The bvp system of ode's I am having for this problem is,
ode1 = f'''[y] + a * f'[y]*f'[y] - f[y]*f''[y] == 0
ode2 = T''[y] + b *f[y]*T'[y] == 0
with boundary conditions
bcs = {f[0] == 0, f'[0] == 1, f'[L] == 0, T[0] == 1, T[L] == 0};
where L=5. Taking some random values for a, b and c,NDSolve report stiffness. The expression which should take values from NDSolve is
SE = f'[y]^2 + b*T[y]^2 + c*f[y]*T[y]; (*at y = 0*)
I want to vary a$\in[0.1,0.5]$, b$\in[0.1,1]$ and c$\in[0,2]$ by 0.1 simultaneously for both the system and the secondary expression SE to create a data table for SE of the form,
a b c SE
Can someone please help?
Edit
The above system is just a test case. So ignore the convergence warnings. I am only interested in generating the table.
test = ParametricNDSolveValue[{ode1, bcs[[1 ;; 2]], f''[0] == ic} /. a -> 5/10, f, {y, 0, L}, ic]; Manipulate[ListLinePlot[test[ic]' // Quiet, PlotRange -> {{0, 5}, {-10, 10}}], {ic, -100, 100, 1/2}], I believe it's a weak solution problem. To overcome the stiffness, I guess a small modification is necessary forode1, but finding it by pure math analysis is somewhat beyond my reach. ) – xzczd Dec 27 '16 at 07:28SE.– zhk Dec 27 '16 at 07:37a=5/10,f'[L]seems to be always<0:te = ParametricNDSolveValue[{ode1, bcs[[;; 2]], f''[0] == ic} /. a -> 5/10, f, {y, 0, L}, ic]; Plot[te[ic]'[L], {ic, -3, 1}]– xzczd Dec 27 '16 at 08:31