Research Article | Open Access
A Lipschitz Stability Estimate for the Inverse Source Problem and the Numerical Scheme
We consider the inverse source problem for heat equation, where the source term has the form . We give a numerical algorithm to compute unknown source term . Also, we give a stability estimate in the case that is a piecewise constant function.
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  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 , Trong et al. , and Shidfar et al. , 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 that is an open set in . We consider the following initial-boundary value problem:
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, represents either temperature or concentration. In this system, is spatial density and unknown is interpreted as either a heat or a material source, respectively, in a chemical or a biochemical application.
When , it means that term is a source. We assume that is a given function and satisfies the following condition:
Our problem is to derive and a conditional stability in the determination of , from the observation where .
2. Some Existing Results
To derive the stability estimate, we need the following inequalities, which can be found in . Here, we just list the result.
Theorem 1. Let , , , and satisfyThen, In particular, forand for , one has
In , by using the above theorem, a Hölder estimate was obtained for the case that and , where
Theorem 2. Let satisfy (2), and set to be defined as above definition.
Then, for arbitrarily given , there exists a constant such that for any .
Corollary 3. In Theorem 2, if one replaces by also the same result can be obtained.
But we are more interested in the case that is a piecewise constant function. In that case, 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
Also from , we can see that fundamental solution satisfies the following conditions:
Therefore, settingwe havewhich is a Volterra integral equation of the first kind with respect to .
Here, we consider the case that .
Theorem 4 (uniqueness). For problem (1), given , then is uniquely determined except for finite point in .
Proof. Suppose that there are two functions satisfying (1). We assume that the discontinuous points of and are and , respectively, and and .
After some resorting procedure, we get points ; then, for , and , so we get .
For , therefore, .
Repeating the above procedure, we can get .
Thus, end the proof.
Suppose that changes its values at , and . Here, we assume that there exists a constant , such that .
Theorem 5. If satisfies (2), .
Then, for , there exists a constant such that for any . Here, is the number of intervals in that .
Proof. Since is positive and bounded and (13), we can see thatChoosing norm from to , we obtainThus, we can see thatwhere is a constant.
For , we can see that Then, we can obtainThen Therefore, in the case of and (20), we get Continuing this argument until , we get the proof of this theorem.
4. Numerical Scheme
For problem (1), if is given, we need to reconstruct function .
Consider an auxiliary problem:
Actually, we can get the solution of problem (25) and suppose that is the solution. Then,
Given data , where , then, for every , we have
We suppose that ; thenThus, we can use the above formula to get the discrete data for function .
Example 6. We choose as the following function:
Choose , , , , and outside interval . Given data . Then, we can get the numerical approximation of using the following formula:
And the numerical result is as shown in Figure 1.
Figure 3 shows the error between the 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 , and
Let and , and the numerical result is as shown in Figure 4.
Figure 6 is the error between exact and the numerical result.
The author declares that there are no competing interests regarding the publication of this paper.
This work was supported by the Natural Science Foundation of China Grant no. 11371231 and Grant no. 11071148 and the Natural Science Foundation of Shandong Province, China, Grant no. ZR2011AQ014.
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