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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.

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  • $\begingroup$ The above inverse is obtained as a series expansion. Under that premise, you just "stupidly" start with the linear term (z) and then construct the quadratic terms $-az^2-bz\overline{z} -c\overline{z}^2 $ such that all quadratic terms will cancel out, which leaves you with cubic terms etc. Again, this is UNDER THE PREMISE of a series expansion. If you want a closed form of the inverse (if it exists), then you have to solve $z = \psi(w)$ for w, which is just the standard inverse function. $\endgroup$ – Andreas Sep 29 '17 at 8:17
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    $\begingroup$ Apart from my answer, an analogous computation is done by Mario Carneiro in this answer. $\endgroup$ – Giuseppe Negro Aug 23 '18 at 13:33
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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 $$ w(z)=w(0)+\left.D_z w\right|_{z=0}(0)\,z+O(|z|^2).$$


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.

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  • $\begingroup$ Thanks! The source is Kuznetsov - Elements of Applied Bifurcation Theory. In your first proof: is this the common way to obtain a inverse transformation or do you know literature which deals with these problems? $\endgroup$ – JDoe Sep 30 '17 at 13:30
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    $\begingroup$ This is the kind of thing everybody seems to know about except yourself. Typically people writes it down like the book you are reading, without further explanation. I have had the chance to speak with experts in these things, and it seems to me that they use the theorems of differential calculus (inverse and implicit function theorem) which they have interiorized to the point that they are obvious to them. That's why they don't tell you what they are exactly using. $\endgroup$ – Giuseppe Negro Oct 1 '17 at 15:21
  • $\begingroup$ One more thing: You set $b=c=0$. Do you do this for simplification or is there something else behind this step? And how do we know that the coefficients are the same? $\endgroup$ – JDoe Oct 9 '17 at 11:37
  • $\begingroup$ Just for laziness. Everything works exactly in the same way with arbitrary $b, c$. There is no differentiability problem, because you only use the real inverse function theorem. (Actually, the inverse function theorem is a real theorem, the only difference with the complex case is that if the functions are complex differentiable then the inverse will be complex differentiable as well). $\endgroup$ – Giuseppe Negro Oct 9 '17 at 11:42
  • $\begingroup$ I see. And the specific form of the coefficients should be obtained by the inverse function theorem? $\endgroup$ – JDoe Oct 9 '17 at 11:44

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