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Is there any real analytic diffeomorphism from two dimensional disk to itself, except to the identity, such that whose restriction to the boundary is identity?

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    If your diffeomorphism restricts to the interior of the disk into itself, then I think the Schwarz lemma applies, and it must be a rotation.2011-11-30
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    Yes Leonardo! In my case they send the interior into itself! Would you please write me a reference for the Schwarz lemma? and explain more?2011-11-30
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    @Leandro: the question specifies the map is only *real* analytic, so it need not be *complex* analytic. So Schwarz isn't relevant unless you can argue the map has to be complex analytic.2011-11-30
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    The common element in Jonas's response and my own is that in the real analytic category there are functions that behave much like bump functions, so you have a fair bit of freedom to manipulate functions, at least at the $C^0$ level.2011-11-30
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    @Ryan: you're right. My bad!2011-12-01

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$$f(x,y) = (x,y) + (-x,-y)(x^2+y^2)(1-x^2-y^2)$$

Doesn't the above map do the job? I'm using the disc in $\mathbb R^2$ given by $x^2+y^2 \leq 1$.

If you want one without a fixed point in the interior,

$$f(x,y) = (x,y) + \left(\frac{1}{10},0\right)(1-x^2-y^2)$$

The fraction $\frac{1}{10}$ just needs to be a positive number strictly smaller than $1/2$.

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    Thanks! I think the example works! Do you know any example without any fixed point in the interior?2011-11-30
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    I'll edit in one.2011-11-30
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$f(z)=ze^{2\pi|z|^2i}$ should do the trick.

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    We perturbed the identity map in different directions.2011-11-30
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Some that spring to mind are $(x,y) \to (x+ c (1 - x^2 - y^2), y)$ where $-1/2 < c < 1/2$, $c \ne 0$.