Heat Equation $u_{t} - \Delta u = \sin(u)$

Let $\Omega \subset \mathbb{R}^n$ bounded domain with smooth boundary.

\begin{equation} \left\{ \begin{aligned} &u_{t} - \Delta u = \sin(u),&&\Omega \times (0, T),\\ & u(x,t) = 0 &&\partial \Omega \times (0, T) \\ & u(x,0) = u_0(x)&&\Omega. \end{aligned} \right. \end{equation}

Show that if $u_0 \leq 1$, then $u(x,t) \leq e^t$, $\forall \ x \in \Omega$, $t > 0$.

Suggestion: Note that $\sin x = \frac{x\sin x}{x}$ and $\frac{\sin x}{x}$ are limited. Try to use the maximum principle

Attempt: I think we have to use the Gronwall Inequality. I don't see where the maximum principle fits.

For example, i tried $uu_t = u\sin(u) + u\Delta u = \frac{u^2\sin(u)}{u} + u\Delta u$. taking the module, we obtain $u_t \leq u^2 + |u\Delta u|$. But I didn't get anywhere

Solution 1:

Let $w(x,t) = u(x,t) - e^t$. Then in $\Omega \times (0,T)$, $$w_t - \Delta w = \sin(u)-e^t<\sin (u)-1 \leqslant 0.$$ On $\partial \Omega \times (0,T)$, $$ w=-e^t<0$$ and on $\Omega \times \{ t=0\}$, $$w=u_0-1\leqslant0. $$ Hence, by the maximum principle $w \leqslant 0$ in $\Omega \times (0,T)$ i.e. $$u(x,t) \leqslant e^t \qquad \text{in }\Omega \times (0,T). $$