Research Article Pascu-Type Harmonic Functions with Positive Coefficients Involving Salagean Operator

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1 International Analysis Article ID pages Research Article Pascu-Type Haronic Functions with Positive Coefficients Involving Salagean Operator K. Vijaya G. Murugusundaraoorthy and M. Kasthuri School of Advanced Sciences VIT University Vellore India Correspondence should be addressed to K. Vijaya; Received 5 Noveber 013; Accepted 1 February 014; Published 6 April 014 Acadeic Editor: Rei Léandre Copyright 014 K. Vijaya et al. This is an open access article distributed under the Creative Coons Attribution License which perits unrestricted use distribution and reproduction in any ediu provided the original work is properly cited. Making use of a Salagean operator we introduce a new class of coplex valued haronic functions which are orientation preserving and univalent in the open unit disc. Aong the results presented in this paper including the coeffcient bounds distortion inequality and covering property extree points certain inclusion results convolution properties and partial susforthisgeneralized class of functions are discussed. 1. Introduction and Preliinaries A continuous function f = u + iv is a coplex-valued haronic function in a coplex doain G if both u and V are real and haronic in G. In any siply connected doain D G wecanwritef=h+g h and g are analytic in D. Wecallh the analytic part and g the coanalytic part of f. A necessary and sufficient condition for f to be locally univalent and orientation preserving in D is that h (z) > g (z) in D (see 1). Denote by H the faily of functions f=h+g (1) which are haronic univalent and orientation preserving in theopenunitdiscu ={z: z <1}so that f is noralized by f(0) = f z (0) = 0.Thusforf=h+g Hthefunctionsh and g are analytic in U and can be expressed in the following fors: h (z) =z+ a n z n g(z) = and f(z) is then given by b n z n (0 b 1 <1) () =z+ a n z n + b n z n (0 b 1 <1). (3) We note that the faily H of orientation preserving noralized haronic univalent functions reduces to the well-known class S of noralized univalent functions if the coanalytic part of f is identically zero; that is g 0. For functions f HJahangiri et al. defined Salagean operator on haronic functions given by D l =D l h (z) + () l D l g (z) (4) D l h (z) =z+ n l a n z n D l g (z) = n l b n z n. In 1975 Silveran 3 introduced a new class T of analytic functions of the for f(z) = z a n z n and opened up a new direction of studies in the theory of univalent functions as well as in haronic functions with negative coefficients 4. Uralegaddi et al. 5 introduced analogous subclasses of star-like convex functions with positive coefficients and opened up a new and interesting direction of research. In fact they considered the functions the coefficients are positive rather than negative real nubers. Motivated by the initial work of Uralegaddi et al. 5 any researchers (see 6 9) introduced and studied various new subclasses of analytic functions with positive coefficients but analogues results on haronic univalent (5)

2 International Analysis functions have not been explored in the literature. Very recentlydixitandporwal10 atteptedtofillthisgapby introducing a new subclass of haronic univalent functions with positive coefficients. Denote by V H the subfaily of H consisting of haronic functions f=h+g of the for A (z) = (1 λ) D l+1 +λd l+ =z+ n l+1 (1 λ+nλ) a n z n (10) =z+ a n z n b n z n (a n 0; b n 00 b 1 <1). Motivated by the earlier works of onthesubject of haronic functions in this paper an attept has been ade to study the class of functions f V H associated with Salagean operator on haronic functions. Further we obtain a sufficient coefficient condition for functions f Hgiven by (3) and also show that this coefficient condition is necessary for functions f V H theclassofharonicfunctionswith positive coefficients. Distortion results and extree points inclusion relations and convolution properties and results on partial sus are discussed extensively. For 0 λ 1 1 < γ 4/3weletP l H (λ γ) be a new subclass of H consisting of all functions of the for (3) satisfying the condition (6) R ( (1 λ) Dl+1 +λd l+ (1 λ) D l +λd l+1 ) <γ (7) () l n l+1 (1 λ nλ) b n z n B (z) = (1 λ) D l +λd l+1 =z+ n l (1 λ+nλ) a n z n + () l n l (1 λ nλ) b n z n. Substituting for A(z) and B(z) in (9) we get A (z) /B (z) A (z) /B (z) () ( n l 1 λ+nλ (n) z n + () l n l 1 λ nλ (n+1) b n z n ) (() n l 1 λ+nλ n () (11) D l f(z) is given by (4) (see). Also let V l H (λ γ) = P l H (λ γ) V H.. Coefficient Bounds In our first theore we obtain a sufficient coefficient condition for haronic functions in P l H (λ γ). Theore 1. Let f=h+g be given by (3).If n l 1 λ+nλ (n γ) + n l 1 λ nλ (n + γ) b n 1 a 1 =1and 1<γ 4/3thenf P l H (λ γ). Proof. We let (8) holdforthecoefficientsoff = h + g. It suffices to show that (8) A (z) /B (z) <1 z U (9) A (z) /B (z) () z n () l n l 1 λ nλ n+() b n z n ) ( n l 1 λ+nλ (n) + () l n l 1 λ nλ (n+1) b n ) (() n l 1 λ+nλ n () () l n l 1 λ nλ n + (γ 1) b n. ) (1) The above expression is bounded above by 1 if n l 1 λ+nλ (n) +n l 1 λ nλ (n+1) b n () (13)

3 International Analysis 3 which is equivalent to n l 1 λ+nλ (n γ) n l 1 λ nλ (n + γ) + b n 1. (14) But (8) is true by hypothesis. Hence (A(z)/B(z) 1)/ (A(z)/B(z) (γ 1)) < 1 z Uandthetheoreisproved for A(z) and B(z) is given by (10)and(11) respectively. Theore. For a 1 = 1 and 1 < γ 4/3 f = h + g (λ γ) if and only if V l H n l 1 λ+nλ (n γ) + n l 1 λ nλ (n + γ) b n 1. (15) Proof. Since V l H (λ γ) Pl H (λ γ)weonlyneedtoprovethe only if part of the theore. To this end for functions f of the for (6) we notice that the condition Equivalently R ( (1 λ) Dl+1 +λd l+ (1 λ) D l +λd l+1 )<γ. (16) The above required condition ust hold for all values of z in U. Upon choosing the values of z on the positive real axis 0 z=r<1weusthave (1 γ n l 1 λ+nλ (n γ) a n r n n l 1 λ nλ (n + γ) b n r n ) (1 n l (1 λ+nλ) a n r n 0. + n l 1 λ nλ b n r n ) (18) If condition (15) does not hold then the nuerator in (18) is negative for r sufficiently close to 1. Hence there exists z 0 = r 0 in (0 1) for which the quotient of (18) isnegative. This contradicts the required condition for f V l H (λ γ). This copletes the proof of the theore. 3. Distortion Bounds and Extree Points By routine procedure (see 10 13) we can easily prove the following results; hence we state the following theores without proof for functions in V l H (λ γ). Theore 3 (distortion bounds). Let f V l H (λ γ).thenfor z = r < 1wehave (1 b 1 )r 1 l (1+λ) ( γ 1+γ γ b 1 )r (1+ b 1 )r+ 1 l (1+λ) ( γ 1+γ ( γ) b 1 )r. (19) Corollary 4 (covering result). If f(z) V l H (λ γ)then {w : w < l+1 (1 + γ) + 1 l (1 + γ) + 1 γ l ( γ) (1+λ) R (((1 γ)z n l (1 λ+nλ)(n γ) a n z n 0. () l n l (1 λ nλ)(n + γ) b n z n ) (z n l (1 λ+nλ) a n z n + () l n l (1 λ nλ) b n z n ) ) (17) l+1 (1 + γ) 1 l (1 + γ) + 1 γ }b l ( γ) 1 (1+λ) f(u). (0) Next we state the extree points of closed convex hulls of V l H (λ γ) denoted by clco Vl H (λ γ). Theore 5. Afunctionf(z) V l H (λ γ) if and only if f(z) = (X nh n (z) + Y n g n (z)) h 1 (z) = z h n (z) = z + ((γ 1)/( l 1 λ + nλ (n γ)))z n (n )andg n (z) = z + ((γ 1)/( l 1 λ nλ (n + γ)))z n (n 1) also (X n +Y n )= 1 X n 0 and Y n 0. In particular the extree points of V l H (λ γ) are {h n} and {g n }.

4 4 International Analysis Theore 6. The faily V l H (λ γ) is closed under convex cobinations. 4. Inclusion Results Now we will exaine the closure properties of the class V l H (λ γ) under the generalized Bernardi-Libera-Livingston integral operator L c (f) which is defined by L c (f) = ((c + 1)/z c ) z 0 tc f(t)dt c>. Theore 7. Let f(z) V l H (λ γ). Then L c(f(z)) V l H (λ γ). Lea 8 (see 15). Let f=h+g be given by (3).If l 1 λ+nλ (n α) 1 α l 1 λ nλ (n+α) + 1 α b n 1 a 1 =1and 0 α<1thenf M H (λ α). (1) Theore 9. Let f=h+g V l H (λ α) be given by (6).Then f V l H (λ (4 3γ)/(3 γ)). Proof. Since f V l H (λ γ)thenbytheore 1 we ust have l 1 λ+nλ (n γ) l 1 λ nλ (n + γ) + b n 1. () To show that f V l H (λ (4 3γ)/(3 γ)) byvirtueof Lea 8 we have to show that l 1 λ+nλ n ((4 3γ) / (3 γ)) 1 ((4 3γ)/(3 γ)) l 1 λ nλ n + ((4 3γ) / (3 γ)) + 1 ((4 3γ)/(3 γ)) b n 1 (3) 0 (4 3γ)/(3 γ) < 1. Forthisitissufficientto prove that 1 λ+nλ (n γ) 1 λ+nλ n ((4 3γ) / (3 γ)) 1 ((4 3γ) / (3 γ)) (n = 3...) 1 λ nλ (n + γ) 1 λ nλ n + ((4 3γ) / (3 γ)) 1 ((4 3γ) / (3 γ)) (n = ) (4) or equivalently (γ 1) 1 λ + nλ (n γ) 0 (n = 3...) and (γ 1) 1 λ nλ (n + γ) 0 (n = ) whichis true and the theore is proved. Corollary 10. V l H (λ γ) Vl H (λ 4/3) Vl H. 5. Convolution Properties For functions f H given by (3)andF H given by F (z) =H(z) + G (z) =z+ A n z n + B n z n (5) we recall the Hadaard product (or convolution) of f and F by (f F)(z) =z+ a n A n z n + b n B n z n (z U). Let F j (z) V l H (λγ) (j=13...p)be given by F j (z) =z+ then the convolution is defined by a nj zn + () l (F 1 F p ) (z) =z a nj zn p j=1 (6) b nj zn (7) + () l b nj zn. p j=1 (8) Theore 11. Let F j (z) V l H (λ γ j) (j = 13...p)then (F 1 F p )(z) V l H (λ β) β=1+ p j=1 (γ j ) l (1+λ) p j=1 ( + γ j) p j=1 (γ j ). (9) Proof. We use the principle of atheatical induction in our proof. Let F 1 V l H (λ γ 1) andf V l H (λ γ ). Byusing Theore we have n l 1 λ+nλ (n γ j) γ j a nj + n l 1 λ nλ (n + γ j) γ j b nj 1; (30)

5 International Analysis 5 then Therefore if ( n l 1 λ+nλ (n γ 1) γ 1 a n1 ) ( n l 1 λ+nλ (n γ ) γ ) 1/ + ( n l 1 λ nλ (n + γ 1) γ 1 b n1 ) 1. ( n l 1 λ nλ (n + γ ) γ b n ) Thus by applying Cauchy-Schwarz inequality we have 1/ (31) n l 1 λ+nλ (n α) α a n1a n n l 1 λ+nλ (n γ 1)(n γ ) a n1a n n l 1 λ nλ (n+α) α b n1b n that is if n l 1 λ nλ (n + γ 1)(n+γ ) b n1b n a n1a n α n α 1 λ+nλ (n γ 1)(n γ ) b n1b n α n+α 1 λ nλ (n + γ 1)(n+γ ) (34) (35) ( n l 1 λ+nλ (n γ 1)(n γ ) a n1a n ) ( n l 1 λ+nλ (n γ 1) γ 1 ( n l 1 λ+nλ (n γ ) γ a n1 ) 1/ ) ( n l 1 λ nλ (n + γ 1)(n+γ ) b n1b n ) ( n l 1 λ+nλ (n + γ 1) γ 1 b n1 ) 1/ 1/ (3) then (F 1 F )(z) V l H (λ α).by(30)wehave Henceweget n 1 λ+nλ (n γ j) l γ j a nj 1 n 1 λ nλ (n + γ j) l γ j b nj 1. a nj γ j 1 λ+nλ (n γ j ) b nj γ j 1 λ nλ (n + γ j ). (36) (37) Then we get ( n l 1 λ+nλ (n + γ ) γ b n ) n l 1 λ+nλ (n γ 1)(n γ ) a n1a n + n l 1 λ nλ (n + γ 1) γ 1 b n1b n 1. 1/. (33) Consequently if (γ 1 )(γ ) n l 1 λ+nλ (n γ 1 )(n γ ) α n α 1 λ+nλ (n γ 1)(n γ ) (γ 1 )(γ ) n l 1 λ nλ (n + γ 1 )(n+γ ) α n+α 1 λ nλ (n + γ 1)(n+γ ). (38)

6 6 International Analysis That is if n α α 1 λ+nλ (n γ 1)(n γ ) n+α α 1 λ nλ (n + γ 1)(n+γ ) then (F 1 F )(z) V l H (λ α).thenweseethat α 1 + =φ(n) α 1 + (n) n l 1 λ+nλ (n + γ 1 )(n+γ )+ (n+1) n l 1 λ nλ (n + γ 1 )(n+γ ) (39) =ψ(n). (40) Since φ(n) for n and ψ(n) for n 1are increasing (γ α 1+ 1 )(γ ) l 1+λ ( + γ 1 )(+γ )+ (41) (γ α 1+ 1 )(γ ) 1 λ (1 + γ 1 )(1+γ ) (4) and also l 1+λ ( + γ 1 )(+γ )+ 1 λ (1 + γ 1 )(1+γ ) then (F 1 F )(z) V l H (λ α) α 1+ l (1+λ) ( + γ 1 )(+γ )+. (43) (44) Next we suppose that (F 1 F F p )(z) V l H (λ β) p j=1 β=1+ (γ j ) l (1+λ) p j=1 ( + γ j) p j=1 (γ j ). (45) Since we have (β 1) (α p+1 ) = p+1 j=1 (γ j ) l (1+λ) p j=1 ( + γ j) p j=1 (γ j ) (β ) (α p+1 ) = =1+ p+1 j=1 (γ j ) l (1+λ) p j=1 ( + γ j) p j=1 (γ j ) (47) p+1 j=1 (γ j ) l (1+λ) p+1 j=1 ( + γ j) p+1 j=1 (γ j ). (48) Corollary 1. Let F j (z) V l H (λ γ)(j = p) then (F 1 F p )(z) V l H (λ β) (γ 1) p β=1+ l (1+λ)( + γ) p p. (49) () 6. Partial Sus Results In 1985 Silvia 16 studied the partial sus of convex functions of order α(0 α < 1).Later onsilveran17 and several researchers studied and generalized the results on partial sus for various classes of analytic functions only but analogues results on haronic functions have not been explored in the literature. Very recently Porwal 18 and Porwal and Dixit 19 filledthisgapbyinvestigating interesting results on the partial sus of star-like haronic univalent functions. Now in this section we discussed the partial sus results for the class of haronic functions with positive coefficients based on Salagean operator of order γ (1 < γ 4/3) on lines siilar to Porwal 18. Let P l H (A n/ B n /) denote the subclass of H consisting of functions f = h + g of the for (3) whichsatisfythe inequality A n + B n b n 1 (50) We can show that (F 1 F F p+1 )(z) V l H (λ ) 1+ (β 1) (γ p+1 ) l (1+λ) ( + β) ( + γ p+1 )+(β)(γ p+1 ). (46) A n = nl 1 λ+nλ (n γ) B n = nl 1 λ nλ (n + γ) and = unless otherwise stated. (51)

7 International Analysis 7 Now we discuss the ratio of a function of the for (6) with b 1 =0being f (z) =z+ a n z n + b n z n f k (z) =z+ a n z n + b n z n k f k (z) =z+ a n z n + b n z n. k We first obtain the sharp bounds for R{f(z)/f (z)}. (5) Theore 13. If f of the for (6) with b 1 = 0 satisfies the condition R { f (z) } (z U) (53) A n { B n if n = 3... if+... if 3... The result (53) issharpwiththefunctiongivenby =z+ Proof. Define the function w(z) by (54) z +1. (55) 1+w(z) 1 w(z) = f(reiθ ) f (re iθ ) =(1+ a n r n e i(n)θ + b n r n e i(n+1)θ ( (1+ a n r n e i(n)θ + a n r n e i(n)θ )) b n r n e i(n+1)θ ). (56) It suffices to show that w(z) 1. Nowfro(56) we can write w (z) =( Henceweobtain w (z) ( a n r n e i(n)θ )) (+( a n r n e i(n)θ + + ( b n r n e i(n+1)θ ) a n r n e i(n)θ )). ( /) ( ) + b n (/) Now w(z) 1 if + b n + Fro condition (50) it is sufficient to show that + A n which is equivalently to ( A n + b n + + ) + B n (57) a. n (58) 1. (59) ( B n ) b n ( A n A n+1 ) 0. (60) (61) To see that the function given by (55) givesthesharpresult we observe that for z=re iπ/n f (z) =1+ z 1 = when r 1. We next deterine bounds for R{f (z)/f(z)}. (6) Theore 14. If f of the for (6) with b 1 =0satisfies condition R { f (z) } (z U) (63) +

8 8 International Analysis A n { B n if n = 3... if+... if 3... The result (63) issharpwiththefunctiongivenby =z+ Proof. Define the function w(z) by (64) z +1. (65) 1+w(z) 1 w(z) = + f (re iθ ) f(re iθ ) + =(1+ a n r n e i(n)θ Theore 15. If f of the for (6) with b 1 =0satisfies condition R { f (z) f (z)} (+1) (z U) (69) A n { B n if n = 3... if +... if 3... (70) The result (69) is sharp with the function given by f(z) = z + (/ )z +1. Proof. Define the function w(z) by 1+w(z) 1 w(z) = (+1) f (z) f (z) (+1) Henceweobtain + b n r n e i(n+1)θ ( (1+ a n r n e i(n)θ + w (z) ( + a n r n e i(n)θ )) b n r n e i(n+1)θ ). ( )) ( + b n (( )/) ) (66) (67) =(1+ na n r n e i(n)θ + nb n r n e i(n+1)θ A +1 (+1) ( na n r n e i(n)θ )) (1+ na n r n e i(n)θ nb n r n e i(n+1)θ ). (71) The result (69) follows by using the techniques as used in Theore 13. Proceeding exactly as in the proof of Theore 14we can prove the following theore. Theore 16. If f of the for (6) with b 1 = 0 satisfies the condition 1. The last inequality is equivalent to l + b n + 1. (68) Making use of (50) and the condition (64) we obtain (61). Finally equality holds in (63)fortheextrealfunctionf(z) given by (65). We next turns to ratios for R{f (z)/f (z)} and R{f (z)/f (z)}. R { f (z) f (z) } (z U). (7) + (+1) The result is sharp with the function given by f(z) = z + (/ )z +1. We next deterine bounds for R{f(z)/f k (z)} and R{f k (z)/f(z)}. Theore 17. If f of the for (6) with b 1 =0satisfies condition R { f k (z) } B k+1 (z U) (73) B k+1

9 International Analysis 9 B n { B k+1 A n if 3...k if n=k+1k+... if 3... (74) The result (73) is sharp with the function given by f(z) = z + (/B k+1 )z k+1. Theore 18. If f of the for (6) with b 1 =0satisfies condition R { f k (z) } B k+1 (z U) (75) B k+1 + B n { if 3...k B k+1 if n=k+1k+... (76) A n if 3... The result (75) is sharp with the function given by f(z) = z + (/B k+1 )z k+1. Proof. Define the function w(z) by 1+w(z) 1 w(z) = B k+1 + f k (re iθ ) f(re iθ ) B k+1 B k+1 + =(1+ a n r n e i(n)θ + b n r n e i(n+1)θ B k+1 k+1 k b B k+1 + n r n e i(n)θ ) (1+ a n r n e i(n)θ k + b n r n e i(n+1)θ ). (77) We oit the details of proof because it runs parallel to that fro Theore 14. Theore 19. If f of the for (6) with b 1 =0satisfies condition R { f k (z) } (z U) (78) A n { if if +... (79) B n { if 3... if +... The result (78) is sharp with the function given by f(z) = z + (/ )z +1. Theore 0. If f of the for (6) with b 1 =0satisfies condition R { f k (z) } B k+1 (z U) (80) B k+1 B n { if 3...k B k+1 if n=k+1k+... (81) A n { if 3...k B k+1 ifn=k+1k+... Theore 1. If f of the for (6) with b 1 =0satisfies condition R { f k (z) } (z U). (8) + Theore. If f of the for (6) with b 1 =0satisfies condition R { f k (z) } B k+1 (z U). (83) B k+1 + The result (83) is sharp with the function given by f(z) = z + (/B k+1 )z k+1. Theore 3. If f of the for (6) with b 1 =0satisfies condition R { f (z) f (z)} (+1) (z U) (84) A n { if n = 3... if +... (85) B n { if 3... if+... The result (84) is sharp with the function given by f(z) = z + (/ )z +1. Theore 4. If f of the for (6) with b 1 =0satisfies condition R { f k (z) f (z) } (z U). (86) + (+1) The result (86) is sharp with the function given by f(z) = z + (/ )z +1. Concluding Rearks. By choosing λ = 0 (or λ = 1)and l=0(or l=1) the various results presented in this paper would provide interesting extensions and generalizations of the subclasses of haronic star-like functions with positive coefficients of order γ (1 < γ 4/3) based on Salagean operator and siilarly for convex functions. The details involved in the derivations of such specializations of the results presented in this paper are fairly straight-forward and hence oitted.

10 10 International Analysis Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. Acknowledgent We record our sincere thanks to the referees for their valuable suggestions. 17 H. Silveran Partial sus of starlike and convex functions Matheatical Analysis and Applicationsvol.09no. 1 pp S. Porwal Partial sus of certain haronic univalent functions Lobachevskii Matheatics vol.3no.4pp S. Porwal and K. K. Dixit Partial sus of starlike haronic univalent functions Kyungpook Matheatical Journalvol.50 no.3pp References 1 J. Clunie and T. Sheil-Sall Haronic univalent functions Annales Acadeiae Scientiaru Fennicae A vol.9pp J.M.JahangiriG.MurugusundaraoorthyandK.Vijaya Salagean-type haronic univalent functions Southwest Journal of Pure and Applied Matheaticsno.pp H. Silveran Univalent functions with negative coefficients Proceedings of the Aerican Matheatical Society vol.51pp H. Silveran Haronic univalent functions with negative coefficients Matheatical Analysis and Applications vol.0no.1pp B.A.UralegaddiM.D.GanigiandS.M.Sarangi Univalent functions with positive coefficients Takang Matheaticsvol.5no.3pp K.K.DixitandV.Chandra Onsubclassofunivalentfunctions with positive coefficients TheAligarhBulletinofMatheatics vol.7no.pp K.K.DixitandA.L.Pathak Anewclassofanalyticfunctions with positive coefficients Indian Pure and Applied Matheaticsvol.34no.pp S. Porwal and K. K. Dixit An application of certain convolution operator involving hypergeoetric functions Rajasthan Acadey of Physical Sciences vol.9no.pp S. Porwal K. K. Dixit V. Kuar and P. Dixit On a subclass of analytic functions defined by convolution General Matheaticsvol.19no.3pp K. K. Dixit and S. Porwal A subclass of haronic univalent functions with positive coefficients Takang Matheaticsvol.41no.3pp J. M. Jahangiri Haronic functions starlike in the unit disk Matheatical Analysis and Applicationsvol.35no. pp G. Murugusundaraoorthy and K. Vijaya A subclass of haronic functions associated with wright hypergeoetric functions Advanced Studies in Conteporary Matheaticsvol. 18no.1pp G. Murugusundaraoorthy K. Vijaya and R. K. Raina A subclass of haronic functions with varying arguents defined by Dziok-Srivastava operator Archivu Matheaticu vol. 45no.1pp S. Porwal and K. K. Dixit New subclasses of haronic starlike and convex functions Kyungpook Matheatical Journal vol. 53no.3pp K. Vijaya Studies on certain subclasses of Haronic functions Ph.D. thesisvituniversityvelloreindia E. M. Silvia On partial sus of convex functions of order α Houston Matheaticsvol.11no.3pp

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