Review / Part V — Fourier Transform / Ch 26

Convolution & the Wave Equation on the Line

Ch 26 — Part V — Fourier Transform

Convolution \((f * g)(x) = \int f(y) g(x-y)\,dy\) encodes the action of a linear, translation-invariant operator as a kernel. For the wave equation, d'Alembert's formula expresses the solution directly.

Convolution & FT

\(\mathcal{F}\{f * g\} = \hat f \cdot \hat g\). Many PDE fundamental solutions act via convolution with a kernel.

d'Alembert's formula (wave on \(\mathbb{R}\))

For \(u_{tt} = c^2 u_{xx}\) on \(\mathbb{R}\), \(u(x,0) = f(x),\; u_t(x,0) = g(x)\): $$u(x,t) = \tfrac{1}{2}\big[f(x - ct) + f(x + ct)\big] + \tfrac{1}{2c} \int_{x-ct}^{x+ct} g(s)\,ds.$$

Interpretation

Information travels at speed \(c\) along the two characteristic families \(x\pm ct = \text{const}\). Sharp features in \(f\) persist (they don't smooth like heat); speed of propagation is finite.

Practice Problems

Problem 1easy
Apply d'Alembert: \(u_{tt} = u_{xx},\; u(x,0) = e^{-x^2},\; u_t(x,0) = 0\).
Hint
Split the initial profile.
Solution

\(u(x,t) = \tfrac{1}{2}[e^{-(x-t)^2} + e^{-(x+t)^2}]\).

Answer: See solution.

Problem 2medium
If \(f\) and \(g\) are Gaussians, describe \(f*g\).
Hint
Variances add.
Solution

Convolution of two Gaussians with variances \(\sigma_1^2\) and \(\sigma_2^2\) is a Gaussian with variance \(\sigma_1^2 + \sigma_2^2\).

Answer: Another Gaussian, variances add.

Problem 3easy
Using \(\mathcal{F}\{f*g\} = \hat f \hat g\), prove the convolution is commutative.
Hint
Multiplication is commutative.
Solution

\(\widehat{f*g} = \hat f \hat g = \hat g \hat f = \widehat{g*f}\). By Fourier inversion, \(f*g = g*f\).

Answer: See solution.