Solving the Heat Equation

Main idea

For a rod with both ends held at 0°, the temperature is a sum of sine modes, and mode $n$ fades at a rate proportional to $n^2$ . The fine wiggles die almost instantly while the smooth first mode lingers, which is why heat smooths everything out.

u(x, t) = \sum_{n=1}^{\infty} {\htmlData{sym=0}{B_n}}\,{\htmlData{sym=1}{e^{-k(n\pi/L)^2t}}}\sin\frac{n\pi x}{L}, \qquad {\htmlData{sym=0}{B_n}} = \frac2L\int_0^L {\htmlData{sym=2}{f(x)}}\sin\frac{n\pi x}{L}\,dx

What the symbols mean

$B_n$
Initial amount of mode n
The sine coefficients of the initial temperature f.
$e^{-k(n\pi/L)^2t}$
Decay factor
Mode n fades at a rate proportional to n², so fine detail vanishes first.
$f(x)$
Initial temperature
The profile at t = 0.

Explanation

The series solution of the heat equation is simply the temperature written as a stack of simple waves, each one fading away on its own schedule. It tells you not just the temperature but how fast each part of the pattern dies out, which is how engineers estimate cooling times. The formula has three pieces: $B_n$ (how much of wave $n$ was there at the start), $e^{-k(n\pi/L)^2t}$ (a fading factor that shrinks that wave as time goes on), and $\sin\frac{n\pi x}{L}$ (the wave's shape, with $n$ humps across the rod). Add them up and you have the temperature everywhere, at every time. It looks heavy, but it's just the recipe from the last lesson with the Fourier coefficients plugged in.

Reading the solution. Each mode keeps its shape and fades. Mode 3 decays 9 times faster than mode 1, mode 10 a hundred times faster. So after a short while only the first mode matters:

u(x, t) \approx B_1e^{-k\pi^2t/L^2}\sin\frac{\pi x}{L}.

There's a graph further down where you can drag time forward and watch the first mode alone catch up with the full solution.

The time scale. The slowest decay rate is $\frac{k\pi^2}{L^2}$ . Because $L$ is squared in the denominator, doubling a rod's length makes it take four times as long to cool.

Smoothing. Even a discontinuous initial temperature becomes perfectly smooth for every $t > 0$ , because the high modes responsible for sharp features are damped by enormous factors. This is the "heat forgets" behavior from the last unit, now coming straight out of the formula. Running time backwards would amplify those modes instead of damping them, which is why the backward heat equation is ill-posed.

See it

t

1.0e-3

The temperature of a rod that starts at 4x(1 − x), with k = 1, next to its first mode by itself. Even at the start they're close: the first mode peaks at 32/π³ ≈ 1.03, against the true 1. Drag t and within a few hundredths of a time unit you can't tell them apart. By t = 0.3 the peak is down to about 0.05.

Worked examples

Example 1

Solve $u_t = u_{xx}$ on $[0, \pi]$ with $u(0, t) = u(\pi, t) = 0$ and $u(x, 0) = 3\sin 2x - \sin 5x$ .

1
The initial condition is already a sine series, so no integrals are needed: $B_2 = 3$ and $B_5 = -1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $u_t = u_{xx}$ on $[0, \pi]$ , zero ends, $u(x, 0) = 3\sin 2x$ .

Problem 2+5 XP

Solve $u_t = 2u_{xx}$ on $[0, \pi]$ , zero ends, $u(x, 0) = \sin x + \frac12\sin 3x$ .

Problem 3+5 XP

Find $B_n$ for $f(x) = x(\pi - x)$ on $[0, \pi]$ .

Problem 4+5 XP

How many times faster does mode 3 decay than mode 1?

Problem 5+5 XP

For large $t$ , which mode dominates the solution?

Problem 6+5 XP

A 1 m steel rod (diffusivity $k = 1.2\times10^{-5}$ m²/s) has its ends held at 0 °C. Its slowest mode decays like $e^{-k\pi^2t}$ . How long does that mode take to fall by a factor of $e$ , in seconds?

Problem 7+5 XP

A rod on $[0, \pi]$ with $k = 1$ and both ends held at 0 starts at $u(x, 0) = 1$ , so $B_1 = \frac4\pi$ . Using only the first term of the series, estimate the temperature at the center, $u\left(\frac\pi2, 2\right)$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A rod on $[0, \pi]$ with $k = 1$ starts at a uniform 100° and is plunged into ice, so both ends are held at 0. Its temperature is $u = \sum_{n \text{ odd}}\frac{400}{n\pi}e^{-n^2t}\sin nx$ . Find the center temperature $u\left(\frac\pi2, 0.5\right)$ to within 0.1°.

Where it's used

Engineers estimate cooling times of rods, walls and electronic components from the slowest mode; the n² decay is why fine-scale temperature fluctuations vanish so quickly.

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