# Reviewing the regularity theory of elliptics PDEs via the Laplace equation [Part III]

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Let us recall Green’s identity, if ${u,v}$ are any functions smooth in ${\bar{\Omega}}$ and ${\Omega}$ is a bounded domain with smooth boundary we have

$\displaystyle \int_\Omega u\Delta v - v \Delta u dx = \int_{\partial \Omega}u\frac{\partial v}{\partial \nu}-v\frac{\partial u}{\partial \nu}dS$

this identity can be obtained with a couple of integration by parts involving the vector fields ${u \nabla v}$ and ${v \nabla u}$.

Lets rewrite the identity as

$\displaystyle \int_\Omega u\Delta vdx = -\int_\Omega v \Delta u dx + \int_{\partial \Omega}u\frac{\partial v}{\partial \nu}-v\frac{\partial u}{\partial \nu}dS$

thus, at least formally, if somehow we could find for every ${x \in \Omega}$ a function ${v_x(y)}$ such that

$\displaystyle \Delta v_x(y)=\delta_x(y)$

$\displaystyle v_x(y) \equiv 0 \mbox{ on } \partial \Omega$

then Green’s identity applied to both ${u}$ and ${v_x}$ in ${\Omega}$ would give us an integral representation formula for harmonic functions

$\displaystyle u(x)=\int_{\partial \Omega} u(y)\frac{\partial v_x(y)}{\partial \nu}dS(Y)$

This can actually be carried out rigorously for all reasonable domains ${\Omega}$, but only in a few cases however, we know a useful expression for the functions ${\left \{ v_x(y) \right \}_x}$. Happily for us, whenever ${\Omega}$ is a ball there is a simple expression, and so, for any function ${u}$ harmonic in a neighborhood of a ball ${B_r(0)}$ we have

Theorem 1

Let ${u}$ be a smooth harmonic function in some neighborhood of the ball ${B_r(0)}$, and let the dimension be ${n\geq2}$, then

$\displaystyle u(x)=\int_{\partial B_r}\frac{1}{n\alpha_n} \frac{r^2-x^2}{|x-y|^n}u(y)dS(y) \;\;\;\; \forall x\in B_r(0)$

Here ${\alpha_n}$ is just the surface area of the ${n-1}$ dimensional sphere, I will not derive it as this can be found in most introductory PDE textbooks (for instance, Evan’s book), suffice it to say that one needs to use the symmetries of Laplace’s equation (in particular under inversions) to manipulate the fundamental solution ${V(x)=|x|^{-n-2}}$ (${n \geq 3}$) and build the function ${v_x(y)}$.

Note that this integral representation for ${u}$ gives us as a corollary the mean value property (just take ${x}$ equal to the center of the ball, in this case ${0}$), and it tells us that for other points ${x \in B_r}$, the value ${u(x)}$ is obtained via a weighted average of ${u}$ on the sphere, the average being balanced according to the position of ${x}$ with respect to the sphere. By the way, this weight function

$\displaystyle K(x,y)=\frac{1}{n\alpha_n}\frac{r^2-|x|^2}{|x-y|^n}, \;\;\;\;\; x \in B_r,\;\;\; y \in \partial B_r$

is known as the Poisson kernel (for the ball). This representation formula also gives us directly another proof of Harnack’s inequality. It tells us even more, it says that if ${u}$ is continuous up to the boundary of ${B_r}$, then ${u}$ is Lipscthiz on the boundary, which can be seen by just looking at the term ${r^2-|x|^2}$. Finally, by differentiating the right hand side of the integral representation formula, we prove again the a priori estimate for the gradient I discussed last time:

$\displaystyle \nabla u(x)=\frac{1}{n\alpha_n}\int_{\partial B_r}\frac{-2x|x-y|^2-(r^2-|x|^2)n(x-y)}{|x-y|^{n+2}}u(y)dy$

$\displaystyle \nabla u(0)=\frac{1}{n\alpha_n}\int_{\partial B_r}\frac{nr^2y}{|y|^{n+2}}u(y)dy=\frac{1}{r}\frac{1}{\alpha_nr^{n-1}}\int_{\partial B_r} y u(y)dy$

taking absolute values we get ${|\nabla u(0)|\leq \frac{1}{r} \frac{1}{\alpha_n r^{n-1}}\int_{\partial B_r}|u(y)|dy}$, so the gradient is bounded by ${\frac{1}{r}}$ times the average of ${u}$ on the sphere of radius ${r}$, which is actually a stronger estimate than what I proved last time, for the average is bounded by the supremum of ${u}$, thus

$\displaystyle |\nabla u(0)|\leq \frac{1}{r}\sup \limits_{\partial B_r} |u|$

(notice I had not specified the constant ${C_n}$ last time I wrote this estimate, here we see that we may take ${C_n=1}$.)