So I have a problem which (in one case) leads me to the following vector Laplace equation:

$\nabla^2 \mathbf{A} = 0$

with $\mathbf{A}$ the magnetic vector potential, whereon I have imposed the Coulomb gauge $\nabla\cdot\mathbf{A} = 0$. The problem contains cylindrical symmetry, which is why I want to solve this in cylindrical coordinates. Boundary conditions are to follow from solving the homogeneous Helmholtz equation and demanding continuity at a cylindrical surface.
However, I need to solve the Laplace equation first, which I don't know how to do. If the coordinates were cartesian, the vector equation would just be equivalent to three scalar equations, but now it's not so easy. Does anyone feel like explaining to me how I would best go about solving this vector Laplace equation in cylindrical coordinates? Greatly appreciated!

  • $\begingroup$ Your vector Laplace equations are three independent equations (without the gauge condition). So I would take the solutions of the scalar Laplace equations and then see what kind of relation you obtain due to the gauge condition. $\endgroup$
    – Fabian
    Dec 7 '12 at 19:44
  • $\begingroup$ I'm pretty sure you can get a nice Fourier series solution using separation of variables. $\endgroup$
    – Matt
    Dec 7 '12 at 21:51
  • $\begingroup$ @Fabian: the gauge condition is necessary to obtain this Laplace equation and I didn't plan on using it anywhere else, I just thought I'd mention it to be complete. Unfortunately, according to this wolfram page (mathworld.wolfram.com/VectorLaplacian.html) the equations for the radial and the angular component of the vector are not independent. They are coupled, which is what's making things difficult for me. $\endgroup$
    – Wouter
    Dec 8 '12 at 3:48
  • $\begingroup$ @Matt: are you sure about that? Again referring to the same wolfram page (mathworld.wolfram.com/VectorLaplacian.html), the equations for the radial and the angular component of the vector are coupled. So separation of variables is not possible, right? $\endgroup$
    – Wouter
    Dec 8 '12 at 3:50
  • $\begingroup$ @Wouter: yes they are coupled by the gauge condition. But still I would argue that you should solve the problem without the gauge condition (which you can do explicitly) and then see what kind of relations between the different modes you get due to the gauge condition. $\endgroup$
    – Fabian
    Dec 8 '12 at 7:13

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