How Do You Solve the Grad-Shafranov Equation for Plasma Physics?

[BacalhauGT]

i need to solve this equation:

R\frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi \right ) + \frac{\partial^2 }{\partial z^2} \Psi = R^2 A_1 - A_2

psi is a function of Z and R. A1 and A2 are constants

can you help me?

 

[fzero]

Have you succeeded in solving the homogeneous version? Separation of variables will work there. The particular solutions to the inhomogenous equation above seem rather easy to guess at.

 

[BacalhauGT]

Have you succeeded in solving the homogeneous version? Separation of variables will work there. The particular solutions to the inhomogenous equation above seem rather easy to guess at.

for particular solutions i have to solve:

first term = A1 R^2

second = -A2

Right?

 

[fzero]

for particular solutions i have to solve:

first term = A1 R^2

second = -A2

Right?

Not necessarily, the equation reads LHS = RHS. You must check whether either or both terms on the LHS can generate either or both terms on the RHS.

 

[BacalhauGT]

Not necessarily, the equation reads LHS = RHS.

Can you explain me why? (btw, i am sorry, but what do you mean by LHS and RHS?)

You must check whether either or both terms on the LHS can generate either or both terms on the RHS.

Can you explain me that?

My knowledge about differential equations is bad. I know the final solution and i just have to found the particular one. With the method i said i can obtain that solution, i just don't know why it works and if it is luck or incorrect :)

Thank you.

 

[fzero]

Can you explain me why? (btw, i am sorry, but what do you mean by LHS and RHS?)

Sorry, I meant left-hand side (LHS) and right-hand side (RHS).

Can you explain me that?

My knowledge about differential equations is bad. I know the final solution and i just have to found the particular one. With the method i said i can obtain that solution, i just don't know why it works and if it is luck or incorrect :)

Thank you.

Sorry again, I should have clarified what you meant, or wrote something differently to avoid confusion. It is true that we can find the particular solution by first finding a particular solution that gives $A_1 R^2$ and then finding another that gives $-A_2$. I was confused and thought that you meant the first term on the LHS gave the first term on the RHS and similarly for the 2nd terms. So I ended up writing something even more confusing.

To summarize, the equation is linear, so the solution is composed of 3 terms. One is the solution to the homogeneous equation, one is the particular solution for $A_1 R^2$ and the last is the particular solution for $-A_2$.

 

[BacalhauGT]

Sorry, I meant left-hand side (LHS) and right-hand side (RHS).

ok

Sorry again, I should have clarified what you meant, or wrote something differently to avoid confusion. It is true that we can find the particular solution by first finding a particular solution that gives $A_1 R^2$ and then finding another that gives $-A_2$. I was confused and thought that you meant the first term on the LHS gave the first term on the RHS and similarly for the 2nd terms. So I ended up writing something even more confusing.

No. i am wrong. I did exactly what you were thinking . So i did:

R\frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi \right ) = R^2 A_1

and

\frac{\partial^2 }{\partial z^2} \Psi = - A_2

I know it is not necessary. But it worked LOL. the particular solution is (A1/8) R^4 - (A2/2) Z^2 and with that method i can obtain (A1/8) R^4 - (A2/2) Z^2 ()

To summarize, the equation is linear, so the solution is composed of 3 terms. One is the solution to the homogeneous equation, one is the particular solution for $A_1 R^2$ and the last is the particular solution for $-A_2$.

Yes, i understand. So you are saying:

\Psi_{particular} = \Psi_{-A_2} + \Psi_{A_1 R^2}, right?

So for \Psi_{A_1 R^2} i have:

R\frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi_{A_1 R^2} \right ) + \frac{\partial^2 }{\partial z^2} \Psi_{A_1 R^2} = R^2 A_1

right?

Thats my problem, i can't solve it. have i to use separation of variables?

Thank you

 

[fzero]

Yes, i understand. So you are saying:

\Psi_{particular} = \Psi_{-A_2} + \Psi_{A_1 R^2}, right?

Yes.

So for \Psi_{A_1 R^2} i have:

R\frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi_{A_1 R^2} \right ) + \frac{\partial^2 }{\partial z^2} \Psi_{A_1 R^2} = R^2 A_1

right?

Thats my problem, i can't solve it. have i to use separation of variables?

Thank you

You can use separation of variables. This will recount some of the solutions to the homogenous equation that you found before, but that's ok. You should get an inhomogenous equation for the R dependence, for which you already found the particular solution with your "wrong" method.

Incidently, your wrong method actually seems to work here because the inhomogenous equation is also separable. In general, it would not. As an example, I don't believe it would work for a term like $R z$ on the right-hand side.

 

[BacalhauGT]

Yes.

You can use separation of variables. This will recount some of the solutions to the homogenous equation that you found before, but that's ok. You should get an inhomogenous equation for the R dependence, for which you already found the particular solution with your "wrong" method.

Incidently, your wrong method actually seems to work here because the inhomogenous equation is also separable. In general, it would not. As an example, I don't believe it would work for a term like $R z$ on the right-hand side.

Yes i understand. But now, with separation:

\Psi_{particular} = \Theta_z \Phi_R

using:

R\frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi_{particular} \right ) + \frac{\partial^2 }{\partial z^2} \Psi_{particular} = R^2 A_1 - A_2

so i have by separation:

\Theta_z R\frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Phi_R \right ) + \Phi_R \frac{\partial^2 }{\partial z^2} \Theta_z = R^2 A_1 - A_2

And now?? that's my problem.

Thank you.

Edit: i don't know if it helps, but the function has to be a pair function in z variable. It helped me when i had terms like const*z with my method, then i did const=0 because const.z is not apair function.

 

[the_wolfman]

What are your boundary conditions?

Its actually pretty easy to find series solutions to the G-S equation when p' = constant and FF' = constant. This was first done by Solov'ev.

One simple solution is
\psi = -S_1/8 R^2 -S_2/2 Z^2 + S_3 + S_4 R^2 + S_5 (R^4 -4 R^2 Z^2)

Freidberg's Ideal MHD textbook talks more about the solution.

There is also a good paper by one of his former students that keeps higher order terms (and logarithmic terms).
A.J. Cerfon and J.P. Freidberg, Physics of Plasmas 17, 2010

 

[fzero]

I'm sorry, my brain isn't up to speed today. Of course separation of variables doesn't work because of the inhomogeneous terms. I would just guess at the form of the particular solutions. You will end up doing something like you were doing in the first place, that I called the "wrong" method . Look at each term on the left separately to guess the particular solution. But you also need to check how the other derivative terms act on the particular solution act. It was leaving out this step that made me say that the method was "wrong." The method is fine as long as you check every term on the left-hand side.

Let me just give an example to explain. Consider finding a solution to

$$\frac{\partial^2}{\partial z^2} \psi = A_1 R^2.$$

We can guess that $\psi = A_1 R^2 Z^2/2$. Now when we check this in the other term

$$ R \frac{\partial}{\partial R} \left( \frac{1}{R} \frac{\partial}{\partial R} \frac{A_1 R^2 Z^2}{2} \right) = 0.$$

So $\psi = A_1 R^2 Z^2/2$ is part of the particular solution.

 

[BacalhauGT]

I'm sorry, my brain isn't up to speed today. Of course separation of variables doesn't work because of the inhomogeneous terms. I would just guess at the form of the particular solutions. You will end up doing something like you were doing in the first place, that I called the "wrong" method . Look at each term on the left separately to guess the particular solution. But you also need to check how the other derivative terms act on the particular solution act. It was leaving out this step that made me say that the method was "wrong." The method is fine as long as you check every term on the left-hand side.

Let me just give an example to explain. Consider finding a solution to

$$\frac{\partial^2}{\partial z^2} \psi = A_1 R^2.$$

We can guess that $\psi = A_1 R^2 Z^2/2$. Now when we check this in the other term

$$ R \frac{\partial}{\partial R} \left( \frac{1}{R} \frac{\partial}{\partial R} \frac{A_1 R^2 Z^2}{2} \right) = 0.$$

So $\psi = A_1 R^2 Z^2/2$ is part of the particular solution.

yes. my solution is:

\begin{split}<br /> R\frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi_1 \right ) = A_1 R^2 &amp; \Leftrightarrow \frac{\partial }{\partial R} \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi_1 \right ) = A_1 R \\ &amp;\Leftrightarrow \left ( \frac{1}{R} \frac{\partial }{\partial R} \Psi_1 \right ) = A_1 \frac{R^2}{2} + E(z) \\<br /> &amp;\Leftrightarrow \frac{\partial }{\partial R} \Psi_1 = A_1 \frac{R^3}{2} + R E(z) \\ &amp;\Leftrightarrow \Psi_1 = A_1 \frac{R^4}{8} + R E(z) + F(z)<br /> \end{split}

and:

<br /> \begin{equation}<br /> \frac{\partial^2 }{\partial z^2} \Psi_1 = - A_2 \Leftrightarrow \frac{\partial }{\partial z} \Psi_1 = - A_2 z + C(R) \Leftrightarrow \Psi_1 = -A_2 \frac{z^2}{2} + z C(R) + D(R)<br /> \end{equation}

The function has to be pair (C(R)=0). So:

\begin{equation}<br /> A_1 \frac{R^4}{8} + R E(z) + F(z) = -A_2 \frac{z^2}{2} + D(R)<br /> \end{equation}

so E(z)=0

then:

\begin{equation}<br /> \Psi_1 = \frac{A_1}{8}R^4 - \frac{A_2}{2}z^2<br /> \end{equation}

Thats my method. And i know from a paper that it is right. But i would like to know if my method can be applied here for some reason or is just luck

what do you think. any idea?
edit: with 1st term = -A2 and sexond = A1 R^2, i can't obtain this solution. just a solution that contains homogenious terms and a function of z that is unknown

 

[BacalhauGT]

What are your boundary conditions?

Its actually pretty easy to find series solutions to the G-S equation when p' = constant and FF' = constant. This was first done by Solov'ev.

One simple solution is
\psi = -S_1/8 R^2 -S_2/2 Z^2 + S_3 + S_4 R^2 + S_5 (R^4 -4 R^2 Z^2)

Freidberg's Ideal MHD textbook talks more about the solution.

There is also a good paper by one of his former students that keeps higher order terms (and logarithmic terms).
A.J. Cerfon and J.P. Freidberg, Physics of Plasmas 17, 2010

Thank you.

i know the solution. i can see it in the books, but never how they solved and that's my problem.

Whats really the name of the book you talked? ideal magnetohydrodynamics?

 

[the_wolfman]

Thats my method. And i know from a paper that it is right. But i would like to know if my method can be applied here for some reason or is just luck

what do you think. any idea?

You're getting the correct particular solutions, but I think you're getting a little lucky. To get the solution I posted earlier and more general solutions assume the flux has the form

\psi = \sum_n \sum_m c_{nm} R^n Z^m.

Then plug that into the Grad-Shafranov equation \Delta^* \psi =A_1 R^2 - A_2 and solve. You'll find that certain combinations of c_{nm} have to sum together in the right way.

It gets a little messy as you keep higher order terms.

Whats really the name of the book you talked? ideal magnetohydrodynamics?

Yes. That is the full title.

 

[BacalhauGT]

You're getting the correct particular solutions, but I think you're getting a little lucky. To get the solution I posted earlier and more general solutions assume the flux has the form

\psi = \sum_n \sum_m c_{nm} R^n Z^m.

Then plug that into the Grad-Shafranov equation \Delta^* \psi =A_1 R^2 - A_2 and solve. You'll find that certain combinations of c_{nm} have to sum together in the right way.

It gets a little messy as you keep higher order terms.

Yes. That is the full title.

thank you.

So, i will assume \psi = \sum_n \sum_m c_{nm} R^n Z^m.

i just want the particular solution.

So, with your method, i have (with GS eq):

\sum_{n} \sum_{m} E_m C_n m (m-2) R^{m-2} Z^n + \sum_{n} \sum_{m} E_m C_n n (n-1) R^{2} Z^{n-2}

then i change n-2 \rightarrow n&#039; \Leftrightarrow n =n&#039; +2 and m-2 \rightarrow m&#039; \Leftrightarrow m =m&#039; +2, right?

so m' will be -2,-1,0,1,2... and n' -2,0,...

So i get:

\sum_{n} \sum_{m&#039;+2} E_{m&#039;+2} C_n m&#039; (m&#039;+2) R^{m&#039;} Z^n + \sum_{n&#039;+2} \sum_{m} E_m C_{n&#039;+2} (n&#039;+2) (n&#039;+1) R^{m} Z^{n&#039;}

sums star in 0. So:

\sum_{n} \sum_{m} E_{m+2} C_n m (m+2) R^{m} Z^n + E_m C_{n+2} (n+2) (n+1) R^{2} Z^{n} = A_1 R^2 + A_2

is this?

Now:

i want for m = 2, n=0. it will become:

E_4 C_0 8 + E_2 C_2 = A_1

but c_2=0.

if a use C_0=1, i will get E_4 = A_1/8, the result i want. But why can't i choose c_0= other value?

for m=0,n=0, itwill become:

2 E_0 C_2 = -A_2

if E_0 = 1, i have c_2 = -A_2/2,the result i want.

But why i have to set E_0 = C_0 = 1?

For example, for n=0,m=0, constant is a homogeneous solution. it is because of that (any constant = constante-1 + 1)?

Thank you!edit: forget Em and Cn. i can write Cmn. right? with that i have got the right solution. is that, right?

:)