Non Linear Differential Equations Asymptotics

Question

I wish to study the asymptotic behaviour of the following equation:

$\frac{d^2 a}{dr^2} = 2 a(r) \phi(r)^2 + B_1 a(r) (1-\phi^2(r) + B_2 a(r)^2)$

$\phi(r)\longrightarrow 1$ as $r\longrightarrow \infty$

Hence my equation boils down to:

$\frac{d^2 a}{dr^2} = 2 a(r) + A a(r)^3$ where I define $A=B_1B_2$.

This should have two solutions. I wish to get both the possible solutions since I'm interested in studying the divergent solution as well. The boundary conditions are not fixed and I wish to study the most general solution. With this context , I have the following two related questions:

1. If I wish to get the divergent solution, then $a^3$ would be leading and $a$ would be subleading and my equation looks like:

$\frac{d^2 a}{dr^2} = A a(r)^3$

It is clear that this equation has $\sqrt{\frac{2}{A}} \frac{1}{r}$ as one solution. However, I tried solving this in Mathematica

DSolve[a''[r] -  A  a [r]^3 == 0, a[r], r]

and Mathematica gave an answer in terms of JacobiSN functions.

{{a[r] -> -((
    2^(1/4) Sqrt[(I Sqrt[A])/Sqrt[C[1]]] Sqrt[C[1]]
      JacobiSN[((-1)^(3/4) Sqrt[
       Sqrt[2] Sqrt[A] r^2 Sqrt[C[1]] + 
        2 Sqrt[2] Sqrt[A] r Sqrt[C[1]] C[2] + 
        Sqrt[2] Sqrt[A] Sqrt[C[1]] C[2]^2])/Sqrt[2], -1])/Sqrt[
    A])}, {a[r] -> (
   2^(1/4) Sqrt[(I Sqrt[A])/Sqrt[C[1]]] Sqrt[C[1]]
     JacobiSN[((-1)^(3/4) Sqrt[
      Sqrt[2] Sqrt[A] r^2 Sqrt[C[1]] + 
       2 Sqrt[2] Sqrt[A] r Sqrt[C[1]] C[2] + 
       Sqrt[2] Sqrt[A] Sqrt[C[1]] C[2]^2])/Sqrt[2], -1])/Sqrt[A]}}

$\sqrt{\frac{2}{A}} \frac{1}{r}$ is not apparent amongst these solutions Mathematica gives.

So I'm not sure why I'm getting an error and how to successfully manipulate this hiccup. The $1/r$ solution also clearly violates the initial working assumption that $a^3(r) >> a(r)$, so I probably need the second solution to understand it better.

2. My second question is regarding solving the original equation as a whole and not ignoring any terms :

$\frac{d^2 a}{dr^2} = 2 a(r) + A a(r)^3$

I did the following :

DSolve[a''[r] - 2 a[r] -  A  a [r]^3 == 0, a[r], r]

and I got the following answer:

Solve[-(((2 + Sqrt[4 - 2 A C[1]]) (1 - (
      A a[r]^2)/(-2 + Sqrt[4 - 2 A C[1]])) (2 + (2 A a[r]^2)/(
      2 + Sqrt[4 - 2 A C[1]])) EllipticF[
     I ArcSinh[a[r] Sqrt[A/(2 + Sqrt[4 - 2 A C[1]])]], -((
      2 + Sqrt[4 - 2 A C[1]])/(-2 + Sqrt[4 - 2 A C[1]]))]^2)/(
   A (4 a[r]^2 + A a[r]^4 + 2 C[1]))) == (r + C[2])^2, a[r]]

This is also a bit unhelpful. Since I'm interested in the asymptotics I also tried AsymptoticDSolveValue to solve the differential equation asymptotically, but Mathematica just gave me back the input command with the equation rearranged.

Any suggestions in getting the asymptotics for these equations, by some other analytic or numerical method which could possibly be better suited would be greatly helpful. Thanks in advance.

Answer 1

Hint.

Considering $\phi(r)\approx 1$ we have

$$ a'' = 2a+A a^3 $$

thus

$$ a'' a' =2 a a'+A a^3 a' $$

and after integration

$$ \frac 12 a'^2=a^2+\frac A4 a^4 + c_0 $$

now considering in the phase plane $(x,y)$ the orbits for

$$ \frac 12 x^2 = y^2+\frac A4 y^4 + c_0\ \ \ \ \ \ (*) $$

we have

$$ y^2=\frac{-2\pm \sqrt{4+2A(x^2-2c_0)}}{A} $$

now as $c_0$ depends on initial conditions, we can argue when $(*)$ represent closed or open orbits. Open orbits having asymptotic behavior and closed orbits representing periodic behavior for the solutions of $a(r)$.

NOTE

Here $c_0 = \frac 12a'(0)^2-a(0)^2-\frac A4 a(0)^4$. Follows some orbits first for $A=1\gt 0$ and $-10\le c_0\le 10$ showing asymptotic behavior

and also for $A=-1\lt 0$ and $-10\le c_0\le 10$ showing periodic behavior.