Possible to solve this coupled system of vector equations?

Question

Let $\gamma, \omega, c$ be positive constants, let $\mathbf{Q}_{a}$ and $\mathbf{Q}_{b}$ be three-dimensional vectors, and let $\mathbf{B}(\mathbf{r})=\mathbf{B}(x,y,z)$ be a vector field. Let $\mathbf{B}_{\circ}= \mathbf{B}(\mathbf{r}_{\circ})$ for some position $\mathbf{r}_{\circ}$ and write $\mathbf{B}=\mathbf{B}_{\circ}+\delta\mathbf{B}$. Consider the following coupled system for the unknown vectors $\mathbf{a}$ and $\mathbf{b}$: $$ -\frac{\omega^2}{\gamma}\mathbf{a}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{a}+\omega\mathbf{b}\times\mathbf{B}_{\circ} \tag{1} $$ $$ -\frac{\omega^2}{\gamma}\mathbf{b}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{b}-\omega\mathbf{a}\times\mathbf{B}_{\circ} \tag{2} $$ If it helps, $\mathbf{Q}_{a}=\int_{0}^{2\pi/\omega}\mathbf{B}\cos(\omega t)\,dt$ and $\mathbf{Q}_{b}=\int_{0}^{2\pi/\omega}\mathbf{B}\sin(\omega t)\,dt$. All the $x, y, z$ are functions of $t$ but $\mathbf{B}$ is (basically) abitrary so those integrals can't be done.

Is it possible to solve $(1)$ and $(2)$ for $\mathbf{a}$ and $\mathbf{b}$ while still keeping everything a vector? I believe that one could simply write out all the components of $(1)$ and $(2)$ to get two linear relationships between $\mathbf{a}$ and $\mathbf{b}$, which is solvable in principle, but I'd really rather not do that. I've tried just subbing one equation into the other, but you end up with the thing to solve for being buried in a double cross product, so I'm not sure.

Answer 1

Hint: as double cross product can be computationally exhausting (a bit), why don't you have a preliminary look at the vector components? If we forget about constants for a second putting $\omega=\gamma=c=1$, then we obtain

$$a=(Q_a-b)\times B, $$ $$b=(Q_b+a)\times B. $$

As $a_i=\epsilon_{ilr}(Q_a-b)_{lr}B_r$, then $b_j=\epsilon_{jik}(Q_b)_iB_k+ \epsilon_{jik}\epsilon_{ilr}(Q_a-b)_{lr}B_r,$ which can be manipulated with

$$\epsilon_{jik}\epsilon_{ilr}=-\delta_{jl}\delta_{kr}+\delta_{jr}\delta_{lk}. $$

Answer 2

to long for a comment, this is not an answer (yet)

Begin with your equations: $$ -\frac{\omega^2}{\gamma}\mathbf{a}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{a}+\omega\mathbf{b}\times\mathbf{B}_{\circ} \tag{1} $$ $$ -\frac{\omega^2}{\gamma}\mathbf{b}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{b}-\omega\mathbf{a}\times\mathbf{B}_{\circ} \tag{2} $$ take the cross product of the first equation with $\mathbf{a}$ and the second with $\mathbf{b}$ to obtain: $$ 0=c\mathbf{a} \times \left( \mathbf{B}_{\circ}\times\mathbf{Q}_{a}\right)+\omega\mathbf{a} \times \left( \mathbf{b}\times\mathbf{B}_{\circ} \right) \tag{3} $$ $$ 0=c\mathbf{b} \times \left( \mathbf{B}_{\circ}\times\mathbf{Q}_{b}\right)-\omega\mathbf{b} \times \left(\mathbf{a}\times\mathbf{B}_{\circ}\right) \tag{4} $$ Now, the Jacobi identity is a close as we get to associativity for the cross-product, $$ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) +\mathbf{b} \times (\mathbf{c} \times \mathbf{a})+\mathbf{c} \times (\mathbf{a} \times \mathbf{b})=0$$ This means we can trade the $-\mathbf{b} \times (\mathbf{a} \times \mathbf{B}_{\circ})$ term for $\mathbf{a} \times (\mathbf{B}_{\circ} \times \mathbf{b})+\mathbf{B}_{\circ} \times (\mathbf{b} \times \mathbf{a})$ to obtain: $$ 0=c\mathbf{b} \times \left( \mathbf{B}_{\circ}\times\mathbf{Q}_{b}\right)+\omega \left(\mathbf{a} \times (\mathbf{B}_{\circ} \times \mathbf{b})+\mathbf{B}_{\circ} \times (\mathbf{b} \times \mathbf{a})\right) \tag{5} $$ then I could add this to (3.) and use $\mathbf{b} \times \mathbf{B}_{\circ} = -\mathbf{B}_{\circ} \times \mathbf{b} $ to cancel a pair of terms leaving...

Well, you get the idea, maybe I'll finish this later, I just want to post this much now to give you an idea, there may be better ways, certainly you need to look up the triple-product formulas.