- #1
tau1777
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I am trying to solve four coupled equations. Three of them are first order differential equations and the fourth is a algebraic one. The equations look something like this:
V[itex]_{l}[/itex](r) = f[itex]_{1}[/itex](r)W'[itex]_{l}[/itex](r) (1)
h''[itex]_{l}[/itex] + f[itex]_{2}[/itex](r)h'[itex]_{l}[/itex] + f[itex]_{3}[/itex](r)h[itex]_{l}[/itex](r) = U[itex]_{l}[/itex](r) (2)
f[itex]_{4}[/itex](r)U'[itex]_{l}[/itex] + f[itex]_{5}[/itex](r)h'[itex]_{l}[/itex] + f[itex]_{6}[/itex]W'[itex]_{l}[/itex] + f[itex]_{7}[/itex]V'[itex]_{l}[/itex]= $ (3)
U[itex]_{l}[/itex]+ h[itex]_{l}[/itex] + f[itex]_{8}[/itex](V[itex]_{l}[/itex] + W[itex]_{l}[/itex]) = 0 (4)
I didn't explicitly write this out for every term but everything is a function of 'r', and they are defined on the domain {0, R}. Also I did lie about having only four equations, the index l runs from l=2, to as large as my computer can handle (lmax). So I have (lmax-2)*4 equations.
I am trying to use a finite difference scheme and the following boundary conditions:
W[itex]_{l}[/itex](R) = 0 , V[itex]_{l}[/itex](0) = U[itex]_{l}[/itex](0) = 0 and
h[itex]_{l}[/itex](R) = houtside[itex]_{l}[/itex](R). What I mean by this is that we know what the function h[itex]_{l}[/itex](r) is outside the domain so the solution inside should match the know solution outside.
The function houtside[itex]_{l}[/itex](R) =[itex]\Sigma \frac{(l+s)!(l-s)!}{(s+2)!}... (2M/R) h_{l,0} [/itex]
The '...' in the equations above means that there are more terms such as (l+s)! before (2M/R) but they are all multiplicative only.
Thanks for reading thus far, so here is the question. How can I determine [itex]h_{l,0}[/itex]?
I know how to implement the other boundary conditions in my code but unless I know [itex]h_{l,0}[/itex], I do not believe I can continue.
I have tried taking the know outside solution and plugging it into equation (2) above, but then the trouble is what is U[itex]_{l}[/itex](R)? I know that U[itex]_{2}[/itex](R) =1 but I'm not sure I can extend this to all values of l.
Thanks for the help.
V[itex]_{l}[/itex](r) = f[itex]_{1}[/itex](r)W'[itex]_{l}[/itex](r) (1)
h''[itex]_{l}[/itex] + f[itex]_{2}[/itex](r)h'[itex]_{l}[/itex] + f[itex]_{3}[/itex](r)h[itex]_{l}[/itex](r) = U[itex]_{l}[/itex](r) (2)
f[itex]_{4}[/itex](r)U'[itex]_{l}[/itex] + f[itex]_{5}[/itex](r)h'[itex]_{l}[/itex] + f[itex]_{6}[/itex]W'[itex]_{l}[/itex] + f[itex]_{7}[/itex]V'[itex]_{l}[/itex]= $ (3)
U[itex]_{l}[/itex]+ h[itex]_{l}[/itex] + f[itex]_{8}[/itex](V[itex]_{l}[/itex] + W[itex]_{l}[/itex]) = 0 (4)
I didn't explicitly write this out for every term but everything is a function of 'r', and they are defined on the domain {0, R}. Also I did lie about having only four equations, the index l runs from l=2, to as large as my computer can handle (lmax). So I have (lmax-2)*4 equations.
I am trying to use a finite difference scheme and the following boundary conditions:
W[itex]_{l}[/itex](R) = 0 , V[itex]_{l}[/itex](0) = U[itex]_{l}[/itex](0) = 0 and
h[itex]_{l}[/itex](R) = houtside[itex]_{l}[/itex](R). What I mean by this is that we know what the function h[itex]_{l}[/itex](r) is outside the domain so the solution inside should match the know solution outside.
The function houtside[itex]_{l}[/itex](R) =[itex]\Sigma \frac{(l+s)!(l-s)!}{(s+2)!}... (2M/R) h_{l,0} [/itex]
The '...' in the equations above means that there are more terms such as (l+s)! before (2M/R) but they are all multiplicative only.
Thanks for reading thus far, so here is the question. How can I determine [itex]h_{l,0}[/itex]?
I know how to implement the other boundary conditions in my code but unless I know [itex]h_{l,0}[/itex], I do not believe I can continue.
I have tried taking the know outside solution and plugging it into equation (2) above, but then the trouble is what is U[itex]_{l}[/itex](R)? I know that U[itex]_{2}[/itex](R) =1 but I'm not sure I can extend this to all values of l.
Thanks for the help.
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