Let $(u,t)$ the $C^2$ solution of the equation $$ u_t=u_{xx}+u, \textrm{ over } [0,a]\times[0,T]\subset \mathbb{R}^2 $$ where $T>0$
Show that $$ \max\limits_{[0,a]\times[0,T]} |u| \leq e^T\max\limits_{\gamma}|u| $$ where $\gamma\dot{=}(\{0\}\times\{[0,T]\})\cup (\{a\}\times[0,T])\cup([0,a]\times\{0\})$.
This is a question of my homework in PDE.
What I have to work: Solutions of the "traditional" heat equations with boundary condition, the dirichlet problem, Green's identities, divergence theorem, Hölder inequality...
My answer: I did some exercises of the heat equation but none is like this... I'm not knowing how to start that.
[UPDATE] My attempt:
$$ u=u_t-u_{xx}\Rightarrow u=\varphi(x)G(t) $$ so $$ u_t=G_t(t)\varphi(x) \textrm{ and } u_xx=G(t)\varphi_{xx}(x) \Rightarrow \varphi(x)G(t)=G_t(t)\varphi(x)-G(t)\varphi_{xx}(x)\Rightarrow $$ $$ \Rightarrow 1=\frac{1}{G(t)}G_t(t)-\frac{1}{\varphi(x)}\varphi_{xx}(x)\Rightarrow\left(1+\frac{1}{\varphi(x)}\varphi_{xx}(x)\right)=\frac{1}{G(t)}G_t(t)=\lambda $$ where $\lambda$ is an arbitrary positive constant. What I have is this:
$$ \frac{1}{G(t)}G_t(t)=\lambda $$ and $$ \left(1+\frac{1}{\varphi(x)}\varphi_{xx}(x)\right)=\lambda $$
So, for the first equation, $G(t)=e^{\lambda t}$ and supposing that we have that $u=\varphi(x)e^{\lambda t}$. But this implies that $u_{xx}=0$ (and this not make sense for me).
Furthermore, if I try solve this by the maximum principle, as $T>0$ and $e^T\geq 1$ $$ \max\limits_{[0,a]\times[0,T]} |u|=\max\limits_{\gamma} |u|\leq e^T\max\limits_{\gamma} |u| $$ and so this confused me because I did not use the hypothesis $u_t=u_{xx}+u$
Here's a hint: Rewrite the PDE to
$$ u = u_t - u_{xx}$$
You probably done the maximum principal (without the forcing term), so what can you say about:
$$\max_{\Omega}|u| = \max_{\Omega} | u_t - u_{xx} | \leq ?$$
Where $\Omega$ is the domain you're working in. If that doesn't help, what if you assume the solution takes the form $u(x,t)=e^t X(x)$?