Time independent perturbation theory for solving coupled differential equations

The eqexact1 and eqexact2 are the coupled differential equation of motion with g lets say a repulsive factor, that I choose.

eqexact1:=4 \[Pi] r^2 (-mu1 p1[r] + r^2 p1[r] + g p1[r] p2[r]^2) - 8 \[Pi] r Derivative[1][p1][r] - 4 \[Pi] r^2 (p1^\[Prime]\[Prime])[r]

eqexact2:= 4 \[Pi] r^2 (-mu2new u u1 p2[r] + r^2 u u1 p2[r] + g p1[r]^2 p2[r]) -8 \[Pi] r u1Derivative[1][p2][r] - 4 \[Pi] r^2 u1 (p2^\[Prime]\[Prime])[r]


In the beginning I solved it for g equal zero with initial condition of p1[0]==1 and p1'[0]==0 and limit p2[r] with r->Inf==0 using power series of eqexact1,2 around r, demanding the terms to be equal to zero and I ended up with the following solutions.

 g := 0
p1[r_] := E^(-(r^2/2)) Hypergeometric1F1[3/4 - mu1/4, 3/2, r^2]
p2[r_] := E^(-(1/2) r^2 Sqrt[u])* HermiteH[1/2 (-1 + mu2new Sqrt[u]), r u^(1/4)])/r


How can I find a solutions with g not equal with zero assuming that I don't expect the form of the solutions to change a lot, with the term g p1[r] p2[r]^2 or g p1[r]^2 p2[r] which can be large with the choice of a large g parameter?

• The boundary condition on p2 determines it only up to a constant multiplier. Also, are there any constraints on the constants in the equations? Oct 14, 2023 at 1:49
• Yes the boundary condition on p2 just gives a constant multiplier . No there no constraints on the constants in the equation. Oct 14, 2023 at 6:13