Laplace equation for u=1/√(x²+y²+z²)

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.

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

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

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

∂²u/∂x² = −r⁻³ − x · (−3 r⁻⁴)(x/r)
        = −1/r³ + 3x²/r⁵
        = (3x² − r²)/r⁵

By symmetry:

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

∂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. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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