Harmonic Functions and Conjugates

Laplace's equation $\nabla^2 u=0$ is the quietest equation in physics and the most universal: it governs the electrostatic potential in a charge-free region, the gravitational potential in empty space, the steady temperature of a body with no internal heat source, and the velocity potential of an incompressible irrotational fluid. Its solutions are the harmonic functions, and their defining trait is a kind of perfect blandness — the value at any point is the average of the values around it, so a harmonic function has no interior peaks, valleys, or dimples. It is the smoothest interpolation of whatever data you clamp onto the boundary. In two dimensions an extraordinary bridge appears. A function $u(x,y)$ is harmonic if and only if it is the real part of an analytic function $f=u+iv$ — and the imaginary part $v$ , its harmonic conjugate, is harmonic too. The link is the Cauchy–Riemann equations, inherited from the prerequisite article: they force $u$ and $v$ to be harmonic, make their level curves cross at right angles, and pair a potential with its stream function. The whole apparatus of complex analysis — Cauchy's integral formula, the maximum principle, conformal mapping — therefore descends onto two-dimensional potential theory. This article builds that dictionary, from Laplace's equation to the complex potential $f=u+iv$ whose derivative $f'$ encodes the field, and back.

Key takeaways

A real function $u(x,y)$ is harmonic on a domain if it is twice continuously differentiable and satisfies Laplace's equation $u_{xx}+u_{yy}=0$ .
If $f=u+iv$ is analytic, the Cauchy–Riemann equations force both $u$ and $v$ to be harmonic — in two dimensions, analyticity is the source of harmonicity.
Given a harmonic $u$ on a simply connected domain, one can construct a harmonic conjugate $v$ (unique up to a constant) so that $u+iv$ is analytic; the level curves of $u$ and $v$ are everywhere orthogonal, since $\nabla u\cdot\nabla v=0$ .
Harmonic functions obey the mean-value property and the maximum principle, and are automatically infinitely differentiable and real-analytic — rigidity inherited from the analytic function.
The pairing turns 2D electrostatics, ideal fluid flow, and steady heat conduction into complex analysis: potential and stream function fuse into the analytic complex potential $f=u+iv$ .
Topology can obstruct it: $u=\tfrac12\ln(x^2+y^2)$ is harmonic on a punctured disc but has no single-valued conjugate there.
Key idea: in two dimensions, harmonic = real part of analytic; the Cauchy–Riemann equations both produce harmonicity and pair each harmonic $u$ with an orthogonal conjugate $v$ , packaging potential and stream function into one analytic complex potential $f=u+iv$ .

The core insight

In two dimensions a function is harmonic exactly when it is the real (or imaginary) part of an analytic function. The Cauchy–Riemann equations both force $u_{xx}+u_{yy}=0$ and pair each harmonic $u$ with a conjugate $v$ whose level curves are orthogonal, so potential and stream function fuse into the single analytic complex potential $f=u+iv$ whose derivative $f'$ gives the field.

Where this lives today

The harmonic-conjugate dictionary is quietly at work far beyond the potential problems that spawned it. In computer graphics, Poisson image editing solves Laplace's equation with prescribed boundary values to paste one photograph into another with no visible seam — the maximum-principle statement that boundary data fix the smooth interior. Conformal maps, analytic functions that preserve the orthogonal net, drive texture mapping and mesh parametrization by flattening curved surfaces with minimal angular distortion. In two-dimensional condensed matter, conformal field theory — the modern descendant of harmonic-conjugate analysis — predicts the exact critical exponents of the Ising model and percolation. Electrical-impedance tomography and airfoil design still solve Dirichlet problems by conformal transplantation, and numerical relativity uses harmonic coordinates to keep Einstein's equations well-posed in black-hole simulations. The line from Laplace's spheroid to a merging-black-hole waveform runs straight through $\nabla^2 u=0$ .

The physical picture

Think of an analytic function as one rigid object with two shadows. Pin down the real part $u$ over a simply connected region and the imaginary part $v$ is determined completely, up to an additive constant — you cannot wiggle one without the other responding in lockstep. The Cauchy–Riemann equations $u_x=v_y,\ u_y=-v_x$ are the gear linkage, and it is rotational: the gradient of $v$ is the gradient of $u$ turned ninety degrees. That single fact is why the level curves of $u$ and of $v$ tile the plane into an orthogonal net of curvilinear squares — if $u$ is an electrostatic potential, the equipotentials and field lines cross at right angles everywhere.

Figure

The orthogonal net of f(z) = z²: equipotentials u = x²−y² = c in blue cross stream lines v = 2xy = k in amber everywhere at 90°, because ∇u·∇v = 0 forces the two families to meet at right angles

Harmonicity itself says a function has no local bumps: the value at a point equals its average over any small circle centred there — the mean-value property — so a harmonic function can never attain a strict interior maximum or minimum; extremes are pushed to the boundary. A stretched soap film obeys the same rule, sagging into the flattest surface its wire frame allows. The second figure makes this a concrete arithmetic fact.

Figure

Mean-value property for the harmonic u = x²−y²: at the centre z₀ = (2, 0) the value is u = 4, and the average of u over the circle of radius 1 equals 4 — even the four cardinal points 9, 3, 1, 3 average to exactly 4

Because $u$ and $v$ are so tightly coupled, boundary data alone reconstruct the entire analytic potential inside — the practical payoff. A hard-looking Dirichlet problem can be transplanted by a conformal map onto a disc or half-plane where the answer is a standard integral, then carried back, angles and orthogonality preserved at every step.

The core equations

The definition is a single second-order equation. A twice-differentiable real function is harmonic on a domain exactly when its two unmixed second partials cancel; the mixed partials never enter. Everything else in the article is a consequence of this line combined with analyticity.

\nabla^2 u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 \qquad (1)

where: $u(x,y)$ — a real function with continuous second partials; $\nabla^2$ — the two-dimensional Laplacian. Reading: the curvature in $x$ must exactly offset the curvature in $y$ at every point, forbidding interior bumps. The engine that manufactures harmonic functions is the Cauchy–Riemann pair, recalled from the prerequisite article. They are the necessary and sufficient conditions for $f=u+iv$ to be complex-differentiable, and they couple the two components rotationally.

u_x=v_y,\qquad u_y=-v_x \qquad (2)

where: $u,v$ — real and imaginary parts of $f=u+iv$ ; subscripts denote first partials. Reading: the gradient of $v$ is that of $u$ rotated ninety degrees, $\nabla v=(-u_y,\,u_x)$ — the geometric heart of the theory. Differentiate the Cauchy–Riemann pair once more and the mixed partials cancel, delivering harmonicity for free. This is the two-way street: analytic parts are automatically harmonic, and on a simply connected domain every harmonic function is the real part of some analytic function.

f=u+iv\ \text{analytic}\ \Longrightarrow\ \nabla^2 u=0\ \ \text{and}\ \ \nabla^2 v=0 \qquad (3)

where: the implication runs both ways locally — $u$ harmonic on a simply connected domain guarantees a conjugate $v$ . In 2D, harmonic and component-of-analytic are the same word. The rotational coupling makes the two families of level curves orthogonal. Substituting Cauchy–Riemann into the dot product of the gradients gives zero identically, so equipotentials and field lines cross at right angles wherever $f'\neq0$ .

\nabla u\cdot\nabla v=u_x v_x+u_y v_y=0 \qquad (4)

where: $\nabla u,\nabla v$ — gradients of the conjugate pair. Reading: the $u=$ const and $v=$ const curves form a grid of curvilinear squares. To build the conjugate of a given harmonic $u$ , integrate its exact differential. The one-form below is closed precisely because $u$ is harmonic, so on a simply connected domain it integrates to a single-valued $v$ .

\mathrm{d}v=v_x\,\mathrm{d}x+v_y\,\mathrm{d}y=-u_y\,\mathrm{d}x+u_x\,\mathrm{d}y \qquad (5)

where: $\mathrm{d}v$ — exact differential of the conjugate, built from derivatives of $u$ . Reading: recover $v$ by a path-independent line integral, fixing the additive constant by one boundary value. The rigidity theorems follow from Cauchy's integral formula applied to $f$ . The centrepiece is the mean-value property: a harmonic function equals its own average over any admissible circle, from which the maximum principle and Dirichlet uniqueness drop out immediately.

u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u\!\left(z_0+re^{i\theta}\right)\mathrm{d}\theta \qquad (6)

where: $z_0$ — an interior point; $r$ — any radius with the closed disc inside the domain. Reading: averaging creates no new extreme, so interior maxima and minima are impossible. Finally, when $u$ is a physical potential the complex derivative packages the whole field into one analytic object of magnitude $|f'|$ . The Milne–Thomson trick recovers $f$ from $u$ alone by formally setting $y=0,\ x=z$ .

f'(z)=u_x+iv_x=u_x-iu_y,\qquad f'(z)=u_x(z,0)-i\,u_y(z,0) \qquad (7)

where: $f'(z)$ — derivative of the complex potential; $|f'|$ is the field strength, $\arg f'$ its direction. Reading: one differentiation holds the entire vector field as a complex function.

Derivation AWhy the parts of an analytic function are harmonic

The goal is to show that analyticity does not merely permit Laplace's equation but forces it: if $f=u+iv$ is analytic, then $u$ and $v$ each satisfy $\nabla^2=0$ , and the same Cauchy–Riemann equations also make their gradients perpendicular. We take the Cauchy–Riemann pair as given from the prerequisite article, differentiate each once more, and use equality of continuous mixed second partials to cancel the cross terms — converting a first-order coupling into a second-order constraint on each component.

Step 1. Start from $u_x=v_y$ and differentiate with respect to $x$ : $u_{xx}=v_{yx}$ . Why: pushing another $x$ -derivative through the first Cauchy–Riemann equation produces the $u_{xx}$ we want on the left and a mixed partial of $v$ on the right.

Step 2. Start from $u_y=-v_x$ and differentiate with respect to $y$ : $u_{yy}=-v_{xy}$ . Why: the second equation carries the complementary $u_{yy}$ term, again at the cost of a mixed partial of $v$ — but with the opposite sign, which is exactly what makes the cancellation work.

Step 3. Add the two results: $u_{xx}+u_{yy}=v_{yx}-v_{xy}$ . Since $f$ analytic makes $v$ twice continuously differentiable (Cauchy's integral formula shows analytic functions are $C^\infty$ ), the mixed partials are equal, $v_{yx}=v_{xy}$ , and the right side vanishes: $\nabla^2 u=0$ . Why: equality of mixed partials is the whole game — the one place the smoothness of analytic functions is spent.

Step 4. By symmetry, differentiating $v_y=u_x$ by $y$ and $v_x=-u_y$ by $x$ and subtracting gives $\nabla^2 v=0$ . Both components are harmonic. Why: the Cauchy–Riemann system treats $u$ and $v$ almost symmetrically, so the argument runs identically for the imaginary part.

Step 5. For orthogonality, form $\nabla u\cdot\nabla v=u_xv_x+u_yv_y$ and substitute $u_x=v_y,\ u_y=-v_x$ to get $v_yv_x-v_xv_y=0$ . Why: the same substitution that produced harmonicity also collapses the gradient dot product, so orthogonal level curves and harmonicity are two faces of one coupling. *The key equalities are $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ , which holds by continuity of second partials, and $\nabla u\cdot\nabla v=0$ , which follows by direct substitution of the Cauchy–Riemann equations.*

Go deeper · Derivation B · 1 minConstructing the harmonic conjugate on a simply connected domain

Derivation BConstructing the harmonic conjugate on a simply connected domain

The goal is the converse: given a harmonic $u$ on a simply connected domain $D$ , manufacture a harmonic conjugate $v$ so that $f=u+iv$ is analytic, and see precisely where Laplace's equation and where the topology of $D$ each enter. The strategy is to write down the differential $v$ must satisfy from Cauchy–Riemann, test whether it is closed, discover that closedness is equivalent to $u$ being harmonic, then invoke simple connectivity to promote closed to exact and integrate.

Step 1. Cauchy–Riemann demands $v_x=-u_y$ and $v_y=u_x$ , so $\mathrm{d}v=-u_y\,\mathrm{d}x+u_x\,\mathrm{d}y$ , built entirely from known derivatives of $u$ . Why: we are not free to choose $v$ — the pairing fixes its gradient pointwise, leaving only a global integration to perform.

Step 2. Write the one-form as $P\,\mathrm{d}x+Q\,\mathrm{d}y$ with $P=-u_y,\ Q=u_x$ . It is closed when $P_y=Q_x$ , i.e. $-u_{yy}=u_{xx}$ . Why: exactness of a differential is the integrability condition that lets a line integral define a single-valued function; closedness is its local test.

Step 3. But $-u_{yy}=u_{xx}$ is exactly $u_{xx}+u_{yy}=0$ — the closedness condition is Laplace's equation. Since $u$ is harmonic by hypothesis, the form is closed. Why: this is the pivot of the whole subject — harmonicity is precisely what is needed for the conjugate's defining differential to be integrable.

Step 4. On a simply connected $D$ , Poincaré's lemma upgrades closed to exact, so $v(x,y)=\int_{(x_0,y_0)}^{(x,y)}(-u_y\,\mathrm{d}x+u_x\,\mathrm{d}y)+C$ is path-independent and defines $v$ up to the constant $C$ . Why: simple connectivity kills the loop integrals that would otherwise make $v$ multivalued — the indispensable topological hypothesis, as the logarithm shows.

Step 5. Differentiating the result reproduces $v_x=-u_y,\ v_y=u_x$ , confirming Cauchy–Riemann and hence that $f=u+iv$ is analytic. Why: closing the loop verifies we really did build an analytic function, not just some antiderivative. *In summary: the differential $-u_y\,\mathrm{d}x+u_x\,\mathrm{d}y$ is closed if and only if $\nabla^2u=0$ , and on a simply connected domain $D$ the resulting line integral is single-valued.*

Go deeper · Derivation C · 1 minThe mean-value property from Cauchy's integral formula

Derivation CThe mean-value property from Cauchy's integral formula

The goal is to derive the mean-value property — that a harmonic function equals its average over any admissible circle — directly from Cauchy's integral formula for the analytic $f$ , and to note how the maximum principle and Dirichlet uniqueness follow as one-line corollaries. The route: evaluate $f$ at the circle's centre by the integral formula, parametrize so the kernel becomes a plain average, then take real parts to reach the harmonic statement.

Step 1. For $f$ analytic on a disc containing the closed disc of radius $r$ about $z_0$ , Cauchy's integral formula gives $f(z_0)=\frac{1}{2\pi i}\oint_{|z-z_0|=r}\frac{f(z)}{z-z_0}\,\mathrm{d}z$ . Why: the integral formula is the single most powerful fact about analytic functions — the value at an interior point is fixed by the values on any enclosing contour.

Step 2. Parametrize the circle as $z=z_0+re^{i\theta}$ , so $\mathrm{d}z=ire^{i\theta}\,\mathrm{d}\theta$ and $z-z_0=re^{i\theta}$ . The factors $r$ , $e^{i\theta}$ , and $i$ cancel, leaving $f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})\,\mathrm{d}\theta$ . Why: on a circle the Cauchy kernel degenerates into a uniform weight, turning the contour integral into an unweighted average around the loop.

Step 3. Take the real part. Since $u=\operatorname{Re}f$ , $u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})\,\mathrm{d}\theta$ — the harmonic mean-value property. Why: the identity holds for the complex $f$ , so it holds component by component; the real part is the harmonic statement we wanted.

Step 4. A strict interior maximum would exceed its own circular average, contradicting Step 3; so harmonic functions attain extremes only on the boundary. Applying this to $u_1-u_2$ for two solutions with equal boundary data forces $u_1\equiv u_2$ . Why: the mean-value property is the whole content of the maximum principle, which is the whole content of Dirichlet uniqueness — three theorems, one derivation. *As a concrete check: for $u=x^2-y^2$ at $z_0=(2,0)$ with $r=1$ , the circular average equals $4=u(z_0)$ , confirming the property numerically. The minimum principle follows immediately by applying the mean-value result to $-u$ .*

Limiting cases

Reading off the elementary complex functions gives the standard conjugate pairs, together with the one case where the construction fails:

Case	Result	Meaning
Power $z^2$	$\operatorname{Re}(z^2)=x^2-y^2,\ \operatorname{Im}(z^2)=2xy$	Both harmonic; $2xy$ is the conjugate of $x^2-y^2$ ; level curves are orthogonal hyperbolae (the net figure)
Power $z^n$	$\operatorname{Re}=r^n\cos n\theta,\ \operatorname{Im}=r^n\sin n\theta$	The multipole potentials of 2D electrostatics; each pair is harmonic on all of the plane
Logarithm	$u=\tfrac12\ln(x^2+y^2)=\ln r,\ v=\arg z=\theta$	Harmonic on the punctured plane but $v$ is multivalued — the topological obstruction to a global conjugate
Exponential	$\operatorname{Re}(e^z)=e^x\cos y,\ \operatorname{Im}(e^z)=e^x\sin y$	A globally single-valued conjugate pair; both entire and harmonic on all of the plane
Linear / constant gradient	$u=ax+by\ \Rightarrow\ v=ay-bx$	Uniform-field potential and its stream function — the simplest non-trivial conjugate pair

Common trapnot every pair of harmonic functions is a conjugate pair, and a conjugate need not exist globally

Two traps hide here. First, being each harmonic is not enough to make $u$ and $v$ a conjugate pair — they must satisfy Cauchy–Riemann together. Both $x^2-y^2$ and $y^2-x^2$ are harmonic, but they are not conjugates; the true conjugate of $x^2-y^2$ is $2xy$ , giving $f=z^2$ . Second, a conjugate need not exist on a domain with holes. The classic case is $u=\ln r=\tfrac12\ln(x^2+y^2)$ , harmonic everywhere but the origin, whose only candidate conjugate is the polar angle $\theta$ — which jumps by $2\pi$ each loop of the origin, so it is multivalued. A single-valued conjugate requires simple connectivity (equivalently, that the period $\oint(-u_y\,\mathrm{d}x+u_x\,\mathrm{d}y)$ around every hole vanish). The physics is real: $\ln r$ is a line charge's potential, and its multivalued conjugate is the flux function whose $2\pi$ jump counts the enclosed charge by Gauss's law.

Worked examples

1 · Building a conjugate from scratch: u = x³ − 3xy² (fully worked)

Given $u=x^3-3xy^2$ , verify it is harmonic, construct its conjugate $v$ , identify $f=u+iv$ , and check orthogonality at $(1,1)$ .

Step 1. Compute the derivatives: $u_x=3x^2-3y^2$ , $u_y=-6xy$ . Then $u_{xx}=6x$ and $u_{yy}=-6x$ , so $u_{xx}+u_{yy}=6x-6x=\mathbf{0}$ — u is harmonic.

Step 2. Use Cauchy–Riemann $v_y=u_x$ and integrate in $y$ : $v=\int(3x^2-3y^2)\,\mathrm{d}y=3x^2y-y^3+g(x)$ , where $g(x)$ is an unknown function of $x$ alone (the "constant" of $y$ -integration).

Step 3. Fix $g$ from the other equation $v_x=-u_y$ . Here $v_x=6xy+g'(x)$ and $-u_y=6xy$ , so $g'(x)=0$ and $g=C$ , a genuine constant. Hence $\mathbf{v=3x^2y-y^3+C}$ .

Step 4. Assemble $f=u+iv=(x^3-3xy^2)+i(3x^2y-y^3)+iC$ . Recognising the binomial expansion $(x+iy)^3=x^3-3xy^2+i(3x^2y-y^3)$ , this is $\mathbf{f(z)=z^3+iC}$ — the construction has reconstructed the analytic function from its real part alone.

Step 5. Check orthogonality at $(1,1)$ : $\nabla u=(3x^2-3y^2,\,-6xy)=(0,-6)$ and $\nabla v=(6xy,\,3x^2-3y^2)=(6,0)$ . Their dot product is $\mathbf{0}$ , and the gradients — one vertical, one horizontal — are visibly perpendicular, so the equipotential and stream line through $(1,1)$ cross at a right angle, exactly as equation (4) demands.

2 · The mean-value property as a numerical check

Confirm the mean-value property for the harmonic $u=x^2-y^2$ on the circle of radius $r=1$ centred at $z_0=(2,0)$ , and explain why even a four-point average is already exact.

Step 1. The centre value is $u(2,0)=4-0=\mathbf{4}$ .

Step 2. Parametrize the circle: $x=2+\cos\theta,\ y=\sin\theta$ . Then $u=(2+\cos\theta)^2-\sin^2\theta=4+4\cos\theta+\cos^2\theta-\sin^2\theta=4+4\cos\theta+\cos2\theta$ .

Step 3. Average over $\theta\in[0,2\pi]$ : the terms $4\cos\theta$ and $\cos2\theta$ integrate to zero, leaving $\frac{1}{2\pi}\int_0^{2\pi}u\,\mathrm{d}\theta=\mathbf{4}=u(z_0)$ . The mean-value property checks out exactly.

Step 4. Sample the four cardinal points: $u(3,0)=9$ , $u(2,1)=3$ , $u(1,0)=1$ , $u(2,-1)=3$ . Their average is $(9+3+1+3)/4=\mathbf{4}$ . The four-point average is already exact because the only angular harmonics on this circle are $\cos\theta$ and $\cos2\theta$ , both annihilated by a symmetric four-point sum — a preview of why the mean-value property forbids any interior extreme.

3 · A line charge as a complex potential

An infinite line charge of density $\lambda=1.0\ \mathrm{nC/m}$ runs into the plane. Write its complex potential, extract the potential and stream functions, and find the field magnitude at radius $r=5.0\ \mathrm{cm}$ .

Step 1. The complex potential is $\Omega(z)=-\dfrac{\lambda}{2\pi\varepsilon_0}\ln z$ . Splitting $\ln z=\ln r+i\theta$ gives the potential $\phi=\operatorname{Re}\Omega=-\dfrac{\lambda}{2\pi\varepsilon_0}\ln r$ and the flux function $\psi=\operatorname{Im}\Omega=-\dfrac{\lambda}{2\pi\varepsilon_0}\theta$ .

Step 2. The two families are the familiar picture: equipotentials $\phi=$ const are circles $r=$ const, and field lines $\psi=$ const are radial rays $\theta=$ const — orthogonal, as every conjugate pair must be. The $2\pi$ jump in $\psi$ around the wire is not a defect but Gauss's law: the flux increment counts the enclosed charge.

Step 3. The complex field is $E_x-iE_y=-\Omega'(z)=\dfrac{\lambda}{2\pi\varepsilon_0}\dfrac{1}{z}$ , so the magnitude is $|E|=\dfrac{\lambda}{2\pi\varepsilon_0\,r}$ — the standard $1/r$ falloff, from one differentiation of an analytic function.

Step 4. Insert numbers with $\varepsilon_0=8.854\times10^{-12}\ \mathrm{F/m}$ : $|E|=\dfrac{1.0\times10^{-9}}{2\pi(8.854\times10^{-12})(0.050)}=\mathbf{3.6\times10^{2}\ V/m}$ (about $360\ \mathrm{V/m}$ ). The whole electrostatics problem reduced to reading off a logarithm and differentiating once.

Physics of the portrait

Harmonic functions carry the name of the man who first wrote their equation for the potential of gravity, long before anyone suspected its two-dimensional marriage to complex analysis.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · potential theory and celestial mechanics

Pierre-Simon Laplace was the great systematizer of Newtonian gravity, and in the course of proving the stability of the Solar System he distilled the potential into the equation that bears his name. Studying the gravitational attraction of a spheroid in his memoirs of the 1780s, he showed that the potential $V$ of an inverse-square force satisfies $\nabla^2 V=0$ in any region free of mass — the first appearance of Laplace's equation, though Lagrange had brushed against it in fluid dynamics earlier. He made it the organizing principle of potential theory across his monumental Mécanique céleste (1799–1825), and the spherical-harmonic expansions he and Legendre developed to solve it remain the working language of gravitation, geodesy, and electrostatics. What Laplace could not have foreseen was the two-dimensional twist: that in the plane his equation is nothing but the real part of the Cauchy–Riemann conditions, so harmonic and analytic became two names for one thing. That synthesis belonged to the next generation — d'Alembert had written the conjugate equations for fluid flow in 1752, Cauchy built the complex integral in the 1820s, and Riemann in 1851 made harmonic functions and conformal maps the foundation of geometric function theory — but the equation, and the idea that a field in empty space is governed by pure smoothness, are Laplace's.

History

1752

— d'Alembert writes the equations later named for Cauchy and Riemann in a memoir on fluid resistance

1785

— Laplace formulates ∇²V = 0 for the gravitational potential in empty space

1828

— Green's essay introduces the potential function and the identities that bear his name

1851

— Riemann's dissertation founds complex function theory on harmonic functions and conformal maps

1890

— Poincaré's balayage method secures existence for the Dirichlet problem on general domains

Going further

Texts. Stone & Goldbart, Mathematics for Physics, complex-analysis chapter (harmonic functions and conjugates) · Arfken & Weber, Mathematical Methods for Physicists, 7th ed., complex-variable chapter · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, ch. 24 · Hassani, Mathematical Physics, complex-analysis chapters · Ahlfors, Complex Analysis, ch. 4 (harmonic functions and the Dirichlet problem). Key self-consistency checks: $\nabla^2 u=0$ follows from Cauchy–Riemann via equality of mixed partials; the worked conjugates are confirmed by expanding $z^2$ , $z^3$ , and $e^z$ directly; the mean-value property for $u=x^2-y^2$ gives centre value $4$ , circular average $4$ , and four-point average $4$ ; the line-charge field satisfies $|E|=\lambda/(2\pi\varepsilon_0 r)=3.6\times10^2\ \mathrm{V/m}$ at $\lambda=1.0\ \mathrm{nC/m},\ r=5.0\ \mathrm{cm}$ . All constants ( $2\pi$ averaging factor, $\varepsilon_0$ , integer exponents) are standard.

Prerequisites. Analytic Functions and the Cauchy–Riemann Equations · Functions of a Complex Variable and Limits · Complex Numbers and the Complex Plane · Elementary Complex Functions. Leads to. Cauchy's Integral Formula and Derivatives · Liouville, Maximum Modulus, Morera · Taylor Series of Analytic Functions.

Ask the tutor: Give me any harmonic $u(x,y)$ and I will construct its conjugate, assemble the complex potential $f(z)$ , and read off the field — step by step.

What exactly is a harmonic conjugate, and is it unique?

Given a harmonic $u$ , its harmonic conjugate is the harmonic $v$ that pairs with it so $f=u+iv$ is analytic — equivalently, $v$ is the function whose gradient is $\nabla u$ rotated ninety degrees. On a simply connected domain it exists and is unique up to an additive real constant, the one freedom left after Cauchy–Riemann pins down every derivative of $v$ ; you fix it by specifying $v$ at a single point. Note the asymmetry of language: $v$ is the conjugate of $u$ , but $-u$ is the conjugate of $v$ , so the roles are not symmetric in sign.

Why does two-dimensionality matter so much? Isn't Laplace's equation the same in 3D?

The equation $\nabla^2u=0$ makes sense in any dimension, but the bridge to complex analysis is special to the plane. Complex numbers are two-dimensional, so only in 2D can you fuse two coordinates into one variable $z=x+iy$ and speak of analytic functions. In three dimensions there is no harmonic conjugate or complex potential — you lose conformal mapping (Liouville's theorem says the only 3D conformal maps are trivial) and with it the toolkit of transplanting solutions. So harmonic functions exist in every dimension, but the machinery of conjugates, orthogonal nets, and $f=u+iv$ is a two-dimensional luxury.

How can I tell whether a given function is harmonic without guessing?

Compute the two unmixed second partials and add them: if $u_{xx}+u_{yy}=0$ identically, $u$ is harmonic. That is the entire test. A faster route is recognition — $e^x\cos y$ is $\operatorname{Re}(e^z)$ , $x^2-y^2$ is $\operatorname{Re}(z^2)$ , $\ln r$ is $\operatorname{Re}(\ln z)$ — and any $\operatorname{Re}(f(z))$ with $f$ analytic is automatically harmonic. Beware the common false positive: $x^2+y^2$ looks harmonic-ish but has Laplacian $2+2=4\neq0$ , so it is not harmonic at all.

What is the complex potential, and why is it so convenient?

The complex potential $f=u+iv$ (often $\Omega=\phi+i\psi$ in physics) bundles a scalar potential $\phi$ and its stream function $\psi$ into one analytic function of $z$ . Its convenience is threefold. One object carries both the equipotentials and the field lines, automatically orthogonal. One differentiation gives the whole field, $E_x-iE_y=-\Omega'(z)$ , so the vector field becomes a complex function you manipulate algebraically. And because $\Omega$ is analytic, you can compose it with conformal maps to solve awkward geometries by transplanting simple ones. It is the most efficient packaging of a 2D potential problem.

Why must the domain be simply connected for the conjugate to exist?

The conjugate is built by integrating the differential $-u_y\,\mathrm{d}x+u_x\,\mathrm{d}y$ , which is closed precisely because $u$ is harmonic. On a simply connected domain a closed differential is exact, so the line integral is path-independent and yields a single-valued $v$ . On a domain with a hole, a loop encircling it can accumulate a nonzero period — the integral around the loop need not vanish — and then $v$ is multivalued. The archetype is $u=\ln r$ : its candidate conjugate is the angle $\theta$ , which increases by $2\pi$ every time you circle the origin. That $2\pi$ jump is the flux of the enclosed line charge, so the obstruction is not a bug but a conservation law.

Are harmonic functions really infinitely differentiable, even if I only assumed two derivatives?

Yes, and this is a striking rigidity result. The definition asks only for continuous second partials, but a harmonic function is automatically $C^\infty$ and in fact real-analytic (locally a convergent power series). The reason is the analytic connection: on any small disc $u$ is the real part of an analytic $f$ , and analytic functions are infinitely differentiable by Cauchy's integral formula, so $u$ inherits that smoothness. You cannot build a merely twice-differentiable harmonic function with a kink in its third derivative — harmonicity forbids it.