Vector and Scalar Helmholtz equation

Question

This is closely related to this recent question

The vector Helmholtz equation is

\begin{align} (\nabla^2 + k^2)\boldsymbol{u} = 0 \end{align}

The scalar Helmholtz equation is

\begin{align} (\nabla^2 + k^2)u = 0 \end{align}

In the linked question and reference therein (Reitz, Milford "Foundations of Electromagnetic Theory") it is claimed that if $u$ satisfies the scalar Helmholtz equation then

\begin{align} \boldsymbol{u}_1 =& \boldsymbol{r}\times \nabla u\\ \boldsymbol{u}_2 =& \nabla \times\left(\boldsymbol{r} \times \nabla u\right) \end{align}

both satisfy the vector Helmholtz equation. I have confirmed this as follows (though I would appreciate a check for correctness)

\begin{align} \left[\left(\nabla^2 + k^2\right)\boldsymbol{u}_1\right]_i =& (\partial_j\partial_j+k^2) \epsilon_{ikl} r_k\partial_lu\\ =&\epsilon_{ijk}r_k\partial_l(\partial_j\partial_j + k^2)u\\ =& 0 \end{align}

and pretty similarly for $\boldsymbol{u}_2$

\begin{align} \left[(\nabla^2 + k^2)\boldsymbol{u}_2\right]_i =& (\partial_j\partial_j + k^2)\epsilon_{ikl}\partial_k \epsilon_{lmn}r_m\partial_nu\\ =& \epsilon_{ikl}\epsilon_{lmn}\partial_kr_m\partial_n(\partial_j\partial_j+k^2)u\\ =& 0 \end{align}

In index notation the proof of the claim seems pretty obvious. In fact, it looks like pretty much any vector quantity derived from $u$ would do the trick.

For example $\boldsymbol{u}_3 = \nabla u$.

\begin{align} \left[(\nabla^2 + k^2)\boldsymbol{u}_3\right]_i =& (\partial_j\partial_j + k^2)\partial_i u\\ =& \partial_i (\partial_j\partial_j + k^2) u\\ =& 0 \end{align}

It seems $\boldsymbol{u}_4 = \boldsymbol{r}u$ would also work.

Here then are my questions:

1) Is it correct that if $u$ satisfies the scalar Helmholtz equation that $\nabla u$ and $\boldsymbol{r}u$ satisfy the vector Helmholtz equation?

2) Are my proofs that $\boldsymbol{u}_1, \boldsymbol{u}_2$ and $\boldsymbol{u}_3$ satisfy the vector Helmholtz equation valid?

3) If the answer to question 1) is yes then why are $\boldsymbol{u}_1$ and $\boldsymbol{u}_2$ natural choices to extend complete families of solutions of the scalar Helmholtz equation to complete families of solutions of the vector Helmholtz equation? Or would other choices be natural and $\boldsymbol{u}_1$ and $\boldsymbol{u}_2$ just happen to be the choices that Reitz and Milford use?

Answer 1

Good question. The answer for question (1) is no. Because $$\nabla\cdot\mathbf{u_{3,4}}\neq 0.$$ They do not satisfy the boundary condition of the electromagnetic wave equation, i.e., $\nabla\cdot\mathbf{E}\neq 0$. That is the reason we have to keep the relation between the scalar solution and the vector solution such that, $\mathbf{u}=\nabla\times(\mathbf{c}u)$, such that $\nabla\cdot(\nabla \times (\mathbf{c}u))=0$.