Separation of variables

Look for solutions of \( u_t = k\,u_{xx} \) on \( 0 < x < L \) with \( u(0,t) = u(L,t) = 0 \) in the form \( u = X(x)\,T(t) \). Substituting and dividing by \( kXT \) gives \[ \frac{T'(t)}{k\,T(t)} = \frac{X''(x)}{X(x)}. \] The left side depends only on \( t \), the right only on \( x \), so both equal a constant, written \( -\lambda \). The PDE splits into two ODEs: \( T' = -k\lambda T \) and the boundary value problem \( X'' + \lambda X = 0 \), \( X(0) = X(L) = 0 \).

The boundary value problem has a non-zero solution only for special \( \lambda \), the eigenvalues. Multiply \( X'' + \lambda X = 0 \) by \( X \) and integrate by parts: \( \lambda\int_0^L X^2\,dx = \int_0^L X'^2\,dx \), so \( \lambda \ge 0 \), and \( \lambda = 0 \) forces \( X \) constant, hence zero. For \( \lambda = \mu^2 > 0 \) the solutions are \( A\cos\mu x + B\sin\mu x \) (the example solves the case \( \mu = 3 \)); \( X(0) = 0 \) kills \( A \), and \( X(L) = 0 \) needs \( \sin\mu L = 0 \). So \[ \lambda_n = \Big(\frac{n\pi}{L}\Big)^2, \quad X_n = \sin\frac{n\pi x}{L}, \quad T_n = e^{-k\lambda_n t}, \quad n = 1, 2, 3, \dots \]

Each product \( X_nT_n \) solves the problem, and because the equation and boundary conditions are linear and homogeneous so does any sum \( \sum b_n e^{-k\lambda_n t}\sin(n\pi x/L) \). To match an initial temperature \( f \) we need \( f(x) = \sum b_n \sin(n\pi x/L) \): an expansion of \( f \) in sines, which is the next lesson.

Picture it: the modes are fixed shapes (one hump, two humps, three) each fading at its own rate. The \( n \)-th decays like \( e^{-kn^2\pi^2 t/L^2} \), so wiggly modes vanish first and the rod soon looks like a single fading hump.

Think it: \( -d^2/dx^2 \) with these boundary conditions is a symmetric operator, \( \int X''Y = \int XY'' \), with positive eigenvalues, and its eigenfunctions are orthogonal. Separation of variables is diagonalisation: in the eigenbasis the PDE becomes independent scalar ODEs, one per mode, exactly as \( \dot{\mathbf v} = -A\mathbf v \) decouples when \( A \) is a symmetric matrix.