8b. Harmonic functions

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To summarize, we have been doing some physics, with the conclusion that both the wave and heat equation involve the Laplace operator. So, let us formulate, as a conclusion, the definition of this operator, that we would like to further understand:

Definition

The Laplace operator in [math]2[/math] dimensions is:

[[math]] \Delta f=\frac{d^2f}{dx^2}+\frac{d^2f}{dy^2} [[/math]]

A function [math]f:\mathbb R^2\to\mathbb C[/math] satisfying [math]\Delta f=0[/math] will be called harmonic.

Here the formula of [math]\Delta[/math] is the one coming from Theorem 8.5 and Theorem 8.6. As for the notion of harmonic function, this is something quite natural. Indeed, we can think of [math]\Delta[/math] as being a linear operator on the space of functions [math]f:\mathbb R^2\to\mathbb C[/math], and previous experience with linear operators, and linear algebra in general, suggests looking first into the eigenvectors of [math]\Delta[/math]. But the simplest such eigenvectors are those corresponding to the eigenvalue [math]\lambda=0[/math], and these are exactly our harmonic functions, satisfying:

[[math]] \Delta f=0 [[/math]]

Getting now to more concrete things, and to some mathematics that we can do, using our knowledge, let us try to find the functions [math]f:\mathbb R^2\to\mathbb C[/math] which are harmonic. And here, as a good surprise, we have an interesting link with the holomorphic functions:

Theorem

Any holomorphic function [math]f:\mathbb C\to\mathbb C[/math], when regarded as real function

[[math]] f:\mathbb R^2\to\mathbb C [[/math]]

is harmonic. Moreover, the conjugates [math]\bar{f}[/math] of holomorphic functions are harmonic too.

Show Proof

The first assertion follows from the following computation, for the power functions [math]f(z)=z^n[/math], with the usual notation [math]z=x+iy[/math]:

[[math]] \begin{eqnarray*} \Delta z^n &=&\frac{d^2z^n}{dx^2}+\frac{d^2z^n}{dy^2}\\ &=&\frac{d(nz^{n-1})}{dx}+\frac{d(inz^{n-1})}{dy}\\ &=&n(n-1)z^{n-2}-n(n-1)z^{n-2}\\ &=&0 \end{eqnarray*} [[/math]]

As for the second assertion, this follows from [math]\Delta\bar{f}=\overline{\Delta f}[/math], which is clear from definitions, and which shows that if [math]f[/math] is harmonic, then so is its conjugate [math]\bar{f}[/math].

■

All this is quite interesting, and the idea in what follows will be that of developing a theory of harmonic functions, as a generalization of the theory that we know for the holomorphic functions, covering as well functions of type [math]\bar{z}[/math]. Them, we will go back to physics, with some applications of this to the wave equation, and the heat equation.

As a first goal, in order to understand the harmonic functions, we can try to find the homogeneous polynomials [math]P\in\mathbb R[x,y][/math] which are harmonic. In order to do so, the most convenient is to use the variable [math]z=x+iy[/math], and think of these polynomials as being homogeneous polynomials [math]P\in\mathbb R[z,\bar{z}][/math]. With this convention, the result is as follows:

Theorem

The degree [math]n[/math] homogeneous polynomials [math]P\in\mathbb R[x,y][/math] which are harmonic are precisely the linear combinations of

[[math]] P=z^n\quad,\quad P=\bar{z}^n [[/math]]

with the usual identification [math]z=x+iy[/math].

Show Proof

As explained above, any homogeneous polynomial [math]P\in\mathbb R[x,y][/math] can be regarded as an homogeneous polynomial [math]P\in\mathbb R[z,\bar{z}][/math], with the change of variables [math]z=x+iy[/math], and in this picture, the degree [math]n[/math] homogeneous polynomials are as follows:

[[math]] P(z)=\sum_{k+l=n}c_{kl}z^k\bar{z}^l [[/math]]

In oder to solve now the Laplace equation [math]\Delta P=0[/math], we must compute the quantities [math]\Delta(z^k\bar{z}^l)[/math], for any [math]k,l[/math]. But the computation here is routine. We first have the following formula, with the derivatives being computed with respect to the variable [math]x[/math]:

[[math]] \begin{eqnarray*} \frac{d(z^k\bar{z}^l)}{dx} &=&(z^k)'\bar{z}^l+z^k(\bar{z}^l)'\\ &=&kz^{k-1}\bar{z}^l+lz^k\bar{z}^{l-1} \end{eqnarray*} [[/math]]

By taking one more time the derivative with respect to [math]x[/math], we obtain:

[[math]] \begin{eqnarray*} \frac{d^2(z^k\bar{z}^l)}{dx^2} &=&k(z^{k-1}\bar{z}^l)'+l(z^k\bar{z}^{l-1})'\\ &=&k\left[(z^{k-1})'\bar{z}^l+z^{k-1}(\bar{z}^l)'\right]+l\left[(z^k)'\bar{z}^{l-1}+z^k(\bar{z}^{l-1})'\right]\\ &=&k\left[(k-1)z^{k-2}\bar{z}^l+lz^{k-1}\bar{z}^{l-1}\right]+l\left[kz^{k-1}\bar{z}^{l-1}+(l-1)z^k\bar{z}^{l-2}\right]\\ &=&k(k-1)z^{k-2}\bar{z}^l+2klz^{k-1}\bar{z}^{l-1}+l(l-1)z^k\bar{z}^{l-2} \end{eqnarray*} [[/math]]

With respect to the variable [math]y[/math], the computations are similar, but some [math]\pm i[/math] factors appear, due to [math]z'=i[/math] and [math]\bar{z}'=-i[/math], coming from [math]z=x+iy[/math]. We first have:

[[math]] \begin{eqnarray*} \frac{d(z^k\bar{z}^l)}{dy} &=&(z^k)'\bar{z}^l+z^k(\bar{z}^l)'\\ &=&ikz^{k-1}\bar{z}^l-ilz^k\bar{z}^{l-1} \end{eqnarray*} [[/math]]

By taking one more time the derivative with respect to [math]y[/math], we obtain:

[[math]] \begin{eqnarray*} \frac{d^2(z^k\bar{z}^l)}{dy^2} &=&ik(z^{k-1}\bar{z}^l)'-il(z^k\bar{z}^{l-1})'\\ &=&ik\left[(z^{k-1})'\bar{z}^l+z^{k-1}(\bar{z}^l)'\right]-il\left[(z^k)'\bar{z}^{l-1}+z^k(\bar{z}^{l-1})'\right]\\ &=&ik\left[i(k-1)z^{k-2}\bar{z}^l-ilz^{k-1}\bar{z}^{l-1}\right]-il\left[ikz^{k-1}\bar{z}^{l-1}-i(l-1)z^k\bar{z}^{l-2}\right]\\ &=&-k(k-1)z^{k-2}\bar{z}^l+2klz^{k-1}\bar{z}^{l-1}-l(l-1)z^k\bar{z}^{l-2} \end{eqnarray*} [[/math]]

We can now sum the formulae that we found, and we obtain:

[[math]] \begin{eqnarray*} \Delta(z^k\bar{z}^l) &=&\frac{d^2(z^k\bar{z}^l)}{dx^2}+\frac{d^2(z^k\bar{z}^l)}{dy^2}\\ &=&k(k-1)z^{k-2}\bar{z}^l+2klz^{k-1}\bar{z}^{l-1}+l(l-1)z^k\bar{z}^{l-2}\\ &&-k(k-1)z^{k-2}\bar{z}^l+2klz^{k-1}\bar{z}^{l-1}-l(l-1)z^k\bar{z}^{l-2}\\ &=&4klz^{k-1}\bar{z}^{l-1} \end{eqnarray*} [[/math]]

In other words, we have reached to the following formula:

[[math]] f=z^k\bar{z}^l\implies\Delta f=\frac{4klf}{|z|^2} [[/math]]

Now let us get back to our homogeneous polynomial [math]P[/math], written as follows:

[[math]] P(z)=\sum_{k+l=n}c_{kl}z^k\bar{z}^l [[/math]]

By using the above formula, the Laplacian of [math]P[/math] is given by:

[[math]] \Delta P(z)=\frac{4}{|z|^2}\sum_{k+l=n}klc_{kl}z^k\bar{z}^l [[/math]]

We conclude that the Laplace equation for [math]P[/math] takes the following form:

[[math]] \begin{eqnarray*} \Delta P=0 &\iff&klc_{kl}=0,\forall k,l\\ &\iff&[k,l\neq0\implies c_{kl}=0]\\ &\iff&P=c_{n0}z^n+c_{0n}\bar{z}^n \end{eqnarray*} [[/math]]

Thus, we are led to the conclusion in the statement. And with the observation that the real formulation of the final result is something quite complicated, and so, for one more time, the use of the complex variable [math]z=x+iy[/math] is something very useful.

■

We know that the holomorphic functions are harmonic, and it follows from this that the real and imaginary parts of the holomorphic functions, as well as any linear combinations of these real and imaginary parts, are harmonic too. That is, if [math]f[/math] is holomorphic, then the following function is harmonic, for any values of the parameters [math]\alpha,\beta\in\mathbb C[/math]:

[[math]] f_{\alpha\beta}=\alpha Re(f)+\beta Im(f) [[/math]]

Observe that this result covers all the examples that we have so far, for instance with the function [math]\bar{z}[/math], that we know to be harmonic, appearing as follows:

[[math]] \bar{z}=Re(z)-iIm(z) [[/math]]

Our main goal in what follows will be that of proving a converse to this, at least locally. For this purpose, let us start with the following definition:

Definition

The Cauchy-Riemann operators are

[[math]] \partial=\frac{1}{2}\left(\frac{d}{dx}-i\frac{d}{dy}\right) \quad,\quad \bar{\partial}=\frac{1}{2}\left(\frac{d}{dx}+i\frac{d}{dy}\right) [[/math]]

where [math]\frac{d}{dx}[/math] and [math]\frac{d}{dy}[/math] are the usual partial derivatives for complex functions.

There are many things that can be said about the Cauchy-Riemann operators [math]\partial,\bar{\partial}[/math], the idea being that in many contexts, these are better to use than the usual partial derivatives [math]\frac{d}{dx},\frac{d}{dy}[/math], and with this being a bit like the usage of the variables [math]z,\bar{z}[/math], instead of the decomposition [math]z=a+ib[/math], for many questions regarding the complex numbers.

We have already seen in fact some instances of this, in our computations above. At the general level, the main properties of [math]\partial,\bar{\partial}[/math] can be summarized as follows:

Proposition

Assume that [math]f:X\to\mathbb C[/math] is differentiable in the real sense.

• [math]f[/math] is holomorphic precisely when [math]\bar{\partial}f=0[/math].
• In this case, its derivative is [math]f'=\partial f[/math].
• The Laplace operator is given by [math]\Delta=4\partial\bar{\partial}[/math].
• [math]f[/math] is harmonic precisely when [math]\partial\bar{\partial}f=0[/math].

Show Proof

We can assume by linearity that we are dealing with differentiability questions at [math]0[/math]. Since our function [math]f:X\to\mathbb C[/math] is differentiable in the real sense, we have a formula as follows, with [math]z=x+iy[/math], and with [math]a,b\in\mathbb C[/math] being the partial derivatives at [math]0[/math]:

[[math]] f(z)=ax+by+o(z) [[/math]]

Now observe that we can write this formula in the following way:

[[math]] \begin{eqnarray*} f(z) &=&a\cdot\frac{z+\bar{z}}{2}+b\cdot\frac{z-\bar{z}}{2i}+o(z)\\ &=&a\cdot\frac{z+\bar{z}}{2}+b\cdot\frac{i\bar{z}-iz}{2}+o(z)\\ &=&\frac{a-ib}{2}\cdot z+\frac{a+ib}{2}\cdot\bar{z}+o(z) \end{eqnarray*} [[/math]]

Now by dividing by [math]z[/math], we obtain from this the following formula:

[[math]] \begin{eqnarray*} \frac{f(z)}{z} &=&\frac{a-ib}{2}+\frac{a+ib}{2}\cdot\frac{\bar{z}}{z}+o(1)\\ &=&\partial f(0)+\bar{\partial}f(0)\cdot\frac{\bar{z}}{z}+o(1) \end{eqnarray*} [[/math]]

But this gives the first two assertions, because in order for the derivative [math]f'(0)[/math] to exist, appearing as the [math]z\to0[/math] limit of the above quantity, the coefficient of [math]\bar{z}/z[/math], which does not converge, must vanish. Regarding now the third assertion, this follows from:

[[math]] \begin{eqnarray*} \Delta &=&\frac{d^2}{dx^2}+\frac{d^2}{dy^2}\\ &=&\left(\frac{d}{dx}-i\frac{d}{dy}\right) \left(\frac{d}{dx}+i\frac{d}{dy}\right)\\ &=&4\partial\bar{\partial} \end{eqnarray*} [[/math]]

As for the last assertion, this is clear from this latter formula of [math]\Delta[/math].

■

In analogy now with the theory of the holomorphic functions, we have:

Theorem

The harmonic functions obey to the same general principles as the holomorphic functions, namely:

• The maximum modulus principle.
• The plain mean value formula.
• The boundary mean value formula.
• The Liouville theorem.

Also, locally, the real harmonic functions are the real parts of holomorphic functions.

Show Proof

This is something quite tricky, the idea being as follows:

(1) Regarding the maximum modulus principle, the statement here is that any harmonic function [math]f:X\to\mathbb C[/math] has the property that the maximum of [math]|f|[/math] over a domain is attained on its boundary. That is, given a domain [math]D[/math], with boundary [math]\gamma[/math], we have:

[[math]] \exists x\in\gamma\quad,\quad |f(x)|=\max_{y\in D}|f(y)| [[/math]]

(2) Regarding the plain mean value formula, here the statement is that given an harmonic function [math]f:X\to\mathbb C[/math], and a disk [math]D[/math], the following happens:

[[math]] f(x)=\int_Df(y)dy [[/math]]

(3) Regarding the boundary mean value formula, here the statement is that given an harmonic function [math]f:X\to\mathbb C[/math], and a disk [math]D[/math], with boundary [math]\gamma[/math], the following happens:

[[math]] f(x)=\int_\gamma f(y)dy [[/math]]

(4) Regarding the Liouville theorem, the statement here is that an entire, bounded harmonic function must be constant:

[[math]] f:\mathbb C\to\mathbb C\quad,\quad |f|\leq M\quad\implies\quad f={\rm constant} [[/math]]

(5) Finally, regarding the proofs, these are not exactly trivial. According to our previous experience with holomorphic functions, the above conditions (2,3) are equivalent, and imply (1,4) via some simple arguments, and the same happens for the harmonic functions. However, establishing (2) and the last assertion is something which is not exactly trivial, and we refer to Rudin for proofs of this, and with the promise that we will be back to this, later in this book, once we will know more calculus.

■

Observe that we are in the process of a tactical retreat from mathematics. As a last objective, however, before completely giving up with all this, and getting back to physics, let us try to find the harmonic functions which are radial, in the following sense:

[[math]] f(z)=\varphi(|z|) [[/math]]

However, things are quite tricky here, involving a blowup phenomenon at the dimension value [math]N=2[/math], which is precisely the one that we are interested in. So, moving now to [math]N[/math] dimensions, with the straightforward definition for [math]\Delta[/math] there, here is the result:

Theorem

The fundamental radial solutions of [math]\Delta f=0[/math] are

[[math]] f(x)=\begin{cases} ||x||^{2-N}&(N\neq 2)\\ \log||x||&(N=2) \end{cases} [[/math]]

with the [math]\log[/math] at [math]N=2[/math] basically coming from [math]\log'=1/x[/math].

Show Proof

Consider indeed a radial function, defined outside the origin [math]x=0[/math]. This function can be written as follows, with [math]\varphi:(0,\infty)\to\mathbb C[/math] being a certain function:

[[math]] f:\mathbb R^N-\{0\}\to\mathbb C\quad,\quad f(x)=\varphi(||x||) [[/math]]

Our first goal will be that of reformulating the Laplace equation [math]\Delta f=0[/math] in terms of the one-variable function [math]\varphi:(0,\infty)\to\mathbb C[/math]. For this purpose, observe that we have:

[[math]] \begin{eqnarray*} \frac{d||x||}{dx_i} &=&\frac{d\sqrt{\sum_{i=1}^Nx_i^2}}{dx_i}\\ &=&\frac{1}{2}\cdot\frac{1}{\sqrt{\sum_{i=1}^Nx_i^2}}\cdot\frac{d\left(\sum_{i=1}^Nx_i^2\right)}{dx_i}\\ &=&\frac{1}{2}\cdot\frac{1}{||x||}\cdot 2x_i\\ &=&\frac{x_i}{||x||} \end{eqnarray*} [[/math]]

By using this formula, we have the following computation:

[[math]] \begin{eqnarray*} \frac{df}{dx_i} &=&\frac{d\varphi(||x||)}{dx_i}\\ &=&\varphi'(||x||)\cdot\frac{d||x||}{dx_i}\\ &=&\varphi'(||x||)\cdot\frac{x_i}{||x||} \end{eqnarray*} [[/math]]

By differentiating one more time, we obtain the following formula:

[[math]] \begin{eqnarray*} \frac{d^2f}{dx_i^2} &=&\frac{d}{dx_i}\left(\varphi'(||x||)\cdot\frac{x_i}{||x||}\right)\\ &=&\frac{d\varphi'(||x||)}{dx_i}\cdot\frac{x_i}{||x||}+\varphi'(||x||)\cdot\frac{d}{dx_i}\left(\frac{x_i}{||x||}\right)\\ &=&\left(\varphi''(||x||)\cdot\frac{x_i}{||x||}\right)\cdot\frac{x_i}{||x||}+\varphi'(||x||)\cdot\frac{||x||-x_i\cdot x_i/||x||}{||x||^2}\\ &=&\varphi''(||x||)\cdot\frac{x_i^2}{||x||^2}+\varphi'(||x||)\cdot\frac{||x||^2-x_i^2}{||x||^3} \end{eqnarray*} [[/math]]

Now by summing over [math]i\in\{1,\ldots,N\}[/math], this gives the following formula:

[[math]] \begin{eqnarray*} \Delta f &=&\sum_{i=1}^N\varphi''(||x||)\cdot\frac{x_i^2}{||x||^2}+\sum_{i=1}^N\varphi'(||x||)\cdot\frac{||x||^2-x_i^2}{||x||^3}\\ &=&\varphi''(||x||)\cdot\frac{||x||^2}{||x||^2}+\varphi'(||x||)\cdot\frac{(N-1)||x||^2}{||x||^3}\\ &=&\varphi''(||x||)+\varphi'(||x||)\cdot\frac{N-1}{||x||} \end{eqnarray*} [[/math]]

Thus, with [math]r=||x||[/math], the Laplace equation [math]\Delta f=0[/math] can be reformulated as follows:

[[math]] \varphi''(r)+\frac{(N-1)\varphi'(r)}{r}=0 [[/math]]

Equivalently, the equation that we want to solve is as follows:

[[math]] r\varphi''+(N-1)\varphi'=0 [[/math]]

Now observe that we have the following formula:

[[math]] \begin{eqnarray*} (r^{N-1}\varphi')' &=&(N-1)r^{N-2}\varphi'+r^{N-1}\varphi''\\ &=&r^{N-2}((N-1)\varphi'+r\varphi'') \end{eqnarray*} [[/math]]

Thus, the equation to be solved can be simply written as follows:

[[math]] (r^{N-1}\varphi')'=0 [[/math]]

We conclude that [math]r^{N-1}\varphi'[/math] must be a constant [math]K[/math], and so, that we must have:

[[math]] \varphi'=Kr^{1-N} [[/math]]

But the fundamental solutions of this latter equation are as follows:

[[math]] \varphi(r)=\begin{cases} r^{2-N}&(N\neq 2)\\ \log r&(N=2) \end{cases} [[/math]]

Thus, we are led to the conclusion in the statement.

■

And good news, that is the end. We have learned many things about the Laplace operator and the harmonic functions, but all this clearly becomes too complicated, spilling sometimes in arbitrary [math]N[/math] dimensions, so time to stop. We will be back to this.

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].