Open Access

An inverse coefficient problem for a quasilinear parabolic equation with nonlocal boundary conditions

Boundary Value Problems20132013:213

DOI: 10.1186/1687-2770-2013-213

Received: 7 June 2013

Accepted: 27 August 2013

Published: 30 September 2013

Abstract

In this paper the inverse problem of finding the time-dependent coefficient of heat capacity together with the nonlocal boundary conditions is considered. Under some natural regularity and consistency conditions on the input data, the existence, uniqueness and continuous dependence upon the data of the solution are shown. Some considerations on the numerical solution for this inverse problem are presented with an example.

1 Introduction

Denote the domain D by
D : = { 0 < x < 1 , 0 < t < T } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equa_HTML.gif
Consider the equation
u t = u x x p ( t ) u + f ( x , t , u ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ1_HTML.gif
(1)
with the initial condition
u ( x , 0 ) = φ ( x ) , x [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ2_HTML.gif
(2)
the nonlocal boundary condition
u ( 0 , t ) = u ( 1 , t ) , u x ( 1 , t ) = 0 , t [ 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ3_HTML.gif
(3)
and the overdetermination data
u x ( 0 , t ) = g ( t ) , t [ 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ4_HTML.gif
(4)

for a quasilinear parabolic equation with the nonlinear source term f = f ( x , t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq1_HTML.gif.

The functions φ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq2_HTML.gif and f ( x , t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq3_HTML.gif are given functions on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq4_HTML.gif and D ¯ × ( , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq5_HTML.gif, respectively.

The problem of finding the pair { p ( t ) , u ( x , t ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq6_HTML.gif in (1)-(4) will be called an inverse problem.

Definition 1 The pair { p ( t ) , u ( x , t ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq6_HTML.gif from the class C [ 0 , T ] × ( C 2 , 1 ( D ) C 1 , 0 ( D ¯ ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq7_HTML.gif, for which conditions (1)-(4) are satisfied and p ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq8_HTML.gif on the interval [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq9_HTML.gif, is called the classical solution of inverse problem (1)-(4).

The problem of identification of a coefficient in a nonlinear parabolic equation is an interesting problem for many scientists [13]. Inverse problems for parabolic equations with nonlocal boundary conditions are investigated in [46]. This kind of conditions arise from many important applications in heat transfer, life sciences, etc. In [7], also the nature of (3) type boundary conditions is demonstrated.

In [1] the boundary conditions are local, the solution is obtained locally and the authors obtained the solution in Holder classes using iteration method. In [5] the boundary condition is nonlocal but the problem is linear and the existence and the uniqueness of the classical solution is obtained locally using a fixed point theorem. In this paper, the existence and uniqueness of the classical solution is obtained locally using the iteration method.

The paper is organized as follows. In Section 2, the existence and uniqueness of the solution of inverse problem (1)-(4) is proved by using the Fourier method and the iteration method. In Section 3, the continuous dependence upon the data of the inverse problem is shown. In Section 4, the numerical procedure for the solution of the inverse problem is given.

2 Existence and uniqueness of the solution of the inverse problem

Consider the following system of functions on the interval [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq10_HTML.gif:
X 0 ( x ) = 2 , X 2 k 1 ( x ) = 4 cos ( 2 π k x ) , X 2 k ( x ) = 4 ( 1 x ) sin ( 2 π k x ) , k = 1 , 2 , , Y 0 ( x ) = x , Y 2 k 1 ( x ) = x cos ( 2 π k x ) , Y 2 k ( x ) = sin ( 2 π k x ) , k = 1 , 2 , . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equb_HTML.gif

The systems of these functions arise in [8] for the solution of a nonlocal boundary value problem in heat conduction. It is easy to verify that the system of functions X k ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq11_HTML.gif and Y k ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq12_HTML.gif, k = 0 , 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq13_HTML.gif , is biorthonormal on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq10_HTML.gif. They are also Riesz bases in L 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq14_HTML.gif (see [5, 6]).

The main result on the existence and uniqueness of the solution of inverse problem (1)-(4) is presented as follows.

We have the following assumptions on the data of problem (1)-(4):

  1. (A1)

    g ( t ) C 1 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq15_HTML.gif, g ( t ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq16_HTML.gif, g ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq17_HTML.gif;

     
  2. (A2)
    φ ( x ) C 3 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq18_HTML.gif,
    1. (1)

      φ ( 0 ) = φ ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq19_HTML.gif, φ ( 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq20_HTML.gif, φ ( 0 ) = φ ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq21_HTML.gif,

       
    2. (2)

      φ 2 k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq22_HTML.gif, k = 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq23_HTML.gif ;

       
     
  3. (A3)
    Let the function f ( x , t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq3_HTML.gif be continuous with respect to all arguments in D ¯ × ( , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq5_HTML.gif and satisfy the following conditions:
    1. (1)
      | f ( n ) ( x , t , u ) f ( n ) ( x , t , u ˜ ) | b ( t , x ) | u u ˜ | , n = 0 , 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equc_HTML.gif

      where b ( x , t ) L 2 ( D ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq24_HTML.gif, b ( x , t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq25_HTML.gif,

       
    2. (2)

      f ( x , t , u ) C 3 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq26_HTML.gif, t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq27_HTML.gif,

       
    3. (3)

      f ( x , t , u ) | x = 0 = f ( x , t , u ) | x = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq28_HTML.gif, f x ( x , t , u ) | x = 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq29_HTML.gif, f x x ( x , t , u ) | x = 0 = f x x ( x , t , u ) | x = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq30_HTML.gif,

       
    4. (4)
      f 2 k ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq31_HTML.gif, f 0 ( t ) + k = 1 ( 2 π k ) 2 ( φ 2 k + 0 T f 2 k ( τ ) d τ ) g ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq32_HTML.gif, t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq33_HTML.gif, where
      φ k = 0 1 φ ( x ) Y k ( x ) d x , f k ( t ) = 0 1 f ( x , t , u ) Y k ( x ) d x , k = 0 , 1 , 2 , . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equd_HTML.gif
       
     
By applying the standard procedure of the Fourier method, we obtain the following representation for the solution of (1)-(3) for arbitrary p ( t ) C [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq34_HTML.gif:
u ( x , t ) = [ φ 0 e 0 t p ( s ) d s + 0 t 0 1 f ( ξ , τ , u ) ξ e τ t p ( s ) d s d ξ d τ ] X 0 ( x ) u ( x , t ) = + k = 1 X 2 k ( x ) [ φ 2 k e ( 2 π k ) 2 t 0 t p ( s ) d s u ( x , t ) = + 0 t 0 1 f ( ξ , τ , u ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d ξ d τ ] u ( x , t ) = + k = 1 X 2 k 1 ( x ) [ ( φ 2 k 1 4 π k t φ 2 k ) e ( 2 π k ) 2 t 0 t p ( s ) d s ] u ( x , t ) = + k = 1 X 2 k 1 ( x ) [ 0 t 0 1 f ( ξ , τ , u ) ξ cos 2 k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d ξ d τ ] u ( x , t ) = k = 1 X 2 k 1 ( x ) u ( x , t ) = × [ 4 π k 0 t 0 1 f ( ξ , τ , u ) ( t τ ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d ξ d τ ] , u 0 ( t ) = φ 0 e 0 t p ( s ) d s + 0 t 0 1 f ( ξ , τ , u ) ξ e τ t p ( s ) d s d ξ d τ , u 2 k ( t ) = φ 2 k e ( 2 π k ) 2 t 0 t p ( s ) d s + 0 t 0 1 f ( ξ , τ , u ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d ξ d τ , u 2 k 1 ( t ) = ( φ 2 k 1 4 π k t φ 2 k ) e ( 2 π k ) 2 t 0 t p ( s ) d s u 2 k 1 ( t ) = + 0 t 0 1 f ( ξ , τ , u ) ξ cos 2 k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d ξ d τ u 2 k 1 ( t ) = 4 π k 0 t 0 1 f ( ξ , τ , u ) ( t τ ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ5_HTML.gif
(5)
Under conditions (A1)-(A3), we obtain
u t x ( 0 , t ) = g ( t ) , 0 t T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ6_HTML.gif
(6)
Equations (5) and (6) yield
p ( t ) = 1 g ( t ) [ g ( t ) + k = 1 ( 8 π k f 2 k 4 ( 2 π k ) 2 φ 2 k e ( 2 π k ) 2 t 0 t p ( s ) d s ) ] 1 g ( t ) k = 1 4 ( 2 π k ) 2 0 t f 2 k ( t ) e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ7_HTML.gif
(7)

Definition 2 Denote the set { u ( t ) } = { u 0 ( t ) , u 2 k ( t ) , u 2 k 1 ( t ) , k = 1 , , n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq35_HTML.gif of continuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq36_HTML.gif functions satisfying the condition 2 max 0 t T | u 0 ( t ) | + 4 k = 1 ( max 0 t T | u 2 k ( t ) | + max 0 t T | u 2 k 1 ( t ) | ) < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq37_HTML.gif by B. Let u ( t ) = 2 max 0 t T | u 0 ( t ) | + 4 k = 1 ( max 0 t T | u 2 k ( t ) | + max 0 t T | u 2 k 1 ( t ) | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq38_HTML.gif be the norm in B. It can be shown that B is the Banach space.

Theorem 3 Let assumptions (A1)-(A3) be satisfied. Then inverse problem (1)-(4) has a unique solution for small T.

Proof An iteration for (5) is defined as follows:
u 0 ( N + 1 ) ( t ) = u 0 ( 0 ) ( t ) + 0 t 0 1 f ( ξ , τ , u ) ξ e τ t p ( N ) ( s ) d s d ξ d τ , u 2 k ( N + 1 ) ( t ) = u 2 k ( 0 ) ( t ) + 0 t 0 1 f ( ξ , τ , u ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( N ) ( s ) d s d ξ d τ , u 2 k 1 ( N + 1 ) ( t ) = u 2 k 1 ( 0 ) ( t ) + 0 t 0 1 f ( ξ , τ , u ) ξ cos 2 k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( N ) ( s ) d s d ξ d τ u 2 k 1 ( N + 1 ) ( t ) = 4 π k 0 t 0 1 f ( ξ , τ , u ) ( t τ ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( N ) ( s ) d s d ξ d τ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ8_HTML.gif
(8)
where N = 0 , 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq39_HTML.gif and
u 0 ( 0 ) ( t ) = φ 0 e 0 t p ( s ) d s , u 2 k ( 0 ) ( t ) = φ 2 k e ( 2 π k ) 2 t 0 t p ( s ) d s , u 2 k 1 ( 0 ) ( t ) = ( φ 2 k 1 4 π k t φ 2 k ) e ( 2 π k ) 2 t 0 t p ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Eque_HTML.gif

From the conditions of the theorem, we have u ( 0 ) ( t ) B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq40_HTML.gif, and let p ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq41_HTML.gif.

Let us write N = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq42_HTML.gif in (8).
u 0 ( 1 ) ( t ) = u 0 ( 0 ) ( t ) + 0 t 0 1 f ( ξ , τ , u ) ξ d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equf_HTML.gif
Adding and subtracting 0 t 0 1 f ( ξ , τ , 0 ) d ξ d τ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq43_HTML.gif on both sides of the last equation, we obtain
u 0 ( 1 ) ( t ) = u 0 ( 0 ) ( t ) + 0 t 0 1 [ f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) f ( ξ , τ , 0 ) ] d ξ d τ + 0 t 0 1 f ( ξ , τ , 0 ) d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equg_HTML.gif
Applying the Cauchy inequality and the Lipschitz condition to the last equation and taking the maximum of both sides of the last inequality yields the following:
max 0 t T | u 0 ( 1 ) ( t ) | | φ 0 | + T b ( x , t ) L 2 ( D ) u ( 0 ) ( t ) + T f ( x , t , 0 ) L 2 ( D ) , u 2 k ( 1 ) ( t ) = φ 2 k e ( 2 π k ) 2 t + 0 t 0 1 [ f ( ξ , τ , u ( 0 ) ) f ( ξ , τ , 0 ) ] sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) d ξ d τ u 2 k ( 1 ) ( t ) = + 0 t 0 1 f ( ξ , τ , 0 ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equh_HTML.gif
Applying Cauchy’s inequality, Hölder’s inequality, Bessel’s inequality, the Lipschitz condition and taking maximum of both sides of the last inequality yields the following:
k = 1 max 0 t T | u 2 k ( 1 ) ( t ) | k = 1 | φ 2 k | + 3 3 b ( x , t ) L 2 ( D ) u ( 0 ) ( t ) + 3 3 f ( x , t , 0 ) L 2 ( D ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equi_HTML.gif
Applying the same estimations, we obtain
k = 1 max 0 t T | u 2 k 1 ( 1 ) ( t ) | k = 1 | φ 2 k 1 | + 2 6 T 3 k = 1 | φ 2 k | + ( 3 3 + 2 2 | T | ) b ( x , t ) L 2 ( D ) u ( 0 ) ( t ) + ( 3 3 + 2 2 | T | ) f ( x , t , 0 ) L 2 ( D ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equj_HTML.gif
Finally, we have the following inequality:
u ( 1 ) ( t ) B = 2 max 0 t T | u 0 ( 1 ) ( t ) | + 4 k = 1 ( max 0 t T | u 2 k ( 1 ) ( t ) | + max 0 t T | u 2 k 1 ( 1 ) ( t ) | ) | φ 0 | + 4 k = 1 ( | φ 2 k | + | φ 2 k 1 | ) + 2 6 T 3 k = 1 | φ 2 k | + ( 2 T + 2 3 3 + 4 2 | T | ) ( b ( x , t ) L 2 ( D ) u ( 0 ) ( t ) B ) + ( 2 T + 2 3 3 + 4 2 | T | ) f ( x , t , 0 ) L 2 ( D ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equk_HTML.gif
Hence u ( 1 ) ( t ) B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq44_HTML.gif. In the same way, for a general value of N, we have
u ( N ) ( t ) B = 2 max 0 t T | u 0 ( N ) ( t ) | + 4 k = 1 ( max 0 t T | u 2 k ( N ) ( t ) | + max 0 t T | u 2 k 1 ( N ) ( t ) | ) | φ 0 | + 4 k = 1 ( | φ c k | + | φ s k | ) + 2 6 T 3 k = 1 | φ 2 k | + ( 2 T + 2 3 3 + 4 2 | T | ) ( b ( x , t ) L 2 ( D ) u ( N 1 ) ( t ) B ) + ( 2 T + 2 3 3 + 4 2 | T | ) f ( x , t , 0 ) L 2 ( D ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equl_HTML.gif
From u ( N 1 ) ( t ) B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq45_HTML.gif we deduce that u ( N ) ( t ) B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq46_HTML.gif,
{ u ( t ) } = { u 0 ( t ) , u 2 k ( t ) , u 2 k 1 ( t ) , k = 1 , 2 , } B . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equm_HTML.gif
An iteration for (7) is defined as follows:
p ( N + 1 ) ( t ) = 1 g ( t ) [ g ( t ) + k = 1 ( 8 π k 0 1 f ( ξ , τ , u ( N ) ) sin 2 π k ξ d ξ 4 ( 2 π k ) 2 φ 2 k e ( 2 π k ) 2 t 0 t p ( N ) ( s ) d s ) ] 1 g ( t ) k = 1 4 ( 2 π k ) 2 0 t 0 1 f ( ξ , τ , u ( N ) ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( N ) ( s ) d s d ξ d τ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equn_HTML.gif
where N = 0 , 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq47_HTML.gif ,
p ( 1 ) ( t ) = 1 g ( t ) [ g ( t ) + k = 1 ( 8 π k 0 1 f ( ξ , τ , u ( 0 ) ) sin 2 π k ξ d ξ 4 ( 2 π k ) 2 φ 2 k e ( 2 π k ) 2 t ) ] 1 g ( t ) k = 1 4 ( 2 π k ) 2 0 t 0 1 f ( ξ , τ , u ( 0 ) ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equo_HTML.gif
For convergence,
p ( 1 ) ( t ) = 1 g ( t ) [ g ( t ) + k = 1 8 π k ( 2 π k ) 2 0 1 f ξ ξ ( ξ , τ , u ( 0 ) ) sin 2 π k ξ d ξ k = 1 4 ( 2 π k ) 2 ( 2 π k ) 3 φ 2 k e ( 2 π k ) 2 t ] 1 g ( t ) k = 1 4 ( 2 π k ) 2 ( 2 π k ) 2 0 t 0 1 f ξ ξ ( ξ , τ , u ( 0 ) ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equp_HTML.gif
Applying Cauchy’s inequality, Hölder’s inequality, Bessel’s inequality, the Lipschitz condition and taking maximum of both sides of the last inequality yields the following:
| p ( 1 ) ( t ) | | g ( t ) g ( t ) | + 6 3 | g ( t ) | k = 1 | φ 2 k | + ( 6 + 3 3 | g ( t ) | ) b ( x , t ) L 2 ( D ) u ( 0 ) ( t ) B + ( 6 + 3 3 | g ( t ) | ) M . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equq_HTML.gif
Hence p ( 1 ) ( t ) B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq48_HTML.gif. In the same way, for a general value of N, we have
| p ( N + 1 ) ( t ) | | g ( t ) g ( t ) | + 6 3 | g ( t ) | k = 1 | φ 2 k | + ( 6 + 3 3 | g ( t ) | ) b ( x , t ) L 2 ( D ) u ( N ) ( t ) B + ( 6 + 3 3 | g ( t ) | ) M . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equr_HTML.gif

We deduce that p ( N ) ( t ) B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq49_HTML.gif.

Now we prove that the iterations u ( N + 1 ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq50_HTML.gif and p ( N + 1 ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq51_HTML.gif converge in B as N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq52_HTML.gif.
u 0 ( 1 ) ( t ) u 0 ( 0 ) ( t ) = 0 t 0 1 [ f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) f ( ξ , τ , 0 ) ] ξ d ξ d τ + 0 t 0 1 f ( ξ , τ , 0 ) ξ d ξ d τ , u 2 k ( 1 ) ( t ) u 2 k ( 0 ) ( t ) = 0 t 0 1 [ f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) f ( ξ , τ , 0 ) ] e ( 2 π k ) 2 ( t τ ) sin 2 π k ξ d ξ d τ + 0 t 0 1 f ( ξ , τ , 0 ) e ( 2 π k ) 2 ( t τ ) sin 2 π k ξ d ξ d τ , u 2 k 1 ( 1 ) ( t ) u 2 k 1 ( 0 ) ( t ) = 0 t 0 1 [ f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) f ( ξ , τ , 0 ) ] e ( 2 π k ) 2 ( t τ ) ξ cos 2 π k ξ d ξ d τ + 0 t 0 1 f ( ξ , τ , 0 ) e ( 2 π k ) 2 ( t τ ) ξ cos 2 π k ξ d ξ d τ 4 π k 0 t 0 1 [ f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) f ( ξ , τ , 0 ) ] ( t τ ) e ( 2 π k ) 2 ( t τ ) sin 2 π k ξ d ξ d τ 4 π k 0 t 0 1 ( t τ ) f ( ξ , τ , 0 ) e ( 2 π k ) 2 ( t τ ) sin 2 π k ξ d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equs_HTML.gif
Applying Cauchy’s inequality, Hölder’s inequality, the Lipschitz condition and Bessel’s inequality to the last equation, we obtain
| u ( 1 ) ( t ) u ( 0 ) ( t ) | ( 2 T + 2 3 3 + 4 2 T ) ( b ( x , t ) L 2 ( D ) u ( 0 ) ( t ) B ) | u ( 1 ) ( t ) u ( 0 ) ( t ) | + ( 2 T + 2 3 3 + 4 2 T ) f ( x , t , 0 ) L 2 ( D ) , K = ( 2 T + 2 3 3 + 4 2 T ) ( b ( x , t ) L 2 ( D ) u ( 0 ) ( t ) B ) K = + ( 2 T + 2 3 3 + 4 2 T ) f ( x , t , 0 ) L 2 ( D ) , u 0 ( 2 ) ( t ) u 0 ( 1 ) ( t ) = 0 t 0 1 [ f ( ξ , τ , u ( 1 ) ( ξ , τ ) ) f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) ] ξ e τ t p ( 1 ) ( s ) d s d ξ d τ + 0 t 0 1 f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) ξ [ e τ t p ( 1 ) ( s ) d s e τ t p ( 0 ) ( s ) d s ] d ξ d τ , u 2 k ( 2 ) ( t ) u 2 k ( 1 ) ( t ) = 0 t 0 1 [ f ( ξ , τ , u ( 1 ) ( ξ , τ ) ) f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) ] e ( 2 π k ) 2 ( t τ ) e τ t p ( 1 ) ( s ) d s sin 2 π k ξ d ξ d τ + 0 t 0 1 f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) sin 2 k ξ e ( 2 π k ) 2 ( t τ ) [ e τ t p ( 1 ) ( s ) d s e τ t p ( 0 ) ( s ) d s ] d ξ d τ , u 2 k 1 ( 2 ) ( t ) u 2 k 1 ( 1 ) ( t ) = 0 t 0 1 [ f ( ξ , τ , u ( 1 ) ( ξ , τ ) ) f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) ] × e ( 2 π k ) 2 ( t τ ) e τ t p ( 1 ) ( s ) d s ξ cos 2 π k ξ d ξ d τ + 0 t 0 1 f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) ξ cos 2 π k ξ e ( 2 π k ) 2 ( t τ ) × [ e τ t p ( 1 ) ( s ) d s e τ t p ( 0 ) ( s ) d s ] d ξ d τ 4 π k 0 t 0 1 ( t τ ) [ f ( ξ , τ , u ( 1 ) ( ξ , τ ) ) f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) ] × e ( 2 π k ) 2 ( t τ ) e τ t p ( 1 ) ( s ) d s sin 2 π k ξ d ξ d τ 4 π k 0 t 0 1 ( t τ ) f ( ξ , τ , u ( 0 ) ( ξ , τ ) ) × e ( 2 π k ) 2 ( t τ ) [ e τ t p ( 1 ) ( s ) d s e τ t p ( 0 ) ( s ) d s ] sin 2 π k ξ d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equt_HTML.gif
Applying Cauchy’s inequality, Hölder’s inequality, the Lipschitz condition and Bessel’s inequality to the last equation, we obtain
| u ( 2 ) ( t ) u ( 1 ) ( t ) | ( 2 T + 2 3 3 + 4 2 T ) ( b ( x , t ) L 2 ( D ) | u ( 1 ) u ( 0 ) | ) | u ( 2 ) ( t ) u ( 1 ) ( t ) | + ( 2 T + 2 3 3 + 4 2 T ) | T | f ( x , t , 0 ) L 2 ( D ) | p ( 1 ) p ( 0 ) | , p ( 1 ) p ( 0 ) = 1 g ( t ) k = 1 ( 8 π k 0 1 [ f ( ξ , τ , u ( 1 ) ) f ( ξ , τ , u ( 0 ) ) ] sin 2 π k ξ d ξ ) p ( 1 ) p ( 0 ) = 1 g ( t ) k = 1 4 ( 2 π k ) 2 φ 2 k e ( 2 π k ) 2 t [ e 0 t p ( 1 ) ( s ) d s e 0 t p ( 0 ) ( s ) d s ] p ( 1 ) p ( 0 ) = 1 g ( t ) k = 1 4 ( 2 π k ) 2 0 t 0 1 [ f ( ξ , τ , u ( 1 ) ) f ( ξ , τ , u ( 0 ) ) ] p ( 1 ) p ( 0 ) = × sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) e 0 t p ( 1 ) ( s ) d s d ξ d τ p ( 1 ) p ( 0 ) = 1 g ( t ) k = 1 4 ( 2 π k ) 2 0 t 0 1 f ( ξ , τ , u ( 0 ) ) p ( 1 ) p ( 0 ) = × sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) [ e 0 t p ( 1 ) ( s ) d s e 0 t p ( 0 ) ( s ) d s ] d ξ d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equu_HTML.gif
Applying Cauchy’s inequality, Hölder’s inequality, the Lipschitz condition and Bessel’s inequality to the last equation, we obtain
| p ( 1 ) p ( 0 ) | ( 6 + 3 3 | g ( t ) | ) b ( x , t ) L 2 ( D ) | u ( 1 ) u ( 0 ) | | p ( 1 ) p ( 0 ) | + ( 3 3 | g ( t ) | M + 6 | T | 3 | g ( t ) | k = 1 | φ 2 k 1 | ) | p ( 1 ) p ( 0 ) | , A = ( 6 + 3 3 | g ( t ) | ) , B = ( 3 3 | g ( t ) | M + 6 T 3 | g ( t ) | k = 1 | φ 2 k 1 | ) , B < 1 , | p ( 1 ) p ( 0 ) | A 1 B b ( x , t ) L 2 ( D ) | u ( 1 ) u ( 0 ) | , | u ( 2 ) ( t ) u ( 1 ) ( t ) | ( 2 T + 2 3 3 + 4 2 T ) ( b ( x , t ) L 2 ( D ) | u ( 1 ) u ( 0 ) | ) | u ( 2 ) ( t ) u ( 1 ) ( t ) | + ( 2 T + 2 3 3 + 4 2 T ) T M A 1 B b ( x , t ) L 2 ( D ) | u ( 1 ) u ( 0 ) | , | u ( 2 ) ( t ) u ( 1 ) ( t ) | ( ( 2 T + 2 3 3 + 4 2 T ) ( 1 + T M A 1 B ) ) b ( x , t ) L 2 ( D ) K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equv_HTML.gif
For N, we have
| p ( N + 1 ) p ( N ) | A 1 B b ( x , t ) L 2 ( D ) | u ( N + 1 ) u ( N ) | , | u ( N + 1 ) ( t ) u ( N ) ( t ) | 1 N ! ( ( 2 T + 2 3 3 + 4 2 T ) ( 1 + M A T 1 B ) ) N b ( x , t ) L 2 ( D ) K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ9_HTML.gif
(9)

It is easy to see that u ( N + 1 ) u ( N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq53_HTML.gif, N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq52_HTML.gif, then p ( N + 1 ) p ( N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq54_HTML.gif, N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq52_HTML.gif.

Therefore u ( N + 1 ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq50_HTML.gif and p ( N + 1 ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq51_HTML.gif converge in B.

Now let us show that there exist u and p such that
lim N u ( N + 1 ) ( t ) = u ( t ) , lim N p ( N + 1 ) ( t ) = p ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equw_HTML.gif
In the same way, we obtain
| u ( t ) u ( N + 1 ) ( t ) | ( 2 T + 2 3 3 + 4 2 T ) b ( x , t ) L 2 ( D ) u ( τ ) u ( N + 1 ) ( τ ) B | u ( t ) u ( N + 1 ) ( t ) | + ( 2 T + 2 3 3 + 4 2 T ) b ( x , t ) L 2 ( D ) u ( N + 1 ) ( τ ) u ( N ) ( τ ) B | u ( t ) u ( N + 1 ) ( t ) | + ( 2 T + 2 3 3 + 4 2 T ) | T | | p ( τ ) p ( N ) ( τ ) | f ( x , t , u ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ10_HTML.gif
(10)
| p p ( N ) | A 1 B ( 0 t 0 1 b 2 ( ξ , τ ) | u ( τ ) u ( N + 1 ) ( τ ) | 2 d ξ d τ ) 1 2 | p p ( N ) | + A 1 B ( 0 t 0 1 b 2 ( ξ , τ ) | u ( N + 1 ) ( τ ) u ( N ) ( τ ) | 2 d ξ d τ ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ11_HTML.gif
(11)
Applying Gronwall’s inequality to (10) and using (9) and (11), we have
u ( t ) u ( N + 1 ) ( t ) B 2 [ K N ! D 2 E 2 b ( x , t ) L 2 ( D ) ] 2 × exp 2 ( D + D 2 | T | M A 1 B ) 2 b ( x , t ) L 2 ( D ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ12_HTML.gif
(12)
Here
D = ( 2 T + 2 3 3 + 4 2 T ) , E = { ( 2 T + 2 3 3 + 4 2 T ) ( 1 + T M A 1 B ) } N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equx_HTML.gif

Then N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq52_HTML.gif, we obtain u ( N + 1 ) u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq55_HTML.gif. Hence p ( N + 1 ) p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq56_HTML.gif.

For the uniqueness, we assume that problem (1)-(4) has two solutions ( p , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq57_HTML.gif, ( q , v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq58_HTML.gif. Applying Cauchy’s inequality, Hölder’s inequality, the Lipschitz condition and Bessel’s inequality to | u ( t ) v ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq59_HTML.gif and | p ( t ) q ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq60_HTML.gif, we obtain
| u ( t ) v ( t ) | ( φ + ( 2 T + 2 3 3 + 4 2 T ) M ) T | p ( t ) q ( t ) | | u ( t ) v ( t ) | + ( 2 T + 2 3 3 + 4 2 T ) ( 0 t 0 π b 2 ( ξ , τ ) | u ( τ ) v ( τ ) | 2 d ξ d τ ) 1 2 , | p ( t ) q ( t ) | A 1 B ( 0 t 0 1 b 2 ( ξ , τ ) | u ( τ ) v ( τ ) | 2 d ξ d τ ) 1 2 , | u ( t ) v ( t ) | [ ( φ + ( 2 T + 2 3 3 + 4 2 T ) M ) | T | A 1 B | u ( t ) v ( t ) | + ( 2 T + 2 3 3 + 4 2 T ) ] | u ( t ) v ( t ) | × ( 0 t 0 1 b 2 ( ξ , τ ) | u ( τ ) v ( τ ) | 2 d ξ d τ ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ13_HTML.gif
(13)

Applying Gronwall’s inequality to (13), we have u ( t ) = v ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq61_HTML.gif. Hence p ( t ) = q ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq62_HTML.gif. □

The theorem is proved.

3 Continuous dependence of ( p , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq63_HTML.gif upon the data

Theorem 4 Under assumptions (A1)-(A3), the solution ( p , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq63_HTML.gifof problem (1)-(4) depends continuously upon the data φ, g.

Proof Let Φ = { φ , g , f } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq64_HTML.gif and Φ ¯ = { φ ¯ , g ¯ , f } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq65_HTML.gif be two sets of the data, which satisfy assumptions (A1)-(A3). Suppose that there exist positive constants M i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq66_HTML.gif, i = 0 , 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq67_HTML.gif, such that
0 < M 0 | g | , 0 < M 0 | g ¯ | , g C 1 [ 0 , T ] M 1 , g ¯ C 1 [ 0 , T ] M 1 , φ C 3 [ 0 , π ] M 2 , φ ¯ C 3 [ 0 , π ] M 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equy_HTML.gif
Let us denote Φ = ( g C 1 [ 0 , T ] + φ C 3 [ 0 , π ] + f C 3 , 0 ( D ¯ ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq68_HTML.gif. Let ( p , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq57_HTML.gif and ( p ¯ , u ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq69_HTML.gif be the solutions of inverse problem (1)-(4) corresponding to the data Φ = { φ , g , f } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq70_HTML.gif and Φ ¯ = { φ ¯ , g ¯ , f } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq65_HTML.gif, respectively. According to (5),
u u ¯ = 2 ( φ 0 φ 0 ¯ ) e 0 t p ¯ ( s ) d s + φ 0 ( e 0 t p ( s ) d s e 0 t p ¯ ( s ) d s ) u u ¯ = + 2 0 t 0 1 [ f ( ξ , τ , u ( ξ , τ ) ) f ( ξ , τ , u ¯ ( ξ , τ ) ) ] ξ e τ t p ¯ ( s ) d s d ξ d τ u u ¯ = + 2 0 t 0 1 f ( ξ , τ , u ( ξ , τ ) ) [ e τ t p ( s ) d s e τ t p ¯ ( s ) d s ] u u ¯ = + 4 k = 1 ( 1 x ) sin 2 π k ξ ( φ 2 k φ 2 k ¯ ) e ( 2 π k ) 2 t [ e τ t p ( s ) d s e τ t p ¯ ( s ) d s ] u u ¯ = + 4 k = 1 ( 1 x ) sin 2 π k ξ φ 2 k e ( 2 π k ) 2 t e τ t p ¯ ( s ) d s u u ¯ = + 4 k = 1 cos 2 π k ξ ( φ 2 k 1 φ 2 k 1 ¯ ) e ( 2 π k ) 2 t [ e τ t p ( s ) d s e τ t p ¯ ( s ) d s ] u u ¯ = + 4 k = 1 cos 2 π k ξ φ 2 k 1 e ( 2 k ) 2 t e τ t p ¯ ( s ) d s u u ¯ = 16 π k = 1 k t cos 2 π k ξ ( φ 2 k φ 2 k ¯ ) e ( 2 π k ) 2 t [ e τ t p ( s ) d s e τ t p ¯ ( s ) d s ] u u ¯ = 16 π k = 1 k t cos 2 π k ξ φ 2 k e ( 2 π k ) 2 t e τ t p ¯ ( s ) d s u u ¯ = + 4 k = 1 0 t 0 1 [ f ( ξ , τ , u ( ξ , τ ) ) f ( ξ , τ , u ¯ ( ξ , τ ) ) ] u u ¯ = × sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ¯ ( s ) d s d ξ d τ u u ¯ = + 4 k = 1 0 t 0 1 f ( ξ , τ , u ( ξ , τ ) ) u u ¯ = × sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) [ e τ t p ( s ) d s e τ t p ¯ ( s ) d s ] d ξ d τ u u ¯ = + 4 k = 1 0 t 0 1 [ f ( ξ , τ , u ( ξ , τ ) ) f ( ξ , τ , u ¯ ( ξ , τ ) ) ] u u ¯ = × ξ cos 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ¯ ( s ) d s d ξ d τ u u ¯ = + 4 k = 1 0 t 0 1 f ( ξ , τ , u ( ξ , τ ) ) u u ¯ = × ξ cos 2 π k ξ e ( 2 π k ) 2 ( t τ ) [ e τ t p ( s ) d s e τ t p ¯ ( s ) d s ] d ξ d τ u u ¯ = 16 π k = 1 k 0 t 0 1 [ f ( ξ , τ , u ( ξ , τ ) ) f ( ξ , τ , u ¯ ( ξ , τ ) ) ] u u ¯ = × ( t τ ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ¯ ( s ) d s d ξ d τ u u ¯ = 16 π k = 1 k 0 t 0 1 f ( ξ , τ , u ( ξ , τ ) ) u u ¯ = × ( t τ ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) [ e τ t p ( s ) d s e τ t p ¯ ( s ) d s ] d ξ d τ , | u u ¯ | ( ( 2 T + 2 3 3 + 4 2 T ) + 8 π 2 6 3 k = 0 | φ 2 k | + 4 k = 0 ( | φ 2 k | + | φ 2 k 1 | ) ) | u u ¯ | × p p ¯ C [ 0 , T ] + ( 1 + 2 6 3 T ) φ φ ¯ C 3 [ 0 , 1 ] | u u ¯ | + ( 2 T + 2 3 3 + 4 2 T ) ( 0 t 0 1 b 2 ( ξ , τ ) | u ( τ ) u ¯ ( τ ) | 2 d ξ d τ ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ14_HTML.gif
(14)
Now, let us estimate the difference p p ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq71_HTML.gif as follows:
p p ¯ = ( g ( t ) g ( t ) + g ( t ) ¯ g ( t ) ¯ ) p p ¯ = + 1 g ( t ) k = 1 8 π k 0 1 f ( ξ , τ , u ) sin 2 π k ξ d ξ p p ¯ = 1 g ¯ ( t ) k = 1 8 π k 0 1 f ( ξ , τ , u ¯ ) sin 2 π k ξ d ξ p p ¯ = 4 1 g ( t ) k = 1 ( 2 π k ) 2 φ 2 k e ( 2 π k ) 2 t e 0 t p ( s ) d s p p ¯ = + 4 1 g ¯ ( t ) k = 1 ( 2 π k ) 2 φ 2 k ¯ e ( 2 π k ) 2 t e 0 t p ( s ) ¯ d s p p ¯ = 4 1 g ( t ) k = 1 ( 2 π k ) 2 0 t 0 1 f ( ξ , τ , u ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) d s d ξ d τ p p ¯ = 4 1 g ¯ ( t ) k = 1 ( 2 π k ) 2 0 t 0 1 f ( ξ , τ , u ¯ ) sin 2 π k ξ e ( 2 π k ) 2 ( t τ ) τ t p ( s ) ¯ d s d ξ d τ , p p ¯ C [ 0 , T ] M 3 g g ¯ C 1 [ 0 , T ] + M 4 φ φ ¯ C 3 [ 0 , 1 ] p p ¯ C [ 0 , T ] + M 5 T p p ¯ C [ 0 , T ] + M 6 ( 0 t 0 π b 2 ( ξ , τ ) | u ( τ ) u ¯ ( τ ) | 2 d ξ d τ ) 1 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equz_HTML.gif
where M k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq72_HTML.gif, k = 3 , 4 , 5 , 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq73_HTML.gif are constants that are determined by M 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq74_HTML.gif, M 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq75_HTML.gif and M 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq76_HTML.gif. Then we obtain M 7 = 1 T M 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq77_HTML.gif, M 8 = max { M 3 , M 4 , M 6 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq78_HTML.gif. The inequality M 5 T < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq79_HTML.gif holds for small T. Finally, we obtain
p p ¯ C [ 0 , T ] M 9 ( E E ¯ C 1 [ 0 , T ] + φ φ ¯ C 3 [ 0 , 1 ] + ( 0 t 0 π b 2 ( ξ , τ ) | u ( τ ) u ¯ ( τ ) | 2 d ξ d τ ) 1 2 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equaa_HTML.gif

where M 9 = M 8 M 7 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq80_HTML.gif.

If we take this estimation in (14)
| u u ¯ | M 12 Φ Φ ¯ + M 13 ( 0 t 0 1 b 2 ( ξ , τ ) | u ( τ ) u ¯ ( τ ) | 2 d ξ d τ ) 1 2 , | u u ¯ | M 12 Φ Φ ¯ + M 13 ( 0 t 0 1 b 2 ( ξ , τ ) | u ( τ ) u ¯ ( τ ) | 2 d ξ d τ ) 1 2 , | u u ¯ | 2 2 M 12 2 Φ Φ ¯ 2 + 2 M 13 2 ( 0 t 0 π b 2 ( ξ , τ ) | u ( τ ) u ¯ ( τ ) | 2 d ξ d τ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equab_HTML.gif
applying Gronwall’s inequality, we obtain
| u u ¯ | 2 2 M 12 2 Φ Φ ¯ 2 x exp 2 M 13 2 ( 0 t 0 1 b 2 ( ξ , τ ) d ξ d τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equac_HTML.gif
taking the maximum of the inequality
u u ¯ B 2 2 M 12 2 Φ Φ ¯ 2 x exp 2 M 13 2 ( 0 t 0 1 b 2 ( ξ , τ ) d ξ d τ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equad_HTML.gif

For Φ Φ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq81_HTML.gif, then u u ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq82_HTML.gif. Hence p p ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq83_HTML.gif. □

4 Numerical procedure for nonlinear problem (1)-(4)

We construct an iteration algorithm for the linearization of problem (1)-(4) as follows:
u ( n ) t = 2 u ( n ) x 2 p ( t ) u ( n ) + f ( x , t , u ( n 1 ) ) , ( x , t ) D , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ15_HTML.gif
(15)
u ( n ) ( 0 , t ) = u ( n ) ( 1 , t ) , t [ 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ16_HTML.gif
(16)
u x ( n ) ( 1 , t ) = 0 , t [ 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ17_HTML.gif
(17)
u ( n ) ( x , 0 ) = φ ( x ) , x [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ18_HTML.gif
(18)
Let u ( n ) ( x , t ) = v ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq84_HTML.gif and f ( x , t , u ( n 1 ) ) = f ˜ ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq85_HTML.gif. Then problem (15)-(18) can be written as a linear problem:
v t = 2 v x 2 p ( t ) v ( x , t ) + f ˜ ( x , t ) , ( x , t ) D , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ19_HTML.gif
(19)
v ( 0 , t ) = v ( 1 , t ) , t [ 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ20_HTML.gif
(20)
v x ( 1 , t ) = 0 , t [ 0 , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ21_HTML.gif
(21)
v ( x , 0 ) = φ ( x ) , x [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ22_HTML.gif
(22)

We use the finite difference method to solve (19)-(22) with a predictor-corrector type approach which was explained in [9].

We subdivide the intervals [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq86_HTML.gif and [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq87_HTML.gif into N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq88_HTML.gif and N t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq89_HTML.gif subintervals of equal lengths h = 1 N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq90_HTML.gif and τ = T N t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq91_HTML.gif, respectively. Then we add two lines x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq92_HTML.gif and x = ( N x + 1 ) h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq93_HTML.gif to generate the fictitious points needed for dealing with the boundary conditions. We choose the implicit scheme, which is absolutely stable and has second-order accuracy in h and first-order accuracy in τ[10]. The implicit scheme for (1)-(4) is as follows:
1 τ ( v i j + 1 v i j ) = 1 h 2 ( v i 1 j + 1 2 v i j + 1 + v i + 1 j + 1 ) p j + 1 v i j + 1 + f ˜ i j + 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ23_HTML.gif
(23)
v i 0 = ϕ i , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ24_HTML.gif
(24)
v 0 j = v N x + 1 j , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ25_HTML.gif
(25)
v N x 1 j = v N x + 1 j , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ26_HTML.gif
(26)

where 1 i N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq94_HTML.gif and 1 j N t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq95_HTML.gif are the indices for the spatial and time steps, respectively, v i j = v ( x i , t j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq96_HTML.gif, ϕ i = φ ( x i ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq97_HTML.gif, f ˜ i j = f ˜ ( x i , t j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq98_HTML.gif, x i = i h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq99_HTML.gif, t j = j τ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq100_HTML.gif. At t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq101_HTML.gif level, adjustment should be made according to the initial condition and the compatibility requirements.

Now, let us construct the predicting-correcting mechanism. First, differentiating equation (1) with respect to x and using (3) and (4), we obtain
p ( t ) = g ( t ) + f ˜ x ( 0 , t ) d x + v x x x ( 0 , t ) g ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ27_HTML.gif
(27)
The finite difference approximation of (27) is
p j = ( ( g j + 1 g j ) / τ ) + ( f ˜ 2 j f ˜ 1 j ) / τ + ( v 0 j + 3 v 1 j 3 v 2 j + v 3 j ) / h 3 g j , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equae_HTML.gif

where g j = g ( t j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq102_HTML.gif, j = 0 , 1 , , N t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq103_HTML.gif.

For j = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq104_HTML.gif,
p 0 = ( ( g 1 g 0 ) / τ ) + ( f ˜ 2 0 f ˜ 1 0 ) / τ + ( ϕ 0 + 3 ϕ 1 3 ϕ 2 + ϕ 3 ) / h 3 g 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equaf_HTML.gif
and the values of ϕ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq105_HTML.gif allow us to start our computation. We denote the values of p j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq106_HTML.gif, v i j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq107_HTML.gif at the s th iteration step p j ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq108_HTML.gif, v i j ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq109_HTML.gif, respectively. In numerical computation, since the time step is very small, we can take p j + 1 ( 0 ) = p j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq110_HTML.gif, v i j + 1 ( 0 ) = v i j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq111_HTML.gif, j = 0 , 1 , 2 , , N t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq112_HTML.gif, i = 1 , 2 , , N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq113_HTML.gif. At each ( s + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq114_HTML.gifth iteration step, we first determine p j + 1 ( s + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq115_HTML.gif from the formula
p j + 1 ( s + 1 ) = ( ( g j + 2 g j + 1 ) / τ ) + ( f ˜ 2 j + 1 f ˜ 1 j + 1 ) / τ + ( v 0 j + 1 ( s ) + 3 v 1 j + 1 ( s ) 3 v 2 j + 1 ( s ) + v 3 j + 1 ( s ) ) / h 3 g j + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equag_HTML.gif
Then from (15)-(18) we obtain
1 τ ( v i j + 1 ( s + 1 ) v i j + 1 ( s ) ) = 1 h 2 ( v i 1 j + 1 ( s + 1 ) 2 v i j + 1 ( s + 1 ) + v i + 1 j + 1 ( s + 1 ) ) p j + 1 ( s + 1 ) v i j + 1 ( s + 1 ) + F i j + 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ28_HTML.gif
(28)
v 0 j + 1 ( s ) = v N x + 1 j + 1 ( s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ29_HTML.gif
(29)
v N x 1 j + 1 ( s ) = v N x + 1 j + 1 ( s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ30_HTML.gif
(30)

The system of equations (28)-(30) can be solved by the Gauss elimination method and v i j + 1 ( s + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq116_HTML.gif is determined. If the difference of values between two iterations reaches the prescribed tolerance, the iteration is stopped, and we accept the corresponding values p j + 1 ( s + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq115_HTML.gif, v i j + 1 ( s + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq116_HTML.gif ( i = 1 , 2 , , N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq113_HTML.gif) as p j + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq117_HTML.gif, v i j + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq118_HTML.gif ( i = 1 , 2 , , N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq113_HTML.gif), at the ( j + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq119_HTML.gifth time step, respectively. By virtue of this iteration, we can move from level j to level j + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq120_HTML.gif.

5 Numerical example

Example 1 Consider inverse problem (1)-(4) with
f ( x , t , u ) = 4 π cos ( 2 π x ) + ( ( 2 π ) 2 + exp ( 2 t ) ) u , φ ( x ) = ( 1 x ) sin ( 2 π x ) , g ( t ) = 2 π exp ( t ) , x [ 0 , 1 ] , t [ 0 , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equah_HTML.gif
It is easy to check that the analytical solution of this problem is
{ p ( t ) , u ( x , t ) } = { 1 + exp ( 2 t ) , ( 1 x ) sin ( 2 π x ) exp ( t ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ31_HTML.gif
(31)

Let us apply the scheme which was explained in the previous section for the step sizes h = 0.05 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq121_HTML.gif, τ = 0.05 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq122_HTML.gif.

In the case when T = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq123_HTML.gif, the comparisons between the analytical solution (31) and the numerical finite difference solution are shown in Figures 1 and 2.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Fig1_HTML.jpg
Figure 1

The analytical and numerical solutions of p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq124_HTML.gifwhen T = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq125_HTML.gif. The analytical solution is shown with dashed line.

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Fig2_HTML.jpg
Figure 2

The analytical and numerical solutions of u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq126_HTML.gifat T = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq125_HTML.gif. The analytical solution is shown with dashed line.

Next, we will illustrate the stability of the numerical solution with respect to the noisy overdetermination data (4) defined by the function
g γ ( t ) = g ( t ) ( 1 + γ θ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Equ32_HTML.gif
(32)
where γ is the percentage of noise and θ are random variables generated from uniform distribution in the interval [ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq127_HTML.gif. Figure 3 shows the exact and the numerical solution of p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq128_HTML.gif when the input data (4) is contaminated by γ = 1 % https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq129_HTML.gif, γ = 3 % https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq130_HTML.gif and 5% noise.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_Fig3_HTML.jpg
Figure 3

The numerical solutions of p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq124_HTML.gif(a) for 1% noisy data, (b) for 3% noisy data, (c) for 5% noisy data. In Figure 3(a)-(c) the analytical solution is shown with dashed line.

It is clear from these results that this method has shown to produce stable and reasonably accurate results for these examples. Numerical differentiation is used to compute the values of g ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq131_HTML.gif and v x x x ( 0 , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq132_HTML.gif in the formula p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-213/MediaObjects/13661_2013_Article_578_IEq133_HTML.gif. It is well known that numerical differentiation is slightly ill-posed and it can cause some numerical difficulties. One can apply the natural cubic spline function technique [11] to get still decent accuracy.

Declarations

Authors’ Affiliations

(1)
Department of Management Information Systems, 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.