Research

# Continuous dependence on data for a solution of the quasilinear parabolic equation with a periodic boundary condition

Fatma Kanca1* and Irem Sakinc Baglan2

Author Affiliations

1 Department of Information Technologies, Kadir Has University, Istanbul, 34083, Turkey

2 Department of Mathematics, Kocaeli University, Kocaeli, 41380, Turkey

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Boundary Value Problems 2013, 2013:28  doi:10.1186/1687-2770-2013-28

 Received: 7 January 2013 Accepted: 29 January 2013 Published: 14 February 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper we consider a parabolic equation with a periodic boundary condition and we prove the stability of a solution on the data. We give a numerical example for the stability of the solution on the data.

### 1 Introduction

Consider the following mixed problem:

(1)

(2)

(3)

(4)

for a quasilinear parabolic equation with the nonlinear source term f = f ( x , t , u ) .

The functions φ ( x ) and f ( x , t , u ) are given functions on [ 0 , π ] and D ¯ × ( , ) respectively. Denote the solution of problem (1)-(4) by u = u ( x , t ) . The existence, uniqueness and convergence of the weak generalized solution of problem (1)-(4) are considered in [1]. The numerical solution of problem (1)-(4) is considered [2].

In this study we prove the continuous dependence of the solution u = u ( x , t ) upon the data φ ( x ) and f ( x , t , u ) . In [3], a similar iteration method is used with this kind of a local boundary condition for a nonlinear inverse coefficient problem for a parabolic equation. Then we give a numerical example for the stability.

### 2 Continuous dependence upon the data

In this section, we will prove the continuous dependence of the solution u = u ( x , t ) using an iteration method. The continuous dependence upon the data for linear problems by different methods is shown in [4,5].

Theorem 1Under the following assumptions, the solution u = u ( x , t ) depends continuously upon the data.

(A1) Let the function f ( x , t , u ) be continuous with respect to all arguments in D ¯ × ( , ) and satisfy the following condition:

| f ( t , x , u ) f ( t , x , u ˜ ) | b ( x , t ) | u u ˜ | ,

where b ( x , t ) L 2 ( D ) , b ( x , t ) 0 ,

(A2) f ( x , t , 0 ) C 2 [ 0 , π ] , t ϵ [ 0 , π ] ,

(A3) φ ( x ) C 2 [ 0 , π ] .

Proof Let ϕ = { φ , f } and ϕ ¯ = { φ ¯ , f ¯ } be two sets of data which satisfy the conditions (A1)-(A3).

Let u = u ( x , t ) and v = v ( x , t ) be the solutions of problem (1)-(4) corresponding to the data ϕ and ϕ ¯ respectively, and

| f ( t , x , 0 ) f ¯ ( t , x , 0 ) | ε for  ε 0 .

The solutions of (1)-(4), u = u ( x , t ) and v = v ( x , t ) , are presented in the following form, respectively:

(5)

Let A u ( ξ , τ ) = u 0 ( τ ) 2 + k = 1 [ u c k ( τ ) cos 2 k ξ + u s k ( τ ) sin 2 k ξ ] .

(6)

Let A v ( ξ , τ ) = v 0 ( τ ) 2 + k = 1 [ v c k ( τ ) cos 2 k ξ + v s k ( τ ) sin 2 k ξ ] .

From the condition of the theorem, we have u ( 0 ) ( t ) and v ( 0 ) ( t ) B . We will prove that the other sequential approximations satisfy this condition.

(7)

(8)

where u 0 ( 0 ) ( t ) = φ 0 , u c k ( 0 ) ( t ) = φ c k e ( 2 k ) 2 t , u s k ( 0 ) ( t ) = φ s k e ( 2 k ) 2 t and v 0 ( 0 ) ( t ) = φ ¯ 0 , v c k ( 0 ) ( t ) = φ ¯ c k e ( 2 k ) 2 t , v s k ( 0 ) ( t ) = φ ¯ s k e ( 2 k ) 2 t .

First of all, we write N = 0 in (6)-(7). We consider u ( 1 ) ( t ) v ( 1 ) ( t )

u ( 1 ) ( t ) v ( 1 ) ( t ) = u 0 ( 1 ) ( t ) v 0 ( 1 ) ( t ) 2 + k = 1 [ ( u c k ( 1 ) ( t ) v c k ( 1 ) ( t ) ) + ( u s k ( 1 ) ( t ) v s k ( 1 ) ( t ) ) ] = ( φ 0 φ 0 ¯ ) + 2 π 0 t 0 π [ f ( ξ , τ , A u ( 0 ) ( ξ , τ ) ) f ¯ ( ξ , τ , A v ( 0 ) ( ξ , τ ) ) ] d ξ d τ + ( φ c k φ c k ¯ ) e ( 2 k ) 2 t + 2 π 0 t 0 π [ f ( ξ , τ , A u ( 0 ) ( ξ , τ ) ) f ¯ ( ξ , τ , A v ( 0 ) ( ξ , τ ) ) ] × e ( 2 π k ) 2 ( t τ ) cos 2 π k ξ d ξ d τ + ( φ s k φ s k ¯ ) e ( 2 k ) 2 t + 2 π 0 t 0 π [ f ( ξ , τ , A u ( 0 ) ( ξ , τ ) ) f ¯ ( ξ , τ , A v ( 0 ) ( ξ , τ ) ) ] × e ( 2 π k ) 2 ( t τ ) sin 2 π k ξ d ξ d τ . (9)

to both sides and applying the Cauchy inequality, Hölder inequality, Lipschitz condition and Bessel inequality to the right-hand side of (8) respectively, we obtain

For N = 1 ,

| u ( 2 ) ( t ) v ( 2 ) ( t ) | | u 0 ( 2 ) ( t ) v 0 ( 2 ) ( t ) | 2 + k = 1 ( | u c k ( 2 ) ( t ) v c k ( 2 ) | + | u s k ( 2 ) ( t ) v s k ( 2 ) ( t ) | ) ( 3 T + π 6 π ) ( 0 t 0 π b 2 ( ξ , τ ) d ξ d τ ) 1 2 A T + ( 3 T + π 6 π ) ( 0 t 0 π b ¯ 2 ( ξ , τ ) d ξ d τ ) 1 2 A T .

For N = 2 ,

In the same way, for a general value of N, we have

| u ( N + 1 ) ( t ) v ( N + 1 ) ( t ) | | u 0 ( N + 1 ) ( t ) v 0 ( N + 1 ) ( t ) | 2 + k = 1 ( | u c k ( N + 1 ) ( t ) v c k ( N + 1 ) ( t ) | + | u s k ( N + 1 ) ( t ) v s k ( N + 1 ) ( t ) | ) A T a N = a N   ( φ φ ¯ + C ( t ) + M 1 f f ¯ ) , (10)

where

a N = ( 3 T + π 6 π ) N A T N ! [ ( 0 t 0 π b 2 ( ξ , τ ) d ξ d τ ) 2 ] N 2 + ( 3 T + π 6 π ) N A T N ! [ ( 0 t 0 π b ¯ 2 ( ξ , τ ) d ξ d τ ) 2 ] N 2

and

M 1 = ( 3 T + π 6 π ) N .

(The sequence a N is convergent, then we can write a N M , ∀N.)

It follows from the estimation ([[1], pp.76-77]) that lim N u ( N + 1 ) ( t ) = u ( t ) .

Then let N for the last equation

| u ( t ) v ( t ) | M φ φ ¯ + M 2 f f ¯ ,

where M 2 = M M 1 .

If f f ¯ ε and φ φ ¯ ε , then | u ( t ) v ( t ) | ε . □

### 3 Numerical example

In this section we consider an example of numerical solution of (1)-(4) to test the stability of this problem. The numerical procedure of (1)-(4) is considered in [2].

Example 1

Consider the problem

(11)

(12)

(13)

It is easy to see that the analytical solution of this problem is

u ( x , t ) = sin 2 x exp ( t ) .

In this example, we take f ( x , t , u ) = f ( x , t , u ) + ε and φ ( x ) = φ ( x ) + ε for different ε values.

The comparisons between the analytical solution and the numerical finite difference solution for ε = 0 , 01 , ε = 0 , 05 values when T = 1 are shown in Figure 1.

Figure 1. The exact and numerical solutions of u ( x , 1 ) . The exact and numerical solutions of u ( x , 1 ) , ( ) for ε = 0 , ( ) for ε = 0.05 , ( . . ) for ε = 0.01 , the exact solution is shown with a dashed line.

The computational results presented are consistent with the theoretical results.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

FK conceived the study, participated in its design and coordination and prepared computing section. ISB participated in the sequence alignment and achieved the estimation.

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

### References

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