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MACQUARIE UNIVERSITY MATH332 (2006D2)

Time-t maps and Poincare maps

This version: 24 November.The 23 November version had an error in Example 3. (See remark following that example).

Consider a set of autonomous first-order ODEs on D Rn, with flow (i.e. general solution) F(x, t),

which means that for any fixed x0

, the map F

(x0, t

) (as a function of t

only) is the particular solutionwith initial condition x(0) =x0. Often, the domain ofFmust be taken to be smaller than D R, butthis will not be important in the examples in this half-unit. For every t R, we define Ft(x) =F(x, t).This is called the time-t map.

Example 1. The general solution to ODE x= x2, with initial condition x(0) =x0= 0 can be obtainedby separation of variables:

1

x2dx=

dt =

1

x=t + c, which implies c=

1

x0.

Ift = 1x0

, this implies

x= 1

t + c=

1

t 1

x0

= x0

1 tx0.

Ifx(0) = 0, then the solution is x(t) = 0 for all t. Since substituting x0 = 0 into the expression x0

1 tx0gives 0, we dont actually have to consider the case x0= 0 separately. The flow is F(x, t) =

x

1 tx.

For any R, the time- map for this system is F(x) = x

1 x. Of course, this is not defined when

= 1x

, and since our initial condition is given at t= 0, it makes sense to assume (, 1x

), ifx >0,or ( 1

x, ), ifx 0 and we consider + 2n to be the same as , for any R and any n Z. Since theequations are decoupled, i.e. each equation contains only one variable, they can be solved separately.The first one is almost the same as Example 1, and the solution to the second one, with initial condition(0) =0, is (t) =t + 0. Hence for any t (subject to certain constraints, as in Example 1), the time-tmap for this flow is

Ft(r, ) =

r 1

1 t(r 1)+ 1, t +

.

Note that, ift= 2

, then t+ = 2+ = (in polar coordinates). If we omit the from expression

forF2

(since the

coordinate doesnt change anyway), we get a map

P(r) = r 1

1 2

(r 1)

+ 1 . (1)

This is an example of a Poincaremap, for which we now give a general definition . . .

Recall that amanifoldis a smooth curve or surface or higher-dimensional analogue.

A Poincare sectionfor F, at a periodic point a , is a manifold S through a, of dimension n 1, thatmakes a nonzero angle with the orbit ofa and doesnt intersect that orbit at any other point.Remark. The dimension ofS, i.e. n 1, may have been omitted in the lecture. You will not be askedfor this definition on the exam.

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Ifx is in Sand is close enough to a then F(x, t) Sfor some t >0, i.e. the orbit ofx returns to S.The Poincare map(or first return map) for Sis defined by P(x) =F(x, t) for the smallest t >0 suchthat F(x, t) S. This is slightly imprecise, since we havent specified the domain ofP. Read on.The following theorem clarifies things.

Theorem. Given any Poincare section S for F at a, there exist U, V S, open relative to S, such thatP :U V, defined as above, is a diffeomorphism.

Remark. Ifxis a periodic point of the ODE then it is a fixed point for the Poincare map at x.

Remark. In the definition ofP, its possible for t to be different for different x values. However, in allof the do-able examples I could come up with, the t is the same for all x values, so P is just a time-tmap, or more precisely the restriction of a time-tmap to S.

Example 2 (continued). In Example 2 above, any point (1, ) is periodic, with period 2

. For any fixed0, let S= {(r, 0) : r >0}. Then Sis a Poincare section for Fat (1, 0). For all r values, F2

(r, 0) S,

and 2

is the smallest t >0 for which this is true. Therefore the restriction of the time-2

map to S isthe Poincare map for S. In symbols,

P(r, 0) =F2

(r, 0) =

r 1

1 2

(r 1)+ 1, 0

.

Since the value never changes, it is usually dropped from the formula, giving Equation 1 above.

Convention: When calculating P in coordinates such that S is defined by setting one coordinate to aconstant value, that coordinate is dropped from the formula.Remark. All of the examples here and in the lecture use polar or cylindrical coordinates, because thissimplifies the calculations.

Example 3. Consider the following system of ODEs in cylindrical coordinates (r,,z),

r= r 2, = 4, z= 3z.

These can be solved separately, giving

r(t) = (r0 2)et + 2, (t) = 4t + 0, z(t) =z0e

3t.

Let a = (2, 0, 0), and note that this is a periodic point with period 2

. Now let S = {(r,,z) : = 0}.This is a Poincare section for a. The time it takes for a solution with initial condition (r0, 0, z0) to returnto S is always

2. Thus the Poincare map for S is

P(r, z) =

(r 2)e

2 + 2, ze3

2

.

Remark. The version of Example 3 from 23 November was badly incorrect, because the system was notautonomous and (2, 0, 0) was not periodic! I simplified the example.

Dont forget to also review the examples in the Week 13 lecture notes.

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