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

  • Fatma Kanca1Email author and

    Affiliated with

    • Irem Sakinc Baglan2

      Affiliated with

      Boundary Value Problems20132013:28

      DOI: 10.1186/1687-2770-2013-28

      Received: 7 January 2013

      Accepted: 29 January 2013

      Published: 14 February 2013

      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:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ1_HTML.gif
      (1)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ2_HTML.gif
      (2)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ3_HTML.gif
      (3)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ4_HTML.gif
      (4)

      for a quasilinear parabolic equation with the nonlinear source term f = f ( x , t , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq1_HTML.gif.

      The functions φ ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq2_HTML.gif and f ( x , t , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq3_HTML.gif are given functions on [ 0 , π ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq4_HTML.gif and D ¯ × ( , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq5_HTML.gif respectively. Denote the solution of problem (1)-(4) by u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq6_HTML.gif. 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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq6_HTML.gif upon the data φ ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq2_HTML.gif and f ( x , t , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq3_HTML.gif. 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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq6_HTML.gif using an iteration method. The continuous dependence upon the data for linear problems by different methods is shown in [4, 5].

      Theorem 1 Under the following assumptions, the solution u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq6_HTML.gif depends continuously upon the data.

      (A1) Let the function f ( x , t , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq3_HTML.gif be continuous with respect to all arguments in D ¯ × ( , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq5_HTML.gif and satisfy the following condition:
      | f ( t , x , u ) f ( t , x , u ˜ ) | b ( x , t ) | u u ˜ | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equa_HTML.gif

      where b ( x , t ) L 2 ( D ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq7_HTML.gif, b ( x , t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq8_HTML.gif,

      (A2) f ( x , t , 0 ) C 2 [ 0 , π ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq9_HTML.gif, t ϵ [ 0 , π ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq10_HTML.gif,

      (A3) φ ( x ) C 2 [ 0 , π ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq11_HTML.gif.

      Proof Let ϕ = { φ , f } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq12_HTML.gif and ϕ ¯ = { φ ¯ , f ¯ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq13_HTML.gif be two sets of data which satisfy the conditions (A1)-(A3).

      Let u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq6_HTML.gif and v = v ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq14_HTML.gif be the solutions of problem (1)-(4) corresponding to the data ϕ and ϕ ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq15_HTML.gif respectively, and
      | f ( t , x , 0 ) f ¯ ( t , x , 0 ) | ε for  ε 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equb_HTML.gif
      The solutions of (1)-(4), u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq6_HTML.gif and v = v ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq14_HTML.gif, are presented in the following form, respectively:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ5_HTML.gif
      (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 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq16_HTML.gif.
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ6_HTML.gif
      (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 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq17_HTML.gif.

      From the condition of the theorem, we have u ( 0 ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq18_HTML.gif and v ( 0 ) ( t ) B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq19_HTML.gif. We will prove that the other sequential approximations satisfy this condition.
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ7_HTML.gif
      (7)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ8_HTML.gif
      (8)

      where u 0 ( 0 ) ( t ) = φ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq20_HTML.gif, u c k ( 0 ) ( t ) = φ c k e ( 2 k ) 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq21_HTML.gif, u s k ( 0 ) ( t ) = φ s k e ( 2 k ) 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq22_HTML.gif and v 0 ( 0 ) ( t ) = φ ¯ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq23_HTML.gif, v c k ( 0 ) ( t ) = φ ¯ c k e ( 2 k ) 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq24_HTML.gif, v s k ( 0 ) ( t ) = φ ¯ s k e ( 2 k ) 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq25_HTML.gif.

      First of all, we write N = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq26_HTML.gif in (6)-(7). We consider u ( 1 ) ( t ) v ( 1 ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq27_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ9_HTML.gif
      (9)
      Adding and subtracting
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equc_HTML.gif
      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
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equd_HTML.gif
      For N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq28_HTML.gif,
      | 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Eque_HTML.gif
      For N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq29_HTML.gif,
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equf_HTML.gif
      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 ¯ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ10_HTML.gif
      (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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equg_HTML.gif
      and
      M 1 = ( 3 T + π 6 π ) N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equh_HTML.gif

      (The sequence a N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq30_HTML.gif is convergent, then we can write a N M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq31_HTML.gif, ∀N.)

      It follows from the estimation ([[1], pp.76-77]) that lim N u ( N + 1 ) ( t ) = u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq32_HTML.gif.

      Then let N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq33_HTML.gif for the last equation
      | u ( t ) v ( t ) | M φ φ ¯ + M 2 f f ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equi_HTML.gif

      where M 2 = M M 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq34_HTML.gif.

      If f f ¯ ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq35_HTML.gif and φ φ ¯ ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq36_HTML.gif, then | u ( t ) v ( t ) | ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq37_HTML.gif. □

      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
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ11_HTML.gif
      (11)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ12_HTML.gif
      (12)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equ13_HTML.gif
      (13)
      It is easy to see that the analytical solution of this problem is
      u ( x , t ) = sin 2 x exp ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Equj_HTML.gif

      In this example, we take f ( x , t , u ) = f ( x , t , u ) + ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq38_HTML.gif and φ ( x ) = φ ( x ) + ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq39_HTML.gif for different ε values.

      The comparisons between the analytical solution and the numerical finite difference solution for ε = 0 , 01 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq40_HTML.gif, ε = 0 , 05 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq41_HTML.gif values when T = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq42_HTML.gif are shown in Figure 1.
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_Fig1_HTML.jpg
      Figure 1

      The exact and numerical solutions of u ( x , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq43_HTML.gif. The exact and numerical solutions of u ( x , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq44_HTML.gif, ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq45_HTML.gif for ε = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq46_HTML.gif, ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq47_HTML.gif for ε = 0.05 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq48_HTML.gif, ( . . ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq49_HTML.gif for ε = 0.01 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-28/MediaObjects/13661_2013_Article_283_IEq50_HTML.gif, the exact solution is shown with a dashed line.

      The computational results presented are consistent with the theoretical results.

      Declarations

      Acknowledgements

      Dedicated to Professor Hari M Srivastava.

      Authors’ Affiliations

      (1)
      Department of Information Technologies, Kadir Has University
      (2)
      Department of Mathematics, Kocaeli University

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      Copyright

      © Kanca and Baglan; licensee Springer. 2013

      This article is published under license to BioMed Central Ltd. 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.