# regularity theorem for the Laplace equation

Warning: This entry is still in the process of being written, hence is not yet .

Let $D$ be an open subset of $\mathbb{R}^{n}$. Suppose that $f\colon D\to\mathbb{R}$ is twice differentiable and satisfies Laplace’s equation. Then $f$ has derivatives of all orders and is, in fact analytic.

Proof:   Let ${\bf p}$ be any point of $D$. We shall show that $f$ is analytic at ${\bf p}$. Since $D$ is an open set, there must exist a real number $r>0$ such that the closed ball of radius $r$ about ${\bf p}$ lies inside of $D$.

Since $f$ satisfies Laplace’s equation, we can express the value of $f$ inside this ball in terms of its values on the boundary of the ball by using Poisson’s formula:

 $f({\bf x})={1\over r^{n-1}A(n-1)}\int_{|{\bf y}-p|=r}f({\bf y}){r^{2}-|{\bf x}% -{\bf p}|^{2}\over|{\bf x}-{\bf y}|^{n}}\,d\Omega({\bf y})$

Here, $A(k)$ denotes the http://planetmath.org/node/4495area of the $k$-dimensional sphere and $d\Omega$ denotes the measure on the sphere of radius $r$ about ${\bf p}$.

We shall show that $f$ is analytic by deriving a convergent power series for $f$. From this, it will automatically follow that $f$ has derivatives of all orders, so a separate proof of this fact will not be necessary.

Since this involves manipulating power series in several variables, we shall make use of multi-index notation to keep the equations from becoming unnecessarily complicated and drowning in a plethora of indices.

First, note that since $f$ is assumed to be twice differentiable in $D$, it is continuous in $D$ and, hence, since the sphere of radius $r$ about $s$ is compact, it attains a maximum on this sphere. Let us denote this maxmum by $M$. Next, let us consider the quantity

 ${1\over|{\bf x}-{\bf y}|^{n}}$

which appears in the integral. We may write this quantity more explicitly as

 $\left(|{\bf y}-{\bf p}|^{2}-2({\bf x}-{\bf p})\cdot({\bf y}-{\bf p})+|{\bf x}-% {\bf p}|^{2}\right)^{-{n\over 2}}.$

Since the values of the variable $y$ has been restricted by the condition $|{\bf y}-{\bf p}|=r$, we may rewrite this as

 ${1\over r^{n}}\left(1+{-2({\bf x}-{\bf p})\cdot({\bf y}-{\bf p})+|{\bf x}-{\bf p% }|^{2}\over r^{2}}\right)^{-{n\over 2}}.$

Assume that $|{\bf x}-{\bf p}|. Then we have

 $\left|{-2({\bf x}-{\bf p})\cdot({\bf y}-{\bf p})+|{\bf x}-{\bf p}|^{2}\over r^% {2}}\right|\leq{2|({\bf x}-{\bf p})\cdot({\bf y}-{\bf p})|\over r^{2}}+{|{\bf x% }-{\bf p}|^{2}\over r^{2}}\leq$
 ${2|{\bf x}-{\bf p}|\>|{\bf y}-{\bf p}|\over r^{2}}+\left({|{\bf x}-{\bf p}|% \over r}\right)^{2}\leq 2\cdot{1\over 4}+\left({1\over 4}\right)^{2}={9\over 1% 6}<1.$

Since this absolute value is less than one, we may apply the binomial theorem to obtain the series

 ${1\over|{\bf x}-{\bf y}|^{n}}={1\over r^{n}}\left(1+{-2({\bf x}-{\bf p})\cdot(% {\bf y}-{\bf p})+|{\bf x}-{\bf p}|^{2}\over r^{2}}\right)^{n\over 2}=$
 $\sum_{m=0}^{\infty}{\left({n\over 2}\right)^{\underline{m}}\over m!}\left({-2(% {\bf x}-{\bf p})\cdot({\bf y}-{\bf p})+|{\bf x}-{\bf p}|^{2}\over r^{2}}\right% )^{m}$

Note that each term in this sum is a polynomial in $x-p$. The powers of the various components of $x-p$ that appear in the $m$-th term range between $m$ and $2m$. Moreover, let us note that we can strengthen the assertion used to show that the binomial series converged by inserting absolute value bars. If we write

 ${-2(x-p)\cdot(y-p)+|x-p|^{2}\over r^{2}}=\sum_{k=0}^{n}c_{k}(y)\>(x-p)_{k}+% \sum_{k_{1},k_{2}=0}^{n}c_{k_{1}k_{2}}(y)\>(x-p)_{k_{1}}(x-p)_{k_{2}},$

(actually, the coefficients $c_{k_{1}k_{2}}$ depend on $y$ trivially, but the dependence on $y$ has been indicated for the sake of uniformity) then

 $\sum_{k=0}^{n}|c_{k}(y)|\>|(x-p)_{k}|+\sum_{k_{1},k_{2}=0}^{n}|c_{k_{1}k_{2}}(% y)|\>|(x-p)_{k_{1}}|\>|(x-p)_{k_{2}}|\leq{9\over 16}.$

Raising this to the $m$-th power, we see that, if we define

 $\left({-2(x-p)\cdot(y-p)+|x-p|^{2}\over r^{2}}\right)^{m}=\sum_{k_{1},k_{2},% \ldots,k_{m}=0}^{n}c_{k_{1}k_{2},\cdots k_{m}}(y)(x-p)_{k_{1}}(x-p)_{k_{2}}% \cdots(x-p)_{k_{m}},$

then we have

 $\sum_{k_{1},k_{2},\ldots,k_{m}=0}^{n}|c_{k_{1}k_{2},\cdots k_{m}}(y)|\>|(x-p)_% {k_{1}}|\>|(x-p)_{k_{2}}|\cdots|(x-p)_{k_{m}}|\leq\left({9\over 16}\right)^{m}$

Because of the fact that one may freely rearrange and regroup the terms in an absolutely convegent series, we may conclude that the expansion of $|x-y|^{-n}$ in powers of $x-p$ converges absolutely. Furthermore, there exist constants $b_{k_{1}k_{2},\cdots k_{m}}$ such that the term involving $|(x-p)_{k_{1}}|\>|(x-p)_{k_{2}}|\cdots|(x-p)_{k_{m}}|$ in the power series is bounded by $b_{k_{1}k_{2},\cdots k_{m}}$.

Title regularity theorem for the Laplace equation RegularityTheoremForTheLaplaceEquation 2013-03-22 14:57:29 2013-03-22 14:57:29 rspuzio (6075) rspuzio (6075) 17 rspuzio (6075) Theorem msc 26B12