Cooling a Rod: Separation of Variables in Practice

The problem

A metal rod of length \(L\) starts at a uniform 100 °C. At \(t = 0\) both ends are clamped to 0 °C. Find \(T(x,t)\), and answer the practical question the series raises: how many terms do you actually need?

\[\frac{\partial T}{\partial t} = \alpha\frac{\partial^2T}{\partial x^2}, \qquad T(0,t) = T(L,t) = 0, \qquad T(x,0) = T_0\]

Step 1: separate

Try \(T = X(x)\tau(t)\). Substituting and dividing by \(X\tau\):

\[\frac{1}{\alpha}\frac{\tau'}{\tau} = \frac{X''}{X} = -\lambda\]

The left side depends only on \(t\), the right only on \(x\), so both must equal a constant. That is the entire separation argument — and it is also where the method's limits live: it works because the domain is a rectangle in \((x,t)\) and the boundary conditions are homogeneous on lines of constant \(x\).

Step 2: solve the spatial eigenvalue problem

\[X'' + \lambda X = 0,\qquad X(0) = X(L) = 0\]

This is a Sturm–Liouville eigenproblem, and it is exactly the finite-dimensional problem of the normal modes lab taken to the continuum. Non-trivial solutions exist only for

\[\lambda_n = \left(\frac{n\pi}{L}\right)^2,\qquad X_n = \sin\frac{n\pi x}{L},\qquad n = 1,2,3,\dots\]

The boundary conditions quantise \(\lambda\) — the same mechanism that quantises energy levels in a box, and it is worth noticing that nothing quantum is involved. Then \(\tau_n \propto e^{-\alpha\lambda_nt}\).

Step 3: match the initial condition with a Fourier series

\[T(x,t) = \sum_n b_n\sin\frac{n\pi x}{L}e^{-n^2\pi^2\alpha t/L^2}\]

At \(t=0\) this must equal the constant \(T_0\), so the \(b_n\) are the Fourier sine coefficients of a constant. Using the orthogonality \(\int_0^L\sin\frac{n\pi x}{L}\sin\frac{m\pi x}{L}dx = \frac{L}{2}\delta_{nm}\):

\[b_n = \frac{2}{L}\int_0^LT_0\sin\frac{n\pi x}{L}dx = \frac{2T_0}{n\pi}\left(1 - \cos n\pi\right) = \begin{cases}4T_0/n\pi & n\ \text{odd}\\ 0 & n\ \text{even}\end{cases}\]

\[\boxed{\;T(x,t) = \frac{4T_0}{\pi}\sum_{n\ \rm odd}\frac{1}{n}\sin\frac{n\pi x}{L}\, e^{-n^2\pi^2\alpha t/L^2}\;}\]

Even harmonics vanish because a constant is symmetric about the midpoint and even sine modes are antisymmetric — their overlap integral is zero. Symmetry doing work, as usual.

Step 4: how many terms?

Take \(L = 1\) m, \(\alpha = 10^{-2}\) m²/s, \(T_0 = 100\) °C, and watch the centre:

\(t\) (s)	full series	one term only
0	99.94	127.32
1	99.92	115.36
5	77.23	77.73
10	47.45	47.455
25	10.798	10.798

Two things stand out.

At \(t = 1\) the one-term approximation gives 115 °C — hotter than the rod ever was. That is not a bug: a single sine cannot represent a flat-topped profile, and truncating a Fourier series near a discontinuity overshoots. It is the Gibbs phenomenon, and at early times you need many terms.

By \(t = 10\) the one-term answer is right to four figures. The \(n\)-th mode decays as \(e^{-n^2\pi^2\alpha t/L^2}\) — quadratically in \(n\) in the exponent — so the third harmonic dies \(e^{-8\pi^2\alpha t/L^2}\) times faster than the first. After one time constant

\[\tau = \frac{L^2}{\pi^2\alpha} = 10.13\ \text{s}\]

everything but the fundamental has been annihilated.

This is the general lesson about diffusion: it is a low-pass filter of extraordinary severity. Fine structure is erased almost instantly, and the long-time behaviour of any initial condition is the slowest-decaying eigenmode with whatever amplitude it happened to start with. Which is why you can predict the late-time state of a diffusive system without knowing much about how it began.

Common mistakes

Expecting even harmonics. They vanish by symmetry here. Change the initial condition to something asymmetric and they return.
Truncating at one term at early times. See the 115 °C above.
Forgetting that separation needs homogeneous boundary conditions. With ends held at 20 °C and 80 °C you must first subtract the steady linear profile, then expand the remainder.
Using the wrong orthogonality weight. For this problem it is 1; for cylindrical or spherical geometry it is not, and the special functions carry their own weights.

Partial differential equations · Fourier series · Eigenvalues & eigenvectors · Inner product spaces — orthogonality is a projection · Special functions · Fourier series builder — watch the Gibbs overshoot