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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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    I don't quite see why the boundary term vanishes.2012-12-21
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    Ok I'll add an explanation to my post.2012-12-21
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    It's OK, I got it. Add an explanation anyway :) Since the boundary terms cancel before you even applied the log convexity argument, it's no wonder that you could prove uniqueness for both cases in one argument.2012-12-21
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    I feel like I must be making an error somewhere, since the proof is totally independent of the value of $\alpha$. I've been trying to find it for a couple days but no luck. :(2012-12-21
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    Maybe the instructor wants you to try two different proof techniques, one being less powerful.2012-12-21
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    Maybe that's the case, but I have a feeling there's probably a mistake in my proof somewhere. It just seems very wierd that the proof is completely independent of the value of $\alpha$, and $\alpha$ does not even have to be a constant for this to work.2012-12-21
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    Indeed, $\alpha$ could be a function of $x$ (boundary with variable permeability). It had to be independent of $t$, though.2012-12-21
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    So the proof is correct then? I'm just confused about why the professor told us to split it into 2 cases if it didnt even matter...2012-12-21
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    @Bartek - remind me why a logarithmically convex non-negative function must be zero.2012-12-21
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    It also needs to satisfy $E(0)=0$ so that we can say $E(t)=0$ by logarithmic convexity, but that just follows from the initial condition. The proof of this result can be done by contradiction and taking $\frac{d}{dt} \log(E)$2012-12-21
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    Yes, the proof is correct. And I understand why the professor wanted you to work the two cases separately: if you used log-convexity for both, you'd miss a chance to practice a proof based on the monotonicity of energy.2012-12-21
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    Ok thanks for checking it! I guess the wording of the question had me thinking there was a problem with the proof.2012-12-21
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    I agree with Pavel.2012-12-21
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    Great, thanks Hans and Pavel for your comments!2012-12-21
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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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