VARIATIONAL METHODS FOR KIRCHHOFF TYPE PROBLEMS WITH TEMPERED FRACTIONAL DERIVATIVE

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1 Electronic Journal of Differential Equations, Vol (2018, No. 34, pp ISSN: UL: or VAIATIONAL METHODS FO KICHHOFF TYPE POBLEMS WITH TEMPEED FACTIONAL DEIVATIVE NEMAT NYAMOADI, YONG ZHOU, BASHI AHMAD, AHMED ALSAEDI Communicated by Paul abinowitz Abstract. In this article, using variational methods, we study the existence of solutions for the Kirchhoff-type problem involving tempered fractional derivatives Z M + u(t 2 dt (Dα, + u(t = f(t, u(t, t, u W α,2 (, where ± u(t are the left and right tempered fractional derivatives of order α (1/2, 1], > 0, W α,2 ( represent the fractional Sobolev space, f C(, and M C( +, Introduction Fractional calculus is a natural extension of ordinary calculus, where integrals and derivatives are defined for arbitrary real orders. Since 17th century, when fractional calculus was born, several kinds of fractional derivatives have been proposed. Examples include iemann-liouville, Hadamard, Grunwald-Letnikov, Caputo, tempered, etc. [4, 5, 9, 11, 12, 13, 14, 17, 19], each of them having its own advantages and disadvantages. The choice of an appropriate fractional derivative, depending on the system under consideration, has led to a variety of researches for fractional differential equations involving different fractional derivatives. For details and examples, we refer the reader to a series of papers [2, 3, 26, 27, 30, 31, 32, 33, 34, 35] and the references cited therein. One of the simplest description of a fractional derivative relies on Fourier transform. If f(x is a function with Fourier transform f(w, then the iemann-liouville fractional derivative D α f(x is the function with Fourier transform (iw α f(w, which is an extension of familiar integer-order formula [9, 17]. The foregoing arguments have motivated the researchers to investigate the tempered fractional derivative f(t, defined in terms of a function having Fourier transform ( + iw α û(w with the tempered fractional integral I α, f(t as its inverse, having Fourier transform (+iw α f(w [4, 5, 12, 13, 19]. In this paper, we apply variational methods to establish the existence of infinitely many solutions 2010 Mathematics Subject Classification. 26A63, 34A38, 35A45. Key words and phrases. Tempered fractional calculus; Kirchhoff type problems; Variational methods. c 2018 Texas State University. Submitted September 6, Published January 24,

2 2 N. NYAMOADI, Y. ZHOU, B. AHMAD, A. ALSAEDI EJDE-2018/34 to the following Kirchhoff-type problem involving tempered fractional derivatives: M + u(t 2 dt ( + u(t = f(t, u(t, t, (1.1 u W α,2 (, where ± u(t denote the left and right tempered fractional derivatives of order α (1/2, 1], > 0, f C(, and M C( +, +. In recent years, there has been a growing interest in the study of fractional differential equations by means of variational methods and critical point theory. One of the pioneering works in this direction was due to Jiao and Zhou [7], who investigated the following fractional boundary value problem by using Mountain Pass Theorem, d dt ( D β t (u (t t D β T (u (t + F (t, u(t = 0, a.e. t [0, T ], u(0 = u(t = 0, where 0 D β t and t D β T are the left and right iemann-liouville fractional integrals of order 0 β < 1 respectively. For more examples, we refer the reader to a series of papers [8, 22, 23, 24, 28, 29, 15, 16, 36] and the references cited therein. In the sequel, we need the following assumptions: (A1 M C( +, + and there exists Υ > 1 such that M(tt Υ M(t for all t [0,, and for all δ > 0 there exists ϱ = ϱ(δ > 0 such that M(t ϱ for all t δ, where M(t = t 0 M(sds; (A2 M C( +, + and there exist three constants 0 < m 1 m 2 < and 1 < β < such that m 1 t β M(t m 2 t β, t + ; (1.2 (A3 f(t, u = o( u as u 0 uniformly for t, (A4 f C(, such that there exist b C(, + with lim t + b(t = 0 and 2 < q < + such that f(t, s b(t s q 1, (t, s ; (A5 There exist µ > 2Υ with Υ > 1 such that 0 < µf (t, ζ ζf(t, ζ, ζ > 0, where F (t, u = u f(t, sds; 0 F (t,ζ (A6 lim ζ 0 ζ = 0 uniformly for a.e. t ; 2Υ (A7 There exist two constants b 1 > 0, 1 < γ 0 < 2 such that F (t, s b 1 s γ0, (t, s ; (A8 f is odd in x, i.e. f(t, x = f(t, x, (t, x. The rest of the paper is organized as follows. In Section 2, we describe some basic concepts related to our main results (Theorems In Section 3, the existence of infinitely many solutions to the problem (1.1 is established.

3 EJDE-2018/34 VAIATIONAL METHODS FO KICHHOFF TYPE POBLEMS 3 2. Preliminaries Let us recall some basic definitions and lemmas that we need in the forthcoming analysis. Definition 2.1. For any > 0, we define the positive tempered fractional integral of a function f L p ( with 1 p < as I α, + f(x = 1 Γ(α x and the negative tempered fractional integral by I α, f(x = 1 Γ(α + x f(ξ(x ξ α 1 e (x ξ dξ, (2.1 f(ξ(ξ x α 1 e (ξ x dξ. (2.2 If = 0, these formulae reduce to the well-known iemann-liouville fractional integrals [9, 14]. Definition 2.2. The positive and negative tempered fractional derivatives of order 0 < α < 1 for a function f : are defined by for any > 0. + f(x = α f(x + f(x = α f(x + Define the fractional space α Γ(1 α α Γ(1 α W α,2 ( = {f L 2 ( : x + x f(x f(ξ (x ξ α+1 e (x ξ dξ, (2.3 f(x f(ξ (ξ x α+1 e (ξ x dξ, (2.4 ( 2 + ω 2 α f(ω } 2 dω <, (2.5 which is a Banach space with the norm 1/2. f α, = ( 2 + ω 2 α f(ω dω 2 (2.6 For any f W α,2 (, let Dα, ± f(x denote the functions with Fourier transform ( ± iω α f(ω ([20], where the Fourier transform of u(x is defined as follows F(u(ξ = Now we state the following known results. e ix ξ u(xdx. Lemma 2.3 (See [20]. (i For any α, > 0 and f L 2 (, we have and for any f W α,2 (, we have (ii For any α, > 0 and f, g W α,2 (, we have ± I α, ± f(x = f(x, (2.7 I α, ± ± f(x = f(x. (2.8 f, + g L2 ( = f, g L2 (. (2.9

4 4 N. NYAMOADI, Y. ZHOU, B. AHMAD, A. ALSAEDI EJDE-2018/34 Lemma 2.4 (See [12]. (i For any α, > 0 and p 1, I α, ± : L p ( L p ( are bounded linear operators with (ii For any α, β, > 0 and f L p (, we have (iii For any α, > 0 and f, g L 2 (, we have I α, ± f Lp ( α f Lp (. (2.10 I α, ± I β, ± f(x = I α+β, ± f(x. (2.11 f, I α, + g L2 ( = I α, f, g L2 (. (2.12 Next, for 0 < α < 1, we define fractional Sobolev space H α ( as follows endowed with the norm u α = ( For 0 < α < 1, we have H α ( = C 0 ( α, u(t 2 dt + ω 2α û(ω 2 dω 1/2. ( α 1 2 u α u α,1 u α, (2.14 u α,1 u α, α u α,1, (2.15 u α, u α,1 α u α,, (2.16 where u α,1 is the norm on W α,2 1 ( and so W α,2 1 ( = H α ( with equivalent norms. Lemma 2.5 (See [1]. Let α > 1/2. Then any u W α,2 ( is uniformly continuous, bounded and there exists a constant C = C α such that sup u(t C u α,. (2.17 t emark 2.6. From Lemma 2.5 and (2.13-(2.15, we have the following implication: if u W α,2 with 1 2 < α < 1, then u Lq ( for all q [2, as u(t q dt u q 2 u 2 L 2 ( 21 α C q 2 u q α,. emark 2.7 ([21]. The imbedding of W α,2 in L q ( T, T is compact for q (2, and any T > Main results Definition 3.1. For every u, v W α,2 (, a weak solution of problem (1.1 is M + u(t 2 dt ( 2 + ω 2 α û(ω v(ωdω = f(t, u(tv(tdt, that is, M + u(t 2 dt + u(t + v(tdt = f(t, u(tv(tdt. Definition 3.2. We say that the functional Φ satisfies the Palais-Smale condition if any sequence {u n } n N X has a convergent subsequence provided {Φ(u n } n N is bounded and Φ (u n 0 as n +.

5 EJDE-2018/34 VAIATIONAL METHODS FO KICHHOFF TYPE POBLEMS 5 Theorem 3.3 ([6, Theorem 2.2]. Let X be a real infinite dimensional Banach space and K C 1 (X be a functional satisfying the Palais-Smale condition and that (i K(0 = 0 and there exist two constants α > 0 and ρ > 0 such that ϕ Bρ α, where B ρ = {u X : u < ρ}; (ii K is even; (iii for all finite dimensional subspace X X, there is = ( X such that ϕ(u 0 on X \ B ( X e. Then the functional K possess an unbounded sequence of critical values characterized by a minimax argument. Consider a functional Φ : W α,2 ( defined by Φ(u = 1 2 M + u(t 2 dt F (t, u(tdt, u W α,2 (, (3.1 where M(t = t M(sds. Obviously, by the conditions (A1 (or (A2 and (A4, 0 Φ C 1 (W α,2 (,, and ( Φ (uv = M + u(t 2 dt + u(t + v(t dt (3.2 f(t, u(tv(tdt, for every u, v W α,2 (. Lemma 3.4. Assume that (A1, (A4-(A6 hold. Then there exist ρ, β > 0 such that Φ(u β, u W α,2 (, u α, = ρ. Proof. In view of (A6, for all ε > 0 there exists δ = δ(ε > 0 such that F (t, u ε u 2Υ, (t, u [0, δ. (3.3 Also, by (A4, for u > δ, there is a T > 0 such that for t > T. Set b T := max t [ T,T ] b(t, then we have F (t, u ε u q, (3.4 F (t, u b ε u q, (3.5 for u > δ, where b ε = max{b T, ε}. As F (t, is even, it follows by (3.3 and (3.5 that for all ε > 0, there is a b ε > 0 such that F (t, u ε u 2Υ + b ε u q, for all (t, u. (3.6 Moreover, from (2.6 and (A1, we have M(t > 0 for all t > 0 and M(t M(1t Υ, for all t [0, 1]. (3.7 From (3.6 and (3.7, for all u W α,2 ( with u α, 1, we get Φ(u 1 2 M( + u(t 2 dt ε u(t 2Υ dt a ε u(t q dt 1 2Υ M(1 u α, ε2 1 α C 2Υ 2 u 2Υ 2 (3.8 α, a ε 2 1 α C q 2 u q α,.

6 6 N. NYAMOADI, Y. ZHOU, B. AHMAD, A. ALSAEDI EJDE-2018/34 Choosing one obtains ε := M( α C 2Υ 2, Φ(u M(1 u 2Υ α, a ε 2 1 α C q 2 u q α, 4 ( M(1 u 2Υ α, 4 a ε 2 1 α C q 2 u q 2Υ α,. Hence, for all u W α,2 ( with u α, = ρ and 0 < ρ < 1 small enough, we have M(1 a ε 2 1 α C q 2 ρ q 2Υ > 0. 4 Therefore, by taking β := ρ 2Υ( M(1 4 a ε 2 1 α C q 2 ρ q 2Υ, we get Φ(u β for all u W α,2 (, u α, = ρ. Thus the conclusion is achieved. Lemma 3.5. Assume that (A1, (A3 (A6 hold. Then, for any finite dimensional subspace E of W α,2, there exists 1 = 1 (E > 0 such that Φ(u 0, u W α,2 ( \ B 1(E, where B 1(E = {u W α,2 : u α, < 1 }. Proof. In a straightforward manner, one can obtain F (t, ζ K ζ µ, where K := 1 r µ inf F (t, ζ > 0. ζ, ζ =r Then, by (A3 (A5, there exists M > 0 such that F (t, ζ K ζ µ M ζ 2, for all (t, ζ. (3.9 Also, by assumption (A1, we have M(t M(1t Υ. (3.10 Let E W α,2 ( be a fixed finite dimensional. Now, for any u E with u α, = 1, by emark 2.6, (3.9 and (3.10, we have Φ(su = 1 2 M(s 2 F (t, su(tdt 1 2 s2υ M(1 Ks µ u(t µ dt + M u(t 2 dt 1 2 s2υ M(1 Ks µ M µ E u µ α, + M21 α u 2 α, = 1 2 s2υ M(1 Ks µ M µ E + M21 α, as s, where M E > 0 such that u L p M E u α, for all u E. As 1, we have sup Φ(u = sup Φ( 1 u. u E, u α, = 1 u E, u α, =1

7 EJDE-2018/34 VAIATIONAL METHODS FO KICHHOFF TYPE POBLEMS 7 Therefore, there exists 0 > 0 large enough such that Φ(u 0 for all u E with u α, = 1 and 1 0. This completes the proof. Lemma 3.6. Assume that (A1, (A4, (A5 hold. Then Φ satisfies Palais-Smale condition. Proof. Assume that {u n } n N W α,2 ( is a sequence such that {Φ(u n} n N is bounded and Φ (u n 0 as n. Then there exists a constant D > 0 such that Φ(u n D and Φ (u n (W α,2 ( D, (3.11 for any n N, where (W α,2 ( is the dual space of W α,2 (. Firstly, we show that {u n } n N is bounded. Without loss of generality, we assume that inf n u n α, = η > 0, denote by ϱ = ϱ(η the number corresponding to δ = η 2 in (A1 such that M( u n 2 α, ϱ for all n. (3.12 In view of (A5 and (3.12, one gets D + D u n α, Φ(u n 1 µ Φ (u n u n = 1 2 M( u n 2 α, 1 µ M( u n 2 α, u n 2 α, 1 (µf (t, u n (t f(t, u n (tu n (tdt µ ( 1 2Υ 1 M( un 2 µ α, u n 2 α, ϱ ( 1 2Υ 1 un 2 µ α,. Since µ > 2Υ, the boundedness of {u n } n N follows directly. subsequence {u n } n N, and u W α,2 such that So, there exist a u n u weakly in W α,2 (, (3.13 which yields ( Φ (u n (u n u = M( u n 2 α, + u n + (u n u dt f(t, u n (u n udt 0 as n. (3.14 Now we show that lim n f(t, u n(u n udt = 0. To this end, by (3.13, there is some constant d > 0 such that u n α, < d and u α, < d, for n N, u n u strongly in L q ( and a.e. in. Moreover, for any ε > 0, (A4 implies that there exists T > 0 such that f(t, u n ε u n q 1, for t > T. (3.15 Then, for n large enough, from (2.17, emark 2.6 and Young inequality, we obtain f(t, u n (u n udt

8 8 N. NYAMOADI, Y. ZHOU, B. AHMAD, A. ALSAEDI EJDE-2018/34 T T f(t, u n u n u dt f(t, u n u n u dt + ε u n + ε t >T εc u n α, + ε t >T t >T u n q 1 u n u dt f(t, u n u n u dt ( q 1 u n q + 1 q µ u n u q dt εc u n α, + q 1 ε2 1 α C q 2 u n q α, q + ε 1 µ 21 α C q 2 u n u q α, εcd + q 1 ε2 1 α C q 2 d q + ε 1 q µ 21 α C q 2 u n u q α,. Then lim f(t, u n (u n udt = 0. n Therefore, by (3.14, we have ( M( u n 2 α, + u n + (u n u dt 0, as n. Thus, by (3.12 and the boundedness of M( u n 2 α,, one can get ( + u n + (u n u dt 0, as n. (3.16 In a similar manner, we can get ( + u + (u n u dt 0, as n. (3.17 Combining (3.16 and (3.17, we obtain ( + (u n u + (u n u dt 0, as n. Hence, u n u α, 0 as n and then Φ satisfies Palais-Smale condition. Theorem 3.7. Assume that (A1, (A3 (A6, (A8 hold. Then problem (1.1 has infinitely many nontrivial solutions. Proof. Assumption (A8 implies that F (t, is even in and so is Φ. Since Φ(0 = 0, it follows from Lemmas and Theorem 3.3 that there exists an unbounded sequence of weak solutions of problem (1.1. To prove our second result, we will use the genus properties. So we recall the following definitions and results (see [18]. Let X be a Banach space, g C 1 (X, and c. Set Σ = {A X \ {0} : A is closed in X and symmetric with respect to 0}, K c = {x X : g(x = c, g (x = 0}, g c = {x X : g(x c}.

9 EJDE-2018/34 VAIATIONAL METHODS FO KICHHOFF TYPE POBLEMS 9 Definition 3.8 ([10]. For A Σ, we say genus of A is j (denoted by γ(a = j if there is an odd map ψ C(A, j \ {0}, and j is the smallest integer with this property. Theorem 3.9. Let g be an even C 1 functional on X which satisfies the Palais- Smale condition. For j N, let Σ j = {A Σ : γ(a j}, c j = inf sup A Σ j u A g(u. (i If Σ j and c j, then c j is a critical value of g. (ii If there exists r N such that c j = c j+1 = = c j+r = c and c g(0, then γ(k c r + 1. Lemma Assume that (A1 and (A4 hold. Then Φ is bounded from below and satisfies the Palais-Smale condition. Proof. By a method similar to the one in [25, Lemma 3.3], for all ε > 0, it follows from (A4 that F (t, u(t ε u(t 2, for all t. (3.18 For any u W α,2 (, by (3.18, we get Φ(u 1 2 M( u 2 α, F (t, u(tdt 1 2 M( u 2 α, ε u(t 2 dt If u α, 1, then by (3.7, we have 1 2 M( u 2 α, ε u 2 α,. Φ(u M(1 u 2Υ α, ε u 2 α, ε. ( If u α, > 1, then by (A1 and ε = ϱ 4, we get Φ(u ϱ 2 u 2 α, ε u 2 α, ϱ 4 u 2 α,. (3.20 Combining (3.19 and (3.20, one can infer that Φ is coercive. Thus Φ is bounded from below and satisfies the Palais-Smale condition. Theorem Assume that (A1, (A4, (A7, (A8 hold. Then problem (1.1 has infinitely many nontrivial solutions. Proof. The assumption (A8 implies that Φ is even and Φ(0 = 0, and by Lemma 3.10, J C 1 (X α, is bounded from below and satisfies the Palais-Smale condition. We make use of Theorem 3.9 to complete the proof. First, we show that there exists ε > 0 such that γ(φ ε n for any n N. (3.21 For each k, we take k disjoint open sets K i such that k K i. For i = 1,..., k, letting u i (W α,2 ( C 0 (K i \ {0} with u i α, = 1, we set X α n = span{u 1,..., u n }, S n = {u X α n : u α, = 1}. (3.22

10 10 N. NYAMOADI, Y. ZHOU, B. AHMAD, A. ALSAEDI EJDE-2018/34 For an u X α n, we can write u(t = for some i, i = 1, 2,..., n. So and u L γ 0 = u 2 α, = i u i (t for t, (3.23 1/γ0 ( n u(t γ0 dt = i γ0 = + u(t 2 dt = 2 i u i 2 α, = 2 i 2 i. u i (t γ0 dt 1/γ0, ( u i (t 2 dt (3.25 Since all norms of a finite dimensional normed space are equivalent, there exists a constant Θ > 0 such that Θ u α, u L γ 0 for u X α n. (3.26 From (A7, for u S n, we can take some Λ 0 such that F (t, u(tdt = ( F t, i u i (t dt b 1 Λ 0 i u i (t γ0 dt := ϱ. (3.27 We claim that ϱ > 0. To this end, suppose otherwise, for any bounded open set Λ, there exists {u k } k N S n such that Λ u k (t γ0 dt = b 1 Λ ik u i (t γ0 dt 0, as k +, where u k = n iku i (t with n 2 ik = 1. Then we have lim ik := i0 [0, 1] k + Thus, for any bounded open set Λ, we get Λ and 2 i0 = 1. i0 u i (t γ0 dt = 0. Since Λ is arbitrary, therefore u 0 = n i0u i (t = 0 a.e. on, which contradicts that u 0 X α = 1. Hence F (t, u(tdt = ( F t, i u i (t dt b 1 Λ 0 i u i (t γ0 dt = ϱ > 0. (3.28

11 EJDE-2018/34 VAIATIONAL METHODS FO KICHHOFF TYPE POBLEMS 11 From (A7, (3.24-(3.26 and (3.28, we have Φ(su = 1 2 M( su 2 α, F (t, su(tdt 1 2 max 0 l 1 M(ls2 1 2 max 0 l 1 M(ls2 b 1 s γ0 F (t, su i (tdt K i i I γ0 u i (t γ0 dt max 0 l 1 M(ls2 b 1 s γ0 u γ0 L γ 0 ( max 0 l 1 M(ls2 b 1 (Θs γ0 u γ0 α, 1 2 max 0 l 1 M(ls2 b 1 (Θs γ0, u S n, 0 < s < δ, which implies that there exist ε > 0 and σ > 0 such that Let Φ(σu < ε u S n. (3.30 S σ n = {σu : u S n }, Ω = { ( 1,..., n n : Then it follows from (3.30 that 2 i < σ 2}. Φ(u < ε u S σ n. (3.31 So, by (3.31 and the fact that Φ C 1 (W α,2 (, and is even, we get S σ n Φ ε Σ. (3.32 On the other hand, in view of (3.23 and (3.25, there exists an odd homeomorphism mapping Ψ C(S σ n, Ω. Using properties of the genus (see [18, 3 of Propositions 7.5 and 7.7 ], one can obtain Hence (3.21 is obtained. Set γ(φ ε γ(s σ n = n. (3.33 c n = inf sup Φ(u. A Σ n As Φ is bounded from below on X α and (3.33 implies that < c n ε < 0, therefore c n (for all n N is a real negative number. Thus it follows from Theorem 3.9 that Φ has infinitely many nontrivial critical points, which correspond to infinitely many nontrivial solutions to system (1.1. The proof is complete. Theorem Assume that (A2, (A4, (A7, (A8 hold. Then, for 2 < q < 2β and 1 < γ 0 < 2β, problem (1.1 has infinitely many nontrivial solutions. Proof. In view of (A2 and (A4 with 2 < q < 2β, one can show that the functional Φ is bounded from below and satisfies the Palais-Smale condition. The rest of the proof is similar to that of Theorem 3.11, so we omit it. u A

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