Heat Equation $u_{t} - \Delta u + u u_x + uu_y = 0$

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?

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.