Shooting Method Help

*To*: mathgroup at smc.vnet.net*Subject*: [mg97456] Shooting Method Help*From*: SK <skhushrushahi at gmail.com>*Date*: Fri, 13 Mar 2009 04:53:52 -0500 (EST)

Hi 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 http://web.mit.edu/~shahriar/Public/shootingmethodhelp.nb 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.