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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 = 
   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 ?

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Calling Export["logb1k5Rm12000.dat", Evaluate[{b1[t]} /. soln2dec], "TSV"] won't give you what you expect... did you checked it? –  mmal Aug 8 '13 at 14:35
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1 Answer 1

First, after running your code, I ran

Plot[Evaluate[{h1[t] /. soln2dec[[1, 1]], h1'[t] /. soln2dec[[1, 4]]}], {t, 0, 10^7}]

Output from Mathematica

They are clearly different functions. Your problem is not at the point where soln2dec was computed.

Second, I ran

Plot[Evaluate[{(h1'[t] /. soln2dec[[1, 1]]) - (h1'[t] /. soln2dec[[1, 4]])}], {t, 0, 10^7}]

The output was was a flat line at zero. This tests shows that computing h1'[t] from the interpolated h1[t] is equivalent to use the 4-th rule in soln2dec. In fact, I would modify the third parameter you passed to NDSolve from

{h1, h2, b1, h1', h2', b1'}


{h1, h2, b1}

Third, I ran a comparison tool on the seemingly same files and I got enter image description here

The first few lines are almost identical, but they are definitely different functions.

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