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Math416Complex Variables (Lecture 17)

Math416 Lecture 17

Continue on Chapter 7

Harmonic conjugates

Theorem 7.18

Existence of harmonic conjugates.

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

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

Proof:

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

ff is holomorphic on Ω\Omega

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

So f=gf=g', fix z0Ωz_0\in \Omega, we can choose q(z0)=u(z0)q(z_0)=u(z_0) and g=u1+iv1g=u_1+iv_1, g=u1x+iv1x=v1yiu1y=uxiuyg'=\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 u1x=ux\frac{\partial u_1}{\partial x}=\frac{\partial u}{\partial x} and u1y=uy\frac{\partial u_1}{\partial y}=\frac{\partial u}{\partial y}

So u1=uu_1=u on Ω\Omega

Re(g)=u1=u\text{Re}(g)=u_1=u on Ω\Omega

If u+ivu+iv is holomorphic, vv is harmonic conjugate of uu

QED

Corollary For Harmonic functions

Theorem 7.19

Harmonic functions are CC^\infty

CC^\infty is a local property.

Theorem 7.20

Mean value property for harmonic functions.

Let uu be harmonic on an open set of Ω\Omega

Then u(z0)=12π02πu(z0+reiθ)dθu(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta

Proof:

Reg(z0)=12π02πReg(z0+reiθ)dθ\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 uu be harmonic on a domain Ω\Omega. If u=0u=0 on some open set GΩG\subset \Omega, then u0u\equiv 0 on Ω\Omega.

If u=vu=v on GΩG\subset \Omega, then u=vu=v on Ω\Omega.

Proof:

We proceed by contradiction.

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

HH is open and nonempty. If HΩH\neq \Omega, then z0HΩ\exists z_0\in \partial H\cap \Omega. Then r>0\exists r>0 such that Br(z0)ΩB_r(z_0)\subset \Omega such that gO(Br(z0))\exists g\in O(B_r(z_0)) such that Reg=u\text{Re}g=u on Br(z0)B_r(z_0)

Since HBr(z0)H\cap B_r(z_0) is nonempty open set, then gg is constant on HBr(z0)H\cap B_r(z_0)

So gg is constant on Br(z0)B_r(z_0)

So uu is constant on Br(z0)B_r(z_0)

So D(z0,r)HD(z_0,r)\subset H. This is a contradiction that z0Hz_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 uu attains a maximum at z0Ωz_0\in \Omega.

For all zz in the neighborhood of z0z_0, u(z)<u(z0)u(z)<u(z_0). We can choose r>0r>0 such that Br(z0)ΩB_r(z_0)\subset \Omega.

By the mean value property, u(z0)=12π02πu(z0+reiθ)dθu(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta

So 0=12π02πu[z0+reiθu(z0)]dθ0= \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 ff is holomorphic on a domain Ω\Omega and attains a maximum on the boundary of Ω\Omega, then ff is constant on Ω\Omega.

Except at z0Ωz_0\in \Omega where f(z0)=0f'(z_0)=0, if ff attains a minimum on the boundary of Ω\Omega, then ff is constant on Ω\Omega.

Dirichlet problem for domain DD

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

We can always solve the problem for the unit disk.

u(z)=12π02πh(eit)Re(eit+zeitz)dtu(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=reiθz=re^{i\theta}

Re(eit+reiθeitreiθ)=1r212rcos(θt)+r2\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(θ,t)>0Pr(\theta, t)>0 and 02πPr(θ,t)dt=1\int_0^{2\pi}Pr(\theta, t)dt=1, r,t\forall r,t

Chapter 8 Laurent series

when n=an(zz0)n\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n converges?

Claim R>0\exists R>0 such that n=an(zz0)n\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n converges if zz0<R|z-z_0|<R and diverges if zz0>R|z-z_0|>R

Proof:

Let u=1zz0u=\frac{1}{z-z_0}

n=0an(zz0)n\sum_{n=0}^{\infty}a_n(z-z_0)^n has radius of convergence 1R\frac{1}{R}

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

So zz0=1u>11R=R|z-z_0|=\frac{1}{|u|}>\frac{1}{\frac{1}{R}}=R

QED

Laurent series

A Laurent series is a series of the form n=an(zz0)n\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n

The series converges in some annulus shape A={z:r1<zz0<r2}A=\{z:r_1<|z-z_0|<r_2\}

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

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