Introduction Preliminaries Main Result Proof of Main Result References
Least energy nodal solution for a quasilinear
biharmonic equations with critical exponent in RN
Haibo Chen
(Joint work with Hongliang Liu)
School of Mathematics and Statistics
Central South University
July 2015 · Changsha
Haibo Chen
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Introduction Preliminaries Main Result Proof of Main Result References
Content
1
Introduction
Haibo Chen
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Introduction Preliminaries Main Result Proof of Main Result References
Content
1
Introduction
2
Preliminaries
Haibo Chen
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Introduction Preliminaries Main Result Proof of Main Result References
Content
1
Introduction
2
Preliminaries
3
Main Result
Haibo Chen
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Introduction Preliminaries Main Result Proof of Main Result References
Content
1
Introduction
2
Preliminaries
3
Main Result
4
Proof of Main Result
Haibo Chen
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Introduction Preliminaries Main Result Proof of Main Result References
Content
1
Introduction
2
Preliminaries
3
Main Result
4
Proof of Main Result
5
References
Haibo Chen
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Introduction Preliminaries Main Result Proof of Main Result References
Model
In this talk, we study the least energy nodal solution for the
following quasilinear biharmonic equations
∆2 u − ∆u + V (x)u −
κ
∆(u2 )u = |u|2∗ −2 u,
2
x ∈ RN ,
(1.1)
where ∆2 is the biharmonic operator, the parameter κ > 0 and
V (x) ∈ C(RN , R), 2∗ = N2N
−4 , (N ≥ 5).
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Physical Background
Most readers have surely seen the dramatic collapse of the
Tacoma Narrows suspension bridge, followed by the collapse of the
structure. There is a usual explanation of the large oscillations of
the bridge. In 1990, Lazer and McKenna [7] first used the
following biharmonic equations
2
∆ u + a∆u = d[(1 + u)+ − 1], in Ω,
(1.2)
u = ∆u = 0,
on ∂Ω,
where u+ = max{u, 0} and d ∈ R, Ω is a smooth bounded domain
of RN , to explain the oscillations of suspension bridges. More
precisely, they studied the traveling waves of problem (1.2).
For a more physical background of the biharmonic equation,
we refer the readers to [7] and references therein.
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Research Status
With the aid of the modern variational methods, the solvability of
the biharmonic equations have been widely studied in recent years.
For example, for the following semilinear (κ = 0 in (1.1))
biharmonic equations
∆2 u − ∆u + λV (x)u = f (x, u),
u ∈ H 2 (RN ).
Many authors pay their attention to study the existence of
nontrivial solutions for problem (1.3).
Haibo Chen
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(1.3)
Introduction Preliminaries Main Result Proof of Main Result References
Research Status
More precisely, Yiwei Ye and Chunlei Tang [17,2013,JMAA]
studied the existence and multiplicity solutions of problem (1.3) by
using mountain pass theorem. Their results unify and sharply
improve the results of Jiu Liu, Shaoxiong Chen and Xian Wu
[10,2012JMAA]. Wen Zhang and Xianhua Tang [21,2013, Taiwan.
J. Math.] also investigated the infinitely many solutions of problem
(1.3) with sign-changing potential V (x) via symmetric mountain
pass theorem. Yinbin Deng and Yi Li [2009,Acta Mathematica
Scientia], Fanglei Wang, Mustafa Avci and Yukun An [2014,JMAA]
studied the existence of nontrivial solution of problem (1.3). For
some other interesting results on biharmonic equations, we refer
readers to [1,5,9,16,17,18,20,22,23] and the references therein.
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Motivation
Very recently, by using mountain pass theorem and the
symmetric mountain pass theorem, Shaoxiong Chen, Jiu Liu and
Xian Wu [3,2014AMC] investigated the existence and multiplicity
results of problem (1.1) with N ≤ 6 and replacing |u|2∗ −2 u by the
generalized form f (x, u). To the best of our knowledge, their
conclusions seem to be the latest results on the quasilinear
biharmonic equation in the entire space up to now although there
are a lot of papers dealing with the semilinear case.
Motivated by the above facts, the aim of the present talk is to
study the least energy nodal solution of problem (1.1) with N ≤ 6.
As far as the authors know, there is no paper handling with this
case.
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Assumption on V (x)
Before stating our main result, we impose the following conditions
on V (x).
(V 1) V ∈ C(RN , R) satisfies inf x∈R3 V (x) ≥ a0 > 0. Moreover,
for every M > 0,
meas({x ∈ RN : V (x) ≤ M }) < ∞,
where meas denotes the Lebesgue measure in RN .
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Working Space
As usual, we let
X :=
Z
u ∈ H (R )
2
V (x)u dx < +∞ .
2
N
RN
Then, by condition (V 1), X is a Hilbert space with the inner
product and norm
Z
1
hu, vi =
[4u4v + ∇u∇u + V (x)uv] dx, kuk = hu, ui 2 .
RN
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Compactness embedding
Moreover, we have the following compactness results from [3].
Lemma 1.1. ([3, Lemma 2.1]) Under the condition (V 1),
X ,→ Lr (RN ) is continuous for 2 ≤ r ≤ 2∗ and X ,→ Lr (RN ) is
compact for 2 ≤ r < 2∗ .
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Variational Setting
Now we define a functional I on X by
Z
1
I(u) =
[|4u|2 + |∇u|2 + V (x)u2 ]dx
2 RN
Z
Z
κ
1
2
2
+
u |∇u| dx −
|u|2∗ dx,
2 RN
2∗ R N
for all u ∈ X. Since N ≤ 6, the Lemma 2.2 in [3] shows that
Z
u2 |∇u|2 dx < ∞,
RN
which implies that I(u) is well defined in X.
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(2.1)
Introduction Preliminaries Main Result Proof of Main Result References
Furthermore, we have
hI 0 (u), ϕi =
Z
[4u4ϕ + ∇u∇ϕ + V (x)uϕ]dx
RNZ
+κ
Z
−
(uϕ|∇u|2 + u2 ∇u∇ϕ)dx
RN
|u|2∗ −2 uϕdx,
RN
for all u, ϕ ∈ X.
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(2.2)
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Definition
To state our main result, we introduce the following notations and
definitions first.
Notation: Throughout this paper, we denote u+ = max{u(x), 0}
and u− = min{u(x), 0}, then u = u+ + u− . C denotes various
positive constants, which may vary from line to line. 2∗ = N2N
−4 for
N ≥ 5 and 2∗ = +∞ for N ≤ 4, is the critical Sobolev exponent
∗
for the embedding H 2 (RN ) ,→ L2 (RN ).
Definition: If u ∈ X is a solution of problem (1.1) with u± 6= 0,
then we call that u is a nodal solution of (1.1). Furthermore, if u
is a nodal solution of problem (1.1) with
I(u) = inf{I(v) : v is the nodal solution of (1.1)}, then we call
that u is the least energy nodal solution of (1.1).
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Main Results
Now, we state our main result as follows:
Theorem 1.1. Suppose that N ≤ 6 and conditions (V 1) holds.
Then problem (1.1) has a least energy nodal solution u with
u+ + u− ∈ N ± and c̄ = inf u∈N ± I(u) > 0, where
N ± := u ∈ X : u± 6= 0, hI 0 (u), u+ i = 0, hI 0 (u), u− i = 0 .
(3.1)
Remark 1.1. Under the conditions of Theorem 1.1, by using
almost the same procedure in [8] (or in [4,14]), we can easily get
that the problem (1.1) has a ground state solution ū with
I(ū) = c = inf u∈N I(u), where N = {u ∈ X : hI 0 (u), ui = 0} is a
Nehari manifold. Therefore, Theorem 1.1 is not only including but
also improving this result.
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Notations
For each u ∈ N ± , we denote
g + (u) := I(u+ ),
g − (u) := I(u− ),
G+ (u) := hI 0 (u), u+ i = hI 0 (u+ ), u+ i = 0,
(3.2)
−
0
−
0 −
−
G (u) := hI (u), u i = hI (u ), u i = 0.
(3.3)
Then we have
1
1
g + (u) = g + (u) − G+ (u) = ku+ k2 +
4
4
1
1
−
4 2∗
1
1
g − (u) = g − (u) − G− (u) = ku− k2 +
4
4
1
1
−
4 2∗
ku+ k22∗∗ ,
(3.4)
ku− k22∗∗ ,
(3.5)
1
I(u) = g + (u) + g − (u) = I(u) − hI 0 (u), ui.
4
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(3.6)
Introduction Preliminaries Main Result Proof of Main Result References
Properties of N ±
Furthermore, we have some properties on N ± which are given by
the following lemmas.
Lemma 2.1. For each u ∈ X with u± 6= 0, there exists a unique
(tu , su ) ∈ R × R with tu , su > 0 such that tu u+ + su u− ∈ N ± with
I(tu u+ + su u− ) = max{I(tu+ + su− ) : t, s ≥ 0},
and Hβ u (tu , su ) is a negative definite matrix, where Hβ u (t, s) is
the Hessian matrix of β u (t, s) := I(tu+ + su− ).
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Proof of Lemma 2.1
Proof: For u ∈ X with u± 6= 0, by the definition of β u (t, s), we
have
β u (t, s) = I(tu+ + su− ) Z= I(tu+ ) + I(su− )
1
κ
1
= t2 ku+ k2 + t4
|u+ |2 |∇u+ |2 dx − |t|2∗ ku+ k22∗∗
2
2 RZN
2∗
1
1 2 − 2
4κ
− 2
− 2
|u | |∇u | dx − |s|2∗ ku− k22∗∗ ,
+ s ku k + s
2
2 RN
2∗
which implies that β u (t, s) > 0 for t, s > 0 small and
β u (t, s) → −∞ as |(t, s)| → ∞. Noting that β u (t, s) = β u (|t|, |s|),
then there exist nonnegative real numbers tu and su such that
β u (tu , su ) = I(tu u+ + su u− ) = max{I(tu+ + su− ) : t, s ≥ 0},
and then we have G± (tu u+ + su u− ) = 0, which means that
tu u+ + su u− ∈ N ± .
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Proof of Lemma 2.1
Now, we show that tu , su > 0. Without loss of generality, we may
assume that su = 0, then if s > 0 is small enough, we have
β u (tu , 0) ≥ β u (tu , s)
= β u (tu , 0) +
κ
s2 − 2
ku k + s4
2
2
Z
|u− |2 |∇u− |2 dx
RN
1 2∗ − 2∗
|s| ku k2∗
2∗
1
s2
≥ β u (tu , 0) + ku− k2 − |s|2∗ ku− k22∗∗
2
2∗
> β u (tu , 0).
−
Obviously, this is a contradiction. Therefore, su > 0. By the similar
argument, we get tu > 0.
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Proof of Lemma 2.1
Furthermore, by a simple computation, we get that
A 0
,
Hβ u (tu , su ) =
0 B
where
κ
2
Z
κ
2κs2u
Z
A = 2κt2u
B=
2
RN
RN
|u+ |2 |∇u+ |2 dx − (2∗ − 2)t2u∗ −2 ku+ k22∗∗ ,
|u− |2 |∇u− |2 dx − (2∗ − 2)s2u∗ −2 ku− k22∗∗ .
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Proof of Lemma 2.1
In view of G+ (tu u+ + su u− ) = 0, we have
Z
κ
2κt2u
|u+ |2 |∇u+ |2 dx − t2u∗ −2 ku+ k22∗∗ < 0.
2 RN
(3.7)
Then it follows that
Z
κ
2κt2u
|u+ |2 |∇u+ |2 dx < t2u∗ −2 ku+ k22∗∗ < (2∗ − 2)t2u∗ −2 ku+ k22∗∗ ,
2 RN
(3.8)
since N ≤ 6. Similarly, we have
Z
2κ
|u− |2 |∇u− |2 dx < s2u∗ −2 ku− k22∗∗ < (2∗ − 2)s2u∗ −2 ku− k22∗∗ .
2κsu
2 RN
(3.9)
Thus, (3.8) and (3.9) mean that A < 0, B < 0 and
det Hβ u (tu , su ) > 0, that is, Hβ u (tu , su ) is a negative definite
matrix.
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Proof of Lemma 2.1
Next, we shall verify the uniqueness of (tu , su ). Suppose, reasoning
by contradiction, that there exists another (t̄u , s̄u ) with t̄u > 0 and
s̄u > 0 such that t̄u u+ + s̄u u− ∈ N ± . Then the Hessian matrix
Hβ u (t̄u , s̄u ) is also negative definite by the almost same procedures
above. Therefore, by the properties of Hessian matrix, (t̄u , s̄u ) is a
local maximum point of β u . Noting that (tu , su ) is a global
maximum point, we have β u (tu , su ) ≥ β u (t̄u , s̄u ) > 0.
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Proof of Lemma 2.1
Let
v + = t̄u u+ ,
v − = s̄u u− ,
tu
teu = ,
t̄u
seu =
su
.
s̄u
Then v = v + + v − = t̄u u+ + s̄u u− ∈ N ± and
e
tu v + + seu v − = tu u+ + su u− ∈ N ± . Moreover, we have
β v (e
tu , seu ) = β u (tu , su ) ≥ β u (t̄u , s̄u ) = β v (1, 1).
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(3.10)
Introduction Preliminaries Main Result Proof of Main Result References
Proof of Lemma 2.1
Without loss of generality, we may assume that e
tu > 0, then it
follows from (3.2) that G+ (v) = 0 and G+ (e
tu v + + seu v − ) = 0.
Hence, we get that
Z
+ 2
2∗ −2
2
2∗ −2
e
e
e
1 − tu
kv k + tu − tu
|v + |2 |∇v + |2 dx = 0,
2κ
RN
which means that e
tu = 1. Similarly, we get seu = 1. It follows from
(3.10) that tu = e
tu and su = seu . Therefore, (tu , su ) is unique. The
proof is completed. Haibo Chen
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Lemma 2.2
Lemma 2.2. For all u ∈ N ± , there exists C > 0 such that
ku± kpp ≥ C > 0. Furthermore, c̄ = inf u∈N ± I(u) > 0.
Proof: Arguing by contradiction, there exists {un } ⊂ N ± such
p
− p
that ku+
n kp → 0 or kun kp → 0 as n → ∞. Without loss of
p
generality, we may assume that ku+
n kp → 0 as n → ∞. It follows
from G+ (un ) = 0 that ku+
n k → 0 as n → ∞.
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Proof of Lemma 2.2.
On the other hand, by using the Sobolev embedding inequality and
G+ (un ) = 0 again, we have
Z
+ 2
2
+ 2
+ 2∗
+ 2∗
kun k + 2κ
|u+
n | |∇un | dx = kun k2∗ ≤ Ckun k .
RN
That is
2
+ 2∗
+ 2∗
ku+
n k ≤ kun k2∗ ≤ Ckun k ,
which implies that there exists C > 0 such that ku+
n k ≥ C since
2∗ > 2 for N ≤ 6. This is a contradiction with ku+
n k → 0 as
±
n → ∞. Hence, there exists C > 0 such that ku kpp > C for all
u ∈ N ± and c > 0 follows from (3.4), (3.5) and (3.6). We
complete the proof. Haibo Chen
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Lemma 2.3
Lemma 2.3. If there exists u ∈ N ± such that I(u) = c̄, where c̄
defined in Lemma 2.2, then u is a critical point of problem (1.1).
Proof: The proof of this lemma is almost the same as to the one
of Lemma 2.5 in [11, J. Liu 2004 Comm. Partial Differential
Equations]. See also in [2,T.Bartsch 2005 Ann. Inst. H. Poincaré
Anal. Non-Linéaire.],[6,Y.Huang 2014 Arch. Math.], [13,Z.Liu2004
Adv. Nonlinear. Stud]. So, we omit it here.
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Proof of Main Result
Proof of theorem 1.1: Let {un } ⊂ N ± be a sequence such that
I(un ) → c > 0 as n → ∞. Up to a subsequence, we may assume
that I(un ) ≤ 2c for all n. Thus, we have
1 0
1
1
1
2
2c ≥ I(un ) = I(un ) − hI (un ), un i = kun k +
−
kun k22∗∗ ,
4
4
4 2∗
which implies that
kun k2 ≤ 8c,
kun k22∗∗ ≤
82∗
c.
2∗ − 4
By Lemma 2.2 there exists C > 0 such that ku±
n k ≥ C and
p
≥
C.
Moreover,
by
Lemma
1.1,
there
exists
u ∈ X such
ku±
k
n p
±
±
±
that un * u and un * u in X as n → ∞ and un → u± in
Lτ (RN ) for τ ∈ [2, 2∗ ) as n → ∞, then u± 6= 0.
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Proof of Main Result
It follows from Lemma 2.1 that there exist t±
∗ > 0 such that
+ + t− u− ∈ N . In what follows, we shall show that t± = 1.
t+
u
±
∗
∗
∗
Noting that {un } ⊂ N ± , then G+ (un ) = 0. Furthermore, we have
Z
Z
+ 2
+ 2
+ 2
ku k + 2κ
|u | |∇u | dx =
|u+ |2∗ dx.
(4.1)
RN
RN
+
− −
On the other hand, it follows from t+
∗ u + t∗ u ∈ N ± that
Z
Z
+ 2
+ 2
+ 2∗
2
+ 2
+ 4
|u
|
|∇u
|
dx
=
|t
|
|u+ |2∗ dx,
|t+
|
ku
k
+
|t
|
2κ
∗
∗
∗
RN
RN
then we have
−2
+ 2
|t+
∗ | ku k + 2κ
Z
RN
2∗ −4
|u+ |2 |∇u+ |2 dx = |t+
∗|
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Z
RN
|u+ |2∗ dx.
(4.2)
Introduction Preliminaries Main Result Proof of Main Result References
Proof of Main Result
From (4.1) and (4.2), we get that
Z
1
2∗ −4
1 − + 2 ku+ k2 = (1 − |t+
|
)
|u+ |2∗ dx,
∗
N
|t∗ |
R
−
which yields that t+
∗ = 1. Similarly, we get that t∗ = 1. It follows
from (3.4), (3.5) and (3.6) that
+
− −
c ≤ I(t+
∗ u + t∗ u )
+ + +
−
− + +
− −
= g (t∗ u + t−
∗ u ) + g (t∗ u + t∗ u )
+
−
= g (u) + g (u)
≤ lim I(un ) = c,
n→∞
+
− −
+
−
which shows that I(t+
∗ u + t∗ u ) = I(u + u ) = c. Therefore,
+
−
by Lemma 2.3, u + u ∈ N ± is a critical point of I. We
complete the proof. Haibo Chen
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Introduction Preliminaries Main Result Proof of Main Result References
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Introduction Preliminaries Main Result Proof of Main Result References
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Introduction Preliminaries Main Result Proof of Main Result References
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Thank you!
Email: math [email protected]
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