Research Article The Intrinsic Structure and Properties of Laplace-Typed Integral Transforms

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1 Hindawi Mahemaical Problems in Engineering Volme 217, Aricle ID , 8 pages hps://doi.org/1.1155/217/ Research Aricle The Inrinsic Srcre and Properies of Laplace-Typed Inegral Transforms Hwajoon Kim Facly of General Edcaion, Kyngdong Universiy, Yangj, Gyeonggi 11458, Repblic of Korea Correspondence shold be addressed o Hwajoon Kim; cellmah@gmail.com Received 14 Febrary 217; Revised 12 April 217; Acceped 1 May 217; Pblished 8 Jne 217 Academic Edior: Salvaore Srano Copyrigh 217 Hwajoon Kim. This is an open access aricle disribed nder he Creaive Commons Aribion License, which permis nresriced se, disribion, and reprodcion in any medim, provided he original work is properly cied. We wold like o esablish he inrinsic srcre and properies of Laplace-yped inegral ransforms. The mehodology of his aricle is done by a consideraion wih respec o he common srcre of kernels of Laplace-yped inegral ransform, and G-ransform, he generalized Laplace-yped inegral ransform, is proposed wih he feare of inclsiveness. The proposed Gransform can provide an adeqae ransform in a nmber of engineering problems. 1. Inrodcion The key moivaion for prsing heories for inegral ransforms is ha i gives a simple ool which is represened by an algebraic problem in he process of solving differenial eqaions [1]. In he mos heories for inegral ransforms, he kernel is doing he imporan role which ransforms one space o he oher space in order o solve he solion. The main reason o ransform is becase i is no easy o solveheeqaioninhegivenspace,oriiseasyofind a characerisic for he special prpose. For example, in comped omography (CT) or magneic resonance imaging (MRI), we obain he projecion daa by inegral ransform and prodce he image wih he inverse ransform. This is he srong poin of inegral ransform. To begin wih, le s see he inrinsic srcre of Laplaceyped inegral ransforms. Of corse, he srcre is dependen on he kernel, and he form of he kernel in Laplaceyped inegral ransforms is as follows. Laplace ransform is defined by Smd one is m (f) = e s f () d, (1) S () = 1 e / f () d, (2) and Elzaki one is E () = e / f () d. (3) SinceheLaplaceransformm(f) canberewrienas e / f () d (4) for s = 1/, we can narally consider ha he form of Laplace-yped inegral ransform G is F () =G(f)= α e / f () d (5) foransiableinegerα: for example, if G (1) = α e / d = α ( ) [e / ] = α+1 (6) for >. Normally, inegral ransforms have a base of exponenial fncion and i ges along wih how o inegrae from o in order o ilize he propery o e s or e / converges o when approaches. Smd/Elzaki ransform is a kind of modified Laplace one inrodced by Wagala [2] in 1993/Elzaki e al. [3] in 212 o solve iniial vale problems in engineering problems [4]. Belgacem e al. [5, 6]

2 2 Mahemaical Problems in Engineering are menioning ha Smd ransform (α = 1)has scale and ni-preserving properies, and i may be sed o solve problems wiho resoring o a new freqency domain. Elzaki e al. [3, 7, 8] insiss ha Elzaki ransform (α = 1) shold be easily applied o he iniial vale problems wih less compaional work and solve he varios examples which are no solved by he Laplace or he Smd ransform. As an applicaion, Agwa e al. [9] deal wih Smd ransform on ime scales and is applicaions, Elayeb and Kilicman [1] have checked some applicaions of Smd one, and Elayeb e al. are highlighing he imporance of fracional operaors of inegral ransform and heir applicaions in [11]. The shifed daa problems, shifing heorems, and he forms of solions of ODEs wih variable coefficiens can be fond in [4, 12, 13]. On he oher hand, Kreyszig [1] says ha Laplace ransform (α = ) has a srong poin in he ransforms of derivaives; ha is, he differeniaion of a fncion f() corresponds o mliplicaion of is ransform m(f) by s. In he oher view, if we wan he inverse case, he ransform giving a simple ool for ransforms of inegrals, hen we can choose a siable form of inegral ransform sch as 1 2 e / f () d. (7) This means ha he ineger α is applicable o 2. Aswe checked above, he comprehensive ransform in Laplaceyped ones is needed, and hs we wold like o propose Gransform, a generalized Laplace-yped inegral ransform, which is more comprehensive and inrinsic han he exising ransforms. This inrinsic srcre in Laplace-yped inegral ransforms has a meaning which can be direcly applied o any siaion by choosing an appropriae ineger α. The mainobjeciveofhispaperisoconsrchegeneralized form of Laplace-yped inegral ransforms and esablish he properies of i, and, o he ahor s knowledge, he proposed G-ransform is he firs aemp o generalize Laplace-yped inegral ransforms. Finally, we wold like o menion ha Laplace ransform gave many consideraions o his aricle. 2. The Properies of Laplace-Typed Inegral Transforms 2.1. The Definiion and he Table of Generalized Inegral Transform G,ShifingTheorems. As menioned before, le s rewrie he definiion of Laplace-yped inegral ransforms, and we wold like o call i G-ransform. Definiion 1. If f() is an inegrable fncion defined for all, is generalized inegral ransform G is he inegral of f() imes α e / from =o. I is a fncion of,say F(), and is denoed by G(f);hs F () =G(f)= α e / f () d. (8) Le s firs check he shifing heorems. Theorem 2. (1) (-shifing) If f() has he ransform F(), hen e a f() has he ransform Tha is, F( ). (9) 1 a G[e a f ()] =F( ). (1) 1 a (2) (-shifing) If f() has he ransform F(), hen he shifed fncion f( a)h( a)has he ransform e a/ F(). In formla, G[f( a) h ( a)] =e a/ F () (11) for h( a) is Heaviside fncion (we wrie h since we need o denoe -space). Addiionally, G[h( a)] = α+1 e a/ holds. Proof. (1) From we obain he resl. (2) G[e a f ()] = α e / e a f () d, e / e a =e (a 1/) =e /(/(1 a)), e a/ F () =e a/ α e τ/ f (τ) dτ = α e (a+τ)/ f (τ) dτ. Sbsiing τ+a=,weobain e a/ F () = α e / f ( a) d a = α e / f ( a) h ( a) d =G[f( a) h ( a)] (12) (13) (14) for h is Heaviside fncion. By he similar way, we have G[h( a)] = α+1 e a/ for h is Heaviside fncion. Using G(f) = α F(1/) for Laplace ransform m(f) = F(s), we can obain he able of generalized inegral ransform G as shown in Table 1. In he able, we regard Laplaceyped inegral ransform G asaransform.however,wecan choose an appropriae consan according o each siaion. For example, he choice of α=has a meri in he ransforms of derivaives, and α= 2has a srong poin in he ransforms of inegrals.

3 Mahemaical Problems in Engineering 3 Table 1: Table of Laplace-yped inegral ransform G. f() G(f) 1 1 α+1 1 α+2 3 n n! n+α+1 4 e a α+1 5 sina 6 cosa 7 sinh a 8 cosha 9 e a cos b 1 e a sin b 1 a a α a 2 α a 2 a α a 2 α a 2 α (1/ a) (1/ a) 2 +b 2 b α (1/ a) 2 +b 2 If f() is defined and is piecewise coninos on and saisfies f() Me k for all,heng(f) exiss for all <1/k.Since G(f) = α e / f () d α = Mα+1 k 1, he saemen is valid. e / f () d α e / Me k d 2.2. Transforms of Derivaives and Inegrals (15) Theorem 3. Le a fncion f be n-h differeniable. Then he ransforms of he firs, second, and n-h derivaives of f() saisfy (1) (2) (3) G(f )= 1 G(f) α f () (16) G (f ) = 1 2 G (f) 1 f () α f () α (17) G(f (n) )= 1 1 G(f) n n 1 f () α 1 n 2 f () α (18) f (n 1) () α for α is an arbirary ineger. (4) Le f() be piecewise coninos for and inegrable. Then G[ f (τ) dτ] = F () (19) holds for F() = G[f()]. Proof. By he inegraion by pars, G(f )= α e / f () d = α [e / f ()] +α 1 e / f () d and, similarly, = 1 G(f) α f () G(f )= 1 G(f ) f () α = 1 [ 1 G(f) α f ()] f () α = 1 2 G(f) 1 f () α f () α (2) (21) follows. Conining his process by sbsiion as he above and sing indcion, (3) follows. Le s minely esablish he validiy of he saemen of (3) by he mahemaical indcion. For n=1, i clearly follows. Nex, we sppose ha G(f (n) )= 1 1 G(f) n n 1 f () α 1 n 2 f () α f (n 1) () α, and we show ha G(f (n+1) ) canbeexpressedby G(f (n+1) )= 1 n+1 G(f) 1 f () α n 1 n 1 f () α f (n) () α. (22) (23) In saemen (1), he proof of saemen (3) follows by applying (1) o f (n). Finally, (4) followsasbelow. Show ha G[ f(τ)dτ] = F() holds for >.Using Theorem 3 and ping g() = f(τ)dτ, G(f()) =G(g ()) = 1 G(g) α g () = 1 G(g). (24) Moreover, F() clearly saisfies a growh resricion.

4 4 Mahemaical Problems in Engineering Le s check an example in [1] by he G-ransform. Theorem 6. Example 4. Solve y y=, y() = 1,andy () = 1. Solion.Taking G-ransform on boh sides, we have 1 2 G (f) 1 y () α y () α G(f) = α+2. (25) Organizing his eqaliy, we have Simplificaion gives Since ( 1 2 1)G(f)=(α+2 + α + α 1 ). (26) G(f)=( α+2 + α + α 1 ) = α α α from Table 1, we have he solion (27) α = α+2 + α+2 1 2, (28) y () = +2sin h + cos h (29) for h is hyperbolic fncions. From he sbsiion, above y() is exacly a solion of he given eqaion. Of corse, by a simple calclaion, he above answer is eqal o he solion of [1] which is e sin h Convolion and Inegral Eqaions for G-Transform. I is a well-known fac ha m(fg) = m(f)m(g) and m(f)m(g) = m(f g) for f g is he convolion of f and g.thismeansha convolion has o do wih he mliplicaion of ransforms in Laplace ransform. Here, we invesigae he change a G-ransform. If wo fncions f and g are inegrable, he following heorem is held. Lemma 5 (Lebesge s dominaed convergence heorem (LDCT) [14]). Le (X, M, μ) be a measre space and sppose ha {f n } is a seqence of exended real-valed measrable fncions defined on X sch ha (a) lim n f n (x) = f(x) exiss μ-a.e. (b) here is an inegrable fncion g so ha, for each n, f n gμ-a.e. Then f is inegrable and lim f n n dμ = f dμ. (3) X X We noe ha he above lemma gives validiy o he following eqaliy: g n dμ = n=1 n=1 for (g n ) is a nondecreasing seqence. g n dμ (31) Proof. Le s p and p Then α G (f g) =G(f) G (g). (32) G(f)= α e τ/ f (τ) dτ (33) G(g)= α e V/ g (V) dv. (34) G(f)G(g)= α e τ/ f (τ) α e V/ g (V) dv dτ. (35) Le s p =V +τ,whereτ is a firs consan. Then V = τ andsowege G(g)= α e ( τ)/ g ( τ) d τ =e τ/ α e / g ( τ) d. τ (36) Since he fncion f is inegrable, we can change he order of inegraion by sing Lebesge s dominaed convergence heorem. Hence G(f)G(g)= α α f (τ) e / g ( τ) d dτ, τ and when varies τ o, τ varies o.hence G(f)G(g)= α α e / f (τ) g ( τ) dτ d = α α e / (f g) () d = α G(f g). (37) (38) I is a well-known fac ha convolion helps s o solve inegral eqaions of cerain ype, mainly Volerra inegral eqaion. Hence, we wold like o check he heorem by means of some examples sing G-ransform. Example 7. Solve he Volerra inegral eqaion of he second kind y () y (τ) sin ( τ) dτ =. (39)

5 Mahemaical Problems in Engineering 5 Solion. This is rewrien as a convolion: y () y sin =. (4) Taking G-ransform on boh sides and applying Theorem 6, we have Y () α Y () α+2 =Y() ( ) (41) = α+2 for Y=G(y).Thesolionis andgivesheanswer by Table 1. Y () =(1+ 2 ) α+2 (42) y () = (43) Example 8. Solve he Volerra inegral eqaion of he second kind y () (1+τ) y ( τ) dτ = 1 sinh. (44) Solion. In a way similar o Example 7, he given eqaion is same as y (1+) y=1 sinh.takingg-ransform, we have Y () α ( α+1 + α+2 ) Y () = α+1 α ; (45) hence, Y () [1 2 ]= α+1 (1 2 ) 1 2. (46) Simplificaion gives Y () = α+1 1 2, (47) and so we obain he answer by Table 1. Example 9. Find he solion of y () = cosh (48) y () + ( τ) y (τ) dτ = 1. (49) Solion. Since his eqaion is y+y = 1,akingGransform on boh sides, we have Y+ α (Y α+2 )= α+1 (5) for Y=G(y).Ths Y= α , (51) andsoweobainhesoliony=cos. Le s check his by he direc calclaion. Expanding he given eqaion, we have y () + y (τ) dτ τy (τ) dτ = 1. (52) Differeniaing boh sides wice wih respec o, wehave y () + y() =.Ths,fromy() = 1 and y () = obained by calclaing corse, we have he solion y=cos. Similarly, we can easily obain he solion of inegral eqaions by sing α= 2. For example, le s consider y () y (τ) dτ = 1. (53) By he G-ransform, we have Y Y= 1 (54) andso,wehavehesoliony=e for Y=G(y).Ofcorse, his resl is he same as he resl y y 1=1 (55) by sing convolion, and he resl of he direc calclaion is so as well. Similarly, since y () y() = (56) G(e a )= α+2 (1 a) 2, (57) he solion of y () +2e e τ y (τ) dτ = e (58) is y=sinh. Here, we noe ha he G-ransform of y+2(y e )=e is for Y=G(y). Y+2 α Y α+1 (1 ) = α+2 (1 ) 2 (59) 3. Differeniaion and Inegraion of Transforms: ODEs wih Variable Coefficiens Theorem 1. Le s p Y=F().Then (1) F () = Y(α/ + / 2 ), (2) F () = Y((α 2 α)/ 2 +2(α 1)/ / 4 ),

6 6 Mahemaical Problems in Engineering (3) F(s)ds = f() sα e /s ds d, (4) G(y) = αy (d/d)g(y), G(y ) = α[(1/)y y() α ] (d/d)g(y ), (5) G(y )= α[(1/ 2 )Y (1/)y() α y () α ] (d/d)g(y ). Proof. (1) Since F () =G(f)= α e / f () d, F () =α α 1 e / f () d for Y=F(). (2) Since we have + α 2 e / f () d = Y ( α + 2 ) (6) F () = d d F () = d d [Y (α + )], (61) 2 F () =Y ( α + 2 )+Y(α + 2 ) Organizing his eqaliy, we have (3) =Y( α + 2 )2 Y( α ). F () =Y( α2 α 2 F (s) ds = = = + 2 (α 1) 3 [s α e /s f () d] ds s α e /s f () d ds f () s α e /s ds d (62) + 2 ). (63) 4 hanks o Lebesge s dominaed convergence heorem. (4) In he definiion of le s p 1/ = s.then (64) F () =G(y)= α e / y () d, (65) F () =G(y)= 1 s α e s y () d, F () = αs α 1 e s y () d 1 s α e s y () d = α G (y) G (y). s (66) Since 1/s is he same as,wehave G (y) = αy F () = αy d G (y) (67) d for Y=G(y). In he above heorem, we noe ha G(y (n) ) can be represened by G(y (n) )= αg(y (n) ) d d G(y(n) ). (68) (5) From (4), he saemen is held for an arbirary ineger n. 4. G-Transforms of Heaviside Fncion and Dirac s Dela Fncion: Shifed Daa Problems 4.1. Heaviside Fncion G [h ( a)] = α e / h ( a) d = α e / d = α+1 [e / ] a a = α+1 e a/, where h is Heaviside fncion. (69) 4.2. Dirac s Dela Fncion. We consider he fncion f k ( a) = 1/k if a a+kand oherwise. Taking G-ransform, we have G[f k ( a)] = α e / f k ( a) d = α+1 k [e / ] a+k a = α+1 k (e (a+k)/ e a/ ) = α+1 k e a/ (e k/ 1). If we denoe he limi of f k as δ( a),hen (7) δ ( a) = lim k f k ( a) = α e a/. (71) 4.3. Shifed Daa Problems. For given differenial eqaions y +ay +by=r(), y( )=c,andy ( )=c 1,where = and a and b are consan, we can se = 1 +.Then= gives 1 =andsowehave y 1 +ay 1 +by 1 =r( 1 + ), y 1 () =c, (72) y 1 () =c 1 for inp r(). Taking he ransform, we can obain he op y().

7 Mahemaical Problems in Engineering 7 5. The Solion of Semi-Infinie Sring by G-Transform Le s check he solion of semi-infinie sring by Gransform in erms of a ypical example as given in [1]. Example 11 (semi-infinie sring). Find he displacemen w(x, ) of an elasic sring sbjec o he following condiions: (a) The sring is iniially a res on x-axis from x=o. (b) For >, he lef end of he sring is moved in a given fashion, namely, according o a single sine wave w(, ) = f() = sin, if 2π and zero oherwise. (c) Frhermore, lim w(x, ) = as x for. Of corse here is no infinie sring, b or model describes a long sring or rope (of negligible weigh) wih is righ end fixed far o on x-axis [1]. Solion. I is a well-known fac ha he eqaion of semiinfinie sring can be expressed by 2 w 2 =c2 2 w x 2 ( ) sbjec o w(, ) = f(),lim w(x, ) = as x, w(x, ) =,andw (x, ) =.TakingG-ransform wih respec o and by Theorem 3, we have G[ 2 w 2 ]= 1 2 W 1 w (x, ) α w (x, ) α = 1 2 W (73) for W(x, ) = G[w(x, )]. On he oher hand, G[ 2 w x 2 ]=α = 2 x 2 α Ths, he eqaion ( ) becomes e / 2 w x 2 d e / w (x, ) d = 2 x 2 G [w (x, )] = 2 W x 2. (74) c 2 2 W x W=. (75) Since his eqaion may be regarded as an ODE for W(x, ) considered as a fncion of x, is general solion can be represened by W (x, ) =A() e x/c +B() e x/c. (76) From he iniial condiions, we have W(, ) = G[w(, )] = G[f()] = F().In[15,16],wehavedealwihhevalidiyon exchangeabiliy of inegral and limi in he solving process of PDEs by sing dominaed convergence heorem. Hence, we have lim W (x, ) = lim x x α e / w (x, ) d (77) = α e / lim w x (x, ) d =, implying ha A() = and so W(, ) = B() = F().Ths, W (x, ) =F() e x/c. (78) By he -shifing heorem (Theorem 2), we obain he inverse ransform w (x, ) =f( x c )h( x c )=sin ( x c ) (79) for x/c < < x/c + 2π and zero oherwise, where h is Heaviside fncion. 6. Conclsion This paper has consrced he generalized form of Laplaceyped inegral ransforms and has esablished he properies of he generalized Laplace-yped inegral ransform, Gransform. The ransform is comprehensive form, and i has been well adaped in a nmber of siaions of engineering problems by choosing adeqae vales α in kernel, and we newly presened he vale α = 2 which is siable for ransforms of inegrals. And he fre work is o find he oher vales of α which are siable for each siaion. The srong poin of his aricle is in he high applicabiliy o engineering problems. Conflics of Ineres The ahor declares ha here are no conflics of ineres regarding he pblicaion of his paper. Acknowledgmens This research was sppored by Kyngdong Universiy Research Gran. References [1] E. Kreyszig, Advanced Engineering Mahemaics, Wiley, Singapore, 213. [2] G. K. Wagala, Smd ransform a new inegral ransform o solve differenial eqaions and conrol engineering problems, Inernaional Mahemaical Edcaion in Science and Technology,vol.24,no.1,pp.35 43,1993. [3] T. M. Elzaki, S. M. Elzaki, and E. M. A. Hilal, Elzaki and Smd ransforms for solving some differenial eqaions, Global Pre & Applied Mahemaics,vol.8,no.2,pp , 212. [4] Hj. Kim, The ime shifing heorem and he convolion for Elzaki ransform, Global Pre and Applied Mahemaics,vol.87,pp ,213.

8 8 Mahemaical Problems in Engineering [5] F. B. Belgacem and S. Sivasndaram, New developmens in compaional echniqes and ransform heory applicaions o nonlinear fracional and sochasic differenial eqaions and sysems, Nonlinear Sdies. The Inernaional Jornal, vol. 22, no. 4, pp , 215. [6] F. B. M. Belgacem and A. Karaballi, Smd ransform fndamenal properies invesigaions and applicaions, Applied Mahemaics and Sochasic Analysis,vol.26,Aricle ID 9183, 23 pages, 26. [7] T. M. Elzaki and J. Biazar, Homoopy perrbaion mehod andelzakiransformforsolvingsysemofnonlinearparial differenial eqaions, World Applied Sciences Jornal, vol.24, no.7,pp ,213. [8] T. M. Elzaki and H. Kim, The solion of radial diffsiviy and shock wave eqaions by Elzaki variaional ieraion mehod, Inernaional Mahemaical Analysis,vol.9,no.21-24, pp , 215. [9] H. A. Agwa, F. M. Ali, and A. Kilicman, A new inegral ransform on ime scales and is applicaions, Advances in Difference Eqaions,vol.6,pp.1 14,212. [1] H. Elayeb and A. Kilicman, On some applicaions of a new inegral ransform, Inernaional Mahemaical Analysis,vol.4,no.1-4,pp ,21. [11] H. Elayeb, A. Kilicman, and M. B. Jleli, Fracional Inegral Transform and Applicaion, Absrac and Applied Analysis,vol. 215, Aricle ID 15863, 1 page, 215. [12] H. Kim, The form of solion of ODEs wih variable coefficiens by means of he inegral and Laplace ransform, Global Pre and Applied Mahemaics, vol. 12, pp , 216. [13] H. Kim, The shifed daa problems by sing ransform of derivaives, Applied Mahemaical Sciences, vol. 8, no , pp , 214. [14] D. L. Cohn, Measre Theory, Birkhäaser, Boson, Mass, USA, 198. [15] H. C. Chae and H. Kim, The validiy checking on he exchange of inegral and limi in he solving process of PDEs, Inernaional Mahemaical Analysis,vol.8,no.22,pp , 214. [16] J. Jang and H. Kim, An applicaion of monoone convergence heorem in pdes and forier analysis, Far Eas Mahemaical Sciences, vol. 98, no. 5, pp , 215.

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