Harmonic Functions and the Poisson Integral — Examples

Differentiating with [L1] gives $u_x=\frac{x}{x^2+y^2},u_y=\frac{y}{x^2+y^2}\mathrm{.}$

Differentiating once more gives $u_{xx}=\frac{y^2-x^2}{(x^2+y^2{)}^2},u_{yy}=\frac{x^2-y^2}{(x^2+y^2{)}^2}\mathrm{.}$

The sum in step 2.1 is $0$, so $u_{xx}+u_{yy}=0$ on $\mathbb{C}\setminus \{0\}$; by Plane harmonic functions, $u$ is harmonic there.

The polynomial $f(z)=z^n$ is entire by [L1], hence holomorphic on all of $\mathbb{C}$.

Its real and imaginary parts are polynomials in $x$ and $y$, so they are of class $C^{\mathrm{\infty }}$ and in particular $C^2$; [L2] therefore makes $\mathrm{Re}(z^n)$ harmonic.

Since $z^2=(x+iy{)}^2=(x^2-y^2)+2ixy$, the real part of $z^2$ is $x^2-y^2$ and the imaginary part is $2xy$.

The polynomial $z^2$ is entire by [L1], so [L2] makes $2xy$ a harmonic conjugate of $x^2-y^2$.

The function $u(r{e}^{i\varphi })=r\cos \varphi$ is the real part of $z$, so [L2] makes it harmonic on $\mathbb{D}$.

On the boundary $\mathrm{\mid }z\mathrm{\mid }=1$, the same formula gives $u(e^{i\theta })=\cos \theta =\phi (e^{i\theta })$. By [L1], the continuous harmonic extension of $\phi$ is unique, so $P[\phi ]=u$.

The function $H(z):=\frac{1+z}{1-z}$ is holomorphic on $\mathrm{\mid }z\mathrm{\mid }<1$ by [L2], and its real part is $\mathrm{Re}H(\rho e^{i\varphi })=\frac{1-{\rho }^2}{1-2\rho \cos \varphi +{\rho }^2}=P(\rho e^{i\varphi },1)$ by [L1]. Therefore $u(z):=P(z,1)$ is harmonic on the unit disc. Since $1-\mathrm{\mid }z{\mathrm{\mid }}^2>0$ and $\mathrm{\mid }1-z{\mathrm{\mid }}^2>0$ for $\mathrm{\mid }z\mathrm{\mid }<1$, it is positive there as well.

Substituting $\varphi =0$ and $\varphi =\pi$ into [L1] gives $u(r{e}^{i0})=\frac{1-r^2}{(1-r{)}^2}=\frac{1+r}{1-r},u(r{e}^{i\pi })=\frac{1-r^2}{(1+r{)}^2}=\frac{1-r}{1+r},$ and $u(0)=P(0,1)=1$.

Fix $R$ with $r<R<1$. The function $u$ is harmonic on a neighbourhood of $\overline{D(0,R)}$, so [L3] gives $\frac{R-r}{R+r}u(0)\le u(r{e}^{i\varphi })\le \frac{R+r}{R-r}u(0)\mathrm{.}$ Letting $R\to 1^{-}$ yields the unit-disc Harnack bounds $\frac{1-r}{1+r}\le u(r{e}^{i\varphi })\le \frac{1+r}{1-r}\mathrm{.}$ Step 1.2 shows equality at $\varphi =\pi$ and $\varphi =0$, respectively. Thus the Poisson kernel realizes both Harnack extremes on the circle $\mathrm{\mid }z\mathrm{\mid }=r$.

Suppose $v$ were a harmonic conjugate of $\log \mathrm{\mid }z\mathrm{\mid }$ on $\mathbb{C}\setminus \{0\}$. Then $F(z):=\log \mathrm{\mid }z\mathrm{\mid }+iv(z)$ would be holomorphic there, and its exponential would satisfy $\exp (F(z))=\mathrm{\mid }z\mathrm{\mid }(\cos v(z)+i\sin v(z))\mathrm{.}$

The function $G(z):=\exp (F(z))\mathrm{/}z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ by [L3], and $\mathrm{\mid }G(z)\mathrm{\mid }=\mathrm{\mid }\exp (F(z))\mathrm{\mid }\mathrm{/}\mathrm{\mid }z\mathrm{\mid }=e^{\mathrm{Re}F(z)}\mathrm{/}\mathrm{\mid }z\mathrm{\mid }=1$ by step 1.1. If $G$ were nonconstant, [L4] would make its image open in $\mathbb{C}$, impossible because $G(\mathbb{C}\setminus \{0\})\subseteq \{\mathrm{\mid }w\mathrm{\mid }=1\}$. Hence $G$ is constant on $\mathbb{C}\setminus \{0\}$.

Since $G$ is constant, for every $z\mathrm{\ne }0$ one has $\exp (F(z)-F(1))=\frac{\exp (F(z))}{\exp (F(1))}=\frac{zG(z)}{G(1)}=z\mathrm{.}$ Thus $L(z):=F(z)-F(1)$ is a continuous logarithm on $\mathbb{C}\setminus \{0\}$, contradicting [L2].

Therefore $\log \mathrm{\mid }z\mathrm{\mid }$ has no global harmonic conjugate on $\mathbb{C}\setminus \{0\}$.

One has $u_{xx}=0$ and $u_{yy}=0$, so $u_{xx}+u_{yy}=0$ and $u$ is harmonic on $\mathbb{C}$ by Plane harmonic functions.

On the line $x=0$, the function $u$ vanishes identically, but $u(1,0)=1\mathrm{\ne }0$. So vanishing on a line does not force vanishing everywhere.

As in the previous example, $u$ and $v$ are harmonic because both have vanishing second partial derivatives.

Their product is $uv=x^2$, and $(x^2{)}_{xx}+(x^2{)}_{yy}=2+0=2\mathrm{\ne }0$. So $uv$ is not harmonic.

By [L1], the function $1\mathrm{/}z$ is holomorphic on $0<\mathrm{\mid }z\mathrm{\mid }<1$, so [L2] makes $u(z)=\mathrm{Re}(1\mathrm{/}z)$ harmonic there.

On the positive real axis, $u(t)=1\mathrm{/}t\to +\mathrm{\infty }$ as $t\downarrow 0$, so $u$ is unbounded near $0$. If $u$ had a harmonic extension across $0$, it would be bounded on some small closed disc around $0$, contradicting [L3].

Therefore $u$ is harmonic on the punctured disc but does not extend harmonically across the puncture.