transactions of the american mathematical society Volume 347, Number 6, June 1995 A CONSTRUCTIVE PROOF OF THE POINCARÉ-BIRKHOFF THEOREM LI YONG AND LIN ZHENGHUA Abstract. In this paper, with the use of

Please download to get full document.

View again

of 16
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.


Publish on:

Views: 32 | Pages: 16

Extension: PDF | Download: 0

transactions of the american mathematical society Volume 347, Number 6, June 1995 A CONSTRUCTIVE PROOF OF THE POINCARÉ-BIRKHOFF THEOREM LI YONG AND LIN ZHENGHUA Abstract. In this paper, with the use of the homotopy method, a constructive proof of the Poincaré-Birkhoff theorem is given. This approach provides a global method for finding fixed points of area-preserving maps and periodic solutions of Duffing equations. 1. Introduction As is well known, the Poincaré-Birkhoff theorem, proposed by Poincaré [38] in 1912 and proven by Birkhoff [7] in 1913, has been generalized by many authors, for example, Birkhoff [8-10], Jacobowitz [29], Ding [20], [21], Conley and Zehnder [12], [13] (about Arnold's conjecture [6]), Franks [24], [25] and Flucher [22]. Today, these notable results have become powerful tools in studying the dynamical systems and periodic solutions for some significant equations, for example Newton equations and Duffing equations. The related works can be found in [14, 16-20, 23, 25, 28, 29, 35] (and [15, 32, 34, 36] utilizing Moser's twist theorem [37]), and the references therein. As an important aspect for applications, it is also significant to have a way of finding fixed points of area-preserving maps, such as Poincaré maps of Newton equations and Duffing equations. Generally, such a map of an annulus which twists the boundary curves in opposite directions has at least two fixed points, and the sum of indices of fixed points on such an annulus equals zero (see [25]). These result in some difficulties in utilizing the classical numerical methods, for example, the Newton method and the continuation method. The main reasons are the local convergence of the former and the strong regularity of the latter, along the manifold of solutions. The main aim of this paper is to present a global method of finding fixed points of area-preserving maps and periodic solutions of Duffing equations. More precisely, with use of the continuous homotopy method, we shall establish constructively the following Poincaré-Birkhoff theorem given by Jacobowitz [29] and Ding [21]. Let A denote an annular region in R2 \ {0}, whose inner boundary Tx and outer boundary T2 are two disjoint closed simple curves. By D we denote Received by the editors February 9, 1994 and, in revised form, August 29, Mathematics Subject Classification. Primary 54H25, 58F05. Key words and phrases. Constructive proof, the Poincaré-Birkhoff theorem, periodic solutions of Duffing equations American Mathematical Society /95 $1.00+ $.25 per page 2112 LI YONG AND LIN ZHENGHUA the bounded open region bounded by T, i = 1, 2. Hence, A = D2\DX, and 0 A c D2. Theorem. Let T:A-* T(A) c R2 \ {0} be an area-preserving homeomorphism. Suppose: (i) T has the polar coordinates r*=f(r,6), e* = e + g(r,6) such that g(r, 6) 0 on Tx and g(r, 6) 0 on T2 where f and g are C2 continuous and 2n-periodic in 6. (ii) There exists a continuous area-preserving map TX:D2 R2 such that TX\A = T, and 0e TX(DX). Then T has at least two fixed points in A. Now let us make some comments. (a) In our result, the inner boundary curve Tx is not necessarily star shaped relative to the orgin. (b) The continuous homotopy method which we use has been proposed originally by Keller [30], Kellogg, Li and Yorke [31] and Smale [39]. This method has played an important role in various problems of finding fixed points or zeros of maps; see, for example, [1-3, 11, 26, 27, 33, 40]. There are three distinct, but interrelated, aspects of the homotopy method: ( 1 ) construction of the right homotopy map, (2) theoretical proof of global convergence for this homotopy map, and (3) tracking the zero curve of this homotopy map. The first aspect is to link the problem considered to a simpler one by considering a single parameter problem. The second aspect is a key to that method, because it guarantees the third aspect, that is, following paths to find the desired solutions. There have been some typical algorithms in the third aspect; see, for example, [2, 3, 27]. Our constructive proof means for almost every point near the set g(r, 6) = 0, there exists a C path passing to that point such that two ends of this path are two distinct fixed points of the area-preserving map considered. Hence, by utilizing the usual path-following methods, one can find numerically two fixed points. In particular, when fixed points of the map are isolated, an end of that C1 path leads to a fixed point with positive index and another end leads to one with negative index. Because the zeros of g(r, 6) = 0 are easier to determine than the fixed points of T (for example, using the Newton method), we provide an effective global method of finding fixed points of area-preserving maps and periodic solutions of the Duffing equations. To our best knowledge, such global methods seem not to have been applied to this problem. Of course, the aspect of path-following can be carried out by utilizing the typical methods. This makes us focus our attention on the theoretical proof of the global convergence for the given homotopy (Newton-type), that is the constructive proof. This paper is organized as follows. Section 2 is the main part, which exhibits such a constructive proof of the Poincaré-Birkhoff theorem. There it will be seen that for the singular case, T has infinite fixed points in the set g(r, 6) = 0, and finding fixed points is generally simpler; but for the nonsingular case, the problem becomes more complicated. In Section 3, we outline a general framework of applications to the Duffing equations. As stated above, we use the path-following algorithm given by Allgower and Georg [3]. Neverthless, we A CONSTRUCTIVE PROOF OF THE POINCARÉ-BIRKHOFF THEOREM 2113 also provide an experimental example, which shows the efficiency of the reduced algorithm. It should be pointed out that recently Alpern and Prasad [4, 5] have established a very interesting conbinational proof of the famous Conley-Zehnder- Franks theorem. This approach is different from ours. In particular, our approach reduces finding fixed points of area-preserving maps into following solutions of suitable ordinary differential equations with initial values. Consequently, it is convenient to implement to applications. 2. Constructive proof In this section, we give a constructive proof of the Poincaré-Birkhoff theorem. From this proof we can obtain a global method of finding two fixed points for area-preserving maps. The following lemmas are vital to our discussion. Lemma 1 (Sard's theorem). If cf : U c Rm R is a Ck map on the open set U with k max{0, m-n}, then the set of singular values of ( has n-dimensional Lebesgue measure zero. Consequently, the set of regular values of j) is dense in Rn. Lemma 2 (the parametrized Sard theorem [11]). Let V c R , U c Rm be open sets, and 4 : V x U Rk a C map, where r max{0, m - k}. If 0 Rk is a regular value of f), then for almost all a V, 0 is a regular value of t a = f(a, ). Lemma 3 (Garcia and Zangwill [26]). Let V c R be an open set and let ( : Vx[0, 1]» R be a Cx function and 0 a regular value on V. Then H~x =: {(x, t) V x [0, l]: f (x, t) = 0} is a finite number of disjoint continuously differentiable paths. Any path is either a loop in V x [0, 1] or starts from a boundary point of V x[0, 1 ] and ends at another boundary point of V x [0, 1 ]. Lemma 4 (Garcia and Zangwill [26]). Let f : V x [0, 1] Rn be a Cx homotopy, and 0 a regular value on V, where V c Rn is an open bounded set. Then each solution x(s) of the initial value problem dx ^ = (-1)'+1 det#, x,(0) =x,o (/= 1,...,«+ 1) determines a Cx path in / _1(0), where s is a parameter and P', = ( t xl,, 0x,_,, ( xm,, t x +t)- Lemma 5. Under the assumptions of the Theorem, there exists an area-preserving map T2: D2 -» R2 such that T2\A = T and T2(0) = 0. Proof. By the assumption (ii), we have P = Trx(0) Dx. Hence there exists a ball BS(P) with center P and the radius ô such that BS(P) C Dx. Let Px db n OP, where OP denotes the line segment with end points 0 and P. Set Ox - \[P + Px]. Choose a polar coordinate system (p, 4 ) in R2 which has Ox as its pole. With this coordinate system, define S: R2» R2 by p* = p, 4 * = cf + h(p), where h(p) is a C function satisfying h(p) = 0, for p 3/2; h(p) = it, for 0 p a/4. Obviously, S is an area-preserving diffeomorphism, and S(0) = P, S\A = id (identity). 2114 LI YONG AND LIN ZHENGHUA Define T2 Tx os. Then T2:D2 - R2 is an area-preserving homeomorphism, because it is a composition of such maps. Moreover, T2\a = TxoS\A = TX\A = T, T2(0) = TxoS(0) = TX(P) = 0, which completes the proof of the lemma. For convenience, we also define a map H R2 \ {0} (R+ \ {0}) x [0, 2n] by U(x,y) Proof of the theorem. Let 'x +y F(/ ) = T(p) -p = Yi-x(f(U(p)), g(u(p))), J = {p A:g(U(p)) = 0}, y o, y 0. where / = f(r, 8) - r. By Lemma 5, without loss of generality, we may assume that the map Tx satisfies Tx(0) = 0. Hence, Tx:D2\{0} - i?2\ {0}. With the polar coordinate system, we can write Tx in the following form: r* = fx(r,6), 8* = 8 + gx(r,d), ind2\{0}, where fx and gx are continuous on D2 \ {0}. Since Ti^ = T, we have (1) fi(r,8) = f(r,8), gx(r, 8) = g(r, 8), on A.. From the property of the polar coordinate system it follows that fx is 2itperiodic in 8, and for some integer k, gx(r, 6) g(r, 6) + 2kn, on D2\ {0}. Since gx(r, 8) is continuous and D2 \ {0} is connected, k is independent of (r, 8). From (1) and the periodicity of g(r, 8), we obtain k = 0, which implies that gx(r, 8) is also 2^-periodic in 8. Define and set f(r, 8) = (Mr, 8),8 + gx(r, 8)) on (R+ \ {0}) x [0, 2tt] F(r, 8) = (Fx(r, 8), F2(r, 8)) = Tx(r, 8) - (r, 8), on (R+ \ {0}) x [0, 2it]. We shall complete the proof by the following two steps. Io. Singular case. For all (r, 8) e UA, the Jacobian F'(r, 8) is singular. Take a simply connected region Q. c D2 such that Dx c Q and dcl c J. Indeed, by (i) there exists a bounded and connected branch Qx c R2 \ J such that 0 Cix and dílx c /. Thus Dx is a region. Let Zi denote the set of all closed simple curves v lying in Qx. By D(v) we denote the bounded region bounded by v. Set Çi = \J{D(v):i lies in Q,}. Then Q is the desired simply»ly connected region. «r 1 We claim: Fr(r, 0) = O, on 9q. A constructive proof of the poincaré-birkhoff THEOREM 2115 By (i), there exists (r0, 80) UA such that rank F'(r0, do) = 1. Hence by Lemma 1, almost every r range (Fx) is a regular value of Fx. First, we prove (2) Fx (r, 8) = c (a constant), on dil. If not, then by the continuity of Fx, Fx(ïldQ) is an interval in R. Let Si denote the set of all regular values in Fx(HdQ). Since dq is bounded and closed, and Fx is C1, Sx is open in R, and consequently, S2 = Fl~x(TlSx) is open in the relative topology of TldÇl. By the inverse function theorem, for each (ro, do) S2, there exists a positive number n = n(ro, do) such that on [8q- n, 80 + n], the equation (3) Fx(r,8) = Fx(r0,8o) has a unique solution (r(8), 8) satisfying r(80) = rq, \r(8)-r0\ n, if FXr(ro, 8q) ^ 0 ; or on [ro - n, r0 + t]], equation (3) has a unique solution (r, 8(r)) satisfying 0(ro) = 0o, Ö(r)-öo f?, if Fxe(r0, 80) t 0. We can choose a small n such that FXr ^ 0, on Z, = [?o - 1, ro + *l]x [do - n, 8o + n]; or FXg / 0 on /,. For definiteness, let us assume it is the former case. Since F' is singular, there exists a continuous function X(r, 8) defined on In such that (4) (F2r,F2e) = k(r,8)(fxr,fxg) on I,. From (4) and _ dfx(r(8),8) =-dd-= dr Flrd8+Fxe it follows that df2(r(8),8) -Jq-= dr dfx(r(8),8) _ rirjq-+ ïie = (r(8), 8)-^-=0, on/,. Hence on InndQ., (5) (Fx(r, 8),F2(r, 8)) = (Fx(r0, 9o),F2(r0, 80)) = (Fx(r0, 80), 0), which implies (r(8), 8) d l. Set I(n(r, 8)) = (r-n(r,8),r + n(r, G)) x (8 - n(r, 8), 8 + n(r, 8)). Since the set of all members in {I(n(r, 8)): (r, 8) S2} disjoint from each other is countable, by (5) Sx is also countable, which contradicts the openess of Sx. Therefore (2) holds. Now we prove that c = 0. If c 0, then from Fx(r, 8) = c on d ï, we have f(r, 8) = r + c, on YldQ,. Note Therefore, g(r, 8) = 0, onodil nf(íí) = {(/-*, 8*):r* = r + c,8* = 8*, (r, 8*) UÇÏ} = G(YIQ), 2116 LI YONG AND LIN ZHENGHUA where G(r, 8) = (r + c, 8). Hence UT(Q.) = G(Ylíí). Given any AS = {(r, 8): 0 rx r r2, dx d d2} c nq, we have G(AS) C YlTx(Cl), meas[c7(as)] = r(02 - dx)[(r2 + c)2 - (r, + c)2] Thus, when c ^ 0, we obtain = ^ñ(82-8x)(r2-rx)(r2 + rx+2c) 360 = ^(82-dx)(r2-rx) + meas(as). measte (fi)] = meas[nri(q)] = meas[c7(nf )] # meas(nq) = meas(q), a contradiction. This proves that c = 0, and consequently, every point in 9Q is a fixed point of T. 2 Regular case. For some «o G IL4, F'(ao) is nonsingular. We claim that for some (r0, d0) UA, f(ro, 0o) - r0 t 0. If not, then f(r, 8)-r = 0, onhl Choose any 8 g(ua) and consider the equation f(r,8)-r = 0, g(r,8)-8 = 0. Denote J(8) = {U~x(r, 8) A:g(r, 6)-0 = 0}. Hence Therefore, (fr-l)dr + fedd = 0, onua, grdr + ggdd = 0, on UJ(8). detffi 1 fe\ =0 onil/(0), V 8r gej which leads to a contradiction, because d is arbitrary. By Lemma 1 and the above claim, almost every point (r, d) (FX(UA), (rji, a2 )) satisfying r / 0 is a regular value of (FX,F2), where ax = maxpgr., g(r(p), d(p)) min^r, g(r(p), d(p)) = o2. For such a regular value (r,8), set TB:r*=f(r,8), 8* = 8 + gx(r, 8) - 8, 8 (ox,o2). Obviously, Tq is an area-preserving homeomorphism defined on Tl(D2 \ {0}). Set Fe(r,d) = (fx(r,d)-r,gx(r,d)-d); and define a homotopy map H: YIÄ x WÄ x \-X(r), 1] R2 by ZZe-(Po, P, A) = F9(p) - (1 - )Fg(po), A CONSTRUCTIVE PROOF OF THE POINCARE-BIRKHOFF THEOREM 2117 where p = (r, 8), Fs(p0) = (r, 0), Po F^x(r, 0), and X(r) is a positive constant such that (6) \\r\m max{r:r = ^x2 + y2, (x,y) AU T(A)}. Notice that if for some po = (r, 8) HA, F'(r, (7) WL8)=-^- )F'^e) 8) is nonsingular, then is also nonsingular for X 1. Set V = {P = (r, 8) F-X(F(YIA)): (r, 8) is a regular value of F on HA and r ^ 0, 8 (ax, a2)}. By Lemma 2, for almost all po V, 0 is a regular value of HPo = H(p0, ), for X [-X(r), 1). Denote by Vx all such regular values. Choose any (r, 8) such that F~x(r, 8) c Vx. Let xx = r, x2 = 8, x- = X and P = (xi, x2). Then by Lemmas 3, 4, for each po = (/ o, öo) F~l(r, 8), the solution (P(s, po), X(s, po)) of the initial value problem dx ^ = (-l)'+idetz/;, 1 = 1,2,3, (P(0),X(0)) = (po,0) determines a C1 path (P(s, po), X(s, po)) in H~x(0), where #i = (//;,#;) = det(^ o) = -^9' zz^ = (/z;i,zz;) = det(/^1 )(=/i (p)); and ZZ-^O) = {/ = (r, 0) e n^:zz(p0, P, A) = 0, (8) Ae[-A(f),l),p06Z='-1(r,ö)} Since d (ox, a2) and we have = {P(s, po) UA:X(s, po) [-X(r), 1), p0 F-X(r, 8)}. H(po,P(s,po),X(s,po)) = 0, gi(r(s,po), d(s,p0)) = gi(r(s,p0), d(s, p0)) - d = (l-x(s,po))-0 = 0, for -X(r) X(s, po) 1. Hence, (9) J[d] = {(r, 0): (r, 0) = (r(s, p0), B(s, p0))} C n^. By Lemma 3, (P(s, po), X(s, po)) is either a loop in TIÄ x [-X(r), 1] or starts from boundary points of YlAx[-X(r), 1] and ends at boundary points of YlAx [-X(f), 1]. We shall prove that the former is impossible. Let (P(s, po), X(s, po)) be a loop. By Q(po) we denote the bounded region bounded by P(s, po) We claim 0 Cl(po) and the orbit {P(s, po)} is star shaped about the origin 0. 2118 LI YONG AND LIN ZHENGHUA First, {P(s, po)} is star shaped about the origin 0. Set P(s) = P(s, po) = (r(s), d(s)), X(s) = X(s, po). If not, then there would exist so R such that d(so) is minimal and dd/ds\s=so = 0. Since dd/ds = grr, we see gr(r(so), 6(so)) = 0. Notice that the line 0 = d(so) is a tangent of (r(s), 0(5)) at (r(so), d(so)) By the area-preserving property of Tg, we have detr (r,0) = det(í l + gg)=r- Hence for any (r, 8) VIA, (10) Ar*A8* = det fr fe \ gr i+gej (r+íar,6+ía6) ArA8 = (r + *Ar)ArA0, whenever Ar, A0 are sufficiently small. Since for sx, s2 R, along the curve {P(s)}, we get (11) Ar*A0* = Ar* Ad = fi(r + Ar, 0 + A0)ArA0. From r*(s) = f(r(s),d(s)) = (l-x(s))r + r(s), (12) d*(s) = g(r(s),d(s))-d = 0, it follows that (13) rank(hr, He, Hx)\{r(s),e{s),x(s)),Ms) \ - 2, gg(r(s0),d(so)) 0, dr*(s) ds s=s0 = fr dr(s) ds dd*(s) ds s=s0 s=s0 dd(s) ds J=J0 = 0, Because P(s) is a loop, we can choose such an Jn to possess the following property: (P) For every e 0, meas{0(s):s [s0 - e, so + e]} 0. Then using (10),(11), and (13) yields 0 r(s0) = fr(r(s0), d(s0))[l + gg(r(s0), d(s0))] = fr(r(s0), d(s0)). Then gg(r(s0),d(so)) = 0. Conseqently, rank(/zr, He, Hx)\(r(s0),8(s0),Us0)) = 1 a contradiction. It is clear that 0 ii(po) By the claim and (12), we have (14) Tg[Çl(p0)]cn(po), if r 0 ; or (15) Te[iî(po)]DCi(pQ), ifr 0. Notice that X(s) 1. Therefore from (14) and (15) it follows that a contradiction. meas[tg(çl(po))] meas(cl(po)), This shows that {(P(s), X(s))} is not a loop. A CONSTRUCTIVE PROOF OF THE POINCARÉ-BIRKHOFF THEOREM 2119 By(6),(9)and(12), we get (16) X(s) -U(r), fors R. Hence T(po) = {(P(s), X(s)):X(s) 1} starts from the points in YIA x {1} and ends at the points in YIA x {1}. Set We claim S± = {q UA: P(s, po) leads to q, as X(s) - l,ie R±}. (17) S+nS- = 0. If not, then S+ n S- ^ 0. By the above arguments, for each s R with X(s) 1, 0(5) is not an extreme value. Otherwise, for some so R, d(so) is an extreme value. Then for any interval I containing so, the property (P) does not hold. Therefore 0(5) = a constant, for X(s) 1. Since rank(zzr, He, H?)\(p{s)^(s))^(s) \ = 2, we have dr(s) d$ 0, fora(5) l. This shows that S+ n _ = 0, a contradiction. Hence when X(s) 1, (18) ^1 0 (or^ 0). ds ~~ ds ~ From (17) and (18) it follows that there exist 8X,82 R with \8X - 02 = 2it such that U I 02, asa(j)-» 1,5CZÎ_. Then F(po) bounds a bounded region l(po) Notice that for each 0 (min{0i, d2}, max{0i, 02}), {(r, d):r 0}r\T(po) contains only one point. Thus by (12), meas(tg(sl(po))) # meas(q(j9o)) This contradiction shows that (17) is true. Using (12), we know that every point p S+liS- is a fixed point of Tg. By (17), Tg has at least two fixed points in A. In particular, when S+ and S- are isolated sets, letting {P+} = S+ and {/ _} = S_, we have index(7g, Z-±) = ±1. Indeed, choose e 0 such that B((P+)nBe(P-) = 0. By Sard's theorem, there exist two sequences of regular values, {X(Sk)}Sk o and {X(tk)}tk o suchthat X(Sk) - 1, X(tk) ^l,k^oo, P(sk) Bt(P+),P(tk) Bf(P-). Hence, for sufficiently large k, deg(id - Tg, B( (P+ ), 0) = deg(id -TB, B((P+), rk) (19) = sgndet(^(^)), where rk = -(1 -X(sk))(o) Notice pe(ià-te)-hrk) P(sk) - P+, X(sk) 1, X(sk) - 1, 5fc 0. 2120 LI YONG AND LIN ZHENGHUA Therefore, from (19) it follows that Similarly, index(rö, P+) = 1. index(rö,zj_) = -l. Now we prove that T has at least two fixed points in A. In the same way as in choosing (r, 8), choose a sequence {(rk, 8k)} such that 8k (ox, a2), 8k -* 0, k oo, and without loss of generality, let fy Fo ^ 0. Set HkiPk, P, A) = Fg (p)-(l- where pk Fgx (rk, 0). Then the Cauchy problem dx (20) 5 Ip*(0),4(0)) = Cp*,0) X)Fg (pk), (-l)l+xdet(hk)'i, i =1,2,3, determines a C1 path (/^(s), Afc(5)) in (//^ '(O). Applying the Arzela-Ascoli theorem and passing to a subsequence if necessary, we may assume Pk P*, and for each compact interval I with Xk(s) 1, (21) Pk(s)-^P(s), Xk(s) ^X(s) uniformly on I. and Set Sk± = {q UA: Pk(s) leads to q, as Xk(s) - l,i oo oo d±=n u^±fc=l =A: Obviously, every point p D+ U Z)_ is a fixed point of T in A. We claim that D+ U Z)_ contains at least two points. Indeed, if not, then D+ U Z)_ = {p}. Since rk - ro ^ 0, we have R±} lim sup{ rí:(5)-^(51) -r- 0/t(5)-0/t(5i) (mod27t) :Afc(í), Afe(ji) 1} 0. fc»oo Notice if D+ U Z)_ = {/?}, then 8'k(s) 0, forí( l, (22) lim meas{8k(s):xk(s) 1} 2n. k-*oo First, we have lim meas{8k(s):xk(s) 1} # 0. fc»oo If not, without loss of generality, we assume lim meas{0^(5):afc(5) 1} = 0. k»oo Hence lim^^meas^^):^^) 1} 0. Since when Xk(s) 1, 8k(s) is monotonie, and if 8'k(s) = 0, then 0^(5) = a constant, and r'k(s) # 0. We have D+uD-=f\\J{rk(s):Xk(s) l}. k=\ i=k A CONSTRUCTIVE PROOF
Related Search
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks