Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.1

1. Let $\Omega=\{(x,y)\in\mathbb R^2\,:x^2+y^2<1\}=\{(r,\theta):\,0\leq r<1,\,0\leq\theta<2\pi\}$, and use separation of the variables $(r,\theta)$ to solve the Dirichlet problem

   $\left\{\begin{aligned} &\Delta u=0\quad\text{in}~\Omega,\\&u(1,\theta)=g(\theta)\quad\text{for}~0\leq\theta<2\pi.\end{aligned}\right.$

Solution

Using the polar coordinates $(r,\theta)$, the Dirichlet problem can be transformed into

$\left\{\begin{aligned} &u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta}=0\quad\text{for}~0\leq r<1,~0\leq\theta<2\pi,\\&u(1,\theta)=g(\theta)\quad\text{for}~0\leq\theta<2\pi.\end{aligned}\right.$

Let $r=e^{-t}$ ($t=-\ln r$) and $u(r,\theta)=X(-\ln r)Y(\theta)=X(t)Y(\theta)$. It is natural to assume that $Y(0)=Y(2\pi)$ and $Y'(0)=Y'(2\pi)$ and $X'(\infty)=0$. The Laplace equaiton becomes

$r^2u_{rr}+ru_r+u_{\theta\theta}=\partial_t^2u+u_{\theta\theta}=X''(t)Y(\theta)+X(t)Y''(\theta)=0$.

From the differential equaiton, we have $X''(t)/X(t)=-Y''(\theta)/Y(\theta)$ for $t\in(0,\infty)$ and $\theta\in(0,2\pi)$, which means there exists a constant $\lambda$ such that $X''(t)+\lambda X(t)=0$ and $Y''(\theta)-\lambda Y(\theta)=0$. To find nontrivial solution $Y$, we divide three cases as follows.

• If $\lambda>0$, we may write $\lambda=\mu^2$ for some $\mu>0$. Then $Y(\theta)=c_1e^{\mu\theta}+c_2e^{-\mu\theta}$. Then by $Y(0)=Y(2\pi)$ and $Y'(0)=Y'(2\pi)$, we find $c_1=c_2=0$ and then $Y\equiv0$, which is a trivial solution.
• If $\lambda=0$, we get $Y(\theta)=c_1+c_2\theta$. By $Y(0)=Y(2\pi)$ and $Y'(0)=Y'(2\pi)$, it is easy to see $c_2=0$ and hence $Y\equiv c_1$. This gives a nonzero solution, provided $c_1\neq0$.
• If $\lambda<0$, we may write $\lambda=-\mu^2$ for some $\mu>0$. Then we have $Y(\theta)=c_1\sin(\mu\theta)+c_2\cos(\mu\theta)$. By $Y(0)=Y(2\pi)$, we get $c_2=c_1\sin(2\mu\pi)+c_2\cos(2\mu\pi)$; by $Y'(0)=Y'(2\pi)$, we have $c_1\mu=c_1\mu\cos(2\mu\pi)-c_2\mu\sin(2\mu\pi)$. Hence $(c_1,c_2)$ is the solution of following linear system

  $\left\{\begin{aligned} &c_1\sin(2\mu\pi)+c_2(\cos(2\mu\pi)-1)=0,\\&c_1(\cos(2\mu\pi)-1)-c_2\sin(2\mu\pi)=0.\end{aligned}\right.$

  To find a nonzero $(c_1,c_2)$, the determinant $\det\begin{bmatrix}\sin(2\mu\pi)&\cos(2\mu\pi)-1\\\cos(2\mu\pi)-1&-\sin(2\mu\pi)\end{bmatrix}=0$, which gives $\cos(2\mu\pi)=1$, i.e., $2\mu\pi=2n\pi$ or $\mu=n$ for $n\in\mathbb N$.

Based on the above discussion, for $\lambda=-n^2$ and $n\in\mathbb N\cup\{0\}$, we have $Y_n(\theta)=a_n\cos(n\theta)+b_n\sin(n\theta)$ for $n\in\mathbb N\cup\{0\}$. For such $\lambda_n$, we can obtain

$X_n(t)=\begin{cases}c_0t+d_0&\text{if}~n=0;\\c_ne^{nt}+d_ne^{-nt}&\text{if}~n\in\mathbb N.\end{cases}$

Thus, we have

$u_n(r,\theta)=\begin{cases}a_0(-c_0\ln r+d_0)&\text{if}~n=0;\\(a_n\cos(n\theta)+b_n\sin(n\theta))(c_nr^{-n}+d_nr^n)&\text{if}~n\in\mathbb N.\end{cases}$

Since $u$ should be finite at $r=0$, so $c_n=0$ for $n\in\mathbb N\cup\{0\}$. Then by superposition, we obtain

$\displaystyle u(r,\theta)=A_0+\sum_{n=1}^\infty r^n(A_n\cos(n\theta)+B_n\sin(n\theta))$,

where $A_n=a_nd_n$ for $n\in\mathbb N\cup\{0\}$ and $B_n=b_nd_n$ for $n\in\mathbb N$. By the Dirichlet boundary condition, we have

$\displaystyle g(\theta)=u(1,\theta)=A_0+\sum_{n=1}^\infty(A_n\cos(n\theta)+B_n\sin(n\theta))\quad\text{for}~0\leq\theta<2\pi$.

According to the Fourier series, the coefficients $A_0$, $A_n$ and $B_n$ can be determined by

$\displaystyle A_0=\frac1{2\pi}\int_0^{2\pi}\!g(\theta)\,\mathrm d\theta,\quad A_n=\frac1\pi\int_0^{2\pi}\!g(\theta)\cos(n\theta)\,\mathrm d\theta,\quad B_n=\frac1\pi\int_0^{2\pi}\!g(\theta)\sin(n\theta)\,\mathrm d\theta\quad\text{for}~n\in\mathbb N$.

Therefore, the Fourier solution $u$ of Laplace equaiton with the Dirichlet boundary condition is given by

$\displaystyle u(r,\theta)=\frac1{2\pi}\int_0^{2\pi}\!g(s)\,\mathrm ds+\frac1\pi\sum_{n=1}^\infty r^n\left[\cos(n\theta)\int_0^{2\pi}\!g(s)\cos(ns)\,\mathrm ds+\sin(n\theta)\int_0^{2\pi}\!g(s)\sin(ns)\,\mathrm ds\right]$.



2. Let $\Omega=(0,\pi)\times(0,\pi)$, and use separation of the variables to solve the mixed boundary value problem

   $\left\{\begin{aligned} &\Delta u=0\quad\text{in}~\Omega,\\&u_x(0,y)=0=u_x(\pi,y)\quad\text{for}~0<y<\pi,\\&u(x,0)=0,\quad u(x,\pi)=g(x)\quad\text{for}~0<x<\pi.\end{aligned}\right.$

Solution

Using the separation of variables, we suppose that the solution $u(x,y)$ is of the form

$\displaystyle u(x,y)=\sum_{n=0}^\infty u_n(x,y)=\sum_{n=0}^\infty X_n(x)Y_n(y)\quad\text{for}~(x,y)\in\Omega$.

Due to the superposition principle, we firstly consider $u(x,y)=X(x)Y(y)$ to fulfill the Laplace equation with the boundary condition $u_x(0,y)=0=u_x(\pi,y)$ for $y\in(0,\pi)$, which gives

$X''(x)Y(y)+X(x)Y''(y)=0\quad\text{and}\quad X'(0)Y(y)=0=X'(\pi)Y(y)$.

Surely, we want to find a general solution so $Y$ is not identically zero and $X'(0)=X'(\pi)=0$. From the differential equation, we have $X''(x)/X(x)=-Y''(y)/Y(y)$ for $(x,y)\in\Omega$, which means there exists a constant $\lambda$ such that $X''(x)-\lambda X(x)=0$ and $Y''(y)+\lambda Y(y)=0$. To find nontrivial solution $X$, we divide three cases as follows.

• If $\lambda>0$, we may write $\lambda=\mu^2$ for some $\mu>0$. Then $X(x)=c_1e^{\mu x}+c_2e^{-\mu x}$. Then by $X'(0)=X'(\pi)=0$, we find $c_1=c_2=0$ and then $X\equiv0$, which is a trivial solution.
• If $\lambda=0$, we get $X(x)=c_1+c_2x$. By $X'(0)=X'(\pi)=0$, it is easy to see $c_2=0$ and hence $X\equiv c_1$. This gives a nonzero solution, provided $c_1\neq0$.
• If $\lambda<0$, we may wrtie $\lambda=-\mu^2$ for some $\mu>0$. Then we have $X(x)=c_1\sin(\mu x)+c_2\cos(\mu x)$. By $X'(0)=0$, we get $c_1=0$. From $X'(\pi)=0$, we have $-c_2\mu\sin(\mu\pi)=0$. Since we want to seek the nontrivial solution, we hope $c_2\neq0$, which means $\sin(\mu\pi)$ should be zero and $\mu$ must be an integer.

Based on the above discussion, for $\lambda=-n^2$ and $n\in\mathbb N\cup\{0\}$, we have $X_n(x)=\cos(nx)$. On the other hand, we can solve

$Y_n(y)=\begin{cases}A_0+yB_0&\text{if}~n=0;\\A_n\cosh(ny)+B_n\sinh(ny)&\text{if}~n\in\mathbb N\end{cases}$.

From the boundary condition $u(x,0)=0$, we have $Y_n(0)=0$ for all $n\in\mathbb N\cup\{0\}$, which shows $A_n=0$ for all $n\in\mathbb N\cup\{0\}$. Thus, the solution $u$ has the following form:

$\displaystyle u(x,y)=yB_0+\sum_{n=1}^\infty B_n\sinh(ny)\cos(nx)$.

Now for the boundary condition $u(x,\pi)=g(x)$, we have

$\displaystyle g(x)=\pi B_0+\sum_{n=1}^\infty B_n\sinh(n\pi)\cos(nx).$

According to the Fourier series, the coefficients $B_n$ can be determined by

$\displaystyle B_0=\frac1{\pi^2}\int_0^\pi\!g(x)\,\mathrm dx,\quad B_n=\frac2{\pi\sinh(n\pi)}\int_0^\pi\!g(x)\cos(nx)\,\mathrm dx\quad\text{for}~n\in\mathbb N.$

Therefore, the solution is given by

$\displaystyle u(x,y)=\frac{y}{\pi^2}\int_0^\pi\!g(s)\,\mathrm ds+\frac2\pi\sum_{n=1}^\infty\frac{\sinh(ny)\cos(nx)}{\sinh(n\pi)}\int_0^\pi\!g(s)\cos(ns)\,\mathrm ds$.



3. Prove that the solution of the Robin or third boundary value problem (5) for the Laplace equation is unique when $\alpha>0$ is a constant.
Solution
Let $\tilde u$ and $u$ be solutions of
$\left\{\begin{aligned} &\Delta\tilde u=0\quad\text{in}~\Omega,\\&\frac{\partial\tilde u}{\partial\nu}+\alpha\tilde u=\beta\quad\text{on}~\partial\Omega;\end{aligned}\right.\quad\left\{\begin{aligned} &\Delta u=0\quad\text{in}~\Omega,\\&\frac{\partial u}{\partial\nu}+\alpha u=\beta\quad\text{on}~\partial\Omega.\end{aligned}\right.$
Define $w=\tilde u-u$ in $\bar\Omega$. Clearly, $w$ satisfies
$\left\{\begin{aligned} &\Delta w=0\quad\text{in}~\Omega,\\&\frac{\partial w}{\partial\nu}+\alpha w=0\quad\text{on}~\partial\Omega.\end{aligned}\right.$
Then by the Green's first identity (6) with $u=v=w$ and using the boundary condition, we have
$\displaystyle\int_\Omega\!|\nabla w|^2\,\mathrm dx=\int_{\partial\Omega}\!w\frac{\partial w}{\partial\nu}\,\mathrm dS-\int_\Omega\!w\Delta w\,\mathrm dx=-\alpha\int_{\partial\Omega}\!w^2\,\mathrm dS\leq0,$
which means $\nabla w\equiv0$ in $\Omega$. By the equality, we have $\displaystyle\int_\Omega\!w^2\,\mathrm dx=0$ and $w\equiv0$ in $\bar\Omega$. Therefore, $\tilde u\equiv u$ in $\bar\Omega$ and we complete the proof of uniqueness.


4. Let $\Omega$ be the unit disk as in Exercise 1.
   1. Solve the Robin problem (5) for the Laplace equation when $\alpha>0$ is a constant.
   2. When $\alpha=-1$, show that uniqueness fails.
Solution
1. As for Exercise 1, the solution $u$ can be expressed as

   $\displaystyle u(r,\theta)=A_0+\sum_{n=1}^\infty r^n(A_n\cos(n\theta)+B_n\sin(n\theta))$.

   In addition, the Robin boundary condition (5) becomes

   $\displaystyle\frac{\partial u}{\partial r}(1,\theta)+\alpha u(1,\theta)=\beta(\theta)\quad\text{for}~0\leq\theta<2\pi$.

   Then by the Robin boundary condition, we get

   $\displaystyle\alpha A_0+\sum_{n=1}^\infty(n+\alpha)[A_n\cos(n\theta)+(n+\alpha)B_n\sin(n\theta)]=\beta(\theta)\quad\text{for}~0\leq\theta<2\pi$.

   According to the Fourier series, the coefficients $A_0$, $A_n$ and $B_n$ can be determined by

   $\displaystyle A_0=\frac1{2\pi\alpha}\int_0^{2\pi}\!\beta(\theta)\,\mathrm d\theta,\quad A_n=\frac1{\pi(n+\alpha)}\int_0^{2\pi}\!\beta(\theta)\cos(n\theta)\,\mathrm d\theta,\quad B_n=\frac1{\pi(n+\alpha)}\int_0^{2\pi}\!\beta(\theta)\sin(n\theta)\,\mathrm d\theta\quad\text{for}~n\in\mathbb N$.

   Therefore, the solution $u$ of Laplace equation with the Robin boundary condition is given by

   $\displaystyle u(r,\theta)=\frac1{2\pi\alpha}\int_0^{2\pi}\!\beta(s)\,\mathrm ds+\frac1\pi\sum_{n=1}^\infty\frac{r^n}{n+\alpha}\left[\cos(n\theta)\int_0^{2\pi}\!\beta(s)\cos(ns)\,\mathrm ds+\sin(n\theta)\int_0^{2\pi}\!\beta(s)\sin(ns)\,\mathrm ds\right].$

2. When $\alpha=-1$ and $\beta=0$, the function $u(r,\theta)=r(a\sin\theta+b\cos\theta)$ solves the Laplace equation with the Robin boundary condition for any $a,b\in\mathbb R$, which shows there are infinite many solutions.


5. Suppose $q(x)\geq0$ for $x\in\Omega$ and consider solutions $u\in C^2(\Omega)\cap C^1(\bar\Omega)$ of $\Delta u-q(x)u=0$ in $\Omega$. Establish uniqueness theorems for (a) Dirichlet problem, and (b) the Neumann problem with $q(x_0)\neq0$ for some $x_0\in\Omega$.
Solution
1. Let $\tilde u$ and $u$ be solutions of

   $\left\{\begin{aligned} &\Delta\tilde u-q(x)\tilde u=0\quad\text{in}~\Omega,\\&\tilde u=g\quad\text{on}~\partial\Omega;\end{aligned}\right.\quad\left\{\begin{aligned} &\Delta u-q(x)u=0\quad\text{in}~\Omega,\\&u=g\quad\text{on}~\partial\Omega.\end{aligned}\right.$

   Define $w=\tilde u-u$ in $\bar\Omega$. Clearly, $w$ satisfies

   $\left\{\begin{aligned} &\Delta w-q(x)w=0\quad\text{in}~\Omega,\\&w=0\quad\text{on}~\partial\Omega.\end{aligned}\right.$

   Then by Green's first identity (6) with $u=v=w$ and using the boundary condition, we have

   $\displaystyle\int_\Omega\!|\nabla w|^2\,\mathrm ds=\int_{\partial\Omega}\!w\frac{\partial w}{\partial\nu}\,\mathrm dS-\int_\Omega\!w\Delta w\,\mathrm ds=-\int_\Omega\!q(x)w^2\,\mathrm dx\leq0$.

   Here we have used the fact that $q(x)\geq0$ in $\Omega$. Hence $\nabla w\equiv0$ and $w\equiv C$ in $\bar\Omega$, where $C$ is a constant. Since $w=0$ on $\partial\Omega$, $C=0$, and we arrive $w\equiv0$ in $\bar\Omega$, which means $\tilde u\equiv u$ in $\bar\Omega$. The proof of uniqueness is complete.
2. Let $\tilde u$ and $u$ be solutions of

   $\left\{\begin{aligned} &\Delta\tilde u-q(x)\tilde u=0\quad\text{in}~\Omega,\\&\frac{\partial\tilde u}{\partial\nu}=h\quad\text{on}~\partial\Omega;\end{aligned}\right.\quad\left\{\begin{aligned} &\Delta u-q(x)u=0\quad\text{in}~\Omega,\\&\frac{\partial u}{\partial\nu}=h\quad\text{on}~\partial\Omega.\end{aligned}\right.$

   Define $w=\tilde u-u$ in $\bar\Omega$. Clearly, $w$ satisfies

   $\left\{\begin{aligned} &\Delta w-q(x)w=0\quad\text{in}~\Omega,\\&\frac{\partial w}{\partial\nu}=0\quad\text{on}~\partial\Omega.\end{aligned}\right.$

   Then by Green's first identity (6) with $u=v=w$ and using the boundary condition, we have

   $\displaystyle\int_\Omega\!|\nabla w|^2\,\mathrm ds=\int_{\partial\Omega}\!w\frac{\partial w}{\partial\nu}\,\mathrm dS-\int_\Omega\!w\Delta w\,\mathrm ds=-\int_\Omega\!q(x)w^2\,\mathrm dx\leq0$.

   Here we have used the fact that $q(x)\geq0$ in $\Omega$. Hence $\nabla w\equiv0$ and $w\equiv C$ in $\bar\Omega$, where $C$ is a constant. From equaiton at $x=x_0$, we have $\Delta w(x_0)-q(x_0)w=0$, which means $C=0$ and hence $w\equiv0$ in $\bar\Omega$. Therefore, we get $\tilde u\equiv u$ in $\bar\Omega$ and complete the proof of uniqueness.
   
   It is clear that the uniqueness fails for $q\equiv0$ because any constant functions are solutions of Laplace equation with the Neumann boundary condition.


6. By direct calculation, show that $v(x)=|x-x_0|^{2-n}$ is harmonic in $\mathbb R^n\backslash\{x_0\}$ for $n\geq3$. Do the same for $v(x)=\log|x-x_0|$ if $n=2$.
Solution

Write $x_0=(x_{01},\dots,x_{0n})$.

• For $n\geq3$, a direct calculation gives

  $\displaystyle\frac{\partial v}{\partial x_i}=(2-n)|x-x_0|^{1-n}\cdot\frac{x_i-x_{0i}}{|x-x_0|}=(2-n)|x-x_0|^{-n}(x_i-x_{0i})\quad\text{for}~x\neq x_0$.

  Moreover, the second order derivative is

  $\begin{aligned}\frac{\partial^2v}{\partial x_i^2}&=-n(2-n)|x-x_0|^{-n-1}\cdot\frac{x_i-x_{0i}}{|x-x_0|}\cdot(x_i-x_{0i})+(2-n)|x-x_0|^{-n}\\&=(n^2-2n)|x-x_0|^{-n-2}(x_i-x_{0i})^2+(2-n)|x-x_0|^{-n}\quad\text{for}~x\neq x_0.\end{aligned}$

  Then we have

  $\begin{aligned}\Delta v&=\sum_{i=1}^n\frac{\partial^2v}{\partial x_i^2}=(n^2-2n)|x-x_0|^{-n-2}\sum_{i=1}^n(x_i-x_{0i})^2+n(2-n)|x-x_0|^{-n}\\&=(n^2-2n)|x-x_0|^{-n-2}\cdot|x-x_0|^2+n(2-n)|x-x_0|^{-n}=0\quad\text{for}~x\neq x_0.\end{aligned}$

• For $n=2$, a direct calculation gives

  $\displaystyle\frac{\partial v}{\partial x_i}=\frac1{|x-x_0|}\cdot\frac{x_i-x_{0i}}{|x-x_0|}=\frac{x_i-x_{0i}}{|x-x_0|^2}\quad\text{for}~x\neq x_0$.

  Moreover, the second order derivative is

  $\displaystyle\frac{\partial^2v}{\partial x_i^2}=\frac1{|x-x_0|^2}-2\frac{x_i-x_{0i}}{|x-x_0|^3}\cdot\frac{x_i-x_{0i}}{|x-x_0|}=\frac{|x-x_0|^2-2(x_i-x_{0i})^2}{|x-x_0|^4}\quad\text{for}~x\neq x_0.$

  Then we have

  $\displaystyle\Delta v=\sum_{i=1}^2\frac{\partial^2v}{\partial x_i^2}=2\frac{|x-x_0|^2-[(x_1-x_{01})^2+(x_2-x_{02})^2]}{|x-x_0|^4}=0\quad\text{for}~x\neq x_0.$



7. 1. If $\Omega$ is a bounded domain and $u\in C^2(\Omega)\cap C(\bar\Omega)$ satisfies (1), then $\max\limits_{\bar\Omega}|u|=\max\limits_{\partial\Omega}|u|$.
   2. If $\Omega=\{x\in\mathbb R^n\,:\,|x|>1\}$ and $u\in C^2(\Omega)\cap C(\bar\Omega)$ satisfies (1) and $\lim\limits_{|x|\to\infty}u(x)=0$, then $\displaystyle\max_{\bar\Omega}|u|=\max_{\partial\Omega}|u|$.
Solution
1. Since $u$ satisfies Laplace equation (1), we know $\Delta u\geq0$ in $\Omega$. Then by (16), we have $\displaystyle\max_{\bar\Omega}u=\max_{\partial\Omega}u$. On the other hand, Laplace equation implies that $\Delta(-u)\geq0$ in $\Omega$. Then by (16), we have $\displaystyle\max_{\bar\Omega}(-u)=\max_{\partial\Omega}(-u)$, which means $\displaystyle\min_{\bar\Omega}u=\min_{\partial\Omega}u$. Therefore, combining them, we find

   $\displaystyle\max_{\bar\Omega}|u|=\max\left\{\max_{\bar\Omega}u,-\min_{\bar\Omega}u\right\}=\max\left\{\max_{\partial\Omega}u,-\min_{\partial\Omega}u\right\}=\max_{\partial\Omega}|u|$,

   which completes the proof.
2. Let $\displaystyle M=\max_{\partial\Omega}|u|=\max_{\partial B_1(0)}|u|$. Clearly, $\displaystyle M\leq\max_{\bar\Omega}|u|$ because $\partial B_1(0)\subsetneq\bar\Omega=\mathbb R^n-B_1(0)$. Since $\displaystyle\lim_{|x|\to\infty}u(x)=0$, there exists $R>1$ such that $|u(x)|\leq M/2$ for $|x|\geq R$, which implies

   $\displaystyle\max_{\mathbb R^n-B_R(0)}|u|\leq M/2<M=\max_{\partial B_1(0)}\leq\max_{\bar\Omega}|u|$.

   Hence $\displaystyle\max_{\bar\Omega}|u|=\max_{\bar B_R(0)-B_1(0)}|u|$. Then by (a), we have

   $\displaystyle\max_{\bar\Omega}|u|=\max_{\partial B_1(0)\cup\partial B_R(0)}|u|=\max\left\{\max_{\partial B_1(0)}|u|,\max_{\partial B_R(0)}|u|\right\}=M=\max_{\partial\Omega}|u|$.

   Here we have used the fact that $\displaystyle\max_{\partial B_R(0)}|u|\leq M/2<M=\max_{\partial B_1(0)}|u|$.


8. Suppose $u\in C^2(\Omega)$ and $\Delta u\geq0$ in $\Omega$ where the boundary $\partial\Omega$ has the following property: For every $x_0\in\partial\Omega$ there is a ball $B_\epsilon(x_1)\subset\Omega$ such that $\partial\Omega\cap\overline{B_\epsilon(x_1)}=\{x_0\}$. Prove that if $u$ is not a constant and for some $x_0\in\partial\Omega$ we have $u(x_0)=\sup\{u(x)\,:\,x\in\Omega\}$, then the outward normal derivative of $u$ is positive at $x_0$: $\displaystyle\frac{\partial u}{\partial\nu}(x_0)>0$, where $\nu$ is the exterior unit normal to the ball.
Solution
This result is called Hopf's lemma whose proof can be found in many PDE textbooks. Let $x_0\in\partial\Omega$ satisfy $u(x_0)=\sup\{u(x)\,:\,x\in\Omega\}$ and $B_\epsilon(x_1)\subset\Omega$ such that $\partial\Omega\cap\overline{B_\epsilon(x_1)}=\{x_0\}$, where $x_1=(x_{11},\dots,x_{1n})$. Then we introduce an auxiliary function $v$ defined by
$v(x)=\exp\left(-\alpha|x-x_1|^2\right)-\exp\left(-\alpha \epsilon^2\right)\quad\text{for}~\epsilon/2<|x-x_1|<\epsilon$,
where $\alpha>2n/\epsilon$ is a constant. Clearly, $v(x)\geq0$ in $B_\epsilon(x_1)-\bar B_{\epsilon/2}(x_1)$. Moreover, $v$ satisfies
$\begin{aligned}\Delta v&=\sum_{i=1}^nv_{x_ix_i}=\sum_{i=1}^n(-2\alpha v(x)+4\alpha^2(x_i-x_{1i})^2v(x))\\&=v(x)[-2n\alpha+4\alpha^2|x-x_1|^2]\geq v(x)[\alpha^2\epsilon^2-2n\alpha]\\&=\alpha v(x)(\alpha \epsilon-2n)\geq0.\end{aligned}$
By Theorem 3 (Maximum Principle) at page 109 in the textbook, we have $u(x)-u(x_0)<0$ on $\partial B_{\epsilon/2}(x_1)$. There is a constant $\kappa>0$ such that $u-u(x_0)+\kappa v\leq0$ on $\partial B_{\epsilon/2}(x_1)$. In addition, we have $u-u(x_0)+\kappa v\leq0$ on $\partial B_\epsilon(x_1)$ because $v\equiv0$ on $\partial B_\epsilon(x_1)$. Now we find
$\begin{aligned} &\Delta(u-u(x_0)+\kappa v)=\Delta u+\kappa\Delta v\geq0\quad\text{in}~B_\epsilon(x_1)-\bar B_{\epsilon/2}(x_1),\\&u-u(x_0)+\kappa v\leq0\quad\text{on}~\partial B_\epsilon(x_1)\cup\partial B_{\epsilon/2}(x_1).\end{aligned}$
Then by Theorem 3 again, we have $u(x)-u(x_0)+\kappa v\leq0$ in $B_\epsilon(x_1)-\bar B_{\epsilon/2}(x_1)$. Taking the normal derivative at $x_0$, we find
$\displaystyle\frac{\partial u}{\partial\nu}(x_0)+\kappa\frac{\partial v}{\partial\nu}(x_0)\geq0$,
which implies
$\displaystyle\frac{\partial u}{\partial\nu}(x_0)\geq-\kappa\frac{\partial v}{\partial\nu}(x_0)=-\kappa v'(R)=2\alpha\epsilon\exp\left(-\alpha\epsilon^2\right)>0$.
The proof of Hopf's lemma is complete.


9. Suppose $u\in C(\Omega)$ satisfies the mean value property in $\Omega$.
   1. Show that $u$ satisfies the maximum principle (15).
   2. Show that boundary values on $\overline{B_r(\xi)}\subset\Omega$ uniquely determine $u$: If $v\in C(\overline{B_r(\xi)})$ satisfies the mean value property in $B_r(\xi)$ and $v=u$ on $\partial B_r(\xi)$, then $v\equiv u$ in $B_r(\xi)$.
Solution
1. Let $\displaystyle\Omega'=\{x\in\Omega:u(x)=\sup_{x\in\Omega}u(x)\}$ be the subset of $\Omega$. Due to the continuity of $u$, the set $u^{-1}(x_0)\cap\Omega=\Omega'$ is relatively closed in $\Omega$. On the other hand, for any $x'\in\Omega'\subseteq\Omega$, by the mean value property of $u$ (see (10)), we have

   $\displaystyle M=u(x')=\frac1{\omega_n}\int_{|\xi|=1}\!u(x'+R\xi)\,\mathrm dS_\xi\leq\frac1{\omega_n}\int_{|\xi|=1}\!M\,\mathrm dS_\xi=M.$,

   where $\overline{B_R(x')}\subsetneq\Omega$. Since the equality holds, we find $u(x)=M$ for $x\in B_R(x')$, which means $B_R(x')\subseteq\Omega'$, and hence $\Omega'$ is relatively open in $\Omega$. Due to the connectness of $\Omega$, $\Omega'$ must be empty set of $\Omega$.
   • If $\Omega'$ is empty, then $\displaystyle u(\xi)<\sup_{x\in\Omega}u(x)$ for all $\xi\in\Omega$.
   • If $\Omega'=\Omega$, then $u$ is a constant function.
   The proof is complete.
2. Since $u$ and $v$ both satisfy the mean value property in $B_r(\xi)$, then functions $w:=u-v$ and $-w=v-u$ both satisfy the mean value property in $B_r(\xi)$ with $w=-w=0$ on $\partial B_r(\xi)$. Suppose by contradiction that $w\not\equiv0$ in $\bar B_r(\xi)$. Then by (a) for $w$ and $-w$, we have

   $\displaystyle0=\min_{\partial B_r(\xi)}w=\inf_{B_r(\xi)}w<w(x)<\sup_{B_r(\xi)}w=\max_{\partial B_r(\xi)}w=0$,

   which leads a contradiction. Thus, $w$ is a constant. By $w=0$ on $\partial B_r(\xi)$, $w\equiv0$ in $\bar B_r(\xi)$, and hence $v\equiv u$ in $\bar B_r(\xi)$. The proof is complete.