜏 𝜋 s i n 𝑠 , 1 ( 𝑘 1 ) ! 𝑐 = 𝜏 𝜋 Γ ( 𝑘 𝑗 1 / 2 ) ( 𝑘 𝑗 1 ) ! Γ ( 𝑘 𝑗 + 1 / 2 ) , 𝑑 ( 𝑘 𝑗 ) ! 𝑛 = 𝑥 s i n 𝑛 𝑥 𝜏 𝜋 s i n 𝑠 Γ ( 𝑘 𝑗 1 / 2 ) ( 𝑘 𝑗 1 ) ! Γ ( 𝑘 𝑗 + 1 / 2 ) , 1 ( 𝑘 𝑗 ) ! 𝑣 = 𝜏 1 𝜏 , 𝑐 𝑛 = 𝑥 s i n 𝑛 𝑥 𝜏 s i n 𝑠 , 𝑤 𝑟 = 𝑤 𝑟 0 𝑤 𝑟 𝑀 ( 𝑀 + 1 ) × 1 𝜑 f o r a n y 𝑟 = 0 , 1 , , 𝑘 , 𝑘 = 0 𝜙 𝑘 1 𝜙 𝑘 𝑀 1 0 ( 𝑀 + 1 ) × 1 , 𝜙 𝑘 𝑛 = 𝜌 𝑡 𝑘 𝑥 s i n 𝑠 𝑥 s i n 𝑛 + 1 𝑥 2 s i n 𝑛 𝑥 + s i n 𝑛 1 2 𝜌 𝑡 𝑘 𝑥 s i n 𝑠 𝑥 s i n 𝑛 1 𝜋 𝑘 𝑚 = 1 Γ ( 𝑘 𝑚 + 1 / 2 ) 𝜏 ( 𝑘 𝑚 ) ! 1 / 2 𝑥 s i n 𝑠 𝑥 s i n 𝑛 𝑡 + 𝑓 𝑘 , 𝑥 𝑛 , ( 3 . 1 4 ) and 𝐷 is ( 𝑀 + 1 ) × ( 𝑀 + 1 ) identity matrix. Using (3.11), we can obtain that 𝑤 𝑘 = 𝐴 1 𝐷 𝜑 𝑘 𝑘 1 𝑗 = 0 𝐵 𝑗 𝑤 𝑗 , 𝑘 = 1 , 2 , , 𝑁 , 𝑤 0 = s i n 𝑥 𝑛 𝑀 𝑛 = 0 . ( 3 . 1 5 ) To solve the resulting difference equations, we apply the method given in (3.15) step by step for 𝑘 = 1 , 2 , , 𝑁 . For the evaluation of 𝑤 𝑟 , 𝑟 = 2 , 3 , , 𝑁 , 𝑤 𝑟 1 is needed. It is obtained in the previous step. Then, the solution pairs ( 𝑢 , 𝑝 ) are obtained by using the last formulas (3.9) and (3.8).

3.2. The Second Order of Accuracy Difference Scheme

For the approximate solution of the problem (3.1), the Crank-Nicholson difference scheme 𝑢 𝑘 𝑛 𝑢 𝑛 𝑘 1 𝜏 = 𝑢 𝑘 𝑛 + 1 2 𝑢 𝑘 𝑛 + 𝑢 𝑘 𝑛 1 2 2 + 𝑢 𝑘 1 𝑛 + 1 2 𝑢 𝑛 𝑘 1 + 𝑢 𝑘 1 𝑛 1 2 2 𝑢 𝑘 𝑛 + 𝑢 𝑛 𝑘 1 2 + 𝑝 𝑘 + 𝑝 𝑘 1 2 𝑞 𝑛 + 𝐷 𝜏 1 / 2 𝑢 𝑡 𝑘 𝜏 2 , 𝑥 𝑛 𝑡 + 𝑓 𝑘 𝜏 2 , 𝑥 𝑛 , 𝑝 𝑘 𝑡 = 𝑝 𝑘 𝑢 , 1 𝑘 𝑁 , 1 𝑛 𝑀 1 0 𝑛 𝑥 = s i n 𝑛 𝑢 , 0 𝑛 𝑀 , 𝑘 0 = 𝑢 𝑘 𝑀 𝑢 = 0 , 0 𝑘 𝑁 , 𝑘 𝑠 + 𝑢 𝑘 𝑠 + 1 𝑢 𝑘 𝑠 𝑥 𝑡 𝑠 = 𝜌 𝑘 𝜋 , 0 𝑘 𝑁 , 𝑠 = 2 ( 3 . 1 6 ) is constructed.

Here, Γ 1 𝑘 𝑟 + 2 = 0 𝑡 𝑘 𝑟 + 1 / 2 𝑒 𝑡 𝑑 𝑡 . ( 3 . 1 7 ) Moreover, applying the second order of approximation formula for 𝐷 𝑡 1 / 2 𝑢 𝑡 𝑘 𝜏 2 = 1 Γ ( 1 / 2 ) 𝑡 𝑘 0 𝜏 / 2 𝑡 𝑘 𝜏 2 𝑠 1 / 2 𝑢 ( 𝑠 ) 𝑑 𝑠 , ( 3 . 1 8 ) it is obtained (see [27]) 𝐷 𝜏 1 / 2 𝑢 = 𝑑 2 3 𝑢 0 + 𝑑 2 3 𝑢 1 + 𝑑 𝜏 3 2 𝑥 s i n 𝑛 , 𝑘 = 1 , 2 𝑑 6 5 𝑢 0 + 𝑑 6 5 𝑢 1 + 𝑑 6 5 𝑢 2 𝑑 𝜏 6 𝑥 1 0 s i n 𝑛 𝑑 , 𝑘 = 2 , 𝑘 1 𝑚 = 2 ( 𝑘 𝑚 ) 𝑏 1 + 𝑏 2 𝑢 𝑚 2 + ( 2 𝑚 2 𝑘 1 ) 𝑏 1 2 𝑏 2 𝑢 𝑚 1 + ( 𝑘 𝑚 + 1 ) 𝑏 1 + 𝑏 2 𝑢 𝑚 + 𝑑 6 2 𝑢 𝑘 2 4 𝑢 𝑘 1 + 5 𝑢 𝑘 , 3 𝑘 𝑁 . ( 3 . 1 9 ) Here and throughout the paper, 𝐷 𝑡 1 / 2 𝑢 = 𝐷 𝜏 1 / 2 𝑢 𝑡 𝑘 𝜏 2 , 𝑥 𝑛 , 2 𝑑 = 𝜋 𝜏 , 𝑏 1 = 1 𝑘 𝑚 + 2 1 𝑘 𝑚 2 , 𝑏 2 1 = 3 1 𝑘 𝑚 + 2 3 / 2 1 𝑘 𝑚 2 3 / 2 . ( 3 . 2 0 ) We search the solution of (3.16) in the following form: 𝑢 𝑘 𝑛 = 𝜂 𝑘 𝑞 𝑛 + 𝑤 𝑘 𝑛 , ( 3 . 2 1 ) where 𝜂 𝑘 = 𝑘 𝑖 = 1 𝑝 𝑖 + 𝑝 𝑖 1 2 𝜏 , 1 𝑘 𝑁 , 𝜂 0 = 0 . ( 3 . 2 2 ) We have that 𝑢 𝑘 𝑠 + 𝑢 𝑘 𝑠 + 1 𝑢 𝑘 𝑠 𝑥 𝑠 = 𝜂 𝑘 𝑥 1 𝑠 𝑞 𝑠 + 𝑥 𝑠 𝑞 𝑠 + 1 + 𝑥 1 𝑠 𝑤 𝑘 𝑠 + 𝑥 𝑠 𝑤 𝑘 𝑠 + 1 𝑡 = 𝜌 𝑘 . ( 3 . 2 3 ) Let us denote 𝑥 𝑦 = 𝑠 = 𝑥 𝑥 , ( 3 . 2 4 ) where 0 𝑦 < 1 . Then, one can write 𝜂 𝑘 = 𝜌 𝑡 𝑘 ( 1 𝑦 ) 𝑤 𝑘 𝑠 𝑦 𝑤 𝑘 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 . ( 3 . 2 5 ) So the values of ( 𝑝 ( 𝑡 𝑘 ) + 𝑝 ( 𝑡 𝑘 1 ) ) / 2 , 1 𝑘 𝑁 can be obtained by the following formula: 𝑝 𝑘 + 𝑝 𝑘 1 2 = 𝜌 𝑡 𝑘 𝑡 𝜌 𝑘 1 𝑤 / 𝜏 ( 1 𝑦 ) 𝑘 𝑠 𝑤 𝑠 𝑘 1 𝑤 / 𝜏 𝑦 𝑘 𝑠 + 1 𝑤 𝑘 1 𝑠 + 1 / 𝜏 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 . ( 3 . 2 6 ) Let 𝑤 𝑟 denote 𝑤 𝑟 = 𝑤 𝑟 0 𝑤 𝑟 𝑀 ( 𝑀 + 1 ) × 1 f o r 𝑟 = 0 , 1 , , 𝑁 . ( 3 . 2 7 ) For 𝑘 = 1 , one can show that 𝑤 1 is the solution of the difference scheme 𝑤 1 𝑛 𝑤 0 𝑛 𝜏 = 𝑤 1 𝑛 + 1 2 𝑤 1 𝑛 + 𝑤 1 𝑛 1 2 2 + 𝑤 0 𝑛 + 1 2 𝑤 0 𝑛 + 𝑤 0 𝑛 1 2 2 𝑤 1 𝑛 + 𝑤 0 𝑛 2 + 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 2 2 𝑞 𝑛 2 × 𝜌 𝑡 1 ( 1 𝑦 ) 𝑤 1 𝑠 𝑦 𝑤 1 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝜌 𝑡 0 ( 1 𝑦 ) 𝑤 0 𝑠 𝑦 𝑤 0 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 2 3 𝜌 𝑡 1 ( 1 𝑦 ) 𝑤 1 𝑠 𝑦 𝑤 1 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑞 ( 𝑛 ) + 𝑤 1 𝑛 𝑑 2 3 𝜌 𝑡 0 ( 1 𝑦 ) 𝑤 0 𝑠 𝑦 𝑤 0 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑞 ( 𝑛 ) + 𝑤 0 𝑛 + 𝑑 𝜏 3 2 𝑡 𝑞 ( 𝑛 ) + 𝑓 1 𝜏 2 , 𝑥 𝑛 𝑤 , 1 𝑛 𝑀 1 , 1 0 = 𝑤 1 𝑀 𝑤 = 0 , 0 𝑛 𝑥 = s i n 𝑛 , 0 𝑛 𝑀 . ( 3 . 2 8 ) We have the system of linear equations and we write them in the matrix form 𝐴 1 𝑤 1 + 𝐵 1 𝑤 0 = 𝐷 𝜑 1 , ( 3 . 2 9 ) where 𝐴 1 = 1 0 0 0 0 0 0 0 𝑎 𝑦 1 𝑎 0 𝑙 1 𝑐 1 0 0 0 0 𝑎 𝑦 1 𝑎 𝑙 2 𝑐 2 0 0 0 0 0 0 𝑎 𝑙 𝑠 + 𝑎 𝑐 𝑠 0 0 0 0 0 0 0 𝑙 𝑠 + 1 + 𝑦 1 𝑐 𝑠 + 1 + 𝑎 0 0 0 0 0 0 0 𝑙 𝑀 1 𝑐 𝑀 1 𝑎 𝑦 1 𝑎 0 0 0 0 0 0 0 0 1 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) , 𝐵 1 = 0 0 0 0 0 0 0 0 𝑎 𝑣 1 𝑎 0 𝑑 1 𝑒 1 0 0 0 0 𝑎 𝑣 1 𝑎 𝑑 2 𝑒 2 0 0 0 0 0 0 𝑎 𝑑 𝑠 + 𝑎 𝑒 𝑠 0 0 0 0 0 0 0 𝑑 𝑠 + 1 + 𝑣 1 𝑒 𝑠 + 1 + 𝑎 0 0 0 0 0 0 0 𝑑 𝑀 1 𝑒 𝑀 1 𝑎 𝑣 1 𝑎 0 0 0 0 0 0 0 0 0 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) . ( 3 . 3 0 ) Here, for any 𝑛 = 1 , 2 , , 𝑀 1 , 1 𝑎 = 2 2 , 𝑦 1 = 1 𝜏 + 1 2 + 1 2 𝑑 2 3 , 𝑣 1 = 1 𝜏 + 1 2 + 1 2 + 𝑑 2 3 , 𝑙 𝑛 = 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 ( 1 𝑦 ) 2 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 𝑛 ( 1 𝑦 ) 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 2 3 𝑞 𝑛 , 𝑐 𝑛 = 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 𝑦 2 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 𝑛 𝑦 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 2 3 𝑞 𝑛 , 𝑑 𝑛 = 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 ( 1 𝑦 ) 2 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 𝑛 ( 1 𝑦 ) 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑑 2 3 𝑞 𝑛 , 𝑒 𝑛 = 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 𝑦 2 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 𝑛 𝑦 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑑 2 3 𝑞 𝑛 , 𝜑 1 = 0 𝜙 1 1 𝜙 1 𝑀 1 0 ( 𝑀 + 1 ) × 1 , 𝜙 1 𝑛 = 𝑥 s i n 𝑛 + 1 𝑥 2 s i n 𝑛 𝑥 + s i n 𝑛 1 2 2 𝑥 s i n 𝑛 2 𝜌 𝑡 1 𝑡 + 𝜌 0 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 2 𝑞 𝑛 3 𝜌 𝑡 1 𝑡 𝜌 0 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 𝜏 3 2 𝑞 𝑛 𝑡 + 𝑓 1 𝜏 2 , 𝑥 𝑛 ( 3 . 3 1 ) and 𝐷 is ( 𝑀 + 1 ) × ( 𝑀 + 1 ) identity matrix. Using (3.29), we can obtain that 𝑤 1 = 𝐴 1 1 𝐷 𝜑 1 𝐵 1 𝑤 0 , 𝑤 0 = s i n 𝑥 𝑛 𝑀 𝑛 = 0 . ( 3 . 3 2 ) For 𝑘 = 2 ,   𝑤 2 is the solution of the difference scheme 𝑤 2 𝑛 𝑤 1 𝑛 𝜏 = 𝑤 2 𝑛 + 1 2 𝑤 2 𝑛 + 𝑤 2 𝑛 1 2 2 𝑤 2 𝑛 + 𝑤 1 𝑛 2 + 𝑤 1 𝑛 + 1 2 𝑤 1 𝑛 + 𝑤 1 𝑛 1 2 2 + 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 2 2 𝑞 𝑛 2 𝜌 𝑡 2 ( 1 𝑦 ) 𝑤 2 𝑠 𝑦 𝑤 2 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝜌 𝑡 1 ( 1 𝑦 ) 𝑤 1 𝑠 𝑦 𝑤 1 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 6 5 𝜌 𝑡 2 ( 1 𝑦 ) 𝑤 2 𝑠 𝑦 𝑤 2 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑞 ( 𝑛 ) + 𝑤 2 𝑛 + 𝑑 6 5 𝜌 𝑡 1 ( 1 𝑦 ) 𝑤 1 𝑠 𝑦 𝑤 1 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑞 ( 𝑛 ) + 𝑤 1 𝑛 2 𝑑 6 5 𝜌 𝑡 0 ( 1 𝑦 ) 𝑤 0 𝑠 𝑦 𝑤 0 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑞 ( 𝑛 ) + 𝑤 0 𝑛 𝑑 𝜏 6 𝑡 1 0 𝑞 ( 𝑛 ) + 𝑓 2 𝜏 2 , 𝑥 𝑛 𝑤 , 1 𝑛 𝑀 1 , 2 0 = 𝑤 2 𝑀 𝑤 = 0 , 0 𝑛 𝑥 = s i n 𝑛 , 0 𝑛 𝑀 . ( 3 . 3 3 ) The system of linear equations given above can be written in the matrix form 𝐴 2 𝑤 2 + 𝐵 2 𝑤 1 + 𝐶 2 𝑤 0 = 𝐷 𝜑 2 , ( 3 . 3 4 ) where 𝐴 2 = 1 0 0 0 0 0 0 0 𝑎 𝑦 2 𝑎 0 𝑔 1 1 0 0 0 0 𝑎 𝑦 2 𝑎 𝑔 2 2 0 0 0 0 0 0 0 𝑔 𝑠 + 𝑎 𝑠 0 0 0 0 0 0 0 𝑔 𝑠 + 1 + 𝑦 2 𝑠 + 1 + 𝑎 0 0 0 0 0 0 0 𝑔 𝑀 1 𝑀 1 𝑎 𝑦 2 𝑎 0 0 0 0 0 0 0 0 1 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) , 𝐵 2 = 0 0 0 0 0 0 0 0 𝑎 𝑣 2 𝑎 0 𝑔 1 1 0 0 0 0 𝑎 𝑣 2 𝑎 𝑔 2 2 0 0 0 0 0 0 0 𝑔 𝑠 + 𝑎 𝑠 0 0 0 0 0 0 0 𝑔 𝑠 + 1 + 𝑣 2 𝑠 + 1 + 𝑎 0 0 0 0 0 0 0 𝑔 𝑀 1 𝑀 1 𝑎 𝑣 2 𝑎 0 0 0 0 0 0 0 0 0 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) , 𝐶 2 = 0 0 0 0 0 0 0 0 0 𝑧 0 0 𝑖 1 𝑗 1 0 0 0 0 0 𝑧 0 𝑖 2 𝑗 2 0 0 0 0 0 0 0 𝑖 𝑠 + 𝑧 𝑗 𝑠 0 0 0 0 0 0 0 𝑖 𝑠 + 1 𝑗 𝑠 + 1 + 𝑧 0 0 0 0 0 0 0 𝑖 𝑀 1 𝑗 𝑀 1 0 𝑧 0 0 0 0 0 0 0 0 0 0 ( 𝑀 + 1 ) × ( 𝑀 + 1 ) . ( 3 . 3 5 ) Here, for any 𝑛 = 1 , 2 , , 𝑀 1 , 1 𝑎 = 2 2 , 𝑦 2 = 1 𝜏 + 1 2 + 1 2 𝑑 6 5 , 𝑣 2 1 = 𝜏 + 1 2 + 1 2 𝑑 6 5 , 𝑔 𝑛 = 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 ( 1 𝑦 ) 2 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 𝑛 ( 1 𝑦 ) 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 6 𝑞 𝑛 ( 1 𝑦 ) 5 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 , i n ( 𝑠 + 1 ) t h c o l u m n , 𝑛 = 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 𝑦 2 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 𝑛 𝑦 2 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 6 𝑞 𝑛 𝑦 5 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑖 , i n ( 𝑠 + 2 ) t h c o l u m n , 𝑛 = 2 𝑑 6 𝑞 𝑛 ( 1 𝑦 ) 5 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 , 𝑗 𝑛 2 = 6 𝑞 𝑛 𝑦 5 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 , 𝑧 = 2 𝑑 6 5 , 𝜑 2 = 0 𝜙 2 1 𝜙 2 𝑀 1 0 ( 𝑀 + 1 ) × 1 , 𝜙 2 𝑛 = 𝑥 s i n 𝑛 + 1 𝑥 2 s i n 𝑛 𝑥 + s i n 𝑛 1 2 2 𝑥 s i n 𝑛 2 + 𝑑 6 𝑞 𝑛 5 × 𝜌 𝑡 2 𝑡 + 𝜌 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 2 𝑑 6 𝑞 𝑛 5 𝜌 𝑡 0 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑑 𝜏 6 𝑞 1 0 𝑛 𝑡 + 𝑓 2 𝜏 2 , 𝑥 𝑛 . ( 3 . 3 6 ) Using (3.34), we can obtain that 𝑤 2 = 𝐴 2 1 𝐷 𝜑 2 𝐵 2 𝑤 1 𝐶 2 𝑤 0 , 𝑤 0 = s i n 𝑥 𝑛 𝑀 𝑛 = 0 . ( 3 . 3 7 ) For 3 𝑘 𝑁 , we can obtain the following difference scheme: 𝑤 𝑘 𝑛 𝑤 𝑛 𝑘 1 𝜏 = 𝑤 𝑘 𝑛 + 1 2 𝑤 𝑘 𝑛 + 𝑤 𝑘 𝑛 1 2 2 + 𝑤 𝑘 1 𝑛 + 1 2 𝑤 𝑛 𝑘 1 + 𝑤 𝑘 1 𝑛 1 2 2 𝑤 𝑘 𝑛 + 𝑤 𝑛 𝑘 1 2 + 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 2 2 𝑞 𝑛 2 × 𝜌 𝑡 𝑘 ( 1 𝑦 ) 𝑤 𝑘 𝑠 𝑦 𝑤 𝑘 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 𝑛 + 1 2 𝑞 𝑛 + 𝑞 𝑛 1 2 2 𝑞 𝑛 2 × 𝜌 𝑡 𝑘 1 ( 1 𝑦 ) 𝑤 𝑠 𝑘 1 𝑦 𝑤 𝑘 1 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑑 𝑘 1 𝑚 = 2 ( 𝑘 𝑚 ) 𝑏 1 + 𝑏 2 × 𝜌 𝑡 𝑚 2 ( 1 𝑦 ) 𝑤 𝑠 𝑚 2 𝑦 𝑤 𝑚 2 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 ( 𝑛 ) ( 2 𝑚 2 𝑘 1 ) 𝑏 1 2 𝑏 2 × 𝜌 𝑡 𝑚 1 ( 1 𝑦 ) 𝑤 𝑠 𝑚 1 𝑦 𝑤 𝑚 1 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 ( 𝑛 ) ( 2 𝑚 2 𝑘 1 ) 𝑏 1 2 𝑏 2 𝑤 𝑛 𝑚 1 × 𝜌 𝑡 𝑚 ( 1 𝑦 ) 𝑤 𝑚 𝑠 𝑦 𝑤 𝑚 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 + 𝑞 ( 𝑛 ) ( 𝑘 𝑚 ) 𝑏 1 + 𝑏 2 𝑤 𝑛 𝑚 2 + ( 𝑘 𝑚 1 ) 𝑏 1 + 𝑏 2 𝑤 𝑚 𝑛 + ( 2 𝑚 2 𝑘 1 ) 𝑏 1 2 𝑏 2 𝑤 𝑛 𝑚 1 𝑑 6 2 𝜌 𝑡 𝑘 2 ( 1 𝑦 ) 𝑤 𝑠 𝑘 2 𝑦 𝑤 𝑘 2 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑞 ( 𝑛 ) + 𝑤 𝑛 𝑘 2 4 𝑑 6 2 𝜌 𝑡 𝑘 1 ( 1 𝑦 ) 𝑤 𝑠 𝑘 1 𝑦 𝑤 𝑘 1 𝑠 + 1 ( 1 𝑦 ) 𝑞 𝑠 + 𝑦 𝑞 𝑠 + 1 𝑞 ( 𝑛 ) + 𝑤 𝑛 𝑘 1 + 5 𝑑 6 2 𝜌 𝑡 𝑘 ( 1 𝑦 ) 𝑤