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Nonsymmetric solutions for some variational problems
Nonlinear Analysis 44 (2001) 705 – 725
Nonsymmetric solutions for some variational problems Lionel Paumond Analyse num erique et EDP, Universit e de Paris-Sud, Bat. ˆ 425, 91405 Orsay Cedex, France Received 14 January 1999; accepted 27 March 1999
Keywords: Kadomtsev–Petviashili equations; Solitary waves; Symmetry; Minimax methods
1. Introduction In this paper we try to -nd nonsymmetric solitary waves for some variational problems. First of all Kadomtsev–Petviashvili equations (KP) are “universal” models for dispersive, weakly nonlinear waves, which are essentialy unidimensional, with weak transverse e5ect. For instance, the works [4,5] of de Bouard and Saut deals with a special case of these equations (KPI), ut + up · ux + uxxx − vy = 0;
u = u(x; y; t);
(x; y) ∈ R2 ; t ¿ 0;
vx = uy ;
(1.1)
in the two-dimensional case, and ut + up · ux + uxxx − vy − wz = 0;
u = u(x; y; z; t);
(x; y; z) ∈ R3 ; t ¿ 0;
vx = uy ;
(1.2)
wx = uz ; in the three-dimensional case, with p = m=n ¿ 0 with m and n relatively prime, n odd. They have shown in [4] the existence of solitary waves for (1.1) and(1.2) where x is the direction of propagation. Next, they have proved in [5] that any ground-state solution of (1.1) and (1.2) is cylindrically symmetric, in the sense that it has radial symmetry with respect to transverse coordinates, up to translation of the origin. E-mail address: [email protected] (L. Paumond). 0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 3 0 1 - 6
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It is legitimate to wonder whether all solitary waves solutions are so. In Sections 3 and 4 we establish that the answer is no for two suitably modi-ed KPI equations in R5 . Thus, we -rst consider the following problem: ut + up · ux1 + @7x1 u − vx2 − wx3 − x4 − x5 = 0; vx1 = ux2 ; (1.3)
wx1 = ux3 ; x1 = ux4 ; x1 = ux5 ;
where p = m=n ¿ 0 with m and n relatively prime, n odd, and with u = u(x1 ; x2 ; x3 ; x4 ; x5 ; t) t ≥ 0 (x1 ; x2 ; x3 ; x4 ; x5 ) ∈ R5 . The equations studied in Section 4 are the same except for a potential in the nonlinearity but the method is somewhat di5erent. Classically we introduce (see [4]): Denition 1.1. Y is the closure of @x1 (C0∞ (R5 )) for the norm 2
2
@x1 Y = (∇L2 (R5 ) + @4x1 L2 (R5 ) )1=2 if ∈ @x1 (C0∞ (R5 )). Remark 1.1. For every u in Y there exists in L10=3 (R5 ) such that u = @x1 . Y is a Hilbert space with the scalar product induced by :Y . The solutions of (1.3) we are interested in are solitary waves of the form u = u(x1 − ct; x2 ; x3 ; x4 ; x5 );
c ¿ 0;
u ∈ Y:
Thus problem (1.3) becomes −c · ux1 + up · ux1 + @7x1 u − vx2 − wx3 − x4 − x5 = 0; vx1 = ux2 ; wx1 = ux3 ;
(1.4)
x1 = ux4 ; x1 = ux5 : Remark 1.2. Note that we may from now on assume that c = 1 by the scale change: u(x ˜ 1 ; x ) = c−1=p · u(x1 =c1=6 ; x =c2=3 ) where x = (x2 ; x3 ; x4 ; x5 ) ∈ R4 . The minimax method we will apply in Section 3 to -nd a solution of (1.4) which is not radial in the transverse direction is an adaptation of that used by Bartsch and Willem in [2]. The key argument is that symmetry allows to recover compactness as it is expounded in [7]. But for topological reasons the used of symmetry seems to be
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more diIcult when transverse coordinates are less than 4. It explains our choice of R5 . Moreover solitary waves existence for (1.4) needs the term @7x1 u. This order is given by Sobolev inequalities in anisotropic Sobolev Spaces (see [3, Section 2]) used when we apply a minimizing method. Finally, consider the following problem: −Ku + b(|x|)u = f(|x|; u);
x ∈ RN ;
(1.5)
u ∈ H 1 (RN ):
The interest of this equation originates from various problems in physics and mathematical physics; for instance, cosmology or constructive -eld theory where (1.5) is called a nonlinear Euclidean scalar -eld equation. A solution of (1.5) can also be interpreted as a kind of solitary waves of nonlinear Klein–Gordon or SchrModinger equation. Bartsch and Willem prove the existence of in-nitely many nonradial solutions of this equation provided some hypotheses on f and b for N = 4 or N ≥ 6. In fact, the modi-cation of the method done in Section 3 allows to obtain the result for N = 5 in the autonomous case where b = 1 and f(|x|; u) = |u|p u (see Section 5). 2. A solution for (1:4) In order to frame problem (1.4) in a variational form, let us consider the fonctional N such that N:
Y →R 1 1 2 p+2 u → · uY − · up+2 : 2 (p + 1)(p + 2)
We can also write 1 2 (u + (@3x1 u)2 N(u) = R5 2 +(Dx−1 ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 u x 5 )2 ) − 1 1 1 1
up+2 d x; (p + 1)(p + 2)
where Dx−1 uxi for i ∈ {1; 2; 3; 4; 5} is the element of Y (the dual of Y in the L2 -duality) 1 such that for any ∈ Y , Dx−1 uxi ; Y ; Y = (u; Dx−1 xi ). 1 1 Remark 2.1. It is easy to check that N ∈ C 2 (Y; R) (see for example [9]) and that weak solutions for (1.4) are critical points for N on Y . The following Lemmas are well known in two- and three-dimensional cases (see for example [4]). Lemma 2.1. There exists C ¿ 0 such that uLq (R5 ) ≤ CuY for any q ∈ [2; 34=11]. Proof. It suIces apply a result of Besov et al. [3, Theorem 15:6, p. 323].
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Remark 2.2. Lemma 2.1 remains valid if we take an open subset P in place of R5 (see [3]). Lemma 2.2. The imbedding Y ,→ Lploc (R5 ) is compact for p ∈ [2; 34=11[: Proof. We follow the same idea as in [4]. Let (un )n be a bounded sequence in Y , so by Sobolev inequality there exists n ∈ L10=3 (R5 ) such that un = @x1 n with @xi n ∈ L2 (R5 ) for i ∈ {1; : : : ; 5}. For now on we will suppose Supp(n ) ⊂ B(O; 2R) since we can multiply n by ∈ D(R5 ) such that 0 ≤ ≤ 1, ≡ 1 on B(O; R) and Supp( ) ⊂ B(O; 2R). Extracting if necessary a subsequence we can assume that un * u = @x1 in Y and replacing n by n − , we may assume that = 0. Thus by Plancherel theorem we have |un |2 = |un |2 ; R5
B(O;2R)
where
u(%) ˆ =
R5
e−2i&x; % u(x) d x
with % = (%1 ; %2 ; %3 ; %4 ; %5 ); x = (x1 ; x2 ; x3 ; x4 ; x5 ) and x; % = %1 x1 + %2 x2 + %3 x3 + %4 x4 + %5 x5 . Let us consider the inclusion R5 ⊂{|%1 | ¿ R1 } ∪ {|%1 | ¡ R1 ; |%i | ¡ R21 ; i ∈ {2; : : : ; 5}} ∪
5
{|%1 | ¡ R1 ; |%i | ¿ R21 }:
i=2
First, we have |un |2 ≤ |%1 |¿R1
|%1 |¿R1
1 ≤ 42 R21 ≤
1 2 |@[ x un | 42 %21 1
R5
2 |@[ x 1 un |
1 2 @x un 2 2 : 42 R21 1 L (R )
Let us now consider the integral 2 |un | = |%1 |¡R1 ;|%2 |¿R21
|%1 |¡R1 ;|%2 |¿R21
≤
1 2 vn L2 (R5 ) : R21
(2.1)
%21 |vn |2 %22 (2.2)
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In the same way, we obtain 1 2 |un |2 ≤ 2 wn L2 (R5 ) ; 2 R1 |%1 |¡R1 ;|%3 |¿R1 |%1 |¡R1 ;|%4 |¿R21
|%1 |¡R1 ;|%5 |¿R21
709
(2.3)
|un |2 ≤
1 2 n L2 (R5 ) ; R21
(2.4)
|un |2 ≤
1 2 n L2 (R5 ) : R21
(2.5)
Thus for all ( we can choose R1 suIciently large to have 5 |un |2 + |un |2 ¡ (: |%1 |¿R1
i=2
|%1 |¡R1 ;|%i |¿R21
Now it remains to evaluate
|%1 |¡R1 ;|%i |¡R21 ; i ∈ {2;:::;5} have un * 0 in L2 (R5 ),
Applying Lemma 2.1 we un (x)e−2i&x; % d x → 0: un =
(2.6)
|un |2 . thus:
n→∞
B(O; 2R)
But by a classical majoration and Lemma 2.1 we obtain |un (%)| ≤ C0 un L2 (R5 ) ≤ C1 ;
∀% ∈ {|%1 | ¡ R1 ; |%i | ¡ R21 ; i ∈ {2; : : : ; 5}}:
We can then apply Lebesgue’s dominated convergence theorem, so |un |2 → 0: |%1 |¡R1 ;|%i |¡R21 ; i ∈ {2;:::;5}
n→∞
To conclude it suIces to remark that 5 |un |2 ≤ |un |2 + R5
|%1 |¿R1
+
i=2
|%1 |¡R1 ;|%i |¿R21
|%1 |¡R1 ;|%i |¡R21 ; i ∈ {2;:::;5}
(2.7)
|un |2
|un |2 :
(2.8)
By (2.1) – (2.8), we obtain un → 0 in L2loc (R5 ). n→∞
The conclusion follows by interpolation using Lemma 2.1. We are now able to give a -rst result: Theorem 2.1. System (1:4) possesses a solution u ∈ Y and u = 0; for p ∈ 0;
12 11
.
Proof. In the spirit of [4] (Theorems 3.1 and 3:2) we consider the constrained minimization problem
2 p+2 u d x d x = * ; * ¿ 0; I* = inf uY ; u ∈ Y; with R5
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which is solved using an application of the concentration–compactness principle of Lions [8]. 3. A solution for (1:4) which is not radial in the transverse direction 3.1. Functional setting As in [9] we shall de-ne a function space X containing no radial function except zero, and then will prove that N has a critical point on X . We -rst consider the group G = {id} × O(2) × O(2) and its action on Y de-ned by ∗:
G×Y →Y (g; u) → g ∗ u(x) = u(g−1 · x):
We get YG = {u ∈ Y=∀g ∈ G g ∗ u = u}. Let / be the involution de-ned on R5 = R × R2 × R2 by /(x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; x4 ; x5 ; x2 ; x3 ). The group H = {id; /} acts on YG as follows: H × YG → YG #:
(h; u) → h # u(x) =
u(x)
if h = id;
−u(h−1 · x) if h = /:
Finally, we denote X = {u ∈ YG =∀u ∈ H h # u = u}. Proposition 3.1. (X; ·; · Y ) is a Hilbert space which is not reduced to {0} and 0 is the only radial function of X . Proof. It is obvious that X is a closed subspace of Y . To construct a nonzero function ∞ 5 in 0 (R ) − {0} such that (x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; r; r ) where r =
X we take ∈ C x22 + x32 and r = x42 + x52 with the following condition: (x1 ; r; r ) = − (x1 ; r ; r). Then u = @x1 is the desired nonzero element of X . The second part of the proposition is left to the reader. 3.2. Some technical lemmas We shall start with a compactness lemma whose proof needs the following variant of a well-known result (see for example [8]). Lemma 3.1. Let (un )n be a bounded sequence in Y such that |un |2 → 0: sup x ∈ R4
Then
B(x1 ; r1 )
B(x1 ; r1 ) R4
B(x ; r )
n→∞
|un |p → 0 for 2 ¡ p ¡ 34 11 : n→∞
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Proof. Let s ∈ ]2; 34=11[ and u ∈ Y . By HMolder inequality we obtain 1=s (1−*)=2 11*=34 2 34=11 s |u| ≤ |u| · |u| V
V
V
with 1=s = (1 − *)=2 + 11*=34 so * = (17s − 34)=6s and where V = B(x1 ; r1 ) × B(x ; r ). Lemma 2.1 implies that (1−*)=2 1=s * |u|2 · uY (V ) ; |u|s ≤C · V
V
where *
uY (V ) =
V
[(@3x1 u)2 + u2 + (Dx−1 ux2 )2 + (Dx−1 u x 3 )2 1 1 1=2
+ (Dx−1 u x 4 )2 1
+
(Dx−1 u x 5 )2 ] 1
Taking * = 2=s, we obtain ((1−*)s)=2 2 s s 2 |u| ≤ C · |u| · uY (V ) : V
(3.1)
V
Now covering R4 by balls B(xi ; r ) in such a way that each point of R4 is contained in at most -ve balls, we -nd s |u| = |u|s B(x1 ; r1 )
R4
∪i B(xi ; r )
B(x1 ; r1 )
= ≤
∪i B(xi ; r )
i
≤C
i
B(x1 ; r1 )
Vi
s
sup
x ∈ R4
But we have 2 uY (Vi ) = i
Cs ·
|u|s
B(xi ; r )
i
≤
B(x1 ; r1 )
|u|2
((1−*)s)=2
+ (Dx−1 u x 2 )2 1
by (3:1)
((1−*)s)=2 2 |u| · uY (Vi ) :
V
B(x1 ; r1 )
+
2
· uY (Vi )
2
B(xi ; r )
|u|s
i
[(@3x1 u)2 + u2
(Dx−1 u x 3 )2 1
+
(Dx−1 u x 4 )2 1
+
(Dx−1 u x 5 )2 ] 1
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≤ 5·
R4
B(x1 ; r1 )
[(@3x1 u)2 + u2
+ (Dx−1 ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 u x 5 )2 ] 1 1 1 1 ≤ 5·
R5
[(@3x1 u)2 + u2
ux4 )2 + (Dx−1 ux5 )2 ] + (Dx−1 ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 1 1 1 1 2
≤ 5 · uY : Finally, we obtain ((1−*)s)=2 2 2 s s sup |u| · uY : |u| ≤ 5C B(x1 ; r1 )
R4
x ∈ R4
V
As s is -xed in ]2; 34=11[ we conclude by interpolating between s and 34=11. As in [9] we de-ne: Denition 3.1. For y ∈ R5 ; and r ¿ 0; let m(y; r; G) = sup{n ∈ N; ∃g1 ; : : : ; gn ∈ G=j = k ⇒ B(gj y; r) ∩ B(gk y; r) = ∅}. Denition 3.2. R5 is compatible with G if for any r ¿ 0 lim|y|→+∞ m(y; r; G) = +∞: Remark 3.1. In our case the diIculty arises from the fact that R5 is not compatible with the group G we have de-ned before. Indeed, if y = (x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; x ), we can -x x and make |x1 | → +∞ with |y| → +∞ and m(y; r; G) -xed. It is then obvious to check that m(y; r; G) → +∞ for any x1 ∈ R5 . |x |→+∞
Now we can prove the following lemma. Lemma 3.2. The imbedding YG ,→ Lploc; x1 (R5 ) is compact for p ∈ ]2; 34=11[. Proof. Let (un )n be a bounded sequence in YG ; and V = B(x1 ; r1 ) × B(x ; r ) be a neighbourhood of y = (x1 ; x ). Then there exists a ball of radius r such that V ⊂ B(y; r). Owing to Remark 3.1 we have 2 supn un 2 |un |2 ≤ |un |2 ≤ : m(y; r; G) V B(y; r)
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But the imbedding YG ,→ Y ,→ L2 (R5 ) are continuous, (see Lemma 2.1), and supn 2 un 2 = C ¡ + ∞ which yields: V |un |2 ≤ C=m(y; r; G). Let ( ¿ 0 and x1 ∈ R be -xed. By Remark 3.1 we can -nd R ¿ 0 such that |un |2 ¡ (: for all n ∈ N sup |x | ≥ R
B(x1 ; r1 )
B(x ; r )
Moreover, the imbedding YG ,→ L2loc (R5 ) is compact and so 2 |un |2 → 0: |un | = B(0; R +r )
This implies sup
|x | ≤ R
B(x1 ; r1 )
B(x1 ; r1 )
B(x ; r )
B(x1 ; r1 )
Finally, we obtain ∃N=∀n ≥ N
x ∈ R4
n→∞
|un |2 →n→∞ 0:
sup
B(0; R +r )
B(x1 ; r1 )
B(x ; r )
|un |2 ¡ 2(:
We conclude by applying Lemma 3.1. We have the obvious corollary: Corollary 3.1. The imbedding X ,→ Lploc; x1 (R5 ) is compact for p ∈ ]2; 34=11[. Let us give another Lemma which permits to recover compactness. Lemma 3.3. Let un be a bounded sequence in Y such that y+r sup |un |6 → 0 for 6 ∈ ]2; 34=11[ y∈R
y−r
n→∞
R4
Then un Lq (R5 ) → 0 for q ∈ ]2; 34=11[. n→∞
Proof. Let 2 ¡ s ¡ 34=11 and u ∈ X . HMolder inequality implies that 1−*
*
uLs (P) ≤ uL6 (P) · uL34=11 (P) with P = ]y − r; y + r[ × R4 and 1=s = 11*=34 + (1 − *)=6. Lemma 2.1 gives 1−*
*
uLs (P) ≤ CuL6 (P) · uY (P) : As in the proof of Lemma 3.2 we take * = 2=s ∈ ]0; 1[. So we obtain (s(1−*))=6 2 s s 6 |u| ≤ C · |u| · yY (P) : P
P
(3.2)
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Consider a covering R of R5 by open sets P(y) = ]y − r; y + r[ × R4 in such a way that each point of R5 is contained in at most two open sets. We have |u|s ≤ |u|s |u|s = R5
∪y ∈ R P(y)
P(y)
y∈R
≤ Cs
y∈R
Finally, R5
s
|u| ≤ C ·
(s(1−*))=6
s
sup
y∈R
P(y)
6
P(y)
|u|
·
|u|6
(s(1−*))=6 ·
P(y)
y∈R
2 uY (P(y))
:
by (3:2)
[(@3x1 u)2 + u2
ux5 )2 ]: ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 + (Dx−1 ux2 )2 + (Dx−1 1 1 1 1 By the choice of the covering we obtain (s(1−*))=6 6 s s |u| |u| ≤ 2 · C · sup · y∈R
R5
R5
P(y)
[(@3x1 u)2 + u2
ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 u x 5 )2 ] + (Dx−1 1 1 1 1 s
= 2·C ·
sup
y∈R
P(y)
(s(1−*))=6 2
6
|u|
· uY :
Since s is -xed in ]2; 34=11[ we conclude by interpolation between s and 34=11. 3.3. The 8rst result Theorem 3.1. (1:4) possesses a solution in X − {0}; for p ∈ 0;
12 11
:
Proof. Observe -rst that N is unbounded from below. Next, N(u) ≥
1 C p+2 2 p+2 · uY − · uY ¿ 0 2 (p + 1)(p + 2) 2
if 0 ¡ uY ¡
(p + 1)(p + 2) 2C p+2
1=p
and there exists r ¿ 0 such that inf u = r; u ∈ X N(u) ¿ 0. Finally, let v0 ∈ X − {0}. Then *v0 ∈ X and there exists * ¿ 0 such that *v0 ¿ r and 0 = N(0) ≥ N(e). Let us de-ne
=e
d = inf max N(8(t)) where U = {8 ∈ C([0; 1]; X ); 8(0) = 0; 8(1) = e}: 8 ∈ U t ∈ [0;1]
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If 8 ∈ U there exists t ∈ [0; 1] such that 8(t) = r, thus b ≤ d. Moreover by Theorem 2:8 in [9] there exists (un )n , a Palais Smale sequence for N in X such that N(un ) → d; n→+∞
N (un ) → 0: n→+∞
Now we will prove that (un )n is bounded in X . Indeed 1 1 1 2 · un Y ≥ 0: N(un ) − N (un ); un = − p+2 2 p+2 Moreover 1 1 N (un ); un ≤ N(un ) + N (un )X · un X : N(un ) − p+2 p+2 By N (un )X →n→+∞ 0 we deduce 1 ∃N=∀n ≥ N N(un ) − N (un ); un ≤ d + 1 + un X : p+2 2
Thus ( 12 − 1=(p + 2)) · un Y ≤ d + 1 + un X , and (un )n is bounded. Suppose by contradiction that y+1 9 = lim sup |un |6 = 0 n→∞ y ∈ R
y−1
R4
with 6 ∈ ]2; 34=11[. Then Lemma 3.3 yields un Lp+2 → 0 with p + 2 ∈ ]2; 34=11[, and n→∞
we get 0 ¡ d = N(un ) − 12 · N (un ); un + o(1). Moreover, 1 1 N(un ) − · N (un ); un = − |un |p+2 (p + 1)(p + 2) R5 2 1 + |un |p+2 2(p + 2) R5 1 1 1 · = |un |p+2 : − 2 p+2 p + 1 R5 We obtain N(un ) −
1 2
· N (un ); un → 0: n→∞
Thus d=0 which is a contradiction, so 9 ¿ 0. Extracting if necessary a subsequence, there exists (yn )n ∈ RN such that yn +1 |un |6 ¿ 9: yn −1
R4
We can now de-ne vn (x1 ; x ) = un (x1 + yn ; x ). The basic fact is that vn still belongs to X . Indeed the actions used to de-ne X , modify only the transverse variables. Thus, we obtain +1 |vn |6 ¿ 9: (3.3) −1
R4
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Moreover, (vn )n is bounded so, after extraction of a subsequence we can assume vn *m→∞ v
weakly in X
But Corollary 3.1 gives vn → v n→∞
in L6loc; x1 (R5 ):
(3.4)
Note that v = 0 by (3.3) and (3.4). On the other hand, for every w in X we have N (v); w = lim N (vn ); w n→∞
= lim N (vn ); wn : n→∞
But we have | N (vn ); wn | ≤ N (un )X · wn X and wn X = wX . Moreover N (un )X → 0 thus N (v) = 0. n→∞
4. “KP” with potential 4.1. Introduction This section is devoted to the problem: up+1 + @7x1 u − vx2 − wx3 − x4 − x5 = 0; ux1 + Q(x) · p + 1 x1 vx1 = ux2 ; wx1 = ux3 ;
(4.1)
x1 = ux4 ; x1 = ux5 with Q ∈ C(R5 ); 1 = lim|x|→+∞ Q(x) = inf R5 Q(x): Our aim is here to -nd a nonradial transverse solution to (4.1). We will use again the mountain pass theorem. The diIculty is that the Palais–Smale condition (P.S.) is 2 not always satis-ed. Let denote Sp = inf uLp+2 =1;u ∈ X uY where p + 2 ∈ ]2; 34=11[. We will solve this minimization problem by a method similar to that of Willem in [9] for the SchrModinger equation. The diIculty is here to -nd a minimizer in X . Let now consider the functional V : X → R, 1 2 u x 2 )2 V(u) = (u + (@3x1 u)2 + (Dx−1 1 R5 2 Q(x) −1 2 −1 2 −1 2 p+2 d x: + (Dx1 ux3 ) + (Dx1 ux4 ) + (Dx1 ux5 ) ) − ·u p+2 Weak solutions for (4.1) in X are critical points for V on X .
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4.2. The critical value 2 Theorem 4.1. There exists v ∈ X such that vY = Sp and vLp+2 = 1; for p ∈ 0;
12 11
:
Proof. Let (un )n be a minimizing sequence for Sp : un Lp+2 = 1;
∀n;
2 un Y → Sp : n→∞
By Lemma 3.3 we can -nd 9 ¿ 0 and (yn )n ∈ RN such that yn +1 |un |6 ¿ 9 ¿ 0: R4
yn −1
Let us set vn (x1 ; x2 ; x3 ; x4 ; x5 ) = un (x1 + yn ; x2 ; x3 ; x4 ; x5 ). Note that vn ∈ X because of the action of G and H . Obviously, +1 (4.2) |vn |6 ¿ 9: −1
R4
vn Lp+2 = un Lp+2 = 1; vn Y = un Y → Sp : n→∞
The sequence (vn )n is bounded in X and since X ,→ L6loc; x1 (R5 ) with compact imbedding for p ∈ ]2; 34=11[ we have after extracting a subsequence: vn * v
weakly in X;
(4.3)
vn → v
in L6loc; x1 (R5 ):
(4.4)
vn → v
almost everywhere in R5 :
(4.5)
By (4.2) and (4.4), v ∈ X − {0}. Moreover, vn Lp+2 = 1 together with (4.5) make possible the use of Brezis–Lieb Lemma [6]. Setting wn = vn − v, we have p+2
p+2
1 = vLp+2 + lim wn Lp+2
(4.6)
n→∞
and 2
2
2
Sq = lim vn Y = vY + lim vn − vY : n→∞
n→∞
(4.7)
By de-nition of Sp we have 2
2
vY ≥ Sp · vLp+2 :
(4.8)
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L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
Thus, p+2 p+2 Sp ≥ Sp · (vLp+2 )2=(p+2) + lim (wn Lp+2 )2=(p+2) n→∞
p+2
p+2
≥ Sp · [(vLp+2 )2=(p+2) + (1 − vLp+2 )2=(p+2) ]
by (4:6):
But since vn Lp+2 = 1 and vn * v weakly in X , we have 0 ≤ vLp+2 ≤ 1. Thus (4.6) yields vLp+2 = 1: 2
2
2
Moreover, vX ≤ limvn X = Sp , and we obtain vX ≥ Sp proving that v is a minimizer. Let us denote un (x1 ; x2 ; x3 ; x4 ; x5 ) = v(x1 + an ; x2 ; x3 ; x4 ; x5 ) where (an )n is a real sequence such that |an | → +∞. n→∞
By the de-nition of V we have 1 2 V((p + 1)1=p Sp1=p un ) = (p + 1)1=p Sp1=p un X 2 1 − Q(x)(p + 1)(p+2)=p Sp(p+2)=p unp+2 (p + 1)(p + 2) R5 1 2 = (p + 1)2=p Sp2=p un X 2 (p+2)=p 2=p Sp − (p + 1) Q(x)unp+2 (p + 2) R5 1 2 = (p + 1)2=p Sp2=p un X 2 Sp(p+2)=p − (p + 1)2=p Q(x1 − an ; x )vp+2 : (p + 2) R5 By Lebesgue’s Theorem
R5
Since
Q(x1 − an ; x )v
R5
p+2
→
n→∞
R5
vp+2 :
2
vp+2 = 1 and vX = Sp we deduce
V((p +
1)1=p Sp1=p un )
→
n→+∞
1 1 − 2 p+2
Sp(p+2)=p (p + 1)2=p :
(4.9)
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719
Let us now consider for h ∈ X , V (p + 1)1=p Sp1=p un ; h X ; X = (p + 1)1=p Sp1=p un ; h X; X −
1 (p + 1)
R5
Q(x)(p + 1)(p+1)=p Sp(p+1)=p unp+1 h
= (p + 1)1=p Sp1=p v; h(x1 − an ; x ) X; X −
1 (p + 1)
R5
Fn (x1 ; x ) d x1 d x ;
where Fn (x1 ; x ) = Q(x1 − an ; x )(p + 1)(p+1)=p Sp(p+1)=p vp+1 h(x1 − an ; x ). Finally, we arrive at V ((p + 1)1=p Sp1=p un ) → 0:
(4.10)
n→+∞
Then, by (4.9) and (4.10), the Palais–Smale condition is not veri-ed at the level c = ( 12 − 1=(p + 2))Sp(p+2)=p (p + 1)2=p . ∗
4.3. Application of the mountain pass theorem We -rst prove that every Palais–Smale sequence is relatively compact in X . Lemma 4.1. Let us consider (un )n ∈ X N a Palais–Smale sequence i.e. V(un ) → d ¡ c∗ ; n→∞
V (un ) → 0: n→∞
Then (un )n countains a convergent subsequence. Proof. The fact that (un ) is bounded in X results from the identity 1 1 1 2 un Y (see Theorem 3:1): V(un ) − V (un ); un = − p+2 2 p+2 One can therefore assume that, up to a subsequence un *n→∞ u weakly in X and by Lemma 3.2 un → u
5 strongly in Lp+2 loc; x1 (R ) for + 2 ∈ ]2; 34=11[
un → u
a:e: in R5 :
n→∞
and n→∞
Brezis–Lieb Lemma yields Q(x)|un |p+2 − Q(x)|un − u|p+2 = R5
R5
Let us denote vn = un − u
R5
Q(x)|u|p+2 + o(1):
(4.11)
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L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
We have
1 1 2 V(un ) = un Y − (p + 1)(p + 2) 2 1 1 2 = un Y − 2 (p + 1)(p + 2) + Q(x)|u|p+2 + o(1) R5
1 1 2 = un Y − (p + 1)(p + 2) 2 1 2 +V(u) − uY + o(1): 2
Since
2 1 2 un Y
2
R5
R5
Q(x)|un |p+2
(4.12)
Q(x)|un − u|p+2 (4.13)
R5
Q(x)|vn |p+2 (4.14)
2
− 12 uY = 12 vn Y + o(1); 2 1 2 vn Y
(4.15)
we obtain V(un ) = V(u) + − 1=((p + 1)(p + 2)) R5 Q(x)|vn |p+2 + o(1) and we deduce 1 1 2 V(u) + vn Y − Q(x)|vn |p+2 → d: (4.16) n→∞ 2 (p + 1)(p + 2) R5 On the other hand, 1 1 2 2 p+2 vn Y − Q(x)|vn | = vn Y − Q(x)unp+2 p + 1 R5 p + 1 R5 1 + Q(x)|u|p+2 + o(1) (4.17) p + 1 R5 1 2 2 = un Y − uY − Q(x)unp+2 p + 1 R5 1 + Q(x)|u|p+2 + o(1) by (4:15) p + 1 R5 = V (un ); un − V (u); u + o(1): But | V (un ; un Y ; Y | ≤ V (un )Y u − nY ≤ M V (un )Y → 0 and V (u); u Y ; Y = 0. Thus we obtain -nally 1 2 vn Y − Q(x)|vn |p+2 → 0: n→∞ p + 1 R5 Since (un )n is bounded in X , so is (vn )n and (4.18) implies 2
vn Y → b; n→∞ 1 Q(x)|vn |p+2 → b: n→∞ p + 1 R5
n→∞
(4.18)
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
Now, we have p+2 (1 − Q(x))|vn | = R5
B(O; R)
p+2
(1 − Q(x))|vn |
+
R5 −B(O;R)
721
(1 − Q(x))|vn |p+2 :
Moreover, Lemma 2.1 yields p+2 (1 − Q(x))|vn |p+2 ≤ (vn Lp+2 (R5 ) ≤ C( R5 −B(O;R)
for R suIciently large. On the other hand, we can write (1 − Q(x))|vn |p+2 ≤ C B(O; R)
B(O; R)
|vn |p+2 → 0 n→∞
because of (4.11). Thus, we have 2
vn Y → b; n→∞ 1 |vn |p+2 → b: n→∞ p + 1 R5 2
(4.19) 2
Now by de-nition of Sp , vn Y ≥ Sp vn Lp+2 (R5 ) and b ≥ Sp [b(p + 1)]2=(p+2) . If b = 0, we conclude by (4.19). If b = 0 then b ≥ Sp(p+2)=p (p + 1)2=p . Now, 1 1 1 1 c∗ = Sp(p+2)=p (p + 1)2=p ≤ b: − − 2 p+2 2 p+2
(4.20)
5 Since V (un ) → 0 and un → u strongly in Lp+2 loc; x1 (R ); u satis-es n→∞
Q(x)
n→∞
p+1
u ux2 + Dx−1 ux3 + Dx−1 ux4 + Dx−1 ux 5 = u − @6x1 u + Dx−1 1 1 1 1 p+1
and we obtain by integration by parts 1 1 V(u) = − uY ≥ 0: 2 p+1 Then by (4.16) we arrive at ( 12 − 1=(p + 1))b ≤ d ¡ c∗ , so we obtain a contradiction with (4.20). Theorem 4.2. Eq. (4:1) possesses a nontrivial solution which is not radial in the transverse direction. Proof. Since V(*u) = *2
1 *p 2 uY − 2 (p + 1)(p + 2)
R5
Q(x)|u|p+2 ;
(4.21)
we have V(*u) → −∞: *→∞
(4.22)
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L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
Moreover, C p+2 C 1 2 p+2 V(u) ≥ uY − uY : 2 (p + 1)(p + 2) It follows that V(u) ¿ 0 if 0 ¡ uY ¡ (((p + 1)(p + 2))=(2C C p+2 ))1=p and there exists r ¿ 0 such that inf uY =r; u ∈ X V(u) ¿ 0: Let v be a minimizing function for Sp . Then *v belongs to X for every * ¿ 0 and by (4.22) there exists * ¿ 0 such that *v = e ¿ r and 0 = V(0) ≥ V(e). Setting d = inf 8 ∈ U maxt ∈ [0; 1] V(8(t)), we have b ≤ d as in Theorem 3.1. Finally, 2 t t p+2 2 vY − max V(tv) = max Q(x)|v|p+2 t≥0 t≥0 2 (p + 1)(p + 2) R5 = ¡
1 1 − 2 p+2 1 1 − 2 p+2
(p + 1)
2=p
(
(p + 1)
2=p
2
R5
vY Q(x)vp+2 )2=(p+2) 2
vY
(p+2)=p
(p+2)=p (4.23)
2
vLp+2 (R5 ) 2
and by (4.23) we have maxt ≥ 0 V(tv) ¿ 0. By the choice of v we get vY = Sp and 2 vLp+2 (R5 ) = 1. Thus 0 ¡ maxt ≥ 0 V(tv) ¡ c∗ and obviously we have d ¡ c∗ . As a consequence of Theorem 2:8 in [9] we can -nd a Palais–Smale sequence at level d. Then the mountain pass Theorem (using Lemma 4.1) yields the existence of a critical value c ∈ [b; c∗ [.
5. A nonsymmetric solution for a semilinear elliptic problem This section is devoted to the problem −Ku + u = |u|p−2 u; u ∈ H 1 (RN );
(5.1)
where N ≥ 2 and 2 ¡ p ¡ 2∗ . For now on we will consider the functional |∇u|2 u2 W(u) = + − F(u) ; 2 2 RN where F(u) = (u+ )p =p. In fact, in this section we will consider only the case N = 5 in order to complete the following result of Bartsch and Willem [1], see also [2]. Theorem 5.1. (Bartsch–Willem). If N = 4 or N ≥ 6 and 2 ¡ p ¡ 2∗ then the problem (5:1) has a nonradial solution.
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
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Our result deals with the case N = 5 which was not treated in [2]. The method is similar to that of the last section and we will only sketch the proof. Theorem 5.2. If N = 5 and 2 ¡ p ¡ 2∗ then problem (5:1) has a nonradial solution. Proof. Let G = {id} × O(2) × O(2). The action of G on H 1 (R5 ) is de-ned by ∗:
G × H 1 (R5 ) → H 1 (R5 ) (g; u) → g ∗ u(x) = u(g−1 · x)
:
We will denote HG1 (R5 ) = {u ∈ H 1 (R5 )=∀g ∈ G g ∗ u = u}. Lemma 5.1. The imbedding HG1 (R5 ) ,→ Lploc; x1 (R5 ) is compact for p ∈ ]2; 10=3[. Proof. Let un * 0 in HG1 (R5 ) and let denote V = ]x1 − r1 ; x1 + r1 [ × B(x ; r ). Since the imbedding H 1 (R5 ) ,→ L2 (R5 ) is continuous, un is bounded in L2 (R5 ). Using the estimate 2 supn un L2 (R5 ) 2 |un | ≤ m(y; r; G) V with m(y; r; G) as in De-nition 3.1 and following the line of Lemma 3.2, we prove that supx ∈ R4 V |un |2 → 0. n→∞ Now we obtain B(x1 ; r1 ) R4 |un |p → 0 for 2 ¡ p ¡ 10=3 using the same arguments n→∞
of that of Lemma 3.1.
We will use later the following Lemma. Lemma 5.2. Let un be a bounded sequence in H 1 (R5 ) such that y+r sup |un |6 → 0 for 6 ∈ ]2; 10=3[: y∈R
y−r
R4
n→∞
Then un Lq (R5 ) → 0 for q ∈ ]2; 10=3[. n→∞
Proof. See Lemma 3.3. Let now / be the involution de-ned on R5 = R × R2 × R2 by /(x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; x4 ; x5 ; x2 ; x3 ). The group H = {id; /}. H acts on HG1 (R5 ) as follows: H × HG1 (R5 ) → HG1 (R5 ) #: u(x) if h = id; (h; u) → h#u(x) = −1 −u(h · x) if h = /:
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L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
Finally, we note X ={u ∈ HG1 (R5 )=∀u ∈ H h#u=u}. We have the immediate property: Proposition 5.1. (X; :; : H 1 (R5 ) ) is a Hilbert space not reduced to {0} and 0 is the only radial function of X . 2
We have W(*u)=*2 ( 12 ·uH 1 (R5 ) −((*p−2 )=p)· W(*u) Since
R5
→
*→+∞
(u+ )p ≤
R5
(u+ )p ) and since p ¿ 2 we obtain
−∞:
R5
(5.2) p
|u|p ≤ CuH 1 (R5 ) for 2 ¡ p ¡ 10=3, we have
C 1 2 p W(u) ≥ uH 1 (R5 ) − uH 1 (R5 ) 2 p
for any u ∈ X:
(5.3)
It follows that there exists r ¿ 0 such that b = inf u=r; u ∈ X W(u) ¿ 0 = W(0). Let u ∈ X , u = 0 then R5 (u+ )p = 0 because u− = (−u)+ and u(x1 ; x2 ; x3 ; x4 ; x5 ) = −u(x1 ; x4 ; x5 ; x2 ; x3 ). Thus by (5.2), there exists e = *u ∈ X such that eH 1 (R5 ) ¿ r and W(e) ≤ 0. We set d = inf max W(8(t))
where U = {8 ∈ C([0; 1]; X ) 8(0) = 0 8(1) = e}:
8 ∈ U t ∈ [0;1]
As in Theorem 3.1 we can prove that b ≤ d and the existence of a Palais–Smale sequence (un )n at the level d. Moreover, such a sequence is bounded in X by the same argument of that of Lemma 1:20 in [9]. We can now prove Theorem 5.2. Suppose by contradiction that y+1 |un |6 = 0 6 ∈ ]2; 10=3[: 9 = lim sup ∞ y∈R
R5
y−1
By Lemma 5.2 un Lp (R5 ) → 0 for p ∈ ]2; 10=3[. But then, n→∞
0 ¡ d = W(un ) − =−
1 p
R5
1 2 W (un ); un
(un+ )p +
1 2
+ o(1)
(5.4)
(un+ )p−1 un + o(1):
(5.5)
R5
(un+ )p−1 un ≤ R5 (un+ )p , 1 1 1 1 + p + p−1 − (u ) + (u ) un ≤ (un+ )p − p R5 n 2 R5 n 2 p R5
Since
R5
≤ and 1 − p
R5
(un+ )p
1 + 2
R5
1 1 − 2 p
(un+ )p−1 un → 0: n→∞
R5
|un |p
(5.6)
(5.7)
(5.8)
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
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So d = 0 and this is a contradiction. Thus 9 ¿ 0. After extracting if necessary a subsequence, there exists (yn )n ∈ RN such that yn +1 |un |6 ¿ 9: (5.9) yn −1
R4
We de-ne vn (x1 ; x ) = un (x1 + yn ; x ). An obvious computation gives vn ∈ X . Moreover, vn H 1 (R5 ) =un H 1 (R5 ) so vn is bounded and up to a subsequence, we have vn * v weakly in X , so by Lemma 5.1 vn → v in Lploc; x1 (R5 ) strongly for 2 ¡ p ¡ 10=3 and n→∞
v = 0 by (5:9). Finally we have for any w ∈ X W (v); w = limn→∞ W (vn ; w = 0 so W (v) = 0. Remark 5.1. The same result for N = 2 or 3 seems to be more diIcult for topological reasons. Acknowledgements
The author wishes to thank Professor Jean-Claude Saut, for helpful support and comments during the work. References [1] T. Bartsch, M. Willem, In-nitely many radial solutions of a semilinear elliptic problem on RN , Arch. Rational Mech. Anal. 124 (1993) 261–276. [2] T. Bartsch, M. Willem, In-nitely many nonradial solutions of a Euclidean scalar -eld equation, J. Funct. Anal. 117 (1993) 447–460. [3] O.V. Besov, V.P. Ilin, S.M. Nikolski, Integral Representation of Functions and Imbeddings Theorems, Vol I, Wiley, New York, 1978. [4] A. de Bouard, J.C. Saut, Solitary waves of generalized Kadomtsev-Petviashvili equations, Ann. Inst. Henri Poincar\e 14 (2) (1997) 211–236. [5] A. de Bouard, J.C. Saut, Symmetries and decay of the generalized Kadomtsev-Petviashvili solitary waves, SIAM J. Math. Anal. 28 (5) (1997) 1064–1085. [6] H. Brezis, E. Lieb, A relation between pointwise convergence of functions and convergence of functionals, Proc. Amer. Math. Soc. 88 (1993) 486–490. [7] P.L. Lions, Sym\etrie et compacit\e dans les espaces de Sobolev, J. Funct. Anal. 49 (1983) 315–334. [8] P.L. Lions, The concentration-Compactness principle in the caculous of variations, The locally compact case, I and II, Ann. I.H.P. Analyse Non Lin\eaire I (1984) 104 –145, 223–283. [9] M. Willem, Minimax Theorems, BirkhMauser, Boston, 1996.
Nonsymmetric solutions for some variational problems Lionel Paumond Analyse num erique et EDP, Universit e de Paris-Sud, Bat. ˆ 425, 91405 Orsay Cedex, France Received 14 January 1999; accepted 27 March 1999
Keywords: Kadomtsev–Petviashili equations; Solitary waves; Symmetry; Minimax methods
1. Introduction In this paper we try to -nd nonsymmetric solitary waves for some variational problems. First of all Kadomtsev–Petviashvili equations (KP) are “universal” models for dispersive, weakly nonlinear waves, which are essentialy unidimensional, with weak transverse e5ect. For instance, the works [4,5] of de Bouard and Saut deals with a special case of these equations (KPI), ut + up · ux + uxxx − vy = 0;
u = u(x; y; t);
(x; y) ∈ R2 ; t ¿ 0;
vx = uy ;
(1.1)
in the two-dimensional case, and ut + up · ux + uxxx − vy − wz = 0;
u = u(x; y; z; t);
(x; y; z) ∈ R3 ; t ¿ 0;
vx = uy ;
(1.2)
wx = uz ; in the three-dimensional case, with p = m=n ¿ 0 with m and n relatively prime, n odd. They have shown in [4] the existence of solitary waves for (1.1) and(1.2) where x is the direction of propagation. Next, they have proved in [5] that any ground-state solution of (1.1) and (1.2) is cylindrically symmetric, in the sense that it has radial symmetry with respect to transverse coordinates, up to translation of the origin. E-mail address: [email protected] (L. Paumond). 0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 3 0 1 - 6
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L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
It is legitimate to wonder whether all solitary waves solutions are so. In Sections 3 and 4 we establish that the answer is no for two suitably modi-ed KPI equations in R5 . Thus, we -rst consider the following problem: ut + up · ux1 + @7x1 u − vx2 − wx3 − x4 − x5 = 0; vx1 = ux2 ; (1.3)
wx1 = ux3 ; x1 = ux4 ; x1 = ux5 ;
where p = m=n ¿ 0 with m and n relatively prime, n odd, and with u = u(x1 ; x2 ; x3 ; x4 ; x5 ; t) t ≥ 0 (x1 ; x2 ; x3 ; x4 ; x5 ) ∈ R5 . The equations studied in Section 4 are the same except for a potential in the nonlinearity but the method is somewhat di5erent. Classically we introduce (see [4]): Denition 1.1. Y is the closure of @x1 (C0∞ (R5 )) for the norm 2
2
@x1 Y = (∇L2 (R5 ) + @4x1 L2 (R5 ) )1=2 if ∈ @x1 (C0∞ (R5 )). Remark 1.1. For every u in Y there exists in L10=3 (R5 ) such that u = @x1 . Y is a Hilbert space with the scalar product induced by :Y . The solutions of (1.3) we are interested in are solitary waves of the form u = u(x1 − ct; x2 ; x3 ; x4 ; x5 );
c ¿ 0;
u ∈ Y:
Thus problem (1.3) becomes −c · ux1 + up · ux1 + @7x1 u − vx2 − wx3 − x4 − x5 = 0; vx1 = ux2 ; wx1 = ux3 ;
(1.4)
x1 = ux4 ; x1 = ux5 : Remark 1.2. Note that we may from now on assume that c = 1 by the scale change: u(x ˜ 1 ; x ) = c−1=p · u(x1 =c1=6 ; x =c2=3 ) where x = (x2 ; x3 ; x4 ; x5 ) ∈ R4 . The minimax method we will apply in Section 3 to -nd a solution of (1.4) which is not radial in the transverse direction is an adaptation of that used by Bartsch and Willem in [2]. The key argument is that symmetry allows to recover compactness as it is expounded in [7]. But for topological reasons the used of symmetry seems to be
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
707
more diIcult when transverse coordinates are less than 4. It explains our choice of R5 . Moreover solitary waves existence for (1.4) needs the term @7x1 u. This order is given by Sobolev inequalities in anisotropic Sobolev Spaces (see [3, Section 2]) used when we apply a minimizing method. Finally, consider the following problem: −Ku + b(|x|)u = f(|x|; u);
x ∈ RN ;
(1.5)
u ∈ H 1 (RN ):
The interest of this equation originates from various problems in physics and mathematical physics; for instance, cosmology or constructive -eld theory where (1.5) is called a nonlinear Euclidean scalar -eld equation. A solution of (1.5) can also be interpreted as a kind of solitary waves of nonlinear Klein–Gordon or SchrModinger equation. Bartsch and Willem prove the existence of in-nitely many nonradial solutions of this equation provided some hypotheses on f and b for N = 4 or N ≥ 6. In fact, the modi-cation of the method done in Section 3 allows to obtain the result for N = 5 in the autonomous case where b = 1 and f(|x|; u) = |u|p u (see Section 5). 2. A solution for (1:4) In order to frame problem (1.4) in a variational form, let us consider the fonctional N such that N:
Y →R 1 1 2 p+2 u → · uY − · up+2 : 2 (p + 1)(p + 2)
We can also write 1 2 (u + (@3x1 u)2 N(u) = R5 2 +(Dx−1 ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 u x 5 )2 ) − 1 1 1 1
up+2 d x; (p + 1)(p + 2)
where Dx−1 uxi for i ∈ {1; 2; 3; 4; 5} is the element of Y (the dual of Y in the L2 -duality) 1 such that for any ∈ Y , Dx−1 uxi ; Y ; Y = (u; Dx−1 xi ). 1 1 Remark 2.1. It is easy to check that N ∈ C 2 (Y; R) (see for example [9]) and that weak solutions for (1.4) are critical points for N on Y . The following Lemmas are well known in two- and three-dimensional cases (see for example [4]). Lemma 2.1. There exists C ¿ 0 such that uLq (R5 ) ≤ CuY for any q ∈ [2; 34=11]. Proof. It suIces apply a result of Besov et al. [3, Theorem 15:6, p. 323].
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L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
Remark 2.2. Lemma 2.1 remains valid if we take an open subset P in place of R5 (see [3]). Lemma 2.2. The imbedding Y ,→ Lploc (R5 ) is compact for p ∈ [2; 34=11[: Proof. We follow the same idea as in [4]. Let (un )n be a bounded sequence in Y , so by Sobolev inequality there exists n ∈ L10=3 (R5 ) such that un = @x1 n with @xi n ∈ L2 (R5 ) for i ∈ {1; : : : ; 5}. For now on we will suppose Supp(n ) ⊂ B(O; 2R) since we can multiply n by ∈ D(R5 ) such that 0 ≤ ≤ 1, ≡ 1 on B(O; R) and Supp( ) ⊂ B(O; 2R). Extracting if necessary a subsequence we can assume that un * u = @x1 in Y and replacing n by n − , we may assume that = 0. Thus by Plancherel theorem we have |un |2 = |un |2 ; R5
B(O;2R)
where
u(%) ˆ =
R5
e−2i&x; % u(x) d x
with % = (%1 ; %2 ; %3 ; %4 ; %5 ); x = (x1 ; x2 ; x3 ; x4 ; x5 ) and x; % = %1 x1 + %2 x2 + %3 x3 + %4 x4 + %5 x5 . Let us consider the inclusion R5 ⊂{|%1 | ¿ R1 } ∪ {|%1 | ¡ R1 ; |%i | ¡ R21 ; i ∈ {2; : : : ; 5}} ∪
5
{|%1 | ¡ R1 ; |%i | ¿ R21 }:
i=2
First, we have |un |2 ≤ |%1 |¿R1
|%1 |¿R1
1 ≤ 42 R21 ≤
1 2 |@[ x un | 42 %21 1
R5
2 |@[ x 1 un |
1 2 @x un 2 2 : 42 R21 1 L (R )
Let us now consider the integral 2 |un | = |%1 |¡R1 ;|%2 |¿R21
|%1 |¡R1 ;|%2 |¿R21
≤
1 2 vn L2 (R5 ) : R21
(2.1)
%21 |vn |2 %22 (2.2)
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
In the same way, we obtain 1 2 |un |2 ≤ 2 wn L2 (R5 ) ; 2 R1 |%1 |¡R1 ;|%3 |¿R1 |%1 |¡R1 ;|%4 |¿R21
|%1 |¡R1 ;|%5 |¿R21
709
(2.3)
|un |2 ≤
1 2 n L2 (R5 ) ; R21
(2.4)
|un |2 ≤
1 2 n L2 (R5 ) : R21
(2.5)
Thus for all ( we can choose R1 suIciently large to have 5 |un |2 + |un |2 ¡ (: |%1 |¿R1
i=2
|%1 |¡R1 ;|%i |¿R21
Now it remains to evaluate
|%1 |¡R1 ;|%i |¡R21 ; i ∈ {2;:::;5} have un * 0 in L2 (R5 ),
Applying Lemma 2.1 we un (x)e−2i&x; % d x → 0: un =
(2.6)
|un |2 . thus:
n→∞
B(O; 2R)
But by a classical majoration and Lemma 2.1 we obtain |un (%)| ≤ C0 un L2 (R5 ) ≤ C1 ;
∀% ∈ {|%1 | ¡ R1 ; |%i | ¡ R21 ; i ∈ {2; : : : ; 5}}:
We can then apply Lebesgue’s dominated convergence theorem, so |un |2 → 0: |%1 |¡R1 ;|%i |¡R21 ; i ∈ {2;:::;5}
n→∞
To conclude it suIces to remark that 5 |un |2 ≤ |un |2 + R5
|%1 |¿R1
+
i=2
|%1 |¡R1 ;|%i |¿R21
|%1 |¡R1 ;|%i |¡R21 ; i ∈ {2;:::;5}
(2.7)
|un |2
|un |2 :
(2.8)
By (2.1) – (2.8), we obtain un → 0 in L2loc (R5 ). n→∞
The conclusion follows by interpolation using Lemma 2.1. We are now able to give a -rst result: Theorem 2.1. System (1:4) possesses a solution u ∈ Y and u = 0; for p ∈ 0;
12 11
.
Proof. In the spirit of [4] (Theorems 3.1 and 3:2) we consider the constrained minimization problem
2 p+2 u d x d x = * ; * ¿ 0; I* = inf uY ; u ∈ Y; with R5
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which is solved using an application of the concentration–compactness principle of Lions [8]. 3. A solution for (1:4) which is not radial in the transverse direction 3.1. Functional setting As in [9] we shall de-ne a function space X containing no radial function except zero, and then will prove that N has a critical point on X . We -rst consider the group G = {id} × O(2) × O(2) and its action on Y de-ned by ∗:
G×Y →Y (g; u) → g ∗ u(x) = u(g−1 · x):
We get YG = {u ∈ Y=∀g ∈ G g ∗ u = u}. Let / be the involution de-ned on R5 = R × R2 × R2 by /(x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; x4 ; x5 ; x2 ; x3 ). The group H = {id; /} acts on YG as follows: H × YG → YG #:
(h; u) → h # u(x) =
u(x)
if h = id;
−u(h−1 · x) if h = /:
Finally, we denote X = {u ∈ YG =∀u ∈ H h # u = u}. Proposition 3.1. (X; ·; · Y ) is a Hilbert space which is not reduced to {0} and 0 is the only radial function of X . Proof. It is obvious that X is a closed subspace of Y . To construct a nonzero function ∞ 5 in 0 (R ) − {0} such that (x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; r; r ) where r =
X we take ∈ C x22 + x32 and r = x42 + x52 with the following condition: (x1 ; r; r ) = − (x1 ; r ; r). Then u = @x1 is the desired nonzero element of X . The second part of the proposition is left to the reader. 3.2. Some technical lemmas We shall start with a compactness lemma whose proof needs the following variant of a well-known result (see for example [8]). Lemma 3.1. Let (un )n be a bounded sequence in Y such that |un |2 → 0: sup x ∈ R4
Then
B(x1 ; r1 )
B(x1 ; r1 ) R4
B(x ; r )
n→∞
|un |p → 0 for 2 ¡ p ¡ 34 11 : n→∞
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Proof. Let s ∈ ]2; 34=11[ and u ∈ Y . By HMolder inequality we obtain 1=s (1−*)=2 11*=34 2 34=11 s |u| ≤ |u| · |u| V
V
V
with 1=s = (1 − *)=2 + 11*=34 so * = (17s − 34)=6s and where V = B(x1 ; r1 ) × B(x ; r ). Lemma 2.1 implies that (1−*)=2 1=s * |u|2 · uY (V ) ; |u|s ≤C · V
V
where *
uY (V ) =
V
[(@3x1 u)2 + u2 + (Dx−1 ux2 )2 + (Dx−1 u x 3 )2 1 1 1=2
+ (Dx−1 u x 4 )2 1
+
(Dx−1 u x 5 )2 ] 1
Taking * = 2=s, we obtain ((1−*)s)=2 2 s s 2 |u| ≤ C · |u| · uY (V ) : V
(3.1)
V
Now covering R4 by balls B(xi ; r ) in such a way that each point of R4 is contained in at most -ve balls, we -nd s |u| = |u|s B(x1 ; r1 )
R4
∪i B(xi ; r )
B(x1 ; r1 )
= ≤
∪i B(xi ; r )
i
≤C
i
B(x1 ; r1 )
Vi
s
sup
x ∈ R4
But we have 2 uY (Vi ) = i
Cs ·
|u|s
B(xi ; r )
i
≤
B(x1 ; r1 )
|u|2
((1−*)s)=2
+ (Dx−1 u x 2 )2 1
by (3:1)
((1−*)s)=2 2 |u| · uY (Vi ) :
V
B(x1 ; r1 )
+
2
· uY (Vi )
2
B(xi ; r )
|u|s
i
[(@3x1 u)2 + u2
(Dx−1 u x 3 )2 1
+
(Dx−1 u x 4 )2 1
+
(Dx−1 u x 5 )2 ] 1
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≤ 5·
R4
B(x1 ; r1 )
[(@3x1 u)2 + u2
+ (Dx−1 ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 u x 5 )2 ] 1 1 1 1 ≤ 5·
R5
[(@3x1 u)2 + u2
ux4 )2 + (Dx−1 ux5 )2 ] + (Dx−1 ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 1 1 1 1 2
≤ 5 · uY : Finally, we obtain ((1−*)s)=2 2 2 s s sup |u| · uY : |u| ≤ 5C B(x1 ; r1 )
R4
x ∈ R4
V
As s is -xed in ]2; 34=11[ we conclude by interpolating between s and 34=11. As in [9] we de-ne: Denition 3.1. For y ∈ R5 ; and r ¿ 0; let m(y; r; G) = sup{n ∈ N; ∃g1 ; : : : ; gn ∈ G=j = k ⇒ B(gj y; r) ∩ B(gk y; r) = ∅}. Denition 3.2. R5 is compatible with G if for any r ¿ 0 lim|y|→+∞ m(y; r; G) = +∞: Remark 3.1. In our case the diIculty arises from the fact that R5 is not compatible with the group G we have de-ned before. Indeed, if y = (x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; x ), we can -x x and make |x1 | → +∞ with |y| → +∞ and m(y; r; G) -xed. It is then obvious to check that m(y; r; G) → +∞ for any x1 ∈ R5 . |x |→+∞
Now we can prove the following lemma. Lemma 3.2. The imbedding YG ,→ Lploc; x1 (R5 ) is compact for p ∈ ]2; 34=11[. Proof. Let (un )n be a bounded sequence in YG ; and V = B(x1 ; r1 ) × B(x ; r ) be a neighbourhood of y = (x1 ; x ). Then there exists a ball of radius r such that V ⊂ B(y; r). Owing to Remark 3.1 we have 2 supn un 2 |un |2 ≤ |un |2 ≤ : m(y; r; G) V B(y; r)
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
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But the imbedding YG ,→ Y ,→ L2 (R5 ) are continuous, (see Lemma 2.1), and supn 2 un 2 = C ¡ + ∞ which yields: V |un |2 ≤ C=m(y; r; G). Let ( ¿ 0 and x1 ∈ R be -xed. By Remark 3.1 we can -nd R ¿ 0 such that |un |2 ¡ (: for all n ∈ N sup |x | ≥ R
B(x1 ; r1 )
B(x ; r )
Moreover, the imbedding YG ,→ L2loc (R5 ) is compact and so 2 |un |2 → 0: |un | = B(0; R +r )
This implies sup
|x | ≤ R
B(x1 ; r1 )
B(x1 ; r1 )
B(x ; r )
B(x1 ; r1 )
Finally, we obtain ∃N=∀n ≥ N
x ∈ R4
n→∞
|un |2 →n→∞ 0:
sup
B(0; R +r )
B(x1 ; r1 )
B(x ; r )
|un |2 ¡ 2(:
We conclude by applying Lemma 3.1. We have the obvious corollary: Corollary 3.1. The imbedding X ,→ Lploc; x1 (R5 ) is compact for p ∈ ]2; 34=11[. Let us give another Lemma which permits to recover compactness. Lemma 3.3. Let un be a bounded sequence in Y such that y+r sup |un |6 → 0 for 6 ∈ ]2; 34=11[ y∈R
y−r
n→∞
R4
Then un Lq (R5 ) → 0 for q ∈ ]2; 34=11[. n→∞
Proof. Let 2 ¡ s ¡ 34=11 and u ∈ X . HMolder inequality implies that 1−*
*
uLs (P) ≤ uL6 (P) · uL34=11 (P) with P = ]y − r; y + r[ × R4 and 1=s = 11*=34 + (1 − *)=6. Lemma 2.1 gives 1−*
*
uLs (P) ≤ CuL6 (P) · uY (P) : As in the proof of Lemma 3.2 we take * = 2=s ∈ ]0; 1[. So we obtain (s(1−*))=6 2 s s 6 |u| ≤ C · |u| · yY (P) : P
P
(3.2)
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Consider a covering R of R5 by open sets P(y) = ]y − r; y + r[ × R4 in such a way that each point of R5 is contained in at most two open sets. We have |u|s ≤ |u|s |u|s = R5
∪y ∈ R P(y)
P(y)
y∈R
≤ Cs
y∈R
Finally, R5
s
|u| ≤ C ·
(s(1−*))=6
s
sup
y∈R
P(y)
6
P(y)
|u|
·
|u|6
(s(1−*))=6 ·
P(y)
y∈R
2 uY (P(y))
:
by (3:2)
[(@3x1 u)2 + u2
ux5 )2 ]: ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 + (Dx−1 ux2 )2 + (Dx−1 1 1 1 1 By the choice of the covering we obtain (s(1−*))=6 6 s s |u| |u| ≤ 2 · C · sup · y∈R
R5
R5
P(y)
[(@3x1 u)2 + u2
ux2 )2 + (Dx−1 ux3 )2 + (Dx−1 ux4 )2 + (Dx−1 u x 5 )2 ] + (Dx−1 1 1 1 1 s
= 2·C ·
sup
y∈R
P(y)
(s(1−*))=6 2
6
|u|
· uY :
Since s is -xed in ]2; 34=11[ we conclude by interpolation between s and 34=11. 3.3. The 8rst result Theorem 3.1. (1:4) possesses a solution in X − {0}; for p ∈ 0;
12 11
:
Proof. Observe -rst that N is unbounded from below. Next, N(u) ≥
1 C p+2 2 p+2 · uY − · uY ¿ 0 2 (p + 1)(p + 2) 2
if 0 ¡ uY ¡
(p + 1)(p + 2) 2C p+2
1=p
and there exists r ¿ 0 such that inf u = r; u ∈ X N(u) ¿ 0. Finally, let v0 ∈ X − {0}. Then *v0 ∈ X and there exists * ¿ 0 such that *v0 ¿ r and 0 = N(0) ≥ N(e). Let us de-ne
=e
d = inf max N(8(t)) where U = {8 ∈ C([0; 1]; X ); 8(0) = 0; 8(1) = e}: 8 ∈ U t ∈ [0;1]
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If 8 ∈ U there exists t ∈ [0; 1] such that 8(t) = r, thus b ≤ d. Moreover by Theorem 2:8 in [9] there exists (un )n , a Palais Smale sequence for N in X such that N(un ) → d; n→+∞
N (un ) → 0: n→+∞
Now we will prove that (un )n is bounded in X . Indeed 1 1 1 2 · un Y ≥ 0: N(un ) − N (un ); un = − p+2 2 p+2 Moreover 1 1 N (un ); un ≤ N(un ) + N (un )X · un X : N(un ) − p+2 p+2 By N (un )X →n→+∞ 0 we deduce 1 ∃N=∀n ≥ N N(un ) − N (un ); un ≤ d + 1 + un X : p+2 2
Thus ( 12 − 1=(p + 2)) · un Y ≤ d + 1 + un X , and (un )n is bounded. Suppose by contradiction that y+1 9 = lim sup |un |6 = 0 n→∞ y ∈ R
y−1
R4
with 6 ∈ ]2; 34=11[. Then Lemma 3.3 yields un Lp+2 → 0 with p + 2 ∈ ]2; 34=11[, and n→∞
we get 0 ¡ d = N(un ) − 12 · N (un ); un + o(1). Moreover, 1 1 N(un ) − · N (un ); un = − |un |p+2 (p + 1)(p + 2) R5 2 1 + |un |p+2 2(p + 2) R5 1 1 1 · = |un |p+2 : − 2 p+2 p + 1 R5 We obtain N(un ) −
1 2
· N (un ); un → 0: n→∞
Thus d=0 which is a contradiction, so 9 ¿ 0. Extracting if necessary a subsequence, there exists (yn )n ∈ RN such that yn +1 |un |6 ¿ 9: yn −1
R4
We can now de-ne vn (x1 ; x ) = un (x1 + yn ; x ). The basic fact is that vn still belongs to X . Indeed the actions used to de-ne X , modify only the transverse variables. Thus, we obtain +1 |vn |6 ¿ 9: (3.3) −1
R4
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Moreover, (vn )n is bounded so, after extraction of a subsequence we can assume vn *m→∞ v
weakly in X
But Corollary 3.1 gives vn → v n→∞
in L6loc; x1 (R5 ):
(3.4)
Note that v = 0 by (3.3) and (3.4). On the other hand, for every w in X we have N (v); w = lim N (vn ); w n→∞
= lim N (vn ); wn : n→∞
But we have | N (vn ); wn | ≤ N (un )X · wn X and wn X = wX . Moreover N (un )X → 0 thus N (v) = 0. n→∞
4. “KP” with potential 4.1. Introduction This section is devoted to the problem: up+1 + @7x1 u − vx2 − wx3 − x4 − x5 = 0; ux1 + Q(x) · p + 1 x1 vx1 = ux2 ; wx1 = ux3 ;
(4.1)
x1 = ux4 ; x1 = ux5 with Q ∈ C(R5 ); 1 = lim|x|→+∞ Q(x) = inf R5 Q(x): Our aim is here to -nd a nonradial transverse solution to (4.1). We will use again the mountain pass theorem. The diIculty is that the Palais–Smale condition (P.S.) is 2 not always satis-ed. Let denote Sp = inf uLp+2 =1;u ∈ X uY where p + 2 ∈ ]2; 34=11[. We will solve this minimization problem by a method similar to that of Willem in [9] for the SchrModinger equation. The diIculty is here to -nd a minimizer in X . Let now consider the functional V : X → R, 1 2 u x 2 )2 V(u) = (u + (@3x1 u)2 + (Dx−1 1 R5 2 Q(x) −1 2 −1 2 −1 2 p+2 d x: + (Dx1 ux3 ) + (Dx1 ux4 ) + (Dx1 ux5 ) ) − ·u p+2 Weak solutions for (4.1) in X are critical points for V on X .
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4.2. The critical value 2 Theorem 4.1. There exists v ∈ X such that vY = Sp and vLp+2 = 1; for p ∈ 0;
12 11
:
Proof. Let (un )n be a minimizing sequence for Sp : un Lp+2 = 1;
∀n;
2 un Y → Sp : n→∞
By Lemma 3.3 we can -nd 9 ¿ 0 and (yn )n ∈ RN such that yn +1 |un |6 ¿ 9 ¿ 0: R4
yn −1
Let us set vn (x1 ; x2 ; x3 ; x4 ; x5 ) = un (x1 + yn ; x2 ; x3 ; x4 ; x5 ). Note that vn ∈ X because of the action of G and H . Obviously, +1 (4.2) |vn |6 ¿ 9: −1
R4
vn Lp+2 = un Lp+2 = 1; vn Y = un Y → Sp : n→∞
The sequence (vn )n is bounded in X and since X ,→ L6loc; x1 (R5 ) with compact imbedding for p ∈ ]2; 34=11[ we have after extracting a subsequence: vn * v
weakly in X;
(4.3)
vn → v
in L6loc; x1 (R5 ):
(4.4)
vn → v
almost everywhere in R5 :
(4.5)
By (4.2) and (4.4), v ∈ X − {0}. Moreover, vn Lp+2 = 1 together with (4.5) make possible the use of Brezis–Lieb Lemma [6]. Setting wn = vn − v, we have p+2
p+2
1 = vLp+2 + lim wn Lp+2
(4.6)
n→∞
and 2
2
2
Sq = lim vn Y = vY + lim vn − vY : n→∞
n→∞
(4.7)
By de-nition of Sp we have 2
2
vY ≥ Sp · vLp+2 :
(4.8)
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Thus, p+2 p+2 Sp ≥ Sp · (vLp+2 )2=(p+2) + lim (wn Lp+2 )2=(p+2) n→∞
p+2
p+2
≥ Sp · [(vLp+2 )2=(p+2) + (1 − vLp+2 )2=(p+2) ]
by (4:6):
But since vn Lp+2 = 1 and vn * v weakly in X , we have 0 ≤ vLp+2 ≤ 1. Thus (4.6) yields vLp+2 = 1: 2
2
2
Moreover, vX ≤ limvn X = Sp , and we obtain vX ≥ Sp proving that v is a minimizer. Let us denote un (x1 ; x2 ; x3 ; x4 ; x5 ) = v(x1 + an ; x2 ; x3 ; x4 ; x5 ) where (an )n is a real sequence such that |an | → +∞. n→∞
By the de-nition of V we have 1 2 V((p + 1)1=p Sp1=p un ) = (p + 1)1=p Sp1=p un X 2 1 − Q(x)(p + 1)(p+2)=p Sp(p+2)=p unp+2 (p + 1)(p + 2) R5 1 2 = (p + 1)2=p Sp2=p un X 2 (p+2)=p 2=p Sp − (p + 1) Q(x)unp+2 (p + 2) R5 1 2 = (p + 1)2=p Sp2=p un X 2 Sp(p+2)=p − (p + 1)2=p Q(x1 − an ; x )vp+2 : (p + 2) R5 By Lebesgue’s Theorem
R5
Since
Q(x1 − an ; x )v
R5
p+2
→
n→∞
R5
vp+2 :
2
vp+2 = 1 and vX = Sp we deduce
V((p +
1)1=p Sp1=p un )
→
n→+∞
1 1 − 2 p+2
Sp(p+2)=p (p + 1)2=p :
(4.9)
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Let us now consider for h ∈ X , V (p + 1)1=p Sp1=p un ; h X ; X = (p + 1)1=p Sp1=p un ; h X; X −
1 (p + 1)
R5
Q(x)(p + 1)(p+1)=p Sp(p+1)=p unp+1 h
= (p + 1)1=p Sp1=p v; h(x1 − an ; x ) X; X −
1 (p + 1)
R5
Fn (x1 ; x ) d x1 d x ;
where Fn (x1 ; x ) = Q(x1 − an ; x )(p + 1)(p+1)=p Sp(p+1)=p vp+1 h(x1 − an ; x ). Finally, we arrive at V ((p + 1)1=p Sp1=p un ) → 0:
(4.10)
n→+∞
Then, by (4.9) and (4.10), the Palais–Smale condition is not veri-ed at the level c = ( 12 − 1=(p + 2))Sp(p+2)=p (p + 1)2=p . ∗
4.3. Application of the mountain pass theorem We -rst prove that every Palais–Smale sequence is relatively compact in X . Lemma 4.1. Let us consider (un )n ∈ X N a Palais–Smale sequence i.e. V(un ) → d ¡ c∗ ; n→∞
V (un ) → 0: n→∞
Then (un )n countains a convergent subsequence. Proof. The fact that (un ) is bounded in X results from the identity 1 1 1 2 un Y (see Theorem 3:1): V(un ) − V (un ); un = − p+2 2 p+2 One can therefore assume that, up to a subsequence un *n→∞ u weakly in X and by Lemma 3.2 un → u
5 strongly in Lp+2 loc; x1 (R ) for + 2 ∈ ]2; 34=11[
un → u
a:e: in R5 :
n→∞
and n→∞
Brezis–Lieb Lemma yields Q(x)|un |p+2 − Q(x)|un − u|p+2 = R5
R5
Let us denote vn = un − u
R5
Q(x)|u|p+2 + o(1):
(4.11)
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We have
1 1 2 V(un ) = un Y − (p + 1)(p + 2) 2 1 1 2 = un Y − 2 (p + 1)(p + 2) + Q(x)|u|p+2 + o(1) R5
1 1 2 = un Y − (p + 1)(p + 2) 2 1 2 +V(u) − uY + o(1): 2
Since
2 1 2 un Y
2
R5
R5
Q(x)|un |p+2
(4.12)
Q(x)|un − u|p+2 (4.13)
R5
Q(x)|vn |p+2 (4.14)
2
− 12 uY = 12 vn Y + o(1); 2 1 2 vn Y
(4.15)
we obtain V(un ) = V(u) + − 1=((p + 1)(p + 2)) R5 Q(x)|vn |p+2 + o(1) and we deduce 1 1 2 V(u) + vn Y − Q(x)|vn |p+2 → d: (4.16) n→∞ 2 (p + 1)(p + 2) R5 On the other hand, 1 1 2 2 p+2 vn Y − Q(x)|vn | = vn Y − Q(x)unp+2 p + 1 R5 p + 1 R5 1 + Q(x)|u|p+2 + o(1) (4.17) p + 1 R5 1 2 2 = un Y − uY − Q(x)unp+2 p + 1 R5 1 + Q(x)|u|p+2 + o(1) by (4:15) p + 1 R5 = V (un ); un − V (u); u + o(1): But | V (un ; un Y ; Y | ≤ V (un )Y u − nY ≤ M V (un )Y → 0 and V (u); u Y ; Y = 0. Thus we obtain -nally 1 2 vn Y − Q(x)|vn |p+2 → 0: n→∞ p + 1 R5 Since (un )n is bounded in X , so is (vn )n and (4.18) implies 2
vn Y → b; n→∞ 1 Q(x)|vn |p+2 → b: n→∞ p + 1 R5
n→∞
(4.18)
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Now, we have p+2 (1 − Q(x))|vn | = R5
B(O; R)
p+2
(1 − Q(x))|vn |
+
R5 −B(O;R)
721
(1 − Q(x))|vn |p+2 :
Moreover, Lemma 2.1 yields p+2 (1 − Q(x))|vn |p+2 ≤ (vn Lp+2 (R5 ) ≤ C( R5 −B(O;R)
for R suIciently large. On the other hand, we can write (1 − Q(x))|vn |p+2 ≤ C B(O; R)
B(O; R)
|vn |p+2 → 0 n→∞
because of (4.11). Thus, we have 2
vn Y → b; n→∞ 1 |vn |p+2 → b: n→∞ p + 1 R5 2
(4.19) 2
Now by de-nition of Sp , vn Y ≥ Sp vn Lp+2 (R5 ) and b ≥ Sp [b(p + 1)]2=(p+2) . If b = 0, we conclude by (4.19). If b = 0 then b ≥ Sp(p+2)=p (p + 1)2=p . Now, 1 1 1 1 c∗ = Sp(p+2)=p (p + 1)2=p ≤ b: − − 2 p+2 2 p+2
(4.20)
5 Since V (un ) → 0 and un → u strongly in Lp+2 loc; x1 (R ); u satis-es n→∞
Q(x)
n→∞
p+1
u ux2 + Dx−1 ux3 + Dx−1 ux4 + Dx−1 ux 5 = u − @6x1 u + Dx−1 1 1 1 1 p+1
and we obtain by integration by parts 1 1 V(u) = − uY ≥ 0: 2 p+1 Then by (4.16) we arrive at ( 12 − 1=(p + 1))b ≤ d ¡ c∗ , so we obtain a contradiction with (4.20). Theorem 4.2. Eq. (4:1) possesses a nontrivial solution which is not radial in the transverse direction. Proof. Since V(*u) = *2
1 *p 2 uY − 2 (p + 1)(p + 2)
R5
Q(x)|u|p+2 ;
(4.21)
we have V(*u) → −∞: *→∞
(4.22)
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Moreover, C p+2 C 1 2 p+2 V(u) ≥ uY − uY : 2 (p + 1)(p + 2) It follows that V(u) ¿ 0 if 0 ¡ uY ¡ (((p + 1)(p + 2))=(2C C p+2 ))1=p and there exists r ¿ 0 such that inf uY =r; u ∈ X V(u) ¿ 0: Let v be a minimizing function for Sp . Then *v belongs to X for every * ¿ 0 and by (4.22) there exists * ¿ 0 such that *v = e ¿ r and 0 = V(0) ≥ V(e). Setting d = inf 8 ∈ U maxt ∈ [0; 1] V(8(t)), we have b ≤ d as in Theorem 3.1. Finally, 2 t t p+2 2 vY − max V(tv) = max Q(x)|v|p+2 t≥0 t≥0 2 (p + 1)(p + 2) R5 = ¡
1 1 − 2 p+2 1 1 − 2 p+2
(p + 1)
2=p
(
(p + 1)
2=p
2
R5
vY Q(x)vp+2 )2=(p+2) 2
vY
(p+2)=p
(p+2)=p (4.23)
2
vLp+2 (R5 ) 2
and by (4.23) we have maxt ≥ 0 V(tv) ¿ 0. By the choice of v we get vY = Sp and 2 vLp+2 (R5 ) = 1. Thus 0 ¡ maxt ≥ 0 V(tv) ¡ c∗ and obviously we have d ¡ c∗ . As a consequence of Theorem 2:8 in [9] we can -nd a Palais–Smale sequence at level d. Then the mountain pass Theorem (using Lemma 4.1) yields the existence of a critical value c ∈ [b; c∗ [.
5. A nonsymmetric solution for a semilinear elliptic problem This section is devoted to the problem −Ku + u = |u|p−2 u; u ∈ H 1 (RN );
(5.1)
where N ≥ 2 and 2 ¡ p ¡ 2∗ . For now on we will consider the functional |∇u|2 u2 W(u) = + − F(u) ; 2 2 RN where F(u) = (u+ )p =p. In fact, in this section we will consider only the case N = 5 in order to complete the following result of Bartsch and Willem [1], see also [2]. Theorem 5.1. (Bartsch–Willem). If N = 4 or N ≥ 6 and 2 ¡ p ¡ 2∗ then the problem (5:1) has a nonradial solution.
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
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Our result deals with the case N = 5 which was not treated in [2]. The method is similar to that of the last section and we will only sketch the proof. Theorem 5.2. If N = 5 and 2 ¡ p ¡ 2∗ then problem (5:1) has a nonradial solution. Proof. Let G = {id} × O(2) × O(2). The action of G on H 1 (R5 ) is de-ned by ∗:
G × H 1 (R5 ) → H 1 (R5 ) (g; u) → g ∗ u(x) = u(g−1 · x)
:
We will denote HG1 (R5 ) = {u ∈ H 1 (R5 )=∀g ∈ G g ∗ u = u}. Lemma 5.1. The imbedding HG1 (R5 ) ,→ Lploc; x1 (R5 ) is compact for p ∈ ]2; 10=3[. Proof. Let un * 0 in HG1 (R5 ) and let denote V = ]x1 − r1 ; x1 + r1 [ × B(x ; r ). Since the imbedding H 1 (R5 ) ,→ L2 (R5 ) is continuous, un is bounded in L2 (R5 ). Using the estimate 2 supn un L2 (R5 ) 2 |un | ≤ m(y; r; G) V with m(y; r; G) as in De-nition 3.1 and following the line of Lemma 3.2, we prove that supx ∈ R4 V |un |2 → 0. n→∞ Now we obtain B(x1 ; r1 ) R4 |un |p → 0 for 2 ¡ p ¡ 10=3 using the same arguments n→∞
of that of Lemma 3.1.
We will use later the following Lemma. Lemma 5.2. Let un be a bounded sequence in H 1 (R5 ) such that y+r sup |un |6 → 0 for 6 ∈ ]2; 10=3[: y∈R
y−r
R4
n→∞
Then un Lq (R5 ) → 0 for q ∈ ]2; 10=3[. n→∞
Proof. See Lemma 3.3. Let now / be the involution de-ned on R5 = R × R2 × R2 by /(x1 ; x2 ; x3 ; x4 ; x5 ) = (x1 ; x4 ; x5 ; x2 ; x3 ). The group H = {id; /}. H acts on HG1 (R5 ) as follows: H × HG1 (R5 ) → HG1 (R5 ) #: u(x) if h = id; (h; u) → h#u(x) = −1 −u(h · x) if h = /:
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L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
Finally, we note X ={u ∈ HG1 (R5 )=∀u ∈ H h#u=u}. We have the immediate property: Proposition 5.1. (X; :; : H 1 (R5 ) ) is a Hilbert space not reduced to {0} and 0 is the only radial function of X . 2
We have W(*u)=*2 ( 12 ·uH 1 (R5 ) −((*p−2 )=p)· W(*u) Since
R5
→
*→+∞
(u+ )p ≤
R5
(u+ )p ) and since p ¿ 2 we obtain
−∞:
R5
(5.2) p
|u|p ≤ CuH 1 (R5 ) for 2 ¡ p ¡ 10=3, we have
C 1 2 p W(u) ≥ uH 1 (R5 ) − uH 1 (R5 ) 2 p
for any u ∈ X:
(5.3)
It follows that there exists r ¿ 0 such that b = inf u=r; u ∈ X W(u) ¿ 0 = W(0). Let u ∈ X , u = 0 then R5 (u+ )p = 0 because u− = (−u)+ and u(x1 ; x2 ; x3 ; x4 ; x5 ) = −u(x1 ; x4 ; x5 ; x2 ; x3 ). Thus by (5.2), there exists e = *u ∈ X such that eH 1 (R5 ) ¿ r and W(e) ≤ 0. We set d = inf max W(8(t))
where U = {8 ∈ C([0; 1]; X ) 8(0) = 0 8(1) = e}:
8 ∈ U t ∈ [0;1]
As in Theorem 3.1 we can prove that b ≤ d and the existence of a Palais–Smale sequence (un )n at the level d. Moreover, such a sequence is bounded in X by the same argument of that of Lemma 1:20 in [9]. We can now prove Theorem 5.2. Suppose by contradiction that y+1 |un |6 = 0 6 ∈ ]2; 10=3[: 9 = lim sup ∞ y∈R
R5
y−1
By Lemma 5.2 un Lp (R5 ) → 0 for p ∈ ]2; 10=3[. But then, n→∞
0 ¡ d = W(un ) − =−
1 p
R5
1 2 W (un ); un
(un+ )p +
1 2
+ o(1)
(5.4)
(un+ )p−1 un + o(1):
(5.5)
R5
(un+ )p−1 un ≤ R5 (un+ )p , 1 1 1 1 + p + p−1 − (u ) + (u ) un ≤ (un+ )p − p R5 n 2 R5 n 2 p R5
Since
R5
≤ and 1 − p
R5
(un+ )p
1 + 2
R5
1 1 − 2 p
(un+ )p−1 un → 0: n→∞
R5
|un |p
(5.6)
(5.7)
(5.8)
L. Paumond / Nonlinear Analysis 44 (2001) 705 – 725
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So d = 0 and this is a contradiction. Thus 9 ¿ 0. After extracting if necessary a subsequence, there exists (yn )n ∈ RN such that yn +1 |un |6 ¿ 9: (5.9) yn −1
R4
We de-ne vn (x1 ; x ) = un (x1 + yn ; x ). An obvious computation gives vn ∈ X . Moreover, vn H 1 (R5 ) =un H 1 (R5 ) so vn is bounded and up to a subsequence, we have vn * v weakly in X , so by Lemma 5.1 vn → v in Lploc; x1 (R5 ) strongly for 2 ¡ p ¡ 10=3 and n→∞
v = 0 by (5:9). Finally we have for any w ∈ X W (v); w = limn→∞ W (vn ; w = 0 so W (v) = 0. Remark 5.1. The same result for N = 2 or 3 seems to be more diIcult for topological reasons. Acknowledgements
The author wishes to thank Professor Jean-Claude Saut, for helpful support and comments during the work. References [1] T. Bartsch, M. Willem, In-nitely many radial solutions of a semilinear elliptic problem on RN , Arch. Rational Mech. Anal. 124 (1993) 261–276. [2] T. Bartsch, M. Willem, In-nitely many nonradial solutions of a Euclidean scalar -eld equation, J. Funct. Anal. 117 (1993) 447–460. [3] O.V. Besov, V.P. Ilin, S.M. Nikolski, Integral Representation of Functions and Imbeddings Theorems, Vol I, Wiley, New York, 1978. [4] A. de Bouard, J.C. Saut, Solitary waves of generalized Kadomtsev-Petviashvili equations, Ann. Inst. Henri Poincar\e 14 (2) (1997) 211–236. [5] A. de Bouard, J.C. Saut, Symmetries and decay of the generalized Kadomtsev-Petviashvili solitary waves, SIAM J. Math. Anal. 28 (5) (1997) 1064–1085. [6] H. Brezis, E. Lieb, A relation between pointwise convergence of functions and convergence of functionals, Proc. Amer. Math. Soc. 88 (1993) 486–490. [7] P.L. Lions, Sym\etrie et compacit\e dans les espaces de Sobolev, J. Funct. Anal. 49 (1983) 315–334. [8] P.L. Lions, The concentration-Compactness principle in the caculous of variations, The locally compact case, I and II, Ann. I.H.P. Analyse Non Lin\eaire I (1984) 104 –145, 223–283. [9] M. Willem, Minimax Theorems, BirkhMauser, Boston, 1996.