TITLE: Solve functional equation $ h(y)+h^{-1}(y)=2y+y^2 $
QUESTION [14 upvotes]: I was trying to solve a certain physics problem, and encountered the functional equation that contains a function $h$ and its inverse $h^{-1}$:
\begin{equation}
h(y)+h^{-1}(y)=2y+y^2.\tag{1}
\end{equation}

Q: Does $(1)$ have a unique solution and is it possible to find it in closed form?

Equation $(1)$ looks quite simple and probably it is simple to analyze too, but I couldn't figured out how.

Background information: The physics problem I was trying to solve is finding the dependence of the current $J_0$ on the voltage $U_0$ in this infinite chain that contains nonlinear elements with quadratic volt-ampere characteristics $I(V)=\alpha V^2$ and ohmic resistors $R$:

According to dimensional analysis one can write
$$
J_0(U_0)=\frac{1}{\alpha R^2}f(\alpha RU_0).
$$
Solving simple system of equations I obtained the following functional equation for the unknown function $f$
$$
f(x)=(x-f(x))^2+f(x-f(x)).\tag{2}
$$
Now introduce another function $h$ according to
$$
x-f(x)=h(x).
$$
Then equation $(2)$ becomes
$$
x-h(x)=h^2(x)+h(x)-h(h(x)).
$$
Let $h(x)=y$, then $x=h^{-1}(y)$ and we get $(1)$.
This is not a textbook problem and I don't even know whether it has a solution. I was studying it out of curiosity.

REPLY [4 votes]: Unfortunately, your proposed functional equation:
$$ h(y)+h^{-1}(y)=2y+y^2 \tag{1} $$ has no
differentiable solutions near $\,y=0.\,$ Suppose that
$$ h(y) = 0 + a_1 y + O(y^2). \tag{2} $$
Then its inverse function is
$$ h^{-1}(y) = 0 + \frac1{a_1} y + O(y^2). \tag{3} $$
Substituting equations $(2)$ and $(3)$ into $(1)$ gives
$$ h(y)\!+\!h^{-1}(y) = \frac{a_1^2+1}{a_1}y \!+\! O(y^2) 
= 2y \!+\! O(y^2). \tag{4}$$
The only solutions for $\,a_1\,$ are $\,a_1=1\,$ and
$\,a_1=-1.\,$ However, we are given that
$$ 0<h(x)<x \tag{5}$$
and thus the only possible value is $\,a_1=1.\,$
Suppose we want more terms in the power series
$$ h(y) = 0 + y + a_2y^2 + O(y^3). \tag{6}$$
The inverse function is now
$$ h^{-1}(y) = 0 + y - a_2y^2 + O(y^3). \tag{7}$$
Adding these two equations together is the equation
$$ h(y) + h^{-1}(y) = 0 + 2y + 0y^2 + O(y^3). \tag{8}$$
This implies that equation $(1)$ can not be satisfied.
There is a possibility that there exists an exponent 
$\,e\,$ not an integer such that
$\, h(y) = 0 + y + c y^e + \cdots \,$ which perhaps
should be studied according to comments.
In fact, define $$ g(x) : = \sqrt{h(x^2)}. \tag{9}$$
Then solving for a power series expansion gives
$$ g(x) = x - \frac{x^2}{\sqrt{6}} + \frac{x^3}6 + O(x^4) \tag{10}$$ which implies that
$$ h(x^2) = (x^2+x^4/2)+\frac1{\sqrt{6}}f(x) \tag{11} $$
where
$$ f(x) \!=\! -2 x^3
\!-\!\frac{11x^5}{24} \!+\! \frac{117x^7}{1280}
\!-\!\frac{5491x^9}{110592} \!+\! \frac{156538363x^{11}}{3715891200} \!+\! O(x)^{13}. \tag{12}$$
A Wolfram Language code to calculate $\,g(x)\,$ is
ClearAll[x, g, gx];
gx[3] = x - 1/Sqrt[6]*x^2 + O[x]^3;
Do[g = Normal[gx[n]] + O[x]^(2+n); gx[n+1] = Simplify[
    g + (x^2 + (Normal[g]/.x -> g)^2 - g^4 - 2*g^2) * 3 /
    ((4+n)*x^2*Sqrt[6])], 
  {n, 3, 6}] 

As some comments suggest, it seems that the power series
for $\,f(x)\,$ has zero radius of convergence. This makes
finding properties of it difficult. Perhaps we need to let
$\,x\,$ approach $\,\infty.$ In this case we find that
$\,h(x) \approx \sqrt{x}\,$ with infinite number of other
terms. Using equation $(1)$ we can find the expansion
$$ h(x) = x^{1/2} - 1 + \frac12 x^{-1/4} + \frac1x s(x) \tag{13} $$
where $$ s(x) :=
\sum_{k=2}^\infty 2^{-k}(x^{-\frac32 2^{-k}} -
  x^{-2^{-k}}  ). \tag{14} $$
This gives moderately good approximations as $\,x\,$ gets large and down to $1.$
One method that leads to equations $(13)$ and $(14)$
is as follows. From equation $(1)$ we immediately get
$$ h^{-1}(y) = y^2 + 2y - h(y) \tag{15} $$
and if $\,y=h(x),\,$ then
$$ x = h(x)^2 + 2h(x) - h(h(x)). \tag{16} $$
We start with an approximation and try to find what
additional term will satisfy equation $(16)$. So we guess
$$ h(x) = x^{1/2} + cx^e + \cdots \tag{17} $$
where $\,\cdots\,$ denotes terms with smaller exponents.
Substitute equation $(17)$ in equation $(16)$ to get
$$ x \!=\! (x \!+\! 2cx^{1/2+e} \!+\! \cdots) \!+\!
 2x^{1/2} \!+\! \cdots \!=\! x \!+\! 2x^{1/2}(cx^e \!+\! 1)
 + \cdots. \tag{18} $$
This implies $\,0 = cx^e + 1.\,$ Solving this we get
our next guess as
$$ \,h(x) = x^{1/2} - 1 + cx^e + \cdots. \tag{19} $$
Repeating this process leads to equations $(13)$ and
 $(14)$.
The series in $(14)$ appears
to converge but I have no proof, only numerical evidence.
I also have no proof that $(13)$ satisfies the functional
equation $(1)$.
An answer to this question by 'Semiclasical' contains
sequence recursions
$$ u_n = u_{n-1} - j_{n-1} \quad \text{ and } \quad
   j_n = j_{n-1} - u_n^2. \tag{20} $$
with the property that $\, u_n = h(u_{n-1})\,$ and
that $\,u_n \to 0^+.\,$ I found that for suitable
starting values of $\,u_1\,$ and $\,j_1\,$ we have
$$ u_n = 6n^{-2} - \frac{15}2n^{-4} + \frac{663}{40}
n^{-6} - \frac{43647}{800}n^{-8} + \cdots. \tag{21} $$
An answer to this question by 'Cesareo' uses recursion to construct a sequence of functions $\,h_n(x)\,$ which seems to converge to a global solution. It would be nice to give a proof of this.