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Solve one dimensional heat equation

Short answer

Separation of variables yields u(x,t) = Σ Bₙ sin(nπx/ℓ) e^(−n²π²t/ℓ²). Applying u(x,0)=u₀ sin(πx/ℓ) gives B₁=u₀ and all other Bₙ=0, so u(x,t) = u₀ e^(−π²t/ℓ²) sin(πx/ℓ).

Solve the one dimensional heat equation ∂u/∂t = ∂²u/∂x² subject to u(0, t) = u(ℓ, t) = 0 and u(x, 0) = u₀ sin(πx/ℓ).

Let u = X(x)T(t). From ∂u/∂t = ∂²u/∂x²: X T′ = X″T, so X″/X = T′/T = −p² (a negative constant, so that the temperature decays).

X″ + p²X = 0   →  X = A cos px + B sin px
T′ + p²T = 0   →  T = C e^(−p²t)
u = (A cos px + B sin px) e^(−p²t)

Boundary conditions

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

Hence u = Σ Bₙ sin(nπx/ℓ) e^(−n²π²t/ℓ²).

Initial condition. u(x, 0) = u₀ sin(πx/ℓ) = Σ Bₙ sin(nπx/ℓ). Comparing both sides, B₁ = u₀ and Bₙ = 0 for n ≥ 2.

Answer:

u(x, t) = u₀ e^(−π²t/ℓ²) sin(πx/ℓ)

✓ Answer checked by Claude Opus 5.5 (high effort)

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