I had a doubt in following theorem of Evans
THEOREM 4 (Strong maximum principle). Suppose $u \in C^2_1(U) \cap C(\bar{U})$ is satisfies $u_t + \Delta u = 0$ within $U_T$. (i) Then $$ \max _{\bar{U_T}} u=\max _{\Gamma_T} u $$ (ii) Furthermore, if $U$ is connected and there exists a point $(x_0, t_0) \in U_T$ such that $$ u\left(x_0,t_0\right)=\max _{\bar{U_T}} u $$ then $u$ is constant within $\bar{U_{t_0}}$.
Assertion (i) is the maximum principle for heat's equation and (ii) is the strong maximum principle.
Proof (as in Evans) -
- Suppose there exists a point $(x_0, t_0) \in U_T$ with $u(x_0,t_0) = M := \max_{\bar{U_T}}u$ Then for all sufficiently small $r > 0$, $E(x_0,t_0;r) \in U_T$; and we employ the mean value property to deduce that $$ M = u(x_0,t_0) = \frac{1}{4r^n}\int\int_{E(x_0,t_0;r)} u(y,s) \frac{|x_0 - y|^2}{(t_0 -s)^2} \leq M $$ since $$ 1 = \frac{1}{4r^n}\int\int_{E(x_0,t_0;r)} \frac{|x_0 - y|^2}{(t_0 -s)^2}$$
Equality holds only if $u$ is identitically equal to $M$ within $E(x_0,t_0;r)$. Consequently $$ u(y,s) = M \,\, \forall (y,s) \in E(x_0,t_0;r)$$
Draw any line segment $L$ in $U_T$ connecting $(x_0,t_0)$ with some other point $(y_0, s_0) \in U_T$ with $s_0 < t_0$. Consider $$ r_0 = \min\{s \geq s_0 : u(x,t) = M \text{for all points } (x,t) \in L \, s\leq t \leq t_0\}$$
Since $u$ is continous the minimum is attained. Assume $r_0 > s_0$. Then $u(z_0, r_0) = M$ for some point $(z_0, r_0)$ on $L \cap U_T$ and so $u \equiv M$ on $E(z_0,r_0;r)$ for all sufficently small $r>0$. Since $E(z_0,r_0;r)$ contains $L \cap \{r_0 - \sigma \leq t \leq r_0 + \sigma\}$ for some small $\sigma >0$, we have a contradiction. Thus $r_0 = s_0$ and hence $u \equiv M$ on $L$.
- Now fix any point $x \in U$ and any time $0 \leq t \leq t_0$. There exists points $\{x_0,x_1, \cdots, x_m =x \}$ such that the line segment in $\mathbb{R}^n$ connecting $x_{i-1}$ to $x_i$ lie in $U$ for $i = 1, \cdots, m.$ Select times $t_0 > t_1 > \cdots > t_m =t$. Then the line segments in $\mathbb{R}^{n+1}$ connecting $(x_{i-1}, t_{i-1})$ to $(x_i, t_i)$ lie in $U_T$. According to step 1, $u \equiv M$ on each such line segment and so $u(x,t) = M$
I can understand the general idea of the proof, which use representation formulae, and then shows that if maximum is attained in the parabolic cylinder then solution has to be constant.
What I do not understand here is why do we need $U$ to be bounded and exactly where in the proof we use this. In time I think since we can do for any $t_0 \leq T$ we can take $T$ to be infinite, So I was trying to do something for space space variable also, but was not able to conclude anything