Journal of Applied Mathematics and Stochastic Analysis
Volume 2009 (2009), Article ID 254720, 18 pages
doi:10.1155/2009/254720
Research Article

Implicit Difference Inequalities Corresponding to First-Order Partial Differential Functional Equations

Institite of Mathematics, University of Gdańsk, Wit Stwosz Street 57, 80-952 Gdańsk, Poland

Received 19 August 2008; Accepted 5 January 2009

Academic Editor: Donal O'Regan

Copyright © 2009 Z. Kamont and K. Kropielnicka. 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

We give a theorem on implicit difference functional inequalities generated by mixed problems for nonlinear systems of first-order partial differential functional equations. We apply this result in the investigations of the stability of difference methods. Classical solutions of mixed problems are approximated in the paper by solutions of suitable implicit difference schemes. The proof of the convergence of difference method is based on comparison technique, and the result on difference functional inequalities is used. Numerical examples are presented.

1. Introduction

The papers [1, 2] initiated the theory of difference inequalities generated by first-order partial differential equations. The results and the methods presented in [1, 2] were extended in [3, 4] on functional differential problems, and they were generalized in [58] on parabolic differential and differential functional equations. Explicit difference schemes were considered in the above papers.

Our purpose is to give a result on implicit difference inequalities corresponding to initial boundary value problems for first-order functional differential equations.

We prove also that that there are implicit difference methods which are convergent. The proof of the convergence is based on a theorem on difference functional inequalities.

We formulate our functional differential problems. For any metric spaces 𝑋 and 𝑌 we denote by 𝐶 ( 𝑋 , 𝑌 ) the class of all continuous functions from 𝑋 into 𝑌 . We will use vectorial inequalities with the understanding that the same inequalities hold between their corresponding components. Write 𝐸 = [ 0 , 𝑎 ] × ( 𝑏 , 𝑏 ) , 𝐷 = 𝑑 0 , 0 × [ 𝑑 , 𝑑 ] , ( 1 . 1 ) where 𝑎 > 0 , 𝑏 = ( 𝑏 1 , , 𝑏 𝑛 ) 𝑛 , 𝑏 𝑖 > 0 for 1 𝑖 𝑛 and 𝑑 = ( 𝑑 1 , , 𝑑 𝑛 ) 𝑛 + , 𝑑 0 + , + = [ 0 , + ) . Let 𝑐 = 𝑏 + 𝑑 and 𝐸 0 = 𝑑 0 𝜕 , 0 × [ 𝑐 , 𝑐 ] , 0 𝐸 = [ 0 , 𝑎 ] × ( [ 𝑐 , 𝑐 ] ( 𝑏 , 𝑏 ) ) , Ω = 𝐸 𝐸 0 𝜕 0 𝐸 . ( 1 . 2 ) For a function 𝑧 Ω 𝑘 , 𝑧 = ( 𝑧 1 , , 𝑧 𝑘 ) , and for a point ( 𝑡 , 𝑥 ) 𝐸 where 𝐸 is the closure of 𝐸 , we define a function 𝑧 ( 𝑡 , 𝑥 ) 𝐷 𝑘 by 𝑧 ( 𝑡 , 𝑥 ) ( 𝜏 , 𝑦 ) = 𝑧 ( 𝑡 + 𝜏 , 𝑥 + 𝑦 ) , ( 𝜏 , 𝑦 ) 𝐷 . Then 𝑧 ( 𝑡 , 𝑥 ) is the restriction of 𝑧 to the set [ 𝑡 𝑑 0 , 𝑡 ] × [ 𝑥 𝑑 , 𝑥 + 𝑑 ] and this restriction is shifted to the set 𝐷 . Write Σ = 𝐸 × 𝐶 ( 𝐷 , 𝑘 ) × 𝑛 and suppose that 𝑓 = ( 𝑓 1 , , 𝑓 𝑘 ) Σ 𝑘 and 𝜑 𝐸 0 𝜕 0 𝐸 𝑘 , 𝜑 = ( 𝜑 1 , , 𝜑 𝑘 ) , are given functions. Let us denote by 𝑧 = ( 𝑧 1 , , 𝑧 𝑘 ) an unknown function of the variables ( 𝑡 , 𝑥 ) , 𝑥 = ( 𝑥 1 , , 𝑥 𝑛 ) . Write 𝑓 𝔽 [ 𝑧 ] ( 𝑡 , 𝑥 ) = 1 𝑡 , 𝑥 , 𝑧 ( 𝑡 , 𝑥 ) , 𝜕 𝑥 𝑧 1 ( 𝑡 , 𝑥 ) , , 𝑓 𝑘 𝑡 , 𝑥 , 𝑧 ( 𝑡 , 𝑥 ) , 𝜕 𝑥 𝑧 𝑘 ( 𝑡 , 𝑥 ) ( 1 . 3 ) and 𝜕 𝑥 𝑧 𝑖 = ( 𝜕 𝑥 1 𝑧 𝑖 , , 𝜕 𝑥 𝑛 𝑧 𝑖 ) , 1 𝑖 𝑘 . We consider the system of functional differential equations 𝜕 𝑡 𝑧 ( 𝑡 , 𝑥 ) = 𝔽 [ 𝑧 ] ( 𝑡 , 𝑥 ) ( 1 . 4 ) with the initial boundary condition 𝑧 ( 𝑡 , 𝑥 ) = 𝜑 ( 𝑡 , 𝑥 ) o n 𝐸 0 𝜕 0 𝐸 . ( 1 . 5 ) In the paper we consider classical solutions of (1.4), (1.5).

We give examples of equations which can be obtained from (1.4) by specializing the operator 𝑓 .

Example 1.1. Suppose that the function 𝛼 𝐸 1 + 𝑛 satisfies the condition: 𝛼 ( 𝑡 , 𝑥 ) ( 𝑡 , 𝑥 ) 𝐷 for ( 𝑡 , 𝑥 ) 𝐸 . For a given 𝑓 𝑓 = ( 1 𝑓 , , 𝑘 ) 𝐸 × 𝑘 × 𝑘 × 𝑛 𝑘 we put 𝑓 𝑓 ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) = 𝑡 , 𝑥 , 𝑤 ( 0 , 𝜃 ) , 𝑤 𝛼 ( 𝑡 , 𝑥 ) ( 𝑡 , 𝑥 ) , 𝑞 o n Σ , ( 1 . 6 ) where 𝜃 = ( 0 , , 0 ) 𝑛 . Then (1.4) is reduced to the system of differential equations with deviated variables 𝜕 𝑡 𝑧 𝑖 𝑓 ( 𝑡 , 𝑥 ) = 𝑖 𝑡 , 𝑥 , 𝑧 ( 𝑡 , 𝑥 ) , 𝑧 𝛼 ( 𝑡 , 𝑥 ) , 𝜕 𝑥 𝑧 𝑖 ( 𝑡 , 𝑥 ) , 𝑖 = 1 , , 𝑘 . ( 1 . 7 )

Example 1.2. For the above 𝑓 we define 𝑓 𝑓 ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) = 𝑡 , 𝑥 , 𝑤 ( 0 , 𝜃 ) , 𝐷 𝑤 ( 𝜏 , 𝑦 ) 𝑑 𝑦 𝑑 𝜏 , 𝑞 o n Σ . ( 1 . 8 ) Then (1.4) is equivalent to the system of differential integral equations 𝜕 𝑡 𝑧 𝑖 𝑓 ( 𝑡 , 𝑥 ) = 𝑖 𝑡 , 𝑥 , 𝑧 ( 𝑡 , 𝑥 ) , 𝐷 𝑧 ( 𝑡 + 𝜏 , 𝑥 + 𝑦 ) 𝑑 𝑦 𝑑 𝜏 , 𝜕 𝑥 𝑧 𝑖 ( 𝑡 , 𝑥 ) , 𝑖 = 1 , , 𝑘 . ( 1 . 9 )

It is clear that more complicated differential systems with deviated variables and differential integral problems can be obtained from (1.4) by a suitable definition of 𝑓 . Sufficient conditions for the existence and uniqueness of classical or generalized solutions of (1.4), (1.5) can be found in [9, 10].

Our motivations for investigations of implicit difference functional inequalities and for the construction of implicit difference schemes are the following. Two types of assumptions are needed in theorems on the stability of difference functional equations generated by (1.4), (1.5). The first type conditions concern regularity of 𝑓 . It is assumed that

(i)the function 𝑓 of the variables ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) , 𝑞 = ( 𝑞 1 , , 𝑞 𝑛 ) , is of class 𝐶 1 with respect to 𝑞 and the functions 𝜕 𝑞 𝑓 𝑖 = ( 𝜕 𝑞 1 𝑓 𝑖 , , 𝜕 𝑞 𝑛 𝑓 𝑖 ) , 1 𝑖 𝑘 , are bounded,(ii) 𝑓 satisfies the Perron type estimates with respect to the functional variable 𝑤 . The second type conditions concern the mesh. It is required that difference schemes generated by (1.4), (1.5) satisfy the condition 1 0 𝑛 𝑗 = 1 1 𝑗 | | 𝜕 𝑞 𝑗 𝑓 𝑖 | | ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) 0 o n Σ f o r 𝑖 = 1 , , 𝑘 , ( 1 . 1 0 ) where 0 and = ( 1 , , 𝑛 ) are steps of the mesh with respect to 𝑡 and ( 𝑥 1 , , 𝑥 𝑛 ) respectively. The above assumption is known as a generalized Courant-Friedrichs-Levy (CFL) condition for (1.4), (1.5) (see [11, Chapter 3] and [10, Chapter 5]). It is clear that strong assumptions on relations between 0 and are required in (1.10). It is important in our considerations that assumption (1.10) is omitted in a theorem on difference inequalities and in a theorem on the convergence of difference schemes.

We show that there are implicit difference methods for (1.4), (1.5) which are convergent while the corresponding explicit difference schemes are not convergent. We give suitable numerical examples.

The paper is organized as follows. A theorem on implicit difference functional inequalities with unknown function of several variables is proved in Section 2. We propose in Section 3 implicit difference schemes for the numerical solving of functional differential equations. Convergence results and error estimates are presented. A theorem on difference inequalities is used in the investigation of the stability of implicit difference methods. Numerical examples are given in the last part of the paper.

We use in the paper general ideas for finite difference equations which were introduced in [1214]. For further bibliographic informations concerning differential and functional differential inequalities and applications see the survey paper [15] and the monographs [16, 17].

2. Functional Difference Inequalities

For any two sets 𝑈 and 𝑊 we denote by 𝐹 ( 𝑈 , 𝑊 ) the class of all functions defined on 𝑈 and taking values in 𝑊 . Let and be the sets of natural numbers and integers, respectively. For 𝑥 = ( 𝑥 1 , , 𝑥 𝑛 ) 𝑛 , 𝑝 = ( 𝑝 1 , , 𝑝 𝑘 ) 𝑘 we put | | 𝑥 𝑥 = 1 | | | | 𝑥 + + 𝑛 | | , 𝑝 | | 𝑝 = m a x 𝑖 | | 1 𝑖 𝑘 . ( 2 . 1 ) We define a mesh on Ω in the following way. Suppose that ( 0 , ) , = ( 1 , , 𝑛 ) , stand for steps of the mesh. For ( 𝑟 , 𝑚 ) 1 + 𝑛 where 𝑚 = ( 𝑚 1 , , 𝑚 𝑛 ) , we define nodal points as follows: 𝑡 ( 𝑟 ) = 𝑟 0 , 𝑥 ( 𝑚 ) = 𝑥 ( 𝑚 1 ) 1 , , 𝑥 ( 𝑚 𝑛 ) 𝑛 = 𝑚 1 1 , , 𝑚 𝑛 𝑛 . ( 2 . 2 ) Let us denote by 𝐻 the set of all = ( 0 , ) such that there are 𝐾 0 and 𝐾 = ( 𝐾 1 , , 𝐾 𝑛 ) 𝑛 satisfying the conditions: 𝐾 0 0 = 𝑑 0 and ( 𝐾 1 1 , , 𝐾 𝑛 𝑛 ) = 𝑑 . Set 1 + 𝑛 = 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) ( 𝑟 , 𝑚 ) 1 + 𝑛 , 𝐷 = 𝐷 1 + 𝑛 , 𝐸 = 𝐸 1 + 𝑛 , 𝐸 0 . = 𝐸 0 1 + 𝑛 , 𝜕 0 𝐸 = 𝜕 0 𝐸 1 + 𝑛 , Ω = 𝐸 𝐸 0 . 𝜕 0 𝐸 . ( 2 . 3 ) Let 𝑁 0 be defined by the relations: 𝑁 0 0 𝑎 < ( 𝑁 0 + 1 ) 0 and 𝐸 = 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) 𝐸 0 𝑟 𝑁 0 1 . ( 2 . 4 ) For functions 𝑤 𝐷 𝑘 and 𝑧 Ω 𝑘 we write 𝑤 ( 𝑟 , 𝑚 ) = 𝑤 ( 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) ) on 𝐷 and 𝑧 ( 𝑟 , 𝑚 ) = 𝑧 ( 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) ) on Ω . We need a discrete version of the operator ( 𝑡 , 𝑥 ) 𝑧 ( 𝑡 , 𝑥 ) . For a function 𝑧 Ω 𝑘 and for a point ( 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) ) 𝐸 we define a function 𝑧 [ 𝑟 , 𝑚 ] 𝐷 𝑘 by 𝑧 [ 𝑟 , 𝑚 ] 𝑡 ( 𝜏 , 𝑦 ) = 𝑧 ( 𝑟 ) + 𝜏 , 𝑥 ( 𝑚 ) + 𝑦 , ( 𝜏 , 𝑦 ) 𝐷 . ( 2 . 5 ) Solutions of difference equations corresponding to (1.4), (1.5) are functions defined on the mesh. On the other hand (1.4) contains the functional variable 𝑧 ( 𝑡 , 𝑥 ) which is an element of the space 𝐶 ( 𝐷 , 𝑘 ) . Then we need an interpolating operator 𝑇 𝐹 ( 𝐷 , 𝑘 ) 𝐶 ( 𝐷 , 𝑘 ) . We define 𝑇 in the following way. Let us denote by ( 𝜗 1 , , 𝜗 𝑛 ) the family of sets defined by 𝜗 𝑖 = { 0 , 1 } i f 𝑑 𝑖 > 0 , 𝜗 𝑖 = { 0 } i f 𝑑 𝑖 = 0 , 1 𝑖 𝑛 . ( 2 . 6 ) Set 𝜐 = ( 𝜐 1 , , 𝜐 𝑛 ) 𝑛 and 𝜐 𝑖 = 0 if 𝑑 𝑖 = 0 , 𝜐 𝑖 = 1 if 𝑑 𝑖 > 0 where 1 𝑖 𝑛 . Write Δ + = 𝜆 𝜆 = 1 , , 𝜆 𝑛 𝜆 𝑖 𝜗 𝑖 f o r 1 𝑖 𝑛 . ( 2 . 7 ) Set 𝑒 𝑖 = ( 0 , , 0 , 1 , 0 , , 0 ) 𝑛 with 1 standing on the 𝑖 th place.

Let 𝑤 𝐹 ( 𝐷 , 𝑘 ) and ( 𝑡 , 𝑥 ) 𝐷 . Suppose that 𝑑 0 > 0 . There exists ( 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) ) 𝐷 such that ( 𝑡 ( 𝑟 + 1 ) , 𝑥 ( 𝑚 + 𝜐 ) ) 𝐷 and 𝑡 ( 𝑟 ) 𝑡 𝑡 ( 𝑟 + 1 ) , 𝑥 ( 𝑚 ) 𝑥 𝑥 ( 𝑚 + 𝜐 ) . Write 𝑇 [ 𝑤 ] ( 𝑡 , 𝑥 ) = 1 𝑡 𝑡 ( 𝑟 ) 0 𝜆 Δ + 𝑤 ( 𝑟 , 𝑚 + 𝜆 ) 𝑥 𝑥 ( 𝑚 ) 𝜆 1 𝑥 𝑥 ( 𝑚 ) 1 𝜆 + 𝑡 𝑡 ( 𝑟 ) 0 𝜆 Δ + 𝑤 ( 𝑟 + 1 , 𝑚 + 𝜆 ) 𝑥 𝑥 ( 𝑚 ) 𝜆 1 𝑥 𝑥 ( 𝑚 ) 1 𝜆 , ( 2 . 8 ) where 𝑥 𝑥 ( 𝑚 ) 𝜆 = 𝑛 𝑖 = 1 𝑥 𝑖 𝑥 ( 𝑚 𝑖 ) 𝑖 𝑖 𝜆 𝑖 , 1 𝑥 𝑥 ( 𝑚 ) 1 𝜆 = 𝑛 𝑖 = 1 𝑥 1 𝑖 𝑥 ( 𝑚 𝑖 ) 𝑖 𝑖 1 𝜆 𝑖 ( 2 . 9 ) and we take 0 0 = 1 in the above formulas. If 𝑑 0 = 0 then we put 𝑇 [ 𝑤 ] ( 𝑡 , 𝑥 ) = 𝜆 Δ + 𝑤 ( 𝑟 , 𝑚 + 𝜆 ) 𝑥 𝑥 ( 𝑚 ) 𝜆 1 𝑥 𝑥 ( 𝑚 ) 1 𝜆 . ( 2 . 1 0 ) Then we have defined 𝑇 [ 𝑤 ] on 𝐷 . It is easy to see that 𝑇 [ 𝑤 ] 𝐶 ( 𝐷 , 𝑘 ) . The above interpolating operator has been first proposed in [10, Chapter 5].

For 𝑤 , 𝑤 𝐹 ( 𝐷 , 𝑘 ) we write 𝑤 𝑤 if 𝑤 ( 𝑟 , 𝑚 ) 𝑤 ( 𝑟 , 𝑚 ) where ( 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) ) 𝐷 . In a similar way we define the relation 𝑤 𝑤 for 𝑤 , 𝑤 𝐶 ( 𝐷 , 𝑘 ) and the relation 𝑧 𝑧 for 𝑧 , 𝑧 𝐹 ( Ω , 𝑘 ) and for 𝑧 , 𝑧 𝐶 ( Ω , 𝑘 ) .

We formulate an implicit difference scheme for (1.4), (1.5). For 𝑥 , 𝑦 𝑛 we write 𝑥 𝑦 = ( 𝑥 1 𝑦 1 , , 𝑥 𝑛 𝑦 𝑛 ) 𝑛 .

Assumption ( 𝐻 [ 𝑓 ] ). The function 𝑓 = ( 𝑓 1 , , 𝑓 𝑘 ) Σ 𝑘 of the variables ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) , 𝑞 = ( 𝑞 1 , , 𝑞 𝑛 ) , is continuous and
(1)the partial derivatives ( 𝜕 𝑞 1 𝑓 𝑖 , , 𝜕 𝑞 𝑛 𝑓 𝑖 ) = 𝜕 𝑞 𝑓 𝑖 , 𝑖 = 1 , , 𝑘 , exist on Σ and the functions 𝜕 𝑞 𝑓 𝑖 , 𝑖 = 1 , , 𝑘 , are continuous and bounded on Σ , (2)there is ̃ 𝑥 ( 𝑏 , 𝑏 ) , ̃ 𝑥 = ( 𝑥 1 , , 𝑥 𝑛 ) , such that ( 𝑥 ̃ 𝑥 ) 𝜕 𝑞 𝑓 𝑖 ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) 𝜃 o n Σ f o r 𝑖 = 1 , , 𝑘 , ( 2 . 1 1 ) (3)there is 𝜀 0 > 0 such that for 0 < 0 < 𝜀 0 and 𝑤 , 𝑤 𝐶 ( 𝐷 , 𝑘 ) , 𝑤 𝑤 , we have 𝑤 ( 0 , 𝜃 ) + 0 𝑓 ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) 𝑤 ( 0 , 𝜃 ) + 0 𝑓 ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) , ( 𝑡 , 𝑥 , 𝑞 ) 𝐸 × 𝑛 . ( 2 . 1 2 )

Remark 2.1. The existence theory of classical or generalized solutions to (1.4), (1.5) is based on a method of bicharacteristics. Suppose that 𝑧 𝐶 ( Ω , 𝑘 ) , 𝑢 𝐶 ( Ω , 𝑛 ) . Let us denote by 𝑔 𝑖 𝑔 [ 𝑧 , 𝑢 ] ( , 𝑡 , 𝑥 ) = 𝑖 . 1 [ 𝑧 , 𝑢 ] ( , 𝑡 , 𝑥 ) , , 𝑔 𝑖 . 𝑛 [ 𝑧 , 𝑢 ] ( , 𝑡 , 𝑥 ) ( 2 . 1 3 ) the 𝑖 th bicharacteristic of (1.4) corresponding to ( 𝑧 , 𝑢 ) . Then 𝑔 𝑖 [ 𝑧 , 𝑢 ] ( , 𝑡 , 𝑥 ) is a solution of the Cauchy problem 𝑦 ( 𝜏 ) = 𝜕 𝑞 𝑓 𝑖 𝜏 , 𝑦 ( 𝜏 ) , 𝑧 ( 𝜏 , 𝑦 ( 𝜏 ) ) , 𝑢 𝜏 , 𝑦 ( 𝜏 ) , 𝑦 ( 𝑡 ) = 𝑥 . ( 2 . 1 4 ) Assumption (2.11) states that the bicharacteristics satisfy the following monotonicity conditions: If 𝑥 𝑗 𝑥 𝑗 0 the function 𝑔 𝑖 𝑗 [ 𝑧 , 𝑢 ] ( , 𝑡 , 𝑥 ) is non increasing. If 𝑥 𝑗 𝑥 𝑗 < 0 then 𝑔 𝑖 𝑗 [ 𝑧 , 𝑢 ] ( , 𝑡 , 𝑥 ) is nondecreasing.
The same property of bicharacteristics is needed in a theorem on the existence and uniqueness of solutions to (1.4), (1.5) see [9]. It is important that our theory of difference methods is consistent with known theorems on the existence of solutions to (1.4), (1.5).

Remark 2.2. Given the function 𝑓 𝑓 = ( 1 𝑓 , , 𝑘 ) 𝐸 × × 𝐶 ( 𝐷 , 𝑘 ) × 𝑛 𝑘 of the variables ( 𝑡 , 𝑥 , 𝑝 , 𝑤 , 𝑞 ) . Write 𝑓 𝑖 𝑓 ( 𝑡 , 𝑥 , 𝑤 , 𝑞 ) = 𝑖 ( 𝑡 , 𝑥 , 𝑤 𝑖 ( 0 , 𝜃 ) , 𝑤 , 𝑞 ) , 𝑖 = 1 , , 𝑘 , on Σ . Then system (1.4) is equivalent to 𝜕 𝑡 𝑧 𝑖 𝑓 ( 𝑡 , 𝑥 ) = 𝑖 𝑡 , 𝑥 , 𝑧 𝑖 ( 𝑡 , 𝑥 ) , 𝑧 ( 𝑡 , 𝑥 ) , 𝜕 𝑥 𝑧 𝑖 ( 𝑡 , 𝑥 ) , 𝑖 = 1 , , 𝑘 . ( 2 . 1 5 ) Note that the dependence of 𝑓 on the classical variable 𝑧 ( 𝑡 , 𝑥 ) is distinguished in (2.15). Suppose that
(1) 𝑓 is nondecreasing with respect to the functional variable,(2)there exists the derivative 𝜕 𝑝 𝑓 = ( 𝜕 𝑝 𝑓 1 , , 𝜕 𝑝 𝑓 𝑘 ) and 𝜕 𝑝 𝑓 𝑖 ( 𝑡 , 𝑥 , 𝑝 , 𝑤 , 𝑞 ) 𝐿 for 𝑖 = 1 , , 𝑘 and 1 + 𝐿 0 0 . Then the monotonicity condition (3) of Assumption ( 𝐻 [ 𝑓 ] )   is satisfied.

Let us denote by 𝐻 the set of all = ( 0 , ) 𝐻 such that 𝑖 𝑏 < m i n 𝑖 𝑥 𝑖 , 𝑥 𝑖 + 𝑏 𝑖 , 𝑖 = 1 , , 𝑛 . ( 2 . 1 6 ) Suppose that 𝜔 Ω . We apply difference operators 𝛿 = ( 𝛿 1 , , 𝛿 𝑛 ) given by i f 𝑥 𝑗 𝑥 ( 𝑚 𝑗 ) 𝑗 < 𝑏 𝑗 t h e n 𝛿 𝑗 𝜔 ( 𝑟 , 𝑚 ) = 1 𝑗 𝜔 ( 𝑟 , 𝑚 + 𝑒 𝑗 ) 𝜔 ( 𝑟 , 𝑚 ) , i f 𝑏 𝑗 < 𝑥 ( 𝑚 𝑗 ) 𝑗 < 𝑥 𝑗 t h e n 𝛿 𝑗 𝜔 ( 𝑟 , 𝑚 ) = 1 𝑗 𝜔 ( 𝑟 , 𝑚 ) 𝜔 ( 𝑟 , 𝑚 𝑒 𝑗 ) , ( 2 . 1 7 ) and we put 𝑗 = 1 , , 𝑛 in (2.17). Let 𝛿 0 be defined by 𝛿 0 𝜔 ( 𝑟 , 𝑚 ) = 1 0 𝜔 ( 𝑟 + 1 , 𝑚 ) 𝜔 ( 𝑟 , 𝑚 ) ( 2 . 1 8 ) and 𝛿 0 𝑧 = ( 𝛿 0 𝑧 1 , , 𝛿 0 𝑧 𝑘 ) . Write 𝔽 [ 𝑧 ] ( 𝑟 , 𝑚 ) = 𝑓 1 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) , 𝑇 𝑧 [ 𝑟 , 𝑚 ] , 𝛿 𝑧 1 ( 𝑟 + 1 , 𝑚 ) , , 𝑓 𝑘 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) , 𝑇 𝑧 [ 𝑟 , 𝑚 ] , 𝛿 𝑧 𝑘 ( 𝑟 + 1 , 𝑚 ) . ( 2 . 1 9 ) Given 𝜑 𝐸 0 . 𝜕 0 𝐸 𝑘 , we consider the functional difference equation 𝛿 0 𝑧 ( 𝑟 , 𝑚 ) = 𝔽 [ 𝑧 ] ( 𝑟 , 𝑚 ) ( 2 . 2 0 ) with the initial boundary condition 𝑧 ( 𝑟 , 𝑚 ) = 𝜑 ( 𝑟 , 𝑚 ) o n 𝐸 0 . 𝜕 0 𝐸 . ( 2 . 2 1 )

The above problem is considered as an implicit difference method for (1.4), (1.5). It is important that the difference expressions ( 𝛿 1 𝑧 𝑖 , , 𝛿 𝑛 𝑧 𝑖 ) , 1 𝑖 𝑘 , are calculated at the point ( 𝑡 ( 𝑟 + 1 ) , 𝑥 ( 𝑚 ) ) and the functional variable 𝑇 𝑧 [ 𝑟 , 𝑚 ] appears in a classical sense.

We prove a theorem on implicit difference inequalities corresponding to (2.20), (2.21). Note that results on implicit difference methods presented in [18] are not applicable to (2.20), (2.21).

Theorem 2.3. Suppose that Assumption ( 𝐻 [ 𝑓 ] )   is satisfied and
(1) 𝐻 , 0 < 𝜀 0 and the functions 𝑢 , 𝑣 Ω 𝑘 satisfy the difference functional inequality 𝛿 0 𝑢 ( 𝑟 , 𝑚 ) 𝔽 [ 𝑢 ] ( 𝑟 , 𝑚 ) 𝛿 0 𝑣 ( 𝑟 , 𝑚 ) 𝔽 [ 𝑣 ] ( 𝑟 , 𝑚 ) o n 𝐸 , ( 2 . 2 2 ) (2)the initial boundary estimate 𝑢 ( 𝑟 , 𝑚 ) 𝑣 ( 𝑟 , 𝑚 ) holds on 𝐸 0 . 𝜕 0 𝐸 . Then 𝑢 ( 𝑟 , 𝑚 ) 𝑣 ( 𝑟 , 𝑚 ) o n 𝐸 . ( 2 . 2 3 )

Proof. We prove (2.23) by induction on 𝑟 . It follows from assumption (2) that estimate (2.23) is satisfied for 𝑟 = 0 and ( 𝑡 ( 0 ) , 𝑥 ( 𝑚 ) ) 𝐸 . Assume that 𝑢 ( 𝑗 , 𝑚 ) 𝑣 ( 𝑗 , 𝑚 ) for ( 𝑡 ( 𝑗 ) , 𝑥 ( 𝑚 ) ) 𝐸 ( [ 0 , 𝑡 ( 𝑟 ) ] × 𝑛 ) . We prove that 𝑢 ( 𝑟 + 1 , 𝑚 ) 𝑣 ( 𝑟 + 1 , 𝑚 ) for ( 𝑡 ( 𝑟 + 1 , 𝑚 ) , 𝑥 ( 𝑚 ) ) 𝐸 . Write 𝑈 𝑖 ( 𝑟 , 𝑚 ) = 𝑢 𝑖 ( 𝑟 , 𝑚 ) + 0 𝑓 𝑖 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) , 𝑇 𝑢 [ 𝑟 , 𝑚 ] , 𝛿 𝑢 𝑖 ( 𝑟 + 1 , 𝑚 ) 𝑣 𝑖 ( 𝑟 , 𝑚 ) 0 𝑓 𝑖 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) , 𝑇 𝑣 [ 𝑟 , 𝑚 ] , 𝛿 𝑢 𝑖 ( 𝑟 + 1 , 𝑚 ) , 𝑖 = 1 , , 𝑘 . ( 2 . 2 4 ) It follows from (2.22) that 𝑢 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) 𝑈 𝑖 ( 𝑟 , 𝑚 ) + 0 𝑓 𝑖 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) , 𝑇 𝑣 [ 𝑟 , 𝑚 ] , 𝛿 𝑢 𝑖 ( 𝑟 + 1 , 𝑚 ) 𝑓 𝑖 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) , 𝑇 𝑣 [ 𝑟 , 𝑚 ] , 𝛿 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) , ( 2 . 2 5 ) where 𝑖 = 1 , , 𝑘 . The monotonicity condition (3) of Assumption ( 𝐻 [ 𝑓 ] )   implies the inequalities 𝑈 𝑖 ( 𝑟 , 𝑚 ) 0 for ( 𝑡 ( 𝑟 ) , 𝑥 ( 𝑚 ) ) 𝐸 , 𝑖 = 1 , , 𝑘 . Then we have 𝑢 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) 0 𝑛 𝑗 = 1 1 0 𝜕 𝑞 𝑗 𝑓 𝑖 𝑄 𝑖 ( 𝑟 , 𝑚 ) [ 𝑣 , 𝜏 ] 𝑑 𝜏 𝛿 𝑗 𝑢 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) , ( 2 . 2 6 ) where 𝑖 = 1 , , 𝑘 and 𝑄 𝑖 ( 𝑟 , 𝑚 ) 𝑡 [ 𝑣 , 𝜏 ] = ( 𝑟 ) , 𝑥 ( 𝑚 ) , 𝑇 𝑣 [ 𝑟 , 𝑚 ] , 𝛿 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) 𝑢 + 𝜏 𝛿 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) . ( 2 . 2 7 ) Write Γ + ( 𝑚 ) = 𝑗 { 1 , , 𝑛 } 𝑥 ( 𝑚 𝑗 ) 𝑗 𝑥 𝑗 , 𝑏 𝑗 , Γ ( 𝑚 ) = { 1 , , 𝑛 } Γ + ( 𝑚 ) . ( 2 . 2 8 ) It follows from (2.11), (2.17) that 𝑢 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) 1 + 0 𝑛 𝑗 = 1 1 𝑗 1 0 | | 𝜕 𝑞 𝑗 𝑓 𝑖 𝑄 𝑖 ( 𝑟 , 𝑚 ) [ 𝑣 , 𝜏 ] | 𝑑 𝜏 0 𝑗 Γ + ( 𝑚 ) 1 𝑗 1 0 𝜕 𝑞 𝑗 𝑓 𝑖 𝑄 𝑖 ( 𝑟 , 𝑚 ) 𝑢 [ 𝑣 , 𝜏 ] 𝑑 𝜏 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 + 𝑒 𝑗 ) 0 𝑗 Γ ( 𝑚 ) 1 𝑗 1 0 𝜕 𝑞 𝑗 𝑓 𝑖 𝑄 𝑖 ( 𝑟 , 𝑚 ) 𝑢 [ 𝑣 , 𝜏 ] 𝑑 𝜏 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 𝑒 𝑗 ) , 𝑖 = 1 , , 𝑘 . ( 2 . 2 9 ) We define 𝑚 𝑛 and 𝜇 , 1 𝜇 𝑘 , as follows: 𝑢 𝜇 𝑣 𝜇 ( 𝑟 + 1 , 𝑚 ) = m a x 1 𝑖 𝑘 𝑢 m a x 𝑖 𝑣 𝑖 ( 𝑟 + 1 , 𝑚 ) 𝑡 ( 𝑟 + 1 ) , 𝑥 ( 𝑚 ) Ω . ( 2 . 3 0 ) If ( 𝑡 ( 𝑟 + 1 ) , 𝑥 ( 𝑚 ) ) 𝜕 0 𝐸 then assumption (2) implies that ( 𝑢 𝜇 𝑣 𝜇 ) ( 𝑟 + 1 , 𝑚 ) 0 . Let us consider the case when ( 𝑡 ( 𝑟 + 1 ) , 𝑥 ( 𝑚 ) ) 𝐸 . Then we have from (2.29) that 𝑢 𝜇 𝑣 𝜇 ( 𝑟 + 1 , 𝑚 ) 1 + 0 𝑛 𝑗 = 1 1 𝑗 1 0 | | 𝜕 𝑞 𝑗 𝑓 𝑖 ( 𝑄 𝑖 ( 𝑟 , 𝑚 ) | | [ 𝑣 , 𝜏 ] ) 𝑑 𝜏 0 𝑢 𝜇 𝑣 𝜇 ( 𝑟 + 1 , 𝑚 ) 𝑗 Γ + ( 𝑚 ) 1 𝑗 1 0 𝜕 𝑞 𝑗 𝑓 𝑖 𝑄 𝑖 ( 𝑟 , 𝑚 ) [ 𝑣 , 𝜏 ] 𝑑 𝜏 𝑗 Γ ( 𝑚 ) 1 𝑗 1 0 𝜕 𝑞 𝑗 𝑓 𝑖 𝑄 ( 𝑟 , 𝑚 𝑖 . [ 𝑣 , 𝜏 ] 𝑑 𝜏 ( 2 . 3 1 ) It follows that ( 𝑢 𝜇 𝑣 𝜇 ) ( 𝑟 + 1 , 𝑚 ) 0 . The the proof of (2.23) is completed by induction.

3. Implicit Difference Schemes

We define 𝑁 = ( 𝑁 1 , , 𝑁 𝑛 ) 𝑁 𝑛 by the relations: 𝑁 1 1 , , 𝑁 𝑛 𝑛 < 𝑏 1 , , 𝑏 𝑛 𝑁 1 + 1 1 𝑁 , , 𝑛 + 1 𝑛 ( 3 . 1 ) and we assume that ( 𝑁 𝑖 + 1 ) 𝑖 = 𝑏 𝑖 if 𝑑 𝑖 = 0 . For 𝑤 𝐶 ( 𝐷 , 𝑘 ) we write 𝑤 𝐷 = m a x 𝑤 ( 𝑡 , 𝑥 ) ( 𝑡 , 𝑥 ) 𝐷 . ( 3 . 2 ) In a similar way we define the norm in the space 𝐹 ( 𝐷 , 𝑘 ) : if 𝑤