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I am interested in how to calculate the Poincare map corresponding to a submanifold.

The vector field $f(x,y)=(-y,x)$ has the periodic solution $(cos(t),sin(t))$

Now I'd like to compute the Poincare map corresponding to $A=\{(x,y)\in\mathbb R^2|x>0, y=0\}$

$P_A(y)=\Phi(\tau(y),y)$ where $\Phi$ is the flow.

$\tau(y)$ is the period if I am correct, so in our case $2\pi$, but what is $\Phi(\tau(y),y)$ ?

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I'm not sure why you're writing the section map as a function of y, when the section you defined has $y=0$. The solution to this ODE is given by A Blumenthal below, where $r$ is along the positive x-axis. The spectrum of this linear system (and the solution) will yield the identity map on the section (or any ray pointed at the origin). –  alancalvitti Jan 29 '13 at 21:48
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up vote 1 down vote accepted

The integral curves for the map are concentric circles $\{(r \cos t, r \sin t) \mid t \in \mathbb{R}\}$ for r > 0. Which direction do they travel in about the origin?

Here's another hint: if you take a Poincare section transversal to a periodic orbit of a flow, then the point where the transversal hits the orbit is a fixed point of the Poincare map.

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Thanks, as alancalvitti pointed out, r is along the pos. x-axis, but what is the IVP ? I understand you hint, I also know a related related lemma which says that: "Suppose $x\in M$ and $T\in I_x$. Let A be a submanifold of codeimension one transversal to f such that $\Phi(T,x)\in A$. Then there exists a neighborhood U of x and $\tau\in C^k(U)$ s.t $\tau(x)=T$ and $\Phi(\tau(y),y)\in A$ for all $y\in U$" But still I do not know how to write down the explicit poincare map –  Alexander Jan 29 '13 at 21:54
    
@Alexander Why work with a technical definition like that? Here is a simpler way to think of it: you have a vector field and you have a small piece of hypersurface (codimension 1) that cuts across the vector field in a neighborhood. The Poincare map freezes the flow precisely at the point when the flow returns to the hypersurface. In this case, draw the concentric circles representing the integral curves and follow them as they depart from the set $A$ and inevitably return. You'll see that the 'departure' point is the same as the 'return' point (since your position on $A$ is precisely... –  A Blumenthal Jan 30 '13 at 18:46
    
(continued) the radius of the circle representing the integral curve. Thus is the action of the return map: it takes a point on $A$ to the same point. –  A Blumenthal Jan 30 '13 at 18:47
    
I'll try not to sound pedantic, but I think I should make a point here: although I cannot speak for other fields, in dynamical systems it's important to have a good feeling or intuition for the objects you're studying, since most of what you'll be doing does have an appealing intuitive interpretation which will serve you well when proving theorems. Reciting a technical definition is a useful exercise once the intuition is 'in your gut', so to speak, but before then will often confuse one to the point of misunderstanding the concept. –  A Blumenthal Jan 30 '13 at 18:51
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