Mathematica Asked by Lele on May 26, 2021
I know that this is a "beginner question", and, in fact, I am. I want to improve my code, as it takes really too much time to run.
I have already read some other discussions, like here,
but I am struggling in translating the simplifications with Table
or Do
into my example.
The For
cycle I want to improve is the following:
zh = 0.4;
list = {};
For[i = 1, i < 10000, i++,
With[{zg = -0.6, dh = i*10^-2},
nsol = Block[{eps = $MachineEpsilon},
NDSolve[{phi''[x] + 2*phi'[x]/x + (2*(zg + zh*Exp[-zh*x/dh])/x + 1)*phi[x] == 0, phi[eps] == 1, phi'[eps] == -(zg + zh)}, phi, {x, eps, 20000},
WorkingPrecision->MachinePrecision, AccuracyGoal->15, PrecisionGoal->8, MaxSteps->Infinity]]];
AppendTo[list, 1/Evaluate[(15000*phi[15000])^2 + ((15000-Pi/2)*phi[15000-Pi/2])^2 /. nsol[[1]]]];]
Clearly, this code, written in this way, is highly inefficient. Also, I need to do more of these, with different values for zg
inside With
, and make some plots out of the lists.
Anyone that can help me with this noob question?
Thanks a lot!
Here is an example that uses Table
(well, ParallelTable
) instead of For
. I've also used ParametricNDSolveValue
instead of With
, mostly to simplify the Table
.
phifunc = ParametricNDSolveValue[
{[Phi]''[x] + 2 [Phi]'[x]/x + (2 (zg + zh Exp[-zh x/dh])/x + 1) [Phi][x] == 0
, [Phi][$MachineEpsilon] == 1
, [Phi]'[$MachineEpsilon] == -(zg + zh)
}
, 1/((15000 [Phi][15000])^2 + ((15000 - [Pi]/2) [Phi][
15000 - [Pi]/2])^2)
, {x, $MachineEpsilon, 20000}
, {dh [Element] Reals, zg [Element] Reals, zh [Element] Reals}
]
list = ParallelTable[phifunc[i 10^-2, -0.6, 0.4], {i, 1, 10000}]
As far as timing goes, it still seems to take a very long time to run 10000 evaluations, so another answer might provide a faster method. I vaguely recall a way to functionalize calls to NDSolve
in a way that stores calculations for faster repeated calls, but I can't find the link.
Answered by Josh Bishop on May 26, 2021
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