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~OO
as ex so 1 is not bounded from below. Furthermore for each kEN, sin kx E E, and 1(sin kx) > '2k 2 ~ '2 -> 00 - 4
4
as k -+ 00 so 1 is not bounded from above. Thus it is not obvious that 1 possesses any critical points other than the trivial one u == 0. Nevertheless we will see later as an application of the Mountain Pass Theorem that 1 possesses positive critical values and the same thing is true for higher dimensional versions of (1.5).
As a second example of an indefinite function,,!, consider the Hamiltc system of ordinary differential equations (1.6)
: dp
dt
-Hq(p, q),
where H: R 2n ~ R is smooth, and P and q are n-tllplcs. \Vc are interest, periodic solutions of (1.6). Taking the period to be 27r and choosing E to t appropriate space of 2if functions. solutions of (1.6) are critical p of
(1.7)
I(p, q) = [ " Ip(t) . q(t) - H(p(I.), q(t))] dl..
(This will be made precise in Chapter 6.) To see the indefinite nature of ( suppose n 1. Taking Pk(t) = sin kt and qdt) = -- cos kt shows I(pk, qk) = bounded term ~ ±oo as k --; ±oo. Thus I is not bounded from above or be Despite this, as we shall see later, minimax methods can be applied tc functional (1.7) to obtain periodic solutions of (1.6). There are at least two sets of methods that have been developed to critical points of functionals: (i) Morse theory and its generalizations am minimax theory. For material on "classical" .tI·!orsc theory, see e.g. [Mi, Ch2]. Generalized Morse theories and the so-called Conley index can be fe in the CBMS monograph of Conley ICC] (see also ISm]). Our lectures focus on minimax theory. This subject originated in work of Ljusternik Schnirelman ILLS] although it certainly had antecedents (see e.g. IBill. What are minimax methods? These are methods that characterize a cri value c of a functional I as a minimax over a suitable class of sets S: ( 1.8)
c = inf max J( u 1. AES uEA
• ,
There is no recipe for choosing S. In any situation the choice must r€ some qualitative change in the topological nature of the level sets of 1, i.e sets I-l(s) for s near c. Thus obtaining and characterizing a critical value in (1.8) is something of an ad hoc process. The Mountain Pass Theorem is the first minimax result that we will study statement involves a useful technical assumption--the Palais-Smale conditie that occurs repeatedly in critical point theory. Suppose E is a real Bal space. Let C1(E, R) denote the set of functionals that are Frechet differenti and whose Frechet derivatives are continuous on E. For IE Cl (E, R), we s satisfies the Palais-Smale condition (henceforth denoted by (PS)) if any sequ (urn) C E for which I(u m ) is bounded and I'(uml --; 0 as m ~ 00 posse a convergent subsequence. The (PS) condition is a convenient way to t some "compactness" into the functional I. Indeed observe that. (PS) im that Kc == {u E EII(u) = c and I'(u) = O}, i.e. the set of critical points ha critical value c, is compact for any c E R. We will see many examples lat, when (PS) is satisfied.
';,
A~
OVER\'IEW
Let Br denote the open ball in E of radius r about 0 and let DBr denote its boundary, ~ow the Mountain Pass Theorem can be stated, THEOREY!. Let E be a real Banach space and IE C1(E.R). Suppose satisfies (PS), 1(0) = 0, (1d there exist constants p, 0: > 0 such that IlaB, 2: 0:, and (h) there is an e E E \ DBp such that I(e) S; 0, Then I possesses a critical value c 2: 0: which can be characterized as
c
= gEl inf
11
I(u),
where
r= {gEC([O,l],E)lg(O) =O,g(l)
e}.
This result is due to Ambrosetti and Rabinowitz [AR]. On a heuristic level, the theorem says if a pair of points in the graph of I are separated by a mountain range, there must be a mountain pass containing a critical point between them. Although the statement of the theorem does not require it, in applications it is generally the case that I has a local minimum at O. A second geometrical example of a minimax result is the following Saddle Point Theorem iR4]: THEOREM. Let E be a real Banach space such that E = V 8 X, where V is finite dimensional. Suppose IE C 1 R), satisfies (PS), and (13) there exists a bounded neighborhood, D, 0/0 in V and a constant 0: such that IlelD S; ct, and (14) there is a constant j3 > ct such that Ilx 2: (3. Then I has a critical value c 2: (3. Moreover c can be characterized as
c
= SEf inf max I (u ) . uES .
where
r = {S = h(J5)lh E C(J5, E)
and h = id on DD}.
Here heuristically c is the minimax of lover all surfaces modelled on D and which share the same boundary. Unlike the Mountain Pass Theorem, in applications of the Saddle Point Theorem generally no critical points of I are known initially. \'ote that (13) and (J,J) are satisfied if I is convex on X, concave on V, and appropriately coercive. Indeed the Saddle Point Theorem was motivated by earlier results of that nature due to Ahmad, Lazer, and Paul [ALP] and Castro and Lazer [CLIo Both of the above theorems, generalizations, and applications will be treated in Chapters 2--B, In particular in a somewhat more restrictive setting both the ).fountain Pass Theorem and Saddle Point Theorem can be interpreted as special cases of a more general critical point theorem which is proved in Chapter 5. yl ueh of the remainder of these lectures will be devoted to the study of variational problems in which symmetries playa role. To be more precise, suppose
AC'J OVEH\'lE\V
E is a real Banach space, G is a group of transformations of E into ;;: IE CI(gR), We say I is invariant under G if I(gu) = I(u) for allg EG lL E g As a first example, consider (L5), It is invariant under G == {id where id denotes the identity map on g Note that we can identify (~ Z2, More generally if p(x, 0 is continuous on [0,7r] X R, is odd in P(x,O Jo(, p(x, t) dt, then (1.9) is invariant under G, As another example consider (1. 7), recalling for th that functions in E are 27r periodic, Let 0 E [0, 27r), z (p, q) E E, (ge~t) ) z(t + 0), and G == {gelO E 10,27r)}, Then it is easy to see that r is im under G, YIoreover G can be identified with Sl, The above examples show that functionals invariant under a group of s;', , tries arise in a natural fashion, It is often the case that such functionals r multiple critical points, Indeed results of this type are among the most faJ ing in minimax theory, The first example of such a theorem goes back t( work of Ljusternik and Schnirelman [LLS], They studied a constrained tional problem, Le, I restricted to a manifold (which must be invariant G) and proved THEOREM, If IE CI(Rn,R) and is even, then distinct pairs of critical points,
115n-1
possesses at
Subsequently other researchers extended this result to an infinite dimer setting, Another multiplicity result is provided by the following Z2 symmetric v of the Mountain Pass Theorem [AR, R2]: THEOREM, Let E be a real Banach space and I E CI(E,R) with I Suppose 1(0) = and I satisfies (PS), (1d, and (12) for all finite dimensional subspace8 E c E, there is an R = R(E that I(u) 2 and r :::: 0 such that for I ~ 1
() < I1 P (X, 0
~ ~p(x,
::::
r,
O·
REMARKS 2.13. (i) Hypothesis (P3) implies that (2.9) possesses the 'It solution" u == O. (ii) Integrating condition (P1) shows that there exist constants (]3,!l4 > 0 that (2.14)
THE \lOUNTAIN PASS THEORE\1
10
for all x E 0 and ~ E R". Thus since Ii > 2. 0 grows at a "superquadratic" rate and by (P4)' P gro\\'s at a "superlillear" rate as I~I ~ 00. (iii) :':ote that if n = 1 and p(x. ~) = i.e. if we are in the setting of example (1.5). (pd(Pl) are satisfied.
e.
THEORE\!
2.15.
1/ p
satisfies (PI) (Pol), (2.9) possesses a nontrivial weak
solutlOn. PROOF. Let E = WJ·2(fl) and 1 be defined by (2.10). The weak solution I of (2.9) will be obtained as a critical point of J with the aid of Theorem 2.2. Proposition B.lO and (pd-(P2) imply 1 E CI(E,R). Clearly 1(0) = O. Thus we must show 1 satisfies (IJ). (I2), and (PS). To verify (I2), note that by (P4) and (2.14), (2.16) for all u E E. where (2.16) implies
J(tu)
(2.17)
Ifll
denotes the measure of
=~
i
fl.
Chosing any u E E \
12 dx ~
In P(x, tu) dx :S ~lluI12~tl'a3 r Ilull'dx-+-04Ifll--> ~OO ~ in 1
'VU
1
as t --> 00. Hence (I2) holds. For (IJ), by (P3), given any E > such that I~I :S 6 implies IP(x, 01 :s: ~EI~12 for all x E
{O}.
O.
there is a 6> 0
0. By (P2) there is a constant A = A(6) > 0 such that IP(x, ~)I :S
AIEls+
I~I
:::: 5 implies
1
for all x EO. Combining these two estimates, for all ~ E R and x EO, (2.18)
IP(x,
01
:S ~IEI2 -+- AI
Consequently,
I) IIul12
(2.19) via the Poincare and Sobolev inequalities. Choosing
Since
osEIlull shows J(u) o(llul1 2 ) IJ(u)! :S
(2.20) E
was arbitrary, (2.20)
J(u)
= !lluil2 ~
J(u)
Ilu!1 :s: (E/2A)1/(s-l)
yields
2
as u
--> O.
Therefore
= ~lIu112 -'-o(lluI!2)
as u -; 0 so (Id holds. Next the verification of (PS) here and in later results is simplified with the aid of the following result whose proof can be found in Appendix B.
PROPOSITIO:\ B 35. Lct.p wt.ls/YiPl) (P2) (wd! hi' 18
is a bounded seqvence m E such thai J'(u m precompllet in E.
) -- () ()8
/)1/ (210
x. then (
m -
By Proposition B.35, to verify (PS), we need only show if(u",)! ) ---+ 0 as m ---+ 00 implies (urn) is a. bounded sequence. For m with 1L Um and T = /clp(x, u)u ~ u) shows (2,21)
J'(u m
'",
iVl -j-Il
_ j ""
Ii 11 II :::
",
l\U) ~11
_1 ' ; ' )
'1111 '1~
:::(~~)
(1 \2
~
l
(2
;' ~ .
1 \ ,. I;) jl1J
-T-, [)i(J:r
TelL,!'
J{xElll ,lI(xlT}
I"'
M
, (.rEO
iu(xli 0 and the second term on the right in (2.21) is posit The third term is bounded by a constant inrlependently of?n, Hence (2 implies (Urn) is bounded in E, Lastly note that 1(0) = 0 while for our critical point u, f(ll) ;::: IX > (], He 11 is a nontrivial weak solution of (2.9), REMARK 2,22, If hypothesis (pd is strengthened to, e,g., (PI) p(x,~) is locally Lipschitz continuous in X R, then (PI )'-(P2) imply any weak solution of (2,9) in E is a classical solutiO! (2,9) (see e.g. [Ag]).
n
COROLLARY 2.23. Under the hypothesis 0/ Theorem 2.15, i/ paisa satis (P'l)' then (2.9) possesses a positive and a ni'9!!tive classical solution. PROOF. Set p(x,
0 = (] for
~ '::: (] and
P(x,
p(x, 0 =
p(x,~) for ~ ;:::
O. Let
~) = fo~ pix, t) dt.
The arguments of Theorem 2.15 show that
(2.24)
7(u) =
~ IIuI12 ~ 10 P(x, 11) dx
satisfy the hypotheses of the Mountain Pass Theorem. Indeed '15 satisfies (p (P3)' Moreover, (P4) holds for ~ > (] while 0 == P = '15 for ~ '::: O. Hypoth( (P4) was required to help verify (PS) and (h). The above weaker version of ( implies P;::: 11- 1'15 for large I~I and this suffices to get (PS). To also satisfy (j note that (2.14) holds for ~ :2: 0. Thus choosing u E E \ {O} in (2.17) to b, nonnegative function, (12) holds. Consequently, by the Mountain Pass Theor (2.25)
~
6.u
=
p(x, u),
u = 0,
x E 0,
x E 80,
has a weak solution, u 'Ie (]. By (pU and Remark 2.22, 11 is a classical soiut of (2.25). Let A == {x E Oiu(x) < O}. Then by the definition of '15,
!:.u = 0,
u =0,
x E A,
x E 8.11.
THE MOUKTAIK PASS THEOREM
12
Consequently the maximum principle shows in n. In fact rewriting (2.25) as -
Ll11 -
11=
- - ' - - - , - ) 11.
(2.26) 11
0 in A so A =:2). Thus
11
xE
0,
xE
11
?: 0
n.
an.
where aT = max(a,O) and a- = min(a,O), and noting that C1p(x, are 0, the strong maximum principle [CHI continuollR if defined to he 0 at ~ implies that l1(X) > 0 for x E nand dl1(X)/aU < 0 for x E dD. u(x) heing the outward pointing normal to aD. The negative solution of (2.9) is produced in a similar fashion. REMARK 2.2i. Theorem 2.15 and Corollary 2.23 can be found in [AR]. Earlier work in this direction was done by Coffman [Col-2] and Hempel [Hel]. RS"lARl':: 2.28. If p = p(O and n > 2, an identity due to Pohozacv [Po] says
f (2nP(11) + (2 -
(2.29)
Jo
n)l1p(l1)) dx =
f
Jao
x· u(x)lvl1I2 dS
for all solutions of (2.9). In (2.29) u(x) is the outward pointing normal to dn. Consequently if n is starshaped with respect to the origin. i.e. x· u(x) ?: 0 for all :r E then
an,
(2.30)
In P(uJdx?: n 2 In 11p(l1)dx.
By imposing additional conditions, e.g. x· u(x) 'Ie 0 and 11 > () in n so Vl1 # 0 on on. the inequality in (2.30) becomes strict. Taking in particular P'(O = (8 + 11 then shows .5 < (n + 2)(n - 2)-1 is a necessary condition for there to exist solutions of (2.9). On the other hand, Brezis and Nirenberg IBN] have established some existence results for P( 0 P' (0 + lower order term when" equals the limit exponent (n + 2)(n 2)--1. In work in progress, Bahri imd Coron have shown if n = 3, and P( 0 P' (0 with s the limit exponent and n noncontractible, then (2.9) has a positive solution. If n is an annulus in Rn it is also known that there exists a positive solution of (2.9) for P( 0 = P'to for all values of s. Just what the relationship between the geometry of nand growth conditions on p need be for there to exist solutions of (2.9) remains an interesting open question. Our next application concerns a nonlinear eigenvalue problem
(2.31)
- 2:,11 11
x En.
= ),p(l1) , = O.
xE
an,
e-e
where), E R. For a class of such problems, being a good model case, we will show (2.31) has at least two positive solutions for all large ),. THEOREM 2.32 [ARj. Suppose p satisfies (p~), (P2) and (Po) there is an r > () such that p(~) > () for ~ E (Oor) and p(r)
O.
THE :"rOUi'\TA li'\ PASS TIlEOTm:,,1
Then there eXIsts a A > claSSIcal solutions wzth are
° such that for all ,\ > III
~. (2.31)
has at lew
ll.
PROOF. As with Corollar y 2.23, the proof begins by redefining p. Set p p( 0 for ~ E [0, rJ and () otherwis e Then 7J satisfies (pi]). (P2), (P:l) ). If 11 is a solution of 6u = AP(U),
(2.33)
U
= 0,
x E 0, x E all,
the argumen t of Corollar y 2.23 shows {x E ll'iU < O} (:3 find U > 0 j Similar reasonin g proves {x E Olu(x) > r} = 0. Hence 0 < nix) :s; r in U satisfies (2.31). By these observat ions and Remark 2.22, to find solutio (2.31), it suffices to produce nontrivi al critical points of (2.34)
dx
Wd
2 on E = (O), £5 being the primitiv e of p. To study the properti es of h" note first that since p satisfies (pd-(p2 )' C 1 (E, R). Moreover (PS) holds for h,: if(u m ) is a sequence in E with I),(u, M, the bounded ness of p implies 1£5(01 :s; a61EI find
(2.35)
M2:I),(um)2:~llumI12
j,\la6!olumldx
Applying the Holder and Poincare inequalities, (2.35) shows (urn) is bour in E. Hence (PS) follows from Proposit ion B,35. ~ote that (2,35) also iill] I A (n) is bounded from below, Hence by Theorem 2.7,
(2.36)
bA == inf h,(u) E
is a critical value of h, for all A E R, It may be the case that b 0 correspo n( A to the trivial critical point u = 0 of fA' Indeed (P2Hp3) and the arguill' centered around (2,18)-(2 .19) show u 0 is a local minimum of fA and satisfies (II!. In fact
(2.37) for 1,\1 :s; AO and Ilull :s; a7 provided that AO and a7 are sufficiently small. On other hand as in (2.35) the linear growth of £5 in E implies
(2.38) for all'\ E Rand u E E. Thus for small A (2,37)-(2,38) show h,(u) 2: 0 h,{ Consequ ently bA = 0, To obtain a nontrivia l solution, let 'P E E \ {O} such that 'Pix) E [0, r) x E O. By (Ps)
,l
P('P(x)) dx > 0,
l4
TilE \!()(::\TAI:\ PASS THEOREM
Thus for A sufficiently large, !),(Y) < 0 ,md bA < 0, Define A =: infp > 01 bA < O}. Then for all A > A, the abo\'(' remarks show IA has a criticial value bA < 0 and a corresponding critical point 1!A which is a posith'c solution of (2,33). \loreover since h(1!AJ < 0, h satisfies (h) of Theorem 2,2, Consequently the \lo11Iltain Pass Theorem yields a second critical point rIA of h such that h(rI A) > 0 > JA(lJ.Al, Clearly rIA is distinct from ]f A and again above remarks is a positive solution of (2.33), The proof is complete, Our final application in this section is to a problem for which one can establish t.he existence of at least three nontrivial solutions, Consider the equation: (2,39)
~ll = II
Aa(x)ll
~
p(X,ll),
xE
= 0,
x
E
n,
an.
The function a is assumed to be positive and e.g. Lipschitz continuous in O. Associated with (2.39) is the Sturm-Liouville eigenvalue problem: ~ ~1J =
(2.40)
IJ
=
/1a(x)1J, 0,
xE
n,
x E an.
As is well known, (2.40) possesses a sequence of eigenvalllefi (A J ) with 0 < Al < A2 ::; ... ::; AJ ::;. . and AJ ---+ 00 as j ---+ 00, (The number of times an eigenvalue appears in the sequence equals its multiplicity,) Concerning solutions of (2,39) we have PROPOSlTIO:" 2,41. Suppose p satisfies (P3) and (pc) (i) () < 0 == p(x, O~-I Jar ~ # 0, (ii) hE C I (0 x R, R) and ~hdx,~) > 0 Jar ~ # 0, (iii) h(x,O ---+ 00 as I~I ---+ 00 llniJormly Jor x E O. Then: (P) Jor A ~ AI, (2,39) possesses only the trivial solution u == 0, (2°) For each A> AI, (2,39) possesses a pair oj solutions with ut (resp, the unique solution oj (2.39) with ut > 0 in n (resp. < 0 in D), In fact the maps A ~ are CI for A> Al and u~ ---+ 0 as A ---> AI·
Since p satisfies (P3), u == 0 is a solution of (2,39) for any value of A, The first assertion of Proposition 2.41 follows simply on multiplying (2,39) by u, integrating by parts, and using (pe)(i) , The second statement can be proved in a variety of different ways~see e,g, [RI, St21~and will be omitted here, We will prove the following result which is essentially due to Struwe !St2], TllEORE!\.{ 2,42. If p satisfies (P3) and (P6), then Jar all A > A2, (2,39) possesses at least three nontrivial solutions,
The proof of Theorem 2.42 will be carried out in a series of steps, First observe that (pe)(iii) implies an a priori bound for solutions of (2,39), Indeed if 11 is a nontrivial solution, either u has a positive (global) maximum or a negative minimum, Assuming the former, if the max is attained at zEn,
o~
~611(Z)
= Aa(z)u(z) -
z)
THE :-'lOUNTAI:\, PASS THEORE:-'l
or
Ii( Z)) :S AO( zj.
(2.43)
The same inequality holds at a negative minimum. Hence by (P6)(iii), the a constant M = M(A) > 0 such that IlulIL= :S 1..,1(,\). In fact since A is fi choosing :\ > A there is an !vi depending on :\ such that Ilnll L= :S M for solution of (2.39) with A :s:\. With this observation, p(x, 0 or actually he can be modified for lei :0: M as in Corollary 2.23 or Theorem 2.32. so hypotheses (p6)(i)-(iii) also hold for the modification 0 c.= 0 a] satisfies
(P2) Ip(x, 01 :S 121
+ 122IEI8-1
for some s as in (P2). Moreover solutions of -
11
p(x, n),
A0.11
611
(2.44)
= 0,
O.
x
ao,
x
for A :S :\ are also solutions of (2.39). It remains to find a third nontrivial solution of (2.44). 2 (rl) and define
Wci·
h.(n) =
il [~lv1112
~an2 -
P(x
11)]
As usual let 1
dx,
where P is the primitive of p. Since (P6)(ii) implies (pd of Proposition B.34 p and (P2) implies (P2), by that result, I), E C2(E.R). Moreover f), satiE (PS) for. suppose
(2.45)
f{
~
:0: h
By (ps)(iii), for any ,8
eE R,
> 0 there
112 is a 1
A
l
= 1(3) :0: 0
P(x, E) :0: l3e
(2.46)
dx -'-
2 ill
l
dx
JIl
such that for all x En
i
1·
Choosing (3 > AllaIIL=/2, (2.45)-(2.46) show (urn) is bounded in E so (I follows from Proposition B.35. We want to find a critical point u), of I), other than the known ones 0, given by Proposition 2.4l. Concerning the latter we have: LEMMA 2.47.
If A> A], then
< 0 and
are strict lowl mini
of h. PROOF. The map A --+
is C 1 for A> Aj. Hence
since I~ (un = O. Moreover by Proposition 2.41. 1;,..(0) = O. Hence I),(n~) < 0 for A> A].
--+
0 as A
--+
Al a
16
THE .\lOl'NTAI:\ PASS THEOhEivl
To show that argument. Let 11.' =
(2.48) [A(11)
are strict local minima of [A' we will lise a comparison or Since h E C'2(£, R), for 11 near w,
= [A (U') +
J~(11')(11 - 10')
+ P;(W)(l1 - W, 11 -tel + 0(1;11
Since J~(w) = 0, to showw is a strict local minimum for Proposition B.34, it suffices to show that
h, by (2.48)
+ (h(]:, w) +
and
dx 2:
for some,) > 0 and all :p E E. The eigenvalue problem
- 6v
(2.50)
h(x,tL')v v
pav, O.
x E \1, xE
an.
U' with v of OIle sign in n. Hence A must be the hiL' iL" a solution p = A ilnd l' "mallcst rigclwalllc to (2.50). ?vloreover A can be characterized as [CHI:
(251)
h(x. w)v 2 ) dx
-r
inf
.\ =
O"'"EE
Comparing (2.51) to
dx shows thilt ! > A. This fact and (2.52) then show
f; (w)
:p)
2: ( 1
(2.53)
2:
(1
~) In ~)
11:p1!2
and the lemma is proved. With the aid of Lemma 2.4 7, we will find another critical point of [A' Consider the larger of h(ll~). (If these numbers are equal either will suffice,) Suppose it occurs at 1l~. Making the change of variables U = and setting 4>A(U) = h(U + ) - JA (l1\), we see that 4>.\ E Cl(£,R), 4>.\(0) = 0, and h satisfies (PS). Moreover Lemma 2.47 and our choice of origin show f.\ satisfies (Jd and (J 2) since h ) 2: h, (Il ~). Thus the Mountain Pass Theorem shows h has a positiH' critical value. Returning to the original coordinate system, it follows that h. hiL'i a critical value CA > fA ), where
1l-1l\
h(g(t)) and
fA = {g E C(lO, Ii. £)lg(O)
= ll~.g(l) =
}.
) by Lemma 2.47 and 0 is a critical point of h., a priori it may be the case that CA = 0 and the corresponding critical point is llA := O. Thus to ensure that 11), is nontrivial it suffices to show that c), < O. To do this,
THE
~10Cj\;TAJ"i
PASS THEOH.EM
the fact that A > A2 will be used. \Ve will construct a curve 9A E l\ sue h is strictly negative on 9A([0.1]). Before doing so, some further obsen about lA are needed. :\Tote first that
l),(tu) S t 2 1),(u)
(2.54)
for all u E E and t E [0, Ii. Indeed by (P6)(ii)
P(x,
==
(2.55)
lZ
0 d~ =
l
p(x, ty)tdy
t 2 r h(x,ty)ydYSt 2 rh(x,y)ydy
./0
./0
for any t E [0, 1J and z E R. Therefore
h,(t11)=t 2
(2.56)
{~(
-A!/(112)d.r+ (P(x,tu)cixSt 2 f:,,(I1)
./n
./n
for all 11 E E. :\Text let V denote t.he span of all eigenfunctions of (2.40) correspond AI and A2 and let fF denote the closure (in E) of the span of the rem eigenvectors of (2.40). Thus E = V Wand it is easy to check that 1/2
(-In
1/2
- Aaw 2 )
cix )
can be taken as equivalent norms on V and W. Abusing notation somewh will write Ilvll, Ilwll for these norms. Thus for u = v + wEE,
(2.57)
h(u)=
~llvI12+~llwI12+ InP(x,v+w)dx.
If h,(u) < 0, (2.57) and (P6)(i) show
(2.58)
+
21n pry, v+ w) dx > IIwl1
Therefore there exists a 0 =
(2.59)
2
°
such that
Ilvll > (1
+ o)ilwll·
wi-
= v~ + Since h < 0, (2.59) shows II> 0. Set Now the path 9A E r,\ can be constructed. The path consists of five pa (a+) The first part is the line segment {tutlt E ,Ii}, where 7+livt and p > is free for the moment. It is clear from (2.56) that h < 01 segment (b+) Homotopy to 7-'-V;: via T+(1't + 81£'T). (c) Join to T-1';' on aBp n V, where TII = (I. !\ote that aB is a connected set since A > A2 implies the dimension of V is at ICilHt 2. (b-) HomotopYTto as in (b-i-). (a-) Join 7-U), to 11), by a straight line segment. As in (a-i-). here h,
E C(u, E) with <J>(u) u-T(u) and T compact, and b E E\ <J>(30). The resulting degree~still denoted by d( <J>, 0, b)~--is called the Leray-Schauder degree (see [LJ8 or 82]). With these preliminaries in hand, we turn to the Saddle Point Theorem [R4]:
°
Tm;OREM 4.6. Let E = V EBX, where E IS a real Banach space and V of {O} and is finite dimensional. Suppose IE CI(E, R), satisfies (PS), and (I3) there is a constant Q and a bounded neighborhood D of 0 in V such that llaD :S Q, and (14) there IS a constant (J > Q such that llx 2: .3. Then 1 possesses a critical value c 2: (J . Moreover c can be characterized as
(4.7)
c = inf maxI(h(u)). hEr uED
where
r = {h
E
C(15,E)lh
id on aD}.
PROOF. Let P denote the projector of E onto V obtained from the given splitting of E. If hEr. Ph E C(D, V). Moreover u E 3D implies Ph(u) = Pu = u of O. Identifying V with Rn for some n, d(Ph, D, 0) is defined and by COfo]I'lfY 45 and 10 of (4.3).
d(Ph, D, 0)
= d(id,
D, 0)
= 1.
Hence by 2° of (4.3), there exists an xED such that Ph(x) = O. Consequently for each hEr, there is an x = x(h) E D such that
(4.8)
h(x)
(id- P)h(x) E X
Hypothesis (1 4 ) now implies
ma:>Sl(h(u)) 2: I(h(x)) 2: (J. uED
THE SADDLE POINT THEOHEM
Thus by (4,7), c 2: ,3, To show that c is a critical value of 1 requires a fami argument, If Kc = Z, W8 set E = ~(p n:) and invoke Theorem A4 obtainin and 1] as usual. Choose hEr such that
1(h(u)) 1L
J)
1L
=
'\a(x)lL + p(x.lL),
0,
x E 0,
x Eon,
where p satisfies (pd-(P4)' This differs from the case treated in Theorem 2.15 due to the presence of the linear term in (5.1). The corresponding functional on E is
(5.2)
1(1L)
=
10 [~iVlLI2
~a1L2 -
P(x, 1L)] dx.
If ,\ < '\1, with '\1 as in (2.39), then (fn(lVlL\2 - '\a1L2) dX)I/2 can be taken as a norm on E. 1 satisfies (h), and one can use the Mountain Pass Theorem to establish the existence of a weak solution to (5.1) (and even a positive solution). However if ), > ), I, (11) no longer holds so our previous existence mechanism fails. The next result from [R5] gives us a tool to treat problems where a milder ver8ion of (Id is satisfied. THEOREM 5.3. Let E be a real Banach space with E = V 9 X, where V IS finite dimensional. Suppose 1 E CI(E, R), satisfies (PS), and (Ii) there (Ire constants p, Ct > 0 slLch that llaB p r-rX ;:: Ct, (lnd (h) there is (In e E oBI n X (lnd R > p such that if Q == (IJ R n V) EEl {relO < r < R}, then llaQ s: O. Then 1 possesses a critical value c ;:: Ct which can be characterized as
(5.4)
c
inf maxl(h(u)),
hEr uEQ
where
r REMARKS 5.5. (il
=
{h E CCQ. E)lh
id on aQ}.
3Q refers to t.he boundary of Q reiative to V@span{e}. 28
SOME GE:'-1ERALIZATIO);S OF THE
~10U"'TAI]\;
PASS THEORE l\l
(ii) If j/ = {O}, then X = E and (I;) reduces to (h). If further 1(0) ) becomes (h). Thus Theorem 5.3 is a generali zation of the Mounta in] Theorem . (iii) Suppose Ilv p such I( Ii) 0 in rl. Theref 11 is a solution of (5.1),
(5.20)
r(A011 + p(x, 11))111 dx ior(-6u)vI dx = r(-6vdudx io =
in
=
Al
r oVludx.
in
Consequently (5.21 )
(AI - A)
r 01LVI dx = .!0p(x, u)vJ dx.
.10
If u is positive in 0, the left-hand side of (5.21) is nonposit.ive while by (ps) right-hand side is nonnegative. Thus there can only be a positive solution if A = Al and p(X,11(X)) == o. An examination of our abstract critical point theorems shows that an im tant ingredient in their proofs is an "intersection theorem" which allows L show that the minimax values defined in these results are indeed critical va E.g. in Theorem 5.3, this is carried out in Proposition 5.9 while in Theor·err it is done in (4.8). Our next goal is to introduce a topological notion of lin which modulo assumptions on the form of the functional will enable us to p a result which contains both Theorems 4.6 and 5.3. It also enables us to we, the splitting assumptions of these theorems. Thus let E be a real Banach space with E = E1 9 E z, where both EI E2 may be infinite dimensionaL Let PI, Pz be the projectors of E onto EI associated with the given splitting of E. Set
S == {1> E G([O, 1) x E, E)[1>(O, u) = u and P21>(t, u) = P2u - K(t, 11), where K: [0,1] x E
~
E2 is compact
Recall K compact means it is continuous and maps bounded sets to relati· compact sets. Let S, Q c E with Q c E, a given subspace of E. Then BQ refer to the boundary of Q in that subspace. We say Sand BQ link if when! 1> E Sand 1>(t,BQ) n S = 0 for all t E [0,1]' then 1>(t, Q) n S of 0 for t E [0, 1). This notion and the examples below are due to Benci and the aut IBR]. For heuristic purposes. one can think of the sets Sand BQ as linkin
32
SOME GE:"ERALIZATlO:"S OF THE :vIOG:"TAl:" PASS THEOREM
every manifold modelled on q and ,hiuing the same boundary intersects S, \\'e will give two examples of such linking corresponding to what occurs in Theorems 4,6 and 5,3, EXAMPLE 5,22, Let q B n E 2 , where B is a neighborhood of 0 in E 2 , E E 2 , q E q, and S q + E 1 , Suppose E Sand
=
(5,23)
=
(t,aq)n(q+Ed
:2
for all tE [0,11,
\Yo claim (t, q) n S # 2 for all t E [0,1]' i,e, there is a w = w(t) E q such that P2(t,w(t)) = q, For u E E 2 , oct W(t,u) P2 (t,u) = u - K(t,u), By (5,23), W(t, u) # q for u E aq, Therefore d(W(t, ,), Q, q) is defined for t E [0,11 and by the infinite dimensional versions of Propositions 4.4 and 4,3, for any t E [0,11,
=
(,'),24)
d(W(O,), q, q) = d(id, q, q) = 1.
d(W(t,), Q, q)
Hence there is a 11'( t) E q as desired and Sand aq link, II EMARK ,5.25, Setting q 0, S = X, and q = D, we are in the setting of the Saddle Point Theorem, EXA:vlPLE 5,26, Let p > 0, S \".E], e E E1 noB 1 , 1'1> p, 1'2> 0, span{e} E 2 , Suppose E Sand Q E iO.rJ]}
=
(,),27)
(t.aQ)
S=2
for all tE [0,1],
\,\'c claim for each t E [0,1]' there is a w(t) E Q such that (t, w(t)) E S, i,e, P 2 ¢(t, (] and ::P1 (t, p, For u E E2 and r E R, set
'11 (t,(r, By (,5,27), d(W(/,),
q, (p, 0))
is well defined and as in (5,24),
d(W(t,(r,u)),q,(p,O)) = 1. Consequently there is a wit) as claimed and Sand aQ link, RE".IARK 5,28, Setting E1 = X, E2 = V, and T1 = R = T2, the relationship between Example 5,26 and Theorem 5,3 becomes clear. .'\ow we are ready to state a critical point theorem which unifies Theorems 4,6 and 5,3, THEORE"1 5,29, Let E be a real Hilbert space with E = EJ ttl E2 and E2 = Et Suppose 1 E C 1 (E,R), satisfies (PS), and (Is) l(u) = HLu, 1l) + btu), where Lu = L 1 Pj u + L 2 P2 U and L,: E, ~ £, is hounded and selfadjoint, i 1. 2, (h) b' is compact, and (17) there exist5 a subspace E c E and sets SeE, q c E and constants o > w such that (i) S c El and lis:::: (x, q is bounded and lidQ 0 such that 0 < ilH(Z) 'S z . Hz(z for all ? r. Then for (my T > 0, (6.2) possesses a nonconstant T periodic solution. PROOF. Making the change of variables
(6.11)
T
= 2'rrtT- 1 ,
(6.2) becomes
dz/dT = ).,JHAz),
where)., = (27T)-IT and z is 27T periodic in T. Since this has the same form as (6.2) with H replaced by )"H, without loss of gcncmlit y w(' can takc T = 27T and work with (6.2). Basically the proof reduces to verifying that. Theorem 5.29 is applicab le here. However there are some technica l complica tions. Since the growth conditio n (6.8) has not been assumed for H, the correspo nding term in (6.9) need not belong to C 1 (E, R). Thus to get a C I function al on E, H wi!! be modified so that it grows like a power of Izl as [z! -+ 00. Let K > 0 and X E COO(R,R) such that x(y) == 1 if y::; K, X(y) == 0 if y ? K + 1, and X'(y) < 0 if y E (K, K + 1), where K is free for now. Set (6.12) where R = R(K) satisfies
R? Then HK E C I (R2n,R) , satisfies (HJ)-(H2) and (6.8) with" = 4. ~loreo\'er a straightf orward computa tion shows (H3) holds with II replaced by v = min(![, 4). Integrati ng this inequali ty then yields
(6.13)
HK(Z) ? a3lzi"
a4
APPLICATIO:-;S TO HA\llLTOI\IAI\
for illl z
c R 2'1.
where
SYSTEMS
> 0 and ilre independent of K.
(lJ .11.j
set
(G.14)
We will ,;how II{ satisfies the hypotheses of Theorem 5.29. This willieild Ilonconstant 27i periodic (weak) solution 21{ (I) of
to
a
(6.15) It will then be shown that ZK is a classical solution of (6.15). Further estimates then prove there is a Ko such that for all K > KO, I ZK I L= ::; Ko. Therefore Jh(ZK) = H(ZK). Hence for such K, ZK satisfies (6.2). Turning to a study of (6.14), by Proposition B.39. IK c CI(E,R). Choosing E 1 == E~ ami E2 == EO ~ E-. we see that I K satisfies (I5) of Theorem 5.29 with L, clefined by for
Z
E E, aIlCI
b(z) =
_12K HK(z) dt.
Proposition B.39 implies that b'(z) is compact. Hence (h) holds. The next three lemmas establish (17).
LE\!\1A 6.16. If H satisfies (H 2 ), (Ii) (i) holds for h· PROOF. Dy (H2), for any E > 0, there is a 6 > 0 such that HK(Z) ::; slzl2 6. Since H K (z)lz!-4 is uniformly bounded as Izl ~ 00, there is an M = M(s. K) such that HK(z) ::; Mlzl4 for Izi 2: 6. Hence
if
Izi ::;
HK(Z)::;
(6.17)
+
for all z E R2n. Therefore by (6.17) and Proposition 6.6,
fTC HK(z) dt::; [llzI11, + Mllzll1, ::; (m2 + Ma411z112)llz112
(6.18)
Consequently for z E El = E+,
h(z) 2: Choose
E
1
!lzil 2 -
(m2 +
Ma411zi12)llz112
and p so that 3M a4p2 = 1. Then for z E fJBp
h(z) 2: 1p2
n E1 ,
a.
Hence IK satisfies (h)(i) with S fJBpnEl' REMARK 6.19. p and a depend on K through M.
LEMMA 6.20. If H satisfies (Hd and (H3J, then PROOF. Let e E BBI
n El
h satisfies
and set
Q == {relO ::; r ::; rl} EB (Be2
n E 2 ),
(I7)(ii).
APPLICATIONS TO HA\llLTO\,IA\, SYSTE'dS
where r1 and r2 are free for the moment. Define Let Z = zO + z~ E Br2 Ii E 2 . Then
Jf{(Z
(6.21)
+ re)
3'
E == span{e}
E2 so Q c
E
= r2 -
'\Tow by simple inequalities and orthogonalit.ies:
(" Hf{(z + re) dt 2:'
a3
"0
2:'
a5
(6.22)
~2" iz + rei" dt -
.10
(1
2 "
27fa4
Iz + rej2 dt) ,,/2
-
a6 v/2
a5
2:'
(1h(lzOI2 + Izj2 +
a7(lzOI" + rV) -
dt )
- at
a6
Combining (6.21)·(6.22) shows
19(z
(6.23)
+ rel
::; r2 .~
Choose r1 so that (6.24) for all r ;::: rj. Set
M = max 'P(r). rE[O,rd
Since
1/J(z) == uniformly as Ilzll -+ 00, in E 2 , 1/J(z) 2:' M if Therefore by (6.23), if Ilzlj ;::: r2,
Ilzll
is large enough, e.g.
19(z + re) ::; M -1/J(z) ::; 0.
(6.25)
Ilzlj ;:::
r2
°
Combining (6.24) and (6.25) with the fact that 19 ::; on E2 via (Hd, we see == won DQ and (h )(ii) holds. REMARK 6.26. Both rj and T2 and therefore Q are independent of K.
19 ::;
°
LEMMA 6.27. If Sand Q are defined as in Lemmas 6.16 and 6.20, then S and DQ link, i.e. If{ satisfies (h)(iii). PROOF. Immediate from the definitions of Sand Q and Example 5.26. The above three lemmas show If{ satisfies (h). '\Tow to be able to use Theorem 5.28, it only remains to verify that 19 satisfies (PS). Thus suppose Ilg(Zm)'1 ::; A1 and Ij.;(zm) --> 0 as m --> 00. Then for large m and z = Zm:
M (6.28)
+ IIzl!
2:lg(z) -1Ij.;(z)z = J02"[~z. Hf{z(z) - Hf{(z)] dt
2:
(2~j - l/~j)
l"
z· Hf{z(z) dt - Mj ;::: M 2 11zlli,
-
M3
10
APPLICATIOi\S TO HAMILTONIAi\ SYSTEMS
via (H3) and the form of H K . In (6.28). both M2 and M3 depend on K. \",'riting zO + z-'- ~ z', (H3) and simple estimates show
z=
"\1
(6.29)
+llzll :: (2- 1
v- 1 )v
12rr
Hg(z) dt - M4
"ote that .\16 and ;\h are independent of K via (6.13). Inequality (6.29) can also be written as
(6.30) \ext taking
Z
= Zm
and
0 with a corresponding critical point ZK = (PI{. qK). It remains to show that Zg is a classicalnonconstant solution of (6.2). Since IK )\. = 0 for all \ E K choosing \ = (',0,0) E W1.2(SI,R2n) shows (6.35)
f2rr[PK'
io
D-
qg . c; -lhz(zl{)' c;J dt = O.
In particular if \' == 1. (6.35) implies
HA~IILTO:;IA:\ ~\,::;TE:,l::;
APPLICATIOt\S TO
Fourier expansion shows that if y E L 2(SI,R 2n), [y] = 0, and ~ E R 2n, ther there exists a unique x E W I ,2(SI,R 2n) such that [x] = E and:i; = y. Th( choice of HK and Proposition 6.6 show that HKz(ZK) E L2(SI,R2n). Conse quently choosing y = lHKz(ZK) and ~ = [ZK], there is a unique Z = (p,q) E W I .2(SI,R 2n) such that !z] = and (6.36) Taking the inner product of (6.36) with and integrating by parts yields
le;. where e;
1
2 1 0, tlwrc l'xins a IlOllconstcmt solution of (6.2) having period T. However T may not he the minimal period of the solution which may Ill'Tk 1 for :;OIlle k E: N, k > 1. Under further hypotheses on II, Ambrosetti and Mancini [AM2], Deng [De], and Ekeland and Hofer [EH] have proved there exists a solution having minimal period T for any T > O. An easy consequence of Theorem 6.10 is COHOLLAH.Y 6.44. Under the hypotheses of Theorem 6.10, there exist inJimtely many (hsUnd T periodic .solutions of (6.2). PROOF. Theorem 6.10 provides one such solution, ZI(t). Suppose its minimal period is Tkjl Apply Theorem 6.10 again with T replaced by T(2ktl- 1 to get a nonconstant T(2ktl- l periodic solution Z2(t). Certainly Z2 is T periodic and it is distinct from ZI since its minimal period is less than tkjl. Repeating this process produces a sequence of distinct nonconstant T periodic solutions of (6.2). RD1AHK 6.45. There is a much stronger version of Theorem 6,10 [R9]: If HE C l (R 2 n,R) and satisfies (H 3 ), then for any R,T > 0, there exists aT periodic solution Z of (6.2) with > R (sec also [Bel]). Simple examples show that T may not be the minimal period of the solution. E.g. suppose n = 1 and H(z) = G(!z:2), where G is smooth and monotone increasing. Then (6,2) hecomes
and
== const for a solution. Setting
¢ = 2iG'(ki
~ 2
= p + iq, , satisfies )\
,0 exp(2iG'
Therefore if T is the minimal period, T = In particular if G' :2 1, T -:: 7f, We conclude this chapter with an application of a seemingly different nature, Consider (6,2) where instead of fixing the period, the energy is prescribed, e,g, H(z) == L 1.
THEORE~1 6.46 [R61. Suppose H E C l (R2n,R) and H-l(l) is the boundary of a compact starshaped neighborhood of 0 with z· Hz # 0 on H-l (1). Then (6.2) has a periodic solution on H-l(l).
This theorem will be obtained with the aid of Theorem 6,10, First a technical result.
APPLICATIONS TO HAMILTO\iL\\i
SYSTI-:\l~
1 PROPOSrTIOT\ 6.47. Suppose H.H E Cl(R2n.R) wl.th H 1(1)= H- (1) and Hz,H z lOon H-l(1). If dt) satisfies ~ = JHz(r;)
(6.48)
and 1(0) E H-l(1), then there exists a reparametrization z(t) of Itt) such that z satisfies (6.2) and z(t) E H-l(l). In particular if \ is periodic, so 18 PROOF. Since H- 1(1) is a level set for Hand Hand Hz,Hz lOon this set, these gradients must be proportional on H-1(1). i.e. there is i1 1/ E C(H-l(l),R \ {O}) such that Hz(z) l/(z)Hz(z) Now dO) E H-1(l) and (6.48) is a Hamiltonian system so ((t) E H-l(1). Set z(t) == \(r(t)) where r(O) = 0 and r is a solution of (6.49) Therefore z satisfies z(O)
i
\(0) E H-l(l) and
= ~~r = 1/(\"(r(t)))JHz(\"(r(t))) = JHz(z(t)),
i.e. (6.2). To justify the assertion about the periodic case, more care must be taken since the solution of (6.49) with r(O) 0 may not be unique if 1/ is merely continuous. Suppose \" is T periodic and without loss of generality assume 1/ > O. Then 1/ > 0 on d[O, Til and (6.49) implies there is a first positive value I of t snch that r(l) = T. Set 3(t) r(t) for t E [0, II, and for J E N, set 3(t) = JT + r(t - JI) for t E [JI, U + I)I]. Then S E C 1 . Indeed we need only check what happens at
t
=
I. lirrlS'(t) tit
= lirrH'(t) = lir!.w(dr(t))) tTt
tTt
= l/(dT)) = v(\(O)) = liJ'!.lI/(\"(r(t
I)))
tll
= liJ'!.lr'(t - I) = liJ'!.l3'(t). tll
ilt
It follows that z(t) = d3(t)) is I periodic. PROOF OF THEOREM 6.46. By Proposition 6.47, it suffices to find an H 1 such that H- (1) = H-l (1) and for which the existence of a periodic solution of (6.48) can be established. The geometrical assumption on H-l(l) leads to the construction of such a function. By the starshaped assumption for each z E R2n, there is a unique w(z) E H-l(l) and a(z) > 0 such that z = a(z)w(z). Indeed a(z) = izi Iw(z)I- 1 Note that a is homogeneous of degree one and is C 1 for z I O. Define H(O) = 0 and H(z) = a(z)4 for z 10. Then H E C 1 (R2n,R), H-1(l) = H-l(I), H 2: 0, H(z) = 0(lzI2) as Izl--. 0, and z·Hz(z) = 4H(z) since H is homogeneous of degree 4. Thus H satisfies all of the hypotheses of Theorem 6.10. Hence for e.g. T = 27f, there exists a nonconstant 27f periodic solution \" of (6.48). It need not be the case that H( 1, then A contains infinitely many distinct points, for if A were finite we could write A = B U (- B) with B as in Example 7.2. But then ,(A) = 1. 45
IT\'CTI()\'.\L:; \\ITll Sy\!:-IETHIES A\,D I\,DEX THEOHlES
46
.,'n
EXA\1PLE ,.j If n ? I iind A is hOlll('oillorphic to • by an odd map. then > 1. Otlwrwisr therr i, a mapping; E ('IA. R {Of) 'xitil ; odd. Choose 'ln~' E A sllch that Plr) > O. Then -.ri < 0 and hy th(' Interlllcdiate \'alll(, Theorelll. ; lllust \'i\nish ,olTll'\\'hrre on ,1m' path in A joining .r and ".1'. i\ contradiction. ~.(A)
The main proper! ies of grnus wrll he listed in thl' nC'xt pl'Oposit ion. For A (- (; and ~ > O. let So(A) denote a uniform 6-neighborhoorJ of A. i.e. X,,(A) ~ {r E £1 IT - A k. Then A n X # 0.
ol1bsJ!ILU:
01 1'; 01
wrizlII.c7I.mJrl
k and
PROOF. Writing E V 1) X with V a k rlimensional complcll]('nt of X, let P denote the projector of E onto V. If AnX = 0, ]' E C(A. V \ {O}). Moreover P is odd. Hence by 2° of Proposition 7.0, ,(A) :S ,(PAl. The radial projection of P A into aE 1 n V is another continuous odd map. Hence ,( A) : 0 such that No(A) E [ and i(No(A)) = itA). REMARK 7.9. If A E [and AnFix9 # 2, then irA) = sUPBEE l(B). Indeed let x E A n Fix::; and define I: A -+ {x} via I (u) = x for all 17 E A. This map is continuous and eCjuivariant. Hence i( A) ::; i ({x}) ::; i( A) via the mapping and monotonicity properties of i. But A can be replaced by for e.g. r < ~ I!xll and any B E [ lies in such a set. The genus, ,. provides us with a simple indcx theory where::; {id. As to other index theories, Benc; has int.roduced an 8 1 versioll of genus [Be]. Cohomological index theories can be found e.g. in iY, FRI-2, FRRI and the references cited there. E.g. an especially general such situation is contained in Fadell-Husseini [FRj. One can also find analogues of Propositions 7.7-7.8 for these theories. In t.he next two chapters we shall see how these tools can be used to obt.ain multiplicity results for symmetric functionals.
8. Multiple Critical Points of Symmetric Functionals: Problems with Constraints
The following two chapters study the existence of multiple critical points of fllnctionals possessing a Z2 symmetry. The same ideas together with an appropriate index theory can be used to treat other kinds of symmetries. SeC' e.g. :AZ, Be2-3, BF, Bg, Brl--2. Pal-2, Sl-2]. Theorem 1.10, the classical ]'('sult of Ljusternik and Schnirelmann ILLS] is for a constmined funct.ional, i.e. i! functional on a manifold. In this chapter we will prove that result. as well as an infinite climensional generalization and give an application to (2.31). Chapter 9 treats unconstrained functionals. For either type of problem, the main clifficuJty is to find an appropriate class of sets with respect to which one can minimax the functional. We will see how this can be done in the constrained setting first, following Ljusternik and Schnirelmann. For convenience we restate their result: THEOREM 8.1. Suppose I E C 1(RTI. R) and is even. Then
1
possesses
at least n distinct pairs of critical points. PROOF. For E = Rn and 1 ~ )
(8.2)
"iJ
~
n, define
= {A E ciA c 5 n -
1
and ,(A) :-,> j}.
This family of sets possesses the following properties: 1c
2° (8.3)
3° 4°
¥ 0. 1 ~ ] ~ n. Alonotonicdy property: ~!l :) ~!2 :) ... :) In' Invariance property: Suppose p E C( 5 n - 1 , 5 n - 1 ) and is odd. Then 'P' -, IJ' i.e. E IJ wheneyer A E Excision property: If A. E ~;J and BE C with I(B) ~ s < ], I]
then Indeed ro follows from Proposition 7.7 with D = sn-l. 2° is trivial, 3° is a consequence of the mapping property of Proposition 7.5, and 4° follows from Remark 7.6. Define
(8.4)
c]
=
inf maxI(u),
AE~I)
uEA
48
1~
J ~ n.
PROBLEMS WITH COl\'STRAINTS From the monotonicity property of the 'J' it is clear that Cl :s: C2 :S:. .:S: Cn . 1 will show that cJ is a critical value of Ilsn-l. This fact in itself is not sufficient prove Theorem S.l since some of the minimax values may coincide with only c corresponding critical point. The following proposition together with Rem, 7.3 shows the Cj'S are critical values and we get enough corresponding criti points. :\ote that 1 (lL) = 1'(lL) AlL, where A = (1'(11),1/). PROPOSITION S. 5. If CJ = ... = CJ + P == c, and and = O}, then ,(ke ) 2: p T 1.
kc
=
{,y E
sn-l
PROOF. Suppose that ,(ke ) :s: p, Then by 5° of Proposition 7,5, then a [; > 0 such that ,(Nt(kc )) :s: p, Hence if .N == No(k e ) n sn-l, by 3° Proposition 7.5, ,U-:V) :s: p. Invoking Theorem A.4 and Remark A.17(iv) w 0== intJV and E = 1, there is all c: E (0,1) and TI E C([O, 1] x sn-l,sn-l) W II(t, u) odd in 11 and satisfying (S.C))
TI(l.
c
\
Chom;c A E
By 4° 01' (S.3), A \ /./ E' ') imel E I)' Therefore by (S.C)) and the definition of c,
I :s: c
c:s:
a contradiction. REMARKS S,7. (i) The minimax values rical characterization, namely (S.S)
CJ
c)
c,
can be given another more geon
= inf{r E Rh(k) 2:]}.
Thus the c 7 's are just those numbers at which the sets A. r change genus. Ind an denote the right-hand side of (S.S) by cJ . If r > c]' ,(k) 2:.7 so Ar E
'J
(S.9)
CJ
:s: max 1= r. A,
~ Then there is an A E Thus (S.9) shows cJ :s: cJ . If c] < cJ ' let c = such that maxA I :s: c. Therefore > 2: j by 3° of (8.3) but c < (' contradiction. (ii) There are other ways to obtain critical values of lis define
bk = sup min I(u),
1 :s: k:S: n,
AE'Ik uEA
Clearly b1 2: b2 2: .. 2: bn and using Theorem A.4 and Remark A.17(iv) s11 the bk's are also critical values of Ilsn-l. Note that Cl = mins"-' I sill! x E sn-l, {x} U { -x} E ,1, :V10reover c 1 bn . To prove this, it suffices to s that = {sn-1}. If not, there is a set A E In such that A # S'1-l Then there is a point y E sn-l \ A. Without loss of generality we can assume y = ( with 6 the origin in R n-1 The projection map P( 11) = (11 1,. ., Un-l, 0) bell to C(A, Rn-1 \ {O}) and is odd. Therefore by the argument of Proposition
'71
PEOBLE\lS \\TfH CO';STHAI';TS
50
~J(A) 'S 11 -- I. a contradiction. Thus Cj = hn · Similarly en = III = maXSn-l f. \Ve do not kno\\' if r J = Ii" -).1 if.J ~ 1. n. However if one used the cohomological index theory of iFRlj instead of genus and defilled corresponding minima:.:: and maximin values it call be shown that = b;, I' Thus ill this sense the cohomological index is a nicer tool to deal with. There are manv infinite dimensional generalizatiolls of Theorem 8.1. E.g.
r;. b;,
r;
THEOHE\l 8,10. Let E be an dImensional Hilbert space and lei f E Cl(E,R) be even Suppose r > 0, f\iJJJ, satisjies (P8), and fliw, 18 bounded from below. Then f,)B, possesses lnjimtely many distmet pairs of entieal points. PHOOF. Define the sets J) iL'i in (8.2) for J E N with 8,,--1 replaced by aE T • These sets still satisfy properties 1CAe (again with 8,,--1 replaced by anT)' !\ow define
(8.11 )
= inf sup
CJ
[(11),
.JE N,
AET] l1EA
Since JjiJlJ, is hounded frolllblll()w, 1:1 / - J(:'. :Vlorccwer (PS) illlplies fe· {11 C anrlf(u) = c and f!~3B,(U) = O} is a compact set for any c E R. With these observatiom3 and Remark A,17(iv), the argument of Proposition 8.5 proves the theorem as earlier. REl\IAHK 8.12. The requirement that fia1J e satisfies (PS) is too stringent a condition for applications. Consider e.g. -
~11
Ap(X.11).
(813) =
11
O.
X
:r E
n.
an,
n as usual. Suppose that p satisfies (pd--(p2) and (ps) ~p(x. 0> 0 if Ef 0, and (pg) p(x,~) is odd in E 2 Let E == \Vci (O) and 11 E E, Set
with
(8,14) Then f E C 1 flaB, we have
flu) == - f P(x, 11) dx. In R) by Proposition B.10 and f is even, At a critical point
(8.
=
for all y E E. Choosing:p =
11
0=
-
f p(x,u)'Pdx
io
11
of
pfYll'Y'Pdx
in
and using (Ps) shows
p=I'(ll)U=- (p(x,u)udx p+ 1.
Ck+J
C
and Kc == {u E EII(u) =
C
anc
PROOF. The proof follows almost the same lines as that of Proposition 8.3 and will be omitted. (Use must be made of the fact that 1(0) = 0 and c, < 0 fOJ 1 SiS k. Therefore 0 rf- KCi so K c, E [ and I(Kc,} < 00 via (PS).) REMARK 9.4. Actually in [C), Clark does not assume that I is bounded frorr below or that there is a K as above but merely that if Ck is defined as in (9.2) then -(X) < Ck < O. The above proof then shows Ck is a critical value of I The form of the special case given in Theorem 9.2 is perhaps more useful fOJ applications. Two such applications will be given next (see e.g. [AR, He2, and R2]). The) are related to problems treated in Chapter 2, namely (2.31) and (2.39) .. QC(nsidm 53
THE (':\(,O:\STH.-'c!:\ED CASE
54
first -
(9.5)
~11 = 11
=
A(a(.r)lJ - plIo u)).
0.
e!fl.
.r
where \2 is a.s usual and p satisfies (P'1)' (P3). (p~) there i" a (l > 0 such that a(.r)(j - p(x. 6) ::; 0 for all x E \2. and (P9).
9. G. Suppose p (p'j). (P:3)' (Ps), (pn), and A> Ak, the kth of (2.40). Then (fl5) possesses at lenst k lilstmet of nontrivial
THEORE,-,!
solutwns. PHOOF. We begin by modifying the problem in a familiar fashion,. For x E \1. set q(I, 0 = a(x)( p(x.O if (E [0. (11. q(x. 0 = q(I. Ed if ( > (l, and let If be odd in (. Consider -
(9.7)
~11
u
= Aq(X, u), x E \2. x E ao. = o.
Arguing ,~s in the proof of Corollary 2.23 shows any solution of (9.7) is a solution of (9.5). Hence to prove Theorem 9.G. it suffices to produce at least k distinct pairs of critical points of
1n[~I\}uI2 -
I(u)
AQ(x.u)ldx,
where Q is the primitive of q and E = WOI2(O) as usual. Since q satisfies (p~) and is a bounded function, I E C l (E, R) via Proposition B.lO. Likewise I is bounded from below and (PS) holds as in the proof of Theorem 2.32. Clearly 1(0) = 0 and I is even. Thus the proof of Theorem 9.G follows from Theorem 9.1 once the existence of a set K as in that theorem has been established. Let Vl, ...• Vk denote the eigenfunctions of (2.40) corresponding to AI, . . , Ak normalized so that
I = 1=
A,
In v~ dx,
1 ::; i ::; k.
Set
(9.8) It is clear that K is homeomorphic to sn-1 by an odd map for any r > O. We claim < 0 if r is sufficiently small. Indeed for small T and u E K, 11[( x) i ::; E1 so Q(u) au 2/2 P(x,u). Therefore by (P3), 1
I(u) =
r - 2'A L }o1J'fdx
k
k
2' L 1=1
1 =-L 2 k
1=1
o(r2)
i=l (
A)
1-~ A,
for small r. Since A > Ab I(u) < 0 and the proof is complete.
THE C"ICONSTHAI"IED CASE
As a second application of Theorem 9.6. we CXiillliIlC agclill the proiJlcIn t real in (2.33) and Theorem 2.32: c,u
(9.9)
= Ap(U),
u = 0,
x EO,
xE
a~.
THEOREM 9.10. Suppose p satisfies (pit), (P3), (P5), and (pg). Then for a J E N, there exists a >:J > 0 such that if A > >:J' (9,lD) possesses at leasl
distinct pairs of solutions. PROOF. As in the proof of Theorem 2.32. it suffices to show that the modifi functional
(9.11) has the appropriate number of critical points. This in turn will follow frc Theorem 9.l. In the proof of Theorem 2.32, it was already established th hE CI(E,R), is bounded from below, satisfies (PS), and IA(O) = O. CleaJ (pg) implies h is even. Thus we need only verify that for any J E N, there a >:) > 0 such that for each A > >:), there is a set K as in Theorem 9,1. Let be as in (9.8). For sufficiently small r, F(u(x)) = P(u(x)) for all u E K and (ps), P(u(x)) > 0 if u(x) f 0, Therefore inf "EK
inr P(u(x)) dx =a
> O.
Choose>: = a- 1r2 Then IAIK::; ~r2 + o(r2) < 0 for small r. The proof complete. Now we turn to a Z2 version of the Mountain Pass Theorem. Such a rest was stated in Chapter l. A more general version will be given here. THEOREM 9.12. Let E be an infinite dimensional Banach space and, IE CI(E, R) be even, satisfy (PS), and 1(0) = O. If E = V Ell X, where V finite dimensional, and I satisfies (I;J there are constants p, a > 0 such that IIDB p 0x ? a, and (12) for each finite dimensional subspace E c E, there is an R = R(E) su that I::; 0 on E \ BR(E)' then I possesses an unbounded sequence of critical values. REMARK 9.13, All of our previous symmetric results have used roughly tJ same class of sets: ik = {A E [h( A) ? k} (or the corresponding class on B to construct critical values. These sets do not suffice for the setting of Theore 9.13. To see why, as a model case consider a functional of the form
a
(9.14)
s
+1
J) dx,
THE lJ;\CO,-;STRAI"ED CASE
56
where 1 < s < (n + 2) (n ). It will be seen later in this chapter that Theorem 9.12 applies to such a functional. Since A E can be chosen to be an arbitrarily large sphere in span{v).. with Vi as in (2.40). (9.14) shows
'k
I(u)
inf
(9.15)
=
-x.
AEik
On the other hand choosing
where] = J(r) is large enough. and using arguments as in Corollary 8.18 we sec minuEA I(u) :::: ~r2 Hence sup minI(u) =
00.
AE""1k uEA
Thus a new family of sets must be produced t.o prove Theorem 9.12. PROOF OF THEOREM 9.12. A sequence of families of sets f m will be introduced and a corresponding sequence (em) of critical values of I will be obtained by taking a minimax of I over each I'm. A separate argument then shows is unbounded. Suppose V' is k dimensional and l' = span{ e) .. .. ,ek}. For m :::: k, inductively choose em+ 1 rt span{e), .... em} == Em· Set Rm == R(Eml and Dm == BRm ",Em· Let
E)ih is odd ar:d h = id on BB Rm
(9.16) ;';Ot8 that id E
em
for all mEN so
Ih
(9.17)
E
em of 2.
Set
em, m :::: J, Y
E
The following proposition shows that the sets f
E, and fry) :; m - J}. J
satisfy conditions like (8.3).
The sets f) possess the following properties: Z for all} E N. 2° (Monotonicity) rJ~l C f J . 3° (Invarwnce) If p E G(E.E) is odd, andp = id on BBRm " Em for all :::: J, then p: r J ---> r J . 4° If BE rj , Z E E, and ,(Z) :; s < J, then B \ Z E r J - 8 · PHOPOSITIO:O: 9.18.
1"
m
r J of
PROOF. Since id E em for all mEN, it follows that r J f/: 0 for all J' E N. If B = h(Dm\Y) E rJ~l. then m :.:: j + 1 :::: J, hE Y E E, and -1(Y) :; m + 1) :; m - J. Therefore B E f J . !'Iext to prove 3°, suppose B =, E f J and p is as above. Then po h is odd, belongs to C( Dm, E), and yO h = id on BBR~ n Em. Therefore yO h E Gm and po h(Dm \ Y) p(E) ~ r). Lastly to get 4°, let B = h(Dm \ Y) E r J and Z E E with ,( Z) :; .) < J \Yc claim
em,
(9.19)
B\Z
THE UNCONSTRAINED CASE Assuming (9.19), note that since h is odd and continuous and Z E [, h- 1 (Z) E [ Therefore Y U h- 1 (Z) E [ and by 4° and 2° of Proposition 7.5,
,(Y U h- 1 (Z)) ::: ,(Y) + ,(h-1(Z)) ::: ,(Y)
+ ,(Z)
:::m-J+s=m-(J-s). Hence B \ Z E r)-s' To prove (9.19), suppose b E h(Dm \(YUh-1(Z))). Then b E h(Dm \Y)\Z B \ Z c B \ Z. Therefore
c
h(Dm \ (Y U h-l(Z))) c B \ Z.
(9.20)
On the other hand if b E B \ Z, then b = h( w) where wE Dm \ Y \ h-1(Z)
c Dm \ (Y U h-l(Z)).
Thus (9.21)
B \ Z
c h(Dm \ (Y U h-l(Z))).
Comparing (9.20)-(9.21) yields (9.19) since h is continuous. Now a consequence of minimax values of I can be defined. Set (9.22)
JE N.
c) = inf maxI(u), BEf, uEB
It will soon be seen that if J > k = dim V, Cj is a critical value of I. The following intersection theorem is needed to provide a key estimate. PROPOSITION 9.23. IJJ>k andBEr J , then (9.24)
BnaBp nX
i= 0.
PROOF. Set B = h(Dm \ Y) where m ~ j and ,(Y) ::: m - J. Let 6 = {x E Dmlh(x) E Bp}. Since h is odd, 0 E 6. Let 0 denote the component of 6 containing O. Since Dm is bounded, 0 is a symmetric (with respect to 0) bounded neighborhood of 0 in Em. Therefore by Proposition 7.7, ,taO) m. We claim
h(aO)
(9.25)
c aB p .
Assuming (9.25) for the moment, set W == {x E Dmlh(x) E aBp}. ThereforE (9.25) implies W :::J ao. Hence by 3° of Proposition 7.5, ,(W) m and by Remark 7.6, ,(W \ Y) ~ m (m - J) J' > k. Thus by 2° of Proposition 7.5, ,(h(W \ Y)) > k. Since co dim X k, h(W \ Y) n X i= (2) by Proposition 7.8. But h(W \ Y) c (B n aBp). Consequently (9.24) holds. It remains to prove (9.25). Note first that by the choice of Hm, (9.26)
Since m > k, aBp (9.27)
n X n Em i=
2. Hence by (1;),
THE UNCO:\STRAI:".'ED CASE
,'i8
Comparing (9,26) and (9,27) shows
Rm >
p, :"ow to verify (9,25), suppose
x E 00 and E B p ' If x E Dm there is a neighborhood /If of x such that hiS) c B p ' But then x (j. 00, Thus x E oDm (with 0 relative to Em), But on oDm' h = id, Consequently if x E oDm and h(x) E B p , ilh(x)l! = = Rm < p contrary' to what we just proved, Thus (9,25) must hold, R E\lARK 9.28. A closer inspection of the above proof shows that
2:) - k, COHOLLIllY CJ
29
JI»k'('l
rx>O
PROOF, If) > k and BE f l , by (9,24) and (1;), maxuEB I(u) 2: 0:, Therefore by (9,22). c) 2: 0:, The next proposition both shows cJ is a critical value of I for J > k and makes an appropriate multipicity statement about degenerate critical values. PROPOSITION 9,30. II)
> k, and c) = ' . , = cJ+p ==
c, then ,(Ke)
2: p + 1.
PROOF, Since 1(0) = 0 while c 2: 0 > 0 via Corollary 9.29, 0 (j. Ke. Therefore Kc E [ and by (PS), Kc is compact. If ,(Ke) 0 such that ,(NIi(Ke )) 0 such that ,(N8(K)) = ,(K) == q, Let s max(q, k + 1). The
THE
U~CONSTRAIl\ED
CASE
Deformation Theorem with c = C, E = C ~ cs , and 0 = N5(K) yields an as usual such that
1/(1, Ac_€
(9.34) Choose
JEN
such that c] > C ~
E
\
0)
C
and B E
,59 E
and 1/
Ac- £ '
r
such that
maxl 0 via Corollary 9.29 Consequently 1/(1, B \ 0) E f J and by (9.34) (9.35) and the choice of c]'
CJ:c:;
m~
I :c:;
c ~ [ < c],
~(l,B\O)
a contradiction, The proof is complete. REMARKS 9.36. (i) If E is finite dimensional, the result of Theorem 9.12 also obtains with the conclusion being that I possesses at least dim X critica points, (ii) There are analogues of Theorem 9.12 when V is infinite dimensional and when we have an SI rather than a Z2 action (see e.g. !FHRj) , Such a result leads to the generalization of Theorem 6,10 mentioned in Remark 6.45. As an application of Theorem 9.12, consider
c.u = p(x, u),
(9,37)
u = 0,
x E {1
x E 30.
THEOREM 9,38. Suppose p satisfies (pd, (P2), (P4), and (pg). Then (9.37)
possesses an unbounded sequence of weak solutions. PROOF, With E = W6,2(0) as usual and
l(u) =
fo[~lvuI2 ~ P(x,u)] dx,
the proof of Theorem 2,15 shows I E Cl(E, R), satisfies (PS), and 1(0) = 0, Moreover the argument in Theorem 2,15 that showed I satisfied (h) equally well yields (I2)' Clearly (pg) implies I is even. Assume for now that I also satisfies (Ii), Then Theorem 9,12 implies that I possesses an unbounded sequence of critical values Ck = I(Uk), where Uk is a weak solution of (9.37). Since I'(Uk)Uk = 0,
(9,39) and it follows that
(9.40)
Ck
=
fo[~P(X,Uk)Uk ~ P(x,uk)]dx -> 00
and k -> 00. Hence by (9,39)·-(9.40) and (P4), (Uk) must be unbounded in E and in LOO(O).
THE C:\CONSTRAINED CASE
60
To verify (I;). choose V = span {Vj, ... ,vd where the functions v) are as in (2.40) and X = V ~. By (P2). for all u EO E.
(9.41)
flu) ::::
By (8.21)(8.22), if u
EO
I, ~
dx
as
in 111ls~l
dx -
a6
BEp n X,
1) - as.
(9.42)
Choose p = p(k) so that the coefficient of p2 in (9.42) is
±
Therefore
(9.43) for 11 E BEp X. Since Ak ~x as k that p2 > 2as. Consequently
i
~
x. p(k)
~
x as k
~ 00.
Choose k so
(9.44) and (I;) holds. As a final example of the ideas used in the proof of Theorem 9.12, consider (9.9) again. In Theorem 9.lD, we proved under appropriate hypotheses on P if A > ),k, (9.9) possesses at least k distinct pairs of weak solutions and these functions correspond to negative critical values of (9.11). Let K be as in (9.8), i.e. K = BET n E k · By our choice of rand ),k, I < 0 on K. Setting Rm = r and Dm = Er n Em, define G m for 1 ::; m ::; k as in (9.16) and define r) as in (9.17) with the further proviso that m ::; k. Proposition 9.18 then holds for j EO [1, k] and m ::; k as does Proposition 9.23. Defining Cj via (9.22), observing that 7)(1..J = id on BET n E k, and noting that Corollary 9.29 holds with k O. the proof of Proposition 9.30 works equally well in this setting. Thus we have proved:
THEORE:,I 9.45 [AR]. Under the hypotheses of Theorem 9.10, for all A> ),k, (9.9) possesses at least 2k distinct pairs of nontrivial solutions, k pairs cor-
responding to negative and k pairs to positive critical values of (9.11).
10. Perturbations from Symmetry
In the last two chapters several examples have been given of the existence of multiple critical points for functionals invariant under a group of symmetries. A natural question to ask is: What happens when such a functional is subjected to a perturbation which destroys the symmetry? Some special cases of this question have been studied and while progress has been made, there are not yet satisfactory general answers. This chapters treats a perturbation question in the setting of Theorem 9.38. Thus consider
- 611. = p(x, 11.),
(10.1)
11. = 0,
xE
x EO,
ao,
where p satisfies (Pl), (P2), (P4), and (pg). By Theorem 9.38 the corresponding functional has an unbounded sequence of critical values and (10.1) has an unbounded sequence of weak solutions. Suppose f E L2(0) and (10.1) is replaced by
611. = p(x, 11.)
(10.2)
11. = 0,
+ f(x), ao.
x EO,
xE
The corresponding functional is
(10.3)
J(11.)
and J is not even if
f
fo[~IV11.12-P(x'11.)-f(X)11.]dX
=f= 0. Nevertheless we have:
L 2(0), then (10.2) possesses an unbounded seq11.ence of weak solutions provided that s in (P2) is further restricted by THEOREM 10.4
(10.5) REMARKS
(10.7)
[RB]. lfp satisfies (pd, (P2), (P4), and (pg) and f
= (n+2)-(n-2)s
/3 _
( ) ns-1
>
E
_/1_
/1-1'
10.6. (il Inequality (10.5) is equivalent to /1n s < 1m
+ (/1 - l)(n + 2) + (/1 - l)(n - 2)'
It is easily checked that if s satisfies (10.7), then 8 < (n + 2)(n 2)--] Also, observing that s = 1 and /1 = 2 satisfies (10.7) shows that (10.5) is nonVaC\lOllS. 61
PERTT:RBATIO:\S FH()\l SY\I\IETRY
62
Iii) Slightly Ie,s general versions of Theorem 10.4 were proved by Bahri and Berestycki :BB. and b:.' Struwc lSI. See also Dong and Li :DL]. Their arguments differ fWIIl the one giv'cn herf'. which is somewhat in the spirit of the s:;nHnetric ~lountain Pil.SS Theorem. For the proof we requirE' an estimate on the deviation from symmetry of I of the form
( 10.8) for 11 E E. Cnfortunately I does not satisty (10.8): however it can be modified in snch a [,t'ihioll that the new functional, J. satisfies (10.8) and critical values and points of J are critical value'S and points of I. To motivate the modified problem, a priori bounds for critical points of I in terms of the corresponding critical values will be obtained. ;'\ote first that by (P4) there are constants a4, a.o > 0 such that
o 2: a51~!!1 ~ a4
(10.9)
for all ~ E R. Therefore there is a constant 1 (10.10) ~(~p(x. 0 + (/3) 2: PIx. 11. for all ~ E R. PROPOSITJO:\ 1 (). 11.
wnstant A depending on
> 0 such that
0 + 01 2:
(151~!!1
Under the hypotheses 0/ Theorem 10.4, there eXlsts a ::/1:1'(0) such that i/u is a critical point 0/ I,
r (P(x, u) +
110.12)
a3
j[]
0.4)
ciJ:
:S A(I(v)2 + 1)1/2
REMARK 10.13. If u is a critical point of (10.3), then (10.10) and (10.12) easily imply a bound for v in E in terms of I (v). PROOF OF PROPOSITION 10.11. Suppose u is a critical point of I. Then by (10.10) and simple estimates,
I(v)
= I(v) 2:
(10.14)
~ ~I'(v)u =
G~ t) l
l [~UP(X'lL) ~
(up(x, u) + (3) dx
2:
a71
2:
°2711(p(x,v)+a4)dx~a9
(P(x, u)
~/v]
P(x,u)
dx
~ ~ IlfIIU(Il) IlvIIL2(O) ~ 0.6
+ (4) dx ~ aslluIIL"([]) ~ ae
and (10.12) follows immediately from (10.14). To introduce the modified problem, let X E C=(R, R) such that X(O E:S 1, X(O := 0 for E 2: 2, and x'(El E (~2,0) for EE (1,2). Set
Q(v) := 2A(J(vJ2 + 1)1/2 and
'I/;(u) := X ( Q(v)-l
l
(P(x, u)
+ 0.4) dX)
.
1 for
PERTURBAT10'iS
FJW~1
SY\I\IETllY
:';ote that by (10.12), if u is a critical point of 1. the arg1ll11(,llt of \ Ii", ill and therefore 11)(11) = 1. Finally set
-1
[1. 2 J(u)= 0.21vul
(10.15)
ro.
~
•
( IJ dx. Px,u)-v(u)fl1
Then J(u) = 1(11) if u is a critical point of I. The following result contains the main technical properties of .1 which we need. PROPOSITION 10.16. Under the hypotheses of Theorem 10.4: 1° JECl(E,R). 2° There exists a constant (jl depending on I f Ii L' (0) such that
(10.17)
!oralluEE. 3° There is a constant Mo > 0 such that if .1(11) ?: Mo and .1'(11) =,0, then J(u) = I(u) and 1'(u) = o. 40 There is a constant M 1 ?: M 0 such that for any c > M I, .1 satisfies (P S) lac at c. PROOF. Hypotheses (Ptl-(p2) imply I E Cl(E,R). Since X is smooth, the same is true for 1/; and therefore J. To prove 2°, note first that if 11 E su pp 10 (the support of 1/;), then
(10.18) where Ctl depends on and (10.10),
lin
II!IIL'(D). Indeed by the Schwarz and Holder inequalities
!UdXI ::; IlfllL'(o)llullL'(D) ::; Ct21Iulb(0)
(10.19)
::; Ct3
(In
(P(x, u) + a4) dX)
III'
If further u E supp 1/;, (10.20)
In
(P(x, 11) + a4) dx ::; 4A(I(u)2 + 1)1/2::; Ct4(II(u)1 + 1)
so (10.18) follows from (10.19)-(10.20). Now to get (10.17), by (10.15) and (P9),
(10.21)
IJ( u) - J( -u)1 ::; (1/J( 11) + 11)( -u))
lin
fu dxl·
To estimate the right-hand side of (10.21), by (10.18),
(10.22)
1/J(u)
lin
fudxl ::; Ctl1/J(u)(II(uWlfL
By (10.3) and (10.15), (10.23)
II(u)l::; IJ(u)j
lin
fUdxl·
+ 1).
PERTCRBATJO:\S FROy! SYW,lETIW
64
Therefore ii/II
(10.24)
\) ~
1
.
Using Young's inequality, the f term on the right-hand side can be absorbed into the left-hand side yielding
(10.25) Combining (10.25) with a similar estimate for the term gives (10.17). To prove 3°. it suffices to show that if A10 is large and 11 is a critical point of J with J(u) 2' ,\10' then
. \ Q(ui
(10.26)
Ifrp. . \
\.1
(X,llj-a4}CLX 0 and k 2: k'. Choose H E '\.k such
that max J(H(u))
( 10.48)
uE[h
s: bk + f
Since D k + 1 = Uk U (-Uk), H can be continuously extended to D k_ 1 as an odd function. Therefore by (10.38),
(10.49)
bk + 1
s: uED max + k
for some
11'
E Dk+ l . If
11'
J(H(u)) = J(H(w)) 1
E Uk, by (10.48)-(10.49),
J(H(w))
( 10.50)
s: bk -1- E.
~uppose
1)' E -[h. Then since bk -> CXl as k -> (Xl via (10.40), (10.49) and (10.17) imply J(-H(w)) > 0 if k is large, e.g. k 2: k. By (10.17), the oddness of H. and (10.48),
J(H(11')) = J( -H(-w))
( 10.51)
s: J(H(-w)) + PI ((J(H( -W)))I/I' -1- 1)
s: bk + E -1- 01((b k + ElI/1' + 1).
Combining (10.49)-( 10.51) yields
bk _
( 10.52) Since
f
1
S:b k
€+3d(b k -+-
-rl).
(10.52) implies
is
( 10.53)
bk+!
s: bk -i- 31(b~/1' -1- 1)
for all k 2: k. It remains to show that (10.53) implies (10.47). This will be done by induction. Suppose (10.47) holds for all k E [k,]] n N. We claim it also holds for j + 1. \Vithout loss of generality, we can assume j 2: 2k and ~
> max, bk+ I - o 0. For our later purposes, the behavior of X' Ilcar 0 must be studied. By (11.8) for zEN and small t E R,
°
(11.15)
(L -
+ t.z)idJx(v
Diffcn'ntiating (11. 1,5) at t ( 111G)
(L-
+ pi H(l' + tz + Xlv + tz))
0 gives
+ p~ H'(l' ~ X(l')))x'(v)z -pLH'(v't'
= 0,
VARIATIO"AL :-'1ETHODS E\ BIFCRCATIOl\ THEORY
73
Similarly (1114) yields
(?}J'(v). z) = ilvll~21(H'(v
(11.17)
- 2(H(1I
+ X(v))(z T
X'(v)z). v)
+ (H(v + X(v)),z)]
+ X(lI)),V)(v, z)llvll~4
Consequently since H'(O) = 0. by (11.16)~(11.17), for v near 0,
Ilx'(v)zll :s: constll(1jJ'(v), z)i"x(V)11 + IIH'(v
-i-
l'
f 0.
xlv)):!
+ Ilx'(v)zll)
:s: const { ( 11.18)
~'-;,-",;---'-'izi'l"x(v)i' -+-
x(v))11
= o(l)llzll + o(1)llx'(v)zil as v
-+
0, v
(11.19)
f
0, or
Ilx'(v)zll
o(I)llzll
as v -+ 0, v f 0. Equation (11.19) shows X'(v) extends continuously to v = 0 and ;((0) = O. With these preliminaries in hand, the proof of Theorem 11.4 can be reduced to the study of a finite dimensional variational problem. Let 1) be a neighborhood of in N in which 1jJ and X are defined and C 1 Let ]v( := {v -i- x(v)lv E 'V}. Then ]v( is a CI manifold of dimension n dim N in E. Let E > and set D, := ]v( naB, so for small E, D, is a compact CI, n 1 manifold. Consider liD,. This functional possesses at least two distinct critical points corresponding to the maximum and minimum of I on D,. We claim that at any critical point U of II D" I'(u) = AU, where u = v -+- xlv) and A = 1jJ(v). Assuming this for the moment, observe that as f -+ 0, if v€ E D€, then v€ -+ O. Therefore v€ -+- x( v,) --+ 0 and 1jJ(v€) -+ fL. Thus the proof of Theorem 11.4 will be complete once the above claim is verified. If u is a critical point of liD" then
°
°
J'(u) 0 > c- such that Q c Br and v E aQ implies either (i) g(f.1.,v) = c+, or (ii) g(f.1., v) = C ,or (iii) 1/;(t, v) E aQ for all t near O.
Aosuming Proposition 11.44 for the moment, we can complete the proof of Case (b). Let T- = g(f.1., .)-I(C-) n S-. The construction of S' and Q shows T- 7= C. Set
f:={hEC([O,l],Q)lh(O)
O,h(1)ET-}.
Define
erA)
(11.45)
= hEr inf max g(.\, h(t)). tEla,l]
Letting A < J1 and choosing p as following (11.38) shows c(A) 2: a(A) > O. We claim c(A) is a critical value of g(A, .). Indeed (A.1) shows -gv(f.1., v) is a pseudogradient vector for -gv (.\, v) for all v E aQ. This permits a modification of the proof of the Deformation Theorem to find TI E C([O, 11 x Q, Q) having the usual properties and the use of TI(1,') in the usual fashion to conclude that erA) is a critical value of g( A, .). If A > f.1., the same argument works on replacing 9 by -g. rt remains to construct the set Q. Let Ac:= {v E Nlg(f.1.,v):S e}. LEMMA 11.46. There is an c > 0 such that if c := maxvEB, g(f.1., v) and x E BE \ S-, then as t -+ -00, the orbit 1/;(t, x) leaves Ac n Br via Ac \ aB r . PROOF. If not, there are sequences such that Zm := 1/;(tm, Xm) E aB r and
Cm -+
0,
Xm
E BEm \ S-, and
tm
0
om
if Itm - Tml is bounded away from O. But our above remarks show -+ 0 as m -+ 00 so It m - Tmi ---t 0 as m -+ 00, i.e. the time it takes the orbit to go from 1/;(Tm' xm) on aBEl to Zm on aBT tends to O. However this cannot happen since gv(fl, v) is bounded away from infinity. Therefore there exists an e as desired. An obvious consequence of Lemma 11.46 is COROLLARY 11.47. There is an
E
> 0 such that if c- = min g(f.1.,v), vEBi;
80
VARIATIONAL METHODS IN BIFURCATIOCI THEORY
and x E B£ \ 5--, as t -00 the orbit 1jJ(t, x) leaves Ac+ n aB via Ac+ and if x E Be \ ST, as t 00 the orbit 1/J( t, x) leaves Ac ,n BT via Ac . -4
T
-4
;-.Jow let E be as given by Corollary 11.47. For each x E BE \ S-, there is a corresponding t- (x) such that g(l1. (x), x)) = c+ and for each x E Be \ S+. there i8 a tT(x) such that g(I1.1jJ(t+(x),x)) = c. If x E S+, set t-(x) = -00: if x E 5-. set t-(x) = 00. Finally define
Q == {'0(t,x)lx E Be and t-(x) < t < t+(x)}. PROOF OF PROPOSlTIOl\ 11.44. Clearly Be C Q. If Z E Q \ Be, Z = '0(t, x) for some x E Be and t E (t-(x),t+(x)). Therefore for 0 small, 1/J(t,BIi(x)) c Q and is a neighborhood of z so Q is a neighborhood of O. Let v E aQ and suppose g(l1,v) cI c T or C-. Consider Ov == {1/J(t,v)lt E R}. Since v = limm~(X)vm, where Vm E Q, and 0,,= n Be cI 0, there exists an x E Ov n BE' Therefore .T= .1')forsomeTERandv= where-TE (x),t+(x)) since 9(11.,11) E (c-. ). Hence x E aBE for otherwise v E Q. Moreover 9(11, 1/J(s, v)) E ,c T ) for s near O. Since v) = '0(s - T,X), it is clear that 1/J(s, v) E DQ. Indeed 1iJ( s, v) = limm~oo 1jJ(s T, where Xm E Be, Xm - 4 x, and therefore s - T E (t- (x m ), t+ (x m )). The proof is complete. REMARK 11.48. If I is even, a much stronger result than Theorem 11.35 holds (see iFR]).
Appendix A
The goal of Appendixes A and B is to prove some results that were required earlier which are rather technical in nature. In particular in this appendix we will prove a fairly general version of the Deformation Theorem. Thus let E be a real Banach space, U c E, and I E Cl (U, R). Then vEE is called a pseudo-gradient vector for I at u E U [P2] if: (A.l)
(i) (ii)
Ilvll::; 211J'(u)ll, 11J'(u)112
J'(u)v ~
In the future pseudo-gradient will be denoted by p.g. for short. :\ote that a p.g. vector is not unique in general and any convex combination of p.g. vectors for I at u is also a p.g. vector for 1 at u. Let I E Cl (E, R) and E == {u E EII'(u) Ie a}. Then \f: E -~ E is called a p.g. vector field on E if \f is locally Lipschit~ continuous and \fIx) is a p.g. vector for I for all x E E. LEMMA
A. 2. If IE C l (E, R), there exists a p.g. vector field for I on
E.
PROOF. For each u E E, we can find a vector 11J E E such that = 1 and J'(u)11J > ~III'(u)ll. Then z = ~III'(u)ll11J is a p.g. vector for I at u with strict inequality in (i) and (ii) of (A.l). The continuity of I' then shows z is a p.g. vector for all v E N u , an open neighborhood of u. Since {Nulu E E} is an open covering of E, it possesses a locally finite refinement which will be denoted by {MJ }. Let PJ(x) denote the distance from x to the complement of MJ . Then PJ(x) is Lipschitz continuous and PJ(x) = 0 if x f/:. M J. Set /
/3J(x) = PJ(x) / I:>k(X). k
The denominator of /3) is only a finite sum since each x E E belongs to only Let z) == finitely many sets M k . Each of the sets M J lies in some ~liF(uJ)II11JJ' a p.g. vector for I in M) and set \fix) == Li (x). Since o ::; /3J(x) ::; 1 and L J /3J(x) = 1, for each x E E, \fix) is 'a convex combination of p.g. vectors for I at x. Moreover \f is locally Lipschitz continuous. The proof is complete. 81
APPE:\DIX A
82
COROLLARY :\.3. If I(x) by an odd W.
1S
even in x, I has a p.g. vector field on j;; given
PROOF. Suppose I is eVeIl. Let \; be as given Lemma A.2 and set W(x) = . Then 11-' is odd. locally Lipschitz continuous, and since l' (x) is odd in T. (i) ::; + I!. (ii) = V(x)\/(x) ~ 2: Recall that A.., =' {11 E ::: s} and Ie == {11 E EII(u) sand 1'(11) = O}. "ow we will prove the following version of the Deformation Theorem (which in particular contains Proposition 2.1 as a very special case). See e.g. [Brl, C, P1, Sl] for earlier such results. ~ (V (x) ~ V (
1(
THEOREM A.4. Let E be a real Banach space and let I E CI(E,R) and satisfy (P8). If C E R, E > 0, and 0 is any neighborhood of K e , then there exists an c: E (0, E) and T) E 0([0.1] x E, E) such that 10 T)(0.11) = 11 for all 11 E E. 2° T)(I.u) = 11 for all tE [0,1) if 1(11) t/: [c E.e +E]. 3° T)(t.11) is a horneornorphl:srn of E onto E for each t E [0, I]. 4° !1T)(t.u) ~ uli ::: 1 for all t E [0,1] and u E E. 5° I(T)(t. ::: I(u) for nil t E [0,1] nnd u E E. 6° T)(l,A o +£ \ 0) C ..1 0 - £ , 7° If Kc = Z, T)(1, C ..1 0 -£ 8° If I(u) is even in 11, T)(t, u) is odd in u. PROOF. The function T) will be constructed as the solution of a suitably modified negative gradient flow for I. A few preliminaries are needed before setting up this differential equation. By (PS), Kc is compact. Set Ns == {11 E Eiliu ~ Kc II < t5}, where liu ~ Ko II denotes the distance from 11 to Ke. Choosing (j suitably small, Ills cO. Therefore it suffices to prove 6° with 0 replaced by N s . ]\'ote also that if Kc = Ns == Z so we get 7° instead. We claim there are constants b, s > such that
°
(A.5)
1I1'(u)11 2: b
for all u EAcH \ (Ae-i U N o/ 8 )·
If not, there are sequences bn -> 0, sn -> 0, and 11n E AeHn \ (Ae-in UNo/ 8 ) such that Il1'(un)11 < bn . By (PS), a subsequence of Un converges to u E Kc \ N o/ s · But this latter set is empty. Hence there are constants b, s as in (A.5). Since (A.5) still holds if E is decreased, it can further be assumed that
(A.6)
o<s<min(E,~~,~,D.
Choosing any c: E (0, s), define A == {u E EII(u)::: c ~ E} U {u E EII(u)
2: c + E}
and
B == {u E
Elc ~ E :::
I(u) :::
c+ c:}.
APPEN])]X ,\
Therefore All B = 0. Set
°
°
Then 9 = on A, 9 = 1 on B. ~ 9 ~ 1, and 9 is Lipschitz continuous on E. Similarly there is a Lipschitz continuous function f on E such that f = 1 on E \ Nb/4 , f = on No/s , and ~ f ~ 1. Note that if 1 is even, A, D, and Nb will be symmetric sets with respect t.o the origin and f and 9 arc even functions. Nextdefineh(s)=lifsE[O,ll andh(s) = 1!8if.s2 1. Since [ C1(F::,R).by Lemma A.2 and Corollary A.3, there exists a p.g. vector field l/ for [ on E with V odd if 1 is even. Finally set W(x) = -f(x)g(x)h(lIV(x)II)ll(x) for x E E and W (x) otherwise. Then by construction, W is locally Lipschitz continuous on E with ~ IIW(x)11 ~ 1 and W is odd if 1 is even. Now we can define the mapping 1/. Consider the Cauchy problem:
°
°
° °
(A.7)
d1/ dt
W(1/),
1/(O,u)
=
u.
The basic existence-uniqueness theorem for ordinary differential equations implies that for each u E E, (A.7) has a unique solution defined for t in a maximal interval (u), t+(u)). We claim t±(u) ±oo. If not, say t+(u) < 00. Let tn -> t+(u) with tn < t+(u). Integrating (A.7) shows (A.S)
- tn !
since !IW(·) II ~ 1. But then 1/(t n, u) is a Cauchy sequence and hence converges to some u as tn -> t+(u). The solution to (A.7) with u as initial data then furnishes a continuation of 1/( t, u) to values of t > t+ (u) contradicting the maximality of t+(u). Similarly t-(u) = -00, The continuous dependence of solutions of (A.7) on the initial data u implies 1/ E C([O, 11 x E, E) and (A.7) implies 10 holds. Since E > E, g(x) = on A so 20 is satisfied. The semigroup property for solutions of (A. 7) gives 3°. Integrating (A.7) and using IIWC)II ~ 1 and 10 gives 40 . Above remarks on the oddness of W when 1 is even yield So. To verify 50, note first that if W(u) = 0, 1/(t, u) == u is the solution of (A.7) (via uniqueness) so 5° is trivially satisfied. If W(u) f= 0, u E E so V(u) is defined as is V(1/(t,u)) and
°
~~...!..!.. (A.9)
=
1'(1/(t,u))~: -1'(1/(t, u))f(1/(t, u))g(1/(t, u)) x h(IIV(1/(t, u)!!)ll(1/(t, u)) ~
°
via (ii) of (A.l) and 5° follows for this case. It remains only to verify 6° or a fortiori, 1/( 1, .1 0 +£ \ Ns) C .1 0 -£, If u E .1 c -" then 1(1/(t, u)) ~ C-f by 5°. Thus we need only proveu E Y == .1C+E \(.1c_,UNs) implies that 1/(l,u) E .1 c - E ' Let u E Y. The reasoning that led to 5° showed
(A.lO)
°
d1(1/(t, u)) < dt -'
APPE:\DIX A
84
Since 9 = implies
°
Oil
A c - c. the orbit 1)(1. u) cannot enter A c -€.
1(1)(0. u))
(A.11)
I(I)(t, u)) 'S
E T E:
Therefore
10)
< 2c'
for all t. 2 0. Suppose that u E }' and 1)(£.u) E Z == A c +€ \
for
U
s E [0, tj. This will certainly be the case for small I. Then for such s, 1)(3. u) E (\'ia (A.5)) anel f(7)(S, u)) = 1 = 9(1)(8, u)). By (A.11). 2E 2 =
i;
fO -I'(rl(.,.1I))h(IIF(1)(s.l1))IIW(1)(s,l1))d,~
r h(!W(I)(s, u))II)['(1)(s. 11))II(1)(s, u)) ds
./0
2 {h(IW(I)('S, u))ill'M'(1)(s, u))11 2 ds . () IA.12)
2 b
11
> -b - 2
> -b
u))rI)!II'(1)(,', 11))11 dx
j'l
- 2 , b 2
h(
ds
0
1t
8.
0
b
W(1)(s,u))ds
u) - ul!,
2
where we ,;u(wssively nseo (A.l)(ii). (A.5). and (A.l)(i). Hence by (A. (A.6).
111) l t. u ) -
"
l1i!
ano
S
4f
'S b < '8
Thus the orbit 1)(t,11) cannot leave Z by entering N O/2' Consequently the only way 1)( t. u) can leave Z is to enter A c _.[· We claim this occurs for some t E (0,1) thereby proving 6°. If not. 1)(t, u) E Z for all t E (0,1) and as in (A.12),
(.'1.13)
-.-'-'--'-_.:.-'.. 'S
-hi IW (1) (1, ullli) Ill' (1)( t, u)) 112
If for sOlIle t E (0.1). IW(1)(t. u)) II 'S 1. hlW (1)(t, 11)) II) = 1 and (A.5) and (A.13) imply
(A.14)
On the other hand, if for some t E (0,1), W(r/(t. u)h' 1 so (A.l)(i) and (A.13) yield
(t,l1m > 1, hlIW(1)(t. u))II)
(A.15) COf],;cql1(,nth' for all t E (0. I). we have
( A.Hi)
dI(1) ( t. u))
- mill (\b
2
1)
'=I .
=
APP~~"D1X
8.)
A
Integrating (A.16l and combining the result with (A.1ll gives
which is contrary to (A.6). The proof of Theorem A.4 is complete. REMARKS A.17. (i) :--Jote that the full strength of (PS) was not required in the above proof. We only needed (PS) to conclude that Kc was compact and that (A.5) holds for some b, E > O. But we can also achieve this by merely assuming (PS)loc: there is a 6 > 0 such that IJ(1171) - ci < Sand J'(lLn) -40 as n 00 implies that (1171) is precompaet. Still weaker forms of (PS) have been found (see e.g. ICe, Chi]). (ii) It is sometimes useful to have a bit more flexibility ill the definition of a p.g. vector. If (A. 1) is replaced by
(A.l')
(il
Ilvll:S o:llI'(l1)II,
(ii)
J'(l1)V ~
1!2,
where 0 < {3 < a, it is easy to check that Lemma A.2 holds with 11' chosell so that J'(u)w > 2{3(a+{3l-11lw!! and z = l\iI'(u)w!!. Moreover the proof of Theorem A.4 also is essentially unchanged aside from replacing (A.6) by
(A.6')
0< f < min
(3bS 62
16a'
2'
(iii) There arc problems where instead of rl(t, u) satisfying 5c 7c well as the other conclusions of Theorem A.4), one wants a function ~(L u) to satisfy J(~(l,u)) ~ flu), and if
As := {u E EIJ(u)
~ s},
then ~(l,Ac_" \ 0) c AC+E (with 0 0 if Kc 0). Such an ~ can be obtained by slightly modifying the proof just given, in particular using a positive rather than a negative gradient flow. (iv) There are versions of Theorem A.4 for a functional J defined on a Banach manifold, M, rather than a Banach space [Brl, Pl-2, 81-21. Using the same sort of ideas as in the proof of Theorem A.4, and in particular constructing a p.g. vector field for M which is tangential to M, we can get an analogue of Theorem A.4 for such a setting. E.g. in Chapter 7 we consider the case of M a sphere, say oBJ[O), in a Hilbert space. For this special case, a tangential p.g. vector field can be constructed via V(x) - (v(x) . where V(x) is it p.g. vector field for Jon E. The next result concerns the effect on the mapping 7)(1, u) of making stronger assumptions on the form of I. Such information is useful in particular for highly indefinite functionals such as arise in treating Hamiltonian systems as ill Chapter 6.
APPS:\DIX A
PHOPO:'ITIO:\ A.18.
w1th 1(II)
~ ~:
L Ii. Il)
-~
Suppose E 1" (L rcal Hilbert space and IE ('I(E.R) ), L 18 selfad)oml. and :;:1 IS compar-t. Thrn 1](1.11) =
lL'here 0
+ h) - P(x, 0 -
)1/8+1
d:r
S
,
1)
)
0)
0, there exists a ::.; =
Ohl
5 Eihi
and ihl 5 :y In particul ar if ;] == (3 and ~i 5
(B.29)
(B.S)
dx 5
Choose E so that 3a7E 5 E. This determin es (B.16), (B.25), (B.27), and (B,29) yields (B,30)
)"'-(S+I )
+ Ilul!S + II'PII
t\'ext since P E ['1 (0 x R, R), given any E, such that (B.2S)
C'C~X)l
Therefor e
a6,(s-t-l -m)!(s+I )(1
in,
"
1
11'PIIL'c) (il,)
10 iii dx 5 2; II'PII +
~i.
(1
Choose , =
+ HullS
Combini ng
I)
+llpi!S)
Finally choose 5 so small that (B.31)
a6,I-m/ (s+I)(2
+ Ilull s )5(m/(8+1))-1 5
~
thereby obtainin g (B,14), To prove that J' (u) is continuo us, let Urn -+ U in E. Then via (B,S). By the Holder inequali ty and (B.S) again, (B.32)
IIJ'(u m )
-
J'(ulll
sup Ii
