# A Regularization of the Backward Problem for Nonlinear Parabolic Equation with Time-Dependent Coefficient

Authors: Pham Hoang Quan; Le Minh Triet; Dang Duc Trong
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We study the backward problem with time-dependent coefficient which is a severely ill-posed problem. We regularize this problem by combining quasi-boundary value method and quasi-reversibility method and then obtain sharp error estimate between the exact solution and the regularized solution. A numerical experiment is given in order to illustrate our results.


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## Body

## 1. Introduction

We consider the inverse time problem for the nonlinear parabolic equation(1.1)ut(x,t)-a(t)uxx(x,t)=f(x,t,u(x,t)),(x,t)∈[0,π]×(0,T],(1.2)u(0,t)=u(π,t)=0,t∈[0,T],(1.3)u(x,T)=g(x),x∈[0,π],
where  a(t)  is thermal conductivity function of (1.1) such that there exist p,  q>0 satisfying
(1.4)0<p≤a(t)≤q,
for all t∈[0,T].In other words, from the temperature distribution at a particular timet=T (final data), we want to retrieve the temperature distribution at any earlier time t<T. This problem is called the backward heat problem (BHP), or final-value problem. As known, this problem is severely ill-posed in Hadamard’s sense; that is, solutions do not always exist, and when they exist, they do not depend continuously on the given data. In practice, datum g is based on (physical) measurements. Hence, there will be measurement errors, and we would actually have datum function gδ such that ∥gδ-g∥L2(0,π)≤δ. Thus, form small error contaminating physical measurements, the solutions corresponding to datum function gδ have large errors. This makes it  difficult to make numerical calculations with perturbed data.In our knowledge, there have been many papers on the linear homogeneous case of the backward problem, but there are a few papers on the nonhomogeneous case and the nonlinear case such as [1–6]; especially, the nonlinear case with time-dependent thermal conductivity coefficient is very scarce. Moreover, the thermal conductivity is the property of a material’s ability to conduct heat. Therefore, the thermal conductivity is not a constant in some cases. In this paper, we extend the result (in [7]) for the case of  the time-dependent thermal conductivity a(t). In future, we will  research the BHP problem for the case of  the time- and space-dependent thermal conductivity a(x,t).In [8], the authors used the quasi-reversibility method to regularize a 1D linear nonhomogeneous backward problem. Very recently, in [9], the methods of integral equations and of Fourier transform have been used to solve a 1D problem in an unbounded region. In recent articles considering the nonlinear backward heat problem, we refer the reader to [10]. In [11], the authors used the quasi-boundary value method to regularize the latter problem. However, in [11], the authors showed that the error between the regularized solution and the exact solution is
(1.5)∥uδ(·,t)-u(·,t)∥≤K(t)δt/T,0<t<T.It is easy to see that the convergence of the error estimate between the regularized solution and the exact solution is very slow whent  is in a neighborhood of zero. For this reason, the error estimate in initial time is given by
(1.6)∥uδ(·,t)-u(·,t)∥≤K(1ln(1/δ))1/4.We can easily find that the exact solution of (1.1)–(1.3) satisfies
(1.7)u(x,t)=∑k=1∞uk(t)sin(kx),
where
(1.8)uk(t)=ek2(λ(T)-λ(t))gk-∫tTek2(λ(s)-λ(t))fk(u)(s)ds,(1.9)fk(u)(s)=2π∫0πf(x,s,u(x,s))sin(kx)dx,(1.10)gk=2π∫0πg(x)sin(kx)dx,(1.11)λ(t)=∫0ta(s)ds.In this paper, we will approximate (1.1)–(1.3) by using the regularization problem:
(1.12)(uδ)t(x,t)-a(t)(uδ)xx(x,t)=∑k=1∞e-k2λ(T)αk(δ)+e-k2λ(T)fk(uδ)(t)sin(kx),uδ(0,t)=uδ(π,t)=0,uδ(x,T)=∑k=1∞e-k2λ(T)αk(δ)+e-k2λ(T)gksin(kx),
where αk(δ)=δk2. Actually, in [12], we also considered the problem (1.1)–(1.3) for the homogeneous case f=0  in  ℝ. Hence, we want to extend for the nonlinear case f(x,t,u) in bounded region  [0,π], and this is the biggest different point in this paper.The remainder of this paper is organized as follows. In Section2, we shall regularize the ill-posed problem (1.1)–(1.3) and give the error estimate between the regularized solution and the exact solution. Then, in Section 3, a numerical example is given.


## 2. Regularization and Error Estimates

For clarity of notation, we denote the solution of (1.1)–(1.3) by u(x,t) and the solution of the regularized problem (1.12) by uδ(x,t). Throughout this paper, we denote λ1=max{1,λ(T)}, and, δ be a positive number such that  0<δ<λ(T). Hereafter, we have a number of inequalities in order to evaluate error estimates.Lemma 2.1.
Leta(t) be a function satisfying (1.4),  αk(δ)=δk2,  Bδ=(δln(λ(T)/δ))-1  and  λ(t) be as in (1.11). Then for k>0  and  0≤t≤s≤T, one gets(i)
ek2(λ(s)-λ(t)-λ(T))/(αk(δ)+e-k2λ(T))≤λ1(δln(λ(T)/δ))(λ(t)-λ(s))/λ(T)=λ1(Bδ)(λ(s)-λ(t))/λ(T),(ii)
e-k2λ(t)/(αk(δ)+e-k2λ(T))≤λ1(δln(λ(T)/δ))(λ(t)-λ(T))/λ(T)=λ1(Bδ)(λ(T)-λ(t))/λ(T).Proof of Lemma2.1.
The proof of Lemma2.1 can be found in [12].Theorem 2.2.
Letg∈L2(0,π)  and  f∈L∞([0,π]×[0,T]×R) such that there exists  L>0  independent of x,t,u,v  satisfying
(2.1)|f(x,t,u)-f(x,t,v)|≤L|u-v|.
Then problem (1.12) has a unique weak solution  uδ∈W=C([0,T];L2(0,π)) satisfying the following equality:
(2.2)uδ(x,t)=∑k=1∞[e-k2λ(t)αk(δ)+e-k2λ(T)gk-∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)fk(uδ)(s)ds]sin(kx),
where
(2.3)gk=2π∫0πg(x)sin(kx)dx,fk(uδ)(s)=2π∫0πf(x,s,uδ)sin(kx)dx.Proof of Theorem2.2.
Step1. The existence and the uniqueness of the solution of the problem (2.2)
Put(2.4)F(u)(x,t)=∑k=1∞(Pk(t)-Kk(u)(t))sin(kx),
where
(2.5)Pk(t)=e-k2λ(t)αk(δ)+e-k2λ(T)gk,Kk(u)(t)=∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)fk(u)(s)ds.
We claim that, for everyu,v∈C([0,T];L2(0,π)),  k≥1, we have
(2.6)∥Fk(u)(·,t)-Fk(v)(·,t)∥2≤(T-t)kk!(Bδ)2kq/pCλ12kL2k∥u-v∥C([0,T];L2(0,π))2,
where C=max{1,T} and ∥·∥C([0,T];L2(0,π)) is supremum norm in C([0,T];L2(0,π)). We shall prove this inequality by induction. For k=1, we have
(2.7)∥F(u)(·,t)-F(v)(·,t)∥2=π2∑k=1∞(Kk(u)(t)-Kk(v)(t))2=π2∑k=1∞[∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)(fk(u)(s)-fk(v)(s))ds]2≤π2∑k=1∞[∫tT(ek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T))2ds∫tT(fk(u)(s)-fk(v)(s))2ds].
From Lemma2.1, we get
(2.8)∥F(u)(·,t)-F(v)(·,t)∥2≤π2∑k=1∞[∫tT(λ1(δln(λ(T)δ))(λ(t)-λ(s))/λ(T))2ds∫tT(fk(u)(s)-fk(v)(s))2ds].
It follows(2.9)∥F(u)(·,t)-F(v)(·,t)∥2≤π2∑k=1∞[∫tTλ12(Bδ)2(λ(s)-λ(t))/λ(T)ds∫tT(fk(u)(s)-fk(v)(s))2ds]≤π2∑k=1∞[∫tTλ12(Bδ)2(qs-pt)/pTds∫tT(fk(u)(s)-fk(v)(s))2ds]≤π2∑k=1∞[∫tTλ12(Bδ)2q/pds∫tT(fk(u)(s)-fk(v)(s))2ds]=λ12Bδ2q/p(T-t)∫tT∫0π(f(x,s,u(x,s))-f(x,s,v(x,s)))2dxds.
Therefore, we have(2.10)∥F(u)(·,t)-F(v)(·,t)∥2≤λ12L2Bδ2q/p(T-t)∫tT∫0π(u(x,s)-v(x,s))2dxds≤(T-t)Bδ2q/pCλ12L2∥u-v∥C([0,T];L2(0,π))2,
where  C=max{1,T}.
Thus, (2.6) holds for k=1. Supposing that (2.6) holds for k=n,  we shall prove that (2.6) holds for  k=n+1.  In fact, we get
(2.11)∥Fn+1(u)(·,t)-Fn+1(v)(·,t)∥2=π2∑k=1∞[Kk(Fn(u))(t)-Kk(Fn(v))(t)]2=π2∑k=1∞(∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)(fk(Fn(u))(s)-fk(Fn(v))(s))ds)2.
Hence, we obtain(2.12)∥Fn+1(u)(x,t)-Fn+1(v)(x,t)∥2≤π2(Bδ)2q/pλ12(T-t)∫tT∑k=1∞(fk(Fn(u))(s)-fk(Fn(v))(s))2ds=(Bδ)2q/pλ12(T-t)∫tT∥f(·,s,Fn(u)(·,s))-fk(·,s,Fn(v)(·,s))∥2ds≤(Bδ)2q/pλ12(T-t)L2∫tT∥Fn(u)(·,s)-Fn(v)(·,s)∥2ds.
Thus, we have(2.13)∥Fn+1(u)(x,t)-Fn+1(v)(x,t)∥2≤(Bδ)2q/pλ12(T-t)L2∫tTL2n(Bδ)2nq/pλ12n(T-s)nn!Cn∥u-v∥C([0,T];L2(0,π))2ds≤Cn(Bδ)2(n+1)q/pλ12n+2(T-t)L2n+2∥u-v∥C([0,T];L2(0,π))2∫tT(T-s)nn!ds≤(T-t)n+1(n+1)!Cn+1(Bδ)2(n+1)q/pλ12n+2(T-t)L2n+2∥u-v∥C([0,T];L2(0,π))2.
Therefore, by the induction principle, we have(2.14)∥Fk(u)(x,t)-Fk(v)(x,t)∥≤(T-t)k/2k!(Bδ)kq/pCk/2λ1kLk∥u-v∥C([0,T];L2(0,π)),
for all  u,v∈C([0,T];L2(0,π)).
We considerF:C([0,T];L2(0,π))→C([0,T];L2(0,π)).  Since
(2.15)(T-t)k/2k!(Bδ)kq/pλ1kCk/2Lk→0,
when k→∞, there exists a positive integer number k0 such that
(2.16)(T-t)k0/2k0!(Bδ)k0q/pλ1k0Ck0/2Lk0<1,
and Fk0 is a contraction. It follows that the equation Fk0(u)=u has a unique solution uδ∈C([0,T];L2(0,π)).We claim that  F(uδ)=uδ. In fact, one has F(Fk0(uδ))=F(uδ). Hence, Fk0(F(uδ))=F(uδ). By the uniqueness of the fixed point of Fk0, one has F(uδ)=uδ; that is, the equation F(u)=u has a unique solution uδ∈C([0,T];L2(0,π)).
Step2. If  uδ∈W  satisfies (2.2), then  uδ  is the solution of the problem (1.12). For  0≤t≤T, we have
(2.17)uδ(x,t)=∑k=1∞[e-k2λ(t)αk(δ)+e-k2λ(T)gk-∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)fk(uδ)(s)ds]sin(kx).
We can verify directly thatuδ∈C([0,T];L2(0,π)∩L2(0,T;H01(0,π))∩C1(0,T;H01(0,π)). In fact, uδ∈C∞([0,T];H01(0,π)). Moreover, one has
(2.18)(uδ)t(x,t)=∑k=1∞-k2a(t)(e-k2λ(t)gk-∫tTek2(λ(s)-λ(t)-λ(T))fk(uδ)(s)ds)αk(δ)+e-k2λ(T)sin(kx)+∑k=1∞e-k2λ(T)αk(δ)+e-k2λ(T)fk(uδ)(t)sin(kx)=-a(t)(uδ)xx(x,t)+∑k=1∞e-k2λ(T)αk(δ)+e-k2λ(T)fk(uδ)(t)sin(kx),uδ(x,T)=∑k=1∞e-k2λ(T)αk(δ)+e-k2λ(T)gksin(kx).
Hence,uδ is the solution of (1.12).
Step3. The problem (1.12) has at most one solution uδ∈C([0,T];L2(0,π)∩L2(0,T;H01(0,π))∩C1(0,T;H01(0,π)).  In fact, let uδ and vδ be two solutions of (1.12) such that uδ,vδ∈W. Putting wδ(x,t)=uδ(x,t)-vδ(x,t),  then wδ satisfies
(2.19)(wδ)t-(wδ)xx=∑k=1∞e-k2λ(T)αk(δ)+e-k2λ(T)(fk(uδ)(t)-fk(vδ)(t))sin(kx).
It follows that(2.20)∥(wδ)t-(wδ)xx∥2≤1δ2∑k=1∞(fk(uδ)(t)-fk(vδ)(t))2≤1δ2∥f(·,t,uδ(·,t))-f(·,t,vδ(·,t))∥2≤L2δ2∥uδ(·,t)-vδ(·,t)∥2=L2δ2∥wδ(·,t)∥2.
By using the result in Lees and Protter [13], we get wδ(·,t)=0. This completes the proof of Step 3.
Finally, by combining three steps, we complete the proof of Theorem2.2.Theorem 2.3 (stability of the modified method).
Let f be as in Theorem2.2, αk(δ)=δk2,g and let  gδ in L2(0,π) satisfy ∥gδ-g∥≤δ. If one supposes that uδ and vδ defined by (2.2) are corresponding to the final values g and gδ in L2(0,π), respectively, then one obtains
(2.21)∥uδ(·,t)-vδ(·,t)∥≤2λ1eL2λ12T(T-t)δλ(t)/λ(T)(ln(λ(T)δ))(λ(t)-λ(T))/λ(T).Proof of Theorem2.3.
Using the inequality(a+b)2≤2(a2+b2)  and Lemma 2.1, we get the estimate
(2.22)∥uδ(·,t)-vδ(·,t)∥2≤π∑k=1∞|e-k2λ(t)αk(δ)+e-k2λ(T)(gk-gk,δ)|2+π∑k=1∞|∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)(fk(uδ)(s)-fk(vδ)(s))ds|2≤π∑k=1∞|e-k2λ(t)αk(δ)+e-k2λ(T)(gk-gk,δ)|2+π(T-t)∑k=1∞∫tT(ek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)(fk(uδ)(s)-fk(vδ)(s)))2ds.
Thus, we get(2.23)∥uδ(·,t)-vδ(·,t)∥2≤πλ12(Bδ)2(λ(T)-λ(t))/λ(T)∑k=1∞|gk-gk,δ|2+π(T-t)∑k=1∞∫tTλ12(Bδ)2(λ(s)-λ(t))/λ(T)|fk(uδ)(s)-fk(vδ)(s)|2ds.
Hence, we obtain(2.24)∥uδ(·,t)-vδ(·,t)∥2≤πλ12(Bδ)2(λ(T)-λ(t))/λ(T)∑k=1∞|gk-gk,δ|2+λ12πT(Bδ)2(λ(T)-λ(t))/λ(T)∑k=1∞∫tT(Bδ)2(λ(s)-λ(T))/λ(T)|fk(uδ)(s)-fk(vδ)(s)|2ds≤πλ12(Bδ)2(λ(T)-λ(t))/λ(T)∑k=1∞|gk-gk,δ|2+λ12πT(Bδ)2(λ(T)-λ(t))/λ(T)∫tT(Bδ)2(λ(s)-λ(T))/λ(T)∑k=1∞|fk(uδ)(s)-fk(vδ)(s)|2ds.
Thus, we get(2.25)(Bδ)2(λ(t)-λ(T))/λ(T)∥uδ(·,t)-vδ(·,t)∥2≤2λ12∥g-gδ∥2+2λ12T∫tT(Bδ)2(λ(s)-λ(T))/λ(T)∥f(·,s,uδ(·,s))-f(·,s,vδ(·,s))∥2≤2λ12∥g-gδ∥2+2L2λ12T∫tT(Bδ)2(λ(s)-λ(T))/λ(T)∥uδ(·,s)-vδ(·,s)∥2.
By using Gronwall’s inequality, we have(2.26)(Bδ)2(λ(t)-λ(T))/λ(T)∥uδ(·,t)-vδ(·,t)∥2≤2λ12e2L2λ12T(T-t)∥gδ-g∥2.
It follows(2.27)∥uδ(·,t)-vδ(·,t)∥≤2λ1eL2λ12T(T-t)(Bδ)(λ(T)-λ(t))/λ(T)∥gδ-g∥≤2λ1eL2λ12T(T-t)δλ(t)/λ(T)(ln(λ(T)δ))(λ(t)-λ(T))/λ(T).
This completes the proof of Theorem2.3.Theorem 2.4.
Let u be the exact solution of problem (1.1)–(1.3) such that
(2.28)Q(t)=3λ12e3L2Tλ12(T-t)∥uxx(·,0)∥2<∞,M(t)=6Tλ12e3L2Tλ12(T-t)∫0T∫0π∑k=1∞|k2ek2λ(s)(us(x,s)-a(s)uxx(x,s))|2dxds<∞,
for all t∈[0,T). Letting αk(δ)=δk2 and vδ(·,t) given by (2.2) corresponding to the perturbed data gδ, then one has for every t∈[0,T)(2.29)∥u(·,t)-vδ(·,t)∥≤C(t)δλ(t)/λ(T)(ln(λ(T)δ))(λ(t)-λ(T))/λ(T),
where C(t)=2λ1eL2λ12T(T-t)+Q(t)+M(t).Proof of Theorem2.4.
From (1.1), we construct the regularized solution corresponding to the exact data and the perturbed data
(2.30)uδ(x,t)=∑k=1∞uk,δ(t)sin(kx),(2.31)vδ(x,t)=∑k=1∞vk,δ(t)sin(kx),
where
(2.32)uk,δ(t)=e-k2λ(t)αk(δ)+e-k2λ(T)gk-∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)fk(uδ)(s)ds,(2.33)vk,δ(t)=e-k2λ(t)αk(δ)+e-k2λ(T)gk,δ-∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)fk(vδ)(s)ds.
Since (1.8) and (2.32), we get
(2.34)|uk(t)-uk,δ(t)|=|αk(δ)e-k2λ(t)αk(δ)+e-k2λ(T)ek2λ(T)gk-∫tTαk(δ)ek2(λ(s)-λ(t)-λ(T))(αk(δ)+e-k2λ(T))ek2λ(T)fk(u)(s)ds+∫tTek2(λ(s)-λ(t)-λ(T))αk(δ)+e-k2λ(T)(fk(uδ)(s)-fk(u)(s))ds|.
Fromαk(δ)=δk2, we get
(2.35)|uk(t)-uk,δ(t)|≤|δe-k2λ(t)δk2+e-k2λ(T)k2ek2λ(T)gk-∫tTδek2(λ(s)-λ(t)-λ(T))δk2+e-k2λ(T)k2ek2λ(T)fk(u)(s)ds|+|∫tTek2(λ(s)-λ(t)-λ(T))δk2+e-k2λ(T)(fk(uδ)(s)-fk(u)(s))ds|≤|δe-k2λ(t)δk2+e-k2λ(T)||k2ek2λ(T)gk-k2∫0Tek2λ(s)fk(u)(s)ds|+|δe-k2λ(t)δk2+e-k2λ(T)|∫0T|k2ek2λ(s)fk(u)(s)|ds+|∫tTek2(λ(s)-λ(t)-λ(T))δk2+e-k2λ(T)(fk(uδ)(s)-fk(u)(s))ds|.
By applying the inequality(a+b+c)2≤3(a2+b2+c2), we get
(2.36)∥u(·,t)-uδ(·,t)∥2=π2∑k=1∞|uk,δ(t)-uk(t)|2≤3π2∑k=1∞|δe-k2λ(t)δk2+e-k2λ(T)|2|k2ek2λ(T)gk-k2∫0Tek2λ(s)fk(u)(s)ds|2+3π2∑k=1∞|δe-k2λ(t)δk2+e-k2λ(T)|2(∫0T|k2ek2λ(s)fk(u)(s)|ds)2+3π2∑k=1∞|∫tTek2(λ(s)-λ(t)-λ(T))δk2+e-k2λ(T)(fk(uδ)(s)-fk(u)(s))ds|2.
Using Lemma2.1, we obtain
(2.37)∥u(·,t)-uδ(·,t)∥2≤3π2λ12(L1(t)+L2(t)+L3(t)),=3π2λ12δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)∑k=1∞|k2uk(0)|2+3π2λ12δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)∑k=1∞(∫0T|k2ek2λ(s)fk(u)(s)|ds)2+3π2λ12Tδ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)×∫tTδ-2λ(s)/λ(T)(ln(λ(T)δ))2(λ(T)-λ(s))/λ(T)∑k=1∞|(fk(uδ)(s)-fk(u)(s))|2ds,
where
(2.38)L1(t)=δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)∑k=1∞|k2uk(0)|2,L2(t)=δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)∑k=1∞(∫0T|k2ek2λ(s)fk(u)(s)|ds)2,L3(t)=Tδ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)×∫tTδ-2λ(s)/λ(T)(ln(λ(T)δ))2(λ(T)-λ(s))/λ(T)∑k=1∞|(fk(uδ)(s)-fk(u)(s))|2ds.
Hence, we get the following estimates(2.39)3π2λ12L1(t)≤3λ12δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)∥uxx(·,0)∥2,3π2λ12L2(t)≤3π2Tλ12δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)∫0T∑k=1∞k4e2k2λ(s)fk2(u)(s)ds,3π2λ12L3(t)≤3L2λ12Tδ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)×∫tTδ-2λ(s)/λ(T)(ln(λ(T)δ))2(λ(T)-λ(s))/λ(T)∥u(·,s)-uδ(·,s)∥2ds.
From the estimate (2.39), we get
(2.40)∥u(·,t)-uδ(·,t)∥2≤3λ12δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)×(∥uxx(·,0)∥2+π2T∫0T∑k=1∞k4e2k2λ(s)fk2(u)(s)ds)+3L2Tλ12δ2λ(t)/λ(T)(ln(λ(T)δ))2(λ(t)-λ(T))/λ(T)×∫tTδ-2λ(s)/λ(T)(ln(λ(T)δ))(2λ(T)-2λ(s))/λ(T)∥u(·,s)-uδ(·,s)∥2ds.
Hence, we have(2.41)δ-2λ(t)/λ(T)(ln(λ(T)δ))2(λ(T)-λ(t))/λ(T)∥u(·,t)-uδ(·,t)∥2≤3λ12(∥uxx(·,0)∥2+π2T∫0T∑k=1∞k4e2k2λ(s)fk2(u)(s)ds)+3L2Tλ12∫tTδ-2λ(s)/λ(T)(ln(λ(T)δ))(2λ(T)-2λ(s))/λ(T)∥u(·,s)-uδ(·,s)∥2ds.
Applying Gronwall’s inequality, we obtain(2.42)δ-2λ(t)/λ(T)(ln(λ(T)δ))(2λ(T)-2λ(t))/λ(T)∥u(·,t)-uδ(·,t)∥2≤3λ12e3L2Tλ12(T-t)(∥uxx(·,0)∥2+π2T∫0T∑k=1∞k4e2k2λ(s)fk2(u)(s)ds).
Hence, we get(2.43)δ-2λ(t)/λ(T)(ln(λ(T)δ))(2λ(T)-2λ(t))/λ(T)∥u(·,t)-uδ(·,t)∥2≤3λ12e3L2Tλ12(T-t)(∥uxx(·,0)∥2+π2T∫0T∑k=1∞k4e2k2λ(s)(fk(u)(s))2ds)=3λ12e3L2Tλ12(T-t)(∥uxx(·,0)∥2+2πT∫0T∑k=1∞k4e2k2λ(s)(∫0πf(x,s,u(x,s))sin(kx)dx)2ds)≤3λ12e3L2Tλ12(T-t)(∥uxx(·,0)∥2+2T∫0T∑k=1∞k4e2k2λ(s)(∫0π|f(x,s,u(x,s))|2dx)ds).
From (1.1), we have
(2.44)δ-2λ(t)/λ(T)(ln(λ(T)δ))(2λ(T)-2λ(t))/λ(T)∥u(·,t)-uδ(·,t)∥2≤3λ12e3L2Tλ12(T-t)(∥uxx(·,0)∥2+2T∫0T∑k=1∞k4e2k2λ(s)(∫0π|f(x,s,u(x,s))|2dx)ds)≤3λ12e3L2Tλ12(T-t)×(∥uxx(·,0)∥2+2T∫0T∫0π∑k=1∞|k2ek2λ(s)(us(x,s)-a(s)uxx(x,s))|2dxds)=Q(t)+M(t),
where
(2.45)Q(t)=3λ12e3L2Tλ12(T-t)∥uxx(·,0)∥2,M(t)=6Tλ12e3L2Tλ12(T-t)∫0T∫0π∑k=1∞|k2ek2λ(s)(us(x,s)-a(s)uxx(x,s))|2dxds.
Therefore, we get the estimate(2.46)∥u(·,t)-uδ(·,t)∥≤Q(t)+M(t)δλ(t)/λ(T)(ln(λ(T)δ))(λ(t)-λ(T))/λ(T).
Letvδ be the solution of (1.12) corresponding to the perturbed data gδ,  and let uδ be the solution of (1.12) corresponding to the exact data g. From Theorem 2.3 and (2.46), we can get
(2.47)∥vδ(·,t)-u(·,t)∥≤∥vδ(·,t)-uδ(·,t)∥+∥uδ(·,t)-u(·,t)∥≤2λ1eL2λ12T(T-t)δλ(t)/λ(T)(ln(λ(T)δ))(λ(t)-λ(T))/λ(T)+Q(t)+M(t)δλ(t)/λ(T)(ln(λ(T)δ))(λ(t)-λ(T))/λ(T)=C(t)δλ(t)/λ(T)(ln(λ(T)δ))(λ(t)-λ(T))/λ(T),
where C(t)=2λ1eL2λ12T(T-t)+Q(t)+M(t).
This completes the proof of Theorem2.4.


## 3. Numerical Experiment

Consider the nonlinear parabolic equation with time-dependent coefficient:(3.1)ut(x,t)-a(t)uxx(x,t)=f(x,t,u(x,t)),(x,t)∈[0,π]×(0,1],u(0,t)=u(π,t)=0,t∈[0,1],u(x,T)=g(x),x∈[0,π],
where
(3.2)a(t)=(2t+1),f(x,t,u(x,t))=u+4etsin(x)cos(x)(2t+1),g(x)=esin(x)cos(x).The exact solution of the equation is(3.3)u(x,t)=etsin(x)cos(x).Lettingt=0, from (3.3), we have
(3.4)u(x,0)=sin(x)cos(x).Consider the measured data(3.5)gδ(x)=(1+δ∥g∥)g(x)=(1+δ1.7034)g(x).Then we have(3.6)∥gδ-g∥=δ.From (2.31) and (3.5), we have the regularized solution for the case t=0 in the form of iteration
(3.7)vδ(n)(ω,0)=∑k=1∞vδ,k(n)(x,0)sin(kx),
where
(3.8)vδ,k(n)(x,0)=1αk(δ)+exp{-2k2}gδ,k-∫t1exp{k2(s2+s-2)}αk(δ)+exp{-2k2}fk(vδ(n-1))(s)ds,gδ,k=2π∫0πgδ(x)sin(kx)dx,vδ(1)(x,0)=(1+δ)sin(x)cos(x),αk(δ)=δk2.We considerδ1=10-1,  δ2=10-2,  δ3=10-3,  δ4=10-4,  δ5=10-5,  δ6=10-10,  δ7=10-20,  δ8=10-50,  and  n=10. Now, we get Table 1 for the case  t=0.Table 1
δ
∥
v
δ
i
(
n
)
(
·
,
0
)
-
u
(
·
,
0
)
∥
2
δ
1
=
10
-
1
3.763414
e
-
001
δ
2
=
10
-
2
3.735245
e
-
001
δ
3
=
10
-
3
3.475127
e
-
001
δ
4
=
10
-
4
2.048545
e
-
001
δ
5
=
10
-
5
4.012725
e
-
002
δ
6
=
10
-
10
4.491189
e
-
007
δ
7
=
10
-
20
2.638466
e
-
013
δ
8
=
10
-
50
2.179771
e
-
013We have in Figure1 the graphs of the regularized solution vδi(n)(·,t),  i=1,2,3  and n=10.The regularized solutions corresponding toδi,  i=1,2,3.
(a)
(b)
(c)We have in Figure2 the graphs of the regularized solution vδi(n)(·,t),  i=4,5,6 and n=10.The regularized solutions corresponding toδi,  i=4,5,6.
(a)
(b)
(c)We have in Figure3 the graphs of the exact solution u(·,t) and of the regularized solution vδi(n)(·,t),  i=7,8  and  n=10.The regularized solutions correspondingδi,  i=7,8  and the exact solution.
(a)
(b)
(c)Now, Figure4 can represent visually the exact solution and regularized solutions corresponding to δi,  i=1,…,8 at initially time t=0.Figure 4
The exact solution and the regularized solutions corresponding toδi,  i=1,…,8 at initial time t=0.Notice that, in Figure4, the curve number 0 expressing the exact solution is indistinguishable from the curve number  i  expressing the regularized solution corresponding to δi,  i=6,7,8.Remark 3.1.
From (1.7) and (3.5), we obtain the exact solution corresponding to the measured data gδ(x):
(3.9)w(x,t)=∑k=1∞wk(t)sin(kx),
where
(3.10)wk(t)=ek2(λ(T)-λ(t))gδ,k-∫tTek2(λ(s)-λ(t))fk(w)(s)ds,fk(w)(s)=2π∫0πf(x,s,w(x,s))sin(kx)dx,gδ,k=2π∫0πgδ(x)sin(kx)dx.
Now, we cannot calculate the formula (3.9) exactly (we need to find fk(w)(s) while we have not known w yet). From Theorem 2.2, we use the iteration for (3.9) at initial time t=0 as follows:
(3.11)w(n)(ω,0)=∑k=1∞wk(n)(x,0)sin(kx),
where
(3.12)wk(n)(x,0)=exp{2k2}gδ,k-∫01exp{k2(s2+s)}fk(w(n-1))(s)ds,gδ,k=2π∫0πgδ(x)sin(kx)dx,w(1)(x,0)=(1+δ)sin(x)cos(x).
Then we get the error in the Table2.
We can see that the error∥wδi(5)(·,0)-u(·,0)∥2  is very large. Therefore, the problem is ill-posed and a regularization is necessary.Table 2
δ
∥
w
δ
i
(
5
)
(
·
,
0
)
-
u
(
·
,
0
)
∥
2
δ
1
=
10
-
1
1.1856
e
+
031
δ
2
=
10
-
2
1.0
e
+
031
δ
3
=
10
-
3
1.0
e
+
031
δ
4
=
10
-
4
1.0
e
+
031
δ
5
=
10
-
5
1.0
e
+
031
δ
6
=
10
-
10
1.0
e
+
031
δ
7
=
10
-
20
1.0
e
+
031
δ
8
=
10
-
50
1.0
e
+
031



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*Source: 943621-2012-10-17.xml*