This is a homework problem. I am looking for an additional hint or reference to theorems/ideas that can help. Some of my thought process is presented below.


\begin{align*} u_t - \Delta u &= 0 \, \text{on} \, \Omega \times [0,T]\\ \Omega &\in R^n, \text{bounded}\\ \frac{\partial u}{\partial n} &=0 \, \text{on} \, \partial \Omega\\ \end{align*}

and $\partial \Omega$ smooth.

$\textbf{Question:}$ Prove that $\max{u}$ occurs when $t = 0$. A hint is given to construct a function $u_{\epsilon,\delta} = u - \epsilon t - \delta \phi(x) $.

My work: Essentially, I need to choose $u_{\epsilon,\delta}$ such that

\begin{align*} \partial_tu_{\epsilon,\delta} - \Delta u < 0\\ \frac{\partial u_{\epsilon,\delta}}{\partial n} <0 \end{align*} The first condition guarantees the weak maximum principle, i.e. that $\max u$ lies either on $\partial \Omega$ or at time $t =0$. The second condition prevent $\max{u}$ from being on $\partial \Omega$.

The first condition is rather easy to satisfy. Essentially we need \begin{align*} -\epsilon + \delta \Delta \phi(x) < 0 \end{align*} Which can be done if $\Delta \phi(x)$ is a constant. This made me choose $\phi(x) = || x||^2$, where $||.||$ is the $l^2$ norm.

The second condition requires:

\begin{align*} \nabla ( u_{\epsilon,\delta}) \cdot \vec{n} = \nabla u \cdot \vec{n} - \delta \nabla(\phi(x)) \cdot \vec{n} < 0 \end{align*}

Which implies that \begin{align*} \nabla \phi(x) \cdot \vec{n} > 0 \end{align*} with $\delta > 0$.

Here is where I am confused. How can I ensure that the dot product is strictly positive? I have no information on the components of $\vec{n}$. Even if I force all the components of $\nabla \phi$ to be positive, the dot product can still be negative.

My suspicion is that I need to use the fact that $\partial \Omega$ is smooth.

  • 1
    $\begingroup$ Your choice of $\phi$ is to blame. You need a function adapted to the geometry of the domain. One possibility: subtract from $||x||^2$ the harmonic function with the same boundary values. This gives a function with constant positive Laplacian and zero boundary values. Hopf lemma implies that normal derivative is positive on the boundary. ... But I don't know if you have these tools at your disposal. $\endgroup$ – user147263 Sep 24 '14 at 2:36

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