R E S E A R C H
Open Access
Existence and multiplicity of solutions for a
class of sublinear Schrödinger-Maxwell
equations
Ying Lv
**Correspondence:
[email protected] School of Mathematics and Statistics, Southwest University, Chongqing, 400715, People’s Republic of China
Abstract
In this paper I consider a class of sublinear Schrödinger-Maxwell equations, and new results about the existence and multiplicity of solutions are obtained by using the minimizing theorem and the dual fountain theorem respectively.
Keywords: Schrödinger-Maxwell equations; sublinear; minimizing theorem; dual fountain theorem
1 Introduction and main result
Consider the following semilinear Schrödinger-Maxwell equations:
⎧ ⎨ ⎩
–u+V(x)u+φu=f(x,u), inR, –φ=u, lim
|x|→∞φ(x) = , inR. ()
Such a system, also known as the nonlinear Schrödinger-Poisson system, arises in an interesting physical context. Indeed, according to a classical model, the interaction of a charge particle with an electromagnetic field can be described by coupling the nonlinear Schrödinger and the Maxwell equations (we refer to [, ] for more details on the physical aspects and on the qualitative properties of the solutions). In particular, if we are looking for electrostatic-type solutions, we just have to solve ().
In recent years, system (), withV(x)≡ or being radially symmetric, has been widely studied under various conditions onf; see, for example, [–]. Since () is set onR, it is well known that the Sobolev embeddingH(R)→Ls(R) (≤s≤∗= ) is not compact, and then it is usually difficult to prove that a minimizing sequence or a sequence that satisfies the (PS) condition, briefly a Palais-Smale sequence, is strongly convergent if we seek solutions of () by variational methods. IfV(x) is radial (for example,V(x)≡), we can avoid the lack of compactness of Sobolev embedding by looking for solutions of () in the subspace of radial functions ofH(R), which is usually denoted byH
r(R), since the
embeddingH
r(R)→Ls(R) ( <s< ) is compact. Specially, Ruiz [] dealt with () under
the assumption thatV(x)≡ andf(u) =up ( <p< ) and got some general existence, nonexistence and multiplicity results.
Moreover, in [] the authors considered system () with periodic potentialV(x), and the existence of infinitely many geometrically distinct solutions was proved by the nonlinear superposition principle established in [].
There are also some papers treating the case with nonradial potentialV(x). More pre-cisely, Wang and Zhou [] got the existence and nonexistence results of () whenf(u) is asymptotically linear at infinity. Chen and Tang [] proved that () has infinitely many high energy solutions under the condition thatf(x,u) is superlinear at infinity inuby the fountain theorem. Soon after, Li, Su and Wei [] improved their results.
Up to now, there have been few works concerning the case thatV(x) is nonradial poten-tial andf(x,u) is sublinear at infinity inu. Very recently, Sun [] treated the above case based on the variant fountain theorem established in Zou [].
Theorem .[] Assume that the following conditions hold:
(V) V∈C(R,R)satisfiesinf
x∈RV(x)≥a> ,wherea> is a constant.For everyM> , meas{x∈R:v(x)≤M}<∞.
(H) F(x,u) =a(x)|u|r,whereF(x,u) =
u
f(x,y)dy,a:R→R+is a positive function such
thata∈L–r (R)and <r< .
Then problem()has infinitely many nontrivial solutions{(uk,φk)}satisfying
R
|∇uk|+V(x)uk
dx–
R|∇
φk|dx+
R
φkukdx–
R
F(x,uk)dx→–
as k→ ∞.
In the present paper, based on the dual fountain theorem, we can prove the same result under a more generic condition, which generalizes the result in []. Our first result can be stated as follows.
Theorem . Assume that V satisfies
(V) V∈C(R,R)andinf
x∈RV(x) > ;
and f satisfies the following conditions.
(W) There exist constantsδ> ,r∈(, )and a functiona∈L–r(R, [, +∞))such that f(x,u) ≤a(x)|u|r–
for allx∈Rand|u| ≤δ;
(W) There exist constantsM> ,r∈(, )and a functiona∈L–r(R, [, +∞))such that
f(x,u) ≤a(x)|u|r–
for allx∈Rand|u| ≥M;
(W) For every m>δ, there exist a constant r ∈(, ) and a function bm ∈L
–r(R,
[, +∞))such that
f(x,u) ≤bm(x)
(W) There exist constantsr∈(, ),η> andζ> such that
F(x,u)≥η|u|r
for allx∈and|u| ≤ζ,wheremeas{}> ,F(x,u) :=uf(x,y)dy; (W) F(x, –u) =F(x,u)for allx∈Randu∈R.
Then problem()has infinitely many nontrivial solutions{(uk,φk)}satisfying
R
|∇uk|+V(x)uk
dx–
R|∇
φk|dx+
R
φkukdx–
RF(x,uk)dx→
–
as k→ ∞.
By Theorem ., we obtain the following corollary.
Corollary . Assume that L satisfies(V)and W satisfies
(W) F(x,u) =a(x)|u|r,whereF(x,u) =u
f(x,y)dy, <r< is a constant anda:R→R
is a function such thata∈L–r (R)anda(x) > forx∈,wheremeas{}> .
Then problem()has infinitely many nontrivial solutions{(uk,φk)}satisfying
R
|∇uk|+V(x)uk
dx–
R|∇
φk|dx+
R
φkukdx–
RF(x,uk)dx→
–
as k→ ∞.
Remark . In Theorem ., infinitely many solutions for problem () are obtained under the symmetry condition (W) by using the dual fountain theorem. As a special case of Theorem ., Corollary . generalizes and improves Theorem .. To show this, it suffices to compare (V) and (V), (H) and (W). Firstly, it is clear that (V) is really weaker than (V). Secondly, in (H)ais assumed to be positive, while in (W) we assume thatais indefinite.
Moreover, under all the conditions of Theorem . except (W) we obtain an existence result.
Theorem . Assume that L satisfies(V)and W satisfies(W), (W), (W), (W).Then problem()possesses a nontrivial solution.
Remark . In Theorem . we obtain the existence of solutions for problem () under the assumption thatf(x,u) is indefinite and without any coercive assumptions respect to Vsuch as (V). There are functionsVandf which satisfy Theorem ., but do not satisfy the corresponding results in [–]. For example,
V(x)≡, f(x,u) =a(x)˜ |u| ()
and
˜ a(x) =
⎧ ⎨ ⎩
(–)nn(|x|–n) forn≤ |x| ≤n+
n,
in whichn≥. It is clear thata˜∈C(R,R) is indefinite. Denoting byπthe area of the unit ball inR, we obtain
Ra˜
(x)dx=
∞
n=
n+
n
n
nr(r–n)dr+
n+
n
n+
n
nr
n+ n –r
dr
π
=π
∞
n= n
n
rdx
=π
∞
n= n–
<∞, ()
which means thata˜∈L
– (R). So, () satisfies our results, but does not satisfy the results
in [–].
2 Preliminary results
In order to establish our results via critical point theory, we firstly describe some properties of the spaceH(R), on which the variational functional associated with problem () is defined. Define the function space
HR:=u∈LR:∇u∈LR
equipped with the norm
uH:=
R
|∇u|+udx
/
and the function space
D,R:=u∈L∗:∇u∈LR
with the norm
uD,=
R|∇u|
dx
/
.
Let
E:=
u∈HR:
RV(x)u
dx< +∞
equipped with the inner product
(u,v) =
R
∇u· ∇v+V(x)uvdx
and the corresponding norm
Note that the following embeddings
E→LsR, ≤s≤∗, D,R→L∗R
are continuous, where ∗= is the critical exponent for the Sobolev embeddings in di-mension . Therefore, there exist constantsCpandC∗such that
uLp≤Cpu, u
L∗≤C∗uD, ()
for allu∈E. HereLp(R) (≤p≤∗) denotes the Banach spaces of a function onRwith values inRunder the norm
uLp=
R
u(x) pdx
/p
.
Let
LraR:=
u:R→R:
Ra(x)|u|
rdx< +∞
,
wherea(x) > for a.e.x∈R. ThenLr
a(R) is a Banach space with the norm
uLra=
Ra(x)|u|
rdx
/r
.
Lemma . Suppose that assumption(V)holds.Then the embedding of E in Lra(R)is
compact,where r∈(, ),a∈L–r (R)is positive for a.e.x∈R.
Proof For any bounded setK⊂E, there exists a positive constantMsuch thatu ≤M for allu∈K. We claim thatKis precompact inLr
a(R). In fact, sincea∈L
–r(R), for any
ε> , there existsTε> such that
|x|≥Tε
a(x)–r dx (–r)/
<ε.
For anyu,v∈K, applying the Hölder inequality forrsuch that r
+ –r
= and the first inequality in (), we have
|x|≥Tε
a(x)|u–v|rdx≤
|x|≥Tε
a(x)–r dx
(–r)/
|x|≥Tε
|u–v|dx
r/
≤ u–vrL
|x|≥Tε
a(x)–r dx (–r)/
≤Cru–vrε
≤CrMrε. ()
Besides, since E(BTε()) ⊂ H (B
Tε()) is compactly embedded in L
r
a(BTε()), where BTε() ={x∈R
:|x| ≤T
ε}, there areu,u, . . . ,um∈Ksuch that for anyu∈K,
|x|≤Tε
Now it follows from () and () thatK is precompact inLr
a(R). Obviously, we haveE
is compact embedded inLr
a(R), wherer∈(, ),a∈L
–r(R) is positive for a.e.x∈R.
Lemma . Assume that assumptions(V), (W), (W)and(W)hold and unu in E.
Then
f(x,un)→f(x,u)
in L(R).
Proof Assume thatunuinE. Then, by Lemma .,
un→u
inLr
a(R), wherer∈(, ),a∈L
–r(R) is positive for a.e.x∈R. Passing to a subsequence
if necessary, it can be assumed that
∞
n=
un–uLra< +∞.
It is clear that
hk(x) := k
n=
un(x) –u(x) ∈Lra
R ()
and
hg–hlLra≤
g
n=l
un–uLra ()
for allg>l∈N+. Since{u
n}is a Cauchy sequence inLra(R), so by () we know that{hk}is
also a Cauchy sequence inLra(R). Therefore, by the completeness ofLra(R), there exists h∈Lr
a(R) such thathk→hinLra(R). Now we show that
hk(x)≤h(x) ()
for all k∈N+ and almost every x∈R. If not, there exist k
∈N+ andS⊂ R, with
meas{S}> , such that
hk(x) >h(x)
for allx∈S. Then there exist a constantc> andS⊂S, withmeas{S}> , such that
for allx∈S. By the definition ofhk, we have
hk(x)≥hk(x)≥h(x) +c
for allk≥kandx∈S. Therefore, one has
Ra(x)|hk–h|
rdx≥
S
a(x)|hk–h|rdx
≥cr
S
a(x)dx.
Lettingk→ ∞, we get
≥cr
S
a(x)dx,
which contradicts the fact thata(x) > for a.e.x∈R. Now we have proved (). It follows from (W) that there existsM> such that
f(x,u) ≤a(x)|u|r– ()
for allx∈Rand|u| ≥M. By (W
), there existsδ> such that
f(x,u) ≤a(x)|u|r– ()
for allx∈Rand|u| ≤δ, which together with (W) shows there existsb
M∈L
–r(R) such
that
f(x,u) ≤a(x)|u|r–+ bM(x)
δr– |u|
r– ()
for allx∈Rand|u| ≤M. Combining () and (), we have
f(x,u) ≤a(x)|u|r–+a(x)|u|r–+ bM
δr–|u|
r– ()
for allx∈Randu∈R. Hence, by () one has
f(x,un) –f(x,u) ≤a(x)
|un|r–+|u|r–
+a(x)|un|r–+|u|r–
+bM(x) δr–
|un|r–+|u|r–
≤a(x)|un–u|r–+ u(x) r–
+a(x)|un–u|r–+ |u|r–
+bM(x) δr–
|un–u|r–+ |u|r–
≤a(x)|h|r–+ |u|r–+a(x)|h|r–+ |u|r– +bM(x)
δr–
for alln∈Nandx∈R. It follows that
f(x,un) –f(x,u)
dx≤a(x)|h|(r–)+ |u|(r–)dx
+ a (x)
|h|(r–)+ |u|(r–)dx +b
M(x)
δ(r–)
|h|(r–)+ |u|(r–)dx
=:(x) ()
for alln∈N. By the Hölder inequality, we have
Ra
(x)|h|(r–)dx≤
Ra(x) –rdx
–r
r
Ra(x)|h|
rdx (r–)
r
=a
r
L
–rh
(r–)
Lr a
<∞. ()
Similarly, we can prove
Ra
(x)|u|(r–)dx<∞,
Ra
(x)|h|(r–)dx<∞,
Ra
(x)|u|(r–)dx<∞,
()
also
Rb
M(x)|h|(r–)dx<∞,
Rb
M(x)|u|(r–)dx<∞. ()
It follows from (), (), () and () that
∈LR,
which together with Lebesgue’s convergence theorem shows
R
f(x,un) –f(x,u)
dx→ ()
asn→ ∞. Now we have proved the lemma.
In the proof of Theorem ., the following lemma is needed.
Lemma . Assume that G⊂R is an open set.Then,for any closed set H ⊂G,there exists a functionϕ∈C∞(R)such thatϕ(x) = for all x∈R\G,ϕ(x) = for all x∈H and ≤φ(x)≤for all x∈G\H.
Proof Letting
˜ α(x) =
⎧ ⎨ ⎩
e
|x|–, |x|< ,
thenα˜∈C∞ (R) andsuppα˜=B(). For any givenε> , definingαandα
εas follows,
α(x) = α(x)˜
Rα(x)˜ dx
, αε(x) =
εα
x ε
,
one hasαε∈C∞ (R),suppαε={x:|x| ≤ε}and
Rαε(x)dx= . Denoting
d= inf
x∈H,y∈∂Gd(x,y)
and
Gθ:=
x∈G,d(x,∂G)≥θ,
it is clear thatd> andH⊂Gd. Lastly, we define
ψ(x) =
⎧ ⎨ ⎩
, x∈Gd
,
, x∈R\Gd
and
ϕ(x) =
Rψ(x–y)α d
(y)dy,
thenϕ(x) = for allx∈Handϕ(x) = for allx∈Gd
. Moreover, by the definition ofαε,
we haveϕ∈C∞(R) and ≤ϕ(x)≤.
SinceEis a Hilbert space, then there exists a basis{vn} ⊂Xsuch thatX=
j≥Xj, where
Xj=span{vj}. Letting Yk =
k
j=Xj, Zk =
j≥kXj, now we show the following lemma,
which will be used in the proof of Theorem ..
Lemma . Suppose r∈(, )and a∈L–r (R),then we have
βk(a,r) := sup u∈Zk,u=
uLra→
as k→ ∞.
Proof It is clear that <βk+(a,r)≤βk(a,r), so there existsβ(a,r)≥ such that
βk(a,r)→β(a,r) ()
ask→ ∞. By the definition ofβk(a,r), there existsuk∈Zkwithuk= such that
ukLra>
βk(a,r)
. ()
Since{uk}k∈N is bounded, then there existsu∈Esuch that
ask→ ∞. Now, since{vj}is a basis ofE, it follows that for allj∈N,
= (uk,vj) ∀k>j
→(u,vj)
ask→ ∞, which shows thatu= . By Lemma . we have
uk→
inLr
a(R) for allr∈(, ) anda∈L
–r(R), which together with () and () implies that
β(a,r) = for allr∈(, ) anda∈L–r (R).
We obtain the existence of a solution for problem () by using the following standard minimizing argument.
Lemma .[] Let E be a real Banach space and∈C(E,R)satisfying the(PS) condi-tion.Ifis bounded from below,
c:=inf
E
is a critical value of.
In order to prove the multiplicity of solutions, we will use the dual fountain theorem. Firstly, we introduce the definition of the (PS)∗ccondition.
Definition . Let∈C(E,R) andc∈R. The functionsatisfies the (PS)∗
ccondition if
any sequence{unj} ∈E, such that
(unj)→c, |
Ynj(unj)→ asnj→ ∞,
contains a subsequence converging to a critical point of.
Now we show the following dual fountain theorem.
Lemma .[] If(–u) =(u)and for every k≥k,there existsρk>γk> such that
(i) ak:=infu∈Zk,u=ρk(u)≥,
(ii) bk:=maxu∈Yk,u=γk(u) < ,
(iii) dk:=infu∈Zk,u=ρk(u)→ask→ ∞.
Moreover,if∈C(X,R)satisfies the(PS)∗
c condition for all c∈[dk, ),thenhas a
sequence of critical points{uk}such that(uk)→–as k→ ∞.
3 Proof of theorems
Define the functionalI:E×D,(R)→Rby
I(u,φ) = u
–
R|∇
φ|dx+
Rφu
dx–
It is easy to know thatIexhibits a strong indefiniteness, namely it is unbounded both from below and from above on an infinitely dimensional subspace. This indefiniteness can be removed using the reduction method described in [], by which we are led to study a variable functional that does not present such a strong indefinite nature.
Now we recall this method. For anyu∈E, consider the linear functionalTu:D,(R)→
Rdefined as
Tu(v) =
Ru
v dx.
By the Hölder inequality and using the second inequality in (), we have
Ru
v dx≤u
L/vL
≤ uL/vL
≤C/C∗uvD,.
So,Tuis continuous onD,(R). Set
μ(u,v) =
R∇u· ∇v dx
for allu,v∈D,(R). Obviously,μ(u,v) is bilinear, bounded and coercive. Hence, the Lax-Milgram theorem implies that for everyu∈E, there exists a uniqueφu∈D,(R) such
that
Tu(v) =μ(φu,v)
for anyv∈D,(R), that is,
Ru
v dx=
R∇
φu· ∇v dx
for anyv∈D,(R). Using integration by parts, we get
R∇φu· ∇v dx= –
Rvφudx
for anyv∈D,(R), therefore
–φu=u ()
in a weak sense. We can write an integral expression forφuin the form
φu=
π
R
u(y) |x–y|dy
It follows from () that
Rφuu
dx=
Rφu(–φu)dx=
R|∇φu|
dx, ()
and by the Hölder inequality, we have
φuD, =
R
φuudx
≤
R
φudx
/
R|u|
/
=C∗φuD,u
L/,
and it follows that
φuD,≤C∗u
L/. ()
Hence,
R
φuudx≤C∗uL/≤C∗C/ u:=Cu. ()
So, we can consider the functional:E→Rdefined by (u) =I(u,φu). By (), the
reduced functional takes the form
(u) = u
+
R
φuudx–
R
F(x,u)dx. ()
By (), we have
F(x,u) ≤a(x) r |u|
r ()
for allx∈Rand|u| ≤δ, wherer∈(, ) anda∈L–r(R). Letu∈E, thenu∈C(R),
the space of continuous functionuonR, such thatu(x)→ as|x| → ∞. Therefore there existsT> such that
u(x) ≤δ ()
for all|x|>T. Hence, one has
|x|>T
F(x,u) dx≤
|x|>T
a(x) r u(x)
r
dx
≤ r
|x|≥T
a(x)–rdx
(–r)/
|x|≥T
u(x) dx
r/
≤ r
|x|≥T
a(x)–rdx (–r)/
ur
L
≤ r
Cr
ura
L
–r
which together with () shows thatis well defined. Furthermore, it is well known that is aCfunctional with derivative given by
(u),v=
R
(∇u· ∇v) +V(x)uv+φuuv–f(x,u)v
dx.
It can be proved that (u,φ)∈E×D,(R) is a solution of problem () if and only ifu∈E is a critical point of the functionalandφ=φu; see, for instance, [].
Lemma . Under conditions(V), (W), (W), (W),satisfies the(PS)∗c condition. Proof Assume that{unj} ⊂Eis a sequence such that
(unj)→c, |
Ynj(unj)→ asnj→ ∞.
Then there existsσ> such that
(unj) ≤σ, |
Ynj(unj) ∗
E≤σ
for allnj∈N.
Firstly, we show that{unj}is bounded. By (), we have F(x,u) ≤a(x)
r |u|
r+a(x)
r |u|
r+ bM(x)
rδr–|u|
r ()
for allu∈Randx∈R, which together with
Rφunjunjdx≥ implies
unj
= (u
nj) –
R
φunjunjdx+
R
F(x,unj)dx
≤σ+ r
Ra(x)|unj|
rdx+
r
Ra(x)|unj|
rdx
+
rδr–
RbM(x)|unj|
rdx
≤σ+ r
Ra(x) –rdx
(–r)/
R|unj|
dx
r/
+ r
Ra(x) –r dx
(–r)/
R|unj|
dx
r/
+
rδr–
RbM(x) –r dx
(–r)/
R|unj|
dx
r/
≤σ+ r
Cr
a
L
–runj
r+
r Cr
a
L
–runj
r
+
rδr–C
r
bM L
–runj
r. ()
Noting thatri< for alli= , , , sounjis bounded.
By the fact that{unj}is bounded inE, there existsu∈Eand a constantd> such that sup
nj∈N
and
unju
inEasnj→ ∞. It is obvious that
(unj) –
(u),u→ ()
and
φuu(unj–u)→ ()
asnj→ ∞. On the other hand, by (V), () and Lemma ., one has
R
f(x,unj) –f(x,u)
unjdx
≤ f(x,unj) –f(x,u)LunjL
≤ Cf(x,unj) –f(x,u)Lunj
≤ Cdf(x,unj) –f(x,u)L
→ ()
asnj→ ∞, which implies
(unj) – (u),u
nj
→ ()
asnj→ ∞. Summing up () and (), we have
(unj) – (u),u
nj–u
→ ()
asnj→ ∞. By the Hölder inequality and (), one gets
R
φunjunj(unj–u)dx≤ φunjunjLunj–uL
≤ φunjLunjLunj–uL
≤C∗φunjD,unjLunj–uL
≤C∗unj
L/unjLunj–uL
≤C∗C/CCunj
u
nj–u
≤C∗C/ CCd
<∞.
Then by Lebesgue’s convergence theorem, we have
R
asnj→ ∞, which together with () implies
R(φunjunj–
φuu)(unj–u)dx→ ()
asnj→ ∞. By Lemma . and (), we get
R
f(x,unj) –f(x,u)
(unj–u)dx
≤ f(x,unj) –f
x,u(x)Lunj–uL
≤ Cf(x,unj) –f(x,u)Lunj–u
≤ Cdf(x,unj) –f(x,u)L
→
asnj→ ∞. Moreover, an easy computation shows that
(unj) –(u),unj–u
=unj–u
+
R(φunjunj–
φuu)(unj–u)dx
–
R
f(x,unj) –f(x,u)
(unj–u)dx.
Consequently,unj–u → asnj→ ∞.satisfies the (PS)∗c condition.
Remark . Under conditions (V), (W), (W), (W),satisfies the (PS) condition. As-sume that{un} ⊂Eis a sequence such thatI(un) is bounded and
I(un)→
asn→ ∞. Then there existsσ> such that
I(un) ≤σ, I(un)∗E≤σ
for alln∈N. The rest of the proof is the same as that of Lemma ..
Proof of Theorem. For anyk∈N, we takekdisjoint open sets{i|i= , . . . ,k}such that k
i=
i⊂.
For anyε> andi, there exist a closed setHiand an open setGisuch thatHi⊂i⊂Gi
and
meas{Gi\i}<ε, meas{i\Hi}<ε.
For everyGi(i= , . . . ,k), by Lemma . there existsϕi∈C∞(Gi,R) such thatϕi|Hi= and
≤ϕi≤. Lettingvi=ϕϕii, can be extended to be a basis{vn} ⊂X. ThereforeX=
j≥Xj,
whereXj=span{vj}. Now we defineYk:=
k
j=Xj,Zk:=
By Lemma .,∈C(E,R) satisfies the (PS)∗
c condition and(u) =(–u). Hence, to
prove Theorem ., we should just show thathas the geometric property (i), (ii) and (iii) in Lemma ..
(i) By Lemma .
βk(a,r) = sup u∈Zk,u=
uLr
a→
ask→ ∞forr∈(, ) anda∈L–r (R). In view of () and the fact that
Rφuudx≥,
we have
(u) = u
+
R
φuudx–
R
F(x,u)dx
≥ |u|
–
RF(x,u)dx
≥ u
– r
Ra(x)|u|
rdx–
r
Ra(x)|u|
rdx
–
rδr–
R
bM(x)|u|rdx
≥ u
– ur
Lra
r – ur
Lra
r – ur
Lra
rδr–
≥ u
–βk(a,r)r r u
r–βk(a,r)
r
r u
r–βk(bM,r)
r
rδr– u
r. ()
Letr:=min{r,r,r}, βk:=max{βk(a,r),βk(a,r),βk(bM,r)}, C:=max{r,r,r
δr–},
thenβk→ ask→ ∞. Hence, we have
(u)≥ u
– Cβr
kur ()
whenu ≤ andβk≤. Now we can chooseρk= (βkrC)/(–r), thenρk→ ask→ ∞.
Whenkis large enough, we haveρk≤,βk≤, which together with () shows
ak:= inf u∈Zk,u=ρk
(u)≥ ρ
k> .
(ii) For anyu∈Yk, there existsλi= , , . . . ,ksuch that
u=
k
i= λivi.
Then we have
ur
Lr =
R u(x) r
dx
=
k
i= |λi|r
i
vi(x) rdx+ k
i= |λi|r
Gi\i
vi(x) rdx
=
k
i= |λi|r
i
vi(x) rdx+ k
i= |λi|r
Gi\i
|ϕi(x)|r
ϕir
≤
k
i= |λi|r
i
vi(x) rdx+ k
i= |λi|r
ϕir
meas{Gi\i}
≤
k
i= |λi|r
i
vi(x) r+ k
i= |λi|r
ϕir
ε ()
and also
u=
R
|∇u|+V(x)udx
=
k
i= λi
Gi
|∇vi|+V(x)vi
dx = k i= λivi
=
k
i=
λi. ()
Since all the norms of a finite dimensional space are equivalent, there is a constantC˜ such that
˜
Cu ≤ uLr
for allu∈Yk. By (), one has
F(x,λivi)≥–
a(x) r |λivi|
r–a(x)
r |λivi|
r– bM(x)
rδr–|λivi|
r.
Therefore, we have
k
i=
Gi\i
F(x,λivi)dx
≥–
k
i=
Gi\i
|λi|r
r a(x)|vi|
rdx–
k
i=
Gi\i
|λi|r
r a(x)|vi|
rdx
–
k
i=
Gi\i
|λi|r
rδr–bM(x)|vi|
rdx
≥–
k
i= |λi|r
r aL
–r
Gi\i
|vi|dx
r/
–
k
i= |λi|r
r aL
–r
Gi\i
|vi|dx
r/
–
k
i= |λi|r
rδr–bML–r
Gi\i
|vi|dx
r/
≥–
k
i= |λi|r
r a L –r
Gi\i
|ϕi|
ϕi
dx
–
k
i= |λi|r
r aL
–r
Gi\i
|ϕi|
ϕi
dx
r/
–
k
i= |λi|r
rδr– bM
L
–r
Gi\i
|ϕi|
ϕi
dx
r/
= – r a L –r k i= |λi|r
ϕir
meas{Gi\i}
r/
– raL
–r
k
i= |λi|r
ϕir
meas{Gi\i}
r/
–
rδr–bML–r
k
i= |λi|r
ϕir
meas{Gi\i}
r/
≥– raL
–r
k
i= |λi|r
ϕir
εr/–
raL
–r
k
i= |λi|r
ϕir
εr/
–
rδr–bML–r
k
i= |λi|r
ϕir
εr/. ()
For anyu∈Ykwithu=
k
i=λi =γk, we can chooseγksmall enough such that|λivi(x)|<
ζ for allx∈Randi= , . . . ,k, which together with (W
) implies
F(x,λivi)≥η|λivi|r ()
for allx∈iandi= , . . . ,k. Combining (), (), (), () and (), we have
(u) = u
+
R
φuudx–
RF(x,u)dx
= u
+C u
– k i= Gi
F(x,λivi)dx
≤ u
–
k
i=
Gi\i
F(x,λivi)dx+
i
F(x,λivi)dx
≤ u
+C u
+ raL
–r
k
i= |λi|r
ϕir
εr/
+ raL
–r
k
i= |λi|r
ϕir
εr/+
rδr–bML–r
k
i= |λi|r
ϕir
εr/
–η
k
i= |λi|r
i
|vi|rdx
= u
+C u
+ raL
–r
k
i= |λi|r
ϕir
εr/
+ r a L –r k i= |λi|r
ϕir
εr/+
rδr– bM L –r k i= |λi|r
ϕir
–η
ur
Lr–
k
i= |λi|r
ϕir
ε
≤ u
+C u
–ηC˜rur+
r a L –r k i= |λi|r
ϕir
εr/
+ r a L –r k i= |λi|r
ϕir
εr/+
rδr– bM L –r k i= |λi|r
ϕir
εr/
+η
k
i= |λi|r
ϕir
ε = k i= λi +C
k
i= λi
–ηC˜r
k
i= λi
r/
+ r a L –r k i= |λi|r
ϕir
εr/
+ r a L –r k i= |λi|r
ϕir
εr/+
rδr– bM L –r k i= |λi|r
ϕir
εr/
+η
k
i= |λi|r
ϕir
ε = γ k + C γ
k –η(Cγ˜ k)r+
raL
–r
k
i= |λi|r
ϕir
εr/
+ raL
–r
k
i= |λi|r
ϕir
εr/+
rδr–bML–r
k
i= |λi|r
ϕir
εr/
+η
k
i= |λi|r
ϕir
ε
≤γk+C γ
k –η(Cγ˜ k)r
for allu∈Ykwithu=γk, whenεandγkare both small enough. Sincer< , we can
chooseγk<ρksmall enough such that
bk:= max u∈Yk,u=γk
(u) < .
(iii) By (), for anyu∈Zkwithu=ρk, we have
(u)≥–Cβkrur. Therefore
≥ inf
u∈Zk,u≤ρk
(u)≥–Cβkrρkr.
Sinceβk,ρk→ ask→ ∞, we have
dk:= inf u∈Zk,u≤ρk
(u)→
Hence, by Lemma ., we obtain that problem () has infinitely many solutions{(uk,φk)}
satisfying
R
|∇uk|+V(x)uk
dx–
R|∇
φk|dx+
R
φkukdx–
RF(x,uk)dx→
–
ask→ ∞.
Proof of Theorem. Similar to (), there exist constantski> ,i= , , , such that
(u)≥ u
–
i=
kiuri ()
for allu∈E. Since <ri< , it follows from () that the functionalis bounded from
below. By Lemma . and Remark .,possesses a critical pointusatisfying
(u) =inf
E ,
(u) = .
It remains to show thatuis nontrivial. For everyε> , there exist an open setGand a closed setHsuch thatH⊂⊂Gand
meas{G\}<ε, meas{\H}<ε.
By Lemma ., there exists a functionϕ∈C∞ (R) such that ≤ϕ(x)≤ andϕ|
H(x) = ,
ϕ|R\G(x) = , thenϕ∈E. Choosing <λ<min{δ,ζ}, then|λϕ(x)|<δfor allx∈R, which
together with () shows
Fx,λϕ(x)≥–a(x) r λϕ(x)
r
for allx∈R. Therefore, one has
G\H
F(x,λϕ)dx≥–
G\H
λr
r a(x)ϕ
rdx
≥–λ
r
r
G\H
a(x)–rdx
(–r)/
G\H
ϕdx
r/
≥–λ
r
r aL
–r
G\H
dx
r/
≥–λ
r
r aL
–r
meas{G\H}r/ ≥–λ
r
r a
L
–r(ε)
r/. ()
In view ofλ<ζ, we have|λϕ(x)|<ζfor allx∈R, which together with (W) implies
for allx∈. It follows from (), (), () that
(λϕ) = λ
ϕ +
R
φλϕ(λϕ)dx–
R
F(x,λϕ)dx
≤λ ϕ
+Cλϕ–
RF(x,λϕ)dx
≤λ ϕ
+Cλϕ–
G
F(x,λϕ)dx
= λ
ϕ
+Cλϕ–
H
F(x,λϕ)dx+
G\H
F(x,λϕ)dx
≤λ ϕ
+Cλϕ–λr
H
η|ϕ|rdx+λ
r
r aL
–r(ε)
r/
≤λϕ+Cλϕ–λrηmeas{H}
<
whenεandλare both small enough. Since() = , thenu= . Hence, (u,φu) is a
non-trivial solution of problem ().
Competing interests
The author declares that she has no competing interests.
Acknowledgements
The author is highly grateful for the referees’ careful reading and comments on this paper. This work is partially supported by the National Natural Science Foundation of China (No. 11071198) and Southwest University Doctoral Fund Project (SWU112107).
Received: 24 September 2012 Accepted: 5 July 2013 Published: 30 July 2013 References
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doi:10.1186/1687-2770-2013-177
Cite this article as:Lv:Existence and multiplicity of solutions for a class of sublinear Schrödinger-Maxwell equations.
