```Date: Apr 10, 2013 3:24 AM
Author: Torsten
Subject: Re: Solving an unusual system of ODES

"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-LindelĂ¶f 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 wishesTorsten.
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