String released from sin³(πx/l)

Problem. u_tt = c² u_xx (here y is the displacement), with y(0, t) = y(l, t) = 0, y(x, 0) = y₀ sin³(πx/l), and zero initial velocity.

Step 1: general solution (ends fixed). Separation of variables gives

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

Step 2: zero initial velocity gives ∂y/∂t = 0 at t = 0, so every Bₙ* (nπc/l) = 0, hence Bₙ* = 0.

y(x, t) = Σ Bₙ cos(nπct/l) sin(nπx/l)

Step 3: initial shape. Use sin 3θ = 3 sin θ − 4 sin³θ, so sin³θ = (3 sin θ − sin 3θ)/4. With θ = πx/l:

y(x, 0) = Σ Bₙ sin(nπx/l)
        = (y₀/4) [ 3 sin(πx/l) − sin(3πx/l) ]

Compare coefficients (the sine functions are orthogonal):

B₁ = 3y₀/4, B₃ = −y₀/4, all other Bₙ = 0.

Answer:

y(x, t) = (y₀/4) [ 3 cos(πct/l) sin(πx/l) − cos(3πct/l) sin(3πx/l) ]

Check: at t = 0 this equals (y₀/4)(3 sin θ − sin 3θ) = y₀ sin³θ ✓.