I need to solve a second order ODE numerically. The ODE depends on two parameters (a,b). Things work fine when 'a' is small, but for large 'a' the solutions are oscillating rapidly and Mathematica takes a long time to solve or eventually just runs out of memory.
I need to integrate over quite a large range (10,000) and that is part of the problem, but I actually only need the value of the InterpolatingFunction produced at the end point. Is there a way to tell Mathematica I just want this last point? And not store the rest (very large) InterpolatingFunction in memory? i.e. just integrate so far, take that point use as ICs for next leg, then integrate to next pt, take that as IC and integrate onwards,etc...
Or just some other strategy for using NDSolve with such highly oscillatory solutions.
Some definitions:
M=1;
rstar[r_] := r + 2 M Log[r/(2 M) - 1];
$MinPrecision = 45;
wp = $MinPrecision;
ac = $MinPrecision - 8;
\[Lambda][l_] = l (l + 1);
rinf = 10000;
rH = 200001/100000;
nH = 200;
The ODE is:
eq[\[Omega]_, l_] := \[CapitalPhi]''[r] + (2 (r - M))/(
r (r - 2 M)) \[CapitalPhi]'[
r] + ((\[Omega]^2 r^2)/(r - 2 M)^2 - \[Lambda][l]/(
r (r - 2 M))) \[CapitalPhi][r] == 0
Without going into detail about why I have these initial conditions, they are :
HorizonICs[l_?IntegerQ, \[Omega]_?NumericQ] :=
Module[{\[CapitalPhi]inrH, d\[CapitalPhi]inrH},
Clear[b];
b[0] = 1; b[-1] = 0; b[-2] = 0;
b[n_] :=
b[n] = Simplify[
1/(2 n (n - 4 I \[Omega])) (-2 I \[Omega] b[-2 + n] +
2 I n \[Omega] b[-2 + n] + l b[-1 + n] + l^2 b[-1 + n] +
n b[-1 + n] - n^2 b[-1 + n] - 4 I \[Omega] b[-1 + n] +
8 I n \[Omega] b[-1 + n])];
uintrunc[r_, n_] := Sum[b[i] (r - 2 M)^i, {i, 0, n}];
\[CapitalPhi]inrH = (Exp[-I \[Omega] rstar[rH]] uintrunc[rH, nH])/(
2 M);
d\[CapitalPhi]inrH =
D[(Exp[-I \[Omega] rstar[r]] uintrunc[r, nH])/(2 M), r] /.
r -> rH;
]
Solve as
\[CapitalPhi]inExt[\[Omega]\[Omega]_ ,
l_] := \[CapitalPhi]inExt[\[Omega]\[Omega], l] = \[CapitalPhi] /.
NDSolve[{eq[\[Omega]\[Omega], l], \[CapitalPhi][rH] ==
N[HorizonICs[l, \[Omega]\[Omega]][[1]], wp], \[CapitalPhi]'[
rH] == N[HorizonICs[l, \[Omega]\[Omega]][[2]],
wp]}, \[CapitalPhi], {r, rH, rinf}, WorkingPrecision -> wp,
AccuracyGoal -> ac, MaxSteps -> \[Infinity],
Method -> "StiffnessSwitching"][[1]];
You can also look for another solution that has the property of being simple near infinity and we set the ICs there. This is the particular solution that seems to be really really slow and causes memory crash.
Some definitions:
M = 1;
r0 = 5/2;
rstar[r_] := r + 2 M Log[r/(2 M) - 1];
$MinPrecision = 45;
wp = $MinPrecision;
ac = $MinPrecision - 8;
\[Lambda][l_] = l (l + 1);
rinf = 10000;
ninfphase =
50;
Set init conditions:
Infinitycs = Module[{n = ninfphase, c},
Clear[c];
veqexp =
CoefficientList[
Series[(-2 - l r - l^2 r + 2 (r + I r^3 \[Omega])
\!\(\*SuperscriptBox["v", "\[Prime]",
MultilineFunction->None]\)[r] + (-2 + r) r^2
\!\(\*SuperscriptBox["v", "\[Prime]",
MultilineFunction->None]\)[r]^2 + (-2 + r) r^2
\!\(\*SuperscriptBox["v", "\[Prime]\[Prime]",
MultilineFunction->None]\)[r])/
r /. {v'[r_] :> Sum[-i c[i]/r^(i + 1), {i, 1, n}],
v''[r_] :>
Sum[i (i + 1) c[i]/r^(i + 2), {i, 1, n}]}, {r, \[Infinity],
n - 1}], r^-1];
Do[c[i] = c[i] /. Simplify[Solve[veqexp[[i]] == 0, c[i]][[1]]];
Print, {i, 1, n}] ;
Table[c[i], {i, 1, n}]];
InfinityICs[ll_?IntegerQ, \[Omega]\[Omega]_?NumericQ] := Module[{c2},
Do[c2[i] =
Infinitycs[[i]] /. {l -> ll, \[Omega] -> \[Omega]\[Omega]}, {i, 1,
ninfphase}];
vtrunc = Sum[c2[i]/r^i, {i, 1, ninfphase}];
init = 1/r Exp[I \[Omega]\[Omega] rstar[r] + vtrunc] /. r -> rinf;
dinit =
D[1/r Exp[I \[Omega]\[Omega] rstar[r] + vtrunc], r] /. r -> rinf;
Clear[c2];
{init, dinit}]
Solve it
\[CapitalPhi]out[\[Omega]\[Omega]_,
l_] := \[CapitalPhi]out[\[Omega]\[Omega], l] = \[CapitalPhi] /.
Block[{$MaxExtraPrecision = 100},
NDSolve[{eq[\[Omega]\[Omega], l], \[CapitalPhi][rinf] ==
N[InfinityICs[l, \[Omega]\[Omega]][[1]], wp], \[CapitalPhi]'[
rinf] ==
N[InfinityICs[l, \[Omega]\[Omega]][[2]],
wp]}, \[CapitalPhi], {r, rinf, r0}, WorkingPrecision -> wp,
AccuracyGoal -> ac, MaxSteps -> \[Infinity]]][[1]];

Burgers equation(although that is a PDE). What does your equation look like? What does your mathematica code look like? I wonder if it is because of your history length in case you are using the%for the previous step. (Try setting $HistoryLength=0) – drN Oct 12 '12 at 21:26omegaetc.. – drN Oct 12 '12 at 22:39NDSolve[(* stuff *), {r, rinf, rinf}, (* more stuff *)]? – 0x4A4D♦ Oct 12 '12 at 22:59