Review / Part V — Fourier Transform / Ch 25

Gaussians and the Heat Equation on the Line

Ch 25 — Part V — Fourier Transform

Gaussians are fixed points (up to rescaling) of the Fourier transform and they are the natural kernel for the heat equation on all of \(\mathbb{R}\).

FT of a Gaussian

$$\mathcal{F}\{e^{-a x^2}\}(\xi) = \sqrt{\frac{\pi}{a}}\, e^{-\xi^2/(4a)}.$$ A Gaussian's Fourier transform is another Gaussian — narrow in \(x\) ↔ wide in \(\xi\).

Heat kernel on \(\mathbb{R}\)

For \(u_t = \alpha u_{xx}\) on \(\mathbb{R}\) with \(u(x,0)=f(x)\): $$u(x,t) = \int_{-\infty}^\infty K(x-y, t) f(y)\,dy, \quad K(x,t) = \frac{1}{\sqrt{4\pi\alpha t}} e^{-x^2/(4\alpha t)}.$$ The kernel \(K\) is a Gaussian that spreads with \(\sqrt t\).

Derivation via FT

Taking FT in \(x\): \(\hat u_t = -\alpha \xi^2 \hat u\Rightarrow \hat u(\xi, t) = \hat f(\xi) e^{-\alpha \xi^2 t}\). Inverse-FT gives the convolution with the Gaussian kernel.

Practice Problems

Problem 1medium
Solve \(u_t = u_{xx}\) on \(\mathbb{R}\) with \(u(x,0) = \delta(x)\).
Hint
Heat kernel.
Solution

\(u(x,t) = \frac{1}{\sqrt{4\pi t}} e^{-x^2/(4t)}\).

Answer: Heat kernel \(K(x,t).\)

Problem 2medium
Compute \(\mathcal{F}\{e^{-x^2/2}\}\).
Hint
Use formula with \(a = 1/2\).
Solution

\(\sqrt{2\pi} e^{-\xi^2/2}\).

Answer: \(\sqrt{2\pi} e^{-\xi^2/2}.\)

Problem 3hard
Interpret "scale invariance" of the heat kernel: if \(u\) solves the heat equation with initial \(\delta\), what is \(u(\lambda x, \lambda^2 t)/u(x,t)\)?
Hint
Substitute and simplify.
Solution

Under \(x\to \lambda x, t\to\lambda^2 t\), the exponent \(-x^2/(4t)\) is invariant and the prefactor scales by \(1/\lambda\). So \(u(\lambda x, \lambda^2 t) = \lambda^{-1} u(x, t)\). This reflects the \(\sqrt t\) spreading law.

Answer: \(1/\lambda.\)