Ngô Quốc Anh

November 9, 2010

Characterization of biharmonic functions

Filed under: Uncategorized — Ngô Quốc Anh @ 16:30

It is well-known that if $B_R(x)$ is a ball with center $x$ and radius $R$ which is completely contained in the open set $\Omega\subset\mathbb R^n$, then the value $u(x)$ of a harmonic function (i.e. $\Delta u =0$) $u :\Omega\to\mathbb R$ at the center of the ball is given by the average value of $u$ on the surface of the ball; this average value is also equal to the average value of $u$ in the interior of the ball. In other words $\displaystyle u(x) = \frac{1}{n\omega_n R^{n-1}}\int_{\partial B_R(x)} ud\sigma = \frac{1}{\omega_n R^n}\int_{B_R(x)} u(y)dy$

where $\omega_n$ is the volume of the unit ball in $n$ dimensions and $\sigma$ is the $n-1$ dimensional surface measure.

For a biharmonic function $u$ (i.e. $\Delta^2u=0$) we have a similar result due to Pizzetti.

Theorem. For any $n \in \mathbb N$, any solution $u$ of $\Delta^2 u =0$ in $B_R(x) \subset \mathbb R^n$

satisfies $\displaystyle u(x)-\frac{1}{\omega_n R^n}\int_{B_R(x)} u(y)dy=\frac{R^2}{2(n+2)}\Delta u(x)$.

Proof. We may assume that $B_R(x)=B_R(0)=B_R$. For $0 let $G_r$ be the fundamental solution of the operator $\Delta^2$ on $B_r$ satisfying $G_r=\Delta G_r=0$ on $\partial B_r$.

Note that $\displaystyle G_r(x)=\frac{1}{r^{n-4}}G_1\left(\frac{x}{r}\right)$.

If $n=4$ we have $\displaystyle G_r(x)=c_0\left(\log\frac{r}{|x|}-\frac{r^2-|x|^2}{4r^2}\right)$.

Applying the mean value formula to the harmonic function $\Delta u$, for some constants $c_1, c_2$ we have $\displaystyle\begin{gathered} 0 = \int_{{B_r}} {{G_r}{\Delta ^2}udx} \hfill \\ \quad= u(0) + \int_{\partial {B_r}} {\left( {\frac{\partial }{{\partial n}}{G_r}\Delta u + \frac{\partial }{{\partial n}}\Delta {G_r}u} \right)d\sigma } \hfill \\ \quad= u(0) - \frac{1}{{n{\omega _n}{r^{n - 1}}}}\int_{\partial {B_r}} {({c_1}{r^2}\Delta u + {c_2}u)d\sigma } \hfill \\ \quad= u(0) - {c_1}{r^2}\Delta u(0) - \frac{{{c_2}}}{{n{\omega _n}{r^{n - 1}}}}\int_{\partial {B_r}} {ud\sigma } \hfill \\ \end{gathered}$

that is, for some constants $c_3, c_4$, $\displaystyle n{r^{n - 1}}u(0) = {c_3}{r^{n + 1}}\Delta u(0) + {c_4}\int_{\partial {B_r}} {ud\sigma }$.

Integrating over $0 and dividing by $R^n$ we obtain the identity $\displaystyle u(0) = {c_5}{R^2}\Delta u(0) + \frac{{{c_6}}}{{{\omega _n}{R^n}}}\int_{{B_R}} {udx}$

with uniform constants $c_5, c_6$ for all biharmonic functions $u$ on $B_R$. Inserting a harmonic function $u$, we obtain the value $c_6=1$, whereas the choice $u(x)=|x|^2$ yields $c_5=\frac{1}{2(n+2)}$.

I haven’t seen such a characterization of triharmonic functions, say function $u$ such that $\Delta^3 u=0$.