3.2 The continuum limit and the 1-D wave equation

Start from the discrete-chain equation we just derived,

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

and take the limit N→∞N \to \infty, a→0a \to 0 with L=NaL = Na fixed and μ≡m/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±a yx+12a2 yxx±16a3 yxxx+⋯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(x−a,t)  =  2y+a2 yxx+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+1−2yn+yn−1  =  a2 ∂2y∂x2  +  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):

μa y¨  =  κ a2 yxx    ⟹    μ y¨  =  (κa) yxx  =  T yxx,\mu a\, \ddot y \;=\; \kappa\, a^2\, y_{xx} \;\;\Longrightarrow\;\; \mu\, \ddot y \;=\; (\kappa a)\, y_{xx} \;=\; T\, y_{xx},

or in canonical form,

    ∂2y∂t2  =  c2 ∂2y∂x2,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 ℓ=Npiece a\ell = N_\text{piece}\, a in tension TT. Extending the piece by δℓ\delta\ell requires force T⋅δℓ/ℓT \cdot \delta\ell / \ell (Hooke for a continuous rod). For the equivalent discrete chain to behave the same way, the total extension δℓ\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 δℓ\delta\ell is κδℓ/Npiece\kappa \delta\ell / N_\text{piece} (because springs in series sum compliances). Equating, T/ℓ=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 t→−tt \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=50 T = 50\,N — about a steel guitar string — the speed is

c  =  T/μ  =  50/5×10−4  ≈  320 m/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.