Theory of natural transform

MATHEMATICS IN ENGINEERING, SCIENCE AND AEROSPACE MESA - www.journalmesa.com Vol. 3, No. 1, pp. 105-135, 2012 c CSP - Cambridge, UK; I&S - Florida, USA, 2012 ⃝

Theory of natural transform ⋆

Fethi Bin Muhammed Belgacem1 , R.Silambarasan 2 1

E-mail :[email protected] E-mail :silambu [email protected] ⋆ Corresponding Author. E-mail: [email protected]

2

Abstract. Nature of the Natural transform is,it converges to Laplace and Sumudu transform.The theme of this paper is to give detailed study of Natural transform.The Natural transform is derived from the Fourier Integral.We showed it to the theoretical dual of Laplace and Sumudu transforms.This work includes Natural-multiple shift theorems,Bromwich contour integral and Heviside’s Expansion formula for Inverse Natural transform.

1 Motivation Differential equations (ordinary,partial) solving by integral transform method is most gifted technique in the mathematics world.For the function f (t) with t in (−∞, ∞) the general integral transform is defined by ∫ I [ f (t)](s) =

∞

−∞

K(s,t) f (t)dt

(1.1)

Here K(s,t) is the kernel of the transform,s the real (or) complex number is independent of t.When the K(s,t) is e−st , tJn (st) and t s−1 (st) the (1.1) gives the respective Laplace,Hankel and Mellin transforms.With these integral transforms various differential equations with initial and boundary conditions are solved.For the same function consider the integral transforms defined by I [ f (t)](u) = I [ f (t)](s, u) =

∫ ∞ −∞

K(t) f (ut)dt

(1.2)

K(s,t) f (ut)dt

(1.3)

∫ ∞ −∞

The kernel K(t) = e−t in (1.2) gives the new integral Sumudu transform,where the parameter s is replaced with u.And (1.3) is the combination of integral transforms (1.1) and (1.2) with the kernels already defined.For the any value of n the generalised Laplace and Sumudu transforms are defined respectively by,for the function f (t) > 0, 2010

Mathematics Subject Classification: Keywords: .

106

Fethi Bin Muhammed Belgacem, R.Silambarasan

L [ f (t)] = F(s) = sn S[ f (t)] = G(u) = un

∫ ∞

∫

0 ∞

e−s

n+1 t

f (snt)dt

(1.4)

e−u t f (tun+1 )dt

(1.5)

n

0

with n = 0 in (1.4) and (1.5) the respective Laplace and Sumudu transform obtained which is in literature.With this interesting group of integral transforms the need for study of some more integral tranforms arised.

2 Introduction Z.H.Khan et al. gave some properties and applications of Natural transform ”in the name of N transform” in [1].Along with the properties they solved the unsteady fluid flow problem over a plane wall.They highlighted,the transform converges to Laplace and Sumudu transform.Elementary properties such as first shift,change of scale,transform of derivatives (first and second) and integrals,and table of N (Natural) transforms of certain functions are given in [1]. Laplace transform is the simplest,well known and widely used integral transform in the mathematical and engineering community.Enough of the about Laplace transform are referred to [14, 15, 16].Enlarged list of Laplace transform of functions are given in [14].Application of Laplace transform in particular to the fractional partial differential equations,summation of series are found in [15]. Sumudu transform new to this century was first introduced by G.K.Watugala in 1993,and the paper here referred was his 1998’s [2],control engineering problems were solved with Sumudu transform.Complex inverse formula for Sumudu transform was given by S.Weerakoon.[3].M.A.Asiru gave application of Sumudu transform to integral equations in [4],discrete dynamical system (difference and differential-difference) equations in [6] and more in [5].Papers on Sumudu transform after 2002 were referred to F.B.M.Belgacem et al.In [7] F.B.M.Belgacem et al. showed Sumudu transform is theoretical dual of Laplace transform (Laplace-Sumudu Duality,abbrevation LSD) and applied Sumudu transform to convolution type integral equation.The wide range of properties,multiple shift theorem,and comprehensive list of Sumudu transform of functions are given in [8].Still more about Sumudu can be seen in [9].The new characterization for Sumudu transform given in [10] by F.B.M.Belgacem and derived the Bilateral Laplace Sumudu Duality (BLSD).More applications of Sumudu transform are seen in A.Kilicman et al.[11, 12, 13]. When the real function f (t) > 0 and f (t) = 0 for t < 0 is sectionwise continuous,exponential order and defined in the set t

A = { f (t)|∃M, τ1 , τ2 > 0, | f (t)| < Me τ j , if t ∈ (−1) j × [0, ∞)}

(2.1)

The N (hereafter Natural transform throughtout the paper) transform of the function f (t) > 0 and f (t) = 0 for t < 0 is defined by [1] N+ [ f (t)] = R(s, u) =

∫ ∞

e−st f (ut)dt; s > 0, u > 0

(2.2)

0

Where s and u are the transform variables.When u ≡ 1 in eqn (2.2) converges to Laplace transform [14, 15, 16] and s ≡ 1 in eqn (2.2) converges to Sumudu transform [7, 8] respectively defined by L [ f (t)] = F(s) = S+ [ f (t)] = G(u) =

∫ ∞

e−st f (t)dt

(2.3)

e−t f (ut)dt; u ∈ (−τ1 , τ2 )

(2.4)

0

∫ ∞ 0

Theory of natural transform

107

G.K.Watugala in [2] defined the Sumudu transform for the function ∞

f (t) =

∑ ant n

n=0

as S+ [ f (t)] = G(u) =

∞

∑ n!an un

(2.5)

n=0

Next we give the systematic derivation of Natural transform from Fourier integral.For the function f (t) that are piecewise continously differentiable in every finite interval and is absolutely integrable on the whole real line,the following integral equations holds true in the domain −∞ < t < ∞ ∫

∞ 1 F [ f (t)] = f (k) = √ e−ikt f (t)dt 2π −∞ ∫ ∞ 1 −1 F [ f (k)] = f (t) = √ eikt f (k)dk 2π −∞

The operators F [ f (t)] and F −1 [ f (k)] are respective Fourier and Inverse Fourier transform.And the product of F [ f (t)] and F −1 [ f (k)] defined by f (x) gives 1 f (x) = 2π

∫ ∞

−∞

∫ ∞

ikt

e dk

−∞

e−ikt f (t)dt

The f (x) is the Fourier integral formula in the domain −∞ < x < ∞.For −∞ < x < 0 , f (x) ≡ 0.And setting for x > 0 f (x) = f (x)H(x)e−cx where H(x) is the Unit step function defined H(x) = 1; x ≥ 0 and 0; x < 0 ∫

∫

∞ 1 ∞ ikt e dk e−(c+ik)t f (t)dt 2π −∞ 0 ∫ ∫ ∞ 1 ∞ (c+ik)t f (x)H(x) = e dk e−(c+ik)t f (t)dt 2π −∞ 0 s ds substituting = c + ik; = idk u u ∫ c+i∞ ∫ st 1 1 ∞ −st f (x)H(x) = e u ds e u f (t)dt 2πi c−i∞ u 0

f (x)H(x)e−cx =

Here N+ [ f (t)] = R(s, u) = N−1 [R(s, u)] = f (t) =

1 u

1 2πi

∫ ∞

e

−st u

0 ∫ c+i∞

f (t)dt st

e u R(s, u)ds

(2.6) (2.7)

c−i∞

The eqn (2.6) is Natural transform (see below theorem 2.1,eqn (2.10)) of function f (t) ∈ A and eqn (2.7) is the Inverse Natural transform R(s,u) and both are linear operators. The LSD given in [7] now be obtained by substituting n = −1 in eqns (1.4) and (1.5) of motivation section ( ) ∫ 1 ∞ −t ( t ) 1 1 L [ f (t)] = F(s) = e f dt = G (2.8) s 0 s s s ( ) ∫ 1 ∞ −t 1 1 + S [ f (t)] = G(u) = e u f (t)dt = F (2.9) u 0 u u

108

Fethi Bin Muhammed Belgacem, R.Silambarasan

Next we derive the relationship between Natural and Laplace,Sumudu transform in successive theorems. Theorem 2.1. Natural-Laplace Duality (NLD).If R(s,u) is Natural transform and F(s) is Laplace transform of function f (t) ∈ A then, N+ [ f (t)] = R(s, u) =

1 u

∫ ∞ 0

st 1 (s) e− u f (t)dt = F u u

(2.10)

Proof. Substituting w = ut in eqn (2.2) and replacing limits and dummy variable the result (2.10) follows. Theorem 2.2. Natural-Sumudu Duality (NSD).If R(s,u) and G(u) are respective Natural and Sumudu transforms of function f (t) ∈ A then, N+ [ f (t)] = R(s, u) =

1 s

∫ ∞

e−t f

( ut ) s

0

1 (u) dt = G s s

(2.11)

Proof. Substituting w = st in eqn (2.2) and performing similar operations as theorem 2.1 the result (2.11) obtained. For any value of n the generalised Natural transform of function f (t) > 0 is defined by N+ [ f (t)] = R(s, u) = un N+ [ f (t)] = R(s, u) = sn

∫ ∞ ∫ 0∞

e−su t f (un+1t)dt

(2.12)

e−s

(2.13)

n

n+1 t

f (sn ut)dt

0

Substituting n = −1 in eqns (2.12) and (2.12) gives the dualities NLD and NSD eqns (2.10) and (2.11) respectively.In the light of eqn (2.11) to the eqn (2.5),For the function f (t) Natural transform may be defined by N+ [ f (t)] = R(s, u) =

∞

n!an un n+1 n=0 s

∑

(2.14)

Throughout this paper we assume the functions are defined in positive axis (0, ∞), thus f (t) > 0 and f (t) = 0 for t < 0,and use the Natural transform definitions eqns (2.2) (2.6) (2.11) and (2.14) for proving various properties wherever necessary.And it is worth to mention t is time variable,s is frequency variable and u is again time variable.Having background to Natural transform certain analytical properties are proved in next section.

3 Analytical properties Theorem 3.1. Weight Shift property.Let the function f (t) belongs to set A be multiplied with weight function e±t then, [ ] s su + ±t N [e f (t)] = R (3.1) (s ∓ u) s ∓ u

Theory of natural transform

109

Proof. From the defining eqn (2.2) ±t

N [e f (t)] = +

Sub w =

(s∓u)t s

t=

s = s∓u

sw s∓u

dt =

∫ ∞

e 0

−sw

∫ ∞

e

−st ±ut

e

∫ ∞

f (ut)dt =

0

e−(s∓u)t f (ut)dt

0

sdw s∓u

[

] [ ] [ ] ∫ ∞ usw s ust s su −st f dw = e f dt = R s∓u s∓u 0 s∓u (s ∓ u) s ∓ u

which ends the proof of theorem 3.1. Theorem 3.2. Change of Scale property.Let the function f (at) in setA,where a is non zero constant then, 1 [s ] N+ [ f (at)] = R , u (3.2) a a Proof. Using the definition,eqn (2.6) 1 N [ f (at)] = u +

∫ ∞

e

−st u

f (at)dt

0

w dw ; dt = a a ∫ dw 1 ∞ −sw e au f (w) = u 0 a ∫ 1 ∞ −st 1 [s ] = e au f (t)dt = R , u au 0 a a √ For the Natural transform of cosh 7t,(entry 13,table 3) we have ( ) √ s 1 1 + N [cosh 7t] = √ N[cosh at]s→ √s = √ 7 7 7 s2 − u2 s→ √s 7 ( s ) √ 1 s 7 =√ = 2 = R(s, u) 2 s 2 s − 7u2 7 −u w = at → t =

7

The similar counterpart of theorem 3.2 for the function f

(t ) a

is

[ ( t )] N+ f = a.R[as, u] a

(3.3)

Theorem 3.3. Natural transform of derivative.If f n (t) is the nth derivative of function f (t) w.r.t.’t’ then its Natural transform is given by N+ [ f n (t)] = Rn (s, u) =

n−1 n−(k+1) sn s R(s, u) − f k (0) ∑ n n−k u u k=0

(3.4)

Proof. For n = 1 and 2 in eqn (3.4) gives the Natural transform of first and second derivatives of f (t) respectively [1].

110

Fethi Bin Muhammed Belgacem, R.Silambarasan ′ s f (0) N+ [ f (t)] = R1 (s, u) = R(s, u) − u u ′ 2 R(s, u) − s f (0) ′′ s f (0) N+ [ f (t)] = R2 (s, u) = − u2 u

(3.5) (3.6)

To proceed the induction process,assuming eqn (3.4) true for n and prove it for n + 1,using eqn (3.5), ′ 1 s N+ [ f n+1 (t)] = N+ [( f n (t)) ] = Rn+1 (s, u) = Rn (s, u) − f n (0) u ] u [ n−1 n−(k+1) n s s s 1 = R(s, u) − ∑ f k (0) − f n (0) n−k u un u u k=0

=

n sn+1 sn−k R(s, u) − ∑ u(n−k)+1 f k (0) un+1 k=0

which is true for n + 1 and when n = 0 and 1 in previous relation,gives eqns (3.5) and (3.6) respectively.Hence the result (3.4) follows. Consider the integration of function f (t) in set A,w.r.t ′t ′ in the interval (0,t) as w(t) and successive integrals as w2 (t) upto wn (t) which is ∫ t

w(t) =

f (t)dt, w2 (t) =

∫ t∫ t

0

0

f (t)(dt)2 , . . . , wn (t) =

0

∫ t

∫ t

... 0

f (t)(dt)n

(3.7)

0

Theorem 3.4. Natural transform of integrals.If wn (t) is given by (3.7) the Natural transform of wn (t) is un (3.8) N+ [wn (t)] = n R(s, u) s Proof. From the Natural transform definition eqn (2.6) ∫

1 ∞ −st n 1 e u w (t)dt = u 0 u Applying the integration by parts N+ [wn (t)] = ∫ t

u=

∫ t

... 0

0

∫ ∞

e 0

−st u

∫ t

∫ t

... 0

f (t)(dt)n dt

0

−st un −st f (t)(dt)n , un = f (t)(dt); v.dv = e u , vn = (−)n n e u s ]∞ [ ∫ ∞ n n −st u 1 u −st = (−)n n e u wn (t) + e u n f (t)dt s u 0 s 0

The first part of the previous equation vanishes and the succeeding integral gives un 1 sn u

∫ ∞

e 0

−st u

f (t)dt =

un R(s, u) = N+ [wn (t)] sn

which ends the proof. At the end of this section Natural transform for T-periodic function was given. Theorem 3.5. The Natural transform of T-periodic function f (t) ∈ A such that f (t + nT ) = f (t), n = 0, 1, 2, . . . is given ∫ T [ ] −sT −1 1 −st N+ [ f (t)] = R(s, u) = 1 − e u (3.9) e u f (t)dt u 0

Theory of natural transform

111

Proof. Writing eqn (2.6) as ∫

∫

1 T −st 1 ∞ −st e u f (t)dt + e u f (t)dt u 0 u T Replacing t = x + T in the second integral

N+ [ f (t)] = R(s, u) =

dt = dx and t = T, x = 0 and t = ∞, x = ∞ thus =

1 u

∫ T

e

−st u

f (t)dt +

0

1 u

∫ ∞

e

−s(x+T ) u

f (x + T )dx

0

Noting for period function f (x + T ) = f (x) 1 = u

∫ T

e

−st u

f (t)dt + e

−sT u

0

1 u

∫ ∞

e

−sx u

f (x)dx

0

Now replacing t for x in the second integral and after rearranging the result (3.9) obtained. At last we can see that constants (also unit step function) have same Natural and Laplace transform and Dirac delta function δ(t) have same Natural and Sumudu transform,given respectively N+ [constant] = L [constant] = N+ [δ(t)] = S+ [δ(t)] =

1 s

1 u

4 Multiple shift and Convolution theorem When the function f (t) in set A is multiplied with some shift t then, ∞

t f (t) =

∑ ant n+1

n=0

The Natural transform of t f (t) gives N+ [t f (t)] = =

∞

(n + 1)!an un+1 u ∞ (n + 1)!an un u ∞ d n!an un+1 = ∑ = ∑ ∑ sn+2 s n=0 sn+1 s n=0 du sn+1 n=0 u d ∞ n!an un u d u ∑ n+1 = uR(s, u). s du n=0 s s du

The generalization of previous result is Theorem 4.1. The function f (t) in set A is multiplied with shift function t n then, N+ [t n f (t)] = Proof. From the defining eqn (2.14)

un d n n u R(s, u) sn dun

(4.1)

112

Fethi Bin Muhammed Belgacem, R.Silambarasan

N+ [t n f (t)] =

(2n)!an u2n un ∞ (2n)!an un ∑ s2n+1 = sn ∑ sn+1 n=0 n=0 ∞

=

un ∞ d n n!an u2n un d n n ∞ n!an un ∑ dun sn+1 = sn dun u ∑ sn+1 sn n=0 n=0

=

un d n n u R(s, u) sn dun

Theorem 4.2. If R(s,u) is Natural transform of function f (t) in set A,and let Rk (s, u) be the kth derivadk tive of R(s,u) w.r.t ′ u′ ie du k R(s, u).Then the Natural transform of function multiplied with shift function t n is given by N+ [t n f (t)] =

un n n k ∑ ak u Rk (s, u) sn k=0

(4.2)

where an0 = n!, ann = 1, an1 = n!n, ann−1 = n2 and for k = 2, 3, . . . , n − 2 n−1 ank = an−1 k−1 + (n − k)ak

(4.3)

Proof. The proof is similar to the theorem given in [8] ([8],theorem 5.2,page 18 and 19) by replacing un to

un and Gk (u) to Rk (s, u) sn

And the coefficents of ank eqn (4.3) for n = 0 through 6 are given by [8] n/k 0 1 2 3 4 5 0 1 1 1 1 2 2 4 1 3 6 18 9 1 4 24 96 72 16 1 5 120 600 600 200 25 1 6 720 4320 5400 2400 450 36 1

(4.4)

Using the coefficients for ank in (4.4) the Natural transform of t 2 f (t) and t 3 f (t) are respectively given by ] u2 [ 2R(s, u) + 4uR1 (s, u) + u2 R2 (s, u) 2 s ] u3 [ N+ [t 3 f (t)] = 3 6R(s, u) + 18uR1 (s, u) + 9u2 R2 (s, u) + u3 R3 (s, u) s

N+ [t 2 f (t)] =

(4.5) (4.6)

which ends the proof. Corollary 4.1. For the function f (t) ∈ A,multiplied with shift function t n its Natural transform is N+ [t n f (t)] = (−u)n

dn R(s, u) dsn

(4.7)

Theory of natural transform

113

Proof. Differentiating eqn (2.2) ’n’ times w.r.t.’s’, ∫

∫

∫

∞ ∂n ∞ dn d n ∞ −st R(s, u) = e f (ut)dt = e−st f (ut)dt = (−t)n e−st f (ut)dt n n n ds ds 0 0 ∂s 0 ∫ 1 ∞ 1 = n (−ut)n e−st f (ut)dt = N+ [t n f (t)] u 0 (−u)n

multiplying both sides by (−u)n gives relation (4.7). Theorem 4.3. If f n (t) is nth derivative of function f (t) w.r.t ′t ′ ,is multiplied with shift function t n then, dn N+ [t n f n (t)] = un n R(s, u) (4.8) du Proof. Differentiating defining eqn (2.11) ’n’ times w.r.t.’u’ gives dn dn R(s, u) = dun dun

∫ ∞ 1 0

s

e−t f

( ut ) s

dt =

∫ ∞ 1 0

∫ ∞ −t ( )n ( ) e t ut = fn dt

s

0

s

s

e−t

∂n ( ut ) f dt ∂un s

s

∫ 1 1 ∞ e−t ( ut )n n ( ut ) f dt = n N+ [t n f n (t)] = n u 0 s s s u

multiplying both sides by un the eqn (4.8) follows. Theorem 4.4. The function f (t) in A is divided with multiple shift function t n its Natural transform defined [ ] ∫ ∫ ∞ 1 ∞ + f (t) N = n ... R(s, u)(ds)n (4.9) tn u s s Proof. Integrating R(s,u) eqn (2.2) w.r.t ′ s′ in the interval (s, ∞),n times ∫ ∞

∫ ∞

... ∫ ∞

=

s

∫ ∞

f (ut)dt 0

n

R(s, u)(ds) = s

∫ ∞

... s

∫ ∞

s

∫ ∞∫ ∞

... s

s

e−st f (ut)dt(ds)n

0

∫ ∞ f (ut) −st −st n e (ds) = e dt n 0

=u

n

t

∫ ∞

e 0

−st

[ ] f (ut) n + f (t) dt = u N (ut)n tn

dividing both sides by un gives the (4.9). Theorem 4.5. The nth derivative of time function f (t) in set A w.r.t ′t ′ is divided with shift function t n ,its N transform given by [ n ] ∫ ∞ ∫ ∞ 1 + f (t) N = ... R(s, u)(du)n (4.10) tn (−u)n u u Proof. Integrating eqn (2.2) R(s,u) w.r.t ′ u′ in the interval (u, ∞),n times

114

Fethi Bin Muhammed Belgacem, R.Silambarasan

∫ ∞

∫ ∞

... u

R(s, u)(du)n =

u

∫ ∞

∫ ∞∫ ∞

... ∫u ∞

u −st

=

e

∫ 0∞

∫ ∞

...

dt

0

∫ ∞

e−st f (ut)dt(du)n f (ut)(du)n

u u f n (ut)

∫ ∞

f n (ut) dt (−t)n (−ut)n 0 0 [ n ] ∫ ∞ n n −st f (ut) n + f (t) = (−u) e dt = (−u) N (ut)n tn 0 e−st dt

=

= un

e−st

dividing both sides by (−u)n ends proof for eqn (4.10). Theorem 4.6. Convolution theorem.If F(s,u) and G(s,u) are the Natural transforms of respective functions f (t) and g(t) both defined in set A then, N+ [( f ∗ g)] = uF(s, u)G(s, u)

(4.11)

where f ∗ g is convolution of two functions defined by ∫ t 0

∫ t

f (a)g(t − a)da =

0

f (t − a)g(a)da

Proof. Expanding the R.H.S of eqn (4.11) ∫ ∞

uF(s, u)G(s, u) = u

−sx

e 0

∫ ∞∫ ∞

=u 0

∫ ∞

f (ux)dx

e−sy g(uy)dy

0

e−s(x+y) f (ux)g(uy)dxdy

0

substituting t = x + y

∫ ∞∫ ∞

= 0

0

e−st f (ux)g(u(t − x))udxdy

setting a = ux and da = udx with ux in [0, ut] and x is in [0,t] thus ∫ ∞

=

−st

∫ t

e 0

∫ ∞

= 0 +

0

e−st

f (ux)g(u(t − x))d(ux)dt

∫ ut

f (a)g(ut − a))d(a)dt

0

= N [( f ∗ g)] Corollary 4.2.

′

N+ [( f ∗ g) ] = sF(s, u)G(s, u) Proof. Using the Natural transform of first derivative and noting ( f ∗ g) = 0 ′ 1 s N+ [( f ∗ g) ] = N[ f ∗ g] − ( f ∗ g)(0) u u s = uF(s, u)G(s, u) u = sF(s, u)G(s, u)

(4.12)

Theory of natural transform

115

We give without proof,Natural transform of certain convolution of functions. Corollary 4.3. ′

N+ [( f ∗ f ) ] = sM 2 (s, u) un N+ [( f ∗ g)n ] = n−2 F(s, u)G(s, u) s N+ [(h1 ∗ h2 ∗ · · · ∗ hn )] = sun−1 H1 (s, u)H2 (s, u) . . . Hn (s, u)

(4.13)

N [( f ∗ g ∗ h)] = su F(s, u)G(s, u)H(s, u)

(4.16)

+

2

(4.14) (4.15)

where ( f ∗ g)n = (h1 ∗ h2 ∗ · · · ∗ hn ) = ( f ∗ g ∗ h) =

∫ t

∫ t

... 0

∫ t 0

∫ t 0

0

f (a)g(t − a)(da)n

h1 (a)h2 (a) . . . hn (t − a)da f (a)g(a)h(t − a)da

Definitions of inverse Natural transform is given in next section.

5 Inverse Natural transform The inverse Natural transform was already derived from the Fourier integral eqn (2.7).Here we give the formal definition with the Cauchy Residue theorem.Before doing so,we give the inverse transform of Laplace [14, 14, 16] and Sumudu transforms [3, 8, 9] in respective succeeding theorems. Theorem 5.1. Complex inverse Laplace transform L −1 [F(s)] = f (t) =

1 2πi

∫ γ+i∞ γ−i∞

est F(s)ds,t > 0

(5.1)

F(s) is the Laplace transform,the integral is taken along s = γ in complex plane s = x + iy.The real number γ is chosen so that s = γ lies on right of all singularities. Theorem 5.2. Complex inverse Sumudu transform ( ) 1. 1s G 1s is meromorphic function with singularities having Re(s) = γ and 2. there exists circular Γ with radius R and positive constants M and K with ( ) G 1 s < MR−k s Then inverse Sumudu transform is given 1 S [G(u)] = f (t) = 2πi −1

∫ γ+i∞ γ−i∞

( ) 1 ds e G s s st

(5.2)

116

Fethi Bin Muhammed Belgacem, R.Silambarasan

Theorem 5.3. Bromwich contour integral for complex inverse Natural transform : R(s,u) is the Natural tranform of function f (t) in A,then its inverse Natural transform is defined by 1 N [R(s, u)] = f (t) = lim T →∞ 2πi −1

∫ γ+iT γ−iT

st

e u R(s, u)ds

(5.3)

c is real constant and the integral is taken along s = γ in complex plane s = x + iy.The real number γ is chosen so that s = γ lies on right of all (finite (or) countably infinite) singularities. Suppose when T → ∞ then the integral over Γ tends to zero,then by Cauchy residue theorem eqn (5.3) is defined by 1 N [R(s, u)] = f (t) = lim T →∞ 2πi −1

∫ γ+iT γ−iT

st

e u R(s, u)ds

st

= sum of residues of e u R(s, u)at the poles of R(s, u) = ∑ sum of residues of e u R(s, u)at the poles of R(s, u) st

Theorem 5.4. Heaviside’s Expansion theorem.F(s,u) and G(s,u) are two polynomials in s and u,with degree of F(s,u) is less than the degree of G(s,u),where F(s,u) and G(s,u) are respective Natural transforms of functions f (t) and g(t) both defined in set A. [ ] n F(αi , u) αi t −1 F(s, u) N = f (t) = ∑ ′ eu (5.4) G(s, u) i=1 G (αi , u) ′

Here G (s, u) is the differentiation of G(s,u) w.r.t ’s’ and αi (i = 1, 2, . . . , n) are the n distinct roots of G(s, u) = 0 Proof. Assume the leading coefficient of G(s,u) (coefficient of highest power of ’s’) is unity.then, n

G(s, u) = ∏(s − αi ) = (s − α1 )(s − α2 ) . . . (s − αn )

(5.5)

i=1

Let P(s, u) =

F(s, u) G(s, u)

(5.6)

The partial fraction decomposition of eqn (5.6) is P(s, u) =

n F(s, u) Ai =∑ G(s, u) i=1 s − ai n

n

i=1 n

i=1

(5.7)

P(s, u) = ∑ Ai ∏(s − αi )

(5.8)

P(s, u) = ∑ Ai (s − α1 )(s − α2 ) . . . (s − αi−1 )(s − αi+1 ) . . . (s − αn )

(5.9)

i=1

Substituting s = αi in eqn (5.9) P(αi , u) = Ai (αi − α1 )(αi − α2 ) . . . (αi − αi+1 ) . . . (αi − αn )

(5.10)

Theory of natural transform

117

Where (i = 1, 2, . . . , n) and differentiating eqn (5.5) and substituting s = αi n

′

G (s, u) = ∑ (s − α1 )(s − α2 ) . . . (s − αi−1 )(s − αi+1 ) . . . (s − αn )

(5.11)

i=1

′

G (αi , u) = (αi − α1 )(αi − α2 ) . . . (αi − αi−1 )(αi − αi+1 ) . . . (αi − αn )

(5.12)

From eqns (5.10) and (5.12) the constant Ai is given by Ai =

F(αi , u) ′ G (αi , u)

(5.13)

Substituting eqn (5.13) in eqn (5.7) for Ai n F(s, u) 1 F(αi , u) =∑ G(s, u) i=1 s − αi G′ (αi , u)

Noting N+ [eat ] =

1 s−au

(5.14)

(entry 5,table 3),so that [ ] ] [ αi t 1 1 −1 −1 u N =N αi u = e s − αi s− u

Hence the inverse Natural transform of eqn (5.14) results [ ] n F(s, u) F(αi , u) αi t N−1 = f (t) = ∑ ′ eu G(s, u) i=1 G (αi , u) where the theorem proved. Theorem 5.5. Lerch Theorem.Let f (t) and g(t) are the two functions in set A.If there exists any constants s0 > 0,such that N[f(t)]=N[g(t)],F(s,u)=G(s,u) for all s0 > 0 and t > 0 except possibly at point of discontinuity.Then the Natural transform is Unique for all s0 > 0,t > 0.

6 Applications Applications of Natural transform of derivatives and Natural multiple shift theorem is our first example. Example 6.1. Consider the Bessel’s differential equation (with polynomial coefficients) and the initial conditions. t

d 2 y dy + + ty(t) = 0 dt 2 dt ′ y(0) = 1, y (0) = c (any constant)

Proof. Applying Natural transform on both sides of eqn (6.1), [ 2 ] [ ] d y dy N+ t 2 + N+ + N+ [ty(t)] = 0 dt dt Using theorem 4.1,eqn (4.1)

(6.1) (6.2)

(6.3)

118

Fethi Bin Muhammed Belgacem, R.Silambarasan ′′ ′ u d u d [u.N+ [y (t)]] + N+ [y (t)] + [u.N+ [y(t)]] = 0 s du s du

(6.4)

The Natural transform of first and second derivatives are respectively from eqns (3.5) and (3.6), [ ( )] ′ u d s2 R(s, u) sy(0) y (0) sR(s, u) y(0) u − 2 − + − 2 s du u u u u u [ ] u d + u R(s, u) + R(s, u) = 0 s du Using initial conditions, [ ] u d s2 R(s, u) s sR(s, u) 1 − −c + − s du u u u u 2 u d u + R(s, u) + R(s, u) = 0 s du s [ 2 ] s R(s, u) s2 d s u sR(s, u) 1 − R(s, u) + 2 + − + 2 s u u du u u u 2 u d u + R(s, u) + R(s, u) = 0 s du s −

sR(s, u) d 1 sR(s, u) 1 + s R(s, u) + + − u du u u u 2 u d u + R(s, u) + R(s, u) = 0 s du s

After cancelling opposite signed terms,the above equation gives, s

dR(s, u) u2 dR(s, u) u + + R(s, u) = 0 du s du s ( ) 2 u dR(s, u) uR(s, u) s+ + =0 s du s s2 + u2 dR(s, u) uR(s, u) =− s du s dR(s, u) udu =− 2 R(s, u) s + u2 udu dR(s, u) + 2 =0 R(s, u) s + u2

(6.5) (6.6) (6.7) (6.8) (6.9)

Integrating ∫

dR(s, u) 1 + R(s, u) 2

∫

2udu =0 s2 + u2

1 log R(s, u) + log(s2 + u2 ) + log constant = 0 = log 1 2 √ log[R(s, u) s2 + u2 constant] = log 1

(6.10) (6.11) (6.12)

Theory of natural transform

Application of anti-logarithm on both sides of eqn (6.12), √ R(s, u) s2 + u2 constant] = 1 constant R(s, u) = √

119

(6.13) 1 s2 + u2

(6.14)

Taking inverse Natural transform constant y(t) = N

−1

[

1 √ 2 s + u2

] (6.15)

gives,(entry 66 (a=1),table 3) constant y(t) = J0 (t)

(6.16)

Using the initial condition eqn (6.2),y(0) = 1 in eqn (6.16) gives the value of constant, constant y(0) = J0 (0) = 1; constant = 1 Thus the eqn (6.16) becomes y(t) = J0 (t)

(6.17)

which is the solution of eqn (6.1). How the Bromwich contour integral of inverse Natural transform works is given in next example. Example 6.2. To find the inverse Natural transform of R(s, u) =

au2 s(s2 + a2 u2 )

(6.18)

Proof. Using the Bromwich contour integral theorem 5.3 eqn (5.3) f (t) = ∑ Residues of e u

st

au2 s(s2 + a2 u2 )

(6.19)

at the poles of eqn (6.18). The three simple poles of eqn (6.18) are s = 0, iau and −iau.And the corresponding residues −iat eiat are,Res at s = 0 is 1a ,Res at s = iau is −2a and Res at s = −iau is e−2a ,whose sum of all residues are 1 a (1 − cos at) = f (t) which is the time function solution of eqn (6.18). The last example is application of Heaviside’s Expansion theorem. Example 6.3. For the inverse Natural transform of F(s, u) 2s2 − 5su − 4u2 = = P(s, u) 3 2 2 s + s u − 2su G(s, u)

(6.20)

120

Fethi Bin Muhammed Belgacem, R.Silambarasan

Proof. Using the theorem 5.4 and noting deg F(s, u) < deg G(s, u), G(s, u) = s3 + s2 u − 2su2 = s(s + 2u)(s − u)

(6.21)

s = 0, −2u, u

(6.22)

are the three simple poles,and differentiating G(s,u) w.r.t ′ s′ ′

G (s, u) = 3s2 + 2su − 2u2

(6.23)

′

applying the poles for ’s’,eqn (6.22),in F(s,u) and G (s, u) respectively given by F(0, u) = −4u2 , F(u, u) = 3u2 , F(−2u, u) = −6u2 ′

′

′

G (0, u) = −2u , G (u, u) = 3u , G (−2u, u) = 6u 2

2

2

(6.24) (6.25)

Now using the eqn (5.4) and αi = (0, u, −2u); (i = 1, 2, 3) 3

F(αi , u) αi t eu ′ G (α , u) i i=1

f (t) = ∑ f (t) =

−4u2 t0 3u2 tu 6u2 −2tu e u + 2 e u − 2 e u = 2 + et − e−2t −2u2 3u 6u

which is the required function. Table 1: Brief theory of Natural transform

(6.26) (6.27)

Theory of natural transform

Property Definition

Derivative

Integration

Definition N [ f (t) ∈ A] = +

∫ ∞

e−st f (ut)dt

0

n−1 n−(k+1) sn s R(s, u) − f k (0) ∑ n n−k u u k=0 [∫ t ∫ t ] un n + ... f (t)(dt) = n R(s, u) N s 0 0

N+ [ f n (t)] =

un d n n u R(s, u) sn dun dn = (−u)n n R(s, u) ds n + n n n d R(s, u) N [t f (t)] = u dun ] [ ∫ ∫ ∞ 1 ∞ + f (t) = N . . . R(s, u)(ds)n tn un s s [ n ] ∫ ∞ ∫ ∞ 1 + f (t) N = ... R(s, u)(du)n tn (−u)n u u N+ [t n f (t)] =

Product shift Product shift and derivative

Division shift

Division shift and derivative Convolution

Complex inverse Natural transform

N+ [( f ∗ g)] = uF(s, u)G(s, u) f (t) =

1 2πi

∫ γ+i∞ γ−i∞ n

Heaviside’s expansion

st

e u R(s, u)ds

F(αi , u) αi t eu ′ i=1 G (αi , u)

f (t) = ∑

121

122

Fethi Bin Muhammed Belgacem, R.Silambarasan

Table 2: Important functions and definitions S.No

Function

1

Gamma function

Γ(n) =

∫ Definition ∞ 0 ∫

un−1 e−u du; n > 0

1

B(m, n) = 2

0

Beta function

um−1 (1 − u)n−1 du

ΓmΓn Γ(m + n) xn x2 Jn (x) = n × {1 − 2 Γ(n + 1) 2(2n + 2) 3 Bessel function 4 x + −...} 2.4(2n + 2)(2n + 4) In (x) = i−n Jn (ix) xn x2 = n × {1 + 4 Modified Bessel function 2 Γ(n + 1) 2(2n + 2) 4 x + −...} 2.4(2n + 2)(2n ∫ t+ 4) 2 2 er f (t) = √ e−u du 5 Error function π 0 ∫ ∞ 2 2 √ e−u du 6 Complementary error function er f c(t) = 1 − er f (t) = π ∫ ∞ −u t e 7 Exponential integral du Ei(t) = ∫tt u sin u 8 Sine integral Si(t) = du ∫ 0t u cos u 9 Cosine integral Ci(t) = du ∫ 0t u 10 Fresnel Sine integral S(t) = sin u2 du =

∫ 0t

11

Fresnel Cosine integral

12

Laguerre polynomial

C(t) = 0

cos u2 du

et d n n −t Ln (t) = t e ; n = 0, 1, 2, . . . n! dt n

Theory of natural transform

Table 3: Special Natural transforms-Entries 1-171 f (t) 1

1

R(s, u) 1 s

2

t

u s2

3

t n−1 ; n = 1, 2, . . . (n − 1)!

un−1 sn

4

t n−1 ;n > 0 Γ(n)

un−1 sn

5

eat

1 (s − au)

6

t n−1 eat ; n = 1, 2, . . . (n − 1)!

un−1 (s − au)n

7

t n−1 eat ; Γ(n)

un−1 (s − au)n

8

sin at a

s2 + a2 u2

9

cos at

s s2 + a2 u2

10

ebt sin at a

u (s − bu)2 + a2 u2

11 ebt cos at

s − bu (s − bu)2 + a2 u2

12

sinh at a

13 cosh at

u

u s2 − a2 u2 s s2 − a2 u2

ebt sinh at a

u (s − bu)2 − a2 u2

15 ebt cosh at

s − bu (s − bu)2 − a2 u2

14

16

ebt − eat ; a ̸= b b−a

u (s − bu)(s − au)

123

124

Fethi Bin Muhammed Belgacem, R.Silambarasan

17

bebt − aeat ; a ̸= b b−a

s (s − bu)(s − au)

18

sin at − at cos at 2a3

u3 (s2 + a2 u2 )2

19

t sin at 2a

su2 (s2 + a2 u2 )2

20

sin at + at cos at 2a

s2 u (s2 + a2 u2 )2

1 21 cos at − at sin at 2

s3 (s2 + a2 u2 )2

22 t cos at

u(s2 − a2 u2 ) (s2 + a2 u2 )2

23

at cosh at − sinh at 2a3

(s2 − a2 u2 )2

24

t sinh at 2a

su2 (s2 − a2 u2 )2

25

sinh at + at cosh at 2a

s2 u (s2 − a2 u2 )2

u3

1 26 cosh at + at sinh at 2

s3 (s2 − a2 u2 )2

27 t cosh at

u(s2 + a2 u2 ) (s2 − a2 u2 )2

28

(3 − a2t 2 ) sin at − 3at cos at 8a5

u5 (s2 + a2 u2 )3

29

t sin at − at 2 cos at 8a3

su4 (s2 + a2 u2 )3

30

(1 + a2t 2 ) sin at − 3at cos at 8a3

(s2 + a2 u2 )3

31

3t sin at + at 2 cos at 8a

s3 u2 (s2 + a2 u2 )3

32

(3 − a2t 2 ) sin at + 5at cos at 8a

s4 u (s2 + a2 u2 )3

s2 u3

Theory of natural transform

33

(8 − a2t 2 ) cos at − 7at sin at 8

s5 (s2 + a2 u2 )3

34

t 2 sin at 2a

u3 (3s2 − a2 u2 ) (s2 + a2 u2 )3

35

t 2 cos at 2

u2 (s2 − a2 u2 ) (s2 + a2 u2 )3

36

t 3 cos at 6

u3 (s4 − 6a2 s2 u2 + a4 u4 ) (s2 + a2 u2 )4

37

t 3 sin at 24a

u4 (s2 − a2 u2 ) (s2 + a2 u2 )3

38

(3 + a2t 2 ) sinh at − 3at cosh at 8a5

u5 (s2 − a2 u2 )3

39

at 2 cosh at − t sinh at 8a3

su4 (s2 − a2 u2 )3

40

at cosh at + (a2t 2 − 1) sinh at 8a3

s2 u3 (s2 − a2 u2 )3

41

3t sinh at + at 2 cosh at 8a

s3 u2 (s2 − a2 u2 )3

42

(3 + a2t 2 ) sinh at + 5at cosh at 8a

s4 u (s2 − a2 u2 )3

43

(8 + a2t 2 ) cosh at + 7at sinh at 8a

s5 (s2 − a2 u2 )3

44

t 2 sinh at 2a

u3 (3s2 + a2 u2 ) (s2 − a2 u2 )3

45

t 2 cosh at 2

su2 (s2 + a2 u2 ) (s2 − a2 u2 )3

46

t 3 cosh at 6

u3 (s4 + 6a2 s2 u2 + a4 u4 ) (s2 − a2 u2 )4

47

48

su4 (s2 + a2 u2 ) t 3 sinh at (s2 − a2 u2 )4 24a [ ] √ √ at −3at e2 √ 3at 3at u2 − cos +e 2 3 sin 2 3a 2 2 s3 + a3 u3

125

126

49

50

51

52

53

Fethi Bin Muhammed Belgacem, R.Silambarasan

[ ] √ √ at −3at e2 3at √ 3at cos + 3 sin −e 2 3a 2 2 [ √ ] at 1 −at 3at e + 2e 2 cos 3 2 [ √ √ ] −at 3at e 2 3at √ 3at e 2 − cos − 3 sin 3a2 2 2 [ ] √ √ −at 3at e 2 √ 3at 3at 3 sin − cos +e 2 3a 2 2 [ √ ] −at 1 at 3at e + 2e 2 cos 3 2

su s3 + a3 u3 s2 s3 + a3 u3 u2 s3 − a3 u3 su s3 − a3 u3 s2 s3 − a3 u3

54

1 [sin at cosh at − cos at sinh at] 4a3

u3 s4 + 4a4 u4

55

sin at sinh at 2a2

su2 s4 + 4a4 u4

56

1 [sin at cosh at + cos at sinh at] 2a

s2 u s4 + 4a4 u4

57 cos at cosh at

s3 s4 + 4a4 u4

58

1 [sinh at − sin at] 2a3

u3 s4 − a4 u4

59

1 [cosh at − cos at] 2a2

s4 − a4 u4

60

1 [sinh at + sin at] 2a

s2 u s4 − a4 u4

61

1 [cosh at + cos at] 2

s3 s4 − a4 u4 ( ) 1 1 √ √ u s + au + s + bu √ u √ s s + au ) √ ( u 1 √ s s − au

62

63

64

e−bt − e−at √ 2(b − a) πt 3 √ er f at √ a √ eat er f at √ a

su2

Theory of natural transform

[ 65 eat

] √ 1 b2 t √ − be er f c(b t) πt

67 I0 (at) 68 an Jn (at); n > −1 69 an In (at); n > −1 ( √ ) 70 J0 a t(t + 2b)

72

√

1 √ s − au + b u

)

1 √ s2 + a2 u2

66 J0 (at)

71

1 √ u

(

√ J0 (a t 2 − b2 );t > b 0;t < b

1 √ 2 s − a2 u2 √ ( s2 + a2 u2 − s)n √ un s2 + a2 u2 √ (s − s2 + a2 u2 )n √ un s2 − a2 u2 ( bs ) √ 1 e u (s − s2 + a2 u2 ) √ s s2 + a2 u2 −b

√

s2 +a2 u2

u e √ s2 + a2 u2

u2

tJ1 (at) a

3

(s2 + a2 u2 ) 2 su

73 tJ0 (at)

3

(s2 + a2 u2 ) 2

74 J0 (at) − atJ1 (at)

75

s2 3

(s2 + a2 u2 ) 2 u2

tI1 (at) a

3

(s2 − a2 u2 ) 2 su

76 tI0 (at)

3

(s2 − a2 u2 ) 2

77 I0 (at) + atI1 (at)

78

f (t) = n, n ≤ t < n + 1, n = 0, 1, 2, . . . [t]

79

f (t) =

∑ rk where

k=1

[t] = greatest integer ≤ t

s2 3

(s2 + a2 u2 ) 2

−s

1 eu = s −s s(e u − 1) s(1 − e u ) −s

1 eu = s −s s(e u − r) s(r − e u )

127

128

80

Fethi Bin Muhammed Belgacem, R.Silambarasan

f (t) = rn , n ≤ t < n + 1, n = 0, 1, 2, . . .

81

√ 1 √ cos 2 at πt

82

√ 1 √ sin 2 at πa

83

84

( t ) n2 a

−au

e s √ su √ −au ue s √ s s

√ Jv (2 at)

e √

( ) √ a er f c b t + √ 2 t

1 πta2n+1 89 ; n > −1 √

90

∫ ∞ 0

−u2 √ un e 4a2 t J2n (2 u)du

e−bt − e−at t

91 Ci(at) 92 Ei(at) 93 logt

94

2(cos at − cos bt) t

95 log2 t

su √

−a2 a √ e 4t 85 2 πt 3 ( ) a √ 86 er f 2 t ( ) a 87 er f c √ 2 t

88 e

−au

un e s sn+1 √ −a √ s u

1 −a2 √ e 4t πt

b(bt+a)

−s

s

1−eu 1−e u = s −s s(e u − r) s(r − e u )

1 −a√u s e u ( √ ) −a 1 √ s 1−e u s ( √) −a 1 √ s e u s   √ −a √ s u 1 e √ √ √  s s+b u un sn+1

( e

√ −a √ u s

)

1 s + au log u s + bu 1 s2 + a2 u2 log 2s a2 u2 1 s + au log s au ] 1[ u log − γ s s s2 + a2 u2 1 log 2 u s + b2 u2 [ ] 1 π2 ( u )2 + γ − log s 6 s

Theory of natural transform

96

logt + γ; γ = euler’s constant = 0.5772156

π2 ; 97 6 γ = euler’s constant = 0.5772156 (logt + γ)2 −

98 t n logt; n > −1

s] 1[ log s u 1 2 (u) log s s un [ sn+1

Γ′(n + 1) + Γ(n + 1) log

sin at t

1 −1 au tan u s

100 Si(at)

1 −1 au tan s s

√ 1 101 √ e−2 at πt

1 au √ e s er f c su

2 2 2a 102 √ e−a t π

( s ) 1 s22 2 e 4a u er f c u 2au

99

103 er f (at) 104 √

1 π(t + a)

( s ) 1 s22 2 e 4a u er f c s 2au (√ ) as eu as √ er f c √ u su as

1 t +a

106

1 2 t + a2 t t 2 + a2

108 tan−1

(√ ) au √ s

( as ) eu √ Ei u su

105

107

u] s

(t ) a

( as )) ( as ) ( as )] as ( π 1 [ cos − Si − sin Ci au u 2 u u u ( as )) ( as ) ( as )] 1 [ as ( π sin − Si + cos Ci u u 2 u u u ( as )) ( as ) ( as )] 1[ as ( π cos − Si − sin Ci s u 2 u u u

( 2 ) 1 t + a2 109 log 2 a2 ( 2 ) 1 t + a2 110 log t a2

( as )) ( as ) ( as )] 1 [ as ( π sin − Si + cos Ci s u 2 u u u [( ] ( as ))2 ( ) u π 2 as − Si +Ci s2 2 u u

111 N(t)

0

129

130

Fethi Bin Muhammed Belgacem, R.Silambarasan

112 δ(t)

1 u

113 δ(t − a)

1 −as e u u

114 U(t − a)

1 −as e u s

x 2 ∞ (−1)n nπx nπt + ∑ sin cos a π n=1 n a a

1 sinh xsu s sinh as u

) ( 4 ∞ (−1)n−1 2n − 1 sin πx ∑ 2n − 1 π n=1 2a 116 ( ) 2n − 1 πt × sin 2a

1 sinh xsu s cosh as u

t 2 ∞ (−1)n nπx nπt cos sin 117 + ∑ a π n=1 n a a

1 cosh xsu s sinh as u

( ) 2n − 1 4 ∞ (−1)n cos πx 1+ ∑ π n=1 2n − 1 2a 118 ( ) 2n − 1 × cos πt 2a

1 cosh xsu s cosh as u

xt 2a ∞ (−1)n nπx nπt + 2 ∑ sin sin 2 a π n=1 n a a

u sinh xsu s2 sinh as u

115

119

8a ∞ (−1)n ∑ (2n − 1)2 π2 n=1 120 ( ) ( ) 2n − 1 2n − 1 πx cos πt × sin 2a 2a

u sinh xsu s2 cosh as u

t 2 2a ∞ (−1)n nπx + 2 ∑ cos 2 2a π n=1 n a 121 ( ) nπt × 1 − cos a

u cosh xsu s2 sinh as u

x+

Theory of natural transform

8a ∞ (−1)n ∑ (2n − 1)2 π2 n=1 122 ( ) ( ) 2n − 1 2n − 1 × cos πx sin πt 2a 2a

u cosh xsu s2 cosh as u

1 2 16a2 ∞ (−1)n (t + x2 − a2 ) − 3 ∑ 2 π n=1 (2n − 1)3 123 ( ) ) ( 2n − 1 2n − 1 × cos πx cos πt 2a 2a

u2 cosh xsu s3 sinh as u

t+

−n2 π2 t 2π ∞ nπx 124 2 ∑ (−1)n ne a2 sin a n=1 a

−(2n−1)2 π2 t π ∞ n−1 4a2 (−1) (2n − 1)e ∑ a2 n=1 125 ) ( 2n − 1 πx × cos 2a

√

s 1 sinh x √u √

u sinh a √ s

u

√

s 1 cosh x √u √

u cosh a √ s

u

−(2n−1)2 π2 t 2 ∞ (−1)n−1 e 4a2 ∑ a 126 n=1 ( ) 2n − 1 × sin πx 2a

s 1 sinh x √u √ √ su cosh a √ s

−n2 π2 t 1 2 ∞ nπx 127 + ∑ (−1)n e a2 cos a a n=1 a

1 cosh x √u √ √ su sinh a √ s

nπx x 2 ∞ (−1)n −n22π2 t e a sin 128 + ∑ a π n=1 n a

s 1 sinh x √u

4 ∞ (−1)n −(2n−1)2 2 π2 t ∑ (2n − 1) e 4a π n=1 129 ( ) 2n − 1 × cos πx 2a 1+

−n2 π2 t xt 2a2 ∞ (−1)n a2 ) + 3 ∑ (1 − e 3 a π n n=1 130 nπx × sin a

√

u

√

s

u

√

√

s sinh a √ s

u

√

s 1 cosh x √u √

s cosh a √ s

u

√

s u sinh x √u √ s2 sinh a √ s

u

131

132

Fethi Bin Muhammed Belgacem, R.Silambarasan

x2 − a2 16a2 ∞ (−1)n +t − 3 ∑ 2 π n=1 (2n − 1)3 131 ( ) −(2n−1)2 π2 t 2n − 1 2 cos × e 4a πx 2a ∞

132

λ2 nt

e a2 J0 (λn x/a) 1−2 ∑ λn J1 (λn ) n=1 whereλ1 , λ2 , . . . are positive roots of

√

s u cosh x √u √ s2 cosh a √ s

u



(√ ) six

√ 1 u (√ ) s J √sia

J0 0

u

J0 (λ) = 0 λ2 nt

133

∞ x2 − a2 e a2 J0 (λn x/a) + t + 2a2 ∑ 4 λ3n J1 (λn ) n=1

whereλ1 , λ2 , . . . are positive roots of J0 (λ) = 0



(√ ) six

√ u u ( √ ) 2 sia s J √

J0 0

u

134 Triangular wave

( as ) u tanh as2 2u

135 Square wave

( as ) 1 tanh s 2u

136 Rectified sine wave

137 Half sine wave rectifier

138 Sawtooth wave 139 Heaviside’s unit step 140 Pulse

141 Step

142 n2 , n ≤ t < n + 1, n = 0, 1, 2, . . . 143 rn , n ≤ t < n + 1, n = 0, 1, 2, . . .

( as ) πau coth s2 a2 + π2 u2 2u [ ] 1 πau s2 a2 + π2 u2 (1 − e −as u ) [ ] −as 1 u e u − s as 1 − e −as u 1 −as e u s ) −εs 1 −as ( e u 1−e u s [ ] 1 1 s 1 − e −as u ] [ −s −2s 1 e u +e u s (1 − e −su )2 [ ] −s 1 1−e u s 1 − re −su

Theory of natural transform

144

sin

( πt ) a

;0 ≤ t ≤ a

0;t > a 145 Laguerre polynomial

146

1 at (e − 1) a

147 (1 + at)eat

) ( −as πau 1 + e u s2 a2 + π2 u2 (s − u)n sn+1 ( ) u 1 s s − au s (s − au)2

148

a sin bt − b sin at a2 − b2

abu3 (s2 + a2 u2 )(s2 + b2 u2 )

149

cos bt − cos at a2 − b2

su2 (s2 + a2 u2 )(s2 + b2 u2 )

150

a sin at − b sin bt a2 − b2

s2 u (s2 + a2 u2 )(s2 + b2 u2 )

151

a2 cos at − b2 cos bt a2 − b2

s3 (s2 + a2 u2 )(s2 + b2 u2 )

152

b sinh at − a sinh bt a2 − b2

abu3 (s2 − a2 u2 )(s2 − b2 u2 )

153

cosh at − cosh bt a2 − b2

su2 (s2 − a2 u2 )(s2 − b2 u2 )

154

a sinh at − b sinh bt a2 − b2

s2 u (s2 − a2 u2 )(s2 − b2 u2 )

155

a2 cosh at − b2 cosh bt a2 − b2

s3 (s2 − a2 u2 )(s2 − b2 u2 )

1 156 t − sin at a 157

1 sinh at − t a

1 158 √ πt e−at 159 √ πt

a2 u3 s2 (s2 + a2 u2 ) a2 u3 s2 (s2 − a2 u2 ) 1 √ su ( ) 1 1 √ √ u s + au

133

134

Fethi Bin Muhammed Belgacem, R.Silambarasan

√ 2 t 160 √ π √ ( )n− 1 2 π t 161 Jn− 1 (at) 2 Γ(n) 2a √ ( )n− 1 2 π t 162 In− 1 (at) 2 Γ(n) 2a 163

nan Jn (at) t

164

nan In (at) t

165

166

169

u2n−1 (s2 − a2 u2 )n

1 [ un+1

0; 0 < t < b

a

u2n−1 (s2 + a2 u2 )n

un+1

ebt − eat √ 2 πt 3 √ I0 (a t 2 − b2 );t > b

[t ]

3

s2

1 [√

2 2 167 √ e−t π

168

√ u

([t]; greatest integer ≤ t)

a[t] − 1 a−1

s−

1 (√ 3

u2 √

−b(

e √

s2 + a2 u2 − s √

s2 − a2 u2

]n ]n

) √ s − au − s − bu

s2 −a2 u2 ) u

s2 − a2 u2 [ 2 ] 1 s2 s 4u e er f c u 2u [ −as ] 1 e u s 1 − e −as u [ ] −s 1 eu s 1 − ae −su −s

170 a[t] ( ) 1 a2 2a ∞ (−1)2 2 2 x +t − − 2 ∑ 2a 3 π n=1 n2 171 nπx nπt × cos cos a a

1−e u

−s

s(1 − ae u ) u cosh xsu s2 sinh as u

7 Conclusion Like the traditional Laplace transform and the new Sumudu transform,we have shown the worth of Natural transform (in solving differential equations) in this work.The results here we obtained using inverse Natural transform and Heaviside’s Expansion formula for inverse Natural transform are exact with the results in literature.

Theory of natural transform

135

References [1] Z.H.Khan and W.A.Khan.N-transform properties and applications. NUST Jour of Engg Sciences.Vol 1 No.1 2008 pp 127-133. [2] G.K.Watugala.Sumudu transform-a new integral transform to solve differential equations and control engineering problems. Math probs in Engg Indust-6 1998 N0.1 pp 319-329. [3] S.Weerakoon.Complex Inversion Formula for Sumudu Transform. Int Jour Math Edu Sci Tech (IJMEST).Vol.29.No.4.1998.pp 618-621. [4] M.A.Asiru.Sumudu Transform and Solution of Integral Equations of Convolution Type. Int Jour of Math Edu Sci Tech vol.32. No.6. 2001. pp 906-910. [5] M.A.Asiru.Further properties of Sumudu Transform and its Applications. Int Jour of Math Edu Sci Tech Vol. 33. No.2. 2002. pp 441-449. [6] M.A.Asiru.Applications of Sumudu Transform to Discrete dynamic system. Int Jour of Math Edu Sci Tech Vol. 34. No.6. 2003.pp 944-949. [7] F.B.M.Belgacem,A.A.Karaballi and S.L.Kalla.Analytical investigations of the Sumudu transform and applications to integral production equations, Mathematical Problems in Engineering (2003), no. 3, pp 103-118. [8] F.B.M.Belgacem and A.A.Karaballi.Sumudu transform fundamental properties investigations and applications. Journal of Applied Mathematics and Stochastic Analysis, 2006. pp 1-23. [9] F.B.M.Belgacem.Introducing and analysing deeper Sumudu properties. Nonlinear Studies Journal,Vol.13,No.1,2006 pp 23-41. [10] F.B.M.Belgacem.Sumudu transform applications to Bessel’s Functions and Equations. Applied Mathematical Sciences.Vol 4.No 74.2010. pp 3665-3686. [11] A.Kilicman and H.Eltayeb.A Note on the Sumudu Transforms and Differential Equations. Applied Mathematical Sciences, Vol.4,2010,no.22,pp 1089 - 1098. [12] A.Kilicman,H.Eltayeb and R.P.Agarwal.On Sumudu transform and System of Differential Equations. Jour of Abstract and Applied Analysis.2010.pp 1-11. [13] A.Kilicman and H.Eltayeb.On a New Integral transform and Differential equations. Math probs in Engg 2010 pp 1-13. [14] M.R.Spiegel.Theory and Problems of Laplace Transforms. Schaums Outline Series,McGraw-Hill,New York,1965. [15] L.Debnath and D.Bhatta.Integral Transforms and their applications. 2nd edition.C.R.C.Press.London.2007. [16] Joel.L.Schiff.Laplace transform Theory and Applications. Auckland, New-Zealand.Springer-2005.