Universal Journal of Mathematics and Applications

Universal Journal of Mathematics and Applications VOLUME I ISSUE III

ISSN 2619-9653 http://dergipark.gov.tr/ujma

VOLUME I ISSUE III ISSN 2619-9653

September 2018 http://dergipark.gov.tr/ujma

UNIVERSAL JOURNAL OF MATHEMATICS AND APPLICATIONS

Honorary Editor-in-Chief

Murat Tosun ¨ ˙ Department of Mathematics, Faculty of Science and Arts, Sakarya University, Sakarya-TURK IYE [email protected]

Editors

Editor in Chief Emrah Evren Kara Department of Mathematics, Faculty of Science and Arts, D¨ uzce University, ¨ ˙ D¨ uzce-TURK IYE [email protected]

Editor in Chief Mahmut Akyi˘ git Department of Mathematics, Faculty of Science and Arts, Sakarya University, ¨ ˙ Sakarya-TURK IYE [email protected] Managing Editor Fuat Usta Department of Mathematics, Faculty of Science and Arts, D¨ uzce University, ¨ ˙ D¨ uzce-TURK IYE [email protected]

Managing Editor Murat Kiri¸sçi Department of Mathematics, ˙ Faculty of Science and Arts, Istanbul University, ˙Istanbul-TURK ¨ ˙IYE [email protected] Editorial Secretariat ˙ Merve Ilkhan Department of Mathematics, Faculty of Science and Arts, D¨ uzce University, ¨ ˙ D¨ uzce-TURK IYE [email protected]

Editorial Board of Universal Journal of Mathematics and Applications

Hari Mohan Srivastava University of Victoria, CANADA

Mohammad Mursaleen Aligarh Muslim University, INDIA

Poom Kumam King Mongkut’s University of Technology Thonburi, THAILAND

Soley Ersoy Sakarya University, ¨ ˙ TURK IYE

Necip S ¸ im¸sek ˙ Istanbul Commerce University, ¨ ˙ TURK IYE

Syed Abdul Mohiuddine Sakarya University, ¨ ˙ TURK IYE

Michael Th. Rassias University of Zurich, SWITZERLAND

Sel¸cuk Topal Bitlis Eren University, ¨ ˙ TURK IYE i

Mujahid Abbas University of Pretoria, SOUTH AFRICA

Feng Qi Henan Polytechnic University, CHINA

Dumitru Baleanu C ¸ ankaya University, ¨ ˙ TURK IYE

Silvestru Sever Dragomir Victoria University, AUSTRALIA

˙ Ishak Altun Kırıkkale University, ¨ ˙ TURK IYE

Yeol Je Cho Gyeongsang National University, KOREA

Changjin Xu Guizhou University of Finance and Economics, CHINA

Ljubisa D. R. Kocinac University of Nis, SERBIA Waqas Nazeer University of Education, PAKISTAN

Sidney Allen Morris Federation University, AUSTRALIA

Lakshmi Narayan Mishra Vellore Institute of Technology (VIT) University, INDIA

Vishnu Narayan Mishra Indira Gandhi National Tribal University, MADHYA PRADESH

ii

Contents 1 On the paranormed binomial sequence spaces Hacer Bilgin Ellidokuzo˘ glu, Serkan Demiriz, Ali K¨ oseo˘ glu

137-147

2 Explicit limit cycles of a class of Kolmogorov system Salah Benyoucef, Ahmed Bendjeddou

148-154

3 A note on hyperbolic quaternions I¸ sıl Arda K¨ osal

155-159

4 Generalized Zagreb index of some dendrimer structures ˙ Prosanta Sarkar, Nilanjan De, Ismail Naci Cang¨ ul, Anita Pal

160-165

5 General helices that lie on the sphere S 2n in Euclidean space E 2n+1 B¨ ulent Altunkaya, Levent Kula

166-170

6 Nonexistence of global solutions to system of semi-linear fractional evolution equations Medjahed Djilali, Ali Hakem 171-177 7 Existence and uniqueness of an inverse problem for a second order hyperbolic equation ˙ Ibrahim Tekin 178-185 8 Multiple solutions for a class of superquadratic fractional Hamiltonian systems Mohsen Timoumi

186-195

9 Geometry of bracket-generating distributions of step 2 on graded manifolds Esmaeil Azizpour, Dordi Mohammad Ataei

196-201

10 On the convergence of a modified superquadratic method for generalized equations Mohammed Harunor Rashid, Md. Zulfiker Ali

202-214

iii

Universal Journal of Mathematics and Applications, 1 (3) (2018) 137-147

Universal Journal of Mathematics and Applications Journal Homepage: www.dergipark.gov.tr/ujma ISSN 2619-9653 DOI: http://dx.doi.org/10.32323/ujma.395247

On the paranormed binomial sequence spaces Hacer Bilgin Ellidokuzo˜glua , Serkan Demirizb* and Ali Köseo˜glua a Recep

Tayyip Erdo˜gan University, Science and Art Faculty, Department of Mathematics, Rize-Turkey University, Science and Art Faculty, Department of Mathematics, Tokat-Turkey * Corresponding author E-mail: [email protected]

b Gaziosmanpas¸a

Article Info

Abstract

Keywords: Binomial sequence spaces, Paranorm, Matrix domain, Matrix transformations 2010 AMS: 46A45, 40C05, 46B20 Received: 15 February 2018 Accepted: 6 March 2018 Available online: 30 September 2018

r,s r,s r,s In this paper the sequence spaces br,s 0 (p), bc (p), b∞ (p) and b (p) which are the generalization of the classical Maddox’s paranormed sequence spaces have been introduced and r,s r,s r,s proved that the spaces br,s 0 (p), bc (p), b∞ (p) and b (p) are linearly isomorphic to spaces c0 (p), c(p), `∞ (p) and `(p), respectively. Besides this, the α−, β − and γ−duals of the r,s r,s spaces br,s 0 (p), bc (p), and b (p) have been computed, their bases have been constructed and some topological properties of these spaces have been studied. Finally, the classes r,s r,s of matrices (br,s 0 (p) : µ), (bc (p) : µ) and (b (p) : µ) have been characterized, where µ is one of the sequence spaces `∞ , c and c0 and derives the other characterizations for the special cases of µ.

1. Introduction We shall denote the space of all real-valued sequences by w as a classical notation. Any vector subspace of w is called a sequence space. The spaces `∞ , c and c0 are the most common and frequently used spaces which are all bounded, convergent and null sequences, respectively. Also bs, cs, `1 and ` p notations are used for the spaces of all bounded, convergent, absolutely and p−absolutely convergent series, respectively, where 1 < p < ∞. First, we point out the concept of a paranorm. A linear topological space X over the real field R is said to be a paranormed space if there is a subadditive function g : X → R such that g(θ ) = 0, g(x) = g(−x) and scalar multiplication is continuous, i.e., |αn − α| → 0 and g(xn − x) → 0 imply g(αn xn − αx) → 0 for all α’s in R and all x’s in X, where θ is the zero vector in the linear space X. Assume here and after that (pk ) be a bounded sequences of strictly positive real numbers with sup pk = H and L = max{1, H}. Then, the linear spaces `∞ (p), c(p), c0 (p) and `(p) were defined by Maddox [19] (see also Simons [21] and Nakano [20]) as follows: `∞ (p) = {x = (xk ) ∈ w : sup |xk | pk < ∞}, k∈N

c(p) = {x = (xk ) ∈ w : lim |xk − l| pk = 0 for some l ∈ R}, k→∞

c0 (p) = {x = (xk ) ∈ w : lim |xk | pk = 0} k→∞

( `(p) =

)

x = (xk ) ∈ w : ∑ |xk |

pk

0 and g2 (x) =

∑ |xk |

pk

,

k

respectively. For convenience in notation, here and in what follows, the summation without limits runs from 0 to ∞. By F and Nk , we shall denote the collection of all finite subsets of N and the set of all n ∈ N such that n ≥ k, respectively. We shall assume throughout that 0 −1 = 1 provided 1 < inf p ≤ H < ∞. p−1 k k + (pk )

Email address: [email protected] (H. B. Ellidokuzo˜glu) [email protected] (S. Demiriz) [email protected] (A. Köseo˜glu)

138

Universal Journal of Mathematics and Applications

Let λ , µ be any two sequence spaces and A = (ank ) be an infinite matrix of real numbers ank , where n, k ∈ N. Then, we say that A defines a matrix mapping from λ into µ, and we denote it by A : λ → µ, if for every sequence x = (xk ) ∈ λ , the sequence Ax = {(Ax)n }, the A−transform of x, is in µ, where (Ax)n = ∑ ank xk , (n ∈ N).

(1.1)

k

By (λ : µ), we denote the class of all matrices A such that A : λ → µ. Thus, A ∈ (λ : µ) if and only if the series on the right-hand side of (1.1) converges for each n ∈ N and every x ∈ λ , and we have Ax = {(Ax)n }n∈N ∈ µ for all x ∈ µ. A sequence x is said to be A−summable to α if Ax converges to α which is called the A−limit of x. r,s r,s r,s 2. The sequence spaces br,s 0 (p), bc (p), b∞ (p) and b (p) r,s r,s r,s r,s r,s r,s r,s In this section, we define the sequence spaces br,s 0 (p), bc (p), b∞ (p) and b (p), and prove that b0 (p), bc (p), b∞ (p) and b (p) are the complete paranormed linear spaces. For a sequence space λ , the matrix domain λA of an infinite matrix A is defined by

λA = {x = (xk ) ∈ w : Ax ∈ λ }.

(2.1)

In [7], Choudhary and Mishra have defined the sequence space `(p) which consists of all sequences such that S−transforms are in `(p) , where S = (snk ) is defined by 1 , 0 ≤ k ≤ n, snk = 0 , k > n. Bas¸ar and Altay [3] have studied the space bs(p) which is formerly defined by Bas¸ar in [4] as the set of all series whose sequences of partial t (p) in [1] and rt (p), rt (p) in [2] which are sums are in `∞ (p). More recently, Altay and Bas¸ar have studied the sequence spaces rt (p), r∞ c 0 derived by the Riesz means from the sequence spaces `(p), `∞ (p), c(p) and c0 (p) of Maddox, respectively. t (p), rt (p) and rt (p) may be redefined by With the notation of (2.1), the spaces `(p), bs(p), rt (p), r∞ c 0 `(p) = [`(p)]S , bs(p) = [`∞ (p)]S , rt (p) = [`(p)]tR t r∞ (p) = [`∞ (p)]tR , rct (p) = [c(p)]tR , r0t (p) = [c0 (p)]tR .

In [8], Demiriz and C ¸ akan have defined the sequence spaces er0 (u, p) and erc (u, p) which consists of all sequences such that E r,u - transforms are in c0 (p) and c(p), respectively E r,u = {ernk (u)} is defined by n n−k rk u , (0 ≤ k ≤ n), k k (1 − r) ernk (u) = 0 , (k > n) for all k, n ∈ N and 0 < r < 1. r,s r,s r,s r,s In [5] and [6], the Binomial sequence spaces br,s the sequence 0 , bc , b∞ and b p , which are the matrix domains of Binomial mean Br,s inr,s r,s spaces c0 , c, `∞ and ` p , respectively, are introduced, some inclusion relations and Schauder basis for the spaces br,s , b c , b∞ and b p are 0 given, and the α−, β − and γ− duals of those spaces are determined. For more papers related to sequence spaces and matrix domains of different infinite matrices one can see [13, 12] and references therein. The main purpose of this paper is to introduce the sequence spaces r,s r,s r,s br,s all sequences whose Br,s −transforms are in the spaces c0 (p), c(p), `∞ (p) and `(p), 0 (p), bc (p), b∞ (p) and b (p) which are the set of r,s respectively; where Br,s denotes the matrix Br,s = {bnk } defined by ( k n−k k 1 r , 0 ≤ k ≤ n, r,s n n s (s+r) bnk = 0 , k > n, where sr > 0. Also, we have constructed the basis and computed the α−, β − and γ−duals and investigated some topological properties of r,s r,s r,s the spaces br,s 0 (p), bc (p), b∞ (p) and b (p). Following Choudhary and Mishra [7], Bas¸ar and Altay [3], Altay and Bas¸ar [1, 2], Demiriz [8], Kiris¸çi [14, 15], Candan and Günes¸ [16] r,s r,s r,s and Ellidokuzo˜glu and Demiriz [9], we define the sequence spaces br,s 0 (p), bc (p), b∞ (p) and b (p), as the sets of all sequences such that Br,s −transforms of them are in the spaces c0 (p), c(p), `∞ (p) and `(p), respectively, that is, ( ) pn n 1 n r,s n−k k b0 (p) = x = (xk ) ∈ w : lim s r xk = 0 , n→∞ (s + r)n ∑ k k=0 (

) pn n n−k k s r xk − l = 0 , ∑ k=0 k pn ( ) 1 n n n−k k r,s s r xk < ∞ , b∞ (p) = x = (xk ) ∈ w : sup n ∑ n∈N (s + r) k=0 k pn ( ) 1 n n n−k k r,s b (p) = x = (xk ) ∈ w : ∑ s r xk < ∞ . n ∑ n (s + r) k=0 k br,s c (p) =

1 x = (xk ) ∈ w : ∃l ∈ C 3 lim n→∞ (s + r)n

n

139


r,s r,s r,s In the case (pn ) = e = (1, 1, 1, ...), the sequence spaces br,s spaces 0 (p), bc (p), b∞ (p) and b (p) are, respectively, reduced tor,sthe sequence r,s r,s r,s r,s br,s , b , b and b which are introduced by Bis ¸ gin [5, 6]. With the notation of (2.1), we may redefine the spaces b0 (p), br,s ∞ c p c (p), b∞ (p) 0 r,s and b (p) as follows: r,s r,s r,s r,s r,s r,s r,s br,s 0 (p) = [c0 (p)]B , bc (p) = [c(p)]B , b∞ (p) = [`∞ (p)]B and b (p) = [`(p)]B .

Define the sequence y = {yn (r, s)}, which will be frequently used, as the Br,s −transform of a sequence x = (xk ), i.e., n 1 n n−k k yn (r, s) := s r xk ; for all k ∈ N. ∑ (s + r)n k=0 k

(2.2)

Now, we may begin with the following theorem which is essential in the text. r,s r,s Theorem 2.1. br,s 0 (p), bc (p) and b∞ (p) are the complete linear metric space paranormed by g, defined by

pn /L 1 n n n−k k g(x) = sup s r xk . n ∑ n∈N (s + r) k=0 k

(2.3)

In addition, br,s (p) is the complete linear metric space paranormed by h, defined by 1 ∑ (s + r)n n=0 ∞

h(x) =

pn !1/M n n−k k s r xk . ∑ k=0 k n

(2.4)

r,s r,s r,s (p), br,s Proof. First, we give the proof for br,s c (p) and b∞ (p). Since the proof is similar for bc (p) and b∞ (p), we give the proof only for r,s0 the space br,s (p). The linearity of b (p) with respect to the co-ordinatewise addition and scalar multiplication follows from the following 0 0 inequalities which are satisfied for x, z ∈ br,s (p) (see Maddox [18, p.30]) 0

pn /L 1 1 n n n−k k s r (x + z ) ≤ k k ∑ (s + r)n k=0 k (s + r)n

pn /L pn /L 1 n n n−k k n n−k k s r xk + ∑ ∑ k s r zk (s + r)n k=0 k=0 k n

(2.5)

and for any α ∈ R (see [21]) |α| pn ≤ max{1, |α|L } = K.

(2.6)

Using (2.6) inequality, we get pn /L pn /L 1 1 n n n n−k k n n−k k pn /L ∑ k s r (αxk ) = |α| (s + r)n ∑ k s r xk (s + r)n k=0 k=0 pn /L 1 n n n−k k 1/L ≤K ∑ k s r xk (s + r)n k=0 r,s for x ∈ br,s 0 (p). This shows the space b0 (p) is a linear space. r,s Now we will see that g is a paranorm on br,s 0 (p). It is clear that g(θ ) = 0 and g(x) = g(−x) for all x ∈ b0 (p). n − x) → 0 and (α ) also be any sequence of scalars such that α → α. Let {xn } be any sequence of the points xn ∈ br,s (p) such that g(x n n 0 Then, since the inequality

g(xn ) ≤ g(x) + g(xn − x) holds by the subadditivity of g, {g(xn )} is bounded and we thus have 1 g(αn x − αx) = sup k k∈N (s + r) n

pk /L k k− j j n ∑ j s r (αn x j − αx j ) j=0 k

≤ |αn − α|g(xn ) + |α|g(xn − x)

(2.7)

which tends to zero as n → ∞. This means that the scalar multiplication is continuous. Hence, g is a paranorm on the space br,s 0 (p). (i)

(i)

(i)

r,s i i It remains to prove the completeness of the space br,s 0 (p). Let {x } be any Cauchy sequence in the space b0 (p), where x = {x0 , x1 , x2 , . . .}. Then, for a given ε > 0 there exists a positive integer n0 (ε) such that

g(xi − x j )
n0 (ε). Using the definition of g we obtain for each fixed k ∈ N that |(Br,s xi )k − (Br,s x j )k | pk /L ≤ sup |(Br,s xi )k − (Br,s x j )k | pk /L < k∈N

ε 2

(2.8)

140


for every i, j > n0 (ε) which leads to the fact that {(Br,s x0 )k , (Br,s x1 )k , (Br,s x2 )k , . . .} is a Cauchy sequence of real numbers for every fixed k ∈ N. Since R is complete, it converges, say (Br,s xi )k → (Br,s x)k as i → ∞. Using these infinitely many limits (Br,s x)0 , (Br,s x)1 , . . ., we define the sequence {(Br,s x)0 , (Br,s x)1 , . . .}. From (2.8) with j → ∞, we have ε (i, j > n0 (ε)) 2

|(Br,s xi )k − (Br,s x)k | pk /L ≤

(2.9)

(i)

for every fixed k ∈ N. Since xi = {xk } ∈ b0r,s (p) for each i ∈ N, there exists k0 (ε) ∈ N such that |(Br,s xi )k | pk /L
n0 (ε) we obtain by (2.9) and (2.10) that |(Br,s x)k | pk /L ≤ |(Br,s x)k − (Br,s xi )k | pk /L + |(Br,s xi )k | pk /L
k0 (ε). This shows that x ∈ br,s 0 (p). Since {x } was an arbitrary Cauchy sequence, the space b0 (p) is complete and this concludes the proof. Now lets show that, br,s (p) is the complete linear metric space paranormed by h defined by (2.4). It is easy to see that the space br,s (p) is linear with respect to the coordinate-wise addition and scalar multiplication. Therefore, we first show that it is a paranormed space with the paranorm h defined by (2.4).

It is clear that h(θ ) = 0 where θ = (0, 0, 0, ...) and h(x) = h(−x) for all x ∈ br,s (p). Let x, y ∈ br,s (p); then by Minkowski’s inequality we have pk !1/M ∞ k k k− j j 1 h(x + y) = ∑ s r (x j + y j ) ∑ k j (s + r) j=0 k=0    pk /M M 1/M 1 ∞ k k k− j j    =  ∑  s r (x j + y j )  k ∑ j (s + r) j=0 k=0 1 ∑ (s + r)k k=0 ∞

≤

pk !1/M k k− j j s r x j + ∑ j=0 j k

∞

∑ k=0

"

1 (s + r)k

pk !1/M k k− j j s r y j ∑ j=0 j k

= h(x) + h(y)

(2.11)

and for any α ∈ R we immediately see that |α| pk ≤ max{1, |α|M }.

(2.12)

{xn }

xn

h(xn − x) → 0

and (λn ) also be any sequence of scalars such that λn → λ . We

h(λn xn − λ x) ≤ h[(λn − λ )(xn − x)] + h[λ (xn − x)] + h[(λn − λ )x].

(2.13)

Let be any sequence of the points observe that

∈ br,s (p)

such that

It follows from λn → λ (n → ∞) that |λn − λ | < 1 for all sufficiently large n; hence lim h[(λn − λ )(xn − x)] ≤ lim h(xn − x) = 0.

n→∞

(2.14)

n→∞

Furthermore, we have lim h[λ (xn − x)] ≤ max{1, |λ |M } lim h(xn − x) = 0.

n→∞

(2.15)

n→∞

Also, we have lim h[(λn − λ )x)] ≤ lim |λn − λ |h(x) = 0.

n→∞

(2.16)

n→∞

Then, we obtain from (2.13), (2.14), (2.15) and (2.16) that h(λn xn − λ x) → 0, as n → ∞. This shows that h is a paranorm on br,s (p). (n) (n) (n) Now, we show that br,s (p) is complete. Let {xn } be any Cauchy sequence in the space br,s (p), where xn = {x0 , x1 , x2 , ...}. Then, for a n m given ε > 0, there exists a positive integer n0 (ε) such that h(x − x ) < ε for all n, m > n0 (ε). Since for each fixed k ∈ N that #1

" r,s n

r,s m

|(B x )k − (B x )k | ≤

M

r,s n

∑ |(B

r,s m

x )k − (B x )k |

pk

= h(xn − xm ) < ε

(2.17)

k

for every n, m > n0 (ε), {(Br,s x0 )k , (Br,s x1 )k , (Br,s x2 )k , ...} is a Cauchy sequence of real numbers for every fixed k ∈ N. Since R is complete, it converges, say (Br,s xn )k → (Br,s x)k as n → ∞. Using these infinitely many limits (Br,s x)0 , (Br,s x)1 , ..., we define the sequence {(Br,s x)0 , (Br,s x)1 , ...}. For each K ∈ N and n, m > n0 (ε) "

K

#1

M

r,s n

∑ |(B k=0

r,s m

x )k − (B x )k |

pk

≤ h(xn − xm ) < ε.

(2.18)

141


By letting m, K → ∞, we have for n > n0 (ε) that " #1

M

h(xn − x) =

∑ |(Br,s xn )k − (Br,s x)k | p

k

< ε.

(2.19)

k

This shows that xn − x ∈ br,s (p). Since br,s (p) is a linear space, we conclude that x ∈ br,s (p); it follows that xn → x, as n → ∞ in br,s (p), thus we have shown that br,s (p) is complete. r,s r,s Note that the absolute property does not hold on the spaces br,s there exists at least one sequence in the spaces 0 (p), bc (p) and b (p), since r,s r,s r,s r,s b0 (p), bc (p) and b (p) and such that g(x) 6= g(|x|), where |x| = (|xk |). This says that br,s (p), br,s c (p) and b (p) are the sequence spaces 0 of non-absolute type. r,s r,s r,s Theorem 2.2. The sequence spaces br,s 0 (p), bc (p), b∞ (p) and b (p) are linearly isomorphic to the spaces c0 (p), c(p), `∞ (p) and `(p), respectively, where 0 < pk ≤ H < ∞.

Proof. To avoid repetition of similar statements, we give the proof only for br,s 0 (p). We should showr,sthe existence of a linear bijection between the spaces br,s (p) and c (p). With the notation of (2.2), define the transformation T from b0 (p) to c0 (p) by x 7→ y = T x. The 0 0 linearity of T is trivial. Furthermore, it is obvious that x = θ whenever T x = θ , and hence T is injective. Let y ∈ c0 (p) and define the sequence 1 k k xk = k ∑ (−s)k− j (s + r) j y j ; (k ∈ N). r j=0 j Then, we have n

n ∑ k sn−k rk xk k=0 n n n−k k k 1 (−s)k− j (s + r) j y j s ∑ ∑ (s + r)n k=0 k j=0 j ! n n 1 n k n−k k− j j ∑ ∑ k j s (−s) (s + r) y j (s + r)n j=0 k= j ! n n 1 n n− j k− j n− j j ∑ ∑ j k − j (−1) s (s + r) y j (s + r)n j=0 k= j ! n n n n− j 1 n− j j k− j s (s + r) (−1) yj ∑ j ∑ (s + r)n j=0 k= j k − j n 1 n n− j s (s + r) j δnk y j ∑ (s + r)n j=0 j n n−n 1 s (s + r)n 1yn (s + r)n n 1 (s + r)n

(Br,s x)n = =

=

=

= = =

= yn . r,s Thus, we have that x ∈ br,s 0 (p) and consequently T is surjective. Hence, T is a linear bijection and this says that the spaces b0 (p) and c0 (p) are linearly isomorphic, as was desired.

r,s r,s 3. The basis for the spaces br,s 0 (p), bc (p) and b (p) Let (λ , g) be a paranormed space. Recall that a sequence (βk ) of the elements of λ is called a basis for λ if and only if, for each x ∈ λ , there exists a unique sequence (αk ) of scalars such that ! n

g x−

∑ αk βk

→ 0 as n → ∞.

k=0

The series ∑ αk βk which has the sum x is then called the expansion of x with respect to (βn ), and written as x = ∑ αk βk . Since it is known that the matrix domain λA of a sequence space λ has a basis if and only if λ has a basis whenever A = (ank ) is a triangle (cf. [11, Remark 2.4]), we have the following. Because of the isomorphism T is onto, defined in the proof of Theorem 2.2, the inverse image of the basis of r,s r,s those spaces c0 (p), c(p) and `(p) are the basis of the new spaces br,s 0 (p), bc (p) and b (p), respectively. Therefore, we have the following: Theorem 3.1. Let λk = (Br,s x)k for all k ∈ N and 0 < pk ≤ H < ∞. Define the sequence b(k) = {b(k) }k∈N of the elements of the space r,s r,s br,s 0 (p), bc (p) and b (p) by (k)

bn =

1 n n−k (s + r)k rn k (−s)

for every fixed k ∈ N. Then

0

, ,

n≥k 0≤k 1 such that p0 k −1 sup ∑ ∑ ank M < ∞. N∈F k n∈N

(4.11)

Lemma 4.2. [17] Let A = (ank ) be an infinite matrix. Then, the following statements hold (i) A ∈ (`(p) : `∞ ) if and only if (a) Let 0 < pk ≤ 1 for all k ∈ N. Then, sup |ank | pk < ∞.

(4.12)

n,k∈N

(b) Let 1 < pk ≤ H < ∞ for all k ∈ N. Then, there exists an integer M > 1 such that p0 k sup ∑ ank M −1 < ∞.

(4.13)

n∈N k

(ii) Let 0 < pk ≤ H < ∞ for all k ∈ N. Then, A = (ank ) ∈ (`(p) : c) if and only if (4.12) and (4.13) hold, and lim ank = βk , ∀k ∈ N.

(4.14)

n→∞

Theorem 4.3. Let K ∈ F and K ∗ = {k ∈ N : n ≥ k} ∩ K for K ∈ F . Define the sets T1r (p), T2r , T3 (p) and T4 (p) as follows: ( ) [ −1/pk T1 (p) = a = (ak ) ∈ w : sup ∑ ∑ cnk M 1 ( ) n T2 = a = (ak ) ∈ w : ∑ ∑ cnk exists for each n ∈ N , n k=0   p0 k  [  T3 (p) = a = (ak ) ∈ w : sup ∑ ∑ cnk M −1 < ∞, ,   N∈F k n∈N M>1 pk ) ( T4 (p) = a = (ak ) ∈ w : sup sup ∑ cnk < ∞ , N∈F k∈N n∈N where the matrix C = (cnk ) defined by 1 n n n−k (s + r)k a n rn ∑k=0 k (−s) cnk = 0

, ,

0 ≤ k ≤ n, k ≥ n.

r,s α α Then, [br,s 0 (p)] = T1 (p), [bc (p)] = T1 (p) ∩ T2 and T3 (p) 1 < pk ≤ H < ∞, ∀k ∈ N, [br,s (p)]α = T4 (p) 0 < pk ≤ 1, ∀k ∈ N.

Proof. We chose the sequence a = (ak ) ∈ w. We can easily derive that with the (2.2) that 1 n n an xn = n ∑ (−s)n−k (s + r)k an yk = (Cy)n , (n ∈ N) r k=0 k

(4.15)

(4.16)

(4.17)

for all k, n ∈ N, where C = (cnk ) defined by (4.15). It follows from (4.17) that ax = (an xn ) ∈ `1 whenever x ∈ br,s 0 (p) if and only if Cy ∈ `1 whenever y ∈ c0 (p). This means that a = (an ) ∈ [br,s (p)]α if and only if C ∈ (c0 (p) : `1 ). Then, we derive by (4.2) with qn = 1 for all n ∈ N 0 α r that [br,s 0 (p)] = T1 (p). α r Using the (4.3) with qn = 1 for all n ∈ N and (4.17), the proof of the [br,s c (p)] = T1 (p) ∩ T2 can also be obtained in a similar way. Also, using the (4.10),(4.11) and (4.17), the proof of the T3 (p) 1 < pk ≤ H < ∞, ∀k ∈ N, [br,s (p)]α = T4 (p) 0 < pk ≤ 1, ∀k ∈ N, can also be obtained in a similar way.

144


Theorem 4.4. The matrix D = (dnk ) is defined by (

j j−k r− j (r + s)k a j k (−s)

∑nj=k

dnk =

(0 ≤ k ≤ n) (k > n)

, ,

0

(4.18)

for all k, n ∈ N. Define the sets T5 (p), T6 , T7 (p), T8 , T9 (p), T10 and T11 (p) as follows: ( [

T5 (p) =

)

n

a = (ak ) ∈ w : sup

∑ |dnk |M

−1/pk

1

o n T6 = a = (ak ) ∈ w : lim |dnk | exists for each k ∈ N , n→∞

( [

T7 (p) =

n

a = (ak ) ∈ w : ∃(αk ) ⊂ R 3 sup

1

( T8 =

∑ |dnk − αk | M

) −1/pk

)

n

∑ |dnk | exists

a = (ak ) ∈ w : lim

n→∞

,

k=0

( [

T9 (p) =

M>1

) p0 −1 k a = (ak ) ∈ w : sup ∑ dnk M < ∞ , n∈N k

n o T10 = a = (ak ) ∈ w : lim dnk exists for each k ∈ N , n→∞

( T11 (p) =

)

a = (ak ) ∈ w : sup |dnk |

pk

0 and L(t)x.x ≥ l(t) |x|2 , ∀(t, x) ∈ R × RN ; t∈R

(L2 ) There exists a constant r0 > 0 such that lim meas({t ∈]s − r0 , s + r0 [/L(t) < bIN }) = 0, ∀b > 0,

|s|−→∞

where meas denotes the Lebesgue’s measure on R. The above conditions on L guarantee the compactness of Sobolev embedding. Besides, in all the above mentioned papers, the potential W is required to be subquadratic or to satisfy the Ambrosetti-Rabinowitz superquadratic condition (A R) at infinity. The aim of this paper is to study the existence of infinitely many solutions for (F H S ), when the function L is unnecessarily positive definite or coercive, and the potential W satisfies some superquadratic conditions at infinity, weaker than the (A R)−condition. More precisely, let W ∈ C1 (R × RN , R) be such that for all r > 0, ∇W is bounded in R × Br (0), we make the following hypotheses: (L) The smallest eigenvalue of L(t) is bounded from below; (W1 )

(W2 )

W (t, 0) = 0 and ∇W (t, x) = o(|x|), as |x| −→ 0 uni f ormly on t ∈ R;

lim

|x|→+∞

|W (t, x)| |x|2

= +∞, ∀t ∈ R,

and (W3 )

W (t, x) ≥ 0, ∀(t, x) ∈ R × RN with |x| ≥ R0 ;W (t, −x) = W (t, x), ∀(t, x) ∈ R × RN ;

(W4 ) There exist g ∈ L1 (R) and constants b0 , c0 > 0 and ν ∈]0, 2[ such that g(t), ∀t ∈ R, |x| ≤ R0 b (t, x) = 1 ∇W (t, x).x −W (t, x) ≥ W b0 |x|ν , ∀t ∈ R, |x| ≥ R0 ; 2 b (t, x), ∀(t, x) ∈ R × RN with |x| ≥ R0 ; |W (t, x)| ≤ c0 |x|2−ν W (W5 ) There exist constants µ > 2 and ρ0 > 0 such that µW (t, x) ≤ ∇W (t, x).x + ρ0 |x|2 , ∀(t, x) ∈ R × RN . Our main results read as follows. Theorem 1.1. Assume that (L), (L2 ) and (W1 ) − (W4 ) are satisfied. Then (F H S ) possesses infinitely many nontrivial solutions. Theorem 1.2. Assume that (L), (L2 ), (W1 ) − (W3 ) and (W5 ) are satisfied. Then (F H S ) possesses infinitely many nontrivial solutions. Remark 1.3. 1. In our results, L(t) is unnecessarily required to be either uniformly positive definite or coercive. For example L(t) = (t 2 sin2 t − 1)IN , where IN is the identity matrix, satisfies (L) and (L2 ), but it does satisfy neither Theorem 1.1 nor Theorem 1.2. 2. Let W (t, x) = a(t) |x|2 ln( 21 + |x|), where a is a continuous bounded function with positive lower bound. Then an easy computation shows that W satisfies the superquadratic conditions (W1 ) − (W4 ). However, W does not satisfy the (A R)−condition. The remainder of this paper is organized as follows. In Section 2, some preliminary results are presented. Section 3 is devoted to the proofs of our results.

2. Preliminaries In this Section, for the reader’s convenience, first we will recall some facts about the fractional calculus on the whole real axis. On the other hand, we will give some preliminaries lemmas for using in the sequel.

188


2.1. Liouville-Weyl fractional calculus The Liouville-Weyl fractional integrals of order 0 < α < 1 on the whole axis R are defined as (see [12], [13], [18]) 1 Γ(α)

α −∞ It u(t) =

Z t

(t − x)α−1 u(x)dx,

−∞

and 1 Γ(α)

α t I∞ u(t) =

Z ∞

(x − t)α−1 u(x)dx.

t

The Liouville-Weyl fractional derivatives of order 0 < α < 1 on the whole axis R are defined as the left-inverse operators of the corresponding Liouville-Weyl fractional integrals (see [12], [13], [18]) α ∞ Dt u(t) =

d (−∞ It1−α u)(t), dt

(2.1)

d 1−α (t I u)(t). dt ∞

(2.2)

and α t D∞ u(t) = −

The definitions of 2.1 and 3.2 may be written in an alternative form as follows α −∞ Dt u(t) =

1 Γ(1 − α)

Z ∞ u(t) − u(t − x)

xα+1

0

dx,

and 1 Γ(1 − α)

α t D∞ u(t) =

Z ∞ u(t) − u(t + x)

xα+1

0

dx.

We establish the Fourier transform properties of the fractional integral and fractional differential operators. Recall that the Fourier transform ub of u is defined by Z ∞

ub(s) =

e−ist u(t)dt.

−∞

Let u be defined on R. Then the Fourier transform of the Liouville-Weyl integrals and differential operators satisfies (see [12,13]) −α α \ ub(s), −∞ It u(s) = (is) −α α d ub(s), t I∞ u(s) = (−is) α α \ b(s), −∞ Dt u(s) = (is) u

Dα∞ u(s) = (−is)α ub(s). t[ Next, we present some properties for Liouville-Weyl fractional integrals and derivatives on the real axis, which were proved in [12]. Denote by L p (R, RN ) (1 ≤ p < ∞), the Banach spaces of functions on R with values in RN under the norms Z

kukL p = ( and

L∞ (R, RN )

1

|u(t)| p dt) p ,

R

the Banach space of essentially bounded functions from R into RN equipped with the norm

kuk∞ = esssup {|u(t)| /t ∈ R} .

2.2. Fractional derivative spaces In order to establish the variational structure which enables us to reduce the existence of solutions of (F H S ) to find critical points of the corresponding functional, it is necessary to construct the appropriate functional spaces. For α > 0, define the semi-norm α |u|I−∞ α = k−∞ Dt uk 2 L

and the norm 1

2 kukI−∞ α = (kukL2 + |u|I α ) 2 , −∞

and let k.k|I α

α I−∞ = C0∞ (R, RN )

−∞

where C0∞ (R, RN ) denotes the space of infinitely differentiable functions from R into RN with vanishing property at infinity. Now, we can define the fractional Sobolev space H α (R, RN ) in terms of the Fourier transform. Choose 0 < α < 1, define the semi-norm

|u| = |s|α ub 2 α

L

189


and the norm 1

kukα = (kukL2 + |u|2α ) 2 , and let k.kα

H α (R, RN ) = C0∞ (R, RN )

.

α if and only if Moreover, we note that a function u ∈ L2 (R, RN ) belongs to I−∞

|s|α ub ∈ L2 (R, RN ). Especially, we have

α b L2 . |u|I−∞ α = |s| u α and H α (R, RN ) are equivalent with equivalent semi-norms and norms. Analogous to I α , we introduce I α . Define the Therefore, I−∞ ∞ −∞ semi-norm

|u|I∞α = kt Dα∞ ukL2 and the norm 1

kukI∞α = (kukL2 + |u|2I∞α ) 2 , and let k.k|I∞α

I∞α = C0∞ (R, RN )

α and I α are equivalent with equivalent semi-norms and norms. Then I−∞ ∞ Let C(R, RN ) denotes the space of continuous functions from R into RN . Then we obtain the following Sobolev lemma.

Lemma 2.1. [[21], Theorem 2.1]. If α > 12 , then H α (R, RN ) ⊂ C(R, RN ), and there exists a constant C = Cα such that kuk∞ = sup |u(t)| ≤ Cα kukα , ∀u ∈ H α (R, RN ). t∈R

Remark 2.2. From Lemma 2.1, we know that if u ∈ H α (R, RN ) with Z R

1 2

< α < 1, then u ∈ L p (R, RN ) for all p ∈]2, ∞[, because

|u(t)| p dt ≤ kuk∞p−2 kuk2L2 .

In this section, we assume the L satisfies the following condition (L0 )

L(t)x.x ≥ |x|2 , ∀(t, x) ∈ R × RN

and we introduce the following fractional space Z X α = u ∈ H α (R, RN )/ L(t)u(t).u(t)dt < ∞ . R

Then X α is a Hilbert space with the inner product Z

< u, v >X α =

[−∞ Dtα u(t).−∞ Dtα v(t) + L(t)u(t).v(t)]dt

R

and the corresponding norm kuk2X α =< u, u >X α . It is easy to see that X α is continuously embedded in H α (R, RN ) and in L2 (R, RN ). In fact, for u ∈ X α , we have kuk2X α =

Z

[|−∞ Dtα u(t)|2 + L(t)u(t).u(t)]dt ≥

R

Z R

[|−∞ Dtα u(t)|2 |u(t)|2 ]dt = kuk2H α ≥ kuk2L2 .

For p ∈]2, ∞[, we have by Remark 2.2 kukLpp =

Z R

|u(t)| p dt ≤ kuk∞p−2 kuk2L2 ≤ Cαp−2 kukXp α .

Hence for all p ∈ [1, ∞], there exists a constant η p > 0 such that kukLpp ≤ η p kukXp α .

(2.3)

The main difficulty in dealing with the existence of infinitely many solutions for (F H S ) is the lack of compactness of the Sobolev embedding. To overcome this difficulty under the assumptions of Theorems 1.1 and 1.2, we employ the following compact embedding lemma.

190


Lemma 2.3. [16] Assume (L0 ) and (L2 ) are satisfied. Then X α is compactly embedded in L2 (R, RN ). Remark 2.4. From Remark 2.2 and Lemma 2.3, it is easy to verify that the embedding of X α in L p (R, RN ) is also compact for p ∈]2, ∞[. To study the critical points of the variational functional associated with (F H S ), we need to recall the Symmetric Mountain Pass Theorem [19]. Definition 2.5. Let X be a Banach space with the norm k.k, we say that Φ ∈ C1 (X, R) satisfies the a) (PS)c -condition, c ∈ R, if any sequence (un ) ⊂ X satisfying Φ(un ) −→ c and Φ0 (un ) −→ 0 as n −→ ∞ possesses a convergent subsequence, b) (C)c −condition, c ∈ R, if any sequence (un ) ⊂ X satisfying

Φ(un ) −→ c and Φ0 (un ) (1 + kun k) −→ 0 as n −→ ∞ possesses a convergent subsequence. Lemma 2.6. Let X be an infinite dimensional Banach space, X = Y ⊕ Z, where Y is finite dimensional space. Suppose that Φ ∈ C1 (X, R) satisfies the Palais-Smale condition and (a)

Φ(0) = 0, Φ(−u) = Φ(u), ∀u ∈ X;

(b)

T hereexistconstantsρ, α > 0suchthatΦ|∂ Bρ ∩Z ≥ α;

(c)

e > 0suchthatΦ(u) ≤ 0onEe \ BR , whereBR = {u ∈ X/ kuk < R} . Forany f initedimensionalsubspaceEe ⊂ X,thereisR = R(E)

Then Φ possesses an unbounded sequence of critical values. Remark 2.7. As shown in [2], a deformation lemma can be proved with (C)c −condition replacing the (PS)c −condition, and it turns out that Lemma 2.3 still holds true with the (C)c −condition instead of the (PS)c −condition.

3. Proof of theorems From (L), (W1 ) and (W2 ), we know that there exists a positive constant d0 such that L(t) + 2d0 IN ≥ IN for all t ∈ R. Let L(t) = L(t) + 2d0 IN and W (t, x) = W (t, x) + d0 |x|2 . Consider the following fractional Hamiltonian system (FHS)

α α t D∞ (−∞ Dt u)(t) + L(t)u(t) = ∇W (t, u(t)), α u ∈ H (R, RN ),

t ∈R

then (F H S ) is equivalent to (F H S ). Moreover, it is easy to check that the hypotheses (W1 ) − (W5 ) still hold for W provided that those hold for W , and L satisfies the conditions (L0 ), (L2 ). Hence, in what follows, we always assume without loss of generality that L satisfies (L0 ) instead of (L). Consider the variational functional Φ associated to (F H S ): Φ(u) =

1 2

Z

[|−∞ Dtα u(t)|2 + L(t)u(t).u(t)]dt −

R

Z

W (t, u)dt R

defined on the space X α introduced in Section 2. In the following, to simplify the notation, we will note the norm k.kX α of X α by k.k. Lemma 3.1. Under assumptions (L0 ), (L2 ) and (W1 ), the functional Z

ψ(u) =

W (t, u)dt R

is continuously differentiable on X α and ψ 0 (u)v =

Z R

∇W (t, u).vdt, ∀u, v ∈ X α .

(3.1)

191


Proof. By (W1 ), there exist constants a0 , R0 > 0 such that |∇W (t, x)| ≤ a0 |x| , ∀(t, x) ∈ R × RN with |x| ≤ R0 .

(3.2)

For any given u ∈ X α , we know that u ∈ H α (R, RN ) and hence there exists a constant T0 > 0 such that R0 , ∀ |t| ≥ T0 . 2

|u(t)| ≤

(3.3)

By (2.3), for any v ∈ X α with kvk ≤

R0 2η∞ ,

we have

R0 . 2

kvkL∞ ≤

(3.4)

Combining (3.2)-(3.4) and (W1 ), by the Mean Value Theorem and Holder’s ¨ inequality, for any T ≥ T0 and v ∈ X α with kvk ≤

R0 2η∞ ,

one has

Z Z Z 1 |t|≥T [W (t, u + v) −W (t, u) − ∇W (t, u).v]dt = |t|≥T 0 [∇W (t, u + sv) − ∇W (t, u)].vdsdt ≤ 2a0

Z

Z

|t|≥T

(|u| + |v|) |v| dt ≤ 2a0 (

Z

≤ 2a0 η2 [( Since u ∈ L2 (R), for any ε > 0, there exist 0 < α1 < Z

2a0 η2 [(

1

|t|≥Tε

|u|2 dt) 2 + η2 kvk] ≤

R0 2η∞

|t|≥T

1

(|u| + |v|)2 dt) 2 kvkL2

1

|t|≥T

|u|2 dt) 2 + η2 kvk] kvk .

and Tε ≥ T0 such that for all v ∈ X α with kvk ≤ α1

ε . 2

(3.5)

It is well known that the functional Z

ψε (u) =

W (t, u)dt

(3.6)

[−Tε ,Tε ]

is continuously differentiable on H 1 ([−Tε , Tε ], RN ). Thus, since X α is compactly embedded in H α (R), there exists a constant α2 > 0 such that for all kvk ≤ α2 Z ε ≤ kvk . [W (t, u + v) −W (t, u) − ∇W (t, u).v]dt (3.7) [−R ,R ] 2 ε ε Taking α = min {α1 , α2 }, then (3.5)-(3.7) imply Z [W (t, u + v) −W (t, u) − ∇W (t, u).v]dt ≤ ε kvk R

for all v ∈ X α with kvk ≤ α. Therefore, ψ is differentiable on X α and satisfies (3.1). It remains to prove that ψ 0 is continuous. Let un −→ u in X α . By Holder’s ¨ inequality, we have

0

ψ (un ) − ψ 0 (u) 0 = sup ψ 0 (un )v − ψ 0 (u)v E kvk=1

Z = sup [∇W (t, un ) − ∇W (t, u)].vdt R kvk=1

Z

≤ sup ( kvk=1

Z

≤ η2 (

R

1

|∇W (t, un ) − ∇W (t, u)|2 dt) 2 kvkL2 1

|∇W (t, un ) − ∇W (t, u)|2 dt) 2 .

(3.8)

R

Lemma 2.3 implies that un −→ u in L2 (R). Let M be a positive constant such that kun kL2 ≤ M for all integer n. By (W1 ), for any ε > 0, there exists a constant 0 < r < R0 such that for all t ∈ R and |x| ≤ r |∇W (t, x)| ≤

ε |x| . 2(M + kukL2 )

(3.9)

Due to (3.9) and the facts that u ∈ H α (R) and un −→ u in L∞ (R), there exists Rε > R0 and N1 ∈ N such that for all |t| ≥ Rε and n ≥ N1 |∇W (t, un (t))| ≤

ε |un (t)| 2(M + kukL2 )

and |∇W (t, u(t))| ≤

ε |u(t)| . 2(M + kukL2 )

192


Thus Z |t|≥Tε

|∇W (t, un ) − ∇W (t, u)|2 dt

1

Z

2

1

≤(

|t|≥Tε

Z

|∇W (t, un )|2 dt) 2 + (

1

|∇W (t, u)|2 dt) 2

|t|≥Tε

ε ≤ (kun kL2 + kukL2 ) 2(M + kukL2 ) ε ≤ . 2

(3.10)

Observing that un −→ u in L∞ (R), then by Lebesgue’s Dominated Convergence Theorem, we have Z [−Tε ,Tε ]

|∇W (t, un ) − ∇W (t, u)|2 dt

1 2

−→ 0 as n −→ ∞.

Hence, there is N2 ∈ N such that for all n ≥ N2 Z

ε , 2

1

( [−Tε ,Tε ]

|∇W (t, un ) − ∇W (t, u)|2 dt) 2 ≤

(3.11)

which together with (3.10) implies that for all n ≥ max {N1 , N2 } Z

(

1

|∇W (t, un ) − ∇W (t, u)|2 dt) 2 ≤ ε.

R

Combining this with (3.8) implies that

ψ 0 (un ) −→ ψ 0 (u)

as n −→ ∞ and then ψ ∈ C1 (X α , R). The proof of Lemma 3.1 is completed.

From Lemma 3.1, we deduce that Φ ∈ C1 (X α , R) and Φ0 (u)v =< u, v > −

Z

∇W (t, u).vdt, ∀u, v ∈ X α .

R

Moreover, if u ∈ X α is a critical point of Φ, we have α α t D∞ (−∞ Dt u)(t) = −L(t)u(t) + ∇W (t, u(t))

which implies that u is a solution of (F H S ). e > 0 such that Lemma 3.2. Under assumptions (L0 ), (L2 ), (W1 ) and (W2 ), for any finite dimensional subspace Ee ⊂ X α , there is R = R(E) e kuk ≥ R. Φ(u) ≤ 0, ∀u ∈ E,

(3.12)

Proof. We will prove the following e Φ(u) −→ −∞ as kuk −→ ∞, u ∈ E.

(3.13)

Arguing indirectly, assume that there exists a sequence (un ) ⊂ Ee with kun k −→ ∞ and a constant M > 0 such that Φ(un ) ≥ −M for all n ∈ N. Set vn = kuun k , then kvn k = 1. Passing to a subsequence if necessary, we may assume that vn * v in X α . Since Ee is finite dimensional, we n have vn −→ v in Ee and vn −→ v a.e. on R. It follows that kvk = 1. For 0 ≤ a < b, let Ωn (a, b) = {t ∈ R/a ≤ |un (t)| < b} A = {t ∈ R/v(t) 6= 0} . Since v 6= 0, then meas(A) > 0. For a.e. t ∈ R, we have limn−→∞ |un (t)| = ∞, hence t ∈ Ωn (R0 , ∞) for n large enough. Since vn (t) −→ v(t) a.e. t ∈ R, we have χΩn (R0 ,∞) (t) |vn (t)| −→ |v(t)| a.e. t ∈ A, where χΩ denotes the characteristic function of Ω. Hence, it follows from (W1 ), (W2 ) and Fatou’s Lemma that 0 = lim

n−→∞

−M

1 = lim [ − n−→∞ ku k n−→∞ 2 n

kun k

1 = lim [ − n−→∞ 2

≤

|un |2

Ωn (0,R0 )

Z

|vn |2 dt −

1 a0 2 + η2 − 2 2

R n−→∞

W (t, un ) |vn |2

Ωn (R0 ,∞)

Ωn (R0 ,∞)

lim inf

|un |2

|un |2

W (t, un ) |vn |2 |un |2

dt]

W (t, un ) |vn |2

Z

|un |2

W (t, un ) |vn |2

Z

Z

R

Ωn (R0 ,∞)

Z

R

1 a0 2 + η2 − lim inf n−→∞ 2 2

dt −

W (t, un ) |vn |2

Z

2

W (t, un ) |vn |2

Z

1 a0 ≤ lim sup[ + 2 2 n−→∞ ≤

Φ(un )

≤ lim

2

|un |2

dt]

dt]

dt

χ|Ωn (R0 ,∞) (t)dt = −∞

which is a contradiction. Hence (3.13) and then (3.12) is verified. The proof of Lemma 3.2 is completed.

(3.14)

193


Let (e j ) j∈N be an orthonormal basis of X α and define X j = Re j Yk = ⊕kj=1 X j , Zk = ⊕∞j=k+1 X j , k ∈ N. Lemma 3.3. Suppose (L0 ) and (L2 ) hold. Then for any p ∈ [2, ∞] l p (k) =

sup u∈Zk ,kuk=1

kukL p −→ 0 as k −→ ∞.

(3.15)

Proof. It is clear that 0 < l p (k + 1) ≤ l p (k), so that l p (k) −→ l p as k −→ ∞. For every k ≥ 1, there exists uk ∈ Zk such that kuk k = 1 and kuk kL p > 12 l p (k). For any v ∈ X α , let v = ∑∞ i=1 vi ei . By the Cauchy-Schwartz inequality, one has ∞ ∞ |< uk , v >| = < uk , ∑ vi ei > = < uk , ∑ vi ei > i=1 i=k+1

∞

∞

(3.16) ≤ kuk k ∑ vi ei ≤ ∑ |vi | kei k −→ 0 as k −→ ∞

i=k+1

i=k+1 which implies that uk * 0. Without loss of generality, Lemma 2.3 implies that uk −→ 0 in L2 (R). Thus we have proved that l p = 0. The proof of Lemma 3.3 is completed. By (3.15), we can choose an integer m ≥ 1 such that kukL2 ≤

1 kuk , ∀u ∈ Zm . 2a0

(3.17)

In the following, we will apply Lemma 3.1 with Y = Ym and Z = Zm . Lemma 3.4. Under assumptions (L0 ), (L2 ) and (W1 ), there exist constants ρ, α > 0 such that Φ|∂ Bρ ∩Z ≥ α. Proof. If kuk =

R0 η∞ ,

W (t, u(t)) ≤

then by (2.3), we have kukL∞ ≤ R0 . Hence, it follows from (W1 ) that a0 R |u(t)|2 , ∀u ∈ X α , kuk = 0 . 2 η∞

(3.18)

Combining (3.16) with (3.17) yields for all u ∈ Z with kuk = Φ(u) =

1 kuk2 − 2

Z

R0 η∞

=ρ

1 a 1 a kuk2 − 0 |u|2 dt ≥ kuk2 − 0 kuk2L2 2 2 R 2 2 1 R 1 0 ≥ kuk2 = ( )2 = α. 4 4 η∞ Z

W (t, u)dt ≥

R

(3.19)

The proof of Lemma 3.4 is completed.

Proof of Theorem 1.1 By assumptions (W1 ) and (W3 ), it is clear that Φ(0) = 0 and Φ(−u) = Φ(u), ∀u ∈ X α .

(3.20)

Thus the condition (a) of Lemma 3.1 is satisfied. Lemmas 3.2, 3.4 imply that conditions (b) and (c) of Lemma 3.1 are satisfied. It remains to prove that Φ satisfies the (C)c −condition. Lemma 3.5. Assume that (L0 ), (L2 ), (W1 ), (W2 ) and (W4 ) are satisfied. Then Φ verifies the (C)c −condition for all c ∈ R. Proof. Let (un ) be a (C)c sequence, that is

Φ(un ) −→ c and Φ0 (un ) (1 + kun k) −→ 0 as n −→ ∞. Firstly, we prove that (un ) is bounded in X α . Arguing by contradiction, suppose that kun k −→ ∞ as n −→ ∞. Let vn = Observe that for n large enough 1 c + 1 ≥ Φ(un ) − Φ0 (un )un = 2

(3.21) un kun k .

Then kvn k = 1.

Z

b (t, un )dt. W

(3.22)

R

It follows from (3.21) that 1 ≤ lim sup 2 n−→∞

Z R

|W (t, un )| kun k2

dt.

(3.23)

194


Passing to a subsequence if necessary, we may assume that vn * v in X α . Then by Lemma 2.3, without loss of generality, we have vn −→ v in L2 (R) and vn −→ v a.e. on R. If v = 0, then vn −→ 0 in L2 (R) and vn −→ 0 a.e. on R. Hence, it follows from (W1 ) that |W (t, un )|

Z

2

kun k

Ωn (0,R0 )

|W (t, un )|

Z

dt =

|vn |2 dt ≤

2

|un |

Ωn (0,R0 )

a0 2

Z Ωn (0,R0 )

|vn |2 dt ≤

a0 kvn k2L2 −→ 0 as n −→ ∞. 2

(3.24)

From (W4 ) and (3.22), one has |W (t, un )|

Z

2

kun k

Ωn (R0 ,∞)

|W (t, un )|

Z

dt =

|un |2

Ωn (R0 ,∞)

≤ ≤ ≤ ≤

c0 kvn k2−ν L∞ kun kν

|vn |2 dt

Z

b (t, un )dt W

Ωn (R0 ,∞)

c0 kvn k2−ν L∞ [1 + c − kun kν c0 η∞2−ν [1 + c − kun kν

Z

c0 η∞2−ν [1 + c + kun kν

Z

Z Ωn (0,R0 )

b (t, un )dt] W

g(t)dt] Ωn (0,R0 )

|g(t)| dt] −→ 0 as n −→ ∞.

(3.25)

R

Combining (3.22) with (3.23) yields Z

|W (t, un )| 2

kun k

R

Z

dt =

|W (t, un )| 2

kun k

Ωn (0,R0 )

|W (t, un )|

Z

dt + Ωn (R0 ,∞)

kun k2

dt −→ 0 as n −→ ∞

which contradicts (3.21). If v 6= 0. By a similar fashion as for (3.14), we can get a contradiction. Therefore, (un ) is bounded in X α . Next, we prove that (un ) possesses a convergent subsequence. Without loss of generality, we can assume by Remark 2.4 that un −→ u in L2 (R). Using Holder’s ¨ inequality, (W1 ) and the fact that ∇W is bounde’d in R × Br (0) for all r > 0, we can show that Z

[∇W (t, un ) − ∇W (t, u)].(un − u)dt −→ 0 as n −→ ∞.

(3.26)

R

Observe that kun − uk2 = (Φ0 (un ) − Φ0 (u))(un − u) +

Z

[∇W (t, un ) − ∇W (t, u)].(un − u)dt.

(3.27)

R

It is clear that (Φ0 (un ) − Φ0 (u))(un − u) −→ 0 as n −→ ∞.

(3.28)

Combining (3.24) − (3.26), we get kun − uk −→ 0 as n −→ ∞. The proof of Lemma 3.5 is completed. Consequently, Lemma 2.3 together with Remark 2.2 imply that Φ possesses an unbounded sequence of critical points. Therefore (F H S ) possesses infinitely many solutions. The proof of Theorem 1.1 is completed. Proof of Theorem 1.2 Lemma 3.6. Under assumptions (L0 ), (L2 ), (W1 ) − (W3 ) and (W5 ), Φ satisfies the (C)c −condition for all c ∈ R. Proof. Let c ∈ R and (un ) ⊂ X α satisfying (3.19). First, we prove that (un ) is bounded in X α . Arguing by contradiction, suppose that kun k −→ ∞ as n −→ ∞. Let vn = kuun k . Then kvn k = 1 and kvn k ≤ η p kvn kL p for 2 ≤ p ≤ ∞. By (W5 ), one has for n large enough n

1 µ −2 1 kun k2 + [ ∇W (t, un ).un −W (t, un )]dt c + 1 ≥ Φ(un ) − Φ0 (un )un = µ 2µ µ R µ −2 ρ0 2 kun k − kun k2L2 , ≥ 2µ µ Z

which implies µ −2 ≤ lim sup kvn k2L2 . 2ρ0 n−→∞

(3.29)

Taking a subsequence if necessary, we may assume that vn −→ v in L2 (R) and vn −→ v a.e. on R. Hence, it follows from (3.27) that v 6= 0. By a similar fashion as for (3.14), we can get a contradiction. Therefore (un ) is bounded in X α . The rest of the proof is the same as that in Lemma 3.5 and the proof of Lemma 3.6 is completed. We conclude as in the proof of Theorem 1.1 that Φ possesses an unbounded sequence of critical points and the proof of Theorem 1.2 is completed.


195

4. Conclusion Using the variational methods and critical point theory, we proved that the fractional Hamiltonian system (F H S ) possesses infinitely many nontrivial solutions, where L is neither uniformly positive definite nor coercive and W does not satisfy the classical superquadratic growth conditions like the well-known Ambrosetti-Rabinowitz superquadratic condition. Recent results in the literature are generalized and significantly improved.

Acknowledgement The author thanks the referee for valuable suggestions.

References [1] O. Agrawal, J. Tenreiro Machado, J. Sabatier, Fractional derivatives and their applications, Springer-Verlag, Berlin, 2004; [2] T. Bartolo, V. Benci, D. Fortunato, Abstract critical point theorems and applications to some nonlinear problems with strong resonance at infinity, Nonlinear Analysis, Vol. 7 , No. 9 (1983) 981-1012; [3] N. Nyamoradi, A. Alsaedi, B. Ahmad, Y. Zou, Multiplicity of homoclinic solutions for fractional Hamiltonian systems with subquadratic potential, Entropy 2017, 19,50,1-24; [4] N. Nyamoradi, A. Alsaedi, B. Ahmad, Y. Zou, Variational approach to homoclinic solutions for fractional Hamiltonian systems, J. Optim. Theory Appl. 2017; [5] Z. Bai, H. Lu, ¨ Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl. (2005), 311, 495-505; [6] Z. Bai, Y. Zhang, The existence of solutions for a fractional multi-point boundary value problem, Computers and Mathematics with Applications 2010, 69, 2364-2372; [7] P. Chen, X. He, X.H. Tang, Infinitely many solutions for a class of Hamiltonian systems via critical point theory, Math. Meth. Appl. Sci. 2016, 39, 1005-1019; [8] Y. Li, B. Dai, Existence and multiplicity of nontrivial solutions for Liouville-Weyl fractional nonlinear Schrodinger ¨ equation, RA SAM (2017); [9] W. Jiang, The existence of solutions for boundary value problems of fractional differential equations at resonance, Nonlinear Analysis (2011), 74, 1987-1994; [10] F. Jiao, Y. Zhou, Existence results for fractional boundary value problem via critical point theory, Intern. Journal of Bif. and Chaos, 22, No. 4 (2012), 1-17; [11] R. Hiffer, Applications of fractional calculus in physics, World Science, Singapore, 2000; [12] S. G. Samko, A.A Kilbas, O.I. Marichev, Fractional integrals and derivatives, Theory and applications, Gordon and Breach, Switzerland 1993; [13] A.A. Kilbas, H.M. Srivastawa, J.J. Trujillo, Theory and applications of fractional differential equations, North-Holland Mathematical Studies; Vol. 204, Singapore 2006; [14] S. Liang, J. Zhang, Positive solutions for boundary value problems of nonlinear fractional differential equations, Nonlinear Analysis, 2009, 71, 5545-5550; [15] J. Mawhin, M. Willem, Critical point theory and Hamiltonian systems, Applied Mathematical Sciences, Springer, Berlin, 1989; [16] A. Mèndez, C. Torres, Multiplicity of solutions for fractional Hamiltonian systems with Liouville-Weyl fractional derivative, arXiv: 1409.0765v1[mathph] 2 Sep. 2014; [17] K. Miller, B. Ross, An introduction to differential equations, Wiley and Sons, New York, 1993; [18] I. Pollubny, Fractional differential equations, Academic Press, 1999; [19] P.H. Rabinowitz, Minimax methods in critical point theory with applications to differential equations, CBMS Reg. Conf. Ser. in Math., Vol. 65, American Mathematical Society, Providence, RI, 1986; [20] K. Tang, Multiple homoclinic solutions for a class of fractional Hamiltonian systems, Progr. Fract. DIff. Appl. 2, , No. 4 (2016), 265-276; [21] C. Torres, Existence of solutions for fractional Hamiltonian systems, Electr. J. DIff. Eq., Vol. 2013 (2013), No. 259, 1-12; [22] C. Torres Ledesma, Existence of solutions for fractional Hamiltonian systems with nonlinear derivative dependence in R, J. Fractional Calculus and Applications; Vol. 7 (2) (2016) 74-87; [23] X. Wu, Z. Zhang, Solutions for perturbed fractional Hamiltonian systems without coercive conditions, Boundary Value Problems (2015) 2015: 149, 1-12; [24] S. Zhang, Existence of solutions for the fractional equations with nonlinear boundary conditions, Computers and Mathematics with Applications (2011), 61, 1202-1208; [25] S. Zhang, Existence of solutions for a boundary value problems of fractional differential equations at resonance, Nonlinear Analysis (2011): 74, 1987-1994; [26] Z. Zhang, R. Yuan, Existence of solutions to fractional Hamiltonian systems with combined nonlinearities, Electr. J. Diff. Eq., Vol. 2016 (2016) No. 40, 1-13; [27] Z. Zhang, R. Yuan, Solutions for subquadratic fractional Hamiltonian systems without coercive conditions, Math. Meth. Appl. Sci. (2014) 37, 2934-2945; [28] Z. Zhang, R. Yuan, Variational approach to solutions for a class of fractional Hamiltonian systems, Math. Meth. Appl. Sci. (2014) 37, 1873-1883;

Universal Journal of Mathematics and Applications, 1 (3) (2018) 196-201

Universal Journal of Mathematics and Applications Journal Homepage: www.dergipark.gov.tr/ujma ISSN 2619-9653 DOI: http://dx.doi.org/10.32323/ujma.416741

Geometry of bracket-generating distributions of step 2 on graded manifolds Esmaeil Azizpoura* and Dordi Mohammad Ataeia a Department

of Pure Mathematics, Faculty of Mathematical Sciences, University of Guilan, Rasht, Iran * Corresponding author E-mail: [email protected]

Article Info

Abstract

Keywords: Graded manifold, Distribution. 2010 AMS: 58A50, 58A30 Received: 18 April 2018 Accepted: 24 September 2018 Available online: 30 September 2018

A Z2 −graded analogue of bracket-generating distribution is given. Let D be a distribution of rank (p, q) on an (m, n)-dimensional graded manifold M , we attach to D a linear map F on D defined by the Lie bracket of graded vector fields of the sections of D. Then D is a bracket-generating distribution of step 2, if and only if F is of constant rank (m − p, n − q) on M .

1. Introduction A smooth distribution D ⊂ T M is said to be bracket-generating if all iterated brackets among its sections generate the whole tangent space to the manifold M, [1, 8]. D is a bracket-generating distribution of step 2 if D2 = T M, where D2 = D + [D, D]. Bejancu showed that a distribution of rank k < m = dimM is a bracket-generating distribution of step 2, if and only if, the curvature of D is of constant rank m − k on M, [1]. In this paper, a Z2 −graded analogue of bracket-generating distribution of step 2 is given. Some differences arise in the graded case due to the presence of odd generators. Given a distribution D of rank (p, q) on an (m, n)-dimensional graded manifold M , we attach to D a linear map F on D defined by the Lie bracket of graded vector fields of the sections of D. Then D is a bracket-generating distribution of step 2, if and only if F is of constant rank (m − p, n − q) on M . In particular, if rankD(z) = (m − 1, n), then for the linear map F = F0 + F1 associated to D, F0 6= 0 and if rankD(z) = (m, n − 1), then F1 6= 0 on M .

2. Preliminaries Let M be a topological space and let OM be a sheaf of super R-algebras with unity. A graded manifold of dimension (m, n) is a ringed space M = (M, OM ) which is locally isomorphic to Rm|n , (see [6]) . Let M and N be graded manifolds. Let φ : M 7→ N be a continuous map such that φ ∗ : ON −→ OM takes ON (V ) into OM (φ −1 (V )) for each open set V ⊂ N, then we say that Φ = (φ , φ ∗ ) : M −→ N is a morphism between M and N . Let A be a super R-algebra, ϕ ∈ EndR A is called a derivation of A, if for all a, b ∈ A, ϕ(ab) = ϕ(a).b + (−1)|ϕ||a| a.ϕ(b),

(2.1)

where for a homogeneous element x of some graded object, |x| ∈ {0, 1} denotes the parity of x (see [6]). A vector field on M is a derivation of the sheaf OM . Let U ⊂ M be an open subset, the OM (U)−super module of derivations of OM (U) is defined by T M (U) := Der(OM (U)). The OM −module T M is locally free of dimension (m, n) and is called the tangent sheaf of M . A vector field is a section of T M . If Ω1 (M ) := T ∗ M be the dual of the tangent sheaf of a graded manifold M , then it is the sheaf of super OM -modules and Ω1 (M ) := Hom(T M , OM ).

Email addresses: [email protected] (E. Azizpour), [email protected] (D. M. Ataei)

(2.2)

197


It is called the cotangent sheaf of a graded manifold M , and the sections of Ω1 (M ) are called super differential 1-forms [2, 6]. Let M = (M, OM ) be an (m, n)-dimensional graded manifold and D be a distribution of rank (p, q) (p < m, q < n) on M . Then for each point x ∈ M there is an open subset U over which any set of generators {Di , Dµ|1 ≤ i ≤ p, 1 ≤ µ ≤ q} of the module D(U) can be enlarged to a set ( ) |D |=1 1≤i≤p 1≤µ≤q |C |=0 Ca , Di , Dµ ,Cα p+1≤a≤m and q+1≤α≤n |Da |=0 and |C µ|=1 i α of free generators of DerOM , [3]. We attach to D a sequence of distributions defined by, D ⊂ D 2 ⊂ . . . ⊂ D r ⊂ . . . ⊂ DerOM , with D 2 = D + [D, D], . . . , D r+1 = D r + [D, D r ], where [D, D r ] = span{[X,Y ] : X ∈ D,Y ∈ D r }. As in the classical case, we say that D is a bracket-generating distribution, if there exists an r ≥ 2 such that D r = DerOM . In this case r is called the step of the distribution D. Suppose that X,Y ∈ D and consider the linear map on D as follows: F(X,Y ) = −(−1)|X||Y | [X,Y ] mod D.

(2.3)

With respect to the above local basis {Di ,Ca , Dµ ,Cα } of DerOM , if [Di , D j ] = Dkij Dk + DdijCd + D˜ νi j Dν + D˜ i jCγ , γ

[Di , Dξ ] = Dkiξ Dk + Ddiξ Cd + D˜ νiξ Dν + D˜ iξ Cγ , γ

γ [Dµ , D j ] = Dkµ j Dk + Ddµ jCd + D˜ νµ j Dν + D˜ µ jCγ ,

[Dµ , Dξ ] = Dkµξ Dk + Ddµξ Cd + D˜ νµξ Dν + D˜ µξ Cγ , γ

then, by using (2.3), we conclude that F(D j , Di ) = DdijCd + D˜ i jCγ mod D, γ

γ F(Dξ , Di ) = Ddiξ Cd + D˜ iξ Cγ mod D,

F(D j , Dµ ) = Ddµ jCd + D˜ µ jCγ mod D, γ

F(Dξ , Dµ ) = Ddµξ Cd

γ + D˜ µξ Cγ

(2.4)

mod D.

Each component Dabc of F is a superfunction on U. Let U be an open subset of M such that U ∩U 6= 0. / If we change the basis of DerOM (U ∩U) to {Di ,Ca , Dµ ,Cα } then we have D j = f ji Di + f j Dµ , µ

Dν = fνi Di + fν Dµ , µ

Cb = fbi Di + gabCa + fb Dµ + gαb Cα , µ

Cβ = fβi Di + gaβ Ca + fβ Dµ + gαβ Cα , µ

where µ

f ji fνi

fj µ fν

gab gaβ

and

gαb , gαβ

are nonsingular supermatrices of smooth functions on U ∩U. Both of these matrices are even. With respect to the basis {D j ,Cb , Dν ,Cβ } on b

β

β

U, if {Dkh , Dkh , . . . , Dξ ρ } are the local components of F, then we have 

b

Dkh  b  Dξ h   Db  kρ b

Dξ ρ

β  Dkh β  Dξ h  ga  ab β D   gβ

kρ β Dξ ρ



j

f  h gαb 0 = j gαβ  fρ 0

0 µ fρ 0 µ fh

µ

fh 0 µ fρ 0

 i fk 0 j fρ   0  0  0 j fξi f h

0 fξν − fkν 0

fkν 0 0 fξν

 0 Daij i  − fξ  Daν µ  fki   Daν j 0 Daiµ

 Dαi j α Dν µ   Dαν j  Dαiµ

(2.5)

198


Since

f ji fνi

i µ fj fj µ is invertible at x ∈ U ∩U, we see that 0 fν

 p+q+1 D  1 2. D(x) =   .. Dm+n 12

p+q+1

D1 3 .. . m+n D1 3

...

p+q+1 p+q

D1

.. . ...

Dm+n 1 p+q

0 µ is invertible and from (2.5) we conclude that if fν

p+q+1

D2 3 .. . m+n D2 3

...

p+q+1 p+q

D2

...

.. . ...

Dm+n 2 p+q

...

p+q+1

D p+q−1 .. . m+n D p+q−1

 p+q

  (x)  p+q

then rank D(x) = rankD(x) . Now we can define the rank of F, which is related to its coefficients matrix. Before doing this, in view of (2.4), we note that the submatrices a Di j (x) D˜ αi j (x)

" Da (x) Daµν (x) and ˜ iµ α ˜ Dµν (x) Dαiµ (x)

# Daµi (x) D˜ αµi (x)

are even and odd respectively. The rank of the first submatrix can be defined but for the second submatrix, since Daµi (x) and D˜ αiµ (x) are even, " # Da (x) Daiµ (x) we consider the matrix ˜ αµi to define its rank. Now set Dµi (x) D˜ αiµ (x) a D (x) r := rank ˜ iαj Di j (x)

Daµν (x) and D˜ αµν (x)

" Da (x) s := rank ˜ αµi D (x) µi

# Daiµ (x) , D˜ α (x) iµ

where i, j = 1, ..., p, a = p + 1, .., m and µ, ν = 1, ..., q, α = q + 1, ..., n. Thus we define rankF(x) = (r, s). If (qa¯ , ξµ¯ ) are local supercoordinates on a coordinate neighborhood U of x ∈ M, (a¯ = 1, ..., m, µ¯ = 1, ..., n), then D is locally given by the graded 1-forms µ¯ φb¯ = φb¯a¯ dqa¯ + φ˜b¯ dξµ¯ = 0, µ¯

φα¯ = φαa¯¯ dqa¯ + φ˜α¯ dξµ¯ = 0,

b¯ = 1, ..., p α¯ = 1, ..., q. µ¯

¯ µ¯ ≤ q are invertible. Let Since D is a distribution of rank (p, q), we may assume that the submatrices (φb¯a¯ ), 1 ≤ a, ¯ b¯ ≤ p, and (φ˜α¯ ), 1 ≤ α, ! µ¯ a ¯ ˜ φb¯ φb¯ ¯ µ¯ ≤ q and suppose the matrix ψ = (ψ•• ) denotes the inverse of the matrix , 1 ≤ a, ¯ b¯ ≤ p, 1 ≤ α, µ¯ φ a¯¯ φ˜ α

¯ µ¯ φ¯a¯ = ψab¯ φb¯ + φã¯ φµ¯ ,

α¯

¯ µ¯ φ¯α¯ = φαb¯ φb¯ + φ˜α¯ φµ¯ .

Therefore, the new notation ya = qa , xi = qi , i = 1, ..., p,

a = p + 1, ..., m,

ζα = ξα , ηµ = ξµ , µ = 1, ..., q,

α = q + 1, ..., n,

for the coordinates, may be performed to bring the local basis of Ω1 (M ) into the form {dxi , dηµ , dya + ria dxi + rµa dηµ , dζα + riα dxi + rµα dηµ }. It is easy to check that δ ∂ ∂ ∂ := − ria − riα , i = 1, . . . , p, δ xi ∂ xi ∂ ya ∂ ζα δ ∂ ∂ ∂ := + rµa − rµα , µ = 1, . . . , q, δ ηµ ∂ ηµ ∂ ya ∂ ζα

(2.6)

are (respectively even and odd) generators of D on U and {δ /δ xi , δ /δ ηµ , ∂ /∂ ya , ∂ /∂ ζα } is a local basis for Der(OM (U)), (see also [4, 5]). With respect to this basis, if we put F(

δ δ ∂ ∂ , ) = Fi aj + F˜i αj mod D, δ x j δ xi ∂ ya ∂ ζα

F(

δ δ ∂ ∂ , ) = Fνaj + F˜ναj mod D, δ x j δ ην ∂ ya ∂ ζα

F(

δ δ ∂ ∂ , ) = Fi aµ + F˜i αµ mod D, δ ηµ δ xi ∂ ya ∂ ζα

F(

δ δ ∂ ∂ , ) = Fνaµ + F˜ναµ mod D, δ ηµ δ ην ∂ ya ∂ ζα

(2.7)


199

then by using (2.3) and (2.6), we deduce that α a δ rα δ r j δ ra δ r j ∂ ∂ δ δ ∂ ∂ + F˜i αj =[ , ]=( i − ) +( i − ) mod D, ∂ ya ∂ ζα δ xi δ x j δ x j δ xi ∂ ya δxj δ xi ∂ ζα

Fi aj

Fνaj

α δ raj ∂ ∂ δ δ δ ra δ rα δ r j ∂ ∂ + F˜ναj =[ , ] = (− ν − ) +( ν − ) mod D, ∂ ya ∂ ζα δ ην δ x j δ x j δ ην ∂ ya δ x j δ ην ∂ ζα

Fi aµ

δ rµa ∂ δ rµα ∂ δ rα δ ra ∂ δ δ ∂ + F˜i αµ =[ , ]=( i + ) +( i − ) mod D, ∂ ya ∂ ζα δ xi δ ηµ δ ηµ δ xi ∂ ya δ ηµ δ xi ∂ ζα

Fνaµ

(2.8)

δ rµa ∂ δ rµα ∂ ∂ δ δ δ ra δ rα ∂ + F˜ναµ =[ , ]=( ν + ) + (− ν − ) mod D. ∂ ya ∂ ζα δ ην δ ηµ δ ηµ δ ην ∂ ya δ ηµ δ ην ∂ ζα

Now let us consider a distribution D of corank one on M . For each z ∈ M, there are two cases. Case1. Let rankD(z) = (m − 1, n). Then there exist a coordinate system (xi ,t, ηµ ), i = 1, ..., m − 1, µ = 1, ..., n, defined in a neighborhood U of z, such that D is locally given by dt + ri dxi + rµ dηµ = 0. Case2. Let rankD(z) = (m, n − 1). Then there exist a coordinate system (x j , ην , θ ), j = 1, ..., m, ν = 1, ..., n − 1 defined in a neighborhood U of z, such that D is locally given by dθ + r j dx j + rν dην = 0. Note that in the first case, (2.8) becomes Fi j

δ δ δ ri δ r j ∂ ∂ =[ mod D, , ]=( − ) ∂t δ xi δ x j δ x j δ xi ∂t

Fν j

δrj ∂ δ δ δ rν ∂ =[ ] = (− ) mod D, , − (−1)|t| ∂t δ ην δ x j δ ην δ x j ∂t

Fiµ

δ rµ ∂ δ δ ∂ δ ri =[ , ) mod D, ]=( + (−1)|t| ∂t δ xi δ ηµ δ ηµ δ xi ∂t

Fν µ

(2.9)

δ rµ ∂ ∂ δ δ δ rν =[ mod D, , ] = ((−1)|t| + (−1)|t| ) ∂t δ ην δ ηµ δ ηµ δ ην ∂t

where Fi j , Fν j , Fiµ and Fν µ are the local components of F with respect to the local basis {δ /δ xi , δ /δ xµ , ∂ /∂t}.

3. Bracket-generating distribution of step 2 In this section, we want to find the conditions under which a distribution D is bracket-generating of step 2. As mentioned in the previous section, we attach to D a linear map F on D defined by the Lie bracket of graded vector fields of the sections of D. We will have several types of possibilities for the rank of F. Using this, we find conditions to describe the problem. Theorem 3.1. Let D be a distribution of rank (p, q) (p < m, q < n) on an (m, n)-dimensional graded manifold M such that m− p ≤

q(q − 1) p(p − 1) q(q − 1) + ,n−q ≤ , 2 2 2

(3.1)

Then D is a bracket-generating distribution of step 2, if and only if, the linear map F associated to D is of constant rank (m − p, n − q) on M. Proof. Let x ∈ M. Suppose D is a bracket-generating distribution of step 2 and let {δ /δ xi , δ /δ ηµ , ∂ /∂ ya , ∂ /∂ ζα } be a basis of DerOM (U) in a coordinate neighborhood U of x. Then rank[D, D](x) = (m − p, n − q). This means that the number of linearly independent graded δ δ δ δ δ δ , ], [ , ], 1 ≤ i, j ≤ p, 1 ≤ µ, ν ≤ q}, (respectively {[ , ], 1 ≤ i ≤ p, 1 ≤ µ ≤ q}) is m − p vector fields of the set {[ δ xi δ x j δ ηµ δ ην δ xi δ ηµ (respectively n − q). Therefore the coefficient matrix, the matrix consisting of the coefficients of the Lie brackets of graded vector fields δ δ δ δ {[ , ], [ , ]} at the point x, denoted by δ xi δ x j δ ηµ δ ην a Di j (x) Daµν (x) a=1,...,m−p , , (modD), D˜ αi j (x) D˜ αµν (x) α=1,...,n−q having the rank m − p, is invertible. Similarly, the coefficient matrix " # Daiµ (x) a=1,...,m−p , , ( mod D), D˜ αiµ (x) α=1,...,n−q δ δ the matrix consisting of the coefficients of the Lie brackets of graded vector fields {[ , ]} at the point x, has rank n − q, (i.e. n − q = δ x δ ηµ i ! a D˜ αiµ (x) Dbc (x) rank , and this matrix is even). Hence associated with F is the graded vector field, represented by the matrix ˜ α , (modD), De f (x) Daiµ (x) relative to the above basis. It is clear that rankF(x) = (m − p, n − q).

200


Conversely, suppose that x ∈ M and rankF(x) = (m − p, n − q) on M . Let {δ /δ xi , δ /δ ηµ , ∂ /∂ ya , ∂ /∂ ζα } be a basis of DerOM (U) in a δ δ δ δ coordinate neighborhood U of x. Consider the coefficient matrix of the graded vector fields F( , ) and F( , ), which is even δ xi δ x j δ ηµ δ ην and denoted by a a (x) Fi j (x) Fµν (3.2) α α (x) . F˜i j (x) F˜µν Note that its rank is m − p, otherwise F would not be a map of the given rank. Thus there are two non-negative integers r and s such that α (x)) = s. Hence we may assume that the submatrices G = (F a0 (x)), 1 ≤ a0 , j 0 − 1 ≤ r, i0 < j 0 r + s = m − p and rank(Fiaj (x)) = r, rank(F˜µν i0 j0 0 α 0 0 0 0 ˜ and J = (F 0 0 (x)), 1 ≤ α , ν − 1 ≤ s, µ < ν , are both invertible. Therefore, the submatrix, µν

G I

" a0 F 0 0 (x) H = ˜iαj0 J Fi0 j0 (x)

# 0 Fµa0 ν 0 (x) , 0 F˜ α0 0 (x)

1≤a0 , j0 −1≤r 1≤α 0 ,ν 0 −1≤s , i0 < j 0 µ 0 0. Let Γ : X ⇒ 2Y be a set-valued mapping. The domain of Γ, denoted by dom Γ, is defined by dom Γ := {x ∈ X : Γ(x) 6= 0}. / The inverse and the graph of Γ, denoted by Γ−1 and gph Γ respectively, are defined by Γ−1 (y) := {x ∈ X : y ∈ Γ(x)} for each y ∈ Y and

gph Γ := {(x, y) ∈ X ×Y : y ∈ Γ(x)}.

204


Let B ⊆ X. The distance from a point x ∈ X to a set B is defined by dist(x, B) := inf kx − bk, b∈B

and the excess from the set A to the set B is defined by e(B, A) = sup{dist(x, A)}. x∈B

The notions of pseudo-Lipschitz and Lipchitz-like set-valued mappings are due to [23]. Aubin [9, 10] introduced these notions and studied extensively. Definition 2.1. Let G : Y ⇒ 2X be a set-valued mapping and let (y, ¯ x) ¯ ∈ gphG. Let rx¯ > 0, ry¯ > 0 and M > 0. Then the mapping G is said to be (a) Lipschitz-like on B(y, ¯ ry¯ ) relative to B(x, ¯ rx¯ ) with constant M if the following inequality holds: e(G(y1 ) ∩ B(x, ¯ rx¯ ), G(y2 )) ≤ Mky1 − y2 k for any y1 , y2 ∈ B(y, ¯ ry¯ ). (b) pseudo-Lipschitz around (y, ¯ x) ¯ if there exist constants a > 0, b > 0 and M 0 > 0 such that G is Lipschitz-like on B(y, ¯ b) relative to 0 B(x, ¯ a) with constant M . The following notion of (L, p)-Hölder continuity property is due to [21]. Definition 2.2. Let f : X → Y be a Fréchet differentiable function on some neighborhood U of x¯ and let ∇2 f be the second Fréchet derivative of f on U. Let p ∈ [0, 1] and L > 0. Then ∇2 f is called (L, p)-Höder continuous on U with constant L if the following condition holds: k∇2 f (x1 ) − ∇2 f (x2 )k ≤ Lkx1 − x2 k p , for any x1 , x2 ∈ U. The following lemma has taken from [23]. This lemma employs a vital role for proving the convergence analysis. ¯ x) ¯ ∈ gph G. Assume that G is Lipschitz-like on B(y, ¯ ry¯ ) relative to B(x, ¯ rx¯ ) Lemma 2.3. Let G : Y ⇒ 2X be a set-valued mapping and let (y, with constant M. Then dist(x, G(y)) ≤ Mdist(y, G−1 (x)) ry¯ . 3 We would like to finish this section with the following lemma that is known in [4]. r

holds for every x ∈ B(x, ¯ rx¯ ) and y ∈ B(y, ¯ 3y¯ ) satisfying dist(y, G−1 (x)) ≤

Lemma 2.4. Let Φ : X ⇒ 2X be a set-valued mapping. Let x¯ ∈ X, c > 0 and 0 < r < 1 be such that dist(x, ¯ Φ(x)) ¯ < c(1 − r);

(2.1)

and e(Φ(x1 ) ∩ B(x, ¯ c), Φ(x2 )) ≤ rkx1 − x2 k

for any x1 , x2 ∈ B(x, ¯ c).

(2.2)

Then Φ has a fixed point in B(x, ¯ c), that is, there exists x ∈ B(x, ¯ c) such that x ∈ Φ(x). Moreover, if Φ is single-valued, then the fixed point of Φ in B(x, ¯ c) is unique.

3. Convergence analysis of MSQM This section is devoted to prove the existence and convergence of the sequences generated by the modified superquadratic method defined by Algorithm 2. To this end, let x ∈ X and let us define the mapping Tx by 1 Tx (·) := f (x) + ∇ f (x)(· − x) + ∇2 f (x)(· − x)2 + F (·). 2 Then for the construction of D(x), we have that n o D(x) = d ∈ X : 0 ∈ Tx (x + d) n o = d ∈ X : x + d ∈ Tx −1 (0) .

(3.1)

Moreover, for any v ∈ X and y ∈ Y , the inclusions 1 v ∈ Tx−1 (y) and y ∈ f (x) + ∇ f (x)(v − x) + ∇2 f (x)(v − x)2 + F (v). 2

(3.2)

are equivalent. In particular, x¯ ∈ Tx¯−1 (y) ¯ for each (x, ¯ y) ¯ ∈ gph ( f + F ). The following result is due to [15]. This result establishes the equivalence relation between ( f + F )−1 and Tx¯−1 . Lemma 3.1. Let f : X → Y be a function and let (x, ¯ y) ¯ ∈ gph ( f + F ). Assume that f is twice differentiable in an open neighborhood Ω of x¯ and that its second-order derivative is continuous at x. ¯ Then the following are equivalent:

205


(i) The mapping ( f + F )−1 is pseudo-Lipschitz at (y, ¯ x); ¯ ¯ x). ¯ (ii) The mapping Tx¯−1 (·) is pseudo-Lipschitz at (y, Let rx¯ > 0, ry¯ > 0 and (x, ¯ y) ¯ ∈ gph ( f + F ). Then, the closed graph property of the set-valued mapping f + F implies that f + F is continuous at x¯ for y, ¯ that is, lim dist(y, ¯ f (x) + F (x)) = 0.

(3.3)

x→x¯

Assume that B(x, ¯ rx¯ ) ⊆ Ω ∩ dom F . Moreover, by Lemma 3.1 we assume that the mapping Tx¯−1 is Lipschitz-like on B(y, ¯ ry¯ ) relative to B(x, ¯ rx¯ ) with constant M, that is, e(Tx¯−1 (y1 ) ∩ B(x, ¯ rx¯ ), Tx¯−1 (y2 )) ≤ Mky1 − y2 k ∀ y1 , y2 ∈ B(y, ¯ ry¯ ).

(3.4)

Let p ∈ (0, 1], L > 0 and setting p+2 p n rx¯ (2 p+1 − 5MLrx¯ ) o L(3 p+2 + 2 p+2 )rx¯ , . α := min ry¯ − p+2 p+2 (p + 1)(p + 2)2 5M2

(3.5)

Then α > 0 if and only if L < min

n 2 p+2 (p + 1)(p + 2)r

y¯

p+2

(3 p+2 + 2 p+2 )rx¯

,

2 p+1 o p . 5Mrx¯

(3.6)

The following lemma plays a vital role for convergence analysis of the modified superquadratic method. The proof is a refinement of the one for [23, Lemma 3.1]. Lemma 3.2. Let Tx¯−1 be a Lipschitz-like mapping on B(y, ¯ ry¯ ) relative to B(x, ¯ rx¯ ) with constant M. Let p ∈ (0, 1] and x ∈ B(x, ¯ r2x¯ ). Assume rx¯ 2 that ∇ f and ∇ f are (L, p)-Höder continuous at x¯ on B(x, ¯ 2 ) with the same constant L defined by (3.6). Let α be defined in (3.5) so that 5M2 p (3.6) is satisfied. Then the mapping Tx−1 is Lipschitz-like on B(y, ¯ α) relative to B(x, ¯ r2x¯ ) with constant p+1 p i.e. 2 − 5MLrx¯ e(Tx−1 (t1 ) ∩ B(x, ¯

5M2 p rx¯ ), Tx−1 (t2 )) ≤ p+1 p kt1 − t2 k 2 2 − 5MLrx¯

for every t1 , t2 ∈ B(y, ¯ α).

Proof. Since α is defined in (3.5) so that (3.6) is satisfied, then it is clear that α > 0. Now let t1 , t2 ∈ B(y, ¯ α) and u0 ∈ Tx−1 (t1 ) ∩ B(x, ¯

rx¯ ). 2

(3.7)

To complete the proof, it is sufficient to show that there exists u00 ∈ Tx−1 (t2 ) such that ku0 − u00 k ≤

5M2 p p 2 p+1 − 5MLrx¯

kt1 − t2 k.

To finish this, we need to verify that there exists a sequence {xk } ⊆ B(x, ¯ rx¯ ) such that t2 ∈ f (x) + ∇ f (x)(xk−1 − x) + ∇ f (x)(x ¯ k − xk−1 ) 1 2 1 + ∇ f (x)(xk−1 − x)2 + ∇2 f (x) ¯ (xk − x) ¯ 2− 2 2

(3.8)

(xk−1 − x) ¯ 2 + F (xk )

and kxk − xk−1 k ≤

5MLr p k−2 5M x¯ kt1 − t2 k 2 2 p+1

(3.9)

hold for each k = 2, 3, 4, . . .. We proceed by induction on k. Write 1 ai := ti − f (x) − ∇ f (x)(u0 − x) − ∇2 f (x)(u0 − x)2 + f (x) ¯ 2 1 + ∇ f (x)(u ¯ 0 − x) ¯ + ∇2 f (x)(u ¯ 0 − x) ¯ 2 for each i = 1, 2. 2

(3.10)

Note by (3.7) that kx − u0 k ≤ kx − xk ¯ + kx¯ − u0 k ≤ rx¯ .

(3.11)

Furthermore, we have, for (3.10), that kai − yk ¯ ≤ kti − yk ¯ + k f (u0 ) − f (x) − ∇ f (x)(u0 − x) 1 − ∇2 f (x)(u0 − x)2 k + k f (u0 ) − f (x) ¯ 2 1 − ∇ f (x)(u ¯ 0 − x) ¯ − ∇2 f (x)(u ¯ 0 − x) ¯ 2 k. 2

(3.12)

206


If ∇ f is (L, p)-Hölder continuous at x¯ with constant L, then we have that k f (x) − f (x) ¯ − ∇ f (x)(x ¯ − x)k ¯ =k

Z 1

[∇ f (x¯ + t(x − x)) ¯ − ∇ f (x)](x ¯ − x)dtk ¯

0

≤

Z 1

k∇ f (x¯ + t(x − x)) ¯ − ∇ f (x)kkx ¯ − xkdt ¯

0

≤ Lkx − xk ¯ p+1

Z 1

t p dt

0

=

L kx − xk ¯ p+1 . p+1

(3.13)

Analogously, if ∇2 f is (L, p)-Hölder continuous at x¯ with constant L, then we have that 1 k f (x) − f (x) ¯ − ∇ f (x)(x ¯ − x) ¯ − ∇2 f (x)(x ¯ − x) ¯ 2k = k 2 ≤

Z 1

[∇ f (x¯ + t(x − x)) ¯ − ∇ f (x) ¯ − ∇2 f (x)( ¯ x¯ + t(x − x) ¯ − x)](x ¯ − x)dtk ¯

0 Z 1

k∇ f (x¯ + t(x − x)) ¯ − ∇ f (x) ¯ − ∇2 f (x)( ¯ x¯ + t(x − x) ¯ − x)kkx ¯ − xkdt ¯

0

Z x

=

k∇ f (u) − ∇ f (x) ¯ − ∇2 f (x)(u ¯ − x)kdu ¯

x¯

=

Z xZ 1n

o k∇2 f (x¯ + s(u − x)) ¯ − ∇2 f (x)kds ¯ ku − xkdu ¯

x¯

≤

0

Z xn

Lku − xk ¯ p+1

x¯

L ≤ p+1 =

Z 1

o s p ds du

0

Z x

ku − xk ¯

p+1

du

x¯

L kx − xk ¯ p+2 . (p + 1)(p + 2) p+2

Then from (3.12), using the relations in (3.7), (3.11) and the relation α ≤ ry¯ −

L(3 p+2 + 2 p+2 )rx¯ by (3.5), we have that (p + 1)(p + 2)2 p+2

L (ku0 − xk p+2 + ku0 − xk ¯ p+2 ) (p + 1)(p + 2) p+2 r L p+2 ≤α+ rx¯ + x¯p+2 (p + 1)(p + 2) 2

kai − yk ¯ ≤α+

p+2

=α+

L(1 + 2 p+2 )rx¯ ≤ ry¯ . (p + 1)(p + 2)2 p+2

That is ai ∈ B(y, ¯ ry¯ ) for each i = 1, 2. Define x1 := u0 . Then x1 ∈ Tx−1 (t1 ) by (3.7) and it follows from (3.2) that 1 t1 ∈ f (x) + ∇ f (x)(x1 − x) + ∇2 f (x)(x1 − x)2 + F (x1 ). 2 This can be written in another form as follows: 1 t1 + f (x) ¯ + ∇ f (x)(x ¯ 1 − x)) ¯ + ∇2 f (x)(x ¯ 1 − x) ¯ 2 ∈ f (x) 2 1 + ∇ f (x)(x1 − x) + ∇2 f (x)(x1 − x)2 + F (x1 ) + f (x) ¯ 2 1 + ∇ f (x)(x ¯ 1 − x) ¯ + ∇2 f (x)(x ¯ 1 − x) ¯ 2. 2 This, by the definition of a1 , implies that 1 a1 ∈ f (x) ¯ + ∇ f (x)(x ¯ 1 − x) ¯ + + ∇2 f (x)(x ¯ 1 − x) ¯ 2 + F (x1 ). 2 Hence x1 ∈ Tx¯−1 (a1 ) by (3.2). This together with (3.7) implies that x1 ∈ Tx¯−1 (a1 ) ∩ B(x, ¯ rx¯ ). Noting that a1 , a2 ∈ B(y, ¯ ry¯ ) and Tx¯−1 is Lipschitz-like by our assumption. Then it follows from (3.4) that there exists x2 ∈ Tx¯−1 (a2 ) such that kx2 − x1 k ≤ Mka1 − a2 k = Mkt1 − t2 k
0 be such that nr 2(p + 1)(p + 2)ry¯ 1285α o x¯ (a) δ ≤ min , , ,1 ; 4 L(2 p+3 + 2 · 4 p+2 + 1) 3 · 2 p p (b) 5(M + 1)L η2 p δ p+1 + 44−p (p + 1)(p + 2)rx¯ ≤ 2 p+1 (p + 1)(p + 2);

209


(c) kyk ¯
0 such that any sequence {xn } generated by ˆ Algorithm 2 with initial point in B(x, ¯ δ ) converges superquadratically to a solution x∗ of (1.1). Proof. According to the continuity of f + F at x¯ for y¯ and assumption (c), we can choose 0 < δˆ ≤ δ be such that dist(0, f (x0 ) + F(x0 )) ≤

Lδ p+2 2(p + 1)(p + 2)

for each x0 ∈ B(x, ¯ δˆ ).

(3.19)

Setting t :=

5ηML2 p δ p+1 p . (p + 1)(p + 2)(2 p+1 − 5MLrx¯ )

It follows, from assumption (b), that p p 5ML η2 p δ p+1 + (p + 1)(p + 2)rx¯ ≤ 5(M + 1)L η2 p δ p+1 + 44−p (p + 1)(p + 2)rx¯ ≤ 2 p+1 (p + 1)(p + 2).

The above inequality implies that t ≤ 1.

(3.20)

Let x0 ∈ B(x, ¯ δˆ ). We use mathematical induction on n to show that Algorithm 2 generates at least one sequence and every sequence {xn } obtained by Algorithm 2 satisfies the following assertions: kxn − xk ¯ ≤ 2δ

(3.21)

and 1 kdn k ≤ t 2

!(p+2)n (3.22)

δ

for each n = 0, 1, 2, .... Now, define 9 MLkx − xk ¯ p+2 + (p + 1)(p + 2)Mkyk ¯ for each x ∈ X. rx := (p + 1)(p + 2) Because η > 1, p ∈ (0, 1] and δ ≤

(3.23)

rx¯ by assumption (a), it follows, from assumption (b), that 4

257(M + 1)Lδ p+1 = (M + 1)L(δ p+1 + 44 δ p+1 ) ≤ (M + 1)L(δ p+1 + 44 δ p ) p

≤ (M + 1)L(2 p ηδ p+1 + 44−p (p + 1)(p + 2)rx¯ ) ≤

2 p+1 (p + 1)(p + 2) , 5

which gives MLδ p+1 ≤

2 p+1 (p + 1)(p + 2) 1285

and Lδ p+1 ≤

2 p+1 (p + 1)(p + 2) . 1285

(3.24)

Thus, by 3 · 2 p δ ≤ 1285α in assumption (a) and second inequality in (3.24), we obtain that kyk ¯
0 by assumption (a). Then we have from (3.5) that p

p

α > 0 ⇒ 2 p+1 − 5MLrx¯ > 0 ⇒ 5MLrx¯ < 2 p+1 .

(3.26)

210


Therefore, with the help of above relation we obtain that p

5 · 4 p MLδ p < 5MLrx¯ < 2 p+1 ⇒ LMδ p
0 such that any sequence {xn } generated by Algorithm 2 with initial point in B(x, ¯ δˆ ) converges superquadratically ∗ to a solution x of (1.1). Proof. By our assumption, Tx¯−1 is pseudo-Lipschitz around (0, x). ¯ Then there exist constants r0 , β and M such that Tx¯−1 is Lipschitz-like on B(0, r0 ) relative to B(x, ¯ β ) with constant M. Then, for each 0 < r ≤ β , one has that e(Tx¯−1 (y1 ) ∩ B(x, ¯ r), Tx¯−1 (y2 ) ≤ Mky1 − y2 k for any y1 , y2 ∈ B(0, r0 ) i.e. Tx¯−1 is Lipschitz-like on B(0, r0 ) relative to B(x, ¯ r) with constant M. Let L ∈ (0, 1) and rx¯ ∈ (0, β ) be such that p+2

and r0 −

rx¯ 2 p+1 p ≤ r, MLrx¯ ≤ 2 5

L(3 p+2 + 2 p+2 )rx¯ > 0. By the (L, p)-Hölder continuous property of ∇2 f , for each x, x0 ∈ B(x, ¯ r2x¯ ), we have that (p + 1)(p + 2)2 p+2

k∇2 f (x) − ∇2 f (x0 )k ≤ Lkx − x0 k p . Choose α so that p+2 p n L(3 p+2 + 2 p+2 )rx¯ rx¯ (2 p+1 − 5MLrx¯ ) o α := min r0 − , > 0, p+2 p+2 (p + 1)(p + 2)2 5M2

and min

2(p + 1)(p + 2)r0 1285α o , > 0. 4 L(2 p+3 + 2 · 4 p+2 + 1) 3 · 2 p

nr

x¯

,

Thus we can choose 0 < δ ≤ 1 such that δ ≤ min

2(p + 1)(p + 2)r0 1285α o , . p+3 p+2 4 L(2 +2·4 + 1) 3 · 2 p

nr

x¯

,

and p 5(M + 1)L η2 p δ p+1 + 44−p (p + 1)(p + 2)rx¯ ≤ 2 p+1 (p + 1)(p + 2).

Now one can easily sees that the assumptions (a)-(c) of Theorem 3.3 are hold. Therefore, to complete the proof of the corollary, we can apply Theorem 3.3 .


213

4. Conclusion The semilocal and local convergence results for the modified superquadratic method are established with η > 1 under the assumptions that Tx¯−1 is Lipschitz-like as well as ∇2 f is (L, p)-Hölder continuous. This result extends and improves the corresponding one [13]. This result seems new for the generalized equation problem (1.1). According to the main result of this study, we have the following conclusions: • If p = 0, then the Fréchet derivative of f satisfies the continuity condition with constant L and we obtain the quadratic convergence of the modified superquadratic method. In this case the result established in the present paper coincides with the result presented in [22, Theorem 3.1, Corollary 3.1]. • If p = 1, then the Fréchet derivative of f satisfies the Lipschitz condition and we obtain the cubic convergence of the modified superquadratic method. In this case the result established in the present paper coincides with the result presented in [22, Theorem 3.2, Corrolary 3.2].

Acknowledgments This work is fully supported by Ministry of Science and Technology, Bangladesh, Grant No. 39.009.002.01.00.057.2015–2016/EAS-326.

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