Multiple positive solutions of fractional elliptic equations involving concave and convex nonlinearities in $R^N$

In this paper, we investigate the existence of positive solutions of the following equation 
\begin{eqnarray} 
(-\Delta)^s v +\lambda v=f(x) v^{p-1}+h(x)v^{q-1}, \ x\in R^N,\\ 
v\in H^s(R^N), 
\end{eqnarray} 
where $1\leq q 2s$ and $\lambda>0$ is a parameter. 
Since the concave and convex nonlinearities are involved, the variational functional 
of the equation has different properties. Via variational method, we show that the equation 
admits a positive ground state solution for all $\lambda>0$ strictly larger than a threshold value. 
Moreover, under certain conditions on $f$ and for sufficiently large $\lambda>0$, we also prove that there are 
at least $k+1$ ($k$ is a positive integer) positive solutions of the equation.


(Communicated by Changfeng Gui)
Abstract. In this paper, we investigate the existence of positive solutions of the following equation where 1 ≤ q < 2 < p < 2 * s = 2N N −2s , 0 < s < 1, N > 2s and λ > 0 is a parameter. Since the concave and convex nonlinearities are involved, the variational functional of the equation has different properties. Via variational method, we show that the equation admits a positive ground state solution for all λ > 0 strictly larger than a threshold value. Moreover, under certain conditions on f and for sufficiently large λ > 0, we also prove that there are at least k + 1 (k is a positive integer) positive solutions of the equation.
Some related problems involving the operator (−∆) s have been studied in [2,6,9,22] and the references therein.
In this paper, we denote the space and equipped with the norm Clearly, · H s is a Hilbertian norm induced by the inner product For the case s = 1, the equation (1) has been studied by many researches. In [27], for q = λ = 1 and f (x) ≡ 1 for all x ∈ R N , Zhu proved that equation admits at least two positive solutions, where it was supposed that h is nonnegative, small, and exponential decay. Without the condition of exponential decay, Cao and Zhou [8] and Hirano [12] proved that equation (1) admits at least two positive solution in R N . For λ = 0, f (x) = h(x) ≡ 1, Ambrosetti et al. [1] considered equation (1) on bounded domain with concave-convex nonlinearity. In [14], Lin studied the equation with s = 1 has a positive ground stste solution and for λ large enough, the equation admits at least k + 1 positive solutions. For more results of the case s = 1, we refer the readers to the papers [14,27,8,25] and the references therein.
Recently, a great attention has been focused on the study of fractional elliptic type problem. For example, using the s-harmonic extension introduced by Caffarelli and Silvestre [7], Yu [26] used the Nehari manifold to establish the existence and multiple solutions for elliptic problems involving the square root of the Laplacian. And Brändle et al. [2] studied the fractional Laplace equation involving concaveconvex nonlinearity. In particular, by the version given by (2) for (−∆) s , Brarrios et al. [3] investigated the fractional Laplace equation involving concave-convex nonlinearity on bounded domain Ω. Any other reference, readers can consider [14,19,21].
Motivated by the results just described, we are interested in the existence and multiplicity of positive solutions of equation (1) in R N . In this paper, since the nonlocal term (−∆) s u is involved, the variational functional of the equation has different properties, we shall deal with it carefully. Moreover, the solution of the basic equation with f (εz) = 1 of (6) is algebra decay which is different from the case of s = 1. In [3], the authors considered the fractional elliptic type problem on bounded domain, however, in our paper, we consider the global space R N , the lack of compactness is another difficult we need to overcome.
In the sequel, we assume f and h satisfy the following conditions: (f 1 ) f is a positive continuous function in R N and lim |x|→+∞ f (x) = f ∞ > 0.
(f 2 ) There exist k points a 1 , · · · , a k in R N such that Associated with equation (3), we consider the C 1 -functional J ε , for u ∈ H s (R N ), where |x−y| N +2s dxdy + R N |u(x)| 2 dx, and u + = max{u, 0} ≥ 0. We know that nonnegative weak solutions of equation are equivalent to the critical points of J ε . Let For the fractional elliptic equation we define the energy functional By the same method of Theorem 4.12 and 4.13 in Wang [24], we have For convenient, we denote Λ := ε 2s(p−q) p−2 and Λ * := ε 2s(p−q) Our main results are as follows.  The article is organized as follows. First of all, we use the argument of Tarantello [23] to divide the Nehari Manifold M ε into the two parts M + ε and M − ε . Next, we prove that the existence of a positive ground state solution u 0 ∈ M + ε . Finally, in section 4, we show there are at least k critical points where Note that M ε contains all nonnegative solutions of equation (1). From the lemma below, we have J ε is bounded from below on M ε .
Lemma 2.1. The energy functional J ε is coercive and bounded from below on M ε .
Proof. For u ∈ M ε , by (8), the Hölder inequality and the Sobolev embedding theorem (5), we get Hence, we have J ε is coercive and bounded from below on M ε . Define Then for u ∈ M ε , we get We apply the method in Tarantello [23], and set By the Hölder inequality and the Sobolev embedding theorem, we get we get a contradiction. Hence, the conclusion is completed.
Proof. It is similar to Theorem 2.3 in Brown and Zhang [4], so we omit it here.
Lemma 2.4. We have the following inequalities Proof. (i) It can be proved by using (10).
(ii) For any u ∈ M + ε ⊂ M ε , by (10), we apply the Hölder inequality to obtain (iii) For any u ∈ M − ε by (10), we have by direct calculation, we get (iii).
Applying Ekeland's variational principle and using the same argument in Cao and Zhou [8] or Tarantello [23], we have the following lemma.
3. Existence of a ground state solution. In order to prove the existence of positive solutions, we claim that J ε satisfies the (P S) β condition in H s (R N ) for and C 0 is defined in the following lemma.
Proof. Since {u n } is a (P S) β -sequence in H s (R N ) for J ε with u n u weakly in H s (R N ), it is easy to check that J ε (u) = 0 in (H s (R N )) −1 and u ≥ 0. Then we have J ε (u), u = 0, that is Similarly, we get Remark 1. By (7), we obtain Lemma 3.2. Assume that f and h satisfy (f 1 ) and (h 1 ).
It follows that {u n } is bounded in H s (R N ). Hence, there exist a subsequence {u n } and a nonnegative u ∈ H s (R N ) such that J ε (u) = 0 in (H s (R N )) −1 , and a.e. in R N .
Using the Brezis-Lieb lemma to get and Next, we claim that For any σ > 0, there exists r > 0 such that [B N (0;r)] c h(εz) p p−q dz < σ. By the Hölder inequality and the Sobolev embedding theorem, we get Then (19) holds.
Applying (f 1 ) and u n → u in L p loc (R N ), we get Let p n = u n − u. Suppose p n 0 strongly in H s (R N ). By (17)-(21), we deduce Then By Theorem 4.3 in Wang [24], there exists a sequence {s n } ⊂ R + such that s n = 1 + o n (1), {s n p n } ⊂ N ∞ and I ∞ (s n p n ) = I ∞ (p n ) + o n (1). By (19) and (21), we obtain which is a contradiction. Hence, u n → u strongly in H s (R N ).
Proof. By lemma 2.7 (i), there is a minimizing sequence there exist a subsequence {u n } and u 0 ∈ H s (R N ) such that u n → u 0 strongly in H s (R N ). It is easy to see that u 0 0 is a solution of equation (1) in R N and J ε (u 0 ) = α ε .
Next, we claim that u 0 ∈ M + ε . On the contrary, we assume that It follows that which contradicts to α ε < 0. By Lemma 2.5 (ii), there exist positive numbers t + <t < t − = 1 such that t which is a contradiction. Hence, u 0 ∈ M + ε and By Lemma 2.2 and the maximum principle [5], then u 0 is a positive solution of equation (1) in R N .

4.
Existence of k+1 solutions. From now on, we assume that f satisfies (f 1 ) − (f 2 ) and h satisfies (h 1 ). Let w ∈ H s (R N ) be the unique, radially symmetric, and positive ground state solution of equation Recall the facts in [10,11] about the equation (27).
There exists a number t 0 > 0 such that for 0 ≤ t ≤ t 0 and any ε > 0 , we have J ε (tw i ε ) < γ max uniformly in i. (ii) There exist positive numbers t 1 and ε 1 such that for any t > t 1 and ε < ε 1 , we have ε } is uniformly bounded in H s (R N ) for any ε > 0, and γ max > 0, then for any ε > 0 and 0 ≤ t ≤ t 0 , we obtain J ε (tw i ε ) < γ max .
(ii) There is an r 0 > 0 such that f (z) ≥ fmax 2 for z ∈ B N (a i ; r 0 ) uniformly in i. Then there exists ε 1 > 0 such that for ε < ε 1 Thus, there is t 1 > 0 such that for any t > t 1 and ε < ε 1 J ε (tw i ε ) < 0 uniformly in i.
Proof. By Lemma 4.1, we only need to show that We know that sup t≥0 I max (tw) = γ max . For t 0 ≤ t ≤ t 1 , we get Applying the results of Lemmas 2.5, 2.6 (ii) and 4.2, we can deduce that Choosing 0 < ρ 0 < 1 such that for any ε < ε 0 and each 1 ≤ i ≤ k.
Lemma 4.4. Suppose there existsδ > 0 such that I ε (u) ≤ γ max +δ for u ∈ N ε , then we obtain Q ε (u) ∈ K ρ 0 2 for any 0 < ε < ε 0 . Proof. On the contrary, there exist sequences {ε n } ⊂ R + and {u n } ⊂ N εn , such that ε n → 0, I εn (u n ) = γ max + o n (1) as n → ∞ and Q εn (u n ) ∈ K ρ 0 2 for all n ∈ N . It is easy to check that {u n } is bounded in H s (R N ). Suppose u n → 0 strongly in L p (R N ). Since and which is a contradiction. Thus u n → 0 strongly in L p (R N ). Applying the concentration-compactness principle [15,16], there exist a constant d 0 > 0 and a sequence {z n } ⊂ R N such that B N (zn;1) Let v n (z) = u n (z +z n ), there are a subsequence {v n } and v ∈ H s (R N ) such that v n v weakly in H s (R N ). Using the similar computation in Lemma 2.5, there is a sequence {s n max } ⊂ R + such thatṽ n = s n max v n ∈ N max and 0 < γ max ≤ I max (ṽ n ) ≤ I εn (s n max u n ) ≤ I εn (u n ) = γ max + o n (1) as n → ∞.
We deduce that a convergent subsequence {s n max } satisfies s n max → s 0 > 0. Then there is subsequenceṽ n ṽ = (s 0 v) weakly in H s (R N ). By (29), thenṽ = 0. Moreover, we can obtain thatṽ n →ṽ strongly in H s (R N ) and I max (ṽ) = γ max . Now, we want to show there exist a subsequence {z n } = {ε nzn } such that z n → z 0 ∈ K.
(i) Claim that the sequence {z n } is bounded in R N . On the contrary, assume that |z n | → ∞, then f (ε n z)(u n ) p + dz = lim inf n→∞ I εn (s n max u n ) ≤ lim inf n→∞ I εn (u n ) = γ max , which is a contradiction.
(ii) Claim that z 0 ∈ K. On the contrary, assume that z 0 ∈ K, that is, f (z 0 ) < f max . Then using the above argument to obtain which is a contradiction.