AA1422 PERIODIC SOLUT1ONS OF PERTURBED SUPERGUADRATIC J/1HAMILTONIAN SYSTEMS(U) WISCONSIN UNIV-MADISON CENTER

I FOR MATHEMATICAL SCIENCES Y LONG DEC 87 CMS-TSR-88-23

UNCLASIID DAAL83-87-K-063FG1/

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CMS Technical Summary Report #88-23

PERIODIC SOLUTIONS OF PERTURBEDSUPERQUADRATIC HAMILTONIAN SYSTFMS

N Yiming Long

Center for the Mathematical SciencesUniversity of Wisconsin-Madison610 Walnut StreetMadison, Wisconsin 53705

December 1987

(Received October 7, 1987)

ITISCSponsored by Approved for public release L _.,Distribution unlimited '

U. S. Army Research Office 9 88P. 0. Box 12211Research Triangle Park tNorth Carolina 27709

Air Force Office of National Science Foundation Office of NavalScientific Research 1800 G Street, N.W. Research

Bolling AFB Washington, DC 20550 Arlington, VAWashington, DC 20332 22217

88

.Acoession ForNTIS GRA&I

UNIVEPSITY OF WISCONSIN-MADISON DTIC TAR

CENTER FOR THE MATHEMATICAL SCIENCES Unannounced-." Just iflea i dr

PERIODIC SOLUTIONS OF PERTURBED SUPERQUADRATIC

HAMILTONIAN SYSTEMS By

Yiming Long istribution/Availability Codes

Technical Summary Report #88-23 Avni' and/or-Dist ,-pecial

December 1987

ABSTRACT

In this paper, we prove the existence of infinitely many distinct T-

periodic solutions of the perturbed Hamiltonian system -

z = J(H'(z) + f(t))

under the conditions that H is of C1, superquadratic, and possesses

exponential or polynomial growth at infinity and that f is of Wl and T-L . -

periodic, via minimax methods. , .,.

AMS (MOS) Subject Classifications: 34C25, 58E05, 58F05

Key Words: Hamiltonian systems, superquadratic, perturbation, monotone

truncations; a priori estimates; minimax methods, multiple

* periodic solutions

.

This research was supported in part by the United State Army under ContractNo. DAAL03-87-K-0043, the Air Force Office of Scientific Research under GrantNo. AFOSR-87-0202, the National Science Foundation under Grant No. MCS-8110556and the Office of Naval Research under Grant No. N00014-88-K-0134.

% % %

- - - - - - - - - - - -

*-*" PERIODIC SOLUTIONS OF PERTURBED SUPERQUADRATIC

HAMILTONIAN SYSTEMS

Yiming Long

-!: !1. Introduction and main results

We consider the existence of periodic solutions of a perturbed

Hamiltonian system

S(.1) z = J(H'(z) + f(t))

where z, f :R R2 N, = dJz 0 (0 ), I is the identity matrix on N,

- dt' T 0

H : R2N + R, H' is its gradient. Let aeb and .*1 denote the usual inner

pdt do o 2N.product and norm on H will be required to satisfy the following

conditions,

(Hi) H E C1 (R2NR)I•

(H2) There exist M > 2, r0 > 0 such that 0 < UH(z) 4 H'(z)-z

vizl > rO.

(H3) There exist 0 < -2-- < q q2 < 2 and ai, Ti > 0, 8i > 0 for

1 = 1,2 such that

TjIslq l T21szq2 2a 1e - 1 H(z) a 2e + 2 V z R2 N

or

(H4) There exist 1 < P1 4 P2 < 2PI + 1, ai > 0, 82 > 0 for i = 1,2

such that

aIzI 1 01 H(z) 42jzz12 + 82 V z R2N

Our main results are

in..

- This research was suppcrted in part by the United States Army under ContractNo. DAAL03-87-K-0043, the Air Force Office of Scientific Research under Grant

No. AFOSR-87-0202, the National Science Foundtion under Grant No. MCS-8110556

and the Office of Naval Research under Grant No. N00014-88-K-0134.

-1-

0N% 2iU

Theorem 1.2. Let H satisfy conditions (HI)-(H3). Then for any given

W1'2(R 2N) I )psessa noneT > 0 and T-periodic function f C Wl 2 (RR ( possesses an unbounded

oc

sequence of T-periodic solutions.

ITheorem 1.3. The conclusion of Theorem 1.2 holds under conditions (Hi),

(H2) and (H4).

- - Such global existence problems have been studied extensively in recent

years. For the autonomous case of (1.1) (i.e. f - 0), the result was proved

by Rabinowitz under the conditions (Hi) and (H2) only ([17, 19]). His proof

is based on a group symmetry possessed by the corresponding variational

formulation. When one considers the forced vibration problem (1.1) (f does

depend on t), such a symmetry breaks down. Bahri and Berestycki ([2])

studied the perturbed problem (1.1) and proved the conclusion of Theorem 1.3

0 by assuming (H2), (H4) and (HI') H £ C2 (R2N,R). Our Theorem 1.3 weakens

their condition on the smoothness of H and Theorem 1.2 allows H to

-4- increase faster at infinity.

Our proof extends Rabinowitz' basic ideas used in [18,191 and ideas used

in (12,13]. In order to get the smoothness and compactness of corresponding

functionals, we introduce a sequence of truncation functions {Hn} of H

in C1 and corresponding modified functionals {Jn} and J of 1, where

(7r (- z.Jz - H(z) - f.z)dt

* We make {Pm} be monotone increasing to H and {On} be monotone decreasing

to J as n increases. These monotonicities also allow us to get LO-

estimates for the critical points of Jn we found. We modify the treatment

* of the Si-action on /2 , 2 (S1,R 2N) by introducing a simpler S1 -action on it

to get upper estimates for certain minimax values. Combining with

applications of Fadell-Rabinowitz cohomological index we get the multiple

existence of periodic solutions of (1.1).

-2-

% A. .

%

*In §2 we define {Hn)i {Jnj and J. With the aid of an auxiliary space

X we define sequences of minimax values {ak(n)), {bk(n)) of Jn,~ {ak}, {bk}'

of J in §3, and discuss their properties in §4. §5 and §6 contain estimates

from above and below for {ak}. We prove the existence of critical values

of Jn in §7. Then in §8 by showing that the critical points of Jn for

large n yields solutions of (1.1), we complete the proofs of our main

theorems. Finally in §9 we discuss more general forced Hamiltonian systems.

Since the proofs of Theorems 1.2 and 1.3 are similar, we shall carry out

the details for the first one only and make some comments on the second in §8.

For the details of the proof of Theorem 1.3, we refer to [13].

Acknowledgement. The author wishes to express his sincere thanks to

Professor Paul H. Rabinowitz for his advise, help and encouragement,

especially for his suggestions on truncation functions. He also thanks

Professor Stephen Wainger and Dr. James Wright for interesting discussions on

imbedding theorems.

4,'

2S

-- 3-

4. %

'p..

.P~'

o43

P. 42. Modified functionals

By rescaling time if necessary we can assume" T = 2w. Let E -

2(S1,R2N). The scalar product in L2 naturally extends as the duality

pairing between E and E' = W 1/ 2(S1,R2N). Thus for z c E, the actional

integral - A(z) is well defined, where2

(2.1) A(z) = 0 z.Jz dt

.,- l. Ik -rkFor k c N, write k [~,k=k -[L], where [k] is the integer part of

k 2Let el...,e2N denote the usual orthonormal basis in R2N . We write

i =-1, and denote for k c N

-5 J = (sin kt)e - (cos kt)e , = (cos kt)e + (sin kt)ek+N k+N

~~(2.2) ^

k= (sin kt)e + (cos kt)e , i4k = (cos kt)e - (sin kt)ek i+N k +N

+ =span{q m. I k 4 n}, E- span m 4 k 4 n) for m,

Let Emn = (k' k kmn {*k'ikm

n~N,~m ~ + =4 E+i E -E +. and E0 -n € N , N 4 m 4 n . E + = E- N E1" " ~ ~~N+ 1 , + w ' E E + , + a n=

span{;klik 1 4 k < NJ. Then E = E+ 0 E- ® E0 and A(z) is positive

definite, negative definite and null on E+, E- and E0 respectively. For

z Z + z- + z0 E E+ 0 E- 0 = E, we take as a norm for E

I z12 A(z) - A(z ) + Iz0l2

* Under this norm, E becomes a Hilbert space and E+, E-, E0 are orthogonal

subspaces of E with respect to the inner product associated with this norm,

as well as with the L2 inner product. (2.2) gives a basis for E.

* A result of Brezis and Wainger [6] implies that E is compactly embedded

" h1into LP(sv,R p<+ and the Orlicz space with M(z) = etiz, _

nk1 IZk T > 0, 0 < q < 2 and n c N, nq > 1, and E C LM. Notek=0

16 that q = 2 is the critical imbedding value (cf. [6,10]). We shall use the

following version of their imbedding theorem,

-4-

N0

- P MII.R

Lemnm 2.3. For T > 0, 0 < q < 2, there exist constants C1 , C2 > 0

%I depending only on T and q such that

f2 ' exp(a~l zlq)dt 4C 1 ex( 2-) V a >0, z E0 (Ll

Proof. We use the notations in [6] and only prove the Lemma for E

Cn mt

For z E. write z(t) =CO+ n -e where Cn c C. Then z=

k*g, where k(t) e 1 L(t,-), (t) n e C L(2,211. By

(8) of [6],

*2 -2

Ujzjq 4 q 12~jz + £2-q

2 2qCl+ jigt)1k 2 22-(CU Iog j)IML(2,-) Igo L(2,2 ) + 2Ca for 0.< t< 1

Coose e -(qClkI 2 , g) L2,2 2 then from EIo 2 ga CIZI and

f2w eaIzqdt = f2wreaIzl dt, we get the lemma.

0 0

For z E and 0 c [0,2w] S, we define an S1-action on E by

(Tez)(t) =z(t+O) V t C [0,2w]

We say a subset B of E is S (E)-invariant if TeZ z B, V z B, 0 c [0,2w].

Note that Fix{T01 [z c E ITez = z, V 8 c [0,2w]) = E0

By Lemma 2.3 and (H13), 1(z) defined in (1.4) is a continuous functional

on E and formally the critical points of I correspond to the solutions of

Without loss of generality we assume ro 1. Set a i= ~ )% 0 1z=r0

-5-0

so 0 61 + max jH(z)J, where 8 1 is given by (H3). Conditions (Hi) and

(H2) imply that 0for some 83 ; 0,

0 fZ I U < H(z) , v IzI > ro

F (2.4)

a I0IzIu H(z) + o (H'(z).z + 83) • (R 2N

Choose a c (0,1) such that pa > 2. We have

Proposition 2.5. Assume conditions (Hi) and (H2). Then there exists a

sequence {Knl C R and a sequence of functions {Hn} such that

10. 0 < K < Kn < Kn+l, V n c N and Kn ++4 as n +4w where K0

somaxf1, -of, ,

0020. Hn C C(R 2N,R) V n c N.

30 . Hn(z) = H(z) V n C N and Yz n K.

40 . Hn (z) H n+1(z) 4 H(z) V n C N and z R2N

5% 0 < UaHn(z) 4 HA(z)-z, V n C N and Izr ) r0 .

6*• For n c N, there exists a constant X > 1 independent of n

and C(n) > 0 such that

H( () (z).z + 1) V z C _2N

Since the proof of this proposition is rather technical and lengthy, we

put it in the Appendix.

Similar to (2.4), there is a constant 84 > 0 independent of n such

a that

{LOIzIo' 4 H (z) V n c N and IzI > )

(2. )1 2N

aoIzt"G < H (z) + 8 - (H'(z).z + 84) V n N and z R Rn 0 l n 4

For n N, we define

(Z) = AC) - r2 0HZ)dt - f2f-zdt V z E

-6-

N N N N N

A-

A- Lemmua 2.7.

I in C E,R), V n £ N.

2'. I(z) In+1(z) 4 In(z), V n c N and z E.

Proof: For 10 we refer to (5,17]. 20 follows from 4° of Proposition 2.5.

From now on in this section, we define T 1 /3pa+10.p4, Lemma 2.8. There is 65 > 0 independent of n such that

S(2T(3L10+2) 3a+2 27r n N d(2.9) 2(T-1) Ifn L2 2 8 0o (Hn(Z) + 80 )dt + 85 n N and z E

Proof. By (2.6)

__2_ T(3a+2)j (Hn(Z) + 80 )dt I 2

. 0- 2(T-) fL

1 2

3a2 a DZ VU T(3ua+2) If (27r) 21ja izi8 0 paU 2(--) L2 I1caLj Lu L o

Since ija > 2, (2.9) holds.

Lemma 2.10. There is 86 > 0 independent of n such that for any n E N,

z C E if I (z) =0 then

3p)+2 2r 1 )2 s 1)/2+(2.11) f (Hn(z) + 80)dt + 8 ((1 A(z) +8 ~0 n 0 5 2 6

Proof. From <I'(z),z> = 0, we get that

A(z) 2 . (H(z)*z + 84 )dt 2 o f.zdt - W841

22

(2.12) ) 27t (Hn(Z) + 80)dt + f 2, f.zdt - wS4 (by (2.6))

32 0 i j-2 ii

> [2 (H (z) + )dt +IZIa f Zn 0 16 0 L a

5'- ElIf2ziL - ir4 - 1.f 2 I 2 4

-7-

%A%

Since pa > 2, (2.11) holds.

. Let X C (R,R) such that X(S) = 1 if s 1 1, X(S) = 0 if s > T,

2and - < x''q) < 0 if 1 < s < T, where T is defined before Lemma 2.8.

For n N and z C E we define

90(z) = ((-! A(z)) + 1)1/2 + 8S2 6

' (z) = f0 (H(z)+8 0)dt + 85 ,n(z) = 8 2w (H (z)+$ )dt + 88 0 0 5 n8 0O n 5

N4. 0pz) = x( l(z)- )z , )ZTq0(z)) n Tn(Z )

J(z) =- A(z) f H(z)dt - -z) f z dt200

J(z) = - A(z) - H(z)dt - i(z) f-z dt

For these functionals on E we have

Lemma 2.13. 10. n E C1 (ER), Jn 4 c1 (ER), V n c N.

.- 20. d C(E,R), J c C(E,R).

Proof. For 10 we refer to [5,17], 20 follows from (H3) and Lemma 2.3.

Lemma 2.14. For any m, n c N, m ) n and z c E,1 ,2' 3 2i

1.. 0 j .0 (Hm(z)-Hn(z))dt 4 J (z) - J (z) . f 0(H (z)-H (z))dt,2 0 m n n m 2 Om n

: 1 21 3 2

20. 0 <1 _ 0 (H(z)-H (z))dt 4 Jn(z) - J(z) r 1 f'0 (H(z)-H (z))dt.2 0 n n 2 0 n

Proof. We only prove 10. The proof of 20 is similar. Let supp *n be the

8.' closure of [z C E j tn(z) O} in E.

If z , supp n U supp *m , then by 40 of Proposition 2.5

* Jn(Z) - Jm(Z) = f21r (Hm(Z)_Hn(Z))dt ) 0

-8-

-

.If z C sUpp Pn U SUpP m, then Tj(z) 4 T 2 0 (z) for j = n or j = m.

By Lemma 2.8,

-9(z) ) t(3pa+2) If' *zNS. 2 (-T-1) 2 2

-~L LSo

2 .3pca+2 IfII Izir-1 8 L2 L2 j(z) 1

2.15) T)o (2() -2

On the other hand

2z-.1r +m (z ) 2 rn n (Z) m (Z) = fo (Hm( z-Hn(z))dt + X'(C, To f.z dt

where we have used the mean value theorem with a number between m

q)(z) Tcp0(z)-- n

and (z)" By the definition of Tm and Wn' we get,

00

V9Since JX'( )j 4 T-2j1, V E R, by (2.15) we get that

XC 311cr+2 r21rfzd

+ '€ ) 8- -- o f3 a

n2 T90 (z) 2

Combining with 40 of Proposition 2.5 and (2.16) we get the proof of 10.

Corollary 2.17. For any n C N and z C E

Jn(z) > Jn+i() > J(z)

Lemma~ 2.18. There exists 87>0independent of n such that for any

MSn , z c E and M > 88, if Jn(z) > M then ro(z) >o

Proof. Since a > 2 and

Jn(z) 4 - A(z) a N0 zI L2 2 Z 2ir80o

-9-

I'-.

.

,p. LeVma 2. 1xith

op, %L.'.

there exists C > 0 independent of n such that

VO (z) = 1 ) + 1)/ + Jn(z) +8 - C2 6 n 6

and this yields the Lemma.

Lemma 2.19. There exists a constant 88 > 0 independent of n such

that for any n C N and z C E, if n(z) ) 68 and <JA(z),z> = 0, then

Jn(z) = In(z) and Jn(z) = In(z).

Proof. For any z, g E, we have that

<J'(z).?> = (1 + T (z))A(z,C)n n,1

C., (2.20)

z)+ W H'(z). dt- ,o(z)o f- dto

where

A(z,C) = 0 + .Jz)dt

9 n (Z) A(z)Tn(Z) 2T 1 (z) = X'( (Z) 2 fez dt

: nr z) 0 (z) ( 0 (z)- 6)

9 (z)T 8,2 (z ) = x'( -/'o 2 .fez dt

2(z) r 0(W) 8Tp 0 (z) '0

If for large enough 88 we have

(2.21) ITn (z)I 16 2 I n,2 <6po

Then from (2.20) with C = z, we get

1 3pa+2 2 ( titA(z) 1 6 (Hn(Z)+80)dt + La L2 Z 84 - 1

. This is (2.12). So 90 (z) > Tn(z), thus $n(z) = 1 and *p'(z) = 0. This

' ' yields the lemma. Therefore we reduce to the proof of (2.21).

If z L supp n' Tn, 1 (z ) = Tn, 2 (z) = 0. If z C supp *n' then

2 •___Iln0 (z) ) n 8 0

NL

-10-

C-%

Thus

~o- I1

T (Z) < T n21 2f L M0 u'-" - 1n,I T-1 L2 2 0 (z) M(y 0 (z))

Similarly

ITn 2 (z)I 4 M(9O(z))Va

for some constant M > 0 independent of n and z, by Lemma 2.18, thisv.

jk%. implies (2.21) and completes the proof of the Lemma.

We say Jn satisfies the Palais-Smale condition (PS) if whenever a

sequence {z} in E satisfies that {Jn(zj)} is bounded and J'(z-) + 0

as j + +w, then (zj) possesses a convergent subsequence.

Lemma 2.22. For any n C N, Jn satisfies (PS) on [Jn]S8 = {z c E

Jn(z) > 881, where the constant 88 > 0 is defined in Lemma 2.19 which is

independent of ni.

Proof. Since J'(zj) + 0, we may assume VJ(z), z E. By

Sn ,<

the assumption 88 4 Jn(zj) 4 M, from (2.20) and (2.21) we get

M + Rz I > J (z ) - 1 -<J'(z.), z.>jlE nj) 2(1+T (z.)) n j

Sn , (n ,1 2p.

1+Tn,2 (zj f2 H'(z ).z.dt - f 0 Hn(z )dt

2 T+T (z z)n(Zj So f , dt

n,1j

0a-2 f2- 1 (Z2)1zdt)2Hnf d

,.

o> H'(z. )zdt +2 (u- 2 ) fo (H(z)+$)dt16U~ 0O n j j ' 0~

2 Eft Iz I 2 -2irL 2 L 2 (4 0 o

-11-

0't

% ~ % %.'~ j .jj'7p 'ALI

W', Therefore there is M, > 0 independent of n and j such that

(2.23) Iz Ipa + (H'(z.).z + 0 )dt M (Iz I + 1)J O f2a(n joz 4 1 JE

+ 0 = Sic f21",Write zj z. + z. +z.. 2ic 0 I 2

Wri zdt, there is M>

independent of n, j such that

(2.24) (zI . 1.. 4 M (Iz if + 1)L U0 2 jE

From (2.20) for <J'(z.), z+>, we getn j J

I iz + 2 4 (1+T (z))A(z ) 4 . f0 ,HI(z Iz+Idt + IfO Iz I + Nz + 2

By 60 of Proposition 2.5 and (2.23) we get

1+1/ 04 M 3(n)(Iz i A + 1)

for some constant M3 (n) > 0 independent of j. Thus there is 4(n) > 0

' independent of j such that

Iz + r M (n)(Izj E 4 jE +1

Similarly":"2 1+ 1/AO

-2 +/X0Iz M 5 (n)(Iz I + 1)

j E 5 j E

for some M5 (n) > 0 independent of J. Combining with (2.24), we get aII constant M6 (n) > 0 independent of j such that

(2.25) Iz jI M (n)

Let Pi E + E - be the orthogonal projections. From (2.20)

P±Jn(Z) = ±(1 + T (z))z t +P(z.)nj n,i))j nj

where Pn is a compact operator by (Hi), (H2) and Proposition 2.5. Since

-12-

AL -2Z % % %

iT n,(z 16

+-Z± n( 1(1 + -1±,± 2. (I T zj)PJn(Zj - (1 + Tn, (j) -n(Z)

By (2.25) this shows that {zt} and {z} are precompact in E. By (2.25)

{z } is also precompact, therefore fzj} is precompact in E, and the proof

is complete.

0

Lemma 2.26: There exists a constant 89 > 0 such that

(2.27) [J(z) - J(T z) < 89(log 1 (IJ(z)l + 1) + 1) V z £ E

where q, is defined in (H3).

Proof: A direct application of H8Ider inequality shows that there

exists c1 > 0 such that

;/1 fo(T J0zjdr)q 1 21re 4- f, e dt + c V z C E

If z C supp 3, IJ(2)7 (3ua-'- 1) f2(H(z) + S )dt _ cf 2wjfzldt - c3

"#.' k 8T 2 1)f

> C 27r 11zi t

a... )c 4 fo edtc

Here we used (H3), and cj's denote positive constants. Therefore

IJ(z) - J(T z)l 2 (z) f0Ifzldt 4 c 6 (z) fo IZ 1

1 /q

S89(log (lJ(z)l + 1) + 1)

for some 69 > 0 and the proof is complete.

.¢

1,.3-

%%%..V

'a.

S3. A minimax structure

For k N, k ) N+1, we define Vk(E) = EN+lk * E- M E0 . By (2.6) and

Corollary 2.17, there exists Rk for k • N+1 such that I < Rk < Rk+1

and J(z) < J (z) 4 0 V n c N, z c Vk(E) with IzIE ) Rk.

Let Dk(OE) = Vk(OE) n Bk(OE), Bk(E) = {z E 3ZIE E R0.

For z = z0 + Z+ + z- E we write

(3.1) z+ k p e k' z = k ake *k where pk'ak ) 0, ak'Sk e [0,2w]k)N+I k>N+1

Then we have

2 0 zO2 -~2a2%Zi = + 2n k(pk+ak), and

E k>N4-1 k

i(ak+kq) A i (T z - z + (Pke 9k + ak e *k) V c [0,2w]8 -k>N+l

We define a new S1-action on E by

0 i(clk+e) k i k+e)T z = z + ) (Pke Vk + ake *k) v e c [0, 2w)]

k>N+ I

for z with expression (3.1) and denote E with T by X. We define an

*' St-action on X x E by

0 (xz) = (Te x,T ez) V (x,z) C X x E and 8 c (0,2w]8e 0 z

We also define sl-invariant sets, equivariant maps, invariant functionals

for X, E, X x E in a usual way (cf. [9]). Let E (or X,F) denote the

family of closed (in E) (or X,XxE) S1-invariant subsets in E\{O} (or

X\{0), X x E\{0}). Then F contains x {0} and {0} xE. For B eX, we

say a map h : B + E is S -equivariant if h(T x) = T h(x) V x c B and

8 e (0,2n]. We also denote by Vk(X), Dk(X), Bk(X) the sets in X

corresponding to Vk(E), DkE), Bk(E), etc. We introduce the Fadell-

-14-

M

Rabinowitz cohomological index theory on F (cf. (9]).

Lemma 3.2: There is an index theory on F i.e. a mapping

y F + (0} U N U {+=} such that if A,B F

10. y(A) < y(B). if there exists h £ C(A,B) with h being S1-

equivariant.

20. y(AUB) 4 y(A) + y(B).

3%" If B C (XxE)\(X0 xE0) and B is compact, then Y(B) < w and there

is a constant 6 > 0 such that y(N6 (B,XxE)) = y(B) where

N6 (B,XxE) = {zCXxE I fz-B XxE 4 6}.

4%• If S C (XxE)\(X0xE0 ) is a 2n-1 dimensional invariant sphere,

then Y(S) = n.0

By identifying X with X x {0}, and E with {0} x E, we may view

that the index theory y is defined on both X and E.

For x C X, z C E, we write z - x if z x 0 Pk(z) = Pk(X)

a k(z) = ak(x), N(z) = c(x) and k(z) = 8k(x) V k > N+1.

" For x x0 + k ke 'k + ake *k) C X, we definek>N+ 1

ik ik k

h(x) = x0 + 7 (pke lk + ike k

k>N+ 1lk

About this map h, we have

Lemma 3.3: 10. h c C(X,E) and is surjective.

20. h is S'-equivariant.

S30. h(B p(X) n Vk(X)) = 3Bp(E)fl Vk(E) V p > 0, k > N+1.

40. h(x) = z if x C and x - z.

5%• If fh(xn)1 is convergent in E, {xn} is precompact in X.bJn

Proof: 1° - 40 are direct consequences of the definition of h. Suppose

'Xnl C X and h(xn) + z in E as n + a. Write xn n; Pk(n), ck(n);

-- 2 2ak(n); 8k(n)). Since 2wk(Pk(n) - 0k(m)) 4 Ih(xn ) - h(x m)E' we get that

* {Pk(n)}, similarly {ak(n)}, is convergent for each k c N. Since { } is

* precompact, and ak(n), 8k(n) c [0,2n], we can choose a sequence {nj} in

N such that {Xnj, {ak(nj)} and {Bk(nj)} are convergent for every fixed

k N. Then x } is convergent in X. This proves 5%

We name the above map h : X + E by "id". With the aid of "id" we can

define minimax structures now.

Definition 3.4. For j c N, j ) N+1, define rj to be the family of

such maps h, which saisfy the following conditions:

1. h c C(D (X),E) and is S equivariant.

20. h = id on (9Bj(X) n vj(X)) u (X0 f Dj(X)) Fj(X).

30- P-h(x) = a(x)P-id(x) + 8(x), V x C Dj(X), where a c C(Dj(X),

(Ia)with 1 4 a < +a* depending on h and is S -invariant,

B c C(D (X), E-) is compact and S1-equivariant, and 8 = 0 on

Fj(X).

40. h(D.(X)) is bounded in E.

Definition 3.5. For j c N, j N+1, define Aj to be the family of

maps h, which satisfy

1o h c C(Dj+I(X) , E) and hlDj(x) E rj.

2. h = id on (aBj+I(X) fl Vj+,(X)) U ((Bj+ 1 (X)\Bj(X)) n Vj(X)) Gj(X).

3'. P-h(x) = a(x)P-id(x) + 8(x), for any x c Dj+ 1 (X) where

a C C(Dj+I(X), [1,o]) with I 4 < +- depending on h,

8 e C(Dj+I(X), E-) is compact and 8 = 0 on Gj(X), and a, 8

0'1

'"9. -.1. 6-

-.

- are extensions of the corresponding maps defined in 10 via the

definition of rj.

4 . h(Dj+1(X)) is bounded in E.

Remark. id E ri r Aj for any j > N+1.

Lemma 3.6. For j ; N+i, any h c rj can be extended to a map in Aj.

Proof. For h c rj, define h = id on Gj(X). This also extends a

and 8 in 30 of Definition 3.4 of h by a = 1 and 8 = 0 on Gj(X). Now

we use Dugundji extension theorem [7) to extend a, 8 to the whole Dj+,(X).

Since by this theorem the image of the extension mapping is contained in the

closed convex hull of the original image, a E C(Dj+1 (X), [1,a]) andft-

8 C C(Dj+M(X), E-) is compact and 40 of Definition 3.5 holds. We use this

theorem again to extend P+h, P0h to the whole Dj+I(X). (p0 : E + E0 is

the orthogonal projector.) Then define h = P+h + P-h + P0h. It is easy to

a..)i check that h c A.

0

Now for k C N, k N+1 we define

Ak = {h(D.(X)\Y) J k, h r., Y £ X with y(Y) 4 j-k}

= {h(D (X)\Y) j > k, h c A., Y c X with y(Y) 4 j-k} ,

ak = inf sup J(z) , bk = inf sup J(z)ACAk zEA B£k zCB

ak(n) = inf sup J n(z) , bk(n) = inf sup J n(z)

AEAk zCA B.Bk zCB

Since id c rj n Aj and 0 4 A n B ' A c Aj, B 4 BL, -< ak , bk , ak(n),

bk(n) < +-.

-17-

.f,

ft.

"K "O

-U-

§4. Properties of sequences of minimax values.

We have

Lemma 4.1. 10. {ak} and {bk) are increasing sequences.

2*. {ak(n)) and {bk(n)} are increasing sequences for fixed n c N.

30- ak ( bk, ak(n) < bk(n) V n, k C N.

4. ak 4 ak(n+l) 4 ak(n), bk < bk(n+l) 4 bk(n) V n, k C N.

. . Proof. 1* and 20 are due to the fact that Ak+1 L__Ak and Bk+1 g_ Bk"

For any B = h(D j+(X)\Y) c Bk' let A = h(D.(X)\Y), then A C B and

A Ak , so 30 holds. 4 follows from Corollary 2.17.

Lemmar 4.2. For any fixed k c N, k ) N+1.

10. lira a(n) = ak

20. lim b k(n) = bk

Proof. We only prove 1%. 20 can be done similarly.

Given any e > 0, by the definition of ak, there is A0 c Ak such

that sup J(z) 4 a + . Sok 3

ZEA0z A0

(4.3) a (n) 4 sup J (z) a.K + +up D(z) VnCNkn ak + u D nZ"z A0 0

* where Dn (Z) = n(z) - J(z). Since A0 is bounded in E, by Lemma 2.3,

sup D (z) < +.. Thus for every n c i4, there exists z n A0 such that.zeA 0 nnnn

(4.4) sup D (z) 4 Dn(zn) + 3z A0

"' "Since B is compactly imbedded into L1 (sI,R 2N), (zn) possesses a

-18-Sic scmatyibde nt SRN,{ osse

subsequence {zn } which converges to some z0 in L I(S,R 2N). From real)

analysis (cf. [14)) {Z nj} has a subsequence which converges to z0 almost

everywhere. We still denote it by {zn}.

Let Q = t [0,27] I Izo(t)I < -} n {t [0,2w] Zn (t) + zo(t) as

j -1, then Q has Lebesgue measure 2w. For any t c Q, there is

N1 (t) > 0 such that IZnj(t)I - jz0(t)I + 1, V j > N1 (t). Choose

N2(t) > NItW such that "n > IzO(t)l + 1, V j > N2 (t) where {Kn} is

defined in Proposition 2.5. Then Hnj (Zn. (t)) = H(zn (t)), V j > N2(t). This

shows that H(znj) - Hnj (zn) + 0 almost everywhere as j + c. Thus H(zn)

-Hnj (z n) + 0 in measure as j + w.

Since {znj} are bounded in E, by (H3) and Lemma 2.3, {H(zn) -* J

Hn (Zn) j c NJ are bounded in L2 (Sl,R 2 N). So by a theorem of)j n

De La Vallde-Poussin (Theorem VI.3.7 [14]) with O(u) = u, {H(zn) - Hn (Znj

j c NJ have equi-absolute continuous integrals.

*'. Now we can apply D. Vitali's theorem (Theorem VI.3.2 [14] ) and get a

constant N 3 > 0 such that

f' 2wH(z ) -H (z ))dt < Vj N0 n n .93

Combining with (4.4) and Lemma 2.14 we get

0 D z(z < - V j Nn n. 3 3

Let N O = nN, then combining with (4.3) yields ak(N 0) 4 ak + c. By Lemma3

4.1 we get

a k ak(n) 4 ak(N 0 ) 0 a k + V n o N0

This completes the proof.

-19-

§5. An upper estimate for the growth rate of {ak)

By (2.27) we get';, 1/q1

(5.1) J(T z) ( J(z) + a9(log (IJ(Z)i + 1) + 1) V z c E and 8 c [0,2w]

So there is M1 > 0 depending only on B9 and ql such that

" (5.2) J(z) + a9 (log (IJ(z)l + 1) + 1) ( 0 if J(z) 4 -M1

We shall prove the following claim in §6: ak + +w as k + c. So there is

a k0 E N, k0 > N+1 such that

(5.3) ak ) M+1, V k > k0

Proposition 5.4. Assume that there is k1 > k0 such that bk = ak,

V k > k1l Then there is M = M(kj) > 0 such that

1/q1 .(5.5) ak < Mk (log k) V k )k I

r

Proof. Assuming the following inequality for a moment

(5.6) inf sup( max J(T0 Z)) 4 inf sup( max J(T0z))cAEk+1 zcA 8c[0, 2 1 BeB k zeB Gd[0,2w]

For k ? k1 we get

a .+ ( inf sup( max J(T 7))BCB k zcB OE[0,2]

For any e > 0, by the definition of bk, there is a B c Bk such that

sup J(z) ( b k + c = ak + C. For this B, using (5.1), (5.2), (5.3) we get

zCB

that

* I/q1J(Tez) 4 ak + C + 89 (log (ak + C + 1) + 1) V z f B and 9 C (0,2w]

Therefore there is M2 > 0 depending only on $9 and q, such that

1/q6 ak+1 ( ak + C + M 2 (log (ak + C) + 1)

Letting E 4 0 yields"; I/qla

(5.7) ak+1 ( ak + M2(log ak + 1)

Write k a k(k logpk) 1, p = 1/ql. Wa need to prove {k} is bounded. Now

-20-

6S%

(5.7) becomes

(5.8) 6 k+1 (k+1)logT(k+1) 4 6k k logpk + M2( (log 6k + log(k logpk))p + 1)

If 6 k+1 > 6k we get

P P(i + log(k log k)

6k log-P6 k 4 M2 log 6 log6 kk(logP(k+1)-logpk)+logp(k+1)

If 6k > e, then

6 k176 ( M2 (1+log k + p log log k)p + 1

log(k+1)

Thus there is M3 > 0 depending only on P and M2 such that 6k 4 M3.

Then from (5.8) it is easy to see that there is a constant M4 > 0 depending

only on M3 and p such that Ak+1 4 M4

Therefore Sk+1 < max{6kM 4 }, V k > k1 . So Sk 4 max{6k1'M4}1 V k > k1 .

Let M = max{6k,M 4 }. This yields (5.5). Therefore we reduce to the proof of

(5.6), i.e.

Lemma 5.9. If L is a continuous Sl-invariant functional on E, then

(5.10) inf sup L(z) 4 inf sup L(z) V k > N+1AcAk+1 zCA bck zCB

Proof. Given any B c 6k, by the definition, there are j > k, h, c Aj.

Y c X with y(Y) 4 j-k such that B = h1 (Dj+1 (X)\Y). Let

AA

U.(X) = {x ( D+(X) I x = x' + Vj+1U+1, x' V (X) , 2 )

and IxIE ( R+I •

-21-

%where and are given by 3 of Definition 3.4. We define

di Q(x) = at(x), 8(x) = B11x) V x U.(x)

' Q(T x) = a1(x), 8(Tsx) = T 8(x) V x £ Uj(X) and 8 [0,2 1

-21-

0

,3 -~ ~

Note that Dj+I(x) = U Te U(X) and for any given y c Dj+,(X).es [0,2w]

T x =y possesses a unique solution (x,e) in Uj(X) x [0,2w). So a and

are well defined, a c C(Dj+I(X), [I,a]) is S 1-invariant, c £ C(Dj+ 1 (X)1 E)

is compact and S1-equivariant.

Define

h (x) = a(x)P id(x) + 8(x) V x C Dj+1 X)+ + ho 0

h (x) = P h (x), h 0x) = P h (x) V x £ U.(x)* 1 1

+ + 0h( 0h (T x) = T h (x), h (T x) = h (x) V x C U.(x) and 0 c [0, 2w)

and

h(x) = h+(x) + h-(x) + h0 (x) V x C D£+I(X).

Then h c rj+ 1, h = h, on Uj(X) and h(T x) = T h(x) V x £ Uj(X) and 6

(0,2w)]. Let A = h(D j+(X)\Y) then A c Ak+i and we have that

sup L(z) = sup ( max L(Tex)) 4 sup L(z)zcA zh 1 (UC(X)\Y) Oc[0,2w] zeB

This completes the proof of Lemma 5.9 and then Proposition 5.4.

Remark. Proposition 5.4 is a variant of Lemma 1.64 [18] in St-setting.

Lemma 5.9 is new. The space X is introduced to get the unique expression

y = T x for given y C Dj+,(X) in terms of (x,O) in Uj(X) x [0, 2w),

which is crucial in the proof of Lemma 5.9.

I%6. A lower estimate for the growth rate of {ak}

In this section, we shall prove the following estimate on {akl.

Proposition 6.1. There are constants X > 0, k0 ) N+1 such that

-22-

2/q2(6.2) ak ) ) k (log k) Vk ) k0

We shall carry out the proof in several steps.

Step 1. We consider a Hamiltonian system

(6.3) =JVF(Izl) F(- zz

and its corresponding Lagrangian functional

*2 1 1 2N. (z) - 70 F(Izl)dt V z c C (S ,RF 2 (-

By direct computation we have

Lemma 6.4. If the function F satisfies

(FI) F c CI ( [0,+-),R).

': F' (t)(F2) Let gF(t) = t" then gF(O) = 0, lim gF(t) = + , and gF(t)

t++mis strictly increasing.

4 (F3) Let hF(t) = F'(t)t - 2F(t), then hF(gF1 (1)) > 0 and hF(t) is

strictly increasing for t gF1(1).

Then

1. The solutions of (6.3) are all in E+ and of the following form

cos(kt)I sinktl\

zk(t) sin(kt) - cos(kt) Vk

for any vk C 2N with Vk1= Yk(r), k c iN and v0 = 0 where Yk(F) = (k)

satisfies Yk(F) + +c as k + +c and 0 = y0 (F) < Yk(F) < yk+1(F) for k c N,

I is the identity matrix on 1.

2 • Let d0 (F) - 0, dk(F) = OF(zk) V k C N, then dk(F) = hF(yk) > 0,

is strictly increasing in k.

. Step 2. We define a function G : [0,+-) + R by

- k q n kG(t)kq ae - tkq

k=n+1 ki k0

where a - 2M2, T ' T2 , q = q 2 ' a2 t T2' q 2 are given by (H3), and n is the

smallest positive integer such that n > + I and Tq 2q L. 4k

q k=n q

-23-

P~~ of j-a of -r -F -' 'r~& 4~ %

* ' Then it is easy to see G satisfies (Fl) and (F2). Since

T t (n + 1 )q k(6.5) hG(t) - G'(t)t - 2G(t) = q n + a(Tqt - 2) [ t

*.. k=n+ I

when t () , h(t) > 0 and is strictly increasing.

4Gt/qWrite Yk= Yk(G) for k C N, we claim that YI > 4q. For

otherwise we have the following contradiction,

S k 2/q

1=g(y) =aq k kq-2 < Tgaq 2/q 1 k <G 1 k=n+l (k-i)! 1 4 ) k=n

Therefore G satisfies (F3), and we also have that

n+ 1 w k(6.6) G'(t)t - 4G(t = (n+l)q q T kq(6.6) GI t~t 4~ aq- t + a(Tqt 4) t > 0

'nI- k-n+1

and is strictly increasing for t 4) 1/q It is also easy to see that' q

there is t I > 0 such that

(6.7) 0 4 G(t) 4 t4 V t [0,t 1

We consider the Hamiltonian system

"" G(IzI)* - (6.8) z=JVG(IzF ) E'

zF J z

and write D = OG' dk = dk(G) V k c {0} U N. Besides properties described in

Lemma 6.4 we also have

Lemma 6.9. There are X, > 0, ki c N such that

S (6.10) dk > k (log k)2/q V k ; k I

Proof. From gG(y k ) = k we have

T" n jta y q-2 k aq jq-2 k

"qke -q 3: (-1)! YkF soTk ) log k + (2-q)log yk - log(aTq)

Thus there is k2 £ N such that2 N

Y (- log k) q V k k2

-24-

e. - -,,

.0

* "T 1 t2q w e

. ,'(t) (n+1)q-2 T 1qSince G(t) at-t we get2 2- Tq

dk - (G'(yk)yk - 2G(Yk) > 2(k 2 2k k 2-q) = k 2 (1 - Y-q

k wGkykkk Tq k k Tqk

So there is k I > k2 such that for any k > k,

d >1 kyk 2 > k (log k)2/q

where = 2/q.w.~ee A 1 2 2 T

For k C N, k > N+1 we define

ck = inf sup O(z)

acA k zCA

Since tc C(E,R), (6.6) and 0 e n A, we get -w <ck < +w. From Ak+1 C Aki ' ACAkWAA'

A we get

S(6.11) ck 4 Cik+1 V k > N+1

" By (H3) and the definition of G, there is a constant > 0 such that

H(z) +4_ I1 2 4 G(IzI) + V z C R2N

: :' For z c E, by H8ider's inequality we get for 2 = C1 2 2

. "-. L1 r2 ,fH1 Zo1 (l2 IzI2 1 f 2

"'' J(z) > 1 A(z) - z) 2 2 • (z) -2 2 J- O L2

So we get

(6.12) ak ) ck - V k > N+1

Step 3 Let m0 = [y1] + 1 and define for m • m0

G(t) , if 0 < t 4 m

m G(m) t4 + G(m) - , if •<

4m4

6.-Then Gm satisfies (Fl) and (F2). For n < t

G'(m) t4 1 G mm-4~)(6.13) G%(t)t - 2Gi(t) =- t 4 1I.. "s 3 2

m

By (6.6) and the properties of G, G. satisfies (F3). So Lemma 6.4 holds

, for Gm . Since

-25-

4C o ; "' "q % ".

-G'(m) 4Gm(t)t - 3 t + (G'(m)m - 4G(m)) , V t > m4Gm m

We get that

0 < Gin(t) 4 Gm+1 (t) 4 G(t) V t 0

and

0 4Gm(t) 4 Git)t V t ;o r I Tq

From the definition of Gm and (6.7), there is a constant am > 0 depending

on m such that

(6.14) Gm(t) 4 amt4 V t > 0

We consider

G'(IzI)(6.15) z =JVG IzI) - Jz

anO.write 1m = Gm' dk(m) = dk(Gm)" Then Om c CI(ER) and satisfies (P.S)

condition. Define

ck(m) = inf sup m (z) V k > N+1, m > m 0A4E% zCA

Then we have - < ck(m) < +w. To get more accurate estimates on ck(m), we

need

Lemma 6.16. Let N+1 < j ( m, 0 < p < Rm , h c rm and

Q {x c Dm(X) I h(x) C DB (E) r) VI (E)}p j-1

* Then Q is compact and y(Q) ) m-j+1.

Proof. This Lemma is a variant of Proposition 1.19 [19). Note that firstly

id(x7) = P-id(xi ) by 50 of Lemma 3.3, {P-id(xi) } being convergent implies that

{x } has a convergent subsequence. This yields that Q is compact. Secondly, by

+ E -4D E0 ,the definition of h c rm, if we let Ek = EN+, k N+l, k

+k = + N+k XX+1k '0 and P E + Ek be the orthogonal projector,

then Pkh(x) = z for z - x X0 Dm(X) and Pkh(aBm(X) n Vm(X) l Xk ) =

-26-

* .. 4w *'144

r

p.Pkid(lBm(X) n vm(x) n xk ) = aEm(E) n Vm(E) ( Ek . This allows us to apply

Borsuk-Ulam theorem (8]. Therefore we can go through the proof of Proposition

1.19 [191. We omit the details here.

°13

Corollary 6.17. Let N+1 4 j 4 m, 0 < p < Rm , h rm. For any Y c X

with y(Y) 4 m-j

h(D(X)\Y) f aB (E) n V I (E) $mp j-1() ,

Lemma 6.18. ck(m) > 0 V k > N+1, m > m0.

®r' Proof. Fix m m0 , k > N+1, by Corollary 6.17 for any A (Ak and

0< p Rk there is a z c An 3B fl E+. Let C denote the embedding

constant from E into L 4 , then by (6.14)

01 ( ( f2wjzl4dt > 2 4 = 1 2(1 cp 2

,d.. Az) - 0 0 ) p - mCp = p (4..I

Choose pm = inl,(2C) /21, we get c (m) > 1 2 > 0.m m k 2 Pm

Lemma 6.19. For any k > N+1, m m 0

I*. ck(m) is a critical value of Im"

20. Any critical point of 0. corresponding to ck(m) lies in E\E 0.

4.. 1' -130

. If Ck+1(i) .... Ck+j(m) c and K = (0J) 1(o) /n (c), then

V,. Y(K) j.S

Proof. 1 and 30 follow from the standard argument, we refer to (19].

"" 20 follows from 10, Lemma 6.4 and Lemma 6.18. We omit details here.

Lemma 6.20. For k c N, k > N+1, m m0 , ck < ck(m+l) • ck(m) and

.lir ck(m) = ckin.

-27-

.- %~~ %.

Proof. The Lemma follows from the proofs of Lemmas 4.1 and 4.2.

Step 4. Proof of Proposition 6.1.

Fix k > N+1, for any m > m0 , by I0 of Lemma 6.19 and Lemma 6.18,

ck(m) = dj(m) for some j > 0. So

m 4 Tn+1U ck(m=) = $m(z) =(Gm(yj)yj - 2Gm(y )) ) G '4 rT (n+1)q

k 3 j m ,3

here we used (6.5) and (6.13). Since by definition of G, G'(t) is strictlyt 3

increasing for t > 0, by Lemma 6.20 we get

G'(m )n+1-0 4 T (n+1)qck(m 0 ) ck(m) > min{r 3 Yj, r aq y

k 0 0

So there exists M > 0 independent of m such that if z is a critical

point of 0m corresponding to ck(m) with m > mo, then IIzi c e M 1 Thus

G.(zI) = G(1zI) for m > mi(k) - max{mo, [MI] + 1}, and then there exists

j(m) c N depending on m such that

(6.21) ck(m) dj(m) V m > m 1(k)

By Lemma 6.20, 0 < dk < dk+1, and (6.10), we get ck = dj for some j c N.

Therefore {ck} is a subset of {dk}.

We claim that c V k ; N+1 If not, by (6.11) we get.-. We claim thatCk+N > ck Vk)NI

c ck =*= Ck+N. By (6.21) there exists m > m0 depending on k such

that c = ck(m) =...= ck+N(m). Let K = (S)-(0) f m1(C). By 30 of Lemma

6.19, y(K) > N+1. But 10 of Lemma 6.4 and 40 of Lemma 3.2 shows that Y(K) N.

This contradiction proves the claim.

Assume cN+1 = dt for some £ N N, then by the above discussion and

(6.10), for k > max{kl,6N}

-28-

k k-N-2 k-N-2 1 N NN+I+( - N I+ fN NIk/q k k .2/q k:

X Ul1 + k - 3)log 2/q(U + '-3) > A' log 2/q -

N N l 2N-2

,. k (log k) 2 / q

for some X > 0. Combining with (6.12) we get (6.2).

" The proof of Proposition 6.1 is complete.

S7. The existence of critical values of Jn

- Fix n, k C N, k > N+1, we have

Proposition 7.1. Suppose bk(n) > ak(n) 0 08. Let Sk(n) c (O,bk(n) -ak(n))

and

B k(n,6k (n)) = {h(D j+I(X)\Y) C8k I J n (h(x)) 4 a k(n) + 6 k(n) for x c Dj(X)\Y.

Let

b k(n,6 k(n)) = inf sup J (z),p"' Bcg(n, 6 k(n)) zcB

*- Then bk(n,hk(n)) is a critical value of Jn"

-,' Remark. Dk(n,Sk(n)) > bkln). By Lemma 3.6, Bk(n,6k0n)) # 0, and

'i bk(n'ak(n)) < +-.

* For the proof of Proposition 7.1, we need the following "Deformation

Theorem", which was proved in (19].

Lemma 7.2. Let Jn be as above, then if b > $8, c > 0 and b is not

a critical value of Jn' there exist c c (0,) and n e C([0,1] x E, E)

such that

-29-

%0

10. n(t,z) = z if z J (b-eb+e)n,. .

20. r(O,z) = z V z c E

3*- n(l, [Jn]b+e) C [Jn]b - c, where [J ]a {z C E Jn(z) 4 al.

4. P-yh(1,z) = al(z)z- + 81(z) V z e E, where al c C(E,[1,e 2 ] ),

$1 (C(E,E-) and 8 is compact.

5*. n(t,-) is a bounded map from E to E for t ( (0,1].

a,

2 (k (n) - akn)) > 0. Ifi Proof of Proposition 7.1I. Let £ = 2 b~)-a~) .I

bk(n,6k(n)) is not a critical value of Jn' then there exist c and n as..

a. in Lemma 7.2. Choose B c Bk(n,Sk(n)) such that

a. sup J(z) 4 b (n,6 k(n)) + czCB

Then there exist j > k, h0 Aj, Y c X with y(Y) < j-k such that B

h (D (X)\Y). Define0 j+1

h(x) = n(1,h 0 (x)) V x c Dj+ 1 (X)\Y E Q1

h(x) = h0 (x) V x C Bj+I(X) n Vj(X) n Y Q2

h(x) = id(x) V x C BBj+I(X) n vj+I(x) Q3 "

Denote Q = Q 1 U Q2 U Q3 "

. For x D (X)\Y, Jn(h0(x)) 4 ak(n) + 6k(n) 4 bk(n) 2e < bk(n,6 k(n)) c.

Thus

(7.3) h(x) -(1,h 0 (x)) h0 (x) V x C Dj(X)\Y

* For x C Q3 U ((Bj+ 1 (X)\Bj(X)) n Vj(X)), Jn(ho(X)) 4 0 < bk(n,6k(n)) - e,.

thus n(1,h 0 (x)) h0 (x) = id(x). So h c C(Q,E).

,A For x (Dj+I(X) ) Vj(X)) U Q3 1 p-(h(x) = P-h0 (x) = a 0(x)P-id(x) +

W0 (x) where a0, 80 are defined for h0 in 30 of Definition 3.5. For

x C Q\Q4

-30-

'' , ' % %. . % % a . . .% . "a ' %a %%. .

P-h(x) = al(h 0 (x))a 0(x)P-id(x) + cl(h 0 (x))O 0 (x) + 01(h 0 (x))

Define

a 0 (x) if x C Q4

a(x){ 1l(ho(x))aO (x) if x C Q\Q 4

. 0(x) , if x C Q4

,e2 0O8x)=

Tn a 1 (h0 (x))(0x(x) + 81 (h0 (x)) , if x Q\Q 4

Then a c C(Q,[[ a 0), 8 C C(Q,E-) is compact and

P-h(x) = a(x)P-id(x) + 8(x) V x C Q

Let W = (Dj+l(X) n Y)\Vj(X), then aW C Q where "a" is taken

within Vj+i(X). Since a, 8, P+h, and P0h are continuously defined on

aW, we may use the Dugundji extension theorem ([7]) to extend them to W,

then define P-h(x) = a(x)P-id(x) + 8(x), and h(x) = P+h(x) + P-h(x) + POh(x).

We have h c A. Thus D E h(D j+(X)\Y) C S k . By (7.3)

Jn(h(x)) = Jn(hO(x)) 4 ak(n) + 6k(n) V x Dj(X)\Y

Thus D c Bk(n,6k(n)). Now 30 of Lemma 7.2 yields

sup J (Z) b k (n , 6 k ( n ) ) - ca,',-. ZED nk kn)-£

S-S- This contradicts to the definition of bk(n,6k(n)). Therefore the proof is

complete.

.%

-31-

a!

K -J8. The proofs of the main theorems

We prove Theorem 1.2 by contradiction. Assume that the functional I is

bounded from above by M1 > 0 on S, the solution set of (1.1).

Since q2 < 2q1, Propositions 5.4 and 6.1 show that there exists k C N

such that

. bk > ak * max{8 8 ,M1 }

.

Let 5 (bk -a). By Lemmas 4.1 and 4.2, there exists n, > 0 such that

bk(n) - ak(n) > 4c and ak(n) < ak+e V n ) n,

S.Let 6k(nl) = e, and 6k(n) = 2e for n > nj. Then by Proposition 7.1,

bk(n,Sk(n)) is a critical value of in for n > n1.

If h(D (X)\Y) e B (nle) then for any x c Dj(X)\Yj+1 k1

Jn (h(x)) < in (h(x)) 4 ak(n1 ) + C < ak + 2e 4 ak(n) + 2e V n > n,

So Bk(nl,e) C Bk(n, 2e) V n > nj. Therefore for n > nj,1 .21r 2 (zdz +d27

b k(n,2e) 4 inf sup ( 2A(z) - +0 0(z Ifd t dt) + b <

Bc&k(nlie) zeB -

where Po(Z) = aolZjVI - So and we used (2.6).

"4 Let zn be a critical point of Jn corresponding to bk(n, 2e) for

n > n1. Using (2.6), f 4 W1 ,2 (S1,R 2N) and the proof of Lemma 5.3 [2], we

a,. get

(8.1) Iz 1 4 M V n > n

L

where the constant M2 > 0 depending on b, but independent of n. Now we

Schoose n2 > n I such thatK2 > M2,where {K is defined in Proposition

2

2.5. From (8.1) and Lemma 2.19 we get that Z) = H'(Z) on

and Zn is a solution of (1.1) i.e. z C S. But% Z2 •Z2

I(Z) n (Zn2) (Zn2) = bkln 2,2 r) > bk(n 2 ) > ak(n2) ) ak >

2 2 2 2 2

-32-

oV

This contradicts to the definition of M1 , and completes the proof cf Theorem

1.2.

The proof of Theorem 1.3 is similar. Instead of (5.5) and (6.2) we shall

(P+I)/Pk 1" (ak y(have "a Mk , V k ) and "a) y (p2+1)/(p 2-1)(H4) they

K". yield "bk > ak for infinitely many k". The proof is rather simpler than

that of Theorem 1.2. For example, the corresponding

-U-.~~~ 11, 2T p+1

O(z) = 1 A(z) - (a2 + -) j0 ,zI dt2 2 2 0O

is C2 and satisfies (P.S.) condition. So the lower estimate for ak is

quite straightforward. For the details we refer to [13].

In (15], Pisani and Tucci gave a result for (1.1),

*Theorem 8.2 (Theorem 1.1 [15]). Let H satisfy (Hi) and the following

- "conditions

"-" S' (z) .z(H5) lim +) .

(H6) There are constants p > 1, ai, 81 > 0 such that

1 Ri+ - 2NH'(z)2z -1 Z R

(H7) There are constants q c [p,p+l) and a2 ' 82 > 0 such that

jz + 2R 2N

Then the conclusion of Theorem 1.3 holds for given T > 0 and T-periodic

function f E LXoc(RR 2 N).

Unfortunately, because of a difficulty caused by the St-action on

1/2 2 1 2NW 2 ' (S ,R ), their proof (Lemma 1.14 (151) is not complete. Using the

minimax idea introduced in §3, this difficulty can be overcome. The condition

(H7) allows us to carry out the proof without doing any truncation on H, so

the function f can be allowed only in L2 and the proof becomes rather

-33-0,

asimpler. We omit the details here.

We also refer readers to a related density result proved earlier.

Theorem 8.3 (Theorem 1.5 [111). Let H satisfy (Hi) and (H5), then for

any T > 0, there exists a dense set D in the space of T-periodic functions

in L2((0,T],R 2N ) such that for every f c D, (1.1) is solvable.

This theorem poses a natrual question whether the condition (H3) or (H4)

is necessary in Theorems 1.2 or 1.3.

§9. Results for general forced systems

In this section we consider the general Hamiltonian system

(9.1) = (t,z)z

-. Firstly we consider (9.1) with bounded perturbations. That is

Theorem 9.2. Let H satisfy the following conditions,'A

-~ (GI) H c C (R x R2,R) and H(t,z) is T-periodic in t.

(G2) There exists H : R2 N + R satisfying (HI), (H2) and constants 0 < q < 2

a, T > 0, B ) 0 such that

10. H(z) 4 ae T l z q + B V z C R2 N.

20. IH(t,z) - H(z)l ( a V (t,z) c R x R2 N.

30. H (t,z) - H (Z)I 4 a(jz p I + 1) V (t,z) E R x R2N where

1 4 p < U and V > 2 is defined in (H2).

then the system (9.1) possesses infinitely many distinct T-periodic solutions.

Remark. Theorem 9.2 weakened conditions of Bahri and Berestycki's

corresponding result, Theorem 10.1 [2], who required H satisfying the

V. following conditions

-34-

0r%% %%

1. H c C (R x R ,R) and is T-periodic in t.

2%* There exists H c C2(R2N,R) satisfying (H2) and constants q > 1,

cL > 0 such that

H(z) 4 (lzl q+l + 1) V z R2N

andA1H. Hn 1 2Nt < -C•

C ( (R 2 N ,R)

In order to prove Theorem 9.2, we let G(t,z) = H(t,z) - H(z), and consider

*functionals

J(z) 1 A(z) 27r H(z)dt - (z) 20 G(t,z)dt

* and

N: J (z) -A(z) - f H n(z) G(t)dnl 2 j0 H~~i n fo

where Hn' " n are defined in §2, and we can go through the proofs in §§2 - 7

with the following estimates for {akl from above.

Lemma 9.3. If bk - ak V k > k1 , then there is M = M(k1 ) > 0 such that

(9.4) ak 4 Mk V k > k I •

Proof. Using 20 of (G2), instead of (2.27), we get that JJ(z) - J(Taz)j

4wa V z C E. So as in the proof of Proposition 5.4, we get that

a +1 4 inf sup( max J(T z)) <inf sup( max J(T z)) (by Lemma 5.9)

AcAk+ 1 z£A eO[0,2w] beBk zcB ec[0,2n]

< inf sup J(z) + 4wa b k + 4a = ak + 4wa V k > k I' BC k zCB

Ii. Let . if I>6k then from the above inequality

(k+1)Sk+l + k6k + 4a +

-35-

k":v, . Sk + 1 • 4 w a

This shows that

6 k+1 4 maxf6k, 4wc} V k ok I

Thus

6k 4 max{6k," 4nc} V k ) k 1

a k

Let M maxk-, 4ll}, we get (9.4), and completes the proof of Lemma 9.3.

11

Now the arguments in §8 yield Theorem 9.2.

Secondly, it is not difficult to get direct generalizations of Theorems 1.2

and 1.3 for (9.1).

Theorem 9.5. Let H satisfy conditions (G1) and

(G3) There exists H : R2N + R satisfying (HI), (H2) and either

2q1i 0 . H satisfying (H3) and there are a > 0 and 1 4 p < min{ q ul

such that

IH(t,z) - H(z)l • azlP+1), IH (t,z) - H(z)l • a(jz1"-+ 1), V(T,z) R x R2N

orV2(p1+1

)

20. H satisfying (H4) and there are a > 0 and 1 4 q < 1 such that

1 -4 a(jzjq-+ 1) 2N

"I-. I+ (t,z) - 1) (t,z) c R x R

Then the system (9.1) possesses infinitely many distinct T-periodic solutions.I,-.. 4 We omit the details here.

-36-

04r

ILp 4

.Ma

Appendix. Monotone truncations of H in CI(R 2 N,R)

".4" In this appendix, we give a proof of Proposition 2.5.

Recall that a c (0,1), pa > 2, and r0 1 1 (§2). Choose ) (a,1) such

.-4 that u(X-a) < 1. Define K1 = K0 + 2, To = 1. For n c N, define inductively

. that

(A.1) r = max{'T + 2, a +1 max H(z)}nn-1i K Il KJ <IlK +1-n n n

(A.2) Kn1 = maxlK + 2, (-0) 1n+1 n

Here a0 minlzl=r H(z) > 0. For K c R, take X(.,K) C C'(R,R) such that'U. 0-,?

'.. X(s,K) = 1 for s < K, X(s,K) = 0 for s > K+1 and X'(s,K) < 0 for s £ (K,K+l)

Then for n C N set

M (z) X(IzIKn)H(z) + (1 - X(IzIKn) )TnIZU) V z E R2 N

This kind of truncation functions was used by Rabinowitz in [17]. Since the

M 's do not satisfy 40 of Proposition 2.5, we need to modify them. Directn

computations (cf. [17]) show that

Lemma A.3. For n E N, Mn C Cl(R 2 N,R) has the following properties,

(A.4) Mn(z) = H(z) if IzI < ,

(A.5) M (z) lzl" , M'(z)z 1, Xjzj UX , if Izi > K + 11'"n n n n n

(A.6) 0 < uXMn (z) < M'(Z).z for jzj > r 0"pn n

Integrating (A.6) we get that

v (A.7) CLoIZI'4 < Mn (z.) for Izi> ro

Lemma A.8. For p > Kn+ 1 we have that

(A.9) max M' (pz)oz ( min H'(pz).z

A zS 2N ZS 2 N

a" and

": -37-o,,

(A.10) max M'(pz).z mrin M'(PZ)oz_ _ F r a s2N-1 n 12N-1 n+

Proof. For any , z S2N- by (H2) and (2.4)

H'(p ) -C >-2 H(p;) > VjLp -

p

) PX - T (by (A.2))

= M'(pz).z . (by (A.5))

This proves (A.9).2N-1

For any ;, z S

M 1 (p)' = X(P,Knl )H'(p )oC + (I - X(p,n+ ))r U - 1

n+1p)n+ n1 n+1UXS+ X'(PK n+I ) (H (p ) T n+1 )

) min{H'(p1)oC, Tn+1 1 -P U )

(by (A.1) and the definition of X)

) M'(pz)oz (by (A.9) and (A.1))n

This proves (A.10).

We now introduce the spherical coordinates (r,e) on R3N. For z =

-.(z, ...,Z2N) 6R2N, write z = rz(6), r = Jz, Z(.) e )

zi = r cos

z2 = r sin 81 cos 82

(A.11) ......

2N-1= r sin 8 1 ... sin 8 2N-2 cos e2N.1

Z2N = r sin 01...sin 02N-2 sin e2N_1

where r > 0, 81 c [0,w], 8i c R for i = 2,...,2N-1. We also write d8 =

-38-

del...d6 2N_1, and doi = de ..d lde ...de for i = 1,...,2N-1.1- i-i i+1 2N-1

Let l = [O,"] x R2N - 2 . For 0 c Q, p > 0 define

2N-1

U(e,p) = ([61 - 7,ir, p'I + "r 1 n [0o,w]) x ,n [0i - "'pai + /-]"i=2

Since H and are uniformly continuous on {Kn 4 IzJ ( K+i1}, there

is a constant 6n C (0,1] such that

(A.12) JH(z) - H( )[ + IM (z) M (z )I 4 (x - ) K

for any z, {Kn IzI Kn 1 } and Iz -ZI 4 Sn " There is a constant

EnE [0,1] such that

: (A.13) Irz() z( )j . n

for any r [Kn,Kn+ I] 0 and C c U(eln). Define..--

(A.14) v (t) = min{it, ven} for t > 0n

For n c N; i = 2,...,2M-1; k = 1,2; 0 c 9; p ) K. define

(elp) = I e lv(p - K, 01 + ."2 (r - 6l)vn(p - K)]

A 1 1

wni(0i,p) = [ v - vn(p - K ), 0. + v (p - Kn)]l,i 1'P 1 2 n n i 2 n n

(k0 ) = _ 2 (p _ K) ' + 2" wv (p - Kn)] n [o,w]'n,1 2 n n 1 2 n n

. (k) (ei,p) [ [ (p - K + k K

* fn,i 1 2 n n i 2 n n

and for j = 1,...,2N-1, define

2N-1

n '=1 n(6,p) = I w .(6.,p)':" j 1 ' J

, (0 p)=1,(9.(2N- I n,

(k) 2N 1(k)( ,p) = f wnj(ejP)n j=l ~

-39-pI .-

(k)l .(0,P) = P~(~ )

1(t42N- 1n

Then direct computations show that

Sl(el,P) C [0,i1 ,

(1) (2)fi n (6,p) C n( (e,p) C an (6,p) C U(e,p - K ),..n n - n n

(A.15) 1W (6,P) v (p - Kn )n1 1 2 n n

I v(p - K ), i = 2,...,2N-1

%V Isn (e,p) V (n(p - K ))2N-1

O n

To simplify the notations we write Vn(p) = Inn(,p)I. These sets satisfy:

Lemma A.16. For n c N, 8 c n, p > Kn ,

10. 8 Qn(0,p) implies 6 C a(I)(0,p).

2*. 8 c Qn(,p) and y ()(8,) ily y QM(2)e

Proof. 10. If 81 C wn,1(eip), we have

1 Vn(p- K) 81 eI +- r )v (p K1 2 1n n 2 - n 1 n n

then

% :81- * V(p - Kn ) 81(1 - V (p - l) 61e1 1

( + 2" 1 )h( - K) 4 81 + 2 Vn(P K)Since 01 (0r1 01~ - K )

n .n

Since 61£ [0,1], :1 £ 4i,1(6lp). Similarly 8i C wn,i(ei,p) implies

Q',8i i ,)for i =2,...,2N-1.Thrfe *ols

20. If C n(OP), Y C (1n)(,p) then

6 - v (p K )'A1 2 1 n n ~ 1 ~1 2 ni n )~~-K

and

-40-

-s'U~~~~NWIR uri- -Wr... WE-- .... .' ...r- nr-~

II.I

• " 81 -- v n(p -K ) ( (1< 8 1 + - vn(p - K ) .

So 1 nT

01 -Wv n(p Kn) - (r + l)vn(p - Kn ) ( 81 - v vn(p - K) y1 n 1 n n 2n n 1

and

iT

Y1Kn) + .1Vnp - Kn) - - elvn(p - Kn) 0 + irv p - K

Since y . [0,n1], we get that yi C wn, 1(6lp). Similarly yi e ii

for i = 2,...,2N-1. Thus 20 holds and the proof is complete.

-- 2N-1For n E N, p ) Kn, z C S define

F (p,z) = min{M'(Pz).z, H'(pz).z}

- 2NNote that Fn is continuous in its arguments. Define for z = rz(6) C R

r = Z, z()=0),r 1 I ~ A

Gn(Z) = K V(p) min F (P,())ddp#, n 1

Yeyn (0,P)

and

Hz (K,n

) n (z) + H(K z(e)), if K < IzI

Note that when r = Izi > Kn+, 1 , by (A.5) and (A.9),

G (z) = f Vn+1 P fa (0) min Fn (p,z(y))dadpn n n (1)

(A.17) (8,P)

+ T (r - KUX )

A1(R2N

Lemma A.18. For n c N, H C (R2,R).

-41-

a.

Proof. Since H, Mn CI(R2 NR) and in the formula of Gn all the

variables r, e only appear linearly in the integration limits, H is C1-t n

continuous on (jz] 4 Kn} and {IzI > Kn }. We bnly need to verify the C I-

nncontinuity of H n at every C c R2N with I Y =1, =

- zFor z = rz() with r = Izi > K, z(8)= Tz7

9. aH (z) BG (z)_____ ____nun F (r,z(y))d$

Dr r V (n(6,r) ( ,r) n(A.19)n

min F (r,z(y))

Y 1 (1)(,r)ydln

for some E n(O,r) by the mean value theorem of integration. Thus

aH (z)n

lim lim min F (r,z(y))z+ r z+ (1) n

Iz~d== jz ~fl= nYE1

Fn(K,) =H'(K

a)r

From the definition of Gn we get that

aG (z)n1 n Vn(P 1(ep)(1 v n(p- K) )[min F (pZ(y))ae0 KnVn() fa n,1eP 2n Yea 1 (Ty)

• ' (A.20)(A.n- n F (p,)(y))]d 1 dp ,

Sn

- ( 1 ) 1_ 1 )where A W (n, (8 + I (Tr- Ol)-(p K ),p) x n, (8,P), A=

1 n,1 1 2 1i n1 A2

k. d "% " mn,(e - 0 V (p - K ),p) x S1(1)(8,p). So for Kn < r n we getn ,1 1 2 1in n n,1 '~ oo nr Kn + Cng e

that6.-

Sn (z) 2M r dp (by (A.15))

a e)l n K nK Kn

=-M r-K + 0 as r + Kw n nn

-42-

4% M,:-;8

where M = max (e,p)X[KnKn+l) Fn(p,z(e)). Thus

3H (Hn z)_aG z) 3H Kn (6))

ur"n = nr,(n + n ___l .z ael z = ael + ael 8

ao1 1e -Z =

lzl>lI=Yn lzl>l;I=Kn

Similarly we heve that

lrn 3H aH() for i = 2,...,2N-1.e z e.

Ic Iz I>lI l=Ixn

This completes the proof.

Lemma A.21. For n C N and z c R2N

V (A.22) H (z) H n+1(z) 4 H(z)

Proof. 1. We prove that H (z) 4 H(z).. n

If IzI 4 , this is true since H (z) = H(z).

If Kn < IzI, write z = rz(e), then

A(z) r 1 ---

(Z) K8 min H'(pz(y))*z(y)dBdp~n 'K Vn(P) n e' ) )

n n( (,P)

n

+ H(K z(e))n

By 1* of Lemma A.16, 8 c Qn(8,p) implies that n ) so

( H(p) f 1 fn(,,P )H'(Pz(e ) )fz( o )dS dp

+ H(Knz(8))

= fr H'(pz(6)).z(O)dp + H(K Z(e))

K n

= H(z) .

20 . We prove that H (Z) 4 H (z).n n +1

-43-.

If jzl 4 +I' this is a consequence of 10 since H n+1(z) = H(z).

If Kn+1 4 jz1, write z = rz(e), then by-the definition of Hn"

A r-

H (z) f rain Fn(p,z(y))d~dpn Kn+ Vn(p) S(n(,, p

n+1 n (1)

An

+ Hn (K n+1Z6))

: : S~+I n+11) S~n~(,O) 6 M(P())(y)d~dp

+ H n (Kn+z 1 )fr 1 n(zy)zydp

K n A + P i + 9P Y

+ H (K +z(8))= K V (p) fa (e,p) YE nApZy).+1d

" " here we used (A.9), (A.10), and 1° of this lemma. Thus

Hp. n n+)1 n+1 ()i + Z

.. and this completes the proof.

?.'.'Lemtma A.23. For every n c N

(A.24) 0 < z0 ( (Z) ) ,lz).z, v(O- > r0 •n n

"-' rH(Kz() ) =,ze

J- Proof. Write z = r= =91 If r 0 4 hze w ue, (A.24) holds by (H2) and H (z) H(z).

he y di oand (A.17) H(z)-z

if Kn+1 n n

In~r So

i -44-

-:.H'(z)oz naH(Z) - Pa CK V (p) (eQ) min Fn(Pz(y))dadp.n n n v p)f1)lp

n n n ye n (8,P)

n

_ jaHK nz(O)) + T n p Xr

) (z - fj MI(pz(6)).z(O)dp

iin

(by 1° of Lemma A.16)

= aHn (Z) + Tnr Xr - VT n r + 1 M n(Knz(e))

uaH(K z(6))

* A

> paH (z)

since X > a and M (Knz(e)) = H(K nZ()).

nn

if yn< Izl < Kn+, by 2 0 of Lemma A. 16,

Hm(Z)-Z( f min F (r,z(y))d,NJ n V (r) (6,r)n, .n n (1)

y e. n ( 0,r )

ii rFmin rFn(r,z())

y (2)(,r)• n

=rFn(r,z(E)),

for some ( 2 (6,r), by the conpactness of Q 2)(e,r). So we get that by

,*5* 10 of Lemma A.16

H'(z)-z > PaH (z) + rF (r,z( )) - Uj fK Fn(P'Z(e))dPn n n n

*(A.25)

a- uH(K z(6)) •

p. We consider two cases:

Case 1. F (r,z( )) = M(rz(E))ez(W).. n

'-45-

N-4

Then from (A.25) and 10 of Lemma A.16

H'(z).z > jjaH (z) + M'(rz( )).rz(V)n n n

- pa r M'(Pz(e)) .()dp - aH(Kn z(e))n

A-

> UaH (Z) + PXM (rz(E)) - naM (rz(O))n n n

(by (A.4) and (A.6))

) licH n(Z) + POX-a)M(rz(E)) - UaJM n(rz(E)) - M (rz(O))I

> aH n(z) + UOL-a)a 0 K' n 1.IOImn (rz(E)) - M n(rz(e))I

(by (A.7))

f"~ •UaH (z).n

In the last inequality, we used (A.12), (A.13), (A.15), and that E (2

Case 2. F (r,z(E)) = H'(rz()).z(E)., ,. n

Then from (A.25) and 10 of Lemma A.16,

A A- rH'(z).z > paH (z) + H'(rz(E)).rz(E) - pa H'(pz(6)).z(6)dpn n

n

=paH(K n T())

> VaH n(z) + UH(rz(E)) - UaH(rz(e)) (by (H2))

>' jaH (Z) + 01l -a)a K~ - UaIH(rz( )) - H(rl(e))In 0 n

(by (2.4))

> UaH (z)0n

here we used (A.12), (A.13), (A.15), and that E C n 2)(e,r).

AA

Ths I(z).z > UaH ()i• ..-'.,Thus H,¢n . (zo. ) if Kn < Izl < n+.,-

Finally from (A.6) and (H2), H (z) > 0 for Izi j r0, and this completes.- n

the proof of (A.24).

S 0

To get 60 of Proposition 2.5, we modify H 's again.n

For n N, z = rz(O), from (A.17) if r > Kn+1, we get that

-46-et

LSO - - - WJT *rr,~rr -W r''W .-.. . . . . . . . . . . . . .--K. .

(z) G (K Z()) + T -(r. + H(Y z(O)n n n+1 n n+1 n

Set

= max IGn(K z) + H(K z) - X jI + 1Cn 2N_ I1 + n n n+lzES

h "-. Then we have that

H (z) T IzlA + c

(A.26)

4 (T n + 1)1z UX vizi > max{K n+,C }

and

Hn+1 +1 -n+ 1

>C-T + l)zIU + - C (by (A.1))(A.27) n n+1

(T + 1)Iz"' viz! > max{K n+1C n1

. Let K = max[Kn+2,Cn,Cn+1} and define

Hn M + 1) Iz'A Vz R

H)n(z) - xjZI, ^n n

Then we have that Hn Cl(R2N,R) possesses the folloiing properties,A

(A.28) H(z) = H (z) for IzI 4 Kn n n

(A.29) H (z) = (T + 1).ZIu for IzI K n + 1"n nn

" (A.30) 0 < uaHn(z) 4 HA(z)-z Viz! > r 0 -

From (A.26) - (A.29) and the definition of Hn we also have that* A A 2

(A.31) H (z) 4 H (z) 4 H (z) V Z R

n n n+1

Now we can give the

Proof of Proposition 2.5. 1 - 30 are true from the definitions of Y,

H , Hn and Lemma A.18. Lemma A.21 and (A.31) yield 40. (A.30) gives 50. 60 isnp

a consequence of (A.29) by letting 0 =

The proof of Proposition 2.5 is complete.

-47-

Us

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