Derive 1D wave equation and solve

Part 1: Derivation of the one-dimensional wave equation

Assumptions. A perfectly flexible, uniform string of mass ρ per unit length is stretched to a constant tension T along the x-axis and fixed at x = 0 and x = l. It vibrates in the xy-plane with small displacements u(x, t); the slope is small, so cos θ ≈ 1 and sin θ ≈ tan θ. Gravity is neglected.

 u
 |            Q ○──→ T (angle θ₂)
 |          ╱
 |   P ○───╱
 |  ↙ T (angle θ₁)
 |_________________________ x
      x        x+δx

Take a small element PQ between x and x + δx. The tensions T at the two ends make small angles θ₁ and θ₂ with the x-axis.

• Horizontal: T cos θ₂ − T cos θ₁ ≈ 0 (no horizontal motion).
• Vertical net force = T sin θ₂ − T sin θ₁ ≈ T (tan θ₂ − tan θ₁).

Since tan θ = ∂u/∂x:

net force = T [ (∂u/∂x)_{x+δx} − (∂u/∂x)_x ]

By Newton's second law, mass × acceleration = force. The mass of the element is ρ δx:

ρ δx ∂²u/∂t² = T [ (∂u/∂x)_{x+δx} − (∂u/∂x)_x ]
∂²u/∂t² = (T/ρ) · [ (∂u/∂x)_{x+δx} − (∂u/∂x)_x ] / δx

Letting δx → 0 and writing c² = T/ρ:

∂²u/∂t² = c² ∂²u/∂x²

Part 2: Solution

Boundary and initial conditions

1. u(0, t) = 0 and u(l, t) = 0 for all t (fixed ends)
2. u(x, 0) = f(x) (initial shape)
3. ∂u/∂t at t = 0 equals g(x) (initial velocity)

Step 1: separation of variables. Put u = X(x) T(t):

X T″ = c² X″ T, so X″/X = T″/(c²T) = −λ² (constant, shown negative below).

X″ + λ² X = 0
T″ + λ²c² T = 0

(A zero or positive separation constant gives only the trivial solution X = 0 once the ends are fixed, so the constant is −λ², a negative value.)

Step 2: apply boundary conditions. X = A cos λx + B sin λx.

• u(0, t) = 0 gives A = 0.
• u(l, t) = 0 gives B sin λl = 0, so λ = nπ/l (n = 1, 2, 3, …) since B ≠ 0.

Xₙ(x) = sin(nπx/l), and T(t) = C cos(λct) + D sin(λct).

Step 3: general solution. By superposition:

u(x, t) = Σ_{n=1}^{∞} [ Bₙ cos(nπct/l) + Bₙ sin(nπct/l) ] sin(nπx/l)*

Step 4: initial conditions.

• u(x, 0) = f(x) gives f(x) = Σ Bₙ sin(nπx/l), a Fourier sine series, so

  Bₙ = (2/l) ∫₀^l f(x) sin(nπx/l) dx

• ∂u/∂t at t = 0: g(x) = Σ Bₙ* (nπc/l) sin(nπx/l), so

  Bₙ* = (2/(nπc)) ∫₀^l g(x) sin(nπx/l) dx

Special case (string released from rest, g = 0): u(x, t) = Σ Bₙ cos(nπct/l) sin(nπx/l).

Answer: u(x, t) as in Step 3, with Bₙ and Bₙ* from Step 4. The n-th term is the n-th normal mode, with frequency nc/(2l).