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Topic: Solving an unusual system of ODES
Replies: 7   Last Post: Apr 10, 2013 3:24 AM

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Torsten

Posts: 1,477
Registered: 11/8/10
Re: Solving an unusual system of ODES
Posted: Apr 10, 2013 3:24 AM
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"Torsten" wrote in message <kk334m$mb3$1@newscl01ah.mathworks.com>...
> "Torsten" wrote in message <kk31o2$iqe$1@newscl01ah.mathworks.com>...
> > D R G <grimesd2@gmail.com> wrote in message <ea631802-c941-4e78-9832-3561e21842bf@googlegroups.com>...
> > >
> > > On Tuesday, April 9, 2013 2:15:07 PM UTC+1, Torsten wrote:

> > > > D R G <grimesd2@gmail.com> wrote in message <5c502161-10e9-4863-9e76-57617f33fb35@googlegroups.com>...
> > > >

> > > > > Yes, that is what I want to solve; however, the case you're referring to is I believe the trivial case; we have already solved for when k = 0, and got a non-trivial solution. As K is small, this will be close to it and there will be non zero solutions.
> > > >
> > > > >
> > > >
> > > > > Regards
> > > >
> > > > >
> > > >
> > > > > DRG
> > > >
> > > > >
> > > >
> > > >
> > > >
> > > > No, y=0 is the only solution that satisfies your differential equation together with the two boundary conditions - also for the case K and/or J not equal to 0.
> > > >
> > > > You will have to change the boundary conditions to get a solution different from y=0.
> > > >
> > > >
> > > >
> > > > Best wishes
> > > >
> > > > Torsten.

> > >
> > > Hmm.. that is quite unexpected! Alright, so can we extend this for the case
> > >
> > > C(RN) = 0
> > > dC/dr (RN) =! 0 ?
> > >
> > > Thanks in advance
> > >
> > > DRG

> >
> > In the case dC/dr (RN) not equal to 0 you will get a nontrivial solution.
> > When you use ODE45 to solve, just interpret the time variable as the space variable:
> >
> > dX(1) = X(2)
> > dX(2) = J*X(1)/(X(1) + K) - 2/t*X(2)
> >
> > Usually stationary second-order ODEs are boundary value problems - i.e. one boundary condition is given at RN and the second at RO. Are you sure the physics of your problem leads to two boundary conditions at RN ?
> >
> > Best wishes
> > Torsten.

>
> I see now that for the special case K=0 and J not equal to 0 you get a nontrivial solution, namely
> C(r)=1/3*J*(r^2/2 + RN^3/r - 1.5*RN^2).
> In the case K not equal to 0 I don't know if a nontrivial solution exists.
> But at least the numerical solver will always produce the trivial solution, that's for sure.
>
> Best wishes
> Torsten.
>


I think the theorem of Picard-Lindelöf guarantees that C=0 is the only solution to your problem if K is not equal to 0.
But you should check this carefully.

Best wishes
Torsten.




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