differential equations - NDSolve runs out of memory

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;
λ[l_] = l (l + 1);
rinf = 10000;
rH = 200001/100000;
nH = 200;

The ODE is:

eq[ω_, l_] := Φ''[r] + (2 (r - M))/(

 r (r - 2 M)) Φ'[
 r] + ((ω^2 r^2)/(r - 2 M)^2 - λ[l]/(
 r (r - 2 M))) Φ[r] == 0

Without going into detail about why I have these initial conditions, they are :

HorizonICs[l_?IntegerQ, ω_?NumericQ] := 
 Module[{ΦinrH, dΦinrH},
 Clear[b];
   b[0] = 1; b[-1] = 0; b[-2] = 0;
   b[n_] := 

  b[n] = Simplify[
 1/(2 n (n - 4 I ω)) (-2 I ω b[-2 + n] + 
    2 I n ω b[-2 + n] + l b[-1 + n] + l^2 b[-1 + n] + 
    n b[-1 + n] - n^2 b[-1 + n] - 4 I ω b[-1 + n] + 
    8 I n ω b[-1 + n])];
  uintrunc[r_, n_] := Sum[b[i] (r - 2 M)^i, {i, 0, n}];
  ΦinrH = (Exp[-I ω rstar[rH]] uintrunc[rH, nH])/(
  2 M);
 dΦinrH = 
  D[(Exp[-I ω rstar[r]] uintrunc[r, nH])/(2 M), r] /. 

   r -> rH;
 ]

Solve as

ΦinExt[ωω_ , 
 l_] := ΦinExt[ωω, l] = Φ /. 
 NDSolve[{eq[ωω, l], Φ[rH] == 
    N[HorizonICs[l, ωω][[1]], wp], Φ'[
     rH] == N[HorizonICs[l, ωω][[2]], 
     wp]}, Φ, {r, rH, rinf}, WorkingPrecision -> wp, 

  AccuracyGoal -> ac, MaxSteps -> ∞, 
  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;
λ[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 ω) 
\!\(\*SuperscriptBox["v", "′",
MultilineFunction->None]\)[r] + (-2 + r) r^2 
\!\(\*SuperscriptBox["v", "′",
MultilineFunction->None]\)[r]^2 + (-2 + r) r^2 
\!\(\*SuperscriptBox["v", "′′",
MultilineFunction->None]\)[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, ∞, 
   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, ωω_?NumericQ] := Module[{c2},
Do[c2[i] = 
Infinitycs[[i]] /. {l -> ll, ω -> ωω}, {i, 1,
 ninfphase}];

vtrunc = Sum[c2[i]/r^i, {i, 1, ninfphase}];
init = 1/r Exp[I ωω rstar[r] + vtrunc] /. r -> rinf;
dinit = 
D[1/r Exp[I ωω rstar[r] + vtrunc], r] /. r -> rinf;
Clear[c2];
{init, dinit}]

Solve it

Φout[ωω_, 
l_] := Φout[ωω, l] = Φ /. 

Block[{$MaxExtraPrecision = 100}, 
  NDSolve[{eq[ωω, l], Φ[rinf] == 
     N[InfinityICs[l, ωω][[1]], wp], Φ'[
      rinf] == 
     N[InfinityICs[l, ωω][[2]], 
      wp]}, Φ, {r, rinf, r0}, WorkingPrecision -> wp, 
   AccuracyGoal -> ac, MaxSteps -> ∞]][[1]];

Answer

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?

One way to go about it is to have the start and end of the integration interval be identical. Consider the following:

y[5] /. First @ NDSolve[{y'[x] == y[x] Cos[x + y[x]], y[0] == 1}, y, {x, 5, 5}]
   0.07731217497157500942

To see that the approach saves space, here's a comparison:

yi = y /. First@NDSolve[{y'[x] == y[x] Cos[x + y[x]], y[0] == 1}, y, {x, 5, 5}];
ByteCount[yi]
   1296


yn = y /. First@NDSolve[{y'[x] == y[x] Cos[x + y[x]], y[0] == 1}, y, {x, 0, 5}];
ByteCount[yn]
   3288