Asymptotic inverse function?

Question

A paper I am reading defines a variable $\theta$ in terms of another variable $\phi$ as an expansion in $u$, where $u$ is understood to be small: $$\theta=\phi-u^2\sin\phi+\mathcal{O}(u^4).$$ They then go on to "invert this relation in the small $u$ limit" and use it to eliminate $\phi$ from another equation. How can I do this in Mathematica?

If I try the obvious guess

Theta[Phi] := Phi - u^2 Sin[Phi]
InverseFunction[Theta][Phi]

The output is just Theta^(-1)[Phi].

If I try

AsymptoticSolve[Phi-u^2 Sin[Phi] == Theta, Phi, {u, 0, 1}]

I get AsymptoticSolve[Phi-u^2 Sin[Phi] == Theta, Phi, {u, 0, 1}] as the output.

Clearly the authors of this paper have managed to find a way to invert this relation, so I am wondering if anyone has any ideas on how to do so. Thanks.

Answer 1

AsymptoticSolve[θ == φ - u^2 Sin[φ], {φ, θ}, {u, 0, 10}]

(*    {{φ -> θ + u^2 Sin[θ] + 
             u^4 Cos[θ] Sin[θ] + 
             1/2 u^6 (2 Cos[θ]^2 Sin[θ] - Sin[θ]^3) + 
             1/3 u^8 (3 Cos[θ]^3 Sin[θ] - 5 Cos[θ] Sin[θ]^3) + 
             1/24 u^10 (24 Cos[θ]^4 Sin[θ] - 88 Cos[θ]^2 Sin[θ]^3 + 13 Sin[θ]^5)}}    *)

Answer 2

@Roman's answer is exactly what you asked for, but as a check, here is a second answer using Bessel's series expansion for the solution of the Kepler equation:

getExpansion[cut_]:=θ+Sum[2/m*BesselJ[m,m*u^2]*Sin[m*θ],
                          {m,1,Ceiling[cut/2]}]//Series[#,{u,0,cut}]&

For example

getExpansion[10]
(* θ
   +Sin[θ] u^2
   +1/2 Sin[2 θ] u^4
   +(-(Sin[θ]/8)+3/8 Sin[3 θ]) u^6
   +(-(1/6) Sin[2 θ]+1/3 Sin[4 θ]) u^8
   +(Sin[θ]/192-27/128 Sin[3 θ]+125/384 Sin[5 θ]) u^10
   +O[u]^11
*)

It coincides with @Roman's answer.