Harmonic Functions and Holomorphy

The harmonic–holomorphic correspondence underlies two-dimensional potential theory: steady-state heat flow, electrostatics in the plane, and incompressible irrotational fluid flow are all modeled by harmonic potentials and their harmonic-conjugate stream functions, letting engineers solve boundary value problems by finding an appropriate conformal map (this is the classical technique of "complex potentials" in aerodynamics, e.g. Joukowski airfoil theory). In electrical engineering, harmonic functions model capacitance and resistance distributions in planar conductors. In pure mathematics, the correspondence is the seed of Hodge theory and the theory of harmonic forms on Riemann surfaces, and of the theory of subharmonic functions used in pluripotential theory and several complex variables. In probability, harmonic functions are exactly the functions annihilated by the generator of planar Brownian motion, giving a probabilistic (martingale) interpretation of the mean value property; this feeds into the modern theory of Brownian motion and Dirichlet problems. In numerical analysis and computer graphics, discrete harmonic functions (solutions of the discrete Laplace equation on graphs/meshes) are used for mesh parameterization and image processing, directly inspired by the continuous theory here. Finally, this article's construction is the essential stepping stone toward Conformal Mappings and the Riemann Mapping Theorem, where harmonic conjugates supply the angle-preserving coordinate systems used to transplant boundary value problems between domains.

Ask the tutor: Can you show me why the maximum principle for harmonic functions follows from the maximum modulus principle for holomorphic functions?

The Cauchy–Riemann equations are two coupled first-order PDEs relating $u$ and $v$. Differentiating the first equation $u_x = v_y$ with respect to $x$, and the second $u_y = -v_x$ with respect to $y$, gives

If $v \in C^2$ so that mixed partials commute ($v_{xy} = v_{yx}$, guaranteed once we know holomorphic functions are $C^\infty$, indeed real-analytic), adding these yields $u_{xx} + u_{yy} = 0$. The same computation with the roles of $u,v$ swapped shows $v$ is harmonic too. So harmonicity of both real and imaginary parts is an automatic corollary of holomorphy — it costs nothing extra. Geometrically, the level curves $u = c_1$ and $v = c_2$ of a holomorphic function $f = u+iv$ with $f' \neq 0$ are orthogonal wherever they cross: the gradient vectors $\nabla u = (u_x,u_y)$ and $\nabla v=(v_x,v_y) = (-u_y, u_x)$ satisfy $\nabla u \cdot \nabla v = -u_x u_y + u_y u_x = 0$. This orthogonal grid of level curves is the seed of Conformal Mappings: holomorphic maps carry orthogonal families to orthogonal families, and $u,v$ can be read as "isothermal coordinates" adapted to $f$. Physically, if $u$ represents a steady-state temperature (or electrostatic potential), Laplace's equation says $u$ has no local maxima or minima in the interior — heat flows without sources or sinks. The conjugate $v$ traces the "streamlines" — level curves along which no heat flows, orthogonal to the isotherms $u = \text{const}$.

Example 1: A non-homogeneous polynomial harmonic function. Let $u(x,y) = x^3 - 3xy^2 + 2x$ on $\Omega = \mathbb{C}$ (simply connected). We verify harmonicity and find the harmonic conjugate. Check harmonicity. Compute

So $u_{xx}+u_{yy} = 6x - 6x = 0$: $u$ is harmonic on all of $\mathbb{C}$. Find $v$ using the Cauchy–Riemann equations. We need $v_y = u_x = 3x^2-3y^2+2$. Integrate in $y$:

for some function $g(x)$ to be determined. Now impose the second Cauchy–Riemann equation $v_x = -u_y = 6xy$:

for a real constant $c$. Take $c=0$ for a canonical choice:

Assemble $f = u + iv$ and identify it. Compute

Group by powers: $x^3 - 3xy^2 + i(3x^2y-y^3) = (x+iy)^3 = z^3$ (this is the standard binomial expansion identity $z^3=x^3+3ix^2y-3xy^2-iy^3$), and $2x+2iy = 2z$. So

manifestly holomorphic (a polynomial), confirming $u = \mathrm{Re}(z^3+2z)$ and $v=\mathrm{Im}(z^3+2z)$ are conjugate harmonic functions, exactly as the theorem predicts. As a sanity check, differentiate directly: $f'(z) = 3z^2+2 = u_x + iv_x = (3x^2-3y^2+2) + i(6xy)$, matching our computed $u_x, v_x$ exactly.

Example 2: Recovering $f$ from $u$ using the Milne-Thomson-style shortcut, with a transcendental $u$. Let $u(x,y) = e^x(x\cos y - y \sin y)$ on $\mathbb{C}$. This looks unpleasant to integrate directly in $y$; we illustrate a faster route that is a direct corollary of the theorem's proof. Harmonicity check first. Write $u = x e^x \cos y - y e^x \sin y$. Compute the partials carefully.

Adding: $u_{xx}+u_{yy} = e^x\cos y(2+x) - ye^x\sin y - e^x\cos y(x+2) + ye^x\sin y = 0$. Good, $u$ is harmonic. Shortcut: express $f'(z)$ directly. Since $f = u+iv$ is holomorphic with $f' = u_x - i u_y$ (using $v_x = -u_y$), we can compute $f'(z)$ purely from $u_x, u_y$ and then re-express the result in terms of $z$ alone by formally substituting $x=z, y=0$ (a standard trick valid because $f'(z)$, being holomorphic, is determined by its restriction to the real axis $y=0$, where $z=x$):

Compute: $u_x(x,0) = e^x(1+x)\cdot 1 - 0 = e^x(1+x)$, and $u_y(x,0) = -e^x\sin(0)(x+1) - 0\cdot e^x\cos 0 = 0$. So along $y=0$:

Integrate: $\displaystyle f(z) = \int (1+z)e^z\,dz = z e^z + C$ (using $\int ze^z\,dz = ze^z - e^z$, so $\int(1+z)e^z\,dz = e^z + ze^z - e^z = ze^z$, plus constant $C$). Verify. Take $C=0$ and expand $f(z) = ze^z$ with $z=x+iy$, using $e^z = e^x(\cos y + i \sin y)$:

So $\mathrm{Re}(f) = e^x(x\cos y - y\sin y) = u$ exactly as given, confirming the computation, and we read off the harmonic conjugate directly:

This example shows the power of the correspondence: differentiating a nasty-looking $u$, restricting to the real axis, and re-integrating in $z$ recovers a clean closed form $f(z) = ze^z$ that would have been hard to guess by inspection.