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<m\leq \min_{x\in\Omega_0} u(t,x).

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

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About baotangquoc

Lecturer School of Applied Mathematics and Informatics Hanoi University of Science and Technology No 1, Dai Co Viet Street, Hanoi
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