# Maximum Principle for Heat Equation Proof (?)

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.

• 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/… – douglares Dec 22 '13 at 20:54
• What theorem by Fermat are you quoting? – Potato Dec 22 '13 at 20:58
• 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 – douglares Dec 22 '13 at 21:01
• 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? – Potato Dec 22 '13 at 21:02
• Or, are you looking at the maximum as a function of $(t,x)$ on $D\times [0,T]$? – Potato Dec 22 '13 at 21:04

Corrections:

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 $u_\varepsilon(x,t)=u(x,t)+\varepsilon|x|^2$.

• 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. – douglares Dec 22 '13 at 21:11