In this paper, we study the following Schrödinger-Poisson equations with double critical exponents:
{−Δu=|u|4u+ϕ|u|3u+λu,inΩ,−Δϕ=|u|5,inΩ,u=ϕ=0,on∂Ω,
where Ω is a bounded domain in R3 with Lipschitz boundary, λ is a real parameter satisfying suitable conditions. Using variational methods, we show the existence and nonexistence of nontrivial solutions for the Schrödinger-Poisson equations.
Citation: Li Cai, Fubao Zhang. The Brezis-Nirenberg type double critical problem for a class of Schrödinger-Poisson equations[J]. Electronic Research Archive, 2021, 29(3): 2475-2488. doi: 10.3934/era.2020125
[1] | Li Cai, Fubao Zhang . The Brezis-Nirenberg type double critical problem for a class of Schrödinger-Poisson equations. Electronic Research Archive, 2021, 29(3): 2475-2488. doi: 10.3934/era.2020125 |
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[9] | Liyan Wang, Jihong Shen, Kun Chi, Bin Ge . On a class of double phase problem with nonlinear boundary conditions. Electronic Research Archive, 2023, 31(1): 386-400. doi: 10.3934/era.2023019 |
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In this paper, we study the following Schrödinger-Poisson equations with double critical exponents:
{−Δu=|u|4u+ϕ|u|3u+λu,inΩ,−Δϕ=|u|5,inΩ,u=ϕ=0,on∂Ω,
where Ω is a bounded domain in R3 with Lipschitz boundary, λ is a real parameter satisfying suitable conditions. Using variational methods, we show the existence and nonexistence of nontrivial solutions for the Schrödinger-Poisson equations.
In this paper, we are concerned with the existence of solutions for the following Schrödinger-Poisson equations
{−Δu=|u|4u+ϕ|u|3u+λu,inΩ,−Δϕ=|u|5,inΩ,u=ϕ=0,on∂Ω, | (1) |
where
−Δu=|u|2∗−2u+λu,inΩ, | (2) |
where
{−Δu=qϕ|u|3u+λu,inBR,−Δϕ=q|u|5,inBR,u=ϕ=0,on∂BR, | (3) |
where
{−Δu+ηϕu=μu−γ,inΩ,−Δϕ=u2,inΩ,u>0,inΩ,u=ϕ=0on∂Ω, | (4) |
where
Further, Fashun Gao and Minbo Yang [12] considered the following general Choquard equation in a bounded domain
−Δu=(Iμ∗|u|2∗μ)|u|2∗μ−2u+λu,inΩ. | (5) |
They proved that if
−Δu+V(x)u=(1|x|μ∗|u|p)|u|p−2u,inR3. | (6) |
A large number of results have been obtained about the existence and qualitative properties of solutions to equation
Also, a great deal of work on elliptic equations can been founded in [4,11,13,14,15,16,17,23,25,28,30].
Recently, in [7], we studied the Brezis-Nirenberg type double critical problem for the Choquard equation
−Δu=α|u|2∗−2u+β(Iμ∗|u|2∗μ)|u|2∗μ−2u+λu,inΩ, | (7) |
where
In the following, we will give some notations which will be used throughout this paper. Let
H1(R3):={u∈L2(R3):∇u∈L2(R3)}. |
D1,2(R3)={u∈L6(R3):∇u∈L2(R3)} |
with the corresponding norm
Define
S:=infD1,2(R3)∖{0}∫R3|∇u|2dx(∫R3|u|6dx)13. |
By [26], we have
S(Ω):=infD1,20(Ω)∖{0}∫Ω|∇u|2dx(∫Ω|u|6dx)13=S. |
Firstly, we observe that by the Lax-Milgram theorem, for given
Lemma 1.1. The following properties hold:
|ϕu|D1,2≤S−12|u|56 |
and
∫Ωϕu|u|5≤S−1|u|106; |
Next we will denote the sequence of eigenvalues of the operator
0<λ1≤⋯≤λi≤λi+1≤⋯ |
and
Ei+1:={u∈H10(Ω):(u,ej)H10=0,∀j=1,2,⋯,i}, | (8) |
while
Now we introduce the energy functional associated to
I(u)=12∫Ω|∇u|2dx−16∫Ω|u|6dx−110∫Ωϕu|u|5dx−λ2∫Ω|u|2dx. |
It is easy to know that
Definition 1.2. Let
Now we state the main results of this paper.
Theorem 1.3.
Remark 1. If
Theorem 1.4. If
This paper is organized as follows. In Section
Lemma 2.1. [27] Let
Lemma 2.2. Let
Proof. Since
|I(un)|≤C,|(I′(un),un‖un‖)|≤C. |
In order to prove the boundedness of
Case 1.
For
C(1+‖un‖)≥I(un)−16(I′(un),un)≥13(‖un‖2−λ|un|22)≥δ13‖un‖2 |
for
Case 2.
Let
C(1+‖un‖)≥I(un)−θ(I′(un),un)≥(12−θ)(‖un‖2−λ|un|22)+(θ−16)|un|66 |
+(θ−110)∫Ωϕu|u|5dx≥(12−θ)(‖xn‖2+‖yn‖2−λ|xn|22−λ|yn|22)+(θ−16)|un|66≥(12−θ)(δ2‖xn‖2+(λ1−λ)|yn|22)+(θ−16)|un|66≥(12−θ)(δ2‖xn‖2+(λ1−λ)|yn|22)+(θ−16)|un|66 |
for
(12−θ)(λ−λ1)|yn|22+C(1+‖un‖)≥(12−θ)δ2‖xn‖2+(θ−16)|un|66≥(12−θ)δ2‖xn‖2+C(θ−16)|un|62≥(12−θ)δ2‖xn‖2+C(θ−16)|yn|62. |
So, from the fact that
∫Ω|∇u0|2dx=λ∫Ωu20dx+∫Ωϕu|u|5dx+∫Ωu60dx. |
And we can deduce that
I(u0)≥25∫Ωϕu|u|5dx+13∫Ωu60dx≥0. |
This completes the proof.
Lemma 2.3. Let
0<c<13−√530(√5−12)12S32, |
then there exists
Proof. From Lemma
c←I(un)=12∫Ω|∇un|2dx−16∫Ωu6ndx−110∫Ωϕun|un|5dx+on(1) | (9) |
and
on(1)=(I′(un),un)=∫Ω|∇un|2dx−∫Ωu6ndx−∫Ωϕun|un|5dx+on(1). | (10) |
From
∫Ω|∇un|2dx→a |
and
∫Ωu6ndx+∫Ωϕun|un|5dx→a, |
as
∫Ω|un|6dx=∫Ω∇ϕun∇undx≤t22∫Ω|∇un|2dx+12t2∫Ω|∇ϕun|2dx=12t2∫Ωϕun|un|5dx+t22∫Ω|∇un|2dx, |
thus, as
b≤12t2l+t22a. |
Taking
l≥3−√52a. |
Then from
c=25l+13b=13a+115l≥13−√530a. |
On the other hand, it follows from Lemma
a≤S−6a5+S−3a3. |
Hence, we have either
Lemma 3.1. If
Proof.
I(u)≥12∫Ω|∇u|2dx−λ2λ1∫Ω|∇u|2dx−16|u|66−110S−1|u|106≥12(1−λλ1)‖u‖2−16S−3‖u‖6−110S−6‖u‖10. |
So we can choose some
I(tu1)≤t22∫Ω|∇u1|2dx−λt22∫Ωu21dx−t66∫Ωu61dx−t1010∫Ωϕu1|u1|5dx<0 |
for
By Lemma
c∗:=infγ∈Γmaxt∈[0,1]I(γ(t))>0, |
where
Γ:={γ∈C([0,1],H10(Ω)):γ(0)=0,I(γ(1))<0}. |
Without loss of generality, we may assume that
uϵ(r)=ψ(r)(ϵ+r2)12. |
Then we have
∫Ω|∇uϵ|2dx=∫BkR|∇uϵ|2dx=∫BR|∇uϵ|2dx=SKϵ12+w∫R0|ψ′(r)|2dr+O(ϵ12), | (11) |
(∫Ωu6ϵdx)13=(∫BkRu6ϵdx)13=(∫Ru6ϵdx)13=Kϵ12+O(ϵ12), | (12) |
and
∫Ωu2ϵdx=∫BkRu2ϵdx=∫BkRu2ϵdx=w∫R0ψ2(r)dr+O(ϵ12), | (13) |
where
∫Ωu6dx≤1√5−1∫Ωϕ|u|5dx+√5−14∫Ω|∇u|2dx, |
hence we can introduce the new functional
J(u):=13−√520∫Ω|∇u|2dx−λ2∫Ωu2dx−2+3√530∫Ωu6dx. |
Obviously we have
Proof of Theorem
0<maxt≥0I(tv)≤maxt≥0J(tv)=maxt≥0{13−√520t2∫Ω|∇v|2dx−λt22∫Ωv2dx−2+3√530t6∫Ωv6dx}, |
Set
y(t):=13−√520t2b1−t22c1−2+3√530t6d1, |
where
b1:=∫Ω|∇v|2dx, |
c1=λ∫Ωv2dx |
d1:=∫Ωv6dx. |
Let
y′=13−√510b1t−c1t−2+3√55d1t5=0, |
which yields that there exists unique
ymax<13−√530(√5−12)12S32. |
Set
t4ϵ=13−√510∫BR|∇uϵ|2dx−λ∫BRu2ϵdx2+3√55∫BRu6ϵdx, |
which yields that
J(tϵuϵ)≤13(13−√510∫BR|∇uϵ|2dx−λ∫BRu2ϵdx)32(2+3√55∫BRu6ϵdx)12. | (14) |
In addition, when
∫R0|ψ′(r)|2dr=π24R2∫R0ψ2(r)dr. | (15) |
Then we take
In this section, we will consider the cross eigenvalue case, i.e., suppose
Theorem 4.1. Let
Then
c=infh∈Γmaxu∈QI(h(u)), |
where
Γ={h∈C(ˉQ,E)|h=idon∂Q}. |
Lemma 4.2. If
Proof.
I(u)≥12∫Ω|∇u|2dx−λ2λi+1∫Ω|∇u|2dx−16S−3‖u‖6−110S−6‖u‖10≥12(1−λλi+1)‖u‖2−16S−3‖u‖6−110S−6‖u‖10. |
We can choose some
∫Ω|∇u|2dx=i∑j=1l2j|∇ej|22,∫Ωu2dx=i∑j=1l2j. |
Then we can get
I(u)=12i∑j=1l2j(|∇ej|22−λ)−16∫Ωu6dx−110∫Ωϕu|u|5dx<12i∑j=1l2j(λj−λ)≤0. |
I(u)=12‖u‖2−λ2|u|22−16|u|66−110∫Ωϕu|u|5dx≤12‖u‖2−C∗‖u‖6, |
where
It is well known that
Without loss of generality, we may assume that
For
Uϵ(x):=ϵ−12U(xϵ),uϵ(x):=φ(x)Uϵ(x), |
then in the view of [12], we have
|∇Uϵ|22=|Uϵ|66=S32, |
and as
∫Ω|∇uϵ|2dx=S32+O(ϵ), | (16) |
∫Ω|uϵ|6dx=S32+O(ϵ3), | (17) |
and
∫Ω|uϵ|2dx≥C0ϵ, | (18) |
where
Define the linear space for any
Gi,ϵ:=span{e1,⋯,ei,uϵ}. | (19) |
Now we give the proof of Theorem 1.3 for the general case.
Proof of Theorem
u=v+tuϵ,∀u∈Gi,ϵ, |
where
Gi,ϵ=Yi⊕Ruϵ. |
From Lemma
infu∈Ei+1,‖u‖=ρ2I(u)≥γ2>0,supu∈Yi∖{0}I(u)<0,supu∈Gi,ϵ,‖u‖≥RI(u)≤0, |
where
ˉc=infγ∈Γmaxu∈VI(γ(u))>0, |
where
Γ:={γ∈C(ˉV,H10(Ω)):γ=idon∂V} |
and
V:=(ˉBR∩Yi)⊕{ruϵ:r∈(0,R)}. |
For any
ˉc≤maxu∈VI(γ(u)) |
and in particular, if we take
ˉc≤maxu∈VI(u). |
Then we have
I(u)=12∫Ω|∇u|2dx−λ2∫Ωu2dx−16∫Ωu6dx−110∫Ωϕu|u|5dx=12∫Ω|∇(v+tuϵ)|2dx−λ2∫Ω|v+tuϵ|2dx−16∫Ω|v+tuϵ|6dx−110∫Ωϕv+tuϵ|v+tuϵ|5dx≤(λi−λ)2|v|22+t22(∫Ω|∇uϵ|2dx−λ∫Ω|uϵ|2dx)+C1t|uϵ|1|v|2−t66∫Ω|uϵ|6dx−∫Ω|tuϵ|5v(x)dx−t1010∫Ωϕuϵ|uϵ|5dx−∫Ωϕtuϵ|tuϵ|4vdx≤(λi−λ)2|v|22+C1t|v|2O(ϵ12)+t22(∫Ω|∇uϵ|2dx−λ∫Ωu2ϵdx)−t66∫Ωu6ϵdx−t1010∫Ωϕuϵ|uϵ|5dx+|v|2t5O(ϵ12)+|v|2t9O(ϵ12)≤C22(λ−λi)O(ϵ)+t22(∫Ω|∇uϵ|2dx−λ∫Ωu2ϵdx)−t66∫Ωu6ϵdx−t1010∫Ωϕuϵ|uϵ|5dx, | (20) |
where in the view of boundedness of
(λi−λ)2|v|22+C1t|v|2O(ϵ12)+|v|2t5O(ϵ12)+|v|2t9O(ϵ12)≤(λi−λ)2|v|22+C|v|2O(ϵ12)≤C22(λ−λi)O(ϵ). |
Then take
∫Ω|∇uϵ|2dx−λ∫Ωu2ϵdx≤S32+O(ϵ)−λC0ϵ. |
Since
∫Ωu6ϵdx=∫Ω∇ϕuϵ∇uϵdx≤12∫Ω|∇ϕuϵ|2dx+12∫Ω|∇uϵ|2dx=12∫Ωϕuϵ|uϵ|5dx+12∫Ω|∇uϵ|2dx. |
For
∫Ωϕuϵ|uϵ|5dx≥2∫Ωu6ϵdx−∫Ω|∇uϵ|2dx=S32+O(ϵ). |
We have
C22(λ−λi)O(ϵ)+t22(∫Ω|∇uϵ|2dx−λ∫Ωu2ϵdx)−t66∫Ωu6ϵdx−t1010∫Ωϕuϵ|uϵ|5dx≤C3ϵ+t22S32−t66S32−t1010S32+t22(O(ϵ)−λC0ϵ)−t66O(ϵ3)+t1010O(ϵ)<t22S32−t66S32−t1010S32≤13−√510(√5−12)12S32 |
for
maxu∈Gi,ϵI(u)<13−√530(√5−12)12S32. |
Taking
In this section, we will prove the nonexistence of nontrivial solutions for
Proof of Theorem
0=(Δu+λu+ϕ|u|3u+|u|4u)(x⋅∇u)=div[(∇u)(x⋅∇u)−x|∇u|22+λ2xu2+15xϕ|u|5+x|u|6]+12|∇u|2−32λu2−35ϕ|u|5−15(x⋅∇ϕ)|u|5−12|u|6 |
and
0=(Δϕ+|u|5)(x⋅∇ϕ)=div[(∇ϕ)(x⋅∇ϕ)−x|∇ϕ|22]+12|∇ϕ|2+(x⋅∇ϕ)|u|5. |
Assume that
−12‖∇u‖22−12∫∂Ω|∂u∂ν|2x⋅ν=−3λ2|u|22−35∫Ωϕ|u|5dx−15∫Ω(x⋅∇ϕ)|u|5−12|u|6 | (21) |
and
−12‖∇ϕ‖22−12∫∂Ω|∂ϕ∂ν|2x⋅ν=∫Ω(x⋅∇ϕ)|u|5. | (22) |
Taking
−12‖∇u‖22−12∫∂Ω|∂u∂ν|2x⋅ν=−3λ2|u|22−35∫Ωϕ|u|5dx+110‖∇ϕ‖22+110∫∂Ω|∂ϕ∂ν|2x⋅ν−12∫Ω|u|6. |
In addition, since
‖∇u‖22=λ|u|22+∫Ωϕ|u|5+|u|66 |
and
−λ|u|22+12∫∂Ω|∂u∂ν|2x⋅ν+110∫∂Ω|∂ϕ∂ν|2x⋅ν=0. |
In the view of
The authors would like to express their sincere gratitude to the referees for careful reading the manuscript and valuable suggestions. The work was supported by the National Natural Science Foundation of China (Nos. 11671077).
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