The Heated Rod
Now the full problem: a rod of length L, its ends held at
0^\circ, starting from a known temperature profile
f(x). We solve
u_t = \alpha\,u_{xx}, \qquad u(0, t) = u(L, t) = 0, \qquad u(x, 0) = f(x),
by
separation of variables.
Each product mode e^{-\alpha\lambda_n t}\sin\frac{n\pi x}{L} solves the
equation and the boundary conditions; the
superposition
of them all is the general solution.
Fitting the start, then letting it decay
The eigenvalues are \lambda_n = (n\pi/L)^2, so the solution is
u(x, t) = \sum_{n=1}^{\infty} b_n\, e^{-\alpha (n\pi/L)^2 t}\,\sin\frac{n\pi x}{L}.
At t = 0 every exponential is 1, so the sum
must equal f(x) — meaning the b_n are exactly
the
Fourier sine coefficients
of the initial profile:
b_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx.
After that, each mode simply decays at its own rate \alpha(n\pi/L)^2 \propto n^2.
High modes (sharp features) vanish fastest, which is why a jagged start rounds off almost
immediately and the whole rod drifts smoothly to 0^\circ.
- Separate to get product modes e^{-\alpha\lambda_n t}\sin(n\pi x/L) with \lambda_n = (n\pi/L)^2.
- Superpose: u(x,t) = \sum b_n e^{-\alpha\lambda_n t}\sin(n\pi x/L).
- Match the start: the b_n are the Fourier sine coefficients of f.
- Each mode decays at rate \propto n^2; the rod cools to zero, smoothest features last.
Cool the rod
The initial profile is a tent (a triangular heat pulse) on [0, \pi],
with \alpha = 1. Slide N to add Fourier sine
modes until the bold curve matches the faint tent at t = 0; then
advance time t and watch the corner round off and the heat drain away
to zero.