# 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 $f:D\to \mathbb{R}$ is twice differentiable^{} and satisfies Laplace’s equation. Then $f$ has derivatives of all orders and is, in fact analytic.

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

$$f(\mathbf{x})=\frac{1}{{r}^{n-1}A(n-1)}{\int}_{|\mathbf{y}-p|=r}f(\mathbf{y})\frac{{r}^{2}-{|\mathbf{x}-\mathbf{p}|}^{2}}{{|\mathbf{x}-\mathbf{y}|}^{n}}\mathit{d}\mathrm{\Omega}(\mathbf{y})$$ |

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

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

$$\frac{1}{{|\mathbf{x}-\mathbf{y}|}^{n}}$$ |

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

$${\left({|\mathbf{y}-\mathbf{p}|}^{2}-2(\mathbf{x}-\mathbf{p})\cdot (\mathbf{y}-\mathbf{p})+{|\mathbf{x}-\mathbf{p}|}^{2}\right)}^{-\frac{n}{2}}.$$ |

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

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

Assume that $$. Then we have

$$\left|\frac{-2(\mathbf{x}-\mathbf{p})\cdot (\mathbf{y}-\mathbf{p})+{|\mathbf{x}-\mathbf{p}|}^{2}}{{r}^{2}}\right|\le \frac{2|(\mathbf{x}-\mathbf{p})\cdot (\mathbf{y}-\mathbf{p})|}{{r}^{2}}+\frac{{|\mathbf{x}-\mathbf{p}|}^{2}}{{r}^{2}}\le $$ |

$$ |

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

$$\frac{1}{{|\mathbf{x}-\mathbf{y}|}^{n}}=\frac{1}{{r}^{n}}{\left(1+\frac{-2(\mathbf{x}-\mathbf{p})\cdot (\mathbf{y}-\mathbf{p})+{|\mathbf{x}-\mathbf{p}|}^{2}}{{r}^{2}}\right)}^{\frac{n}{2}}=$$ |

$$\sum _{m=0}^{\mathrm{\infty}}\frac{{\left(\frac{n}{2}\right)}^{\underset{\xaf}{m}}}{m!}{\left(\frac{-2(\mathbf{x}-\mathbf{p})\cdot (\mathbf{y}-\mathbf{p})+{|\mathbf{x}-\mathbf{p}|}^{2}}{{r}^{2}}\right)}^{m}$$ |

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

$$\frac{-2(x-p)\cdot (y-p)+{|x-p|}^{2}}{{r}^{2}}=\sum _{k=0}^{n}{c}_{k}(y){(x-p)}_{k}+\sum _{{k}_{1},{k}_{2}=0}^{n}{c}_{{k}_{1}{k}_{2}}(y){(x-p)}_{{k}_{1}}{(x-p)}_{{k}_{2}},$$ |

(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

$$\sum _{k=0}^{n}|{c}_{k}(y)||{(x-p)}_{k}|+\sum _{{k}_{1},{k}_{2}=0}^{n}|{c}_{{k}_{1}{k}_{2}}(y)||{(x-p)}_{{k}_{1}}||{(x-p)}_{{k}_{2}}|\le \frac{9}{16}.$$ |

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

$${\left(\frac{-2(x-p)\cdot (y-p)+{|x-p|}^{2}}{{r}^{2}}\right)}^{m}=\sum _{{k}_{1},{k}_{2},\mathrm{\dots},{k}_{m}=0}^{n}{c}_{{k}_{1}{k}_{2},\mathrm{\cdots}{k}_{m}}(y){(x-p)}_{{k}_{1}}{(x-p)}_{{k}_{2}}\mathrm{\cdots}{(x-p)}_{{k}_{m}},$$ |

then we have

$$\sum _{{k}_{1},{k}_{2},\mathrm{\dots},{k}_{m}=0}^{n}|{c}_{{k}_{1}{k}_{2},\mathrm{\cdots}{k}_{m}}(y)||{(x-p)}_{{k}_{1}}||{(x-p)}_{{k}_{2}}|\mathrm{\cdots}|{(x-p)}_{{k}_{m}}|\le {\left(\frac{9}{16}\right)}^{m}$$ |

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 ${b}_{{k}_{1}{k}_{2},\mathrm{\cdots}{k}_{m}}$ such that the term involving $|{(x-p)}_{{k}_{1}}||{(x-p)}_{{k}_{2}}|\mathrm{\cdots}|{(x-p)}_{{k}_{m}}|$ in the power series is bounded by ${b}_{{k}_{1}{k}_{2},\mathrm{\cdots}{k}_{m}}$.

Title | regularity theorem for the Laplace equation |
---|---|

Canonical name | RegularityTheoremForTheLaplaceEquation |

Date of creation | 2013-03-22 14:57:29 |

Last modified on | 2013-03-22 14:57:29 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 17 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 26B12 |