Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 12, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SEMILINEAR ELLIPTIC PROBLEMS INVOLVING
HARDY-SOBOLEV-MAZ’YA POTENTIAL AND
HARDY-SOBOLEV CRITICAL EXPONENTS
RUI-TING JIANG, CHUN-LEI TANG
Abstract. In this article, we study a class of semilinear elliptic equations
involving Hardy-Sobolev critical exponents and Hardy-Sobolev-Maz’ya potential in a bounded domain. We obtain the existence of positive solutions using
the Mountain Pass Lemma.
1. Introduction and statement of main results
The main purpose of this article is to investigate the existence of positive solution
to the following semilinear elliptic problem with Dirichlet boundary value conditions
∗
−∆u − µ
|u|2 (s)−2 u
u
=
+ λf (x, u),
|y|2
|y|s
u > 0, in Ω,
u = 0,
in Ω,
(1.1)
on ∂Ω,
where f ∈ C(Ω × R, R), Ω is a smooth bounded domain in RN = Rk × RN −k with
N ≥ 3 and 2 ≤ k < N , a point x ∈ RN is denoted as x = (y, z) ∈ Rk × RN −k
2
and (0, z 0 ) ∈ Ω, 0 ≤ µ < µ̄ = (k−2)
for k > 2, µ = 0 for k = 2. The so-called
4
−s)
Hardy-Sobolev critical exponents are denoted by 2∗ (s) = 2(N
N −2 where 0 ≤ s < 2,
2∗ = 2∗ (0) = N2N
−2 are the Sobolev critical exponents. F (x, t) is a primitive function
Rt
of f (x, t) defined by F (x, t) = 0 f (x, s)ds. H01 (Ω) is the Sobolev space with norm
Z
1/2
u2
kuk =
(|∇u|2 − µ 2 )dx
,
|y|
Ω
which is equivalent to its general norm due to the Hardy inequality
Z
Z
u2
Ck
dx
≤
|∇u|2 dx, ∀u ∈ D1,2 (RN ),
2
RN
RN |y|
2010 Mathematics Subject Classification. 35B09, 35B33, 35J75.
Key words and phrases. Hardy-Sobolev-Maz’ya potential; Hardy-Sobolev critical exponents;
positive solution; Mountain Pass Lemma; local Palais-Smale condition.
c
2016
Texas State University.
Submitted July 14, 2015. Published January 7, 2016.
1
2
R.-T. JIANG, C.-L. TANG
EJDE-2016/12
2
where Ck = k−2
is the best constant and is not attained. Let Sµ be the best
2
Sobolev constant, namely
R
u2
|∇u|2 − µ |y|
2 dx
RN
Sµ =
inf
(1.2)
R |u|2∗ (s) 2∗2(s) .
u∈D 1,2 (RN \(0,z 0 )),u6=0
dx
s
N
|y|
R
When k = N , problem (1.1) becomes
∗
−∆u − µ
|u|2 (s)−2 u
u
=
+ λf (x, u),
2
|x|
|x|s
u > 0, in Ω,
u = 0,
in Ω,
(1.3)
on ∂Ω,
2
where 0 ≤ µ < µ̄ = (N −2)
. There are many papers concerning the Dirichlet
4
problem with critical exponents (see [1, 4, 6, 8, 11]) after the work of Breizis and
Nirenberg [3]. When µ = 0 and s = 0, problem (1.3) becomes the well-known
Brezis-Nirenberg problem, and is studied extensively in [11]. When µ 6= 0, the
problem has its singularity at 0 and attracts much attention. Ding and Tang in
[5] studied the existence of positive solutions with N ≥ 3, 0 ≤ s < 2 and f (x, t)
satisfying the (AR) condition in the case λ = 1. Kang in [7] showed the existence
of positive solutions replacing f (x, u) by |u|q−2 u with q > 2 for 0 ≤ s < 2.
When 2 ≤ k < N , the problem has stronger singularity. Bhakta and Sandeep [2]
studied the regularity, P.S. characterization and existence of solutions with λ = 0.
Wang and Wang in [12] showed that the existence of infinitely many solutions
replacing f (x, u) by u for N > 6 + s. In [13], Yang studied (1.1) with Neumann
boundary condition and obtained the existence of positive solution by the Mountain
Pass Lemma. In order to estimate the mountain pass energy, the author use the
extremal function of Sµ ([2]), which is achieved when
s=2−
N −k+
N −2
p
.
(k − 2)2 − 4µ
In this article, we estimate the mountain pass energy for 0 ≤ s < 2 through λ large
enough instead of the extremal function, which makes the results more extensive
and interested. Here is our main result.
Theorem 1.1. Suppose N ≥ 3, 2 ≤ k < N and 0 ≤ µ < µ̄. f ∈ C(Ω × R+ , R)
satisfies
(A1) f (x, t) = 0 for t ≤ 0 uniformly for x ∈ Ω. There exists a nonempty open
subset Ω0 ⊂ Ω with (0, z 0 ) ∈ Ω0 , such that f (x, t) ≥ 0 for almost everywhere
x ∈ Ω and all t > 0, and f (x, t) > 0 for almost x ∈ Ω0 and all t > 0,
(A2) limt→0+ f (x,t)
= 0 and limt→+∞ t2f∗(x,t)
(s)−1 = 0 uniformly for x ∈ Ω.
t
Then there exists Λ∗ > 0, such that problem (1.1) admits at least one positive
solution for all λ ≥ Λ∗ .
Remark 1.2. First, there are many functions satisfying our assumptions of Theorem 1.1. For instance, f (x, t) = tq with 1 < q < 2∗ (s) − 1. Second, it is worth
noting that, when µ = s = 0, problem (1.1) reduces to the classical semilinear elliptic problem with critical exponents, [3] proved the existence of positive solution
for λ > 0 large enough. In this paper, we obtain the similiar result as in [3] when
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SEMILINEAR ELLIPTIC PROBLEMS
3
2 ≤ k < N and 0 ≤ s < 2. Our results complete the existence of positive solutions
for elliptic problem with Hardy-Sobolev critical exponents.
2. Proof of Theorem 1.1
In this article, we use the following notation:
• The dual space of a Banach space E will be denoted by E 0 .
• Lp (Ω, |y|−s dx) denotes the weighted Sobolev space.
• → denotes the strong convergence, while * denotes the weak convergence.
• C, Ci (i = 0, 1, 2 . . . ) will denote various positive constants and their values can
vary from line to line.
To study the positive solutions of problem (1.1), we first consider the existence
of nontrivial solutions to the problem
∗
−∆u − µ
(u+ )2 (s)−1
u
=
+ λf (x, u+ ),
2
|y|
|y|s
u > 0, in Ω,
u = 0,
in Ω,
(2.1)
on ∂Ω,
+
where u = max{u, 0}. The energy functional corresponding to (2.1) is
Z Z
Z
∗
u2 (u+ )2 (s)
1
1
|∇u|2 − µ 2 dx − ∗
dx
−
λ
F (x, u+ )dx, (2.2)
I(u) =
2 Ω
|y|
2 (s) Ω
|y|s
Ω
for u ∈ H01 (Ω). Clearly, I is well defined and is C 1 smooth thanks to the HardySobolev-Maz’ya inequality [9]
Z Z |u|2∗ (s) 2∗2(s)
u2 2
−1
|∇u|
−
µ
dx
≤
S
dx,
(2.3)
µ
|y|2
|y|2
RN
RN
where Sµ = S(µ, N, k, s) is the best constant defined in (1.2). By the existence
of the one to one correspondence between the critical points of I and the weak
solutions of problem (2.1), we know that if u is a weak solution of problem (2.1),
there holds
Z Z
Z
∗
uv (u+ )2 (s)−1 v
hI 0 (u), vi =
(∇u, ∇v) − µ 2 dx −
dx
−
λ
f (x, u+ )vdx = 0,
s
|y|
|y|
Ω
Ω
Ω
for any v ∈ H01 (Ω).
Before proving Theorem 1.1, it is necessary to prove the following lemmas.
Lemma 2.1. Assume that conditions (A1), (A2) hold. Then for 0 ≤ µ < µ and
λ > 0, we can deduce that
(1) there exist r, α > 0 such that I(u) ≥ α when kuk = r,
(2) there exists u1 ∈ H01 (Ω) such that ku1 k > r and I(u1 ) < 0.
Proof. (1) From the continuity of embeddings
H01 (Ω) ,→ Lq (Ω)(1 ≤ q ≤ 2∗ ),
∗
H01 (Ω) ,→ L2
(s)
(Ω, |y|−s dx),
there exist C1 > 0 and C2 > 0 such that
Z
Z
∗
∗
|u|2 (s)
|u|q dx ≤ C1 kukq ,
dx ≤ C2 kuk2 (s) .
s
|y|
Ω
Ω
(2.4)
It follows from (A2) that for ε > 0, there exists C3 > 0, such that
∗
|F (x, t)| ≤ ε|t|2 + C3 |t|2
(s)
,
(2.5)
4
R.-T. JIANG, C.-L. TANG
EJDE-2016/12
for all t ∈ R+ and x ∈ Ω. Combining (2.4) and (2.5), one has
Z
Z
Z
∗
∗
1
1
(u+ )2 (s)
2
I(u) ≥ kuk2 − ∗
dx
−
λε
|u|
dx
−
λC
|u|2 (s) dx
3
s
2
2 (s) Ω
|y|
Ω
Ω
∗
∗
C
1
4
≥ kuk2 − ∗ kuk2 (s) − λεC5 kuk2 − λC6 kuk2 (s) .
2
2 (s)
Therefore for a fixed λ > 0, there exists α > 0 such that I(u) ≥ α > 0 for all
kuk = r, where r > 0 small enough.
(2) Fix v0 ∈ C0∞ (Ω) \ {0} with v0 ≥ 0 in Ω and kv0 k = 1. From (A1), one has
Z
Z
∗
∗
1
1
(v0+ )2 (s)
dx
−
I(tv0 ) = t2 kv0 k2 − ∗ t2 (s)
F (x, tv0 )dx
2
2 (s)
|y|s
Ω
Ω
Z
∗
∗
1
(v0+ )2 (s)
1
dx,
≤ t2 kv0 k2 − ∗ t2 (s)
2
2 (s)
|y|s
Ω
then limt→+∞ I(tv0 ) → −∞. Thus we can find t0 > 0 such that I(t0 v0 ) < 0 when
kt0 v0 k > r. Set u1 = t0 v0 , the proof is complete.
According to the Mountain Pass Lemma without (PS) condition (see [10]), there
exists a sequence {un } ⊂ H01 (Ω), such that
0
I(un ) → cλ > α > 0, I 0 (un ) → 0 in H01 (Ω) ,
as n → ∞, where
cλ = inf max I(γ(t)),
γ∈Γ t∈[0,1]
Γ = {γ ∈ C([0, 1], H01 (Ω))|γ(0) = 0, γ(1) = u1 }.
Lemma 2.2. Suppose that (A1) holds, then limλ→+∞ cλ = 0.
Proof. If v0 is the function given in the proof of Lemma 2.1, then one deduces that
I(tv0 ) > 0 for t > 0 small enough and I(tv0 ) → −∞ as t → +∞. Thus there exists
tλ > 0 such that I(tλ v0 ) = maxt≥0 I(tv0 ). Hence
Z
Z
∗
(v0+ )2 (s)
2∗ (s)
2
2
tλ kv0 k = tλ
dx + λ
f (x, tλ v0+ )tλ v0+ dx.
|y|s
Ω
Ω
By (A1), one has
∗
(v0+ )2 (s)
dx,
|y|s
Ω
which implies that {tλ } is bounded. Therefore there exist a sequence {λn } and
t0 ≥ 0, such that λn → +∞ and tλn → t0 as n → ∞. Consequently, there is C6 > 0
such that t2λn kv0 k2 ≤ C6 for all n ∈ N, and thus
Z
Z
∗
(v0+ )2 (s)
2∗ (s)
λn
f (x, tλn v0+ )tλn v0+ dx + tλn
dx ≤ C6 ,
(2.6)
|y|s
Ω
Ω
2∗ (s)
t2λ kv0 k2 ≥ tλ
Z
for all n ∈ N. If t0 > 0, by (A1), one obtains
Z
Z
2∗ (s)
+
+
lim λn
f (x, tλn v0 )tλn v0 dx + tλn
n→∞
Ω0
Ω0
then
Z
lim λn
n→∞
Ω
∗
f (x, tλn v0+ )tλn v0+ dx
+
2∗ (s)
tλn
Z
Ω
(v0+ )2 (s)
dx = +∞,
|y|s
∗
(v0+ )2 (s)
dx = +∞,
|y|s
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SEMILINEAR ELLIPTIC PROBLEMS
5
which contradicts (2.6). Thus we conclude that t0 = 0. Now, let us consider the
path γ∗ (t) = tu1 for t ∈ [0, 1], which belongs to Γ, then we get the following estimate
1
0 < cλ ≤ max I(γ∗ (t)) ≤ I(tλ v0 ) ≤ t2λ kv0 k2 .
2
t∈[0,1]
As λ → +∞, from the above inequality, we get cλ → 0.
Proof of Theorem 1.1. By Lemma 2.1, there exists a sequence {un } ⊂ H01 (Ω) such
that I(un ) → cλ and I 0 (un ) → 0 as n → ∞. According to Lemma 2.2, one deduces
that there exists Λ∗ > 0 such that
N −s
2−s
Sµ2−s
0 < cλ <
2(N − s)
as λ ≥ Λ∗ . First, we prove that {un } is bounded. Indeed, by (A2) and the
boundedness of Ω, for any ε > 0, there exists M > 0, such that
∗
|F (x, t)| ≤ ε|t|2
(s)
,
x ∈ Ω, t ≥ M ;
|F (x, t)| ≤ C1 (ε),
t ∈ (0, M ];
2 (s)
,
x ∈ Ω, t ≥ M ;
|f (x, t)t| ≤ C2 (ε),
t ∈ (0, M ].
|f (x, t)t| ≤ ε|t|
∗
Thus, we have
∗
|F (x, t)| ≤ C1 (ε) + ε|t|2
(s)
,
∗
|f (x, t)t| ≤ C2 (ε) + ε|t|2
(s)
,
for any (x, t) ∈ Ω × R+ . Then for ξ ∈ (2, 2∗ (s)), one has
∗
1
1
F (x, t) − f (x, t)t ≤ F (x, t) − f (x, t)t ≤ C3 (ε) + ε|t|2 (s) ,
2
ξ
(2.7)
(2.8)
∗
for any (x, t) ∈ Ω × R+ . Set l(x, t) := |y|−s |t|2 (s)−1 + λf (x, t), we claim that l(x, t)
satisfies the (AR) condition. By (2.8), one easily gets
ξ
∗
ξL(x, t) − l(x, t)t = ∗
− 1 |y|−s |t|2 (s) + λ ξF (x, t) − f (x, t)t
2 (s)
ξ
∗
∗
≤ ∗
− 1 |y|−s |t|2 (s) + ξλC4 (ε) + ξλε|t|2 (s)
2 (s)
∗
ξ
− 1)|y|−s + λξε |t|2 (s) + ξλC4 (ε).
= ( ∗
2 (s)
Thus for a fixed λ > 0 and ε > 0 sufficiently small, there exists Mλ0 > 0, such that
0 ≤ ξL(x, t) ≤ l(x, t)t,
where L(x, t) =
t ≥ Mλ0 ,
Rt
l(x, s)ds. Moreover, by (A2), we obtain
1
1
L(x, t) − l(x, t)t ≤
max
F (x, t) − f (x, t)t := Mλ .
ξ
ξ
x∈Ω,0≤t≤Mλ0
0
It follows from the inequalities above that
1
L(x, t) − l(x, t)t ≤ Mλ , for all x ∈ Ω \ {(0, z 0 )}, t ≥ 0.
ξ
Then, one has
1
c + 1 + o(1)kun k ≥ I(un ) − hI 0 (un ), un i
ξ
Z
1
2∗ (s)
1 1
1 (u+
n)
2
≥
− kun k +
− ∗
dx
2 ξ
ξ
2 (s) Ω
|y|s
(2.9)
6
R.-T. JIANG, C.-L. TANG
1
2
1
≥
2
Z 1
+ +
F (x, u+
)
−
f
(x,
u
)u
n
n
n dx
ξ
Ω
Z 1
1
+ +
− kun k2 −
L(x, u+
n ) − l(x, un )un dx
ξ
ξ
Ω
1
− kun k2 − Mλ |Ω|.
ξ
−λ
≥
EJDE-2016/12
∗
Thus, {unR } is bounded. Due to the continuity of embedding H01 (Ω) ,→ L2 (s) (Ω),
∗
we have Ω |un |2 (s) dx ≤ C7 < ∞. Up to a subsequence, still denoted by {un },
there exists u0 ∈ H01 (Ω) satisfying
un * u0 ,
un → u0 ,
weakly in H01 (Ω),
strongly in Lp (Ω), 1 ≤ p < 2∗ ,
un (x) → u0 (x),
∗
u2n
(s)−1
2∗ (s)−1
* u0
,
a.e. in Ω,
∗
weakly in L2
(s)
(2.10)
0
(Ω, |y|−s dx) ,
as n → ∞. By (A2), for any ε > 0 there exists a(ε) > 0 such that
|F (x, t)| ≤
∗
1
ε|t|2 (s) + a(ε)
2C7
for (x, t) ∈ Ω × R+ .
ε
2a(ε)
> 0. When E ⊂ Ω, meas E < δ, one gets
Z
Z
+
F (x, un )dx ≤
|F (x, u+
n )|dx
E
E
Z
Z
∗
1
≤
a(ε)dx +
ε
|un |2 (s) dx
2C7 E
E
1
≤ a(ε) meas E +
εC7 ≤ ε.
2C7
R
Hence
F (x, u+
is equi-absolutely-continuous. It follows from Vin )dx, n ∈ N
Ω
tali’s Convergence Theorem that
Z
Z
F (x, u+
)dx
→
F (x, u+
(2.11)
n
0 )dx,
Set δ =
Ω
Ω
as n → ∞. Applying the same method, one has
Z
Z
+
f (x, un )un dx →
f (x, u+
0 )u0 dx,
Ω
(2.12)
Ω
as n → ∞. By (2.10) and (2.12) we have
Z Z
2∗ (s)−1
u0 v (u+
v
0
0)
lim hI (un ), vi =
(∇u0 , ∇v) − µ 2 dx −
dx
s
n→+∞
|y|
|y|
Ω
Ω
Z
−
f (x, u+
0 )vdx = 0,
(2.13)
Ω
for all v ∈ H01 (Ω). That is, hI 0 (u0 ), vi = 0 for any v ∈ H01 (Ω). Then u0 is a critical
point of I, thus u0 is a solution of problem (2.1). Now we verify that u0 6≡ 0. Let
v = u0 in (2.13), we get
Z
Z
2∗ (s)
(u+
2
0)
dx −
f (x, u+
(2.14)
ku0 k −
0 )u0 dx = 0.
|y|s
Ω
Ω
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SEMILINEAR ELLIPTIC PROBLEMS
Set wn = un − u0 , then we have
Z
Z
Z
2
2
|∇un | dx =
|∇u0 | dx +
|∇wn |2 dx + o(1).
Ω
Ω
7
(2.15)
Ω
And from Brézis-Lieb’s lemma [3], it follows that
Z
Z
Z
u20
wn2
u2n
dx
=
dx
+
dx + o(1),
2
2
2
Ω |y|
Ω |y|
Ω |y|
Z
Z
Z
∗
2∗ (s)
2∗ (s)
(u+
(u+
(wn+ )2 (s)
n)
0)
dx =
dx +
dx + o(1).
|y|s
|y|s
|y|s
Ω
Ω
Ω
(2.16)
(2.17)
By (2.11) and (2.15)-(2.17), one has
1
1
I(un ) = I(u0 ) + kwn k2 − ∗
2
2 (s)
∗
Z
Ω
(wn+ )2 (s)
dx = cλ + o(1).
|y|s
(2.18)
Since hI 0 (un ), un i = o(1), combining with (2.12) and (2.14), one has
Z
2
kwn k −
Ω
∗
(wn+ )2 (s)
dx = o(1).
|y|s
We may assume that kwn k2 → b and
Z
Ω
∗
(wn+ )2 (s)
dx → b
|y|s
as n → ∞. Clearly, b ≥ 0. We now suppose that u0 ≡ 0. On the one hand, if b = 0,
together with (2.18), we get cλ = I(0) = 0, which contradicts with cλ > 0. On the
other hand, if b 6= 0, we have from the definition of Sµ that
kwn k2 =
Z Z (w+ )2∗ (s) 2∗2(s)
w2 n
|∇wn |2 − µ n2 dx ≥ Sµ
dx
,
|y|
|y|s
Ω
Ω
N −s
then b ≥ Sµ2−s , together with (2.18) we deduce
1
1
cλ + o(1) = I(u0 ) + kwn k2 − ∗
2
2 (s)
N −s
2−s
≥
Sµ2−s + o(1),
2(N − s)
Z
Ω
∗
(wn+ )2 (s)
dx + o(1)
|y|s
N −s
2−s
2−s
which contradicts cλ < 2(N
. Therefore u0 6≡ 0 and u0 is a nontrivial
−s) Sµ
−
solution of problem (2.1). Then by hI 0 (u0 ), u−
0 i=0 where u0 R = min{u0 , 0}, one has
−
ku0 k = 0, which implies that u0 ≥ 0. From (2.13), we get Ω (∇u0 , ∇v)dx ≥ 0 for
any v ∈ H01 (Ω), which means −4u0 ≥ 0 in Ω. By the strong maximum principle, we
know u0 is a positive solution of problem (1.1). Therefore Theorem 1.1 holds. Acknowledgments. This research was supported by the National Natural Science
Foundation of China (No. 11471267).
8
R.-T. JIANG, C.-L. TANG
EJDE-2016/12
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Rui-Ting Jiang
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: 1152373321@qq.com
Chun-Lei Tang (corresponding author)
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: tangcl@swu.edu.cn, phone +86 23 68253135, fax +86 23 68253135