MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 fixed spelling and grammar in title.

# Does reactiondiffusionequaitonwithfocusingnonlinearitythere exist a global solution in L^2 forreactiondiffusionequationwithfocusingnonlinearity?

2 added 56 characters in body

It is known that the solution of equation $$u_t - \triangle u = \kappa|u|^{\sigma}u, u(0) = u_0$$ blow up in finite times if $\sigma > 0$. That is, the $L^{\infty}$ norm of solution $u$ will goes to $\infty$ as $t$ goes to $t_0$ for some finite time $t_0$. My question is what happens if we consider $L^2$ norm instead of $L^\infty$? Dose the $L^2$ norm of solution also blows up in finite time?

In my opinion, it may be exist a global solution in $L^2$. We consider $\sigma = \frac{4}{N}$ for brief. It is known that the above problem is local well posed in $L^2$ under this growing condition. On the other hand, multiplying the equation by $u$ and integrating on $R^N$ we find $$\frac{1}{2}\frac{d}{dt}\int frac{1}{2}\frac{d}{dt} \|u\|^2 + \|\nabla u\|^2 = \int |u|^{2 + \frac{4}{N}}. frac{4}{N}}$$ where $\|\cdot\|$ denotes the norm of $L^2$. And since the Lieb-Thirring inequality $\int |u|^{2 + \frac{4}{N}} \leq C \|\nabla u\|^2$, which implies that $\|u\| \leq \|u_0\|$ for $t \leq T$ if we choose $\kappa$ small enough. That is , the $L^2$ norm of solution $u$ is finite for any given initial data $u_0$.

Is my conclusion correct?

1

# Does reaction diffusion equaiton with focusing nonlinearity exist a global solution in L^2?

It is known that the solution of equation $$u_t - \triangle u = \kappa|u|^{\sigma}u, u(0) = u_0$$ blow up in finite times if $\sigma > 0$. That is, the $L^{\infty}$ norm of solution $u$ will goes to $\infty$ as $t$ goes to $t_0$ for some finite time $t_0$. My question is what happens if we consider $L^2$ norm instead of $L^\infty$? Dose the $L^2$ norm of solution also blows up in finite time?

In my opinion, it may be exist a global solution in $L^2$. We consider $\sigma = \frac{4}{N}$ for brief. It is known that the above problem is local well posed in $L^2$ under this growing condition. On the other hand, multiplying the equation by $u$ and integrating on $R^N$ we find $$\frac{1}{2}\frac{d}{dt}\int \|u\|^2 + \|\nabla u\|^2 = \int |u|^{2 + \frac{4}{N}}.$$ And since the Lieb-Thirring inequality $\int |u|^{2 + \frac{4}{N}} \leq C \|\nabla u\|^2$, which implies that $\|u\| \leq \|u_0\|$ for $t \leq T$ if we choose $\kappa$ small enough. That is , the $L^2$ norm is finite for any given initial data $u_0$.

Is my conclusion correct?