# Uniqueness for 3-dimensional heat equation initial Robin boundary value problem (SOLVED)

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.

• 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. Commented Dec 21, 2012 at 0:05
• Maybe the instructor wants you to try two different proof techniques, one being less powerful. Commented Dec 21, 2012 at 1:35
• Indeed, $\alpha$ could be a function of $x$ (boundary with variable permeability). It had to be independent of $t$, though.
– user53153
Commented Dec 21, 2012 at 4:03
• @Bartek - remind me why a logarithmically convex non-negative function must be zero. Commented Dec 21, 2012 at 4:18
• 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
Commented Dec 21, 2012 at 4:34