A study on Some Properties of Legendre Polynomials and Integral

Int. J. Open Problems Compt. Math., Vol. 5, No. 2, June 2012 c ISSN 1998-6262; Copyright ICSRS Publication, 2012 www.i-csrs.org

A study on Some Properties of Legendre Polynomials and Integral Transforms Ghazi S. Khammash and A. Shehata Department of Mathematics, Al-Aqsa University, Gaza Strip, Palestine. e-mail: [email protected] Department of Mathematics, Faculty of Science, Assiut University, Assiut 71516, Egypt. e-mail:[email protected]

Abstract The main aim of this paper is to use the integral representation method to derive the properties of Legendre polynomials. Their recurrence relations, differential equations, relationship with Hermite polynomials have been obtained. Keywords and phrases: Hermite polynomials; Legendre polynomials; Differential operators; Integral transforms. 2000 Mathematics Subject Classification: 35A22, 45P05, 47G10, 65R10.

1 Introduction Orthogonal polynomials is an emergent field whose development has reached important results from both the theoretical and practical points of view. This paper deals with the introduction and the study of Legendre polynomials taking advantage of the Hermite polynomials recently treated in [9]. The Legendre polynomials are introduced and their connections with differential equations are established. The Legendre polynomials of the second kind are defined by the series 1

Pn (x) =

[ 2 n] X k=0

(−1)k (2n − 2k)! xn−2k n 2 k!(n − k)!(n − 2k)!

(1)

2

Ghazi S. Khammash and A. Shehata

can be slightly generalized as 1

Pn (x, y) =

[ 2 n] X k=0

(−1)k (2n − 2k)! y k xn−2k 2n k!(n − k)!(n − 2k)!

in addition, that the Hermite polynomials are generated by ∞ X tn 2 Hn (x, y) = exp 2xt − yt n! n=0

(2)

(3)

where 1

Hn (x, y) = n!

[ 2 n] X k=0

(−1)k y k (2x)n−2k k!(n − 2k)!

by means of the integral transform Z ∞ 2 Pn (x, y) = √ exp(−t2 )tn Hn (xt, y)dt. n! π 0

(4)

(5)

It is clear that P−1 (x, y) = 0, P0 (x, y) = 1, P1 (x, y) = x and Pn (−x, y) = (−1)n Pn (x, y). It has already been shown that most of the properties of the Pn (x, y) polynomials, linked to the ordinary case by n x Pn (x, y) = y 2 Pn ( √ ) and Pn (x, 1) = Pn (x) (6) y can be directly inferred from those of the Hermite polynomials and from the integral representation given in (5). By recalling that the Hn (x, y) satisfies the differential equation of the second order [9] 2 ∂ ∂ (7) + 2n Hn (x, y) = 0 y 2 − 2x ∂x ∂x and the recurrences properties can be derived either from (3) or (4). It is indeed easy to prove that ∂ Hn (x, y) = 2nHn−1 (x, y), ∂x ∂ Hn (x, y) = −n(n − 1)Hn−2 (x, y), ∂y ∂ Hn (x, y), Hn+1 (x, y) = 2x − y ∂x Hn+1 (x, y) = 2xHn (x, y) − 2nyHn−1 (x, y)

(8)

A study on Some Properties of Legendre Polynomials and Integral Transforms

3

this once combined yield ∂2 ∂ Hn (x, y) = 0 H (x, y) + 4 n ∂x2 ∂y and, in general, we get ∂r 2r n! Hn−r (x, y), 0 ≤ r ≤ n H (x, y) = n ∂xr (n − r)! ∂r 1 (−1)r n! Hn−2r (x, y), 0 ≤ r ≤ [ n]. H (x, y) = n r ∂y (n − 2r)! 2

(9)

(10)

According (9), it is clear that the Hn (x, y) is the natural solution of the heat partial differential equation. We find the recurrences properties from (8) and by using the integral transform (5) ∂ ∂ − n Pn (x, y), y Pn−1 (x, y) = x ∂x ∂x ∂ ∂ Pn+1 (x, y) − y Pn−1 (x, y), (2n + 1)Pn (x, y) = ∂x ∂x ∂ ∂ (11) x Pn (x, y) = Pn+1 (x, y) − (n + 1)Pn (x, y), ∂x ∂x ∂ 2 x −y Pn (x, y) = nxPn (x, y) − nyPn−1 (x, y), ∂x (n + 1)Pn+1 (x, y) = (2n + 1)xPn (x, y) − nyPn−1 (x, y). In the following section, we will derive further consequences from the above relations and show how the method of the integral transform may be useful for the study of the properties of generalized forms of Legendre polynomials. 2 Operational identities for Legendre polynomials The recurrences given in (11) can be exploited to define rising and lowerˆ −1 ing operators for Legendre polynomials, indeed introducing the operator D x denoting a kind of inverse derivative, we can write ∂ 1 ˆ −1 − n Pn (x, y), Pn−1 (x, y) = Dx x y ∂x (12) 1 ∂ −1 ˆx x Pn+1 (x, y) = (2n + 1)x − nD − n Pn (x, y). n+1 ∂x These last relations allow the introduction of the rising and lowering operators 1 ∂ −1 ˆ− = D ˆ M x −n ˆ , y x ∂x (13) ∂ 1 −1 ˆ ˆ M+ = (2n + 1)x − nDx x −n ˆ . n+1 ∂x

4


where n ˆ is a number operator in the sense n ˆ Ps (x, y) = sPs (x, y), which act on Pn (x, y) according to the rules ˆ − Pn (x, y) = Pn−1 (x, y), M ˆ + Pn (x, y) = Pn+1 (x, y). M

(14)

Equation (14) can be exploited to derive the differential equation satisfied by Pn (x, y), namely ˆ +M ˆ − Pn (x, y) = Pn (x, y) M

(15)

namely ∂ 1 ∂ −1 −1 ˆ ˆ D x −n ˆ (2n + 1)x − nDx x −n ˆ Pn (x, y) = Pn (x, y). y(n + 1) x ∂x ∂x (16) which after some algebra and the use of the obvious identity ˆ −1 = ˆ1 ∂x D x yields the differential equation of second order in the form 2 ∂ 2 ∂ + n(n + 1) Pn (x, y) = 0. (y − x ) 2 − 2x ∂x ∂x

(17)

(18)

In the forthcoming section, we will touch on the problem of considering the whole family of Legendre polynomials, using the integral representation method and study their properties in terms of those of Hermite polynomials. 3 Connections between Legendre with Hermite polynomials We will try to understand more deeply the role played by the above integral transform connecting Legendre polynomials and Hermite polynomials and whether it can be understood in terms of ordinary integral transforms. By noting that both Pn (x, y) and Hn (x, y) are reduced to ordinary forms for √ n 1 Hn (x, ) = t− 2 Hn (x t). t We find that 2 Pn (x) = √ n! π

Z 0

∞

exp(−t2 )tn Hn (xt)dt.

(19)

A study on Some Properties of Legendre Polynomials and Integral Transforms

5

After a suitable change of variable, the equation (3.1) yields (Putting u = xt) Z ∞ 2 u 2 n+1 √ Pn (x) = exp − ( ) u Hn (u)du (20) x n!xn+1 π 0 which ensures that the identity (19) can be viewed as a kind of Mellin transform in [6], associated with the function ξ 2 (21) F (ξ) = exp − ( ) Hn (ξ). x Let us now consider the problem from an operational point of view. By recalling that the so called dilatation operator acts on a generic function f (x) as ∂ exp λx f (x) = f x exp (λ) . (22) ∂x Since xt = xeln t it follows that ∂

∂

f (xt) = f (xeln t ) = eln t (x ∂x ) f (x) = tx ∂x f (x), ∂

∂

Hn (xt) = Hn (xeln t ) = eln t (x ∂x ) Hn (x) = tx ∂x Hn (x).

(23)

We obtain from (19) 2 Pn (x) = √ n! π

Z

∞

n exp − t t 2 e 2

∂ ln t (x ∂x )

Hn (x)dt

(24)

0

or Z ∞ ∂ 2 Pn (x) = √ exp − t2 tn tx ∂x Hn (x)dt n! π 0 Z ∞ ∂ 2 exp − t2 t(n+x ∂x ) Hn (x)dt. = √ n! π 0 Using the well known definition of the Gamma function Z ∞ 1 exp(−t2 )t2µ dt. Γ(µ + ) = 2 2 0

(25)

(26)

Using (26), we can recast (27) in the operational form ∂ Γ( 21 (n + 1 + x ∂x )) √ Pn (x) = Hn (x). n! π

(27)

In this paper we have shown that the use of integral transforms and operational methods is a fairly useful tool to deal with old and new families of polynomials.

6


Further applications will be discussed in a forthcoming paper.

4 Open problem One can use the same class of integral representation for some other polynomials of two variables. Hence, new results and further applications can be obtained.

References [1] M. Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Washington, D.C., 1970. [2] G.E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 2000. [3]G. Dattoli, Integral transforms and Chebyshev-like polynomials, Appl. Math. Comput., 148 (2004), 225-234. [4] G. Dattoli, A. Torre and M. Carpanese, Operational rules and arbitrary order Hermite generating functions, J. Math. Analysis Appl., 227 (1998), 98-111. [5] G. Dattoli, A. Torre and S. Lorenzutta, Operational identities and properties of ordinary and generalized special functions, J. Math. Analysis Appl. 236 (1999), 399-414. [6] V.A. Ditkin, A. Prudnikov, Integral Transforms and Operational Calculus, Pergamon-Press, Oxford, 1965. [7] I.S. Gradshteyn and I.M. Ryzhik, Table of Integral, Series and Products, Academic Press, New York, 1980. [8] A. P. Prudnikov, Yu.A. Brychkov, and O.I. Marichev, Integrals and Series, More Special Functions, Vol. 3, Gordon and Breach, New York, (1990). [9] E.D. Rainville, Special Functions, The Macmillan Company, New York, 1962. [10] H.M. Srivastava and H.L. Manocha, A Treatise on Generating Functions, Ellis Horwood, New York, 1984.


Some conditions under which prime near-rings are commutative rings Abdelkarim Boua Université Moulay Isma¨ıl, Faculté des Sciences et Techniques Département de Mathématiques, Groupe d’Algèbre et Applications B. P. 509 Boutalamine, Errachidia; Maroc

[email protected]

Abstract Let N be a prime left near-ring, and I be a nonzero semigroup ideal of N . We prove that if N admits a derivation d satisfying any one of the following properties: (i) d([x, y]) = [x, y], (ii) d([x, y]) = [d(x), y], (iii) [d(x), y] = [x, y], (iv) d(x◦y) = x◦y, (v) d(x)◦y = x◦y and (vi) d(x)◦y = x◦y for all x, y ∈ I, then N is a commutative ring. Moreover, example proving the necessity of the primeness condition is given. Keywords: prime near-rings, derivations, commutativity. 2010 MSC: 16W25, 16Y30.

1

Introduction

A left near-ring is a set N with two operations + and · such that (N, +) is a group and (N, ·) is a semigroup satisfying the left distributive law x · (y + z) = x · y + x · z for all x, y, z ∈ N. A left near-ring N is called Zero symmetric if 0 · x = 0 for all x ∈ N (recall that left distributivity yields x · 0 = 0). Throughout this paper, unless otherwise specified, we will use the word nearring to mean zero symmetric left near-ring and denote xy instead of x · y. A nonempty subset I of N will be called a semigroup ideal if IN ⊂ I and N I ⊂ I. An additive mapping d : N −→ N is said to be a derivation if d(xy) = xd(y) + d(x)y for all x, y ∈ N, or equivalently, as noted in [7], that d(xy) = d(x)y + xd(y) for all x, y ∈ N. According to [4], a near-ring N is said

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Abdelkarim Boua

to be prime if xN y = 0 for x, y ∈ N implies x = 0 or y = 0. For any x, y ∈ N, the symbol [x, y] stands for the commutator xy − yx, while the symbol x ◦ y will denote xy + yx. The symbol Z(N ) will represent the multiplicative center of N, that is, Z(N ) = {x ∈ N | xy = yx for all y ∈ N }. Recall that N is called 2-torsion free if 2x = 0 implies x = 0 for all x ∈ N. There is an increasing body of evidence that prime near-rings with derivations have ring like behavior, indeed, there are several results (see for example [2], [3], [4], [5], [6]) asserting that the existence of a suitably-constrained derivation on a prime near-ring forces the near-ring to be a ring. In this paper we continue the line of investigation regarding the study of prime near-rings with derivations. More precisely, we shall prove that a prime near-ring which admits a nonzero derivation satisfying certain differential identities must be a commutative ring.

2

Main Results

In order to prove our main theorems, we shall need the following lemmas. Lemma 2.1 ([3], Theorem 2.1) Let N be a prime near-ring, and I be a nonzero semigroup ideal of N . If N admits a nonzero derivation d for which d(I) ⊂ Z(N ), then N is a commutative ring. Lemma 2.2 ([4], Lemma 1) Let d be an arbitrary derivation on the nearring N . Then N satisfies the following partial distributive law: (xd(y) + d(x)y)z = xd(y)z + d(x)yz for all x, y, z ∈ N. Lemma 2.3 ([3], Lemma 1.4(i)) Let N be a prime near-ring, and I a nonzero semigroup ideal of N . Let d be a nonzero derivation on N . If x, y ∈ N and xIy = {0}, then x = 0 or y = 0. Lemma 2.4 Let N be a 2-torsion free prime near-ring, and I be a nonzero semigroup ideal of N . If N admits a nonzero derivation d for which d(−I) ⊂ Z(N ), then N is a commutative ring. Proof. By hypothesis given we have d(−x) ∈ Z(N ) for all x ∈ I.

(1)

Replacing x by tx in (1), where t ∈ Z(N ), we get d(t)(−x) + td(−x) ∈ Z(N ) for all x ∈ I, t ∈ Z(N ). This implies that d(t)(−x) ∈ Z(N ) for all x ∈ I, t ∈ Z(N ).

(2)

9

Some conditions under which prime near-rings are commutative rings

Hence d(t)N [−x, u] = {0} for all x ∈ I, t ∈ Z(N ), u ∈ N.

(3)

Since N is prime and using (3), we obtain d(Z(N )) = 0 or − x ∈ Z(N ) for all x ∈ I.

(4)

(i) If d(Z(N )) = {0}, then 0 = d2 (u(−x)) = d2 (u)(−x)+2d(u)d(−x)+ud2 (−x) for all x, ∈ I, u ∈ N which, because of d2 (−x) = 0, forces d2 (u)(−x) + 2d(u)d(−x) = 0 for all x ∈ I, u ∈ N.

(5)

Applying d again, we get d3 (u)(−x) + 3d2 (u)d(−x) = 0 for all x ∈ I, u ∈ N.

(6)

Taking d(u) instead of u in (6) gives d3 (u(−x)) + 2d2 (u)d(−x) = 0 Combining the last equation with (6) we obtain d2 (u)d(−x) = 0 so that d2 (u)N d(−x) = 0 for all x ∈ I, u ∈ N.

(7)

By primeness of N , equation (7) gives either d2 (u) = 0 or d(−x) = 0. Hence d2 (u) = 0 or d(x) = 0 for all x ∈ I, u ∈ N.

(8)

If d(x) = 0 for all x ∈ I, then it is easy to find that d = 0 which contradicts our hypothesis. If d2 (u) = 0 for all u ∈ N , then ([4], Lemma 3) assures that d = 0 which is impossible. (ii) Assume that −x ∈ Z(N ) for all x ∈ I. Replacing x by vx, where v ∈ N, in the above equation we have v(−x) ∈ Z(N ) and thus (−x)N [y, v] = 0 for all x ∈ I, y, v ∈ N. (9) Since N is prime and I 6= {0} so that y ∈ Z(N ) for all y ∈ N . Therefore, d(y) ∈ Z(N ), then d(N ) ⊂ Z(N ). Using Lemma 2.1, we conclude that N is a commutative ring. This completes the proof of our theorem. Theorem 2.5 Let N be a prime near-ring, and I be a nonzero semigroup ideal of N . If N admits a nonzero derivation d such that d([x, y]) = [x, y] for all x, y ∈ I, then N is a commutative ring.

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Abdelkarim Boua

Proof. Assume that d([x, y]) = [x, y] for all x, y ∈ I.

(10)

Replacing y by xy in (10), because of [x, xy] = x[x, y], we get x[x, y] = d(x[x, y]) forall x, y ∈ I. Since d(x[x, y]) = xd([x, y]) + d(x)[x, y], then according to (10) we obtain x[x, y] = x[x, y] + d(x)[x, y] and therefore d(x)[x, y] = 0. Hence d(x)xy = d(x)yx for all x, y ∈ I.

(11)

Substituting yz for y in (11), where z ∈ N , because of d(x)xyz = d(x)yxz, we obtain d(x)y[x, z] = 0 for all x, y ∈ I, z ∈ N which leads to d(x)I[x, z] = 0 for all x ∈ I, z ∈ N.

(12)

Using Lemma 1, equation (12) reduces to d(x) = 0 or [x, z] = 0 for all x ∈ I, z ∈ N.

(13)

From (13) it follows that for each fixed x ∈ I we have d(x) = 0 or x ∈ Z(N ).

(14)

But x ∈ Z(N ) also implies that d(x) ∈ Z(N ) and equation (14) forces d(x) ∈ Z(N ) for all x ∈ I.

(15)

In the light of (15), d(I) ⊂ Z(N ) and using Lemma 2.1 we conclude that N is a commutative ring. This completes the proof of our theorem. Theorem 2.6 Let N be a prime near-ring, and I be a nonzero semigroup ideal of N . If N admits a nonzero derivation d such that either d([x, y]) = [d(x), y] for all x, y ∈ I or [x, y] = [d(x), y] for all x, y ∈ I, then N is a commutative ring. Proof. Assume that d([x, y]) = [d(x), y] for all x, y ∈ I.

(16)

Replacing y by xy in (16) we get [d(x), xy] = d(x[x, y]) for all x, y ∈ I.

(17)


11

Since d(x[x, y]) = d(x)[x, y] + xd([x, y]), in light of (16), equation (17) reduces to d(x)yx = xd(x)y for all x, y ∈ I. (18) Substituting yz for y in (18), where z ∈ N and using xd(x)yz = d(x)yxz, we obtain d(x)y[x, z] = 0. Hence d(x)I[x, z] = 0 for all x ∈ I, z ∈ N.

(19)

Since equation (19) is the same as equation (12), arguing as in the proof of Theorem 2.5, we conclude that N is a commutative ring. Now suppose that [d(x), y] = [x, y] for all x, y ∈ I.

(20)

Replacing x by yx in (20), because of [yx, y] = y[x, y], we get [d(yx), y] = y[x, y] = y([d(x), y]) for all x, y ∈ I. Since [d(yx), y] = d(yx)y − yd(yx), then according to Lemma 2.2 we obtain yd(x)y + d(y)xy − yd(y)x − y 2 d(x) = yd(x)y − y 2 d(x), so that d(y)xy = yd(y)x for all x, y ∈ I.

(21)

Since equation (21) is the same as equation (12), arguing as in the first case we find that N is a commutative ring. The conclusion of Theorems 2.4 and 2.5 no remains valid if we replace the product [x, y] by x ◦ y. In fact, we obtain the following result: Theorem 2.7 Let N be a 2-torsion free prime near-ring, and I be a nonzero semigroup ideal of N . Then there is no derivation d such that d(x ◦ y) = x ◦ y for all x, y ∈ I. Proof. If there exists a nonzero d such that d(x ◦ y) = xy + yx for all x, y ∈ I.

(22)

Then, replacing y by xy in (22), we get d(x ◦ (xy)) = x2 y + xyx for all x, y ∈ I.

(23)

Since x ◦ (xy) = x(x ◦ y), then (22) yields d(x ◦ (xy)) = x(x ◦ y) + d(x)(x ◦ y). Hence equation (23) reduces to x(x ◦ y) + d(x)(x ◦ y) = x2 y + xyx for all x, y ∈ I.

(24)

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Abdelkarim Boua

As x2 y + xyx = x(x ◦ y), then (24) assures that d(x)(x ◦ y) = 0 for all x, y ∈ I, which leads to d(x)xy = −d(x)yx for all x, y ∈ I.

(25)

Substituting yz for y in (25), where z ∈ N, we find that −d(x)yzx = d(x)xyz = (−d(x)yx)z = d(x)y(−x)z for all x, y ∈ I, z ∈ N. (26) Since −d(x)yzx = d(x)yz(−x), then (26) becomes d(x)yz(−x) = d(x)y(−x)z for all x, y ∈ I, z ∈ N.

(27)

d(x)I[−x, z] = 0 for all x, y ∈ I, z ∈ N.

(28)

Then By Lemma 2.3, equation (28) assures that for each x ∈ I, either −x ∈ Z(N ) or d(x) = 0. Accordingly, d(−x) = 0 or − x ∈ Z(N ) for all x ∈ I.

(29)

In light of d(Z(N )) ⊂ Z(N ), equation (29) yields then d(−x) ∈ Z(N ) for all x ∈ I.

(30)

Hence d(−I) ⊂ Z(N ), by Lemma 2.4, equation (30) assures that N is a commutative ring. Use the fact that N is a 2-torsion free, the hypothesis d(x◦y) = x◦y for all x, y ∈ I, becomes d(xy) = xy for all x, y ∈ I. So that d(x)y + xd(y) = xy for all x, y ∈ I. Replacing x by xz, then the last equation can be written as xzd(y) = 0 for all x, y, z ∈ I, thus xId(y) = 0 for all x, y ∈ I. Since I 6= {0}, then Lemma 2.3 shows that d = 0 on I, then it is easy to see that d = o on N ; a contradiction. If there exists a zero derivation d such that d(x ◦ y) = x ◦ y for all x, y ∈ I, then we can easily see that x = 0 for all x ∈ I; a contradiction. Theorem 2.8 Let N be a 2-torsion free prime near-ring, and I be a nonzero semigroup ideal of N . Then there is no derivation d such that d(x) ◦ y = x ◦ y for all x, y ∈ I.


13

Proof. Suppose there exists a nonzero derivation d such that d(x ◦ y) = d(x)y + yd(x) for all x, y ∈ I.

(31)

Then, replacing y by xy in (31), we get d(x ◦ (xy)) = d(x)xy + xyd(x) for all x, y ∈ I.

(32)

Since x ◦ (xy) = x(x ◦ y), then d(x ◦ (xy)) = d(x)(x ◦ y) + xd(x ◦ y). As d(x ◦ y) = d(x) ◦ y by hypothesis, then d(x ◦ (xy)) = d(x)(x ◦ y) + x(d(x) ◦ y). Hence equation (32) reduces to d(x)yx = −xd(x)y for all x, y ∈ I.

(33)

Substituting yz for y in (33), where z ∈ N we find that d(x)yzx = −xd(x)yz = xd(x)y(−z) = d(x)y(−x)(−z) = −d(x)y(−x)z. (34) Since −d(x)yzx = d(x)yz(−x), then (34) becomes d(x)yz(−x) = d(x)y(−x)z for all x, y ∈ I, z ∈ N.

(35)

Since equation (35) is the same as equation (27), arguing as in the proof of Theorem 2.7 we conclude that N is a commutative ring. Use the fact that N is a 2-torsion free and the hypothesis of Theorem we arrive at d(x)y = xy for all x, y ∈ I. Replacing x par xz, we get d(x)zy + xd(z)y = xzy for all x, y, z ∈ I, then d(x)zy = 0 for all x, y, z ∈ I. Since I 6= 0, then Lemma 2.3 assure that d = 0; a contradiction. If there exists a zero derivation d such that d(x) ◦ y = x ◦ y for all x, y ∈ I, then we can easily see that x = 0 for all x ∈ I; a contradiction. Theorem 2.9 Let N be a 2-torsion free prime near-ring, and I be a nonzero semigroup ideal of N . If N admits a derivation d such that d(x ◦ y) = d(x) ◦ y for all x, y ∈ I, then d = 0. Proof.Suppose that d(x) ◦ y = x ◦ y for all x, y ∈ I.

(36)

Replacing x by yx in (36) we obtain d(yx) ◦ y = y(x ◦ y) = y(d(x) ◦ y) for all x, y ∈ I.

(37)

Since d(yx) ◦ y = d(yx)y + yd(yx), then according to Lemma 2.2 we obtain yd(x)y + d(y)xy + yd(y)x + y 2 d(x) = yd(x)y + y 2 d(x),

14

Abdelkarim Boua

this implies that d(y)xy = −yd(y)x for all x, y ∈ I.

(38)

As equation (38) is the same as equation (33), arguing as above we conclude that N is a commutative ring. Using the hypothesis of Theorem we have xd(y) = 0 for all x, y ∈ I. Hence xId(y) = 0 for all x, y ∈ I. Since I 6= {0}, then Lemma 2.3 shows that d = 0 on I, then it is easy to see d = 0 on N . The following example demonstrate that the primeness hypothesis in Theorems 2.5, 2.6, 2.7, 2.8 and 2.9 cannot be omitted. x y Example. Let S be a commutative ring. Set N = | x, y ∈ S 0 0 0 x and I = | x ∈ S . It is clear that N is not prime. Moreover, 0 0 x y 0 x d = , d is a nonzero derivation of N and I is a semigroup 0 0 0 0 ideal of N such that: (i) d([A, B]) = [A, B], (ii) d([A, B]) = [d(A), B], (iii) [d(A), B] = [A, B], (iv) d(A ◦ B) = A ◦ B, (v) d(A) ◦ B = d(A) ◦ B and (vi) d(A) ◦ B = A ◦ B for all A, B ∈ I, but N is a noncommutative ring.

3

Open Problem

In this section we introduce the following open question: (i) Does the results remain valid for I a left semigroup ideal? (ii) Does the results remain valid for I a right semigroup ideal? ACKNOWLEDGEMENTS. The author would like to thank the referee for providing very helpful comments and suggestions.

References [1] M. Ashraf and A. Shakir, On (σ, τ )-derivations of prime near-rings-II , Sarajevo J. Math. 4 (16) (2008), 23-30.


15

[2] K. I. Beidar, Y. Fong and X. K. Wang, Posner and Herstein theorems for derivations of 3-prime near-rings, Comm. Algebra 24 (5) (1996), 15811589. [3] H. E. Bell, On derivations in near-rings II, Kluwer Academic Publishers Netherlands (1997), 191-197. [4] H. E. Bell and G. Mason, On derivations in near-rings, North-Holand Mathematics Studies 137 (1987), 31-35. [5] H. E. Bell and G. Mason, On derivations in near-rings and rings, Math. J. Okayama Univ. 34 (1992), 135-144. [6] A. Boua and L. Oukhtite, Derivations on prime near-rings, Int. J. Open Problems Compt. Math. 4 (2011), 162-167. [7] X. K. Wang, Derivations in prime near-rings, Proc. Amer. Math. Soc. 121 (1994), 361-366.


On Fixed-point theorems in Intuitionistic Fuzzy metric Space T.K. Samanta , Sumit Mohinta and Iqbal H. Jebril Department of Mathematics, Uluberia College, India-711315. e-mail: mumpu− [email protected] e-mail: [email protected] Department of Mathematics, Taibah University,Saudi Arabia. e-mail: [email protected] Abstract

In this paper, first we have established two sets of sufficient conditions for a mapping to have unique fixed point in a intuitionistic fuzzy metric space and then we have redefined the contraction mapping in a intuitionistic fuzzy metric space and thereafter we have proved the Banach Fixed Point theorem. Keywords: Fuzzy Sets, Intuitionistic Fuzzy Set, Intuitionistic Fuzzy Metric, Contraction Mapping, Contractive Sequence, Cauchy sequence. 2010 Mathematics Subject Classification: 03F55, 46S40.

1

Introduction

Fuzzy set theory was first introduce by Zadeh[13] in 1965 to describe the situation in which data are imprecise or vague or uncertain. Thereafter the concept of fuzzy set was generalized as intuitionistic fuzzy set by K. Atanassov[7, 8] in 1984. It has a wide range of application in the field of population dynamics , chaos control , computer programming , medicine , etc. The concept of fuzzy metric was first introduced by Kramosil and Michalek[10] but using the idea of intuitionistic fuzzy set, Park[6] introduced the notion of intuitionistic fuzzy metric spaces with the help of continuous t-norms and continuous t-conorms, which is a generalization of fuzzy metric space due to George and Veeramani[2, 3].

On Fixed-point theorems in Intuitionistic Fuzzy metric Space

17

In intuitionistic fuzzy metric space, Mohamad[1] proved Banach Fixed Point theorem. But in his paper[1], the definition of contractive mapping is not natural and to prove the iterative sequence is a Cauchy sequence, he first proved that every iterative sequence is a contractive sequence and then assumed that every contractive sequences are Cauchy. Several authors[9, 11] have tried to established the Banach fixed point theorem in fuzzy metric space by introducing the contraction mapping with the help of the membership function for the fuzzy metric. But these contraction mapping do not bear the intension of the contraction mapping with respect to a fuzzy metric, when a fuzzy metric gives the degree of nearness of two points with respect to a parameter t. Considering this meaning of fuzzy metric, in our paper, we have redefined the notion of contraction mapping in a intuitionistic fuzzy metric space and then directly, it has been proved that the every iterative sequence is a Cauchy sequence, that is, we don’t need to assume that every contractive sequences are Cauchy sequences. Thereafter we have established the Banach Fixed Point theorem. In this paper, also, we have established another two sets of sufficient conditions for a mapping to have unique fixed point.

2

Preliminaries

We quote some definitions and statements of a few theorems which will be needed in the sequel. Definition 2.1 [4]. A binary operation ∗ : [ 0 , 1 ] × [ 0 , 1 ] −→ [ 0 , 1 ] is continuous t - norm if ∗ satisfies the following conditions : ( i ) ∗ is commutative and associative , ( ii ) ∗ is continuous , ( iii ) a ∗ 1 = a ∀ a ε [0 , 1] , ( iv ) a ∗ b ≤ c ∗ d whenever a ≤ c , b ≤ d and a , b , c , d ε [ 0 , 1 ]. Definition 2.2 [4]. A binary operation : [ 0 , 1 ] × [ 0 , 1 ] −→ [ 0 , 1 ] is continuous t-conorm if satisfies the following conditions : ( i ) is commutative and associative , ( ii ) is continuous , ( iii ) a 0 = a ∀ a ∈ [0 , 1] , ( iv ) a b ≤ c d whenever a ≤ c , b ≤ d and a , b , c , d ∈ [ 0 , 1 ]. Result 2.3 [5]. ( a ) For any r 1 , r 2 ∈ ( 0 , 1 ) with r 1 > r 2 , there exist r 3 , r 4 ∈ ( 0 , 1 ) such that r 1 ∗ r 3 > r 2 and r 1 > r 4 r 2 . ( b ) For any r 5 ∈ ( 0 , 1 ) , there exist r 6 , r 7 ∈ ( 0 , 1 ) such that r 6 ∗ r 6 ≥ r 5 and r 7 r 7 ≤ r 5 .

18

T.K. Samanta , Sumit Mohinta , and Iqbal H. Jebril

Definition 2.4 [6] Let ∗ be a continuous t-norm , be a continuous tconorm and X be any non-empty set. An intuitionistic fuzzy metric or in short IFM on X is an object of the form A = { ((x , y , t) , µ(x , y , t) , ν (x , y , t) ) : (x , y , t) ∈ X 2 × (0 , ∞)} where µ , ν are fuzzy sets on X 2 × (0 , ∞) , µ denotes the degree of nearness and ν denotes the degree of non−nearness of x and y relative to t satisfying the following conditions : for all x, y, z ∈ X, s, t > 0 (i) µ(x ,y ,t) + ν (x ,y ,t) ≤ 1 ∀ ( x , y , t ) ∈ X 2 × (0, ∞); ( ii ) µ ( x , y , t ) > 0 ; ( iii ) µ ( x , y , t ) = 1 if and only if x = y ( iv ) µ ( x , y , t ) = µ ( y , x , t ); (v) µ(x , y , s) ∗ µ(y , z , t) ≤ µ(x , z , s + t ); ( vi ) µ ( x , y , · ) : (0 , ∞ ) → (0 , 1] is continuous; ( vii ) ν ( x , y , t ) > 0 ; ( viii ) ν ( x , y , t ) = 0 if and only if x = y ; ( ix ) ν ( x , y , t ) = ν ( y , x , t ); (x) ν (x , y , s) µ(y , z , t) ≥ ν (x , z , s + t ); ( xi ) ν ( x , y , · ) : (0 , ∞ ) → (0 , 1] is continuous. If A is a IFM on X , the pair ( X , A ) will be called a intuitionistic fuzzy metric space or in short IFMS. We further assume that ( X , A ) is a IFMS with the property ( xii ) For all a ∈ (0 , 1), a ∗ a = a and a a = a Remark 2.5 [6] In intuitionistic fuzzy metric space X, µ ( x , y , · ) non-decreasing and ν ( x , y , · ) is non-increasing for all x , y ∈ X.

is

Definition 2.6 [2] A sequence { x n } n in a intuitionistic fuzzy metric space is said to be a Cauchy sequence if and only if for each r ∈ ( 0 , 1 ) and t > 0 there exists n 0 ∈ N such that µ ( xn , xm , t ) > 1 − r and ν ( xn , xm , t ) < r for all n , m ≥ n 0 . A sequence { x n } in a intuitionistic fuzzy metric space is said to converge to x ∈ X if and only if for each r ∈ ( 0 , 1 ) and t > 0 there exists n 0 ∈ N such that µ ( xn , x , t ) > 1 − r and ν ( xn , x , t ) < r for all n , m ≥ n 0 . Note 2.7 [12] A sequence { x n } n in an intuitionistic fuzzy metric space is a Cauchy sequence if and only if lim µ ( xn , xn + p , t ) = 1 and

n→∞

lim ν ( xn , xn + p , t ) = 0

n→∞

A sequence { x n } n in an intuitionistic fuzzy metric space converges to x ∈ X if and only if lim µ ( xn , x , t ) = 1 and

n→∞

lim ν ( xn , x , t ) = 0

n→∞

19


Definition 2.8 [1] An intuitionistic fuzzy metric space (X, A) has the property ♠ if lim µ ( x , y , t ) = 1 and lim ν ( x , y , t ) = 0 for all x, y ∈ X t → ∞

t → ∞

Definition 2.9 [1] Let ( X , A ) be a intuitionistic fuzzy metric space. We will say the mapping f : X → X is t-uniformly continuous if for each ε, with 0 < ε < 1, there exists 0 < r < 1, such that µ ( x , y , t ) ≥ 1 − r and ν ( x , y , t ) ≤ r implies µ ( f (x) , f (y) , t ) ≥ 1 − ε and ν ( f (x) , f (y) , t ) ≤ ε for each x, y ∈ X and t > 0. Definition 2.10 [1] Let ( X , A ) be a intuitionistic fuzzy metric space. A mapping f : X → X is intuitionistic fuzzy contractive if there exists k ∈ (0 , 1) such that 1 −1 ≤ k µ ( f (x) , f (y) , t ) 1 1 and −1 ≥ ν ( f (x) , f (y) , t ) k

1 −1 µ(x , y , t)

!

1 −1 ν (x , y , t)

!

for each x , y ∈ X and t > 0. ( k is called the contractive constant of f .) Proposition 2.11 [1] Let ( X , A ) be a intuitionistic fuzzy metric space. If f : X → X is fuzzy contractive then f is t-uniformly continuous. Definition 2.12 [1] Let ( X , A ) be an intuitionistic fuzzy metric space.We will say that the sequence { xn } in X is intuitionistic fuzzy contractive if there k ∈ (0 , 1) such that

and

µ ( xn+1

1 −1 ≤ k , xn+2 , t )

ν ( xn+1

1 1 −1 ≥ , xn+2 , t ) k

1 −1 µ ( xn , xn+1 , t )

!

1 −1 ν ( xn , xn+1 , t )

!

for all t > 0, n ∈ N. Theorem 2.13 [1] Let ( X , A ) be a complete intuitionistic fuzzy metric space with the property ♠ in which intuitionistic fuzzy contractive sequences are Cauchy. Let T : X → X be a intuitionistic fuzzy contractive mapping such that k is the contractive constant. Then T has a unique fixed point.

20

3


Fixed-point theorems

Definition 3.1 Let ( X , A ) be an IFMS , x ∈ X, r ∈ (0 , 1) , t > 0, B( x , r , t ) = { y ∈ X / µ( x , y , t ) > 1 − r , ν( x , y , t ) < r }. Then B( x , r , t ) is called an open ball centered at x of radius r w.r.t. t. Definition 3.2 Let ( X , A ) be an IFMS and P ⊆ X . P is said to be a closed set in ( X , A ) if and only if any sequence { xn } in P converges to x ∈ P i.e, iff. nlim µ ( xn , x , t ) = 1 and nlim ν ( xn , x , t ) = 0 ⇒ →∞ →∞ x ∈ P. Definition 3.3 Let ( X , A ) be an IFMS, x ∈ X, r ∈ (0 , 1) , t > 0, S( x , r , t ) = { y ∈ X / µ( x , y , t ) > 1 − r , ν( x , y , t ) < r }. Hence S( x , r , t ) is said to be a closed ball centered at x of radius r w.r.t. t iff. any sequence { xn } in S( x , r , t ) converges to y then y ∈ S( x , r , t ). Theorem 3.4 (Contraction on a closed ball) :- Suppose ( X , A ) is a complete IFMS with the property ♠ in which IF contractive sequences are Cauchy. Let T : X → X be IF contractive mapping on S( x 0 , r , t ) with contractive constant k. Moreover, assume that 1 − 1 < (1 − k) µ ( x0 , T (x0 ) , t ) and

1 1 −1 > ν ( x0 , T (x0 ) , t ) 1−k

1 −1 1−r

1 −1 r

Then T has unique fixed point in S( x 0 , r , t ) . Proof. Let x1 = T (x0 ) , x2 = T (x1 ) = T 2 (x0 ) , · · · , xn = T (xn−1 ) i.e, xn = T n (x0 ) for all n ∈ N. Now 1 − 1 < (1 − k) µ ( x0 , x 1 , t ) i.e,

1 < (1 − k) µ ( x0 , x 1 , t )

1 −1 1−r

r 1−r

= (1 − k)

+ 1

r − rk + 1 − r 1 − rk = 1−r 1−r 1−r ⇒ µ ( x0 , x 1 , t ) > > 1−r 1 − rk ⇒ µ ( x0 , x 1 , t ) > 1 − r ··· (i) =

Again, 1 1 −1 > ν ( x0 , x 1 , t ) 1−k

1−r r

r 1−r


i.e. ,

21

1 1−r > + 1 ν ( x0 , x 1 , t ) r (1 − k) =

⇒ ν ( x0 , x 1 , t )
1 − r ] 1−r rk rk + 1 − r 1 1−1+r < k + 1 = + 1 = ⇒ µ ( x1 , x 2 , t ) 1−r 1−r 1−r

i.e. ,

1 r(k − 1) + 1 < µ ( x1 , x 2 , t ) 1−r

⇒ µ ( x1 , x 2 , t ) >

1−r 1−r = > 1−r 1 + r(k − 1) 1 − r(1 − k)

⇒ µ ( x1 , x 2 , t ) > 1 − r Again, 1 1 −1 = −1 ν ( x1 , x 2 , t ) ν ( T (x0 ) , T (x1 ) , t ) 1 ≥ k

1 −1 ν ( x0 , x 1 , t )

!

1 1 i.e. , −1 ≥ ν ( x1 , x 2 , t ) k

1 −1 ν ( x0 , x 1 , t )

!

1 > k ⇒

1 −1 r

1 = k

1−r r

1 1−r 1 − r + rk 1 − r(1 − k) > +1 = = ν ( x1 , x 2 , t ) rk rk rk

22


i.e. ,

ν ( x1 , x 2 , t )
1 − r , ν ( x2 , x 3 , t ) < r , · · · , µ ( xn − 1 , x n , t ) > 1 − r and ν ( xn − 1 , xn , t ) < r. Thus, we see that , µ ( x0 , x n , t ) ≥ µ

t x0 , x 1 , n

∗µ

t x1 , x 2 , n

t ∗ · · · ∗ µ xn − 1 , x n , n

> (1 − r) ∗ (1 − r) ∗ · · · ∗ (1 − r) = 1 − r i.e. ,

µ ( x0 , x n , t ) > 1 − r

t t ν x1 , x 2 , ν ( x 0 , x n , t ) ≤ ν x0 , x 1 , n n < r r ··· r = r

··· ν

t xn − 1 , x n , n

Thus , µ ( x0 , xn , t ) > 1 − r and ν ( x0 , xn , t ) < r ⇒ xn ∈ S( x 0 , r , t ) Hence, by the theorem 22[1], T has unique fixed point in S( x 0 , r , t ). Note 3.5 It follows from the proof of Theorem 2.13[1] that for any x ∈ X the sequence of iterates { T n (x)} converges to the fixed point of T. Lemma 3.6 Let ( X , A ) be IFMS and T : X → X be t-uniformly continuous on X. If xn → x as n → ∞ in ( X , A ) then T (xn ) → T (x) as n → ∞ in ( X , A ) . Proof. Proof directly follows from the definitions of t-uniformly continuity and convergence of a sequence in a IFMS. Lemma 3.7 Let ( X , A ) be IFMS. If xn → x and yn → y in ( X , A ) then µ ( xn , yn , t ) → µ ( x , y , t ) and ν ( xn , yn , t ) → ν ( x , y , t ) as n → ∞ f or all t > 0 in R . Proof. We have, lim

µ ( xn , x , t ) = 1 ,

lim

µ ( yn , y , t ) = 1 ,

n → ∞

and

n → ∞

µ ( xn , y n , t ) ≥ µ

xn , x ,

lim

ν ( xn , x , t ) = 0

lim

ν ( yn , y , t ) = 0

n → ∞

n → ∞

t 2

∗ µ

x , yn ,

t 2

23


≥ µ ⇒

lim

n → ∞

t xn , x , 2

∗µ

t x, y, 4

∗µ

t y , yn , 4

µ ( xn , y n , t ) ≥ µ ( x , y , t )

µ(x, y, t) ≥ µ

t x , xn , 2

∗ µ

t xn , y , 2

t t ∗ µ xn , y n , ∗ µ ≥ µ x , xn , 2 4 ⇒ µ ( x , y , t ) ≥ lim µ ( xn , yn , t ) ∀ t > 0. n → ∞ Then, lim µ ( xn , yn , t ) = µ ( x , y , t ) for all t > 0,

t yn , y , 4

n → ∞

Similarly,

lim

n → ∞

ν ( xn , yn , t ) = ν ( x , y , t ) for all t > 0.

Theorem 3.8 ( X , A ) be a complete IFMS with the property ♠ in which IF contractive sequences are Cauchy sequences and T : X → X be a tuniformly continuous on X. If for same positive integer m, T m is a IF contractive mapping with k its contractive constant then T has a unique fixed point. Proof.Let B = T m , n be a arbitrary but fixed positive integer and x0 ∈ X. we now show that B n T (x0 ) → B n (x0 ) in ( X , A ) . Now, µ

( B n T (x

1 1 −1 = −1 n n − 1 µ ( B(B T (x0 )) , B(B n − 1 (x0 )) , t ) 0 ) , B (x0 ) , t ) ≤ k ≤ kn

=⇒ 0 ≤ =⇒

lim

n → ∞

1 − 1 n − 1 µ(B T (x0 ) , B n − 1 (x0 ) , t ) 1 − 1 µ ( T (x0 ) , x0 , t )

1 − 1 n µ ( B T (x0 ) , B n (x0 ) , t )

!

!

!

≤ 0

lim µ ( B n T (x0 ) , B n (x0 ) , t ) = 1, f or all t > 0.

n → ∞

Similarly, lim ν ( B n T (x0 ) , B n (x0 ) , t ) = 0, for all t > 0. n → ∞ Thus, B n T (x0 ) → B n (x0 ) in ( X , A ) . Again, by the theorem 22[1], we see that B has a unique fixed point x(say),

≤ ···

24


and from the note [3.5], it follows that B n (x0 ) → x as n → ∞ in ( X , A ) . Since T is t-uniformly continuous on X, it follows from the above lemma[3.6] that B n T (x0 ) = T B n (x0 ) → T (x) as n → ∞ in ( X , A ) . Again since n lim µ ( B n T (x0 ) , B n (x0 ) , t ) = 1 and n lim ν ( B n T (x0 ) , B n (x0 ) , t ) → ∞ → ∞ = 0, we have by the lemma [3.7], lim µ ( T (x) , x , t ) = 1 and lim ν ( T (x) , x , t ) = 0, for all t > 0, n → ∞ n → ∞ i.e. , µ ( T (x) , x , t ) = 1 and ν ( T (x) , x , t ) = 0, for all t > 0. ⇒ T (x) = x ⇒ x is a fixed point of T. 0 0 0 0 0 If x is a fixed point of T, i.e. , T (x ) = x , then T m (x ) = T m − 1 (T (x )) = 0 0 0 0 0 T m − 1 (x ) = · · · = x ⇒ B(x ) = x ⇒ x is a fixed point of B. 0 But x is the unique fixed point of B, therefore x = x which implies that x is the unique fixed point of T. Definition 3.9 Let ( X , A ) be IFMS and T : X → X. T is said to be TS-IF contractive mapping if there exists k ∈ (0 , 1) such that k µ ( T (x) , T (y) , t ) ≥ µ ( x , y , t ) and

1 ν ( T (x) , T (y) , t ) ≤ ν ( x , y , t ) ∀ t > 0. k

Theorem 3.10 Let ( X , A ) be a complete IFMS and T : X → X be TSIF contractive mapping with k its contraction constant. Then T has a unique fixed point. Proof. Let x ∈ X and x n = T n (x) for all n ∈ N . Now for each t > 0, k µ ( x2 , x1 , t ) = k µ ( T (x1 ) , T (x) , t ) ≥ µ ( x1 , x , t ) i.e. , k µ ( x2 , x1 , t ) ≥ µ ( x1 , x , t ) and 1 1 ν ( x2 , x 1 , t ) = ν ( T (x1 ) , T (x) , t ) k k ≤ ν ( x1 , x , t ) i.e. ,

1 ν ( x2 , x 1 , t ) ≤ ν ( x1 , x , t ) k

Again, k µ ( x3 , x2 , t ) = k µ ( T (x2 ) , T (x1 ) , t )

25


≥ µ ( x2 , x 1 , t ) ⇒ k 2 µ ( x3 , x 2 , t ) ≥ k µ ( x2 , x 1 , t ) ≥ µ ( x1 , x , t ) i.e. , k 2 µ ( x3 , x2 , t ) ≥ µ ( x1 , x , t ) and 1 1 ν ( x3 , x 2 , t ) = ν ( T (x2 ) , T (x1 ) , t ) k k ≤ ν ( x2 , x 1 , t ) 1 1 ⇒ ν ( x2 , x 1 , t ) ≤ ν ( x1 , x , t ) ν ( x , x , t ) ≤ 3 2 k2 k 1 i.e. , 2 ν ( x3 , x2 , t ) ≤ ν ( x1 , x , t ) k By Mathematical induction, we have, k n µ ( x n + 1 , x n , t ) ≥ µ ( x1 , x , t ) and 1 ν ( x n + 1 , x n , t ) ≤ ν ( x1 , x , t ), f or all t > 0. kn We now verify that { x n } is a cauchy sequence in ( X , A ). Let t1 = µ( xn , xn+p , t)

t . p

≥ µ ( xn , xn + 1 , t1 ) ∗ µ( xn + 1 , xn + 2 , t1 ) ∗ · · · ∗ µ ( xn + p − 1 , xn + p , t1 )

=

1 n k µ ( xn , x n + 1 , t 1 ) kn

∗

1

kn+1 ∗

k n + 1 µ ( xn + 1 , x n + 2 , t 1 ) 1

kn+p−1

∗ ···

k n + p − 1 µ ( x n + p − 1 , x n + p , t1 )

1 1 1 µ ( x1 , x , t1 ) ∗ µ ( x1 , x , t1 ) ∗ · · · ∗ µ ( x1 , x , t1 ) n n + 1 n + k k k p−1 1 1 1 ≥ µ ( x , x , t ) ∗ · · · ∗ µ ( x , x , t ) = µ ( x , x , t ) 1 1 1 1 1 1 kn kn kn 1 ⇒ 1 < lim µ ( x , x , t ) ≤ lim µ ( xn , xn + p , t ) ≤ 1 1 1 n → ∞ n → ∞ kn ≥

⇒

lim µ ( xn , xn + p , t ) = 1

n → ∞

In the similar way, we have, ⇒ lim ν ( xn , xn + p , t ) = 0 n → ∞

Hence { x n }n is a cauchy sequence in ( X , A ).

26


So, ∃ y ∈ X such that xn → y as n → ∞ in ( X , A ). Now, k µ ( T (xn ) , T (y) , t ) ≥ µ ( xn , y , t ) 1 µ ( xn , y , t ) k 1 1 ⇒ nlim µ ( T (xn ) , T (y) , t ) ≥ nlim µ ( xn , y , t ) = > 1 →∞ →∞ k k ⇒ 1 < nlim µ ( T (xn ) , T (x) , t ) ≤ 1 →∞ i.e. , µ ( T (xn ) , T (y) , t ) ≥

⇒

lim µ ( T (xn ) , T (x) , t ) = 1

n→∞

Again, ν ( T (xn ) , T (y) , t ) ≤ k ν ( xn , y , t ) ⇒ ⇒

lim ν ( T (xn ) , T (y) , t ) ≤ lim k ν ( xn , y , t ) = 0

n→∞

n→∞

lim ν ( T (xn ) , T (x) , t ) = 0

n→∞

Thus we see that lim µ ( T (xn ) , T (y) , t ) = 1 and lim ν ( T (xn ) , T (y) , t ) = 0, f or all t > 0.

n→∞

n→∞

⇒ nlim T (xn ) = T (y) in ( X , A ) =⇒ →∞

lim xn+1 = T (y) in ( X , A ).

n→∞

i.e. , y = T (y) ⇒ y is a fixed point of T. To prove the uniqueness, assume T (z) = z for some z ∈ X. Then for t > 0, we have µ ( y , z , t ) = µ ( T (y) , T (z) , t ) ≥ = ≥ ≥ =⇒ 1 < lim

n→∞

1 µ(y, z, t) k 1 µ ( T (y) , T (z) , t ) k 1 µ(y, z, t) ≥ ··· k2 1 µ ( y , z , t ) −→ ∞ as n → ∞ kn 1 µ(y, z, t) ≤ µ(y, z, t) ≤ 1 kn

⇒ µ(y, z, t) = 1 ν ( y , z , t ) = ν ( T (y) , T (z) , t )


27

≤ k µ(y, z, t) = k ν ( T (y) , T (z) , t ) ≤ k2 ν (y , z , t) ≤ ··· ≤ k n µ ( y , z , t ) −→ 0 as n → ∞ =⇒ 0 ≤ µ ( y , z , t ) ≤ lim k n µ ( y , z , t ) < 0 n→∞

⇒ ν (y, z, t) = 0 ⇒ y = z This completes the proof.

4

Open Problem

One can further try to prove the Kannan’s fixed point theorem and Chatterjee’s fixed point theorem with the help of TS-IF contractive mapping and also with respect to the contractive mapping of [1].

References [1] Abdul Mohamad, Fixed-point theorems in Intuitionistic fuzzy metric spaces, Chaos, Solitons and Fractals, 34 ( 2007 ) 1689−1695. [2] A. George, P. Veeramani, On Some result in fuzzy metric spaces, Fuzzy Sets Syst Vol. 64 ( 1994 ) 395−399. [3] A. George, P. Veeramani, On Some results of analysis for fuzzy metric spaces, Fuzzy Sets Syst Vol. 90 ( 1997 ) 365−368. [4] B. Schweizer , A. Sklar, Statistical metric space, Pacific journal of mathematics 10 ( 1960 ) 314−334. [5] E. P. Klement, R. Mesiar, E. Pap, Triangular norms , Kluwer , Dordrecht ( 2000 ). [6] Jin Han Park, Intuitionistic fuzzy metric spaces, Chaos, Solitons and Fractals 22 ( 2004 ) 1039−1046. [7] K. Atanassov, Intuitionistic fuzzy sets , In : Sgurev V. editor. VII ITKR’s Session, Sofia June, 1983 ( Central Sci. and Technol Library, Bulgarian Academy of Sciences, 1984 ). [8] K. Atanassov, Intuitionistic fuzzy sets , Fuzzy Sets and Systems Vol.20 , 1986, pp 87−96

28


[9] M. Grabiec, Fixed points in fuzzy metric spaces , Fuzzy Sets and Systems 27 ( 1988 ) 385−389. [10] O. Kramosil, J. Michalek, Fuzzy metric and statisticalmetric spaces, Kybernetica 11 ( 1975 ) 326−334. [11] R. Vasuki, P. Veeramani, Fixed point theorems and Cauchy sequences in fuzzy metric spaces, Fuzzy Sets Syst Vol. 135 ( 2003 ) 415−417. [12] T. K. Samanta and Iqbal H. Jebril, Finite dimentional intuitionistic fuzzy normed linear space, Int. J. Open Problems Compt. Math., Vol 2, No. 4 ( 2009 ) 574−591. [13] L. A. Zadeh, Fuzzy sets, Information and control 8 ( 1965 ) 338−353.


Even-order Magic Squares with Special Properties Saleem Al-ashhab Department of Mathematics, Al-albayt University, Jordan. e-mail: [email protected]

Abstract

In this paper we introduce new concepts in the field of magic squares. We focus on special types of magic squares of order six. We list enumerations of the squares of this type. We prove that the nullity of some even-order magic squares is at least one. Keywords: Magic Squares, nullspace.

1

Introduction

A magic square is a square matrix, where the sum of all entries in each row, column or both main diagonals yields the same number. This number is called the magic constant. A natural magic square of order n is a matrix such that its entries consist of all integers from one to n2 . The magic constant in this case 2 is n(n2+1) . A pandiagonal magic square is a magic square such that the sum of all entries in all broken diagonals equals the magic constant. A symmetric magic square is a natural magic square of order n such that the sum of all opposite entries equals n2 + 1, i. e. the following relations hold

aij + an+1−i,n+1−j = n2 + 1 for all 1 ≤ i, j ≤ n The following square is a natural symmetric magic square

30

Saleem Al-ashhab



 15 14 1 18 17  19 16 3 21 6     2 22 13 4 24     20 5 23 10 7  9 8 25 12 11 The sum of all opposite entries is 26. We emphasize here that a symmetric magic square is not a symmetric matrix in the sense that its identical to its transpose. A natural magic square can not be a symmetric matrix since all entries are distinct. The number of natural magic squares of order five is known. Schroeppel computed this number in 1971 (see [11]). It is well-known that there are pandiagonal magic squares and symmetric squares of order five (see [6]). The number of natural magic squares of order six is til now unknown. We give here the number of a subset of such squares. One possible form of a symmetric magic square with magic sum 3s is

       

a f b g h r q D E p A B

C s−F s−B s−A k s−j s−p s−E o s−v s−D s−q v s−o s−r s−h j s−k s−g s−b F s−C s−f s−a

       

where A = j − 2b − g − h − a − k + p − q + 3s, B = q − g − h − o − p − f − 2r + 3s + v, C = 92 s − b − f − g − h − k − o − r − a, D = h + o − q + r − v, E = b + g − j + k − p, F = a + b + f + g + h − j + r − 32 s − v. We obtain this form by solving the equations resulting from the definition of such squares. Due to the fraction in the assignment for C we see that noninteger numbers might appear. In particular, a natural symmetric magic square can not exist since the magic constant for natural magic squares is 111. The same problem occurs when we solve the equations resulting from the definition of the pandiagonal magic square. Hence, there are also no natural pandiagonal magic squares. It is well-known that the following structure

31

Even-order Magic Squares with Special Properties



 A B C 2s − A − B − C   E 2s − A − B − E A + E − C B+C −E     s−C A+B+C −s s−A s−B s−A−E+C s−B−C +E s−E A+B+E−s is the general structure of the pandiagonal magic square 4 × 4 (see [11]). Here, the magic constant is 2s. We note that the sum in each pair of antipodals is s, i. e. aij + ai+2,j+2 = s, for all 1 ≤ i, j ≤ 2 aij + ai+2,j−2 = s, for all 1 ≤ i ≤ 2, 3 ≤ j ≤ 4

We define next classes of magic squares of order six, which possesses similar properties for the antipodals.

2

Quasi Pandiagonal Magic Squares

We can generalize idea of the structure of 4 × 4 pandiagonal magic squares in order to obtain new types of magic squares of even order. For example, the 6 × 6 square having the structure        

a b c d i j e f g h k l m n o p q r s−d s−i s−j s−a s−b s−c s−h s−k s−l s−e s−f s−g s−p s−q s−r s−m s−n s−o

       

with the following restrictions

a + b + c + d + i + j = 3s, e + f + g + h + k + l = 3s, m + n + o + p + q + r = 3s,

a + e + m = d + h + p, b + f + n = i + k + q, c + g + o = j + l + r, is a magic square with magic constant 3s. This structure is called a quasi pandiagonal magic square of order six. The previous system of linear equations has the following augmented coefficient matrix:

32

       

Saleem Al-ashhab

1 0 0 1 0 0

1 0 0 0 1 0

1 1 1 1 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 −1 0 1 0 0 −1

0 1 0 1 0 0

0 1 0 0 1 0

0 0 0 0 1 1 1 1 0 0 0 0 0 −1 0 0 0 0 −1 0 1 0 0 −1

0 0 1 1 0 0

0 0 1 0 1 0

 0 0 0 0 −3 0 0 0 0 −3   1 1 1 1 −3   0 −1 0 0 0   0 0 −1 0 0  1 0 0 −1 0

It has the following squared submatrix        

1 0 0 1 0 0 0 0 0 0 −1 −1

0 0 1 0 1 0

0 0 0 0 0 0 1 1 1 0 −1 0 0 0 −1 1 0 0

       

of determinant two. Hence, the general solution of these equations can be given as: j = 3s − a − b − c − d − e, l = 3s − g − h − i − j − k, 3 n = c + d + g + h + i + k − m − r − s, 2 o = 6s − a − b − 2c − d − e − f − 2g − h − i − k + r, p = a − d + e − h + m, 3 q =b+c+d+f +g+h−m−r− s 2 Let Λ be a quasi pandiagonal magic square of order six. Then, the nullity of Λ is at least one. Note that the square Λ has the form A B S−B S−A


33

where all entries of the matrix S are s. We seek the vectors in the nullspace of this matrix having the form x for some x ∈ R3 −x The product between the square Λ and this vector leads   0 Ax − Bx  ..  = .  −Bx + Ax 0 and, hence, we look for vectors x ∈ R3 satisfying the following equation: Ax − Bx = 0 (1) This equation has nonzero vectors as solution because the matrix A − B does not have full rank. This is due to the following reason: replacing the last row with the summation of all rows yields the following row in the matrix: (a + e + m − d − h − p, b + f + n − i − k − q, c + g + o − j − l − r) According to our requirement this is a row of zeros. Hence, the rank of A − B is two and, there exists a nontrivial solution of (1). In general the rank of this type of squares is five. The following square is an example:   1 2 0 4 1 −2  0 0 1 1 0 4     1  3 1 −3 4 0    −2 1  4 1 0 2    1 2 −2 2 2 1  5 −2 2 1 −1 1 We can define certain types of Θ × Θ magic squares, where Θ is an even natural number greater than three. For example, let us consider the case Θ = 8. The square has the following structure:   a b c d A B C D  e f g h E F G H     i j k l I J K L     m  n o p M N O P    s−A s−B s−C s−D s−a s−b s−c s−d     s−E s−F s−G s−H s−e s−f s−g s−h     s−I s−J s−K s−L s−i s−j s−k s−l  s−M s−N s−O s−P s−m s−n s−o s−p

34

Saleem Al-ashhab

where a + b + c + d + A + B + C + D = 4s, e + f + g + h + E + F + G + H = 4s, i + j + k + l + I + J + K + L = 4s, m + n + o + p + M + N + O + P = 4s,

a + e + i + m = A + E + I + M, b + f + j + n = B + F + J + N,

c + g + k + o = C + G + K + O, d + h + l + p = D + H + L + P. The previous system of linear equations has the following squared submatrix in the augmented coefficient matrix:   1 0 0 0 0 0 0 0  0 1 0 0 0 0 0 0     0 0 1 0 0 0 0 0     0  0 0 1 1 1 1 1    0  0 0 0 0 0 −1 0    −1 −1 0 1 0 0 0 −1     0 0 0 0 1 0 0 0  −1 −1 −1 0 0 1 0 0 of determinant different than zero. Hence, the system of linear equations possess a nontrivial solution. The generated square is called quasi magic square. We can generalize these results. Let θ = Θ2 . The magic square having the following structure   a11 . . a1θ a1(θ+1) . . a1Θ   . . . . . . . .     . . . . . . . .     a . . a a . . a θ1 θθ θΘ θ(θ+1)    s − a1(θ+1) . . s − a1Θ s − a11 . . s − a1θ      . . . . . . . .     . . . . . . . . s − aθ(θ+1) . . s − aθΘ s − aθ1 . . s − aθθ

35


is called a quasi pandiagonal magic square of order Θ. The requirement that the matrix is a magic square yields to a solvable linear system. This is because we can following the above reasoning to find a nonsingular submatrix in the augmented coefficient matrix of the system. Hence, there exists quasi pandiagonal magic squares of order Θ, where Θ is an even natural number greater than three. Let Λ be a quasi pandiagonal magic square of order Θ. Then, the nullity of Λ is at least one. Recall that the square Λ of order Θ can be rewritten as A B S−B S−A where A, B and S are square matrices of order θ. Now, we seek the vectors in the nullspace of this matrix having the form x for some x ∈ Rθ −x The product between the square Λ and this vector leads   0 Ax − Bx  ..  = .  −Bx + Ax 0 The matrix A − B does not have full rank. This is due to the following reason: replacing the last row with the summation of all rows yields the following row in the matrix: (a11 + ... + aθ1 − a1(θ+1) − ... − aθ(θ+1) , ..., a1θ + ... + aθθ − a1Θ − ... − aθΘ ) Since the square is magic, we can assure that this row is a row of zeros. Hence, there exists a nontrivial solution of the equation Ax − Bx = 0 Using this nontrivial solution we determine a nonzero vector in the nullspace of Λ.

3

Four-corner Magic Squares

A four-corner magic square of order 6 is a magic square (aij ) i=1,...,6 with magic j=1,...,6

constant 3s such that a33 + a44 + a34 + a43 = 2s and ai,j + a(i+3),(j+3) + ai,(j+3) + a(i+3),j = 2s for all i = 1, 2, 3 and j = 1, 2, 3. The entries of a four-corner magic square of order 6 satisfy

36

Saleem Al-ashhab

a14 + a25 + a36 + a41 + a52 + a63 = 3s, a13 + a22 + a31 + a61 + a55 + a64 = 3s

These two conditions represent the sum of the entries of two broken diagonals. If the magic square is pandiagonal, then we have to consider all broken diagonals. To see the validity of the first equation we know from the definition that a11 + a44 + a14 + a41 = 2s, a22 + a55 + a25 + a52 = 2s, a33 + a66 + a36 + a63 = 2s. Adding up these equations and subtracting from the addition the following equation a11 + a22 + a33 + a44 + a55 + a66 = 3s we obtain the desired equation. A four-corner magic square of order 6 can be written as        

x f g t G z h n j q w H e a m 2s − b − t − x k 2s − b − a − e b D 2s − o − j − z p d o 2s − p − q − h B L A 3s − b − j − o − a − t E

where A = a + b − d − g − n + s, B = b + j + o − s + t − w, D = g − j − k − o − p − q + s + w + x + e, E = f + h + k − m + p − s, F = p − b + q + s − x − e, G = j − g − f + o + p + q + s − w − x − e, H = b + j + o − s + t − w, J = d − a + g + n − p − q + x, L = b + j + o − s + t − w, M = 2s − o − p − q − j − t + w + e, R = a + b − g + j + o + p + q − 2s + t − w, T = h − d + j + q − s + z, Y = 3s − j − n − q − h − z.

M Y J R T F

       

37


We are interested in a subclass of the four-corner magic squares. A four-corner magic squares with semi-symmetric center is a four-corner magic square of order 6 such that a33 + a43 = s and a34 + a44 = s. A four-corner magic squares with semi-symmetric center of order 6 can be written as

       

x f g t G z h n j q w H e a m s+a−t−x k s−e s−a D 2s − o − j − z p d o 2s − p − q − h B L 2s − g − n − d 2s − o − j − t E

M Y J R T F

     ......(3)   

where the capital letters are dependent variables. We can see that the square has seventeen independent variables. This can be checked with computers by solving the system of equations included in the definition of four-corner magic squares with semi-symmetric center.

3.1

Property preserving transformations

There are seven classical transformations, which take a magic square into another magic square. They are the composition of rotations with angles π2 , π, 3π 2 and reflection about the main diagonals. Now, a four-corner magic squares with semi-symmetric center can be transformed into other of the same class. To illustrate this we transform the square in (3) into        

x f g t z 2s − p − q − h d o w m e a s+a−t−x D s−e s−a 2s − o − j − z q n j B L 2s − g − n − d 2s − o − j − t

 G M p Y   H J   k R   h T  E F

In order to eliminate the effect of the previous transformations we compute all natural four-corner magic squares with semi-symmetric center for which the following conditions hold:

38

Saleem Al-ashhab

p < q, a < e < s − a, 1 ≤ a ≤ 17......(4) The condition p < q eliminates the effect of the last mentioned transformation since it does not change the center. Hence, the total number of all natural four-corner magic squares with semi-symmetric center is the computed number multiplied with sixteen. The four-corner magic squares with semi-symmetric center possesses another property preserving transformations besides the previous one. The square in (3) can be transformed into

       

x f g t G M z p+q+h−s n j s−p Y w H e a m J s+a−t−x k s−e s−a D R 2s − o − j − z s−q d o s−h T B L 2s − g − n − d 2s − o − j − t E F

       

We see that the 2 × 2 center also remains unchanged. Besides, any square satisfying the conditions (4) will be transformed into another square satisfying (4). Hence, the number of natural four-corner magic squares with semisymmetric center satisfying (4) will be even. Unfortunately, we can not eliminate the effect of the last transformation by requiring a comparison relation.

3.2

Number of squares

We used pentium IV computers core 2 duo CPU (3 GHz) to count the fourcorner magic squares with semi-symmetric center. It took about three months to finish. The C code is presented in the appendix. It is based on the choice of a specific center in each run. The values of the other independent variables will be assigned using some nested loops. The value of the dependent variables will be sequentially computed. At each stage we test the computed value for being in the range from 1 to 36 and for being different from other existing values. The number of centers is 306. We list the number for all different values of a tabulated according to e in the following tables: a=1

39


e number 2 116511842 3 132987568 4 201772222 5 195213542 6 247269840 7 249849504 8 279043734 9 282858168 10 314225850

e number 11 295406900 12 325569546 13 314724780 14 330356970 15 319871530 16 338570808 17 319583422 18 334466160 19 334597154

e number 20 332716362 21 332269470 22 334768630 23 343668656 24 353535258 25 351413848 26 354646898 27 356452980 28 358076270

e 29 30 31 32 33 34 35

number 350383960 360519204 351628294 352098982 347005202 346545330 336033008

a=2 e number e number 11 331287390 19 326515794 12 321640070 20 345399018 13 350613710 21 335036204 14 332114584 22 333026268 15 336104578 23 356959948 16 331537886 24 360229516 17 350953660 25 353925968 18 311139312 26 358846606

e 27 28 29 30 31 32 33 34

number 356484732 361179738 374039976 344502648 353814694 363592188 362587560 347566238

a=3 e number e number e number 11 306028538 19 320541794 4 251739982 12 336328382 20 336807852 5 240505110 13 324869294 21 336809760 6 273691912 14 329469256 22 339051514 7 275204588 15 324711266 23 344044096 8 299899266 16 337436298 24 355226184 9 294890678 17 328147564 25 362975754 10 308909490 18 323876138 26 353891828

e 27 28 29 30 31 32 33

number 360898736 363278990 346910646 366423556 352598114 354459440 361775374

a=4 e number e number 12 329152950 19 335007752 13 341948432 20 326708492 14 327143618 21 341404426 15 341003048 22 346408234 16 329377468 23 354547734 17 322104494 24 358998042 18 323967196 25 358784510

e 26 27 28 29 30 31 32

number 353888706 366826140 354933908 372809534 354855806 365478056 354459440

e number 3 206302610 4 210992494 5 251163058 6 245105736 7 274508468 8 287150618 9 296599162 10 302170466

e number 5 277262012 6 278106484 7 305137672 8 286925428 9 318200762 10 326734318 11 325879172

a=5

40

Saleem Al-ashhab

e number 6 291627212 7 289665812 8 319609974 9 303484650 10 319630542 11 331302656

e number 7 307831520 8 307220566 9 326632592 10 315214080 11 329922646 12 337240842

e number 8 318913908 9 313804886 10 332965950 11 312416444 12 339820720

e number 9 322651726 10 313931550 11 350259448 12 328728052 13 331352496

e number 10 332885406 11 325241400 12 342083312 13 321193778

e number e 12 329152040 19 13 335041250 20 14 386116304 21 15 332832472 22 16 341768070 23 17 320454620 24 18 328627682 25 a=6 e number e 13 335408322 19 14 323632092 20 15 343900408 21 16 317359492 22 17 329088296 23 18 324929256 24 a=7 e number e 13 326680272 19 14 338270640 20 15 321017534 21 16 331596514 22 17 315920958 23 18 333891060 24 a=8 e number e 14 336499586 19 15 327023778 20 16 321879182 21 17 349000474 22 18 332648602 23 a=9 e number e 14 336616920 19 15 312280764 20 16 327204636 21 17 322272062 22 18 331971266 23

a = 10

number 323827718 346739100 329511150 343736302 373008254 358755138 368793648

e number 26 381023514 27 355241148 28 353360938 29 362792908 30 367872592 31 356279070

number 327838034 325045442 346521984 346273390 366929874 357798606

e number 25 362278616 26 344707584 27 357310716 28 352453064 29 364671418 30 354408080

number 334337068 343814008 338858100 361348270 344125594 350598284

e number 25 353600518 26 353855988 27 352947394 28 369485474 29 361325404

number 326604342 351929926 342614872 338372718 355184854

e number 24 344790314 25 345471054 26 371263102 27 368869600 28 358386792

number 340019774 335943324 326267006 338824226 342484208

e number 24 359800942 25 363693102 26 353751236 27 369395860

41


e number 11 346174266 12 331299312 13 343459744 14 316652498 e number 12 319612194 13 311520480 14 334495182 e number 13 333424860 14 307147362 15 315379026 e number 14 323015258 15 308092470 e number 15 315555354 16 313515758 e number 16 342594820 e number 17 339798226

e number e 15 320420772 19 16 314295184 20 17 329404430 21 18 315451068 22 a = 11 e number e 15 307807314 19 16 327835916 20 17 318252788 21 18 318965356 22 a = 12 e number e 16 310667578 19 17 328751622 20 18 315313592 21 a = 13 e number e 16 326470472 19 17 312747912 20 18 340473770 21 a = 14 e number e 17 372360568 19 18 322125932 20 a = 15 e number e 17 321828952 19 18 351825242 20 a = 16 e number e 18 342610906 19 a = 17 e number e 18 398369256 19

number 333407464 317775590 334963576 324426154

e number 23 351659296 24 348631644 25 360576518 26 346157032

number 315707450 330765638 355567308 339579470

e number 23 345843756 24 359195928 25 368711314

number 330888746 321671738 353451262

e number 22 335530678 23 355516716 24 346475818

number e number 331125814 22 353459532 341748002 23 355448444 331096842 number e number 334305526 21 347812602 370518284 22 326982930 number e number 318003462 21 340531500 341764226 number e number 324696620 20 328137776 number 344164516

The total number of squares is 101, 425, 060, 998 = 50, 712 , 530, 499 ∗ 2 Hence, the total number of natural squares is 101425060998 ∗ 16 = 1, 622 , 800 , 975 , 968

42

4

Saleem Al-ashhab

Open Problem

The number of four-corner magic squares of order 6. We want here to estimate the number of four-corner magic squares of order 6. By computing the number of squares with semi-symmetric center we noticed that the number of different possible values for a and e is 2 + 4 + 8 + ... + 34 = 306 The average of the squares per one possible pattern for a and e is 101425060998∗16 306

= 5. 303 3 × 109

The number of all different possible values for a, b and e by computing the number of four-corner magic squares is 3429. Hence we estimate the number of the four-corner magic squares 5. 303 3 × 109 ∗ 3429 = 1. 818 5 × 1013 We noticed that the number for some possible values for a, b and e, which do not lead to a symmetric center is less that the largest observed number (398369256). This motivates the following The number of four-corner magic squares of order 6 does not exceed 398369256 ∗ 16 ∗ 3429 = 21 , 856 , 130 , 861 , 184.

Acknowledgement Thanks are due to the Deanship of the Academic Research at Al-Albayt University for funding this work (Grant number 5016). We thank also K. Al Share’o for his contributions to the research, which lead to this work. Finally, Harry White contributed to preparing the source code.

References [1] Ahmed, M. [2004] “Algebraic Combinatorics of Magic Squares,” Ph. D. Thesis, Mathematics Dep., University of California, Davis. [2] Al-Ashhab, S. [2011] “Special semi magic squares,” Acta Universitatis Apulensis 25, 195–205. [3] Al-Ashhab, S. & Mueller, W. [2003] “Semi Pandiagonal Magic 4x4 Squares,” Results in Mathematics 44, 25–28.


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[4] Andress, W. [1960] “Basic Properties of Pandiagonal Magic Squares,” The American Mathematical Monthly 67: 2, 143–152. [5] Bellew, S. [1997] “Counting the Number of Compound and Nasik Magic Squares,” Mathematics Today 33:4, 111–118. [6] Benson, A. & Jacoby, S. [1976] New Recreations With Magic Squares, Dover Publications. [7] Duijvestijn, A. & Federico, P. [1982] “Compound Perfect Squares,” The American Mathematical Monthly 89:1, 15–32. [8] Hendricks, J. [1989] “The Determinant of a Pandiagonal Magic Square of Order 4 is Zero,” Journal of Recreational Mathematics 21, 179-181. [9] Lehmer, D. [1933] “A Complete Census of 4x4 Magic Squares,” The American Mathematical Society Bulletin 39, 764–767. [10] Nakkar, H., Nasr Al - Din, I., Al-Ashhab, S. & Elhamale, M. [2011] “Properties of Nested 6x6 Semi Magic Squares,” Res. J. aleppo university 75, 43–54. [11] The order 5 magic squares, Written by Michael Beeler with assistance from Schroeppel—see M. Gardner’s column on mathematical games, Sci. Am. 234, 1976, pp. 118–123.

Appendix The C-code #include ¡assert.h¿ #include ¡errno.h¿ #include ¡io.h¿ #include ¡stdio.h¿ #include ¡stdlib.h¿ #include ¡string.h¿ #include ¡time.h¿ const int N = 6; const int NN = N*N;const int Sum2 = NN - 1; const int Sum4 = Sum2 + Sum2; const int Msum = Sum2 + Sum4; struct bools {bool used[NN];}; struct bools allFree; #define Uint unsigned int void writeSquare(int *p, FILE *wfp) {char squareString[120], *s = squareString; int cells = 0; {int i; for (i = 0; i ¡ NN; ++i) {int x = p[i] + 1; if (x ¡ 10) { *s++ = ’ ’; *s++ = ’0’ + x; } else if (x ¡ 20) { *s++ = ’1’; *s++ = ’0’ - 10 + x; }

44

Saleem Al-ashhab

else if (x ¡ 30) { *s++ = ’2’; *s++ = ’0’ - 20 + x; } else { *s++ = ’3’; *s++ = ’0’ - 30 + x; } if (++cells == N) { *s++ = ’\n0 ; cells = 0; }else ∗ s + + =0 0 ; }} *s++ = ’\n0 ; ∗s + + =0 \00 ; f puts(squareString, wf p); } Uint makeSquares(int a, int b, int e, int J, FILE *wfp) {Uint count = 0, pcount = 0; bools v = allFree; int Z[NN]; Z[20]=J; v.used[e] = true; v.used[a] = true; v.used[b] = true; v.used[J] = true; {int t; for (t = 1; t ¡ 2 ; ++t) if (!v.used[t]) {v.used[t] = true; {int x; for (x = 21; x ¡ 22; ++x) if (!v.used[x]) {Z[18] = Sum4-b-t-x; if ((Z[18] ¡ 0) ||(Z[18]> = N N )||v.used[Z[18]]||(Z[18] == x))continue; v.used[x] = true; v.used[Z[18]] = true;{int j; for (j = 0; j ¡ NN; ++j) if (!v.used[j]) { v.used[j] = true;{int o; for (o = 0; o ¡ NN; ++o) if (!v.used[o]) { Z[33]=Msum-j-b-o-a-t; if ((Z[33] ¡ 0) ||(Z[33]> = N N )||v.used[Z[33]]||(Z[33] == o))continue; v.used[o] = true; v.used[Z[33]] = true; {int z; for (z = 0; z ¡ NN; ++z) if (!v.used[z]) {Z[24]=Sum4-j-z-o; if ((Z[24] ¡ 0) ||(Z[24]> = N N )||v.used[Z[24]]||(Z[24] == z))continue; v.used[z] = true; v.used[Z[24]] = true; {int w; for (w = 0; w ¡ NN; ++w) if (!v.used[w]) {Z[30]=b+j+o+t-w-Sum2; if ((Z[30] ¡ 0) ||(Z[30]> = N N )||v.used[Z[30]]||(Z[30] == w))continue; v.used[w] = true; v.used[Z[30]] = true; {int p; for (p = 0; p ¡ (NN-1); ++p) if (!v.used[p]) {v.used[p] = true; {int q; for (q = p+1; q ¡ NN; ++q) if (!v.used[q]) { Z[5]=Sum4-o-p-q-j-t+w+e; if ((Z[5] ¡ 0) ||(Z[5]> = N N )||v.used[Z[5]]||(Z[5] == q))continue; Z[35]=Sum2-b+p+q-x-e;if ((Z[35] ¡ 0) ||(Z[35]> = N N )||v.used[Z[35]]|| (Z[35] == q) ||(Z[35] == Z[5]))continue; v.used[q] = true; v.used[Z[5]] = true; v.used[Z[35]] = true; {int h; for (h = 2; h ¡ 3; ++h) if (!v.used[h]) {Z[28]=Sum4-p-q-h; if ((Z[28] ¡ 0) ||(Z[28]> = N N )||v.used[Z[28]]||(Z[28] == h))continue; v.used[h] = true; v.used[Z[28]] = true; {int n; for (n = 0; n ¡ NN; ++n) if (!v.used[n]) {Z[11]=Msum-j-z-n-q-h; if ((Z[11] ¡ 0) ||(Z[11]> = N N )||v.used[Z[11]]||(Z[11] == n)) continue; v.used[n] = true; v.used[Z[11]] = true; {int d; for (d = 0; d ¡ NN; ++d) if (!v.used[d]) { Z[29]=Msum-Z[24]-p-d-o-Z[28]; if ((Z[29] ¡ 0) ||(Z[29]> = N N )||v.used[Z[29]]||(Z[29] == d)) continue; v.used[d] = true; v.used[Z[29]] = true; {int g; for (g = 0; g ¡ NN; ++g) if (!v.used[g]) {Z[32]=a+b-d-g-n+Sum2; if ((Z[32] ¡ 0) ||(Z[32]> = N N )||v.used[Z[32]]||(Z[32] == g))


45

continue; Z[17]=d-a+g+n-p-q+x; if ((Z[17] ¡ 0) ||(Z[17]> = N N )||v.used[Z[17]]|| (Z[17] == g) ||(Z[17] == Z[32]))continue; Z[23]=a+b-g+j+o+p+q+t-w-Sum4; if ((Z[23] ¡ 0) ||(Z[23]> = N N )||v.used[Z[23]]|| (Z[23] == g) ||(Z[23] == Z[32])||(Z[23] == Z[17]))continue; v.used[g] = true; v.used[Z[32]] = true; v.used[Z[17]] = true; v.used[Z[23]] = true; {int f; for (f = 34; f ¡ 35; ++f) if (!v.used[f]) {Z[4]=Msum-x-g-t-f-Z[5]; if ((Z[4] ¡ 0) ||(Z[4]> = N N )||v.used[Z[4]]||(Z[4] == f ))continue; v.used[f] = true; v.used[Z[4]] = true; {int m; for (m = 0; m ¡NN; ++m) if (!v.used[m]) { Z[13]=Msum-m-w-e-a-Z[17]; if ((Z[13] ¡ 0) ||(Z[13]> = N N )||v.used[Z[13]]||(Z[13] == m))continue; v.used[m] = true; v.used[Z[13]] = true; {int k; for (k = 0; k ¡ NN; ++k) if (!v.used[k]) { v.used[k] = true; Z[22]=Msum-k-b-Z[18]-Z[23]-Z[20]; if ((Z[22] ¿= 0) && (Z[22] ¡ NN) && !v.used[Z[22]]) { v.used[Z[22]] = true; Z[34]=Msum-m-q-Z[4]-Z[22]-Z[28]; if ((Z[34] ¿= 0) && (Z[34] ¡ NN) && !v.used[Z[34]]) { v.used[Z[34]] = true; Z[31]=Msum-f-h-k-p-Z[13]; if ((Z[31] ¿= 0) && (Z[31] ¡ NN) && !v.used[Z[31]]) { Z[0]=x; Z[1]=f; Z[2]=g; Z[3]=t; Z[6]=z; Z[7]=h; Z[8]=n; Z[9]=j; Z[10]=q; Z[12]=w; Z[14]=e; Z[15]=a; Z[16]=m; Z[19]=k; Z[21]=b; Z[25]=p; Z[26]=d; Z[27]=o; ++count; writeSquare(Z, wfp); if (++pcount == 1000000) {printf(”count %lu\n”, count); pcount = 0; f f lush(wf p); }} v.used[Z[34]] = false;}v.used[Z[22]] = false;}v.used[k] = false;}} v.used[m] = false; v.used[Z[13]] = false;}} v.used[f] = false; v.used[Z[4]] = false;}} v.used[g] = false; v.used[Z[17]] = false;v.used[Z[23]] = false; v.used[Z[32]] = false;}} v.used[d] = false; v.used[Z[29]] = false;}} v.used[n] = false; v.used[Z[11]] = false;}} v.used[h] = false; v.used[Z[28]] = false;}} v.used[q] = false; v.used[Z[5]] = false; v.used[Z[35]] = false;}} v.used[p] = false;}} v.used[w] = false; v.used[Z[30]] = false;}} v.used[z] = false; v.used[Z[24]] = false;}}v.used[o] = false; v.used[Z[33]] = false;}} v.used[j] = false;}} v.used[x] = false; v.used[Z[18]] = false;}} v.used[t] = false;}} printf(”number of squares %d\n”, count); returncount; } void get rest of line(int c) { if (c != ’\n0 )do{c = getchar(); }while(c! =0 \n0 ); }

46

Saleem Al-ashhab

void get abe(int *a, int *b, int *e) { int unused = scanf(”%d %d %d”, a, b, e); int c = getchar(); get rest of line(c);} void getNumPatterns(int *num) {int unused = scanf(”%d”, num); int c = getchar(); get rest of line(c);} bool check abe(int a, int b, int e) {bool rv = true;if ((a ¡= 0) || (a ¿ NN) ||(b< = 0)||(b>N N )||(e< = 0)||(e>N N )){ printf(”\aV aluerangeis1to%d.\n\n”, N N ); rv = f alse; }returnrv; } bool checkNum(int num) { bool rv = true;if (num ¡= 0) {printf(”\aN umbermustbeapositiveinteger\n\n”); rv = false;}return rv;} const int bufSize = 128; void openOutput(int a, int b, int e, char *wfpName, FILE **wfp) { const int defSize = 15; char buf[bufSize], buf1[bufSize], defaultName[defSize]; if (a == 0) strcpy(defaultName, ”s6 counts”); else sprintf(defaultName, ”s6 a%ib%ie%i”, a, b, e); strcpy(buf, defaultName); strcat(buf, ”.txt”); {int sub = 0; do {if ((fopen(buf, ”r”) == NULL) && (errno == ENOENT)) {break;} else { strcpy(buf, defaultName); sprintf(buf1, ” %i”, ++sub); strcat(buf, buf1); strcat(buf, ”.txt”);} } while (true); if ( (*wfp = fopen(buf, ”w”)) != NULL) { printf(”\n%sf ileis%s\n”, a == 0?”Data” : ”Squares”, buf ); strcpy(wfpName, buf); } else { strcpy(buf1, ”\a\nCan0 topenf orwrite”); strcat(buf1, buf); perror(buf1);}}} int getSquares(int a0, int b0, int e0, int num, FILE *wfpc) { char wfpsName[bufSize]; FILE *wfps = NULL; int linecount = 0; Uint count = 0; int patterns = 0; {int a; for (a = –a0; a ¡ NN; ++a) { {int b; for (b = –b0; b ¡ NN; ++b) if (a != b) { {int e; for (e = –e0; e ¡ NN; ++e) if ((a ¡ e) && (e ¡ b)) { int J = Sum4-e-a-b; if ((J ¡= a) ||(J == b)||(J == e)||(J> = N N ))continue; openOutput(a+1, b+1, e+1, wfpsName, &wfps); if (wfps != NULL) { time t startTime; startTime = time(NULL); count = makeSquares(a, b, e, J, wfps); fprintf(wfpc, ”a: %2d b: %2d e: %2d number of squares: %10lu time: ”, a+1, b+1, e+1, count); { int elapsed t = (int)(time(NULL) - startTime); int hr = elapsed t/3600; elapsed t %= 3600; { int min = elapsed t/60, sec = elapsed t%60; { char *fmt = ”%3d:%02d:%02d\n”; printf(fmt, hr, min, sec); fprintf(wfpc, fmt, hr, min, sec);


47

if (++linecount == 5) { fprintf(wfpc, ”\n”); linecount = 0; } fflush(wfpc); fclose(wfps); if (++patterns == num) return patterns; }}}}}}}} }} return patterns;} int main() { int a, b, e; printf(”\nInputstartvaluesf orabe : ”); get abe(&a, &b, &e); if (check abe(a, b, e)) { int num = -1; printf(”\nInputnumberof abepatternstorun : ”); getNumPatterns(&num); if (checkNum(num)) { char wfpcName[bufSize]; FILE *wfpc = NULL; openOutput(0, 0, 0, wfpcName, &wfpc); if (wfpc != NULL) { time t startTime = time(NULL); int patterns = getSquares(a, b, e, num, wfpc); int elapsed t = (int)(time(NULL) - startTime); int hr = elapsed t/3600; elapsed t %= 3600; { int min = elapsed t/60, sec = elapsed t%60; { char *fmt = ”\na, b, epatterns : %delapsedtime : %d : %02d : %02d\n”; printf(fmt, patterns, hr, min, sec); fprintf(wfpc, fmt, patterns, hr, min, sec); fclose(wfpc); } } } }} { int unused = getchar(); } return 0;}


Positive solutions for a higher order p-Laplacian boundary value problem with even derivatives Youzheng Dinga, b , Zhongli Weia, b , Jiafa Xub a

Department of Mathematics, Shandong Jianzhu University, Jinan, Shandong, China. b School of Mathematics, Shandong University, Jinan, Shandong, China.

Abstract In this work, we study the existence and multiplicity of positive solutions for the higher order p-Laplacian boundary value problem with even derivatives ( (ϕp ((−1)n−1 x(2n) ))00 = f (t, x, −x00 , . . . , (−1)n−2 x(2n−4) ), αx(2i) (0) − βx(2i+1) (0) = 0, γx(2i) (1) + δx(2i+1) (1) = 0, where t ∈ [0, 1], n ≥ 2, i = 0, 1, . . . , n, α, β, γ, δ ≥ 0 with ρ := αγ + αδ + βγ > 0 and f ∈ C([0, 1] × Rn−1 + , R+ ) (R+ := [0, ∞)). By virtue of some properties of concave functions and Jensen’s integral inequality, we adopt fixed point index theory to establish our main results. Moreover, our nonlinear term f is allowed to grow superlinear and sublinear.

Keywords: p-Laplacian equation; positive solution; fixed point index; cone; Jensen’s inequality. MSC : 34B18, 47H07, 47H11, 45M20, 26D15

1

Introduction

The paper mainly concerns with the existence and multiplicity of positive solutions for the following higher order boundary value problem with p-Laplacian operator and the nonlinear term f involving even derivatives ( (ϕp ((−1)n−1 x(2n) ))00 = f (t, x, −x00 , . . . , (−1)n−2 x(2n−4) ), (1.1) αx(2i) (0) − βx(2i+1) (0) = 0, γx(2i) (1) + δx(2i+1) (1) = 0, where t ∈ [0, 1], n ≥ 2, i = 0, 1, . . . , n, α, β, γ, δ ≥ 0 with ρ := αγ + αδ + βγ > 0 and 2n+2 f ∈ C([0, 1] × Rn−1 [0, 1] + , R+ ). Here, by a positive solution (1.1) we mean a function x ∈ C that solves (1.1) and satisfies x(t) > 0 for all t ∈ (0, 1).

49

Positive solutions

We are here interested in the case where f depends explicitly on derivatives. There are a very large number of papers dealing with higher order boundary value problems when f is independent of derivatives, see for example [11–13, 20, 26, 32, 33, 37–39, 41] and references cited therein. In [20], Graef and Yang applied Krasnosel’skii’s fixed point theorem to derive intervals for λ in which the boundary value problem consisting of the equation u(n) + λa(t)f (u) = 0, t ∈ (0, 1),

(1.2)

and the boundary condition (i)

(n−2)

u (0) = 0, i = 0, . . . , n − 2, u

(1) =

m X

aj u(n−2) (tj )

(1.3)

j=1

has a positive solution, Pang et al. [33] used a fixed point index argument to obtain existence criteria for the boundary value problem consisting of Eq. (1.2) with λ = 1 and the boundary condition m X (i) u (0) = 0, i = 0, . . . , n − 2, u(1) = aj (tj ). (1.4) j=1

When f involves all even derivatives explicitly, many authors [1–5,7,14–16,18,31,34] studied the following Lidstone boundary value problem (n ≥ 2) ( (−1)n u(2n) = f (t, u, −u00 , . . . , (−1)n−1 u(2n−2) ), (1.5) u(2i) (0) = u(2i) (1) = 0, i = 0, 1, 2, . . . , n − 1. In [42], Yang considered the existence and uniqueness of positive solutions for the following generalized Lidstone boundary value problem  n (2n)  = f (t, u, −u00 , . . . , (−1)n−1 u(2n−2) ), (−1) u (1.6) α0 u(2i) (0) − β0 u(2i+1) (0) = 0(i = 0, 1, 2, . . . , n − 1),   (2i) (2i+1) α1 u (1) − β1 u (1) = 0(i = 0, 1, 2, . . . , n − 1), where αj ≥ 0, βj ≥ 0 (j = 0, 1) and α0 α1 + α0 β1 + α1 β0 > 0. The results obtained here are similar with that of second (n = 1) boundary value problem [27, 28]. In view of symmetry, these results demonstrate that problem (1.5) and (1.6) are essentially identical with Dirichlet problem and Sturm-Liouville problem of second order ordinary differential equations, respectively. In recent years, due to mathematical and physical background [8, 17, 36], the existence of positive solutions for nonlinear boundary value problems with p-Laplacian operator has received wide attention. There exist a very large number of papers devoted to the existence of solutions for differential equations with p-Laplacian, see, for instance, [6, 9, 10, 21–23, 29, 30, 35, 43–46] and references therein. However, the existence of positive solutions for p-Laplacian equation with Lidstone boundary value problems has not been extensively studied yet. To the best of our knowledge, only [24] is devoted to this direction. In [24], Guo and Ge considered the following boundary value problems ( (Φ(y (2n−1) ))0 = f (t, y, y 00 , . . . , y (2n−2) ), 0 ≤ t ≤ 1, (1.7) y (2i) (0) = y (2i) (1) = 0, i = 0, 1, 2, . . . , n − 1,

50

Youzheng Ding et al.

where f ∈ C([0, 1] × Rn , R)(R := (−∞, +∞)). Some growth conditions are imposed on f which yield the existence of at least two symmetric positive solutions by using a fixed point theorem on cones. An interesting feature in [24] is that the nonlinearity f may be sign-changing. Motivated by the works mentioned above, in this paper, we discuss the positive solutions for (1.1). To overcome the difficulty resulting from even derivatives, we first transform (1.1) into a boundary value problem for an associated fourth-order integro-ordinary differential equation. Then, using fixed point index theory, combined with a priori estimates of positive solutions, we obtain some results on the existence and multiplicity of positive solutions for (1.1). Nevertheless, our methodology and results in this paper are new and entirely different from those in the papers cited above. We observe that if p = 2, then (1.1) reduces to problem ( (−1)n−1 x(2n+2) = f (t, x, −x00 , . . . , (−1)n−2 x(2n−4) ), (1.8) αx(2i) (0) − βx(2i+1) (0) = 0, γx(2i) (1) + δx(2i+1) (1) = 0, (i = 0, 1, . . . , n). We will show some connections between them by repeatedly invoking Jensen’s integral inequality in our proofs. Our main tool in the proofs is fixed point index theory based on a priori estimates achieved by utilizing some properties of concave functions and Jensen’s integral inequality. The idea stems from [40], but we here study the higher order p-Laplacian boundary value problem with even derivatives, while the nonlinearity of [40] is independent of derivatives. Thus our main results here improve and extend the corresponding ones in [40]. This paper is organized as follows. Section 2 contains some preliminary results. Section 3 is devoted to the existence and multiplicity of positive solutions for (1.1). In Section 4, we introduce an open problem involving Lidstone problem by Eloe [19].

2

Preliminaries

The basic space used in this paper is E := C[0, 1]. It is well known that E is a real Banach space with the norm k · k defined by kuk := maxt∈[0,1] |u(t)|. Put P := {u ∈ E : u(t) ≥ 0, ∀t ∈ [0, 1]}, then P is a cone on E. We denote Bρ := {u ∈ E : kuk < ρ} for ρ > 0 in the sequel. Let Z 1 n−1 (2n−2) Gn (t, s)u(s)ds, u := (−1) x , (Bn u)(t) := 0

where 1 G1 (t, s) = ρ Z Gn (t, s) =

( (β + αs)(γ + δ − γt), 0 ≤ s ≤ t ≤ 1, (β + αt)(γ + δ − γs), 0 ≤ t ≤ s ≤ 1, 1

Gn−1 (t, τ )G1 (τ, s)dτ, t, s ∈ [0, 1], n ∈ N+ .

0

It is easy to see that problem (1.1) is equivalent to the following p-Laplacian boundary value problem of fourth-order integro-ordinary differential equation  00 00  (ϕp (u )) = f (t, (Bn−1 u)(t), (Bn−2 u)(t), . . . , (B1 u)(t)) , (2.1) αu(0) − βu0 (0) = 0, γu(1) + δu0 (1) = 0,   00 αu (0) − βu000 (0) = 0, γu00 (1) + δu000 (1) = 0,

51

Positive solutions

which can be written in the form (see [40]) Z 1 Z 1 G1 (s, τ )f (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ ds G1 (t, s)ϕq u(t) = 0

0

(2.2)

:= (Au)(t). Note that if f ∈ C([0, 1] × Rn−1 + , R+ ), then A : P → P is a completely continuous operator, and the existence of positive solutions for (1.1) is equivalent to that of positive fixed points of A. By the definition of Bn , we see Bn (n = 1, 2, . . .) : E → E are completely continuous linear operators and they are also positive operators, i.e. Bn (P ) ⊂ P . Let λ1 > 0 be the first eigenvalue and ψ ∈ C 2 [0, 1] ∩ P the associated eigenfunction of ( −u00 = λu, αu(0) − βu0 (0) = 0, γu(1) + δu0 (1) = 0, R1 with 0 ψ(t)dt = 1, which can be written in the form Z 1 ψ(s) = λ1 G1 (t, s)ψ(t)dt = λ1 (B1 ψ)(s). (2.3) 0

Therefore λ1 = 1/r(L). Moreover, the symmetry of G1 (t, s) implies that Z 1 i ψ(s) = λ1 Gi (t, s)ψ(t)dt = λi1 (Bi ψ)(s), for i = 1, 2, . . . .

(2.4)

0

Lemma 2.1 Let κ1 := maxt,s∈[0,1] {G1 (t, s)} > 0, and κ :=

κn 1 −1 κ1 −1

− 1, then

n−1

1X G (t, s) := Gi (t, s) ∈ [0, 1], ∀t, s ∈ [0, 1]. κ i=1 Proof. From the definition Gi (i = 2, 3, . . .), we have Z 1 Gi (t, s) = Gi−1 (t, τ )G1 (τ, s)dτ ≤ κi1 , 0

and thus

n−1 X

Gi (t, s) ≤ κ1 + κ21 + . . . + κ1n−1 =

i=1

κn1 − 1 − 1 := κ > 0. κ1 − 1

Therefore, we claim G (t, s) ∈ [0, 1], ∀t, s ∈ [0, 1], as required. This completes the proof. Lemma 2.2 If u ∈ P is concave on [0, 1]. Then there is a ω > 0 such that Z 1 u(t)ψ(t)dt ≥ ωkuk.

(2.5)

0

Proof. We first prove that u is concave on [0, 1], where u is determined by (2.2). Indeed, by simple computation, we have Z 1 00 u = −ϕq G1 (t, s)f (s, (Bn−1 u)(s), (Bn−2 u)(s), . . . , (B1 u)(s))ds ≤ 0, 0

52


and then u is concave on [0, 1]. In what follows, we divide three cases. Case 1. If kuk = u(0). Then from the concavity and nonnegativity of u, we find 1

Z

Z

1

u(t · 1 + (1 − t) · 0)ψ(t)dt ≥ u(0)

u(t)ψ(t)dt = 0

1

Z

(1 − t)ψ(t)dt.

0

0

Case 2. If kuk = u(1). Similar to Case 1, we have Z

1

Z

Z

1

u(t · 1 + (1 − t) · 0)ψ(t)dt ≥ u(1)

u(t)ψ(t)dt = 0

1

0

tψ(t)dt. 0

Case 3. If there is a t0 ∈ (0, 1) such that u(t0 ) = kuk, then we arrive at Z t0 Z 1 Z t0 Z 1 t t0 − t u u(t)ψ(t) dt = u(t)ψ(t) dt + u(t)ψ(t) dt = · t0 + · 0 ψ(t) dt t0 t0 0 t0 0 0 Z t0 Z 1 Z 1 1−t t − t0 + u tψ(t) dt + (1 − t)ψ(t) dt . · t0 + · 1 ψ(t) dt ≥ u(t0 ) 1 − t0 1 − t0 t0 t0 0 Combining the above three cases and taking Z 1 Z 1 Z (1 − t)ψ(t)dt, tψ(t)dt, ω := min 0

0

t0

Z

1

tψ(t) dt +

0

(1 − t)ψ(t) dt > 0,

t0

R1 then we have 0 u(t)ψ(t)dt ≥ ωkuk, as claimed. This completes the proof. Lemma 2.3( [25]) Let Ω ⊂ E be a bounded open set and A : Ω ∩ P → P is a completely continuous operator. If there exists v0 ∈ P \ {0} such that v − Av 6= λv0 for all v ∈ ∂Ω ∩ P and λ ≥ 0, then i(A, Ω ∩ P, P ) = 0. Lemma 2.4( [25]) Let Ω ⊂ E be a bounded open set with 0 ∈ Ω. Suppose A : Ω ∩ P → P is a completely continuous operator. If v 6= λAv for all v ∈ ∂Ω ∩ P and 0 ≤ λ ≤ 1, then i(A, Ω ∩ P, P ) = 1. Lemma 2.5 (Jensen’s integral inequalities) Let θ > 0 and f ∈ C([0, 1], R+ ). Then Z

1

Z θ Z 1 θ (f (t)) dt, θ ≥ 1, f (t)dt ≤ 0

0

3

1

θ Z 1 f (t)dt ≥ (f (t))θ dt, 0 < θ ≤ 1. 0

0

Main Results

For the reason of notational brevity, we denote by y = (y1 , y2 , . . . , yn−1 ) ∈ Rn−1 + , p∗ := min{1, p − 1}, p∗ := max{1, p − 1}, " βp :=

κp∗ −1

We now list our hypotheses. (H1) f ∈ C([0, 1] × Rn−1 + , R+ ).

1 Pn+1

−i i=3 λ1

# p−1 p

"

∗

, αp :=

κp∗ −1

1 Pn+1 i=3

# p−1 p∗ λ−i 1

.

53

Positive solutions

P n−1 p−1 (H2) There exist a1 > βp and c > 0 such that f (t, y) ≥ a1 ( n−1 − c for all y ∈ R+ and i=1 yi ) t ∈ [0, 1]. Pn−1 p−1 (H3) There exist b1 ∈ (0, αp ) and r > 0 such that f (t, y) ≤ b1 ( i=1 yi ) for all y ∈ [0, r]n−1 and t ∈ [0, 1]. Pn−1 p−1 (H4) There exist a2 > βp and r > 0 such that f (t, y) ≥ a2 ( i=1 yi ) for all y ∈ [0, r]n−1 and t ∈ [0, 1]. P p−1 (H5) There exist b2 ∈ (0, αp ) and c > 0 such that f (t, y) ≤ b2 ( n−1 + c for all y ∈ Rn−1 + i=1 yi ) and t ∈ [0, 1]. (H6) There are ζ > 0 and ς ∈

− p−1 p

such that the inequality f (t, y) ≤ ς p−1 ζ p−1 holds for

0, κ1

all y ∈ [0, ζ]n−1 and t ∈ [0, 1]. Theorem 3.1 Suppose that (H1)-(H3) are satisfied. Then (1.1) has at least one positive solution. Proof. Let M1 := {u ∈ P : u = Au + λψ for some λ ≥ 0}, where ψ(t) is determined by (2.3) and (2.4). We claim M1 is bounded. Indeed, u ∈ M1 implies u is concave and u(t) ≥ (Au)(t). Now, by Jensen’s inequality and (H2), we find 1

Z

p∗

Z

1

G1 (t, s)ϕq G1 (s, τ )f (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ u (t) ≥ 0 0 Z 1Z 1 p∗ G1 (t, s)G1 (s, τ )f p−1 (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ ds ≥ 0

ds

0 1

Z

p ∗

Z



1

≥

G1 (t, s)G1 (s, τ ) a1 0

0

p∗ p−1

Z

i=1 1

1

Z

≥ a1

0 p∗ p−1

Z

!p ∗ Z n−1 X p∗ p−1 dτ ds − c G1 (t, s)G1 (s, τ ) (Bi u)(τ )

0

0 p∗ p−1

1

0

i=1

1

Z

G1 (t, s)G1 (s, τ )dτ ds 0

Z 1 p∗ Z 1Z 1 p∗ dτ ds − c p−1 G1 (t, s)G1 (s, τ )dτ ds G1 (t, s)G1 (s, τ ) κ G (τ, z)u(z)dz 0 0 0 0 Z 1Z 1 Z 1Z 1Z 1 p∗ p∗ p−1 G1 (t, s)G1 (s, τ )dτ ds G1 (t, s)G1 (s, τ )G (s, τ )u (z)dzdτ ds − c

1

≥ a1

≥ κp∗ a1

p∗  p−1 !p−1 n−1 X − c dτ ds (Bi u)(τ )

1

Z

0 p∗

= κp∗ −1 a1p−1

0 0 Z n+1 1 X

0 p∗

Gi (t, s)up∗ (s)ds − c p−1

0

i=3

1

Z

0

1

Z

G1 (t, s)G1 (s, τ )dτ ds. 0

0

(3.1) Multiply the both sides of the above by ψ(t) and integrate over [0,1], and use (2.4) to obtain Z

1 p∗ −1

p∗

u (t)ψ(t)dt ≥ κ

p∗ p−1

a1

0

n+1 X

λ−i 1

and thus 0

p∗

p−1 up∗ (t)ψ(t)dt − λ−2 1 c

0

i=3

Z

1

Z

p∗

1

up∗ (t)ψ(t)dt ≤

p−1 λ−2 1 c p∗ Pn+1 p−1

κp∗ −1 a1

i=3

λ−i 1 −1

:= N1 .

54


Recall that every u ∈ M1 is concave and increasing on [0,1]. So is up∗ with p∗ ∈ (0, 1]. Now Lemma 2.2 yields kup∗ k ≤ ω −1 N1 (3.2) for all u ∈ M1 , which implies the boundedness of M1 , as claimed. Taking R > sup{kuk : u ∈ M1 }, we have u − Au 6= λψ, ∀u ∈ ∂BR ∩ P, λ ≥ 0. Now by virtue of Lemma 2.3, we obtain i(A, BR ∩ P, P ) = 0.

(3.3)

Let M2 := {u ∈ B r ∩ P : u = λAu for some λ ∈ [0, 1]}. We shall prove M2 = {0}. Indeed, if u ∈ M2 , we have for any u ∈ B r ∩ P Z 1 Z 1 G1 (t, s)ϕq G1 (s, τ )f (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ ds. u(t) ≤ 0

(3.4)

0

Now by (H3) and Jensen’s inequality, we obtain Z

p∗

u (t) ≤

1

Z

G1 (t, s)ϕq 0 1Z 1

Z ≤

1

p∗

G1 (s, τ )f (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ

ds

0 p∗

G1 (t, s)G1 (s, τ )f p−1 (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ ds 0

0

p∗

≤ b1p−1

1

Z

G1 (t, s)G1 (s, τ ) 0

p∗

1

Z

p∗ p−1

= κ b1

0

Z

∗

p∗

(Bi u)(τ )

dτ ds

1

Z

p ∗

1

G (τ, z)u(z)dz dτ ds G1 (t, s)G1 (s, τ ) 0 Z 1Z 1Z 1 ∗ G1 (t, s)G1 (s, τ )G (τ, z)up (z)dzdτ ds 0

≤ κp b1p−1

#p∗

i=1

1

Z

" n−1 X

0

0

= κp

∗ −1

p∗ p−1

b1

0 0 Z n+1 1 X i=3

∗

Gi (t, s)up (s)ds.

0

Multiply the both sides of the above by ψ(t) and integrate over [0, 1] and use (2.4) to obtain Z

1

p∗

p∗ −1

u (t)ψ(t)dt ≤ κ 0

p∗ p−1

b1

n+1 X

λ−i 1

Z

i=3

1

∗

up (t)ψ(t)dt.

0

R1 ∗ Therefore, 0 up (t)ψ(t)dt = 0, whence u(t) ≡ 0, ∀u ∈ M2 . As a result, M2 = {0}, as claimed. Consequently, u 6= λAu, ∀u ∈ ∂Br ∩ P, λ ∈ [0, 1]. Now Lemma 2.4 yields i(A, Br ∩ P, P ) = 1.

(3.5)

55

Positive solutions

Combining this with (3.3) gives i(A, (BR \B r ) ∩ P, P ) = 0 − 1 = −1. Hence the operator A has at least one fixed point on (BR \ B r ) ∩ P and therefore (1.1) has at least one positive solution. This completes the proof. Theorem 3.2 Suppose that (H1), (H4) and (H5) are satisfied. Then (1.1) has at least one positive solution. Proof. Let M3 := {u ∈ B r ∩ P : u = Au + λψ for some λ ≥ 0}, where ψ(t) is determined by (2.3) and (2.4). We claim M3 ⊂ {0}. Indeed, if u ∈ M3 , then we have u ≥ Au by definition. Now by (H4) and Jensen’s inequality, we obtain Z

p∗

u (t) ≥

1

G1 (t, s)ϕq 0 1Z 1

Z ≥ 0

p∗

≥ a2p−1

p∗


ds

0 p∗

G1 (t, s)G1 (s, τ )f p−1 (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ ds 0 " n−1 #p∗ Z 1Z 1 X G1 (t, s)G1 (s, τ ) (Bi u)(τ ) dτ ds 0

p∗

1

Z

0

= κ a2

i=1

1

Z

p∗ p−1

p∗

Z

p∗

1

G (τ, z)u(z)dz dτ ds G1 (t, s)G1 (s, τ ) 0 Z 1Z 1Z 1 G1 (t, s)G1 (s, τ )G (τ, z)up∗ (z)dzdτ ds 0

≥ κp∗ a2p−1

1

Z

0

0

0 0 Z n+1 1 X

p∗

= κp∗ −1 a2p−1

i=3

Gi (t, s)up∗ (s)ds.

0

Multiply the both sides of the above by ψ(t) and integrate over [0, 1], and use (2.4) to obtain Z

1 p∗

p∗ −1

u (t)ψ(t)dt ≥ κ 0

p∗ p−1

a2

n+1 X

λ−i 1

i=3

Z

1

up∗ (t)ψ(t)dt,

0

R1 so that 0 up∗ (t)ψ(t)dt = 0, whence u(t) ≡ 0, ∀u ∈ M3 . Therefore, we claim M3 ⊂ {0}. As a result of this, we have u − Au 6= λψ, ∀u ∈ ∂Br ∩ P, λ ≥ 0. Now Lemma 2.3 gives i(A, Br ∩ P, P ) = 0. Let M4 := {u ∈ P : u = λAu for some λ ∈ [0, 1]}.

(3.6)

56


We assert M4 is bounded. Indeed, if u ∈ M4 , then u is concave and u ≤ Au by definition. Now by (H5) and Jensen’s inequality, we obtain Z

p∗

u (t) ≤

1

Z

G1 (t, s)ϕq 0 1Z 1

Z ≤

1

p ∗


ds

0 p∗

G1 (t, s)G1 (s, τ )f p−1 (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ ds 0

0



1

1Z

Z ≤

n−1 X

G1 (t, s)G1 (s, τ ) b2 0 p∗ p−1

0 1

Z

n−1 X

1

Z

G1 (t, s)G1 (s, τ ) 0 p∗ p−1

= κ b3 p∗

(Bi u)(τ )

dτ ds

+ c

i=1

≤ b3

p∗

p∗  p−1

!p−1

p∗ p−1

≤ κ b3

0 1

Z

!p∗

p∗ p−1

(Bi u)(τ )

G1 (t, s)G1 (s, τ )dτ ds 0

Z

p∗

1

0 p∗ p−1

G (τ, z)u(z)dz dτ ds + c1 G1 (t, s)G1 (s, τ ) 0 0 0 Z 1Z 1Z 1 Z p∗ p−1 p∗ G1 (t, s)G1 (s, τ )G (τ, z)u (z)dzdτ ds + c1 0

p∗ −1

=κ

1

Z

dτ ds + c1

i=1 1

Z

1

Z

p∗ p−1

b3

0 0 n+1 Z 1 X

p∗ p−1

Z

Z

1

1

G1 (t, s)G1 (s, τ )dτ ds 0 1

Z

G1 (t, s)G1 (s, τ )dτ ds 0

1

G1 (t, s)G1 (s, τ )dτ ds

Gi (t, s)u (s)ds + c1

0

0

i=3

1

Z

0

0 p∗

1

Z

0

(3.7) for all u ∈ M4 , b3 ∈ (b2 , αp ) and c1 > 0 being chosen so that (b2 z + c)

p∗ p−1

p∗ p−1

≤ b3 z

p∗ p−1

p∗ p−1

+ c1 , ∀z ≥ 0.

Multiply the both sides of (3.7) by ψ(t) and integrate over [0, 1], and use (2.4) to obtain Z

1

p∗

p∗ −1

u (t)ψ(t)dt ≤ κ

p∗ p−1

b3

n+1 X

0

λ−i 1

Z

u (t)ψ(t)dt +

p−1 λ−2 1 c1

∗

up (t)ψ(t)dt ≤

0

p∗

1−

This, together with Jensen’s inequality and Z u(t)ψ(t)dt ≤ κ2 0

p−1 λ−2 1 c1

p∗

1

Z

1

p∗

p∗

0

i=3

and thus

Z

1

1

p∗

u (t) 0

ψ(t) κ2

κp∗ −1 b3p−1

ψ(t) κ2

dt

−i i=3 λ1

∈ [0, 1](κ2 := kψk), leads to

! p1∗

p∗

:= N2 .

Pn+1

Z ≤ κ2

1

p∗

u (t) 0

ψ(t) κ2

p1∗ 1 1− 1∗ ∗ dt ≤ κ2 p N2p (3.8)

for all u ∈ M4 . From Lemma 2.2, we find 1− p1∗

kuk ≤ ω −1 κ2

1 ∗

N2p , ∀u ∈ M4 .

57

Positive solutions

Now the boundedness of M4 , as asserted. Taking R > sup{kuk : u ∈ M4 }, we have u 6= λAu, ∀u ∈ ∂BR ∩ P, λ ∈ [0, 1]. Now Lemma 2.4 yields i(A, BR ∩ P, P ) = 1.

(3.9)

Combining this with (3.6) gives i(A, (BR \B r ) ∩ P, P ) = 1 − 0 = 1. Hence the operator A has at least one fixed point on (BR \ B r ) ∩ P and therefore (1.1) has at least one positive solution. This completes the proof. Theorem 3.3 Suppose that (H1), (H2), (H4) and (H6) are satisfied. Then (1.1) has at least two positive solutions. Proof. By (H6), we have Z 1 Z 1 kAuk = max G1 (t, s)ϕq G1 (s, τ )f (τ, (Bn−1 u)(τ ), (Bn−2 u)(τ ), . . . , (B1 u)(τ ))dτ ds t∈[0,1] 0 0 Z 1 Z 1 p−1 p−1 ≤ κ1 ϕq κ1 ς ζ dτ ds < ζ 0

0

(3.10) and thus kAuk < kuk for all u ∈ Bζ ∩ P , so that u 6= λAu, ∀u ∈ ∂Bζ ∩ P, λ ∈ [0, 1]. Now Lemma 2.4 yields i(A, Bζ ∩ P, P ) = 1.

(3.11)

On the other hand, in view of (H2) and (H4), we may choose R > ζ and r ∈ (0, ζ) so that (3.3) and (3.6) hold (see the proofs of Theorem 3.1 and 3.2). Combining (3.3), (3.6) and (3.11), we obtain i(A, (BR \B ζ ) ∩ P, P ) = 0 − 1 = −1, i(A, (Bζ \B r ) ∩ P, P ) = 1 − 0 = 1. Hence A has at least two fixed points, one on(BR \ B ζ ) ∩ P and the other on (Bζ \ B r ) ∩ P . This proves that (1.1) has at least two positive solutions. The proof is completed.

4

Open Problem

In [19], Eloe considered the nonlinear Lidstone boundary value problem (with a(t) continuous and nonnegative) ( (−1)n u(2n) = λa(t)f (t, u, −u00 , . . . , (−1)n−1 u(2n−2) ), u(2i) (0) = u(2i) (1) = 0, i = 0, 1, 2, . . . , n − 1, where f ∈ C([0, 1] × Rn+ , R+ ) and λ > 0 is a real parameter. He posed an open question in this way: ”Can the methods employed here apply to a Lidstone BVP with nonlinear dependence on

58


odd order derivatives of the unknown function ?” That question is completely open. The problem is that large in norm does not imply large componentwise; by exploiting the nested feature of Lidstone BVPs in [19], large in norm will, in fact, imply large in the appropriate components. In this paper, we only answer the question partly by considering the simple case a(t) := 1. However, the resulting problem is the Lidstone problems with p-Laplacian operator, furthermore, some connections between (1.1) and (1.5) are established by repeatedly invoking Jensen’s integral inequality. This, together with the fact that our nonlinearity f may be of distinct growth, means that our methodology and results in this paper are entirely new in the existing literature.

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[13] F. Cong, Existence of periodic solutions of (2n + 1)th-order ordinary differential equations, Appl. Math. Lett. 17 (2004) 727-732. [14] J. Davis, P. Eloe, J. Henderson, Triple positive solutions and dependence on higher order derivatives, J. Math. Anal. Appl. 237 (1999) 710-720. [15] M. Del Pino, R. Manasevich, Existence for a fourth-order nonlinear boundary problem under a twoparameter nonresonance condition, Proc. Amer. Math. Soc. 112 (1991) 81-86. [16] C. De Coster, C. Fabry, F. Munyamarere, Nonresonance conditions for fourth-order nonlinear boundary value problems, Int. J. Math. Math. Sci., 17 (1994) 725-740. [17] K. Deimling, Nonlinear Functional Analysis, Spring-Verlag, Berlin, 1980. [18] J. Ehme, J. Henderson, Existence and local uniqueness for nonlinear Lidstone boundary value problems, J. Inequ. Pure and Appl. Math. 1 (2000), Article 8. [19] P. Eloe, Nonlinear eigenvalue problems for higher order Lidstone boundary value problems, E. J. Qualitative Theory of Diff. Equ. 2 (2000) 1-8. [20] J. Graef, B. Yang, Positive solutions to a multi-point higher order boundary value problems, J. Math. Anal. Appl. 316 (2006) 409-421. [21] J. Graef, L. Kong, Necessary and sufficient conditions for the existence of symmetric positive solutions of singular boundary value problems, J. Math. Anal. Appl. 331 (2007) 1467-1484. [22] Y. Guo, W. Ge, Three positive solutions for the one-dimensional p-Laplacian, J. Math. Anal. Appl. 286 (2003) 491-508. [23] Z. Guo, J. Yin, Y. Ke, Multiplicity of positive solutions for a fourth-order quasilinear singular differential equation, E. J. Qualitative Theory of Diff. Equ. 27 (2010) 1-15. [24] Y. Guo, W. Ge, Twin positive symmetric solutions for Lidstone boundary value problems, Taiwanese Journal of Mathematics, 8 (2004) 271-283. [25] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Orlando, 1988. [26] F. Li, Y. Li, Z. Liang, Existence and multiplicity of solutions to 2mth-order ordinary differential equations, J. Math. Anal. Appl. 331 (2007) 958-977. [27] F. Li, Z. Liu, Multiple positive solutions of some operators and applications, Acta Mathematica Sinica, Chinese Series, 41 (1998) 97-102. [28] Z. Liu, F. Li, Multiple positive solutions of nonlinear two-point value problems, J. Math. Anal. Appl. 203 (1996) 610-625. [29] J. Li, J. Shen, Existence of three positive solutions for boundary value problems with pLaplacian, J. Math. Anal. Appl. 311 (2005) 457-465.

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[30] A. Lakmeche, A. Hammoudi, Multiple positive solutions of the one-dimensional p-Laplacian, J. Math. Anal. Appl. 317 (2006) 43-49. [31] Y. Ma, Existence of positive solutions of Lidstone boundary value problems, J. Math. Aanal. Appl. 314 (2006) 97-108. [32] S. Odda, Positive solutions for boundary value problems of higher order differential equations, International Mathematical Forum, 5 (2010) 2131-2136. [33] C. Pang, W. Dong, Z. Wei, Green’s function and positive solutions of nth order m-point boundary value problem, Appl. Math. Comput. 182 (2006) 1231-1239. [34] Y. Wang, On 2nth-order Lidstone boundary value problems, J. Math. Anal. Appl. 312 (2005) 383-400. [35] Y. Wang, C. Hou, Existence of multiple positive solutions for one-dimensional p-Laplacian, J. Math. Anal. Appl. 315 (2006) 144-153. [36] H. Wang, On the existence of positive solutions for nonlinear equations in the annulus, J. Differential Equation, 109 (1994) 1-7. [37] J. Xu, Z. Yang, Positive solutions of boundary value problem for system of nonlinear nth order ordinary differential equations, J. Sys. Sci. & Math. Scis. 30 (5) (2010) 633-641. [38] J. Xu, Z. Yang, Three positive solutions for a system of singular generalized Lidstone problems, Electron. J. Diff. Equ. 163 (2009) 1-9. [39] J. Xu, Z. Yang, Positive solutions for a system of generalized Lidstone problems, J. Appl. Math. Comput. 37 (2011) 13-35. [40] J. Xu, Z. Yang, Positive solutions for a fourth order p-Laplacian boundary value problem, Nonlinear Anal. 74 (2011) 2612-2623. [41] D. Xie, C. Bai, Y. Liu, C. Wang, Positive solutions for nonlinear semipositone nth-order boundary value problems, E. J. Qualitative Theory of Diff. Equ. 7 (2008) 1-12. [42] Z. Yang, Existence and uniqueness of psoitive solutions for a higher order boundary value problem, Comput. Math. Appl. 54 (2007) 220-228. [43] J. Yang, Z. Wei, Existence of positive solutions for fourth-order m-point boundary value problems with a one-dimensional p-Laplacian operator, Nonlinear Anal. 71 (2009) 29852996. [44] X. Zhang, L. Liu, Positive solutions of fourth-order four-point boundary value problems with p-Laplacian operator, J. Math. Anal. Appl. 336 (2007) 1414-1423. [45] X. Zhang, L. Liu, A necessary and sufficient condition for positive solutions for fourthorder multi-point boundary value problems with p-Laplacian, Nonlinear Anal. 68 (2008) 3127-3137.

Positive solutions

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The Existence of Intrinsic Set Properties Implies Cantor’s Theorem. The concept of Cardinal Revisited. Juan-Esteban Palomar Taranc´ on Dep. Math. Institut Jaume I, Burriana - (Castellón) Spain e-mail: [email protected]

Abstract The author introduces the concept of intrinsic set property, by means of which the well-known Cantor’s Theorem can be deduced. As a natural consequence of this fact, it is proved that Cantor’s Theorem need not imply the existence of a tower of different-size infinities, because the impossibility of defining a bijection between any infinite countable set and its power can be a consequence of the existence of any intrinsic property which does not depend on size.

Keywords: Cardinal, actual infinity, computability 1 .

1

Introduction

Roughly speaking, what we term intrinsic set properties are those predicates, being satisfied by some sets, which are preserved under bijections. Thus, if an intrinsic set property P is satisfied by a set X, then it is also satisfied by any other set Y having the same cardinal as X. Recall that two sets have the same cardinal if and only if there is a bijection between them. Thus, cardinality is the most natural intrinsic set property, because of it is a concept that cannot be separated from the bijection notion. Other intrinsic property of any set is its size. Indeed, size is preserved under bijections. It is worth mentioning that, although size and cardinality are frequently regarded as synonymous, in this paper cannot be identified. 1

Mathematical Subject classification: 03E10,03E65

Intrinsic Properties

63

Cardinals are defined by means of bijective maps; therefore the concept of cardinal is comparative. In fact, cardinals are equivalence-classes, and the corresponding equivalence relation C is cardinality, which is defined as follows. For every couple of sets X and Y , the pair (X, Y ) lies in C if and only if there is a bijection f : X → Y . Usually this relation is denoted writing #(X) = #(Y ). However, assuming the existence of at least one couple of sets X and Y such that the existence of a bijection between them is an undecidable question, then the cardinals of both X and Y cannot be compared, but we can suppose that, although there were no possibility of discerning whether or not the equality #(X) = #(Y ) holds, each of these sets has a certain size. Of course, for ordinary finite sets cardinality and size can be always identified, but in our context we are assuming that there are sets the cardinals of which cannot be compared. Nevertheless, even when it is a non-accessible attribute, size is an intrinsic set property which cannot be disregarded without forgetting the identity of the considered set. To analyze these concepts deeply, consider the following statements. A1: “There is a bijection between X and Y ”. A2: “Both sets X and Y have the same size”. The first predicate is equivalent to the relation #(X) = #(Y ), since bijectionexistence and cardinal-equality are synonymous. With these statements, the well-known Hume’s principle can be stated as follows. A3: There is a bijection between X and Y if and only if both sets are equal in size. This equivalence can be split into the following implications. A3a: If there is a bijection between X and Y , then both sets are equal in size. A3b: If both sets X and Y are equal in size, then there is a bijection between them. The later is equivalent to the following one. A3bb: If there is no bijections between X and Y , then both sets are different in size. Statement A3a is evident; but, in general, both Statement A3b and Statement A3bb cannot be assumed without proving them. On the one hand, suppose that each member of a set X = {x1 , x2 . . . xn } is defined by means of an infinite collection of predicates and there is no algorithm handling them. Under these circumstances it is not clear that a bijection can be defined between X and In = {1, 2 . . . n}, since to define a bijection it is required to discern

64

J. E. Palomar

whether or not two members of X are the same. If there is no algorithm by means of which they can be distinguished, it is not clear that a bijection can be defined by some finite procedure. Even the existence of a bijection γ : X → In can be an undecidable question. In fact, the possibility of defining a bijection only is guaranteed whenever the members of the involved sets can be determined or approached arbitrarily, by means of some finite procedure or algorithm. If a general bijection theory dealing only with bijection existence is consistent, then, according to Gödel’s theorem [6], such a theory cannot be complete, that is to say, it must contain at least one undecidable proposition. Such an undecidability can involve the existence of some bijection, and consequently the cardinals of the corresponding sets cannot be compared. If this is the case, the nonexistence of a bijection need not imply size-inequality, that is to say, Statement A3bb need not be true. To illustrate this fact, consider a real number r lying in the unit interval [0, 1], and suppose that the decimal notation of r is 0.c1 c2 . . . cn . . . . If no previous constrain is imposed, then, in order to construct r, the figures cn can be chosen at random; however to obtain a rational number lying in the unit interval the method is not admissible. Thus, the possibility of constructing a real number by this method is a property that no rational one can satisfy. It is not difficult to see, that the randomness nature is preserved under bijections; consequently, no bijection can translate random constructions from R ∩ [0, 1] into Q ∩ [0, 1], and proofs and inferences can be based upon this construction method. By contrast, for other purposes any result can be biased depending on the assumed construction method, for instance see [9]. On the other hand, consider that bijections are structure-free set isomorphisms. An isomorphism between two structured sets X and Y can only be defined provided that both X and Y satisfy the same structure properties. For instance, if G4 = (Z4 , +) is the cyclic group of order 4, and P4 = (Z2 , +)×(Z2 , +) is the direct product of two instances of (Z2 , +), then both G4 and P4 consist of 4 members, but there is no isomorphism between them. However, if the algebraic structure of both groups is forgotten, then there is a bijection, that is, a set-isomorphism between the corresponding underlying sets. Accordingly, bijection existence is possible when structures are forgotten. However, set-size is an unforgettable property, and this is why bijection existence depends on size, and set-size is an intrinsic set property. The aim of this paper consists of proving the existence of another intrinsic or unforgettable set property, namely, T -representability, and consequently the nonexistence of a bijection between two arbitrary sets need not be always a consequence of size-difference. At least, to assume Statement A3b a proof must be required, and this is an open problem. After these considerations the following definition is adequate.


65

Definition 1.1 A property P of a set X is forgettable, provided that there is at least one set Y which does not satisfy P together with a bijection f : X → Y . From now on, let us say every non-forgettable property P of any set X to be intrinsic. Lemma 1.2 If a set property P is preserved under bijections, then it is intrinsic. Proof. If P is a property of a class of sets C which is preserved under bijections, then for every set Y and any member X of C, the existence of a bijection f : X → Y implies that Y satisfies P; consequently, P cannot be forgettable. Lemma 1.3 If {Pi |i ∈ I} is a class of intrinsic set-properties, then their conjunction P = ∧i∈I Pi is again an intrinsic one. Proof. It is a straightforward consequence of Definition 1.1. In Theorem 2.3 it is shown the existence of an intrinsic set property, namely, T -representability, being satisfied by N that its power P(N) does not satisfy. Likewise, in Corollary 2.4 Cantor’ Theorem is derived from T -representability; consequently the existence of a tower of different infinities need not be interpreted as the existence of different-size infinities. Perhaps the only difference consists of T -representability. Discerning which is the correct case is an open problem which is proposed in section 3.

2


Henceforth, say a set T = {Ti | i ∈ I} of Turing machines, indexed by a no-void subset I of N, to be regular provided that the following axioms hold. Axiom 1: All members of T have the same finite tape-alphabet Γ, the same blank ∅ ∈ Γ, the same initial state q0 and the same one-element final state set F = {Halt}. In addition, the tape of each member of T is infinite. Axiom 2: The tape alphabet Γ contains the n figures Bn = {0, 1, 2 . . . n} of a base-n numeral system. Notation. To denote a tape-configuration a1 a2 a3 . . . an together with the underlying state q of any Turing machine T , write (a1 a2 a3 . . . an ; q). Recall that, in general, blanks are disregarded, hence only those blanks lying between two non-blank symbols will be considered. In addition, denote as Γ∗ the freemonoid generated by tape-alphabet Γ; and likewise, denote as ΓN the collection of all sequences in Γ.

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Definition. With the same notations as in the preceding axioms, given a regular set of Turing machines T , say a nonempty set X to be T -representable, provided that there are two maps ϕ : X → Γ∗ and τ : X → T together with an injective one Λ : X → ΓN satisfying the following condition. Condition 1: For each x ∈ X, if ϕ(x) = ax,1 ax,2 . . . ax,k , then for every integer j ∈ N, the base-n expression of which is c1 c2 . . . cm , and from the initial tape-configuration (c1 c2 . . . cm ∅ax,1 ax,2 . . . ax,k ; q0 ), after a finite step sequence the Turing machine τ (x) = Tix gets the final state Halt together with a tapeconfiguration (e1 e2 . . . el ∅bx,1 bx,2 . . . bx,s ; Halt) such that, if Λ(x) = r1 r2 r3 · · · ∈ ΓN then ∀m ≤ j : bx,m = rm . It is worth pointing out, that T -representability for a set X implies that there is some algorithm, which is performed by a member T of T , by means of which, for every positive integer j, the member rj of the sequence Λ(x) = √ r1 r2 r3 . . . can be computed. For instance, the set X = { n | n ∈ N} is T -representable, whenever there is T ∈ T the corresponding algorithm of which computes the digits of the square root of any integer, in any base-m numeral √ system. In this case, the injective mapping ϕ : X → Γ∗ sends each expression s1 s2 s3 . . . denoting x in the member x of X into the corresponding √ base-m √ numeral system, while Λ ( n) = r1 r2 r3 . . . is the numeric expression for n in the base-m numeral system. Analogously, the set A of all algebraic numbers is T -representable with respect to any regular Turing-machine set T such that some member T of which can perform the following actions. On the one hand, T can compute arbitrary numeric approaches for every solution x of any algebraic equation. On the other hand, for each positive integer j, T can compute an approach for x containing at least j-digits. Indeed, at the initial state, the tape-configuration of T contains the expression for the integer j together with the equation coefficients. Lemma 2.1 Let T be a regular set of Turing machines. Every nonempty subset Y of each T -representable one X is again T -representable. Proof. It is a straightforward consequence of the preceding definition. Lemma 2.2 For every regular set T of Turing machines, T -representability is preserved under bijections; consequently, by virtue of Lemma 1.2, it is an intrinsic property. Proof. Let X be a T -representable set and f : X → Y a bijection. By hypothesis, there are two maps ϕ : X → Γ∗ , and τ : X → T together with an injective one Λ : X → ΓN satisfying Condition 1. It is not difficult to see,

67


that the three maps ϕ ◦ f −1 : Y → Γ∗ , Λ ◦ f −1 : Y → ΓN and τ ◦ f −1 : Y → T satisfy Condition 1 too. Remark. Lemma 2.1 implies that the concept of T -representability does not depend on set-size, since it is a property inherited by all proper nonempty subsets. Likewise, the preceding lemma shows that it is an intrinsic property. Thus, it is an instance of intrinsic property which does not depend on size. Theorem 2.3 The following statements are true. 1. There is a regular set of Turing machines T such that the set N of all natural numbers is T -representable. 2. There is a regular set of Turing machines T such that the set Q of all rational numbers is T -representable. 3. If a set X is T -representable, then there is an injective map γ : X → N, for every regular set of Turing machines T . 4. There is no regular Turing-machine set T , such that the powerset P(N) is T -representable. 5. There is no regular Turing-machine set T , such that the unit interval [0, 1] is T -representable. Proof. 1. Let B = {c1 c2 . . . cn } be the digits of a base-n numeral system. For every m ∈ N, let ϕ(m) = c1 c2 . . . ck ∈ Γ∗ be the expression for m in the numeral system B, and let Λ(m) = c1 c2 . . . ck ∅∅ · · · ∈ ΓN . If T is a Turing machine such that from the initial state q0 changes to the final one Halt, remaining its tape unaltered, then Condition 1 is satisfied. Thus, Statement 1) holds for any regular Turing-machine set T containing T . 2. Let B2 = {0, 1} ⊆ Γ the figures of the binary numeral system, and Λ the n into its binary expression map sending each rational number m n Λ = r1 r2 . . . rn .rn+1 rn+2 . . . m Let ϕ

n

= am,1 am,2 . . . am,k ∅an,k+1 an,k+2 . . . an,k+j m where the sequences am,1 am,2 . . . am,k and an,k+1 an,k+2 . . . an,k+j are the n be a Turing binary expressions for n and m respectively. Let T = τ m machine performing the ordinary division algorithm working with binary digits. Assume that the algorithm T can determine the j-th figure of

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J. E. Palomar

every quotient, for every positive integer j. With these assumptions, from the initial configuration (c1 c2 . . . cs ∅am,1 am,2 . . . am,k ∅an,k+1 an,k+2 . . . an,k+j ; q0 ) where c1 c2 . . . c2 is the binary expression for j, by means of the division algorithm performed by T , it is possible to obtain the j-th digit of n Λ m , for every positive integer j ∈ N, and Condition 1 can be satisfied. Accordingly, for every regular Turing machine set T containing such an algorithm T , the set Q of rational numbers is T -representable. 3. Since it is assumed X to be T -representable, then, by definition, for every x ∈ X there are a Turing machine τ (x) = Tix ∈ T and a word ϕ(x) = ax,1 ax,2 . . . ax,k ∈ Γ∗ satisfying Condition 1. Likewise, since T is indexed by a subset I of N, there is an injective map ξ : T → {pi | i ∈ N}; where {pi | i ∈ N} is the collection of all primes. With these assumptions, the injective map γ : X → N can be defined as follows. β(a )β(a )...β(ax,k ) γ(x) = pix x,1 x,2 ∈N where pix = ξ(Tix ) and β : Γ → C is any bijection onto the figure set C of a base-n numeral system. Recall, that since Γ is assumed to be finite, then there is such a bijection, provided that #(C) = #(Γ). It remains to be shown that γ is injective. To this end, consider that for each couple x and y of members of X, the equality β(ax,1 )β(ax,2 )...β(ax,k )

γ(x) = pix

β(ay,1 )β(ay,2 )...β(ay,k )

= piy

= γ(y)

(1)

implies pix = piy , and because of ξ is injective, then Tix = Tiy ; accordingly τ (x) = τ (y). In addition, since both pix and piy are primes, then equation (1) implies that both exponents must be the same; therefore ∀j ≤ k : β(ax,j ) = β(ay,j ), and because β is bijective, then it follows that ∀j ≤ k : ax,j = ay,j ; hence ϕ(x) = ϕ(y). Since Tix = Tiy , by virtue of Condition 1, if Λ(x) = a1 a2 a3 . . . and Λ(y) = b1 b2 b3 . . . , then for every j ∈ N: aj = bj ; consequently Λ(x) = Λ(y). Now, taking into account, that by definition, the map Λ is injective; it follows that x = y, and γ is also injective. 4. By virtue of the preceding statement, the existence of a regular Turing machine set T , such that the powerset P(N) is T -representable, implies the existence of an injective map γ : P(N) → N. Thus, if λ : img(γ) → N is the canonical inclusion, since γ is injective, then the map γ ∗ : P(N) → img(γ) determined by γ = λ ◦ γ ∗ is bijective, consequently there is a bijection between P(N) and a subset of N which contradicts Cantor’s Theorem. Recall that since λ is the canonical inclusion, then it is injective; hence γ ∗ is uniquely specified by γ = λ ◦ γ ∗ .


69

5. First, we prove that there is a bijection δ : [0, 1] → P(N). For every x ∈ [0, 1], let 0, cx,1 cx,2 . . . cx,n . . . be the expression for x in the binary numeral system; hence ∀xj ∈ N : cx,j ∈ {0, 1}. Now, letting δ(x) = {n ∈ N | cx,n = 1} the mapping δ is bijective; hence, by virtue of Lemma 2.2, if [0, 1] is T -representable, then so is P(N), which contradicts the former statement. Remark. Both Statement 3) and Statement 4) in the preceding theorem state some relationship between countability and T -representability, and such a relationship can be understood. To this end, consider any infinite subset X of the unit interval [0, 1]. If X can be defined by means of a finite set of conditions, in general, its members can be constructed through some procedure, which involves some effective algorithm. However, it is possible to build members of [0, 1] at random. For instance, one can write the binary expression for a member x of [0, 1] tossing a coin in an endless process, that is to say, the n-th digit of the binary expression for x is 0 or 1 depending on the obtained result. For any member x = 0.c1 c2 c3 . . . of an infinite subset of [0, 1] being built at random, in general, it is not possible to define an algorithm satisfying Condition 1 in order to determine its digits, because, there is no predictable pattern in the corresponding digit sequence 0.c1 c2 c3 . . . . By contrast, Condition 1 implies that every member x of X is determined by a finite set of initial data a1 a2 . . . an ; hence the expression for x in terms of tape-alphabet symbols cannot be obtained at random; consequently Condition 1 cannot be satisfied by [0, 1]. In addition, to define a bijection by means of a finite procedure it is required to discern whether or not two members of [0, 1] are the same by means of a finite method, because under bijections different points must have different images. Notice, that for each couple of rational numbers x1 = 0.c1 c2 . . . cn . . . and x2 = 0.d1 d2 . . . dn . . . there is a positive integer Nx1 ,x2 such that the predicate ∀n ≤ Nx1 ,x2 : cn = dn implies x1 = x2 . By contrast, for all members of [0, 1] this criterion does not work properly, since the possibility of defining the figures c1 , c2 , c3 . . . at random excludes the existence of any predictable pattern. Corollary 2.4 Cantor’s Theorem can be derived from the conjunction of Statement 1) and Statement 4) in the preceding theorem. Proof. Let us assume that both Statement 1) Statement 4) in the preceding theorem are true for some regular Turing machine set T . If there were a bijection f : N → P(N), by virtue of Lemma 2.2 together with Statement 1) in the preceding theorem, P(N) would be T -representable, which contradicts Statement 4); consequently there is no bijection between N and P(N), and Cantor’s Theorem holds.

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Open Problem

From Aristotle to Gauss and Pointcaré, the existence of actual infinity is frequently rejected. For instance consider the following quotation from Henri Poincaré (1854-1912). Actual infinity does not exist. What we call infinite is only the endless possibility of creating new objects no matter how many exist already. However, Cantor’s Theorem introduces a tower of different infinities. Nevertheless, the axiomatic set theorists state that all existing definitions of “actual infinity” are unsatisfactory and have no role to play in serious logical and mathematical analysis, for instance see [1] or [11]. In fact, there is some opposition against the negation of the uniqueness of infinity, or more accurately, against the assumption that an infinity could be surpassed by another. To clarify these ideas, Corollary 2.4 provides a version of Cantor’s Theorem deduced from the existence of an intrinsic property, namely, T -representability. If Cantor’s Theorem is induced by any intrinsic property, this fact need not imply the existence of two infinities being different in size, but it implies one of the following statements. 1. Both N and P(N) have the same size, and for only one of them there is a regular Turing-machine set T with respect to which is T -representable. 2. N and P(N) have different size, and for only one of them there is a regular Turing-machine set T with respect to which is T -representable. Thus, Corollary 2.4 implies that the existence of two infinities being different in size is not an unavoidable consequence of Cantor’s Theorem, but in order to solve this question it must be proved which of the preceding statements holds, and this is an open problem. Finally, since there are other cardinals apart from #(N) and # (P(N)), for instance, the Woodin or Shelah cardinals [10], it must be proved whether or not the existence of each of which can be a consequence of a specific intrinsic set property, and this is another open problem.

References [1] A. A. Fraenkel and Y. Bar-Hillel, Foundation of set theory, North-Holland Publishing Company, Amsterdam, (1958). [2] Frege, G. and J.L. Austin (trans.), The Foundations of Arithmetic, Northwestern University Press, (1884).


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[3] Krzysztof, C. Set theory for the working mathematician, London Mathematical Society Student Texts, 39, Cambridge University Press, (1997). [4] Halmos, P., (1974).

Naive set theory. Princeton, Springer-Verlag, New York,

[5] Mayberry, J. P. The foundations of mathematics in the theory of sets, Encyclopedia of Mathematics and its Applications, 82, Cambridge University Press, (2000). [6] Smorynski, C., The incompleteness theorems, Handbook of Mathematical Logic, North-Holland, pp. 821-866, (1982). [7] Sridevi, B. and Nadarajan, R., “Fuzzy Similarity Measure for Generalized Fuzzy Numbers”, Int. J. Open Problems Compt. Math., Vol. 2, No. 2, pp. 242-253, (2009). [8] Turing, A.M “On Computable Numbers, with an Application to the Entscheidungsproblem”, Proceedings of the London Mathematical Society. 42: 230-265. (1937). [9] Wang, J. and Jackson, R., “Resolving Bertrand’s Probability Paradox”, Int. J. Open Problems Compt. Math., Vol. 4, No. 3, pp. 73-103, (2011). [10] Schimmerling, E., “Woodin cardinals, Shelah cardinals and the MitchellSteel core model”, Proceedings of the American Mathematical Society 130/11, pp. 3385-3391, (2002). [11] Zenkin, A. A., “Logic of actual infinity and G. Cantor’s diagonal proof of the uncountability of the continuum”, Rev. Mod. Log., (9),(2004).


Existence of Multiple Positive Solutions for 3nth Order Three-Point Boundary Value Problems on Time Scales K. R. Prasad1 and N. Sreedhar2 1

Department of Applied Mathematics, Andhra University, Visakhapatnam, 530 003, India. e-mail:[email protected] 2 Department of Mathematics, GITAM University, Visakhapatnam, 530 045, India. e-mail:[email protected]

Abstract We establish the existence of at least three positive solutions for the 3nth order three-point boundary value problem on time scales by using LeggettWilliams fixed point theorem. We also establish the existence of at least 2m − 1 positive solutions for an arbitrary positive integer m. Keywords: Boundary value problem, Cone, Green’s function, Positive solution, Time scale. AMS Subject Classification: 39A10, 34B05.

1

Introduction

In this paper, we establish the existence of multiple positive solutions for the 3nth order boundary value problem on time scales, (−1)n y ∆

(3n)

(t) = f (t, y(t)),

t ∈ [t1 , σ(t3 )],

(1.1)

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Existence of Multiple Positive Solutions

satisfying the general three-point boundary conditions, α3i−2,1 y ∆

(3i−3)

(t1 ) + α3i−2,2 y ∆

(3i−2)

(t1 ) + α3i−2,3 y ∆

(3i−1)

α3i−1,1 y ∆

(3i−3)

(t2 ) + α3i−1,2 y ∆

(3i−2)

(t2 ) + α3i−1,3 y ∆

(3i−1)

α3i,1 y

∆(3i−3)

(σ(t3 )) + α3i,2 y

∆(3i−2)

(σ(t3 )) + α3i,3 y

∆(3i−1)

 (t1 ) = 0,  

(t2 ) = 0,    (σ(t3 )) = 0,

(1.2)

for 1 ≤ i ≤ n, where n ≥ 1, α3i−2,j , α3i−1,j , α3i,j , for j = 1, 2, 3, are real constants, t1 < t2 < σ(t3 ). We assume that f : [t1 , σ(t3 )]×R+ → R+ is continuous and f (t, ·) does not vanish identically on any subset of [t1 , σ(t3 )]. The study of the existence of positive solutions of the higher order boundary value problems (BVPs) arises in various fields of applied mathematics and physics. BVPs describe many phenomena in the applied mathematical sciences, which can be found in the theory of nonlinear diffusion generated by nonlinear sources, in thermal ignition of gases and in chemical or biological problems. In these applied settings only positive solutions are meaningful. Recently, there has been much attention focussed on existence of positive solutions to the BVPs on time scales due to their striking applications to almost all area of science, engineering and technology. Studying BVPs on time scales will unify the theory of differential and difference equations and provide an accurate information of phenomena that manifest themselves partly in continuous time and partly in discrete time. The existence of positive solutions are studied by many authors. A few papers along these lines are Agarwal and Regan [1], Anderson [4], Anderson and Avery [6], F. M. Atici and G. Sh.Guseinov [7], Kaufmann [19] and Sun [20]. For convenience, we use the following notations. For 1 ≤ i ≤ n, let us denote, βij = α3i−3+j,1 tj + α3i−3+j,2 , γij = α3i−3+j,1 t2j + α3i−3+j,2 (tj + σ(tj )) + 2α3i−3+j,3 , for j = 1, 2; βi3 = α3i,1 σ(t3 ) + α3i,2 and γi3 = α3i,1 (σ(t3 ))2 + α3i,2 (σ(t3 ) + σ 2 (t3 )) + 2α3i,3 . Also, for 1 ≤ i ≤ n, we define, mijk =

α3i−3+j,1 γik − α3i−3+k,1 γij , 2(α3i−3+j,1 βik − α3i−3+k,1 βij )

Mijk =

βij γik − βik γij , α3i−3+j,1 βik − α3i−3+k,1 βij

for j, k =1, 2, 3 and let pi = max{mi12 , mi13 , mi23 }, q q qi = min mi23 + m2i23 − Mi23 , mi13 + m2i13 − Mi13 , di = α3i−2,1 (βi2 γi3 − βi3 γi2 ) − βi1 (α3i−1,1 γi3 − α3i,1 γi2 ) + γi1 (α3i−1,1 βi3 − α3i,1 βi2 ) and lij = α3i−3+j,1 σ(s)σ 2 (s) − βij (σ(s) + σ 2 (s)) + γij , where j = 1, 2, 3. We assume the following conditions throughout this paper: (A1) α3i−2,1 > 0, α3i−1,1 > 0, α3i,1 > 0 and for all 1 ≤ i ≤ n,

α3i,2 α3i,1

>

α3i−1,2 α3i−1,1

>

α3i−2,2 , α3i−2,1

2 (A2) pi ≤ t1 < t2 < σ(t3 ) ≤ qi and 2α3i−2,3 α3i−2,1 > α3i−2,2 , 2 2 2α3i−1,3 α3i−1,1 < α3i−1,2 , 2α3i,3 α3i,1 > α3i,2 , for all 1 ≤ i ≤ n,

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K. R. Prasad and N. Sreedhar

(A3) m2i23 > Mi23 , m2i12 < Mi12 , m2i13 > Mi13 and di > 0, for all 1 ≤ i ≤ n, (A4) The point t ∈ [t1 , σ(t3 )] is not left dense and right scattered at the same time. This paper is organized as follows. In Section 2, we establish certain lemmas which are needed in our main results. In Section 3, we establish the existence of at least three positive solutions of the BVP (1.1)-(1.2) by using the LeggettWilliams fixed point theorem. We also establish the existence of at least 2m−1 positive solutions of the BVP (1.1)-(1.2) for an arbitrary positive integer m. Finally as an application, we give an example to illustrate our result.

2

The Green’s Function and Bounds

In this section, we construct the Green’s function for the homogeneous problem corresponding to (1.1)-(1.2) and estimate bounds for the Green’s function. Let Gi (t, s) be the Green’s function for the homogeneous BVP, 3

−y ∆ (t) = 0,

t ∈ [t1 , σ(t3 )],

(2.1)

satisfying the general three-point boundary conditions, ∆

α3i−2,1 y(t1 ) + α3i−2,2 y (t1 ) + α3i−2,3 y

∆2 2

 (t1 ) = 0,  

α3i−1,1 y(t2 ) + α3i−1,2 y ∆ (t2 ) + α3i−1,3 y ∆ (t2 ) = 0,   2 α3i,1 y(σ(t3 )) + α3i,2 y ∆ (σ(t3 )) + α3i,3 y ∆ (σ(t3 )) = 0,

(2.2)

for 1 ≤ i ≤ n. Lemma 2.1 For 1 ≤ i ≤ n, the Green’s function Gi (t, s) for the homogeneous BVP (2.1)-(2.2) is given by    Gi1 (t, s), t1 < σ(s) < t ≤ t2 < σ(t3 )    Gi (t,s)  Gi (t, s), t1 ≤ t < s < t2 < σ(t3 )  t∈[t1 ,t2 ] =   2   Gi3 (t, s), t1 ≤ t < t2 < s < σ(t3 )  Gi (t, s) = (2.3)    G (t, s), t < t < σ(s) < t ≤ σ(t )   i4 1 2 3   Gi (t,s)  G (t, s), t < t ≤ t < s < σ(t ) =  1 2 3   t∈[t2 ,σ(t3 )]  i5 Gi6 (t, s), t1 ≤ σ(s) < t2 < t < σ(t3 ) where 1 [−(βi2 γi3 − βi3 γi2 ) + t(α3i−1,1 γi3 − α3i,1 γi2 ) − t2 (α3i−1,1 βi3 − 2di α3i,1 βi2 )]li1 ,

Gi1 (t, s) =


75

1 {[−(βi1 γi3 − βi3 γi1 ) + t(α3i−2,1 γi3 − α3i,1 γi1 ) − t2 (α3i−2,1 βi3 − 2di α3i,1 βi1 )]li2 + [(βi1 γi2 − βi2 γi1 ) − t(α3i−2,1 γi2 − α3i−1,1 γi1 )+ t2 (α3i−2,1 βi2 − α3i−1,1 βi1 )]li3 },

Gi2 (t, s) =

1 [(βi1 γi2 − βi2 γi1 ) − t(α3i−2,1 γi2 − α3i−1,1 γi1 ) + t2 (α3i−2,1 βi2 − 2di α3i−1,1 βi1 )]li3 ,

Gi3 (t, s) =

1 {[−(βi2 γi3 − βi3 γi2 ) + t(α3i−1,1 γi3 − α3i,1 γi2 ) − t2 (α3i−1,1 βi3 − 2di α3i,1 βi2 )]li1 + [(βi1 γi3 − βi3 γi1 ) − t(α3i−2,1 γi3 − α3i,1 γi1 )+ t2 (α3i−2,1 βi3 − α3i,1 βi1 )]li2 },

Gi4 (t, s) =

1 [(βi1 γi2 − βi2 γi1 ) − t(α3i−2,1 γi2 − α3i−1,1 γi1 ) + t2 (α3i−2,1 βi2 − 2di α3i−1,1 βi1 )]li3 ,

Gi5 (t, s) =

1 [−(βi2 γi3 − βi3 γi2 ) + t(α3i−1,1 γi3 − α3i,1 γi2 ) − t2 (α3i−1,1 βi3 − 2di α3i,1 βi2 )]li1 .

Gi6 (t, s) =

Lemma 2.2 Assume that the conditions (A1)-(A4) are satisfied. Then, for 1 ≤ i ≤ n, the Green’s function Gi (t, s) of (2.1)-(2.2) is positive, for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ]. Proof: For 1 ≤ i ≤ n, the Green’s function Gi (t, s) is given in (2.3). By using the conditions (A1)-(A4), we obtain Gi1 (t, s) > 0, for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ]. Similarly, we can establish the positivity of the Green’s function in the remaining cases. 2 Theorem 2.3 Assume that the conditions (A1)-(A4) are satisfied. Then, for 1 ≤ i ≤ n, the Green’s function Gi (t, s) satisfies the following inequality, mi Gi (σ(s), s) ≤ Gi (t, s) ≤ Gi (σ(s), s), for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ], (2.4) where ) ( Gi2 (t1 , s) Gi4 (σ(t3 ), s) Gi1 (σ(t3 ), s) Gi3 (t1 , s) , , , < 1. 0 < mi = min Gi1 (t1 , s) Gi3 (σ(t3 ), s) Gi2 (σ(t3 ), s) Gi4 (t1 , s)

76


Proof: For 1 ≤ i ≤ n, the Green’s function Gi (t, s) is given (2.3) in six different cases. In each case we prove the inequality as in (2.4). Case 1. For t1 < σ(s) < t ≤ t2 < σ(t3 ). G 1 (t,s) Gi (t,s) = Gi i(σ(s),s) Gi (σ(s),s) 1

=

[−(βi2 γi3 − βi3 γi2 ) + t(α3i−1,1 γi3 − α3i,1 γi2 ) − t2 (α3i−1,1 βi3 − α3i,1 βi2 )] . [−(βi2 γi3 − βi3 γi2 ) + σ(s)(α3i−1,1 γi3 − α3i,1 γi2 ) − (σ(s))2 (α3i−1,1 βi3 − α3i,1 βi2 )]

From (A1)-(A4), we have Gi1 (t, s) ≤ Gi1 (σ(s), s) and also Gi (t, s) Gi1 (t, s) Gi1 (t, s) Gi (σ(t3 ), s) = ≥ ≥ 1 . Gi (σ(s), s) Gi1 (σ(s), s) Gi1 (t1 , s) Gi1 (t1 , s) Therefore, Gi (t, s) ≤ Gi (σ(s), s) and Gi (t, s) ≥ (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ].

Gi1 (σ(t3 ),s) Gi1 (t1 ,s)

Gi (σ(s), s), for all

Case 2. For t1 ≤ t < t2 < s < σ(t3 ). G 3 (t,s) Gi (t,s) = Gi i(σ(s),s) Gi (σ(s),s) 3

=

[(βi1 γi2 − βi2 γi1 ) − t(α3i−2,1 γi2 − α3i−1,1 γi1 ) + t2 (α3i−2,1 βi2 − α3i−1,1 βi1 )] . [(βi1 γi2 − βi2 γi1 ) − σ(s)(α3i−2,1 γi2 − α3i−1,1 γi1 ) + (σ(s))2 (α3i−2,1 βi2 − α3i−1,1 βi1 )]

From (A1)-(A4), we have Gi3 (t, s) ≤ Gi3 (σ(s), s) and also Gi3 (t, s) Gi3 (t, s) Gi3 (t1 , s) Gi (t, s) = ≥ ≥ . Gi (σ(s), s) Gi3 (σ(s), s) Gi3 (σ(t3 ), s) Gi3 (σ(t3 ), s) Therefore, Gi (t, s) ≤ Gi (σ(s), s) and Gi (t, s) ≥ (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ].

Gi3 (t1 ,s) Gi3 (σ(t3 ),s)

Gi (σ(s), s), for all

Case 3. For t1 ≤ t < s < t2 < σ(t3 ). From (A1)-(A4) and case 2, we have Gi2 (t, s) ≤ Gi2 (σ(s), s) and also ( ) Gi (t, s) Gi3 (t1 , s) Gi2 (t1 , s) ≥ min , . Gi (σ(s), s) Gi3 (σ(t3 ), s) Gi2 (σ(t3 ), s) Therefore, Gi (t, s) ≤ Gi (σ(s), s) and ( ) Gi3 (t1 , s) Gi2 (t1 , s) Gi (t, s) ≥ min , Gi (σ(s), s), Gi3 (σ(t3 ), s) Gi2 (σ(t3 ), s)

77


for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ]. Case 4. For t1 < t2 < σ(s) < t ≤ σ(t3 ). From (A1)-(A4) and case 1, we have Gi4 (t, s) ≤ Gi4 (σ(s), s) and ( ) Gi (t, s) Gi1 (σ(t3 ), s) Gi4 (σ(t3 ), s) ≥ min , . Gi (σ(s), s) Gi1 (t1 , s) Gi4 (t1 , s) Therefore, Gi (t, s) ≤ Gi (σ(s), s) and ( ) Gi1 (σ(t3 ), s) Gi4 (σ(t3 ), s) Gi (t, s) ≥ min , Gi (σ(s), s), Gi1 (t1 , s) Gi4 (t1 , s) for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ]. Case 5. For t1 < t2 ≤ t < s < σ(t3 ). From case 2, we have Gi (t, s) ≤ Gi (σ(s), s) and Gi (t, s) ≥ for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ].

Gi3 (t1 ,s) Gi3 (σ(t3 ),s)

Gi (σ(s), s),

Case 6. For t1 ≤ σ(s) < t2 < t < σ(t3 ). From case 1, we have Gi (t, s) ≤ Gi (σ(s), s) and Gi (t, s) ≥ for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ].

Gi1 (σ(t3 ),s) Gi1 (t1 ,s)

Gi (σ(s), s),

From all above cases, for 1 ≤ i ≤ n, we have mi Gi (σ(s), s) ≤ Gi (t, s) ≤ Gi (σ(s), s), for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ], where ( 0 < mi = min

Gi1 (σ(t3 ), s) Gi3 (t1 , s) Gi2 (t1 , s) Gi4 (σ(t3 ), s) , , , Gi1 (t1 , s) Gi3 (σ(t3 ), s) Gi2 (σ(t3 ), s) Gi4 (t1 , s)

) < 1. 2

Lemma 2.4 Assume that the conditions (A1)-(A4) are satisfied and Gi (t, s) is defined as in (2.3). Take H1 (t, s) = G1 (t, s) and recursively define Z

σ(t3 )

Hj−1 (t, r)Gj (r, s)∆r, f or 2 ≤ j ≤ n.

Hj (t, s) = t1

Then Hn (t, s) is the Green’s function for the homogeneous BVP corresponding to (1.1)-(1.2).

78


Lemma 2.5 Assume that the conditions (A1)-(A4) hold. If we define K=

n−1 Y

Kj and L =

j=1

n−1 Y

mj Lj ,

j=1

then the Green’s function Hn (t, s) in Lemma 2.4 satisfies 0 ≤ Hn (t, s) ≤ KkGn (·, s)k, for all (t, s) ∈ [t1 , σ(t3 )] × [t1 , t3 ] and Hn (t, s) ≥ mn LkGn (·, s)k, for all (t, s) ∈ [t2 , σ(t3 )] × [t1 , t3 ], where mn is given as in Theorem 2.3, σ(t3 )

Z

kGj (·, s)k∆s > 0, for 1 ≤ j ≤ n,

Kj = t1

Z

σ(t3 )

kGj (·, s)k∆s > 0, for 1 ≤ j ≤ n

Lj = t2

and k · k is defined by kxk =

max

|x(t)|.

t∈[t1 ,σ(t3 )]

3


In this section, we establish the existence of at least three positive solutions to the BVP (1.1)-(1.2) by using the Leggett-Williams fixed point theorem. We also establish the existence of at least 2m − 1 positive solutions for the BVP (1.1)-(1.2) for an arbitrary positive integer m. Let E be a real Banach space with cone P . A map S : P → [0, ∞) is said to be a nonnegative continuous concave functional on P , if S is continuous and S(λx + (1 − λ)y) ≥ λS(x) + (1 − λ)S(y), for all x, y ∈ P and λ ∈ [0, 1]. Let α and β be two real numbers such that 0 < α < β and S be a nonnegative continuous concave functional on P . We define the following convex sets Pα = {y ∈ P : kyk < α} and P (S, α, β) = {y ∈ P : α ≤ S(y), kyk ≤ β}.

79


Theorem 3.1 [Leggett-Williams fixed point theorem] Let T : P c → P c be completely continuous and S be a nonnegative continuous concave functional on P such that S(y) ≤ kyk, for all y ∈ P c . Suppose that there exist a, b, c and d with 0 < d < a < b ≤ c such that (i) {y ∈ P (S, a, b) : S(y) > a} = 6 ∅ and S(T y) > a, for y ∈ P (S, a, b), (ii) kT yk < d, for kyk ≤ d, (iii) S(T y) > a, for y ∈ P (S, a, c) with kT yk > b. Then T has at least three fixed points y1 , y2 , y3 in P c satisfying ky1 k < d, a < S(y2 ), ky3 k > d, S(y3 ) < a. Let M = mn

n−1 Y j=1

mj Lj . Kj

Theorem 3.2 Assume that the conditions (A1)-(A4) are satisfied and also a1 < a2 assume that there exist real numbers a0 , a1 and a2 with 0 < a0 < a1 < M such that a0 f (t, y(t)) < Qn j=1

Kj

, for t ∈ [t1 , σ(t3 )] and y ∈ [0, a0 ],

a1 a1 , for t ∈ [t2 , σ(t3 )] and y ∈ [a1 , ], M j=1 mj Lj a2 , for t ∈ [t1 , σ(t3 )] and y ∈ [0, a2 ]. f (t, y(t)) < Qn j=1 Kj

f (t, y(t)) > Qn

(3.1) (3.2) (3.3)

Then the BVP (1.1)-(1.2) has at least three positive solutions. Proof: Let the Banach Space E = C[t1 , σ(t3 )] be equipped with the norm kyk =

|y(t)|.

max t∈[t1 ,σ(t3 )]

We denote P = {y ∈ E : y(t) ≥ 0, t ∈ [t1 , σ(t3 )]}. Then, it is obvious that P is a cone in E. For y ∈ P , we define S(y) =

min

|y(t)| and

t∈[t2 ,σ(t3 )]

Z

σ(t3 )

Hn (t, s)f (s, y(s))∆s, t ∈ [t1 , σ(t3 )].

T y(t) = t1

It is easy to see that S is a nonnegative continuous concave functional on P with S(y) ≤ kyk, for y ∈ P and that T : P → P is completely continuous and

80


fixed points of T are solutions of the BVP (1.1)-(1.2). First, we prove that, if there exists a positive number r such that f (t, y(t)) < Qn r Kj , for t ∈ [t1 , σ(t3 )] j=1

and y ∈ [0, r], then T : P r → Pr . Indeed, if y ∈ P r , then for t ∈ [t1 , σ(t3 )], we have Z

σ(t3 )

T y(t) =

Hn (t, s)f (s, y(s))∆s t1

< Qn

Hn (t, s)∆s

Kj

j=1

≤ Qn

σ(t3 )

Z

r

t1 σ(t3 )

Z

r Kj

j=1

kGn (·, s)k∆s = r.

K t1

Thus, kT yk < r, that is, T y ∈ Pr . Hence, we have shown that if (3.1) and (3.3) hold, then T maps P a0 into Pa0 and P a2 into Pa2 . Next, we show that a1 a1 ) : S(y) > a1 } 6= ∅ and S(T y) > a1 , for all y ∈ P (S, a1 , M ). In {y ∈ P (S, a1 , M fact, the constant function a1 a1 + M a1 ∈ {y ∈ P (S, a1 , ) : S(y) > a1 }. 2 M a1 Moreover, for y ∈ P (S, a1 , M ), we have

a1 ≥ kyk ≥ y(t) ≥ min y(t) = S(y) ≥ a1 , t∈[t2 ,σ(t3 )] M for all t ∈ [t2 , σ(t3 )]. Thus, in view of (3.2), we see that Z σ(t3 ) S(T y) = min Hn (t, s)f (s, y(s))∆s t∈[t2 ,σ(t3 )]

t1 σ(t3 )

Z ≥

min t∈[t2 ,σ(t3 )]


a1 mn L j=1 mj Lj

> Qn

Z

σ(t3 )

kGn (·, s)k∆s = a1 t2

a1 as required. Finally, we show that, if y ∈ P (S, a1 , a2 ) and kT yk > M , then a1 S(T y) > a1 . To see this, we suppose that y ∈ P (S, a1 , a2 ) and kT yk > M , then, by Lemma 2.5, we have Z σ(t3 ) S(T y) = min Hn (t, s)f (s, y(s))∆s t∈[t2 ,σ(t3 )]

t1

Z ≥

min t∈[t2 ,σ(t3 )]

Z

σ(t3 )

kGn (·, s)kf (s, y(s))∆s

mn L t1

σ(t3 )

≥ mn L

kGn (·, s)kf (s, y(s))∆s t2

81


for all t ∈ [t1 , σ(t3 )]. Thus mn L S(T y) ≥ max K t∈[t1 ,σ(t3 )] mn L = kT yk K m n L a1 = a1 . > K M

Z

σ(t3 )


To sum up, all the hypotheses of Theorem 3.1 are satisfied. Hence T has at least three fixed points, that is, the BVP (1.1)-(1.2) has at least three positive solutions y1 , y2 and y3 such that ky1 k < a0 , a1
a0 ,

min t∈[t2 ,σ(t3 )]

y3 (t) < a1 . 2

Theorem 3.3 Let m be an arbitrary positive integer. Assume that there exist b1 numbers ai (i = 1, 2, · · ·, m) and bj (j = 1, 2, · · ·, m − 1) with 0 < a1 < b1 < M < bm−1 b2 a2 < b2 < M < · · · < am−1 < bm−1 < M < am such that f (t, y(t)) < Qn

ai

j=1

Kj

, f or t ∈ [t1 , σ(t3 )] and y ∈ [0, ai ], i = 1, 2, · · ·, m, (3.4)

bj bj , f or t ∈ [t2 , σ(t3 )] and y ∈ [bj , ], j = 1, 2, · · ·, m − 1. M j=1 mj Lj (3.5) Then the BVP (1.1)-(1.2) has at least 2m − 1 positive solutions in P am . f (t, y(t)) > Qn

Proof: We use induction on m. First, for m = 1, we know from (3.4) that T : P a1 → Pa1 , then, it follows from Schauder fixed point theorem that the BVP (1.1)-(1.2) has at least one positive solution in P a1 . Next, we assume that this conclusion holds for m = k. In order to prove that this conclusion holds for m = k + 1, we suppose that there exist numbers ai (i = 1, 2, · · ·, k + 1) b2 b1 and bj (j = 1, 2, · · ·, k) with 0 < a1 < b1 < M < a2 < b2 < M < · · · < ak < bk bk < M < ak+1 such that f (t, y(t)) < Qn

ai

j=1

Kj

, for t ∈ [t1 , σ(t3 )] and y ∈ [0, ai ], i = 1, 2, · · ·, k + 1,

(3.6) bj bj f (t, y(t)) > Qn , for t ∈ [t2 , σ(t3 )] and y ∈ [bj , ], j = 1, 2, · · ·, k. M j=1 mj Lj (3.7)

82


By assumption, the BVP (1.1)-(1.2) has at least 2k − 1 positive solutions ui (i = 1, 2, · · ·, 2k − 1) in P ak . At the same time, it follows from Theorem 3.2, (3.6) and (3.7) that the BVP (1.1)-(1.2) has at least three positive solutions u, v and w in P ak+1 such that kuk < ak , bk
ak ,

min t∈[t2 ,σ(t3 )]

w(t) < bk .

Obviously, v and w are different from ui (i = 1, 2, · · ·, 2k − 1). Therefore, the BVP (1.1)-(1.2) has at least 2k + 1 positive solutions in P ak+1 , which shows that this conclusion holds for m = k + 1. 2

4

Example

Let us consider an example to illustrate the usage of the Theorem 3.2. Let 1 : n ∈ N} ∪ [ 12 , 23 ]. Now, consider the following BVP, n = 2 and T = {0} ∪ { 2n+1 6

y ∆ (t) = f (t, y), t ∈ [0, σ(1)] ∩ T

(4.1)

subject to the boundary conditions,  1 ∆ ∆2  y(0) − y (0) + 2y (0) = 0,   2   1 1   1 2 ∆ ∆  − 3y + 2y = 0, 2y   2 2 2    1 ∆ 1 ∆2   y(σ(1)) + y (σ(1)) + y (σ(1)) = 0, 2 3 3 ∆3 4 5  y (0) − 2y ∆ (0) + 3y ∆ (0) = 0,    4    1 1 1 3 4 5  ∆ ∆ ∆  y − 2y +y = 0,   2 2 2    1 1 4 5 3 ∆ ∆ ∆  y (σ(1)) + y (σ(1)) + y (σ(1)) = 0, 2 2 and ( f (t, y) =

sint 100 sint 100

+ +

13 10 y , 50 6656 , 25

y ≤ 2, y ≥ 2.

(4.2)


83

Then the conditions (A1)-(A4) are satisfied. The Green’s function G1 (t, s) in Lemma 2.1 is    G11 (t, s), 0 < σ(s) < t ≤ 12 < σ(1)    G1 (t,s)  G12 (t, s), 0 ≤ t < s < 21 < σ(1) =   t∈[0, 21 ]    G13 (t, s), 0 ≤ t < 12 < s < σ(1)  G1 (t, s) =      G14 (t, s), 0 < 12 < σ(s) < t ≤ σ(1)   G (t,s) 1  G (t, s), 0 < 12 ≤ t < s < σ(1) =    t∈[ 12 ,σ(1)]  15 G16 (t, s), 0 ≤ σ(s) < 21 < t < σ(1) where ih 1 i 12 h 91 23 + t − 5t2 σ(s)σ 2 (s) + (σ(s) + σ 2 (s)) + 4 , 481 12 6 2 12 nh 26 8 7 ih 3i G12 (t, s) = − t − t2 2σ(s)σ 2 (s) + 2(σ(s) + σ 2 (s)) + 481 3 3 4 2 h 13 29 ih io 3 8 + + t + t2 σ(s)σ 2 (s) − (σ(s) + σ 2 (s)) + , 2 4 2 3 ih 3 8i 12 h 13 29 2 2 2 + t + t σ(s)σ (s) − (σ(s) + σ (s)) + , G13 (t, s) = 481 2 4 2 3 nh ih i 12 91 23 1 2 2 2 G14 (t, s) = + t − 5t σ(s)σ (s) + (σ(s) + σ (s)) + 4 481 12 6 2 h 26 8 ih 7 2 3 io 2 2 + − + t + t 2σ(s)σ (s) + 2(σ(s) + σ (s)) + , 3 3 4 2 ih 12 h 13 29 3 8i G15 (t, s) = + t + t2 σ(s)σ 2 (s) − (σ(s) + σ 2 (s)) + , 481 2 4 2 3 h ih i 1 12 91 23 + t − 5t2 σ(s)σ 2 (s) + (σ(s) + σ 2 (s)) + 4 . G16 (t, s) = 481 12 6 2 The Green’s function G2 (t, s) in Lemma 2.1 is    G21 (t, s), 0 < σ(s) < t ≤ 12 < σ(1)    G (t,s) 2  G22 (t, s), 0 ≤ t < s < 21 < σ(1)  1 =  t∈[0, ]   2  G23 (t, s), 0 ≤ t < 12 < s < σ(1)  G2 (t, s) =      G24 (t, s), 0 < 12 < σ(s) < t ≤ σ(1)   G2 (t,s)  G (t, s), 0 < 12 ≤ t < s < σ(1) =    t∈[ 12 ,σ(1)]  25 G26 (t, s), 0 ≤ σ(s) < 21 < t < σ(1) G11 (t, s) =

where G21 (t, s) =

ih 3 i 16 h 39 11 + t − 3t2 σ(s)σ 2 (s) + 2(σ(s) + σ 2 (s)) + 6 , 635 8 4 4

84


16 nh 15 25 ih 3 1i 15 − t − t2 σ(s)σ 2 (s) + (σ(s) + σ 2 (s)) + 635 4 8 2 4 h 17 93 io 7 2 ih 3 + + t + t σ(s)σ 2 (s) − (σ(s) + σ 2 (s)) + 3 , 2 16 8 2 i 16 h 17 93 7 ih 3 G23 (t, s) = + t + t2 σ(s)σ 2 (s) − (σ(s) + σ 2 (s)) + 3 , 635 2 16 8 2 ih 3 i 16 nh 39 11 G24 (t, s) = + t − 3t2 σ(s)σ 2 (s) + 2(σ(s) + σ 2 (s)) + 6 635 8 4 4 h 15 25 2 ih 3 1 io + − 15 + t + t σ(s)σ 2 (s) + (σ(s) + σ 2 (s)) + , 4 8 2 4 i 16 h 17 93 7 ih 3 G25 (t, s) = + t + t2 σ(s)σ 2 (s) − (σ(s) + σ 2 (s)) + 3 , 635 2 16 8 2 i ih 3 16 h 39 11 2 2 2 + t − 3t σ(s)σ (s) + 2(σ(s) + σ (s)) + 6 . G26 (t, s) = 635 8 4 4 G22 (t, s) =

From Theorem 2.3 and Lemma 2.5, we get m1 = 0.4406779661, K1 = 0.6552328771, L1 = 0.183991684, m2 = 0.5596707819, K2 = 0.7449516076, L2 = 0.2551181102. Therefore, K = 0.6552328771, L = 0.08108108108 and M = 0.06925585335. Clearly f is continuous and increasing on [0, ∞). If we choose a0 = 0.25, a1 = 2 a1 < a2 and f satisfies and a2 = 150 then 0 < a0 < a1 < M (i) f (t, y) < 0.512172512 = (ii) f (t, y) > 172.757353 =

a0 , Π2j=1 Kj

for t ∈ [0, σ(1)] and y ∈ [0, 0.25],

a1 , Π2j=1 mj Lj

(iii) f (t, y) < 307.3035072 =

a2 , Π2j=1 Kj

for t ∈ [ 21 , σ(1)] and y ∈ [2, 28.87842548], for t ∈ [0, σ(1)] and y ∈ [0, 150].

Then all the conditions of Theorem 3.2 are satisfied. Thus, by Theorem 3.2, the BVP (4.1)-(4.2) has at least three positive solutions y1 , y2 and y3 satisfying k y1 k< 0.25, 2
0.25,

t∈[ 21 ,σ(1)]

min y3 (t) < 2.

t∈[ 21 ,σ(1)]

Remark: If f involves the derivatives of y in the equation (1.1), we can establish the existence of positive solutions by defining a suitable cone in Banach space with suitable norm or by applying the method used by Xu and Yang [22].

85


5

Open Problem

In this paper, we established the existence of at least three positive solutions for 3nth order three-point boundary value problem on time scales by using Leggett-Williams fixed point theorem. It will be interesting to obtain multiple positive solutions for the 3nth order general boundary value problem on time scales, (3n) (3n−1) (−1)n y ∆ (t) = f (t, y(t), y ∆ (t), · · ·, y ∆ (t)) satisfying the general three-point boundary conditions, 3 X

[α3i−3+j,k y ∆

(3i−4+k)

(t1 )+β3i−3+j,k y ∆

(3i−4+k)

(t2 )+γ3i−3+j,k y ∆

(3i−4+k)

(σ(t3 ))] = 0,

k=1

for j = 1, 2, 3, and 1 ≤ i ≤ n, where n ≥ 1. ACKNOWLEDGEMENTS: The authors thank the referees for their valuable suggestions and comments.

References [1] Agarwal, R. P. & O’Regan, D. [2000] “Triple solutions to boundary value problems on time scales,” Appl. Math. Lett. 13(4), 7-14. [2] Agarwal, R. P., O’Regan, D. & Wong, P. J. Y. [1999] Positive Solutions of Differential, Difference and Integral Equations, Kluwer Academic Publishers, Dodrecht, The Netherlands. [3] Anderson, D. R. [1998] “Multiple positive solutions for a three-point boundary value problem,” Math. Comp. Modelling 27(6), 49-57. [4] Anderson, D. R. [2002] “Solutions to second order three-point problems on time scales,” J. Diff. Eqns. Appl. 8, 673-688. [5] Anderson, D. R. & Avery, R. I. [2001] “Multiple positive solutions to a third order discrete focal boundary value problem,” J. Comp. Math. Appl. 42, 333-340. [6] Anderson, D. R. & Avery, R. I. [2002] “Existence of three positive solutions to a second order boundary value problem on a measure chain, Dynamic equations on time scales,” J. Comp. Appl. Math. 141(1 and 2), 65-73.

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[7] Atici, F. M. & Guseinov, G. Sh. [2002] “On Green’s functions and positive solutions for boundary value problems on time scales,” J. Comp. Appl. Math. 141, 75-99. [8] Avery, R. I. [1999] “A generalization of the Leggett-Williams fixed point theorem,” Math. Sci. Res. Hot-Line 3(7), 9-14. [9] Avery, R. I. & Peterson, A. C. [1998] “Multiple positive solutions of a discrete second order conjugate problem,” Panamer. Math. J. 8(3), 1-12. [10] Bohner, M. & Peterson, A. C. [2001] Dynamic Equations on Time Scales, An Introduction with Applications, Birkhauser, Boston, MA. [11] Bohner, M. & Peterson, A. C. [2003] Advances in Dynamic Equations on Time Scales, Birkhauser, Boston. [12] Eloe, P. W. & Henderson, J. [1997] “Positive solutions for (n-1, 1) conjugate boundary value problems,” Nonlinear Anal. 28, 1669-1680. [13] Eloe, P. W. & Henderson, J. [1997] “Positive solutions and nonlinear multipoint conjugate eigenvalue problems,” Elec. J. Diff. Eqns. 1997(3), 1-11. [14] Eloe, P. W. & Henderson, J. [1998] “Positive solutions and nonlinear (k, n-k) conjugate eigenvalue problems,” Diff. Eqns. Dyn. Sys. 6, 309-317. [15] Erbe, L. H. & Peterson, A. C. [2000] “Positive solutions for a nonlinear differential equation on a measure chain,” Math. Comp. Modelling 32, 571-585. [16] Erbe, L. H. & Wang, H. [1994] “On the existence of positive solutions of ordinary differential equations,” Proc. Amer. Math. Soc. 120, 743-748. [17] Henderson, J. & Kaufmann, E. R. [1997] “Multiple positive solutions for focal boundary value problems,” Comm. Appl. Anal. 1, 53-60. [18] Henderson, J. & Wang, H. [1997] “Positive solutions for nonlinear eigenvalue problems,” J. Math. Anal. Appl. 208, 252-259. [19] Kaufmann, E. R. [2003] “Positive solutions of a three-point boundary value problem on time scales,” Elec. J. Diff. Eqns. 2003(82), 1-11. [20] Sun, J. P. [2006] “Existence of solutions and positive solutions of boundary value problems for nonlinear third order dynamic equation,” Nonlinear Anal. 64(1), 165-176.


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[21] Sun, H. R. & Li, W. T. [2004] “Positive solutions for nonlinear three point boundary value problems on time scales,” J. Math. Anal. Appl. 299, 508524. [22] Xu, J. & Yang, Z. [2009] “Three positive solutions for a system of singular generalized Lidstone problems,” Elec. J. Diff. Eqns. 2009(163), 1-9. [23] Xu, J. & Yang, Z. [2010] “Triple positive solutions for singular integral boundary value problems,” Int. J. Open Problems Compt. Math. 3(4), 455-467. [24] Yao, Q. [2003] “The existence and multiplicity of Positive solutions for a third order three point boundary value problem,” Acta. Math. Appl. Sinica 19, 117-122.


Some Open Problems On a Class of Finite Groups Marius T˘ arn˘ auceanu Faculty of Mathematics, ”Al.I. Cuza” University, Ia¸si, Romania e-mail: [email protected] Abstract In this note we introduce and study a class of finite groups for which the exponents of subgroups satisfy a certain inequality. It is closely connected to some well-known arithmetic classes of natural numbers.

Keywords: finite groups, subgroup lattices, exponents, deficient numbers, perfect numbers. MSC (2010): Primary 20D60, 20D30; Secondary 11A25, 11A99.

1

Introduction

Let n be a natural number and σ(n) be the sum of all divisors of n. We say that n is a deficient number if σ(n) < 2n and a perfect number if σ(n) = 2n (for more details on these numbers, see [3]). Thus, the set consisting of both the deficient numbers and the perfect numbers can be characterized by the inequality X d ≤ 2n , d∈Ln

where Ln = {d ∈ N | d|n}. Now, let G be a finite group. Then the set L(G) of all subgroups of G forms a complete lattice with respect to set inclusion, called the subgroup lattice of G. Two remarkable subsets of L(G) are constituted by all cyclic subgroups and by all normal subgroups of G, respectively. They are called the poset of cyclic subgroups and the normal subgroup lattice of G and will be denoted by C(G) and N (G). Notice that N (G) is a sublattice of L(G), while C(G) is in general only a subposet of L(G). If the group G is cyclic of order n, then L(G) = C(G) = N (G)

Some open problems on a class of finite groups

89

and these are isomorphic to the lattice Ln . So, the fact that n is deficient or perfect can be written in the following equivalent three ways: X |H| ≤ 2|G| , (1) H∈L(G)

X

|H| ≤ 2|G| ,

(2)

|H| ≤ 2|G| .

(3)

H∈C(G)

X H∈N (G)

In [1] and [8] we have studied the classes Ci , i = 1, 2, 3, consisting of all finite groups G which satisfy the above inequalities, respectively. Recall only that the unique groups contained in C1 are the cyclic groups of deficient or perfect orders and that C1 is properly included in C2 . The starting point for our discussion is given by the remark that (1)-(3) also equivalent for finite cyclic groups to the inequality X exp(H) ≤ 2 exp(G) , (4) H∈L(G)

where exp(H) denotes the exponent of the subgroup H of G, that is the lower common multiple of the orders of elements of G. Hence, in a similar manner, one can introduce the class C consisting of all finite groups G which satisfy (4). Its investigation is the main goal of our paper. The paper is organized as follows. In Section 2 we study some basic properties of C and the connections with Ci , i = 1, 2, 3. The containment of some important classes of finite groups, as p-groups, nilpotent groups, dihedral groups, ZM-groups and symmetric groups to C is also discussed. In the final section we propose several problems with respect to this subject. Most of our notation is standard and will not be repeated here. Basic definitions and results on groups can be found in [2] and [5]. For subgroup lattice concepts we refer the reader to [4] and [6].

2

Main results

For a finite group G let us denote σ(G) =

X 1 exp(H) . exp(G) H∈L(G)

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Marius T˘arn˘auceanu

In this way, C is the class of all finite groups G for which σ(G) ≤ 2. First of all, we remark that σ is a multiplicative function, that is if G and G0 are two finite groups satisfying gcd(|G|, |G0 |) = 1, then σ(G × G0 ) = σ(G)σ(G0 ) . By a standard induction argument, it follows that if Gi , i = 1, 2, . . . , k, are finite groups of coprime orders, then k

k

×G ) = Y σ(G ) .

σ(

i

i=1

i

i=1

Obviously, C contains the finite cyclic groups of prime order. On the other hand, we easily obtain σ(Zp × Zp ) =

1 + 2p + p2 > 2, for any prime p . p

This relation shows that C is not closed under direct products or extensions. Observe next that for every finite group G we have X 1 X 1 exp(H) ≥ |H| σ(G) ≥ exp(G) |G| H∈C(G)

H∈C(G)

and thus if G does not belong to C2 , then σ(G) > 2. Consequently, the following result holds. Theorem 1. The class C is contained in the class C2 . In the following we will focus on characterizing some particular classes of groups in C. The simplest case is constituted by finite p-groups. Lemma 2. A finite p-group is contained in C if and only if it is cyclic. Proof. Let G be a finite p-group of order pn which is contained in C, set pm = exp(G) and suppose that G is not cyclic. Then m < n. It is wellknown that G possesses at least an element a of order pm . One obtains that exp(hai) = pm and therefore σ(G) ≥

exp(1) + exp(hai) + exp(G) 1 + 2pm = > 2. pm pm

This shows that G does not belong to C, a contradiction. Conversely, for a finite cyclic p-group G of order pn , we obviously have σ(G) =

1 + p + ... + pn pn+1 − 1 = ≤ 2. pn pn+1 − pn

91


The above lemma leads to a precise characterization of finite nilpotent groups contained in C. It shows that the finite cyclic groups of deficient or perfect order are in fact the unique such groups. Theorem 3. Let G be a finite nilpotent group. Then G is contained in C if and only if it is cyclic and its order is a deficient or perfect number. k

Proof. Assume that G belongs to C and let

×G

i

be its decomposition as a

i=1

direct product of Sylow subgroups. Since Gi , i = 1, 2, ..., k, are of coprime orders, one obtains k Y σ(G) = σ(Gi ) ≤ 2. i=1

This inequality implies that σ(Gi ) ≤ 2, for all i = 1, k. So, each Gi is contained in C and it must be cyclic, by Lemma 2. Therefore G itself is cyclic and | G | is a deficient or perfect number. The converse is obvious. Theorem 3 shows that in order to find examples of noncyclic groups contained in C, we must look to some classes of nonnilpotent groups. Three such classes are investigated in the following theorem. Theorem 4. The following finite groups are not contained in C: 1. dihedral groups; 2. ZM-groups; 3. symmetric groups. Proof.

1. The exponent of the dihedral group D2n = hx, y | xn = y 2 = 1, yxy = x−1 i, n ≥ 2,

can be easily computed: exp(D2n ) =

  2n, n ≡ 1 (mod 2) 

n,

n ≡ 0 (mod 2) .

For every divisor d of n, D2n has a unique cyclic subgroup of order d and n subgroups isomorphic to D2d . Denote by τ (n) the number of divisors d of n. We infer that for n odd we have   X X 1  n  1 σ(D2n ) = d+ 2d = (σ(n) + 2nτ (n)) > τ (n) ≥ 2, 2n d 2n d|n

d|n

92


while for n even we have   X X 1 n  1 σ(D2n ) ≥  d = (σ(n) + nτ (n)) > τ (n) ≥ 2. d+ n d n d|n

d|n

Hence D2n does not belong to C. 2. Recall that a ZM-group is a finite nonabelian group with all Sylow subgroups cyclic. By [2], such a group is of type ZM(m, n, r) = ha, b | am = bn = 1, b−1 ab = ar i, where the triple (m, n, r) satisfies the conditions gcd(m, n) = gcd(m, r − 1) = 1 and rn ≡ 1 (mod m). Obviously, we have exp(ZM(m, n, r) = mn. Set rn − 1 3 L = (m1 , n1 , s) ∈ N | m1 | m, n1 | n, s < m1 , m1 | s n1 . r −1 Then it is well-known that there is a bijection between L and the subgroup lattice of ZM(m, n, r), namely the function that maps a triple (m1 , n1 , s) ∈ L into the subgroup H(m1 ,n1 ,s) defined by n

H(m1 ,n1 ,s) =

n1 [

α(n1 , s)k ham1 i = ham1 , α(n1 , s)i,

k=1

where α(x, y) = bx ay , for all 0 ≤ x < n and 0 ≤ y < m. Since exp(H(m1 ,n1 ,s) ) = mmn , we infer that 1 n1 rn − 1 1 X X mn ≥ σ(ZM(m, n, r)) = gcd m1 , n1 mn m1 n1 r −1 m1 | m n1 | n

1 (x + y) , mn X mn rn − 1 x= gcd 1, n1 = m σ(n) n1 r −1 ≥

where

n1 | n

and

X n rn − 1 y= gcd m, n1 ≥ mn. n1 r −1 n1 | n


93

The above inequalities lead to σ(ZM(m, n, r)) ≥

1 1 (m σ(n) + mn) > (mn + mn) = 2, mn mn

which proves that ZM(m, n, r) is not contained in C. 3. The exponent of the symmetric group Sn is given by exp(Sn ) = lcm(1, 2, . . . , n) . Clearly, we have exp(Sn ) = lcm(exp(Sn−1 ), n) ≤ nexp(Sn−1 ) . On the other hand, Sn has at least n subgroups isomorphic to Sn−1 , namely Hi = {σ ∈ Sn | σ(i) = i}, i = 1, 2, ..., n. One obtains ! n X 1 1+ exp(Hi ) + exp(Sn ) = σ(Sn ) ≥ exp(Sn ) i=1 1 (1 + nexp(Sn−1 ) + exp(Sn )) ≥ exp(Sn ) 1 ≥ (1 + 2exp(Sn )) > 2, exp(Sn ) =

implying that Sn does not belong to C. The containment to C can be also studied for other important classes of finite groups. Unfortunately, for the moment we are not able to decide whether C consists only of cyclic groups of deficient or perfect orders.

3

Open problems

Problem 1. Determine all finite groups contained in the class C. Is it true that C = C1 ? Problem 2. Study if C is closed under subgroups and homomorphic images, i.e. if all subgroups and all quotients of a group in C also belong to C. Problem 3. Investigate the class C 0 consisting of all finite groups G for which X 1 exp(H) ≤ 2 . σ 0 (G) = exp(G) H∈N (G)

94


References [1] De Medts, T., T˘arn˘auceanu, M., Finite groups determined by an inequality of the orders of their subgroups, Bull. Belg. Math. Soc. Simon Stevin 15 (2008), 699-704. [2] Huppert, B., Endliche Gruppen, I, Springer Verlag, Berlin, 1967. [3] Sándor, J., Crstici, B., Handbook of number theory, II, Kluwer Academic Publishers, Dordrecht, 2004. [4] Schmidt, R., Subgroup lattices of groups, de Gruyter Expositions in Mathematics 14, de Gruyter, Berlin, 1994. [5] Suzuki, M., Group theory, I, II, Springer Verlag, Berlin, 1982, 1986. [6] T˘arn˘auceanu, M., Groups determined by posets of subgroups, Ed. Matrix Rom, Bucure¸sti, 2006. [7] T˘arn˘auceanu, M., On the poset of subhypergroups of a hypergroup, Int. J. Open Problems Compt. Math. 3 (2010), no. 2, 115-122. [8] T˘arn˘auceanu, M., Finite groups determined by an inequality of the orders of their normal subgroups, Sci. An. Univ. ”Al.I. Cuza” Ia¸si, Math., 57 (2011), 229-238.


Newton-Raphson Type Methods Mircea I. Cˆırnu Department of Mathematics, Faculty of Applied Sciences Polytechnic University of Bucharest, Romania e-mail: [email protected] Abstract The most famous iteration scheme for solving algebraic equations is Newton-Raphson method. It is derived by the first order Taylor expansion and gives a recurrence formula for the iterations that approximates an exact root of the equation. Methods have appeared recently, based on certain recurrent polynomial equations, called in this paper resolvent equations, from which the variations of iterations are obtained, instead of a recurrence formula for iterations. Thus, J. H. He proposed in 2003 a method based on a quadratic resolvent equation, deduced by the second-order Taylor expansion. Unfortunately, his method is not so effective, because the resolvent equation have not real roots for any initial data. Using third-order Taylor expansion, D. Wei, J. Wu and M. Mei obtained in 2008 an iteration scheme based on solving a cubic resolvent equation, which has always a real root and therefore the method has no restriction regarding the choice of the initial data. The trouble with the resolvent polynomial equations obtained both by He and Wei-Wu-Mei is the constant term that has a complicated expression. In this paper we give two types of methods for finding the approximate values (iterations) of a real root of an algebraic equation, based on the determination of the resolvent polynomial equation of general order, one of its roots being the variation from an iteration to the next. The two types of methods will be obtained in accordance with the type of conditions that will be considered. The resolvent equation of the first type methods is the Taylor polynomial of general order in the considered iteration. Particularly, the first-order case gives the initial Newton-Raphston method. The resolvent equations of the second type methods will differ from those of first type only by the constant term. In the second-order and

96

Mircea I. Cˆırnu third-order cases, He and respective Wei-Wu-Mei methods are obtained. A simple algebraic calculation, based on a telescopic procedure, will produce an improvement of the constant term of the resolvent equations of second type. We give several examples and two applications to some extremum problems, namely to calculation of the minimum distance from the origin to the graphs of two functions inverse to each other, the exponential and the natural logarithm. The algebraic equations to which these problems are reduced, will be solved by the methods given in the present paper.

Keywords: algebraic equations, Newton-Raphson iterative method, Taylor expansion. 2000 Mathematics Subject Classification: 65H05, 41A25.

1

Introduction We consider the algebraic equation f (x) = 0,

(1)

where f (x) is a function of real variable with derivatives up to necessary order. The iterative methods of Newton-Raphson type consists in determining a sequence of iterations xn , n = 0, 1, 2, . . . , that approximate an exact real root of the equation (1), starting with conveniently chosen initial datum. In the new methods of this type, the variations of iterations tn = xn+1 − xn , n = 0, 1, 2, . . . ,

(2)

will be obtained, starting with the initial data x0 and x1 , hence with the initial variation t0 = x1 − x0 , from a recurrent polynomial equation, deduced by Taylor expansion of the function f (x) and named resolvent equation. From these variations, the iterations are determined from the relation xn+1 = xn + tn , n = 1, 2, . . . ,

(3)

starting with x1 . We consider the value of the function f at xn+1 given by the Taylor expansion of order m = 1, 2, 3 . . . , at the point xn , with the remainder Rm,n , f (xn+1 ) = Pm,n (tn ) + Rm,n , n = 0, 1, 2, . . . , where Pm,n (tn ) =

m X

f (k) (xn ) k tn , n = 0, 1, 2, . . . , k! k=0

is the Taylor polynomial of order m in xn with the variable tn .

(4)

(5)

Newton-Raphson Type Methods

2

97

Newton-Raphson Type Methods of First Kind

2.1

Methods of General Order

For an arbitrary natural number n ≥ 1, we require to have approximately f (xn+1 ) = 0, Rm,n = 0.

(6)

Then the Taylor expansion (4) takes the form Pm,n (tn ) =

m X

f (k) (xn ) k tn = 0, n = 1, 2, . . . , k! k=0

(7)

this being the resolvent equation of the method. Starting with x0 and x1 , we have t0 = x1 − x0 and the resolvent equation (7) gives recurrently the variation tn , which is one of the roots of the equation (7), conveniently chosen. Then the iterations xn are obtained by relation (3). Remark 2.1 (i) The formula (7) gives the simplest form of the resolvent equation, from which can be determined tn . It is the Taylor polynomial in xn . (ii) Because in the conditions (6) and (13) below we have approximate equality, we obtain approximate values of the variations tn and iterations xn .

2.2

First Order Method: Initial Newton-Raphson Method

For m = 1, the resolvent equation (7) takes for n = 1, 2, . . . , the form f (xn ) , hence according (3), the f 0 (xn )tn + f (xn ) = 0, with the root tn = − 0 f (xn ) iterations are given by the well-known recurrence relation f (xn ) xn+1 = xn − 0 , n = 1, 2, . . . , (8) f (xn ) starting with x1 . Here we consider x0 = 0. Remark 2.2 Newton used a variant of this formula to solve the polynomial equation presented in example 2.4 below, in his work Method of Fluxions, written in 1671 but published only in 1736. In turn, J. Raphson gave the formula in his book Analysis Aequationum Universalis, published in 1690.

2.3

Second and Third Order Methods

For m = 2, the resolvent equation (7) is the quadratic equation 1 00 f (xn )t2n + f 0 (xn )tn + f (xn ) = 0, n = 1, 2, . . . , 2 and for m = 3 is the cubic equation 1 000 1 f (xn )t3n + f 00 (xn )t2n + f 0 (xn )tn + f (xn ) = 0, n = 1, 2, . . . · 6 2

(9)

(10)

98

Mircea I. Cˆırnu

2.4

Example: Newton’s Polynomial Equation

The polynomial equation f (x) = x3 − 2x − 5 = 0, was solved initially by x3 − 2xn − 5 Newton using his recurrence relation (8), which gives xn+1 = xn − n 2 3xn − 2 2x3n + 5 = 2 , n = 1, 2, . . . · Taking x1 = 2, we obtain x2 = 2.1 and x3 = 2.0946. 3xn − 2 The second order resolvent equation is 3xn t2n +(3x2n −2)tn +x3n −2xn −5 = 0, n = 1, 2, . . . · Taking x0 = 0, x1 = 2 and n = 1, we obtain the equation 6t21 + 10t1 − 1 = 0, with the roots −1.7613 which not agree and t1 = 0.0946. Hence we obtain the iteration x2 = x1 + t1 = 2, 0946.

2.5

Calculation of a Radical

√ We calculate by the first type method the radical 3 2, so we solve the equation f (x) = x3 − 2 = 0. The quadratic resolvent equation is 3xn t2n + 3x2n tn + x3n − 2 = 0, n = 1, 2, . . . · Taking x0 = x1 = 1, hence t0 = 0, we obtain the equations 3t21 + 3t1 − 1 = 0, with the roots −1.2638, inconvenient and t1 = 0.2638, hence x2 = x1 + t1 = 1.2638 and 3.7914t22 + 4.7916t2 + 0.0185 = 0, with the roots −1.2599, inconvenient and t2 = −0.0039, hence x3 = x2 + t2 = 1.2599.

2.6

Example

We consider the algebraic equation f (x) = x − e−x = 0. For m = 1, xn − e−xn 1 + xn the recurrent relation (8) is xn+1 = xn − = , n = 1, 2, . . . · −x n 1+e 1 + e xn Taking x1 = 0.5, we obtain x2 = 0.5663 and x3 = 0.5671. In the methods based on resolvent equation we will take x0 = x1 = 0, hence t0 = 0. For m = 2, the resolvent equation (9) takes the form 1 − e−xn t2n + (1 + e−xn )tn + xn − e−xn = 0, n = 1, , 2, . . . · 2

(11)

For n = 1, the resolvent equation (11) reduces to −0.5t21 + 2t1 − 1 = 0 from which one obtains t1 = 0.5858 = x2 and for n = 2, to equation −0.2783t22 + 1.5567t2 + 0.0291 = 0, which gives t2 = −0.0186 and x3 = x2 + t2 = 0.5672. For m = 3, the resolvent equation (10) takes the form 1 −xn 3 1 −xn 2 e tn − e tn + (1 + e−xn )tn + xn − e−xn = 0, n = 1, 2, . . . · 6 2

(12)

From (12) we obtain the equations 0.1667t31 − 0.5t21 + 2t1 − 1 = 0 from which one obtains t1 = 0.5647 = x2 and 0.0948t32 − 0.2843t22 + 1.5685t2 − 0.0038 = 0, which gives t2 = 0.0024, hence x3 = 0.5671.


3

99

Newton-Raphson Type Methods of Second Kind

3.1

Methods of General Order

We now suppose m ≥ 2. For an arbitrary natural number n ≥ 1, we require to have approximately f (xn+1 ) = 0, Rm,n = Rm,n−1 .

(13)

Then the Taylor expansion (4) takes the form Pm,n (tn ) + Rm,n = 0.

(14)

According to (13) and (4), the last relation for n substituted by n − 1, the rest Rm,n is given by formula Rm,n = Rm,n−1 = f (xn ) − Pm,n−1 (tn−1 ) and the equation (14) becomes Pm,n (tn ) + f (xn ) − Pm,n−1 (tn−1 ) = 0, n = 1, 2, . . . ·

(15)

According to the notation (5), the equation (15) has the form m X

m X f (k) (xn ) k f (k) (xn−1 ) k tn + f (xn ) − tn−1 = 0, n = 1, 2, . . . , k! k! k=0 k=0

(16)

it being the resolvent equation from which can be recursively determined tn , starting with the value t0 = x1 − x0 , given by the initial data x0 and x1 . Remark 3.1 The resolvent equation (16) was given by J. H. He, [3], for m = 2 and by D. Wei, J. Wu and M. Mei, [4], for m = 3. The method of resolvent polynomial equation introdused by Ji-Huan He in the above mentioned paper must be added to other methods of solving various types of equations proposed by the same author, such as homotopy perturbation technique, [2] and variational iteration method, [3].

3.2

Improved Methods

Now we give a new form of the resolvent equation (16), more convenient for its writing. Replacing in the form (15) of the resolvent equation (16) the index n by n − 1, n − 2, . . . , 2, 1, we obtain the relations Pm,n−1 (tn−1 ) + f (xn−1 ) − Pm,n−2 (tn−2 ) = 0, Pm,n−2 (tn−2 ) + f (xn−2 ) − Pm,n−3 (tn−3 ) = 0, .................................... Pm,2 (t2 ) + f (x2 ) − Pm,1 (t1 ) = 0, Pm,1 (t1 ) + f (x1 ) − Pm,0 (t0 ) = 0.

100

Mircea I. Cˆırnu

Adding these relations to (15), we obtain the following improved form of this equation, Pm,n (tn ) +

n X

f (xj ) − Pm,0 (t0 ) = 0, n = 1, 2, . . . ,

(17)

j=1

hence, according with the notation (5), it results the resolvent equation n m X f (k) (x0 ) k f (k) (xn ) k X tn + f (xj ) − t0 = 0, n = 1, 2, . . . · k! k! j=1 k=0 k=0 m X

3.3

(18)

Simplified Methods

If we choose x0 = x1 , then t0 = 0 and the equation (18) reduces to the equations m X f (k) (x1 ) k t1 = 0, (19) k! k=0 that is the same with (7), for n = 1 and n f (k) (xn ) k X f (xj ) = 0, tn + k! j=2 k=0 m X

(20)

that is different from (7), for n ≥ 2. Remark 3.2 (i) For m = 2, respectively m = 3, we call the methods obtained in this Section and based on the resolvent equations (19) and (20), the improved He, respectively improved Wei-Wu-Mei method. (ii) In both methods of first and second type, the second-order resolvent equation cannot be used if it has complex roots. See example 3.5 below.

3.4

Example

We solve here by second type methods, namely by improved He and WeiWu-Mei methods, the same equation f (x) = x−e−x = 0, solved in example 2.6 by methods of first type. We start with x0 = x1 = 0, hence t0 = 0. As result of (19), the resolvent equation for t1 is the same as in example 2.6. Therefore, for m = 2, we have t1 = 0.5858 = x2 . For n ≥ 2 the resolvent equation is n X 1 − e−xn t2n + (1 + e−xn )tn + xn − e−xn + (xj − e−xj ) = 0. 2 j=2

For n = 2 this equation is −0.2783t22 + 1.5567t2 + 0.0582 = 0, resulting t2 = −0.0371, hence x3 = 0.5487.


101

For n = 3, the equation is −0.2889t23 + 1.5777t3 − 0.0289 = 0, resulting t3 = 0.0184, hence x4 = 0.5671. For m = 3, we have t1 = 0.5647 = x2 , in conformity with example 2.6. For n ≥ 2 the resolvent equation is n X 1 −xn 3 1 −xn 2 e tn − e tn + (1 + e−xn )tn + xn − e−xn + (xj − e−xj ) = 0. 6 2 j=2

For n = 2 this equation is 0.0948t32 − 0.2843t22 + 1.5685t2 − 0.0076 = 0, resulting t2 = 0.0048, hence x3 = 0.5695. For n = 3 the equation is 0.0943t33 − 0.2829t23 + 1.5658t3 + 0.0037 = 0, resulting t3 = −0.0024, hence x4 = 0, 5671.

3.5

Example

We give an example, taken from [6], in which He’s method does not apply. Here we use first type and improved second type methods, i.e. the resolvent equations (7) and (20), while in [6] the usual He and Wei-Wu-Mei methods were used, i.e. the resolvent equation (16), for m = 2 and m = 3. Consider equation f (x) = x3 −e−x = 0. Newton-Raphson formula (8) gives x3 − e−xn , n = 1, 2, . . . · the recurrence relation xn+1 = xn − n2 3xn + e−xn Taking x1 = 1, we obtain x2 = 0.8123, x3 = 0.7743 and x4 = 0.7729. For the methods based on resolvent equation, we take x0 = x1 = t0 = 0. For m = 2, both the first and second type methods give the same quadratic equation t21 − 2t1 + 2 = 0, which has complex roots, therefore these methods are not applicable. For m = 3, both the first and second type methods give the same cubic equation 1.1667t31 −0.5t21 +t1 −1 = 0, from which one obtains t1 = 0.7673 = x2 . The first method gives the cubic equation 1.0774t32 + 2.3019t22 + 2.2305t2 − 0.0125 = 0, from which one obtains t2 = 0.0056, hence x3 = 0.7729. By the second method, the improved Wei-Wu-Mei method gives the following recurrence process 1.0774t32 + 2.0698t22 + 2.2305t2 − 0.025 = 0, t2 = 0.0111, x3 = 0.7784 and 1.0765t33 + 2.1056t23 + 2.2769t3 + 0.0125 = 0, t3 = −0.0055, x4 = 0.7729.

4

Applications

In the applications below, the algebraic equations will be solved approximately by the Newton-Raphson method and also by the new methods, based on the resolvent equations, given in the present work.

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4.1. Determine the point on the graph of the function y = ex that is closest to the origin of the coordinate axes in plan and the distance from the origin to graph. √ We should minimize the distance d(x) = x2 + e2x from origin to an arbitrary point on the graph, hence its square, namely the function d2 (x) = x2 +e2x . With this end in view, it is necessary to have (d2 (x))0 = 2x + 2e2x = 0. Therefore we have to solve the equation f (x) = x + e2x = 0, what we do by the above methods. For those based on the resolvent equations are used the initial data x0 = x1 = 0, t0 = 0. The Newton-Raphson method is based on recurrence relation xn+1 =

(2xn − 1)e2xn , n = 1, 2, . . . · . 1 + 2e2xn

Taking x1 = 0, we have x2 = −0.3333, x3 = −0.4222, and x4 = −0.4263. By first type method of second-order, the resolvent equations are 2t21 + 3t1 + 1 = 0, from which one obtains t1 = −0.5 = x2 and 0.7358t22 + 1.7358t2 − 0.1321 = 0, from which one obtains t2 = 0.0738, hence x3 = −0.4262. By improved He’s method we have the same first resolvent equation, with the root t1 = −0.5 = x2 , then 0.7358t22 + 1.7358t2 − 0.2642 = 0, with the root t2 = 0.1435, hence x3 = x2 + t2 = −0.3565, 0.9804t23 + 1.9803t3 + 0.1353 = 0, with the root t3 = −0.0708, hence x4 = x3 + t3 = −0.4273 and 0.8509t24 + 1.8509t4 − 0.0021 = 0 with the root t4 = 0.0011, hence x5 = x4 + t4 = −0.4262. By first type method of third-order, we obtain the resolvent equations 1.3333t31 + 2t21 + 3t1 + 1 = 0, with the root t1 = −0.417 = x2 and 0.5791t32 + 0.8686t22 + 1.8686t2 + 0.0173 = 0, hence t2 = −0.0093 and x3 = −0.4263. By improved Wei, Wu and Mei method we have the same first resolvent equation, with the root t1 = −0.417 = x2 , then 0.5791t32 +0.8686t22 +1.8686t2 + 0.0346 = 0, from which one obtains t2 = −0.0187, hence x3 = x2 +t2 = −0.4357 and 0.5578t33 + 0.8367t23 + 1.8367t3 − 0.0174 = 0, from which one obtains t3 = 0.0094, hence x4 = x3 + t3 = −0.4263. It follows that the point on graph, closest of the origin of coordinates is (x, ex ) = ( − 0.426, 0.653) √ the minimum distance being d(x) = x2 + e2x = 0.78.

4.2. Determine the point on the graph of the function y = ln x that is closest to the origin of the coordinate axes in plan and the distance from the origin to graph. q

We should minimize the distance d(x) = x2 + ln2 x from origin to an arbitrary point on graph, hence its square, namely the function d2 (x) = x2 + 2 2 ln2 x. For this, is necessary to have (d2 (x))0 = 2x + ln x = (x2 + ln x) = 0. x x

103


Therefore we must solve the equation f (x) = x2 + ln x = 0, what we do by the above methods. For the methods based on resolvent equations will be used the initial data x0 = x1 = 0.5, t0 = 0. The Newton-Raphson method is based on recurrence relation xn+1 = xn − x3n + xn ln xn . Taking x1 = 0.5, we have x2 = 0.6477 and x3 = 0.6529. 2x2n + 1 By first type method of second-order, we have the resolvent equations −t21 + 3t1 − 0.4431 = 0, with the roots 2.8442 and t1 = 0.1558, hence x2 = 0.6558, and −0.1626t22 + 2.8365t2 + 0.0082 = 0, from which one obtains t2 = −0.0029 and x3 = 0.6529. By improved He’s method we obtain the same first resolvent equation from which one obtains t1 = 0.1558, hence x2 = x1 + t1 = 0.6558, and the second resolvent equation −0.1626t22 + 2.8365t2 + 0.0164 = 0, with the roots 17.4504 and t2 = −0.0058, hence x3 = x2 + t2 = 0.65. By first type method of third-order, we have the resolvent equations 2.6667t31 − t21 + 3t1 − 0.4431 = 0, with real root t1 = 0.1523, hence x2 = x1 + t1 = 0.6523, then 0.0417t32 − 0.1626t22 + 2.8365t2 + 0.0082 = 0, from which t2 = −0.0029, hence x3 = 0.6494 and 0.0913t33 − 0.1856t23 + 2.8387t3 − 0.01 = 0, from which t3 = 0.0035, hence x4 = 0.6529. By improved Wei, Wu and Mei method we obtain by the same first resolvent equation the root t1 = 0.1523, hence x2 = x1 + t1 = 0.6523 and 1.201t32 − 0.1751t22 + 2.8376t2 − 0.0036 = 0, with real root t2 = 0.0013, hence x3 = x2 + t2 = 0.6536. It follows that the point on graph, closest to the origin, is (x, ln x) = (0.653, −0.426), the minimum distance being d(x) =

q

x2 + ln2 x = 0.78.

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Remark 4.1 The points obtained in the two applications are symmetrical about the bisetrix of the first quadrant and the obtained minimum distances are the same, because the two considered functions are inverse to each other.

5

Conclusions and open problems

Following the direction initiated by J. H. He of finding new methods of Newton-Raphson type for solving algebraic equations, based on resolvent polynomial equations, in this paper we present methods whose resolvent equations are simpler than those previously given by He and Wei-Wu-Mei. Taking into consideration the importance of finding such new methods and determining their theoretical foundation, we propose the following open problems related to the methods presented in this paper: 5.1. Demonstrate convergence of the methods of Newton-Raphson type based on resolvent equations and determine their speed of convergence. Compare different methods presented in this paper in terms of their speed of convergence. 5.2. Expand the methods based on resolvent equations to systems of algebraic equations. 5.3 Establish criteria to indicate the real root of the resolvent polynomial equation that can be taken as variation of iterations of the exact real root of an algebraic equation.

References [1] R. L. Burden, J. D. Faires, Numerical Analysis, 8th edition, Brook/Cole, 2004. [2] J. H. He, Homotopy Perturbation Technique, Comput. Methods Appl. Mech. Engrg, 178 (1999) 257-262. [3] J. H. He, Variational Iteration Method a Kind of Nonlinear Analytical Technique: Some Examples, International J. of Nonlin. Mech. 344 (1999) 699-708. [4] J. H. He, Improvement of Newton Iteration Method, Int. J. Nonlinear Sci. Numer. Simulation, 1(2000), 239-240. [5] J. H. He, A New Iteration Method for Solving Algebraic Equations, Appl. Math. Comput., 135 (2003), 81-84. [6] D. Wei, J. Wu, M. Mei, A More Effective Iteration Method for Solving Algebraic Equations, Appl. Math. Sci., 2(2008), 28, 1387-1391.


Diffusion Processes With Generalized Beta Density Functions Mario Lefebvre ´ Ecole Polytechnique de Montréal, Canada e-mail: [email protected]

Abstract We consider two time-inhomogeneous (one-dimensional) diffusion processes having a particular generalized beta density function. The initial state of the processes also has a generalized beta distribution. We show that the classical beta distribution is stationary for two transformations of the original diffusion processes. Finally, we find stationary density functions for other diffusion processes.

Mathematics Subject Classification: 60G99. Keywords: Infinitesimal parameters, Kolmogorov forward equation, stationary density functions, time-inhomogeneous processes.

1

Introduction

The main diffusion processes, such as the Wiener and Ornstein-Uhlenbeck processes, are Gaussian processes, so that they can take on any value in R. Similarly, the geometric Brownian motion, which is very important in mathematical finance, takes its values in an infinite interval, namely (0, ∞). Suppose that we want to find a model for a certain random variable X(t), which varies with t, but which remains in a bounded domain for a fixed value of t. For instance, suppose that X(t) always belongs to the interval (0, 1). We could of course consider a diffusion process like the Wiener process (or Brownian motion), if we assume that the boundaries at 0 and 1 are reflecting. Then, the process {X(t), t ≥ 0} will indeed evolve between 0 and 1. However, it would be nice to have a diffusion process for which its first-order density

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function

P [X(t) ∈ (x, x + dx)] dx↓0 dx

f (x; t) := lim

is that of a variable that is intrinsically bounded, rather than being a truncated Gaussian random variable, for example. One such distribution is the beta distribution. Remember that we say that the random variable X has a beta distribution with parameters α and β (both positive) if fX (x) =

xα−1 (1 − x)β−1 B(α, β)

where B(α, β) :=

for 0 < x < 1,

(1)

Γ(α)Γ(β) Γ(α + β)

(and Γ(·) denotes the gamma function). Moreover, if a < b and if we let Y := a + (b − a)X, then Y has a generalized beta distribution on (a, b): fY (y) =

(y − a)α−1 (b − y)β−1 1 (b − a)α+β−1 B(α, β)

for a < y < b.

In the next section, we will find two time-inhomogeneous diffusion processes {X(t), t ≥ 0} for which X(t) has a (t-dependent) generalized beta distribution with parameters α = β = 2. Next, we will see that, in each case, a simple transformation Y (t) of X(t) leads to a diffusion process having a stationary beta distribution with parameters α = β = 2, if we assume that Y (0) also has this distribution. Then, in Section 3, we will find stationary density functions for the Wiener and Ornstein-Uhlenbeck processes, as well as for the geometric Brownian motion.

2

Processes having generalized beta density functions

The first time-inhomogeneous diffusion process that we consider is defined by its infinitesimal mean m(x; t) =

1 x + 1 + t x − (1 + t)

(2)

and infinitesimal variance v(x; t) ≡ 1.

(3)

Diffusion Processes With Generalized Beta Density Functions

107

The density function of X(t) satisfies the Kolmogorov forward equation (see Lefebvre (2007, p. 64), for instance) ∂f ∂ 1 ∂2 + [m(x; t)f (x; t)] − [v(x; t)f (x; t)] = 0. ∂t ∂x 2 ∂x2

(4)

The boundaries at x = 0 and x = t + 1 are assumed to be reflecting. Hence, this equation is considered in the interval (0, t + 1). Remark. Notice that the functions m and v are continuous for all t ≥ 0 and x ∈ (0, t + 1). One can check the validity of the following proposition. Proposition 2.1 The function f (x; t) that satisfies Eq. (4), subject to the initial condition f (x; 0) = is f (x; t) =

x(1 − x) = 6x(1 − x) B(2, 2)

6 x(1 + t − x) (1 + t)3

for 0 < x < 1,

(5)

for 0 < x < t + 1.

Remark. The proposition means that if X(0) has a beta distribution with parameters α = β = 2, then we find that X(t) has a generalized beta distribution (also with parameters α = β = 2), such that a = 0 and b = t + 1. Next, assume that m(x; t) =

x x − (1 + t)

and v(x; t) = 1 + t

(6)

for t ≥ 0 and x ∈ (t, t + 1) (that is, the boundaries at x = t and x = t + 1 are reflecting). As above, the functions m and v are continuous in the intervals considered. We must solve the partial differential equation ∂f 1 ∂2 ∂ x + f (x; t) − [(1 + t)f (x; t)] = 0. (7) ∂t ∂x x − (1 + t) 2 ∂x2 Proposition 2.2 The function f (x; t) = 6(x − t)(1 + t − x) for t < x < t + 1 solves Eq. (7), subject to the initial condition (5). Thus, X(t) has a generalized beta distribution with parameters α = β = 2, and a = t and b = t + 1.

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The two diffusion processes considered above take on their values in a bounded interval, which depends on t. In the case of the process with infinitesimal parameters (2) and (3), let Y (t) :=

X(t) t+1

∀ t ≥ 0.

The stochastic process {Y (t), t ≥ 0} will always remain in the interval (0, 1). Moreover, we calculate f (y; t) = 6y(1 − y) for 0 < y < 1, so that Y (t) has a beta distribution for all t ≥ 0. Hence, the probability density function f (y; t) is stationary for the stochastic process {Y (t), t ≥ 0}. Let us compute the infinitesimal parameters of {Y (t), t ≥ 0}. We have [see Lefebvre (2007, p. 181)]: 1 E [Y (t + ) − Y (t) | Y (t) = y] ↓0 1 X(t + ) X(t) = lim E − X(t) = (1 + t)y . ↓0 1+t+ 1+t

m(y; t) := lim

We find, after some calculation, that 1 m(y; t) = (1 + t)2

1 y−1

.

(8)

Next, we may write that 1 E {Y (t + ) − Y (t)}2 | Y (t) = y ↓0 " # 2 1 X(t + ) X(t) X(t) = (1 + t)y . = lim E − ↓0 1+t+ 1+t

v(y; t) := lim

We obtain that v(y; t) =

1 . (1 + t)2

(9)

We can now state the following proposition. Proposition 2.3 Let {Y (t), t ≥ 0} be the diffusion process having infinitesimal mean (8) and variance (9). If the initial state Y (0) has a beta distribution with parameters α = β = 2, and if the boundaries at 0 and 1 are reflecting, then so has Y (t) for all t > 0.


109

In the case of the diffusion process {X(t), t ≥ 0} having the infinitesimal parameters given in (6), we define Z(t) = X(t) − t for t ≥ 0. Like Y (t), the random variable Z(t) takes on its values in the interval (0, 1). Furthermore, we also obtain that f (z; t) = 6z(1 − z) for 0 < z < 1. Proceeding as above, we can calculate the infinitesimal parameters of {Z(t), t ≥ 0}. Proposition 2.4 Suppose that the diffusion process {Z(t), t ≥ 0} has infinitesimal parameters t+1 and v(z; t) = 1 + t. m(z; t) = z−1 Then, given that Z(0) has a beta distribution with parameters α = β = 2 and that the boundaries at 0 and 1 are reflecting, Z(t) has the same beta distribution ∀ t > 0. In the next section, we will consider the problem of obtaining stationary density functions for important diffusion processes.

3

Stationary probability density functions

If m(x; t) = m(x) and v(x; t) = v(x), and the function f (x; t) can be written as f (x; t) = f (x) ∀ t ≥ 0, then the Kolmogorov forward equation (4) reduces to 1 d2 d [m(x)f (x)] − [v(x)f (x)] = 0, dx 2 dx2 which implies that 1 d [v(x)f (x)] = c, (10) 2 dx where c is a constant. We will find interesting solutions to this ordinary differential equation (o.d.e.) for important diffusion processes. m(x)f (x) −

I) Assume that m(x) ≡ 0 and v(x) ≡ 1, so that {X(t), t ≥ 0} is a standard Brownian motion. It follows that we have: f 0 (x) = c

=⇒

f (x) = c1 x + c0 ,

where c0 and c1 are constants. Let us choose c0 = 0 and c1 = 2. Then, f (x) is a valid density function on the interval (0, 1).

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Proposition 3.1 If {X(t), t ≥ 0} is a standard Brownian motion, considered between reflecting boundaries at x = 0 and x = 1, for which X(0) is a random variable such that fX(0) (x) = 2x for 0 < x < 1, then this probability density function (p.d.f.) is also the p.d.f. of X(t) for t > 0. Remarks. i) This result is also valid if v(x) ≡ v0 > 0. ii) By choosing c1 = 0 and c0 = 1 instead, we obtain that X(t) has a uniform distribution on the interval (0, 1) for all t ≥ 0. II) When m(x) ≡ m0 6= 0 and v(x) ≡ v0 > 0, the diffusion process {X(t), t ≥ 0} is a Wiener process with drift m0 and diffusion coefficient v0 . We must solve v0 m0 f (x) − f 0 (x) = c. 2 We find at once that c 2m0 x f (x) = + c1 exp . m0 v0 As in the previous case, we see that by choosing c1 = 0 and c = m0 , we get a uniform distribution on the interval (0, 1) for all random variables X(t). If we assume that m0 < 0, and if we take c = 0 and c1 = −2m0 /v0 , we obtain the following proposition. Proposition 3.2 When {X(t), t ≥ 0} is a Wiener process with infinitesimal parameters m0 < 0 and v0 > 0 and there is a reflecting boundary at x = 0, if X(0) ∼ Exp(−2m0 /v0 ), that is, if X(0) has an exponential distribution with parameter −2m0 /v0 , then X(t) ∼ Exp(−2m0 /v0 ) for any t > 0 as well. III) Next, let us consider the Ornstein-Uhlenbeck process. That is, {X(t), t ≥ 0} is the diffusion process for which m(x) = −αx, where α > 0, and v(x) ≡ v0 > 0. The o.d.e. that we must solve is −αxf (x) −

v0 0 f (x) = c. 2

The general solution to this differential equation involves the error function and does not lead to a classical distribution for X(t), apart from the Gaussian distribution. Actually, it is well known that the distribution of the stationary Ornstein-Uhlenbeck process is Gaussian. Proposition 3.3 Suppose that X(0) has the probability density function √ α α 2 fX(0) (x) = √ exp − x for x ∈ R, πv0 v0


111

where X(0) is the initial state of the Ornstein-Uhlenbeck process {X(t), t ≥ 0} having infinitesimal mean −αx and infinitesimal variance v0 . Then, fX(t) (x) = fX(0) (x)

for all t > 0.

Remark. The function fX(0) (x) is the p.d.f. of a Gaussian random variable with mean µ = 0 and variance σ 2 = v0 /(2α). By choosing a different constant in front of the exponential function, the random variable X(t) could have a Gaussian distribution, conditioned to be positive or to remain in the interval (0, 1) (if we assume that the boundary at x = 0 (or the boundaries at x = 0 and x = 1) is (are) reflecting), etc. IV) Finally, if {X(t), t ≥ 0} is a geometric Brownian motion, we can write that (11) m(x) = αx and v(x) = x2 , where α ∈ R. Remark. Actually, m(x) = (µ + 12 σ 2 )x and v(x) = σ 2 x2 for a geometric Brownian motion, where µ ∈ R and σ > 0. However, to solve the differential equation (10), we may assume, without loss of generality, that σ = 1 and α = µ + 21 . The general solution of αxf (x) − is f (x) =

d 2 x f (x) = c dx

 2c   + c0 x2(α−1) if α 6= 1/2,   x(2α − 1)    

−2c ln(x) + c0 x

if α = 1/2.

Proposition 3.4 Consider a geometric Brownian motion for which α in (11) is greater than 1/2. If X(0) has the following probability density function: fX(0) (x) = (2α − 1)x2(α−1)

for 0 < x < 1,

then X(t) has the same p.d.f. as X(0) for all t > 0 (assuming that the boundary at x = 1 is reflecting). Remarks. i) Remember that the origin is a natural boundary for the geometric Brownian motion. ii) With α = 3/2, we obtain that fX(0) (x) = 2x, for 0 < x < 1, as in the case of the standard Brownian motion.

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To conclude, it is worth mentioning yet another interesting case. Consider the time-homogeneous diffusion process {X(t), t ≥ 0} characterized by m(x) =

α β (1 − x) + x and v(x) = x(1 − x) 2 2

for 0 < x < 1, where α and β are positive constants. This process has applications in population genetics [see Tavaré and Zeitouni (2004), and also Karlin and Taylor (1981, section 15.2)]. We find that as t tends to infinity, the p.d.f. f (x; t) of X(t) tends to that of a random variable having a beta distribution with parameters α and β. Therefore, if we assume that X(0) also has a beta distribution with parameters α and β, then we can assert that the p.d.f. defined in (1) is stationary for the diffusion process considered. However, we cannot state that X(t) has a generalized beta distribution for any finite (and positive) t.

4

Conclusion and Open Problems

In this work, we have shown that two particular time-inhomogeneous diffusion processes, considered between two reflecting boundaries, have a generalized beta distribution (with parameters α = β = 2) which depends on t. Next, by considering simple transformations of the original processes, we have obtained two diffusion processes for which the beta density function (with the same parameters) is stationary. Finally, in Section 3, we have found various interesting stationary density functions for important diffusion processes in (generally) bounded intervals. This type of work could be carried out in two or more dimensions. In Section 3, other one-dimensional diffusion processes, such as the Bessel process, could also be considered. Finally, it would be nice to find a diffusion process having a t-dependent generalized beta distribution for which X(0) is deterministic; that is, fX(0) (x) would be a Dirac delta function.

References [1] S. Karlin and H. Taylor, A Second Course in Stochastic Processes, Academic Press, New York, (1981). [2] M. Lefebvre, Applied Stochastic Processes, Springer, New York, (2007). [3] S. Tavaré and O. Zeitouni, Lectures on Probability Theory and Statistics. Ecole d’Eté de Probabilités de Saint-Flour XXXI-2001. Ed. J. Picard. Lecture Notes in Mathematics, 1837, Springer, New York, (2004).


On Sequence Spaces of Invariant Means and Lacunary Defined by a Sequence of Orlicz Functions C ¸ i˘ gdem A. Bekta¸s1 and G¨ ulcan Atıci2 Department of Mathematics, Firat University, Elazig, 23119, TURKEY. e-mail:[email protected] Department of Mathematics, Mu¸s Alparslan University, Mu¸s, 49100, TURKEY. e-mail: [email protected]

Abstract The purpose of this paper is to introduce and study some sequence spaces which are defined by combining the concepts of a sequence of Orlicz functions, invariant mean and lacunary convergence. We establish some inclusion relations between these spaces under some conditions. Keywords: Difference sequence, sequence of Moduli, lacunary sequence.

1

Introduction

One hundred years ago mathematics was undergoing a revolution. The Kantian dictate that Euclidean Geometry is the only rationally conceivable basis for the physical universe had been debunked. Numerous alternative geometries, each self-consistent, were being discovered, axiomatized, and developed. Felix Klein found a unifying principle for relating and classifying the various geometries Invariant Theory. The key idea is to classify mathematical structures by the transformations under which they are invariant. Invariant Theory has achieved wide influence in mathematics, physics (including relativity and quantum mechanics), and computer science. The calculus developed here is based upon relatively simple aspects of Invariant Theory. From a mathematical point of view, transition from classical mechanics to quantum mechanics amounts to, among other things, passing from the commutative algebra of classical observables to the non-commutative algebra of

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quantum mechanical observables. Recall that in classical mechanics an observable (e.g. energy, position, momentum, etc.) is a function on a manifold called the phase space of the system. A little more than 50 years after these developments, Alain Connes realized that a similar procedure can in fact be applied to areas of mathematics where the classical notions of space (e.g. measure space, locally compact space, or a smooth space) loses its applicability and pertinence and can be replaced by a new idea of space, represented by a non-commutative algebra. Conne’s theory, which is generally known as noncommutative geometry, is a rapidly growing new area of mathematics that interacts with and contributes to many disciplines in mathematics and physics. For a recent survey, see Conne’s article [3]. Examples of such interactions and contributions include: theory of operator algebras, index theory of elliptic operators, algebraic and differential topology, number theory, standard model of elementary particles, quantum Hall effect, renormalization in quantum field theory and string theory. As cited above operator algebras are presently one of the dynamic areas of mathematics. Invariant means on amenable groups are an important tool in many parts of mathematics, especially in harmonic analysis invariant means and their generalizations for vector-valued functions play also an important role in the stability of functional equations and selections of set-valued functions (see, for example, [8,9]). Thus it seems natural to ask what are possible limitations of the use of invariant means. We will show that invariant means are, in some sense, naturally restricted to reflexive Banach spaces. (see [9]) In this paper, by introducing some sequence spaces which are related to a sequence of Orlicz functions, invariant mean, lacunary convergence, we establish some inclusion relations between these spaces under some conditions. Nowadays operator algebra, operator theory and lacunary convergence play an important role in different areas of mathematics, and its applications, particularly in Mathematics, Physics and Numerical analysis. It is hoped that this study about operator theory serves for researchers who carry research in various fields of science. Let `∞ and c denote the Banach spaces of bounded and convergent sequences x = (xk ), with xk ∈ R or C, normed by kxk = supk |xk |, respectively. The difference sequence spaces was first introduced by Kızmaz [11] and then the concept was generalized by Et and C ¸ olak [5]. Later on, Et and Esi [6] extended the difference sequence spaces to the sequence spaces m X (∆m v ) = {x = (xk ) : (∆v x) ∈ X} ,

for X = `∞ , c or c0 , where v = (vk ) be any fixed sequence of non-zerocomplex i m m−1 m numbers and (∆m xk − ∆m−1 xk+1 ), ∆m vk+i xk+i v xk ) = (∆v v v xk =i=0 (−1) i for all k ∈ N.

Some New Type of Lacunary Generalized Difference...

115

m m The sequence spaces ∆m v (`∞ ), ∆v (c) and ∆v (c0 ) are Banach spaces normed

by m kxk∆ =m i=1 |vi xi | + k∆v xk∞ .

Let σ be a mapping of the set of positive integers into itself. A continuous linear functional φ on `∞ , is said to be an invariant mean or σ−mean if and only if (i) φ(x) ≥ 0 when the sequence x = (xn ) has xn ≥ 0 for all n, (ii) φ(e) = 1, e = (1, 1, ...) (iii) φ(xσ(n) ) = φ(x) for all x ∈ `∞ . If x = (xk ), write T x = (T xk ) = (xσ(k) ). It can be shown that n o Vσ = x ∈ `∞ : limtkn (x) = l, uniformly in n , k

l = σ − lim x where tkn (x) =

xn + xσ1 (n) + xσ2 (n) + ... + xσk (n) [14]. k+1

In the case σ is the translation mapping n → n + 1, σ−mean is often called a Banach limit and Vσ , the set of bounded sequences all of whose invariant means are equal, is the set of almost convergent sequence (see [12]). By a lacunary sequence θ = (kr ); r = 0, 1, 2, ... where k0 = 0, we will mean an increasing sequence of nonnegative integers with kr − kr−1 → ∞. The intervals determined by θ will be denoted by Ir = (kr−1 , kr ] and we let hr = kr − kr−1 . The ratio kr /kr−1 will be denoted by qr . The space of lacunary strongly convergent sequences Nθ was defined by Freedman et al. [7] as 1 |xk − l| = 0, for some l . Nθ = x = (xk ) : lim r hr k∈I r An Orlicz function is a function M : [0, ∞) → [0, ∞) which is continuous, non-decreasing and convex with M (0) = 0, M (x) > 0 for x > 0 and M (x) → ∞ as x → ∞. It is well known that if M is a convex function and M (0) = 0, then M (λx) ≤ λM (x) for all λ with 0 ≤ λ ≤ 1. Lindenstrauss and Tzafriri [13] used the idea of Orlicz function to define what is called an Orlicz sequence space |xk | ∞ `M = x ∈ w :k=1 M < ∞, for some ρ > 0 ρ which is a Banach space with the norm |xk | ∞ kxk = inf ρ > 0 :k=1 M ≤1 . ρ

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Definition 1.1 Any two Orlicz functions M1 and M2 are said to be equivalent if there are positive constant α and β, and x0 such that M1 (αx) ≤ M2 (x) ≤ M1 (βx) for all x with 0 ≤ x ≤ x0 , [10]. Definition 1.2 A sequence space E is said to be solid or normal if (αk xk ) ∈ E whenever (xk ) ∈ E and for all sequences of scalars (αk ) with |αk | ≤ 1, [10]. Definition 1.3 A sequence space E is said to be monotone if it contains the canonical pre-images of all its step spaces, [10]. Remark 1. From the two above definitions it is clear that ”A sequence space E is solid implies that E is monotone”, [2]. The following inequality will be used throughout the article. Let p = (pk ) be a positive sequence of real numbers with 0 < pk ≤ sup pk = G, D = max(1, 2G−1 ). Then for all ak , bk ∈ C for all k ∈ N , we have |ak + bk |pk ≤ D{|ak |pk + |bk |pk }.

2

(1)

Main Results

Definition 2.1 Let M = (Mk ) be a sequence of Orlicz functions, p = (pk ) be any sequence of strictly positive real numbers and u = (uk ) be a sequence of positive real numbers . Then we define the following sequence spaces: pk |tkn (∆m 1 v xk )| θ ∞ m [w , M, p, u]σ (∆v ) = x = (xk ) : sup uk Mk 0 and uniformly in n. Some well-known spaces are obtained by specializing Mk , u, v and m: m θ ∞ m If u = (uk ) = (1, 1, ...) for all k ∈ N, then [wθ , M, p, u]∞ σ (∆v ) = [w , M, p]σ (∆v ), θ m θ 0 m θ 0 m [wθ , M, p, u]σ (∆m v ) = [w , M, p]σ (∆v ) and [w , M, p, u]σ (∆v ) = [w , M, p]σ (∆v ). If Mk = M for all k ∈ N and u = (uk ) = (1, 1, ...) for all k ∈ N, then we have m θ m θ 0 m the sequence spaces [wθ , M, p]∞ σ (∆v ), [w , M, p]σ (∆v ) and [w , M, p]σ (∆v ). If Mk (x) = x for all k ∈ N and u = (uk ) = (1, 1, ...) for all k ∈ N, then we m θ m θ 0 m have the sequence spaces [wθ , p]∞ σ (∆v ), [w , p]σ (∆v ) and [w , p]σ (∆v ). If m = 0 and v = (vk ) = (1, 1, ...) for all k ∈ N, then we obtain [wθ , M, p, u]∞ σ , θ m θ m [w , M, p, u]σ and [wθ , M, p, u]0σ instead of [wθ , M, p, u]∞ (∆ ), [w , M, p, u] (∆ σ σ v v ) θ 0 m and [w , M, p, u]σ (∆v ), respectively.


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m θ 0 θ m m Theorem 2.2 [wθ , M, p, u]∞ σ (∆v ), [w , M, p, u]σ (∆v ) and [w , M, p, u]σ (∆v ) are linear space over the field of complex numbers.

Proof is trivial. Theorem 2.3 [wθ , M, p, u]0σ (∆m v ) is a topological linear space, total paranormed by ( ) pk 1/H m 1 |t (∆ x )| kn v k g (x) = inf ρpr /H : uk Mk ≤ 1, r ≥ 1, m = 1, 2, ... hr k∈Ir ρ where H = max (1, sup pk ). The proof is routine verification by using standard arguments and therefore omitted. θ m Theorem 2.4 Let M be an Orlicz function, then [wθ , M, p]0σ (∆m v ) ⊂ [w , M, p]σ (∆v ) m ⊂ [wθ , M, p]∞ σ (∆v ).

Proof. The first inclusion is obvious. We establish the second inclusion. Let x ∈ [wθ , M, p]σ (∆m v ). Then there exists some positive number ρ1 such that pk 1 |tkn (∆m v xk − le)| M →0 hr k∈Ir ρ1 as r → ∞ uniformly in n. Define ρ = 2ρ1 . Since M is non-decreasing and convex, we have pk 1 |tkn (∆m v xk )| M hr k∈Ir ρ pk pk D |tkn (∆m D |le| v xk − le)| ≤ M + M hr k∈Ir ρ1 hr k∈Ir ρ ( 1 G ) pk m D |tkn (∆v xk − le)| |le| ≤ M + D max 1, M hr k∈Ir ρ1 ρ1 m by (1). Thus x ∈ [wθ , M, p]∞ σ (∆v ).

Theorem 2.5 Let M = (Mk ) be a sequence of Orlicz functions. If supk [Mk (z)]pk < ∞ for all z > 0, then θ ∞ m [wθ , M, p]σ (∆m v ) ⊂ [w , M, p]σ (∆v ).

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Proof. Let x ∈ [wθ , M, p]σ (∆m v ). By using (1), we have pk pk 1 |tkn (∆m D |tkn (∆m v xk )| v xk − le)| Mk ≤ Mk hr k∈Ir ρ hr k∈Ir ρ pk |le| D + Mk . hr k∈Ir ρ Since supk [Mk (z)]pk < ∞, we can take that supk [Mk (z)]pk = K. Hence we m get x ∈ [wθ , M, p]∞ σ (∆v ). This completes the proof. The proofs of the following theorems are obtained by using the same technique of Bekta¸s [1], therefore we give it without proof. Theorem 2.6 Let M = (Mk ) be a sequence of Orlicz functions. Then the following statements are equivalent: m θ ∞ m (i) [wθ , p]∞ σ (∆v ) ⊂ [w , M, p]σ (∆v ), θ ∞ m θ 0 (ii) [w , p]σ (∆v ) ⊂ [w , M, p]σ (∆m v ), 1 pk (iii) supr hr k∈I [Mk (z /ρ )] < ∞ for all z, ρ > 0. r

Theorem 2.7 Let M = (Mk ) be a sequence of Orlicz functions. Then the following statements are equivalent: m θ 0 (i) [wθ , M, p]0σ (∆m v ) ⊂ [w , p]σ (∆v ), θ ∞ m θ 0 (ii) [w , M, p]σ (∆v ) ⊂ [w , p]σ (∆m v ), 1 pk (iii) inf r hr k∈I [Mk (z /ρ )] > 0 for all z, ρ > 0. r

m Theorem 2.8 Let M = (Mk ) be a sequence of Orlicz functions. [wθ , M, p]∞ σ (∆v ) ⊂ [wθ , p]0σ (∆m v ) if and only if

1 [Mk (z /ρ )]pk = ∞ (z, ρ > 0). r→∞ hr k∈I r lim

Theorem 2.9 Let m ≥ 1 be a fixed integer, then m−1 m (i) [wθ , M, p, u]∞ ) ⊂ [wθ , M, p, u]∞ σ (∆v σ (∆v ), (ii) [wθ , M, p, u]σ (∆m−1 ) ⊂ [wθ , M, p, u]σ (∆m v v ), θ 0 m−1 θ 0 (iii) [w , M, p, u]σ (∆v ) ⊂ [w , M, p, u]σ (∆m v ). Proof. The proof of the inclusions follows from the following inequality pk pk |tkn (∆m D |tkn (∆m−1 xk )| 1 v xk )| v uk Mk ≤ uk Mk hr k∈Ir ρ hr k∈Ir ρ pk D |tkn (∆m−1 xk+1 )| v + uk Mk . hr k∈Ir ρ


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Theorem 2.10 Let M = (Mk ) and T = (Tk ) be any two sequence of Orlicz functions. Then we have m θ ∞ m θ ∞ m (i) [wθ , M, p, u]∞ σ (∆v ) ∩ [w , T , p, u]σ (∆v ) ⊂ [w , M + T , p, u]σ (∆v ), θ m θ m (ii) [wθ , M, p, u]σ (∆m v ) ∩ [w , T , p, u]σ (∆v ) ⊂ [w , M + T , p, u]σ (∆v ), m θ 0 m θ 0 (iii) [wθ , M, p, u]0σ (∆m v ) ∩ [w , T , p, u]σ (∆v ) ⊂ [w , M + T, p, u]σ (∆v ). m θ ∞ m Proof. (i) Let x ∈ [wθ , M, p, u]∞ σ (∆v ) ∩ [w , T , p, u]σ (∆v ). Then pk 1 |tkn (∆m v xk )| sup uk Mk 1, then there exists δ > 0 such that kr ) ≥ 1 + δ for all r ≥ 1. Furthermore we have hkrr ≤ (1+δ) and qr = ( kr−1 δ kr−1 hr

≤ 1δ , for all r ≥ 1. Then we may write pj |tjn (∆m 1 v xj )| uj Mj hr j∈Ir ρ pj pj 1 kr |tjn (∆m 1 kr−1 |tjn (∆m v xj )| v xj )| = uj Mj − uj Mj hr j=1 ρ hr j=1 ρ pj m kr |tjn (∆v xj )| r = kr−1 kj=1 uj Mj hr ρ pj |tjn (∆m kr−1 v xj )| −1 kr−1 kr−1 j=1 uj Mj . − hr ρ Now suppose that lim supr qr < ∞ and let ε > 0 be given. Then there exists s0 such that for every s ≥ s0 pk 1 |tkn (∆m v xk )| As = uk Mk < ε. hs k∈Is ρ We can also choose a number K > 0 such that As < K for all s. If lim supr qr < ∞, then there exists a number β > 0 such that qr < β for all r. Now let i be any integer with kr−1 < i ≤ kr , where r > L. Then pj |tjn (∆m v xj )| −1 i ij=1 uj Mj ρ pj |tjn (∆m v xj )| −1 kr ≤ kr−1 j=1 uj Mj ρ pj pj |tjn (∆m |tjn (∆m v xj )| v xj )| −1 = kr−1 j∈I1 uj Mj +j∈I2 uj Mj ρ ρ pj m |tjn (∆v xj )| +... +j∈Ir uj Mj ρ pj pj |tjn (∆m |tjn (∆m s0 v xj )| v xj )| −1 r = kr−1 s=1 j∈Is uj Mj +s=s0 +1 j∈Is uj Mj ρ ρ pj m |tjn (∆v xj )| s0 −1 −1 = kr−1 + ε(kr − ks0 )kr−1 s=1 j∈Is uj Mj ρ −1 −1 ≤ kr−1 {h1 A1 + h2 A2 + ... + hs0 As0 } + ε(kr − ks0 )kr−1 −1 −1 ≤ kr−1 sup As ks0 + ε(kr − ks0 )kr−1