Rod ends changed from 0/100 to 15/85

Idea. Find the initial steady state, the new steady state, and then solve for the transient part that decays.

Step 1: initial steady state (ends 0° and 100°). In steady state u_xx = 0, so u is linear: u(x, 0) = 100x/L.

Step 2: new boundary conditions u(0, t) = 15 and u(L, t) = 85 for t > 0. The new steady state is linear:

uₛ(x) = 15 + (85 − 15)x/L = 15 + 70x/L

Step 3: transient part. Write u = uₛ(x) + v(x, t). Then v satisfies v_t = c² v_xx with

• v(0, t) = v(L, t) = 0
• v(x, 0) = 100x/L − (15 + 70x/L) = 30x/L − 15

Step 4: solve for v.

v(x, t) = Σ_{n=1}^{∞} bₙ sin(nπx/L) e^{−c²n²π²t/L²}

bₙ = (2/L) ∫₀^L (30x/L − 15) sin(nπx/L) dx

Evaluate the two parts:

• (2/L)(30/L) ∫₀^L x sin(nπx/L) dx = (60/L²)(−L²(−1)ⁿ/(nπ)) = −60(−1)ⁿ/(nπ)
• (2/L)(−15) ∫₀^L sin(nπx/L) dx = −30[1 − (−1)ⁿ]/(nπ)

Add: bₙ = −30[1 + (−1)ⁿ]/(nπ). This is 0 for odd n, and −60/(nπ) for even n. Writing n = 2m: b₂ₘ = −30/(mπ).

Answer:

u(x, t) = 15 + 70x/L − (30/π) Σ_{m=1}^{∞} (1/m) sin(2mπx/L) e^{−4c²m²π²t/L²}

As t → ∞, u → 15 + 70x/L, the new steady state.

(Numerical check, L = 2, x = 0.7: the series at t = 0 gives 35.0005 ≈ 100x/L = 35 ✓.)