k1 = 1;
k2 = 5;
t0 = 0;
ts = 10000000;
t01 = ts - ts/3;
t02 = 7000;
tfnh = 500;
t0nh01 = tfnh/3;
tf = ts;
bv2 = 0;
bb2 = 0;
fh = 0.99;
b10 = 1.0
h10 = fh*b10*(k2/k1)
h20 = 0
Rm2 = 12000;
soln2dec =
NDSolve[{
h1'[t] == (2/3)*h2[t]*b1[t] - (2/3) (k1/k2)^2*h1[t] - 2*h1[t] (k1/k2)^2/Rm2,
h2'[t] == -(2/3)*h2[t]*b1[t] + (2/3) (k1/k2)^2*h1[t] - 2*h2[t]/Rm2,
b1'[t] == (2/3) h1[t]*h2[t] (k1/k2)^2 - (2/3)*
b1[t] (k1/k2)^2 - (2/Rm2) b1[t] (k1/k2)^2, h1[0] == h10,
b1[0] == b10, h2[0] == h20}, {h1, h2, b1, h1', h2', b1'}, {t, t0, ts},
MaxSteps -> 15000000]
Export["logb1k5Rm12000.dat", Evaluate[{b1[t]} /. soln2dec], "TSV"]
Export["logh2k5Rm12000.dat", Evaluate[{h2[t]} /. soln2dec], "TSV"]
Export["logh1k5Rm12000.dat", Evaluate[{h1[t]} /. soln2dec], "TSV"]
Export["logh1dk5Rm12000.dat", Evaluate[{h1'[t]} /. soln2dec], "TSV"]
Export["logh2dk5Rm12000.dat", Evaluate[{h2'[t]} /. soln2dec], "TSV"]
Export["logb1dk5Rm12000.dat", Evaluate[{b1'[t]} /. soln2dec], "TSV"]
The first three exports rightly retrieve the data. But the last three seem to retrieve the function itself and not the derivatives.
Is there a way out ?
Export["logb1k5Rm12000.dat", Evaluate[{b1[t]} /. soln2dec], "TSV"]
won't give you what you expect... did you checked it? $\endgroup$