---
title: "Laplace equation in polar form"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:laplace-eq-polar-form"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/laplace-eq-polar-form
site: BCT Engineering
---

# Laplace equation in polar form

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2069 Poush; 2068 Magh, 10 marks (asked 2 times)
- Video: [The Laplacian in Polar Coordinates](https://www.youtube.com/watch?v=wlbZxa66ncg) — Mike, the Mathematician
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/laplace-eq-polar-form
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

Change/deduce the two dimensional Laplace equation ∂²u/∂x² + ∂²u/∂y² = 0 into polar form.

## Short answer

Use x = r cos θ, y = r sin θ and chain rule. Compute u_xx + u_yy = u_rr + (1/r)u_r + (1/r²)u_θθ. Thus the polar form is ∂²u/∂r² + (1/r)∂u/∂r + (1/r²)∂²u/∂θ² = 0.

## Answer

**Aim.** Transform u_xx + u_yy = 0 using x = r cos θ, y = r sin θ.

**Step 1: relations.**

r² = x² + y², θ = tan⁻¹(y/x)

```text
∂r/∂x = x/r = cos θ        ∂r/∂y = y/r = sin θ
∂θ/∂x = −y/r² = −sin θ/r   ∂θ/∂y = x/r² = cos θ/r
```

**Step 2: first derivatives (chain rule), u = u(r, θ).**

```text
u_x = u_r cos θ − u_θ sin θ / r
u_y = u_r sin θ + u_θ cos θ / r
```

**Step 3: second derivatives.** Apply the same operator ∂/∂x = cos θ ∂/∂r − (sin θ / r) ∂/∂θ to u_x:

```text
u_xx = cos θ ∂/∂r (u_r cos θ − u_θ sin θ / r)
       − (sin θ / r) ∂/∂θ (u_r cos θ − u_θ sin θ / r)

     = cos²θ u_rr − 2 sinθ cosθ ( u_rθ / r − u_θ / r² )
       + (sin²θ / r²) u_θθ + (sin²θ / r) u_r
```

Similarly, with ∂/∂y = sin θ ∂/∂r + (cos θ / r) ∂/∂θ:

```text
u_yy = sin²θ u_rr + 2 sinθ cosθ ( u_rθ / r − u_θ / r² )
       + (cos²θ / r²) u_θθ + (cos²θ / r) u_r
```

**Step 4: add.** The cross terms cancel and sin²θ + cos²θ = 1:

u_xx + u_yy = u_rr + (1/r) u_r + (1/r²) u_θθ

**Answer:** Laplace's equation in polar form is

**∂²u/∂r² + (1/r) ∂u/∂r + (1/r²) ∂²u/∂θ² = 0**

or equivalently r²u_rr + r u_r + u_θθ = 0.

## Check yourself

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

1. partial theta / partial x =
   A. -sin theta / r
   B. cos theta
   C. sin theta/r
   D. -y/r
   Answer: A. -y/r^2. (Key point: Relations)

2. u_x =
   A. u_r cos theta + u_theta sin theta/r
   B. u_r cos theta - u_theta sin theta/r
   C. u_r sin theta
   D. u_r cos theta
   Answer: B. Chain rule. (Key point: Chain rule)

3. Adding u_xx and u_yy,
   A. cross terms double
   B. all vanish
   C. cross terms cancel and sin^2+cos^2=1
   D. r terms cancel
   Answer: C. Equal and opposite. (Key point: Cross terms)

4. Polar form is
   A. u_rr + u_theta theta = 0
   B. u_rr + r u_r + u_theta theta = 0
   C. u_rr + u_r + u_theta theta/r = 0
   D. u_rr + u_r/r + u_theta theta/r^2 = 0
   Answer: D. Standard. (Key point: Result)

## Related in Applied Mathematics

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