How to obtain the inverse transformation

Question

I found the following complex coordinate transformation in a proof: \begin{equation*} z:= \psi(w) = w+aw^2+bw\overline{w}+c\overline{w}^2. \end{equation*} A little bit later the inverse transformation is given without any information how to obtain it: \begin{equation*}w = \psi^{-1}(z) = z-az^2-bz\overline{z} -c\overline{z}^2 + O(|z|^3).\end{equation*} I have tried to obtain the inverse transformation but I don't manage it. Does someone know if there is a general way how to get $\psi^{-1}$?

EDIT: The source can be found in Kuznetsov - Elements of Applied Bifurcation Theory.

Answer 1

Can you point the source?

Anyway, here's how you can do it with some rigor. Let us set $b=c=0$, so we are given the equation $$\tag{1} z=w + aw^2. $$ Presumably this relation is of interest in a neighborhood of the solution $z=w=0$ (you should really cite the source). For all $z$ in such a neighborhood, (1) can be solved for $w$ by the inverse function theorem, yielding a differentiable map $w=w(z)$.

We will prove the following property, which is the key step: $$\tag{2}w=z+O(|z|^2).$$

Once (2) is proven, we will have $w^2=z^2+O(|z|^3)$. Therefore, we can rewrite (1) as $$w=z-aw^2=z-az^2+O(|z|^3).$$

It remains to prove (2). This is another consequence of the inverse function theorem, because we have that $$\left.D_w z\right|_{w=0}= ( \left. D_z w\right|_{z=0})^{-1}=I.$$ Here $I\colon \mathbb C\to \mathbb C$ denoted the identity mapping, and this relation can be easily obtained by differentiating (1) termwise at the solution $(z, w)=(0,0)$. Therefore, (2) follows by the Taylor expansion $$ w(z)=w(0)+\left.D_z w\right|_{z=0}(0)\,z+O(|z|^2).$$

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NOTE. Here's another proof of (2) which is less dependent on the differentiability, and so might be useful for less regular equations. We need to prove that a constant $C\ge 0$ exists such that $|w(z)-z|\le C|z|^2,$ for all $z$ in a small neighborhood of the origin. From (1) we know that $$|w(z)-z|=|a| |w(z)|^2, $$ so it will be enough to prove that $|w(z)|\le C|z|$. Again from (1) we have that $$\tag{3}|w(z)|\le |z| + |a| |w(z)|^2, $$ and we also know that $w(0)=0$, so, by continuity, we have that $|a||w(z)|^2\le \frac12 |w(z)|$ if $|z|$ is sufficiently small. (In short: the square of a small number is an even smaller number). Inserting this into (3) we obtain $\frac12 |w(z)|\le |z|$, and the proof is complete.

Incidentally, let me note that this strategy of improving bounds by iterating the bounds themselves is sometimes called bootstrapping.