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Existence and multiplicity of nontrivial solutions for semilinear elliptic equations involving Hardy-Sobolev critical exponents

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Abstract
A class of semi-linear elliptic equations with critical Hardy-Sobolev exponent has been considered. This model is widely used in hydrodynamics and glaciology, gas combustion in thermodynamics, quantum field theory and statistical mechanics, as well as gravity balance problems in galaxies. The $\left( PS\right) _{c}$ sequence of energy functional has been investigated, and then the mountain pass lemma was used to prove the existence of at least one nontrivial solution. Also a multiplicity result has been obtained. Some known results have been generalized.
Keywords: 
Subject: Computer Science and Mathematics  -   Mathematics

1. INTRODUCTION

Consider the problem
Δ u μ u x 2 = u 2 * s 2 x s u + f ( x , u ) , x Ω 0 , u = 0 , x Ω .
Here Ω is an open bounded domain with smooth boundary Ω in R N N 3 , and 0 Ω . 0 μ < μ ¯ : = N 2 2 2 , 2 * s = 2 N s / N 2 ( 0 < s < 2 ) is the Hardy-Sobolev critical exponent and 2 * = 2 * 0 is the Sobolev critical exponent. f C Ω × R , R , F x , t = 0 t f x , s d s . We point out that (1) is related to the application of hydrodynamics and glaciology ([1]). And it is also used in some physical or mathematical problems, such as the theory of gas combustion in thermodynamics ([2]), quantum field theory and statistical mechanics ([3,4,5]), as well as gravity balance problems in galaxies ([2,6]). For more investigations on solutions for nonlinear equations with Hardy potential, one can see [7,8,9] etc.
The modern variational method ([10,11,12,13]) plays a significant role in studying PDEs (see [14,15,16,17,18]). In 1973, the mountain pass lemma was proposed by A. Ambrosetti and P. Rabinwitz in [14], it is a milestone in the history of the development of critical point theory. However, in the process of studying the properties for certain equations, there are a lot of phenomena that lose compactness conditions, such as semilinear elliptic equations that involving Sobolev critical exponent or Hardy-Sobolev critical exponent on bounded domain. In 1983, H. Brezis and L. Nirenberg first chose special mountain pass and selected energy estimates to prove the existence of a critical point if the energy functional satisfies the local P S condition (see [15]), they investigated the problem
Δ u = u 2 * 2 u + λ u , x Ω , u = 0 , x Ω .
and obtained that there exists a λ * 0 , λ 1 such that for any λ λ * , λ 1 , problem (2) admits a positive solution. It is a special case of equation (1) ( s = 0 , μ = 0 and f x , u = λ u ). Since then, many excellent results based on the above methods (see [11,19,20,21]) appeared.
In the past decades, the semilinear elliptic equation with Hardy term and Sobolev critical exponent (i.e. when s = 0 and μ 0 ) has been investigated by many mathematicians, one can refer to [22,23,24,25] etc. For example, the following elliptic problem
Δ u μ u x 2 = u 2 * 2 u + λ u , x Ω , u = 0 , x Ω .
is considered in [22,23,24].
For simplicity, in the following, we denote the condition (H1) and (H2) as follows:
(H1) 0 < λ < λ 1 μ and 0 μ μ ¯ 1 ;
(H2) μ ¯ 1 < μ < μ ¯ and λ * μ < λ < λ 1 μ .
In [24], by the variational method, E. Jannelli proved that: If (H1) or (H2) holds, then (3) has at least one positive solution in H 0 1 Ω . Later in [26], the authors investigated problem (1) with f x , u = λ u q 2 u or f x , u = λ u . And obtained the following conclusion.
Theorem A ([26]).
Assume 0 s < 2 ,   q = 2 , and β = 2 μ ¯ + μ ¯ μ . If (H1) or (H2) holds, then problem (1) has a positive solution u in H 0 1 Ω .
Also there are some results dealing with the case μ 0 ,   s 0 and the general form f x , u (see [27,28]). In [27], M.C. Wang and Q. Zhang showed that problem (1) has at least one nonnegative solution. In [28], L. Ding and C.L. Tang also investigated problem (1) and obtained the existence result. Inspired by [26,27,28], we study the existence of nontrivial solutions for problem (1). Our main conclusions are
Theorem 1.
Suppose that N 3 , 0 μ < μ ¯ , β = μ ¯ + μ ¯ μ ,   0 < s < 2 , f x , t satisfies
f 1   f C Ω ¯ × R + , R + and lim t 0 + f x , t t = λ ,   lim t + f x , t t 2 * s 1 = η , uniformly for x Ω ¯ , where λ , η > 0 .
f 2 There exists 2 < ρ 2 * s , such that 1 ρ f x , t t F x , t 1 2 1 ρ λ t 2 , for any x Ω ¯ , t R + .
If (H1) or (H2) holds, then (1) has a positive solution u in H 0 1 Ω , where
λ * μ = min φ H 0 1 Ω 0 Ω φ 2 / x 2 β d x Ω φ 2 / x 2 β d x ,
and
λ 1 μ = inf u H 0 1 Ω 0 Ω u 2 μ u 2 x 2 d x Ω u 2 d x .
Theorem 2.
Suppose that N 3 , 0 μ < μ ¯ , β = μ ¯ + μ ¯ μ ,   0 < s < 2 . f x , t satisfies f 2 and
f 3   f C Ω ¯ × R , R + ,   lim t 0 + f x , t t = λ ,   lim t + f x , t t 2 * s 1 = η , uniformly for x Ω ¯ , where λ , η > 0 .
If (H1) or (H2) holds, then (1) has at least two distinct nontrivial solutions in H 0 1 Ω .
Remark 1.
i Let f x , u = λ u , we can get 1 ρ f x , u u F x , u = 1 2 1 ρ λ u 2 , thus 1 2 1 ρ λ is the best constant.
i i Comparing with [27] and [28], the restrictions on the nonlinear term f x , u are weaken.
i i i If f x , u = λ u + η u 2 * s 1 , then it is easy to verify that f x , u satisfies f 1 - f 3 .
Remark 2.
To prove Theorem A, when (H1) holds, the authors used the analytical techniques as that in [24]. In this paper, by accurate estimates of u ε 2 and Ω u ε 2 * s / x s d x , we obtain c < 2 s 2 N s A μ , s N s / 2 s , thus in this case, the mountain pass lemma could also be used. We unify the methods for proving the existence of solutions of equation (1) for both cases (H1) and (H2). The results in this paper integrally contain all the cases of Theorem A in [26].

2. PROOF OF THEOREMS

Obviously, in Theorem 1, the values of f x , t are irrelevant for t < 0 , therefore, we define
f x , t = 0 for x Ω ¯ and t 0 .
By Hardy inequality and Hardy-Sobolev inequality (see [29]), we define equivalent norm and inner product in H 0 1 Ω
u : = Ω u 2 μ u 2 x 2 d x 1 2 , u , v : = Ω u v μ u v x 2 d x , u , v H 0 1 Ω .
Let
u + : = max 0 , u , F + x , t : = 0 t f + x , s d s , f + x , t : = f x , t , t 0 , 0 , t < 0 .
The energy functional J : H 0 1 Ω R to (1) is given by
J u = 1 2 u 2 1 2 * s Ω u + 2 * s x s d x Ω F + x , u d x , u H 0 1 Ω .
We can easily obtain that J u is well defined with J C 1 H 0 1 Ω , R and
J u , v = u , v Ω u + 2 * s 1 x s v d x Ω f + x , u v d x , u , v H 0 1 Ω .
When 0 μ < μ ¯ , the best constant can be defined as follows (see [30])
A μ , s : = inf u H 0 1 Ω 0 Ω u 2 μ u 2 x 2 d x Ω u 2 * s x s d x 2 2 * s .
Lemma 1.
Suppose f 1 hold. For any 0 < ε 1 < min { λ , η } and α 1 , α 2 1 , 2 * s 1 , there exists ξ > 0 such that
f x , t λ ε 1 t + η ε 1 t 2 * s 1 ξ t α 1 , for t 0 and x Ω ¯ ,
and
f x , t λ + ε 1 t + η + ε 1 t 2 * s 1 + ξ t α 2 , for t 0 and x Ω ¯ .
Proof. 
It follows from f 1 that ε 1 > 0 , δ > 0 and M 1 > 0 ,
f x , t t λ ε 1 , for t , x 0 , δ × Ω ¯ ,
and
f x , t t 2 * s 1 η ε 1 , for t , x M 1 , + × Ω ¯ ,
from (7), we get
f x , t λ ε 1 t , for t , x 0 , δ × Ω ¯ ,
for α 1 1 , 2 * s 1 , if we take ξ max 0 , η ε 1 δ 2 * s 1 α 1 , then for any t 0 , δ , we have η ε 1 t 2 * s 1 ξ t α 1 0 , thus
f x , t λ ε 1 t + η ε 1 t 2 * s 1 ξ t α 1 , for t , x 0 , δ × Ω ¯ .
From (8), we know
f x , t η ε 1 t , for t , x M 1 , + × Ω ¯ ,
for α 2 1 , 2 * s 1 , if we take ξ max 0 , λ ε 1 M 1 1 α 1 , then for any t M 1 , + , we have λ ε 1 t ξ t α 1 0 , thus
f x , t λ ε 1 t + η ε 1 t 2 * s 1 ξ t α 1 , t , x M 1 , + × Ω ¯ .
When t δ , M 1 , taking ξ max 0 , max t δ , M 1 { λ ε 1 t 1 α 1 + η ε 1 t 2 * s 1 α 1 } , we have
f x , t λ ε 1 t + η ε 1 t 2 * s 1 ξ t α 1 , t , x δ , M 1 × Ω ¯ .
As mentioned above, if we take
ξ max 0 , λ ε 1 M 1 1 α 1 , η ε 1 δ 2 * s 1 α 1 , max t δ , M 1 { λ ε 1 t 1 α 1 + η ε 1 t 2 * s 1 α 1 } ,
then
f x , t λ ε 1 t + η ε 1 t 2 * s 1 ξ t α 1 , for t 0 and x Ω ¯ .
Similarly, we may obtain that there exists ξ > 0 such that
f x , t λ + ε 1 t + η + ε 1 t 2 * s 1 + ξ t α 2 , for t 0 and x Ω ¯ .
The conclusion is proved.
Now we introduced the extremal functions. Let
C ε = 2 ε μ ¯ μ N s μ ¯ μ ¯ / 2 s , U ε x = y ε x C ε ,
define a cut-off function φ C 0 Ω such that
φ x = 1 , x R , 0 , x 2 R ,
where B 2 R 0 Ω , 0 φ x 1 , for R < x < 2 R , set
u ε x = φ x U ε x , v ε x = u ε x Ω u ε x 2 * s x s d x 1 2 * s .
Lemma 2 
([26]). Let v ε x be defined as above, then v ε x satisfies
v ε ( x ) 2 = A μ , s + O ε N 2 2 s ,
Ω v ε ( x ) q d x = O ε μ ¯ q 2 s , 1 q < N μ ¯ + μ ¯ μ , O ε μ ¯ q 2 s ln ε , q = N μ ¯ + μ ¯ μ , O ε μ ¯ N q μ ¯ 2 s μ ¯ μ , N μ ¯ + μ ¯ μ < q < 2 * .
Lemma 3 ([31]).
Let u ε ( x ) , U ε x ,   A μ , s , C ε be defined as above, then the exact estimates of u ε 2 and Ω u ε 2 * s x s d x are as follows:
u ε ( x ) 2 = C ε 2 A μ , s N s 2 s + D , Ω u ε ( x ) 2 * s x s d x = C ε 2 * s A μ , s N s 2 s + E ,
where
D = R x 2 R u ε ( x ) 2 μ u ε 2 ( x ) x 2 d x x R U ε ( x ) 2 μ U ε 2 ( x ) x 2 d x ,
E = x R U ε ( x ) 2 * s x s d x + R x 2 R u ε ( x ) 2 * s x s d x .
Moreover, ∃ R 0 > 0 such that for any R R 0 ,
lim ε 0 + D < Ω φ x 2 x 2 β d x .
Lemma 4.
Suppose f 1 ,   f 2 and λ < λ 1 μ hold. Assume u n H 0 1 Ω is a (PS)   c sequences, that is,
J u n c 0 , 2 s 2 N s A μ , s N s 2 s ,
and
J u n 0 , in H 0 1 Ω 1 .
Then there exists u H 0 1 Ω such that u n u weakly in H 0 1 Ω , or a subsequence u n k u weakly in H 0 1 Ω , moreover, J u = 0 and u is a nontrivial solution of (1).
Proof.
First, we claim that if f 1 ,   f 2 and λ < λ 1 μ hold, then any (PS)c sequence u n is bounded in H 0 1 Ω . Otherwise, suppose that u n , since J u n c , there exists N 1 such that when n > N 1 ,   J u n < c + 1 . J u n 0 implies 1 ρ J u n , u n < o 1 u n , thus for any ε 1 0 , λ 1 μ λ , when n > N 1 ,
c + 1 + o 1 u n J u n 1 ρ J u n , u n = 1 2 1 ρ u n 2 + Ω 1 ρ f + x , u n u n F + x , u n d x + 1 ρ 1 2 * s Ω u n + 2 * s x s d x 1 2 1 ρ u n 2 1 2 1 ρ λ + ε 1 Ω u n 2 d x 1 2 1 ρ u n 2 1 2 1 ρ λ + ε 1 λ 1 μ u n 2 = 1 2 1 ρ 1 λ + ε 1 λ 1 μ u n 2 .
Which shows that u n is a bounded sequence in H 0 1 Ω . By the reflexivity of H 0 1 Ω , we know that there exists u such that u n u (or a subsequence of u n convergence to u). Furthermore, J u = 0 by the weak continuity of J . From u n H 0 1 Ω , u n u , by the compactness of the embedding, we have u n u in L γ Ω for any 1 < γ < 2 * s . Let f 1 x , u = f x , u u , from f 1 , we have f 1 x , u n a + b u n 2 * s , by the definition of Uryson operator, we know f 1 : L 2 * s Ω L 1 Ω is a continuous operator. Thus
lim n Ω f 1 x , u n f 1 x , u d x = 0 ,
that is,
lim n Ω f x , u n u n d x = Ω f x , u u d x .
Similarly,
lim n Ω F x , u n d x = Ω F x , u d x .
In addition, by the convergence of u n , u n u in H 0 1 Ω .
Assume that u 0 in Ω , from J u n , u n = o 1 and (15) we know
u n 2 Ω u n + 2 * s x s d x = o 1 .
By (4),
u n 2 A μ , s Ω u n + 2 * s x s d x 2 2 * s .
From (16) and (17), we have
o 1 u n 2 1 A μ , s 2 * s 2 u n 2 * s 2 .
If u n 0 , then (16) implies that J u n 0 , while J u n c , which contradicts c > 0 . Hence
u n 2 A μ , s N s 2 s + o 1 .
By (12), (16) and (18), we obtain
J u n = 1 2 u n 2 1 2 * s Ω u n + 2 * s x s d x + o 1 = 2 s 2 N s u n 2 + o 1 2 s 2 N s A μ , s N s 2 s + o 1 ,
which contradicts c < 2 s 2 N s A μ , s N s 2 s . Thus, u is not constantly equal to 0 and u is a nontrivial solution of problem (1).
Lemma 5.
If f 1 ,   f 2 and λ < λ 1 μ hold, then the functional J admits a P S sequence at level
c = inf γ Γ max t 0 , 1 J γ t ,
where
Γ = γ C 0 , 1 , H 0 1 Ω ; γ 0 = 0 , J γ 1 < 0 .
Proof.
We need to prove J satisfy all assumptions of the mountain pass lemma except for the P S condition. Obviously, J 0 = 0 . Moreover, from the Hardy-Sobolev inequality and the Hardy inequality, we can easily get
Ω u 2 * s x s d x C 1 u 2 * s , u q q C 2 u q for 1 q 2 * , u H 0 1 Ω .
Then, by (6), (19) and Lemma 1, we have
J u = 1 2 u 2 1 2 * s Ω u + 2 * s x s d x Ω F + x , u d x 1 2 u 2 C 1 2 * s u 2 * s η + ε 1 2 * s u 2 * s 2 * s λ + ε 1 2 u 2 2 ξ α 2 + 1 u α 2 + 1 α 2 + 1 1 λ + ε 1 / λ 1 μ 2 u 2 C 1 2 * s u 2 * s C 2 2 * s u 2 * s ξ α 2 + 1 u α 2 + 1 ,
which implies that α , ρ > 0 such that
J u α > 0 , u u H 0 1 Ω | u = ρ .
Taking u 0   H 0 1 Ω 0 , such that Ω u 0 + 2 * s x s d x C 3 > 0 , for any t > 0 , we have
J t u 0 = t 2 2 u 0 2 t 2 * s 2 * s Ω u 0 + 2 * s x s d x Ω F + x , t u 0 d x t 2 2 u 0 2 t 2 * s 2 * s C 3 ,
notice lim t + J t u 0 = , then there exists t 0 > 0 such that t 0 u 0 > ρ and J t 0 u 0 0 . By mountain pass theorem without the (PS) condition (see Theorem 2.2 in [15]), we know that J admits a P S sequence at the c level.
Lemma 6.
Assume f 1 ,   f 2 and 0 < s < 2 , if (H1) or (H2) holds, then
0 < c < 2 s 2 N s A μ , s N s 2 s .
Proof.
Define
g t : = J t v ε = t 2 2 v ε 2 t 2 * s 2 * s Ω F + x , t v ε d x
and
g ¯ t : = t 2 2 v ε 2 t 2 * s 2 * s .
It is easy to see that lim t + g t = , g 0 = 0 and g t > 0 when t is small enough, so there exists some t ε > 0 , such that g t ε = sup t 0 g t > 0 , which shows that c > 0 . Obviously g t ε = 0 , that is,
0 = g t ε = t ε v ε 2 t ε 2 * s 1 Ω f + x , t ε v ε v ε d x ,
thus
v ε 2 = t ε 2 * s 2 + 1 t ε Ω f + x , t ε v ε v ε d x t ε 2 * s 2 ,
therefore,
t ¯ ε : = v ε 2 2 * s 2 t ε .
From (7), we know
Ω f x , t ε v ε v ε d x λ + ε 1 t ε Ω v ε 2 d x + η + ε 1 t ε 2 * s 1 Ω v ε 2 * s d x + ξ t ε α 2 Ω v ε α 2 + 1 d x .
Hence
v ε 2 t ε 2 * s 2 + λ + ε 1 Ω v ε 2 d x + η + ε 1 t ε 2 * s 2 Ω v ε 2 * s d x + ξ t ε α 2 1 Ω v ε α 2 + 1 d x .
Moreover, from Lemma 2, we have
t ε 2 * s 2 A μ , s ε 2 .
On the other hand, g ¯ t   g ¯ t ¯ ε for any t 0 , t ¯ ε . From (5), (11), (12), (21), (22) and Lemma 2, we get
g t ε = g ¯ t ¯ ε Ω F x , t ε v ε d x 2 s 2 N s v ε 2 N s 2 s λ ε 1 2 t ε 2 Ω v ε 2 d x + ξ α + 1 t ε α 1 + 1 Ω v ε α 1 + 1 d x η ε 1 2 * s t ε 2 * s Ω v ε 2 * s d x 2 s 2 N s v ε 2 N s 2 s λ ε 1 A μ , s ε 2 2 2 * s 2 2 Ω v ε 2 d x + ξ α 1 + 1 A μ , s ε 2 α 1 + 1 2 * s 2 Ω v ε α 1 + 1 d x η ε 1 2 * s A μ , s ε 2 2 * s 2 * s 2 Ω v ε 2 * s d x .
If (H1) holds, notice that 2 N μ ¯ + μ ¯ μ , then when ε is sufficiently small, the sign of λ ε 1 2 t ε 2 Ω v ε 2 d x + ξ α + 1 t ε α 1 + 1 Ω v ε α 1 + 1 d x η ε 1 2 * s t ε 2 * s Ω v ε 2 * s d x is decided by the sign of Ω v ε 2 * s d x . Thus, when ε is small enough, (20) holds true.
If (H2) holds, since α 1 > 1 is arbitrary, we can choose α 1 > N μ ¯ + μ ¯ μ , then by Lemma 2, we know that when ε is sufficiently small, the sign of ξ α + 1 t ε α 1 + 1 Ω v ε α 1 + 1 d x η ε 1 2 * s t ε 2 * s Ω v ε 2 * s d x is decided by the sign of Ω v ε 2 * s d x . Thus from (23), when ε ,   ε 1 and ε 2 are small enough,
g t ε < 2 s 2 N s v ε 2 N s 2 s λ * μ 2 A μ , s 2 2 * s 2 Ω v ε 2 d x .
From (13) we know
lim ε 0 + v ε 2 N s 2 s A μ , s N s 2 s A μ , s N 2 2 s Ω v ε 2 d x = lim ε 0 + u ε 2 N s 2 s A μ , s N s 2 s Ω u ε 2 * s x s d x N s 2 s A μ , s N 2 2 s Ω u ε 2 * s x s d x N 2 2 s 2 2 * s Ω u ε 2 d x = lim ε 0 + D + C ε 2 A μ , s N s 2 s N s 2 s A μ , s N s 2 s E + C ε 2 * s A μ , s N s 2 s N 2 2 s A μ , s N 2 2 s E + C ε 2 * s A μ , s N s 2 s N 2 2 s 2 2 * s Ω u ε 2 d x = lim ε 0 + C ε 2 D + A μ , s N s 2 s N s 2 s A μ , s N s 2 s E C ε 2 * s + A μ , s N s 2 s N 2 2 s C ε 2 A μ , s N 2 2 s E C ε 2 * s + A μ , s N s 2 s N 2 2 s 2 2 * s Ω u ε 2 d x = lim ε 0 0 C 2 D ε 0 + A μ , s N s 2 s N s 2 s A μ , s N s 2 s E C 2 * s ε 0 N s N 2 + A μ , s N s 2 s N 2 2 s C 2 ε 0 A μ , s N s N 2 2 s 2 Ω u ε 2 d x = N s 2 s lim ε 0 0 + P C 2 D + ε 0 D ε 0 A μ , s N s 2 s Q C 2 * s E ε 0 2 s N 2 + E ε 0 ε 0 N s N 2 C 2 A μ , s N s N 2 2 s 2 Ω u ε 2 d x = N s 2 s D Ω φ x 2 x 2 β d x ,
where C = 2 μ ¯ μ N s μ ¯ μ 2 s ,   ε 0 = ε μ ¯ 2 s ,   P = C 2 D ε 0 + A μ , s N s 2 s N 2 2 s ,   Q = E C 2 * s ε 0 N s N 2 + A μ , s N s 2 s N 4 + s 2 s . By (14), (13) and (24), we have
lim ε 0 + v ε 2 N s 2 s A μ , s N s 2 s A μ , s N 2 2 s Ω v ε 2 d x < N s 2 s Ω φ x 2 x 2 β d x Ω φ x 2 x 2 β d x ,
so, if ε is small enough, then c g t ε < 2 s 2 N s A μ , s N s 2 s .
Proof of Theorem 1.
By Lemmas 4, 5 and 6, we can get that equation (1) has a nonnegative solution u H 0 1 Ω , by the maximum principle, this solution is positive. Which completes the proof.
Proof of Theorem 2.
Since f 3 contains f 1 , Theorem 1 implies the existence of a positive solution u 1 for equation (1). Let f x , t = h x , t for t R , h x , u satisfies f 1 and f 2 , then
Δ u μ u x 2 = u 2 * s 2 x s u + h x , u
has at least one nonnegative solution v. Let u 2 = v , then u 2 is a solution of
Δ u μ u x 2 = u 2 * s 2 x s u + f x , u .
Clearly, u 1 , u 2 0 . So problem (1) has at least two distinct nontrivial solutions.

Acknowledgments

This work is partially supported by NNSF of China (11201213), NSF of Shandong Province (ZR2015AM026).

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