D R G <firstname.lastname@example.org> wrote in message <email@example.com>... > > On Tuesday, April 9, 2013 2:15:07 PM UTC+1, Torsten wrote: > > D R G <firstname.lastname@example.org> wrote in message <email@example.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 ?