Understanding and inverting Kirchhoff's integral for the diffraction theory

Question

The Kirchhoff diffraction integral has been discussed in the science society pretty often and appears to be a non consistent theory. Never the less it's applicable and "gives" great results. Kirchhoff's integral can be summed up as follows (in my Case it's 2D):

Imagine an area (green) that contains all the sources in our system. This area is bounded by the curve $B1$. The Boundary $B2$ is considered to be very far away (infinity) and represents the boundary of the whole system. The $n$ is an inward directed normal vector to $B1$ which is important for the integral. According to Kirchhoff's diffraction theory, in order to calculate the Electric or Magnetic field at any point $P$ outside the green region and bounded by $B2$, it is sufficient to know the electric and magnetic fields on the boundary $B1$ (as written, the $B2$ is considered to be very far away).

I have successfully applied this integral for an analytically known case and now I would like to invert the system. My desired configuration is as follows:

As you can see, now the sources are outside and the normal vector is also directed towards the green region. The problem is, that due to my undestanding this should not be possible with Kirchhoff's integral, since the boundary $B2$ is simply missing (and this shouldn't be the case, since it's part of the whole integral). In the first case, one can neglect it's value, since it's very far away (at infinity) but here it is not.

Does anyone have an idea how the configuration should look like if I wish to "invert" the first example as I have tried to explain?

Please let me know if something is not very clear

Answer 1

The answer is you do exactly the same thing as before.

Kirchoff's integral works whenever you want to solve the wave equation in a closed volume. One simple case is the first case you presented - the enclosing surface is B1. In the second case, the enclosing surface is the union of B1 & B2 - integrate over both surfaces to integrate over the entire enclosing surface. (If B2 in the second case is infinitely far away - I'm not fully clear on your problem - then you would need to make an argument that the surface integral over this surface is 0, or deal with problems with infinity).

When doing diffraction in 2D, also make sure you use the 2D Kirchhoff integral

$u(r)=-\frac{1}{4}\left[\int_{B_1+B_2} u \partial_n Y_0(k r)-Y_0(k r) \partial_n u\right]$

where $\partial_n$ is the normal derivative, with respect to the inwards pointing normal vector, and $Y_0$ is the Bessel function of the second kind. The integral for a 3D geometry is different.