Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Torsten
Posts:
1,717
Registered:
11/8/10


Re: Solving an unusual system of ODES
Posted:
Apr 10, 2013 3:24 AM


"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 <ea631802c9414e7898323561e21842bf@googlegroups.com>... > > > > > > On Tuesday, April 9, 2013 2:15:07 PM UTC+1, Torsten wrote: > > > > D R G <grimesd2@gmail.com> wrote in message <5c50216110e948639e7657617f33fb35@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 nontrivial 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 secondorder 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 PicardLindelöf 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.



