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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 15, Issue 1

# Determination of a diffusion coefficient in a quasilinear parabolic equation

Fatma Kanca
Published Online: 2017-02-13 | DOI: https://doi.org/10.1515/math-2017-0003

## Abstract

This paper investigates the inverse problem of finding the time-dependent diffusion coefficient in a quasilinear parabolic equation with the nonlocal boundary and integral overdetermination conditions. Under some natural regularity and consistency conditions on the input data the existence, uniqueness and continuously dependence upon the data of the solution are shown. Finally, some numerical experiments are presented.

MSC 2010: 35K59; 35R30

## 1 Introduction

In this paper, an inverse problem of determining of the diffusion coefficient a(t) has been considered with extra integral condition ${\int }_{0}^{1}u\left(x,\phantom{\rule{thinmathspace}{0ex}}t\right)dx$ which has appeared in various applications in industry and engineering [1]. The mathematical model of this problem is as follows: $ut=a(t)uxx+f(x,t,u),(x,t)∈DT:=(0,1)×(0,T)$(1) $u(x,0)=φ(x),x∈[0,1],$(2) $u(0,t)=u(1,t),ux(1,t)=0,t∈[0,T],$(3) $E(t)=∫01u(x,t)dx,0≤t≤T,$(4) The functions φ(x) and f (x, t, u) are given functions.

The problem of a coefficient identification in nonlinear parabolic equation is an interesting problem for many scientists [25]. In [6] the nature of (3)-type conditions is demonstrated.

In this study, we consider the inverse problem (1)-(4) with nonlocal boundary conditions and integral overdetermination condition. We prove the existence, uniqueness and continuous dependence on the data of the solution by applying the generalized Fourier method and we construct an iteration algorithm for the numerical solution of this problem.

The plan of this paper is 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 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

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

(A1)E(t) ∈ C1[0, T], E′(t)≤ 0,

(A2)

1. φ(x) ∈ C4[0, 1], φ(0) = φ(1), φ′(1) = 0, φ″(0) = φ″(1),

2. φ2k ≥ 0, k = 1, 2, …

(A3)

1. Let the function f(x, t, u) be continuous with respect to all arguments in ${\overline{D}}_{T}×\left(-\mathrm{\infty },\mathrm{\infty }\right)$ and satisfy the following condition $∂(n)f(x,t,u)∂xn−∂(n)f(x,t,u~)∂xn≤b(x,t)|u−u~|,n=0,1,2,$ where b(x, t) ∈ L2(DT), b(x, t) ≥ 0,

2. $f\left(x,t,u\right)\in {C}^{4}\left[0,1\right],t\in \left[0,T\right],f\left(x,t,u\right){|}_{x=0}=f\left(x,t,u\right){|}_{x=1},{f}_{x}\left(x,t,u\right){|}_{x=1}=0,{f}_{xx}\left(x,t,u\right){|}_{x=0}={f}_{xx}\left(x,t,u\right){|}_{x=1},$

3. f2k(t) ≥ 0, f0(t) > 0, ∀ t ∈ [0, T], where $φk=∫01φ(x)Yk(x)dx,fk(t)=∫01f(x,t,u)Yk(x)dx,k=0,1,2,…$ $X0(x)=2,X2k−1(x)=4cos⁡2πkx,X2k(x)=4(1−x)sin⁡2πkx,k=1,2,….Y0(x)=x,Y2k−1(x)=xcos⁡2πkx,Y2k(x)=sin⁡2πkx,k=1,2,….$

The systems of functions Xk(x) and Yk(x), k = 0, 1, 2,… are biorthonormal on [0, 1]. They are also Riesz bases in L2[0, 1] (see [7]).

We obtain the following representation for the solution of (1) - (3) for arbitrary a(t) by using the Fourier method: $u(x,t)=φ0+∫0tf0(τ)dτX0(x)+∑k=1∞φ2ke−(2πk)2∫0ta(s)ds+∫0tf2k(τ)dτe−(2πk)2∫τta(s)dsdτX2k(x)+∑k=1∞(φ2k−1−4πkφ2kt)e−(2πk)2∫0ta(s)dsX2k−1(x)+∑k=1∞∫0t(f2k−1(τ)−4πkf2k(τ)(t−τ))e−(2πk)2∫τta(s)dsdτX2k−1(x)$(5) Differentiating (5) we obtain $∫01ut(x,t)dx=E′(t),0≤t≤T.$(6)

(5) and (6) yield $a(t)=−E′(t)+2f0(t)+∑k=1∞2πkf2k(t)∑k=1∞8πkφ2ke−(2πk)2∫0ta(s)ds+∫0tf2k(τ)e−(2πk)2∫τta(s)dsdτ$(7)

#### Definition 2.1

$\left\{u\left(t\right)\right\}=\left\{{u}_{0}\left(t\right),{u}_{2k}\left(t\right),{u}_{2k-1}\left(t\right),k=1,\dots ,n\right\},$ are continuous functions on [0, T] and satisfying the condition $\underset{0\le t\le T}{max}|{u}_{0}\left(t\right)|+\sum _{k=1}^{\mathrm{\infty }}\left(\underset{0\le t\le T}{max}|{u}_{2k}\left(t\right)|+\underset{0\le t\le T}{max}|{u}_{2k-1}\left(t\right)|\right)<\mathrm{\infty }.$ The set of these functions is denoted by B1 and the norm in B1 is $\parallel u\left(t\right)\parallel =\underset{0\le t\le T}{max}|{u}_{0}\left(t\right)|+\sum _{k=1}^{\mathrm{\infty }}\left(\underset{0\le t\le T}{max}|{u}_{2k}\left(t\right)|+\underset{0\le t\le T}{max}|{u}_{2k-1}\left(t\right)|\right).$ It can be shown that B1 is the Banach space.

#### Theorem 2.2

If the assumptions (A1)–(A3) are satisfied, then the inverse coefficient problem (1) - (4) has at most one solution for small T.

#### Proof

We define an iteration for Fourier coefficient of (5) as follows: $u0(N+1)(t)=u0(0)(t)+∫0t∫01f(ξ,τ,u(N)(ξ,τ))ξdξdτu2k(N+1)(t)=u2k(0)(t)+∫0t∫01f(ξ,τ,u(N)(ξ,τ))sin⁡2πkξe−(2πk)2∫τta(N)(s)dsdξdτu2k−1(N+1)(t)=u2k−1(0)(t)+∫0t∫01f(ξ,τ,u(N)(ξ,τ))ξcos2πkξe−(2πk)2∫τta(N)(s)dsdξdτ−4πk∫0t∫01(t−τ)f(ξ,τ,u(N)(ξ,τ))sin⁡2πkξe−(2πk)2∫τta(N)(s)dsdξdτ$(8) where N = 0, 1, 2, … and $u0(0)(t)=φ0,u2k(0)(t)=φ2ke−(2πk)2∫0ta(s)ds,u2k−1(0)(t)=(φ2k−4πktφ2k−1)e−(2πk)2∫0ta(s)ds.$

It is obvious that u(0)(t)∈ B1 and a(0)C[0, T].

For N = 0, $u0(1)(t)=u0(0)(t)+∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]ξdξdτ+∫0t∫01f(ξ,τ,0)ξdξdτ.$

Let us apply Cauchy inequality, $u0(1)(t)≤|φ0|+∫0tdτ12∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]ξdξ2dτ12+∫0tdτ12∫0t∫01f(ξ,τ,0)ξdξ2dτ12.$ and with Lipschitz condition we obtain $u0(1)(t)≤|φ0|+t∫0t∫01b(ξ,τ)u(0)(ξ,τ)dξ2dτ12+t∫0t∫01f(ξ,τ,0)dξ2dτ12.$

If we take the maximum of the last inequality, we get the following estimation for ${u}_{0}^{\left(1\right)}\left(t\right).$ $max0≤t≤Tu0(1)(t)≤|φ0|+T∥b(x,t)∥L2(DT)u(0)(t)B1+T∥f(x,t,0)∥L2(DT).$ $u2k(1)(t)=φ2ke−(2πk)2∫0ta(s)ds+∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]sin⁡2πkξe−(2πk)2∫τta(s)dsdξdτ+∫0t∫01f(ξ,τ,0)sin⁡2πkξe−(2πk)2∫τta(s)dsdξdτ.$

Let us apply Cauchy inequality, $u2k(1)(t)≤|φ2k|+∫0tdτ12∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]sin⁡2πkξdξ2dτ12+∫0tdτ12∫0t∫01f(ξ,τ,0)sin⁡2πkξdξ2dτ12.$ and take the sum of the last inequality and partial derivative of f with respect to ξ and apply Hölder inequality, $∑k=1∞u2k(1)(t)≤∑k=1∞|φ2k|+12π∑k=1∞1k212.∫0t∑k=1∞∫01[fξ(ξ,τ,u(0)(ξ,τ))−fξ(ξ,τ,0)]cos⁡2πkξdξ2dτ12+12π∑k=1∞1k212∫Ot∑k=1∞∫O1fξ(ξ,τ,0)cos⁡2πkξdξ2dτ12.$

By applying Bessel inequality we obtain $∑k=1∞u2k(1)(t)≤∑k=1∞|φ2k|+6T12∫0t∑k=1∞∫01[fξ(ξ,τ,u(0)(ξ,τ))−fξ(ξ,τ,0)]dξ2dτ12+6T12∫0t∑k=1∞∫01fξ(ξ,τ,0)dξ2dτ12.$

If we use Lipschitzs condition and take the maximum of the last inequality, we get the following estimation for $\sum _{k=1}^{\mathrm{\infty }}\left|{u}_{2k}^{\left(1\right)}\left(t\right)\right|.$ $∑k=1∞max0≤t≤Tu2k(1)(t)≤∑k=1∞|φ2k|+6T12∥b(x,t)∥L2(D)u(0)(t)+6T12∥fx(x,t,0)∥L2(D).∑k=1∞max0≤t≤Tu2k(1)(t)≤∑k=1∞|φ2k|+6T12∥b(x,t)∥L2(DT)u(0)(t)B1+6T12M.u2k−1(1)(t)=u2k−1(0)(t)+∫0t∫01f(ξ,τ,u(0)(ξ,τ))ξcos⁡2πkξe−(2πk)2∫τta(0)(s)(ds)dξdτ−4πk∫0t∫01(t−τ)f(ξ,τ,u(0)(ξ,τ))sin⁡2πkξe−(2πk)2∫τta(0)(s)(ds)dξdτ.$

Similarly, let us apply Cauchy inequality, $u2k−1(1)(t)≤|φ2k−1|+4πkt|φ2k|+∫0tdτ12∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]cos⁡2πkξdξ2dτ12+∫0tdτ12∫0t∫01fξ,τ,0cos⁡2πkξdξ2dτ12+4πkt∫0tdτ12∫0t∫01[fξ,τ,u0ξ,τ−fξ,τ,0]sin⁡2πkξdξ2dτ12+4πkt∫0tdτ12∫0t∫01fξ,τ,0sin⁡2πkξdξ2dτ12,$

and take the sum of the last inequality and partial derivative of f with respect to ξ and apply Hölder inequality and Bessel inequality, $∑k=1∞u2k−1(1)(t)≤∑k=1∞|φ2k−1|+t6∑k=1∞φ2k″+∑k=1∞t2πk∫0t∫01[fξ(ξ,τ,u(0)(ξ,τ))−fξ(ξ,τ,0)]dξ2dτ12+∑k=1∞t2πk∫0t∫01fξ(ξ,τ,0)dξ2dτ12+∑k=1∞4πktt(2πk)2∫0t∫01[fξξ(ξ,τ,u(0)(ξ,τ))−fξξ(ξ,τ,0)]dξ2dτ12+∑k=1∞4πktt(2πk)2∫0t∫01fξξ(ξ,τ,0)dξ2dτ12.$

If we use Lipschitzs condition and take the maximum of the last inequality, we get the following estimation for $\sum _{k=1}^{\mathrm{\infty }}\left|{u}_{2k-1}^{\left(1\right)}\left(t\right)\right|.$ $∑k=1∞max0≤t≤Tu2k−1(1)(t)≤∑k=1∞|φ2k−1|+6T6∑k=1∞φ2k″+6T12+6TT6∥b(x,t)∥L2(DT)u(0)(t)B1+6Tπ12+6TT6M.$

Finally we obtain the following inequality: $∥u(1)(t)∥B1=max0≤t≤Tu0(1)(t)+∑k=1∞max0≤t≤Tu2k(1)(t)+max0≤t≤Tu2k−1(1)(t)≤∥φ∥+6T6∑k=1∞φ2k″+T+6T6+6TT3∥b(x,t)∥L2(DT)u(0)(t)B1+T+6T6+6TT3M.$

where $\parallel \phi \parallel =|{\phi }_{0}|+4\left[|{\phi }_{2k}|+|{\phi }_{2k-1}|\right].$ Hence u(1)(t) ∈ B1. In the same way, for N we have $u(N)(t)B1=max0≤t≤Tu0(N)(t)+∑k=1∞max0≤t≤Tu2k(N)(t)+max0≤t≤Tu2k−1(N)(t)≤∥φ∥+6T6∑k=1∞φ2k″+T+6T6+6TT3∥b(x,t)∥L2(DT)u(N−1)(t)B1+T+6T6+6TT3M.$

Since u(N − 1)(t) ∈ B1, we have u(N)(t) ∈ B1, ${u(t)}={u0(t),u2k(t),u2k−1(t),k=1,2,…}∈B1.$

We define an iteration for (7) as follows: $a(N+1)(t)=−E′(t)+∫01f(ξ,τ,u(N))dx∑k=1∞8πkφ2ke−(2πk)2∫0ta(N)(s)ds+∫0t∫01f(ξ,τ,u(N))sin2πkξe−(2πk)2∫0ta(N)(s)dsdξdτ$

It is clear that $\underset{0}{\overset{1}{\int }}f\left(\xi ,\phantom{\rule{thinmathspace}{0ex}}\tau ,\phantom{\rule{thinmathspace}{0ex}}u\right)dx=2{f}_{0}\left(t\right)+\sum _{k=1}^{\mathrm{\infty }}\frac{2}{\pi k}{f}_{2k}\left(t\right).$ For N = 0, $a(1)(t)=−E′(t)+∫01f(ξ,τ,u(0))dx∑k=1∞8πkφ2ke−(2πk)2∫0ta(0)(s)ds+∫0t∫01f(ξ,τ,u(0))sin2πkξe−(2πk)2∫0ta(0)(s)dsdξdτ$

Let us add and subtract $\underset{0}{\overset{1}{\int }}f\left(\xi ,\phantom{\rule{thinmathspace}{0ex}}\tau ,\phantom{\rule{thinmathspace}{0ex}}0\right)d\phantom{\rule{thinmathspace}{0ex}}\xi \phantom{\rule{thinmathspace}{0ex}}d\phantom{\rule{thinmathspace}{0ex}}\tau$ to the last equation and use the Cauchy inequality and take the maximum to obtain: $a(1)(t)C[0,T]<_E′(t)C2+1C2∥b(x,t)∥L2(DT)u(0)(t)B1+1C2M$

where $C2=E(T)−2φ0−2∫0Tf0(τ)dτ.$

Hence a(1)(t) ∈ C[0, T]. In the same way, for N, we have $a(N)(t)C[0,T]<_E′(t)C2+1C2b(x,t)L2(DT)u(N−1)(t)B1+1C2M$

Since u(N − 1)(t) ∈ B1, we have a(N)(t) ∈ C[0, T].

Now let us prove that the iterations u(N + 1)(t) and a(N + 1)(t) converge in B1 and C[0, T], respectively, as N → ∞.

$u(1)(t)−u(0)(t)=(u0(1)(t)−u0(0)(t))+∑k=1∞[(u2k(1)(t)−u2k(0)(t))+(u2k−1(1)(t)−u2k−1(0)(t))]=(∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]ξdξdτ)+∫0t∫01f(ξ,τ,0)ξdξdτ+∑k=1∞∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]e−(2πk)2∫τta(0)(s)dssin⁡2πkξdξdτ+∑k=1∞∫0t∫01f(ξ,τ,0)e−(2πk)2∫τta(0)(s)dssin⁡2πkξdξdτ+∑k=1∞∫0t∫01[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]e−(2πk)2∫τta(0)(s)dsξcos⁡2πkξdξdτ+∑k=1∞∫0t∫01f(ξ,τ,0)e−(2πk)2∫τta(0)(s)dsξcos⁡2πkξdξdτ−16πk∑k=1∞∫0t∫01(t−τ)[f(ξ,τ,u(0)(ξ,τ))−f(ξ,τ,0)]e−(2πk)2∫τta(0)(s)dssin⁡2πkξdξdτ+16πk∑k=1∞∫0t∫01(t−τ)f(ξ,τ,0)e−(2πk)2∫τta(0)(s)dssin⁡2πkξdξdτ$

Applying Cauchy inequality, Hölder inequality, Lipshitzs condition and Bessel inequality to the last equation, we obtain: $u(1)(t)−u(0)(t)B1<_T+6T6+6TT3∥b(x,t)∥L2(DT)u(0)(t)B1+T+6T6+6TT3M.K=T+6T6+6TT3∥b(x,t)∥L2(DT)u(0)(t)B1+T+6T6+6TT3M.u(2)(t)−u(1)(t)=u0(2)(t)−u0(1)(t)+∑k=1∞[(u2k(2)(t)−u2k(1)(t))+(u2k−1(2)(t)−u2k−1(1)(t))]=(∫0t∫01[f(ξ,τ,u(1)(ξ,τ))−f(ξ,τ,u(0)(ξ,τ))]ξdξdτ)+∑κ=1∞∫0t∫01[f(ξ,τ,u(1)(ξ,τ))−f(ξ,τ,u(0)(ξ,τ))]e−(2πk)2∫τta(1)(s)dssin⁡2πkξdξdτ+∑κ=1∞∫0t∫01f(ξ,τ,u(0)(ξ,τ))(e−(2πk)2∫τta(1)(s)ds−e−(2πk)2∫τta(0)(s)ds)sin⁡2πkξdξdτ+∑κ=1∞∫0t∫01[f(ξ,τ,u(1)(ξ,τ))−f(ξ,τ,u(0)(ξ,τ))]e−(2πk)2∫τta(1)(s)dsξcos⁡2πkξdξdτ+∑κ=1∞∫0t∫01f(ξ,τ,u(0)(ξ,τ))(e−(2πk)2∫τta(1)(s)ds−e−(2πk)2∫τta(0)(s)ds)ξcos⁡2πkξdξdτ−16πk∑κ=1∞∫0t∫01(t−τ)[f(ξ,τ,u(1)(ξ,τ))−f(ξ,τ,u(0)(ξ,τ))]e−(2πk)2∫τta(1)(s)dssin⁡2πkξdξdτ−16πk∑κ=1∞∫0t∫01(t−τ)f(ξ,τ,u(0)(ξ,τ))(e−(2πk)2∫τta(1)(s)ds−e−(2πk)2∫τta(0)(s)ds)sin⁡2πkξdξdτ$

Applying the same estimations we obtain: $u(2)(t)−u(1)(t)B1<_T+6T6+6TT3∥b(x,t)∥L2(DT)u(1)−u(0)B1+6T6+6TT3TMa(1)−a(0)B2.$ $a(1)−a(0)=E′(t)+∫01f(ξ,τ,u(1))dξ∑k=1∞8πkφ2ke−(2πk)2∫0ta(1)(s)ds+∫0t∫01f(ξ,τ,u(1))sin⁡2πkξe−(2πk)2∫0ta(1)(s)dsdξdτ−E′(t)+∫01f(ξ,τ,u(0))dξ∑k=1∞8πkφske−(2πk)2∫0ta(0)(s)ds+∫0t∫01f(ξ,τ,u(0))sin⁡2πkξe−(2πk)2∫0ta(0)(s)dsdξdτ$

If we apply the Cauchy inequality, the Hölder Inequality, the Lipschitz condition and the Bessel inequality to the last equation, we obtain: $∥a(1)−a(0)∥C[0,T]<_(|E′(t)|π26C22∑k=1∞|φ2k(4)|+|E′(t)|M26C22+M26C22∑k=1∞|φ2k(4)|+M2)T∥a(1)−a(0)∥C[0,T]+(2|E′(t)|6C22+26C22∑k=1∞|φ2k″|+M)∥b(x,t)∥L2(DT)∥u(1)−u(0)∥B1A=(2|E′(t)|6C22+26C22∑k=1∞|φ2k″|+M),B=E′(t)π26C22∑k=1∞φ2k(4)+E′(t)M26C22+M26C22∑k=1∞φ2k(4)+M2a(1)−a(0)C[0,T]<_A1−BT∥b(x,t)∥L2(DT)u(1)−u(0)B1∥u(2)(t)−u(1)(t)∥B1<_[(T+6T6+6TT3)+(6T6+6TT3)MAT1−BT]∥b(x,t)∥L2(DT)KC=T+6T6+6TT3D=6T6+6TT3u(2)(t)−u(1)(t)B1<_C+DMAT1−BT∥b(x,t)∥L2(DT)K$

If we use the same estimations, we get $u(3)(t)−u(2)(t)B1<_12C+DMAT1−BT2∥b(x,t)∥L2(DT)2K$

For N : $a(N+1)−a(N)C[0,T]<_A1−BT∥b(x,t)∥L2(DT)u(N+1)−u(N)B1u(N+1)(t)−u(N)(t)B1<_KN!C+DMAT1−BTN∥b(x,t)∥L2(DT)N$(9)

It is easy to see that if u(N + 1)u(N), N → ∞, then a(N + 1)a(N), N → ∞. Therefore u(N + 1)(t) and a(N + 1)(t) convergence in B1 and C[0, T], respectively.

Now let us show that there exist u and a such that $limN→∞u(N+1)(t)=u(t),limN→∞a(N+1)(t)=a(t).$

If we apply the Cauchy inequality, the Hölder Inequality, the Lipshitzs condition and the Bessel inequality to |uu(N + 1)| and |aa(N)| $|u−u(N+1)|<_C(∫0t∫01b2(x,t)|u(τ)−u(N+1)(τ)|2dξdτ)12+C(∫0t∫01b2(x,t)|u(N+1)(τ)−u(N)(τ)|2dξdτ)12+D(∫0t∫01|a(τ)−a(N)(τ)|2dξdτ)12|a−a(N)|<_A1−BT(∫0t∫01b2(x,t)|u(τ)−u(N+1)(τ)|2dξdτ)12+A1−BT(∫0t∫01b2(x,t)|u(N+1)(τ)−u(N)(τ)|2dξdτ)12$

and the Gronwall inequality to the last inequality and using inequality (9), we have $u(t)−u(N+1)(t)B12<_2C+DMAT1−BTN+1KN!∥b(x,t)∥L2(DT)N+12×exp⁡2C+DMAT1−BT2∥b(x,t)∥L2(DT)$

Then N → ∞ we obtain u(N+1)u, a(N+1) → a.

Let us prove the uniqueness of these solutions. Assume that problem (1)-(4) has two solution pair (a,u), (b,v) . Applying the Cauchy inequality, the Hölder Inequality, the Lipshitzs condition and the Bessel inequality to |u(t)-v(t)| and |a(t)-b(t)|, we obtain: $|u(t)−v(t)|<_63∑κ=1∞φ2k‴+φ2k−1‴+6T6M∑κ=1∞φ2k−1ιv+6TM3+26TM3×T∫0t∫0π|a(τ)−b(τ)|2dξdτ12+T+6T6+6TT3∫0t∫01b2(ξ,τ)|u(τ)−v(τ)|2dξdτ12,|a(t)−b(t)|<_A1−BT∫0t∫01b2(ξ,τ)|u(τ)−v(τ)|2dξdτ12,$

and applying the Gronwall inequality to the last inequality we have u(t) = v(t) . Hence a(t) = b(t), here $T<\frac{1}{B}.$

The theorem is proved.

## 3 Continuous dependence of solution upon the data

#### Theorem 3.1

If the assumptions (A1) – (A3) are satisfied, the solution (a,u) of problem (1)-(4) depends continuously upon the data φ, E.

#### proof

Let $\mathrm{\Phi }=\left\{\phi ,E,f\right\}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\overline{\mathrm{\Phi }}=\left\{\overline{\phi },\overline{E},f\right\}$ be two sets of the data, which satisfy the assumptions (A1)-(A3) . Suppose that there exist positive constants Mi, i = 0, 1, 2 such that $∥E∥C1[0,T]<_M1,∥E¯∥C1[0,T]<_M1,∥φ∥C4[0,1]<_M2,∥φ¯∥C4[0,1]<_M2.$

Let us denote $\parallel \mathrm{\Phi }\parallel =\left(\parallel E{\parallel }_{{C}^{1}\left[0,T\right]}+\parallel \phi {\parallel }_{{C}^{4}\left[0,1\right]}+\parallel f{\parallel }_{{C}^{4.0}\left({\overline{D}}_{T}\right)}\right).$ Let (a,u) and $\left(\overline{a},\overline{u}\right)$ be solutions of (1)-(4) corresponding to the data $\mathrm{\Phi }=\left\{\phi ,E,f\right\}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\overline{\mathrm{\Phi }}=\left\{\overline{\phi },\overline{E},f\right\}$ respectively. According to (5), we have $|u−u¯|<_∥φ−φ¯∥C4[0,1]+26T3∑κ=1∞|φ2k(4)|+|φ2k−1(4)|+4∑κ=1∞|φ2k′′|+26TM3+26T3⋅T∫0t∫01|a(τ)−a¯(τ)|2dξdτ12+2T+26Tπ3+6TT3∫0t∫01b2(ξ,τ)|u(τ)−u¯(τ)|2dξdτ12|a−a¯|<_M3∥φ−φ¯∥C4[0,1]+M4∥E(t)−E′(t)∥C1[0,T]+M5∫0t∫01b2(ξ,τ)|u(τ)−u¯(τ)|2dξdτ12+M6T|a−a¯|(1−TM6)|a−a¯|<_M7∥Φ−Φ¯∥+∫0t∫01b2(ξ,τ)|u(τ)−u¯(τ)|2dξdτ12|a−a¯|<_M8|Φ−Φ¯|+∫0t∫01b2(ξ,τ)|u(τ)−u¯(τ)|2dξdτ12|u−u¯|2<_2M92∥Φ−Φ¯∥2+2M102∫0t∫01b2(ξ,τ)|u(τ)−u¯(τ)|2dξdτ$

where M7 = max(M3, M4, M6), ${M}_{8}=\frac{{M}_{7}}{1-T{M}_{6}},{M}_{10}={M}_{8}+\left(2\sqrt{T}+\frac{2\sqrt{6T}\pi }{3}+\frac{\sqrt{6T}\tau }{3}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}T<\frac{1}{{M}_{6}}.$

Applying the Gronwall inequality, $∥u−u¯∥B12<_2M92∥Φ−Φ¯∥2×exp⁡2M102∫0t∫01b2(ξ,τ)dξdτ2.$

$\text{For}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\Phi }\to \overline{\mathrm{\Phi }}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}u\to \overline{u}.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{Hence}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}a\to \overline{a}.$

## 4 Numerical method for the problem (1)-(4)

In order to solve problem (1)-(4) numerically, we need the linearization of the nonlinear terms: $ut(n)=a(t)uxx(n)+f(x,t,u(n−1)),(x,t)∈DT$(10)

$u(n)(0,t)=u(n)(1,t),t∈[0,T]$(11)

$ux(n)(1,t)=0,t∈[0,T]$(12)

$u(n)(x,0)=φ(x),x∈[0,1].$(13)

Let u(n)(x,t) = v(x,t) and $f\left(x,t,{u}^{\left(n-1\right)}\right)=\stackrel{~}{f}\left(x,t\right).$ Then we obtain a linear problem: $vt(n)=a(t)vxx(n)+f~(x,t)(x,t)∈DT$(14)

$v(0,t)=v(1,t),t∈[0,T]$(15)

$vx(1,t)=0,t∈[0,T]$(16)

$v(x,0)=φ(x),x∈[0,1].$(17)

In this step, we use the implicit finite difference approximation for the discretizing problem (14)-(17) : $1τvij+1−vij=aj+11h2vi−1j+1−2vij+1+vi+1j+1+f~ij+1,$

$vi0=ϕi,$(18)

$v0j=vNx+1j,$(19)

$vNx−1j=vNx+1j,$(20)

where x = ih, t = , 1 ≤ iNX and 1 ≤ jNt, ${v}_{i}^{j}=v\left({x}_{i},{t}_{j}\right),{\varphi }_{i}=\phi \left({x}_{i}\right),{\stackrel{~}{f}}_{i}^{j}=\stackrel{~}{f}\left({x}_{i},{t}_{j}\right),{x}_{i}=ih,{t}_{j}=j\tau .$

Let us integrate the equation (1) with respect to x and use (3) and (4) to obtain $a(t)=−E′(t)+∫01f~(x,t)dxvx(0,t).$(21)

The finite difference approximation of (21) is $aj=[((Ej+1−Ej)/τ)−(Fin)j]hv1j−v0j,$

where ${E}^{j}=E\left({t}_{j}\right),\left(Fin{\right)}^{j}={\int }_{0}^{1}\stackrel{~}{f}\left(x,{t}_{j}\right)dx,j=0,1,\dots ,{N}_{t}.$ We mention that ${\int }_{0}^{1}\stackrel{~}{f}\left(x,{t}_{j}\right)dx$ is numerically calculated using Simpson’s rule of integration.

${a}^{j\left(s\right)},{v}_{i}^{j\left(s\right)}$ are the values of ${a}^{j},{v}_{i}^{j}$ at the s-th iteration step, respectively. At each (s+1)-th iteration step, aj+1(s+1) is as follows $aj+1(s+1)=[((Ej+2−Ej+1)/τ)−(Fin)j+1]hv1j+1(s)−v0j+1(s).$

The iteration of (18)-(20) is $1τvij+1(s+1)−vij+1(s)=1h2aj+1(s+1)vi−1j+1(s+1)−2vij+1(s+1)+vi+1j+1(s+1)+f~ij+1,$(22)

$v0j+1(s)=vNxj+1(s),$(23)

$v1j+1(s)=vNx+1j+1(s),s=0,1,2,….$(24)

The system of equations (22)-(24) is solved by the Gauss elimination method and ${v}_{i}^{j+1\left(s+1\right)}$ is determined. If the difference of values between two iterations reaches the prescribed tolerance, the iteration is stopped and we accept the corresponding values ${a}^{j+1\left(s+1\right)},{v}_{i}^{j+1\left(s+1\right)}\left(i=1,2,\dots ,{N}_{x}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}^{j+1},{v}_{i}^{j+1}\left(i=1,2,\dots ,{N}_{x}\right),$ on the (j+1)-th time step, respectively.

#### Example 4.1

(smooth diffusion coefficient). The first example investigates finding the exact solution ${a(t),u(x,t)}=(t2+2),x3−2x2+x+5exp⁡(−t).$

for the given functions $φ(x)=x3−2x2+x+5,E(t)=6112exp⁡(−t),F(x,t)=−u−(t2+2)(6x−4)exp⁡(−t).$

The step sizes are h = 0.01, τ = 0.005.

The comparisons between the exact solution and the numerical finite difference solution are shown in Figures 1 and 2 when T = 2.

Figure 1

The analytical and numerical solutions of a(t) when T = 2 The analytical solution is shown with dashed line.

Figure 2

The analytical and numerical solutions of u(x,t) when T = 2 The analytical solution is shown with dashed line.

In order to investigate the stability of the numerical solution, noise is added to the overdetermination data (4) as follows $Eγ(t)=E(t)(1+γθ),$(25)

where γ is the percentage of noise and θ are random variables generated from a uniform distribution in the interval [−1, 1].

Figure 3 shows the exact and numerical solutions of a(t) when the input data (4) are contaminated by γ = 1%, 5% and 10% noise. From these figures it can be seen that the numerical solution becomes unstable as the input data is contaminated with noise. We use wavelet decomposition and thresholding to remove noise and we obtain Figure 4.

Figure 3

The exact and approximate solutions of a(t),(a) for 1% noisy data, (b) for 5% noisy data, (c) for 10% noisy data. In figure (a)-(c) the exact solution is shown with dashed line.

Figure 4

The exact and approximate solutions of a(t), after thresholding, (a) for 3% noisy data, (b) for 5% noisy data, (c) for 10% noisy data. In figure (a)-(c) the exact solution is shown with dashed line.

#### Example 4.2

(discontinuous diffusion coefficient). In the previous Example 4.1, a smooth function given by a(t)= t2 + 1 is considered. In Example 4.2, a more severe discontinuous test function is given: $a(t)=−(t2+2),t∈[0,1)(t2+2),t∈[1,2]$

Let us apply the scheme above for the step sizes h = 0.01, τ = 0.005. Figure 5 shows the exact and the numerical solutions of a(t) when T = 2.

Figure 5

The analytical and numerical solutions of a(t) when T = 2 The analytical solution is shown with dashed line.

## Some discussions

In the previous section, in Example 4.1, the man-made noise in the measured output data is added to show the stability of the numerical method. Unstable numerical solution is obtained and wavelet decomposition and thresholding are used to remove noise. Also in Example 4.2, discontinuous source function is given to show the efficiency of the present method. From Figure 5 it can be seen that the agreement between the numerical and exact solutions for a(t) is excellent.

In future the fractional problem of this inverse problem can be studied [810].

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Accepted: 2016-10-21

Published Online: 2017-02-13

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 77–91, ISSN (Online) 2391-5455,

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