Tag Archives: harmonic functions

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.

Counterexample of regularity of newtonian potential

Define the Newtonial potential in a bounded domain $\Omega$ by

$\displaystyle N(x)=\int_{\Omega}\Gamma(x-y)f(y)dy$

If $f$ is bounded and integrable on $\Omega$, then $N\in C^1(\mathbb{R}^n)$. And if $f$ is bounded and locally holder continuous in $\Omega$ then $N\in C^2(\Omega)$ and $\Delta N=f$. But when $f$ is only continuous, $N$ is not necessarily secondly differentiable. Here is a counterexample.

$\mathbf{Problem:}$ Let $P$ be a homogeneous harmonic polynomial of degree 2. Suppose $D^\alpha P\neq 0$ for some multi-index $\alpha=2$, for instance $P=x_1x_2, D_{12}P\neq 0$. Choose a cut-off function $\eta\in C^\infty_0(\{x||x|<2\})$ with $\eta=1$ when $|x|<1$. Denote $t_k=2^k$, and let $c_k\to 0$ as $k\to \infty$ and $\sum c_k$ divergent. Define

$\displaystyle f(x)=\sum\limits_0^\infty c_k\left(\Delta(\eta P)\right)(t_kx)$

Prove $f$ is a continuous function but $\Delta u=f$ has no $C^2$ solution near the origin.

$\mathbf{Proof:}$ If $x=0$, then $f=0$. If $x\neq 0$, then there exists only one $k_0$ such that $1\leq |t_{k_0}x|\leq 2$. Since $P$ is a harmonic polynomial and $\eta$ has compact support,

$f(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)\leq Const.c_{k_0}$

So $f\in C^\infty(\mathbb{R}^n\backslash \{0\})$. As $x\to 0$, $c_k\to 0$, so $f(x)\to 0$, which means $f$ is continuous at 0.

Let $\displaystyle w(x)=\sum\limits_{0}^\infty t^{-2}_kc_k(\eta P)(t_kx)$, then it is easy to prove $w$ is well defined and for $x\neq 0$ and the unique $k_0$ such that $1\leq |t_{k_0}x|\leq 2$,

$\displaystyle D^\alpha w(x)=\sum\limits_0^{k_0}c_k+c_{k_0}D^\alpha(\eta P)(t_{k_0}x)\quad (1)$

$\Delta w(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)=f(x) \quad (2)$

(1) means $w\in C^2(\mathbb{R}^n\backslash\{0\})$ but $w\not\in C^2(\mathbb{R}^n)$, otherwise $D^\alpha w(0)=\sum c_k$ which does not exist.

Suppose there exists $u$ is a $C^2$ solution at $B_{\epsilon}(0)$, (2) means that $\Delta (u-w)=0$ on $B_{\epsilon}\backslash\{0\}$. Since $u-w$ is bounded, by the removable singularity theorem,  $u-w$ is a harmonic function in $B_\epsilon (0)$, thus an analytic function. However this means $w=u-(u-w)$ is a $C^2$ function on $B_\epsilon (0)$. Contradiction.

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

$\mathbf{Remark:}$ Gilbarg Trudinger’s book. Chapter 4. Exercise 4.9

$\mathbf{Erratum:}$ $u-w$ is not bounded in general, because $w$ can not be proved bounded directly. But we have $w(x)=o(\log r)$. In fact suppose $2^{-l}\leq|x|< 2^{-l+1}$, some $l\geq 1$, then

$\displaystyle w(x)=\sum_{k=0}^{l-1} c_k+c_l\frac{(\eta P)(t_lx)}{t_l^2}=\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)$

Because $c_k\to 0$,

$\displaystyle \left|\frac{w(x)}{\log |x|}\right|\leq \frac{|\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)|}{l\log 2}\to 0\text{ as }l\to \infty$

So $u-w=o(\log r)$ as $r\to 0$. Then one can use the Removable singularity theorem(Bochner), or the conclusion of 3.7 in GT’s book.

A bilinear integral identity for harmonic functions

$\mathbf{Thm:}$ Suppose $\phi_1$ and $\phi_2$ are harmonic function in $\Omega_1$ and $\Omega_2$ respectively. For any $x_1\in\Omega_1,x_2\in\Omega_2$ and $0\leq a,b,c,d\leq \min\{dist(x_1,\Omega_1), dist(x_2,\Omega_2)\}$ with $ab=cd$, we have

$\displaystyle \int_{|w|=1}\phi_1(x_1+aw)\phi_2(x_2+bw)dw=\int_{|w|=1}\phi_{1}(x_1+cw)\phi_2(x_2+dw)dw$

$\mathbf{Proof:}$ We give two different proofs. The first is using possion integral formula

$\displaystyle u(x)=\int_{\partial B} \frac{R^2-|x|^2}{n\omega_nR|x-y|^n}\phi(y)ds_y=\int_{|y|=1} \frac{R^{n-2}(R^2-|x|^2)}{n\omega_n|x-y|^n}\phi(y)dy$

Here $\displaystyle R$ is the radius of $B$

WLOG, assume $x_1,x_2$ are the origin. And $a, $a=\lambda d$, $c=\lambda b$. Define

$\displaystyle L(a,b)=\int_{|w|=1}\psi_1(aw)\psi_2(bw)dw$

We have to prove $L(a,b)=L(c,d)$. Apply the possion integral formula

$\displaystyle \psi_1(aw)=\int_{|y|=1}P(aw,cy)\psi_1(y)dy$

where $\displaystyle P(aw,cy)=\frac{c^{n-2}(c^2-a^2)}{n\omega_n|cy-aw|^n}$. Noticing that

$\displaystyle P(aw,cy)=P(\lambda dw,\lambda b y)=P(dw,by)=P(dy,bw)$

So $\displaystyle L(a,b)=\int_{|w|=1}\psi_1(aw)\psi_2(bw)dw=\int_{|w|=1}\int_{|y|=1}P(aw,by)\psi_1(cy)dy\psi_2(bw)dw$

$\displaystyle =\int_{|y|=1}\int_{|w|=1}P(dy,bw)\psi_2(bw)dw\psi_1(cy)dy=\int_{|y|=1}\psi_2(dy)\psi_1(cy)dy=L(c,d)$

The second proof relies on Green formula. Let $\rho=\sqrt{ab}$, we will prove $L(\lambda \rho,\lambda^{-1} \rho)$ is independent of $\lambda$. Here we require $\lambda \rho,\lambda^{-1}\rho$ are still inside the domain of $\Omega_1$ and $\Omega_2$

$\displaystyle \frac{dL(\lambda \rho,\lambda^{-1} \rho)}{d\lambda}=\int_{|w|=1}\left[\frac{\partial\psi_1(\lambda\rho w)}{\partial\lambda}\psi_2(\lambda^{-1}\rho w)+\psi_1(\lambda\rho w)\frac{\partial\psi_2(\lambda^{-1}\rho w)}{\partial\lambda}\right]dw$

Since $\displaystyle \frac{\partial \psi_1(\lambda\rho w)}{\partial \lambda}=\frac{\rho}{\lambda}\frac{\partial \psi_1(\lambda\rho w)}{\partial\rho}$, $\displaystyle \frac{\partial \psi_2(\lambda^{-1}\rho w)}{\partial \lambda}=-\frac{\rho}{\lambda}\frac{\partial \psi_1 (\lambda^{-1}\rho w)}{\partial\rho}$, plug in this fact

$\displaystyle \frac{dL(\lambda \rho,\lambda^{-1} \rho)}{d\lambda}=\frac{\rho}{\lambda}\int_{|w|=1}\left[\frac{\partial\psi_1(\lambda\rho w)}{\partial\rho} \psi_2(\lambda^{-1}\rho w)-\psi_1(\lambda\rho w)\frac{\partial\psi_2(\lambda^{-1}\rho w)}{\partial\rho} \right]dw$

Define $u_1(r\theta)=\psi_1(\lambda r\theta)$,  $u_2(r\theta)=\psi_2(\lambda^{-1} r\theta)$ for any $0\leq r\leq \rho$, $|\theta|=1$, then $u_1,u_2$ are harmonic in the disc $B_\rho(0)$

$\displaystyle \frac{\partial u_1}{\partial \nu}=\frac{\partial u_1}{\partial r}=\frac{\partial \psi_1}{\partial \rho}$, $\displaystyle \frac{\partial u_2}{\partial \nu}=\frac{\partial u_2}{\partial r}=\frac{\partial \psi_2}{\partial \rho}$ on $\partial B_\rho(0)$.

$\displaystyle \frac{dL(\lambda \rho,\lambda^{-1} \rho)}{d\lambda}=\frac{\rho}{\lambda}\int_{|\theta|=1}\left[\frac{\partial u_1}{\partial\nu} u_2-u_1\frac{\partial u_2}{\partial\nu} \right]d\theta=\frac{1}{\lambda\rho^{n-2}}\int_{\partial B_\rho(0)}\left[\frac{\partial u_1}{\partial\nu} u_2-u_1\frac{\partial u_2}{\partial\nu} \right]ds$

By the Green second identity,  we know that the end of last equation is 0.

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

$\mathbf{Remark:}$ See Gustin’s paper: A bilinear integral identity for harmonic functions. And Gilbarg Trudinger’s book, chapter 2,2.18

Monotone formula of harmonic functions

Suppose $B_1\in \mathbb{R}^n$, and $u$ is a harmonic function is $B_1$

$\Delta u=0$

For $r\in (0,1)$, we can define

$\displaystyle E(r)=\int_{B_r}|\nabla u|^2dx$,
$\displaystyle V(r)=\int_{\partial B_r}u^2ds$
$\displaystyle F(r)=\frac{rE(r)}{V(r)}$

In general $F(r)$ is called the frequency of $u$. If $u$ is a homogeneous harmonic polynomial of degree $k$, then $F(r)=k$
Suppose $u=r^k\phi(\theta)$, where $\phi$ is a spherical harmonic polynomial functions on $\mathbb{S}^{n-1}$, then $u_n=kr^{k-1}\phi(\theta)$,

$\displaystyle F(r)=\frac{rE(r)}{V(r)}=k$.

$\mathbf{Thm:}$ $F(r)$ is a non-decreasing function in $r\in(0,1)$

$\mathbf{Proof:}$
Since $\Delta u=0$, then $\Delta u^2=2|\nabla u|^2$. From the Green formula
$\displaystyle E(r)=\frac 12\int_{B_r}\Delta u^2=\int_{\partial B_r}u u_n$
$\displaystyle F'(r)=\frac{E(r)}{V(r)}+\frac{rE'(r)}{V(r)}-\frac{rV'(r)E(r)}{V(r)}=F(r)\left(\frac 1r +\frac{E'(r)}{E(r)}-\frac{V'(r)}{V(r)}\right)$
We will prove that $F'(r)\geq 0$ for $\displaystyle r\in (0,1)$.
$\displaystyle V(r)=\int_{\partial B_r}u^2ds=\int_{\partial B_1}u^2(rw)r^{n-1}dw$
$\displaystyle V'(r)=\int_{\partial B_1}(n-1)r^{n-2}u^2(rw)dw+\int_{\partial B_1}2u(rw)\frac{\partial u}{\partial r}(rw)r^{n-1}dw=\frac{n-1}{r}H(r)+2\int_{\partial B_1}u u_nds$

$\displaystyle \frac{V'(r)}{V(r)}=\frac{n-1}{r}+\frac{2\int_{\partial B_r}u u_n}{\int_{\partial B}u^2}\quad (1)$
$\displaystyle E'(r)=\int_{\partial B_r}|\nabla u|^2ds$

By the Green formula

$\displaystyle \int_{\partial B_r}u_i^2\cdot x^2_jds=r\int_{\partial B_r}u_i^2x_j\nu_jds=r\int_{B_r}2u_iu_{ij}x_j dx+\int_{B_r}u^2_idx\quad (2)$

where $\displaystyle \nu_j=\frac{x_j}{r}$ is the j-th component of outer normal vector of $\displaystyle \partial B_r$.
Again apply green formula

$\displaystyle \int_{ B_r}u_i u_{ij}x_jdx+\int_{B_r}u_{ii}u_jx_jdx+\int_{B_r}u_iu_j\delta_{ij}=\int_{\partial B_r}u_iu_jx_j\nu_i \quad (3)$

Plug (3) into (2), we get
$\displaystyle \int_{\partial B_r}u^2_i x_j^2=r\int_{B_r}u_i^2dx+2\int_{\partial B_r}u_iu_jx_ix_j-2r\int_{B_r}u_{ii}u_jx_j-2r\int_{B_r}u_iu_j\delta_{ij}$
Sum over j and i, noticing $u$ is haromic,

$\displaystyle r^2E(r)=rnD(r)+2r^2\int_{\partial B_r}u_n^2-2rD(r)$

So

$\displaystyle \frac{E'(r)}{E(r)}=\frac{n-2}{r}+\frac{2\int_{\partial B_r}u_n^2}{\int_{\partial B_r}uu_n}\quad (4)$

Plugging (1) and (4) to $F'(r)$, we get

$\displaystyle F'(r)=F(r)\left(\frac{\int_{\partial B_r}u_n^2}{\int_{\partial B_r}uu_n}-\frac{\int_{\partial B_r }uu_n}{\int_{\partial B_r}u^2}\right)\geq 0$

since the cauchy inequality.