3.2 The continuum limit and the 1-D wave equation

Start from the discrete-chain equation we just derived,

my¨n  =  κ(yn+12yn+yn1),m\, \ddot y_n \;=\; \kappa\, (y_{n+1} - 2 y_n + y_{n-1}),

and take the limit NN \to \infty, a0a \to 0 with L=NaL = Na fixed and μm/a\mu \equiv m/a fixed. The displacement yn(t)y_n(t) becomes a continuous field y(x,t)y(x, t) with x=nax = n a. Watch what happens.

Spatial second difference becomes second derivative

Taylor-expand yn±1=y(x±a,t)y_{n \pm 1} = y(x \pm a, t) around xx:

y(x±a,t)  =  y±ayx+12a2yxx±16a3yxxx+y(x \pm a, t) \;=\; y \pm a\, y_x + \tfrac12 a^2\, y_{xx} \pm \tfrac16 a^3\, y_{xxx} + \cdots

Adding the ++ and - versions:

y(x+a,t)+y(xa,t)  =  2y+a2yxx+O(a4).y(x + a, t) + y(x - a, t) \;=\; 2 y + a^2\, y_{xx} + O(a^4).

So the second difference is

yn+12yn+yn1  =  a22yx2  +  O(a4).y_{n+1} - 2 y_n + y_{n-1} \;=\; a^2\, \frac{\partial^2 y}{\partial x^2} \;+\; O(a^4).

Substituting into the equation of motion and using m=μam = \mu a and κa=T\kappa a = T (where TT is the tension — see derivation below):

μay¨  =  κa2yxx        μy¨  =  (κa)yxx  =  Tyxx,\mu a\, \ddot y \;=\; \kappa\, a^2\, y_{xx} \;\;\Longrightarrow\;\; \mu\, \ddot y \;=\; (\kappa a)\, y_{xx} \;=\; T\, y_{xx},

or in canonical form,

    2yt2  =  c22yx2,c2  =  T/μ.    \boxed{\;\;\frac{\partial^2 y}{\partial t^2} \;=\; c^2\, \frac{\partial^2 y}{\partial x^2}, \qquad c^2 \;=\; T/\mu.\;\;}

This is the one-dimensional wave equation. The propagation speed is c=T/μc = \sqrt{T/\mu} — the square root of tension over mass per unit length. It depends only on the properties of the medium, not on the amplitude or the shape of the disturbance.

Why κa=T\kappa a = T in the continuum limit Derivation

The springs in the discrete chain have stiffness κ\kappa. In the continuum limit, the same physical material is described by a tension TT — the force per unit cross-section needed to stretch it. To extract the right scaling, hold a small piece of string of length =Npiecea\ell = N_\text{piece}\, a in tension TT. Extending the piece by δ\delta\ell requires force Tδ/T \cdot \delta\ell / \ell (Hooke for a continuous rod). For the equivalent discrete chain to behave the same way, the total extension δ\delta\ell is the sum of extensions of NpieceN_\text{piece} springs in series, each with stiffness κ\kappa. The force to extend the chain by δ\delta\ell is κδ/Npiece\kappa \delta\ell / N_\text{piece} (because springs in series sum compliances). Equating, T/=κ/NpieceT / \ell = \kappa / N_\text{piece}, i.e. T=κaT = \kappa a.

The shortcut: a discrete-chain stiffness κ\kappa between masses spaced by aa corresponds to a continuum tension T=κaT = \kappa a. Both have units of force (cross-section dropped because we’re 1-D).

A real wave equation

The 1-D wave equation has two spatial derivatives and two time derivatives. The two-ness on both sides is crucial: it is what makes the equation invariant under ttt \to -t (waves travel either direction) and what allows the existence of two independent travelling-wave families (next lesson). A “wave equation” with one time derivative and two spatial derivatives is the heat equation — and behaves entirely differently (diffusion, no propagation).

For a string of mass density μ=0.5\mu = 0.5\,g/m under tension T=50T = 50\,N — about a steel guitar string — the speed is

c  =  T/μ  =  50/5×104    320m/s.c \;=\; \sqrt{T/\mu} \;=\; \sqrt{50 / 5 \times 10^{-4}} \;\approx\; 320\,\text{m/s}.

Suggestively close to the speed of sound in air. This is also no accident: both are square roots of restoring-force-per-unit-extension over mass-per-unit-length, with the relevant moduli computed differently.

The two interpretations

The same equation can be read two ways:

The next lesson formalises the second interpretation through d’Alembert’s solution.