The Existence of Positive Solution for Semilinear Elliptic Equations with Multiple an Inverse Square Potential and Hardy-Sobolev Critical Exponents

and Applied Analysis 3 where b(μ) = √μ+√μ − μ and a(μ) = √μ−√μ − μ, (see [13] for details). Moreover, 0 ≤ a (μ) < N − 2 2 < b (μ) ≤ N − 2. (14) We consider ρ > 0 such that B(a, 2ρ) ⊂ Ω and define a cut function φ ∈ C∞ 0 (Ω) such that 0 ≤ φ ≤ 1, |∇φ| ≤ C, φ = 1 for |x − a| ≤ ρ and φ = 0 for |x − a| > 2ρ. Set ua μ,ε (x) = φ (x)Ua μ,ε (x) , Va μ,ε (x) = ua μ,ε (x) (∫Ω (󵄨󵄨󵄨󵄨󵄨ua μ,ε (x)󵄨󵄨󵄨󵄨󵄨2∗(s) / |x − a|s)dx)∗ , (15)

We suppose the following: (H 1 ) 0 <   < , for every  = 1, 2, ⋅ ⋅ ⋅ ,  and ∑  =1   <  = (( − 2)/2) 2 .(H 2 ) There is an  0 , 1 ≤  0 ≤ , such that min ⋅ (  0 (  0 )) and (H 3 ) 0 <   0 ≤  − 1 where  0 is given in (H 2 ).The function   () is a positive bounded on Ω, for every (1 ≤  ≤ ).Furthermore, The reason why we investigate (1) is the presence of the Hardy-Sobolev exponent and the so-called inverse square potential in the linear part, which cause the loss of compactness of embedding  1 0 (Ω) →  2 * (Ω),  1 0 (Ω) →  2 (|| −2 , Ω) and  1 0 (Ω) →  2 * () (|| − , Ω)).Hence, we face a type of triple loss of compactness whose interacting with each other will result in some new difficulties.In last two decades, loss of compactness leads to many interesting existence and nonexistence phenomena for elliptic equations.Many important results on the singular problems with Hardy-Sobolev critical exponents (the case that  ̸ = 0 and   =  were obtained such as the existence and multiplicity of solutions in these works and these results give us very good insight into the problem; see, for example, [1][2][3][4][5][6][7] and references therein.In the present paper, we use a variational method to deal with problem (1) with general form and generalize the results in [8].As  ≥ 2 to our knowledge, there are no results on the existence of nontrivial solutions for (1).It is therefore significant for us to study the problem (1) deeply.However, because of the singularities caused by the terms | − | − ( = 1, 2, . . ., ), our problem becomes more complicated to deal with than [8] and therefore we have to face more difficulties.Despite the multiple terms of hardy and the coefficients of the critical nonlinearity, but we will see how, they will play an important role in the search for the bubble whose energy is below the level of local compactness (PS).The existence result is obtained via constructing a minimax level within this range and the Mountain Pass Lemma due to A. Ambrosetti and P.H. Rabinowitz (see also [9]).
Our main result is the following.
Then problem (1) has at least one positive solution.
The paper is organized as follows: in Section 2, preliminary results about Palais-Smale condition for  in a suitable interval and construct some auxiliary functions and estimate their norms.In Section 3, fill the conditions of Mountain Pass Theorem and we establish our result.
Let  be a Banach space and  −1 be the dual space of .The functional  ∈  1 (, R) is said to satisfy the Palais-Smale condition at level  (()  in short), if any sequence {  } ⊂  satisfying (  ) → ,   () → 0 strongly in  −1 as  → +∞ contains a subsequence converging in  to a critical point of the functional .
Lemma 3. The functional  satisfies ()  condition for any where   is the Dirac mass at   ∈ R  .
Arguing by contradiction, let us suppose that there exist  0 such that Thus, Letting  → +∞, we get so, by (61), we obtain which contradicts the assumption that Hence, up to a subsequence, we obtain that   →  strongly in  1 0 (Ω).

Proof of Main Result 1
We verify that the functional  satisfies the mountain pass geometry.To this end, we consider the energy level.
for all  ∈  1 0 (Ω) with ‖‖ =  0 .Let V ∈  1 0 (Ω) given in Lemma 4. Since lim →+∞ (V) = −∞, hence there exists  0 > 0 such that ‖ 0 V‖ ≥  0 and ( 0 V) < 0, by Lemma 4, we obtain Moreover, by the Mountain Pass Theorem [9] and Lemma 3, we obtain that  1 is critical value of  at point  and thus is a solution of problem (1).Then the rest of the proof follows exactly the lines as that in [3].In order to find the positive solution of (1), we replace () with  + () defined as follows: + () = where  + = max {, 0}.Repeating the above arguments, we find a critical point of  + and by applying the maximum principle we obtain a positive solution.So, the proof of Theorem 1 is therefore completed.