Partial Differential Equations 2 | Laplace's Equation [dark version]

TL;DR

Laplace’s operator is Δu = Σ_{j=1}^n ∂²u/∂x_j², and Laplace’s equation is Δu = 0.

Briefing Cornell Notes

Briefing

Laplace’s equation—written as Δu = 0—turns up whenever a scalar “potential” has no sources nearby, such as the electric potential in regions with no charge. The Laplace operator Δ acts on a function u by summing its pure second derivatives: Δu = ∂²u/∂x₁² + … + ∂²u/∂xₙ². Solutions on an open set Ω are typically required to be twice continuously differentiable (u ∈ C²), and any such solution is called a harmonic function. Constant functions automatically qualify, but the more interesting question is what nontrivial harmonic functions look like.

A key simplification comes from radial symmetry. If u is harmonic on Ω {0} and depends only on the distance to the origin, then u(x) can be written as u(x) = φ(|x|), where |x| is the Euclidean norm. In that setting, the multivariable Laplace equation collapses into an ordinary differential equation for φ as a function of r = |x|. Carrying out the chain rule and product rule for the derivatives of φ(|x|) and summing over the n spatial directions yields the radial ODE

φ''(r) + (n − 1)/r · φ'(r) = 0,

valid for every r > 0. This reduction matters because it converts a PDE problem into a solvable ODE problem, letting one classify entire families of radial harmonic functions on punctured space.

Solving the ODE is straightforward once the equation is rewritten in terms of f(r) = φ'(r). The factor (n − 1)/r leads to an integrating factor whose exponent is the logarithm of r, producing r^{n−1}. After multiplying through, the equation becomes a single derivative: d/dr [r^{n−1} f(r)] = 0. That forces r^{n−1} f(r) to be constant, so φ'(r) = c₁ r^{1−n}. Integrating once more introduces a second constant c₂ and produces two cases.

For n = 2, the integral of 1/r gives a logarithm, so φ(r) = c₂ + c₁ ln r. For n ≠ 2, integrating r^{1−n} yields a power law: φ(r) = c₂ + c₁ r^{2−n} (with constants absorbing any prefactors). Therefore, every radial symmetric harmonic function on Rⁿ {0} has the form u(x) = c₂ + c₁ |x|^{2−n} for n ≠ 2, and u(x) = c₂ + c₁ ln|x| for n = 2.

In applications, the three-dimensional case (n = 3) is especially important: the non-constant term behaves like 1/|x|, which is the prototype of a “fundamental solution” for Laplace’s equation. That connection sets up what comes next: using these radial solutions to build the core singular potential used throughout physics and PDE theory.

Cornell Notes

Laplace’s equation, Δu = 0, describes harmonic functions—twice continuously differentiable solutions whose pure second derivatives sum to zero. Imposing radial symmetry u(x) = φ(|x|) on the punctured domain Rⁿ {0} reduces the PDE to the ODE φ''(r) + (n−1)/r · φ'(r) = 0 for r > 0. Solving via an integrating factor shows φ'(r) = c₁ r^{1−n}, and integrating again gives two cases: for n = 2, φ(r) = c₂ + c₁ ln r; for n ≠ 2, φ(r) = c₂ + c₁ r^{2−n}. These formulas classify all radial harmonic functions away from the origin and explain why the 3D term looks like 1/r, a key ingredient for fundamental solutions in physics.

Why does assuming radial symmetry turn Laplace’s equation into an ODE?

With u(x) = φ(|x|) and r = |x|, every point on a sphere of radius r shares the same value of u. When computing Δu, the chain rule and product rule express the second derivatives in terms of φ'(r) and φ''(r). After summing over the n coordinate directions, the PDE collapses to φ''(r) + (n−1)/r · φ'(r) = 0 for r > 0.

What role does the punctured domain Rⁿ \ {0} play?

The derivation assumes r = |x| is nonzero so expressions like (n−1)/r are well-defined. That’s why the harmonic functions are classified on Rⁿ without the origin, allowing singular behavior at r = 0 (e.g., terms like r^{2−n} or ln r) while still satisfying Δu = 0 for every r > 0.

How does the integrating factor method work here?

Let f(r) = φ'(r). The ODE becomes f'(r) + (n−1)/r · f(r) = 0. The integrating factor is exp(∫(n−1)/r dr) = exp((n−1) ln r) = r^{n−1}. Multiplying through gives d/dr [r^{n−1} f(r)] = 0, so r^{n−1} f(r) is constant and f(r) = c₁ r^{1−n}.

Why is n = 2 a special case?

When n = 2, φ'(r) = c₁ r^{1−n} becomes c₁/r. Integrating 1/r produces ln r, so φ(r) = c₂ + c₁ ln r. For n ≠ 2, integrating r^{1−n} yields a power r^{2−n}, giving φ(r) = c₂ + c₁ r^{2−n}.

What does the n = 3 case imply for physics?

For n = 3, the non-constant radial term is r^{2−3} = 1/r. That 1/|x| behavior matches the familiar singular potential outside a point source, foreshadowing the “fundamental solution” used to model potentials in three dimensions.

Review Questions

Starting from u(x)=φ(|x|), derive the radial ODE for φ(r) and identify where the factor (n−1)/r comes from.
Solve φ''(r) + (n−1)/r · φ'(r) = 0 by setting f(r)=φ'(r). What are φ(r) for n=2 and for n≠2?
Why is it valid to classify solutions only on Rⁿ \ {0} rather than all of Rⁿ? What kinds of singularities does this allow?

Key Points

1
Laplace’s operator is Δu = Σ_{j=1}^n ∂²u/∂x_j², and Laplace’s equation is Δu = 0.
2
Harmonic functions are twice continuously differentiable (u ∈ C²) solutions of Δu = 0 on an open set Ω.
3
Imposing radial symmetry u(x)=φ(|x|) on Rⁿ \ {0} reduces the PDE to φ''(r) + (n−1)/r · φ'(r) = 0 for r>0.
4
Setting f(r)=φ'(r) turns the radial ODE into a first-order linear equation solvable with integrating factor r^{n−1}.
5
All radial harmonic functions on Rⁿ \ {0} take the form φ(r)=c₂+c₁ ln r when n=2.
6
For n≠2, radial harmonic functions take the form φ(r)=c₂+c₁ r^{2−n}, giving u(x)=c₂+c₁|x|^{2−n}.
7
In three dimensions (n=3), the singular term behaves like 1/|x|, matching the structure of a fundamental solution outside a point source.

Highlights

Radial symmetry collapses Laplace’s PDE into the ODE φ''(r) + (n−1)/r · φ'(r) = 0.

The integrating factor is r^{n−1}, leading directly to φ'(r)=c₁ r^{1−n}.

The n=2 solution uniquely produces a logarithm: φ(r)=c₂+c₁ ln r.

For n=3, the key term becomes 1/r, foreshadowing the fundamental solution used in physics.

Topics

Laplace Operator
Harmonic Functions
Radial Symmetry
Integrating Factor
Fundamental Solution