# Vector Helmholtz Equation

In my recent exercise book I've derived the following equation that needs solving: $$\nabla^2\vec{u} + k^2\vec{u} = 0.$$ The deformation vectors points only in the $$\hat{e}_r$$ direction. I didn't want to write out the Laplace in spherical coordinates, so I tried using what I learned in my PDE course the previous semester. It turns out, the vector Helmholtz equation is quite different from scalar one we've studied.

Suppose I have basic knowledge in solving scalar Helmholtz in spherical (and other coordinate systems). Is there any analogy that translates over to the vector version? In other words, should I be able to solve vector Helmholtz if I can solve scalar versions?

• Would Mathematics be a better home for this question? Mar 13, 2020 at 5:00

Yes, indeed you can use your knowledge of the scalar Helmholtz equation. The difficulty with the vectorial Helmholtz equation is that the basis vectors $$\mathbf{e}_i$$ also vary from point to point in any other coordinate system other than the cartesian one, so when you act $$\nabla^2$$ on $$\mathbf{u}$$ the basis vectors also get differentiated. This forces you to calculate $$\nabla^2 \mathbf{u}$$ through the identity $$\nabla^2 \mathbf{u} = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) - \boldsymbol{\nabla}\times (\boldsymbol{\nabla}\times \mathbf{u}) \tag{1}$$ which is really cumbersome to deal with by brute force. A smart way to avoid all the hassle is by using the ansatz $$\mathbf{u} = \mathbf{r} \times (\boldsymbol{\nabla} \psi) \tag{2}$$ where $$\psi$$ satisfies the scalar Helmholtz equation $$(\nabla^2 + k^2) \psi = 0.$$
To check that $$(\nabla^2 + k^2) \mathbf{u} = 0$$ yourself you have to plug the ansatz $$(2)$$ on $$(1)$$ and make use of many vector identities and the scalar Helmholtz equation. The calculation is quite involved, so I'll point you to check Reitz, Milford & Christy's Foundations of Electromagnetic Theory, there they do the full calculation. With ansatz $$(2)$$ proven, it's just a matter of plugging the relevant mode $$\psi_{lm}$$ in eq. $$(2)$$ that you get your solution $$\mathbf{u}_{lm}$$.