Multiplicity of positive solutions for a class of singular elliptic equations with critical Sobolev exponent and Kirchhoff-type nonlocal term

We study a class of singular elliptic equations involving critical Sobolev exponent and Kirchhoff-type nonlocal term − ( a + b ∫ Ω |∇u| 2dx ) ∆u = u5 + g(x, u) + λu−γ, x ∈ Ω, u > 0, x ∈ Ω, u = 0, x ∈ ∂Ω, where Ω ⊂ R3 is a bounded domain, a, b, λ > 0, 0 < γ < 1 and g ∈ C(Ω×R) satisfies some conditions. By the perturbation method, variational method and some analysis techniques, we establish a multiplicity theorem.


Introduction
In this paper, we consider the following singular elliptic equation with critical Sobolev exponent and Kirchhoff-type nonlocal term where Ω ⊂ R 3 is a bounded domain, a, b, λ > 0, 0 < γ < 1 and 2 * = 6 is the well-known critical Sobolev exponent.The nonlinear term g(x, s) : Ω × R → R satisfies the following conditions.
Because of the presence of the term b Ω |∇u| 2 dx, which implies that the equation is no longer a pointwise identity, problem (1.1) is called the nonlocal problem.This phenomenon provokes some mathematical difficulties, which makes the study of such a class of problem particularly interesting.Its physical motivation about the operator Ω |∇u| 2 dx ∆u, which appears in the Kirchhoff equation.Thus, problem (1.1) is always called Kirchhoff-type problem.The Kirchhoff equation is related to the following stationary analogue of the equation proposed by Kirchhoff [14] in 1883 as an extension of the classical D'Alembert's wave equation for free vibration of elastic strings.Kirchhoff's model takes into account the changes in length of the string produced by transverse vibrations.In problem (1.2), u denotes the displacement, f (x, u) the external force and b the initial tension while a is related to the intrinsic properties of the string (such as Young's modulus).It is worth pointing out that problem (1.2) received much attention only after the work of Lions [23] where a function analysis framework was proposed to the problem.After that, the Kirchhoff-type problem has been extensively investigated, for examples [1, 4, 9-13, 15, 17-22, 24, 25, 27-37].
To our best knowledge, the pioneer work on the Kirchhoff-type problem with critical Sobolev exponent is Alves, Corrêa and Figueiredo [1], they considered the following critical Sobolev exponent problem where ds is superquadratic at the origin and subcritical at infinity.By using the variational method, under appropriate conditions, they obtained that problem (1.3) has a positive solution for all λ > 0 large enough.After that, the Kirchhoff-type problem with critical exponent has been extensively studied, and some important and interesting results have been obtained, see [4, 8-13, 15-21, 24, 27-29, 33-37].
Based on [16] and [24], the mountain-pass level value is the most obstacle in proving the existence of the second solution of problem (1.4).This obstacle stems from the local term b Ω |∇u| 2 dx, which shows that the difference between the classic elliptic problem(that is, b = 0) and the Kirchhoff-type problem.In this paper, we give another way to overcome this obstacle.We add a supperlinear term g(x, u) in problem (1.4), that is problem (1.1).Combining with the perturbation method and variational method, we obtain two positive solutions for problem (1.1).
For all u ∈ H 1 0 (Ω), we define Obviously, the energy functional I does not belong to C 1 (H 1 0 (Ω), R).Note that a function u is called a weak solution of problem (1.1) for all ϕ ∈ H 1 0 (Ω).Let S be the best Sobolev constant, namely . (1.7) Our main results can be described as follows.
Remark 1.2.To the best of our knowledge, our result is up to date.As we known, [16] is the first paper which considered the singular Kirchhoff-type problem with critical exponent, that is, problem (1.4).However, there exists a small gap in the proof of the second positive solution, that is, the estimation of B(t ε v ε ) in Page 533 of [16].Indeed, when using the inequality of (3.14) in Page 532 of [16] to estimate B(t ε v ε ), they need check α t ε v ε is small enough for |x| ≤ ε 1−γ 16 and ε small.However, it maybe is not true.Obviously, if |x| ≤ ε 1−γ 16 and ε → 0, for some C > 0, , which does not implies that α t ε v ε is small.So far there is no way to correct it.In [24], the authors avoided the similar question by multiplying |x| −β in front of the singular term λ u γ , see problem (1.5).In here, in order to arrive at the same effect, we add a continuous subcritical function g in the right hand side of equation (1.4).
Comparing with Theorem 1.1 in [28], our problem (1.1) is a singular perturbing problem of that paper.Thanks to this perturbation, we get another solution.Moreover, our condition (g 3 ) is more general than the following condition (g 3 ) in [28], (g 3 ) There exists a constant θ ∈ (4, 6) such that g(x, s)s − θG(x, s) ≥ 0 for all x ∈ Ω and s ≥ 0, where G(x, s) = s 0 g(x, t)dt.We should point out that [28] generalized a part of Brézis-Nirenberg's result in [7] to the Kirchhoff-type problem.
Our condition (g 4 ) is first given by [7], which is used to estimate the level of the mountainpass value.Thanks to (g 4 ), we obtain the second positive solution of problem (1.1).
In view of the typical power nonlinearities, Theorem 1.1 allows us to ensure the following corollary.
Remark 1.4.When g(x, u) = u p−1 (4 < p < 6), then clearly g satisfies (g 1 )-(g 4 ).For the proof, we can consider instead with g(x, u) = (u + ) p−1 .This paper is organized as following: in Section 2, we consider an auxiliary problem, and in Section 3, we give the proof of Theorem 1.1.For the convenience of writing, we denote C, C 1 , C 2 , . . .as various positive constants in the following.

The auxiliary problem
In order to overcome the difficulty of the singular term, for every > 0, we study the following perturbation problem where u + = max{u, 0}.The energy functional corresponding to problem (2.1) is Obviously, the energy functional I is of class C 1 on H 1 0 (Ω).As well known that there exists a one-to-one correspondence between all solutions of problem (2.1) and the critical points of I on H 1 0 (Ω).We mean a function u is called a weak solution of problem (2.1 for all ϕ ∈ H 1 0 (Ω).First, we prove that I satisfies the local (PS) c condition.Lemma 2.1.Suppose that a, b, λ > 0, 0 < γ < 1 and g satisfies (g 1 ) − (g 3 ), then I satisfies the (PS) c condition, where c < Θ − Dλ 2 1+γ with D = D(γ, S, κ, Ω) is a positive constant and ), that is, as n → +∞.We claim that {u n } is bounded in H 1 0 (Ω).In fact, from (g 1 ) and (g 2 ), there exists Note that the subadditivity of t 1−γ , one has Consequently, combining with the Sobolev inequality, it follows from (2.3) and (2.5) that by the dominated convergence theorem and (2.7), one has Moreover, for every > 0, by (2.7), one gets Therefore, it follows from the dominated convergence theorem that (2.9) From (2.7), one also has ) By (g 2 ) and (2.7), one has ) ) As usually, letting w n = u n − u, we need prove that w n → 0 as n → ∞.Let lim n→∞ w n = l ≥ 0. If l = 0, our conclusion is true.Suppose that l > 0. By the Brézis-Lieb Lemma (see [6]), one has From (2.3), (2.7), (2.9) and (2.13), one obtains consequently, it follows from (2.10), (2.11) and (2.15) that Moreover, by (2.3), for any ϕ ∈ H 1 0 (Ω), one has lim n→∞ I (u n ), ϕ = 0, that is, where d = u n 2 + o(1) is a positive constant.Particularly, choosing ϕ = u − in (2.18), one has u − = 0. Thus, we have u ≥ 0 in Ω.On the one hand, from (2.5), (2.17) and (g 3 ), by the Hölder inequality, Sobolev inequality and Poincaré inequality, we have where the last inequality is obtained by the Young inequality and On the other hand, it follows from (2.14), (2.16) and (2.17) that and From (1.7), one has consequently, it follows from (2.20) that which implies that which contradicts (2.19).Hence, l ≡ 0, that is, . This completes the proof of Lemma 2.1.
As well known, the function is an extremal function for the minimum problem (1.7), that is, it is a positive solution of the following problem −∆u = u 5 , ∀x ∈ R 3 .Now, we estimate the level value of functional I and obtain the following lemma.
Proof.We divide three steps to prove Theorem 2.3.
Step 1.We prove that there exists a positive local minimizer solution of problem (2.1).
First, we claim that there exist λ * > 0 and R, ρ > 0 such that In fact, by (g 1 ) and (g 2 ), we infer that for all x ∈ Ω and s ∈ R. Consequently, by the Hölder inequality and (1.7) and (2.5), we have where C 12 = 4C 11 and B = 4|Ω| 5+γ 6 and R = t max , according to (2.35), there exists Thus there exists u ∈ H 1 0 (Ω) with u small enough such that I (u) < 0. Thus, we have inf u∈B R I (u) < 0. Therefore, our claim is true.
Step 2. We prove that there exists a positive mountain-pass type solution of problem (2.1).
Step 3. We prove that there exists a positive ground state solution of problem (2.1).
Let N = {u ∈ H 1 0 (Ω) : I (u) = 0} and m 0 = inf u∈N I (u).For all u ∈ N , by the Sobolev inequality, it follows from (2.4) and (2.5) that since 0 < γ < 1, which implies that m 0 is well defined.According to Step 1 and Step 2, one has u , v ∈ N .Thus m 0 = inf u∈N I (u) ≤ I (u ) < 0. From (2.36), we can easy obtain that m 0 > −∞.Therefore, for the minimization problem m 0 , we can get a (PS) m 0 sequence.By Lemma 2.1 and Lemma 2.2, there exists u ∈ H 1 0 (Ω) such that I (u) = m 0 and I (u) = 0. Similar to u in Step 1, by the strong maximum principle, one obtains that u is a positive ground state solution of problem (2.1).This completes the proof of Theorem 2.3.

The proof of Theorem 1.1
According to Section 2, we know that Θ, Λ, D are independent of .Therefore, there exist two sequences of positive solutions {u n } and {v n } of the auxiliary problem (2.1) with I n (u n ) < 0 and I n (v n ) > 0, where n → 0 + as n → +∞.Now, we will prove the limits of the two sequences of positive solutions are two different positive solutions of problem (1.1).Now, we give the proof of Theorem 1.1.
Proof of Theorem 1.1.First, we prove that {u n } and {v n } have a convergent subsequence in H 1 0 (Ω), respectively.Without loss of generality, we only need prove that {u n } has a convergent subsequence in H 1 0 (Ω).Similarly, we can also obtain {v n } has a convergent subsequence.Since u n is the positive solution of problem (2.1), one has Let e be a positive weak solution of the following problem and for every Ω 0 ⊂⊂ Ω, there exists e 0 > 0 such that e| Ω 0 ≥ e 0 .Therefore, by the comparison principle, we get In particular, from e| Ω 0 ≥ e 0 > 0, one has Similar to (2.6), we can easy obtain that Up to a subsequence, combining with (3.1), there exists As usually, letting w n = u n − u * , we need prove that w n → 0 as n → ∞.Let lim n→∞ w n = l ≥ 0. For every n > 0, since and similar to (2.3) in [22], one has