Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.3
- Show that if $u\in C^2(\Omega)$ is subharmonic then $\Delta u\geq0$ in $\Omega$.
- Show that if $\Omega$ satisfies an exterior sphere condition at $z\in\partial\Omega$, then $z$ is a regular point.
- When $n\geq3$, it is easy to verify that
$\begin{aligned} &\frac{\partial Q_z}{\partial x_i}(x)=(2-n)|x-\xi|^{-n}\cdot(x_i-\xi_i),\\&\frac{\partial^2Q_z}{\partial x_i^2}(x)=(2-n)|x-\xi|^{-n}+n(n-2)|x-\xi|^{-n-2}(x_i-\xi_i)^2\quad\text{for}~x\in\Omega.\end{aligned}$
Thus, we have$\displaystyle\Delta Q_z(x)=n(2-n)|x-\xi|^{-n}+n(n-2)|x-\xi|^{-n-2}\sum_{i=1}^n(x_i-\xi_i)^2=0\quad\text{for}~x\in\Omega$,
which means $Q_z$ is harmonic and hence subharmonic in $\Omega$. Here we have used the fact that $\xi\notin\Omega$. On the other hand, since $|z-\xi|=R$, we have $Q_z(z)=0$. For $x\in\bar\Omega-\{z\}$, we have $|x-\xi|>R$ (since $\bar\Omega\cap\partial B_R(\xi)=\{z\}$), implies $Q_z(x)=|x-\xi|^{2-n}-R^{2-n}<0$. Therefore, $Q_z$ is indeed a barrier function at $z$. - When $n=2$, it is easy to verify that
$\displaystyle\frac{\partial Q_z}{\partial x_i}(x)=-\frac{x_i-\xi_i}{|x-\xi|^2},\quad\frac{\partial^2Q_z}{\partial x_i^2}=-\frac1{|x-\xi|^2}+\frac{2(x_i-\xi_i)^2}{|x-\xi|^4}\quad\text{for}~x\in\Omega.$
Thus, we have$\displaystyle\Delta Q_z(x)=-\frac2{|x-\xi|^2}+\frac{2|x-\xi|^2}{|x-\xi|^4}=0\quad\text{for}~x\in\Omega,$
which means $Q_z$ is harmonic and hence subharmonic in $\Omega$. Here we have used the fact that $\xi\notin\Omega$. On the other hand, since $|z-\xi|=R$, we have $Q_z(z)=0$. For $x\in\bar\Omega-\{z\}$, we have $|x-\xi|>R$ (since $\bar\Omega\cap\partial B_R(\xi)=\{z\}$), which implies $Q_z(x)=\log R-\log|x-y|<0$. Therefore, $Q_z$ is indeed a barrier function at $z$. - Show that a bounded domain $\Omega$ with $C^2$ boundary $\partial\Omega$ satisfies an exterior sphere condition and hence is regular.
- If $n=2$ and $\Omega$ satisfies an exterior cone condition at $z\in\partial\Omega$, that is, there is a cone $C=\{x\in\mathbb R^2\,:\,|x-z|<\epsilon,\,|\text{arg}(x-z)-\theta_0|<\delta\}$ such that $C\cap\bar\Omega=\{z\}$, then $z$ is a regular point. (Ths is similarly true for $n\geq3$.)
- In the textbook "Partial Differential Equations: Methods and Applications", a barrier function at $z\in\partial\Omega$ is a function $Q_z\in C(\bar\Omega)$ that is subharmonic in $\Omega$ with $Q_z(z)=0$ and $Q_z(x)<0$ for $x\in\partial\Omega-\{z\}$. This definition is too weaker to control the sign of barrier in the domain $\Omega$. In many books (for example, "Elliptic Partial Differential Equations of Second Order" and "Potential Theory"), they call a function $Q_z$ as a barrier function at $z\in\partial\Omega$ if $Q_z$ is subharmonic in $\Omega$, $Q_z(z)=0$ and $Q_z(x)<0$ for $x\in\bar\Omega-\{z\}$. It is clear that such barrier function still fullfil the McOwen's condition.
- The existence of barrier function is a local property: Let $\mathcal N_z$ be a neighborhood of the boundary point $z$. If function $Q_z(x)$ is subharmonic in $\Omega\cap\mathcal N_z$ and satisfies $Q_z(z)=0$ and $Q_z(x)<0$ in $\bar\Omega\cap\mathcal N_z-\{z\}$, then we call such $Q_z$ is a local barrier at $z\in\partial\Omega$. Then we construct a barrier function $\tilde Q_z$ with respect to $z\in\partial\Omega$ as follows. Let $B$ be an open ball satisfying $z\in B\subset\subset\mathcal N$ and $\displaystyle M=\sup_{\mathcal N-B}Q_z<0$. We define
$\tilde Q_z(x)=\begin{cases}\max\{Q_z(x),M\}&\text{if}~x\in\bar\Omega\cap B;\\M&\text{if}~x\in\bar\Omega-B.\end{cases}$
Hence $\tilde Q_z$ is a barrier at $z$ relative to $\Omega$ because $\tilde Q_z$ is continuous on $\bar\Omega$, subharmonic in $\Omega$ (by Lemma 2 at page 129) and $\tilde Q_z(x)<0$ for $x\in\bar\Omega-\{z\}$.
Solution
For any $x\in\Omega$, there exists a positive constant $R$ such that $\overline{B_R(x)}\subsetneq\Omega$. Since $u\in C^2(\Omega)$ is subharmonic, we have$\displaystyle u(x)\leq\frac1{\omega_n}\int_{|\xi|=1}\!u(x+r\xi)\,\mathrm dS_\xi\quad\text{for}~0<r\leq R$,
which implies$\begin{aligned}u(x)&=\frac n{r^n}\int_0^r\!u(x)\rho^{n-1}\,\mathrm d\rho\leq\frac n{\omega_nr^n}\int_0^r\!\rho^{n-1}\int_{|\xi|=1}\!u(x+\rho\xi)\,\mathrm dS_\xi\,\mathrm d\rho\\&=\frac{n}{\omega_nr^n}\int_0^r\!\rho^{n-1}\cdot\frac1{\rho^{n-1}}\int_{\partial B_\rho(x)}\!u(y)\,\mathrm dS_y\,\mathrm d\rho=\frac n{\omega_nr^n}\int_0^r\!\int_{\partial B_\rho(x)}\!u(y)\,\mathrm dS_y\,\mathrm d\rho\\&=\frac{n}{\omega_nr^n}\int_{B_r(x)}\!u(y)\,\mathrm dy=\frac n{\omega_n}\int_{|\xi|\leq1}\!u(x+r\xi)\,\mathrm d\xi\quad\text{for}~0<r\leq R.\end{aligned}$
Suppose by contradiction that $\Delta u(x_0)<0$ for some $x_0\in\Omega$. Since $u\in C^2(\Omega)$, there exists $R_0>0$ such that $\overline{B_{R_0}(x_0)}\subsetneq\Omega$ and $\Delta u(x)<0$ for $x\in\overline{B_{R_0}(x_0)}$. Consider the function $\phi:(0,\infty)\to\mathbb R$ defined by$\displaystyle\phi(r)=\frac1{\omega_n}\int_{|\xi|=1}\!u(x_0+r\xi)\,\mathrm dS_\xi$.
Since $u$ is continuous, we have $\displaystyle\lim_{r\to0^+}\phi(r)=u(x_0)$. Moreover, we find$\begin{aligned}\phi'(r)&=\frac1{\omega_n}\int_{|\xi|=1}\!\nabla u(x_0+r\xi)\cdot\xi\,\mathrm dS_\xi=\frac1{\omega_nr^{n-1}}\int_{\partial B_r(x_0)}\!\nabla u(y)\cdot\frac{y-x_0}r\,\mathrm dS_y\\&=\frac1{\omega_nr^{n-1}}\int_{B_r(x_0)}\!\Delta u(y)\,\mathrm dy<0\quad\text{for}~0<r\leq R_0.\end{aligned}$
This means $\phi$ is strictly decreasing for $r\in(0,R_0]$, i.e.,$u(x_0)=\lim_{r\to0^+}\phi(r)>\phi(R_0)=\frac1{\omega_n}\int_{|\xi|=1}\!u(x+R_0\xi)\,\mathrm dS_\xi$,
which leads a contradiction. Hence $\Delta u(x_0)\geq0$ for all $x_0\in\Omega$ and we complete the proof.Solution
Let $B_R(\xi)$ be the exterior ball with respect to the boundary point $z\in\partial\Omega$, i.e., $B_R(\xi)\subset\mathbb R^n-\bar\Omega$ and $\partial B_R(\xi)\cap\bar\Omega=\{z\}$. Then we claim$Q_z(x)=\begin{cases}|x-\xi|^{2-n}-R^{2-n}&\text{if}~n\geq3;\\\log R-\log|x-\xi|&\text{if}~n=2\end{cases}$
is a barrier function at $z$ so $z$ is a regular point (cf. Definitions at page 130).Solution
By translation and rotation, we may assume that $0\in\partial\Omega$ and $e_n=(0,\dots,0,1)$ is the exterior unit normal vector at $0$. Moreover, we write $\partial\Omega\cap B_\delta(0)=\{(x',x_n)\in\mathbb R^n\,:\,x_n=f(x'),\,|x'|<\delta\}$ for some small positive $\delta$. Since $\partial\Omega$ is $C^2$, function $f$ is also $C^2$ in $B_\delta'(0)=\{x'\in\mathbb R^{n-1}\,:\,|x'|<\delta\}$. Note that$\Omega\cap B_\delta(0)=\{(x',x_n)\in\mathbb R^n\,:\,x_n<f(x'),\,|x'|<\delta,\,|x_n|<\delta\}$.
The Hessian matrix of $f$ at the origin is a $(n-1)\times(n-1)$ matrix given by$Hf=[f_{x_ix_j}(0)]_{1\leq i,j\leq n}$,
which is symmetric and hence diagonalizable with real eigenvalues $\lambda_1,\dots,\lambda_{n-1}$. Let $C=\max\{\lambda_1,\dots,\lambda_{n-1}\}$ (which may be negative). Then for sufficiently small $\delta>0$, we have$f(x')\leq C|x'|^2\quad\text{for}~|x'|<\delta$.
Then we pick $r>0$ with $r^{-1}>\max\{2C,\delta^{-1}\}$, then we can prove that $B_r(0,\dots,0,r)$ lies in the region $x_n\geq C|x'|^2$ and $\bar B_r(0,\dots,0,r)\cap\bar\Omega=\{0\}$. For $x=(x',x_n)\in B_r(0,\dots,0,r)$, we have $|x'|^2+(x_n-r)^2\leq r^2$, which implies $|x'|^2\leq2x_nr$, and $C|x'|^2<(2r)^{-1}|x'|^2\leq x_n$. Thus, one may see that$\begin{aligned}\partial B_r(0,\dots,0,r)\cap\bar\Omega&\subseteq\{(x',x_n)\in\mathbb R^n\,:\,x_n\geq C|x'|^2\}\cap(\bar\Omega\cap\bar B_\delta(0))\\&=\{(x',x_n)\in\mathbb R^n\,:\,x_n\geq C|x'|^2\}\cap\{(x',x_n)\in\mathbb R^n\,:\,x_n\leq C|x'|^2\}\\&=(x',x_n)\in\mathbb R^n\,:\,x_n=C|x'|^2\}.\end{aligned}$
By $r^{-1}>\max\{2C,\delta^{-1}\}$, we can use $|x'|^2=2x_nr$ (the boundary of $B_r(0,\dots,0,r)$ to get $(x',x_n)=0$. Thus, $\bar B_r(0,\dots,0,r)\cap\bar\Omega=\{0\}$. Therefore, $B_r(0,\dots,0,r)$ is an exterior ball with respect to the boundary $\partial\Omega$ at the origin.Solution
We firstly remark the notion of barrier function at a boundary point $z\in\partial\Omega$.
Let $\Omega_0=\Omega\cup(B_\epsilon(z)-C)$. Then $z$ is also a boundary point of $\Omega_0$. Since $\Omega\subset\Omega_0$ and $z\in\partial\Omega'\cap\partial\Omega$, it suffices to show that there is a barrier function at $z\in\partial\Omega_0$ with respect to $\Omega_0$. Since a barrier at $z$ on $B_r(z)-C$ is a local barrier at $z$ on $\Omega_0$, it is enough to construct a barrier at $z$ on $B_\epsilon(z)-C$. Henceforth, by the translation and rotation, we may assume that $z=0$, $\theta_0=0$ and $\Omega=B_\epsilon(0)-C$.
Due to the exterior cone condition, all boundary points of $\Omega$, except possibly for $0$, are regular points. Let $\Omega'$ be a magnification of $\Omega$ by a factor of $2$, i.e., $\Omega'=\{2x\in\mathbb R^2\,:\,x\in\Omega\}$. Let $r(x)=-|x|$ for $x\in\bar\Omega'$. Now let $H_r$ be the Dirichlet solution on $\Omega'$ with the boundary data $r$ on $\partial\Omega'$ by Perron's method. Since $H_r$ is harmonic on $\Omega'$, by the strong maximum principle, $H_r(x)\leq-|x|<0$ for all $x\in\Omega'$. Then it remains to show that $\displaystyle\lim_{x\to0,\,x\in\Omega}H_r(x)=0$ so that $H_r$ is a barrier function with respect to $0$ relative to $\Omega$. Define functions $u$ in $\bar\Omega'-\{0\}$ and $v$ in $\bar\Omega-\{0\}$ by$\begin{aligned} &u(x)=\begin{cases}H_r(x)&\text{if}~x\in\Omega,\\r(x)&\text{if}~\partial\Omega'-\{0\}.\end{cases}\\&v(x)=u(2x).\end{aligned}$
It is clear that functions $u$ and $v$ are continuous on $\bar\Omega'-\{0\}$ and $\bar\Omega-\{0\}$, respectively. Then we show that $u>v$ on $\partial\Omega\cap\partial B_\epsilon(0)$. Since $r=-|x|\geq-2\epsilon$ on $\partial\Omega'$, we find $-2\epsilon\in S_r=\{u\in C(\bar\Omega')\,:\,u~\text{is harmonic in}~\Omega'~\text{and}~u\leq r~\text{on}~\partial\Omega'\}$. Hence $u=H_r\geq-2\epsilon$ in $\Omega'$ by the definition of Perron's method. Since $u$ is harmonic on $\Omega'$ and not a constant in $\bar\Omega'$, it cannot attain its minimum value at any point of $\Omega'$ by the maximum principle. Thus $u>-2\epsilon$ in $\Omega'$. In particular, $u>-2\epsilon$ in $\Omega'\cap\partial\Omega\cap\partial B_\epsilon(0)$. On the other hand, since $v=-2\epsilon$, we obtain $u>v$ on $\Omega_2\cap\partial\Omega\cap\partial B_\epsilon(0)$. For $y\in\partial\Omega'\cap\partial\Omega\cap\partial B_\epsilon(0)$, $y$ is a regular boundary with respect to $\Omega'$, which implies $\displaystyle\lim_{x\to y}u(x)=-\epsilon$ and hence $u(y)=-\epsilon>-2\epsilon=v(y)$. Thus, $u>v$ on $\partial\Omega\cap\partial B_\epsilon(0)$. Since $u$ and $v$ are continuous on the compact set $\partial\Omega\cap B_\epsilon(0)$, there is a constant $\alpha\in(1/2,\,1)$ such that $u>\alpha v$ on $\partial\Omega\cap\partial B_\epsilon(0)$. On $\partial\Omega-(\partial\Omega\cap\partial B_\epsilon(0))-\{0\}$, $u=v/2$. Therefore, $u>\alpha v$ on $\partial\Omega-\{0\}$. Since $u$ and $v$ are continuous on $\bar\Omega-\{x\}$, we have $\displaystyle\lim_{x\to y,\,x\in\Omega}(\alpha v(x)-u(x))\leq0$ for all $y\in\partial\Omega-\{0\}$. Now we define the function $U(x)=\alpha v(x)-u(x)-\rho(K(|x|)+c)$ where $\rho>0$ and $c$ can be chosen so that $K(|x|)+c\geq0$ on $\bar\Omega$, where $K$ is the fundament solution. Then we see that$\displaystyle\limsup_{x\to y}U(x)\leq\begin{cases}0&\text{if}~y\in\partial\Omega-\{0\},\\-\infty&\text{if}~y=0.\end{cases}$
Since $U$ is subharmonic in $\Omega$, we get $U\leq0$ in $\Omega$. Then for any closed ball $B_\delta(x)\subseteq\Omega$, we have$\displaystyle\alpha\frac1{|B_\delta(x)|}\int_{B_\delta(x)}\!v(y)\,\mathrm dy-\frac1{|B_\delta(x)|}\int_{B_\delta(x)}\!u(y)\,\mathrm dy-\rho\left(\frac1{|B_\delta(x)|}\int_{B_\delta(x)}\!K(|y|)\,\mathrm dy+c\right)\leq0.$
Letting $\rho\to0$ and then $\delta\to0$, we get $\alpha v(x)-u(x)\leq0$ for any $x\in\Omega$. Therefore, we obtain$\displaystyle\liminf_{x\to0,\,x\in\Omega}u(x)\geq\alpha\liminf_{x\to0,\,x\in\Omega}v(x)=\alpha\lim_{x\to0,\,x\in\Omega}u(2x).$
Note that $\alpha\neq1$ and two extreme limits are the same, we get$\displaystyle\liminf_{x\to0,\,x\in\Omega}H_r(x)=\lim_{x\to0,\,x\in\Omega}u(x)=0.$
Since $H_r(x)\leq0$, we get $\displaystyle\lim_{x\to0,\,x\in\Omega}H_r(x)=0$ and $H_r$ is a barrier function. The proof is complete.This proof is due to Zaremba.
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