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Let $\Omega \subset \mathbb{R}^3$ be a bounded domain. Using an energy argument, show that the IBVP \begin{align} u_t &= \Delta u ~~~~~~~~~~x \in \Omega, ~t>0\\ \frac{\partial u}{\partial \nu} + \alpha u &= h(x) ~~~~~~~~x \in \partial\Omega, ~t>0\\ u(x,0)&=g(x) ~~~~~~~~~x \in \Omega \end{align} where $\nu$ is the exterior unit normal and $\alpha$ is a constant has at most one solution. Treat the cases $\alpha \geq 0$ and $\alpha<0$ separately. Use logarithmic convexity for the second case.

My attempted solution: By contradiction, suppose that there are two solutions $u_1$ and $u_2$ and define $v = u_1 - u_2$. Then $v$ satisfies \begin{align} v_t &= \Delta v ~~~~~~~~~~x \in \Omega, ~t>0\\ \frac{\partial v}{\partial \nu} + \alpha v &= 0 ~~~~~~~~~~~~~x \in \partial\Omega, ~t>0\\ v(x,0)&=0 ~~~~~~~~~~~~~x \in \Omega \end{align} Define the energy functional to be $E(t)= \frac{1}{2}\int_\Omega v^2 \,dx.$ The case $\alpha \geq 0$ is trivial. I just showed that $\frac{dE}{dt}=\int_\Omega v \, v_t \,dx \leq 0$ using Green's first identity and the conditions on $v$. Then since $E(0)=0$ we must have $E(t)=0$, and hence $v=0$.

For the $\alpha<0$ case I want to show that $E\frac{d^2E}{d^2t} - \left( \frac{dE}{dt} \right)^2 \geq 0\,.$ Since $E \geq 0$ then by logarithmic convexity we would have $E=0$.

However, I'm running into some problems. I take \begin{align} \frac{d^2E}{dt^2}=\int_\Omega v_t^2 \,dx + \int_\Omega v \, v_{tt} \,dx\,. \end{align} Then, for the second term I write \begin{align} \int_\Omega v \, v_{tt} \,dx &=\int_\Omega v \, \Delta v_t \, dx\\ &= \int_\Omega v_t \, \Delta v \, dx \\ &= \int_\Omega (v_t)^2 \, dx \end{align}

where I used Green's second identity and the boundary term vanished due to the boundary condition on $v$. Explicitly:

\begin{align} \int_{\partial \Omega} v\frac{\partial v_t}{\partial \nu} - v_t \frac{\partial v}{\partial \nu} dS= \alpha \int_{\partial \Omega} -v v_t + v_t v \, dS = 0 \end{align} by the homogeneous Robin condition.

So I get $E\frac{d^2E}{d^2t} - \left( \frac{dE}{dt} \right)^2 = \frac{1}{2}\int_\Omega v^2 dx \cdot 2 \int_\Omega v_t^2 dx - \left( \int_\Omega v v_t \, dx \right)^2 \geq 0$ by the Cauchy-Schwarz inequality.

So I did the proof without even using the assumption $\alpha < 0$, which seems very strange. Did I make a mistake somewhere?

EDIT: Looks like my proof is actually correct. I guess the wording of the question just had me thinking there was an issue.

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    @Hans, Pavel, Bartek: It would be nice if one of you posted an answer so that it can be "officially" marked as accepted. Editing the title of the question to say "(SOLVED)" is insufficient, as the system will still consider it unanswered and periodically bump it to the front page.2012-12-21

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My original proof is actually correct. I was just confused by the wording of the problem and thought I had made a mistake. Thanks to Hans and Pavel for checking it.