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On Multiple Solutions of Quasilinear Equations Involving the Critical Sobolev Exponent
Journal of Mathematical Analysis and Applications 231, 142]160 Ž1999. Article ID jmaa.1998.6230, available online at http:rrwww.idealibrary.com on
On Multiple Solutions of Quasilinear Equations Involving the Critical Sobolev Exponent Yisheng Huang Department of Mathematics, Stockholm Uni¨ ersity, S-106 91 Stockholm, Sweden Submitted by R. E. Showalter Received March 2, 1998
1. INTRODUCTION In this paper we consider the existence of multiple solutions for the following quasilinear elliptic equation:
½
yD p u s a k Ž x . < u < qy 2 u q b h Ž x . < u < p
U
y2
x g RN
u,
Ž 1.
u g D 1, p Ž R N . ,
where 1 - p - N, pU [ NprŽ N y p . is the critical Sobolev exponent, 1 - q - pU , k Ž x . g Ls ŽR N . with s s pU rŽ pU y q ., hŽ x . g C ŽR N . l L`ŽR N ., a and b are real parameters, D 1, p ŽR N . is the completion of C0`ŽR N . with respect to the norm 5 u 5 s HR N N =u N p 41r p and D p u [ divŽN =u N py 2 =u. is the p-Laplacian. We say u g D 1, p ŽR N . is a Žweak. solution of Ž1. if ; ¨ g D 1, p ŽR N .
HR
N
< =u < py 2 =u ? =¨ s a
HR
N
k Ž x . < u < qy 2 u¨ q b
HR
N
hŽ x . < u< p
U
y2
u¨ .
Notice that Ž1. always has the trivial solution u s 0. Solutions of Ž1. correspond to critical points of the ‘‘energy’’ functional I : D 1, p ŽR N . ª R given by I Ž u. [
1
H p R
N
< =u < p y
a
H q R
N
k Ž x . < u< q y
b U
p
HR U
U
N
hŽ x . < u< p .
Ž 2.
Since D 1, p ŽR N . is not compactly embedded in L p ŽR N ., I does not in general satisfy the Palais]Smale condition. This causes serious difficulties in the search of critical points of I by standard variational methods. The 142 0022-247Xr99 $30.00 Copyright Q 1999 by Academic Press All rights of reproduction in any form reserved.
MULTIPLE SOLUTIONS
143
existence of nontrivial solutions to equations like Ž1. with u g D01, p Ž V ., where V is a domain in R N, has been studied since the celebrated paper of Brezis ´ and Nirenberg w6x. We refer to the recent paper w14x, the book w19x, and the references therein for the case p s 2 in both bounded and unbounded domains. When p / 2, p F q - pU , we refer to w1, 7]10x for the case of bounded V and to w3, 15, 17x for the case of V unbounded or equal to R N . When p ) 1, 1 - q - p and k Ž x . s b hŽ x . ' 1, Azorero et al. have shown in w1, theorem 4.5x that if a ) 0 is small enough and V is bounded, then there exist infinitely many solutions of Ž1. in D01, p Ž V . which satisfy I Ž u. - 0. In this paper we show a similar result for Ž1. with V s R N; see Theorem 1 for the details. It is not clear whether this result is true for large a even if the domain is bounded because of the problem with the Palais]Smale condition Žsee w1x.. On the other hand, inspired by w5x, we prove that if k Ž x . and hŽ x . satisfy suitable symmetry conditions, then the Palais]Smale condition holds for all a ) 0. Consequently, under such assumptions the result mentioned above remains true for every a ) 0 and b g R Žsee Theorem 2.; we also show the existence of infinitely many solutions for Ž1. with positive energy for every b ) 0 and a g R Žsee Theorem 3.. These two results were showed by Bartsch and Willem w2, Theorems 1 and 2x under the assumptions that V is bounded, p s 2, k Ž x . s hŽ x . ' 1 and the nonlinearities are subcritical. Recently Tshinanga w18x obtained results similar to w2x for p s 2, unbounded domains, special k Ž x ., hŽ x . and subcritical nonlinearities. Let us finally remark that in the symmetric case our result on the existence of solutions with positive energy holds both for 1 - q - p and for p - q - pU .
2. PRELIMINARIES AND THE ŽPS.c-CONDITION Throughout this paper we always denote D 1, p ŽR N . by X and weak convergence by ‘‘© .’’ Denote 5 u 5 r [ HR N N u N r 41r r and r X [ rrŽ r y 1.. For a function f, let fq Ž x . [ max 0, f Ž x .4 and f Ž`. [ lim sup < x < ª` f Ž x .. PROPOSITION 1. Ži. The functional F Ž u. [ HR N k Ž x . N u N q is well defined and weakly continuous on X. Moreo¨ er, F is continuously differentiable, its deri¨ ati¨ e F X : X ª X U is gi¨ en by
² F X Ž u . , ¨: s qH k Ž x . < u < qy 2 u¨ N R
and maps weakly con¨ ergent sequences to weakly con¨ ergent ones.
Ž 3.
144
YISHENG HUANG U
Žii. The functional G Ž u. [ HR N hŽ x . N u N p is well defined on X. Moreo¨ er, G is continuously differentiable, its deri¨ ati¨ e G X : X ª X U is gi¨ en by
² G X Ž u . , ¨: s pUH h Ž x . < u < p N
U
y2
u¨
Ž 4.
R
and maps weakly con¨ ergent sequences to weakly con¨ ergent ones. Proof. It is clear that F and G are well defined on X. In order to prove F , G g C 1 Ž X, R. it suffices to show both F and G have continuous Gateaux derivative on X. We consider only F since the proof for G is ˆ simpler. Our argument is similar to that in w19, Lemma 3.10x. Let uŽ x ., ¨ Ž x . g X and 0 -N t N- 1. Using the mean value theorem we obtain l g Ž0, 1. such that u Ž x . q t¨ Ž x .
q
y uŽ x .
q
r< t < s q u Ž x . q l t¨ Ž x .
qy1
¨ Ž x.
qy 1
F q uŽ x . q ¨ Ž x .
¨ Ž x. .
Ž 5.
By Holder’s inequality, we have ¨
HR
N
k Ž x . w < u < q < ¨
qy 1
< ¨ < F 5 k 5 s 5 < u < q < ¨ < 5 qy1 pU 5 ¨ 5 pU ,
where s s pU rŽ pU y q . s Ž pU rq .X . It follows from Ž5. and the Lebesgue dominated convergence theorem that F is Gateaux differentiable and Ž3. ˆ holds. Assume that u ª u in X. By the continuity of the embedding n U U X ¨ L p ŽR N ., uUn ª u in ULX p ŽR N .. Let us define f Ž u. [ qk Ž x . N u N qy 2 u. Ž L p ŽR N ., LŽ p . ŽR N .. Žcf. w11, p. 30x.. It follows that f Ž u n . ª Then f g C U X Ž p . ŽR N .. Using Holder’s f Ž u. in L and Sobolev’s inequalities, we get ¨
² F X Ž u n . y F X Ž u . , ¨:
F f Ž un . y f Ž u.
Ž pU .X 5 ¨ 5 pU
F c1 f Ž u n . y f Ž u .
Ž pU .X 5 ¨ 5
for all ¨ g X. Hence, 5 F X Ž u n . y F X Ž u.5 ª 0 and F g C 1 Ž X, R.. U U Let u n © u. Since u n g L p ŽR N ., NU uX n N q is bounded in L Up r q ŽR N ., Ž k xU. XN u n N qy 2 u n is bounded in LŽ p . ŽR N . and hŽ x . N u n N p y2 u n in LŽ p . ŽR N .. Hence, going if necessary to a subsequence, we can assume < u n < q © U1
in
k Ž x . < u n < qy 2 u n © U2
Lp
U
rq
ŽRN . , U X
in
LŽ p . Ž R N .
in
LŽ p . Ž R N . .
and hŽ x . < un < p
U
y2
u n © U3
U X
145
MULTIPLE SOLUTIONS
The proof of the weak convergence part will be complete if we show that U U1 sN u N q, U2 s k Ž x . N u N qy 2 u and U3 s hŽ x . N u N p y2 u. By the compact r Ž N .Ž embedding theorem, u n ª u in L loc R 1 - r - pU .. So going to a subsequence once more, we may assume that u nŽ x . ª uŽ x . a.e. in R N . Consequently, we have q
unŽ x . k Ž x . unŽ x .
qy 2
q
ª uŽ x .
a.e. in R N , qy2
un Ž x . ª k Ž x . uŽ x .
u Ž x . a.e. in R N
and hŽ x . unŽ x .
pU y2
unŽ x . ª hŽ x . uŽ x .
pU y2
u Ž x . a.e. in R N .
Therefore, U1 sN u N q, U2 s k Ž x . N u N qy2 u and U3 s hŽ x . N u N p
U
y2
u.
It follows from Proposition 1 that the functional I is well defined on X and has a continuous derivative given by
² I X Ž u . , ¨: s H < =u < py 2 =u ? =¨ y aH k Ž x . < u < qy 2 u¨ N N R
R
yb
HR
N
hŽ x . < u< p
U
y2
u¨ .
Thus any critical point of I is a Žweak. solution of equation Ž1.. Recall that I is said to satisfy the Palais]Smale condition at the level c g R ŽŽPS.c in short. if any sequence u n4 ; X such that I Ž u n . ª c and I X Ž u n . ª 0 has a convergent subsequence. In order to find critical points of I we will establish conditions under which ŽPS.c holds. We will need the following result due to P. L. Lions w13x. PROPOSITION U 2. Let u n4 ; X be a sequence such that u n © u and suppose N u n N p © n , N =u n N p © m in the sense of measures. Then there exists an at most countable index set J such that Ža. m GN =u N p q Ý j g J m j d x , m j ) 0, j U Žb. n sN u N p q Ý j g J n j d x , n j ) 0, j U Žc. m j G Sn jp r p , where x j g R N, j g J, d x j are the Dirac measures at x j and S is the best Sobole¨ constant. Recall that S [ inf
½H
RN
< =u < p ; u g X ,
HR
U
N
< u< p s 1 .
5
146
YISHENG HUANG
Hence 5 u 5 pU F Sy1 r p 5 u 5 , ; u g X .
Ž 6.
It is known from w13x that S is attained by the function u« Ž x . s K
ž
Ž Nyp .rp
« 1rŽ py1. « p rŽ py1. q < x < p rŽ py1.
, x g RN
/
as well as all translates of u« . Here « ) 0 is arbitrary and K is a constant such that 5 u« 5 s 1. Let 1 - q - p. Then
PROPOSITION 3.
Ži. ;b ) 0 there exists a 1 ) 0 such that if 0 - a - a 1 and c - 0, then I satisfies ŽPS.c . Žii. ;a ) 0 there exists b 1 ) 0 such that if 0 - b - b 1 and c - 0, then I satisfies ŽPC.c . Proof. Let u n4 be a sequence in X such that I Ž u n . ª c - 0 and
I X Ž u n . ª 0.
Ž 7.
Then for n large enough, we have c q 1 q 5 un 5 G I Ž un . y
1 U
p
² I X Ž u n . , u n: s
ž
1 p
y G
1 N
5 un 5 p y
ž
a q
y
a U
p
/
y
ž
a q
1 pU y
/
5 un 5 p
a pU
/H
RN
k Ž x . < un < q
Syq r p 5 k 5 s 5 u n 5 q .
It follows that u n4 is bounded. We can assume, going if necessary to a subsequence, that u n © u in X. There exist measures m and n such that Ža., Žb., and Žc. of Proposition 2 hold. Let x k be a singular point of the measures m and n . As in the paper w1x we define a function f g C0`ŽR N , w0, 1x. such that f Ž x . s 1 in B Ž x k , « ., f Ž x . s 0 in R N _ B Ž x k , 2 « . and N =f Ž x . NF 2r« in R N . Then f u n4 is bounded in X. Indeed, using Holder’s inequality, we have ¨
HR
< u n=f < F p
N
žH žH
F c2
BŽ xk , 2 « .
< un <
BŽ xk , 2 « .
pU
< un <
prpU
prN
/ žH /
RN
pU
prpU
,
< =f < N
/
147
MULTIPLE SOLUTIONS
where c 2 is a constant independent of « , n, and k. Hence
HR
N
=Ž u n f .
p
F2
py1
c2
žH
RN
< un <
pU
prpU
q
/
HR
N
< =u n < p ,
and this implies that f u n4 is bounded in X. It follows from Ž7. that ² I X Ž u n ., f u n : ª 0, i.e.,
a
HR
N
k Ž x . f < u< q q b
HR
s lim
H
nª` R N
N
h Ž x . f dn y
HR
N
f dm
< =u n < py 2 u n=u n ? =f .
By Holder’s inequality, ¨
lim nª`
HR
N
< =u n < py2 u n=u n ? =f F lim
žH
RN
nª`
< =u n < p
1rp
Fc 3
žH
RN
< u=f < p
/
F c4
žH
BŽ xk , 2 « .
< u< p
U
Ž py1 .rp
/
1rp
žH
RN
< u n=f < p
/
1rpU
/
ª 0 as « ª 0
Ž 8. and we deduce 0 s lim a «ª0
žH
R
N
k Ž x . f < u< q q b
HR
N
h Ž x . f dn y
HR
N
f dm
/
s b hŽ xk . n k y m Ž xk 4 . ; hence
b h Ž x k . n k s m Ž x k 4 . G mk .
Ž 9.
This inequality says that the concentration of n cannot occur at points where hŽ x k . F 0; that is, if hŽ x k . F 0, then m k s n k s 0. If hŽ x k . ) 0, we get from Žc. of Proposition 2 that either Ži. Žii.
n k s 0 or n k G S N r p w b hŽ x k .xyN r p G S N r p Ž b 5 hq 5 ` .yN r p.
148
YISHENG HUANG
In order to obtain information about possible concentration at infinity we define, as in w5x,
m` [ lim lim
H
Rª` nª` < x <)R
< =u n < p ,
H
n` [ lim lim
Ž 10 .
U
n
Ž 11 .
Now we define a function x g C`ŽR N , w0, 1x. such that x Ž x . s 1 on < x < G 2, x Ž x . s 0 on < x < F 1 and we set x R Ž x . s x Ž xrR.. As in w5x or w3x, we see that
m` s lim lim
H
Rª` nª` R N
n` s lim lim
H Rª` nª` R
< =u n < px Rp s lim lim U
N
H
Rª` nª` R N
U
< u n < p x Rp s lim lim
H Rª` nª` R
< =u n < px R , U
N
< u n < p xR
and U
m` G Sn`p r p .
Ž 12 .
Since
HR
N
=Ž u n xR .
p
F2
py1
F 2 py1
HR HR
< =u n < q p
N
N
ž
HR
< =u n < p q c5
N
pU
< un <
žH
RN
prpU
prN
/ ž /
< un < p
U
HR
N
< =x R < N
/
prpU
for some constant c5 independent of R and n, u n x R 4 is bounded in X. It follows from Ž7. that lim nª`² I X Ž u n ., u n x R : s 0, i.e., 0 s lim
nª`
HR
N
< =u n < py 2 u n=u n ? =x R y a
HR
yb
HR
U
N
N
h Ž x . < u n < p xR q
k Ž x . < u n < qx R
HR
N
< =u n < px R .
149
MULTIPLE SOLUTIONS
Notice that 1rp
lim
H nª` R
N
< =u n < py2 u n=u n ? =x R F lim 5 u n 5 py1
ž
nª`
F c6
žH
R- < x <-2 R
HR
< u<
N
pU
< u n=x R < p
/
1rpU
/
ª 0 as R ª `
Žcf. Ž8.., lim lim
H Rª` nª` R
U
h Ž x . < u n < p x R F 5 hq 5 ` n`
N
Ž 13 .
and using the weak continuity of F , lim
H nª` R
N
k Ž x . < u n < qx R F
H< x <)R
k Ž x . < u < q ª 0 as R ª `.
Therefore 0 F b 5 hq 5 ` n` y lim lim
H
Rª` nª` R N
< =u n < px R s b 5 hq 5 ` n` y m` .
Ž 14 .
Hence we obtain from Ž12. that either Žiii. n` s 0 or Živ. n` G S N r p Ž b 5 hq 5 ` .yN r p. We claim that Žii. and Živ. are impossible if a 1 or b 1 are chosen small enough. Indeed, it follows from the weak lower semicontinuity of the norm, weak continuity of F , Ž6. and Ž7. that 0 ) c s lim
nª`
H N R
nª`
1
H N R
I Ž un . y
1
s lim G
ž
N
N
1 pU
² I X Ž u n . , u n:
< =u n < p y a
< =u < p y a
ž
1 q
y
1
ž
q
1 U
p
y
/
1 pU
/
/H
R
N
k Ž x . < un < q
5 k 5 s 5 u 5 qpU G
S N
5 u 5 ppU y a A 5 u 5 qpU , Ž 15 .
where A s Ž1rq y 1rpU .5 k 5 s . This yields 5 u 5 qpU F c 7 a q rŽ pyq.
Ž 16 .
150
YISHENG HUANG
for some constant c 7 . If Živ. occurs, we obtain by Ž12., Ž15. and Ž16. that 0 ) c s lim
nª`
s lim
nª`
G G
1
ž
I Ž un . y
1
H N R
N
lim lim
N
pU
² I X Ž u n . , u n:
< =u n < p y a
H N Rª` nª` R S
1
N
ž
1 q
y
1 pU
/
/H
R
N
k Ž x . < un < q
< =u n < px R y a A 5 u 5 qpU G 1
U
n`p r p y c8 a p rŽ pyq. G
N
1 N
m` y c8 a p rŽ pyq.
S N r p 5 hq 5 `1y N r pb 1yN r p y c8 a p rŽ pyq. .
However, if b ) 0 is given, we can choose a 1 ) 0 so small that for every 0 - a - a 1 the last term on the right-hand side above is greater than 0 which is a contradiction. Similarly, if a ) 0 is given, we can take b 1 ) 0 so small that for every 0 - b - b 1 the last term on the right-hand side above is greater than 0. By a similar argument, with x R replaced by the function f introduced earlier, we see that Žii. cannot occur if a 1 or b 1 are chosen properly. Hence n k s n` s 0 for all k g J and this implies that lim
H nª` R
U
N
< u n < p s lim lim
Rª` nª`
HR
U
N
< u n < p Ž 1 y xR . q
HR
U
U
N
< u n < p xR s
HR
U
N
< u< p .
U
By the uniform convexity of L p ŽR N . we see that u n ª u in L p ŽR N .. On the other hand, it follows from Clarkson’s inequality < j 2 < p G < j 1 < p q p < j 1 < py2j 1 ? Ž j 2 y j 1 . q C Ž p .
< j2 y j1< s
Ž < j 2 < q < j 1 <.
syp
,
where j 1 , j 2 g R N , s s 2 if p g Ž1, 2. and s s p if p G 2 Žcf. w12, Lemma 4.2x., that < =u n y =u < p F c9
½ Ž < =u < n
py2
=u n y < =u < py2 =u . ? = Ž u n y u .
? Ž < =u n < p q < =u < p .
1y tr2
,
where t s p if p g Ž1, 2. and t s 2 if p G 2.
5
tr2
Ž 17 .
151
MULTIPLE SOLUTIONS
In the case 1 - p - 2, it follows from Ž17. and Holder’s inequality that ¨ pr2
5 u n y u 5 p F c9 ?
Ž < =u n < py2 =u n y < =u < py2 =u . ? =Ž u n y u .
½H žH ½H Ž ½² Ž RN
RN
F c10
1y pr2
< =u n < p q
RN
5
HR
N
< =u < p
/ pr2
< =u n < py2 =u n y < =u < py2 =u . ? = Ž u n y u .
5
I X u n . y I X Ž u . , u n y u:
s c10
qa
HR
qb
HR
N
k Ž x . Ž < u n < qy 2 u n y < u < qy 2 u . Ž u n y u .
N
hŽ x . Ž < un < p
U
y2
un y < u< p
U
y2
pr2
u. Ž un y u.
5
.
Since
HR
N
k Ž x . Ž < u n < qy 2 u n y < u < qy2 u . Ž u n y u . 1 qy 1 F 5 k 5 s Ž 5 u n 5 qy pU q 5 u 5 p U . 5 u n y u 5 pU ,
HR
N
hŽ x . Ž < un < p
U
y2
un y < u< p
F 5 h 5` Ž 5 un 5 p
U
y1
U
y2
q 5 u5 p
u. Ž un y u.
U
y1
. 5 un y u5 p
U
U
and u n © u in X, u n ª u in L p ŽR N ., it follows that u n ª u in X. The argument via Ž17. in the case p G 2 is similar but simpler. This completes the proof of Proposition 3.
3. EXISTENCE OF INFINITELY MANY SOLUTIONS Let X be the Banach space in Section 2 and let E denote the class of sets A ; X _ 04 such that A is closed in X and symmetric with respect to the origin. For A g E , recall the genus g Ž A. is defined by
g Ž A . [ min k g N; 'w g C Ž A, R k _ 0 4 . , w Ž x . s yw Ž yx . 4 .
152
YISHENG HUANG
If there is no mapping w as above for any k g N, then g Ž A. [ q`. Below we list the main properties of the genus Žcf. w16x.. Let A, B g E . Then
PROPOSITION 4.
18 If there exists an odd map f g C Ž A, B ., then g Ž A. F g Ž B .. 28 If A ; B, then g Ž A. F g Ž B .. 38 g Ž A j B . F g Ž A. q g Ž B .. 48 If S is a sphere centered at the origin in R N , then g Ž S . s N. 58 If A is compact, then g Ž A. - q` and there exists d ) 0 such that Ž Nd A. g E and g Ž Nd Ž A.. s g Ž A., where Nd Ž A. s x g X; 5 x y A 5 F d 4 . Let I be the functional defined by Ž2.. Assume that 1 - q - p and a ) 0, b ) 0. Since I Ž u. s G
1
H p R
1 p
N
< =u < p y
a
H q R
N
k Ž x . < u< q y
b U
p
HR
N
hŽ x . < u< p
U
U
5 u 5 p y a c11 5 u 5 q y b c12 5 u 5 p ,
I Ž u. G QŽ5 u 5., where QŽ t . [
1 p
U
t p y a c11 t q y b c12 t p .
It is not difficult to see that given b ) 0 there exists a 2 ) 0 so small that for every 0 - a - a 2 we can find 0 - T0 - T1 such that QŽ t . - 0 for 0 - t - T0 , QŽ t . ) 0 for T0 - t - T1 , QŽ t . - 0 for t ) T1. Similarly, given a ) 0, we can choose b 2 ) 0 with the property that T0 , T1 as above exist for each 0 - b - b 2 . Clearly, QŽT0 . s QŽT1 . s 0. Using the same idea as in w1x, we consider the truncated functional I˜Ž u . [
1
H p R
N
< =u < p y
a
H q R
N
k Ž x . < u< q y
b U
p
c Ž u.
HR
U
N
hŽ x . < u< p ,
where c Ž u. s t Ž5 u 5. and t : Rqª w0, 1x is a nonincreasing C` function such that t Ž t . s 1 if t F T0 and t Ž t . s 0 if t G T1. Thus we get
˜Ž 5 u 5 . , I˜Ž u . G Q
Ž 18 .
where
˜Ž t . [ Q
1 p
U
t p y a c11 t q y b c12 t p c Ž t . .
153
MULTIPLE SOLUTIONS
It is clear that I˜g C 1 Ž X, R. and I˜ is bounded from below. Moreover, by Ž18. and Proposition 3, we have PROPOSITION 5. Ž1. If I˜Ž u. - 0, then 5 u 5 - T0 and I˜Ž u. s I Ž u.. Ž2. ;b ) 0 there exists a 0 ) 0 Ž a 0 s min a 1 , a 2 4. such that if 0 - a - a 0 and c - 0, then I˜ satisfies ŽPS.c . Ž3. ;a ) 0 there exists b 0 ) 0 Ž b 0 s min b 1 , b 2 4. such that if 0 - b - b 0 and c - 0, then I˜ satisfies ŽPS.c . Remark 1. Denote K c [ u g X; I˜X Ž u. s 0, I˜Ž u. s c4 . If a , b are as in Ž2. or Ž3. above, then it follows from ŽPS.c that K c Ž c - 0. is compact. Denote I˜a [ u g X; I˜Ž u. F a4 . Suppose that k Ž x . ) 0 on some open subset V ; R N . Then ;k g N'« k - 0 such that g Ž I˜« k . G k. Indeed, denote by D01, p Ž V . the closure of C0`Ž V . with respect to the norm 5 u 5 s HV N =u N p 41r p. Extending functions in D01, p Ž V . by 0 outside V we can assume that D01, p Ž V . ; X Žrecall X s D 1, p ŽR N ... Let X k be a k-dimensional subspace of D01, p Ž V .. It is clear that there exists a constant d k ) 0 such that
HV k Ž x . < ¨ <
q
G dk
for all ¨ g X k with 5 ¨ 5 s 1. If u g X k , u / 0, we write u s r k¨ with 5 ¨ 5 s 1 and r k s 5 u 5. Thus, for 0 - r k - T0 , we have I˜Ž u . s I Ž u . s F F
1 p 1 p
1
H < =u < p V
5 u5 p y r kp y
a
p
y
a
H k Ž x . < u< q V
H k Ž x . < u< q V
a dk q
q
q
y
q b c12 5 u 5 p
b U
p
HV h Ž x . < u <
pU
U
U
r kq q b c12 r kp [ « k ;
therefore we can choose r k g Ž0, T0 . so small that I˜Ž u. F « k - 0. Let Sr k s u g D01, p Ž V .; 5 u 5 s r k 4 , then S r k l X k ; I˜« k . Hence, by Proposition 4, g Ž I˜« k . G g Ž Sr k l X k . s k. Thus, if we denote Tk [ A g E ; g Ž A. G k 4 and c k [ inf sup I˜Ž u . , AgGk ugA
Ž 19 .
then y` - c k F « k ) 0,
kgN
because I˜« k g Gk and I˜ is bounded from below.
Ž 20 .
154
YISHENG HUANG
PROPOSITION 6. Let a , b be as in Ž2. or Ž3. of Proposition 5 and let k Ž x . ) 0 on an open subset of R N . Then all c k Ž gi¨ en by Ž19.. are critical ¨ alues of I˜ and c k ª 0. Proof. It is clear that c k F c kq1. By Ž20., c k - 0. Hence c k ª c F 0. Moreover, since ŽPS.c is satisfied, it follows from a standard argument Žcf. ˜ w1, Lemma 4.4x or w16, Proposition 9.29x. that all c k are critical values of I. We claim that c s 0. If c - 0, then by Remark 1 K c is compact and K c g E . It follows from 58 of Proposition 4 that g Ž K c . s m - q` and there exists d ) 0 such that
g Ž K c . s g Ž Nd Ž K c . . s m. By the deformation lemma Žcf. w16, Theorem A.4x. there exist « ) 0Ž c q « - 0. and an odd homeomorphism h : X ª X such that
h Ž I˜cq « _ Nd Ž K c . . ; I˜cy « .
Ž 21 .
Since c k p c, 'k g N such that c k ) c y « and c kqm F c. There exists A g Gkq m such that sup ug A I˜Ž u. - c q « , i.e., A ; I˜cq « .
Ž 22 .
It follows from 18, 28 and 38 of Proposition 4 that
g Ž A _ Nd Ž K c . . G g Ž A . y g Ž Nd Ž K c . . G k, g Ž h Ž A _ Nd Ž K c . . . G k ; therefore h Ž A _ Nd Ž K c . . g Gk . Consequently, sup
I˜Ž u . G c k ) c y « .
Ž 23 .
ug h Ž A_Nd Ž K c ..
On the other hand, by Ž21. and Ž22.,
h Ž A _ Nd Ž K c . . ; h Ž I˜cq « _ Nd Ž K c . . ; I˜cy « which contradicts Ž23.. Hence c k ª 0. By Ž1. of Proposition 5, I˜Ž u. s I Ž u. if I˜Ž u. - 0. This and Proposition 6 give the following result: THEOREM 1. Let I Ž u. be defined as in Ž2.. Assume that 1 - q - p and k Ž x . ) 0 on an open subset of R N . Ži. ;b ) 0 there exists a 0 ) 0 such that if 0 - a - a 0 , then equation Ž1. has a sequence of solutions ¨ k 4 with I Ž ¨ k . - 0 and I Ž ¨ k . ª 0 as k ª `;
155
MULTIPLE SOLUTIONS
Žii. ;a ) 0 there exists b 0 ) 0 such that if 0 - b - b 0 , then equation Ž1. has a sequence of solutions ¨ k 4 with I Ž ¨ k . - 0 and I Ž ¨ k . ª 0 as k ª `. In the following we will discuss equation Ž1. for k Ž x . and hŽ x . satisfying some symmetry conditions. First of all, let us introduce some notation. Let O Ž N . be the group of orthogonal linear transformations in R N and let G ; O Ž N . be a subgroup. For x / 0 we denote the cardinality of Gx [ gx; g g G4 by N Gx N and set N G N[ inf x g R N , x / 0 N Gx N . We say that a function f : R N ª R is G-invariant if f Ž g x . s f Ž x . for all g g G and x g R N. We denote XG [ DG1, p ŽR N . the subspace of D 1, p ŽR N . consisting of all G-invariant functions and XGU [ DG1, p ŽR N .U its dual space. Now we consider equation Ž1. on XG , i.e.,
½
yD p u s a k Ž x . < u < qy 2 u q b h Ž x . < u < p
U
y2
u,
x g RN
u g XG ,
Ž 24 .
where 1 - p - N, 1 - q - pU , k Ž x . and hŽ x . are G-invariant functions such that hŽ x . gC ŽR N . lL`ŽR N . and k Ž x . gLs ŽR N . with sspU rŽ pU yq .. Let u g XG ; we use the same notation I Ž u. given by Ž2. for the functional associated with equation Ž24.. As in Proposition 1, I is well defined and has a continuous derivative I X : XG ª XGU given by
² I X Ž u . , ¨: s H < =u < py 2 =u ? =u y aH k Ž x . < u < qy 2 u¨ N N R
R
yb
HR
N
hŽ x . < u< p
U
y2
u¨ .
Then any critical point of I is a Žweak. solution of equation Ž1. by the following principle of symmetric criticality: PROPOSITION 7.
If I X Ž u. s 0 in XGU , then I X Ž u. s 0 in X U .
This is a special case of Theorem 1.28 in w19x. PROPOSITION 8. If 1 - q - p, N G Ns `, hŽ0. s hŽ`. s 0, then ŽPS.c holds in XG for all c g R. Proof. Let u n4 be a sequence in XG such that I Ž u n . ª c with c g R and I X Ž u n . ª 0 in XGU . As in Proposition 3 we see u n4 is bounded. There exist measures m and n such that Ža., Žb., and Žc. of Proposition 2 hold. We claim that concentration of n cannot occur at any x / 0. For if x k / 0 is a singular point of n , then n k s n Ž x k . ) 0 and by G-invariance of n , n Ž gx k . s n k ) 0 for all g g G. Since N G Ns `, the sum in Žb. of Proposition 2 is infinite. Hence n k s 0. It follows from Ž9. and hŽ0. s 0 that
156
YISHENG HUANG
n 0 s 0 Žwhere n 0 corresponds to x s 0.. So the concentration of n cannot occur at 0. Next we prove that the concentration cannot occur at infinity. Let m` and n` be as in Ž10. and Ž11.. Given « ) 0, by the definition of hŽ`. we can find R 0 Ž « . such that
H< x <)R h Ž x . < u
U
n
H< x <)R < u
n
U
for all R ) R 0 Ž « .; therefore
H hŽ x . < u Rª` nª` < x <)R lim lim
U
n
< p s 0.
Using Ž12. it follows by the same argument as in Ž13., Ž14. that U
Sn`p r p F m` s 0; U
this implies that n` s m` s 0. Hence u n ª u in LGp ŽR N . and the argument at the end of the proof of Proposition 3 shows that u k ª u in XG . Remark 2. Suppose that hŽ x . ) 0 a.e. in R N , hŽ0. s hŽ`. s 0 and N G Ns `. Let u n4 be a sequence such that I Ž u n . ª c and I X Ž u n . ª 0. If p - q - pU and b ) 0, then for n large enough, c q 1 q 5 u n 5 G I Ž u n . y ² I X Ž u n ., u n :rq G Ž1rp y 1rq .5 u n 5 p which gives the boundedness of u n4 . Consequently, the conclusion of Proposition 8 remains true in this case. Assume that k Ž x . ) 0 on an open subset V ; R N . Since k Ž x . is G-invariant, V is G-invariant, i.e., g V s V for every g g G. Let D0,1, Gp Ž V . be the subspace of D01, p Ž V . consisting of all G-invariant functions. By extending functions in D0,1, Gp Ž V . by 0 outside V we see that D0,1, Gp Ž V . ; XG . Consequently, by Proposition 8 and the same methods used in the proof of Theorem 1, we have THEOREM 2. Let k, h be G-in¨ ariant, k Ž x . ) 0 on an open subset of R N , 1 - q - p, N G Ns ` and hŽ0. s hŽ`. s 0. Then for e¨ ery a ) 0 and b g R, equation Ž1. has a sequence of solutions ¨ k 4 such that I Ž ¨ k . - 0 and I Ž ¨ k . ª 0 as k ª `. Moreover, from Proposition 8 and Remark 2 we can obtain existence of infinitely many solutions of equation Ž1. with positive energy: THEOREM 3. Let k, h be G-in¨ ariant, hŽ x . ) 0 a.e. in R N and hŽ0. s hŽ`. s 0. If p, q g Ž1, pU ., p / q and N G Ns `, then for e¨ ery b ) 0, a g R, equation Ž1. has a sequence of solutions u n4 such that I Ž u n . ª ` as k ª `. Since XG is a separable Banach space, there is a linearly independent sequence e j 4 such that XG s [`js1 X j ,
157
MULTIPLE SOLUTIONS
where X j [ R e j Žsee, e.g., w4, p. 44x.. Denote Yk [ [j F k X j and Zk [ [j G k X j. In order to prove Theorem 3 we will use the following result which can be found in w19, Theorem 3.6x: PROPOSITION 9. Let w g C 1 Ž XG , R. be an e¨ en functional satisfying ŽPS.c for e¨ ery c ) 0. If for e¨ ery k g N there exist r k ) r k ) 0 such that
Ž B1 .
a k [ max w Ž u . F 0,
Ž B2 .
bk [ inf w Ž u . ª ` as k ª `,
ugYk 5 u 5s r k ugZ k 5 u 5sr k
then w has a sequence of critical ¨ alues tending to `. Proof of Theorem 3. It is clear that the functional I associated with Ž24. is even and I g C 1 Ž XG , R.. By Proposition 8 and Remark 2, I satisfies ŽPS.c for every c g R. Since Yk is a finite dimensional subspace of XG for each k g N and hŽ x . ) 0 a.e. in R N , it follows that there exists a constant « k ) 0 such that
HR
U
N
hŽ x . < ¨ < p G « k
for all ¨ g Yk with 5 ¨ 5 s 1. If u g Yk , u / 0, we write u s r k¨ , with r k s 5 u 5 and 5 ¨ 5 s 1, and obtain I Ž u. s F
1
H p R
1 p
N
< =u < p y
a
H q R
N
r kp q < a < c11 r kq y
k Ž x . < u< q y
b« k U
p
b U
p
HR
N
hŽ x . < u< p
U
U
r kp F 0
for r k sufficiently large. This proves ŽB 1 . of Proposition 9. Define
g k [ sup ugZ k u/0
žH
RN
hŽ x . < u< 5 u5
pU
1rpU
/
s sup ugZ k 5 u 5s1
žH
RN
hŽ x . < u< p
U
1rpU
/
158
YISHENG HUANG
and 1rq
u k [ sup
k Ž x . < u< q
žH
RN
5 u5
ugZ k u/0
/
1rq
s sup ugZ k 5 u 5s1
ž
HR
k Ž x . < u< q
N
/
.
Under the assumptions of this theorem we will see that
l k ª 0 and
u k ª 0 as
k ª `.
Ž 25 .
In fact, it is clear that 0 F l kq 1 F l k , so that l k ª l 0 G 0. Then for UeveryU k G 1 there exists u k g Zk such that 5 u k 5 s 1 and Ž HR N hŽ x . N u k N p .1r p ) l0r2. Since u k © 0, there exist measures m and n such that relations Ža., Žb. and Žc. of Proposition 2 hold. By the assumption that N G Ns ` and the arguments of Propositions 3 and 8, a concentration of the measure n U can only occur at 0 and `. Hence u k ª 0 in L p Ž r -N x N- R . for each 0 - r - R. Since also hUŽ0. s hŽ`. s 0, ;« ) 0 we can choose r and R so U U U Ž H< x < ) R hŽ x . N u k N p .1r p - «r3. that Ž H< x < - r hŽ x . N u k N p .1r p - «Ur3 and U It follows that Ž HR N hŽ x . N u k N p .1r p ª 0; hence l0 s 0 and l k ª 0. Since u k © 0, by the weak continuity of the functional u ¬ HR N N k Ž x . N N u N q Žcf. Proposition 1. we get u k ª 0. In order to prove ŽB 2 . we distinguish two cases. In the case p ) q, there exists R ) 0 such that < a < c11 q
1
5 u5 q F
2p
5 u5 p
for 5 u 5 G R. Then, for u g Zk , we have I Ž u. G G
1 p
5 u5 p y
1 2p U
< a < c11
5 u5 y p
q
bl kp U
p U
5 u5 q y
bl kp pU
U
5 u5 p
U
U U
5 u5 p .
Choosing r k [ Ž4 pbl kp rpU .1rŽ pyp . and using Ž25., we get r k ª ` as k ª ` and ŽB 2 .. U In the case p - q, we take r k s minŽ «rl k .1rŽ p yp ., Ž «ru k .1rŽ qyp.4 , U where « is a constant with ŽN a Nrq q brp . « - 1rŽ2 p .. It follows from Ž25. that r k ª ` as k ª `. Let u g Zk and 5 u 5 s r k ; then ŽB 2 . follows
159
MULTIPLE SOLUTIONS
from I Ž u. G
1 p
r kp y
< a
r kq y
bl k U
p
U
r kp G
1 2p
r kp .
Since we can choose r k ) r k , it follows from Proposition 9 that the conclusion of Theorem 3 holds.
ACKNOWLEDGMENT The author thanks his adviser Andrzej Szulkin for guidance and encouragement.
REFERENCES 1. J. G. Azorero and I. P. Alonso, Multiplicity of solutions for elliptic problems with critical exponent or with a nonsymmetric term, Trans. Amer. Math. Soc. 323 Ž1991., 877]895. 2. T. Bartsch and M. Willem, On an elliptic equation with concave and convex nonlinearities, Proc. Amer. Math. Soc. 123 Ž1995., 3555]3561. 3. A. K. Ben-Naoum, C. Troestler, and M. Willem, Extrema problems with critical Sobolev exponents on unbounded domains, Nonlinear Analysis, T. M. A. 26 Ž1996., 823]833. 4. C. Bessaga and A. Pełczynski, ´ ‘‘Selected Topics in Infinite-Dimensional Topology,’’ PWN, Warszawa, 1975. 5. G. Bianchi, J. Chabrowski, and A. Szulkin, On symmetric solutions of an elliptic equation with a nonlinearity involving critical Sobolev exponent, Nonlinear Analysis, T. M. A. 25 Ž1995., 41]59. 6. H. Brezis ´ and L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math. 36 Ž1983., 437]477. 7. D. M. Cao and G. B. Li, On a variational problem proposed by H. Brezis, Nonlinear ´ Analysis, T. M. A. 20 Ž1993., 1145]1156. 8. J. Chabrowski, On multiple solutions for the nonhomogeneous p-Laplacian with a critical Sobolev exponent, Diff. Int. Eq. 8 Ž1995., 705]716. 9. H. Egnell, Existence and nonexistence results for m-Laplace equations involving critical Sobolev exponents, Arch. Rat. Mech. Anal. 104 Ž1988., 57]77. 10. M. Guedda and L. Veron, Quasilinear elliptic equations involving critical Sobolev exponents, Nonlinear Analysis, T. M. A. 13 Ž1989., 879]902. 11. M. A. Krasnoselskii, ‘‘Topological Methods in the Theory of Nonlinear Integral Equations,’’ Pergamon, Elmsford, NY, 1964. 12. P. Lindqvist, On the equation divŽN =u N py 2 =u. q l N u N py 2 u s 0, Proc. Amer. Math. Soc. 109 Ž1990., 157]164. 13. P. L. Lions, The concentration-compactness principle in the calculus of variations. The limit case, Re¨ ista Mat. Iberoamer. 1 Ž1985., 145]201; 1, Ž1985., 45]120. 14. O. H. Miyagaki, On a class of semilinear elliptic problems in R N with critical growth, Nonlinear Analysis, T. M. A. 29 Ž1997., 773]781. 15. E. S. Noussair, C. A. Swanson, and J. F. Yang, Quasilinear elliptic problems with critical exponents, Nonlinear Analysis, T. M. A. 20 Ž1993., 285]301.
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16. P. H. Rabinowitz, ‘‘Minimax Methods in Critical Point Theory with Applications to Differential Equations,’’ CBMS Regional Conf. Ser. in Math. No. 65, Amer. Math. Soc., Providence, RI, 1986. 17. C. A. Swanson and L. S. Yu, Critical p-Laplacian problems in R N, Annali Mat. Pura Appl. 169 Ž1995., 233]250. 18. S. B. Tshinanga, On multiple solutions of semilinear elliptic equation on unbounded domains with concave and convex nonlinearities, Nonlinear Analysis, T. M. A. 28 Ž1997., 809]814. 19. M. Willem, ‘‘Minimax Theorems,’’ Birkhauser, Basel, 1996. ¨
On Multiple Solutions of Quasilinear Equations Involving the Critical Sobolev Exponent Yisheng Huang Department of Mathematics, Stockholm Uni¨ ersity, S-106 91 Stockholm, Sweden Submitted by R. E. Showalter Received March 2, 1998
1. INTRODUCTION In this paper we consider the existence of multiple solutions for the following quasilinear elliptic equation:
½
yD p u s a k Ž x . < u < qy 2 u q b h Ž x . < u < p
U
y2
x g RN
u,
Ž 1.
u g D 1, p Ž R N . ,
where 1 - p - N, pU [ NprŽ N y p . is the critical Sobolev exponent, 1 - q - pU , k Ž x . g Ls ŽR N . with s s pU rŽ pU y q ., hŽ x . g C ŽR N . l L`ŽR N ., a and b are real parameters, D 1, p ŽR N . is the completion of C0`ŽR N . with respect to the norm 5 u 5 s HR N N =u N p 41r p and D p u [ divŽN =u N py 2 =u. is the p-Laplacian. We say u g D 1, p ŽR N . is a Žweak. solution of Ž1. if ; ¨ g D 1, p ŽR N .
HR
N
< =u < py 2 =u ? =¨ s a
HR
N
k Ž x . < u < qy 2 u¨ q b
HR
N
hŽ x . < u< p
U
y2
u¨ .
Notice that Ž1. always has the trivial solution u s 0. Solutions of Ž1. correspond to critical points of the ‘‘energy’’ functional I : D 1, p ŽR N . ª R given by I Ž u. [
1
H p R
N
< =u < p y
a
H q R
N
k Ž x . < u< q y
b U
p
HR U
U
N
hŽ x . < u< p .
Ž 2.
Since D 1, p ŽR N . is not compactly embedded in L p ŽR N ., I does not in general satisfy the Palais]Smale condition. This causes serious difficulties in the search of critical points of I by standard variational methods. The 142 0022-247Xr99 $30.00 Copyright Q 1999 by Academic Press All rights of reproduction in any form reserved.
MULTIPLE SOLUTIONS
143
existence of nontrivial solutions to equations like Ž1. with u g D01, p Ž V ., where V is a domain in R N, has been studied since the celebrated paper of Brezis ´ and Nirenberg w6x. We refer to the recent paper w14x, the book w19x, and the references therein for the case p s 2 in both bounded and unbounded domains. When p / 2, p F q - pU , we refer to w1, 7]10x for the case of bounded V and to w3, 15, 17x for the case of V unbounded or equal to R N . When p ) 1, 1 - q - p and k Ž x . s b hŽ x . ' 1, Azorero et al. have shown in w1, theorem 4.5x that if a ) 0 is small enough and V is bounded, then there exist infinitely many solutions of Ž1. in D01, p Ž V . which satisfy I Ž u. - 0. In this paper we show a similar result for Ž1. with V s R N; see Theorem 1 for the details. It is not clear whether this result is true for large a even if the domain is bounded because of the problem with the Palais]Smale condition Žsee w1x.. On the other hand, inspired by w5x, we prove that if k Ž x . and hŽ x . satisfy suitable symmetry conditions, then the Palais]Smale condition holds for all a ) 0. Consequently, under such assumptions the result mentioned above remains true for every a ) 0 and b g R Žsee Theorem 2.; we also show the existence of infinitely many solutions for Ž1. with positive energy for every b ) 0 and a g R Žsee Theorem 3.. These two results were showed by Bartsch and Willem w2, Theorems 1 and 2x under the assumptions that V is bounded, p s 2, k Ž x . s hŽ x . ' 1 and the nonlinearities are subcritical. Recently Tshinanga w18x obtained results similar to w2x for p s 2, unbounded domains, special k Ž x ., hŽ x . and subcritical nonlinearities. Let us finally remark that in the symmetric case our result on the existence of solutions with positive energy holds both for 1 - q - p and for p - q - pU .
2. PRELIMINARIES AND THE ŽPS.c-CONDITION Throughout this paper we always denote D 1, p ŽR N . by X and weak convergence by ‘‘© .’’ Denote 5 u 5 r [ HR N N u N r 41r r and r X [ rrŽ r y 1.. For a function f, let fq Ž x . [ max 0, f Ž x .4 and f Ž`. [ lim sup < x < ª` f Ž x .. PROPOSITION 1. Ži. The functional F Ž u. [ HR N k Ž x . N u N q is well defined and weakly continuous on X. Moreo¨ er, F is continuously differentiable, its deri¨ ati¨ e F X : X ª X U is gi¨ en by
² F X Ž u . , ¨: s qH k Ž x . < u < qy 2 u¨ N R
and maps weakly con¨ ergent sequences to weakly con¨ ergent ones.
Ž 3.
144
YISHENG HUANG U
Žii. The functional G Ž u. [ HR N hŽ x . N u N p is well defined on X. Moreo¨ er, G is continuously differentiable, its deri¨ ati¨ e G X : X ª X U is gi¨ en by
² G X Ž u . , ¨: s pUH h Ž x . < u < p N
U
y2
u¨
Ž 4.
R
and maps weakly con¨ ergent sequences to weakly con¨ ergent ones. Proof. It is clear that F and G are well defined on X. In order to prove F , G g C 1 Ž X, R. it suffices to show both F and G have continuous Gateaux derivative on X. We consider only F since the proof for G is ˆ simpler. Our argument is similar to that in w19, Lemma 3.10x. Let uŽ x ., ¨ Ž x . g X and 0 -N t N- 1. Using the mean value theorem we obtain l g Ž0, 1. such that u Ž x . q t¨ Ž x .
q
y uŽ x .
q
r< t < s q u Ž x . q l t¨ Ž x .
qy1
¨ Ž x.
qy 1
F q uŽ x . q ¨ Ž x .
¨ Ž x. .
Ž 5.
By Holder’s inequality, we have ¨
HR
N
k Ž x . w < u < q < ¨
qy 1
< ¨ < F 5 k 5 s 5 < u < q < ¨ < 5 qy1 pU 5 ¨ 5 pU ,
where s s pU rŽ pU y q . s Ž pU rq .X . It follows from Ž5. and the Lebesgue dominated convergence theorem that F is Gateaux differentiable and Ž3. ˆ holds. Assume that u ª u in X. By the continuity of the embedding n U U X ¨ L p ŽR N ., uUn ª u in ULX p ŽR N .. Let us define f Ž u. [ qk Ž x . N u N qy 2 u. Ž L p ŽR N ., LŽ p . ŽR N .. Žcf. w11, p. 30x.. It follows that f Ž u n . ª Then f g C U X Ž p . ŽR N .. Using Holder’s f Ž u. in L and Sobolev’s inequalities, we get ¨
² F X Ž u n . y F X Ž u . , ¨:
F f Ž un . y f Ž u.
Ž pU .X 5 ¨ 5 pU
F c1 f Ž u n . y f Ž u .
Ž pU .X 5 ¨ 5
for all ¨ g X. Hence, 5 F X Ž u n . y F X Ž u.5 ª 0 and F g C 1 Ž X, R.. U U Let u n © u. Since u n g L p ŽR N ., NU uX n N q is bounded in L Up r q ŽR N ., Ž k xU. XN u n N qy 2 u n is bounded in LŽ p . ŽR N . and hŽ x . N u n N p y2 u n in LŽ p . ŽR N .. Hence, going if necessary to a subsequence, we can assume < u n < q © U1
in
k Ž x . < u n < qy 2 u n © U2
Lp
U
rq
ŽRN . , U X
in
LŽ p . Ž R N .
in
LŽ p . Ž R N . .
and hŽ x . < un < p
U
y2
u n © U3
U X
145
MULTIPLE SOLUTIONS
The proof of the weak convergence part will be complete if we show that U U1 sN u N q, U2 s k Ž x . N u N qy 2 u and U3 s hŽ x . N u N p y2 u. By the compact r Ž N .Ž embedding theorem, u n ª u in L loc R 1 - r - pU .. So going to a subsequence once more, we may assume that u nŽ x . ª uŽ x . a.e. in R N . Consequently, we have q
unŽ x . k Ž x . unŽ x .
qy 2
q
ª uŽ x .
a.e. in R N , qy2
un Ž x . ª k Ž x . uŽ x .
u Ž x . a.e. in R N
and hŽ x . unŽ x .
pU y2
unŽ x . ª hŽ x . uŽ x .
pU y2
u Ž x . a.e. in R N .
Therefore, U1 sN u N q, U2 s k Ž x . N u N qy2 u and U3 s hŽ x . N u N p
U
y2
u.
It follows from Proposition 1 that the functional I is well defined on X and has a continuous derivative given by
² I X Ž u . , ¨: s H < =u < py 2 =u ? =¨ y aH k Ž x . < u < qy 2 u¨ N N R
R
yb
HR
N
hŽ x . < u< p
U
y2
u¨ .
Thus any critical point of I is a Žweak. solution of equation Ž1.. Recall that I is said to satisfy the Palais]Smale condition at the level c g R ŽŽPS.c in short. if any sequence u n4 ; X such that I Ž u n . ª c and I X Ž u n . ª 0 has a convergent subsequence. In order to find critical points of I we will establish conditions under which ŽPS.c holds. We will need the following result due to P. L. Lions w13x. PROPOSITION U 2. Let u n4 ; X be a sequence such that u n © u and suppose N u n N p © n , N =u n N p © m in the sense of measures. Then there exists an at most countable index set J such that Ža. m GN =u N p q Ý j g J m j d x , m j ) 0, j U Žb. n sN u N p q Ý j g J n j d x , n j ) 0, j U Žc. m j G Sn jp r p , where x j g R N, j g J, d x j are the Dirac measures at x j and S is the best Sobole¨ constant. Recall that S [ inf
½H
RN
< =u < p ; u g X ,
HR
U
N
< u< p s 1 .
5
146
YISHENG HUANG
Hence 5 u 5 pU F Sy1 r p 5 u 5 , ; u g X .
Ž 6.
It is known from w13x that S is attained by the function u« Ž x . s K
ž
Ž Nyp .rp
« 1rŽ py1. « p rŽ py1. q < x < p rŽ py1.
, x g RN
/
as well as all translates of u« . Here « ) 0 is arbitrary and K is a constant such that 5 u« 5 s 1. Let 1 - q - p. Then
PROPOSITION 3.
Ži. ;b ) 0 there exists a 1 ) 0 such that if 0 - a - a 1 and c - 0, then I satisfies ŽPS.c . Žii. ;a ) 0 there exists b 1 ) 0 such that if 0 - b - b 1 and c - 0, then I satisfies ŽPC.c . Proof. Let u n4 be a sequence in X such that I Ž u n . ª c - 0 and
I X Ž u n . ª 0.
Ž 7.
Then for n large enough, we have c q 1 q 5 un 5 G I Ž un . y
1 U
p
² I X Ž u n . , u n: s
ž
1 p
y G
1 N
5 un 5 p y
ž
a q
y
a U
p
/
y
ž
a q
1 pU y
/
5 un 5 p
a pU
/H
RN
k Ž x . < un < q
Syq r p 5 k 5 s 5 u n 5 q .
It follows that u n4 is bounded. We can assume, going if necessary to a subsequence, that u n © u in X. There exist measures m and n such that Ža., Žb., and Žc. of Proposition 2 hold. Let x k be a singular point of the measures m and n . As in the paper w1x we define a function f g C0`ŽR N , w0, 1x. such that f Ž x . s 1 in B Ž x k , « ., f Ž x . s 0 in R N _ B Ž x k , 2 « . and N =f Ž x . NF 2r« in R N . Then f u n4 is bounded in X. Indeed, using Holder’s inequality, we have ¨
HR
< u n=f < F p
N
žH žH
F c2
BŽ xk , 2 « .
< un <
BŽ xk , 2 « .
pU
< un <
prpU
prN
/ žH /
RN
pU
prpU
,
< =f < N
/
147
MULTIPLE SOLUTIONS
where c 2 is a constant independent of « , n, and k. Hence
HR
N
=Ž u n f .
p
F2
py1
c2
žH
RN
< un <
pU
prpU
q
/
HR
N
< =u n < p ,
and this implies that f u n4 is bounded in X. It follows from Ž7. that ² I X Ž u n ., f u n : ª 0, i.e.,
a
HR
N
k Ž x . f < u< q q b
HR
s lim
H
nª` R N
N
h Ž x . f dn y
HR
N
f dm
< =u n < py 2 u n=u n ? =f .
By Holder’s inequality, ¨
lim nª`
HR
N
< =u n < py2 u n=u n ? =f F lim
žH
RN
nª`
< =u n < p
1rp
Fc 3
žH
RN
< u=f < p
/
F c4
žH
BŽ xk , 2 « .
< u< p
U
Ž py1 .rp
/
1rp
žH
RN
< u n=f < p
/
1rpU
/
ª 0 as « ª 0
Ž 8. and we deduce 0 s lim a «ª0
žH
R
N
k Ž x . f < u< q q b
HR
N
h Ž x . f dn y
HR
N
f dm
/
s b hŽ xk . n k y m Ž xk 4 . ; hence
b h Ž x k . n k s m Ž x k 4 . G mk .
Ž 9.
This inequality says that the concentration of n cannot occur at points where hŽ x k . F 0; that is, if hŽ x k . F 0, then m k s n k s 0. If hŽ x k . ) 0, we get from Žc. of Proposition 2 that either Ži. Žii.
n k s 0 or n k G S N r p w b hŽ x k .xyN r p G S N r p Ž b 5 hq 5 ` .yN r p.
148
YISHENG HUANG
In order to obtain information about possible concentration at infinity we define, as in w5x,
m` [ lim lim
H
Rª` nª` < x <)R
< =u n < p ,
H
n` [ lim lim
Ž 10 .
U
n
Ž 11 .
Now we define a function x g C`ŽR N , w0, 1x. such that x Ž x . s 1 on < x < G 2, x Ž x . s 0 on < x < F 1 and we set x R Ž x . s x Ž xrR.. As in w5x or w3x, we see that
m` s lim lim
H
Rª` nª` R N
n` s lim lim
H Rª` nª` R
< =u n < px Rp s lim lim U
N
H
Rª` nª` R N
U
< u n < p x Rp s lim lim
H Rª` nª` R
< =u n < px R , U
N
< u n < p xR
and U
m` G Sn`p r p .
Ž 12 .
Since
HR
N
=Ž u n xR .
p
F2
py1
F 2 py1
HR HR
< =u n < q p
N
N
ž
HR
< =u n < p q c5
N
pU
< un <
žH
RN
prpU
prN
/ ž /
< un < p
U
HR
N
< =x R < N
/
prpU
for some constant c5 independent of R and n, u n x R 4 is bounded in X. It follows from Ž7. that lim nª`² I X Ž u n ., u n x R : s 0, i.e., 0 s lim
nª`
HR
N
< =u n < py 2 u n=u n ? =x R y a
HR
yb
HR
U
N
N
h Ž x . < u n < p xR q
k Ž x . < u n < qx R
HR
N
< =u n < px R .
149
MULTIPLE SOLUTIONS
Notice that 1rp
lim
H nª` R
N
< =u n < py2 u n=u n ? =x R F lim 5 u n 5 py1
ž
nª`
F c6
žH
R- < x <-2 R
HR
< u<
N
pU
< u n=x R < p
/
1rpU
/
ª 0 as R ª `
Žcf. Ž8.., lim lim
H Rª` nª` R
U
h Ž x . < u n < p x R F 5 hq 5 ` n`
N
Ž 13 .
and using the weak continuity of F , lim
H nª` R
N
k Ž x . < u n < qx R F
H< x <)R
k Ž x . < u < q ª 0 as R ª `.
Therefore 0 F b 5 hq 5 ` n` y lim lim
H
Rª` nª` R N
< =u n < px R s b 5 hq 5 ` n` y m` .
Ž 14 .
Hence we obtain from Ž12. that either Žiii. n` s 0 or Živ. n` G S N r p Ž b 5 hq 5 ` .yN r p. We claim that Žii. and Živ. are impossible if a 1 or b 1 are chosen small enough. Indeed, it follows from the weak lower semicontinuity of the norm, weak continuity of F , Ž6. and Ž7. that 0 ) c s lim
nª`
H N R
nª`
1
H N R
I Ž un . y
1
s lim G
ž
N
N
1 pU
² I X Ž u n . , u n:
< =u n < p y a
< =u < p y a
ž
1 q
y
1
ž
q
1 U
p
y
/
1 pU
/
/H
R
N
k Ž x . < un < q
5 k 5 s 5 u 5 qpU G
S N
5 u 5 ppU y a A 5 u 5 qpU , Ž 15 .
where A s Ž1rq y 1rpU .5 k 5 s . This yields 5 u 5 qpU F c 7 a q rŽ pyq.
Ž 16 .
150
YISHENG HUANG
for some constant c 7 . If Živ. occurs, we obtain by Ž12., Ž15. and Ž16. that 0 ) c s lim
nª`
s lim
nª`
G G
1
ž
I Ž un . y
1
H N R
N
lim lim
N
pU
² I X Ž u n . , u n:
< =u n < p y a
H N Rª` nª` R S
1
N
ž
1 q
y
1 pU
/
/H
R
N
k Ž x . < un < q
< =u n < px R y a A 5 u 5 qpU G 1
U
n`p r p y c8 a p rŽ pyq. G
N
1 N
m` y c8 a p rŽ pyq.
S N r p 5 hq 5 `1y N r pb 1yN r p y c8 a p rŽ pyq. .
However, if b ) 0 is given, we can choose a 1 ) 0 so small that for every 0 - a - a 1 the last term on the right-hand side above is greater than 0 which is a contradiction. Similarly, if a ) 0 is given, we can take b 1 ) 0 so small that for every 0 - b - b 1 the last term on the right-hand side above is greater than 0. By a similar argument, with x R replaced by the function f introduced earlier, we see that Žii. cannot occur if a 1 or b 1 are chosen properly. Hence n k s n` s 0 for all k g J and this implies that lim
H nª` R
U
N
< u n < p s lim lim
Rª` nª`
HR
U
N
< u n < p Ž 1 y xR . q
HR
U
U
N
< u n < p xR s
HR
U
N
< u< p .
U
By the uniform convexity of L p ŽR N . we see that u n ª u in L p ŽR N .. On the other hand, it follows from Clarkson’s inequality < j 2 < p G < j 1 < p q p < j 1 < py2j 1 ? Ž j 2 y j 1 . q C Ž p .
< j2 y j1< s
Ž < j 2 < q < j 1 <.
syp
,
where j 1 , j 2 g R N , s s 2 if p g Ž1, 2. and s s p if p G 2 Žcf. w12, Lemma 4.2x., that < =u n y =u < p F c9
½ Ž < =u < n
py2
=u n y < =u < py2 =u . ? = Ž u n y u .
? Ž < =u n < p q < =u < p .
1y tr2
,
where t s p if p g Ž1, 2. and t s 2 if p G 2.
5
tr2
Ž 17 .
151
MULTIPLE SOLUTIONS
In the case 1 - p - 2, it follows from Ž17. and Holder’s inequality that ¨ pr2
5 u n y u 5 p F c9 ?
Ž < =u n < py2 =u n y < =u < py2 =u . ? =Ž u n y u .
½H žH ½H Ž ½² Ž RN
RN
F c10
1y pr2
< =u n < p q
RN
5
HR
N
< =u < p
/ pr2
< =u n < py2 =u n y < =u < py2 =u . ? = Ž u n y u .
5
I X u n . y I X Ž u . , u n y u:
s c10
qa
HR
qb
HR
N
k Ž x . Ž < u n < qy 2 u n y < u < qy 2 u . Ž u n y u .
N
hŽ x . Ž < un < p
U
y2
un y < u< p
U
y2
pr2
u. Ž un y u.
5
.
Since
HR
N
k Ž x . Ž < u n < qy 2 u n y < u < qy2 u . Ž u n y u . 1 qy 1 F 5 k 5 s Ž 5 u n 5 qy pU q 5 u 5 p U . 5 u n y u 5 pU ,
HR
N
hŽ x . Ž < un < p
U
y2
un y < u< p
F 5 h 5` Ž 5 un 5 p
U
y1
U
y2
q 5 u5 p
u. Ž un y u.
U
y1
. 5 un y u5 p
U
U
and u n © u in X, u n ª u in L p ŽR N ., it follows that u n ª u in X. The argument via Ž17. in the case p G 2 is similar but simpler. This completes the proof of Proposition 3.
3. EXISTENCE OF INFINITELY MANY SOLUTIONS Let X be the Banach space in Section 2 and let E denote the class of sets A ; X _ 04 such that A is closed in X and symmetric with respect to the origin. For A g E , recall the genus g Ž A. is defined by
g Ž A . [ min k g N; 'w g C Ž A, R k _ 0 4 . , w Ž x . s yw Ž yx . 4 .
152
YISHENG HUANG
If there is no mapping w as above for any k g N, then g Ž A. [ q`. Below we list the main properties of the genus Žcf. w16x.. Let A, B g E . Then
PROPOSITION 4.
18 If there exists an odd map f g C Ž A, B ., then g Ž A. F g Ž B .. 28 If A ; B, then g Ž A. F g Ž B .. 38 g Ž A j B . F g Ž A. q g Ž B .. 48 If S is a sphere centered at the origin in R N , then g Ž S . s N. 58 If A is compact, then g Ž A. - q` and there exists d ) 0 such that Ž Nd A. g E and g Ž Nd Ž A.. s g Ž A., where Nd Ž A. s x g X; 5 x y A 5 F d 4 . Let I be the functional defined by Ž2.. Assume that 1 - q - p and a ) 0, b ) 0. Since I Ž u. s G
1
H p R
1 p
N
< =u < p y
a
H q R
N
k Ž x . < u< q y
b U
p
HR
N
hŽ x . < u< p
U
U
5 u 5 p y a c11 5 u 5 q y b c12 5 u 5 p ,
I Ž u. G QŽ5 u 5., where QŽ t . [
1 p
U
t p y a c11 t q y b c12 t p .
It is not difficult to see that given b ) 0 there exists a 2 ) 0 so small that for every 0 - a - a 2 we can find 0 - T0 - T1 such that QŽ t . - 0 for 0 - t - T0 , QŽ t . ) 0 for T0 - t - T1 , QŽ t . - 0 for t ) T1. Similarly, given a ) 0, we can choose b 2 ) 0 with the property that T0 , T1 as above exist for each 0 - b - b 2 . Clearly, QŽT0 . s QŽT1 . s 0. Using the same idea as in w1x, we consider the truncated functional I˜Ž u . [
1
H p R
N
< =u < p y
a
H q R
N
k Ž x . < u< q y
b U
p
c Ž u.
HR
U
N
hŽ x . < u< p ,
where c Ž u. s t Ž5 u 5. and t : Rqª w0, 1x is a nonincreasing C` function such that t Ž t . s 1 if t F T0 and t Ž t . s 0 if t G T1. Thus we get
˜Ž 5 u 5 . , I˜Ž u . G Q
Ž 18 .
where
˜Ž t . [ Q
1 p
U
t p y a c11 t q y b c12 t p c Ž t . .
153
MULTIPLE SOLUTIONS
It is clear that I˜g C 1 Ž X, R. and I˜ is bounded from below. Moreover, by Ž18. and Proposition 3, we have PROPOSITION 5. Ž1. If I˜Ž u. - 0, then 5 u 5 - T0 and I˜Ž u. s I Ž u.. Ž2. ;b ) 0 there exists a 0 ) 0 Ž a 0 s min a 1 , a 2 4. such that if 0 - a - a 0 and c - 0, then I˜ satisfies ŽPS.c . Ž3. ;a ) 0 there exists b 0 ) 0 Ž b 0 s min b 1 , b 2 4. such that if 0 - b - b 0 and c - 0, then I˜ satisfies ŽPS.c . Remark 1. Denote K c [ u g X; I˜X Ž u. s 0, I˜Ž u. s c4 . If a , b are as in Ž2. or Ž3. above, then it follows from ŽPS.c that K c Ž c - 0. is compact. Denote I˜a [ u g X; I˜Ž u. F a4 . Suppose that k Ž x . ) 0 on some open subset V ; R N . Then ;k g N'« k - 0 such that g Ž I˜« k . G k. Indeed, denote by D01, p Ž V . the closure of C0`Ž V . with respect to the norm 5 u 5 s HV N =u N p 41r p. Extending functions in D01, p Ž V . by 0 outside V we can assume that D01, p Ž V . ; X Žrecall X s D 1, p ŽR N ... Let X k be a k-dimensional subspace of D01, p Ž V .. It is clear that there exists a constant d k ) 0 such that
HV k Ž x . < ¨ <
q
G dk
for all ¨ g X k with 5 ¨ 5 s 1. If u g X k , u / 0, we write u s r k¨ with 5 ¨ 5 s 1 and r k s 5 u 5. Thus, for 0 - r k - T0 , we have I˜Ž u . s I Ž u . s F F
1 p 1 p
1
H < =u < p V
5 u5 p y r kp y
a
p
y
a
H k Ž x . < u< q V
H k Ž x . < u< q V
a dk q
q
q
y
q b c12 5 u 5 p
b U
p
HV h Ž x . < u <
pU
U
U
r kq q b c12 r kp [ « k ;
therefore we can choose r k g Ž0, T0 . so small that I˜Ž u. F « k - 0. Let Sr k s u g D01, p Ž V .; 5 u 5 s r k 4 , then S r k l X k ; I˜« k . Hence, by Proposition 4, g Ž I˜« k . G g Ž Sr k l X k . s k. Thus, if we denote Tk [ A g E ; g Ž A. G k 4 and c k [ inf sup I˜Ž u . , AgGk ugA
Ž 19 .
then y` - c k F « k ) 0,
kgN
because I˜« k g Gk and I˜ is bounded from below.
Ž 20 .
154
YISHENG HUANG
PROPOSITION 6. Let a , b be as in Ž2. or Ž3. of Proposition 5 and let k Ž x . ) 0 on an open subset of R N . Then all c k Ž gi¨ en by Ž19.. are critical ¨ alues of I˜ and c k ª 0. Proof. It is clear that c k F c kq1. By Ž20., c k - 0. Hence c k ª c F 0. Moreover, since ŽPS.c is satisfied, it follows from a standard argument Žcf. ˜ w1, Lemma 4.4x or w16, Proposition 9.29x. that all c k are critical values of I. We claim that c s 0. If c - 0, then by Remark 1 K c is compact and K c g E . It follows from 58 of Proposition 4 that g Ž K c . s m - q` and there exists d ) 0 such that
g Ž K c . s g Ž Nd Ž K c . . s m. By the deformation lemma Žcf. w16, Theorem A.4x. there exist « ) 0Ž c q « - 0. and an odd homeomorphism h : X ª X such that
h Ž I˜cq « _ Nd Ž K c . . ; I˜cy « .
Ž 21 .
Since c k p c, 'k g N such that c k ) c y « and c kqm F c. There exists A g Gkq m such that sup ug A I˜Ž u. - c q « , i.e., A ; I˜cq « .
Ž 22 .
It follows from 18, 28 and 38 of Proposition 4 that
g Ž A _ Nd Ž K c . . G g Ž A . y g Ž Nd Ž K c . . G k, g Ž h Ž A _ Nd Ž K c . . . G k ; therefore h Ž A _ Nd Ž K c . . g Gk . Consequently, sup
I˜Ž u . G c k ) c y « .
Ž 23 .
ug h Ž A_Nd Ž K c ..
On the other hand, by Ž21. and Ž22.,
h Ž A _ Nd Ž K c . . ; h Ž I˜cq « _ Nd Ž K c . . ; I˜cy « which contradicts Ž23.. Hence c k ª 0. By Ž1. of Proposition 5, I˜Ž u. s I Ž u. if I˜Ž u. - 0. This and Proposition 6 give the following result: THEOREM 1. Let I Ž u. be defined as in Ž2.. Assume that 1 - q - p and k Ž x . ) 0 on an open subset of R N . Ži. ;b ) 0 there exists a 0 ) 0 such that if 0 - a - a 0 , then equation Ž1. has a sequence of solutions ¨ k 4 with I Ž ¨ k . - 0 and I Ž ¨ k . ª 0 as k ª `;
155
MULTIPLE SOLUTIONS
Žii. ;a ) 0 there exists b 0 ) 0 such that if 0 - b - b 0 , then equation Ž1. has a sequence of solutions ¨ k 4 with I Ž ¨ k . - 0 and I Ž ¨ k . ª 0 as k ª `. In the following we will discuss equation Ž1. for k Ž x . and hŽ x . satisfying some symmetry conditions. First of all, let us introduce some notation. Let O Ž N . be the group of orthogonal linear transformations in R N and let G ; O Ž N . be a subgroup. For x / 0 we denote the cardinality of Gx [ gx; g g G4 by N Gx N and set N G N[ inf x g R N , x / 0 N Gx N . We say that a function f : R N ª R is G-invariant if f Ž g x . s f Ž x . for all g g G and x g R N. We denote XG [ DG1, p ŽR N . the subspace of D 1, p ŽR N . consisting of all G-invariant functions and XGU [ DG1, p ŽR N .U its dual space. Now we consider equation Ž1. on XG , i.e.,
½
yD p u s a k Ž x . < u < qy 2 u q b h Ž x . < u < p
U
y2
u,
x g RN
u g XG ,
Ž 24 .
where 1 - p - N, 1 - q - pU , k Ž x . and hŽ x . are G-invariant functions such that hŽ x . gC ŽR N . lL`ŽR N . and k Ž x . gLs ŽR N . with sspU rŽ pU yq .. Let u g XG ; we use the same notation I Ž u. given by Ž2. for the functional associated with equation Ž24.. As in Proposition 1, I is well defined and has a continuous derivative I X : XG ª XGU given by
² I X Ž u . , ¨: s H < =u < py 2 =u ? =u y aH k Ž x . < u < qy 2 u¨ N N R
R
yb
HR
N
hŽ x . < u< p
U
y2
u¨ .
Then any critical point of I is a Žweak. solution of equation Ž1. by the following principle of symmetric criticality: PROPOSITION 7.
If I X Ž u. s 0 in XGU , then I X Ž u. s 0 in X U .
This is a special case of Theorem 1.28 in w19x. PROPOSITION 8. If 1 - q - p, N G Ns `, hŽ0. s hŽ`. s 0, then ŽPS.c holds in XG for all c g R. Proof. Let u n4 be a sequence in XG such that I Ž u n . ª c with c g R and I X Ž u n . ª 0 in XGU . As in Proposition 3 we see u n4 is bounded. There exist measures m and n such that Ža., Žb., and Žc. of Proposition 2 hold. We claim that concentration of n cannot occur at any x / 0. For if x k / 0 is a singular point of n , then n k s n Ž x k . ) 0 and by G-invariance of n , n Ž gx k . s n k ) 0 for all g g G. Since N G Ns `, the sum in Žb. of Proposition 2 is infinite. Hence n k s 0. It follows from Ž9. and hŽ0. s 0 that
156
YISHENG HUANG
n 0 s 0 Žwhere n 0 corresponds to x s 0.. So the concentration of n cannot occur at 0. Next we prove that the concentration cannot occur at infinity. Let m` and n` be as in Ž10. and Ž11.. Given « ) 0, by the definition of hŽ`. we can find R 0 Ž « . such that
H< x <)R h Ž x . < u
U
n
H< x <)R < u
n
U
for all R ) R 0 Ž « .; therefore
H hŽ x . < u Rª` nª` < x <)R lim lim
U
n
< p s 0.
Using Ž12. it follows by the same argument as in Ž13., Ž14. that U
Sn`p r p F m` s 0; U
this implies that n` s m` s 0. Hence u n ª u in LGp ŽR N . and the argument at the end of the proof of Proposition 3 shows that u k ª u in XG . Remark 2. Suppose that hŽ x . ) 0 a.e. in R N , hŽ0. s hŽ`. s 0 and N G Ns `. Let u n4 be a sequence such that I Ž u n . ª c and I X Ž u n . ª 0. If p - q - pU and b ) 0, then for n large enough, c q 1 q 5 u n 5 G I Ž u n . y ² I X Ž u n ., u n :rq G Ž1rp y 1rq .5 u n 5 p which gives the boundedness of u n4 . Consequently, the conclusion of Proposition 8 remains true in this case. Assume that k Ž x . ) 0 on an open subset V ; R N . Since k Ž x . is G-invariant, V is G-invariant, i.e., g V s V for every g g G. Let D0,1, Gp Ž V . be the subspace of D01, p Ž V . consisting of all G-invariant functions. By extending functions in D0,1, Gp Ž V . by 0 outside V we see that D0,1, Gp Ž V . ; XG . Consequently, by Proposition 8 and the same methods used in the proof of Theorem 1, we have THEOREM 2. Let k, h be G-in¨ ariant, k Ž x . ) 0 on an open subset of R N , 1 - q - p, N G Ns ` and hŽ0. s hŽ`. s 0. Then for e¨ ery a ) 0 and b g R, equation Ž1. has a sequence of solutions ¨ k 4 such that I Ž ¨ k . - 0 and I Ž ¨ k . ª 0 as k ª `. Moreover, from Proposition 8 and Remark 2 we can obtain existence of infinitely many solutions of equation Ž1. with positive energy: THEOREM 3. Let k, h be G-in¨ ariant, hŽ x . ) 0 a.e. in R N and hŽ0. s hŽ`. s 0. If p, q g Ž1, pU ., p / q and N G Ns `, then for e¨ ery b ) 0, a g R, equation Ž1. has a sequence of solutions u n4 such that I Ž u n . ª ` as k ª `. Since XG is a separable Banach space, there is a linearly independent sequence e j 4 such that XG s [`js1 X j ,
157
MULTIPLE SOLUTIONS
where X j [ R e j Žsee, e.g., w4, p. 44x.. Denote Yk [ [j F k X j and Zk [ [j G k X j. In order to prove Theorem 3 we will use the following result which can be found in w19, Theorem 3.6x: PROPOSITION 9. Let w g C 1 Ž XG , R. be an e¨ en functional satisfying ŽPS.c for e¨ ery c ) 0. If for e¨ ery k g N there exist r k ) r k ) 0 such that
Ž B1 .
a k [ max w Ž u . F 0,
Ž B2 .
bk [ inf w Ž u . ª ` as k ª `,
ugYk 5 u 5s r k ugZ k 5 u 5sr k
then w has a sequence of critical ¨ alues tending to `. Proof of Theorem 3. It is clear that the functional I associated with Ž24. is even and I g C 1 Ž XG , R.. By Proposition 8 and Remark 2, I satisfies ŽPS.c for every c g R. Since Yk is a finite dimensional subspace of XG for each k g N and hŽ x . ) 0 a.e. in R N , it follows that there exists a constant « k ) 0 such that
HR
U
N
hŽ x . < ¨ < p G « k
for all ¨ g Yk with 5 ¨ 5 s 1. If u g Yk , u / 0, we write u s r k¨ , with r k s 5 u 5 and 5 ¨ 5 s 1, and obtain I Ž u. s F
1
H p R
1 p
N
< =u < p y
a
H q R
N
r kp q < a < c11 r kq y
k Ž x . < u< q y
b« k U
p
b U
p
HR
N
hŽ x . < u< p
U
U
r kp F 0
for r k sufficiently large. This proves ŽB 1 . of Proposition 9. Define
g k [ sup ugZ k u/0
žH
RN
hŽ x . < u< 5 u5
pU
1rpU
/
s sup ugZ k 5 u 5s1
žH
RN
hŽ x . < u< p
U
1rpU
/
158
YISHENG HUANG
and 1rq
u k [ sup
k Ž x . < u< q
žH
RN
5 u5
ugZ k u/0
/
1rq
s sup ugZ k 5 u 5s1
ž
HR
k Ž x . < u< q
N
/
.
Under the assumptions of this theorem we will see that
l k ª 0 and
u k ª 0 as
k ª `.
Ž 25 .
In fact, it is clear that 0 F l kq 1 F l k , so that l k ª l 0 G 0. Then for UeveryU k G 1 there exists u k g Zk such that 5 u k 5 s 1 and Ž HR N hŽ x . N u k N p .1r p ) l0r2. Since u k © 0, there exist measures m and n such that relations Ža., Žb. and Žc. of Proposition 2 hold. By the assumption that N G Ns ` and the arguments of Propositions 3 and 8, a concentration of the measure n U can only occur at 0 and `. Hence u k ª 0 in L p Ž r -N x N- R . for each 0 - r - R. Since also hUŽ0. s hŽ`. s 0, ;« ) 0 we can choose r and R so U U U Ž H< x < ) R hŽ x . N u k N p .1r p - «r3. that Ž H< x < - r hŽ x . N u k N p .1r p - «Ur3 and U It follows that Ž HR N hŽ x . N u k N p .1r p ª 0; hence l0 s 0 and l k ª 0. Since u k © 0, by the weak continuity of the functional u ¬ HR N N k Ž x . N N u N q Žcf. Proposition 1. we get u k ª 0. In order to prove ŽB 2 . we distinguish two cases. In the case p ) q, there exists R ) 0 such that < a < c11 q
1
5 u5 q F
2p
5 u5 p
for 5 u 5 G R. Then, for u g Zk , we have I Ž u. G G
1 p
5 u5 p y
1 2p U
< a < c11
5 u5 y p
q
bl kp U
p U
5 u5 q y
bl kp pU
U
5 u5 p
U
U U
5 u5 p .
Choosing r k [ Ž4 pbl kp rpU .1rŽ pyp . and using Ž25., we get r k ª ` as k ª ` and ŽB 2 .. U In the case p - q, we take r k s minŽ «rl k .1rŽ p yp ., Ž «ru k .1rŽ qyp.4 , U where « is a constant with ŽN a Nrq q brp . « - 1rŽ2 p .. It follows from Ž25. that r k ª ` as k ª `. Let u g Zk and 5 u 5 s r k ; then ŽB 2 . follows
159
MULTIPLE SOLUTIONS
from I Ž u. G
1 p
r kp y
< a
r kq y
bl k U
p
U
r kp G
1 2p
r kp .
Since we can choose r k ) r k , it follows from Proposition 9 that the conclusion of Theorem 3 holds.
ACKNOWLEDGMENT The author thanks his adviser Andrzej Szulkin for guidance and encouragement.
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