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 $u_t=u_{xx}+u$

  • $\begingroup$ This is a problem in Evans, isn't it? I feel like I've done something like this before, but with hints, and I don't remember anything else. $\endgroup$
    – abnry
    Commented May 16, 2014 at 0:03
  • $\begingroup$ Have you consider some sort of Grunwald inequality? $\endgroup$
    – abnry
    Commented May 16, 2014 at 0:06
  • $\begingroup$ @nayrb I don't know if this is a problem in Evans, but I will verify that. Please don't consider this inequality the teacher don't explained this in the class. $\endgroup$
    – Felipe
    Commented May 16, 2014 at 0:10

1 Answer 1


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 $u(x,t)=e^t X(x)$?

  • $\begingroup$ The maximum principle gives me that $ \max\limits_{\Omega\times[0,T]}|u|$ $=\max\limits_{\Omega\times[0,T]}|u_t-u_{xx}|=\max\limits_{\gamma}|u_t-u_{xx}|. $ But, who is $\leq$ that the last part? Must I use the triangular inequality? Can I assume that solution for $u$? $\endgroup$
    – Felipe
    Commented May 16, 2014 at 13:54
  • $\begingroup$ If $u(x,t)=e^tX(x)$ so $u_t=u$ and $u_{xx}=0$, this make sense? $\endgroup$
    – Felipe
    Commented May 16, 2014 at 14:35

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