## Control the boundary value

Consider the problem

$u_t - Lu = f(u)$                  (1)

in an open, convex, bounded domain $\Omega$ with the boundary condition

$\alpha \partial u/\partial \nu + \beta u = h(t,x)$ on $\partial \Omega$

and intial data $u(0,x) = u_0(x)$.

We assume that $\alpha, \beta, h(t,x), u_0(x) \geq 0$.

By maximum principle, we know that if $u(t,x)$ is a classical solution to (1) then $u(t,x)\geq 0$ for all $t\geq 0, x\in \Omega$.

Sometimes we need to divide by $u(t,x)$. This leads to the a difficulty that $u(t,x)$ can be zero some where. The following lemma gives us a very useful result to avoid zero-value of $u$.

Lemma. (Moving planes)

Assume that $u$ is a solution to (1). There is a compact subdomain $\Omega_0$ of $\Omega$ and a positive constant $\gamma >0$ such that

$\int_\Omega u(t,x)dx \leq (\gamma +1)\int_{\Omega_0}u(t,x)dx$ for all $t\geq 0$.

Moreover, if $f$ is a increasing function, then

$\int_\Omega f(u(t,x))dx \leq (\gamma +1)\int_{\Omega_0}f(u(t,x))$ for all $t\geq 0$.

Proof: It’s too long. May be given in another post.

Remark: By maximum principle, $u(t,x)$ can be zero only on $\partial \Omega$. This implies that

$0.

Hence, we can divide $u(t,x)$ in the domain $\Omega_0$.