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String with initial deflection lx - x²

Short answer

With fixed ends and zero initial velocity, the solution is a sine series in x with cosine time dependence. Fourier coefficients are Bₙ = (2/l)∫₀^l (lx−x²) sin(nπx/l)dx, non-zero only for odd n. y(x,t) = (8l²/π³) Σ_{n odd} (1/n³) cos(nπct/l) sin(nπx/l).

Solve the wave equation for a tightly stretched string of length l fixed at both ends if the initial deflection is y(x, 0) = lx - x² and the initial velocity is zero.

Problem. y_tt = c² y_xx on 0 < x < l, with y(0, t) = y(l, t) = 0, y(x, 0) = lx − x², and zero initial velocity.

Step 1: general solution (separation of variables with fixed ends):

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

Step 2: zero initial velocity. y_t(x, 0) = Σ Bₙ* (nπc/l) sin(nπx/l) = 0, so Bₙ* = 0.

Step 3: initial deflection. y(x, 0) = lx − x² = Σ Bₙ sin(nπx/l), so

Bₙ = (2/l) ∫₀^l (lx − x²) sin(nπx/l) dx

Integrating by parts twice gives ∫₀^l (lx − x²) sin(nπx/l) dx = 2l³[1 − (−1)ⁿ]/(n³π³). Hence

Bₙ = 4l²[1 − (−1)ⁿ]/(n³π³)
   = 8l²/(n³π³)   for n odd
   = 0            for n even

Answer:

y(x, t) = (8l²/π³) Σ_{n = 1,3,5,…} (1/n³) cos(nπct/l) sin(nπx/l)

Check: at t = 0 and l = 2, x = 0.7, the series sums to 0.9100 = lx − x² ✓ (computed numerically).

✓ Answer checked by Claude Opus 5.5 (high effort)

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