Prove the functions are unique in a volume, vector calculus problem

[moonkey]

Homework Statement

In a volume V, enclosed by a surface S, the vector fields X and Y satisfy the coupled equations

∇×∇×X=X+Y

∇×∇×Y=Y−X

If the values of ∇×X and ∇×Y are given on S, show that X and Y are unique in V.

Homework Equations

∇.(A×B)=B.(∇×A)−A.(∇×B)

∇×(A×B)=∇(∇.A)−∇²A

The Attempt at a Solution

I am assuming that I need to show that ∇²X and ∇²Y are equal to zero to satisfy the uniqueness theorem for Poisson's equation. But am unsure of a good way to get there, so before I write my scribbles if someone could point me in the right direction it would be great.

 

[eoinmurray]

We need to prove two things, first that ∇ ^2 X = 0 and ∇ ^2 Y = 0, and also that X and Y are zero on S.

It is straight forward to show that

2X = ∇×∇×X - ∇×∇×Y
2Y = ∇×∇×X + ∇×∇×Y

Thus from the second identity above

2X = ∇ (∇.X) - ∇^2 X - ∇ (∇.Y) + ∇^2
2Y = ∇(∇.X) - ∇^2 X + ∇ (∇.Y) - ∇^2 Y

Can we somehow now show from this using the divergence theorem that the above conditions are satisfied?