Problem with too many inputs into bvp5c

Hello, I want to test whether bvp4c, bvp5c and bvp6c solvers produce the same solution. The problem I face is that bvp4c and bvp6c allow my ode and bc functions to have additional inputs. Namely, these solvers have the following syntax: function sol = bvp4c(ode, bc, solinit, options, varargin) function sol = bvp6c(ode, bc, solinit, options, varargin) The solver bvp5c however, does not allow for varargin: function sol = bvp5c(ode, bc, solinit, options)

So, does anyone know how to go about this issue? Is there a way to manually edit bvp5c to allow for additional inputs?

Thanks, Artem.

Answer 1

[Jan]

You should not use additional inputs at all, even some integrators still support this ancient method. See http://www.mathworks.com/matlabcentral/answers/1971: Use anonymous functions to provide parameters. This is explained in the documentation also: https://www.mathworks.com/help/matlab/ref/bvp5c.html:

 Parameterizing Functions explains how to provide additional 
 parameters to the function odefun, ...

with a link to: https://www.mathworks.com/help/matlab/math/parameterizing-functions.html.

[Artem Korzhenevych]

Thanks for the fast reaction, Jan, but I guess this is beyond my programming skills. This is my bvp call:

Solution = bvp4c(@odefun,@bc,Solution_guess,OPTIONS,par);

This is my ode function with endogenous parameters end_par and exogenous parameters par that I need to get around:

function xdot=odefun(t,x,end_par,par)

And this is my boundary condition:

function lhs=bc(x0,xt,end_par,par)

Now, at which place do I need to change smth? I see that I need to use @ handles, but the examples in the documentation are not matching my problem (more than one x as function argument).

[Artem Korzhenevych]

OK, seems I figured it out. Just needed to change the bvp call:

Solution = bvp4c(@(t,x,end_par) odefun(t,x,end_par,par), @(x0,xt,end_par) bc(x0,xt,end_par,par), Solution_guess, OPTIONS);