String with initial velocity 3(lx-x²)

Problem. ∂²u/∂t² = c² ∂²u/∂x², with

• u(0, t) = u(l, t) = 0
• u(x, 0) = 0
• ∂u/∂t at t = 0 equals g(x) = 3(lx − x²)

Step 1: separation of variables. Put u = X(x)T(t). Then X″ + p²X = 0 and T″ + c²p²T = 0.

The ends being fixed gives X = sin(nπx/l), p = nπ/l (n = 1, 2, …).

T(t) = C cos(nπct/l) + D sin(nπct/l)

Step 2: u(x, 0) = 0 means every cosine coefficient is zero, so the solution has only sine terms in t:

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

Step 3: velocity condition. Differentiate at t = 0:

∂u/∂t |_{t=0} = Σ Bₙ (nπc/l) sin(nπx/l) = 3(lx − x²)

This is a Fourier sine series, so

Bₙ (nπc/l) = (2/l) ∫₀^l 3(lx − x²) sin(nπx/l) dx

Step 4: the integral. Integrate by parts twice:

∫₀^l (lx − x²) sin(nπx/l) dx = 2l³[1 − (−1)ⁿ] / (n³π³)

which equals 4l³/(n³π³) for odd n and 0 for even n.

Bₙ (nπc/l) = (2/l) · 3 · 4l³/(n³π³) = 24 l² /(n³π³)   (n odd)
Bₙ = 24 l³ / (c π⁴ n⁴)                                (n odd)

Answer:

u(x, t) = (24 l³ / (c π⁴)) Σ_{n = 1,3,5,…} (1/n⁴) sin(nπct/l) sin(nπx/l)

or, writing n = 2m − 1: u(x, t) = (24l³/(cπ⁴)) Σ_{m=1}^{∞} sin((2m−1)πct/l) sin((2m−1)πx/l)/(2m−1)⁴.

(Numerical check: l = 2, c = 1.3, x = 0.7: the series for ∂u/∂t at t = 0 sums to 2.7300, equal to 3(lx − x²) = 2.73.)