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Is it possible to use a functional limit as an initial condition in NDsolve

Posted 10 years ago
Want to get this to work, but it's saying the system is overdetermined.

m = 7/34169*10^(-6)*mp;
mp = 2.435*10^(18);
k = 8*10^(-41);

NDSolve[{u'' + (-2/n^2 + k^2 + m^2/H^2/n^2)*u == 0,
   Normal[Series[u, {n, Infinity, 0}]] == Exp/(2*k)^(1/2)},
  u, {n, 8*10^(38), 1.13*10^(42)}]
POSTED BY: Jessica Cook
2 Replies
Posted 10 years ago
It's a differential equation. I figured you couldn't put the limit for NDsolve to Infinity. Besides, I don't care what the function looks like all the way to infinity. I just care that it has that functional form for n -> Infinity.
POSTED BY: Jessica Cook
 m = 7/34169*10^(-6)*mp;
 
 mp = 2.435*10^(18);
 k = 8*10^(-41);
 H = 1; (* was missing *)
 
 NDSolve[{u''[n] + (-2/n^2 + k^2 + m^2/H^2/n^2)*u[n] == 0,
   Normal[Series[u[n], {n, Infinity, 0}]] == Exp[(2*k)^(1/2)]}, u, {n,
   8*10^(38), 1.13*10^(42)}]

NDSolve::overdet: There are fewer dependent variables, {u[n]}, than equations, so the system is overdetermined. >>
is this meant to be a differential equation (why then there are conditions {n, Infinity, 0} outside the range {n, 8 10^38, 1.13 10^42) or is this meant to be a recurrence equation (why then are the derivatives)?
POSTED BY: Udo Krause
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