Lecture 7: the Poisson integral formula

1.  The Poisson integral formula

Last time, we stated (but did not prove) that harmonic functions satisfy the mean value property. We’ll see the proof of that in a week or two. Today, our main result will be that this actually completely characterizes harmonic functions: any function satisfying the mean value formula is harmonic.

Our central tool will be the Poisson integral formula, a precursor to Cauchy’s integral formula which we’ll see in a few weeks, and which will be one of the central results of the class. Our perspective on the Poisson integral formula will be that it provides a solution to the Dirichlet problem on the unit disk .

What does this mean? Last time, we talked about harmonic functions, which are solutions to the differential equation $\mathrm{\Delta }u=0$. We saw in fact that there are many of these, e.g. the real and imaginary parts of any analytic function. This is not necessarily what we expect from differential equations: often, we hope to get a unique solution.

But in order for this to be possible, we have to fix some “boundary data”: for example, $f^{\prime }=f$ does not have a unique solution in continuously differentiable functions $ℝ\to ℝ$, but if we add the condition that $f(0)=1$ then it does have a unique solution, namely $f(x)=e^x$.

The Dirichlet problem is the analogous question in the complex plane for Laplace’s equation. If $D$ is a domain and $u:D\to ℂ$ is a smooth function (i.e. all its partial derivatives are defined and continuous on $D$), asking that $u$ is harmonic does not uniquely define a solution: there are many analytic functions $D\to ℂ$, hence many harmonic functions $D\to ℝ$ (and hence $D\to ℂ$). We need to add some boundary condition. What this means is: we start with a continuous function $h:\partial D\to ℝ$, where $\partial D=\overline{D}\setminus D$ is the boundary of $D$. Then we ask for a harmonic function $\tilde{h}:D\to ℂ$ such that for $z_0\in \partial D$, as $z\to z_0$ inside $D$ we have ${lim}_{z\to z_0}\tilde{h}(z)=h(z_0)$. This is the Dirichlet problem on $D$.

For today, we are interested in the case $D=𝔻$, the open unit disk. In this case, the Dirichlet problem is as follows: suppose we have a continuous function $h:\partial 𝔻\to ℂ$, i.e. a function $h(e^{i\theta })$. The Dirichlet problem is the problem of extending $h$ to a harmonic function $\tilde{h}$ on $𝔻$.

Any solution to the Dirichlet problem on the disk is unique: if we have two solutions ${\tilde{h}}_1,{\tilde{h}}_2$, then ${\tilde{h}}_1-{\tilde{h}}_2$ is harmonic on $𝔻$, and on $\partial 𝔻$ we have ${\tilde{h}}_1(e^{i\theta })-{\tilde{h}}_2(e^{i\theta })=h(e^{i\theta })-h(e^{i\theta })=0$, so by the maximum principle ${\tilde{h}}_1-{\tilde{h}}_2=0$ on all of $𝔻$, so ${\tilde{h}}_1={\tilde{h}}_2$. Therefore it suffices to find a formula for $\tilde{h}$ on $𝔻$; if it’s harmonic and recovers the right values on the boundary, it must be the only solution.

Showing that a solution exists is harder. We will write down a formula without justification: if you’re interested in pursuing this further, this could be a good paper topic for the final unit of the class.

For $z\in 𝔻$, we define the Poisson kernel

$$P(z)=\frac{\overline{z}}{1-\overline{z}}+1+\frac{z}{1-z}.$$

This has many different forms: for example, if $z=r{e}^{i\theta }$ (with , for $z\in 𝔻$) then

$$P(z)=\frac{1-r^2}{1-2r\cos \theta +r^2},$$

which shows that it is real-valued. For $h:\partial 𝔻\to ℂ$ a continuous function, we define its Poisson integral to be the function $\tilde{h}:𝔻\to ℂ$ given by

$$\tilde{h}(r{e}^{i\theta })=\frac{1}{2\pi }{\int }_0^{2\pi }h(e^{i\varphi })P(r{e}^{\theta -\varphi })𝑑\varphi .$$

Theorem.

Let $h(e^{i\theta })$ be a continuous function on $\partial 𝔻$. Then its Poisson integral $\tilde{h}(z)$ is a harmonic function on $𝔻$ with boundary values $h$, i.e. ${lim}_{z\to \zeta }\tilde{h}(z)=h(\zeta )$ for $\zeta =e^{i\theta }\in 𝔻$.

In particular, $\tilde{h}$ solves the Dirichlet problem on $𝔻$.

We will not prove this theorem, but if we have time at the end of today’s lecture we’ll come back and try to give some intuition, at least through some worked examples.

As a consequence, we obtain the Poisson integral formula.

Corollary.

Let $h:𝔻\to ℂ$ be a harmonic function extending continuously to $\partial 𝔻$. Then for any $z_0=r{e}^{i\theta }\in 𝔻$,

$$h(z_0)=\frac{1}{2\pi }{\int }_0^{2\pi }h(e^{i\varphi })P(r{e}^{\theta -\varphi })𝑑\varphi .$$

That is, we can recover the values of a harmonic function on the disk from its values on the boundary. This is quite remarkable!

Proof.

The given integral is the Poisson integral of the boundary values for $h$, so it gives a harmonic extension to $𝔻$; since $h$ itself is such an extension and these are unique, the claim follows. ∎

2.  Characterizing harmonic functions

The Poisson integral formula is interesting and powerful, in that it lets us apply tools to harmonic functions otherwise only available for analytic functions such as recovering a function on a domain from its values on the boundary. However, it is also very specialized: it holds only on the unit disk $𝔻$ centered at the origin.

It is not too difficult to generalize this to other disks, by scaling and translation. In other words, for any disk $D\subset ℂ$, we can solve the Dirichlet problem on $D$: given a continuous function $h:\partial D\to ℂ$, there is a unique harmonic function $\tilde{h}:D\to ℂ$ with boundary values $h$, which we can write explicitly via an integral formula.

While this is a simple consequence of the Poisson integral formula it lets us fully characterize harmonic functions on any domain $\mathrm{\Omega }$.

Recall that a harmonic function $h$ on $\mathrm{\Omega }$ satisfies the mean value property: for any $z_0\in \mathrm{\Omega }$, on any sufficiently small disk $D\subset \mathrm{\Omega }$ of radius $r$ centered at $z_0$ we have

$$h(z_0)=\frac{1}{2\pi }{\int }_0^{2\pi }h(z_0+r{e}^{i\theta })𝑑\theta .$$

We claim that this property actually characterizes harmonic functions:

Theorem.

Let $h:\mathrm{\Omega }\to ℂ$ be a continuous function. Then $h$ satisfies the mean value property if and only if it is harmonic.

Proof.

We have seen that a harmonic function satisfies the mean value property, so it remains to see the converse: if $h$ satisfies the mean value property, it is harmonic.

Suppose that $h$ satisfies the mean value property. Fix $z_0\in \mathrm{\Omega }$, and let $D$ be a sufficiently small disk of radius $r$ around $z_0$. Restricting $h$ to $\partial D$ gives a continuous function $\partial D\to ℂ$, so since we can solve the Dirichlet problem on $D$ there exists a unique harmonic function $g:D\to ℂ$ with boundary values $h$.

Since $h$ itself is a continuous function on $D$ with boundary values $h$, $h-g$ is a continuous function on $D$ with boundary values $0$. Since $g$ is harmonic, it satisfies the mean value property, so since $h$ does by hypothesis, so does $h-g$.

Now, recall that when we proved the maximum principle for harmonic functions in the last lecture, we did it using the mean value property; and that was the only place where we used that our function is harmonic. In particular, we actually don’t need to know that $u$ is harmonic to know that it satisfies the maximum principle: we only need to know that it satisfies the mean value property. (Of course, these are actually the same so there’s no real distinction, but the difference is important for this proof specifically!)

So since $h-g$ satisfies the mean value property, it satisfies the maximum principle as well. Since its boundary values are $0$, it follows that $h-g$ must also vanish, i.e. $h=g$. Since $g$ is harmonic, so is $h$, i.e. $h$ is the unique extension of its boundary values to a harmonic function on $D$.

Since $h$ is therefore harmonic on a neighborhood of every point $z_0\in \mathrm{\Omega }$, it is harmonic on $\mathrm{\Omega }$. ∎

This is remarkable in that we started with the assumption that $h$ is continuous, and concluded that it is harmonic and therefore smooth, by assuming only that $h$ satisfies the mean value property which a priori doesn’t involve any differential information.

We mentioned above that the Poisson integral formula has an analogue for analytic functions, Cauchy’s integral formula (which is actually significantly more powerful, since we now get to assume analyticity). The characterization of harmonic functions by the mean value property also has an analogue for analytic functions, Morera’s theorem, though it doesn’t quite look the same; it’s similarly found as an application of Cauchy’s formula. We’ll talk more about this in the next unit.

3.  Motivating the Poisson kernel

Let’s return to Dirichlet’s problem on the disk, which is the question of finding, for a given continuous function $h:\partial 𝔻\to ℂ$, a harmonic function $\tilde{h}:𝔻\to ℂ$ with boundary values $h$.

Perhaps the simplest case is when $h(e^{i\theta })=c$ is a constant. Then $\tilde{h}(r{e}^{i\theta })=c$ is a harmonic function, and it has boundary values $h$, so it solves the Dirichlet problem; and since any solution is unique, it is the only one.

The next simplest might be something like $h(e^{i\theta })=e^{i\theta }$. This naturally extends to $\tilde{h}(z)=z$, which is analytic and therefore harmonic. More generally, for any integer $n\ge 0$ if $h(e^{i\theta })=e^{in\theta }$ then $\tilde{h}(z)=z^n$ solves the Dirichlet problem with boundary value $h$.

For , this doesn’t work, since $h(z)=z^n$ is not analytic (or even defined) on $𝔻$ if ; it has a pole at $z=0$. However, we can use the following trick: $e^{-i\theta }=\overline{e^{i\theta }}$, so for we have $e^{in\theta }={\overline{z}}^{-n}=\overline{z^{-n}}$ if $z=e^{i\theta }$. Since , $-n>0$, so this is well-defined on $𝔻$. It is not analytic, since $\overline{z}$ is not analytic, but it is still harmonic: for any analytic function $f=u+iv:D\to ℂ$ on any domain $D$, $\overline{f}=u-iv$ is the sum of harmonic functions, hence harmonic, so taking $f(z)=z^{-n}$ (with ) we find that ${\overline{z}}^{-n}$ is harmonic. (Later in the course, we’ll see some more elegant ways to understand this.) Therefore for $h(e^{i\theta })=e^{in\theta }$ for , we have a solution $\tilde{h}(z)={\overline{z}}^{-n}$ to the Dirichlet problem with boundary value $h$.

We can now solve the Dirichlet problem for any monomial $h(e^{i\theta })=e^{in\theta }$, for any integer $n$. Since linear combinations of harmonic functions are harmonic, we can generalize to “trigonometric polynomials”: if

$$h(e^{i\theta })=\sum _n{a}_n{e}^{in\theta }$$

where $n$ ranges over some (finite) set of integers (positive or negative), then we can find some $\tilde{h}$ solving the Dirichlet problem for $h$, by scaling and adding up the solutions for each term. To find the resulting expression, we rewrite the solutions in polar coordinates: if $z=r{e}^{i\theta }$, for $n\ge 0$ the solution is $z^n=r^n{e}^{in\theta }$, and for the solution is ${\overline{z}}^{-n}=r^{-n}{e}^{in\theta }$, so for all $n$ we can write the corresponding term as $r^{|n|}{e}^{in\theta }$. Therefore for $h$ as above the Dirichlet problem has solution

$$\tilde{h}(r{e}^{i\theta })=\sum _n{a}_n{r}^{|n|}{e}^{in\theta }.$$

Taking the limit as $r\to 1$ visibly recovers $h$.

Now, this is not the same thing as our Poisson integral; and what we’ve derived so far only applies to these trigonometric polynomials, which are certainly not the only continuous functions one can define on $\partial 𝔻$. However, I want to argue that for $h$ as above, we actually recover the same $\tilde{h}$. (By uniqueness, this had certainly better be true!)

Suppose $h(e^{i\theta })=e^{in\theta }$ for some integer $n$. I claim that

$$P(r{e}^{i\theta })=\sum _{n\in \mathbb{Z}}{r}^{|n|}{e}^{in\theta }.$$

(We haven’t talked about series in this class yet, but this will converge (absolutely) for , i.e. on $𝔻$.) You can verify that this formula agrees with the one introduced above by splitting the sum into $n>0$, $n=0$, and and taking the geometric series.

Given this claim, we can evaluate the Poisson integral by exchanging the order of summation and integration:

${\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}h(e^{i\varphi })P(r{e}^{\theta -\varphi })𝑑\varphi$ $={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}{e}^{i\varphi n}{\displaystyle \sum _{m\in \mathbb{Z}}}{r}^{|m|}{e}^{im(\theta -\varphi )}d\varphi$

$={\displaystyle \frac{1}{2\pi }}{\displaystyle \sum _{m\in \mathbb{Z}}}{r}^{|m|}{e}^{im\theta }{\displaystyle {\int }_0^{2\pi }}{e}^{i(n-m)\varphi }𝑑\varphi .$

For any integer $k\ne 0$, ${\int }_0^{2\pi }{e}^{ik\varphi }𝑑\varphi ={e^{ik\varphi }ik|}_0^{2\pi }=0$, while for $k=0$ it is ${\int }_0^{2\pi }1𝑑\varphi =2\pi$; so the integral above is $0$ unless $m=n$, in which case it is $2\pi$. Therefore the only term which contributes is from $m=n$, which gives

$$r^{|n|}{e}^{in\theta },$$

which was our prediction for $\tilde{h}(r{e}^{i\theta })$.

If we had a linear combination of such terms for $h$, by the linearity of integrals we would get a linear combination of the $r^{|n|}{e}^{in\theta }$ for $\tilde{h}$, just as claimed above. So the Poisson integral formula is right for all trigonometric polynomials.

One can formally show that it also holds for other continuous functions (by carefully verifying the harmonicity and boundary values), but instead here is the sketch of a heuristic argument. It turns out that (almost) any function $h:\partial 𝔻\to ℂ$ (equivalently, any periodic function $f:ℝ\to ℂ$ of period $2\pi$, by $h(e^{i\theta })=f(\theta )$) can be written as an infinite linear combination of $e^{in\theta }$-terms. This is called the Fourier series of $h$ (or $f$). Then the argument above adapts, with a little more care, to show that the Poisson integral for $h$ recovers the Fourier series, incorporating an $r$-factor. That is: if

$$h(e^{i\theta })=\sum _n{a}_n{e}^{in\theta },$$

then

$$\tilde{h}(r{e}^{i\theta })=\sum _n{a}_n{r}^{|n|}{e}^{in\theta }.$$