Laplace's Equation

\nabla^2 u = 0

The operator \( \nabla^2 \) is the Laplacian. It compares the value of a function at a point to the average of its neighbors. If \( \nabla^2 u > 0 \), then \(u\) at that point is less than the average of its neighbors (a “dip” in 3D); if \( \nabla^2 u < 0 \), it’s greater (a “peak”). If \( \nabla^2 u = 0 \), the region looks locally “flat.”

In two and three spatial dimensions:

\nabla^2 u \;=\; \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}

\nabla^2 u \;=\; \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}

Laplace’s equation models steady-state fields (no explicit time dependence), e.g., electrostatic or magnetostatic potentials. Once at equilibrium, nothing changes in time, but values still vary spatially.

Laplace's Equation in Polar Coordinates

For geometries with circular symmetry (disks, annuli, cylinders), rewrite the Laplacian in polar coordinates. Starting from \( \nabla^2 u = 0 \), we obtain

u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 0.

Recall the transformations

r^2 = x^2 + y^2,\qquad \theta = \tan^{-1}\!\Big(\frac{y}{x}\Big),\qquad x = r\cos\theta,\quad y = r\sin\theta,

and then apply the chain rule (tedious but straightforward). We will typically use boundary data that is a function of \( \theta \) (radial symmetry at the boundary).

Associated Boundary Conditions

Dirichlet: Value of \(u\) given on the boundary (often a function of \(\theta\)).
Neumann: Normal derivative (flux) given on the boundary (e.g., \(u_r\) on \(r=1\)).
Robin: Linear combination of value and flux.

We will focus on Dirichlet conditions and solve the Interior Dirichlet Problem on the unit disk.

Interior Dirichlet Problem

Consider the BVP

\text{PDE: } \quad u_{rr}+\frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta} = 0,\quad 0 < r < 1

\text{BC: }\; u(1,\theta) = g(\theta),\quad 0\le \theta < 2\pi.

Use separation of variables: assume \(u(r,\theta)=R(r)\Theta(\theta)\). Then

u_r = R'(r)\Theta(\theta),\quad u_{rr} = R''(r)\Theta(\theta),\quad u_{\theta\theta} = R(r)\Theta''(\theta).

Substitute into the PDE:

R''\Theta + \frac{1}{r}R'\Theta + \frac{1}{r^2}R\Theta'' = 0.

Divide by \(R\Theta\) and separate variables:

\frac{R'' + \tfrac{1}{r}R'}{R} + \frac{1}{r^2}\frac{\Theta''}{\Theta} = 0 \;\;\Longrightarrow\;\; r^2\frac{R'' + \tfrac{1}{r}R'}{R} = -\frac{\Theta''}{\Theta} = k.

Hence two ODEs:

r^2 R'' + r R' - k R = 0, \qquad \Theta'' + k\,\Theta = 0.

Solving \(\Theta'' + k\Theta = 0\) with \(2\pi\)-periodicity

Case analysis:

\(k<0\): \(\Theta = Ae^{\sqrt{-k}\,\theta}+Be^{-\sqrt{-k}\,\theta}\) is not \(2\pi\)-periodic unless \(\sqrt{-k}=0\), which contradicts \(k<0\). Discard.
\(k=0\): \(\Theta = A\theta + B\). Periodicity forces \(A=0\), so \(\Theta=C\).
\(k>0\): \(\Theta = A\cos(\sqrt{k}\,\theta) + B\sin(\sqrt{k}\,\theta)\). \(2\pi\)-periodicity requires \(\sqrt{k}\in\mathbb{Z}\), i.e., \(k=n^2\) with \(n\in\{0,1,2,\dots\}\).

Thus the eigenfunctions are

\Theta_n(\theta) = a_n\cos(n\theta) + b_n\sin(n\theta),\quad n=0,1,2,\dots

Solving the radial Cauchy–Euler equation

For \(k=n^2\ge 0\), the radial ODE \(r^2 R''+rR' - n^2 R = 0\) has solutions

R(r)= \begin{cases} a_0 + b_0\ln r, & n=0,\\[4pt] a_n r^{n} + b_n r^{-n}, & n\ge 1. \end{cases}

Boundedness at the origin (\(r\to 0^+\)) forces \(b_0=0\) and \(b_n=0\) for \(n\ge 1\). Hence

R_0(r)=a_0,\qquad R_n(r)=a_n r^{n}\quad (n\ge 1).

Separated solutions and superposition

The separated solutions are \(u_n(r,\theta)=R_n(r)\Theta_n(\theta)\), so the general bounded solution on the disk is

u(r,\theta) = a_0 + \sum_{n=1}^{\infty} r^{n}\bigl[a_n\cos(n\theta)+b_n\sin(n\theta)\bigr].

Fit the boundary data \(u(1,\theta)=g(\theta)\)

At \(r=1\):

g(\theta) = u(1,\theta) = a_0 + \sum_{n=1}^{\infty}\bigl[a_n\cos(n\theta)+b_n\sin(n\theta)\bigr].

Use orthogonality on \([0,2\pi]\):

\int_0^{2\pi}\!\sin(n\theta)\sin(m\theta)\,d\theta = \begin{cases} \pi, & n=m,\\ 0, & n\ne m, \end{cases}

\int_0^{2\pi}\!\cos(n\theta)\cos(m\theta)\,d\theta = \begin{cases} \pi, & n=m,\\ 0, & n\ne m, \end{cases}

\int_0^{2\pi}\!\sin(n\theta)\cos(m\theta)\,d\theta = 0 \quad \text{for all } n,m\in\mathbb{Z}_{\ge 0}.

Projecting \(g\) onto sines and cosines gives

a_n = \frac{1}{\pi}\int_0^{2\pi} g(\theta)\cos(n\theta)\,d\theta,\qquad b_n = \frac{1}{\pi}\int_0^{2\pi} g(\theta)\sin(n\theta)\,d\theta\quad (n\ge 1),

and the \(n=0\) term matches the Fourier convention with \(\frac{a_0}{2}\) in front:

u(r,\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} r^{n}\bigl[a_n\cos(n\theta)+b_n\sin(n\theta)\bigr],

a_0 = \frac{1}{\pi}\int_0^{2\pi} g(\theta)\,d\theta,\qquad a_n = \frac{1}{\pi}\int_0^{2\pi} g(\theta)\cos(n\theta)\,d\theta,\qquad b_n = \frac{1}{\pi}\int_0^{2\pi} g(\theta)\sin(n\theta)\,d\theta.

This solves the Interior Dirichlet Problem for Laplace’s equation on the unit disk (e.g., the electrostatic potential in a circular domain with prescribed boundary potential).