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Part IA Michaelmas, Lent, Easter Term

Laplace's Equation

The Equation

∇2ϕ=∂2ϕ∂x2+∂2ϕ∂y2=0\nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0

In 3D: ∇2ϕ=∂2ϕ∂x2+∂2ϕ∂y2+∂2ϕ∂z2=0\nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = 0


Physical Context

  • Steady-state temperature distribution
  • Electrostatic potential (no charges)
  • Gravitational potential (no masses)
  • Incompressible, irrotational fluid flow

Temperature distribution in a rectangular plate with boundary conditions, showing isotherms (level curves) of potential

Key property: Solutions are harmonic functions — smooth, with no local maxima/minima (maximum principle).


Separation of Variables

Problem: Find ϕ(x,y)\phi(x, y) in rectangle 0<x<a0 < x < a, 0<y<b0 < y < b with:

Rectangle with boundary conditions on each edge - showing Dirichlet conditions

  • ϕ(0,y)=0\phi(0, y) = 0 (left edge)
  • ϕ(x,0)=0\phi(x, 0) = 0 (bottom edge)
  • ϕ(x,b)=0\phi(x, b) = 0 (top edge)
  • ϕ(a,y)=f(y)\phi(a, y) = f(y) (right edge)

Ansatz: ϕ(x,y)=X(x)Y(y)\phi(x, y) = X(x)Y(y)

Substituting into Laplace equation:

X′′Y+XY′′=0X''Y + XY'' = 0

X′′X=−Y′′Y=k\frac{X''}{X} = -\frac{Y''}{Y} = k


Solution

For homogeneous Dirichlet conditions on yy edges:

The separation constant must be negative: k=−γ2<0k = -\gamma^2 < 0.

YY equation: Y′′+γ2Y=0Y'' + \gamma^2 Y = 0

Y(y)=Asin⁡(γy)+Bcos⁡(γy)Y(y) = A\sin(\gamma y) + B\cos(\gamma y)

Boundary condition Y(0)=0⇒B=0Y(0) = 0 \Rightarrow B = 0.

Y(b)=0⇒sin⁡(γb)=0Y(b) = 0 \Rightarrow \sin(\gamma b) = 0, so γ=nπb\gamma = \frac{n\pi}{b} for n=1,2,3,…n = 1, 2, 3, \ldots

XX equation: X′′−γ2X=0X'' - \gamma^2 X = 0

X(x)=Csinh⁡(γx)+Dcosh⁡(γx)X(x) = C\sinh(\gamma x) + D\cosh(\gamma x)

Boundary condition X(0)=0⇒D=0X(0) = 0 \Rightarrow D = 0.

Combined solution:

ϕ(x,y)=∑n=1∞Cnsin⁡(nπyb)sinh⁡(nπxb)\phi(x, y) = \sum_{n=1}^{\infty} C_n \sin\left(\frac{n\pi y}{b}\right) \sinh\left(\frac{n\pi x}{b}\right)


Applying the Inhomogeneous Boundary Condition

At x=ax = a: ϕ(a,y)=f(y)\phi(a, y) = f(y)

This is a Fourier sine series, so:

f(y)=∑n=1∞Cnsin⁡(nπyb)sinh⁡(nπab)f(y) = \sum_{n=1}^{\infty} C_n \sin\left(\frac{n\pi y}{b}\right) \sinh\left(\frac{n\pi a}{b}\right)

Fourier coefficients:

Cn=2bsinh⁡(nπa/b)∫0bf(y)sin⁡(nπyb)dyC_n = \frac{2}{b\sinh(n\pi a/b)} \int_0^b f(y)\sin\left(\frac{n\pi y}{b}\right) dy


Example: Constant f(y)=1f(y) = 1

Cn=2bsinh⁡(nπa/b)∫0bsin⁡(nπyb)dyC_n = \frac{2}{b\sinh(n\pi a/b)} \int_0^b \sin\left(\frac{n\pi y}{b}\right) dy

=2bsinh⁡(nπa/b)⋅bnπ[1−(−1)n]= \frac{2}{b\sinh(n\pi a/b)} \cdot \frac{b}{n\pi}[1 - (-1)^n]

=2nπsinh⁡(nπa/b)×{2n odd0n even= \frac{2}{n\pi\sinh(n\pi a/b)} \times \begin{cases} 2 & n \text{ odd} \\ 0 & n \text{ even} \end{cases}


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\phi_1 and ϕ2\phi_2 are solutions, then ψ=ϕ1−ϕ2\psi = \phi_1 - \phi_2 is harmonic with zero boundary conditions. By maximum principle, ψ=0\psi = 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