Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.2

1. If $n=2$ and $a=1$, show that (44) is equivalent to
  $\displaystyle u(r,\theta)=\frac{1-r^2}{2\pi}\int_0^{2\pi}\!\frac{g(\phi)}{1+r^2-2r\cos(\theta-\phi)}\,\mathrm d\phi$.
2. Use (a) and Exercise 1 in Section 4.1 to verify the formula
  $\displaystyle r^k\cos k\theta=\frac{1-r^2}{2\pi}\int_0^{2\pi}\!\frac{\cos(k\phi)}{1+r^2-2r\cos(\theta-\phi)}\,\mathrm d\phi$,
  where $k$ is an integer and $0\leq r<1$.

Solution

When $n=2$ and $a=1$, the scircumference of the unit disk is $2\pi$, and (44) becomes
$\displaystyle u(\xi_1,\xi_2)=\frac{1-(\xi_1^2+\xi_2^2)}{2\pi}\int_{x^2+y^2=1}\!\frac{g(x,y)}{(x-\xi_1)^2+(y-\xi_2)^2}\,\mathrm dS_{x,y}.$
Using the polar coordinate $x=\cos\phi$ and $y=\sin\phi$ for $0\leq\phi<2\pi$; $\xi_1=r\cos\theta$ and $\xi_2=r\sin\theta$ for $0\leq r\leq1$ and $0\leq\theta<2\pi$, we define $\tilde u(r,\theta)=u(r\cos\theta,r\sin\theta)=u(\xi_1,\xi_2)$ and $\tilde g(\phi)=g(\cos\phi,\sin\phi)=g(x,y)$. Then we have $\mathrm dS_{x,y}=\mathrm d\phi$ and
$\begin{aligned}\tilde u(r,\theta)&=\frac{1-r^2}{2\pi}\int_0^{2\pi}\!\frac{\tilde g(\phi)}{(\cos\phi-r\cos\theta)^2+(\sin\phi-r\sin\theta)^2}\,\mathrm d\phi\\&=\frac{1-r^2}{2\pi}\int_0^{2\pi}\!\frac{\tilde g(\phi)}{1+r^2-2r\cos\theta\cos\phi-2r\sin\theta\sin\phi}\,\mathrm d\phi\\&=\frac{1-r^2}{2\pi}\int_0^{2\pi}\!\frac{\tilde g(\phi)}{1+r^2-2r\cos(\theta-\phi)}\,\mathrm d\phi.\end{aligned}$
The proof is complete.
Consider the following Dirichlet problem
$\left\{\begin{aligned} &\Delta u=0\quad\text{in}~B_1(0),\\&u(1,\theta)=\cos(k\theta)\quad\text{for}~0\leq\theta<2\pi.\end{aligned}\right.$
Then using Exercise 1 in Section 4.1, we find
$\begin{aligned}u(r,\theta)&=\frac1{2\pi}\int_0^{2\pi}\!g(s)\,\mathrm ds+\frac1\pi\sum_{n=1}^\infty r^n\left[\cos(n\theta)\int_0^{2\pi}\!g(s)\cos(ns)\,\mathrm ds+\sin(n\theta)\int_0^{2\pi}\!g(s)\sin(ns)\,\mathrm ds\right]\\&=\frac1{2\pi}\int_0^{2\pi}\!\cos(ks)\,\mathrm ds+\frac1\pi\sum_{n=1}^\infty r^n\left[\cos(n\theta)\int_0^{2\pi}\cos(ks)\cos(ns)\,\mathrm ds+\sin(n\theta)\int_0^{2\pi}\!\cos(ks)\sin(ns)\,\mathrm ds\right]\\&=r^k\cos(k\theta).\end{aligned}$
Here we have used the fact that $\displaystyle\int_0^{2\pi}\!\cos(ks)\,\mathrm ds=\int_0^{2\pi}\!\cos(ks)\sin(ns)\,\mathrm ds=0$ and $\displaystyle\int_0^{2\pi}\!\cos(ks)\cos(ns)\,\mathrm ds=\delta_{ks}$ for $k\in\mathbb N$, where $\delta_{kn}=1$ if $k=n$; $0$ if $k\neq n$. On the other hand, by (a), we obtain the formula
$\displaystyle r^k\cos k\theta=u(r,\theta)=\frac{1-r^2}{2\pi}\int_0^{2\pi}\!\frac{\cos(k\phi)}{1+r^2-2r\cos(\theta-\phi)}\,\mathrm d\phi.$
The proof is complete.

Let $\Omega$ be a bounded domain and $f\in C^k(\bar\Omega)$. Show that in fact the domain potential (29a) satisfies $u\in C^{k+1}(\Omega)$. Conclude that $f\in C^\infty(\bar\Omega)$ implies $u\in C^\infty(\Omega)$. [In Section 8.2 we show that $f\in C^\infty(\bar\Omega)$ implies $u\in C^\infty(\bar\Omega)$, provided $\partial\Omega$ is $C^\infty$.]

Solution

Clearly, the statement holds true for $k=1$ by (iii) of Proposition at page 114 in the textbook. We firstly show that the statement holds true for $k=2$, i.e., $f\in C^2(\bar\Omega)$ implies $u\in C^3(\Omega)$. For any $i\in\{1,\dots,n\}$, from the argument of Proposition at page 114, we have

$\displaystyle\partial_iu(x)=\int_\Omega\!(\partial_iK)(x-y)f(y)\,\mathrm dy\quad\text{for}~x\in\Omega$.

It suffices to show $\partial_iu\in C^2(\Omega)$. To achieve this, we consider the Poisson equation $-\Delta\phi=\partial_if$ in $\Omega$. Since $\partial_if\in C^1(\bar\Omega)$ (from $f\in C^2(\bar\Omega)$), then by the Proposition at page 114, we know the solution $\phi\in C^2(\Omega)$ and

$\displaystyle\phi(x)=\int_\Omega\!K(x-y)(\partial_if)(y)\,\mathrm dy$.

Using the integration by parts, we have

$\displaystyle\phi(x)=\int_{\partial\Omega}\!K(x-y)f(y)\nu_i(y)\,\mathrm dS_y+\int_\Omega\!(\partial_iK)(x-y)f(y)\,\mathrm dy.$

For any $y\in\partial\Omega$, the function $\int_{\partial\Omega}\!K(\cdot-y)f(y)\nu_i(y)\,\mathrm dS_y$ is smooth. Hence, we find

$\displaystyle\partial u_i(x)=\phi(x)-\int_{\partial\Omega}\!K(x-y)f(y)\nu_i(y)\,\mathrm dS_y\quad\text{for}~x\in\Omega$,

which means $\partial_iu\in C^2(\Omega)$ and $u\in C^3(\Omega)$.

Now we prove the statement by mathematical induction. Suppose the statement holds true for some positive integer $m\geq2$, i.e., $f\in C^m(\bar\Omega)$ implies $u\in C^{m+1}(\Omega)$. Since $f\in C^{m+1}(\bar\Omega)$, we have $\partial f_i\in C^m$ for any $i\in\{1,\dots,n\}$. Note that $\partial_iu\in C^m(\Omega)$ and satisfies $\Delta(\partial_iu)=\partial_if$ in $\Omega$. Here we have used the fact that $m\geq2$. Then by the inductive hypothesis, $\partial_iu\in C^{m+1}(\Omega)$. Due to the arbitrariness of $i$, we know $u\in C^{m+2}(\Omega)$. Therefore, by mathematical induction, the statement holds true for all $n\in\mathbb N$.

When $f\in C^\infty(\bar\Omega)$, we have $f\in C^m(\bar\Omega)$ for all $m\in\mathbb N$, which gives $u\in C^{m+1}(\Omega)$ for all $m\in\mathbb N$. Thus, we know $u\in C^\infty(\Omega)$. The proof is complete.

The symmetry of the Green's function [i.e., $G(x,\xi)=G(\xi,x)$ for all $x,\xi\in\Omega$] is an important fact connected with the self-adjointness of $\Delta$.
1. Verify the symmetry of $G(x,\xi)$ by direct calculation when $\Omega=\mathbb R_+^n$ and $\Omega=B_a(0)$.
2. Prove the symmetry of $G(x,\xi)$ when $\Omega$ is any smooth, bounded domain.

Solution

Recall that $K(x)=\begin{cases}\displaystyle\frac1{2\pi}\log|x|&\text{if}~n=2;\\\displaystyle\frac1{(2-n)\omega_n}|x|^{2-n}&\text{if}~n\geq3.\end{cases}$ (cf. (23)).
- When $\Omega=\mathbb R_+^n$, the Green's function $G$ is defined in (35), i.e.,
  $\displaystyle G(x,\xi)=K(x-\xi)-K(x-\xi^*)=\begin{cases}\displaystyle\frac{\log|x-\xi|-\log|x-\xi^*|}{2\pi}&\text{if}~n=2;\\\displaystyle\frac{|x-\xi|^{2-n}-|x-\xi^*|^{2-n}}{(2-n)\omega_n}&\text{if}~n\geq3.\end{cases}$
  Here $\xi^*=(\xi_1,\dots,\xi_{n-1},-\xi_n)$ denotes the reflection point of $\xi=(\xi_1,\dots,\xi_{n-1},\xi_n)$ with respect to the plane $\partial\mathbb R_+^n$. Clearly, $|x-\xi|=|\xi-x|$ and
  $\displaystyle|x-\xi^*|=\sqrt{\sum_{i=1}^{n-1}(x_i-\xi_i)^2+[x_n-(-\xi_n)]^2}=\sqrt{\sum_{i=1}^{n-1}(\xi_i-x_i)^2+[\xi_n-(-x_n)]^2}=|\xi-x^*|$,
  where $x^*=(x_1,\dots,x_{n-1},-x_n)$ denotes the reflection of $x$ with respect to the plane $\partial\mathbb R_+^n$. Therefore, we verify that
  $G(x,\xi)=K(x-\xi)-K(x-\xi^*)=K(\xi-x)-K(\xi-x^*)=G(\xi,x)$ for all $x,\xi\in\mathbb R_+^n$ and $x\neq\xi$.
- When $\Omega=B_a(0)$, the Green's function $G$ is defined in (42), i.e.,
  $\displaystyle G(x,\xi)=K(x-\xi)-K\left(\frac{|\xi|}a(x-\xi^*)\right)=\begin{cases}\displaystyle\frac1{2\pi}\left[\log|x-\xi|-\log\left(\frac{|\xi|}a|x-\xi^*|\right)\right]&\text{if}~n=2;\\\displaystyle\frac1{(2-n)\omega_n}\left[|x-\xi|^{2-n}-\left(\frac{|\xi|}a|x-\xi^*|\right)^{2-n}\right]&\text{if}~n=3.\end{cases}$
  Here $\displaystyle\xi^*=a^2\frac{\xi}{|\xi|^2}$ denotes the reflection of $\xi\neq0$ with respect to the sphere $\partial B_a(0)$. Clearly, $|x-\xi|=|\xi-x|$ and
  $\begin{aligned}\left|\frac{|\xi|}a(x-\xi^*)\right|^2&=\frac{\displaystyle\sum_{i=1}^n\xi_i^2}{a^2}\left(\sum_{i=1}^nx_i^2-\frac{\displaystyle2a^2\sum_{i=1}^nx_i\xi_i}{\displaystyle\sum_{j=1}^n\xi_j^2}+a^4\frac{\displaystyle\sum_{i=1}^n\xi_i^2}{\displaystyle\left(\sum_{j=1}^n\xi_j^2\right)^2}\right)\\&=\frac{\displaystyle\sum_{i=1}^nx_i^2}{a^2}\left(\sum_{i=1}^n\xi_i^2-\frac{\displaystyle2a^2\sum_{i=1}^nx_i\xi_i}{\displaystyle\sum_{j=1}^nx_j^2}+a^4\frac{\displaystyle\sum_{i=1}^nx_i^2}{\displaystyle\left(\sum_{j=1}^nx_i^2\right)^2}\right)=\left|\frac{|x|}a(\xi-x^*)\right|^2,\end{aligned}$
  where $\displaystyle x^*=a^2\frac{x}{|x|^2}$ for $x\neq0$. Therefore, we verify that
  $\begin{aligned}G(x,\xi)&=K(x-\xi)-K\left(\frac{|\xi|}a(x-\xi^*)\right)\\&=K(\xi-x)-K\left(\frac{|x|}a(\xi-x^*)\right)=G(\xi,x)\quad\text{for all}~x,\xi\in B_a(0)-\{0\}~\text{and}~x\neq\xi.\end{aligned}$
Recall that the notion of Green's function $G$ is defined at page 117: $G(x,\xi)=K(x-\xi)+\omega(x)$ for $x\neq\xi$. Hence $G$ satisfies $\Delta_xG(x,\xi)=0$ for $x\in\Omega-\{\xi\}$ with $G(x,\xi)=0$ for $x\in\partial\Omega$ (cf. (32)). Fix $x,\xi\in\Omega$ arbitrarily with $x\neq\xi$. We want to prove $G(x,\xi)=G(\xi,x)$. For this purpose, we define $u:\Omega-\{x\}\to\mathbb R$ and $v:\Omega-\{\xi\}\to\mathbb R$ by
$\begin{aligned} &u(z)=G(z,x)\quad\text{for}~z\in\Omega-\{x\},\\&v(z)=G(z,\xi)\quad\text{for}~z\in\Omega-\{\xi\}.\end{aligned}$
It suffices to show $u(\xi)=v(x)$. Since $x$ and $\xi$ are fixed, there exists a positive constant $\epsilon>0$ such that $B_\epsilon(x)\subsetneq\Omega$, $B_\epsilon(\xi)\subsetneq\Omega$ and $\overline{B_\epsilon(x)}\cap\overline{B_\epsilon(\xi)}=\emptyset$. Let $\Omega_\epsilon=\Omega-(B_\epsilon(x)\cup B_\epsilon(\xi))$. By the definition of Green's function, $\Delta u(z)=0$ for $z\in\Omega-\{x\}$ and $\Delta v(z)=0$ for $z\in\Omega-\{\xi\}$, which implies
$\displaystyle\int_{\Omega_\epsilon}(v\Delta u-u\Delta v)\,\mathrm dz=0$.
Using the Green's second identity (7) at page 107, we have
$\begin{aligned}0=\int_{\Omega_\epsilon}\!(v\Delta u-u\Delta v)\,\mathrm dz&=\int_{\partial\Omega_\epsilon}\!\left(v\frac{\partial u}{\partial\nu}-u\frac{\partial v}{\partial\nu}\right)\,\mathrm dS_z\\&=\left(\int_{\partial\Omega}+\int_{\partial B_\epsilon(x)}+\int_{\partial B_\epsilon(\xi)}\right)\left(v\frac{\partial u}{\partial\nu}-u\frac{\partial v}{\partial\nu}\right)\,\mathrm dS_z.\end{aligned}$
In addition, we note the fact that $u(z)=G(z,x)=0$ for $z\in\partial\Omega$ and $v(z)=G(z,\xi)=0$ for $z\in\partial\Omega$ (from (32)). Hence we get
$\displaystyle\int_{\partial B_\epsilon(x)}\left(v\frac{\partial u}{\partial\nu}-u\frac{\partial v}{\partial\nu}\right)\,\mathrm dS_z=\int_{\partial B_\epsilon(\xi)}\left(u\frac{\partial v}{\partial\nu}-v\frac{\partial u}{\partial\nu}\right)\,\mathrm dS_z,$
where $\nu$ is the unit vector pointing the inward on $\partial B_\epsilon(x)\cup\partial B_\epsilon(\xi)$. Since $v$ is smooth near $x$ and $|u(z)|=|G(z,x)|\leq|K(z-x)|+|\omega(z)|$, we find
$\displaystyle\left|\int_{\partial B_\epsilon(x)}\!u\frac{\partial v}{\partial\nu}\,\mathrm dS_z\right|\leq n\omega_n\epsilon^{n-1}\cdot\max_{z\in\overline{B_1(x)}}|\nabla v(z)|\cdot\left(\max_{z\in\partial B_\epsilon(x)}|K(z-x)|+\max_{z\in\overline{B_1(x)}}|\omega(z)|\right)\to0\quad\text{as}~\epsilon\to0$.
Here we have used the fact that
$\displaystyle\lim_{\epsilon\to0^+}\epsilon^{n-1}K(|\epsilon|)=\begin{cases}\displaystyle\lim_{\epsilon\to0^+}\frac{\epsilon\log\epsilon}{2\pi}=0&\text{if}~n=2;\\\displaystyle\lim_{\epsilon\to0^+}\frac{\epsilon^{n-1}\epsilon^{2-n}}{(2-n)\omega_n}=0&\text{if}~n\geq3.\end{cases}$
On the other hand, by the smoothness of $\omega$ and (22)-(23), we have
$\displaystyle\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(x)}\!v\frac{\partial u}{\partial\nu}\,\mathrm dS_z=\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(x)}v(z)\frac{\partial K}{\partial\nu}(z-x)\,\mathrm dS_z=v(x)$.
Hence we see $\displaystyle\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(x)}\left(v\frac{\partial u}{\partial\nu}-u\frac{\partial v}{\partial\nu}\right)\,\mathrm dS_z=v(x)$. Similarly, one may check that $\displaystyle\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(\xi)}\left(u\frac{\partial v}{\partial\nu}-v\frac{\partial u}{\partial\nu}\right)\,\mathrm dS_z=u(\xi)$. Therefore, we arrive at $u(\xi)=v(x)$ and complete the proof.

1. Use the weak maximum principle (16) to prove that $G(x,\xi)\leq0$ for $x,\xi\in\Omega$ with $x\neq\xi$.
2. Use the strong maximum principle (15) to prove that $G(x,\xi)<0$ for $x,\xi\in\Omega$ with $x\neq\xi$.

Solution

Fix $\xi\in\Omega$. From the definition of Green's function $G$ at page 117, we write $G(x,\xi)=K(x-\xi)+\omega(x)$, where $\omega$ satisfies $\Delta\omega=0$ in $\Omega$ and $\omega(x)=-K(x-\xi)$ for $x\in\partial\Omega$. Since $\omega$ is bounded in $\bar\Omega$ and $\displaystyle\lim_{x\to\xi}K(x-\xi)=-\infty$, there exists $\epsilon>0$ such that $\overline{B_\epsilon(\xi)}\subsetneq\Omega$ and $G(x,\xi)<0$ for $x\in\overline{B_\epsilon(\xi)}$ and $x\neq\xi$. Let $\Omega_{\epsilon,\xi}=\Omega-\overline{B_\epsilon(\xi)}$. Then we find
$\begin{aligned} &\Delta_x(x,\xi)=0\quad\text{in}~\Omega_{\epsilon,\xi},\\&G(x,\xi)\leq0\quad\text{for}~x\in\partial\Omega_{\epsilon,\xi}=\partial\Omega\cup\partial B_\epsilon(\xi).\end{aligned}$
Then by the weak maximum principle (16), we obtain
$\displaystyle\max_{x\in\bar\Omega_{\epsilon,\xi}}G(x,\xi)=\max_{x\in\partial\Omega_{\epsilon,\xi}}G(x,\xi)=\max\left\{\max_{x\in\partial\Omega}G(x,\xi),\max_{x\in\partial B_\epsilon(\xi)}G(x,\xi)\right\}=0$,
which implies $G(x,\xi)\leq0$ for $x\in\bar\Omega_{\epsilon,\xi}$. Since $G(x,\xi)<0$ for $x\in\overline{B_\epsilon(\xi)}-\{\xi\}$, we arrive at $G(x,\xi)<0$ for $x\neq\xi$.
Following the process of part (a), $G$ is not constant because $G(x,\xi)=0$ for $x\in\partial\Omega$ and $G(x,\xi)<0$ for $x\in\partial B_\epsilon(\xi)$. Then by Theorem 3 (Maximun Principle) at page 109, we obtain
$\displaystyle G(x,\xi)<\sup_{z\in\Omega}G(z,\xi)=\max_{z\in\bar\Omega}G(z,\xi)=0\quad\text{for}~x\in\Omega~\text{but}~x\neq\xi$.
The proof is complete.

Use (33) to prove (38).

Solution

Let $\Omega_R=B_R(0)\cap\mathbb R_+^n=\{x=(x_1,\dots,x_n)\in\mathbb R^n\,:\,|x|<R,\,x_n>0\}$ for any $R>0$ and $u\equiv1$ in $\bar{\mathbb R}_+^n$. Using (31) and Green's function $G$ defined by (35), we have

$\begin{aligned}1&=\int_{\Omega_R}\!G(x,\xi)\Delta u\,\mathrm dx+\int_{\partial\Omega_R}\!\left(u(x)\frac{\partial G(x,\xi)}{\partial\nu_x}-G(x,\xi)\frac{\partial u(x)}{\partial\nu}\right)\,\mathrm dS_x\\&=\int_{\partial\Omega_R}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=\left(\int_{\partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\})}+\int_{\partial\Omega_R\cap\mathbb R_+^n}\right)\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x\quad\text{for}~\xi\in\Omega_R.\end{aligned}$

For $x\in\partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\})$, we have

$\displaystyle\frac{\partial G(x,\xi)}{\partial\nu_x}=\left.-\frac{\partial G}{\partial x_n}(x,\xi)\right|_{x_n=0}=\frac{2\xi_n}{\omega_n}|x'-\xi|^{-n}$,

where $x'=(x_1,\dots,x_{n-1})\in\mathbb R^{n-1}$ (cf. (36)). For $x\in\partial\Omega_R\cap\mathbb R_+^n$, we have

$\begin{aligned}\frac{\partial G(x,\xi)}{\partial\nu_x}&=\nabla_xG(x,\xi)\cdot\frac{x}{|x|}=\frac1{\omega_n}\left(|x-\xi|^{-n}(x-\xi)-|x-\xi^*|^{-n}(x-\xi^*)\right)\cdot\frac{x}R\\&=\frac1{\omega_nR}\left(\frac{R^2-\xi\cdot x}{|x-\xi|^n}-\frac{R^2-\xi^*\cdot x}{|x-\xi^*|^n}\right)\\&=\left(\frac{R}{\omega_n}-\sum_{i=1}^{n-1}\frac{\xi_ix_i}{\omega_nR}\right)\left(\frac1{|x-\xi|^n}-\frac1{|x-\xi^*|^n}\right)-\frac{\xi_nx_n}{\omega_nR}\left(\frac1{|x-\xi|^n}+\frac1{|x-\xi^*|^n}\right).\end{aligned}$

A direct estimate gives

$\displaystyle\left|\int_{\partial\Omega_R\cap\mathbb R_+^n}\frac{\xi_nx_n}{\omega_nR}\left(\frac1{|x-\xi|^n}+\frac1{|x-\xi^*|^n}\right)\,\mathrm dS_x\right|\leq\frac{|\xi_n|}{\omega_n}\cdot\frac2{(R-|\xi|)^n}\cdot\frac{\omega_nR^{n-1}}2=\frac{|\xi|R^{n-1}}{R-|\xi|^n}\to0\quad\text{as}~R\to\infty.$

Here we have used the fact that $|x-\xi|\geq|x|-|\xi|=R-|\xi|$ and $|x-\xi^*|\geq|x|-|\xi^*|=R-|\xi|$ for $x\in\partial B_R(0)$. Moreoever, by the trinagle inequality and $x\in\partial B_R(0)\cap\mathbb R_+^n$, we have $R-|\xi|\leq|x-\xi|\leq|x-\xi^*|\leq R+|\xi|$, which implies

$\displaystyle\frac1{|x-\xi|^n}-\frac1{|x-\xi^*|^n}\leq\frac1{(R-|\xi|)^n}-\frac1{(R+|\xi|)^n}=\frac{2nR^{n-1}|\xi|+\cdots}{(R^2-|\xi|^2)^n}.$

Hence we get

$\displaystyle\left|\int_{\partial\Omega_R\cap\mathbb R_+^n}\!\left(\frac R{\omega_n}-\sum_{i=1}^{n-1}\frac{\xi_ix_i}{\omega_nR}\right)\left(\frac1{|x-\xi|^n}-\frac1{|x-\xi^*|^n}\right)\,\mathrm dS_x\right|\leq\left(\frac R{\omega_n}+\sum_{i=1}^{n-1}\frac{|\xi_i|}{\omega_n}\right)\cdot\frac{2nR^{n-1}|\xi|+\cdots}{(R^2-|\xi|^2)^n}\cdot\frac{\omega_nR^{n-1}}2$

Combining these results, it follows that $\displaystyle\lim_{R\to\infty}\int_{\partial\Omega_R\cap\mathbb R_+^n}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=0$. Therefore, we obtain

$\displaystyle1=\lim_{R\to\infty}\int_{\partial\Omega_R}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=\lim_{R\to\infty}\int_{\partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\})}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=\int_{\mathbb R^{n-1}\times\{0\}}H(x',\xi)\,\mathrm dx'\quad\text{for}~x\in\mathbb R_+^n.$

The proof is complete.

Warning: Note that we should not use (33) to prove (38) because Green's function $G$ defined in (35) is for half-plane $\mathbb R_+^n$ but not $B_R(0)\cap\mathbb R_+^n$. Particularly, $G\equiv0$ on $\partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\})$ but $G$ may not becomes zero on $\partial\Omega_R\cap\mathbb R_+^n$. However, since $u\equiv1$, the normal derivative of $u$ becomes zero so the boundary value of function $G$ is not important.

For $n=2$, use the method of reflections to find the Green's function for the first quadrant $\Omega=\{(x,y)\,:\,x,y>0\}$.

Solution

Let $\xi=(\xi_1,\xi_2)\in\Omega$, i.e., $\xi_1$ and $\xi_2$ are positive. Define $\xi^*=(-\xi_1,\xi_2)$. Then $\xi^*$, $-\xi$ and $-\xi^*$ are in the second, the third and the fourth quadrant, respectively. Now we claim

$\begin{aligned}G(x,\xi)&=G(x_1,x_2,\xi_1,\xi_2)=K(x-\xi)-K\left(\frac{|x+\xi^*||x-\xi^*|}{|x+\xi|}\right)\\&=\frac{\log|x-\xi|-\log|x-\xi^*|-\log|x+\xi^*|+\log|x+\xi|}{2\pi}\\&=\frac1{2\pi}\log\frac{|x-\xi||x+\xi|}{|x-\xi^*||x+\xi^*|}\quad\text{for}~x=(x_1,x_2)\in\Omega~\text{but}~x\neq\xi\end{aligned}$

is a Green's function for the first quadrant $\Omega$. Since $\xi^*$, $-\xi$ and $-\xi^*$ are not in the first quadrant, $\Delta_xG(x,\xi)=0$ for $x=(x_1,x_2)\in\Omega$ except $\xi$. In addition, for $x_1=0$, we have $|x-\xi|=|x-\xi^*|$ and $|x+\xi|=|x+\xi^*|$; for $x_2=0$, we have $|x-\xi|=|x+\xi^*|$ and $|x+\xi|=|x-\xi^*|$. Thus, for $x\in\partial\Omega$, it is necessary that $\displaystyle\frac{|x-\xi||x+\xi|}{|x-\xi^*||x+\xi^*|}=1$, which implies $G(x,\xi)=0$. Therefore, by the definition, $G$ is a Green's function for the first quadrant $\Omega$.

If $u\in C(\Omega)$ satisfies the mean value property of Section 4.1.d, then $u$ is harmonic in $\Omega$.

Solution

Let $\xi\in\Omega$. Then there exists a positive constant $r$ such that $\overline{B_r(\xi)}\subsetneq\Omega$. Since $u\in C(\Omega)$ satisfies the mean value property, we have

$\displaystyle u(\xi)=\frac1{\omega_n}\int_{|x|=1}\!u(\xi+rx)\,\mathrm dS_x$.

Consider the Laplace equation $\Delta v=0$ in $B_r(\xi)$ with the Dirichlet boundary condition $v=u\Big|_{\partial B_r(\xi)}$. Clearly, $u\Big|_{\partial B_r(\xi)}$ is continuous on $\partial B_r(\xi)$. Then by Theorem 4 at page 122, $v$ can be solved by Poisson integral formula (44) and $v\in C^2(B_r(\xi))\cap C(\overline{B_r(\xi)})$ is harmonic. Thus, $v$ also satisfies the mean value property. Using the Exercise 9 in Section 4.1, we obtain $v\equiv u$ in $\overline{B_r(\xi)}$. Thus, $u\in C^2(B_r(\xi))\cap C(\overline{B_r(\xi)})$ and $\Delta u=0$ in $B_r(\xi)$. Due to arbitrariness of $\xi$, $\Delta u=0$ in $\Omega$. The proof is complete.

Let $\Omega=B_a(0)$, $\Omega_+=\Omega\cap\mathbb R_+^n$, and $\Omega_0=\{x\in\Omega\,:\,x_n=0\}$. If $u\in C^2(\Omega_+)\cap C(\Omega_+\cup\Omega_0)$ is harmonic in $\Omega_+$, and $u=0$ on $\Omega_0$, prove that $u$ may be extended ot a harmonic function on all of $\Omega$. (This is called a reflection principle.)

Solution

Define the extension $\tilde u:\Omega\to\mathbb R$ of $u$ by

$\tilde u(x)=\tilde u(x',x_n)=\begin{cases}u(x',x_n)&\text{if}~x_n\geq0;\\-u(x',-x_n)&\text{if}~x_n<0.\end{cases}$

Here $x'$ denotes $(x_1,\dots,x_{n-1})$. Since $u\equiv0$ on $\Omega_0$, function $\tilde u$ is continuous on $\Omega$. By Exercise 7, it suffices to show that $\tilde u$ satisfies the mean value property: For any $\xi\in\Omega$, there exists a positive constant $r$ (which may be dependent on $\xi$) such that $\overline{B_r(\xi)}\subsetneq\Omega=B_a(0)$ and

$\displaystyle\tilde u(\xi)=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+rx)\,\mathrm dS_x$.

Since $\tilde u\equiv u$ in $\Omega_+$, for $\xi\in\Omega_+$, there exists $r>0$ such that $\overline{B_r(\xi)}\subsetneq\Omega_+\subsetneq\Omega$, then we have

$\displaystyle\tilde u(\xi)=u(\xi)=\frac1{\omega_n}\int_{|x|=1}\!u(\xi+|\xi_n|x)\,\mathrm dS_x=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+|\xi_n|x)\,\mathrm dS_x$.

Similarly, for $\xi\in\Omega-(\Omega_+\cup\Omega_0)$, there exists $r>0$ such that $\overline{B_r(\xi)}\subsetneq\Omega-(\Omega_+\cup\Omega_0)\subsetneq\Omega$, we have

$\begin{aligned}\tilde u(\xi)&=-u(\xi',-\xi_n)=-\frac1{\omega_n}\int_{|x|=1}\!u((\xi',-\xi_n)+|\xi_n|x)\,\mathrm dS_x=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+|\xi_n|(x',-x_n))\,\mathrm dS_x\\&=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+|\xi_n|(x',-x_n))\,\mathrm dS_x.\end{aligned}$

Here we have used the symmetry of the integral over the surface $|x|=1$. For $\xi\in\Omega_0$ and $0<r<a-|\xi|$, we observe that

$\begin{aligned}\frac1{\omega_n}\int_{|x|=1}\!\tilde u((\xi',0)+rx)\,\mathrm dS_x&=\frac1{\omega_n}\int_{|x|=1, x_n>0}\!u(\xi'+rx',rx_n)\,\mathrm dS_x-\frac1{\omega_n}\int_{|x|=1,\,x_n<0}\!u(\xi'+rx',-rx_n)\,\mathrm dS_x\\&=0=\tilde u(\xi',0)\quad\text{(by substitution $y_n=-x_n$, the two integrals are same)}.\end{aligned}$

Therefore, continuous function $\tilde u$ satisfies the mean value property, which implies $\tilde u$ is harmonic in $\Omega$. The proof is complete.

Remark. If the condition is replaced by a stronger condition: $u\in C^2(\Omega_+\cup\Omega_0)$, then we can compute the second derivatives $\displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x)$ at each point $x\in\Omega$, and then sum up to get $\Delta\tilde u(x)=0$ for all $x\in\Omega$.

If $x\in\Omega_+$, then $\displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x)=\frac{\partial^2u}{\partial x_i^2}(x)$, which implies $\Delta\tilde u(x)=\Delta u(x)=0$.
If $x\in\Omega-(\Omega_+\cup\Omega_0)$, then $\displaystyle\frac{\partial\tilde u}{\partial x_i}(x)=-\frac{\partial u}{\partial x_i}(x',-x_n)$ for $i=1,\dots,n-1$ and $\displaystyle\frac{\partial\tilde u}{\partial x_n}(x)=\frac{\partial u}{\partial x_n}(x',-x_n)$. Moreover, we get $\displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x)=-\frac{\partial^2u}{\partial x_i^2}(x',-x_n)$ for $i=1,\dots,n-1$ and $\displaystyle\frac{\partial^2\tilde u}{\partial x_n^2}(x)=-\frac{\partial^2u}{\partial x_n^2}(x',-x_n)$. Thus, we get $\Delta\tilde u(x)=-\Delta u(x',-x_n)=0$.
If $x\in\Omega_0$, then by $\tilde u\equiv u\equiv0$ on $\Omega_0$, we have $\displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x',0)=0$ for $i=1,\dots,n-1$. Moreorever, we use the definition of partial derivative to find
$\begin{aligned} \frac{\partial\tilde u}{\partial x_n}(x',0^-)&=\lim_{h\to0^-}\frac{\tilde u(x',h)-\tilde u(x',0)}h=\lim_{h\to0^-}\frac{u(x',-h)}{-h}=\frac{\partial u}{\partial x_n}(x',0)\\&=\lim_{h\to0^+}\frac{u(x',h)}h=\lim_{h\to0^+}\frac{\tilde u(x',h)-\tilde u(x',0)}h=\frac{\partial\tilde u}{\partial x_n}(x',0^+).\end{aligned}$
Thus $\displaystyle\frac{\partial\tilde u}{\partial x_n}(x',0)$ exists. Then the second derivative gives
$\begin{aligned} &\frac{\partial^2\tilde u}{\partial x_n^2}(x',0^+)=\lim_{h\to0^+}\frac{\displaystyle\frac{\partial\tilde u}{\partial x_n}(x',h)-\frac{\partial\tilde u}{\partial x_n}(x',0)}h=\lim_{h\to0^+}\frac{\displaystyle\frac{\partial u}{\partial x_n}(x',h)-\frac{\partial u}{\partial x_n}(x',0)}h=\frac{\partial^2u}{\partial x_n^2}(x',0)=0,\\&\frac{\partial^2\tilde u}{\partial x_n^2}(x',0^-)=\lim_{h\to0^-}\frac{\displaystyle\frac{\partial\tilde u}{\partial x_n}(x',h)-\frac{\partial\tilde u}{\partial x_n}(x',0)}{h}=-\lim_{h\to0^-}\frac{\displaystyle\frac{\partial u}{\partial x_n}(x',-h)-\frac{\partial u}{\partial x_n}(x',0)}{h}=-\frac{\partial^2u}{\partial x_n^2}(x',0)=0.\end{aligned}$
Thus, $\displaystyle\frac{\partial^2\tilde u}{\partial x_n^2}(x',0)=0$. Therefore $\Delta\tilde u(x)=0$.

Warning: The original statement is that "If $u\in C^2(\Omega)\cap C(\Omega_+\cup\Omega_0)$ ...". It should be modified as "If $u\in C^2(\Omega_+)\cap C(\Omega_+\cup\Omega_0)$ ...".

Show that the bounded solution of the Dirichlet problem in a half-space is unique. Give unbounded counterexamples.

Solution

Let $\tilde u$ and $u$ be the bounded solutions of the Dirichlet problem in a half-space. Then $v:=\tilde u-u$ satisfies

$\left\{\begin{aligned} &\Delta v=0\quad\text{in}~\mathbb R_+^n,\\&v=0\quad\text{on}~\partial\mathbb R_+^n.\end{aligned}\right.$

Define $\tilde v:\mathbb R^n\to\mathbb R$ by the reflection

$\tilde v(x)=\tilde v(x',x_n)=\begin{cases}v(x',x_n)&\text{if}~x_n\geq0;\\-v(x',-x_n)&\text{if}~x_n<0.\end{cases}$

For any $x\in\mathbb R^n$, there exists a sufficiently large $a>0$ such that $x\in B_a(0)$. Then by Exercise 8, $\tilde v$ is harmonic in $B_a(0)$ and hence $\Delta\tilde v(x)=0$. Thus, $\tilde v$ is harmonic in $\mathbb R^n$. Since $\tilde u$ and $u$ are bounded in $\mathbb R_+^n$, $\tilde v$ is also bounded. By Theorem 7 (Liouville's Theorem) at page 123, $\tilde v$ must be a constant, and hence $\tilde v(x)=\tilde v(0)=0$ for all $x\in\mathbb R^n$. This implies $\tilde u\equiv u$ in $\mathbb R_+^n$ and the uniqueness of bounded solution of the Dirichlet problem in the half-space follows.

It is clear that the function $u(x',x_n)=x_n$ is a unbounded harmonic solution to the zero Dirichlet problem in the half-space. (The bounded harmonic solution to the zero Dirichlet problem in the half-space is identically zero.)

1. Suppose $u\in C^2(\Omega)$ is harmonic, $u\geq0$, and $\overline{B_a(0)}\subset\Omega$. Use (44) to show
  $\displaystyle\frac{a^{n-2}(a-|\xi|)}{(a+|\xi|)^{n-1}}u(0)\leq u(\xi)\leq\frac{a^{n-2}(a+|\xi|)}{(a-|\xi|)^{n-1}}u(0)\quad\text{for}~|\xi|<a$.
2. Prove (45).

Solution

In $B_a(0)$, we use the Poisson integral formula (44):
$\displaystyle u(\xi)=\frac{a^2-|\xi|^2}{a\omega_n}\int_{|x|=a}\!\frac{u(x)}{|x-\xi|^n}\,\mathrm dS_x\quad\text{for}~|\xi|<a$.
By the triangle inequality, we have $a-|\xi|=|x|-|\xi|\leq|x-\xi|\leq|x|+|\xi|=a+|\xi|$ for $|x|=a$, which gives $\displaystyle\frac1{(a+|\xi|)^n}\leq\frac1{|x-\xi|^n}\leq\frac1{(a-|\xi|)^n}$. Then by $u(x)\geq0$, we have
$\begin{aligned}u(\xi)&\geq\frac{a^2-|\xi|^2}{a\omega_n}\int_{|x|=a}\!\frac{u(x)}{(a+|\xi|)^n}\,\mathrm dS_x=\frac{a^n(a^2-|\xi|^2)}{a\omega_n(a+|\xi|)^n}\int_{|x|=a}\!\frac{u(x)}{a^n}\,\mathrm dS_x\\&=\frac{a^{n-1}(a-|\xi|)}{\omega_n(a+|\xi|)^{n-1}}\int_{|x|=a}\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a-|\xi|)}{(a+|\xi|)^{n-1}}\cdot\frac{a}{\omega_n}\int_{|x|=a}\!\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a-|\xi|)}{(a+|\xi|)^{n-1}}u(0).\end{aligned}$
Simiarly, from $u(x)\geq0$, we have
$\begin{aligned}u(\xi)&\leq\frac{a^2-|\xi|^2}{a\omega_n}\int_{|x|=a}\!\frac{u(x)}{(a-|\xi|)^n}\,\mathrm dS_x=\frac{a^n(a^2-|\xi|^2)}{a\omega_n(a-|\xi|)^n}\int_{|x|=a}\!\frac{u(x)}{a^n}\,\mathrm dS_x\\&=\frac{a^{n-1}(a+|\xi|)}{\omega_n(a-|\xi|)^{n-1}}\int_{|x|=a}\!\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a+|\xi|)}{(a-|\xi|)^{n-1}}\cdot\frac{a}{\omega_n}\int_{|x|=a}\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a+|\xi|)}{(a-|\xi|)^{n-1}}u(0).\end{aligned}$
Therefore, we complete the proof of Harnack's inequality on a ball.
Let $r=\frac13\text{dist}(\bar\Omega_1,\partial\Omega)$. Then for any $\xi,\eta\in\bar\Omega_1$, and $|\xi-\eta|\leq r$, we use the mean value property (12) to find
$\begin{aligned}u(\xi)&=\frac n{\omega_n}\int_{|x|\leq1}\!u(\xi+2rx)\,\mathrm dx=\frac n{\omega_n(2r)^n}\int_{B_{2r}(\xi)}\!u(y)\,\mathrm dy\\&\geq\frac n{\omega_n(2r)^n}\int_{B_r(\eta)}\!u(y)\,\mathrm dy=\frac n{2^n\omega_n}\int_{|x|\leq1}\!u(\eta+rx)\,\mathrm dx=\frac{u(\eta)}{2^n}.\end{aligned}$
Since $\Omega_1$ is a bounded domain, $\Omega_1$ is connected and $\bar\Omega_1$ is compact. Then there exists a finite open converging $\{B_i\}_{i=1}^N$ such that the radius of $B_i$ is $r$ and $B_i\cap B_{i-1}\neq\emptyset$. In addition, by the extreme value theorem, there exists $x\in\bar\Omega_1$ and $y\in\bar\Omega_1$ such that
$\displaystyle u(x)=\max_{\bar\Omega_1}u=\sup_{\Omega_1}u,\quad u(y)=\min_{\bar\Omega_1}u=\inf_{\Omega_1}u.$
Therefore, we have
$\displaystyle u(y)\geq\frac1{2^{nN}}u(x)$,
which implies (45) and $C_1$ depends on $N$, i.e., depends on $\Omega_1$.

Use (46) to prove Liouville's theorem.

Solution

Let $u$ be the bounded harmonic function in $\mathbb R^n$ and $\xi\in\mathbb R^n$. Define $v(x)=u(x+\xi)$ for $x\in\mathbb R^n$. Clearly, $v$ is also bounded and harmonic. By (46), we have

$\displaystyle\left|\frac{\partial u}{\partial\xi_i}(\xi)\right|=\left|\frac{\partial v}{\partial\xi_i}(0)\right|\leq\frac na\max_{|x|=a}|v(x)|=\frac na\max_{|x|=a}|u(x+\xi)|\to0\quad\text{as}~a\to\infty$.

This shows $\nabla u(\xi)=0$ for all $\xi\in\mathbb R^n$, and hence $u$ is constant.

Use Stirling's formula $k!\sim(2\pi k)^{1/2}k^ke^{-k}$ as $k\to\infty$ to show there is a constant $C$ so that $|\alpha|^{|\alpha|}\leq C\alpha!e^{|\alpha|}n^{|\alpha|}$ holds for all multi-indices $\alpha$.

Solution

The Stirling's formula implies

$\displaystyle\lim_{k\to\infty}\frac{k^k}{k!e^kn^k}=\lim_{k\to\infty}\frac{(2\pi k)^{1/2}k^ke^{-k}}{k!}\cdot\frac{k^k}{(2\pi k)^{1/2}k^ke^{-k}e^kn^k}=1\cdot\lim_{k\to\infty}\frac1{n^k\sqrt{2\pi k}}=0.$

There exists a positive constant $K$ such that $k^k\leq k!e^kn^k$ for $k\geq K$. Hence we take

$\displaystyle C=\max\left\{1,\max_{1\leq|\alpha|<K}\frac{|\alpha|^{|\alpha|}}{\alpha!e^{|\alpha|}n^{|\alpha|}}\right\},$

which is desired.

艾利歐領域

2024年2月29日星期四