A priori estimates to $u_t - \Delta u = u^2$ [closed]

Question

My research is now considering the a priori estimates on the equation $$ \begin{cases}u_t - \Delta u = u \min(u,c) \\ u(0,y) = u(1,y)\\ u_x(0,y) = u_x(1,y)\\ \partial_n u(x,0) = \partial_n u(x,1) = 0 \end{cases}$$ on a 2-dimensional periodic domain $[0, 1]^2$, where $c > 0$ is some fixed given constant. I try to do inner product of the whole equation with $u$: $$ \frac{1}{2} \frac{d}{dt}|u|_{L^2}^2 + \|u\|^2_{H^1} = \langle u\min(u,c) , u \rangle. $$ It seems that I have no way to bound the right hand side in the following form: $$ \bigg|\int u^2\min(u,c)\bigg| \leq \epsilon \|u\|^2_{H^1} + C|u|^2_{L^2} + C,$$ because the integral looks like $\int |u|^3 = |u|_{L^3}^3$ and it has exponent to the 3; while the RHS has only 2. Is there a way to fix it?

Answer 1

I figured out a way to show it, basically by separating the space and consider where $A^+ := \{u>0\}$ and $A^- := \{u<0\}$. Then the RHS reads \begin{equation} \int u^2 \min(u,c) = \int_{\{u\geq 0\}} + \int_{\{u<0\}} \\ = (I) + (II) \end{equation} $(I)$ term is a good term because when $u \geq 0$, $|u^2\min(u, c)| \leq c|u|^2$ and so $|(I)| \leq c|u|^2_{L^2}$. $(II)$ term looks bad but notice that the integrand $u^3$ is negative when $u < 0$, and so we can safely move it to the left: \begin{equation} \frac{1}{2} \frac{d}{dt} |u|_{L^2}^2 + \|u\|_{H^1}^2 - (II) \leq c|u|_{L^2}^2, \text{where } (II) < 0. \end{equation} In particular, \begin{equation} \frac{1}{2} \frac{d}{dt} |u|_{L^2}^2 \leq c|u|_{L^2}^2. \end{equation} We are good with the $L_t^\infty(0,T; L_x^2)$ bound for $u$.