Lecture 4

Let $\mathbb D$ denote the unit disk, $C$ denote the unit circle, $C(r)$ denote the circle of radius $r$, let $\mathbb{D}(r)$ denote the disk of radius $r$.

In this lecture we study two questions:

• Find representation of $u$ on $\mathbb{D}$ by φ.
• Can we reproduce $\varphi(re^{i\theta})$ by $\lim_{r\to 1}u(re^{i\theta}))$?

Some complex analysis¶

The Cauchy integral formula can be rewritten as follows:

Let $z = re^{it}$, then $f(z) = f(re^{it})$. Let $f_r(\theta) = f(e^{i\theta})$, we obtain a function $f_r$ on the circle, then

where $C_r(t)=\frac{1}{1-re^{it}}$, called the Cauchy kernel.

The Poisson kernel¶

On the unit disk $\mathbb{D}$, conjugate harmonic functions exists, so $u(z)$ is harmonic if and only if $u = Ref$ for some complex analytic function $f(z)$. We want to find similar reproducing formula for $u$ from the analytic case.

Let $\sum_{n=0}^{+\infty} a_nz^n$ be the power series expansion of $f$, then $f_1(\theta) = f(e^{i\theta}) = \sum_{n=0}^{+\infty}a_n e^{in\theta}$, this is the Fourier series of $f_1$. Note that the Fourier coefficients of $f$ is not symmetric because $f$ is not real valued (otherwise has to be zero).

Similarly, we have the Fourer series expansion $f_r(\theta)=\sum_{n=0}^{+\infty} a_n r^ne^{in\theta}$. Here I write $''=''$ because the series does converge uniformly.

By the convolution theorem, the Fourier coefficients of $C_r$ should be $r^n$, so $C_r(\theta) = \sum_{n=0}^{+\infty}r^ne^{in\theta}$. One can directly verify this by the geometric series $\frac{1}{1-z} = \sum_{n=0}^{+\infty} z^n$.

Now let $u$ be a harmonic function on $\mathbb{D}$. Then we can write $u(z) = f(z)+\overline{f(z)}$ for an analytic function $f$ on $\mathbb{D}$. Let $f(z) = \sum_{n=0}^{+\infty} a_n z^n$ be the power series of $f$. Then $u(re^{i\theta}) = 2Re a_0 + \sum_{n = 1}^{+\infty} a_n r^n e^{in\theta} + \sum_{n = 1}^{+\infty} \overline{a_n}r^n e^{-in\theta} = \sum_{n = -\infty}^{+\infty} c_n r^{|n|}e^{in\theta}$, where we take

Then we have $u_r(\theta) = u_1 * P_r(\theta)$. From the convolution theorem we see that the Fourier coefficients of $P_r(\theta)$ should be $r^{|n|}$, so $P_r(\theta) = \sum_{n=-\infty}^{+\infty}r^{|n|}e^{in\theta}$. The series converges absolutely (hence uniformly) so the equation is justified. We call it the Poisson kernel on the circle.

Expression of Poisson kernel¶

Next we find a simple formula for the Poisson kernel. We can rewrite the Poisson kernel as

Lemma. Let φ be an integrable function on the circle, then $\varphi*P_r$ is a harmonic function on $\mathbb{D}$.

Proof. Let $H(z) = \frac{1+z}{1-z}$, then $P_r(\theta) = ReH(re^{i\theta})$. Then $u_{\varphi}(re^{i\theta})$ is the real part of $g(re^{i\theta}) = \frac{1}{2\pi}\int_{-\pi}^\pi \varphi(t)\frac{1+re^{i(\theta - t)}}{1-re^{i(\theta - t)}}dt=\frac{1}{2\pi}\int_{-\pi}^\pi \varphi(t)\frac{e^{it}+z}{e^{it}-z}dt$. Then $\frac{\partial}{\partial \bar{z}}g(z) = \frac{1}{2\pi}\int_{-\pi}^\pi \varphi(t)\frac{\partial}{\partial \bar{z}}\frac{e^{it}+z}{e^{it} -z}dt = 0$ because $\frac{e^{it}+z}{e^{it} - z}$ is analytic on $\mathbb{D}$. To justify the switch of integral and derivative, use $\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y})$ and the following tip.

Alternative proof. The Poisson formula and convolution thoerem tells us that the Fourier series of $u_\varphi(z)$ is $\sum_{n = -\infty}^{+\infty} c_n r^{|n|}e^{in\theta}$, where $c_n$ is the Fourier coefficients of φ. Since φ is integrable, $|c_n|$ is bounded by $\|\varphi\|_{L^1}$. Then for $r<1$ the series converges uniformly. Moreover, if we take derivatives on $r$ or θ term by term, the resulted series still converges uniformly, because its absolute value is controlled by series of the form $\sum_{n\in \mathbb{Z}}n^{q}r^{|n|}$, which converges, this tells us we are allowed to take partial derivatives term by term. Each term in the series looks like either $c_n z^n$ or $c_{-n}\overline{z^n}$, which are analytic or anti-analytic, so harmonic. Then $\Delta u_{\varphi}(z) = \sum_{n = 1}^{+\infty}\Delta (c_nz^{n}) + \sum_{n = -\infty}^{-1}\Delta(c_n\overline{z^n}) + \Delta(c_0) = 0$.

Boundary value of harmonic functions¶

From the lemma we know from a (for example, integrable) function φ on $\mathbb{D}$, we can produce a harmonic function $u_\varphi$ on $\mathbb{D}$. But it is not clear so far whether the $u_\varphi$ we get are compatible with φ on the boundary. In this section we study boundary behavior of harmonic functions on the disk.

Theorem¶

The Poisson kernel is an approximate identity when $r\to 1$.

Proof.

• To show $\frac{1}{2\pi}\int_{-\pi}^{\pi}P_r(\theta)d\theta = 1$, use its Fourier series expansion term by term integrate. You can change integral and summation because the series converges uniformly.

• Note that $P_r$ is positive. For $\delta > 0$, $\sup_{\theta \in [-\pi,\pi]\setminus (-\delta,\delta)}P_r(\theta)\leq C_{\delta}\cdot (1-r^2)$ for some constant $C_\delta$ depending on δ. Then $P_r(\theta)$ converges to 0 uniformly when $r\to 1$. Then $\int_{(-\pi,\pi)\setminus (\delta,\delta)}P_r(\theta)d\theta \to 0$ when $r\to 1$.

On the Dirichlet problem¶

Theorem. Let φ be an integrable function on the unit cirlcle.

(1) $u_{\varphi}(re^{i\theta}) = \frac{1}{2\pi} \int_{-\pi}^\pi \varphi(t) \frac{1-r^2}{1-2r\cos(\theta-t)+r^2}dt$ is a harmonic function on $\mathbb{D}$, such that $\lim_{r\to 1}u_\varphi(re^{i\theta}) = \varphi(\theta)$ when θ is continuous point of φ.

(2) Let φ be a continuous function on the unit circle. Then $u_{\varphi}(re^{i\theta}) = \frac{1}{2\pi} \int_{-\pi}^\pi \varphi(t) \frac{1-r^2}{1-2r\cos(\theta-t)+r^2}dt$ is the unique solution to the Dirichlet problem, and $u_r\to \varphi$ uniformly when $r\to 1$.

Proof. Just need to prove uniqueness. Suppose $\tilde u$ is another solution, let $v = u_\varphi - \tilde u$. Then $v$ is harmonic in $\mathbb{D}$ and continuous on $\bar{\mathbb{D}}$ (the closed unit disk), and $v = 0$ on the unit circle. Now suppose $u_\varphi(z_0)\neq \tilde{u}(z_0)$ for some $z_0\in \mathbb{D}$, then $|v(z_0)|> 0$. By maximum principle, on every $C(r)$ for $|z_0|<r<1$, there exists some points $w_r$ such that $|v(w_r)|>|v(z_0)|$. Then there must be a subsequence of $(w_{r_n})$ such that $w_{r_n}\to w'\in C$. Since $v$ is continuous, $v(w') \geq |v(z_0)|> 0$. Contradicts to $v$ being zero on $C$.

Appendix: Heat flow from differential forms¶

In the note we assume all functions involved are smooth, though most results holds for merely differentiable functions.

Understanding vector fields¶

A vector field on $\mathbb{R}^3$ is a a vector valued function $\mathbf{v}(\mathbf{x}) = (p(\mathbf{x}),q(\mathbf{x}),r(\mathbb{x}))$. There are two types of vector fields:

In the 2-dimensional case, a “surface” becomes a curve, the flux is given by the integral $\int_{C}\mathbf{E}\cdot d\mathbf{n}$, where $\mathbf n$ is the outwards normal to the curve $C$. Let $\gamma(t)=(x(t),y(t))$ be a parametrization of the curve $C$ on $[0,1]$, then the surface integral is given by $\int_{0}^1 p(t)y'(t)-q(t)x'(t)dt$. We also write it in the form $\int_C -pdy+qdx$.

On the other hand, if we want to calculate the change of electronic potentials along a curve $C$, then it is given by the line integral $\int_{C}\mathbf{E}\cdot d\mathbf{r} = \int_{0}^1 pdx+qdy$.

Differential forms on $\mathbb{R}^2$¶

Here is a breif summery of differential forms in 2-dimensions.

Let $D$ be a domain with smooth boundary $C = \partial D$. Then the Green theorem, says $\int_{\partial D} p dx + qdy = \int_{D}(\frac{\partial q}{\partial x} - \frac{\partial p}{\partial y})dxdy$.

Let ω be the one-form $pdx + qdy$, then $d\omega = \frac{\partial p}{\partial y} dy\wedge dx + \frac{\partial q}{\partial x}dx \wedge dy = (\frac{\partial q}{\partial x} - \frac{\partial p}{\partial y})dx\wedge dy$. Then the Green theorem can be written as $\int_{\partial D}\omega = \int_{D} d\omega$ for a one form ω.

Intepreting vector fields¶

Quite naturally, we can identify the gradient field $\mathbf{E} = \nabla U$ as the 1-form $dU = \frac{\partial U}{\partial x}dx+\frac{\partial U}{\partial y}dy$. Then when we calculate the change of potential along curve $C$, it is just given by $\int_C dU$. Since we have $d(dU) = 0$ by direct calculation, when $C = \partial D$ is a closed curve, $\int_C dU = \int_{D}d(dU) = 0$, this explains the nature that change of potential does not depend on path chosen.

On the other hand, when we try to calculate the electronic flux outside $D$, we do $\int_C -pdy+qdx$, so it seems we should identify $\mathbf{E}$ with the 1-form $-pdy+qdx$ in this case. This motivates us the following operator

Let $\omega = pdx+qdy$ be a 1-form. Define the $*$-operator by $*\omega = pdy-qdx$. Then we have $**\omega = -\omega$.

The $*$ is a spacial case of the Hodge * operator.

In particular,

The Fourier’s law¶

The Fourier’s law says the heat flux is proportional to the gradient of temperature. In terms of vector fields, this is simply $\nabla u = -\kappa F$, where $F$ is the heat flux. But from the point of view of differential forms, if we denote ω by the heat flux so that the amout of heat flowing outside $D$ is given by $\int_{\partial D}\omega$, then the Fourier law should be intepreted as $\omega = -\kappa *du$.

By first law of thermaldynamics, increasing of temperature on $D = \int_D\frac{\partial }{\partial t}u(\mathbf{x},t)dm\mathbf({x}) = -c \int_{\partial D}\omega$. Plug in Fourier law we get $\int_{D}\frac{\partial}{\partial t}u(\mathbf{x},t)dt = \kappa c \int_{D}d*du = \kappa c \int_{D}\Delta u dx\wedge dy$. Then we get the heat equation.

Polar coordinate¶

Using differential forms we can give a simple calculation of Laplacian in polar coordiantes. The exterior differential is coordinate invariant, so we always have $du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy = \frac{\partial u}{\partial r} dr + \frac{\partial u}{\partial \theta}d\theta$. And we know that $d*d u = \Delta u(x,y) dx\wedge dy = \Delta u (r,\theta)\cdot rdrd\theta$. So it sufficies to calculate $*$ in polar coordinates. From the Cartesian coordinate case, we have $*dx = dy$, $*dy = -dx$, so $**dx = *(dy) = -dx$, $**dy = *(-dx) = -*dx=-dy$, we conclude that

Lemma. $**\omega = -\omega$.

Note that this is independent of coordinates because we can define the $*$ operator in a coordinate-free way.

Now we calculate $d*d u$ in polar coordinates. $du =\frac{\partial u}{\partial r}dr + \frac{\partial u}{\partial \theta}d\theta$, the main question is what is $*dr$ and what is $*d\theta$.

Let’s examine the function $f = \ln(x^2+y^2)=\ln r^2$. Then $df = \frac{\partial f}{\partial r}dr + \frac{\partial }{\partial \theta}d\theta= \frac{2}{r}dr+0$. On the other hand, in Cartesian coordinate $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = \frac{2x}{x^2+y^2}dx + \frac{2y}{x^2+y^2}dy$. $*df = -\frac{\partial f}{\partial y}dx + \frac{\partial f}{\partial x}dy = -\frac{2r\sin\theta}{r^2}d(r\cos\theta)+\frac{2r\cos\theta}{r^2}d(r\cos\theta)=2(\cos^2\theta+\sin^2\theta)d\theta = 2d\theta.$

We see from this example that it should hold that $2d\theta = *df = *(\frac{2}{r}dr)$, so

Apply $** = -1$ we get $*d\theta = *\frac{1}{r}*dr = \frac{1}{r}**dr = -\frac{1}{r}dr$.

We are almost done.

In polar coordinates,

We conclude that $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} +\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2\theta}{\partial\theta^2}$.