Let $k$ be a field of characteristic zero.

It is well-known, see for example [Corollary 10.2.21][1], that if $(x,y) \mapsto (p,q) \in k[x,y]^2$ is a counterexample to the two-dimensional Jacobian Conjecture, then there exists an automorphism $g$ of $k[x,y]$ such that the support of $g(p)$ is contained in some rectangular 
$\{ (i,j) | 0 \leq i \leq a, 0 \leq j \leq b \}$ with $(a,b)$ in the support of $g(p)$;
such $g(p)$ is called *subrectangular*.
By [Lemma B][5] or [Theorem 3.4][6], we obtain that in this case 
$g(q)$ is also subrectangular (since $g(p)$ and $g(q)$ are similar).

Interestingly, the same result holds in the non-commutative case, namely, in the first Weyl algebra, see [Theorem 5.12][3].

In both the commutative and non-commutative cases:
> Is it true that
we can further obtain that a possible counterexample has the form: $g(p)$ is as above and, in addition, $\{(a,v)_{0 \leq v \leq b-1}\}$ are *not* in the support of $g(p)$ and
$\{(u,b)_{0 \leq u \leq a-1}\}$ are *not* in the support of $g(p)$?
Something like the picture of $(p_0,q_0)$ on [page 50][2].

I think that I have once seen a half positive answer to my question, see the picture on [page 21][4] (in that picture both $P$ and $\phi(P)$ are 'half good' for me). More precisely, I think that I have seen that  
the support of $g(p)$ is contained in the quadrangle with vertices
$\{(0,0),(a,0),(a,b),(b-a)\}$, when $a \leq b$.
(Perhaps this is a result of S. Abhyankar).
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Thank you very much!

  [1]: https://books.google.co.il/books?redir_esc=y&id=wKJqqd8t8KUC&q=proposition%2010.2.6#v=snippet&q=proposition%2010.2.6&f=false
  [2]: https://arxiv.org/pdf/1401.1784.pdf
  [3]: https://arxiv.org/pdf/1111.6100v1.pdf
  [4]: https://arxiv.org/pdf/1111.6100.pdf
  [5]: https://www.researchgate.net/publication/265368034_On_Appelgate-Onishi's_Lemmas
  [6]: https://www.sciencedirect.com/science/article/pii/002240499190128O