---
title: "Derive one-dimensional heat equation"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:heat-eq-derive"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/heat-eq-derive
site: BCT Engineering
---

# Derive one-dimensional heat equation

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2071 Bhadra
- Video: [Derivation of one dimensional heat equation -  | VTU Mathematics | Module 3](https://www.youtube.com/watch?v=06N2o3U7z_c) — Dr Dhananjaya M
- 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-derive
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

Derive one-dimensional heat equation.

## Short answer

Apply Fourier's law of conduction and conservation of energy to a thin slice of the rod. Equate net heat inflow to heat absorbed, take limit as slice thickness goes to zero, obtaining ∂u/∂t = c² ∂²u/∂x² with c² = K/(σρ).

## Answer

**Setting.** A thin homogeneous bar of uniform cross-section A has its lateral surface insulated, so heat flows only along its length (the x-axis). Let u(x, t) be the temperature; ρ = density, σ = specific heat, K = thermal conductivity (all constants).

**Law of conduction.** The rate of heat flow across a cross-section in the +x direction is −K A ∂u/∂x. (The minus sign shows that heat flows from hot to cold.)

**Element.** Consider the slice between x and x + δx.

```text
  heat in →  ┌────────────┐  → heat out
   at x      │  mass      │    at x+δx
             │  ρ A δx    │
             └────────────┘
```

- Heat entering at x per unit time: Q₁ = −K A (∂u/∂x)ₓ
- Heat leaving at x + δx per unit time: Q₂ = −K A (∂u/∂x)_{x+δx}

Net heat gained per unit time:

Q₁ − Q₂ = K A [ (∂u/∂x)_{x+δx} − (∂u/∂x)ₓ ]

**Heat absorbed.** To warm a mass ρ A δx at the rate ∂u/∂t requires

σ ρ A δx ∂u/∂t

**Equating:**

```text
σρ A δx ∂u/∂t = K A [ (∂u/∂x)_{x+δx} − (∂u/∂x)ₓ ]

∂u/∂t = (K/σρ) [ (∂u/∂x)_{x+δx} − (∂u/∂x)ₓ ] / δx
```

As δx → 0, the bracket divided by δx tends to ∂²u/∂x². Writing c² = K/(σρ) (the diffusivity of the material):

**∂u/∂t = c² ∂²u/∂x²**

This is the one-dimensional heat (diffusion) equation.

## Check yourself

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

1. What is the rate of heat flow across a cross-section in the +x direction?
   A. -K A du/dx
   B. K A u
   C. -K A d2u/dx2
   D. K A du/dt
   Answer: A. Heat flow is proportional to the temperature gradient; the minus sign means heat flows from hot to cold. (Key point: Heat flow rate)

2. How much heat per unit time is needed to warm the slice of mass rho A dx at rate du/dt?
   A. K A dx du/dt
   B. sigma rho A dx du/dt
   C. rho A dx u
   D. sigma K A dx du/dx
   Answer: B. Heat = mass x specific heat x rate of temperature change. (Key point: Heat absorbed)

3. How does the term d2u/dx2 appear in the derivation?
   A. It is assumed from the start
   B. (u at x+dx - u at x)/dx tends to u_xx
   C. [(u_x at x+dx) - (u_x at x)]/dx tends to u_xx as dx tends to 0
   D. The slice mass is divided by dx
   Answer: C. The difference quotient of the gradient is the definition of the second derivative. (Key point: Limit step)

4. What does c^2 stand for in u_t = c^2 u_xx?
   A. sigma rho/K
   B. K sigma rho
   C. K/A
   D. K/(sigma rho)
   Answer: D. Dividing by sigma rho A dx leaves c^2 = K/(sigma rho), the diffusivity. (Key point: Diffusivity)

## Close variants set in other exams

- [Derive two-dimensional heat equation](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Aheat-eq-2d-derive) — Applied Mathematics, chapter 3, asked 2×
- [Derive 1D heat equation and solve](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Aheat-eq-derive-and-solve) — Applied Mathematics, chapter 3, asked 4×

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