Bounded state solutions of Kirchhoff type problems with a critical exponent in high dimension

In the present paper, we consider the following Kirchhoff type problem \begin{document}$\begin{cases}-\Big(a+λ∈t_{\mathbb R^N} | \nabla u|^2dx\Big) Δ u+V(x)u = |u|^{2^*-2}u \;\;\;{\rm in}\ \mathbb{R}^N,\\u∈ D^{1,2}(\mathbb R^N),\end{cases}$ \end{document} where \begin{document}$a$ \end{document} is a positive constant, \begin{document}$λ$ \end{document} is a positive parameter, \begin{document}$V∈ L^{\frac{N}{2}}(\mathbb{R}^N)$ \end{document} is a given nonnegative function and \begin{document}$2^*$ \end{document} is the critical exponent. The existence of bounded state solutions for Kirchhoff type problem with critical exponents in the whole \begin{document}$\mathbb R^N$ \end{document} ( \begin{document}$N≥5$ \end{document} ) has never been considered so far. We obtain sufficient conditions on the existence of bounded state solutions in high dimension \begin{document}$N≥4$ \end{document} , and especially it is the fist time to consider the case when \begin{document}$N≥5$ \end{document} in the literature.

1. Introduction and main results. In the present paper, we consider the existence of bounded state solutions for the Kirchhoff type problem − a + λ R N |∇u| 2 dx ∆u + V (x)u = |u| 2 * −2 u in R N , u ∈ D 1,2 (R N ), where a is a positive constant, λ is a positive parameter, V ∈ L N 2 (R N ) is a given nonnegative function and 2 * = 2N N −2 , (N ≥ 4), is the critical exponent. When λ = 0, this problem reduces to the scalar field equation. A large number of papers have been published about the scalar field equation both in H 1 (R N ) and in D 1,2 (R N ). Felmer et al. [6] obtained some monotonicity properties for ground states of the scalar field equation with subcritical exponent. Benci and Cerami [2] studied the existence of bounded state solutions for the scalar field equation with critical exponents.
However, quite a few papers have been published about this problem in 4 dimension. In Naimen [18], the author considered the following Kirchhoff type problem − a + b Ω |∇u| 2 dx ∆u = νu q + µu 3 , u > 0 in Ω, where Ω ⊂ R 4 is a bounded domain with smooth boundary ∂Ω. The author obtained some existence results with some suitable conditions on a, ν, µ > 0, b ≥ 0 and 1 ≤ q < 3. Naimen [18] is the first author investigating the Kirchhoff type problem with critical exponent in 4 dimension. We refer the readers to [9,10,14,15,16] for related results.
As we all know, the existence of bounded state solutions for Kirchhoff type problems with critical exponents in the whole R N (N ≥ 4) has never been considered so far. To state our result, we denote the best Sobolev constant for the embedding D 1,2 (R N ) → L 2 * (R N ) by S. We also denote the norm in L p (R N ) by | · | p .
When N = 4, we have the following result. then problem (SK * ) has at least one bounded state solution.
Remark 1.2. It should be mentioned that the basic idea in the proof follows from those in Benci and Cerami [2] and Xie et al. [25]. For problem (SK * ) with a = 1, λ = 0 in 4 dimension, Benci and Cerami [2] obtained a positive solution under the following assumptions: V (x) ≥ 0 for any x ∈ R 4 , there exist two positive constants p 1 < 2 < p 2 such that and 0 < |V | 2 < ( √ 2 − 1)S. Clearly, our theorem improves the main result in Benci and Cerami [2] when N = 4.
For N ≥ 5, if we set then we have the following result.
Theorem 1.3. Let a > 0, N ≥ 5 and V ∈ L N 2 (R N ) be a nonnegative function. There exist two positive constants µ := µ(a, N ) < 1 2 Λ 0 and ν := ν(a, N, λ) such that problem (SK * ) has at least two bounded state solutions for Λ 0 − µ < λ < Λ 0 and 0 < |V | N 2 < ν. |∇u| ∈ L 2 (R N )} with the inner product and norm, respectively, As we know, the following classical Schrödinger equation has been well studied. On the one hand, positive solutions must be the form On the other hand, infinitely many sign-changing solutions are obtained by Ding [5] when N ≥ 3.
(SK * ∞ ) Now the functionals I and I ∞ related to (SK * ) and (SK * ∞ ) are introduced, respectively, Firstly, we will consider the existence of positive and sign-changing solutions for this limit equation (SK * ∞ ) with N ≥ 4, which is heavily dependent on the dimension N and the parameter λ. What should be mentioned is that the case when N = 3 has been considered in Xie et al. [25]. Before that, we need the following lemma.
Then g(t) = 0 has a unique solution K − if and only if either N = 4 and λ < S −2 or N ≥ 5 and λ = Λ 0 . Moreover, g(t) = 0 has two different solutions K − and K + , (0 < K − < K + ) if and only if N ≥ 5 and λ < Λ 0 . These results are also valid for Proof. We only prove our results for g. When N = 4, one obtains It is easy to prove that g(t) = 0 has a unique positive solution if and only if λ < S −2 . When N ≥ 5, we know 0 < 4 N −2 ≤ 4 3 . Moreover, g(0) = a and g(t) → ∞ as t → ∞. It is easy to prove that is the unique global minima point of g. Thus min t≥0 g(t) = g(K min ) = 0 if and only if λ = Λ 0 . Moreover, This completes the proof. Proof. (a) On the one hand, if w is a positive solution of problem (S * ), then there exists at least one positive constant t such that u = tw solves problem (SK * ∞ ). Actually, if u = tw solves problem (SK * ∞ ), then one obtains −(a + λt 2 w 2 D 1,2 )t∆w = t 2 * −1 w 2 * −1 . Combining the uniqueness of positive solutions for problem (S * ) and w 2 N −2 + a = 0; namely g(t) = 0. If N = 4 and λ < S −2 or N ≥ 5 and λ ≤ Λ 0 hold, then we get the existence of the constant t by Lemma 2.1.
On the other hand, if u is a positive solution of problem (SK * ∞ ), then is also a positive solution of problem (S * ). It follows from the uniqueness of positive solutions for problem (S * ) and w 2 that there exists at least a ξ = u D 1,2 > 0 solving the following problem In other words, f (t) = 0 has at least a positive solution ξ. By Lemma 2.1, we have N = 4 and λ < S −2 or N ≥ 5 and λ ≤ Λ 0 .

QILIN XIE AND JIANSHE YU
(b) Arguing indirectly, if u is a sign-changing solution of problem (SK * ∞ ), then w = a + λ u 2 u is also a sign-changing solution of problem (S * ). With the help of w 2 If N = 4 holds, then which contradicts with (2.7).
Either ϕ − or ϕ + must be the least energy solution. Through out this paper, we only consider the case where ϕ + is the least energy solution. Since the other case can be dealt with similarly. It should be mentioned that both of them can be the least energy solution for some λ 0 at the same time and we only avoid those λ 0 .
Thus, for N ≥ 5, we know that ϕ + is the least energy solution. We define As Naimen pointed out in Remark 4.4 of [18], the least energy level c a,λ,N + can be negative. For example, when N = 6, by direct calculations, we can obtain the following results Obviously, for some suitable λ, then one can easily see c a,λ,6 Remark 2.5. As to problem (SK * ∞ ) with N = 3, from Xie et al. [25], we obtain the following branching diagram, in which the blue one is the positive solution and the red one is a sign-changing solution of (SK * ∞ ). When N = 4, we will have a similar bifurcation result from Proposition 2.2 and Remark 2.3. Note that u s is a sign-changing solution of (S * ).   We set the Nehari manifolds as follows, (2.17) and N N,± := u ∈ N N : ± a u 2 It is easy to see that which implies that has a unique positive solution This completes the proof.
is suitably small, then This completes the proof.
From the above two lemmas, we can define When N = 4, we can easily obtain the following result. This proof is similar to that of the next Proposition 2.9 or Proposition 2.5 in Xie et al. [25]. So we omit it here. When N ≥ 5, the following two results can be proved.
which implies that t ∞ − = 1 and contradicting with (2.26). This completes the proof.
be a nonegative function. Then there exist two positive constants µ 1 := µ 1 (a, N ) and ν 1 := ν 1 (a, N, λ) such that the relation m N, Proof. This proof is mainly similar to that of Proposition 2.9. Firstly, we show Hence, m N,+ ∞ ≤ c a,λ,N + . For any u ∈ N N,+ ∞ , one has On the one hand, by Lemma 2.7, for any The arbitrariness of u from N N, On the other hand, m N,+ ≤ m N,+ ∞ needs to be proved. By Lemma 2.7, consider the sequence u n = t n ϕ + δ,zn ∈ N N,+ , with ϕ + δ,zn defined by (2.13), {z n } ⊂ R N such that |z n | → ∞ as n → ∞ and t n = t + ϕ δ,zn . Applying a similar argument as that of (2.24), we can obtain lim Then m N, which implies that t ∞ + = 1 and R N V (x)|v 0 | 2 dx = 0. Similarly to the proof of Proposition 2.9, we can get a contradiction. This completes the proof.
Proof of Theorem 1.5. If either N = 4 and λ > S −2 or N ≥ 5 and λ > Λ 0 hold, then with the help of Lemma 2.1, one obtains which is impossible. The proof is completed.
3. Compactness results. In this section, we investigate the behavior of Palais-Smale sequences of I. The following proposition is a description of Palais-Smale sequences of I, which comes from Proprosition 5.1 in Xie et al. [25].
2) such that, up to subsequences, there hold

4)
as n → ∞ and we define that As a consequence, we have the following compactness result in N = 4.
Assume that there exist l (≥ 1) and k ≤ min{[ 1 λS 2 ], l} such that In what follows, we will estimate every term in the right hand side of the above equality.
Proof. We only prove the result on N N,− for N ≥ 5. The following methods are similar to the proof of Theorem 2.1 in [3] or the proof of Theorem 1 in [19]. By Ekeland's Variational Principle, there exists a minimising subsequence {u n } ⊂ N N,− of the minimisation problem such that I(u n ) < c a,λ,N − + 1 n and Since {u n } ⊂ N N,− , similar to (2.23), we get where ξ i , i = 1, 4, come from Lemma 5.2. It follows from With the help of (5.9), (3.23) and (3.24), one obtains Now we will show I (u n ) → 0.
By the continuity of t n (w) and t n (0) = 1, without loss of generality, we can assume ε n is small enough such that 1 2 ≤ t n (w) ≤ 3 2 for w D 1,2 < ε n . By (3.22), we obtain that is, Then it follows that By the choice of ε n , we obtain If we prove that | t n (0), w | ≤ C w D 1,2 , (3.28) then by (3.27), we get Hence, for any 0 < ε < ε n , we have for some C > 0 independent of ε and n. Taking ε → 0, we obtain (3.26). We now turn to proving (3.28). Indeed, by (5.12), we have To prove (3.28), we only need to show that for some positive constant δ and n large. By the definition of f , we get By the continuity of f (t), there exists M > 0 such that A n := 2 N −4 a− 1 n λ 1 2 satisfying f (A n ) < 0 for n > M . Combining with f (ξ 1 ) = 0, we get that ξ 1 < A n .
Arguing indirectly, assume that, for a subsequence (still denoted by {u n }), we have, for n > M , which implies that n which implies u n 2 D 1,2 ≥ A n > ξ 1 . This is impossible because of (3.25). Therefore, we conclude that I (u n ) → 0 as n → ∞ in the dual space of D 1,2 (R N ). What has been proved is that the minimizing sequence {u n } for I on N N,− is a Palais-Smale sequence of I at level c a,λ,N − . This result follows from Proposition 2.9 and Proposition 3.3 immediately. 4. Existence of bounded state solutions. In this section, we build a suitable min-max scheme for our problem. Firstly, it should be recalled the definition of the barycenter of a function. Setting where With the help of Lemmas 2.6 and 2.7, we can define the map θ ± : (y, δ) ∈ R N ×R + → D 1,2 (R N ) by where ϕ ± δ,y is defined in (2.8) or (2.13). Proof. It follows from t − ϕ ϕ − δ,y ∈ N 4,− that A direct computation gives This completes the proof. (4.4) Proof. Actually, By Lemma 2.7, we have t + ϕ = 1 + o(1) as |V | N 2 → 0. Thus, there exists ν 2 = ν 2 (a, N, λ) such that, for |V | N 2 < ν 2 , Hence, the result (4.4) holds.
Proof. By the compactness of the interval [δ 1 , δ 2 ] and Lemma 4.6, there exists a positive constant R such that I(θ ± (y, δ)) < c a,λ,N ± + κ for any δ ∈ [δ 1 , δ 2 ] and y ∈ R N with |y| ≥ R. Noting δ 1 , δ 2 and R from Lemma 4.7, Lemma 4.8, and Lemma 4.9, respectively, we define a bounded domain D ⊂ R N by  where D • is the interior of the set D.