## Tag Archives: Dirichlet problem

### Rigidity of upper half plane

Thm: Suppose ${\mathbb{R}^n_+}$ is the upper half plane. ${u\in C^0(\overline{\mathbb{R}^n_+})}$ is non-negative harmonic function in ${\mathbb{R}^n_+}$ and ${u\equiv 0}$ on ${\partial \mathbb{R}^n_+}$. Then ${u=ax_n}$ for some ${a\geq 0}$. Note that there is no requirement on the growth of $u$ at infinity.

Lemma: Let ${u}$, ${v}$ be positive solutions to ${Lw=0}$ in ${B^+_1}$ continuously vanishing on ${\{x_n=0\}}$ with

$\displaystyle u(\frac{1}{2}e_n)=v(\frac{1}{2}e_n)=1$

Then, in ${\bar{B}^+_{1/2}}$,

$\displaystyle \frac{v}{u}\text{ is of class }C^\alpha$

and

$\displaystyle ||\frac{v}{u}||_{L^\infty(B^+_{1/2})},\quad ||\frac{v}{u}||_{C^\alpha(B^+_{1/2})}\leq C(n,\lambda)$

Remark: Refer to Caffarelli’s book, A Geometric Approach to Free Boundary Problems, this lemma is related to boundary harnack inequality or Carleson estimate.

Proof: For ${x=(x_1,\cdots,x_{n})}$, denote ${x'=(x_1,\cdots,x_{n-1})}$, then ${x=(x',x_n)}$. Fix ${\forall \,t>0}$, define ${\tilde{u}(x)=u(2xt)/u(0,t)}$, ${\tilde{v}(x)=2x_n}$ for ${x\in B^+_1}$. Apply the lemma, we can get

$\displaystyle \tilde{u}(x)\leq C\tilde{v}(x)\quad \forall \, x\in B^+_{1/2}$

$\displaystyle C^{-1}\tilde{v}(x)\leq \tilde{u}(x)\quad \forall \, x\in B^+_{1/2}$

where ${C=C(n)\geq 1}$. These are equivalent to

$\displaystyle u(x)\leq Cx_n\frac{u(0,t)}{t} \quad \forall \,x=(x',x_n)\in B^+_{t}\quad (1)$

$\displaystyle C^{-1}x_n\frac{u(0,t)}{t}\leq u(x)\quad \forall \, x=(x',x_n)\in B^+_{t}\quad (2)$

Suppose

$\displaystyle \lim\limits_{\overline{t\rightarrow+\infty}}\frac{u(0,t)}{t}=a$

(i) If ${a=0}$, then ${\forall \epsilon>0}$, ${\exists\, t>0}$ large enough such that ${(1)}$ implies

$\displaystyle u(x)\leq \epsilon Cx_n\quad \forall \,x=(x',x_n)\in B^+_1\subset B^+_{t}\quad$

This means ${u\equiv 0}$ in ${B^+_1}$. Hence ${u\equiv 0}$ on ${\mathbb{R}^n_+}$.

(ii) If ${a>0}$, then ${\forall\, \epsilon>0}$, ${\exists\, t>r}$ large enough such that ${(2)}$ implies that

$\displaystyle u(x)\geq (a-\epsilon)C^{-1}x_n \quad \forall \,x\in B^+_r\subset B^+_{t}$

Since ${r,\epsilon>0}$ are arbitrary, then

$\displaystyle u(x)\geq aC^{-1}x_n\text{ in }\mathbb{R}^n_+$

Define

$\displaystyle v_1=u(x)-aC^{-1}x_n$

Then ${v_1}$ is non-negative and harmonic in ${\mathbb{R}^n_+}$, however

$\displaystyle \lim\limits_{\overline{t\rightarrow+\infty}}\frac{v_1(0,t)}{t}=a(1-C^{-1})$

Applying the above lemma and repeating the above procedure, we get

$\displaystyle v_1\geq a(1-C^{-1})C^{-1}x_n$

So we construct

$\displaystyle v_2=v_1-a(1-C^{-1})C^{-1}x_n$

Then ${v_2}$ is non-negative and harmonic in ${\mathbb{R}^n_+}$ with ${\lim\limits_{\overline{t\rightarrow+\infty}}\frac{v_2(0,t)}{t}=a(1-C^{-1})^{2}}$. Inductively, we construct

$\displaystyle v_k=v_{k-1}-a(1-C^{-1})^{k-1}C^{-1}x_n=u-\sum_{i=0}^{k-1}a(1-C^{-1})^{i}C^{-1}x_n$

Then ${v_\infty=u-ax_n}$ is non-negative and harmonic in ${\mathbb{R}^n_+}$ with

$\displaystyle \lim\limits_{\overline{t\rightarrow+\infty}}\frac{v_\infty(0,t)}{t}=0$

From (i), we know $v_\infty\equiv 0$, that is $u=ax_n$.

Remark: This problem was presented to me by Tianling Jin. As he said, there should have a very elementary proof, which do not require such advanced theorem.

### Mixed boundary condition and uniqueness of solution

$\bf{Problem: }$ Let $L=a^{ij}D_{ij}u+b^i(x)D_iu+c(x)u$ is uniformly elliptic in a bounded domain $\Omega$, $c\leq 0$, $c/\lambda$ is bounded and $Lu=0$ in $\Omega$
(1) Let $\partial \Omega=S_1\cup S_2$ ($S_1$ non-empty) and assume an interior sphere condition at each point of $S_2$. Suppose $u\in C^2(\Omega)\cap C^1(\Omega\cup S_2)\cap C^0(\overline{\Omega})$ satisfies the mixed boundary condition
$\displaystyle u=0$ on $S_1$, $\displaystyle \sum \beta_iD_i u=0$ on $S_2$
where the vector $\boldsymbol{\beta}(x)=(\beta_1(x),\cdots,\beta_n(x))$ has a non-zero normal component(to the interior sphere) at each point $x\in S_2$, then $u\equiv 0$.

(2) Let $\partial \Omega$ satisfy an interior sphere condition, and assume that $u\in C^2(\Omega)\cap C^1(\overline{\Omega})$ satisfies the regular oblique derivative boundary condition
$\displaystyle \alpha(x)u+\sum \beta(x)D_iu=0$ on $\partial \omega$
where $\alpha(\vec{\beta}\cdot \vec{\nu}>;;;;;;;;;0$, $\nu$ is the outer normal vector. Then $u\equiv 0$.

$\mathbf{Proof: }$ (1) Suppose $u\not\equiv$ constant. If there exists $x\in \Omega$ such that $u(x)>;;;;;;;;0$, then $\exists\, x_0\in\partial S_2$ such that $\sup _{x\in \Omega}u=u(x_0)$. WLOG, assume $x_0$ is the origin and $\vec{\nu}$ is pointing to the negative $x_n-$axis. And $B$ is the interior ball at $x_0\in S_2$. Since $u$ is not constant, we know that $u(x)0$. This means $D_nu(x-0)>;0$
Since $u\in C^1(\Omega\cup S_2)$, then $u\in C^1(B\cup x_0)$. So $\displaystyle D_iu(x_0)=0$ for $i=1,2,\cdots,n-1$. So $\vec{beta}\cdot \vec{Du}=0$ means that $\beta_n=0$, this contradicts to the fact that $\vec{\beta}$ has non-zero component on $\vec{nu}$.
So $u\leq 0$ in $\Omega$. Similarly, $-u\leq 0$. So $u\equiv 0$.
(2) Use the same technique.

$\bf{Remark:}$ Gilbarg, Trudinger’ book. Chapter 3, exercise 3.1.

### Boundary condition and uniqueness

$\mathbf{Problem:}$ Prove that if $\Delta u=0$ in $\Omega\subset \mathbb{R}^n$ and $u=\partial u/\partial\nu=0$ on an open smooth portion of $\partial \Omega$, then $u$ is identically zero.

Remark: I saw this nice proof from my friend and also someone anonymous on the internet notified me he/she has a similar idea.

$\mathbf{Proof:}$ Extend $u$ to be zero near the smooth part outside the domain(suppose it is $\Omega\cup B_r(x_0)$). Then we get a $C^1$ function which vanishes on an open set. If we can show $u$ is a harmonic function, then by the analyticity, $u\equiv 0$.

To do that we only need to show that $u$ satisfies $\int_{B_{\varepsilon}(z)}\frac{\partial u}{\partial \nu}ds=0$ for any $\varepsilon$ small enough(more or less like the local mean value property) where $\nu$ is the outer unit normal of $B_\varepsilon(z)$. For any point $z\in \Omega$ and $z\in B_r(x_0)\backslash \Omega$, mean value property holds locally. For any point $a\in \partial \Omega\cap B_r(x_0)$, we need to show for any $\varepsilon$ small enough

$\int_{B_\varepsilon(z)}\frac{\partial u}{\partial \nu}ds=0$

However, $B_{\varepsilon}(z)\cap \Omega$ is a domain with piecewise smooth boundary when $\varepsilon$ is small enough, therefore

$\int_{B_\varepsilon(z)}\frac{\partial u}{\partial \nu}ds=\int_{B_\varepsilon(z)\cap \Omega}\frac{\partial u}{\partial \nu}ds=\int_{B_\varepsilon(z)\cap \Omega}\Delta u(y)dy=0.$

where we have used $u=\partial u/\partial\nu=0$ on $B_\varepsilon(z)\cap \partial\Omega$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Proof:}$ Suppose $\Gamma\subset \partial\Omega$ is this portion. Choose $x_0\in \Gamma$ and a ball $B$ centered at $x_0$ small enough.

Define $\displaystyle w=u(x)$ when $x\in B\cap \Omega$ and $w=0$ when $x\in \overline{B}\backslash\Omega$. Then $w$ is harmonic on $B\cap \Omega$ and $B\backslash\overline{\Omega}$. If we can prove $w$ has continuous second partial derivatives, then $w$ will be harmonic on whole $B$, thus $w$ must be identically zero since harmonic functions are analytic ones.

For any point $x\in B\cap\Gamma$. Since Laplace equation is invariant with respect to rotation and translation, we can assume $x$ is the origin and the outer normal vector is negative $x_n$ axis. And $\displaystyle \phi(x_1,x_2,\cdots,x_{n-1})$ is the boundary $B\cap \Gamma$

Since $\Gamma$ is smooth and $\displaystyle u|_{\Gamma}$ is smooth along $\Gamma$, we can assume $u\in C^2(\overline{B\cap \Omega})$. Then we have

$u_{x_i}(0)=\left(u|_{\Gamma}\right)_{x_i}(0)=0$, for $i=1,2,\cdots, n-1$

$u_{x_ix_i}(0)=\left(u|_{\Gamma}\right)_{x_ix_i}(0)=0$, for $i=1,2,\cdots, n-1$.

$\displaystyle u_{x_n}(0)=-\frac{\partial u}{\partial \nu}(0)=0$ and $\displaystyle u_{x_nx_n}(0)=-\sum\limits_{i=1}^{n-1}u_{x_ix_i}=0$

Noticing $u(x_1,x_2,\cdots,x_{n-1}, \phi(x_1,x_2,\cdots,x_{n-1}))=0\quad (1)$

Differentiate (1) twice

$\displaystyle u_{x_ix_i}+u_{x_ix_n}\phi_{x_i}+u_{x_nx_n}\phi^2_{x_i}+u_{x_n}\phi_{x_ix_i}=0$ for $i=1,2,\cdots,n-1$.

Using the previous equalities, at origin we have $u_{x_ix_n}=0$ for $i=1,2,\cdots,n-1$. Thus we have proved all the second partial derivatives of $u$ are zeros.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Gilbarg Trudinger Book exercise 2.2. Up to now I am not sure this proof is right. Its proof shouldn’t be very difficult. I spent two weeks to realize that $u_{x_i}(0)=\left(u|_{\Gamma}\right)_{x_i}(0)$. It is such a trivial fact that I never notice. I would like to thank the insightful help of Jingang Xiong.

$\mathbf{Remark:}$ See a new proof

Suppose ${\Gamma\subset \partial\Omega}$ is this portion. Choose ${x_0\in \Gamma}$ and a ball ${B}$ centered at ${x_0}$ small enough. Define ${\displaystyle w=u(x)}$ when ${x\in B\cap \Omega }$ and ${w=0}$ when ${x\in \overline{B}\backslash\Omega}$. Then ${w}$ is harmonic on ${B\cap \Omega }$ and ${B\backslash\overline{\Omega}}$. If we can prove ${w}$ has continuous second partial derivatives, then ${w}$ will be harmonic on whole ${B}$, thus ${w}$ must be identically zero since harmonic functions are analytic ones.

Since Laplace equation is invariant with respect to rotation and translation, we can assume ${\Gamma\cap B}$ can be represented locally ${x_n=\phi(x_1,x_2,\cdots,x_{n-1})}$, ${x'=(x_1,\cdots,x_{n-1})\in V\subset\mathbb{R}^{n-1}}$. Since ${\Gamma}$ is smooth, and ${u=0}$ is ${C^\infty}$ on ${\Gamma}$, we can assume ${u\in C^2(\overline{B\cap \Omega})}$. We will prove ${u_{x_i}|_{\Gamma\cap B}=0}$ and ${u_{x_ix_j}|_{\Gamma\cap B}=0}$.

Since ${u(x',\phi(x'))=0}$, ${x'\in V}$, differentiating with respect to ${x_i}$, we get

$\displaystyle u_{x_i}+u_{x_n}f_{x_i}=0\text{ for } i=1,2,\cdots, n-1\quad (1)$

Also ${\displaystyle \frac{\partial u}{\partial \nu}=0}$ on ${\Gamma\cap B}$ implies

$\displaystyle \sum\limits_{i=1}^{n-1} u_{x_i}f_{x_i}-u_{x_n}=0\quad (2)$

Multiply ${(1)}$ by ${f_{x_i}}$, sum ${i}$ from ${1}$ to ${n-1}$ and then subtract ${(2)}$ we get

$\displaystyle u_{x_n}=0\text{ hence } u_{x_i}=0, i=1,\cdots,n-1.$

Differentiating ${(1)}$ furthermore with respect to ${x_j}$, use ${u_{x_n}=0}$ on ${B\cap \Gamma}$ we get

$\displaystyle u_{x_ix_j}+u_{x_ix_n}f_{x_j}+u_{x_ix_j}f_{x_i}+u_{x_nx_n}f_{x_j}f_{x_i}=0, \forall\, i,j=1,2\cdots, n-1\quad (3)$

Differentiating ${(2)}$ furmore with respect to ${x_j}$, use ${u_{x_i}=0}$ on ${B\cap \Gamma}$ we get

$\displaystyle \sum_{i=1}^{n-1}u_{x_ix_j}f_{x_i}+u_{x_ix_n}f_{x_i}f_{x_j}-u_{x_nx_j}-u_{x_nx_n}f_{x_j}=0, \forall\, j=1,2,\cdots,n-1\quad (4)$

Multiplying ${(3)}$ by ${f_{x_i}}$ and sum ${i}$, then subtract ${(4)}$ we get

$\displaystyle u_{x_nx_j}+u_{x_nx_n}f_{x_j}=0\quad (5)$

From this ${(3)}$ is equivalent to

$\displaystyle u_{x_ix_j}+u_{x_ix_n}f_{x_j}=0\quad (6)$

Let ${j=i}$ in ${(6)}$ and sum ${i}$, we get

$\displaystyle \sum\limits_{i}^{n-1}u_{x_ix_i}=-\sum\limits_{i=1}^{n-1}u_{x_ix_n}f_{x_i}=\sum\limits_{i=1}^{n-1}u_{x_nx_n}f^2_{x_i}$

the last equality is obtained from ${(5)}$. Note that ${\Delta u=0}$, the above equality means that

$\displaystyle \sum\limits_{i=1}^{n-1}u_{x_nx_n}f^2_{x_i}=-u_{x_nx_n}$

this implies that ${u_{x_nx_n}=0}$. From ${(5)}$, ${u_{x_ix_n}=0}$, ${i=1,2,\cdots,n-1}$. Then ${(6)}$ implies ${u_{x_ix_j}=0}$, ${\forall\, i,j=1,2\cdots, n-1}$.