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
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