---
title: "Rod with ends at 0°C, u(x,0)=3sin(πx/l)"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:heat-eq-rod-0c-sin"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/heat-eq-rod-0c-sin
site: BCT Engineering
---

# Rod with ends at 0°C, u(x,0)=3sin(πx/l)

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2074 Bhadra, 10 marks
- Video: [Heat Equation | Separation of Variables Method in PDE | Example & Concepts by GP Sir](https://www.youtube.com/watch?v=NGz3Qo0rGiI) — Dr.Gajendra Purohit
- 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-0c-sin
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

A uniform rod of length l has its end maintained at a temperature 0°C and the initial temperature of the rod is: u(x, 0) = 3 sin(πx/l) for 0 < x < l. Find the temperature u(x, t).

## Short answer

The initial condition matches a single sine mode. By separation of variables, u(x,t) = 3 sin(πx/l) e^{−c²π²t/l²}. Temperature decays exponentially to zero.

## Answer

**Problem.** u_t = c² u_xx on 0 < x < l with both ends at 0°C: u(0, t) = u(l, t) = 0, and u(x, 0) = 3 sin(πx/l).

**Step 1: general solution** (separation of variables, ends at 0°C):

u(x, t) = Σ_{n=1}^{∞} Bₙ sin(nπx/l) e^{−c²n²π²t/l²}

**Step 2: initial condition.**

3 sin(πx/l) = Σ Bₙ sin(nπx/l)

Comparing both sides directly (no integration is needed because the initial profile is already a single sine mode): B₁ = 3 and Bₙ = 0 for n ≥ 2.

**Answer:**

**u(x, t) = 3 sin(πx/l) e^{−c²π²t/l²}**

Check: u(0, t) = u(l, t) = 0 ✓, u(x, 0) = 3 sin(πx/l) ✓, and u_t = −(c²π²/l²) u = c² u_xx ✓. The temperature decays exponentially to 0°C.

## Check yourself

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

1. Why does the general solution contain only sine terms?
   A. Both ends are held at 0 degrees, so u(0,t)=u(l,t)=0 leaves sin(n pi x/l)
   B. The rod is insulated at both ends
   C. The initial temperature is a sine
   D. Heat flow is proportional to sine of position
   Answer: A. u(0,t)=0 kills cosine and u(l,t)=0 gives p = n pi/l. (Key point: General solution form)

2. How are the coefficients found here?
   A. By integrating 3 sin(pi x/l) x from 0 to l for every n
   B. By comparing both sides, since the initial profile is already a single sine mode: B1 = 3, others 0
   C. By setting t = 0 and x = 0
   D. By assuming B_n = 3 for all n
   Answer: B. Orthogonality of sine modes means only B1 is nonzero. (Key point: Finding coefficients)

3. What is u(x,t)?
   A. 3 sin(pi x/l) e^{-c^2 t/l^2}
   B. 3 sin(pi x/l) e^{c^2 pi^2 t/l^2}
   C. 3 sin(pi x/l) e^{-c^2 pi^2 t/l^2}
   D. 3 sin(pi x/l) cos(c pi t/l)
   Answer: C. The n = 1 mode has decay exponent c^2 pi^2 t/l^2. (Key point: Final answer)

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