A doubt on Heat equation I'm studing the Maximum Principle of Heat equation:
Let $u\in C(\overline(U_T)\cap C^2_1(U_T))$ in $\mathbb{R}^n$ satisfy $u_t=c\Delta u$ on $(x,t)\in U_T$. Then $\displaystyle\max_{\overline{U}_T}u=\max_{\partial U_T}u$.
The proof begin as follow: Suposse $u$ has a local maximum at $(x,t)\in U_T$. Then $\color{red}{u_t=u_x=0}$ and $\color{red}{\Delta u\leq 0}$. Why is this truth (the red things)?
Thanks!
 A: If a $C^2$ function $u$ has a local maximum at $u(x,t)$, then


*

*$\nabla u = 0$, i.e. all of the partial derivatives of $u$ are zero (otherwise, you are not at a critical point: by moving "uphill" you increase $u$).

*$\nabla^2 u$, the Hessian of $u$, has no positive eigenvalues. Suppose otherwise, that $(v_x, v_t)$ is an eigenvector of $\nabla^2 u$ with $\lambda>0$ as an eigenvalue. Then consider $u(x+h v_x, t+h v_t)$ for small $h$. By Taylor's theorem


$$u(x+h v_x, t+h v_t) = u(x,t) + h\nabla u(x,t)\cdot v + \frac{h^2}{2} v^T \nabla^2 u v + h^2 R(h)$$
$$u(x+h v_x, t+h v_t) = u(x,t) + 0 + \frac{h^2}{2}\lambda + h^2 R(h),$$
with $\lim_{h\to 0} R = 0$. Therefore for sufficiently small $h$,
$$u(x+h v_x, t+h v_t) > u(x,t)$$
and $u$ does not have a maximum at $x,t$, a contradiction. 
Since all of the eigenvalues of $\nabla^2 u$ are nonpositive, $\Delta u = \operatorname{tr}\nabla^2 u \leq 0.$
A: Supose $u$ has a local maximum at $(x,t)\in U_{T}$, for the calculo, if the function $f$ has a maximum or minimun at $x_{0}$, then $\frac{d}{dx}f(x_{0})=0$ ,still more, if you have a maximum in $x_{0}$, you have $\frac{d^{2}}{dx^{2}}f(x_{0})<0$ , in your case, since the function depends on two parameters $(x,t)$, then partial derivatives are zero  and $\Delta u<0$, at which point $(x,t)$
