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Let $U\subset\mathbb{R}^{n}$ be open, $q(x)\geq0$ continuous, and suppose $u\in\mathscr{C}^{2}(U\times[0,T])$ solves $$\left\{\begin{array}{rl} u_{tt}-\Delta u=q(x)u&\text{in}\;U_{T}\\ u=f&\text{on}\;\partial U\times[0,T]\\ u=g&\text{on}\;U\times\{t=0\}\\ u_{t}=h&\text{on}\;U\times\{t=0\}.\end{array}\right. $$ Prove that $u$ is uniquely determined by its initial/boundary data $f,g,h$.

The only way I know how to prove uniqueness results for PDE which do not obey a maximum principle is by basic energy arguments (variational methods). To find an appropriate energy function I multiply as is usual the PDE by $u_{t}$ and after some basic computations find $$E'(t):=\frac{1}{2}\frac{d}{dt}\int_{U}u_{t}^{2}+|\nabla u|^{2}-qu^{2}\;dx=\int_{\partial U}u_{t}\frac{\partial u}{\partial\nu}\;dS.$$ So if $u_{1},u_{2}$ are two solutions and I define $u$ to be their difference, the boundary term on the RHS vanishes and I have for all $t>0$ $$E'(t)\equiv0\Longrightarrow E(t)=E(0)=0,$$ the last equality holding because of the boundary/initial data of $u:=u_{1}-u_{2}$ being zero.

Consequently, $$|u_{t}(x,t)|^{2}+|\nabla u(x,t)|^{2}=q(x)|u(x,t)|^{2}$$ for every $(x,t)\in U_{T}:=U\times(0,T].$ Now, if $q(x)$ were nonpositive, then I could easily deduce $u\equiv0$ in $U$ since the only way a sum of squares is $0$ is if each term is $0$. But $q(x)\geq0$, and so things aren't so simple, and this is where I am having trouble.

I have tried to mess around with the two relations for $u$ that I have from above: $$ \left\{\begin{array}{c} q(x)u=u_{tt}-\Delta u\\ q(x)u^{2}=u_{t}^{2}+|\nabla u|^{2},\end{array}\right. $$ but could not find a proof and would appreciate any hints in this direction. One thing I noticed though is that if we take $q=1$ and $n=1$, we have $u=u_{tt}-u_{xx}$ and $u^{2}=u_{t}^{2}+u_{x}^{2},$ the ladder condition being being a Pythagorean relation. Are these two relations compatible? Is it possible for a non-trivial solution to exist that satisfies these two equations, in addition to the zero boundary/initial data? I figured once I could establish the answer to that for this simplified case, it would be easy to extend for the general case.

(Current Progress I)

Define energies (since I'm not sure which one will be useful yet) $$E_{1}(t):=\frac{1}{2}\int_{U}u_{t}^{2}+|\nabla u|^{2}\;dx\;\;\;\;\text{and}\;\;\;\;E_{2}:=\frac{1}{2}\int_{U}u_{t}^{2}+|\nabla u|^{2}+qu^{2}\;dx.$$

Then for $E_{1}$ we estimate \begin{align*} \frac{dE_{1}}{dt} &=\int_{U}u_{t}u_{tt}+\nabla u\cdot\nabla u_{t}\;dx\\ &=\int_{U}u_{t}(qu+\Delta u)+\nabla u\cdot\nabla u_{t}\;dx\\ &=\underbrace{\int_{U}\nabla\cdot(u_{t}\nabla u)\;dx}_{\int_{\partial U}u_{t}\frac{\partial u}{\partial\nu}\;dS\;=\;0}+\int_{U}quu_{t}\;dx\\ &\leq\int_{U}\frac{1}{2}u_{t}^{2}+\frac{1}{2}(qu)^{2}\;dx\\ &\leq\frac{1}{2}\int_{U}u_{t}^{2}+|\nabla u|^{2}\;dx+\underbrace{\frac{1}{2}\int_{U}(qu)^{2}\;dx}_{U(t)}\\ &=E_{1}(t)+U(t). \end{align*} Since $E_{1}$ and $U$ are nonnegative, we can apply Gronwall's inequality to obtain $$E_{1}(t)\leq e^{\int_{0}^{t}\;ds}\left(\underbrace{E_{1}(0)}_{0}+\int_{0}^{t}U(s)\;ds\right)=e^{t}\int_{0}^{t}\frac{1}{2}\int_{U}(qu)^{2}\;dxds.$$

An almost identical computation for $E_{2}$ gives the same estimate times a factor of $2$: $$E_{2}(t)\leq e^{t}\int_{0}^{t}\int_{U}(qu)^{2}\;dxds.$$

Since the RHS side of the $E_{1}$ estimate is nonnegative, so we can remove the $\frac{1}{2}$ without affecting the inequality in order to obtain $$0\leq E_{1}(t),E_{2}(t)\leq e^{t}\int_{0}^{t}\int_{U}(qu)^{2}\;dxds.$$

But as pointed out already, we cannot subtract these inequalities since they both go the same direction.

(Current Progress II - Solution!)

$U$ is bounded and $q\geq0$ continuous, so $$1\leq\sup\limits_{x\in U}q(x)+1:=M<\infty.$$ Using $E_{2}$ from above, we then have $$E_{2}'(t)\leq\int_{U}qu^{2}q+u_{t}^{2}\;dx\leq\int_{U}(q+1)u_{t}^{2}+(q+1)|\nabla u|^{2}+(q+1)qu^{2}\;dx\leq ME_{2}(t).$$ Then Gronwall's inequality and the initial conditions on $u$ give $$E_{2}(t)\leq e^{Mt}E_{2}(0)=0.$$ Consequently, the integrand defiing $E_{2}(t)$ being at least continuous, we find for all $(x,t)\in U_{T}$ that $$u_{t}^{2}+|\nabla u|^{2}+qu^{2}=0$$ which implies, being the sum of squares, each term above vanishes in $U_{T}$. In particular, the derivatives of $u$ vanish and so $u$ is a constant $C=0$ due to the boundary conditions. Thus, solutions to the above PDE IVP/BVP are unique.

Kind of ironic that in my original attempt at solution I had $$u_{t}^{2}+|\nabla u|^{2}-qu^{2}=0$$ in $U_{T}.$ :p

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I don't see how you can subtract the two energies. Just use $E_2$ and try to bound $E_2'$ by a constant multiple of $E_2$. – timur Mar 5 '13 at 2:37
Nevermind, you're right -- you can't subtract them. I would need a lower bound on one and an upper bound on the other. – Taylor Martin Mar 5 '13 at 2:45
Can you please write what estimate you get for $E_2'$? – timur Mar 5 '13 at 2:57
I posted a correct (I hope) solution with the estimate. I was applying Gronwall's inequality to an estimate of the form $E_{2}'\leq \phi E_{2}+\psi$ intead of $E_{2}'\leq\phi E_{2}.$ In the former, you get an extra term which is hard to work with (in particular, to show it vanishes), but in the ladder you only get a product with one factor being $E_{2}(0)$ which of course vanishes due to the boundary conditions on $u:=u_{1}-u_{2}.$ – Taylor Martin Mar 5 '13 at 3:12
up vote 1 down vote accepted

Do not include the minus $qu^2$ term in the energy. Just define the energy as usual, or if it fails, include $qu^2$ with positive sign. You might find the Gronwall inequality handy.

The reason why you don't want negative $qu^2$ in the energy is that you want the energy to control the solution, in the sense that if the energy is 0 then you want to be able to say that the solution also vanishes.

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I'm not sure what you're trying to get at here. All the computations that I've done all lead to what has already been recorded in my above answer, namely if we define the energy as usual that $$E'(t)=\int_{U}q(x)u(x,t)u_{t}(x,t)\;dx=\frac{1}{2}\frac{d}{dt}\int_{U}u^{2}(x,‌​t)q(x)\;dx.$$ I feel like I keep going around and around in circles with the integration by parts, substituting terms from the equation, writing expressions as derivatives, etc. – Taylor Martin Mar 4 '13 at 5:13
I'll toy around a little more and see if I can't get $E$ and $E'$ to both appear in the same expression in order to apply Gronwall's inequality. – Taylor Martin Mar 4 '13 at 5:19
Try something like $E=\int u_t^2+|\nabla u|^2+qu^2$. – timur Mar 4 '13 at 15:45
With that energy you get $E'(t)=4\int_{U}uu_{t}q\;dx$ instead of $E'(t)=0$! – Taylor Martin Mar 4 '13 at 19:37
@JTian: You just need something like $E'\leq 2E$ to use Gronwall. The condition $E'=0$ is not necessary! – timur Mar 5 '13 at 0:28

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