Some uniqueness theorems for inverse spacewise dependent source problems in nonlinear PDEs

Abstract

We investigate the uniqueness of a solution for an inverse problem of determining a spacewise-dependent source in some non-linear parabolic and hyperbolic problems. The material coefficients appearing in the governing equations may depend on both space and time. The aim is to identify a spacewise-dependent source from the usual initial and boundary conditions and the final-time over-determination. We formulate conditions on coefficients which guarantee the uniqueness of a solution of the inverse problem.

Keywords:

• semilinear parabolic problem
• wave equation with non-linear damping
• inverse problem
• uniqueness

AMS Subject Classification:

• 35R30

Introduction

Transient problems of parabolic and hyperbolic types have been intensively studied in the last decades. The theory for direct problems is nicely elaborated in many mathematical papers and books. One of the classical inverse problems (IPs) is the source determination. This can depend both on space and on time. This general setting is up to now too hard to attack. Therefore, most of the papers in the literature deal with linear partial differential equations (PDEs) and the unknown source term depends either on space or on time. The aim of this paper is to address uniqueness of a solution to the IP of the space-dependent source determination from a final-time measurement. We consider two classes of non-linear model problems: parabolic and hyperbolic.

The first part of this paper deals with a semilinear parabolic problem with applications in reactive contaminant transport in the saturated zone, cf. [1, Chap. 15]. The governing PDE is nothing else than the continuity equation. Adopting the Darcy law for diffusion in a saturated zone one can get a linear parabolic PDE for the contaminant concentration d, which depends on the time variable $t\in [0,T]$. Reaction transformation can be formally captured via augmenting the source by a generalized mass loss rate ${\mathit{\partial }}_{t}s$, which depends on the sorbed concentration s. The usual linear first-order form is$$\begin{array}{c}\hfill {\mathit{\partial }}_{t}s=K_r(K_{d}d-s)\end{array}$$with some given constants $K_r$ and $K_d$. This can be formally resolved as$$\begin{array}{c}\hfill s(t)=e^{-K_{r}t}s(0)+K_r{K}_d{\int }_0^t{e}^{-K_r(t-\mathit{\xi })}d(\mathit{\xi })\mathit{\xi }.\end{array}$$We can see that the right-hand side of the governing PDE for d will depend on the time integral of $d$. Considering a more general relation between contaminant and its sorption, one can even get a time integral of a non-linear function of concentration.

We study a problem of source identification from given final over-determination for the non-linear parabolic equation with Dirichlet boundary conditions. To formulate this situation mathematically, we assume that we have a non-homogeneous and non-isotropic body, denoted by $\mathrm{\Omega }$, occupying a bounded domain in ${\mathbb{R}}^n$, where $n\ge 1$. The unknown functions $u$ and $f$ obey(1) $$\begin{array}{c}\hfill \{\begin{array}{cc}{u}_t(x,t)+Lu(x,t)=f(x)+h(x,t)+{\int }_0^{t}g(u(x,s))\mathit{ds}\hfill & \text{in}\mathrm{\Omega }\times (0,T),\hfill \\ u(x,t)=\mathit{\alpha }(x,t)\hfill & \text{on}\mathit{\partial }\mathrm{\Omega }\times (0,T),\hfill \\ u(x,0)=u_0(x)\hfill & \text{for}x\in \mathrm{\Omega },\hfill \end{array}\end{array}$$(1) where the final time $T>0$. Here, $L$ is a symmetric linear elliptic operator of the second-order with coefficients depending on both space and time. The precise form of $L$ will be specified later depending on the situation under consideration. The data functions $h,g,\mathit{\alpha }\mathrm{and}{u}_0$ as well as the coefficients appearing in the operator $L$, are given. We show that the final data(2) $$\begin{array}{c}\hfill u(x,T)={\mathit{\psi }}_T(x).\end{array}$$(2) together with appropriate conditions on data uniquely determine the solution.

The second part of this paper is devoted to source identification from given final over-determination for the wave equation with a non-linear damping subject to the Dirichlet boundary conditions. We consider the following model problem with unknown functions $u$ and $f$ (the other data functions appearing in the problem setting are known)(3) $$\begin{array}{c}\hfill \{\begin{array}{cc}{u}_{\mathit{tt}}(x,t)+g(u_t(x,t))+Lu(x,t)=f(x)+h(x,t)\hfill & \text{in}\mathrm{\Omega }\times (0,T),\hfill \\ u(x,t)=\mathit{\alpha }(x,t)\hfill & \text{on}\mathit{\partial }\mathrm{\Omega }\times (0,T),\hfill \\ u(x,0)=u_0(x)\hfill & \text{for}x\in \mathrm{\Omega },\hfill \\ u_t(x,0)=v(x)\hfill & \text{for}x\in \mathrm{\Omega }.\hfill \end{array}\end{array}$$(3) We show that the final over-determination () together with suitable conditions on data uniquely determines the solution.

Parabolic problem

This IP for a linear parabolic PDE has mainly been considered in a few theoretical papers, for example, [2–9]. Recently, see for example [10], an iterative procedure was proposed and analysed for finding the source given from () in the case of time-independent operators $L$. Very nice and simple proof technique for uniqueness has been developed in [11]. Our goal is to apply this technique to the non-linear problems () and ().

We shall work in a variational framework; therefore, we introduce some standard notations, first. We assume that $\mathrm{\Omega }$ has a Lipschitz boundary $\mathit{\partial }\mathrm{\Omega }$. The space $L^2(\mathrm{\Omega })$ consists of square integrable functions on $\mathrm{\Omega }$ with the usual norm $\parallel ·\parallel$ and scalar product $(·,·)$. The space $H^1(\mathrm{\Omega })$ denotes the standard Sobolev space on $\mathrm{\Omega }$, i.e. the space of functions with generalized derivatives in $L^2(\mathrm{\Omega })$. Due to the smoothness of the boundary of $\mathrm{\Omega }$, the trace of functions in $H^1(\mathrm{\Omega })$ to the boundary is well defined and $H_0^1(\mathrm{\Omega })$ consists of functions with $u{|}_{\mathit{\partial }\mathrm{\Omega }}=0$.

The first theorem deals with a steady-state differential operator $L$.

Theorem 2.1

Consider a linear differential operator$$\mathit{Lu}(x,t)=\nabla ·(-\mathit{A}(x)\nabla u(x,t))+c(x)u(x,t),$$with bounded (discontinuous) coefficient obeying $\mathbf{A}(x)={\mathit{A}}^{\mathit{tr}}(x)$ and$$\begin{array}{c}\hfill {\mathit{\xi }}^{\mathit{tr}}·\mathit{A}(x)\mathit{\xi }=\sum _{i,j=1}^n{a}_{i,j}(x){\mathit{\xi }}_i{\mathit{\xi }}_j\ge C|\mathit{\xi }{|}^2\forall \mathit{\xi }\in {\mathbb{R}}^n.\end{array}$$Let $u_0,{\mathit{\psi }}_T\in L^2(\mathrm{\Omega })$ and $g^{\prime }\ge 0$. Then, there exists at most one spacewise-dependent source $f\in L^2(\mathrm{\Omega })$ such that () together with the condition () hold.

 

Proof

The paper [8] assumed that $c\ge 0$, so $(\mathit{Lu},u)\ge C_0\parallel u{\parallel }_{H^1(\mathrm{\Omega })}^2$. Therefore, $L$ was strictly accretive and $-L$ was dissipative. Following the Lumer–Phillips theorem cf. [12, Section 1.4], we see that $L$ generates a contractive semigroup, which was the crucial point in the proof of uniqueness in [8].

Please note that we do not assume that $c\ge 0$. If $c$ is bounded, then the operator $L$ obeys the Garding inequality:$$(\mathit{Lu},u)={\int }_{\mathrm{\Omega }}\mathit{Lu}u\ge C_0{\parallel \nabla u\parallel }^2-C\parallel u{\parallel }^2,\forall u\in H_0^1(\mathrm{\Omega }).$$Let us have two solutions $〈u_1,f_1〉$ and $〈u_2,f_2〉$ to () and (). Set $u=u_1-u_2$ and $f=f_1-f_2$. Then, we see that $u(x,0)=0$ and $u(x,T)=0$. For $\mathit{\phi }\in H_0^1(\mathrm{\Omega })$ we get the following weak formulation(4) $$\begin{array}{c}\hfill (u_t,\mathit{\phi })+(\mathit{A}\nabla u,\nabla \mathit{\phi })+(\mathit{cu},\mathit{\phi })=(f,\mathit{\phi })+({\int }_0^t[g(u_1(s))-g(u_2(s))]\text{d}s,\mathit{\phi }).\end{array}$$(4) We have to show that $u=0$ and $f=0$. First, we show that $u=0$ and later that $f=0$. We have to get rid of $f$ at the first stage. The main idea is very simple, namely$${\int }_0^{T}f(x)u_t(x,t)\text{d}t=f(x)u(x,T)-f(x)u(x,0)=0.$$Thus, we set $\mathit{\phi }=u_t(x,t)$ into () and integrate in time over $(0,T)$ to get$$\begin{array}{c}\hfill {\int }_0^T{\parallel u_t\parallel }^2\text{d}t+{\int }_0^T(\mathit{A}\nabla u,\nabla u_t)\text{d}t+{\int }_0^T(\mathit{cu},u_t)\text{d}t\hfill \\ \hfill ={\int }_0^T({\int }_0^t[g(u_1)-g(u_2)],u_t(t))\text{d}t.\hfill \end{array}$$The coefficients $\mathit{A}={\mathit{A}}^{\mathit{tr}}$ and $c$ are time-independent! Simple calculations give$$\begin{array}{cc}\hfill {\int }_0^T(\mathit{A}\nabla u,\nabla u_t)\text{d}t& ={\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}{\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,\nabla u_t)\text{d}t={\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,{\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u_t)\text{d}t\hfill \\ \hfill & ={\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u\right))\text{d}t=\frac{1}{2}{\int }_0^T{\mathit{\partial }}_t{\parallel {\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u\parallel }^2\text{d}t=0\hfill \end{array}$$because of $u(x,0)=0$ and $u(x,T)=0$.

Analogously, we may write$${\int }_0^T(\mathit{cu},u_t)\text{d}t=\frac{1}{2}{\int }_0^T(c,{\mathit{\partial }}_t{u}^2)\text{d}t=0.$$Integration by parts yields$$\begin{array}{ccc}\hfill {\int }_0^T({\int }_0^t[g(u_1)-g(u_2)],u_t(t))\text{d}t& =\hfill & {({\int }_0^t[g(u_1)-g(u_2)],u(t))|}_0^T\hfill \\ \hfill & \hfill & -{\int }_0^T(g(u_1)-g(u_2),u_1-u_2)\text{d}t\hfill \\ \hfill & =\hfill & -{\int }_0^T(g(u_1)-g(u_2),u_1-u_2)\text{d}t.\hfill \end{array}$$Collecting all the results above, we arrive at$${\int }_0^T{\parallel u_t\parallel }^2\text{d}t+{\int }_0^T(g(u_1)-g(u_2),u_1-u_2)\text{d}t=0.$$Due to the fact that $g$ is monotonically increasing, we deduce that$${\int }_0^T{\parallel u_t\parallel }^2\text{d}t=0⟹u_t=0\text{a.e.}\text{in}\mathrm{\Omega }\times [0,T]$$and $u$ is a constant in time. Therefore,$$u(x,0)=0⟹u(x,t)=0.$$Putting this information into (), we obtain$$(f,\mathit{\phi })=0\forall \mathit{\phi }\in H_0^1(\mathrm{\Omega }),$$from which we conclude that $f=0$. $\square$

Next theorem generalizes Theorem 2.1 to a transient differential operator $L$.

Theorem 2.2

Consider a linear differential operator$$\mathit{Lu}(x,t)=\nabla ·(-\mathit{A}(x,t)\nabla u(x,t))+c(x,t)u(x,t)$$with ${\mathit{A}}^{\mathit{tr}}=\mathit{A}$ and(5) $$\begin{array}{c}\hfill {\mathit{\xi }}^{\mathit{tr}}·\mathit{A}\mathit{\xi }=\sum _{i,j=1}^n{a}_{i,j}(x,t){\mathit{\xi }}_i{\mathit{\xi }}_j\ge C|\mathit{\xi }{|}^2,\forall \mathit{\xi }\in {\mathbb{R}}^n\end{array}$$(5) All the coefficients appearing in the operator $L$ are assumed to be bounded for any $(x,t)$. Moreover, assume that$${\mathit{\xi }}^{\mathit{tr}}·{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right){\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathit{\xi }\le 0,\forall \mathit{\xi }\in {\mathbb{R}}^n\text{and}{\mathit{\partial }}_{t}c(·,t)\le 0,\forall t\in [0,T].$$Let $u_0,{\mathit{\psi }}_T\in L^2(\mathrm{\Omega })$ and $g^{\prime }\ge 0$. Then, there exists at most one spacewise-dependent source $f\in L^2(\mathrm{\Omega })$ such that () together with the condition () hold.

 

Proof

The proof follows the same line as in Theorem 2.1. The main difference is in handling of the time-dependent coefficients. Assume that we have two solutions $〈u_1,f_1〉$ and $〈u_2,f_2〉$ to () and (). We denote $u=u_1-u_2$ and $f=f_1-f_2$ and recall that $u(x,0)=0$ and $u(x,T)=0$. We have to show that $u=0$ and $f=0$.

For $\mathit{\phi }\in H_0^1(\mathrm{\Omega })$, we get (). Setting $\mathit{\phi }=u_t(t)$ and integrating in time over $(0,T)$ to see that$$\begin{array}{c}\hfill {\int }_0^T{\parallel u_t\parallel }^2\text{d}t+{\int }_0^T(\mathit{A}\nabla u,\nabla u_t)\text{d}t+{\int }_0^T(\mathit{cu},u_t)\text{d}t\hfill \\ \hfill ={\int }_0^T({\int }_0^t[g(u_1(s))-g(u_2(s))]\text{d}s,u_t(t))\text{d}t.\hfill \end{array}$$Monotonicity of $g$ which is found in the proof of Theorem 2.1 gives$${\int }_0^T({\int }_0^t[g(u_1(s))-g(u_2(s))]\text{d}s,u_t(t))\text{d}t\le 0.$$Integrating by parts and using $u(x,0)=0$ and $u(x,T)=0$, we have$${\int }_0^T(\mathit{cu},u_t)\text{d}t=\frac{1}{2}{\int }_0^T(c,{\mathit{\partial }}_t{u}^2)\text{d}t=-\frac{1}{2}{\int }_0^T({\mathit{\partial }}_{t}c,u^2)\text{d}t\ge 0$$and$$\begin{array}{cc}\hfill {\int }_0^T(\mathit{A}\nabla u,\nabla u_t)\text{d}t& ={\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}{\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,\nabla u_t)\text{d}\mathit{tMYAMP}]={\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,{\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u_t)\text{d}\mathit{tMYAMP}]={\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u\right))\text{d}t-{\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right)\nabla u)\text{d}\mathit{tMYAMP}]=\frac{1}{2}{\int }_0^T{\mathit{\partial }}_t{\parallel {\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u\parallel }^2\text{d}t-{\int }_0^T({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right)\nabla u)\text{d}\mathit{tMYAMP}]=-{\int }_0^T({\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right){\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\nabla u,\nabla u)\text{d}\mathit{tMYAMP}]\ge 0.\hfill \end{array}$$Collecting the results above, we deduce that$${\int }_0^T{\parallel u_t\parallel }^2\text{d}t=0⟹u_t=0$$and$$u(x,0)=0⟹u(x,t)=0.$$Now, returning back to (), we conclude that $f=0$. $\square$

The matrix $\mathit{A}$ obeys (), hence it is positive definite with respect to both space and time-dependent entries. Thus, there exists a unique positive-definite square root ${\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}$. We have to be careful because generally$${\mathit{\partial }}_t\mathit{A}={\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}{\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right)={\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right){\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}+{\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right)\ne 2{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right){\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}.$$This equality holds true for e.g. diagonal matrices. More generally, ${\mathit{\partial }}_t\mathit{A}=2{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right){\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}$ iff the matrix commutes with its derivative which is equivalent to $\mathit{A}(x,t)\mathit{A}(x,s)=\mathit{A}(x,s)$$\mathit{A}(x,t)$ for all $t,s$ and $x$, see [13, 14].

If we want to express conditions of Theorem 2.2 for scalar coefficients, we will have () and both coefficients A(x, t) and c(x, t) must decrease in time.

Hyperbolic problem

The IP with the final over-determination for parabolic problems admits at most one solution, but the situation for hyperbolic equations can be quite different. One can also find some non-uniqueness examples, cf. [5, Section 7.2] In this section, we address the uniqueness results for the IPs () and (). We will show that the damping term $g(u_t)$ helps to establish the uniqueness results for hyperbolic case. We will use a similar proof technique as in the parabolic event. The first theorem deals with a steady-state differential operator L.

Theorem 3.1

Consider a linear differential operator$$\begin{array}{c}\hfill \mathit{Lu}(x,t)=\nabla ·(-\mathit{A}(x)\nabla u(x,t))+c(x)u(x,t),\end{array}$$with bounded (discontinuous) coefficients obeying $\mathit{A}(x)={\mathit{A}}^{\mathit{tr}}(x)$ and$$\begin{array}{c}\hfill {\mathit{\xi }}^{\mathit{tr}}·\mathit{A}\mathit{\xi }=\sum _{i,j=1}^n{a}_{i,j}{\mathit{\xi }}_i{\mathit{\xi }}_j\ge C|\mathit{\xi }{|}^2\forall \mathit{\xi }\in {\mathbb{R}}^n.\end{array}$$Let $u_0,{\mathit{\psi }}_T\in L^2(\mathrm{\Omega })$ and $g^{\prime }>0$. Then, there exists at most one spacewise-dependent source $f\in L^2(\mathrm{\Omega })$ such that () together with the condition () holds.

 

Proof

Suppose that we have two solutions $〈u_1,f_1〉$ and $〈u_2,f_2〉$ to () and (). Set $u=u_1-u_2$ and $f=f_1-f_2$. Please note that $u(x,0)=0,u_t(x,0)=0$ and $u(x,T)=0$. We have to show that $u=0$ and $f=0$.

For $\mathit{\phi }\in H_0^1(\mathrm{\Omega })$, we get(6) $$\begin{array}{c}\hfill (u_{\mathit{tt}},\mathit{\phi })+(g({\mathit{\partial }}_t{u}_1)-g({\mathit{\partial }}_t{u}_2),\mathit{\phi })+(\mathit{A}\nabla u,\nabla \mathit{\phi })+(\mathit{cu},\mathit{\phi })=(f,\mathit{\phi }).\end{array}$$(6) We put $\mathit{\phi }=u_t(x,t)$ and integrate in time over $(0,T)$ to get(7) $$\begin{array}{c}\hfill \frac{1}{2}{\parallel u_t(T)\parallel }^2+{\int }_0^T(g({\mathit{\partial }}_t{u}_1)-g({\mathit{\partial }}_t{u}_2),u_t)\text{d}t+{\int }_0^T(\mathit{A}\nabla u,\nabla u_t)\text{d}t+{\int }_0^T(\mathit{cu},u_t)\text{d}t=0.\end{array}$$(7) The last two terms on the left-hand side vanish due to the same reasoning as in Theorem 2.1. Therefore, we may write$$\begin{array}{c}\hfill \frac{1}{2}{\parallel u_t(T)\parallel }^2+{\int }_0^T(g({\mathit{\partial }}_t{u}_1)-g({\mathit{\partial }}_t{u}_2),{\mathit{\partial }}_t{u}_1-{\mathit{\partial }}_t{u}_2)\text{d}t=0.\end{array}$$Here, we can see why we do need the damping term. Without it, we would have that $\parallel u_t(T)\parallel =0$, which gives no guarantee that $u=0$.

Employing the fact that the function g is strictly monotonically increasing, we get that $u_t=0$, i.e. u is constant in time. Therefore,$$\begin{array}{c}\hfill u(x,0)=0⟹u(x,t)=0.\end{array}$$Returning back to (), we conclude that $f=0$. $\square$

Next theorem addresses the uniqueness result for a transient operator L in a hyperbolic problem.

Theorem 3.2

Consider a linear differential operator$$\begin{array}{c}\hfill \mathit{Lu}(x,t)=\nabla ·(-\mathit{A}(x,t)\nabla u(x,t))+c(x,t)u(x,t)\end{array}$$with ${\mathit{A}}^{\mathit{tr}}=\mathit{A}$ and$$\begin{array}{c}\hfill {\mathit{\xi }}^{\mathit{tr}}·\mathit{A}\mathit{\xi }=\sum _{i,j=1}^n{a}_{i,j}(x,t){\mathit{\xi }}_i{\mathit{\xi }}_j\ge C|\mathit{\xi }{|}^2\end{array}$$All the coefficients appearing in the operator L are assumed to be bounded for any (x,t). Moreover, assume that$$\begin{array}{c}\hfill {\mathit{\xi }}^{\mathit{tr}}·{\mathit{\partial }}_t\left({\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\right){\mathit{A}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathit{\xi }\le 0,\forall \mathit{\xi }\in {\mathbb{R}}^n\text{and}{\mathit{\partial }}_{t}c(·,t)\le 0\forall t\in [0,T].\end{array}$$Let $u_0,{\mathit{\psi }}_T\in L^2(\mathrm{\Omega })$ and $g^{\prime }>0$. Then, there exists at most one spacewise-dependent source $f\in L^2(\mathrm{\Omega })$ such that () together with the condition () hold.

 

Proof

Let us have two solutions $〈u_1,f_1〉$ and $〈u_2,f_2〉$ to () and (). Put $u=u_1-u_2$ and $f=f_1-f_2$. Please note that $u(x,0)=0,u_t(x,0)=0$ and $u(x,T)=0$. We will show that $u=0$ and $f=0$.

The relation () holds true. The last two terms on the left-hand side of () are non-negative, due to the same reasoning as in Theorem 2.2. Thus, we see that$$\begin{array}{c}\hfill \frac{1}{2}{\parallel u_t(T)\parallel }^2+{\int }_0^T(g({\mathit{\partial }}_t{u}_1)-g({\mathit{\partial }}_t{u}_2),{\mathit{\partial }}_t{u}_1-{\mathit{\partial }}_t{u}_2)\text{d}t\le 0.\end{array}$$According to the strict monotonicity of the function g, we obtain that $u_t=0$, i.e. u is constant in time. Looking at $u(x,0)=0$ we get $u(x,t)=0$, Inspecting the relation (), we conclude that $f=0$. $\square$

 

Conclusions

In this work, we have studied the uniqueness of two inverse source problems: parabolic () and (); and hyperbolic () and (). We have showed that under reasonable physical conditions on the data functions, the final over-determination is sufficient for the uniqueness of a space dependent source.

Acknowledgments

This work was supported by the BOF/GOA-project No. 01G006B7 of Ghent University.

Notes

The symbol ${\mathit{A}}^{\mathit{tr}}$ denotes the transpose of $\mathit{A}$.