# $L^\infty$ bound on solutions of linear parabolic equations

We work on a closed Riemannian manifold $M$. Let $u$ and $v$ be the non-negative weak solutions of $$au_t - 2d\,\Delta au = cv - f$$ $$bv_t - d\,\Delta bv = f$$ $$u(0)=u_0, \quad v(0)=v_0$$ where $f$ is a given non-negative function in $L^2(0,T;H^1(M))$ and $a$, $b$ , $c$ and $d$ are positive constants, and $u_0$ and $v_0$ are given non-negative functions that are bounded on $M$.

These solutions lie in the space $L^2(0,T;H^1(M))$ with time derivative in $L^2(0,T;L^2(M))$.

Are $u$ and $v$ in $L^\infty(0,T;L^\infty(M))$?

I'd like to prove this using weak/variational techniques and not recourse to more "classical" theory if possible.

Note that $f$ is not bounded. The problem is that the diffusion coefficients are not equal. If they were, we can add the two equations and test with $(au+bv-C)^+$ where $C$ depends on the $L^\infty$ bounds on the initial data. (The term $cv$ on the right hand side of the first equation can be neglected when we add the equations because we can rescale $v$ by an expontential function and so get rid of it).

• What is $\dim M$? – Fan Zheng Feb 6 '16 at 22:08
• @FanZheng Just suppose it is an embedded hypersurface in $\mathbb{R}^{n+1}$ $n \geq 2$. – Kies Feb 6 '16 at 23:50
• so you mean $L^\infty(M)$ for a fixed $t>0$, with the constant blowing up when $t\to0$? – Fan Zheng Feb 7 '16 at 1:01
• @FanZheng Oops sorry, no, I mean they should be bounded in space and time. I edited the post. – Kies Feb 7 '16 at 9:25
• also $f$ is non-negative. – Kies Feb 7 '16 at 11:43