Hyperbolicity in a class of one-dimensional maps

Publicacions Matemátiques, Vol 34 (1990, 93-105 . Abstract HYPERBOLICITY IN A CLASS OF ONE-DIMENSIONAL MAPS GREGORY J . DAVIs In this paper we provide a direct proof of hyperbolicity for a class of onedimensional maps on the unit interval . The maps studied are degenerate forms of the standard quadratic map on the interval . These maps are important in understanding the Newhouse theory of infinitely many sinks due to homoclinic tangencies in two dimensions . Introduction In the theory of infinitely many sinks, two-dimensional invariant sets are formed when homoclinic tangencies between stable and unstable manifolds of a hyperbolic periodic point are formed . In order to show that infinitely many sinks occur in this situation, we must show that these invariant sets are hyperbolic ([1], [4], or [6]), which is a major undertaking . When the homoclinic tangency is quadratic in nature, the two-dimensional problem has been thought of as a perturbation of the one-dimensional map ([3], [5]) . In [5], a more complete and elegant proof was obtained by conjucating the hyperbolic invariant set for the quadratic map fb(x) = bx(1 - x) to the twodimensional invariant set in the two-dimensional infinitely many sinks problem . In the case where the homoclinic tangency is degenerate (Le ., of order r, r = 4,6 . . . . ), the present proof is very long and involved [1] . If a conjucacy between the one and two-dimensional degenerate problems can be determined, then it may be possible to treat the higher order tangencies in two dimensions using the same type of ideas as presented in [5] . With the above motivation, we will examine the hyperbolicity of the following family of one-dimensional maps on the unit interval . The one-dimensional maps that we are concerned with are of the form r fb(x) = b[2 r - Cx - 2) 1, b>2r, r=4,6,8, ... where b is a real parameter and r is a fixed positive even integer . By studying these one-dimensional maps we will gain insight as to how the hyperbolicity of an invariant set near a degenerate homoclinic tangency in two-dimensions is justified . We will restrict our attention to the cases where r > 2, due to the fact that when r = 2 we obtain the well known and studied quadratic map fb(x) = bx(1 - x) ; see [2], [3] or [5] for more information about this map . 94  G .J . DAvis Statements of results It is easily seen that fb(x) = 0 for x = 0,1 and that fb(x) achieves its maximum value of b/2'r when x = 1/2 . In order for I = [0,1] to be covered by itself under the map fb(x), bmust be larger than 2' . We are interested in the set A contained in I that is invariant for fb(x) ; Le ., fb(A) =A . Explicitly, the invariant set which we are interested in for this máp is given by A = (1°_ ° o fb 2 (I) . (The set A is the analogue of A  (t) in the two-dimensional problem [1]) . It is our goal to show that A is a hyperbolic set for fb(x) . A hyperbolic set for fb is a closed, bounded, invariant set A for which there exists an K > 0 such that for all x E A we have j(fb)'(x)j > 1 for all k > K . If b is large enough, the hyperbolicity of A is relatively easy to establish . Set fb 1(I) = I 1 U 12, where I 1  = [0, c], c E [0,1/2], and 12 = [d, l], dE [1/2,1], so that fb(Ii) = I, i = 1, 2 . Then we have Proposition 1 . Leí fb(x)=b[2 -(x-2)rJ, b>2 r , r=4,6,8, ... thenforb>_2''(3), f b '(x)1 > 1 for al' x E fn 1 (I) . The proof of proposition 1 and the remaining propositions in this section will be deferred to the next section . Continuing with our discussion of the hyperbolicity of fb(x), if 2 r <b<2 r C 3r+2 1 3r ' then not all points in the pre-image of I, fn 1(I), have 1 fb(x) 1 > 1 . Fortunately, the absolute value of fb(p) is greater than one where p is the fixed point of fb which is an element of 1 2 . (See Figure 1) . Figure 1 Proposition 2 . HYPERBOLICITY IN ONE-DIMENSIONAL MAPS  95 The fixed point p of fb(x) which líes in ¡he interval 12 is a repellor, that is l fb(P)1 > 1 . Let xi be the poinis such that jfb(ij j = 1 ; 2i E Ii, i = 1,2 .  Define J I = [0,¡,) and J 2 = (i2,1] . Clearly all xEJ I U J 2have ¡he property lfb(x)1 > 1 . Proposition 3 . If x o E A(1[(h UI 2 )-(J I UJ 2 )], then there is a positive integer k = k(xo) > 1 such that j(fti y(x o )j > 1 . The idea behind Proposition 3 is that the values of x o that do not Nave j(fb)'(xo)j > 1 are mapped close to one and then near zero . When a point is mapped near zero, the point stays in the interval J l , where the derivative is greater than one, for its next few iterates . The accumulative result is that the iterates for which the derivative is greater than one overcome the initial contraction of the point xo . We now have, by Propositions 1-3, the following result : Main result . Let fb(x) = b [2 - (x - 2)1-  , b > 2 1- , r = 4, 6, 8 . . . . fixed .  The invariant set A = fl°_= o fn '(I) is a hyperbolic set for fb(x) . Proof of Proposition 1 : Set a=331-2, and b 1 =2 1- ( 3 ) =2 1- « . It is sufFicient to show Jfb,(x)j > 1 since Jfb(x)j > Jfb,(x)j for b > b l . Consider the function Differentiating fb,(x) with respect to x, we obtain fb, (x) = -2ra(2x - 1)1--1 . Therefore, Jfb l (x)j = I2ra(2x - 1)''-1 . Let xi E I, where i = 1, 2, be the points such that fb,(xi) = 1 ; Le ., a[1 - (2xi - 1) 1- ] = 1 . Solving for xi we obtain, Proofs of results fb,(x) - b [2 1- - (x - 2/ 1-J = a[1 - (2x - 1)1-] . 1_(2xi_1)r1 a (2xi _ 1) 1- = 1 - 1 a 96  G .J . DAvis or and For all x E f~,i(I), xi = 2[± .(l1)  +ll . 1fbi(x)1 ~ 1fbi(xi)1, l fb,(xi)l = I2ra Define the function lab(z) to be = 2r = 2r 2rf (112 ) -11 __ IL 3r 2 3r - 3r 4 +2 (3r+2~ - Therefore Jfb l ( x)j > 3r+2 ( 3 )' > 1 . Thus Jfb l (x)1 > 1 for all xE f¿l l (I), and Proposition 1 is proven . . " Proof of Proposition 2 : The idea of this proof is to show that l a < p < 1 which implies jb(p)j > 1 . Consider gb(x) = fb(x) - x, and note that gb(p) = 0 . Set z = (x - 1/2), then 96(x) _ .fb Cz + ~) - ( z + 2/ = b (2 r - z r ) - (z + 2/ r i _ -b (Zr + b + 2 2rb b l . ri hb(z) = zr + b + 2 2rb b . and Evaluating hb(z) at z = (p - 1/2) and z = 0 yields the following : since b > 2r . Because hb(y) has only one variation in sign, Descartes's rule of signs implies that hb(y) has at most one positive real root ; however, we already know that z = (p - 1/2) is a positive root of hb(z) . Therefore, to prove that x2 < p, that is p E J 2 , it is sufficient to show that hb(z) < 0 where z = (xl - 1/2) . First it is necessary to calculate z . Recall that and The point x2 satisfies the relation fb(-¡2) = -1 ; that is 1)r-1 _ -1 . Replacing (i 2 - 1/2) with z we obtain We now proceed to show that hb(z) < 0 . hb(z) = zr + z/b -f- (2b) -1 - 2-r _ (rb) 11 + b-1(rb) 11 -}- (2b) -1 - 2-r < (r2r) r-1 + 2r (r2r) r-1 + 2 12r - 2-r' - (r2 r ) r-1 + 2r (r2 r ) r-1  2 +1 2r [(rr22r) r11 + (r2r) r11 - 2 HYPERBOLICITY IN ONE-DIMENSIONAL MAPS  97 hb(p - 1/2) = 0, 2r-1 - b hb(0) =  2rb  < 0 fb(x) = b (1- (x - 1)r1 2r  2 ' r-1 fb(x) = -rb (x - 2) -rbzr -1 = -1, or z = (rb) r=1 . 1+ r -2 -1 1 r -1 2r(2r )=1 since b > 2r 9 8  G .J . DAvis Set A= <0 . then hb(2) < A (1 -}- r - r(2r) r 11 < A (1 -Ir - r [1 -fr 1 l ln(2r) J = A (1 - rr 1 ln(2r)1 - =A (1 - (1 + r 1 1 ) ln(2r) I ln(2r) 1 < A  -  , since 1n(2r) > 1 for r >_ 2 r-1 Thus, we will have shown that hb(z) < 0 as soon as equation (*) is verified ; however, equation (*) is true due to the fact that (2r)~'r = e7'r ln(2r) Proof of Proposition 3 : where (higher order terms) > 0 since rl il ln(2r) > 0 for r > 2 . Therefore, hb(2) < 0, which implies pE J2 or Jfb(p)j > 1, and the proof of Proposition 2 is complete . Let 1 + gbe the maximum value obtained by fb(x) where x EI ; i .e ., let 1+g = fb(1/2) . Choose any x o E Afl[I l UI2)-(J, UJ2)] . Define 6 = S(xo) > 0 so that the distance between fb(xo) and 1 -fg is b6', and define y = -y(X0) > 0 to be the distance between fb(xo) and 1 . Then fb(xo) = rb6r -i , and y < bbr . = 1 + 1 ln(2r) + (higher order terms) r-1 < 1 + 1 r-1 ln(2r) (See Figure 2) . but HYPERBOLICITY IN ONE-DIMENSIONAL MAPS  99 Let ~max be the maximum value of f¿(x) where x E I ; that is, let Define k = k(x o ) >2 to be the integer such that for j < k - 1 . We now have that Then Figure 2 Amax = fb(O) = rb(1/2)r-1 . ,xk-1 Y  1 max 21 ~maxy 1 2 , \max -  > 2bbr' y _ Define 0, where 0 < A < 1/2, to be the value of ó for which bAr = 2 + 9 bA r =2+9=fb(1/2)=2 -2 . Solving for Qr we find that O r  ___ __ 1 b  1 _ 1 b-2 r 1 b (2r  2)  b2 r ( 10 0 Therefore, r amex From this relation we have that Therefore, for which y = y(x o ) has the property G .J . DAvis Let a,nin be the minimum value of fb(x) for x E [0,12 - 0l, = f b (112 -,á) = rbár-1 . Using our formula for Ar we see that Or -1 is given by Qr-1 = (Or) r - i = ,~ b2r (b - 2r-1)1 - Amin = rb (b2r (b_  2r Combining the formulas for An,ax and Amin we obtain the following ratio : Amin  i 2-1-r -1 - (b - 2r-1)--r r b-2- > 1 . '/ r i \ k-1 ,\ min 1 in r ax  (b-2 r r~b Z r r 2 l ~ r We will now show that (fñ),(xo) > 1 (f6 )i(xo) i rbór-lñmin (fb),(xo) ~ : (_\max)rrl(b-2r-1)_rrirrb2- .2ir_  r > rbór -1 -1 1 1  (b-2r - 1)rr l r :b rr2 l rr 2rr b= ar-1 2-2r 1+, 3 - r  -, >2 -r  b-(b-2 r -1 )= 2-2r  +i «b-2r-1)r-1 \ =2 -r br_3 - r  C (b - 2r - 1 )r-1 \ r-1 br-3 4- >1 . It should be noted that the above argument is valid only for the values of x o , \max?' > 2, but ñm ax y < 2 Retan that that is, xi satisfies HYPERBOLICITY IN ONE-DIMENSIONAL MAPS  101 where j < k - 1 and k = k(x o) >_ 2 . In addition, the above argument is only legitimate in the case when a < 11/2 - fb 1 (1 - yma .)I, where -%,,,a>, is the value of y for which ñmaxymax = 1/2 . In Lemma 4 we will prove the existente of certain parameter values b for which there exist values of x o that do not satisfy the above condition . These values of b are shown to also have the property of J(fb)'(x o )j > 1, where k = 2, but an argument different from the one above is required to show this fact . This argument will be presented in Lemma 5 . However, before we prove Lemma 4 and Lemma 5, it will be necessary to provide additional notation . Set ama . = 1 1 /2 - fb 1(1 - ymax)j and define the intervals L l and L 2 to be L l = I l r1 [1/2 - amax, 1 /2] ; L2 = 12 n [1/2, 1/2 + bmax] . Lemma 4 . There exist intervals of parameter values M(r) C ( 2r , u2 r ] such that L ; U Ji I i, i = 1, 2, where a= (3r±2\ . 3Jr Proof .. Define xi, where i = 1, 2, to be the values of x for which fb(~i) = 1 - ymax, and define ii, where i = 1, 2, to be the values of x for which jfb( .¡ i )1 = 1 . To prove Lemma 4 it is sufficient to show that there is an interval M(r) of parameter values b for which 11/2 - x i l < 11/2 - .¡¡l . Solving ñmaxymax = 1/2 for ymax we obtain ymax = 2 Amax rb Amax = 2r-1' Therefore, ymax =  b, and hence, - ymax  4rb' The above equation implies that the value xi satisfies _ 2r fb(xé) = 1  4rb' 1)r1 2 . . 77z