Is the following a correct proof? It is easier than a proof I have been provided with, but I feel like it is wrong.

Prop: If $ u$ satisfies $ u_{t} = \sum_{i=1}^{n} u_{x_{i} x_{i} } $ on $\ D \times [0,T] $ where $ D $ is some open domain , then u attains its maximum on $\ \partial D$

Proof: By Heine-Borel, $ u $ attains its maximum on $\partial D \cup D $ .

Suppose $u$ attains its maximum at $ x_0 \in D $. At $x_0$ then $u_t = 0 $ (by Fermat), and $ u_{x_{i} x_{i}} < 0$ (as the Hessian is negative definite).

This contradicts our heat equation and therefore the maximum must be attained in $ \partial D $.

I think the mistake is where I claim $u_{x_{i} x_{i}} < 0$, so maybe I am confused about the Hessian.

  • $\begingroup$ I was using this a bit. Its possible I have misunderstood it. thanks for helping and I apologize for my awful spelling. en.wikipedia.org/wiki/… $\endgroup$ – douglares Dec 22 '13 at 20:54
  • $\begingroup$ What theorem by Fermat are you quoting? $\endgroup$ – Potato Dec 22 '13 at 20:58
  • $\begingroup$ i might have mixed it up. the theorem i was using was: at a local minima/maxima NOT on the boundary we have first derivative is zero $\endgroup$ – douglares Dec 22 '13 at 21:01
  • $\begingroup$ So, it's not clear to me what you are trying to prove. I think what you are trying to show is that if you fix a time $t$, then the resulting function of $x$ (on $D$) attains its maximum on the boundary $\partial D$. Is this right? $\endgroup$ – Potato Dec 22 '13 at 21:02
  • $\begingroup$ Or, are you looking at the maximum as a function of $(t,x)$ on $D\times [0,T]$? $\endgroup$ – Potato Dec 22 '13 at 21:04


  1. $u$ attains maximum in $K=(D\cup\partial D)\times [0,T]$, if $D$ is bounded.

  2. $u$ does not attain maximum at $x_0$ but at a point of the form $(x_0,t_0)$.

  3. It $(x_0,t_0)$ is an interior point, i.e., $(x_0,t_0)\in D\times (0,T)$, then $u_t=0$ and $u_{x_ix_i}\le 0$.

So, you do not have sufficient amount of evidence to show that $u$ can attain maximum on the boundary.

Hint. You need to consider $u_\varepsilon(x,t)=u(x,t)+\varepsilon|x|^2$.

  • $\begingroup$ thanks! I have proven it this way with the epsilon, in my tiredness i thought i had an easier proof but i was being stupid haha. $\endgroup$ – douglares Dec 22 '13 at 21:11

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