---
title: "Derive two-dimensional heat equation"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:heat-eq-2d-derive"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/heat-eq-2d-derive
site: BCT Engineering
---

# Derive two-dimensional heat equation

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2075 Baisakh, 10 marks
- Video: [Concept of Two Dimensional Heat Flow Equation](https://www.youtube.com/watch?v=uMtkvE0bL3w) — Dr Sadik
- 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-2d-derive
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

Derive two dimensional heat equation.

## Short answer

In a thin plate, heat flows in x and y directions. For an element, heat gain through faces equals K h (∂²u/∂x²+∂²u/∂y²) δx δy. Equate to ρσ h δx δy ∂u/∂t to get ∂u/∂t = c² (∂²u/∂x²+∂²u/∂y²) with c² = K/(ρσ).

## Answer

**Setting.** A thin homogeneous rectangular plate (thickness h, density ρ, specific heat σ, thermal conductivity K) has its two faces insulated, so heat flows only in the xy-plane. Let u(x, y, t) be the temperature.

**Law of conduction.** Heat flows from hot to cold at a rate proportional to the temperature gradient: the heat crossing an area A per unit time is −K A ∂u/∂n.

**Element.** Take a small rectangle with sides δx and δy.

```text
   y+δy  ┌──────────────┐
         │              │
         │   element    │
   y     └──────────────┘
        x             x+δx
```

**Heat flow in the x-direction.** Area of the faces = h δy.

- Gain at x: −K h δy (∂u/∂x)ₓ
- Loss at x + δx: −K h δy (∂u/∂x)_{x+δx}

Net gain in x-direction per unit time = K h δy [ (∂u/∂x)_{x+δx} − (∂u/∂x)ₓ ] ≈ K h δy δx ∂²u/∂x²

**Heat flow in the y-direction.** Area of faces = h δx. Similarly

Net gain = K h δx [ (∂u/∂y)_{y+δy} − (∂u/∂y)_y ] ≈ K h δx δy ∂²u/∂y²

**Total net gain** = K h δx δy (∂²u/∂x² + ∂²u/∂y²)

**Heat absorbed.** The element has mass ρ h δx δy, and its temperature changes at rate ∂u/∂t:

heat absorbed per unit time = σ ρ h δx δy ∂u/∂t

**Equate:**

σρ h δx δy ∂u/∂t = K h δx δy (u_xx + u_yy)

Cancel h δx δy and put c² = K/(σρ), the diffusivity:

**∂u/∂t = c² (∂²u/∂x² + ∂²u/∂y²)**

This is the two-dimensional heat equation. In steady state (∂u/∂t = 0) it reduces to Laplace's equation u_xx + u_yy = 0.

## Check yourself

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

1. Which law is used for the heat crossing an area A per unit time in the derivation?
   A. Heat crossed = -K A du/dn (proportional to temperature gradient, flowing hot to cold)
   B. Heat crossed = K A u (proportional to temperature itself)
   C. Heat crossed = -K A d2u/dn2
   D. Heat crossed = K A du/dt
   Answer: A. Heat flow is proportional to the temperature gradient, with a minus sign because heat flows from hot to cold. (Key point: Conduction law)

2. For flow in the x-direction through the element, what is the area of each face?
   A. h dx
   B. h dy
   C. dx dy
   D. h dx dy
   Answer: B. Flow along x crosses faces of thickness h and width dy, so the area is h dy. (Key point: Face areas)

3. What gives the net heat gain in the x-direction as dx becomes small?
   A. The sum of du/dx at x and x+dx, giving K h u_x
   B. The difference in u itself, giving K h dy u
   C. The difference of du/dx at x+dx and x, which becomes K h dx dy u_xx
   D. Only the heat entering at x, giving -K h dy u_x
   Answer: C. The net gain is the difference of the gradients at the two faces, which tends to the second derivative u_xx times dx. (Key point: Net gain)

4. Which equation is equated to get the heat equation?
   A. sigma rho h dx dy u_t = K h dx dy (u_x + u_y)
   B. K u_t = sigma rho (u_xx+u_yy)
   C. sigma rho u_xx = K u_t
   D. Heat absorbed (sigma rho h dx dy u_t) = net heat gained (K h dx dy (u_xx+u_yy))
   Answer: D. The heat that raises the temperature of the element equals the net heat flowing in from both directions. (Key point: Equating heat)

## Close variants set in other exams

- [Derive one-dimensional heat equation](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Aheat-eq-derive) — Applied Mathematics, chapter 3, asked 5×

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- [Derive 1D wave equation and solve](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Awave-eq-derive-and-solve) — Applied Mathematics, chapter 3, asked 5×

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