Reaction diffusion equations with grow up solutions

Last time, I’ve written a paper which considered the following Reaction diffusion equations on \mathbb R^n

(1)            \begin{cases} u_t - \Delta u + f(x,u) + \lambda u = g(t,x)\\ u(0,x) = u_0(x)\end{cases}

Under condition \lambda >0 and suitable assumptions on nonlinearity f and external force g, I’ve been able to prove the existence of a uniform attractor for the corresponding process.

Now I get a question: Why \lambda has to be positive? What happens if \lambda <0.

To answer this question, I should first state that “I need a boundedness uniformly in time t on solutions u(t,x)

Now, I go back to where I need \lambda >0. Multiply (1) by u then integrating over \mathbb R^n, after some computations, we get

\frac{d}{dt}\|u\|^2 + 2\|\nabla u\|^2 + \lambda \|u\|^2 \leq C(1+\|g\|^2).

If \lambda >0, then we easily get the boundness of u(t,x) in L^2 by Gronwall’s inequality, that means, \|u(t,x)\|_{L^2(\mathbb R^n)} \leq C for all t\geq 0.

If \lambda \leq 0 we don’t see any way to do that (at least by standard way).

I also state here a counter-example in one dimensional case: Consider

\begin{cases} u_t - u_{xx} + u - (2\pi^2+1)u = 0,\\ u(t,0) = u(t,1) = 0, \\u(0,x) = \sin(\pi x)\end{cases}

Here we have f(x,u) \equiv u, g(t,x) \equiv 0 and \lambda = -(2\pi^2+1) < 0. It’s easy to check that u(t,x) = e^{\pi^2 t}\sin(\pi x) is a solution of the equation, and

\|u(t)\|_{L^2(0,1)}^2 = e^{2\pi^2t}\int_0^1|\sin(\pi x)|^2dx \rightarrow +\infty as t\rightarrow +\infty.

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 Attractors. 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