# A Lipschitz Stability Estimate for the Inverse Source Problem and the Numerical Scheme

Authors: Xianzheng Jia
Journal: Advances in Mathematical Physics
 (2016)

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## Abstract

We consider the inverse source problem for heat equation, where the source term has the formf(t)ϕ(x). We give a numerical algorithm to compute unknown source term f(t). Also, we give a stability estimate in the case that f(t) is a piecewise constant function.


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## Body

## 1. Introduction to the Problem

The inverse source problems in heat equation are solving the heat equation with unknown heat sources. They are well known to be ill-posed. There were many studies to identify different types of heat sources. For example, Cannon and DuChateau [1] estimated the nonlinear temperature dependent heat source. In [2, 3], the method of fundamental solutions has been presented for an inverse time-dependent heat source problem. In [4–10], several numerical schemes have been proposed to determine a space-dependent heat source. In [11–13], these numerical methods were considered to solve some two- and three-dimensional inverse heat source problems. In Savateev [14], Trong et al. [15], and Shidfar et al. [16], the authors considered the source as a function of both space and time, but it was additive or separable.Throughout this paper, we will assume thatΩ is an open set in Rn. We consider the following initial-boundary value problem:(1)ut=Δu+ftϕxx,t∈Ω×0,T,ux,0=0x∈Ω,ux,t=0x,t∈∂Ω×0,T.Problems of the above equations include inverse problems, heat conduction processes, hydrology, material sciences, and heat transfer problems. In the context of heat conduction and diffusion,u represents either temperature or concentration. In this system, ϕ(x) is spatial density and unknown f(t) is interpreted as either a heat or a material source, respectively, in a chemical or a biochemical application.Whenf(t)>0, it means that term f(t)ϕ(x) is a source. We assume that ϕ is a given function and satisfies the following condition:(2)ϕ≥0,≢0inΩϕhas  compact  support  inΩ.Our problem is to derivef(t) and a conditional stability in the determination of f(t),0≤t≤T, from the observation (3)ux0,t,0<t<T,where x0∈Ω/supp⁡ϕ¯.


## 2. Some Existing Results

To derive the stability estimate, we need the following inequalities, which can be found in [17]. Here, we just list the result.Theorem 1.
Letp≥1, δ>0, 0≤α<T, and f,g∈L∞(0,T) satisfy(4)0≤f,g≤M<∞,0<t<T.
Then,(5)fLpα,TgLp0,δ≤M2p-2/p∫αT+δ∫αtfsgt-sdsdt1/p.
In particular, for(6)f∗gt=∫0tft-sgsds,0<t<Tand for α=0, one has (7)fLpα,TgLp0,δ≤M2p-2/pf∗gL10,T+δ1/p.In [17], by using the above theorem, a Hölder estimate was obtained for the case that Ω=Rn and f∈U, where (8)U=f∈C0,T;fC0,T≤M,fchanges  the  signs  at  mostN-times.Theorem 2.
Letϕ satisfy (2), and set U to be defined as above definition.
Then, for arbitrarily givenδ>0, there exists a constant C=Cx0,ϕ,T,p,δ,U>0 such that (9)fLp0,T≤Cux0,·L10,T+δ1/pN,for any f∈U.Corollary 3.
In Theorem2, if one replaces U by (10)PN=f;fisapolynomialwhoseorderisatmostN,fC0,T≤M,
also the same result can be obtained.But we are more interested in the case thatf(t) is a piecewise constant function. In that case, f(t) can have infinity number of zeros. For that case, we obtained the similar result, and we will claim the result and the proof in the next section.


## 3. The Estimate

For problem (1), we can see that solution u(x,t) can be expressed in the following form: (11)ux,t=∫0t∫ΩKx,y,t-sfsϕydyds,x∈Ω,t>0,where K(x,y,t) is the fundamental function for problem (1). See [18].Also from [18], we can see that fundamental solution K(x,y,t) satisfies the following conditions: (12)Kx,y,t>0,t>0.Therefore, setting(13)μx0t=∫ΩKx0,y,tϕydy,t>0,we have(14)ux0,t=∫0tμx0t-sfsds,0<t<T,which is a Volterra integral equation of the first kind with respect to f.Here, we consider the case thatf∈F={fis  a  piecewise  constant  function  in[0,T]}.Theorem 4 (uniqueness).
For problem (1), given u(x0,t), then f∈F is uniquely determined except for finite point in (0,T).Proof.
Suppose that there are two functionsf1(t),f2(t)∈F satisfying (1). We assume that the discontinuous points of f1 and f2 are 0=t01<t11<⋯<tI11=T and 0=t02<t12<⋯<tI22=T, respectively, and f1(t)=fj1,t∈(tj-11,tj1] and f2(t)=fj2,t∈(tj-12,tj2].
After some resorting procedure, we get points0=τ0<τ1<⋯<τI3=T; then, for 0<t<τ1, (15)∫0tμx0t-sf1s-f2sds=ux0,t-ux0,t=0f11-f12∫0tμx0t-sds=0and ∫0tμx0(t-s)ds>0,0<t<τ1, so we get f1(t)=f2(t),0<t<τ1.
Forτ1<t<τ2, (16)∫0tμx0t-sf1s-f2sds=∫τ1tμx0t-sf1s-f2sds=0,therefore, f1(t)=f2(t),t∈(τ1,τ2).
Repeating the above procedure, we can getf1(t)=f2(t),t∈(τj-1,τj),j=1,2,…,I3.
Thus, end the proof.Suppose thatf changes its values at 0=t0<t1<t2<⋯<tI=T,I≤N, and f(t)=fj,t∈[tj-1,tj),j=1,2,…,I. Here, we assume that there exists a constant c>0, such that t(i)-t(i-1)≥c,i=1,2,…,I.Theorem 5.
Ifϕ satisfies (2), x0∈Ω/supp⁡ϕ¯.
Then, forp>1, there exists a constant C=C(x0,ϕ,T,p,F,N)>0 such that (17)fLp0,T≤Cux0,·Lp0,T,for any f∈F. Here, N is the number of intervals in [0,T] that f≠0.Proof.
Sinceμx0(t) is positive and bounded and (13), we can see that(18)ux0,t=f1∫0tμx0t-sds,0≤t≤t1.Choosing Lp norm from 0 to t1, we obtain(19)uLp0,t1=f1∫0tμx0t-sdsLp0,t1.Thus, we can see that(20)f1≤C1uLp0,t1,where C1>(∫0tμx0(t-s)dsLp(0,t1))-1 is a constant.
Fort1≤t≤t2, we can see that (21)ux0,t=∫0tμx0t-sfsds=f1∫0t1μx0t-sds+f2∫t1tμx0t-sds,ux0,t-f1∫0t1μx0t-sds=f2∫t1tμx0t-sds.Then, we can obtain(22)f2·∫t1tμx0t-sdsLpt1,t2=ux0,t-f1∫0t1μx0t-sdsLpt1,t2≤ux0,tLpt1,t2+f1∫0t1μx0t-sdsLpt1,t2≤ux0,tLpt1,t2+C2ux0,tLp0,t1≤C3ux0,tLp0,t2.Then (23)f2≤Cux0,tLp0,t2.Therefore, in the case of ti-ti-1≥c>0,i=1,2,…,I and (20), we get (24)fLp0,t2≤Cux0,tLp0,t2.
Continuing this argument untilti=T, we get the proof of this theorem.


## 4. Numerical Scheme

For problem (1), if ux0,t is given, we need to reconstruct function ft.Consider an auxiliary problem:(25)ut=Δu+ϕxx,t∈Ω×0,T,ux,0=0x∈Ω,ux,t=0x,t∈∂Ω×0,T.Actually, we can get the solution of problem (25) and suppose that u~ is the solution. Then,(26)u~x,t=∫0t∫ΩKx,y,t-sϕydyds.Given datau(x0,ti),i=0,1,2,…,N, where 0=t0<t1<⋯<tN=T, then, for every ti, we have(27)ux0,ti=∫0ti∫ΩKx0,y,ti-sfsϕydyds,i=0,1,2,…,N.We suppose thatf(t)=fi,t∈[ti-1,ti],i=1,2,…,N; then(28)ux0,ti=∫0ti∫ΩKx0,y,ti-sfsϕydyds=∑j=1ifj∫tj-1tj∫ΩKx0,y,ti-sϕydyds=f1∫0ti∫ΩKx0,y,ti-sϕydyds+∑j=2ifj-f1∫tj-1tj∫ΩKx0,y,ti-sϕydyds=f1u~x0,ti+f2-f1∫t1ti∫ΩKx0,y,ti-sϕydyds+∑j=3ifj-f1-f2-f1∫tj-1tj∫ΩKx0,y,ti-sϕydyds=f1u~x0,ti+f2-f1∫0ti-t1∫ΩKx0,y,ti-t1-sϕydyds+∑j=3ifj-f2∫tj-1tj∫ΩKx0,y,ti-sϕydyds=⋯=f1u~x0,ti+∑j=2ifj-fj-1u~x0,ti-tj-1.Thus, we can use the above formula to get the discrete data for function f(t).Example 6.
We choosef(t) as the following function:(29)ft=1,t∈0,1-1,t∈1,2.3-0.5,t∈2.3,3.50.5,t∈3.5,4.ChooseΩ=(0,2), x0=1.5, ϕ(x)=1, x∈(0,1), and ϕ(x)≡0 outside interval (0,1). Given data u(x0,ti),i=1,2,…,N. Then, we can get the numerical approximation of f(t) using the following formula: (30)ux0,ti=f1u~x0,ti+∑j=2ifj-f1u~x0,ti-tj-1,i=1,2,…,N.And the numerical result is as shown in Figure1.Figure 1
Numerical result.Figure1 is the numerical result and Figure 2 is the exact function.Figure 2
Exact solution.Figure3 shows the error between the numerical result and exact solution.Figure 3
The error between numerical result and exact solution.Example 7.
Now, we will give a numerical result for the 2D case. Suppose the problem is also (1), and the domain Ω=(-1,1)×(-1,1), and(31)ϕx=1,x≤0.50,x∈Ω,x>0.5ft=1,t∈0,0.80,t∈0.8,2.2-2,t∈2.2,3.Letx0=(0.7,0.8) and T=3, and the numerical result is as shown in Figure 4.Figure 4
Numerical result.Figure4 is the numerical result and Figure 5 is the exact function.Figure 5
Exact solution.Figure6 is the error between exact f and the numerical result.Figure 6
The error between numerical result and exact solution.



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*Source: 9504829-2016-07-20.xml*