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Shooting Method Help

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  • Subject: [mg97456] Shooting Method Help
  • From: SK <skhushrushahi at>
  • Date: Fri, 13 Mar 2009 04:53:52 -0500 (EST)


Im sure if anyone has done the shooting method on two variables
instead of one would find this problem easy to tackle.
I have two coupled differential equations one in the variable named
vphi[Rr] and the other wz[Rr] both are a function of Rr. My notebook
file can be accessed at

I only have two boundary conditions to satisfy. They are vphi[1]==0
and vphi[0]==0.

Since eqn1 has vphi''[Rr] as second order. It needs two initial
conditions, one is vphi[1]=0 and the other I decided to implement the
shooting method on to get vphi[0]==0.

So the initial condition vphi'[1]=A but since I have eqn2 with wz'[Rr]
in it I need an initial condition on wz as well. I decided to use the
shooting method on it as well so I kept wz[1]==B.

I essentially am doing the shooting method with two variables, A and B
but since I dont have any other boundary condition to satisfy on wz,
when it comes time to solve for the roots in the shooting method I
need two equations to solve for A and B

One equation I use to set vphi[0]==0 but I dont know what to set the
second one to for it to find a root for both A and B. So my findroot
expression looks like this

rooteqn =
 FindRoot[{fpA[A, B] == 0,
   fpA[A, B] == 0}, {{A, {Al, Au}}, {B, {Bl, Bu}}}]

where I repeat the function twice over

fpend[A_, B_] := vphi[0] /. NDSolve[{Eqnstosolve, vphi[1] == 0,
\!\(\*SuperscriptBox["vphi", "\[Prime]",
MultilineFunction->None]\)[1] == A, \[Omega]z[1] ==
     B}, {vphi, \[Omega]z}, {Rr, 0, xend}, MaxSteps -> Infinity]

This results in a result for vphi'[0]==A but it doesnt actually solve
for values of B. It gives me the same result for whatever bounds I
select for B.

How do I solve for both A and B? Is there another way to implement
this so I can solve for vphi[Rr] and wz[Rr]
Any help on this matter is greatly appreciated.

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