In my class we were given a proof for the weak maximum principle of the heat PDE. However, there are certain parts of the argument that I don't understand. The proof is below:
I don't understand the parts underlined in blue. My questions are:
- Why does the derivative of $u$ with respect to $t$ differ based on whether $t_0 \leq T -\epsilon$. Surely from basic calculus, if we have a min/max at $(t_0,x_0)$ then the derivatives there should be equal to $0$?
2)How do we know that this Taylor expansion is non-positive. I know that if $u$ has a maximum at the point $(t_0,x_0)$ then the second derivative there should be negative. I assumed that because $(t_0,x_{\*})$ is near $(t_0,x_0)$, then by continuity if $u_{xx}(t_0,x_0) \leq 0$ then it follows that $u_{xx}(t_0,x_{\*}) \leq 0$. However, from the next line of the proof, it seems like they already knew that the Taylor expansion was negative and that they use this to justify that $u_xx(t_0,x_{\*}) \leq 0$, rather than the argument I said regarding continuity. The next line of the proof is
Best Answer
If the maximum of a function on a closed interval occurs at an endpoint, then the derivative doesn't have to be zero there.
What it means for $u(t_0,x_0)$ to be the greatest value is that all other values are smaller (or equal): $$u(t,x) \le u(t_0,x_0) \quad \iff \quad u(t,x) - u(t_0,x_0) \le 0.$$