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pages/Math416/Math416_L17.md

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+# 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.
+

pages/Math416/_meta.js

@@ -20,4 +20,5 @@ export default {
     Math416_L14: "Complex Variables (Lecture 14)",
     Math416_L15: "Complex Variables (Lecture 15)",
     Math416_L16: "Complex Variables (Lecture 16)",
+    Math416_L17: "Complex Variables (Lecture 17)",
 }