The Equation
∇2ϕ=∂x2∂2ϕ+∂y2∂2ϕ=0
In 3D: ∇2ϕ=∂x2∂2ϕ+∂y2∂2ϕ+∂z2∂2ϕ=0
Physical Context
- Steady-state temperature distribution
- Electrostatic potential (no charges)
- Gravitational potential (no masses)
- Incompressible, irrotational fluid flow

Key property: Solutions are harmonic functions — smooth, with no local maxima/minima (maximum principle).
Separation of Variables
Problem: Find ϕ(x,y) in rectangle 0<x<a, 0<y<b with:

- ϕ(0,y)=0 (left edge)
- ϕ(x,0)=0 (bottom edge)
- ϕ(x,b)=0 (top edge)
- ϕ(a,y)=f(y) (right edge)
Ansatz: ϕ(x,y)=X(x)Y(y)
Substituting into Laplace equation:
X′′Y+XY′′=0
XX′′=−YY′′=k
Solution
For homogeneous Dirichlet conditions on y edges:
The separation constant must be negative: k=−γ2<0.
Y equation: Y′′+γ2Y=0
Y(y)=Asin(γy)+Bcos(γy)
Boundary condition Y(0)=0⇒B=0.
Y(b)=0⇒sin(γb)=0, so γ=bnπ for n=1,2,3,…
X equation: X′′−γ2X=0
X(x)=Csinh(γx)+Dcosh(γx)
Boundary condition X(0)=0⇒D=0.
Combined solution:
ϕ(x,y)=∑n=1∞Cnsin(bnπy)sinh(bnπx)
Applying the Inhomogeneous Boundary Condition
At x=a: ϕ(a,y)=f(y)
This is a Fourier sine series, so:
f(y)=∑n=1∞Cnsin(bnπy)sinh(bnπa)
Fourier coefficients:
Cn=bsinh(nπa/b)2∫0bf(y)sin(bnπy)dy
Example: Constant f(y)=1
Cn=bsinh(nπa/b)2∫0bsin(bnπy)dy
=bsinh(nπa/b)2⋅nπb[1−(−1)n]
=nπsinh(nπa/b)2×{20n oddn even
Maximum Principle
Statement: A harmonic function on a bounded domain attains its maximum and minimum on the boundary.
Physical interpretation: Steady-state temperature cannot be hotter than the hottest boundary point.
Uniqueness
For given boundary conditions, Laplace’s equation has a unique solution.
Proof: If ϕ1 and ϕ2 are solutions, then ψ=ϕ1−ϕ2 is harmonic with zero boundary conditions. By maximum principle, ψ=0.
Why Laplace’s Equation Matters
- Describes equilibrium in many physical systems
- Solved analytically using separation of variables
- Basis for understanding more complex PDEs
- Important in electrostatics, fluid dynamics, and heat transfer