An Existence Result for Fractional Kirchhoff-Type

Zeitschrift f¨ ur Analysis und ihre Anwendungen Journal of Analysis and its Applications Volume 35 (2016), 181–197 DOI: 10.4171/ZAA/1561

c European Mathematical Society

An Existence Result for Fractional Kirchhoff-Type Equations Giovanni Molica Bisci and Francesco Tulone Dedicated to Anna and Sandro Abstract. The aim of this paper is to study a class of nonlocal fractional Laplacian equations of Kirchhoff-type. More precisely, by using an appropriate analytical context on fractional Sobolev spaces, we establish the existence of one non-trivial weak solution for nonlocal fractional problems exploiting suitable variational methods. Keywords. Fractional equations, variational methods, critical point results Mathematics Subject Classification (2010). Primary 35J62, 35J92, 35J20, secondary 35J15, 47J30

1. Introduction This paper is devoted to the following nonlocal problem ( −(a + bkuk2X0 )LK u = f (x, u) in Ω u=0 in Rn \ Ω.

(PfK )

Here and in the sequel Ω is a bounded domain in (Rn , | · |), where 2s < n < 4s and s ∈ (0, 1), with continuous boundary ∂Ω, and f : Ω×R → R is a continuous function verifying the conditions stated in the sequel. Moreover, a, b denote two positive real constants and Z 2 kukX0 := |u(x) − u(y)|2 K(x − y) dxdy. Rn ×Rn

Finally, LK is a nonlocal operator defined as follows: Z LK u(x) := (u(x + y) + u(x − y) − 2u(x)) K(y) dy,

(x ∈ Rn )

Rn

where K : Rn \ {0} → (0, +∞) is a kernel function with the properties that: G. Molica Bisci: Dipartimento P.A.U. - University of Reggio Calabria, Feo di Vito, 89124 Reggio Calabria, Italy; [email protected] F. Tulone: Department of Mathematics - University of Palermo, Via Archirafi 34, 90123 Palermo, Italy; [email protected]

182

G. Molica Bisci and F. Tulone

(k1 ) γK ∈ L1 (Rn ), where γ(x) := min{|x|2 , 1}; (k2 ) there exists θ > 0 such that K(x) ≥ θ|x|−(n+2s) , for any x ∈ Rn \ {0}. A typical example of the kernel K is given by K(x) := |x|−(n+2s) . In this case LK is the fractional Laplace operator defined as Z u(x + y) + u(x − y) − 2u(x) s −(−∆) u(x) := dy, x ∈ Rn . n+2s |y| n R Aim of this paper is to get the existence of weak solutions for problem (PfK ). By a weak solution for (PfK ), we mean a function u : Rn → R such that u ∈ X0 and Z 2 (a + bkukX0 ) (u(x) − u(y))(ϕ(x) − ϕ(y))K(x − y) dxdy Rn ×Rn Z = f (x, u(x))ϕ(x) dx ∀ ϕ ∈ X0 Ω

Here and in the sequel we set X0 := u ∈ X : u = 0 a.e. in CΩ , where the functional space X denotes the linear space of Lebesgue measurable functions from Rn to R such that the restriction to Ω of any function u in X belongs to L2 (Ω) and p ((x, y) 7→ (u(x) − u(y)) K(x − y)) ∈ L2 (Rn × Rn ) \ (CΩ × CΩ), dxdy , with CΩ := Rn \ Ω. Setting Z t f (x, τ ) dτ, F (x, t) :=

and G(x, t) := f (x, t)t − 4F (x, t),

∀ (x, t) ∈ Ω × R

0

the main result reads as follows. Theorem 1.1. Let Ω be a bounded domain in (Rn , | · |), where 2s < n < 4s and s ∈ (0, 1), with continuous boundary ∂Ω. Further, let K : Rn \ {0} → (0, +∞) be a function satisfying hypotheses (k1 ) and (k2 ). Finally, let f ∈ C 0 (Ω × R) such that the following conditions hold: (f0 ) There exists a positive constant C such that |f (x, t)| ≤ C(1 + |t|q−1 ), 2n for some q ∈ 4, n−2s ;

∀ (x, t) ∈ Ω × R

Fractional Kirchhoff-Type Equations

183

(f1 ) tf (x, t) ≥ 0 in Ω × R; (f2 ) For some σ ≥ 1, one has G(x, t) ≥

G(x, ζt) , σ

∀ (x, t) ∈ Ω × R

and every ζ ∈ [0, 1]; (f3 ) There is δ > 0 such that F (x, t) ≤ a

λ1 2 t, 2

for every x ∈ Ω and t ∈ (−δ, δ), where λ1 is the first eigenvalue of −LK in X0 ; = +∞, uniformly in x ∈ Ω. (f4 ) lim|t|→+∞ f (x,t) t3 Then, problem (PfK ) has at least one non-trivial weak solution. The above result represents a non-local version of an existence result obtained by Sun and Tang for Kirchhoff-type equations defined on bounded domains of the n-dimensional Euclidean space, with n < 4 (see [34, Theorem 1]). This dimensional restriction is replaced in the fractional setting by 2s < n < 4s. This assumption is essential in our technical approach in order to guarantee the embedding of the working space X0 in the Lebesgue space Lq (Rn ), where 4 0 and u1 ∈ E with ku1 k > ρ. Let Γ := {γ ∈ C 0 ([0, 1], E) : γ(0) = 0, γ(1) = u1 }. and c := inf max J(γ(τ )). γ∈Γ τ ∈[0,1]

Then c ≥ β > 0 and there exists a sequence {uj }j∈N ⊂ E such that J(uj ) → c,

and

(1 + kuj k)J 0 (uj ) → 0.

Moreover, if J satisfies the (C)c condition, then c is a critical value of J. For the sake of completeness, we also recall that the C 1 -functional J : E → R satisfies the Cerami condition at level c ∈ R (briefly (C)c condition) when


187

(C)c Every sequence {uj }j∈N ⊂ E such that J(uj ) → c and n o (1 + kuj k) sup |hJ 0 (uj ), ϕi| : ϕ ∈ E, kϕk = 1 → 0 as j → ∞, possesses a convergent subsequence in E. Such a sequence is then called a Cerami sequence of the functional J. Finally, J satisfies the compactness Cerami condition ((C) condition for short) if (C)c holds for every c ∈ R.

3. Technical results Among others, two notions of fractional operators are well-known and widely studied in the literature in connection with elliptic problems of fractional type, namely the integral one (which reduces to the classical fractional Laplacian), and the spectral one (that is sometimes called the local, fractional Laplacian). We would like to note that, as pointed out in [32], these two fractional operators are different. Indeed, the spectral operator depends on the domain Ω considered (since its eigenfunctions and eigenvalues depend on Ω), while the integral one (−∆)s evaluated at some point is independent on the domain in which the equation is set. Further, it is easily seen that the eigenvalues of the spectral Laplacian are the s-th power of the eigenvalues of the classical Laplacian. On the contrary, our abstract framework is more delicate and our approach is based on a careful analysis of the linear problem −LK u = λu in Ω (4) u=0 in Rn \ Ω, related to the operator −LK . A spectral theory for general integrodifferential nonlocal operators was developed in [30, Proposition 9 and Appendix A]. See also [27] for further properties of the spectrum of −LK and of its eigenfunctions. With respect to the eigenvalue problem (4), we recall that it possesses a divergent sequence of eigenvalues 0 < λ1 < λ2 ≤ · · · ≤ λk ≤ λk+1 ≤ · · · . To avoid possible confusions, we stress the fact that the eigenvalues that we consider, even in the model case of the fractional Laplacian, are not the s-th power of the eigenvalues of the standard Laplacian. As usual, we denote by ek the eigenfunction related to the λk , eigenvalue k ∈ N. From [30, Proposition 9], we know that we can choose ek k∈N normalized in such a way that this sequence provides an orthonormal basis in L2 (Ω) and an orthogonal basis in X0 , so that for any k, i ∈ N with k 6= i Z hek , ei iX0 = ek (x)ei (x) dx = 0 and kek k2X0 = λk kek k2L2 (Ω) = λk . Ω

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Furthermore, by [30, Proposition 9 and Appendix A], we have the following characterization of the eigenvalue λ1 : Z |u(x) − u(y)|2 K(x − y) dxdy Rn ×Rn Z . λ1 = min u∈X0 \{0} |u(x)|2 dx Ω

Finally, the first eigenfunction e1 ∈ X0 is non-negative in Ω, see [30, Proposition 9 and Appendix A]. Denote by A the class of all continuous functions f : Ω × R → R such that |f (x, t)| < +∞, q−1 (x,t)∈Ω×R 1 + |t| sup

for some q ∈ (4, 2∗ ).

For the proof of our result, we observe that problem (PfK ) has a variational structure, indeed it is the Euler-Lagrange equation of the functional JK : X0 → R defined in (2). Note that the functional JK is Fréchet differentiable in u ∈ X0 and one has Z 0 2 (u(x) − u(y))(ϕ(x) − ϕ(y))K(x − y) dxdy hJK (u), ϕi = (a + bkukX0 ) Rn ×Rn Z − f (x, u(x))ϕ(x) dx, Ω

for every ϕ ∈ X0 . Thus, critical points of JK are solutions to problem (PfK ). Lemma 3.1. Let f ∈ A and assume that conditions (f1 ), (f2 ) are verified in addition to (f4 ). Then, every Cerami sequence {uj }j∈N ⊂ X0 of the functional JK is bounded in X0 . Proof. Let {uj }j∈N be a Cerami sequence, i.e. for some c ∈ R, one has Z b a 2 4 JK (uj ) = kuj kX0 + kuj kX0 − F (x, uj (x)) dx → c, 2 4 Ω and

o n 0 (1 + kuj kX0 ) sup |hJK (uj ), ϕi| : ϕ ∈ X0 , kϕkX0 = 1 → 0,

as j → ∞. Hence c = JK (uj ) + o(1), (5) n o and, since |hJK0 (uj ), uj i| ≤ kuj kX0 sup |hJK0 (uj ), ϕi| : ϕ ∈ X0 , kϕkX0 = 1 , we also have that hJK0 (uj ), uj i = o(1), (6)


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where o(1) → 0, as j → ∞.Thus, from (5) and (6), for j large enough, it follows that 1 1 + c ≥ JK (uj ) − hJK0 (uj ), uj i Z4 (7) a 1 2 = kuj k + f (x, uj (x))uj (x) − F (x, uj (x)) dx. 4 4 Ω We claim that {uj }j∈N is bounded in X0 . If the assertion were false, up to a subsequence, we could suppose kuj kX0 → ∞, as j → ∞. Set uj wj := , ∀ j ∈ N. kuj kX0 Clearly kwj kX0 = 1, so that {wj }j∈N is bounded. Hence, since X0 is a reflexive space, bearing in mind that for every r ∈ [1, 2∗ ) the embedding X0 ,→ Lr (Ω) is compact, we may assume (up to a subsequence) that wj * w in X0 , wj → w in Lr (Ω), for every 1 ≤ r < 2∗ , wj (x) → w(x) a.e. x ∈ Ω, for some w ∈ X0 . Now, we divide the proof in two cases. Case 1. If w ≡ 0, we choose a sequence {tj }j∈N ⊂ [0, 1] such that JK (tj uj ) = max JK (tuj ). t∈[0,1]

Now, for any m > 0, let r 4 8m wj , vj,m := b

(note that b > 0) for every j ∈ N.

At this point, owing to vj,m → 0 in Lq (Ω), by (f0 ) one has that Z q |F (x, vj,m (x))| dx ≤ C1 kvj,m kL1 (Ω) + kvj,m kLq (Ω) → 0,

(C1 > 0)

Ω

as j → ∞. Thus Z lim F (x, vj,m (x)) dx = 0. j→∞ Ω √ 4 8m So, for j sufficiently large, kuj kXb ∈ (0, 1), and 0

kvj,m k2X0

r

r

r

8m 2

8m u 2 8m

4

4

j = wj = ,

=

b b kuj kX0 b X0

as well as kvj,m k4X0 =

X0

8m . b

Hence, there exists j0 ∈ N such that Z JK (tj uj ) ≥ JK (vj,m ) ≥ 2m − F (x, vj,m (x)) dx ≥ m, Ω

190


for every j ≥ j0 . Then, we have JK (tj uj ) → +∞,

as j → ∞.

(8)

Now, since JK (0) = 0 and JK (uj ) → c, we deduce that tj ∈ (0, 1) and Z 2 2 (a + bktj uj kX0 )ktj uj kX0 − f (x, tj uj (x))tj uj (x) dx Ω

=

hJK0 (tj uj ), tj uj i

(9)

dJK (twj ) = tj dt t=tj = 0. Therefore, by using (f2 ), it follows that, G(x, tj uj (x)) ≤ σG(x, uj (x)),

∀x ∈ Ω

(10)

and tj ∈ (0, 1). Hence, by (9) and (10) one has 4JK (tj uj ) = 4JK (tj uj ) − hJK0 (tj uj ), tj uj i Z 2 = aktj uj kX0 + G(x, tj uj (x)) dx ΩZ ≤ aktj uj k2X0 + σ G(x, uj (x)) dx.

(11)

Ω

Moreover, since aktj uj k2X0 ≤ aσkuj k2X0

(σ ≥ 1),

Z

f (x, uj (x))uj (x) dx = akuj k2X0 + bkuj k4X0 + o(1), Ω Z b a F (x, uj (x)) dx = kuj k2X0 + kuj k4X0 − c + o(1), 2 4 Ω

by (11), it follows that JK (tj uj ) ≤ cσ + o(1), which contradicts (7). Case 2. The function w ∈ X0 is not identically zero in Ω. Hence, let us denote Ω1 := {x ∈ Ω : w(x) 6= 0},

and Ω2 := {x ∈ Ω : w(x) = 0}.

Clearly, one has that |Ω1 | > 0,

Ω = Ω1 ∪ Ω2 ,

and Ω1 ∩ Ω2 = ∅.


191

Since |uj (x)| = |wj (x)|kuj kX0 , we have that |uj (x)| → ∞ for every x ∈ Ω1 , and, thanks to (f4 ), we also have f (x, uj (x)) = +∞, j→∞ uj (x)3 lim

uniformly in Ω1 . Hence, the Fatou’s Lemma, implies that Z f (x, uj (x)) lim |wj (x)|4 dx → +∞, as j → ∞. 3 j→∞ Ω uj (x) 1

(12)

On the other hand, taking into account that f is a continuous function, it is easy to see that Z f (x, uj (x)) C2 |Ω2 |, (13) |wj (x)|4 dx ≥ − 3 uj (x) kuj k4X0 Ω2 for some constant C2 > 0. Then, relations (12) and (13) imply that Z f (x, uj (x))uj (x) dx = +∞. lim j→∞ Ω kuj k4X0 Now, by (6) it follows that

R Ω

f (x,uj (x))uj (x) kuj k4X 0

Z lim inf j→∞

Ω

(14)

dx = b+ kujak2 − kuo(1) 4 . Consequently jk X0

f (x, uj (x))uj (x) dx = b, kuj k4X0

X0

(15)

that contradicts (14). In conclusion, in any case, the sequence {uj }j∈N is bounded in X0 . Lemma 3.2. Let f ∈ A and assume that conditions (f1 ), (f2 ) are verified in addition to (f4 ). Then, the functional JK satisfies the (C) compactness condition. Proof. Let {uj }j∈N ⊂ X0 be a Cerami sequence. By Lemma 3.1, the sequence {uj }j∈N is necessarily bounded in X0 . Since X0 is reflexive, we can extract a subsequence which for simplicity we shall call again {uj }j∈N , such that uj * u∞ in X0 . This means that Z uj (x) − uj (y) ϕ(x) − ϕ(y) K(x − y) dxdy Q Z (16) → u∞ (x) − u∞ (y) ϕ(x) − ϕ(y) K(x − y) dxdy, Q

for any ϕ ∈ X0 , as j → ∞.

192


We will prove that {uj }j∈N strongly converges to u∞ ∈ X0 . Exploiting the derivative JK0 (uj )(uj − u∞ ), we obtain Z 0 ha(uj ), uj − u∞ i = hJK (uj ), uj − u∞ i + f (x, uj (x))(uj − u∞ )(x) dx, (17) Ω

where we set Z ha(uj ), uj − u∞ i :=

|uj (x) − uj (y)|2 K(x − y) dxdy

Q

Z

− uj (x) − uj (y) u∞ (x) − u∞ (y) K(x − y) dxdy Q Z 2 × a + b |uj (x) − uj (y)| K(x − y) dxdy .

Q

n o Since (1 + kuj kX0 ) sup |hJK0 (uj ), ϕi| : ϕ ∈ X0 , kϕkX0 = 1 → 0, and taking into account that the sequence {uj − u∞ }j∈N is bounded in X0 , one gets hJK0 (uj ), uj − u∞ i → 0,

as j → ∞.

(18)

Since the embedding X0 ,→ Lq (Ω) is compact, clearly uj → u∞ strongly in Lq (Ω). So, by condition (f0 ), standard computations ensure that Z |f (x, uj (x))||uj (x) − u∞ (x)| dx → 0. (19) Ω

By (17) relations (18) and (19) yield ha(uj ), uj − u∞ i → 0,

as j → ∞.

(20)

Now, observe that Z 0 0, such that JK (u) ≥ β for every u ∈ X0 with kukX0 = ρ. Proof. By (f0 ) and (f3 ), there exists C1 > 0 such that a F (x, t) ≤ λ1 t2 + C1 |t|q , 2

(23)

for every (x, t) ∈ Ω × R. Thus, by (23), we have Z b a 2 4 JK (u) = kukX0 + kukX0 − F (x, u(x)) dx 2 4 Ω Z Z a b a 2 4 2 ≥ kukX0 + kukX0 − λ1 |u(x)| dx − C1 |u(x)|q dx, 2 4 2 Ω Ω for every u ∈ X0 . Now, since X0 ,→ Lq (Ω) continuously, the above inequality becomes JK (u) ≥ 4b kuk4X0 − C2 kukqX0 , where C2 := C1 cqq . Since q > 4, choosing ρ
0, 4 for every u ∈ X0 with kukX0 = ρ. This concludes the proof. Lemma 4.2. There exists e ∈ X0 with kekX0 > ρ such that JK (e) < 0. Proof. By (f4 ), for every x ∈ Ω, one has f (x, t) = +∞. t→+∞ t3 lim

Then, for any M > 0, there exists δM > 0 such that f (x, t) 1 ≥ , t3 M for every t > δM and x ∈ Ω. Setting cM := for every t ≥ 0 and x ∈ Ω. Then

√ 3

δM , ε

it follows that f (x, t) ≥

z 3 t4 f (x, st)t ≥ − cM t, M

t3 −cM , ε

(24)

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for every (x, t) ∈ Ω × [0, +∞), and z ∈ [0, 1]. Integrating both sides of the inequality (24) on [0, 1] with respect to z, we obtain t4 − cM t, (25) 4M for every (x, t) ∈ Ω × [0, +∞). Now, since the first eigenfunction e1 of the operator −LK in X0 , is not negative in Ω (see Section 3), by (25) it follows that F (x, t) ≥

t4 e1 (x)4 − cM te1 (x), 4M for every (x, t) ∈ Ω × [0, +∞). Hence, for every j ∈ N, one has F (x, te1 (x)) ≥

(26)

F (x, je1 (x)) e1 (x)4 cM e1 (x) − ≥ , j4 4M j3 for every x ∈ Ω. Consequently Z Z e1 (x)4 cM e1 (x) F (x, je1 (x)) − dx. dx ≥ j4 4M j3 Ω Ω

(27)

By (27) the Fatou’s lemma, immediately yields Z ke1 k4L4 (Ω) F (x, je1 (x)) lim inf dx ≥ , j→∞ j4 4M Ω for every M > 0. Hence, passing to the limit for M → 0, one has Z F (x, je1 (x)) lim dx = +∞. j→∞ Ω j4 Thus JK (je1 ) a b = 2 ke1 k2X0 + ke1 k4X0 − 4 j 2j 4

Z Ω

F (x, je1 (x)) dx → −∞, j4

as j → ∞. Finally, the above relation ensures that there exists ν0 ∈ N such that, setting e := ν0 e1 ∈ X0 , it follows that kekX0 > ρ and JK (e) < 0. The conclusion is achieved. Proof of Theorem 1.1. Take E = X0 and the functional J = JK in Proposition 2.1. Let us define Γ := {γ ∈ C 0 ([0, 1], X0 ) : γ(0) = 0, γ(1) = e},

and c := inf max JK (γ(τ )), γ∈Γ τ ∈[0,1]

where e ∈ X0 is given in Lemma 4.2. Now, since JK (0) = 0, by Lemma 4.2 one has max{JK (0), JK (e)} = 0. Moreover, Lemmas 4.1 and 4.2 ensure that 0 0 and e ∈ X0 , with kekX0 > ρ. Then, by Proposition 2.1, it


195

follows that c ≥ β > 0. Finally, by Lemma 3.1, since the Cerami compactness condition holds at level c, there exists u0 ∈ X0 such that JK0 (u0 ) = 0 and JK (u0 ) = c ≥ β > 0. Thus u0 ∈ X0 is a non-trivial critical point of JK . This completes the proof of Theorem 1.1. Remark 4.3. Let f : Ω × R → R be a function such that (f03 ) limt→0 f (x,t) = 0, uniformly in Ω. t 0 Assumption (f3 ) yields F (x, t) lim = 0, t→0 t2 uniformly in Ω. Consequently, condition (f3 ) immediately holds. In conclusion, we present a direct application of Theorem 1.1 and Remark 4.3. Example 4.4. Let s ∈ ( 34 , 1) and let Ω be an open bounded set of R3 with continuous boundary ∂Ω. Moreover, let f : R → R be the continuous function defined by f (t) := t3 log(1 + |t|), ∀ t ∈ R. Simple and direct computations ensure that F (t) =

(t4 − 1) (t2 + 2) 2 (t2 + 3) |t| + log(|t| + 1) − t, 12 4 16

and

(t2 + 3) (t2 + 2) 2 |t| + log(|t| + 1) + t, 3 4 for every t ∈ R. Further, one has |f (t)| ≤ 1 + |t|4 , as well as G(t) := f (t)t − 4F (t) = −

f (t) = 0, t→0 t

lim

and

f (t) = +∞. |t|→∞ t3 lim

Finally G(t) ≥ G(st),

∀ (t, s) ∈ R × [0, 1].

Then, owing to Theorem 1.1, the following problem Z  |u(x) − u(y)|2  a+b dxdy (−∆)s u = u3 log(1 + |u|) 3+2s |x − y| 3 3 R ×R  u|R3 \Ω = 0,

in Ω

admits one non-trivial weak solution in the fractional Sobolev space H0 := u ∈ H s (R3 ) : u = 0 a.e. in CΩ , for every real constants a, b > 0. Acknowledgement. The authors warmly thank the anonymous referees for their useful and nice comments on the paper. The manuscript was realized within the auspices of the INdAM - GNAMPA Project 2015 titled Modelli ed equazioni non-locali di tipo frazionario.

196


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Received May 5, 2015; revised October 23, 2015