Proof of $ \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)} - k^2 \mathbf{A} = \mathbf{0}$

Question

I have this exercise and I don't know how to do it.

These are the instructions:

Show that if the vector A satisfy the equation:

1.$ \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)} - k^2 \mathbf{A} = \mathbf{0}$

It will also satisfy:

2. $ \nabla^2 \mathbf{A} + k^2 \mathbf{A} = 0$

and

3. $ \nabla \cdot \mathbf{A} = 0$

Notice that $k$ is a constant.

Hint: apply $ \nabla \cdot $ to the first equation.

So I tried to apply the $ \nabla \cdot $ to the first equation and I had this:

$ \nabla\cdot \mathbf{(} \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)}- k^2 \mathbf{A} \mathbf{)}$

$\delta_{ij}$ $\partial_i \mathbf{(} \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)}- k^2 \mathbf{A} \mathbf{)}_j$

$\partial_i \epsilon_{ijk} \partial_j (\nabla \times \mathbf{A})_k - \partial_ik^2 \mathbf{A}_i$

$\epsilon_{ijk} \epsilon_{klm}\partial_i \partial_j \partial_l \mathbf{A}_m - \partial_ik^2 \mathbf{A}_i$

$(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})$$(\partial_i \partial_j \partial_l \mathbf{A}_m) - \partial_ik^2 \mathbf{A}_i$

$(\delta_{il}\delta_{jm} \partial_i \partial_j \partial_l \mathbf{A}_m-\delta_{im}\delta_{jl}\partial_i \partial_j \partial_l \mathbf{A}_m) - \partial_ik^2 \mathbf{A}_i$

$(\partial_i \partial_j \partial_i \mathbf{A}_j - \partial_i \partial_j \partial_j \mathbf{A}_i) - \partial_ik^2 \mathbf{A}_i$

So from here I just don't know how to change the index notation to vector notation, also, I can't see how this satisfies the equations 2 and 3. So please let me know if I made a mistake in any of the previous steps and help me telling me what I should do next.

Thank you very much.

Answer 1

For the first one, it holds for all vector $\mathbf{X}$ that $$\nabla\cdot (\nabla \times \mathbf{X})=0$$ Applying this to your equation, we get $$0=\nabla\cdot 0=\nabla\cdot[\nabla \times(\nabla \times \mathbf{A})]-k^2\nabla \cdot\mathbf{A}=0-k^2\nabla \cdot\mathbf{A},$$ so if $k\neq 0$ then you have that $\nabla\cdot \mathbf{A}=0$.

Now, another well-known identity is that $$\nabla \times (\nabla\times \mathbf{X})=\nabla(\nabla\cdot \mathbf{X})-\nabla^2\mathbf{X}.$$ For our case, and provided that $\nabla\cdot \mathbf{A}=0$, then $$0=\nabla \times (\nabla\times \mathbf{A})-k^2\mathbf{A}=\nabla(\nabla\cdot \mathbf{A})-\nabla^2\mathbf{A}-k^2\mathbf{A}=0-\nabla^2\mathbf{A}-k^2\mathbf{A}$$ and you get that $$\nabla^2\mathbf{A}+k^2\mathbf{A}=0.$$