Harmonic Functions

In this lesson you’ll learn what harmonic means, how every harmonic function is the real part of an analytic one, how to construct harmonic conjugates, and how the Dirichlet problem is solved on a disc and transported everywhere else.

A real function $u(x,y)$ with continuous second partials is harmonic on a domain if it satisfies Laplace’s equation

$$\nabla^2u = u_{xx}+u_{yy} = 0$$

Laplace’s equation is arguably the most important PDE in physics. It governs steady-state temperature, electrostatic and gravitational potential in empty space, incompressible irrotational fluid flow, and the equilibrium shape of a stretched membrane. All of those are the same equation, so all of them are the same mathematics.

Two facts, and they run in both directions.

If $f = u+iv$ is analytic, then $u$ and $v$ are both harmonic.

This came out of the Cauchy–Riemann equations in lesson 5: differentiate $u_x=v_y$ with respect to $x$, differentiate $u_y=-v_x$ with respect to $y$, and add.

Conversely, if $u$ is harmonic on a simply connected domain, there is a harmonic $v$, unique up to a constant, with $u+iv$ analytic. That $v$ is the harmonic conjugate of $u$.

Together these say that harmonic functions and analytic functions are two views of the same objects. Real two-dimensional potential theory is complex analysis in different clothing, which is why every result in this section has a physical reading.

The simple connectivity matters. On an annulus, $u = \ln|z|$ is harmonic but its conjugate would have to be $\arg z$, which cannot be made single-valued there. That is the same obstruction as the branch cut for the logarithm.

Because they are real parts of analytic functions, harmonic functions inherit the strong behaviour of the section.

Mean value property. The value at a point is the average over any circle about it inside the domain:

$$u(z_0) = \frac{1}{2\pi}\int_0^{2\pi}u\big(z_0+re^{i\theta}\big)\,d\theta$$

The figure tests this. For $u = x^2-y^2$, the real part of $z^2$, every circular average equals the centre value exactly. For $u = x^2+y^2 = |z|^2$, which has $\nabla^2u = 4 \ne 0$, the averages exceed the centre value and grow with the radius - in fact by exactly $r^2$, which is the Laplacian times $r^2/4$.

Maximum principle. A harmonic function on a bounded domain attains its maximum and minimum on the boundary, and if an interior maximum is attained the function is constant. This is immediate from the mean value property: a strict interior maximum could not equal the average of its neighbours.

Uniqueness. Two harmonic functions with the same boundary values are identical, since their difference is harmonic with zero boundary values, hence zero by the maximum principle. This is what makes physics well posed: specify the temperature on the boundary and the interior temperature is determined.

Smoothness. Harmonic functions are automatically infinitely differentiable, and even real-analytic, inherited from the corresponding property of analytic functions.

Liouville. A harmonic function bounded on all of $\mathbb{R}^2$ is constant.

Dirichlet problem. Given a domain $D$ and a continuous function on its boundary, find the harmonic function on $D$ matching those boundary values.

On the unit disc there is an explicit answer, the Poisson integral formula:

$$u(re^{i\theta}) = \frac{1}{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\varphi)+r^2}\,u(e^{i\varphi})\,d\varphi$$

The kernel is a weighted average that concentrates near the nearest boundary point as $r\to1$. Setting $r=0$ recovers the mean value property, so the formula is its refinement to off-centre points.

For any other simply connected region the strategy is the one from the conformal mapping lesson:

1. Map the region conformally to the disc.
2. Carry the boundary data along.
3. Apply the Poisson formula.
4. Pull back.

Because harmonicity survives conformal maps, this solves the problem on every region you can map - and by the Riemann mapping theorem, that is every simply connected region.

For $f=u+iv$ analytic, the level curves of $u$ and $v$ are orthogonal, which we saw in lesson 6. Physically:

$u$	$v$	Setting
equipotentials	field lines	electrostatics
isotherms	heat flow lines	conduction
velocity potential	streamlines	fluid flow

The pair $f = u+iv$ is called the complex potential, and $f'(z) = u_x + iv_x$ gives the field directly: in fluid dynamics $\overline{f'(z)}$ is the velocity vector. One analytic function encodes an entire two-dimensional field, which is the compression that makes the method powerful.

Example 1: Verify $u = x^2-y^2$ is harmonic and find its conjugate.

Solution. $u_{xx}=2$ and $u_{yy}=-2$, so $u_{xx}+u_{yy}=0$ ✓

For the conjugate, use $v_y = u_x = 2x$, so $v = 2xy+g(x)$. Then $v_x = 2y+g'(x)$, and this must equal $-u_y = 2y$, forcing $g'=0$.

$$v = 2xy+C, \qquad f = z^2 + iC$$

∎

Example 2: Show $u = x^2+y^2$ is not harmonic.

Solution. $u_{xx}+u_{yy} = 2+2 = 4 \ne 0$.

$$\text{not harmonic}$$

Consequently it has no harmonic conjugate, and indeed $|z|^2 = z\bar z$ involves $\bar z$ and is not analytic. The figure shows the failure quantitatively: circular averages exceed the centre value. ∎

Example 3: Verify $u = \ln|z|$ is harmonic away from 0.

Solution. With $r = \sqrt{x^2+y^2}$, $u = \frac12\ln(x^2+y^2)$. Then

$$u_x = \frac{x}{x^2+y^2}, \qquad u_{xx} = \frac{y^2-x^2}{(x^2+y^2)^2}$$

and by symmetry $u_{yy} = \frac{x^2-y^2}{(x^2+y^2)^2}$. They cancel, so $\nabla^2u=0$ on $\mathbb{C}\setminus\{0\}$. ∎

Its conjugate is $\arg z$, and $\operatorname{Log}z = \ln|z|+i\arg z$. On any annulus around 0 the conjugate cannot be single-valued, which shows the simple connectivity hypothesis is not removable. Physically this $u$ is the potential of a point charge, whose field lines wind around the charge and never close.

Example 4: Find the harmonic conjugate of $u = e^x\cos y$.

Solution. Check first: $u_{xx} = e^x\cos y$ and $u_{yy} = -e^x\cos y$, which sum to 0 ✓

From $v_y = u_x = e^x\cos y$, integrate in $y$:

$$v = e^x\sin y + g(x)$$

Then $v_x = e^x\sin y+g'(x)$ must equal $-u_y = e^x\sin y$, so $g'=0$.

$$v = e^x\sin y+C, \qquad f = e^z+iC$$

∎

Example 5: Use the maximum principle.

$u$ is harmonic on $|z|\le1$ with $u = 3$ on the boundary circle. What is $u$ inside?

Solution. The constant function 3 is harmonic and matches the boundary data. By uniqueness there is only one such function, so

$$u \equiv 3$$

∎

Example 6: Solve a Dirichlet problem on the disc.

$u$ is harmonic on $|z|<1$ with boundary values $u(e^{i\theta}) = \cos\theta$. Find $u$.

Solution. Guess a low-order candidate: $\operatorname{Re}(z) = x = r\cos\theta$ is harmonic and equals $\cos\theta$ when $r=1$.

$$u = x = r\cos\theta$$

By uniqueness this is the answer, with no integration needed. ✓

At the centre, $u(0)=0$, which is the average of $\cos\theta$ over the circle. Consistent with the mean value property, and a good check to run on any such solution.

Example 7: Use the complex potential.

The complex potential of uniform flow past a cylinder of radius $a$ is $f(z) = U\left(z+\frac{a^2}{z}\right)$. Find the streamlines and the velocity.

Solution. With $z = re^{i\theta}$,

$$f = U\left(re^{i\theta}+\frac{a^2}{r}e^{-i\theta}\right)$$

so the stream function is

$$\psi = \operatorname{Im}f = U\left(r-\frac{a^2}{r}\right)\sin\theta$$

On $r=a$ this is identically 0, so the circle is a streamline - the flow does not cross the cylinder, which is exactly the physical boundary condition.

The velocity comes from the derivative:

$$f'(z) = U\left(1-\frac{a^2}{z^2}\right)$$

which vanishes at $z=\pm a$, the two stagnation points at the front and back of the cylinder, and tends to $U$ far away as it should. ∎

One analytic function delivered the entire flow field, boundary condition included.

Steady heat conduction. Temperature in a body with fixed boundary temperatures and no internal sources is harmonic. Solving for it is the Dirichlet problem, and conformal mapping handles awkward cross-sections.

Electrostatics. Potential in a charge-free region is harmonic, and field lines are the conjugate’s level curves. Capacitance calculations for irregular electrodes were built on this before numerical solvers.

Groundwater and porous flow. Darcy flow has a harmonic pressure head, so aquifer models and seepage under dams are solved as Dirichlet problems. Flow nets drawn by hand in civil engineering are exactly these orthogonal families.

Aerodynamics. Potential flow around bodies is the complex potential of Example 7, and adding circulation gives lift. The stagnation points are physically observable.

Membrane deflection. A stretched membrane with a fixed boundary takes a harmonic shape under small deflection, which is why soap-film experiments were once used as analog computers for Laplace’s equation.

Image processing. Harmonic inpainting fills a missing region by solving Laplace’s equation with the surrounding pixels as boundary data. The maximum principle guarantees no new extremes appear inside the patch, which is why the result looks smooth.