# Green's Function Boundary Conditions

• BOAS
In summary: I think that there is a reason for the Green's function to satisfy the boundary conditions. If you don't impose the boundary conditions before finding the particular solution , you are not going to get the general solution of the equation.
BOAS

## Homework Statement

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

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.

## 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

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.

Orodruin said:
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.

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)##.

BOAS said:

## Homework Statement

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

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.

## 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.

Last edited:
Ray Vickson said:
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).

BOAS said:

## Homework Statement

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

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.

## 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.

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).

pasmith said:
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)##

Last edited:

## 1. What are Green's function boundary conditions?

Green's function boundary conditions are a set of mathematical equations used to describe the behavior of a physical system at its boundaries. They are derived from Green's function, which is a mathematical tool used to solve differential equations.

## 2. How are Green's function boundary conditions used in science?

Green's function boundary conditions are used in many areas of science, including physics, engineering, and mathematics. They are particularly useful in solving problems involving partial differential equations, such as those found in electromagnetism, fluid dynamics, and quantum mechanics.

## 3. What is the significance of Green's function boundary conditions?

Green's function boundary conditions are significant because they allow scientists to model and understand the behavior of physical systems at their boundaries. This is important in many fields of science, as boundary conditions can greatly influence the overall behavior and properties of a system.

## 4. How are Green's function boundary conditions different from other boundary conditions?

Green's function boundary conditions are different from other boundary conditions in that they are based on Green's function, which is a specific mathematical tool. Other boundary conditions may be based on different mathematical methods or physical principles.

## 5. Can Green's function boundary conditions be applied to all physical systems?

No, Green's function boundary conditions may not be applicable to all physical systems. They are most commonly used in systems that can be described by partial differential equations, and may not be suitable for systems with different types of equations or behaviors.

Replies
1
Views
617
Replies
4
Views
438
Replies
4
Views
2K
Replies
1
Views
2K
Replies
7
Views
1K
Replies
4
Views
1K
Replies
1
Views
916
Replies
8
Views
1K
Replies
1
Views
1K
Replies
4
Views
2K