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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)}]
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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 10 years ago
 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)?