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[parent] regularity theorem for the Laplace equation (Theorem)

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

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:

$\displaystyle f({\bf x}) = {1 \over r^{n-1} A(n-1)} \int_{\vert{\bf y} - p\vert... ...bf x} - {\bf p}\vert^2 \over \vert{\bf x} - {\bf y}\vert^n} \, d\Omega({\bf y})$
Here, $ A(k)$ denotes the area 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

$\displaystyle {1 \over \vert{\bf x} - {\bf y}\vert^n}$
which appears in the integral. We may write this quantity more explicitly as
$\displaystyle \left( \vert{\bf y} - {\bf p}\vert^2 - 2 ({\bf x} - {\bf p}) \cdot ({\bf y} - {\bf p}) + \vert{\bf x} - {\bf p}\vert^2 \right)^{-{n \over 2}}.$
Since the values of the variable $ y$ has been restricted by the condition $ \vert{\bf y} - {\bf p}\vert = r$, we may rewrite this as
$\displaystyle {1 \over r^n} \left( 1 + {- 2({\bf x} - {\bf p}) \cdot ({\bf y} - {\bf p}) + \vert{\bf x} - {\bf p}\vert^2 \over r^2} \right)^{-{n \over 2}}.$

Assume that $ \vert{\bf x} - {\bf p}\vert < r/4$. Then we have

$\displaystyle \left\vert {- 2({\bf x} - {\bf p}) \cdot ({\bf y} - {\bf p}) + \v... ... y} - {\bf p})\vert \over r^2} + {\vert{\bf x} - {\bf p}\vert^2 \over r^2} \le $
$\displaystyle {2 \vert{\bf x} - {\bf p}\vert \> \vert{\bf y} - {\bf p}\vert \ov... ...t)^2 \le 2 \cdot {1 \over 4} + \left( {1 \over 4} \right)^2 = {9 \over 16} < 1.$

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

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

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

$\displaystyle {- 2(x - p) \cdot (y - p) + \vert x - p\vert^2 \over r^2} = \sum_... ... - p)_k + \sum_{k_1, k_2 = 0}^n c_{k_1 k_2} (y) \> (x - p)_{k_1} (x - p)_{k_2},$
(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
$\displaystyle \sum_{k = 0}^n \vert c_k (y)\vert \> \vert(x - p)_k\vert + \sum_{... ...y)\vert \> \vert(x - p)_{k_1}\vert \> \vert(x - p)_{k_2}\vert \le {9 \over 16}.$
Raising this to the $ m$-th power, we see that, if we define
$\displaystyle \left( {- 2(x - p) \cdot (y - p) + \vert x - p\vert^2 \over r^2} ... ...n c_{k_1 k_2, \cdots k_m} (y) (x - p)_{k_1} (x - p)_{k_2} \cdots (x - p)_{k_m},$
then we have
$\displaystyle \sum_{k_1, k_2, \ldots, k_m = 0}^n \vert c_{k_1 k_2, \cdots k_m} ... ... p)_{k_2}\vert \cdots \vert(x - p)_{k_m}\vert \le \left( {9 \over 16} \right)^m$
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 $ \vert x - y\vert^{-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 $ \vert(x - p)_{k_1}\vert \> \vert(x - p)_{k_2}\vert \cdots \vert(x - p)_{k_m}\vert$ in the power series is bounded by $ b_{k_1 k_2, \cdots k_m}$.



"regularity theorem for the Laplace equation" is owned by rspuzio. [ full author list (2) ]
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Cross-references: bounded, converges absolutely, uniformity, coefficients, binomial, range, components, powers, polynomial, sum, series, binomial theorem, absolute value, restricted, integral, compact, continuous, indices, equations, multi-index notation, variables, necessary, power series, convergent, sphere, measure, Poisson's formula, boundary, terms, radius, closed ball, real number, point, proof, analytic, orders, derivatives, Laplace's equation, differentiable, open subset
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This is version 14 of regularity theorem for the Laplace equation, born on 2005-01-21, modified 2007-06-02.
Object id is 6655, canonical name is RegularityTheoremForTheLaplaceEquation.
Accessed 2546 times total.

Classification:
AMS MSC: 26B12 (Real functions :: Functions of several variables :: Calculus of vector functions)
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question for the regularity theorem by mathforever on 2005-01-23 11:53:10
When I first saw the formulation of the theorem I was quite suprised. Ussually one considers Laplace equation together with boundary conditions. Then in this situation the following factors influence the regularity of solution, namely:

- regularity of the boundary
- regularity of the boundary data

and all of this are missing here. So, am I understanding right, that INDEPENDANTLY of the boudary conditions, and INDEPENDANTLY of the boundary (it could be even not Lipshitz continuous), the function sattisfing Laplace equation inside domain is ANALYTIC? Is it just a feature Laplace equation, or the same is true for some other PDE (partial diff.eq.), like Helmholz equation?

Thanks in advance.
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(hopefully not infinite) descent by rspuzio on 2005-01-22 16:02:06
A good part of the reason why this entry is taking so long to complete is that I keep finding myself backtracking recursively to add background material. Since it makes an interesting and amusing story and comments on the state of the website, I think I'll relate this tale of recursive descent. (It's nothing too original since I'm sure most of you could tell similar stories.) To begin with, I wrote the current entry while revising my proof of the Riemann mapping theorem when I realized that the proof used the regularity theorem, which had had not yet been entered in the encyclopedia. So I went off to write this entry. In the course of writing it, I used the Poisson formula, which is also not defined. Oh, but that uses the measure on the sphere, which isn't documented either. Hence, I find myself about to add an entry on something as basic as solid angle and measure on the sphere!

Don't get me wrong. This isn't a complaint. I'm not at all upset, just a bit amused. As I see it, this is an opportunity to fill in some basic gaps. If it means taking a week to finish the original proof so be it, because when I'm done not only will I have a really nice proof, but will have added a bunch of related material which was somehow overlooked.

I guess the moral of this story is something like this. A few days ago, Aaron congratulated us on 4000 entries. That certainly is an achievement to be proud of. Only let's be careful not to rest on our laurels! It'll probably take another 4000 entries to fill all the little gaps in the knowledge base so that one could type in a topic at, say an advanced graduate level, and be 99% sure that all the more basic terms, definitions, and theorems it refers to have already been entered. To the best of my knowledge, Planet Math is already the most comprehensive collection of math on the web (and, on top of that, it's GNU free, so you don't have to worry about a publishing company suing your pants off) and, when we get to the point of 99% completeness for standard topics, I can easily see it becoming a standard reference like Math reviews --- if you need to track down a reference, you go to Math Reviews, whilst if its a definition or a theorem that you need, you go to Planet Math.
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