---
title: "Rod ends changed from 0/100 to 15/85"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:heat-eq-rod-steady-state-100"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/heat-eq-rod-steady-state-100
site: BCT Engineering
---

# Rod ends changed from 0/100 to 15/85

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2070 Magh, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/heat-eq-rod-steady-state-100
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

A rod of length L has its ends A and B maintained at 0° and 100° respectively until steady state prevails. If the changes are made by reducing the temperature of end B to 85° and increasing that of end A to 15°, then find the temperature distribution in the rod at a time t.

## Short answer

Find initial steady state 100x/L, new steady state 15+70x/L, then solve transient v with zero ends. The solution is u(x,t) = 15 + 70x/L − (30/π) Σ_{m=1}∞ (1/m) sin(2mπx/L) e^{−4c²m²π²t/L²}.

## Answer

**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}

```text
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 ✓.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the initial steady state u(x,0) = 100x/L?
   A. Because u_t = 100 in steady state
   B. Because u is sine-shaped in steady state
   C. Because the ends were at the same temperature
   D. In steady state u_xx = 0, so u is linear and fits 0 at x=0 and 100 at x=L
   Answer: D. Steady state means u_t = 0, so u_xx = 0 and u is a straight line. (Key point: Initial steady state)

2. What is the new steady state?
   A. 15 + 70x/L
   B. 85 + 15x/L
   C. 15 + 85x/L
   D. 100x/L
   Answer: A. A linear profile from 15 at x=0 to 85 at x=L gives 15 + (85-15)x/L. (Key point: New steady state)

3. What is v(x,0) for the transient part v = u - us?
   A. 100x/L, with v = 0 at both ends
   B. 100x/L - (15 + 70x/L) = 30x/L - 15, with v = 0 at both ends
   C. 30x/L - 15, with v(0)=15 and v(L)=85
   D. 15 + 70x/L, with v = 0 at both ends
   Answer: B. Subtracting the new steady state makes the end conditions homogeneous. (Key point: Transient initial condition)

4. Which statement about the result is correct?
   A. Only odd n terms survive, and u tends to 100x/L
   B. All n terms survive and u tends to 0
   C. Only even n terms survive, and u tends to 15 + 70x/L as t tends to infinity
   D. Only even n terms survive and u tends to 50
   Answer: C. b_n = -30[1+(-1)^n]/(n pi) vanishes for odd n; the transient decays leaving the new steady state. (Key point: Final behaviour)

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