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.
Suppose
\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.