Solving PDEs via SoV + Fourier Series
Ch 21–22 — Part IV — PDEs & Fourier Series
Combining separation of variables with Fourier series solves a wide class of linear PDEs on bounded domains with homogeneous BCs.
Heat equation with general IC
For \(u_t = \alpha u_{xx}\) on \([0,L]\), Dirichlet zero BCs, \(u(x,0) = f(x)\): $$u(x,t) = \sum_{n=1}^\infty b_n \sin\!\left(\frac{n\pi x}{L}\right) e^{-\alpha(n\pi/L)^2 t},$$ $$b_n = \frac{2}{L}\int_0^L f(x) \sin\!\frac{n\pi x}{L}\,dx.$$
Wave equation on a string
For \(u_{tt} = c^2 u_{xx}\) on \([0,L]\), \(u(0,t)=u(L,t)=0\), \(u(x,0)=f(x)\), \(u_t(x,0)=g(x)\): $$u(x,t) = \sum_n \sin\!\frac{n\pi x}{L}\left[b_n \cos\omega_n t + d_n \sin\omega_n t\right],\quad \omega_n = \frac{n\pi c}{L}.$$
Inhomogeneous BCs
Split \(u = u_s + w\) where \(u_s\) is a steady-state satisfying the BCs (and, for heat, Laplace's equation); \(w\) then solves a homogeneous-BC problem.
Practice Problems
Hint
Solution
\(u(x,t) = 2\sin(\pi x) e^{-\pi^2 t} - \sin(4\pi x) e^{-16\pi^2 t}\).
Answer: See solution.
Hint
Solution
\(\omega_n = n\). \(d_n = (2/n)\cdot\) (sine coefficient of \(u_t(x,0)\)). Only \(n=2\) mode: \(d_2 = 1/2\). So \(u(x,t) = \tfrac{1}{2}\sin(2x)\sin(2t)\).
Answer: \(u = \tfrac{1}{2}\sin(2x)\sin(2t).\)
Hint
Solution
Let \(u_s(x) = T_1 x / L\) (solves steady state and BCs). Then \(w = u - u_s\) satisfies heat equation with zero BCs and initial data \(f(x) - u_s(x)\).
Answer: Split off steady state.