Strong maximum principle for heat equation; If $u(x,t)$ is constant for times $t\leq t_0$, is $u$ also constant at times $t\geq t_0$? I am learning the strong maximum principle by following Evan's Partial differential equations. It is stated as follows: Let $U$ be an open subset of $\mathbb{R}^n$ and $T>0$. Let $u=u(x,t)$, where $x=(x_1,\ldots,x_n)\in\Omega$ and $0\leq t\leq T$. Denote $\partial_t$ the partial derivative wrt $t$ and $\Delta_x=\sum_{i=1}^n\partial_{x_i}^2$ the Laplacian on the variables $x_i$.
Theorem 4(b) (page 55) of Evan's book states, in equivalent terms

Theorem (strong maximum principle for heat equation): Suppose that
  
  
*
  
*$U$ is connected;
  
*$u$ is $C^2$ on the variables $x_i$, and $C^1$ on the variable $t$, on $U\times(0,T]$, and continuous on all of $\overline{U}\times [0,T]$;
  
*(the heat equation) $$\partial_t u=\Delta_x u\quad\text{on}\quad U\times (0,T)$$
  
*There exists $x_0\in U$ such that $u(x_0,T)=\sup u(U\times([0,T]))$
Then $u$ is contant of $\overline{U}\times[0,T]$.

My question is: Does the analogous result hold if we assume instead that $u$ attains its maximum at a point $(x_0,t_0)$, where $0<t_0<T$?
My guess is that this is not the case: If there is a nonzero solution $v$ of the heat equation on some square $U\times[0,T]$ which satisfies $v=0$ when $t=0$, then we can glue the constant function $0$ on $U\times[0,t_0]$ with $w(x,t)=v(x,t-t_0)$ when $t\geq t_0$, and hope that the resulting function is sufficiently regular. However we should not have any good boundedness assumptions on $v$ to avoid all those uniqueness theorems for solutions of the heat equation..

You are right. If $u$ attains its maximum at $(x_0,t_0)$, then you can only conclude that $u$ must be constant up to that time-point, i.e. in $\bar{U}_{t_0}$. Beyond $t_0$, the function may change value before reaching the final time $T$.
