by 3940-07 notes 5d Details of the Construction of the Smale horseshoe map. Let f (x; y) = ( 4 x + 8 ; 5y ). Then f maps D onto the rectangle R de ned 8 x 3 8 y 9 : In particular, it squeezes D horizontally by a factor of and stretches it vertically 4 by a factor of 5 Next, choose a matrix A and a two dimensional vector b such that such x x that the mapping! A + b is invertible and maps the following points y y onto the given images: () 8 ; 9 7! 8 ; 8 ; 5 7! 8 ; 3 3 8 ; 9 5! 8 ; 3 8 ; 5 5! 8 ; 3 : This gives a set of eight linear equations in six unknowns, but if we write out the resulting augmented matrix and compute its row reduced echelon form using Maple, we nd that there is a unique A and b; namely, A = 0 0 ; b = 4 With this choice, the map takes the rectangle de ned by R \ y 5 ; :
where R is de ned by () ;onto the rectangle R de ned by 5 8 x 7 8 y 3 : In doing so it maps the left side of R onto the right side of R ; and maps the top of R onto the bottom of R and the bottom of R onto the top of R : It maps the smaller rectangle R \ f3 y 4g onto R \ D:: We now de ne a function f of the plane into itself. 8 u Let >< f (u; v) = >: if v v 3 u A + b if v v 5 g (u; v) if 3 v 5 where g (u; v) is a smooth function which maps the vertical strip x 3; 3 y 8 8 onto that semicircular portion of the annular region 5 64 x + y 3 9 64 which lies above y = 3: In addition, g should be chosen so that f is smooth. The question which arises is the existence of such a function g: This something I have not found addressed in any text, and I won t address it here! We then consider the map S = f f restricted to the set D: It is clear that the image S (D) of this map is two strips ~V = (x; y) j 8 x 3 8 ; y 3 and ~V (= R ) = (x; y) j 5 8 x 7 8 ; y 3
plus the set. Of more interest is the intersection D \ S (D) : This consists of exactly the two strips: V = (x; y) j 8 x 38 ; 0 y and V (= R \ D) = (x; y) j 58 x 78 ; 0 y and We also wish to consider the inverse images of these two strips, that is, H = f(x; y) D j S (x; y) V g H = f(x; y) D j S (x; y) V g : H is easiest to determine. Since f is the identity map on the strip V ~ ; we just have to determine the inverse image of this strip under f : Looking at the formula for f ; we see that H = (x; y) D j 8 4 x + 8 3 8 ; 0 5y = (x; y) j 0 x ; 0 y 3 0 This is a horizontal strip in D of width ; height : With a little more work we can 5 determine that 7 H = (x; y) j 0 x ; 0 y 9 : 0 Observe that the inverse image under S of any point in V [ V is unique. Also, Sj H is the composition of two linear maps, as is Sj H : Each of these maps is linear so the compositions are linear. However Sj H is not the same linear map as Sj H the formulas di er, not just the domains. In fact, Sj H = f j H but on H the e ect of f is non-trivial. Let us look a little more carefully at the action of S on the two sets H and H : The map f expands the y direction by a factor of 5: Any vertical line in H of length (which is the vertical dimension of H 5 ) is expanded by f to a line of length : Since H is the inverse image of V ; f does not a ect S on H so f : H! V ; and this image line of length spans the whole length of V ; from y = 0 to y =. 3
Since S : H! V D; the second iterate S = S S is de ned on H. We wish to determine the forward image S (H ) = S (V ) : S (V ) looks like S (D) only thinner. It includes points outside of D: A point p = S (q) in S (V ) lines in D only if it lies in V or V : This means that p H [H : For example, p could lie in H \V : From the de nition of H ; S (H \ V ) V : Also, some points of V lie in H, which is another horizontal strip spreading all across D: Clearly S (H \ V ) V : Similar considerations to those above show that S (H \ V ) ; denoted by V is again a vertical strip, also of width and extending the whole length of V 6. Points of V which are not in H or H are not mapped into V or V ; because H and H are the inverse images of V and V. They must therefore be mapped into the interior of the annular region ; and that is not part of the domain of S; so no further iterates of such points are de ned. We can then consider S (H \ V ) ; denoted by V and S (H \ V ) ; :denoted by V We nd that that S applied to (x; y) j S (x; y) is de ned is a set which extends outside of D;but whose intersection with D is a total of four vertical strips, two within V and two within V ; all of height and width 6 : Continuing in this fashion, we see that for each k ; S k applied to D k = (x; y) j S k (x; y) is de ned intersects D in a set of k vertical strips, all of height and width 4 k : It is convenient to write S k (D) where what we mean is S k (D k ), since S k is not de ned on all of D: We will need a more detailed notation to keep track of these various vertical strips. We have seen that Further, S (D) \ D = V [ V : S (V [ V ) \ D = S (H \ V ) [ S (H \ V ) [ S (H \ V ) [ S (H \ V ) 4
is the union of four strips, which we have denoted by V ; V ; V ; and V ;. The notation will be that V and V are both contained in V, while V and V are contained in V : Also, V is the image of H \ V We then have that the eight sets created at the next stage are V ; V ; V ; V ; and so forth, where for any i; j, V ij and V ij are both contained in V ij : Also, V ijk = S (H i \ V jk ) ; and so forth to higher order iterates. We now consider the inverse mapping S : This is not de ned on all of D; but only on the strips V and V : Still, we will write S (D) where we mean, S (V [ V ) : Note that S j V and S j V are both linear maps (di erent linear maps), found from the inverse of of f and the inverse of f :On V ; S = f f : Then, S (D) = H [ H ; and (S ) (D) = S (D) will be the union of four horizontal strips, which we denote by H ; ::; H ; where H ; and H ; are both contained in H ; etc. The domain of S is actually V [ V [ V [ V : Similarly, the domain of S is actually [ i;j=h ij. Recall that the width of H and H was : It is similarly seen that the width of each H 5 ij is : 5 So for each k > 0 the domain of S k is the union of k vertical strips of width ; while the domain of S k is k horizontal strips of width : 4 k 5 k We therefore see that domain S j \ domain S k is the intersection of k vertical strips with j horizontal strips. Furthermore, obviously, domain S k+ domain S k : We now ask: What is an invariant set for the mapping S? This will consist of the set of all points of D for which every S k and every S k are de ned. This will be an in nite intersection of nested compact sets. It is a theorem from topology that 5
such an intersection is non-empty. each of which can be labeled as p In fact, it will have uncountably many points, where is some doubly in nite sequence = f i g j <i< of s and s which de nes the various V sets and H sets which the point p s is contained in. For example, the point p f:::;;;^;;;:::g denotes a point which lies in V and H : The ^ denotes the zero point of the sequence. Consider more carefully the meaning of the label p : As a speci c example, consider a point with label p :::^;;;:::. The meaning of this is that p lies in the set V ; which is a subset of V and hence a subset of V : Also S (p) lies in H \V : In particular, S (p) lies in V ; a subset of V : But S (p) V implies that S (p) lies in H \ V V, Hence p V implies that there is a q such that q V ; S (q) V and S (q) = p V. In this way we prove the following: Theorem: If = f i g j < i < ; then there is a q D such that S j (q) V j ; for any integer j > 0 and S j (q) H j for any j < 0:. We therefore have a correspondence between any bi-in nite sequence and a set of points in D: We now ask if the point q in this theorem is unique. It is unique, and this follows from the expansion and contraction properties of the map S: We have seen that the width of a strip V 3 k is : Similarly the 4 k backward iterates of the horizontal strips decrease in height. Suppose that there are two points q 6= p corresponding to the same sequence. Each is in the invariant set : But for large enough j and k the intersections of the k th iterate vertical strips and j th iterate horizontal strips must have smaller diameter than the distance jp qj : This means that p and q must lie in di erent intersections, at some stage, and so correspond to di erent sequences. This is an important point, which is only guaranteed because of the constant expansion and contraction ratios of the map. A map with constant expansion and 6
contraction ratios (> and < respectively) is called hyperbolic. It is possible to de ne a horseshoe type map without such a strong requirement. We can still have the desired intersections of horizontal and vertical strips, but the widths of these strips might not go to zero. In this case, di erent points might correspond to the same sequence. This kind of map is sometimes called a topological horseshoe. It is a weaker concept than the standard Smale horseshoe, which is assumed to be hyperbolic. This has an important consequence concerning possible de nitions of chaos. One of the properties often assumed is so-called sensitive dependence on initial conditions. In the language of the horseshoe map, this means that any two points for which all iterates of the map are de ned correspond to di erent sequences. There is a : correspondence between the invariant set and the set of all sequences. If there is a point in the set corresponding to each sequence, but this correspondence is not :, then there is a weaker sensitivity to initial conditions. We now wish to discuss in more detail the connection between orbits of the map S and operations on biin nte sequences. Let f = Sj. Thus, f is a : map de ned only on the Cantor-like set ; and f :! :, so that f k x is de ned for every integer k; positive and negative, and every x : Consider as well, the space of all bi-in nite sequence of s and s. We wish to place a topology on this space, by de ning a metric, or a distance between two sequences. We let d (; ) = k= k jkj ; where k = 0 if k = k if k 6= k : We have seen that every point p of the invariant set corresponds to a sequence, namely, the sequence of V s and H s which forward and backward iterates of f follow starting at p: We denote the correspondence by : We then have the following result: Theorem: The map :! is a homeomorphism, meaning that it is :, onto, continuous, and its inverse is continuous. Little proof is required. If a sequence fp i g of points in converges to a point p ; then the set of iterates of f applied to the p i follow successively more and 7
more of the same sequence of V s and H s as the iterates of p: This means that the corresponding sequences agree for larger and larger jkj : It is easy to see that this implies that d ( (p i ) ; (p))! 0; which is the de nition of continuity. We have already discussed the facts that is : and onto. Further, the inverse is continuous by essentially the same argument. Now we consider a special map s :! : This is the shift map s ( ; ; ; ^; ; ; ) = ; ; ; ; ^; ; 3 ::: In other words, we can say that s shifts the decimal point one place to the right, or that s shifts every j one place to the left. We compare this with the action of f on : Suppose that p V ; f (p) V ; f (p) V 3 : This means that (p) = ; ^; ; ; ;. But q = f (p) starts in V, and then f k (q) follows the same sequence as f k+ (p). That is, (f (p)) = ; ; ^:; ;. So, we have f = s Because is a homeomorphism, we say that f is topologically equivalent to the shift map s on. Remark: Many authors use 0 and as labels, rather than and. You can see pictures of the process on pages 3 and 3 of GH (but with a di erent convention on how the strips are numbered, though the labels are and ) and on pages 8, 83, and 84 of Global bifurcations and Chaos by S. Wiggins where the naming convention is like the one I used but the labels are 0 and. 8