Solved: Green's Functions - Kirchhoff Diffraction Theory

[NeoDevin]

[SOLVED] Green's Functions

Homework Statement

The basic equation of the scalar Kirchhoff diffraction theory is:

$$\psi(\vec r_1) = \frac{1}{4\pi}\int_{S_2}\left[\frac{e^{ikr}}{r}\nabla \psi(\vec r_2) - \psi(\vec r_2)\nabla (\frac{e^{ikr}}{r})\right] \cdot d\vec S_2$$

where $\psi$ satisfies the homogeneous (three-dimensional) Helmholtz equation and $r = |\vec r_1 - \vec r_2|$. Derive the above equation, assuming that $\vec r_1$ is interior to the closed surface $S_2$

Homework Equations

Helmholtz Equation

$$(\nabla^2 + k^2)\psi = 0$$

Modified Helmholtz Equation

$$(\nabla^2 - k^2)\psi = 0$$

Has Green's Function

$$G(\vec r_1, \vec r_2) = \frac{\mathrm{exp}(-k|\vec r_1 - \vec r_2|)}{4\pi|\vec r_1 - \vec r_2|}$$

Green's Function of three-dimensional Laplacian ($\nabla^2$) is proportional to

$$G(\vec r_1, \vec r_2) \propto \int \mathrm{exp}(i\vec k\cdot(\vec r_1 - \vec r_2))\frac{d^3k}{k^2}$$

The Attempt at a Solution

We have $\psi$ as a solution of the equation

$$(\nabla_1{}^2 + k^2)\psi(\vec r_1) = (\nabla_1{}^2 - k^2)\psi(\vec r_1) + 2k^2\psi(\vec r_1) = 0$$

or

$$\nabla_1{}^2 \psi(\vec r_1) = -k^2\psi(\vec r_1)$$

Which gives that

$$(\nabla_1{}^2 - k^2)\psi(\vec r_1) = -2k^2\psi(\vec r_1) = 2\nabla_1{}^2 \psi(\vec r_1)$$

So we can write the solution in terms of the Green's function for the modified Helmholtz Equation

$$\psi(\vec r_1) & = & \int_{V_2}G(\vec r_1, \vec r_2)2\nabla_1{}^2\psi(\vec r_1) dV_2$$

And then I don't know how to take it from there to the solution. It doesn't look like it should be too hard, but I don't know how to do it.

Any help would be greatly appreciated.

Thanks in advance,
Devin

 

[NeoDevin]

Noone can help?

 

[NeoDevin]

Anyone, this is due tomorrow :(

 

[NeoDevin]

Ok, I got a little further, still not sure how to finish it though...

I then apply Green's theorem to the RHS and get

$$\psi(\vec r_1) = 2\int_{S_2}(G(\vec r_1, \vec r_2)\nabla_2\psi(\vec r_2) - \psi(\vec r_2)\nabla_2G(\vec r_2, \vec r_2))dS_2 + 2\int_{V_2}\psi(\vec r_2)\nabla_2{}^2G(\vec r_1, \vec r_2)dV_2$$

But we know that

$$\nabla_2{}^2G(\vec r_1, \vec r_2) = k^2G(\vec r_1, \vec r_2) - \delta(\vec r_1, \vec r_2)$$

So the equation becomes

$$3\psi(\vec r_1) = 2\int_{S_2}(G(\vec r_1, \vec r_2)\nabla_2\psi(\vec r_2) - \psi(\vec r_2)\nabla_2G(\vec r_2, \vec r_2))dS_2 + 2k^2\int_{V_2}\psi(\vec r_2)G(\vec r_1, \vec r_2)dV_2$$

The first term is close to what I need, it is

$$\frac{1}{2\pi}\int_{S_2}\left[\frac{e^{-kr}}{r}\nabla\psi(\vec r_2) - \psi(\vec r_2)\nabla\frac{e^{-kr}}{r}\right]\cdot dS_2$$

But I don't know how to change that negative to an $i$, or how to get rid of the extra term, or how to get rid of the extra factor of 2/3.

 

[NeoDevin]

Ok, for the second term, we have

$$k^2\psi(\vec r_2) = -\nabla_2{}^2\psi(\vec r_2)$$

And then

$$\int_{V_2}2\nabla_2{}^2\psi(\vec r_2)G(\vec r_1, \vec r_2)dV_2 = \psi(\vec r_1)$$

So we bring it over to the other side, and divide by 4 to get

$$\psi(\vec r_1) = \frac{1}{8\pi}\int_{S_2}\left[\frac{e^{-kr}}{r}\nabla\psi(\vec r_2) - \psi(\vec r_2)\nabla\frac{e^{-kr}}{r}\right]\cdot d\vec S_2$$

Which is only off by a factor of 1/2, and has those negatives instead of $i$'s in the exponent.

 

[NeoDevin]

Figured it out, the method I was using was the wrong one.

 

[NeoDevin]

I still can't mark my threads as solved...