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Mathematical Methods

Partial differential equations: waves, heat and separation of variables

The three classic equations, d'Alembert's solution, the vibrating string with normal modes and Fourier series, the heat equation and diffusion times, Laplace's equation.

The differential equations of Analysis 2 had a single independent variable, time. But a signal travelling along a cable, heat spreading in a heat sink, the potential inside a semiconductor device depend on time and space. They call for partial differential equations (PDEs). In this lesson we look at the three classic equations, the method of separation of variables with the vibrating string and the heat equation, and the link with Fourier series.

The three classic equations

TypeEquationPhenomena
hyperbolicu_{tt} = c^2 u_{xx} (waves)vibrating strings, sound, electromagnetic waves, signals on transmission lines
parabolicu_t = \alpha\,u_{xx} (heat)heat conduction, carrier diffusion in semiconductors
ellipticu_{xx} + u_{yy} = 0 (Laplace)electrostatic potential, steady-state temperature

A unique solution needs boundary conditions (what happens at the edges of the domain) and, for equations involving time, initial conditions.

The wave equation and d’Alembert’s solution

For an infinite string, the equation u_{tt} = c^2u_{xx} has the general solution

    \[ u(x, t) = F(x - ct) + G(x + ct) \]

for any functions F and G (differentiate twice to check). F(x - ct) is a shape travelling right at speed c without distortion, G(x + ct) one travelling left. With initial conditions u(x, 0) = f(x) and u_t(x, 0) = 0 one gets u = \tfrac12[f(x - ct) + f(x + ct)]: the initial shape splits into two halves moving in opposite directions.

On transmission lines the voltage obeys the same equation, with c = 1/\sqrt{LC} (inductance and capacitance per unit length): that is why fast signals on printed circuit boards must be treated as waves, with reflections at the ends if the impedance is not matched.

Separation of variables: the vibrating string

A string of length L fixed at both ends: u(0, t) = u(L, t) = 0. Look for solutions of the form u(x, t) = X(x)\,T(t). Substituting into the equation and dividing by XT:

    \[ \frac{T''}{c^2\,T} = \frac{X''}{X} = -\lambda \]

The left side depends only on t, the right side only on x: they can be equal for every x and t only if both equal a constant. The PDE splits into two ordinary equations.

The eigenvalues

X'' + \lambda X = 0 with X(0) = X(L) = 0. For \lambda \le 0 the only solution is zero. For \lambda = k^2 \gt 0, X = A\sin kx + B\cos kx: X(0) = 0 gives B = 0, X(L) = 0 gives \sin kL = 0, that is

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

Only a discrete sequence of shapes is allowed: the normal modes.

The first three vibration modes of a string fixed at both ends
The first three normal modes of a string of length L fixed at both ends: mode n has n antinodes and n − 1 interior nodes, and vibrates at n times the fundamental frequency.

For each mode, T'' + c^2k_n^2T = 0 gives a harmonic oscillation with angular frequency \omega_n = c\,n\pi/L. The frequencies are integer multiples of the fundamental f_1 = \frac{c}{2L}: they are the harmonics of the lesson on Fourier series. It is the physics of every string instrument: halving the length (pressing the fret halfway up a guitar neck) doubles the frequency, one octave higher.

The general solution

By linearity, a sum of solutions is still a solution:

    \[ u(x, t) = \sum_{n=1}^{\infty}\sin\frac{n\pi x}{L}\Big(a_n\cos\omega_n t + b_n\sin\omega_n t\Big) \]

The coefficients come from the initial conditions, and here the Fourier series returns: with the string held still in an initial shape f(x), we need f(x) = \sum a_n\sin\frac{n\pi x}{L}, so the a_n are the coefficients of the sine series of f:

    \[ a_n = \frac2L\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx \]

Plucking the string at a different point changes the coefficients, that is the weight of the harmonics, and so the timbre: that is why a guitar sounds different plucked near the bridge or in the middle.

The heat equation

A bar of length L with its ends held at zero temperature and an initial distribution f(x): u_t = \alpha u_{xx}. Separation of variables gives the same spatial modes, but the time equation is now first order: T' = -\alpha k_n^2T, with a decaying exponential solution. So

    \[ u(x, t) = \sum_{n=1}^{\infty}a_n\,e^{-\alpha(n\pi/L)^2t}\sin\frac{n\pi x}{L} \]

Two important remarks:

  • there are no oscillations: the temperature decays towards equilibrium;
  • the higher modes die out much faster (the factor is n^2): after a short time only the first mode remains, half a sine wave. Heat smooths out the irregularities of the initial distribution.

The characteristic time is \tau = \frac{L^2}{\pi^2\alpha}, proportional to the square of the length. For aluminium \alpha \approx 9.7\times 10^{-5} m²/s: a 10 cm bar settles in about 0.01/(9.87\cdot 9.7\times 10^{-5}) \approx 10 s, a 1 m bar in about 17 minutes. Doubling the distance heat must travel quadruples the time: one of the golden rules of thermal design.

Laplace’s equation

In steady state the heat equation becomes u_{xx} + u_{yy} = 0. Separation of variables on a rectangle gives solutions like \sin(kx)\sinh(ky); on a disc, in polar coordinates, r^n\cos n\theta. The solutions are the harmonic functions of the lesson on complex functions, and this is where conformal mappings become useful: they take a complicated domain to a simple one where the solution is known.

When the geometry is too complicated for any analytical method, one turns to numerical methods, finite differences or finite elements, the subject of the Numerical Analysis course.

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