Heat equation, insulated ends, u(x,0)=lx-x²

Problem. u_t = c² u_xx with (i) u bounded as t → ∞, (ii) u_x = 0 at x = 0 and x = l (ends insulated), (iii) u(x, 0) = lx − x².

Step 1: separate variables. u = X(x)T(t): X″/X = T′/(c²T) = k.

Step 2: choose the sign of k. Condition (i) rules out k > 0, because T = e^{kc²t} would blow up as t → ∞.

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

Step 3: apply (ii) u_x = 0 at both ends.

• k = 0: X′ = B = 0, so X = constant (allowed).
• k = −p²: X′ = −Ap sin px + Bp cos px. X′(0) = 0 gives B = 0. X′(l) = 0 gives A p sin pl = 0, so p = nπ/l.

Step 4: general solution.

u(x, t) = a₀ + Σ_{n=1}^{∞} aₙ cos(nπx/l) e^{−c²n²π²t/l²}

Step 5: initial condition (iii). lx − x² = a₀ + Σ aₙ cos(nπx/l) is a Fourier cosine series on (0, l):

a₀ = (1/l) ∫₀^l (lx − x²) dx = (1/l)(l³/2 − l³/3) = l²/6

aₙ = (2/l) ∫₀^l (lx − x²) cos(nπx/l) dx

Integrating by parts twice: ∫₀^l (lx − x²) cos(nπx/l) dx = −l³[1 + (−1)ⁿ]/(n²π²).

So aₙ = −2l²[1 + (−1)ⁿ]/(n²π²), which is 0 for odd n. For even n = 2m: aₙ = −4l²/(4m²π²) = −l²/(m²π²).

Answer:

u(x, t) = l²/6 − (l²/π²) Σ_{m=1}^{∞} (1/m²) cos(2mπx/l) e^{−4c²m²π²t/l²}

As t → ∞, u → l²/6, the mean initial temperature (no heat escapes through insulated ends).

(Numerical check: l = 2, x = 0.7: the series at t = 0 gives 0.91000 = lx − x² ✓.)