Prove Eqn. 1 (below) using Eqns. 2-4. [Suggestion: I'd set up Cartesian coordinates at the surface, with z perpendicular to the surface and x parallel to the current.]
I used ϑ for partial derivatives.
Eqn. 1: ϑAabove/ϑn - ϑAbelow/ϑn = -μ0K
Eqn. 2: ∇ ⋅ A = 0
Eqn. 3: Babove - Bbelow = μ0(K × n-hat)
Eqn. 4: Aabove = Abelow
The Attempt at a Solution
Conceptually, I'm mostly stuck at the partial derivatives with respect to n. n is just a normal vector to a plane surface. It will flip completely as soon as you go from looking at points below the surface to points above the surface.
I've taken Eqn. 3 and plugged in B = ∇ × A to get:
∇ × Aabove - ∇ × Abelow = μ0(K × n-hat)
It looks pretty close, but by Eqn. 4, the two terms on the left should be equal and thus everything is zero. That's hardly going to help.
The usefulness of Eqn. 2 seems dubious to me, but it would be useful if I need find A using Poisson's equation, which is only possible by Eqn. 2.
∇2A = -μ0J
But then again, the surface is 2D so J doesn't really fit.
I need a nudge in the right direction. Help?