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Solve ∂²u/∂x² = ∂u/∂t with u = 0 at ends

Short answer

Separation of variables with k=−p² gives u(x,t) = Σ_{n=1}∞ B_n sin(nπx/l) e^{−n²π²t/l²}. The constants B_n are determined by the initial condition via Fourier sine series.

Solve ∂²u/∂x² = ∂u/∂t given that u = 0 as t → ∞ as well as u = 0 at x = 0 and x = l.

Problem. ∂²u/∂x² = ∂u/∂t (so c² = 1), with u = 0 as t → ∞, and u = 0 at x = 0 and x = l.

Step 1: separate variables. u = X(x)T(t): X″/X = T′/T = k.

Step 2: sign of k. The condition u → 0 as t → ∞ needs T → 0, so k = −p² < 0 (for k > 0, T = e^{kt} grows; for k = 0, the end conditions force u ≡ 0).

X″ + p² X = 0     →  X = A cos px + B sin px
T′ + p² T = 0     →  T = C e^{−p²t}

Step 3: end conditions.

  • u(0, t) = 0 gives A = 0.
  • u(l, t) = 0 gives B sin pl = 0, so p = nπ/l.

Step 4: general solution by superposition:

u(x, t) = Σ_{n=1}^{∞} Bₙ sin(nπx/l) e^{−n²π²t/l²}

If a heat equation with diffusivity c² is intended, replace t by c²t in the exponent.

Step 5: constants. The initial condition is not given. If u(x, 0) = f(x), then

Bₙ = (2/l) ∫₀^l f(x) sin(nπx/l) dx

(For example, with f(x) = sin(πx/l) only B₁ = 1 survives: u = sin(πx/l) e^{−π²t/l²}.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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