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} ba positive solution to {Lw=0} in {B^+_1} continuously vanishing on {\{x_n=0\}} with

\displaystyle u(\frac{1}{2}x_n)=v(\frac{1}{2}x_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<c, 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.