On A Nonlocal Schrödinger-Poisson System With Critical Exponent
Mohammed Massar
yReceived 26 January 2020
Abstract
This work is concerned with a class of Kirchho¤-Schrödinger-Poisson systems involving the critical Sobolev exponent. By means of the variational method and the concentration compact argument, for each positive integerk;the existence ofkpairs of nontrivial solutions is established.
1 Introduction
Let be a bounded smooth domain ofR3:Consider the system 8>
<
>:
M R
jruj2dx u+ u= juj4u+f(x; u) in ;
=u2 in ;
u= = 0 on@ ;
(1)
where >0andM; f are continuous functions satisfying some conditions which will be given later.
The presence of the nonlocal term M R
jruj2dx in (1) causes some mathematical di¢ culties and so the study of such a class of problems is of much interest. This type of problems are closely related to the following hyperbolic equation
@2u
@t2
0
h + E 2L
Z L 0
@u
@x
2
dx
!@2u
@x2 = 0; (2)
which was proposed by Kirchho¤ [13] as a model to describe the transversal vibrations of a stretched string by considering the subsequent change in string length during the vibrations. Recently, the Kirchho¤ type problems with or without critical growth have been investigated by many researchers, we cite here [1, 3,8, 9, 10,11,12,16,21].
WhenM 1;system (1) reduces to Schrödinger-Poisson system. Due to its importance in many physical applications (see[5,18]), it has received much attention in the recent years. Many interesting papers have been devoted to the existence or multiplicity of solutions for (1), when 2 f0;1gsee for instance [7,14,15, 20].
In [4], Batkam and Júnior studied the following Kirchho¤-Schrödinger-Poisson system 8>
<
>:
a+bR
jruj2dx u+ u=f(x; u) in ;
=u2 in ;
u= = 0 on@ ;
(3)
with = 1: The authors proved that (3) has at least three solutions. Furthermore, in case f is odd with respect tou;they established the existence of unbounded sequence of solutions. Under the general singular assumptions on f and by using the variational arguments, Li et al. [17], proved the existence and the uniqueness of solutions for (3). In the critical case, to my knowledge, the existence of multiple solutions for
Mathematics Sub ject Classi…cations: 35J20, 35J50, 35J60.
yDepartment of Mathematics, Facculty of Sciences & Techniques, Al Hoceima, Abdelmalek Essaadi University, Morocco
44
system (1) has not investigated until now. Motivated by the above results, in this note we are interested in
…nding multiple solutions by using the variational method and the concentration compact argument.
Throughout the paper, we assume the following conditions on the Kirchho¤ function and the nonlinearity:
(m1) M : [0;+1)![M0;+1)is continuous for someM0>0;
(m2) 2Mc(t) M(t)tfor allt 0;whereMc(t) =Rt
0M(s)ds;
(m3) There exista >0andb 0such thatM(t) a+btfor allt 0;
(f1) f(x; t)2C( R;R)is odd int;
(f2) f(x; t) =o(jtj5)asjtj ! 1;uniformly in ;
(f3) There exists an open 0 of positive measure such that lim infjtj!1F(x;t)t4 = 0; uniformly in 0; where F(x; t) :=Rt
0f(x; s)ds;
(f4) There existq2[0;2)anda1; a2>0such that F(x; t) 1
4f(x; t)t a1+a2jtjq for allx2 andt2R; (f5) There existr2(2;6) andb1; b2>0 such that
F(x; t) b1+b2jtjr for allx2 andt2R: (f6) (x) := lim supt!0F(x;t)t2 is such thatmaxf0; (x)g=: +(x)2L1( ):
The main result is the following theorem.
Theorem 1 Suppose that(m1) (m3); and(f1) (f4)hold, furthermore, one of conditions(f5)or(f6)is veri…ed. Then for each positive integerk; there exists k >0such that system (1) admits at least kpairs of nontrivial solutions for all 2(0; k):
2 Auxiliary Results
We look for solutions in the Sobolev spaceH01( ) with the normjjujj2=R
jruj2dx: Denote
jujp= Z
jujpdx
1 p
foru2Lp( ):
Thanks to the Lax-Milgram theorem, for any u2H01( )the Poisson problem =u2 in ; = 0on
@ ;has a unique solution u2H01( ):Moreover, we recall the following lemma (see [15]).
Lemma 1 Let u2H01( ): Then 1. u(x) 0; x2 ;
2. For allt 0; tu=t2 u; 3. There existsA >0 such that R
uu2dx=R
jr uj2dx A jjujj4; 4. Ifun* uinH01( );then un! u inH01( ):
By substituting u;system (1) reduces to following problem
( M R
jruj2dx u+ uu= juj4u+f(x; u) in ;
u= 0 on@ : (4)
The energy functional associated to (4) is given by I (u) =1
2Mc Z
jruj2dx +1 4
Z
uu2dx 6
Z
juj6dx Z
F(x; u)dx:
By assumptions of Theorem1,I 2C1(H01( )) and for allu; v2H01( ) hI0(u); vi=M
Z
jruj2dx Z
rurvdx+ Z
uuvdx Z
juj4uvdx Z
f(x; u)vdx: (5)
Lemma 2 Suppose that(m1) (m2); (f1) (f2); (f4) and one of conditions (f5) or(f6) hold. Then, for any >0;there is >0 such that for all 2(0; );the functionalI satis…es the(P S)c condition at every level c < :
Proof. Letfung H01( )be a sequence such thatI (un)!c < andI0(un)!0:
We claim thatfungis bounded. Indeed, we have jttj6q !0asjtj ! 1;thus for given" >0;there existsC">
such that
jtjq "t6+C" for allt2R: (6) Therefore assumptions(m2)and(f4)yield
c+on(1) +on(1)jjunjj I (un) 1
4hI0(un); uni 12junj66
Z
F(x; un) 1
4f(x; un)un dx 12junj66 a1j j a2
Z
junjqdx
12 "a2 junj66 (a1+a2C")j j: Thus, choose" < 12a
2;for some c1; c2>0 and fornlarge enough
junj66 c1+c2jjunjj: (7)
On the other hand, by(m1); (f5)and Lemma 1we have M0
2 jjunjj2 1
2Mc(jjunjj2)
6junj66+b1j j+b2junjrr+c+on(1):
Sincer2(2;6);we can …ndC"0 >0such that jtjr "t6+C"0 for allt2R:Then M0
2 jjunjj2 6 +"b2 junj66+ (b1+b2C"0)j j+c+on(1):
It follows from (7) that for somec3; c4>and fornlarge enough M0
2 jjunjj2 c3jjunjj+c4:
This shows thatfungis bounded inH01( ):
Now the hypothesis (f5)will be replaced by condition(f6):By(f2)we have lim sup
jtj!1
jF(x; t)j
t6 = 0 uniformly in : This and(f6)imply that for any" >0;there existsC">such that
F(x; t) (j +j1+")t2+C"t6: (8) Therefore, by similar arguments as above, we show thatfungis bounded. Then, up to subsequence for some u2H01( );
un* u inH01( );
un!u a.e. in ;
un!u inLs( )for alls2[1;6);
jrunj2* weakly in the sense of measures, u6n* weakly in the sense of measures,
(9)
where and are nonnegative bounded measures on :Applying concentration compact result due to Lions [19], we can …nd at most countable index setJ and elements fxjgj2J of such that
8>
<
>:
=juj6+P
j2J j xj; j>0;
jruj2+P
j2J j xj; j>0
S 1=3j j for allj2J:
(10)
where
S := inf
u2H01( )nf0g
jjujj2 juj
1 3
6
: We claim that j M0S 32
for all j 2J: Let j 2 J be …xed and for an arbitrary" > 0; choosing " of C01(R3)such that0 " 1;
"(x) = 1 ifx2B(xj; ");
0 ifx =2B(xj;2");
andjr "j1 2
". ClearlyhI0(un); "uni=on(1);that is M
Z
jrunj2dx Z
unrunr "dx+ Z
"jrunj2dx + Z
unu2n "dx
= Z
junj6 "dx+ Z
f(x; un)un "dx+on(1): (11) By the Hölder inequality, we have
lim sup
n!1
Z
unrunr "dx lim sup
n!1
Z
jrunj2
1
2 Z
junj2jr "j2dx
1 2
C Z
juj2jr "j2dx
1 2
C Z
B(xj;")jr "j3dx
!13 Z
B(xj;")juj6dx
!16
Cjr "j1w
1 3
6"
Z
B(xj;")juj6dx
!16
!0 as"!0; (12)
wherew6 is the volume ofB(0;1):In view of Lemma 1, un ! uin H01( ); thus
nlim!1
Z
unu2ndx= Z
uu2dx:
Since " is bounded in ;
nlim!1
Z
unu2n "dx= Z
uu2 "dx!0 as"!0: (13)
Using(f1)and thatf(x; un)un "!f(x; u)u " a.e. in ;we have by compactness Lemma of Strauss [6]
nlim!1
Z
f(x; un)un "dx= Z
f(x; u)u "dx!0 as "!0: (14) Lettingn! 1and"!0in (11), from (9) and (12)–(14) we obtainM0 j j:Therefore (10) implies
j
M0S 32 :
Now we prove thatJ is empty. Suppose by contradiction that there is j2J:Then +on(1) > c+on(1) =I (un) 1
4hI0(un); uni 12junj66 a1j j a2
Z
junjqdx
12junj66 a1j j a2j j66q Z
u6ndx
q 6
therefore lettingn!+1and using (9);we get
12 ( ) +a1j j+a2j j66q ( )q6: (15) If ( )>1;from the last inequality, we can write
( ) 12
+a1j j+a2j j66q ( )q6; hence
( ) 0
@12 +a1j j+a2j j66q 1 A
6 6 q
!0 as !0+:
If ( ) 1;then
( )
12 +a1j j+a2j j66q
as !0+: Sincemax 1;66q < 32; in both above cases, there exists >0 such
j ( )< M0S 32
for all 2(0; ):
which is impossible and henceJ =;for all 2(0; ):It follows thatun !uin L6( );thus
nlim!1
Z
u5n(un u)dx= 0: (16)
On the other hand, we see that
nlim!1
Z
f(x; un)(un u)dx= 0: (17)
According to Lemma1, we have
nlim!1
Z
unun(un u)dx= 0: (18)
SincehI0(un); un ui=on(1); by(m1)and (16)-(18) we deduce that
nlim!1
Z
runr(un u)dx= 0:
Similarly, we have
nlim!1
Z
rur(un u)dx= 0:
So thatun!uinH01( ):
3 Proof of Theorem 1
To this end, we need to ensure thatI satis…es the conditions of the following version of Symmetric Mountain Pass theorem [2].
Theorem 2 LetH =V W be a real Banach space withdimV <1:Assume that I2C1(H;R)is an even functional verifyingI(0) = 0and
(i) there exist ; >0 such that
u2@Binf(0)\WI(u) ; (ii) there exists a subspaceE H such that dimV < dimE and
maxu2EI(u) for some >0;
(iii) the functionalI satis…es(P S)c for every c2(0; ).
ThenI admits at leastdimE dimV nontrivial critical points.
Letfe1; e2; ::::gbe a system of the normalized eigenfunctions of( ; H01( )):Consider thek-subspace Vk =fe1; e2; :::; ekg:
Then H01( ) = Vk Vk?: Moreover, for any s 2 [2;6) and > 0 there exists ks; 2 N such that for all k ks;
jujss jjujjsfor allu2Vk?: (19)
Lemma 3 Assume that (m1); (f1) (f2)and one of conditions(f5) or(f6)hold. Then, there exist ; = ( ); ; >0 such that8k kr; and8 2(0; );
inf
u2@B (0)\Vk?I (u) :
Proof. By (f5);(m1)and (19), for allu2Vk?;
I (u) M0
2 jjujj2 6juj66 b1j j b2jujrr
M0
2 b2 jjujjr 2 jjujj2 c5
6 jjujj6 b1j j: Letjjujj= = ( ) := 4bM0
2 1 r 2
:Then
I (u) M0 4
2 c5 6
6 b1j j:
Since ( )!+1as !0+;we can …nd >0and so >0 such thatb1j j M80 2:Therefore I (u) M0
8
2 c5 6
6
and hence there exists such that for all 2(0; )
I (u) >0 for allu2@B(0; )\Vk?: (20)
Now assume that(f6)holds. Then, from(m1)and (8)we have I (u) M0
2 jjujj2
6 +C" juj66 (j +j1+")juj22: Arguing as in above, also (20) follows in this case.
Lemma 4 Assume that (m3); (f1) and (f3) hold. Then, for each positive integer k; there is a k-subspace Vk0 H01( ) and k >0 such that for all >0:
max
u2Vk0I (u)< k:
Proof. Letfe01; e02; :::g be a system of eigenfunctions of( ; H01( 0))and consider thek-subspace Vk0=fe01; e02; :::; e0kg H01( ) withe0i = 0in n 0:
SincedimVk0<1;there existsCk>0 such that
juj44 Ckjjujj4 for allu2Vk0: (21) By(f3)and the continuity ofF on R;for" >b+A4C
k there existsC">0 such that F(x; t) "t4 C" for all(x; t)2 0 R:
Combining this last inequality with(m3);(21)and Lemma 1, we get I (u) a
2jjujj2 "Ck b+A
4 jjujj4+C"j j ! 1 asjjujj ! 1: Therefore, there exists k>0such that
max
u2Vk0I (u) k for all >0:
Proof of Theorem 1. ObviouslyI is an even functional and I (0) = 0:Letk be a positive integer. In view of Lemma3, we can choosek02Nsu¢ ciently large,V =Vk0 andW =Vk?0 such thatH01( ) =V W and condition Theorem2 (i)holds for all 2 (0; ): By Lemma 4, for the positive integer k+k0; there exists a subspaceVk+k0 0 H01( ) withdimVk+k0 0 =k+k0 and k >0such that condition Theorem2 (ii):
follows. According to lemma2, there exists k such that for all 2(0; k); I veri…es(P S)c for allc < k: Let k := min( k; ):Then, by applying Theorem2,I hask+k0 k0=knontrivial critical points, for all 2(0; k):
Acknowledgment. The author would like to thank the referees and editors for the careful review and helpful suggestions which led to an improvement of the original manuscript.
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