# Heat equation, maximum principle

For a solution of $v(x,t)$ in $C^2([-2,2]\times[0,T])$ of the heat eqution $v_t-v_{xx}=0$, if there is a point $(x_0,t_0)$ with $v_x(x_0,t_0)=v_t(x_0,t_0)=0$ and $v_{xx}(x_0,t_0)<0$ then $(x_0,t_0)\in([-2,2]\times \{0\})\cup(\{-2\}\times [0,T])\cup(\{2\}\times [0,T])$

For the heat equation maximum principle holds, and the given point looks like a maximum, so it should be on the boundary, but why is the top of the rectangle, $([-2,2]\times \{T\})$ doesn't belong to the boundary ?

Suppose $v_{xx}(x_0, t_0) < 0$ for some $t_0 > 0$ and $x_0$ inside the spatial domain. This implies that $v_t(x_0, t_0) < 0$. Since $v$ is $C^2$, this means that $v(x_0, t_0 - \delta) > v(x_0, t_0)$ for a sufficiently small $\delta > 0$. Thus $t_0 = T$ is not possible for a maximum.