# MATH3401

Lecture 28

### Harmonic functions

##### Precise definition

$$u: \Omega\subset \R^n \ra \R^n$$ harmonic says:

1. $$u$$ and partials of order 1 and 2 are continuous;
2. $$\Delta u = 0$$.

### Scale factor

Last remarks on conformal mappings

$$f: z \mapsto w$$ conformal (i. e. analytic, $$f'(z_0) \neq 0$$).

For $$z$$ near $$z_0, z\neq z_0$$: $\frac{|f(z)-f(z_0)|}{|z-z_0|} \approx |\,f'(z_0)|.$

$$\implies |f(z)-f(z_0)| \approx |\,f'(z_0)| \cdot |z-z_0|$$

### Scale factor

Last remarks on conformal mappings

$$\implies |\,f'(z_0)|$$ represents the scaling factor or dilatation factor. That is

• stretching, if $$\,\left|\,f'(z_0)\right| \gt 1$$;
• shrinking, if $$\,\left|\,f'(z_0)\right| \lt 1$$.

### Scale factor

Example: $$f(z) = z^2$$ at $$z_0=1+i$$, ($$z=x+iy,$$ $$w = u+iv$$)

$$u = x^2-y^2$$,   $$v= 2xy$$

Scaling factor = $$|\,f'(z_0)|$$ $$= 2 |z_0|$$ $$=2 \sqrt{2}$$.

### Poisson's integral formula

Let $$C_0$$ be a positively oriented circle of radius $$r_0$$, centred at the origin.

Recall Cauchy: if $$f$$ is analytic in and on $$C_0$$, then for $$z_0 \in \text{Int} \,C_0$$ $f(z) = \frac{1}{2\pi i}\int_{C_0} \frac{f\left( \zeta \right)}{\zeta - z}d \zeta \qquad (1)$

Recall for $$z = r e^{i\theta}$$, $$r\gt 0$$ the inverse point to $$z$$ relative to the circle $$C_0$$ is $$r^* e^{i \theta}$$ with $$r^*$$ such that $$r^* \cdot r = r^2_0$$.

### Poisson's integral formula

Recall $$(z^*)^*=z$$. Note also:

$$\qquad z^* = r^* e^{i \theta}$$ $$= \dfrac{r_0^2}{r} e^{i \theta}$$ $$= \dfrac{r_0^2}{re^{-i\theta}}$$ $$= \dfrac{r_0^2}{\conj{z}}$$ $$=\dfrac{\zeta \conj{\zeta}}{\conj{z}}$$ $$\;(2)\;$$

for any $$\zeta\in C_0$$. Now fix $$z \in \text{Int}\, C_0, \; z_0\neq 0$$ and note $\ds \int_{C_0} \frac{f\left( \zeta \right)}{\zeta - z}d \zeta = 0 \quad(\text{Cauchy}),$ since $$\zeta\mapsto \dfrac{f(\zeta)}{\zeta -z^*}$$ is analytic in and on $$C_0$$ ($$z^*\in \text{Ext}\, C_0$$).

### Poisson's integral formula

Combine this with $$(1)$$ implies

$$f(z) = \dfrac{1}{2 \pi i} \ds \int_{C_0} \left( \frac{1}{\zeta -z} - \frac{1}{\zeta - z^*} \right) f(\zeta) d \zeta$$ $$\quad (3)$$

$$\quad I = \ds \left( \frac{\zeta}{\zeta -z} - \frac{\zeta}{\zeta - z^*} \right) \frac{1}{\zeta}$$ $$= \ds \left( \frac{\zeta}{\zeta -z} - \frac{1}{1 - \frac{\conj{\zeta}}{\conj{z}}} \right) \frac{1}{\zeta}$$

$$\;\;\quad = \ds \left( \frac{\zeta}{\zeta -z} - \frac{\conj{z}}{\conj{z} - \conj{\zeta}} \right) \frac{1}{\zeta}$$ $$= \ds \frac{\zeta \conj{\zeta} - z \conj{z}}{|\zeta - z|^2} \cdot \frac{1}{\zeta}$$ $$\quad (4)$$

### Poisson's integral formula

Now for $$z = r e^{i \theta}$$ put $$\zeta = r_0 e^{i \phi}$$, $$0\leq \phi \leq 2 \pi$$.

Then $$d\zeta = r_0i e^{i \phi}d\phi$$ and $I = \frac{r_0^2 - r^2}{|\zeta - z|^2} \cdot \frac{1}{r_0 e^{i \phi}} \qquad (5)$

### Poisson's integral formula

Using the cosine rule: $|\zeta - z|^2 = r_0^2+r^2-2r_0r \cos \left(\phi - \theta\right)\quad (5)'$

### Poisson's integral formula

$$(3)$$ implies $f\left(re^{i \theta}\right)= \frac{1}{2\pi i} \int_{0}^{2 \pi}\frac{r_0^2-r^2}{|\zeta - z|^2}\cdot \frac{f\left(r_0e^{i\phi}\right)r_0 i e^{i\phi}d \phi}{r_0 e^{i\phi}}\quad \quad$

$$\qquad \quad= \ds \frac{r_0^2-r^2}{2 \pi } \int_{0}^{2 \pi} \frac{f\left(r_0e^{i\phi}\right) d\phi}{r_0^2-2r_0r \cos \left(\phi - \theta\right) +r^2}$$. 😀