Let a $\Omega = \{(x,y); -1 < x < 1, -1 < y < 1\}$ and suppose that u is a smooth solution of
\begin{equation}
\left\{
\begin{aligned}
&u_{t} – \Delta u + u u_x + uu_y = 0, \ \ ,&&\Omega \times (0, T),\\
& u(x,y,0) = f(x,y) && \Omega\\
\end{aligned}
\right.
\end{equation}
Show that
\begin{equation}
\max_{\Omega \times [0,T]}u(x,y,t) \leq max\{\max_{0 \leq t \leq T}u(\pm 1, \pm 1, t), \max_{\overline{\Omega}} f(x,y)\}
\end{equation}
Suggestion: Write $u = v + \epsilon t$ and making $\epsilon \rightarrow 0$
Attempt
\begin{equation}
u_t – \Delta u + uu_x + uu_y = v_t – \Delta v + (v + \epsilon)v_x + (v + \epsilon) = 0
\end{equation}
Making $\epsilon \rightarrow 0$, we obtain that $v$ satisfies the heat problem. How then to use the maximum principle?
Best Answer
This follows directly from the maximum principle. Let $ b (x) = u(x)(1,1)$ then $Lu=-\Delta u + b \cdot \nabla u$ is elliptic and $u_t+Lu= 0$ in $\Omega_T:=\Omega \times (0,T)$. Hence, \begin{align*}\max_{\overline{\Omega}_T} u &= \max_{\Gamma_T} u \leqslant \max \{ \max_{\partial \Omega \times [0,T]} u , \max_{\overline{\Omega}}f \} \end{align*} where $\Gamma_T = \overline{\Omega}_T \setminus \Omega_T$ is the parabolic boundary.