Partial Differential Equations 2

TL;DR

Laplace’s equation is ∆u = 0, and its solutions are harmonic functions u ∈ C².

Briefing Cornell Notes

Briefing

Laplace’s equation—defined by setting the Laplacian of a function to zero—becomes especially tractable when the function is assumed to be radially symmetric. In physics, that translates directly to electrostatic potential in regions with no charge: away from sources, the potential must satisfy ∆u = 0. Mathematically, the focus is on harmonic functions, meaning twice continuously differentiable functions u (in C²) on an open set Ω whose second derivatives combine to give zero.

To find nontrivial harmonic functions, the discussion narrows to domains with a “puncture,” removing the origin (Ω = Rⁿ \ {0}). Under radial symmetry, u(x) depends only on the distance r = ||x||, so u(x) can be written as u(x) = φ(r) for some single-variable function φ defined on (0, ∞). Plugging this form into the Laplacian converts the PDE into an ordinary differential equation. Carrying out the chain rule and product rule for derivatives of φ(||x||) yields a radial ODE:

φ''(r) + (n − 1)/r · φ'(r) = 0 for all r > 0.

This step matters because it turns a multi-variable second-order PDE into a solvable one-variable equation, letting standard ODE techniques take over.

The ODE is linear and is handled by introducing f(r) = φ'(r). The equation then becomes a first-order relation for f(r), and an integrating factor simplifies it. The integrating factor comes from the coefficient (n − 1)/r, whose antiderivative is (n − 1) ln r, producing an integrating factor r^{n−1}. After multiplying through, the equation collapses to a single derivative:

d/dr [ r^{n−1} f(r) ] = 0.

So r^{n−1} f(r) must equal a constant, giving f(r) = c₁ r^{1−n}. Integrating once more produces φ(r), introducing a second constant c₂. The final form depends on dimension:

- For n = 2, φ(r) = c₂ + c₁ ln r. - For n ≠ 2, φ(r) = c₂ + c₁ r^{2−n} (with constants absorbing factors).

Therefore, every radially symmetric harmonic function on Rⁿ \ {0} has the form u(x) = φ(||x||) with φ as above. The three-dimensional case (n = 3) yields a 1/r behavior, which is highlighted as the shape of the fundamental solution to Laplace’s equation—an ingredient expected to play a central role in what comes next.

Cornell Notes

Laplace’s equation requires the Laplacian of a function to vanish: ∆u = 0. Harmonic functions are those u ∈ C² whose second derivatives combine to give zero, and radial symmetry makes the problem solvable. On the punctured domain Rⁿ \ {0}, assuming u(x) = φ(r) with r = ||x|| converts the PDE into the ODE φ''(r) + (n−1)/r · φ'(r) = 0. Solving via an integrating factor r^{n−1} gives φ'(r) = c₁ r^{1−n}, and integrating again yields φ(r) = c₂ + c₁ ln r when n = 2, or φ(r) = c₂ + c₁ r^{2−n} when n ≠ 2. This produces explicit families of radially symmetric harmonic functions, including the 1/r form in three dimensions.

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

With radial symmetry on Rⁿ \ {0}, the function has the form u(x) = φ(r) where r = ||x||. The Laplacian ∆u involves second derivatives with respect to the coordinates, but after applying the chain rule to φ(||x||) and summing over the n coordinate directions, all dependence collapses to r. The result is the radial ODE φ''(r) + (n−1)/r · φ'(r) = 0 for r > 0.

What is the exact radial ODE that φ(r) must satisfy?

The derived condition for a radially symmetric harmonic function is: φ''(r) + (n−1)/r · φ'(r) = 0 for every r in (0, ∞). This comes from substituting u(x)=φ(||x||) into ∆u = 0 and simplifying the chain-rule and product-rule terms.

How does the integrating factor method solve the ODE?

Let f(r) = φ'(r). The ODE becomes a first-order equation for f with a coefficient (n−1)/r. The integrating factor uses the antiderivative of (n−1)/r, which is (n−1) ln r, so the integrating factor is e^{(n−1)ln r} = r^{n−1}. Multiplying through yields d/dr [ r^{n−1} f(r) ] = 0, so r^{n−1} f(r) = constant and f(r) = c₁ r^{1−n}.

Why is n = 2 a special case?

When n = 2, the formula for φ'(r) becomes proportional to r^{1−n} = r^{-1}. Integrating r^{-1} produces ln r, so φ(r) = c₂ + c₁ ln r. For n ≠ 2, integrating r^{1−n} gives a power law r^{2−n} instead.

What do the solutions look like in three dimensions, and why is that important?

In n = 3, the non-log solution gives φ(r) proportional to r^{2−3} = 1/r (plus an additive constant). That 1/r behavior is singled out as matching the fundamental solution shape for Laplace’s equation, which is expected to matter for applications like potential fields.

Review Questions

Derive the radial ODE starting from u(x)=φ(||x||) and ∆u=0: what terms combine to produce (n−1)/r · φ'(r)?
Solve φ''(r) + (n−1)/r · φ'(r) = 0 using f(r)=φ'(r). What integrating factor do you use and what is the resulting expression for φ'(r)?
Explain why the solution involves ln r specifically when n = 2, and give the power-law form for n ≠ 2.

Key Points

1
Laplace’s equation is ∆u = 0, and its solutions are harmonic functions u ∈ C².
2
In charge-free physics regions, electrostatic potential satisfies Laplace’s equation.
3
Assuming radial symmetry on Rⁿ \ {0} lets write u(x)=φ(r) with r=||x||.
4
Substituting u(x)=φ(r) into ∆u=0 yields the radial ODE φ''(r) + (n−1)/r · φ'(r) = 0.
5
Introducing f(r)=φ'(r) reduces the ODE to first order and is solved using integrating factor r^{n−1}.
6
The resulting radially symmetric solutions are φ(r)=c₂+c₁ ln r for n=2 and φ(r)=c₂+c₁ r^{2−n} for n≠2.
7
In three dimensions, the non-constant term behaves like 1/r, foreshadowing the fundamental solution.

Highlights

Radial symmetry converts a multi-variable PDE into the single ODE φ''(r) + (n−1)/r · φ'(r) = 0.

The integrating factor is r^{n−1}, turning the equation into d/dr [ r^{n−1} φ'(r) ] = 0.

Dimension controls the functional form: ln r appears only at n = 2; otherwise the solution is a power r^{2−n}.

The three-dimensional case produces the characteristic 1/r dependence tied to Laplace’s fundamental solution.

Topics

Laplace's Equation
Harmonic Functions
Radial Symmetry
Integrating Factor
Fundamental Solution

Mentioned

PTEES
PTE
ODE
C2

Partial Differential Equations 2 | Laplace's Equation