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I S S N 2provided 3 4 7 -by1921 KHALSA PUBLICATIONS Volume 12 Number 06 J o u r n a l of Advances in Mathematics
Leibniz’s rule and Fubini’s theorem associated with Hahn difference operators
å Alaa E. Hamza † , S. D. Makharesh † † Jeddah University, Department of Mathematics, Saudi, Arabia † Cairo University, Department of Mathematics, Giza, Egypt E-mail: [email protected] å † Cairo University, Department of Mathematics, Giza, Egypt E-mail: [email protected] ABSTRACT
In 1945 , Wolfgang Hahn introduced his difference operator Dq, , which is defined by f (qt ) f (t) D f (t) = , t , q, (qt ) t 0 where = with 0 < q <1, > 0. In this paper, we establish Leibniz’s rule and Fubini’s theorem associated with 0 1 q this Hahn difference operator.
Keywords. q, -difference operator; q, –Integral; q, –Leibniz Rule; q, –Fubini’s Theorem. 1Introduction The Hahn difference operator is defined by f (qt ) f (t) D f (t) = , t , (1) q, (qt ) t 0
where q(0,1) and > 0 are fixed, see [2]. This operator unifies and generalizes two well known difference operators. The first is the Jackson q -difference operator which is defined by f (qt) f (t) D f (t) = , t 0, (2) q qt t
see [3, 4, 5, 6]. The second difference operator which Hahn’s operator generalizes is the forward difference operator f (t ) f (t) f (t) = , t R, (3) (t ) t where is a fixed positive number, see [9, 10, 13, 14]. The associated integral of (3) is the well known Nörlund sum
x f (t)t = f (x k), (4) k=0 see [12, 13, 15]. In some literature Nörlund sums are called the indefinite sums, cf. [14]. Then we can define
b f (t)t = f (a k) f (b k), (5) a k=0 whenever the convergence of the series is guaranteed. Note that, under appropriate conditions,
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lim Dq, f (t) = f (t), q1
lim Dq, f (t) = Dq f (t), (6) 0
lim Dq, f (t) = f (t). q1,0
In [1], A.Hamza et al. gave a rigorous analysis of the calculus associated with Dq, . They stated and proved some basic properties of such a calculus. For instance, they defined the inverse of Dq, which contains the right inverse of Dq and the right inverse of . Then, they proved a fundamental lemma of Hahn calculus. This paper is devoted to establishing Leibniz’s rule and Fubini’s theorem associated with the q, - difference operator. We organize this paper as follows. Section 2 gives an introduction to q, - difference calculus. In Section 3, we prove Leibniz’s rule which is concerning with differentiating under the integral sign. Some related results are obtained. Also, we prove Fubini’s theorem in Hahn difference operator setting, that is, we prove that the iterated integrals are equal. 2 Preliminaries
Let N be the set of natural numbers and N0 := N{0}. For k N0 and 0 < q <1, we define the q -numbers 1 qk [k] := . q 1 q
Let I is an interval of R containing 0 , where 0 := /(1 q), and h denote the transformation
h(t):= qt , t I.
One can see that
> t, fort < 0 , h(t) = t, fort = 0 , < t, fort > 0.
1 The transformation h has the inverse h (t) = (t )/q,t I . The k th order iteration of h is given by
k k h (t) := hhh(t) = q t [k]q , t I, (1) ktimes
k 1 k 1 1 1 t [k]q (h (t)) := h (t) := hhh(t) = k , t I. (2) ktimes q
k Furthermore, {h (t)}k=1 is a decreasing (an increasing) sequence in k when t > 0 ( t < 0 ) with
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k inf h (t), t > 0 , kN 0 = (3) k suph (t), t < 0. kN
k The sequence {h (t)}k=1 is increasing (decreasing), t > 0 ( t < 0 ) with
k suph (t), t > 0 , kN = (4) k inf h (t)), t < 0. kN Let f be a function defined on I . The Hahn difference operator is defined in [2] by f (qt ) f (t) D f (t) := , if t , (5) q, (qt ) t 0
' and Dq, f (0 ) = f (0 ) , provided that f is differentiable at0 , where q(0,1) and > 0. In this case, we call Dq, f , the q, -derivative of f . Finally, we say that f is q, -differentiable, i.e. throughout I , if Dq, f (0 ) exists. One can easily check that if f , g are q, -differentiable at t I, then
Dq, (f g)(t) = Dq, f (t) Dq, g(t), , C,
Dq, ( fg)(t) = Dq, ( f (t))g(t) f (qt )Dq, g(t),
f Dq, ( f (t))g(t) f (t)Dq, g(t) Dq, (t) = , g g(t)g(qt ) provided in the last identity g(t)g(qt ) 0, cf. [1]. The right inverse for Dq, is defined in [1] in terms of Jackson-Nörlund sums as follows. Let a,b I the q, -integral of f from a to b is defined to be
b b a f (t)dq,t := f (t)dq,t f (t)dq,t, (6) a 0 0
x k k f (t)dq,t := x(1 q) q f (xq [k]q ), x I, (7) 0 k=0 provided that the series converges at x = a and x = b. It is known that if f is continuous at 0 , then the series in (7) is uniformly convergent.
In the folloing we present some needed results from [1] concerning the calculus associated with Dq. . Corollary 1 The series
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qk (t(1 q) ), k=0 is uniformly convergent to | t 0 | on interval I = [a,b] containing 0 .
Theorem 2Let X be a Banach space endowed with a norm⃦. .⃦. Assume f : I X is continuous at 0 . Then the following statements are true
k • { f ((sq ) [k]q )}kN converges uniformly to f (0 ) on I.
k k • q f (sq [k] ) is uniformly convergent on I and consequently f is q, -integrable over I . k=0 q • The function
x F(x) = f (t)dq,t, x I, 0 is continuous at0 . Furthermore, Dq, F(x) exists for every x I and
Dq, F(x) = f (x).
Conversely,
b Dq, f (t)dq,t = f (b) f (a) for all a,b I. a
Consequently, the q, - integration by parts for continuous function f , g is giving in by
b b b f (t)Dq, g(t)dq,t = f (t)g(t) |a Dq, ( f (t))g(qt )dq,t, a,b I. (8) a a
Theorem 3 Let f : I R be continuous function at 0 , then for t I
t f (s)dq, s = (t h(t)) f (t). h(t)
We will apply the time scales calculus tools, see [7, 8], to obtain a q, -analog of the chain rule.
Theorem 4 Let g : I R be continuous and q, -differentiable and f :R R be continuously differentiable. Then, there exists c between qt and t such that,
' Dq, ( f g)(t) = f (g(c))Dq, g(t).(9)
3 q, -Differentiation Under The Integral Sign
In this section we study the continuity and the q, -differentiation of the integral
(t) f (t, s)dq, s. (t)
We establish Libnitz’s rule. Finally, we prove that the iterated integrals are equal (this theorem is known by Fubini’s Theorem). Let f : I I R . We begin by the following definitions.
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Definition 1
(i) We say that f (t,s) is continuous at t = t0 uniformly with respect to s[a,b] I if f (t,s) = f (t ,s)uniformly with respect to s[a,b] I. limtt0 0
(ii) The q, -partial derivative of f with respect to t is defined by f (t, s) f (h(t), s) t I \ , t h(t) 0 Dq,,t f (t, s) = f (t, s) f (0 , s) lim t = 0 , t 0 t 0 whenever the limit exists.
(iii) We say that f (t,s) is uniformly partially differentiable at t = 0 I , with respect to s[a,b] I if f (t, s) f ( , s) lim 0 t 0 t 0 exists uniformly with respect to s[a,b] .
Definition 2 Let 0 [a,b] I . We define the q, -interval by
k k [a,b]q, ={aq [k]q : k N0}{bq [k]q : k N0}{0}.
For any point c I , we denote by
k [c]q, ={cq [k]q : k N0}{0}.
(t) From now on, we assume some appropriate conditions that imply the integrals of the form g(t, s)dq, s := F(t) exist, (t) where g : I I R and , : I I .
Lemma 3 The following statements are true
(i) Assume that f (t,s) is continuous at t = 0 uniformly with respect to s[b]q, . Then
b F(t) = f (t, s)dq, s 0 is continuous at t = 0 .
b (i) Dq, F(t) = Dq,,t f (t,s)dq, s, t 0 . 0
(ii) If Dq,,t f (t,s) exists uniformly at t = 0 with respect to s[b]q, , then Dq, F(t) exists at t = 0 and
b D F( ) = D f ( , s)d s. q,0 0 q,,t 0 q, 0 Proof.
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(i) Let > 0 . There exists > 0 such that
k k t I(| t 0 |< | f (t,bq [k]q ) f (0 ,bq [k]q ) |< ,k). | b 0 |
In view of corollary 1, for t I]0 , 0[ , we have
| F(t) F( ) | =| qk (b(1 q) )[ f (t,bqk [k] ) f ( ,bqk [k] )]| 0 q 0 q k=0 < .
(ii) For t 0 , we have
k q (b(1 q) ) k k Dq, F(t) = [ f (t,bq [k]q ) f (h(t),bq [k]q )] k=0 t h(t) k k = q (b(1 q) )Dq,,t f (t,bq [k]q ) k=0 b = Dq,,t f (t, s)dq, s. 0
(iii) Assume that Dq,,t f (t,s) exists uniformly at t = 0 with respect to s[b]q, . We conclude that
b F(t) F(0 ) | Dq,,t f (0 , s)dq, s | t 0 0 k q (b(1 q) ) k =| f (t,bq [k]q ) k=0 t 0 k q (b(1 q) ) k f (0 ,bq [k]q ) k =0 t 0 k k q (b(1 q) )Dq,,t f (0 ,bq [k]q ) | k =0 f (t,bqk [k] ) f ( ,bqk [k] ) =| (qk (b(1 q) )[ q 0 q k=0 t 0 k Dq,,t f (0 ,bq [k]q )]|.
For > 0, there exists > 0 such that for all t I we have 0 <| t 0 |<
k k f (t,bq [k]q ) f (0 ,bq [k]q ) k | Dq,,t f (0 ,bq [k]q ) |< . t 0 | b 0 |
In view of corollary 1, for t I such that 0 <|t 0 |< , we see that
b F(t) F(0 ) | Dq,,t f (0 , s)dq, s |< . t 0 0
Corollary 4 If Dq,,t f (t,s) exists uniformly at t = 0 with respect to s[a]q, and s[b]q, , then
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b b Dq,,t f (t,s)dq, s = Dq,,t f (t,s)dq, s, t I. a a Proof. By lemma 3, we get the desired result from the following inequality
b b a Dq,,t f (t,s)dq, s = Dq,,t [ f (t,s)dq, s f (t,s)dq, s]. a 0 0 Theorem 5 (Leibniz's integral rule) Define the function F by
t F(t) := f (t, s)dq, s. 0 The following statements are true
(i) For t 0 , we have
t Dq, F(t) = f (h(t),t) Dq,,t f (t, s)dq, s. (1) 0
(ii) If f (t,s) is continuous at (0 ,0 ) , then F is differentiable at t = 0 and ' F (0 ) = f (0 ,0 ) . Proof.
(i) At t 0 , we have
k q (t(1 q) ) k Dq, F(t) = [ f (t,tq [k]q ) k=0 t h(t)
k k f (h(t),tq [k]q ) f (h(t),tq [k]q )]
k1 q (t(1 q) ) k1 f (h(t),tq [k 1]q ) k=0 t h(t)
k q (t(1 q) ) k k = [ f (t,tq [k]q ) f (h(t),tq [k]q )] k=0 t h(t)
k q (t(1 q) ) k f (h(t),tq [k]q ) k=0 t h(t)
k1 q (t(1 q) ) k1 f (h(t),tq [k 1]q ) k=0 t h(t)
1 h(t) t = [ f (h(t), s)dq, s f (h(t), s)dq, s] t h(t) 0 0
t f (h(t), s) f (t, s) dq, s 0 t h(t)
1 h(t) t = f (h(t), s)dq, s Dq,,t f (t, s)dq, s. t h(t) t 0
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By theorem 3, we get
t Dq, F(t) = f (h(t),t) Dq,,t f (t,s)dq, s. 0 (ii) Using corollary 1, we see that
F(s) F(0 ) | f (0 ,0 ) | s 0 1 k k = | q (s(1 q) )[ f (s, sq [k]q ) f (0 ,0 )]|. (s 0 ) k=0
The continuity of f (t,s) at (0 ,0 ) implies that given > 0 , there exists > 0 such that k k | s 0 |< , ( and consequently sq [k]q < ), implies | f (s,sq [k]q ) f (0 ,0 ) |< . Thus
F(s) F(0 ) | f (0 ,0 ) |< , s 0 whenever 0 <| s 0 |< , which completes the proof. Theorem 6 Let : I I be bounded. Define the function F by
(t) F(t) := f (t, s)dq, s. 0 The following statements are true
(i) The function F is q, -differentiable at t 0 and
1 (h(t)) (t) Dq, F(t) = f (h(t), s)dq, s Dq,,t f (t,s)dq, s, t 0. (2) t h(t) (t) 0
(ii) Assume that the following conditions hold - is q, -differentiable.
- f (t,s) is uniformly partially differentiable at t = 0 with respect to s I such that Dq,,t f (0 ,s) is continuous at s = 0 .
t - The function H(t) = f (0 , s)dq, s is differentiable at (0 ) . 0
Then Dq, F(t) exists at t = 0 and
( ) ' 0 Dq, F(0 ) = f (0 ,(0 )) (0 ) Dq,,t f (0 ,s)dq, s. 0 Proof.
(i) For t 0 we have
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k q ((t)(1 q) ) k Dq, F(t) = [ f (t,(t)q [k]q ) k=0 t h(t)
k f (h(t),(t)q [k]q )]
k q ((t)(1 q) ) k f (h(t),(t)q [k]q ) k=0 t h(t)
k1 q ((t)(1 q) ) k1 f (h(t),(t)q [k 1]q ) k=0 t h(t)
(t) f (h(t), s) f (t, s) 1 (t) = dq, s [ f (t, s)dq, s 0 t h(t) t h(t) 0
(h(t)) f (t, s)dq, s] 0
1 (t) (t) = f (t, s)dq, s Dq,,t f (t, s)dq, s. t h(t) (h(t)) 0
(ii) To ensure the differentiability at t = 0 , we write F as follows F(t) = P(t) (H )(t), where
(t) P(t) = { f (t, s) f (0 , s)}dq, s. 0
(0 ) First, we show that Dq, P(0 ) = Dq,,t f (0 ,s)dq, s . Indeed, one can see that 0
( ) P(t) P(0 ) 0 | Dq,,t f (0 , s)dq, s | t 0 0 (t) (t) f (t, s) f (0 , s) =| [ Dq,,t f (0 , s)]dq, s Dq,,t f (0 , s)dq, s 0 t 0 0 (0 ) Dq,,t f (0 , s)dq, s | 0 (t) (t) f (t, s) f (0 , s) |[ Dq,,t f (0 , s)]|dq, s | Dq,,t f (0 , s)dq, s 0 t 0 0 (0 ) Dq,,t f (0 , s)dq, s |. 0
Since f (t,s) is uniformly partially differentiable at t = 0 with respect to s I , and is bounded, then
(t) f (t,s) f (0 ,s) | Dq,,t f (0 ,s) |dq, s 0as t 0. 0 t 0
t The continuity of Dq,,t f (0 ,s) at s=0 implies that K(t)= Dq,,t f (0 , s)dq, s is continuous at 0 t = 0 which in turn implies that K((t)) is continuous at t = 0 , that is
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(t) (0 ) | Dq,,t f (0 ,s)dq, s Dq,,t f (0 ,s)dq, s | 0ast 0. 0 0 Consequently, we conclude that
( ) P(t) P(0 ) 0 | Dq,,t f (0 , s)dq, s | 0ast 0. (3) t 0 0
Since H is differentiable at (0 ) and is differentiable at 0 , then H is differentiable at ' ' ' t = 0 and Dq, (H )(0 ) = (H ) (0 ) = H ((0 )) (0 ) = f (0 ,(0 ))(0 ). Therefore, we get the desired result. Corollary 7 Let , : I I be bounded functions. Define the function F by
(t) F(t) := f (t, s)dq, s. (t) Then, the following statements are true
(i) For t 0 , we have
1 (t) (t) Dq, F(t) = [ f (h(t), s)dq, s f (h(t), s)dq, s] (h(t)) (h(t) t h(t) (t) Dq,,t f (t, s)dq, s, t 0. (t)
(ii) Assume that the following conditions hold - and are q, -differentiable.
- f (t,s) is uniformly partially differentiable at t = 0 with respect to s I such that Dq,,t f (0 ,s) is continuous at s = 0 .
t - The function H(t) = f (0 , s)dq, s is differentiable at (0 ) and (0 ) . 0
Then Dq, F(t) exists at t = 0 and
( ) ' ' 0 Dq, F(0 ) = f (0 , (0 )) (0 ) f (0 ,(0 )) (0 ) Dq,,t f (0 ,s)dq, s. (0 ) Proof. By theorem 6 and using the definition
(t) (t) (t) Dq,,t f (t,s)dq, s = Dq,,t [ f (t,s)dq, s f (t,s)dq, s], (t) 0 0 we get the desired result.
Theorem 8 (Fubini's theorem)
Let f be defined on the closed rectangle R = [0 ,a][0 ,b] I I . Assume that f (t,s) is
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continuous at t = 0 uniformly with respect to s[a]q, and continuous at s = 0 uniformly with respect to t [b]q, . Then, the double q, -integrals
b a a b f (t,s)dq,t dq, s and f (t,s)dq, s dq,t 0 0 0 0 exist and they are equal, that is
b a a b f (t, s)dq,t dq, s = f (t,s)dq, s dq,t. 0 0 0 0
a b Proof. By assumptions, lemma 3 tells us that f (t, s)dq,t and f (t, s)dq, s are continuous at 0 0
s = 0 and at t = 0 respectively. Therefore both double q, -integrals above exist and we have
b a b j j f (t, s)dq,tdq,ø s = [ q (a(1 q) ) f (aq [ j]q , s)]dq, s 0 0 0 j=0 k j j k =q (b(1 q) )(a(1 q) ) f (aq [ j]q ,bq [k]q ) k =0 j=0 jk j k =q (a(1 q) )(b(1 q) ) f (aq [ j]q ,bq [k]q ) j=0 k =0 a a b k k = [ q (b(1 q) ) f (t,bq [k]q )]dq,t = f (t, s)dq, sdq,t. 0 k =0 0 0 References [1] M. H. Annaby, A. E. Hamza and K. A. Aldwoah, Hahn difference operator and associated Jackson-Norlund integrals, Journal of optimization theory and applications, 154 (2012), 133-153. [2] W. Hahn, Uber orthogonalpolynoe, die q -differenzenlgleichunge geungeu, Math. Nachr 2(1949). src: 888 nnoo.tex
[3] G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, Cambridge, 1999. [4] R. D. Carmichael, Linear difference equations and their analytic solutions, Trans. Amer. Math. Soc., 12 (1911), 99-134. [5] M. E. Ismail, Classical and Quantum Orthogonal Polynomials in One variable, Encyclopedia Math. Appl. Cambridge Univ.Press, Cambridge, 98 (2005). [6] F. H. Jackson, On q -functions and a certain difference operator, Trans. Roy. Soc. Edin., 46 (1908), 253-281.
[7] M. Bohner, A. Peterson, Advanced in Dynamic Equations on Time Scales. Birkhäuser, Basel, 2003. [8] M. Bohner, A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications. Birkhäuser, Basel, 2001. [9] M. T. Bird, On generalizations of sum formulas of the Euler-Maclaurin type, Amer. J. Math, 58(1936), 487-503. [10] G. D. Birkhoff, General theory of linear difference equations, Trans. Amer. J.Math. Soc., 12(1911), 243-284. [11] R. D. Carmichael, Linear difference equations and their analytic solutions, Trans. Amer. Math. Soc., 12(1911), 99-134. [12] N. Nörlund, Vorlesungen über Differencenrechnung, Springer Verlag, Berlin, 1924. [13] D. L. Jagerman, Difference Equations with Applications to Queues, Marcel Dekker, New York, 2000. [14] C. Jordan, Calculus of Finite Differences, Chelsea, New York, 1965. [15] T. Fort, Finite Differences and Difference Equations in Real Domain, Oxford University Press, Oxford, 1948.
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