This paper is concerned with the following Schrödinger-Poisson system
(Pμ):−Δu+u+K(x)ϕu=|u|p−1u+μh(x)u, −Δϕ=K(x)u2, x∈R3,
where p∈(3,5), K(x) and h(x) are nonnegative functions, and μ is a positive parameter. Let μ1>0 be an isolated first eigenvalue of the eigenvalue problem −Δu+u=μh(x)u, u∈H1(R3). As 0<μ≤μ1, we prove that (Pμ) has at least one nonnegative bound state with positive energy. As μ>μ1, there is δ>0 such that for any μ∈(μ1,μ1+δ), (Pμ) has a nonnegative ground state u0,μ with negative energy, and u0,μ(n)→0 in H1(R3) as μ(n)↓μ1. Besides, (Pμ) has another nonnegative bound state u2,μ with positive energy, and u2,μ(n)→uμ1 in H1(R3) as μ(n)↓μ1, where uμ1 is a bound state of (Pμ1).
Citation: Lirong Huang, Jianqing Chen. Existence and asymptotic behavior of bound states for a class of nonautonomous Schrödinger-Poisson system[J]. Electronic Research Archive, 2020, 28(1): 383-404. doi: 10.3934/era.2020022
[1] | Lirong Huang, Jianqing Chen . Existence and asymptotic behavior of bound states for a class of nonautonomous Schrödinger-Poisson system. Electronic Research Archive, 2020, 28(1): 383-404. doi: 10.3934/era.2020022 |
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This paper is concerned with the following Schrödinger-Poisson system
(Pμ):−Δu+u+K(x)ϕu=|u|p−1u+μh(x)u, −Δϕ=K(x)u2, x∈R3,
where p∈(3,5), K(x) and h(x) are nonnegative functions, and μ is a positive parameter. Let μ1>0 be an isolated first eigenvalue of the eigenvalue problem −Δu+u=μh(x)u, u∈H1(R3). As 0<μ≤μ1, we prove that (Pμ) has at least one nonnegative bound state with positive energy. As μ>μ1, there is δ>0 such that for any μ∈(μ1,μ1+δ), (Pμ) has a nonnegative ground state u0,μ with negative energy, and u0,μ(n)→0 in H1(R3) as μ(n)↓μ1. Besides, (Pμ) has another nonnegative bound state u2,μ with positive energy, and u2,μ(n)→uμ1 in H1(R3) as μ(n)↓μ1, where uμ1 is a bound state of (Pμ1).
In this paper, we study a class of Schrödinger-Poisson system with the following version
{−Δu+u+K(x)ϕu=|u|p−1u+μh(x)u in R3,−Δϕ=K(x)u2 in R3, | (1) |
where
{−Δu+u+ϕu=f(u) in R3,−Δϕ=u2 in R3, | (2) |
which has been derived from finding standing waves of the Schrödinger-Poisson system
{iψt−Δψ+ϕψ=f(ψ) in R3,−Δϕ=|ψ|2 in R3. |
A starting point of studying system (1) is the following fact. For any
ϕu(x)=14π∫R3K(y)|u(y)|2|x−y|dy |
such that
−Δu+u+K(x)ϕuu=|u|p−1u+μh(x)u,u∈H1(R3). | (3) |
Problem (3) can be also looked on as a usual semilinear elliptic equation with an additional nonlocal perturbation
From Lemma 2.1, we know that under the condition (A1), the following eigenvalue problem
−Δu+u=μh(x)u,u∈ H1(R3) |
has a first eigenvalue
F(u):=∫R3K(x)ϕu(x)|u(x)|2dx |
and introduce the energy functional
Iμ(u)=12‖u‖2+14F(u)−∫R3(1p+1|u|p+1+μ2h(x)u2)dx, |
where
⟨I′μ(u),v⟩=∫R3(∇u∇v+uv+K(x)ϕuuv−|u|p−1uv+μh(x)uv)dx. |
It is known that there is a one to one correspondence between solutions of (3) and critical points of
Theorem 1.1. Suppose that the assumptions of
The second result is about
Theorem 1.2. Under the assumptions of
The proofs of Theorem 1.1 and Theorem 1.2 are based on critical point theory. There are several difficulties in the road of getting critical points of
While for nonautonomous version of Schrödinger-Poisson system, only a few results are known in the literature. Jiang et.al.[21] have studied the following Schrödinger-Poisson system with non constant coefficient
{−Δu+(1+λg(x))u+θϕ(x)u=|u|p−2u in R3,−Δϕ=u2in R3,lim|x|→∞ϕ(x)=0, |
in which the authors prove the existence of ground state solution and its asymptotic behavior depending on
{−Δu+u+ϕu=V(x)|u|4u+μP(x)|u|q−2uin R3,−Δϕ=u2in R3,2<q<6,μ>0 |
has been studied by Zhao et al. [31]. Besides some other conditions, Zhao et. al. [31] assume that
{−Δu+u+L(x)ϕu=g(x,u) in R3,−Δϕ=L(x)u2 in R3. | (4) |
Besides some other conditions and the assumption
−Δu=a(x)|u|p−2u+˜μk(x)u, in RN, | (5) |
Costa et.al.[14] have proven the mountain pass geometry for the related functional of (5) when
This paper is organized as follows. In Section 2, we give some preliminaries. Special attentions are focused on several lemmas analyzing the Palais-Smale conditions of the functional
Notations. Throughout this paper,
Sp+1=infu∈H1(R3)∖{0}∫R3(|∇u|2+|u|2)dx(∫R3|u|p+1dx)2p+1. |
For any
In this section, we give some preliminary lemmas, which will be helpful to analyze the (PS) conditions for the functional
Lu(v)=∫R3K(x)u2vdx,v∈D1,2(R3), |
one may deduce from the Hölder and the Sobolev inequalities that
|Lu(v)|≤C‖u‖2L125‖v‖L6≤C‖u‖2L125‖v‖D1,2. | (6) |
Hence, for any
ϕu(x)=14π∫R3K(y)u2(y)|x−y|dy. |
Clearly
‖ϕu‖2D1,2=∫R3|∇ϕu|2dx=∫R3K(x)ϕuu2dx. | (7) |
Using (6) and (7), we obtain that
‖ϕu‖L6≤C‖ϕu‖D1,2≤C‖u‖2L125≤C‖u‖2. | (8) |
Then we deduce that
∫R3K(x)ϕu(x)u2(x)dx≤C‖u‖4. | (9) |
Hence on
F(u)=∫R3K(x)ϕu(x)u2(x)dx | (10) |
and
Iμ(u)=12‖u‖2+14F(u)−∫R3(1p+1|u|p+1+μ2h(x)u2)dx | (11) |
are well defined and
⟨I′μ(u),v⟩=∫R3(∇u∇v+uv+K(x)ϕuuv−|u|p−1uv−μh(x)uv)dx. |
The following Lemma 2.1 is a direct consequence of [28,Lemma 2.13].
Lemma 2.1. Assume that the hypothesis
Using the spectral theory of compact symmetric operators on Hilbert space, the above lemma implies the existence of a sequence of eigenvalues
−Δu+u=μh(x)u,inH1(R3) |
with
μ1=infu∈H1(R3)∖{0}‖u‖2∫R3h(x)u2dx,μn=infu∈S⊥n−1∖{0}‖u‖2∫R3h(x)u2dx, |
where
Next we analyze the
Definition 2.2. For
Lemma 2.3. Let
Proof. For
d+1+o(1)‖un‖=Iμ(un)−14⟨I′μ(un),un⟩=14‖un‖2−μ4∫R3h(x)u2ndx+p−34(p+1)∫R3|un|p+1dx. | (12) |
Note that
∫R3h(x)u2ndx≤(∫R3|un|p+1dx)2p+1(∫R3|h(x)|p+1p−1dx)p−1p+1≤2ϑp+1∫R3|un|p+1dx+p−1p+1ϑ−2p−1∫R3|h(x)|p+1p−1dx. |
Choosing
d+1+o(1)‖un‖≥14‖un‖2−D(p,h)μp+1p−1, | (13) |
where
The following lemma is a variant of Brezis-Lieb lemma. One may find the proof in [20].
Lemma 2.4. [20] If a sequence
limn→∞F(un)=F(u0)+limn→∞F(un−u0). |
Lemma 2.5. There is a
Iμ(u)>−p−12(p+1)Sp+1p−1p+1. |
Proof. Since
Iμ(u)=12(‖u‖2−μ∫R3h(x)u2dx)+14F(u)−1p+1∫R3|u|p+1dx=p−12(p+1)(‖u‖2−μ∫R3h(x)u2dx)+p−34(p+1)F(u). |
Noticing that
Iμ1(u)≥p−34(p+1)F(u)>0. |
Next, we claim: there is a
Iμ(u)>−p−12(p+1)Sp+1p−1p+1. |
Suppose this claim is not true, then there is a sequence
Iμ(n)(uμ(n))≤−p−12(p+1)Sp+1p−1p+1. |
Note that
Iμ(n)(uμ(n))+o(1)‖uμ(n)‖≥Iμ(n)(uμ(n))−14⟨I′μ(n)(uμ(n)),uμ(n)⟩≥14‖uμ(n)‖2−D(p,h)(μ(n))p+1p−1. |
This implies that
‖uμ(n)‖2−μ(n)∫R3h(x)(uμ(n))2dx≥(1−μ(n)μ1)‖uμ(n)‖2→0 |
because
Iμ(n)(uμ(n))=p−12(p+1)(‖uμ(n)‖2−μ(n)∫R3h(x)(uμ(n))2dx)+p−34(p+1)F(uμ(n)), |
we deduce that
lim infn→∞Iμ(n)(uμ(n))≥p−34(p+1)lim infn→∞F(uμ(n))≥0, |
which contradicts to the
Iμ(n)(uμ(n))≤−p−12(p+1)Sp+1p−1p+1. |
This proves the claim and the proof of Lemma 2.5 is complete.
Lemma 2.6. If
Proof. Let
d+o(1)=12‖un‖2−μ2∫R3h(x)u2ndx+14F(un)−1p+1∫R3|un|p+1dx |
and
⟨I′μ(un),un⟩=‖un‖2−μ∫R3h(x)u2ndx+F(un)−∫R3|un|p+1dx. |
Then we can prove that
‖un‖2=‖u0‖2+‖wn‖2+o(1), |
F(un)=F(u0)+F(wn)+o(1) |
and
‖un‖p+1Lp+1=‖u0‖p+1Lp+1+‖wn‖p+1Lp+1+o(1). |
Using Lemma 2.1, we also have that
d+o(1)=Iμ(un)=Iμ(u0)+12‖wn‖2+14F(wn)−1p+1∫R3|wn|p+1dx. | (14) |
Noticing
‖u0‖2−μ∫R3h(x)u20dx+F(u0)=∫R3|u0|p+1dx. | (15) |
Since
o(1)=‖un‖2−μ∫R3h(x)u2ndx+F(un)−∫R3|un|p+1dx. |
Combining this with (15) as well as Lemma 2.1, we obtain that
o(1)=‖wn‖2+F(wn)−∫R3|wn|p+1dx. | (16) |
Recalling the definition of
Suppose that the case (ⅰ) occurs. We may obtain from (16) that
‖wn‖2≥Sp+1(‖wn‖2+F(wn)−o(1))2p+1. |
Hence we get that for
‖wn‖2≥Sp+1p−1p+1+o(1). | (17) |
Therefore using (14), (16) and (17), we deduce that for
d+o(1)=Iμ(un)=Iμ(u0)+12‖wn‖2+14F(wn)−1p+1∫R3|wn|p+1dx.=Iμ(u0)+p−12(p+1)‖wn‖2+p−34(p+1)F(wn)>−p−12(p+1)Sp+1p−1p+1+p−12(p+1)‖wn‖2+p−34(p+1)F(wn)>0, | (18) |
which contradicts to the condition
Next we give a mountain pass geometry for the functional
Lemma 2.7. There exist
Proof. For any
u=te1+v,where∫R3(∇v∇e1+ve1)dx=0. | (19) |
Hence we deduce that
‖u‖2=‖∇(te1+v)‖2L2+‖te1+v‖2L2=t2+‖v‖2, | (20) |
μ2∫R3h(x)v2dx≤‖v‖2,μ1∫R3h(x)e21dx=‖e1‖2=1 | (21) |
and
μ1∫R3h(x)e1vdx=∫R3(∇v∇e1+ve1)dx=0. | (22) |
We first consider the case of
Iμ1(u)=12‖u‖2+14F(u)−μ12∫R3h(x)u2dx−1p+1∫R3|u|p+1dx=12‖te1+v‖2+14F(te1+v)−μ12∫R3h(x)(te1+v)2dx−1p+1∫R3|te1+v|p+1dx≥12(1−μ1μ2)‖v‖2+14F(te1+v)−1p+1∫R3|te1+v|p+1dx≥θ1‖v‖2+14F(te1+v)−C1|t|p+1−C2‖v‖p+1. |
Next we estimate the term
F(te1+v)=14π∫R3×R3K(x)K(y)(te1(y)+v(y))2(te1(x)+v(x))2|x−y|dydx. |
Since
(te1(y)+v(y))2(te1(x)+v(x))2=t4(e1(y))2(e1(x))2+(v(y))2(v(x))2+2t3(e1(y)(e1(x))2v(y)+e1(x)(e1(y))2v(x))+2t(e1(x)v(x)(v(y))2+e1(y)v(y)(v(x))2)+t2((e1(x))2(v(y))2+4e1(y)e1(x)v(y)v(x)+(e1(y))2(v(x))2), |
we know that
|∫R3×R3K(x)K(y)(e1(y)(e1(x))2v(y)+e1(x)(e1(y))2v(x))|x−y|dydx|≤C‖v‖; | (23) |
|∫R3×R3K(x)K(y)(2(e1(x))2(v(y))2+4e1(y)e1(x)v(y)v(x))|x−y|dydx|≤C‖v‖2 | (24) |
and
|∫R3×R3K(x)K(y)(e1(x)v(x)(v(y))2+e1(y)v(y)(v(x))2)|x−y|dydx|≤C‖v‖3. | (25) |
Hence
Iμ1(u)≥θ1‖v‖2+θ2|t|4−C1|t|p+1−C2‖v‖p+1−C3|t|3‖v‖−C4|t|2‖v‖2−C5|t|‖v‖3+14F(v), |
where
t2‖v‖2≤2p+1|t|p+1+p−1p+1‖v‖2(p+1)p−1, |
|t|‖v‖3≤1p+1|t|p+1+pp+1‖v‖3(p+1)p |
and for some
|t|3‖v‖≤1q0‖v‖q0+q0−1q0|t|3q0q0−1. |
Therefore we deduce that
Iμ1(u)≥θ1‖v‖2+θ2|t|4−C3q0‖v‖q0−C3(q0−1)q0|t|3q0q0−1−2C4p+1|t|p+1−(p−1)C4p+1‖v‖2(p+1)p−1−C5p+1|t|p+1−pC5p+1‖v‖3(p+1)p−C|t|p+1−C‖v‖p+1. | (26) |
From
Iμ1(u)≥θ3‖v‖2+θ4|t|4 |
provided that
Iμ1(u)≥θ5‖u‖4for‖u‖2≤˜θ25. | (27) |
Set
Iμ(u)=Iμ1(u)+12(μ1−μ)∫Rh(x)u2dx≥θ5‖u‖4−μ−μ12μ1‖u‖2=‖u‖2(θ5‖u‖2−μ−μ12μ1)≥‖u‖2(12θ5˜θ25−14θ5˜θ25)=14θ5˜θ25‖u‖2 |
for
In this section, our aim is to prove Theorem 1.1. For
Proposition 3.1. Let the assumptions
dμ1=infγ∈Γ1supt∈[0,1]Iμ1(γ(t)) |
with
Γ1={γ∈C([0,1],H1(R3)) : γ(0)=0, Iμ1(γ(1))<0}. |
Then
Before proving Proposition 3.1, we analyze the
Lemma 3.2. If the assumptions
Proof. It suffices to find a path
supt∈[0,1]Iμ1(γ(t))<p−12(p+1)Sp+1p−1p+1. |
Define
‖UR‖2−μ1∫h(x)U2Rdx+T2RF(UR)−Tp−1R∫Up+1Rdx=0. |
If
1T2R(‖UR‖2−μ1∫h(x)U2Rdx)+F(UR)=Tp−3R∫Up+1Rdx→∞, |
which is impossible either. Hence we only need to estimate
Iμ1(tUR)≤g(t)+CF(UR), |
where
g(t)=t22(‖UR‖2−μ1∫R3h(x)U2Rdx)−|t|p+1p+1∫R3Up+1Rdx. |
Noting that under the assumptions
F(UR)≤(∫R3K(x)65U125Rdx)56(∫R3ϕ6URdx)16≤C(∫R3e−65a|x+Rθ|(U(x))125dx)56≤C(∫R3e−65aRe(65a−125(1−ε))|x|dx)56≤Ce−aR | (28) |
since
∫R3h(x)U2Rdx=∫R3h(x+Rθ)U2(x)dx≥C∫R3e−b|x+Rθ|U2(x)dx≥C∫R3e−b|x|−bRU2(x)dx≥Ce−bR∫R3e−b|x|U2(x)dx≥Ce−bR. | (29) |
It is now deduced from (28) and (29) that
supt>0Iμ1(tUR)≤supt>0g(t)+Ce−aR≤p−12(p+1)(‖UR‖2−μ1∫R3h(x)U2Rdx)p+1p−1(‖UR‖−2Lp+1)p+1p−1+Ce−aR≤p−12(p+1)Sp+1p−1p+1−Ce−bR+o(e−bR)+Ce−aR<p−12(p+1)Sp+1p−1p+1 |
for
Lemma 3.3. Under the assumptions
Proof. Let
d+o(1)=12‖un‖2−μ12∫R3h(x)u2ndx+14F(un)−1p+1∫R3|un|p+1dx |
and
⟨I′μ1(un),un⟩=‖un‖2−μ1∫R3h(x)u2ndx+F(un)−∫R3|un|p+1dx. |
Hence we can deduce that
‖un‖2=‖u0‖2+‖wn‖2+o(1),F(un)=F(u0)+F(wn)+o(1) |
and
‖un‖p+1Lp+1=‖u0‖p+1Lp+1+‖wn‖p+1Lp+1+o(1). |
Since
d+o(1)=Iμ1(un)=Iμ1(u0)+12‖wn‖2+14F(wn)−1p+1∫R3|wn|p+1dx. | (30) |
From
‖u0‖2−μ1∫R3h(x)u20dx+F(u0)=∫R3|u0|p+1dx |
and then
Iμ1(u0)≥p−12(p+1)(‖u0‖2−μ1∫R3h(x)u20dx)+p−34(p+1)F(u0)≥0. |
Now using an argument similar to the proof of (16), we obtain that
o(1)=‖wn‖2+F(wn)−∫R3|wn|P+1dx. | (31) |
By the relation
Suppose that the case (I) occurs. Then up to a sbusequence, we may obtain from (31) that
‖wn‖2≥Sp+1(‖wn‖2+F(wn)−o(1))2p+1, |
which implies that for
‖wn‖2≥Sp+1p−1p+1+o(1). |
It is deduced from this and (30) that
Proof of Proposition 3.1. Since
Proof of Theorem 1.1. By Proposition 3.1, the
dμ1≤maxt∈[0,1]Iμ1(γn(t))<dμ1+1n. | (32) |
By Ekeland's variational principle [5], there exists
{dμ1≤maxt∈[0,1]Iμ1(γ∗n(t))≤maxt∈[0,1]Iμ1(γn(t))<dμ1+1n;maxt∈[0,1]‖γn(t)−γ∗n(t)‖<1√n; there existstn∈[0,1]such thatzn=γ∗n(tn) satisfies:Iμ1(zn)=maxt∈[0,1]Iμ1(γ∗n(t)),and‖I′μ1(zn)‖≤1√n. | (33) |
By Lemma 3.2 and Lemma 3.3 we get a convergent subsequence (still denoted by
In this section, we always assume the conditions
inf{Iμ(u) : u∈M},M={u∈H1(R3) : ⟨I′μ(u),u⟩=0} |
to get a ground state solution. But for
N={u∈H1(R3)∖{0}:I′μ(u)=0}. |
And then we consider the following minimization problem
c0,μ=inf{Iμ(u):u∈N}. | (34) |
Lemma 4.1. Let
d0,μ=inf‖u‖<ρIμ(u). |
Then the
Proof. Firstly, we prove that
Iμ(u)=12‖u‖2−μ2∫R3h(x)u2dx+14F(u)−1p+1∫R3|u|p+1dx≥12‖u‖2−μ2μ1‖u‖2−C‖u‖p+1>−∞ |
as
Iμ(te1)=t22‖e1‖2−μt22∫R3h(x)e21dx+t44F(e1)−tp+1p+1∫R3|e1|p+1dx. |
It is now deduced from
Iμ(te1)=t22(1−μμ1)‖e1‖2+t44F(e1)−tp+1p+1∫R3|e1|p+1dx. |
Since
Secondly, let
Iμ(un)→d0,μandI′μ(un)→0. |
Then we can prove that
We emphasize that the above lemma does NOT mean that
Lemma 4.2. For
Proof. By Lemma 4.1, we know that
For any
Iμ(u)=Iμ(u)−14⟨I′μ(u),u⟩≥14‖u‖2−D(p,h)μp+1p−1. |
Therefore the
Now let
Iμ(un)→c0,μandI′μ(un)=0. |
Since
Next, to analyze further the
Lemma 4.3. There exists
Proof. The proof is divided into two steps. In the first place, for
‖u‖2−μ1∫R3h(x)u2dx+F(u)=∫R3|u|p+1dx |
and hence
Iμ1(u)=p−12(p+1)(‖u‖2−μ1∫R3h(x)u2dx)+p−34(p+1)F(u). |
Since
Iμ1(u)≥p−34(p+1)F(u)>0. |
In the second place, denoted by
I′μ(n)(u0,μ(n))=0 |
and we also have that
c0,μ(n)=Iμ(n)(u0,μ(n))<0. |
Hence we deduce that
Iμ(n)(u0,μ(n))=p−12(p+1)(‖u0,μ(n)‖2−μ(n)∫R3h(x)(u0,μ(n))2dx)+p−34(p+1)F(u0,μ(n)). |
Using the definition of
‖u0,μ(n)‖2−μ(n)∫R3h(x)(u0,μ(n))2dx≥(1−μ(n)μ1)‖u0,μ(n)‖2→0 |
because
Claim. As
Proof of the Claim. From
o(1)+Iμ(n)(u0,μ(n))=Iμ(n)(˜u0)+12‖˜wn‖2+14F(˜wn)−1p+1∫R3|˜wn|p+1dx, | (35) |
where
Now we distinguish two cases:
Suppose that the case (ⅰ) occurs. We may deduce from a proof similar to Lemma 2.6 that
Iμ(n)(u0,μ(n))+o(1)≥Iμ1(˜u0)+p−12(p+1)Sp+1p−1p+1, |
which is a contradiction because
Next we prove that
lim infn→∞Iμ(n)(u0,μ(n))≥p−34(p+1)F(˜u0)>0, |
which is also a contradiction since
Hence there is
Remark 4.4. The proof of Lemma 4.3 implies that (1) of Theorem 1.2 holds.
In the following, we are going to prove the existence of another nonnegative bound state solution of (3). To obtain this goal, we have to analyze further the
Lemma 4.5. Under the assumptions of
Proof. Let
d+o(1)=12‖un‖2−μ2∫R3h(x)u2ndx+14F(un)−1p+1∫R3|un|p+1dx |
and
⟨I′μ(un),un⟩=‖un‖2−μ∫R3h(x)u2ndx+F(un)−∫R3|un|p+1dx. |
Similar to the proof in Lemma 2.3, we can deduce that
‖un‖2=‖u0‖2+‖wn‖2+o(1), |
F(un)=F(u0)+F(wn)+o(1) |
and
‖un‖p+1Lp+1=‖u0‖p+1Lp+1+‖wn‖p+1Lp+1+o(1). |
Using Lemma 2.1, we also have that
d+o(1)=Iμ(un)=Iμ(u0)+12‖wn‖2+14F(wn)−1p+1∫R3|wn|p+1dx. | (36) |
Since
Iμ(u0)≥c0,μ |
and
‖u0‖2−μ∫R3h(x)u20dx+∫R3ϕu0u20=∫R3|u0|p+1dx. |
Note that
o(1)=‖un‖2−μ∫R3h(x)u2ndx+F(un)−∫R3|un|p+1dx |
imply that
o(1)=‖wn‖2+F(wn)−∫R3|wn|p+1dx. | (37) |
Using
Suppose (I) occurs. Up to a subsequence, we may obtain from (37) that
‖wn‖2≥Sp+1(‖wn‖2+F(wn)−o(1))2p+1. |
Hence we get that for
‖wn‖2≥Sp+1p−1p+1+o(1). | (38) |
Therefore using (36) and (38), we deduce that for
d+o(1)=Iμ(un)=Iμ(u0)+12‖wn‖2+14F(wn)−1p+1∫R3|wn|p+1dx=Iμ(u0)+p−12(p+1)‖wn‖2+p−34(p+1)F(wn)≥c0,μ+p−12(p+1)‖wn‖2+p−34(p+1)F(wn)>c0,μ+p−12(p+1)Sp+1p−1p+1, | (39) |
which contradicts to the assumption
Next, for the
d2,μ=infγ∈Γ2supt∈[0,1]Iμ(γ(t)) |
with
Γ2={γ∈C([0,1],H1(R3)) : γ(0)=w0,μ, Iμ(γ(1))<c0,μ}. |
Lemma 4.6. Suppose that the conditions
d2,μ<c0,μ+p−12(p+1)Sp+1p−1p+1. |
Proof. It suffices to find a path starting from
Iμ(w0+tUR)=12(‖w0+tUR‖2−μ∫R3h(x)|w0+tUR|2dx)+14F(w0+tUR)−1p+1∫R3|w0+tUR|p+1dx=Iμ(w0)+A1+A2+A3+t22‖UR‖2−μ2∫R3h(x)U2Rdx, |
where
A1=⟨w0,tUR⟩−μt∫R3h(x)w0URdx, |
A2=14(F(w0+tUR)−F(w0)) |
and
A3=1p+1∫R3(|w0|p+1−|w0+tUR|p+1)dx. |
Since
From an elementary inequality:
we deduce that
For the estimate of
Since
Similarly we can deduce that for
Since
for
Proposition 4.7. Under the conditions (A1)-(A4), if
Proof. Since for
Proof of Theorem 1.2. The conclusion (1) of Theorem 1.2 follows from Lemma 4.3 and Remark 4.4. It remains to prove (2) of Theorem 1.2. By Proposition 4.7, the
(40) |
By Ekeland's variational principle, there exists
(41) |
By Lemma 4.6 we get a convergent subsequence (still denoted by
Next, let
and
(42) |
Next, similar to the proof in Lemma 2.3, we can deduce that
which contradicts to (42). Hence
The author thanks the unknown referee for helpful comments.
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