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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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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. –  Hans Engler Dec 21 '12 at 0:05
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Maybe the instructor wants you to try two different proof techniques, one being less powerful. –  Hans Engler Dec 21 '12 at 1:35
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Indeed, $\alpha$ could be a function of $x$ (boundary with variable permeability). It had to be independent of $t$, though. –  user53153 Dec 21 '12 at 4:03
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@Bartek - remind me why a logarithmically convex non-negative function must be zero. –  Hans Engler Dec 21 '12 at 4:18
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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. –  user53153 Dec 21 '12 at 4:34

1 Answer 1

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.

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