Review / Part IV — PDEs & Fourier Series / Ch 19

Heat Equation: Separation of Variables

Ch 19 — Part IV — PDEs & Fourier Series

Separation of variables assumes \(u(x,t) = X(x) T(t)\), reducing a PDE to ODEs. It is the entry point to Fourier series.

Derivation

For \(u_t = \alpha u_{xx}\) on \(0 < x < L\) with \(u(0,t) = u(L,t) = 0\):

Substitute \(u = X(x) T(t)\): $$X T' = \alpha X'' T\Rightarrow \frac{T'}{\alpha T} = \frac{X''}{X} = -\lambda \quad (\text{constant}).$$ The BCs force \(X(0) = X(L) = 0\), leading to eigenvalues \(\lambda_n = (n\pi/L)^2\) and eigenfunctions \(X_n(x) = \sin(n\pi x / L)\).

Time part

\(T_n' = -\alpha \lambda_n T_n\Rightarrow T_n(t) = e^{-\alpha \lambda_n t}\).

Superposition and initial data

$$u(x,t) = \sum_{n=1}^\infty b_n \sin\!\left(\frac{n\pi x}{L}\right) e^{-\alpha (n\pi/L)^2 t},$$ with coefficients \(b_n\) fixed by expanding the initial data \(u(x,0) = f(x)\) in a sine Fourier series (next concept).

x u t = 0.01 t = 0.1 t = 0.3 t = 0.8 t = 2.0
Heat diffusion on \([0,\pi]\): profile at increasing times.

Practice Problems

Problem 1medium
Solve \(u_t = u_{xx}\) on \([0,\pi]\), \(u(0,t)=u(\pi,t)=0,\; u(x,0)=\sin(3x)\).
Hint
Only one mode in the IC.
Solution

\(u(x,t) = \sin(3x) e^{-9t}\).

Answer: \(\sin(3x) e^{-9t}.\)

Problem 2medium
What are the eigenvalues of \(-X''\) on \([0,L]\) with Neumann BCs \(X'(0) = X'(L) = 0\)?
Hint
Try \(X = \cos(n\pi x/L)\).
Solution

\(\lambda_n = (n\pi/L)^2\), \(n=0,1,2,\ldots\) (including zero with constant eigenfunction).

Answer: \(\lambda_n = (n\pi/L)^2,\; n \ge 0.\)

Problem 3easy
Explain why high-frequency modes decay faster in the heat equation.
Hint
Decay rate \(\propto n^2\).
Solution

Mode \(n\) decays like \(e^{-\alpha (n\pi/L)^2 t}\). Larger \(n\) ⇒ faster decay. High frequencies smoothed away, explaining the smoothing effect of diffusion.

Answer: \(n^2\) dependence in exponent.