I seek to prove a solution in power series in y
or by numerical methods for the second order nonlinear differential equation:
nlde = {Sqrt[1 - f[y]^2] f''[y] + y == 0, f'[0] == 0, f[0] == 1};
It seems that there is an analytical answer in power series for this ODE (see appendix in this paper):
My failed attempts to prove the analytical solution with Mathematica:
Power series:
AsymptoticDSolveValue[nlde, f, {y, 0, 4}]
(* Indeterminate expression 0 ComplexInfinity encountered*)
Series[DifferentialRoot[Function @@ {{f, y}, nlde}][y], {y, 0, 4}]
(* supplied equation is not a
linear differential equation with polynomial coefficient *)
Numerical:
NDSolveValue[nlde, f, {y, 0, 4}]
(* Infinite expression encountered*)
Anyone has any suggestions on how to solve the equation? Thanks!
t=0
! $\endgroup$SolveAlways[With[{f = C[0] + Sum[C[k] y^k, {k, 1, 6}] + O[y]^7}, Sqrt[1 - f^2] D[f, {y, 2}] + y == 0], y]
, you might get an idea on what constraints your series coefficients have to follow for a power series to satisfy your nonlinear ODE. $\endgroup$SolveAlways[]
with formula A6:1 + Sum[C[k] y^k, {k, 2, 10}] /. With[{g = -Sum[C[k] y^(k - 2), {k, 2, 10}] + O[y]^(11 - 2)}, Last[SolveAlways[g (2 - y^2 g) (y^2 D[g, {y, 2}] + 4 y D[g, y] + 2 g)^2 == 1, y]]]
$\endgroup$SolveAlways
is picking even the other solutions for the imaginary coefficients! $\endgroup$