The Laplace Equation

⚡ Key Takeaways

Laplace's equation is the second-order partial differential equation ∇2u=0\nabla^2 u = 0, stating that a function's value at each point equals the average of its surrounding values. Its solutions, called harmonic functions, model steady-state phenomena: equilibrium temperature, electrostatic potential in charge-free regions, incompressible irrotational fluid flow, and gravitational potential in vacuum. Named for Pierre-Simon Laplace (1749–1827), it is the canonical elliptic PDE and the foundation of Potential Theory.

💡 The Core Insight

A harmonic function has no local extrema in the interior: at every point its value equals the average over any surrounding sphere. There are no "peaks" or "valleys" — heat has flowed until everything is as smooth and balanced as possible. This single averaging property explains uniqueness, the maximum principle, smoothness, and why boundary data alone determines the entire interior.

Formal Statement

DefinitionLaplace's Equation and the Laplacian

For a scalar function uu defined on an open domain Ω⊆Rn\Omega \subseteq \mathbb{R}^n, Laplace's equation is

1
∇2u  ≡  Δu  =  ∑i=1n∂2u∂xi2  =  0.\quad \nabla^2 u \;\equiv\; \Delta u \;=\; \sum_{i=1}^{n} \frac{\partial^2 u}{\partial x_i^2} \;=\; 0.

Derivation hint: The Laplacian Δ\Delta is the divergence of the gradient, ∇⋅(∇u)\nabla\cdot(\nabla u); expand in Cartesian coordinates to obtain the sum of pure second derivatives. Here u:Ω→Ru:\Omega\to\mathbb{R} is the unknown field, xix_i are Cartesian coordinates, and Δ=∇⋅∇\Delta=\nabla\cdot\nabla is the Laplace operator. A function satisfying [1] on Ω\Omega is called harmonic on Ω\Omega. The inhomogeneous version Δu=f\Delta u = f is Poisson's Equation.

In two and three dimensions the equation reads explicitly:

2
uxx+uyy=0(n=2),uxx+uyy+uzz=0(n=3).\quad u_{xx}+u_{yy}=0 \quad(n=2), \qquad u_{xx}+u_{yy}+u_{zz}=0 \quad(n=3).

Derivation hint: Specialize equation [1] to n=2n=2 and n=3n=3, writing uxx=∂2u/∂x2u_{xx}=\partial^2 u/\partial x^2, etc. In polar coordinates (r,θ)(r,\theta) the Laplacian becomes

3
Δu=1r∂∂r ⁣(r∂u∂r)+1r2∂2u∂θ2=urr+1rur+1r2uθθ.\quad \Delta u = \frac{1}{r}\frac{\partial}{\partial r}\!\left(r\frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}.

Derivation hint: Apply the chain rule to x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta, expressing ∂x,∂y\partial_x,\partial_y in terms of ∂r,∂θ\partial_r,\partial_\theta, and sum the second derivatives — the cross terms cancel.

TheoremMean Value Property

If uu is harmonic on Ω\Omega and the closed ball B‾(x0,R)⊂Ω\overline{B}(x_0,R)\subset\Omega, then the value at the centre equals the average over the bounding sphere ∂B\partial B and over the solid ball:

4
u(x0)=1∣∂B(x0,R)∣∫∂B(x0,R)u dS=1∣B(x0,R)∣∫B(x0,R)u dV.\quad u(x_0)=\frac{1}{|\partial B(x_0,R)|}\int_{\partial B(x_0,R)} u \, dS = \frac{1}{|B(x_0,R)|}\int_{B(x_0,R)} u \, dV.

Derivation hint: Define ϕ(r)=\fint∂B(x0,r)u dS\phi(r)=\fint_{\partial B(x_0,r)}u\,dS; differentiate, apply the divergence theorem to get ϕ′(r)=rn\fintBΔu dV=0\phi'(r)=\frac{r}{n}\fint_B \Delta u\,dV = 0, so ϕ\phi is constant and equals its r→0r\to0 limit u(x0)u(x_0). Remarkably, this property characterizes harmonic functions: a continuous function satisfying the mean value property everywhere is automatically harmonic and smooth.

TheoremMaximum Principle

A non-constant harmonic function on a connected domain Ω\Omega attains neither a maximum nor a minimum at any interior point. If Ω\Omega is bounded and uu is continuous on Ω‾\overline{\Omega},

5
max⁡Ω‾u=max⁡∂Ωu,min⁡Ω‾u=min⁡∂Ωu.\quad \max_{\overline{\Omega}} u = \max_{\partial\Omega} u, \qquad \min_{\overline{\Omega}} u = \min_{\partial\Omega} u.

Derivation hint: If an interior max were attained at x0x_0, the mean value property [4] would force u≡u(x0)u\equiv u(x_0) on a small sphere, hence (by a connectedness argument) on all of Ω\Omega, contradicting non-constancy.

Why It Holds / Derivation

ProofUniqueness of the Dirichlet Problem via the Maximum Principle

We prove that the Dirichlet problem — solve Δu=0\Delta u=0 in Ω\Omega with u=gu=g on ∂Ω\partial\Omega — has at most one solution.

Step 1. Suppose u1u_1 and u2u_2 are both harmonic in Ω\Omega, continuous on Ω‾\overline{\Omega}, with u1=u2=gu_1=u_2=g on ∂Ω\partial\Omega.

Step 2. Let w=u1−u2w=u_1-u_2. Since the Laplacian is linear, Δw=Δu1−Δu2=0\Delta w=\Delta u_1-\Delta u_2=0, so ww is harmonic.

Step 3. On the boundary, w=g−g=0w=g-g=0.

Step 4. By the maximum principle [5], max⁡Ω‾w=max⁡∂Ωw=0\max_{\overline\Omega} w = \max_{\partial\Omega} w = 0 and min⁡Ω‾w=min⁡∂Ωw=0\min_{\overline\Omega} w = \min_{\partial\Omega} w = 0.

Step 5. Therefore 0≤w≤00\le w\le 0 everywhere, i.e. w≡0w\equiv 0, so u1=u2u_1=u_2. ■\blacksquare

∎
ProofHarmonicity of Real Parts of Analytic Functions

We show that if f=u+ivf=u+iv is holomorphic, then uu and vv are harmonic.

Step 1. Holomorphy implies the Cauchy–Riemann equations:

6
ux=vy,uy=−vx.\quad u_x = v_y, \qquad u_y = -v_x.

Derivation hint: These follow from requiring the complex derivative f′(z)f'(z) to be independent of direction of approach.

Step 2. Differentiate the first equation with respect to xx: uxx=vyxu_{xx}=v_{yx}.

Step 3. Differentiate the second with respect to yy: uyy=−vxyu_{yy}=-v_{xy}.

Step 4. Since vv is smooth, mixed partials commute (vyx=vxyv_{yx}=v_{xy}). Adding the two results:

7
uxx+uyy=vyx−vxy=0.\quad u_{xx}+u_{yy}=v_{yx}-v_{xy}=0.

Derivation hint: Add the results of Steps 2–3 and invoke Clairaut's theorem on equality of mixed partials. An identical computation shows vxx+vyy=0v_{xx}+v_{yy}=0. Thus both components are harmonic, and vv is the harmonic conjugate of uu. ■\blacksquare

∎
PL
Pierre-Simon Laplace
1749–1827

The French mathematician and astronomer who, in his monumental Mécanique céleste, used the operator now bearing his name to analyze the gravitational potential of planetary bodies. Laplace argued that the gravitational potential of a mass distribution satisfies ΔV=0\Delta V = 0 in empty space, reducing the study of forces to a single scalar function. He famously claimed celestial mechanics had no need of the "hypothesis" of divine intervention — the equations sufficed. Though the equation appears earlier in Euler's and Lagrange's fluid work, Laplace's systematic exploitation of potential theory cemented his name to it.

Building Intuition

🧠 Intuition

Imagine a thin metal plate whose edge temperature you fix arbitrarily, then wait for equilibrium. Heat flows from hot to cold until no point can change — meaning every interior point sits exactly at the average temperature of its neighbours. That balance condition is Laplace's equation. Now stretch a rubber membrane (a soap film) over a bent wire loop: the film minimizes its area, and for small slopes its height is harmonic. There are no bumps an ant could climb to find a local summit; the highest and lowest points are always on the wire. This is the maximum principle made tactile. Finally, in 2D, picture the flow lines of a frictionless incompressible fluid: the velocity potential is harmonic, and the streamlines are the level curves of its harmonic conjugate — the same magic that lets Conformal Mapping reshape one flow into another.

Figure
Steady-state temperature on a square plate: harmonic interior, fixed boundary, isotherms shown in gold
u = 100° (top edge) u = 0° (bottom edge) u = 0° u = 0° 75° 50° 25° 10° center = avg of neighbours Isotherms (gold) curve toward the hot edge; no interior peak exists

Worked Examples

ExampleSteady-State Temperature in a Disk — Poisson Integral

Given: A circular plate of radius R=2R=2 m. The boundary temperature is g(θ)=50+30cos⁡θg(\theta)=50 + 30\cos\theta degrees Celsius. Find the interior temperature u(r,θ)u(r,\theta), then evaluate at the center and at (r,θ)=(1,0)(r,\theta)=(1,0).

Step 1 — General solution. By separation of variables in [3], the bounded harmonic functions on a disk are

8
u(r,θ)=a02+∑n=1∞(rR)n ⁣(ancos⁡nθ+bnsin⁡nθ).\quad u(r,\theta)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(\frac{r}{R}\right)^n\!\big(a_n\cos n\theta + b_n\sin n\theta\big).

Derivation hint: Set u=P(r)Θ(θ)u=P(r)\Theta(\theta) in [3]; the angular part gives Θ′′=−n2Θ\Theta''=-n^2\Theta, and boundedness at r=0r=0 selects the rnr^n radial solution over r−nr^{-n}.

Step 2 — Match boundary data. On r=Rr=R: u(R,θ)=a02+∑(ancos⁡nθ+bnsin⁡nθ)=50+30cos⁡θu(R,\theta)=\frac{a_0}{2}+\sum (a_n\cos n\theta + b_n\sin n\theta)=50+30\cos\theta. Matching Fourier coefficients: a0/2=50a_0/2=50, a1=30a_1=30, all others zero.

Step 3 — Assemble. With R=2R=2:

u(r,θ)=50+30(r2)cos⁡θ=50+15 rcos⁡θ.u(r,\theta)=50 + 30\left(\frac{r}{2}\right)\cos\theta = 50 + 15\,r\cos\theta.

Step 4 — Evaluate. Center (r=0)(r=0): u=50u=50°C — exactly the average of the boundary data, confirming the mean value property. At (r,θ)=(1,0)(r,\theta)=(1,0): u=50+15(1)(1)=65u=50+15(1)(1)=65°C.

u(r,θ)=50+15 rcos⁡θ;u(0)=50∘C,u(1,0)=65∘C\boxed{u(r,\theta)=50+15\,r\cos\theta;\quad u(0)=50^\circ\text{C},\quad u(1,0)=65^\circ\text{C}}
ExampleElectrostatic Potential Between Concentric Spheres

Given: Two concentric conducting spheres, inner radius a=0.10a=0.10 m at potential Va=100V_a=100 V, outer radius b=0.30b=0.30 m at Vb=0V_b=0 V. The region between them is charge-free, so VV is harmonic and depends only on rr (spherical symmetry). Find V(r)V(r) and the field at r=0.20r=0.20 m.

Step 1 — Radial Laplacian. With spherical symmetry, ΔV=1r2ddr ⁣(r2dVdr)=0\Delta V=\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dV}{dr}\right)=0, so

9
r2dVdr=C1  ⇒  V(r)=−C1r+C2.\quad r^2\frac{dV}{dr}=C_1 \;\Rightarrow\; V(r)=-\frac{C_1}{r}+C_2.

Derivation hint: Integrate ddr(r2V′)=0\frac{d}{dr}(r^2 V')=0 once to get r2V′=C1r^2V'=C_1, then integrate V′=C1/r2V'=C_1/r^2 again.

Step 2 — Apply boundary conditions. V(a)=−C1/a+C2=100V(a)=-C_1/a + C_2 = 100 and V(b)=−C1/b+C2=0V(b)=-C_1/b + C_2 = 0. Subtracting: −C1(1/a−1/b)=100⇒−C1(10−3.333)=100⇒C1=−15.0-C_1(1/a-1/b)=100 \Rightarrow -C_1(10-3.333)=100 \Rightarrow C_1=-15.0 V·m. Then C2=C1/b=−15.0/0.30=−50.0C_2 = C_1/b = -15.0/0.30 = -50.0 V.

Step 3 — Potential.

V(r)=15.0r−50.0 (V, with r in metres).V(r)=\frac{15.0}{r}-50.0 \ \text{(V, with } r \text{ in metres)}.

Check: V(0.10)=150−50=100V(0.10)=150-50=100 V ✓; V(0.30)=50−50=0V(0.30)=50-50=0 V ✓.

Step 4 — Field at r=0.20r=0.20 m. E=−dVdr=15.0r2=15.00.04=375E=-\dfrac{dV}{dr}=\dfrac{15.0}{r^2}=\dfrac{15.0}{0.04}=375 V/m, directed radially outward.

V(r)=15.0r−50.0 V,E(0.20)=375 V/m outward\boxed{V(r)=\frac{15.0}{r}-50.0\ \text{V},\qquad E(0.20)=375\ \text{V/m outward}}

Common Misconceptions

⚠️ Common mistake

"Harmonic functions can have interior maxima." They cannot (unless constant). If your computed solution has a bump in the interior, you have made an error or your domain contains a hidden source — which means you are really solving Poisson's Equation, not Laplace's. Likewise, do not assume superposition of products gives every solution: separation of variables only works in specially shaped domains (rectangles, disks, spheres) where the boundary aligns with a coordinate surface.

⚠️ Common mistake

Boundary data must be self-consistent. For the Neumann problem (specifying the normal derivative ∂u/∂n=h\partial u/\partial n=h on ∂Ω\partial\Omega), no solution exists unless the compatibility condition ∮∂Ωh dS=0\oint_{\partial\Omega} h\,dS=0 holds — physically, net heat flux must vanish at steady state. And Neumann solutions are unique only up to an additive constant.

Counterexample

The mean value property fails for Δu≠0\Delta u \ne 0. Consider u(x,y)=x2+y2u(x,y)=x^2+y^2 on the unit disk. Then Δu=2+2=4≠0\Delta u = 2+2=4\ne 0, so uu is not harmonic. At the origin u(0,0)=0u(0,0)=0, but the average over the unit circle is \fint02π(cos⁡2θ+sin⁡2θ) dθ2π=1≠0\fint_{0}^{2\pi}(\cos^2\theta+\sin^2\theta)\,\frac{d\theta}{2\pi}=1\ne 0. The center value undershoots the boundary average — a hallmark of subharmonic functions (Δu≥0\Delta u\ge 0), which lie below their boundary averages, the opposite of the harmonic balance.

Connections

connection

Laplace's equation sits at the crossroads of several theories. It is the steady-state limit of the Heat Equation ut=Δuu_t=\Delta u (set ut=0u_t=0) and of the diffusion equation. Adding a source term yields Poisson's Equation Δu=f\Delta u = f, the workhorse of electrostatics. In two dimensions, harmonic functions are exactly the real/imaginary parts of holomorphic functions (Complex Analysis), so Conformal Mapping transplants solutions between domains. The mean value and maximum principles connect it to Brownian Motion: u(x)u(x) equals the expected boundary value reached by a random walker started at xx. Finally, harmonic functions are the equilibrium states minimizing the Dirichlet energy ∫Ω∣∇u∣2 dV\int_\Omega |\nabla u|^2\,dV, linking the topic to the Calculus of Variations and Euler–Lagrange theory.

Real-World Applications

🔬 Where this appears

Historical Development

⏳ History
- 1752 — Euler derives the equation for the velocity potential of incompressible fluid flow, the first appearance of Δϕ=0\Delta\phi=0 in physics.
- 1762 — Lagrange systematizes the velocity-potential approach, reinforcing the role of harmonic functions in fluid dynamics.
- 1782–1785 — Laplace introduces the operator in his memoirs on planetary attraction, proving the gravitational potential satisfies ΔV=0\Delta V=0 in empty space; the name becomes attached.
- 1813 — Poisson generalizes to ΔV=−4πρ\Delta V=-4\pi\rho inside matter, correcting Laplace's equation where mass density is nonzero.
- 1828 — George Green publishes his Essay, introducing Green's functions and Green's identities, the analytic backbone of boundary-value theory.
- 1840s — Gauss & Thomson (Kelvin) develop potential theory rigorously, including the Gauss mean value theorem and energy methods.
- 1850s — Dirichlet & Riemann formulate the Dirichlet principle, asserting existence of the energy-minimizing harmonic function — used freely until its rigor is questioned.
- 1870 — Weierstrass exhibits a counterexample casting doubt on Dirichlet's principle, sparking a search for rigorous existence proofs.
- 1900–1904 — Hilbert restores the Dirichlet principle on firm footing; Perron (1923) later gives a robust subharmonic-function method for existence under general boundary conditions.

Subtleties & Limits

open-problem

Existence of solutions can fail at irregular boundary points: Lebesgue's spine (a sharp inward cusp) is a domain where the Dirichlet problem has no continuous solution attaining the prescribed boundary value at the cusp tip — motivating the theory of regular boundary points and Wiener's criterion. More broadly, the regularity theory for elliptic equations (how smooth solutions are, given rough coefficients or data) remains active: De Giorgi–Nash–Moser theory addresses divergence-form equations, but sharp regularity for general nonlinear and degenerate elliptic operators, and for free-boundary problems built on the Laplacian (e.g. the obstacle problem and optimal regularity of free boundaries), continues to be researched. The interplay between harmonic measure, potential theory, and the geometry of rough boundaries (rectifiability) is a vibrant modern frontier.

See Also

Poisson's Equation · Harmonic Functions · Potential Theory · Complex Analysis · Conformal Mapping · Heat Equation · Spherical Harmonics · Green's Functions · Calculus of Variations · Brownian Motion