# Maximum Principles in Laplace Equation

I am currently practising some past year exams questions on intro to PDE in my school and I have problem doing the following:

Let $\Omega$ be a bounded domain and let $u$ satisfies: $-\Delta u+ c(x)u=f(x)$ for $x\in \Omega$ and $u=0$ on $\partial\Omega$

Prove that:

1. If $c(x)\geq c>0$ with a positive constant $c$, then $\max_{\Omega} |u(x)| \leq \frac{1}{c} \sup_{\Omega} |f(x)|$.

2. If $0\leq c(x)\leq d<\infty$ with a constant $d$, then $\max_{\Omega} |u(x)| \leq M\, \sup_{\Omega} |f(x)|$, where $M$ depends only on $d$.

I tried to multiply the equation by $u$ and integrate it, so I get something like $\int_\Omega c(x) u-f(x) u^2\;dx <0$, but I don't think I am getting closer to the ans.

Any help would be greatly appreciated. Thanks!

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For the second question, are you sure you copied the question correctly? I get the feeling that $M$ should at least depend on $\Omega$. – Willie Wong Nov 24 '11 at 15:49
yes I copied the question correctly. – digiboy1 Nov 26 '11 at 14:54
Take $c = 0$, $f \equiv 2$, and set $\Omega_R\subset \mathbb{R}$ to be $[-R,R]$. The solution is $u = (x-R)(x+R)$, which takes its maximum at $R^2$. If $M$ is asserted to be independent on $\Omega$, the above gives a counterexample to the second question. (Note that my argument below shows that for the first question the estimate is indeed independent of $\Omega$.) – Willie Wong Nov 28 '11 at 8:45
In 2. $d$ must depend on the width of $\Omega$. – timur Dec 6 '11 at 5:49

For the first question. The quantity $\max |u|$ is attained either at $\max u > 0$ or $\min u < 0$. In the former case, you have that the Hessian of a function at its maximum must be negative. In the latter case, the Hessian at the minimum must be positive. So, in the first case

$$\sup f(x) \geq f(x_M) = c(x_M) u(x_M) - \triangle u(x_M) \geq c(x_M) u(x_M) \geq c u(x_M) \geq 0$$

using the assumptions. In the second case

$$\inf f(x) \leq f(x_m) = c(x_m) u(x_m) - \triangle u(x_m) \leq c(x_m) u(x_m) \leq c u(x_m) \leq 0$$

Either case leads to the conclusion you seek.

(I'll look at the second one later.)

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