Is there a diffeomorphism of the disk with constant sum of singular values?

Question

This question is a relaxed version of this question.

Let $D \subseteq \mathbb{R}^2$ be the closed unit disk, and let $c \ge 2$.

Does there exist a diffeomorphism $f:D \to D$ with constant sum of singular values $ \sigma_1(df)+\sigma_2(df)=c $?

The necessary condition $c \ge 2$ comes from the AM-GM inequality. For $c=2$ there is the identity map, so the question is really about $c>2$. If I am not mistaken this is an "additive" version of the Beltrami equation.

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Some more context:

If we remove the origin, the answer is positive for every $c>2$; choose fixed numbers $\sigma_1,\sigma_2$ such that $\sigma_1+\sigma_2=c,\sigma_1\sigma_2=1$. Then, by the answer to this question, there exist a diffeomorphism $f:D\setminus \{0\} \to D \setminus \{0\}$ with the constant singular values $\sigma_1,\sigma_2$.

At the moment, however, I don't know how to construct such a diffeomorphism with everywhere constant singular values on the entire disk. I hope that by relaxing the requirement from fixed singular values, to a fixed sum, we will be able to use this added freedom to build a diffeomorphism of the whole disk.

The fact that I don't require $f$ to have constant Jacobian also means that we are in a different context from this previous question of mine.

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A possible simplification:

Let $f$ be the map described in polar coordinates by

$$ \big(r,\theta\big )\mapsto \big(\psi(r),\theta+\phi(r)\big). \tag{1}$$

For such maps $f$, the PDE $ \sigma_1(df)+\sigma_2(df)=c $ reduces to the following ODE:

We have $$ [df]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} \psi' & 0 \\\ \phi'\psi & \frac{\psi}{r}\end{pmatrix}. $$

For a matrix $A=\begin{pmatrix} a & 0 \\\ b & e\end{pmatrix}$ with positive determinant, $$\sigma_1(A)+\sigma_2(A)=c \iff c^2=\sigma_1(A)^2+\sigma_2(A)^2+2\sigma_1(A)\sigma_2(A)=|A|^2+2\det A.$$

So, $\sigma_1(A)+\sigma_2(A)=c \iff (a+e)^2+b^2=c^2.$

Thus, for $f$ given by the form $(1)$,
$ \sigma_1(df)+\sigma_2(df)=c $ if and only if

$$ (\psi'+\frac{\psi}{r})^2+(\phi'\psi)^2=c^2. \tag{2}$$

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Answer 1

Take a map $f$ of the form that you propose, so that the equation reduces to the ODE $$(\psi' + r^{-1}\psi)^2 + (\phi'\psi)^2 = c^2.$$ If we fix $c := 12/5 > 2$ and $$\psi(r) := \frac{6}{5}r\left(1-\frac{1}{6}r^4\right),$$ the equation reduces further to $$\phi'(r) = 2r(1-r^4/4)^{1/2}(1-r^4/6)^{-1}.$$ This has an analytic solution for $r < \sqrt{2}$ of the form $$\phi(r) = r^2\left(1 + \sum_{k \geq 1} a_kr^{4k}\right).$$ The corresponding map $$f(z) = \left(|z|^{-1}\psi(|z|) e^{i \phi(|z|)}\right) \cdot z = \frac{6}{5}z(1-|z|^4/6)e^{i |z|^2\left(1 + \sum_{k \geq 1} a_k|z|^{4k}\right)}$$ is an analytic diffeomorphism from $D$ onto itself and satisfies the desired equation.

(Alternatively, one can choose $\psi$ to be any concave, smooth increasing function with $\psi(0) = 0$ and $\psi(1) = 1$ such that $\psi$ is linear near the origin, and take $c = 2\psi'(0)$. Then $f$ is a just a dilation near the origin and we don't need to worry about singularities there).