$\nabla \times \mathbf{A} = (K_{0}(|\mathbf{r}-\mathbf{a}|)+K_{0}(|\mathbf{r}-\mathbf{b}|)) \hat{z}$, what is $\mathbf{A} ?$

Question

1) SIMPLE CASE

We have the following equation:

$\nabla \times \mathbf{A} = K_{0}(|\mathbf{r}|) \hat{z}$

where $K_{0}$ is a Modified Bessel Function of the Second Kind. What is $\mathbf{A}$ ? This case is very simple. In polar coordinates we have that

$\mathbf{A}(r, \varphi)= \varphi r K_{0}(|\mathbf{r}|) \hat{r}$

does the job, since $(\nabla \times \mathbf{A})_z = \frac{1}{r} \Big ( \frac{\partial(r A_{\varphi})}{\partial r} - \frac{\partial A_r}{\partial \varphi} \Big )$.

2) COMPLICATED CASE

We have the following equation:

$\nabla \times \mathbf{A} = (K_{0}(|\mathbf{r}-\mathbf{a}|)+K_{0}(|\mathbf{r}-\mathbf{b}|)) \hat{z}$.

Polar coordinates do not seem to facilitate the task. So probably Cartesian coordinates is the way to go. Therefore we have:

$\partial_x A_y - \partial_y A_x=K_0 \Big (\sqrt{(x-a_x)^2 + (y-a_y)^2} \Big )+K_0 \Big (\sqrt{(x-b_x)^2 + (y-b_y)^2} \Big)$.

Let's choose $A_x =0$, then we end up with

$\partial_x A_y =K_0 \Big (\sqrt{(x-a_x)^2 + (y-a_y)^2} \Big )+K_0 \Big (\sqrt{(x-b_x)^2 + (y-b_y)^2} \Big)$.

Integrating both sides we get

$A_y (x,y) = \int dx \; \Big ( K_0 \Big (\sqrt{(x-a_x)^2 + (y-a_y)^2} \Big )+K_0 \Big (\sqrt{(x-b_x)^2 + (y-b_y)^2} \Big) \Big ) +f(y)$,

where $f(y)$ is any function of $y$. I am not able to continue, any help please?

Thank you.

Answer 1

Those two problem have the same level of difficulty really

$$\nabla\times{A}=(K_{0}(|\vec{r}-\vec{a}|)+K_{0}(|\vec{r}-\vec{b}|))\hat{z}$$ Let $$A=A_{a}+A_{b}$$ Such that $$\nabla\times{A}_{a}=K_{0}(|\vec{r}-\vec{a}|)\hat{z}$$ and $$\nabla\times{A}_{b}=K_{0}(|\vec{r}-\vec{b}|)\hat{z}$$ In the first equation we introduce the set of cyllindrical polar coordiantes $$x-a=r_{a}\cos(\varphi_{a})$$ $$y-a=r_{a}\sin(\varphi_{a})$$ $$z-a=Z_{a}$$ And for the second one $$x-b=r_{b}\cos(\varphi_{b})$$ $$y-b=r_{b}\sin(\varphi_{b})$$ $$z-b=Z_{b}$$ The solutions are $$A_{a}=\varphi_{a}r_{a}K_{0}(r_{a})\hat{r}_{a}$$ $$A_{b}=\varphi_{b}r_{b}K_{0}(r_{b})\hat{r}_{b}$$ Using $$\varphi_{a}=\arctan\Big(\frac{y-a}{x-a}\Big)$$ $$\varphi_{b}=\arctan\Big(\frac{y-b}{x-b}\Big)$$ and $$\hat{r}_{a}=\frac{\vec{r}-\vec{a}}{|\vec{r}-\vec{a}|}$$ $$\hat{r}_{b}=\frac{\vec{r}-\vec{b}}{|\vec{r}-\vec{b}|}$$ You get $$A_{a}=\arctan\Big(\frac{y-a}{x-a}\Big)K_{0}(|\vec{r}-\vec{a}|)(\vec{r}-\vec{a})$$ $$A_{b}=\arctan\Big(\frac{y-b}{x-b}\Big)K_{0}(|\vec{r}-\vec{b}|)(\vec{r}-\vec{b})$$