Gene
Posts:
35
Registered:
11/2/07


Re: Third order nonlinear boundary value problem
Posted:
Mar 28, 2014 9:59 AM


"Rachel " <racheldore20@hotmail.com> wrote in message <lh1o34$9lh$1@newscl01ah.mathworks.com>... > Hey guys, > > I'm having a lot of trouble trying to get a numerical solution the the following boundary value problem, > > f'''+1/3ff''+1/3f'^2=0 > > with the following boundary conditions, > > eta=0, f=0 & f''=0 > eta=infintiy, f'=0 > > I am new to Matlab and I'm completely lost so your help would be much appreciated. I think this is how I set it up but I'm not sure?? > > function dfdeta = mat4ode(eta,f) > dfdeta = [ f(2) > f(3) > 1/3*f(1)*f(3)1/3*Y(2)*Y(2) ]; > > and for the boundary conditions, > > function res = mat4bc(ya,yb) > res = [ ya(3) > ya(1) > yb(2)]; > > And this is an attempt at the rest the code, > > function mat4bvp(solver) > > if nargin < 1 > solver = 'bvp4c'; > end > bvpsolver = fcnchk(solver); > > infinity = 3; > > solinit = bvpinit(linspace(0,infinity,5),[0,0,0]); > > sol = bvpsolver(@mat4ode,@mat4bc,solinit); > > eta = sol.x; > f = sol.y; > > figure(1) > plot(eta,f) > legend('F_1', 'F_2', 'F_3', 3) > grid > > end > > I would be very grateful for the help. Thanks a mill. > > Rachel
Hi Rachel:
I believe your ODE can be rewritten as f''' + (1/3) (f f')' = 0 which integrates to f'' + (1/3) (f f') = C Your b.c. imply C = 0 The 2nd order system also admits a first integral f' + (1/6) f^2 = C; again the b.c. imply C = 0 The result is now 1st order and separable.
gene

