Consider the following system defined on the open interval (-1, 1):

$y_1' = c y_3 \\ y_2' = c y_4 \\ y_3' = -f(y_1, y_2)y_2 \\ y_4' = f(y_1, y_2)y_1 $

given $ y_3(-1) = 0 = y_3(1) \\ y_4(-1) = 1, y_4(1) = -1 $

Note that the system cannot be reduced to the standard form of two point BVP in $y_3, y_4$ by eliminating $y_1, y_2$.

What methods/packages can I use to solve this system ?

  • 2
    $\begingroup$ What do you mean by 'standard form'? Matlab's bvp4c or any collocation implementation should well work for you. $\endgroup$ – Jan Jan 8 '15 at 11:28
  • $\begingroup$ By standard form, I meant a system of two-point BVPs in $y_3, y_4$ eliminating $y_1,y_2$. Thank you for your comment, I looked at bvp4c and it does do the job. $\endgroup$ – me10240 Jan 9 '15 at 15:38

It's true that you can't eliminate $y_1,y_2$ to obtain something where every variable you have has appropriate boundary values. However, you can take derivatives on both sides of the first two equations to get $$y_1'' = c y_3' \\ y_2'' = c y_4' \\ y_3' = -f(y_1, y_2)y_2 \\ y_4' = f(y_1, y_2)y_1 $$ and then eliminate $y_3',y_4'$: $$ y_1'' = -c f(y_1, y_2)y_2 \\ y_2'' = c f(y_1, y_2)y_1 $$ Now you just need boundary values. If you recall that $y_1'=cy_3$ and similarly $y_2'=cy_v$, then they read $$ y_1'(-1) = 0 = y_1'(1) \\ y_2'(-1) = 1, \\ y_2'(1) = -1. $$ In other words, you have a coupled nonlinear second-order system in $y_1,y_2$ with Neumann boundary values.

  • $\begingroup$ I had thought along similar lines before posting the question. However I started having doubts about the validity of the boundary condition generated this way because, the given ODE system is defined on the open interval (-1,1), and boundary conditions are independently supplied at the end points -1 and 1. Given the boundary condition $y_3(1)$, is it valid to assume that $y_1'(1) = cy_3(1)$ ? Please correct me if I am wrong. $\endgroup$ – me10240 Jan 9 '15 at 15:36
  • 2
    $\begingroup$ Sure. If your solution is sufficiently smooth, then this condition can be extended from the interior to the boundary. $\endgroup$ – Wolfgang Bangerth Jan 9 '15 at 15:41

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