Green's Function Boundary Conditions

[BOAS]

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

 

[Orodruin]

Your $y_1$ and $y_2$ are not linearly independent.

There is no reason to assume the function value at 0 and 2pi to be zero.

 

[BOAS]

Your $y_1$ and $y_2$ are not linearly independent.

There is no reason to assume the function value at 0 and 2pi to be zero.

I thought that the conditions on the Green's function meant that for any second order linear differential operator on [a,b], y(a) = y(b) = 0.

 

[Orodruin]

That is not true. It depends on the boundary conditions of the problem you are trying to solve. In your case, you are trying to solve an equation with a periodic boundary condition so your Green's function needs to satisfy periodic boundary conditions, nothing more. Note that the choice of what angle corresponds to $\phi = 0$ or $2\pi$ is completely arbitrary and your result should not depend on an arbitrary assignment of coordinates. Furthermore, you could just as well take the interval to be $[-\pi,\pi)$.

 

[Ray Vickson]

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

Back in the Stone Age when I was learning this material we did not worry about boundary conditions on Green's functions. The idea of using a Green's function was just to get some particular solution to the non-homogeneous DE. The homogeneous part of the solution could then be used to impose boundary conditions. So, we can write the solution as
$$\tilde \psi = c_1 \tilde \psi_1 + c_2 \tilde \psi_2 + \tilde \psi_p,$$
where the $c_i$ are constants, $\tilde \psi_1, \tilde \psi_2$ are homogeneous solutions and $\tilde \psi_p$ is any particular solution. If you have boundary conditions on the final solution $\tilde \psi$, these can be handled by adjusting the constants $c_1$ and $c_2.$

However, I don't know: maybe things are taught differently nowadays.

 

[Orodruin]

Back in the Stone Age when I was learning this material we did not worry about boundary conditions on Green's functions. The idea of using a Green's function was just to get some particular solution to the non-homogeneous DE. The homogeneous part of the solution could then be used to impose boundary conditions. So, we can write the solution as
$$\tilde \psi = c_1 \tilde \psi_1 + c_2 \tilde \psi_2 + \tilde \psi_p,$$
where the $c_i$ are constants, $\tilde \psi_1, \tilde \psi_2$ are homogeneous solutions and $\tilde \psi_p$ is any particular solution. If you have boundary conditions on the final solution $\tilde \psi$, these can be handled by adjusting the constants $c_1$ and $c_2.$

However, I don't know: maybe things are taught differently nowadays.

I don't like this way of teaching it. It completely obscures the fact that a Green's function with appropriate boundary conditions can be used to take care also of inhomogeneous boundary conditions in many cases.

Edit: Also, in this case we are dealing with a periodic function so the Green's function must be periodic. You cannot just ignore this condition and solve it with boundary conditions (there is no boundary).

 

[Ray Vickson]

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

If you want to satisfy $G(0; \xi) = G(2 \pi; \xi) = 0$ as well as continuity of $G(\varphi;\xi)$ and the jump condition on $G_\varphi(\varphi; \xi)$ at $\varphi = \xi$, you can do it by keeping more constants. For $\xi \in (0,2\pi)$, let
$$G(\varphi;\xi) = \begin{cases} A_1 \cos(\varphi) + B_1 \sin(\varphi) & \text{if} \; 0 \leq \varphi < \xi\\
A_2 \cos(\varphi) + B_2 \sin(\varphi) & \text{if} \; \xi < \varphi \leq 2 \pi
\end{cases} $$
You have four conditions to be satisfied, and four constants to apply, so it is do-able.

 

[pasmith]

I think @BOAS has the right of it: there is no such Green's function.

If you're looking for a linear combination of sines and cosines then you can write it as $$G(\varphi;\xi) = \begin{cases} C_1\cos(\varphi - \xi) + D_1\sin(\varphi - \xi) & \varphi \in [0,\xi) \\ G_\xi & \varphi = \xi \\ C_2\cos(\varphi - \xi) + D_2\sin(\varphi - \xi) & \varphi \in (\xi,2\pi) \end{cases}$$ and the conditions to be satisfied at $\varphi = \xi$ are then $C_1 = G_\xi = C_2$ and $D_2 = D_1 + 1$.

Now this function is $2\pi$ periodic in the sense that $G(\varphi;\xi) = G(\varphi + 2\pi; \xi)$ for all $\varphi \in [0,2\pi)$. However we also require continuity: $$\lim_{\varphi \to 2\pi^{-}} G(\varphi; \xi) = G(0,\xi).$$ But we don't have this: instead $$G(0;\xi) - \lim_{\varphi \to 2\pi^{-}}G(\varphi;\xi) = (D_2 - D_1)\sin\xi = \sin\xi$$ which does not hold for all $\xi \in (0,2\pi)$. (The cases of $\xi \in \{0, 2\pi\}$ - which should yield the same result - need separate treatment anyway.)

Explicitly imposing $G(0;\xi) = G(2\pi;\xi) = 0$ doesn't fix this problem: In @Ray Vickson's notation, you are immediately imposing $A_1 = A_2 = 0$ and are then left with $$(B_1 - B_2) \begin{pmatrix} \sin \xi \\ \cos \xi \end{pmatrix} = \begin{pmatrix} 0 \\ -1\end{pmatrix}$$ which has a solution only for $\xi = \pi$ - indeed infinitely many solutions in that case.

I hope I've missed something, because otherwise there seems to be a serious error in a published paper (which I have not had the opportunity to read myself).

 

[Ray Vickson]

I think @BOAS has the right of it: there is no such Green's function.

If you're looking for a linear combination of sines and cosines then you can write it as $$G(\varphi;\xi) = \begin{cases} C_1\cos(\varphi - \xi) + D_1\sin(\varphi - \xi) & \varphi \in [0,\xi) \\ G_\xi & \varphi = \xi \\ C_2\cos(\varphi - \xi) + D_2\sin(\varphi - \xi) & \varphi \in (\xi,2\pi) \end{cases}$$ and the conditions to be satisfied at $\varphi = \xi$ are then $C_1 = G_\xi = C_2$ and $D_2 = D_1 + 1$.

Now this function is $2\pi$ periodic in the sense that $G(\varphi;\xi) = G(\varphi + 2\pi; \xi)$ for all $\varphi \in [0,2\pi)$. However we also require continuity: $$\lim_{\varphi \to 2\pi^{-}} G(\varphi; \xi) = G(0,\xi).$$ But we don't have this: instead $$G(0;\xi) - \lim_{\varphi \to 2\pi^{-}}G(\varphi;\xi) = (D_2 - D_1)\sin\xi = \sin\xi$$ which does not hold for all $\xi \in (0,2\pi)$. (The cases of $\xi \in \{0, 2\pi\}$ - which should yield the same result - need separate treatment anyway.)

Explicitly imposing $G(0;\xi) = G(2\pi;\xi) = 0$ doesn't fix this problem: In @Ray Vickson's notation, you are immediately imposing $A_1 = A_2 = 0$ and are then left with $$(B_1 - B_2) \begin{pmatrix} \sin \xi \\ \cos \xi \end{pmatrix} = \begin{pmatrix} 0 \\ -1\end{pmatrix}$$ which has a solution only for $\xi = \pi$ - indeed infinitely many solutions in that case.

I hope I've missed something, because otherwise there seems to be a serious error in a published paper (which I have not had the opportunity to read myself).

If your first sentence means that there cannot be a periodic Green's function, then I agree. If $\xi \in (0,2\pi)$ implies that
$$G(\varphi; \xi) = \begin{cases} G_1(\varphi), & \varphi < \xi\\
G_2(\varphi), & \varphi > \xi
\end{cases}$$
then the form is pinned down by the anchor at $\xi$, so that $G(\varphi + 2 \pi n; \xi) = G_2(\varphi+2 \pi n)$ and $G(\varphi-2 \pi n; \xi) = G_1(\varphi-2 \pi n)$ for all $n \geq 1$. By adding multiples of $2 \pi$ to $\varphi$ we eliminate the "switching" from $G_1$ to $G_2.$ Of course, if we change both arguments we can have periodicity, so that $G(\varphi \pm 2 \pi n; \xi \pm 2 \pi n) = G(\varphi; \xi)$