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.

About baotangquoc

Lecturer School of Applied Mathematics and Informatics Hanoi University of Science and Technology No 1, Dai Co Viet Street, Hanoi
This entry was posted in Entropy. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s