The image of a disc under a diffeomorphism. This question is a simpler version of this.
Suppose that you have a diffeomorphism $f:\mathbb{R}^2\to \mathbb{R}^2$ acting on the unit disc $D=\{(x,y):x^2+y^2\leq 1\}$. Let $D_xf$ denote the total derivative of $f$ at point $x$. Assume also that the biggest eigenvalue of $D_xf$ at $x$, let us call it $\lambda_x$, satisfies $|\lambda_x|>M$, for all $x\in D$ and some $M>0$. Take now $S$ to be a square with center $f(0)$ and side length $2$.
Question: Is it true that the image of the unit disc under $f$ will intersect $S$ assuming $M$ big enough? In other words is it true that $f(D)\cap S\not=\emptyset$ for big enough $M$?
The idea is that $f$ stretches locally the disc in one direction by a lot but of course not at the same direction at every point. So  the answer to my question might be no but it seems very hard to believe. Does anyone have any ideas?

 A: No, if you are given $M$ and $\epsilon>0$, and the condition is that $\lambda_x(f)>M$ for all $x\in D$ then you may find a diffeo that maps your disc into a disk of radius $\epsilon$. To construct it explicitly is difficult but here is a deformation argument. Map you disk by a linear map into an ellipse of major axis greater than 2M and minor axis smaller than $\frac{\epsilon^2}{100M}$. Now wrap this long but very narrow ellipse around as a spiral inside the disk $D_\epsilon$ of radius $\epsilon$ by a map which is close to a local isometry on the ellipse. This can be done as its area is much smaller than that of the disk $D_\epsilon$. Make sure that in the process the spiral branches are at distances which has a uniform lower bound; then as the topology (also from a differential point of view) are disks both in the source and the image, you may fill in the rest of the plane and get a diffeo of the whole plane verifying the stated conditions.
This construction seems to fail if you assume that  $\lambda_x(f)>M$ in the whole plane (as the filling part above may have strong contractions at some places). I do not know the answer for that case.  
A: I interpret your question as follows: Does there exist an $M>0$ such that every diffeomorphism $f$ with $\lambda_x(f)\ge M$ for all $x \in \mathbb{R}^2$ satisfies $f(D)\cap S \neq \emptyset$?
Edit: I replaced my previous answer (which misunderstood the question) by something which is not an answer, but might still be helpful.
We may assume without loss of generality that $f(0)=0$ and ask whether $f(D)\subset (0,1)^2$ (i.e. $f(D)\cap S= \emptyset$) implies some upper bound on $M$. 
Suppose $\gamma:[0,\ell]\rightarrow D$ is a unit-speed curve, connecting $\gamma(0)=0$ with $\gamma(\ell)\in \partial D$. If we can find such a curve with $d f_{\gamma(t)}(\dot\gamma(t))=\lambda_{\gamma(t)} \dot \gamma(t)$, then for $i=1,2$ we have $$
1>\vert f(\gamma(\ell))\vert_\infty =\vert \int_0^\ell d f_{\gamma(t)}(\dot\gamma(t)) d t \vert_{\infty} \ge M \vert \int_0^\ell \dot\gamma_i(t) dt\vert = M \vert \gamma_i(\ell) \vert
$$
and thus 
$$
2> M^2 (\vert \gamma_1(\ell) \vert^2 + \vert \gamma_2(\ell) \vert^2) = M^2,
$$
which means that $M$ cannot be arbitrarily large.
How to construct such a curve? I suspect we can define a smooth vector field $\xi$ on $\mathbb{R}^2$ that at every point $x\in \mathbb{R}^2$ is an eigenvector of $d f_x$ with eigenvalue $\lambda_x$. (One has to think more carefully about what happens when both eigenvalues are the same and there is no distinct direction  associated to the highest eigenvalue). Then $\gamma$ should be an integral curve of $\xi$ and we only have to make sure that, starting at $0$, it eventually leaves $D$. I have not figure out how to do this though.
