Let $T>0$, $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open and $u\in C([0,T)\times\overline\Omega)\cap C^{1,\:2}((0,T)\times\overline\Omega)$ be nonnegative with $$u=0\;\;\;\text{on }(0,T)\times\partial\Omega.\tag1$$
Now assume that we are in the simple situation, where $\Omega=(-R,R)$ for some $R>0$ and $$u(t,x)=u(t,-x)\;\;\;\text{for all }(t,x)\in(0,T)\times\Omega\tag2.$$
Question 1: Why does it follow that $$\frac{\partial u}{\partial x}(t,0)=0\tag3$$ for all $t\in(0,T)$?
Remark: If the assumptions are not enough to ensure $(3)$, the particular situation I'm interested in also entails the assumption that $$\frac{\partial u}{\partial t}=\Delta_xu+u^p\;\;\;\text{in }(0,T)\times\Omega\tag4$$ for some $p>1$ and $$u(0,\;\cdot\;)=u_0\;\;\;\text{in }\Omega\tag5$$ for some nonnegative $u_0\in C^{1+\alpha}(\overline\Omega)$, $\alpha>0$, satisfying $$u_0'(x)\le0\;\;\;\text{for all }x\in[0,R)\tag6$$ and $$u_0(x)=u_0(-x)\;\;\;\text{for all }x\in\Omega.\tag7$$
EDIT: Ah, isn't the simple argument that $0$ is the maximum value of $u$?
