- #1

BOAS

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## Homework Statement

I am trying to fill in the gaps of a calculation (computing the deflection potential ##\psi##) in this paper:

http://adsabs.harvard.edu/abs/1994A&A...284..285K

We have the Poisson equation:

##\frac{1}{x}\frac{\partial}{\partial x} \left( x \frac{\partial \psi}{\partial x} \right) + \frac{1}{x^2} \frac{\partial^2 \psi}{\partial \varphi^2} = \frac{\sqrt{f}}{x \Delta(\varphi)}##,

where ##\Delta(\varphi) := \sqrt{\cos^2 \varphi + f^2 \sin^2 \varphi}##.

This is reduced to a nonhomogeneous ODE by the ansatz ##\psi(x, \varphi) = x \tilde \psi(\varphi)##, resulting in

##\tilde \psi + \frac{d^2 \tilde \psi}{d \varphi^2} = \frac{\sqrt{f}}{\Delta(\varphi)}##

The author states that we may solve this equation using Green's method.

## Homework Equations

## The Attempt at a Solution

(i'm new to this and trying to follow the description here: http://www.damtp.cam.ac.uk/user/dbs26/1BMethods/GreensODE.pdf)

[/B]

So, to begin with, I want to solve the homogeneous equation

##\tilde \psi + \frac{d^2 \tilde \psi}{d \varphi^2} = 0##

However, I am thoroughly confused about the boundary conditions I should employ here. If I use the interval ##[0, 2 \pi]##, which would seem to make sense given ##\varphi## is the polar angle, I end up with nonsense when I compute the homogeneous solutions satisfying the boundary conditions.

To illustrate what I mean:

If ##[0, 2 \pi]## is my interval, then my boundary conditions must be that ##\tilde \psi (0) = \tilde \psi ( 2 \pi) = 0##.

The general homogeneous solution to my equation is ##\tilde \psi = c_1 \sin \varphi + c_2 \cos \varphi##

so I say that ##y_1(\varphi) = \sin \varphi## and ##y_2 = \sin (2 \pi - \varphi)## satisfying the boundary conditions at ##\varphi = 0## and ##\varphi = 2 \pi## respectively.

So my Green's function is

##G(\varphi; \xi) = A(\xi) \sin \varphi## for ##0 \leq \varphi \leq \xi##

and

##G(\varphi; \xi) = -B(\xi) \sin \varphi## for ##\xi \leq \varphi \leq 2\pi##

Applying the continuity condition gives me that ##A(\xi) = - B(\xi)##

and the jump condition ##A(\xi)y'_1(\xi) - B(\xi)y'_2(\xi) = \frac{1}{\alpha(\xi)}##

leads me to the conclusion that ##0 = 1##...

Something is going horribly wrong but I don't know what