## Tag Archives: Neumman problem

### 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 ${u_{x_ix_j}=0}$, ${\forall\, i,j=1,2\cdots, n-1}$.