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

I am working through the following problem, but finding it hard to know where to go.

Using the Divergence theorem and the following identities

$\nabla \cdot (A \times B) = B\cdot(\nabla \times A) - A\cdot(\nabla \times B)$

$\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla ^2 A$

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

$\nabla \times \nabla \times X = X+Y$

$\nabla \times \nabla \times Y = Y-X$

If the values of $\nabla \times X$ and $\nabla \times Y$ are given on S, show that X and Y are unique in V.

I am assuming that I need to show that $\nabla ^2 X$ and $\nabla ^2 Y$ are equal to zero and that X and Y are zero on S 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 write direction it would be great.

Any help, pointing in the right direction would be very helpful.

EDIT: Fixed the second expression, original didn't make sense.

-

With this I am not even sure what the question is
but here is the answer I give to prove $X \neq Y$:

Sum the two equations $$2Y = \nabla \times \nabla \times X + \nabla \times \nabla \times Y$$ Subtract the two equations $$2X = \nabla \times \nabla \times X - \nabla \times \nabla \times Y$$

Using $\nabla \times \nabla \times A = \nabla(\nabla \cdot A) - \nabla^2 A$ \begin{align} 2Y &= \nabla(\nabla \cdot X) − \nabla ^2 X + \nabla(\nabla \cdot Y) − \nabla ^2 Y\\ 2Y &= \nabla\big(\nabla \cdot(X+Y)\big) − \nabla ^2 (X+Y) \end{align} and then resub $\nabla \times \nabla \times X = X+Y$ $$2Y = \nabla\big(\nabla \cdot (\nabla \times \nabla \times X)\big)− \nabla ^2 (\nabla \times \nabla \times X)$$ similarly $$-2X = \nabla\big(\nabla \cdot (\nabla \times \nabla \times Y)\big) − \nabla ^2 (\nabla \times \nabla \times Y)$$

Therefore if $X=Y$ $$\nabla \times X = -\nabla \times Y$$ which implys $Y=0$ and $X=0$.

Therefore $X = Y$ only in trivial case.

-
The question says nothing about whether $X=Y$. And of course if $X=Y$, from the right sides of the two equations we'd have $X+Y=Y-X$, which implies $X=0$. – Robert Israel Nov 13 '12 at 22:34
$\large X + Y = \nabla{\rm f}$. – Felix Marin Sep 22 '13 at 1:43