This question is a relaxed version of [this question](https://mathoverflow.net/questions/352982/is-there-a-volume-preserving-diffeomorphism-of-the-disk-with-prescribed-singular).


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 $?
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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](https://mathoverflow.net/questions/352982/is-there-a-volume-preserving-diffeomorphism-of-the-disk-with-prescribed-singular), 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.
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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](https://mathoverflow.net/questions/354451/is-there-an-area-preserving-diffeomorphism-of-the-disk-which-is-nowhere-conforma?noredirect=1&lq=1).




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


