RegularityTheoremForTheLaplaceEquation 2005-01-21 02:20:07 2007-06-02 23:07:10 Theorem Laplace equation % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here {\bf Warning: This entry is still in the process of being written, hence is not yet \PMlinkescapetext{complete}.} 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. {\it 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 \PMlinkid{area of the $k$-dimensional sphere}{4495} 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 $|{\bf x} - {\bf p}| < r/4$. Then we have $$\left| {- 2({\bf x} - {\bf p}) \cdot ({\bf y} - {\bf p}) + |{\bf x} - {\bf p}|^2 \over r^2} \right| \le {2 |({\bf x} - {\bf p}) \cdot ({\bf y} - {\bf p})| \over r^2} + {|{\bf x} - {\bf p}|^2 \over r^2} \le $$ $${2 |{\bf x} - {\bf p}| \> |{\bf y} - {\bf p}| \over r^2} + \left( {|{\bf x} - {\bf p}| \over r} \right)^2 \le 2 \cdot {1 \over 4} + \left( {1 \over 4} \right)^2 = {9 \over 16} < 1.$$ 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 $$\sum_{k = 0}^n |c_k (y)| \> |(x - p)_k| + \sum_{k_1, k_2 = 0}^n |c_{k_1 k_2} (y)| \> |(x - p)_{k_1}| \> |(x - p)_{k_2}| \le {9 \over 16}.$$ 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 $$\sum_{k_1, k_2, \ldots, k_m = 0}^n |c_{k_1 k_2, \cdots k_m} (y)| \> |(x - p)_{k_1}| \> |(x - p)_{k_2}| \cdots |(x - p)_{k_m}| \le \left( {9 \over 16} \right)^m$$ 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 $b_{k_1 k_2, \cdots k_m}$.