---
title: "Laplace equation for u=1/√(x²+y²+z²)"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:laplace-equation-u-1-sqrt"
source: https://www.bctengineer.com/q/semester_2/mathematics/laplace-equation-u-1-sqrt
site: BCT Engineering
---

# Laplace equation for u=1/√(x²+y²+z²)

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2072 Magh, 5 marks
- Video: [Partial Differentiation | Multivariable Calculus | u = 1/√(x² + y² + z²), ∂²u/∂x²+∂²u/∂y²+∂²u/∂z²= 0](https://www.youtube.com/watch?v=FTqtP4hg3L8) — Mathematics Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/laplace-equation-u-1-sqrt
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

If u = 1/√(x²+y²+z²), prove that ∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² = 0.

## Short answer

Let r = √(x²+y²+z²). Then u = 1/r. Second derivatives: ∂²u/∂x² = (3x²−r²)/r⁵, similarly for y,z. Summing gives (3(x²+y²+z²)−3r²)/r⁵ = 0, so ∇²u=0.

## Answer

Let r = √(x² + y² + z²), so u = 1/r = r⁻¹ and ∂r/∂x = x/r, ∂r/∂y = y/r, ∂r/∂z = z/r.

#### First derivative
∂u/∂x = −r⁻² · ∂r/∂x = −r⁻² · (x/r) = −x r⁻³.

#### Second derivative
```text
∂²u/∂x² = −r⁻³ − x · (−3 r⁻⁴)(x/r)
        = −1/r³ + 3x²/r⁵
        = (3x² − r²)/r⁵
```

By symmetry:

```text
∂²u/∂y² = (3y² − r²)/r⁵
∂²u/∂z² = (3z² − r²)/r⁵
```

#### Add the three

```text
∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² = [3(x² + y² + z²) − 3r²]/r⁵
                              = (3r² − 3r²)/r⁵ = 0
```

So u = 1/√(x² + y² + z²) satisfies Laplace's equation ∇²u = 0. ∎

## Check yourself

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

1. With r = √(x²+y²+z²), what is ∂u/∂x for u = 1/r?
   A. x/r³
   B. −x/r³
   C. −1/r²
   D. −x/r²
   Answer: B. ∂u/∂x = -r⁻²(x/r) = -x r⁻³. (Key point: First derivative)

2. What is ∂²u/∂x²?
   A. (r²−3x²)/r⁵
   B. (3x²−r²)/r⁵
   C. (x²−3r²)/r⁵
   D. −3x²/r⁵
   Answer: B. Differentiating -x r⁻³ gives -1/r³ + 3x²/r⁵ = (3x²−r²)/r⁵. (Key point: Second derivative)

3. Why does the sum of the three second derivatives vanish?
   A. Each term is zero
   B. x, y, z cancel pairwise
   C. r = 0
   D. 3(x²+y²+z²) − 3r² = 0
   Answer: D. Numerator is 3(x²+y²+z²) − 3r² = 3r² − 3r² = 0. (Key point: Sum)

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