📂Partial Differential Equations

The Mean Value Theorem for Laplace's Equation

Theorem¹

Let us assume that an open set $\Omega \subset \mathbb{R}^{n}$ is given. Also, let’s assume that $u \in C^2(\Omega)$ satisfies the Laplace equation. Then, for each open ball $B(x,r)\subset \subset \Omega$, the following holds.

$$ \begin{align*} u(x) &= \dfrac{1}{n \alpha (n)r^{n-1}} \int _{\partial B(x,r)} udS =: -\!\!\!\!\!\! \int_{\partial B(x,r)} udS \\ &= \dfrac{1}{\alpha (n)r^n}\int_{B(x,r)}udy =: -\!\!\!\!\!\! \int _{B(x,r)} udy \end{align*} $$

• It is denoted as $V\subset \subset U$ when $V \subset \bar V \subset U$ and $\bar V$ are compact.
• $\bar V$ is the closure of $V$
• $B(x,r)=\left\{ y \in \mathbb{R}^n \ \big|\ |y-x|<r \right\}$
• $\partial B(x,r)=\left\{ y \in \mathbb{R}^n \ \big|\ |y-x|=r \right\}$= Boundary of $B(x,r)$
• $- \!\!\!\!\! \int _{\partial B(x,r)} udS =$ The average value of $u$ on the boundary of open ball $B(x,r)$
• $- \!\!\!\!\! \int _{B(x,r)} udy=$ The average value of $u$ in open ball $B(x,r)$

The converse also holds.

Let’s say $u \in C^2(\Omega)$ satisfies the average value property as follows in each open ball $B(x,r) \subset \Omega$.

$$ u(x)=-\!\!\!\!\!\! \int_{\partial B(x,r)} u dS $$

Then, $u$ is harmonic.

Proof

• Part 1. $\int \!\!\!\!\!-_{\partial B(x,r)}u(y)dS(y)=u(x)$

  There is a fixed point $x \in \Omega$. And let’s denote $d_{x}$ as follows.

  $$ d_{x} = \mathrm{dist}(x, \partial \Omega) := \inf \limits_{y \in \partial \Omega} |x-y| >0 $$

  In other words, $d_{x}$ denotes the shortest distance from a point $x$ inside $\Omega$ to the boundary of $\Omega$. And let’s define $\varphi(r)$ as follows.

  $$ \begin{align*} \varphi(r) &:= \dfrac{1}{n \alpha (n)r^{n-1}} \int_{\partial B(x,r)} u(y) dS(y) \quad \mathrm{for}\ 0< r <d_{x} \\ &= -\!\!\!\!\!\!\int_{\partial B(x,r)} u(y)S(y) \end{align*} $$

  Here, the objective is to show that $\varphi(r)$ is independent of $r$, and its value is $u(x)$. Therefore, first, it needs to be shown that $\dfrac{d \phi}{dr}=0$. To obtain the desired result, let’s do a variable transformation to $y=x+rz$. Then $y \in \partial B(x,r)$, $z \in \partial B(0,1)$, and since $dS(y)=r^{n-1}dS(z)$, the following holds.

  $$ \begin{align} \varphi(r) &= \dfrac{1}{n \alpha (n) r^{n-1} } \int_{\partial B(x,r)} u(y)dS(y) \nonumber \\ &= \dfrac{1}{n \alpha (n) r^{n-1} } \int_{\partial B(0,1)} u(x+rz)r^{n-1}dS(z) \nonumber \\ &= \dfrac{1}{n \alpha (n) } \int_{\partial B(0,1)} u(x+rz)dS(z) \label{eq1} \end{align} $$

  If we say $f(r)=g(x+rz)$, the total differential is as follows.

  $$ df=\dfrac{\partial g}{\partial (x_{1}+rz_{1})}d(x_{1}+rz_{1})+\cdots + \dfrac{\partial g}{\partial (x_{1}+rz_{1})}d(x_{1}+rz_{1}) $$

  Then, $\dfrac{d f}{d r}$ is as follows.

  $$ \begin{align*} f^{\prime}(r) &= \dfrac{df(r)}{dr} \\ &= \frac{\partial g}{\partial (x_{1}+rz_{1})}\dfrac{ d(x_{1}+rz_{1})}{dr}+\cdots + \frac{\partial g}{\partial (x_{1}+rz_{1})}\dfrac{ d(x_{1}+rz_{1}) }{dr} \\ &= \frac{\partial g(x+rz)}{\partial (x_{1}+rz_{1})}z_{1}+\cdots + \frac{\partial g(x+rz)}{\partial (x_{1}+rz_{1})}z_{n} \\ &= \left( \frac{\partial g(x+rz)}{\partial (x_{1}+rz_{1})}, \cdots , \frac{\partial g(x+rz)}{\partial (x_{1}+rz_{1})}\right) \cdot(z_{1},\cdots,z_{n}) \\ &= Dg(x+rz)\cdot z \end{align*} $$

  Applying this to $\eqref{eq1}$, we obtain the following.

  $$ \begin{align*} \varphi^{\prime}(r) &= \dfrac{1}{n\alpha (n)} \int_{\partial B(0,1) } Du(x+rz)\cdot zdS(z) \\ &= \dfrac{1}{n\alpha (n)} \int_{\partial B(x,r)} Du(y) \cdot \dfrac{y-x}{r} \dfrac{1}{r^{n-1}}dS(y) \\ &= \dfrac{1}{n\alpha (n)r^{n-1}} \int_{\partial B(x,r)} Du(y)\cdot \dfrac{y-x}{r}dS(y) \\ &= \dfrac{1}{n\alpha (n)r^{n-1}} \int_{\partial B(x,r)} Du(y)\cdot \boldsymbol{\nu} dS(y) \\ &= \dfrac{1}{n\alpha (n)r^{n-1}} \int_{\partial B(x,r)} \dfrac{ \partial u(y)}{\partial \nu}dS(y) \\ &= \dfrac{1}{n \alpha (n) r^{n-1} } \int_{B(x,r)} \Delta u dy \\ &= 0 \end{align*} $$

  Here, $\boldsymbol{\nu}$ is the outward unit normal vector. The fourth and fifth equalities hold by the definition of the outward unit normal vector. $\dfrac{y-x}{r}$ is outward from $\partial B(x,r)$ and its magnitude is $1$, so $\dfrac{y-x}{r}=\boldsymbol{\nu}$. The sixth equality holds by Green’s theorem(i). The last equality holds by the assumption that $u$ satisfies $\Delta u=0$.

  Now, since $\varphi^{\prime}(r)=0$, $\phi (r)$ is constant for $r$, which is $0<r<d_{x}$. Therefore, the following holds.

  $$ \begin{align*} -\!\!\!\!\!\! \int_{\partial B(x,r)} u(y)dS(y) &= \varphi(r) = \lim \limits_{t \rightarrow 0^+} \varphi (t) \\ &= \lim \limits_{t \rightarrow 0^+} \dfrac{1}{n\alpha (n) r^{n-1}}\int _{\partial B(x,t)} u(y)dS(y) \\ &= \lim \limits_{t \rightarrow 0^+} -\!\!\!\!\!\! \int_{\partial B(x,t)} u(y)dS(y)
  \\ &= u(x) \end{align*} $$

  If it is $t \rightarrow 0^+$, the diameter of the ball gradually decreases, so its average value becomes $u(x)$.

• Part 2. $\int\!\!\!\!\!\!- _{B(x,r)} udx=u(x)$

  Let’s assume $x, d_{x}$ is the same as in Part 1. Then, the following holds for $0 < r < d_{x}$.

  $$ \begin{align*} \int_{B(x,r)} u(y)dy &= \int_{0}^r \left( \int_{\partial B(x,s)} u(y)dS(y) \right) ds \\ &= \int_{0}^r n\alpha (n)s^{n-1}\left( \dfrac{1}{n\alpha (n)s^{n-1}}\int_{\partial B(x,s)} u(y)dS(y) \right) ds \\ &= \int_{0}^r n\alpha (n)s^{n-1}\left( -\!\!\!\!\!\! \int_{\partial B(x,s)} u(y)dS(y) \right) ds
  \\ &= \int_{0}^rn\alpha (n)s^{n-1}u(x) ds
  \\ &= n\alpha (n) u(x) \int_{0}^r s^{n-1} ds \\ &= n\alpha (n) u(x) \dfrac{r^n}{n} \\ &= \alpha (n)r^n u(x) \end{align*} $$

  Therefore, if we leave only $u$ on the right side and organize, it is as follows.

  $$ -\!\!\!\!\!\! \int_{B(x,r)} udy = \dfrac{1}{\alpha (n) r^n} \int_{B(x,r)} udy = u(x) $$

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