## 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 $t\rightarrow +\infty$.