NoteNextra-origin/content/Math416/Math416_L17.md

Math416 Lecture 17

Continue on Chapter 7

Harmonic conjugates

Theorem 7.18

Existence of harmonic conjugates.

Let u be a harmonic function on \Omega a convex open subset in \mathbb{C}. Then there exists g\in O(\Omega) such that \text{Re}(g)=u on \Omega.

Moreover, g is unique up to an imaginary additive constant.

Proof:

Let f=2\frac{\partial u}{\partial z}=\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}

f is holomorphic on \Omega

Since \frac{\partial u}{\partial \overline{z}}=0 on \Omega, f is holomorphic on \Omega

So f=g', fix z_0\in \Omega, we can choose q(z_0)=u(z_0) and g=u_1+iv_1, g'=\frac{\partial u_1}{\partial x}+i\frac{\partial v_1}{\partial x}=\frac{\partial v_1}{\partial y}-i\frac{\partial u_1}{\partial y}=\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}, given that \frac{\partial u_1}{\partial x}=\frac{\partial u}{\partial x} and \frac{\partial u_1}{\partial y}=\frac{\partial u}{\partial y}

So u_1=u on \Omega

\text{Re}(g)=u_1=u on \Omega

If u+iv is holomorphic, v is harmonic conjugate of u

QED

Corollary For Harmonic functions

Theorem 7.19

Harmonic functions are C^\infty

C^\infty is a local property.

Theorem 7.20

Mean value property for harmonic functions.

Let u be harmonic on an open set of \Omega

Then u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta

Proof:

\text{Re}g(z_0)=\frac{1}{2\pi}\int_0^{2\pi}\text{Re}g(z_0+re^{i\theta})d\theta

QED

Theorem 7.21

Identity theorem for harmonic functions.

Let u be harmonic on a domain \Omega. If u=0 on some open set G\subset \Omega, then u\equiv 0 on \Omega.

If u=v on G\subset \Omega, then u=v on \Omega.

Proof:

We proceed by contradiction.

Let H=\{z\in \Omega:u(z)=0\} be the interior of G

H is open and nonempty. If H\neq \Omega, then \exists z_0\in \partial H\cap \Omega. Then \exists r>0 such that B_r(z_0)\subset \Omega such that \exists g\in O(B_r(z_0)) such that \text{Re}g=u on B_r(z_0)

Since H\cap B_r(z_0) is nonempty open set, then g is constant on H\cap B_r(z_0)

So g is constant on B_r(z_0)

So u is constant on B_r(z_0)

So D(z_0,r)\subset H. This is a contradiction that z_0\in \partial H

QED

Theorem 7.22

Maximum principle for harmonic functions.

A non-constant harmonic function on a domain cannot attain a maximum or minimum on the interior of the domain.

Proof:

We proceed by contradiction.

Suppose u attains a maximum at z_0\in \Omega.

For all z in the neighborhood of z_0, u(z)<u(z_0). We can choose r>0 such that B_r(z_0)\subset \Omega.

By the mean value property, u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta

So 0= \frac{1}{2\pi}\int_0^{2\pi}u[z_0+re^{i\theta}-u(z_0)]d\theta

We can prove the minimum is similar.

QED

Maximum/minimum (modulus) principle for holomorphic functions.

If f is holomorphic on a domain \Omega and attains a maximum on the boundary of \Omega, then f is constant on \Omega.

Except at z_0\in \Omega where f'(z_0)=0, if f attains a minimum on the boundary of \Omega, then f is constant on \Omega.

Dirichlet problem for domain D

Let h: \partial D\to \mathbb{R} be a continuous function. Is there a harmonic function u on D such that u is continuous on \overline{D} and u|_{\partial D}=h?

We can always solve the problem for the unit disk.

u(z)=\frac{1}{2\pi}\int_0^{2\pi}h(e^{i t})\text{Re}\left(\frac{e^{it}+z}{e^{it}-z}\right)dt

Let z=re^{i\theta}

\text{Re}\left(\frac{e^{it}+re^{i\theta}}{e^{it}-re^{i\theta}}\right)=\frac {1-r^2}{1-2r\cos(\theta-t)+r^2}

This is called Poisson kernel.

Pr(\theta, t)>0 and \int_0^{2\pi}Pr(\theta, t)dt=1, \forall r,t

Chapter 8 Laurent series

when \sum_{n=-\infty}^{\infty}a_n(z-z_0)^n converges?

Claim \exists R>0 such that \sum_{n=-\infty}^{\infty}a_n(z-z_0)^n converges if |z-z_0|<R and diverges if |z-z_0|>R

Proof:

Let u=\frac{1}{z-z_0}

\sum_{n=0}^{\infty}a_n(z-z_0)^n has radius of convergence \frac{1}{R}

So the series converges if |u|<\frac{1}{R}

So |z-z_0|=\frac{1}{|u|}>\frac{1}{\frac{1}{R}}=R

QED

Laurent series

A Laurent series is a series of the form \sum_{n=-\infty}^{\infty}a_n(z-z_0)^n

The series converges in some annulus shape A=\{z:r_1<|z-z_0|<r_2\}

The annulus is called the region of convergence of the Laurent series.