Heat equation, insulated ends, u(x,0)=x

Problem. u_t = c² u_xx on 0 < x < L, with u_x(0, t) = u_x(L, t) = 0 and u(x, 0) = x.

Step 1: separate variables. u = X(x)T(t): X″/X = T′/(c²T) = k. For u to stay bounded as t → ∞, k ≤ 0.

• k = 0: X = A + Bx, T = constant
• k = −p²: X = A cos px + B sin px, T = e^{−c²p²t}

Step 2: insulated ends.

• k = 0: X′ = B = 0, so X = constant (accepted).
• k = −p²: X′(0) = 0 gives B = 0; X′(L) = 0 gives sin pL = 0, so p = nπ/L.

Step 3: general solution.

u(x, t) = a₀ + Σ_{n=1}^{∞} aₙ cos(nπx/L) e^{−c²n²π²t/L²}

Step 4: initial condition u(x, 0) = x (cosine series on (0, L)):

a₀ = (1/L) ∫₀^L x dx = L/2

aₙ = (2/L) ∫₀^L x cos(nπx/L) dx
   = (2/L) [ x (L/nπ) sin(nπx/L) + (L/nπ)² cos(nπx/L) ]₀^L
   = (2/L) (L²/n²π²) [ (−1)ⁿ − 1 ]
   = 2L [ (−1)ⁿ − 1 ] / (n²π²)

So aₙ = −4L/(n²π²) for odd n, and aₙ = 0 for even n.

Answer:

u(x, t) = L/2 − (4L/π²) Σ_{n = 1,3,5,…} (1/n²) cos(nπx/L) e^{−c²n²π²t/L²}

As t → ∞, u → L/2, the average of the initial temperature.

(Numerical check: L = 2, x = 0.7: the series at t = 0 gives 0.70000 ✓.)