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Divergence and curl identity

Short answer

Divergence: ∇·V = ∂V₁/∂x + ∂V₂/∂y + ∂V₃/∂z. Curl: ∇×V = (∂V₃/∂y−∂V₂/∂z)i + (∂V₁/∂z−∂V₃/∂x)j + (∂V₂/∂x−∂V₁/∂y)k. Then div(curl V) = ∂²V₃/∂x∂y − ∂²V₂/∂x∂z + … = 0 because mixed partial derivatives are equal.

Define divergence and curl of V. Prove that div(Curl V) = 0.

Divergence

For a vector function V = V₁ i + V₂ j + V₃ k, the divergence is the scalar

div V = ∇·V = ∂V₁/∂x + ∂V₂/∂y + ∂V₃/∂z.

It measures the net outflow of the field per unit volume at a point.

Curl

curl V = ∇ × V = | i    j    k  |
                 | ∂x   ∂y   ∂z |
                 | V₁   V₂   V₃ |
       = (∂V₃/∂y − ∂V₂/∂z) i + (∂V₁/∂z − ∂V₃/∂x) j + (∂V₂/∂x − ∂V₁/∂y) k

It measures the rotation of the field at a point.

Proof that div(curl V) = 0

Let W = curl V, with components W₁ = ∂V₃/∂y − ∂V₂/∂z, W₂ = ∂V₁/∂z − ∂V₃/∂x, W₃ = ∂V₂/∂x − ∂V₁/∂y.

div W = ∂W₁/∂x + ∂W₂/∂y + ∂W₃/∂z
      = ∂²V₃/∂x∂y − ∂²V₂/∂x∂z
      + ∂²V₁/∂y∂z − ∂²V₃/∂y∂x
      + ∂²V₂/∂z∂x − ∂²V₁/∂z∂y

If V has continuous second partial derivatives, the mixed partials are equal (∂²V₃/∂x∂y = ∂²V₃/∂y∂x, etc.). Each pair cancels:

(V₃: xy − yx) + (V₂: zx − xz) + (V₁: yz − zy) = 0

Hence div(curl V) = ∇·(∇ × V) = 0. (Also, ∇·(∇ × V) is the scalar triple product [∇, ∇, V], which vanishes because two rows are identical.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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