Bound of solution of pde with a distance function

Question

I would like to solve the PDE $\Delta u=-K$ in $\Omega$ and $u=0$ on the boundary, where $K$ is some positive constant. I read a paper which stated that $u(x)$ is no less than the distance from $x$ to the boundary of $\Omega$. How can I obtain this result? Is it by a Schauder estimate? Could you please give me some details?

Answer 1

Let $d(x) = \mathrm{dist} (x, \partial \Omega)$. We want $u \geq \alpha d$ for some small $\alpha(\Omega,K) > 0$. By maximum principle, it is enough to show this on some neighborhood of $\partial \Omega$.

Consider the neighborhood $\Gamma_{\mu} = \{ x \in \Omega : d(x) < \mu\}$. For $\mu > 0$ small enough we have $d \in C^2(\Gamma_{\mu})$. Choose a cut-off function $\eta \in C^2(\Omega)$ such that $\eta = 1$ on $\Gamma_{\mu/2}$ and $\eta = 0$ outside of $\Gamma_\mu$. Then $- \Delta (\eta d) \leq C$ for some constant $C$. (See e.g. Elliptic partial differential equations of second order by Gilbarg and Trudinger).

Now choose any $0 < \alpha < K/C$, then the function $v = u - \alpha \eta d$ satisfies $- \Delta v \geq K-\alpha C > 0$ in $\Omega$. Since $v = 0$ on $\partial \Omega$, by maximum principle $v \ge 0$ in $\Omega$. This proves that $u \geq \alpha d$ on $\Gamma_{\mu/2}$.

Answer 2

This is in fact quite elementary: if $\Omega$ is smooth, then there is $r > 0$ such that at every boundary point $y$ of $\Omega$ there is a tangent ball $B(z, r)$ of radius $r$ which is contained in $\Omega$. Observe that $v(x) = \tfrac{K}{2n} (r^2-|x-z|^2)_+$ satisfies $\Delta v(x) = -K$ in $B(z, r)$ and $v(x) = 0$ in $\Omega \setminus B(z, r)$. Hence, $u - v$ is continuous in $\Omega$, harmonic in $B(z, r)$ and nonnegative in $\Omega \setminus B(z, r)$. By the maximum principle, $u - v$ is nonnegative in $\Omega$, and so $u \geqslant v$ in $B(z, r)$.

If $x \in \Omega$ and $d(x, \partial \Omega) < r$, then choose $y \in \partial \Omega$ so that $|x - y| = r$, and consider a ball $B(z, r)$ as described above. Then $$u(x) \geqslant v(x) = r^2 - |x - z|^2.$$ Smoothness of $\partial \Omega$ implies that $x, y, z$ are co-linear, and $|x - z| = r - d(x, \partial \Omega)$. (A formal proof of this statement requires a bit of work, but to convince oneself, it is enough to make a picture). We conclude that $$u(x) \geqslant r^2 - (r - d(x, \partial \Omega))^2 = d(x, \partial \Omega) (2 r - d(x, \partial \Omega)) \geqslant r d(x, \partial \Omega).$$