The Wave Equation

Chem 3240 · Lecture 2.2

Davit Potoyan

The Classical Wave Equation

  • A second-order PDE for the displacement \(u(x,t)\)

\[\frac{\partial^2 u(x,t)}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 u(x,t)}{\partial t^2}\]

  • Governs evolution in space and time
  • Given initial conditions, predicts any complicated wave

Why a Guitar String?

  • A plucked string is the classic 1D wave system
  • Specify initial conditions, predict evolution over space and time
  • Linear PDE: any combination of solutions is a solution

The Big Picture: Three Ingredients

  • 1. Boundary conditions (string fixed at both ends):

\[u(0, t) = 0 \quad \text{and} \quad u(L, t) = 0\]

  • 2. Separation of variables (assume \(x\), \(t\) vary independently):

\[u(x, t) = X(x) \cdot T(t)\]

  • 3. Superposition (linearity lets us sum solutions):

\[u = c_1 u_1+c_2u_2\]

Step 1: Separate the Variables

  • Substitute \(u(x,t) = X(x)T(t)\) into the wave equation
  • Divide through: each side depends on one variable only, so both equal a constant \(K\)

\[\frac{1}{T(t)v^2}\frac{\partial^2 T(t)}{\partial t^2} = \frac{1}{X(x)}\frac{\partial^2 X(x)}{\partial x^2} = K\]

  • One PDE becomes two ODEs:

\[\frac{\partial^2 T(t)}{\partial t^2} - K T(t) v^2 = 0 \qquad \frac{\partial^2 X(x)}{\partial x^2} - K X(x) = 0\]

Step 2: The Sign of K Decides

  • Recipe for linear, homogeneous ODEs: plug in \(y \to e^{kx}\), solve the algebraic equation for \(k\)
  • \(K > 0\): real exponentials, boundary conditions force \(X(x) = 0\) (no music!)
  • \(K < 0\) (set \(K = -\beta^2\)): oscillatory solution

\[X(x) = A \cos(\beta x) + B \sin(\beta x)\]

  • Boundary conditions: \(A = 0\), and \(\sin(\beta L) = 0\) for a non-trivial solution

Normal Modes Emerge

  • \(\sin(\beta L) = 0\) quantizes \(\beta\):

\[\beta = \frac{n \pi}{L}, \quad n = 1, 2, 3, \ldots\]

  • Infinite set of normal modes:

\[X(x) = B \sin \left(\frac{n \pi}{L} x \right)\]

Modes Combine: Superposition in Motion

  • Any string motion = sum of normal modes, each oscillating at its own frequency

The Temporal Part

  • Time has no boundary conditions: it marches forward freely
  • Solution is oscillatory with mode frequency \(\omega_n = \beta v = \frac{n \pi v}{L}\)

\[T(t) = D_n \cos(\omega_n t) + E_n \sin(\omega_n t) = A_n \cos (\omega_n t + \phi_n)\]

  • Constants \(A_n\) and \(\phi_n\) are fixed by initial conditions (how the string is plucked)

Step 3: The Full Solution

  • Combine spatial and temporal parts, sum over all modes:

\[u(x, t) = \sum_n A_n \sin \left(\frac{n \pi}{L} x \right) \cdot \cos (\omega_n t + \phi_n)\]

  • Nodes (points fixed at zero) grow with \(n\):
    • \(n=1\): 0 nodes, fundamental (first harmonic)
    • \(n=2\): 1 node, second harmonic
    • \(n=3\): 2 nodes, third harmonic

Extending to 2D Membranes

  • Separate three functions: \(u(x, y, t) = X(x) Y(y) T(t)\)
  • 2D mode is a product of 1D modes, indexed by \(n\) and \(m\)

\[\omega_{nm} = v\pi \left(\frac{n^2}{a^2} + \frac{m^2}{b^2}\right)^{1/2}\]

The Sound of Music

  • Musical sound is a superposition of normal modes (harmonics, overtones)
  • Instrument size sets the frequency range: small = high, large = low

Takeaway

Separation of variables plus boundary conditions turns the wave equation into quantized normal modes, and any vibration is a superposition of them.