Review / Part IV — PDEs & Fourier Series / Ch 18

Introduction to PDEs

Ch 18 — Part IV — PDEs & Fourier Series

A partial differential equation (PDE) involves an unknown function of several variables and its partial derivatives. We meet three classical linear second-order PDEs.

The three classics

Heat (parabolic): \(u_t = \alpha u_{xx}\) — diffusion of heat, concentration. Wave (hyperbolic): \(u_{tt} = c^2 u_{xx}\) — waves on a string, sound, light. Laplace (elliptic): \(u_{xx} + u_{yy} = 0\) — steady-state temperature, electrostatic potential.

Boundary vs initial data

PDEs on bounded spatial domains require boundary conditions (Dirichlet \(u = g\), Neumann \(\partial u/\partial n = g\), or mixed) and often initial conditions. These data dramatically shape the solution.

Linear superposition

All three classical PDEs above are linear; solutions add and scale. This is what makes Fourier methods so powerful.

Practice Problems

Problem 1easy
Verify that \(u(x,t) = \sin(x) e^{-t}\) solves \(u_t = u_{xx}\).
Hint
Compute \(u_t\) and \(u_{xx}\).
Solution

\(u_t = -\sin(x) e^{-t}\); \(u_{xx} = -\sin(x) e^{-t}\). Equal.

Answer: Yes.

Problem 2easy
Verify that \(u(x,t) = \sin(x - ct)\) solves \(u_{tt} = c^2 u_{xx}\).
Hint
Compute both derivatives.
Solution

\(u_{tt} = -c^2 \sin(x-ct)\); \(u_{xx} = -\sin(x-ct)\). Equal after multiplying by \(c^2\).

Answer: Yes.

Problem 3easy
Classify as hyperbolic / parabolic / elliptic: \(u_{tt} - u_{xx} + u = 0\).
Hint
Compare top-derivative signs.
Solution

\(u_{tt}\) and \(-u_{xx}\) have opposite signs, so hyperbolic (wave-like).

Answer: Hyperbolic.