---
title: "Heat equation, insulated ends, u(x,0)=x"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:heat-eq-insulated-ends-u-x"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/heat-eq-insulated-ends-u-x
site: BCT Engineering
---

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

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2070 Bhadra, 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-insulated-ends-u-x
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

Solve one dimensional heat equation ∂u/∂t = c² ∂²u/∂x² under the boundary condition ∂u/∂x = 0 when x = 0 and x = L and initial condition u(x, 0) = x for 0 < x < L.

## Short answer

Separate variables and use insulated boundary conditions to get a cosine series. The solution is u(x,t) = L/2 − (4L/π²) Σ_{n odd} (1/n²) cos(nπx/L) e^{−c²n²π²t/L²}. As t→∞, u→L/2.

## Answer

**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)):

```text
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 ✓.)

## Check yourself

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

1. Why is the separation constant taken as k <= 0?
   A. So that the solution stays bounded as t tends to infinity
   B. So that X is oscillatory at t = 0
   C. So that the initial condition is satisfied
   D. Because heat flows from hot to cold in the x direction
   Answer: A. For k > 0 the factor e^{c^2 k t} grows without bound. (Key point: Sign of separation constant)

2. What do the conditions u_x = 0 at x = 0 and x = L give?
   A. A = 0 and cos pL = 0, leaving sine terms
   B. B = 0 and sin pL = 0, so p = n pi/L, leaving cosine terms
   C. B = 0 and p = (2n-1) pi/(2L)
   D. A = B = 0, so u = 0
   Answer: B. X' = -Ap sin px + Bp cos px; X'(0)=0 gives B=0 and X'(L)=0 gives sin pL=0. (Key point: Insulated ends)

3. Which series is used to fit u(x,0) = x?
   A. A Fourier sine series on (0, L)
   B. A Taylor series about x = 0
   C. A Fourier cosine series on (0, L)
   D. A full series with both sine and cosine terms and a 2L period
   Answer: C. The solution has cos(n pi x/L) terms, so the initial data is expanded as a cosine series. (Key point: Series type)

4. What is a0 and what is the temperature as t tends to infinity?
   A. a0 = L, u tends to L
   B. a0 = 0, u tends to 0
   C. a0 = L/2, but u tends to 0
   D. a0 = L/2, the average of the initial temperature; u tends to L/2
   Answer: D. With insulated ends no heat escapes, so the temperature tends to the mean of x over (0, L), which is L/2. (Key point: Constant term)

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