International Journal of Analysis and Applications

Volume 17, Number 2 (2019), 167-190

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-17-2019-167

NEW INTEGRAL TRANSFORM: SHEHU TRANSFORM A GENERALIZATION OF

SUMUDU AND LAPLACE TRANSFORM FOR SOLVING DIFFERENTIAL

EQUATIONS

SHEHU MAITAMA∗, WEIDONG ZHAO

School of Mathematics, Shandong University, Jinan, Shandong 250100, P.R. China

∗Corresponding author: smusman12@sci.just.edu.jo

Abstract. In this paper, we introduce a Laplace-type integral transform called the Shehu transform which

is a generalization of the Laplace and the Sumudu integral transforms for solving differential equations in

the time domain. The proposed integral transform is successfully derived from the classical Fourier integral

transform and is applied to both ordinary and partial differential equations to show its simplicity, efficiency,

and the high accuracy.

1. Introduction

Historically, the origin of the integral transforms can be traced back to the work of P. S. Laplace in

1780s and Joseph Fourier in 1822. In recent years, differential and integral equations have been solved

using many integral transforms ( [1]- [11]). The Laplace transform, and Fourier integral transforms are the

most commonly used in the literature. The Fourier integral transform [12] was named after the French

mathematician Joseph Fourier. Mathematically, Fourier integral transform is defined as:

z[f(t)] = f(ω) =
1
√

2π

∫ ∞
−∞

exp (−iωt) f(t)dt. (1.1)

Received 2018-07-27; accepted 2018-10-06; published 2019-03-01.

2010 Mathematics Subject Classification. 44A10, 44A15, 44A20, 44A30, 44A35.

Key words and phrases. Shehu transform; Fourier integral transform; Laplace transform; natural transform; Sumudu trans-

form; ordinary and partial differential equations.

c©2019 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

167

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-17-2019-167


Int. J. Anal. Appl. 17 (2) (2019) 168

The Fourier transform have many applications in physics and engineering processes [13]. The Laplace integral

transform is similar with the Fourier transform and is defined as:

£[f(t)] = F(s) =

∫ ∞
−∞

exp (−st) f(t)dt. (1.2)

The Laplace transform is highly efficient for solving some class of ordinary and partial differential equations

[14]. By replacing the variable iω with the variable s in Equ.(1.1), the well-known Fourier transform will

become a Laplace transform and the vice-versa. The only difference between the Laplace transform, and the

Fourier transform is that the Laplace transform can be defined for both stable and unstable system while

the Fourier transform can only be defined on a stable system. In mathematical literature, the discrete-time

equivalent of the Laplace transform called z-transform [15] converts a discrete-time signal into a complex

frequency-domain representation. The basic idea of the z-transform was known to Laplace and later it was

re-introduced by the Jewish-Polish mathematician Witold Hurewicz to treat a sampled-data control systems

used with radar in 1947 ( [16]- [17]). In mathematics and signal processing, the bilateral or two-sided

z-transform of a discrete-time signal x[n] is the normal power series X(z) which is defined as:

X(z) = Z{x[n]} =
∞∑

n=−∞
x[n]z−n, (1.3)

where n is an integer and z is in general a complex number [18].

The multiplicative version of the two-sided Laplace transform called the Mellin integral transform is

defined as [19]:

M[f(s); s] = f∗(s) =

∫ ∞
0

xs−1f(x)dx. (1.4)

The Mellin integral transform is similar with the Laplace transform and Fourier transform and is widely

applied in computer science and number theory due to its invariant property [20, 21]. In railway engineering,

the Laplace-Carson transform [22] which is a Laplace-type integral transform named after Pierre Simon

Laplace and John Renshaw Carson is defined as:

f̂C(p) = p

∫ ∞
0

exp (−pt) f(t)dt, t ≥ 0. (1.5)

The Laplace-Carson integral transform have many applications in physics and engineering and can easily be

converted into a Mellin deconvolution problem, see [23, 24]. In mathematics, the Hankel’s integral transform

[25] which is similar to the Fourier transform was first introduced by the German mathematician Hermann

Hankel and was widely used in physical science and engineering [26]. The Hankel’s transform is defined as:

Fv(s) = Hv[f(r)] =

∫ ∞
0

rf(r)Jv(sr)dr, r ≥ 0, (1.6)

where Jv is the Bessel function of the first kind of order v with v ≥−12 .



Int. J. Anal. Appl. 17 (2) (2019) 169

In 1993, Watugala introduced a Laplace-like integral transform called the Sumudu integral transform [27].

In recent years, Sumudu transform has been applied to many real-life problems because of its scale and unit

preserving properties ( [28]- [31]). The mathematical definition of the Sumudu transform is given by:

S[f(t)](u) = G(u) =
1

u

∫ ∞
0

exp

(
−t
u

)
f(t)dt, (1.7)

provided the integral exists for some u. Based on the basic idea of the Laplace and the Sumudu integral

transform, the Elzaki transform was proposed in 2011. The Elzaki transform is closely related with the

Laplace transform, Sumudu transform, and the natural transform. Elzaki transform is defined as [32]:

E[f(t)] = T(u) = u

∫ ∞
0

exp

(
−t
u

)
f(t)dt, (1.8)

provided the integral exists for some u.

The natural transform [33] which is similar to Laplace and Sumudu integral transform was introduced in

2008. In recent years, natural transform was successfully applied to many applications (see [34, 35]). The

natural transform is defined by the following integral:

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

u

∫ ∞
0

exp

(
−st
u

)
f(t)dt, s > 0, u > 0, (1.9)

provided the integral exists for some variables u and s. Recently, a new integral transform called the M-

transform which is also similar to natural transform is introduced by Srivastava et al. in 2015. Mathemat-

ically speaking, M-transform is closely connected with the well-known Laplace transform and the Sumudu

integral transform. M-transform was successfully applied to first order initial-boundary value problem (see

Srivastava et al. [36]). The M-transform is defined as:

Mρ,m[f(t)](u,v) =
∫ ∞

0

exp (−ut) f(vt)
(tm + vm)

ρ dt, (1.10)

(ρ ∈ C; <(ρ) ≥ 0, m ∈ Z+ = 1, 2, 3, · · ·) , where both u ∈ C and v ∈ R+ are the M-transform variables.

In 2013, Atangana and Kilicman introduced a novel integral transform called the Abdon-Kilicman integral

transform [37] for solving some differential equations with some kind of singularities. The novel integral

transform is defined as:

Mn(s) = Mn[f(x)](s) =

∫ ∞
0

xn exp (−xs) f(x)dx. (1.11)

The Atangana-Kilicman integral becomes Laplace transform when n = 0. Recently, a Laplace-type integral

transform called the Yang transform ( [38]- [40]) for solving steady heat transfer problems was introduced

in 2016. The integral transform is defined as:

Y [φ(τ)] = φ(ω) =

∫ ∞
0

exp

(
−τ
ω

)
φ(τ)dτ, (1.12)

provided the integral exists for some ω.



Int. J. Anal. Appl. 17 (2) (2019) 170

Due to the rapid development in the physical science and engineering models, there are many other

integral transforms in the literature. However, most of the existing integral transforms have some limitations

and cannot be used directly to solved nonlinear problems or many complex mathematical models. As a

result, many authors became highly interested to come up with the alternative approach for solving many

real-life problems. In 2016, Atangana and Alkaltani introduced a new double integral equation and their

properties based on the Laplace transform and decomposition method. The double integral transform was

successfully applied to second order partial differential equation with singularity called the two-dimensional

Mboctara equation [41]. Recently, Eltayeb applied double Laplace decomposition method to nonlinear partial

differential equations [42]. In 2017, Belgacem el at. extended the applications of the natural and the Sumudu

transforms to fractional diffusion and Stokes fluid flow realms [43].

Motivated by the above-mentioned researches, in this paper we proposed a Laplace-type integral transform

called Shehu transform for solving both ordinary and partial differential equations. The Laplace-type integral

transform converges to Laplace transform when u = 1, and to Yang integral transform when s = 1. The

proposed integral transform is successfully applied to both ordinary and partial differential equations. All

the results obtained in the applications section can easily be verified using the Laplace or Fourier integral

transforms. Throughout this paper, the Shehu transform is denoted by an operator S[.].

2. Main result

Definition 1. The Shehu transform of the function v(t) of exponential order is defined over the set of

functions,

A =
{
v(t) : ∃ N, η1, η2 > 0, |v(t)| < N exp

(
|t|
ηi

)
, if t ∈ (−1)i × [0,∞)

}
,

by the following integral

S [v(t)] = V (s,u) =
∫ ∞

0

exp

(
−st
u

)
v(t)dt

= lim
α→∞

∫ α
0

exp

(
−st
u

)
v(t)dt; s > 0, u > 0. (2.1)

It converges if the limit of the integral exists, and diverges if not.

The inverse Shehu transform is given by

S−1 [V (s,u)] = v(t), for t ≥ 0. (2.2)

Equivalently

v(t) = S−1 [V (s,u)] =
1

2πi

∫ α+i∞
α−i∞

1

u
exp

(
st

u

)
V (s,u) ds, (2.3)



Int. J. Anal. Appl. 17 (2) (2019) 171

where s and u are the Shehu transform variables, and α is a real constant and the integral in Equ.(2.3) is

taken along s = α in the complex plane s = x + iy.

Theorem 1. The sufficient condition for the existence of Shehu transform. If the function v(t) is

piecewise continues in every finite interval 0 ≤ t ≤ β and of exponential order α for t > β. Then its Shehu

transform V (s,u) exists.

Proof. For any positive number β, we algebraically deduce∫ ∞
0

exp

(
−
st

u

)
v(t)dt =

∫ β
0

exp

(
−
st

u

)
v(t)dt +

∫ ∞
β

exp

(
−
st

u

)
v(t)dt. (2.4)

Since the function v(t) is piecewise continues in every finite interval 0 ≤ t ≤ β, then the first integral on the

right hand side exists. Besides, the second integral on the right hand side exists, since the function v(t) is of

exponential order α for t > α. To verify this claim, we consider the following case∣∣∣∣
∫ ∞
α

exp

(
−
st

u

)
v(t)dt

∣∣∣∣ ≤
∫ ∞
α

∣∣∣∣exp
(
−
st

u

)
v(t)

∣∣∣∣dt
≤

∫ ∞
0

exp

(
−
st

u

)
|v(t)|dt

≤
∫ ∞
α

exp

(
−
st

u

)
Ne exp (βt) dt

= N

∫ ∞
α

exp

(
−

(s−βu)t
u

)
dt

= −
uN

(s−βu)
lim
γ→∞

[
exp

(
−

(s−βu)t
u

)]γ
0

=
uN

s−βu
.

The proof is complete. �

Property 1. Linearity property of Shehu transform. Let the functions αv(t) and βw(t) be in set A,

then (αv(t) + βw(t)) ∈A, where α and β are nonzero arbitrary constants, and

S [αv(t) + βw(t)] = αS [v(t)] + βS [w(t)] . (2.5)

Proof. Using the Definition 1 of Shehu transform, we get

S [αv(t) + βw(t)] =
∫ ∞

0

exp

(
−st
u

)
(αv(t) + βw(t))dt

=

∫ ∞
0

exp

(
−st
u

)
(αv(t)) dt +

∫ ∞
0

exp

(
−st
u

)
(βw(t))dt

= α

∫ ∞
0

exp

(
−st
u

)
v(t)dt + β

∫ ∞
0

exp

(
−st
u

)
w(t)dt

= αu

∫ ∞
0

exp (−st) v(ut)dt + βu
∫ ∞

0

exp (−st) w(ut)dt

= αS [v(t)] + βS [w(t)] .



Int. J. Anal. Appl. 17 (2) (2019) 172

The proof is complete. �

In particular, using the Definition 1 and Property 1, we obtain

S [3 cos(t) + 5 sin(2t)] = 3S [cos(t)] + 5S [sin(2t)]

=
3us

s2 + u2
+

5u2

s2 + (2u)2
,

see entries of table 1.

Property 2. Change of scale property of Shehu transform. Let the function v(βt) be in set A, where

β is an arbitrary constant. Then

S [v(βt)] =
u

β
V

(
s

β
,u

)
. (2.6)

Proof. Using the Definition 1 of Shehu transform, we deduce

S [v(βt)] =
∫ ∞

0

exp

(
−st
u

)
v(βt)dt (2.7)

Substituting η = βt which implies t = η
β

and dt = dη
β

in Equ.(2.7) yields

S [v(βt)] =
1

β

∫ ∞
0

exp

(
−sη
uβ

)
v(η)dη

=
1

β

∫ ∞
0

exp

(
−st
uβ

)
v (t) dt

=
u

β

∫ ∞
0

exp

(
−st
β

)
v(ut)dt

=
u

β
V

(
s

β
,u

)
.

The proof is complete. �

Theorem 2. Derivative of Shehu transform. If the function v(n)(t) is the nth derivative of the function

v(t) ∈ A with respect to ′t′, then its Shehu transform is defined by

S
[
v(n)(t)

]
=
sn

un
V (s,u) −

n−1∑
k=0

(s
u

)n−(k+1)
v(k)(0). (2.8)

When n=1, 2, and 3 in Equ. (2.8) above, we obtain the following derivatives with respect to t.

S [v′(t)] =
s

u
V (s,u) −v(0). (2.9)

S [v′′(t)] =
s2

u2
V (s,u) −

s

u
v(0) −v′(0). (2.10)

S [v′′′(t)] =
s3

u3
V (s,u) −

s2

u2
v(0) −

s

u
v′(0) −v′′(0). (2.11)



Int. J. Anal. Appl. 17 (2) (2019) 173

Proof. Now suppose Equ. (2.8) is true for n = k. Then using Equ. (2.9) and the induction hypothesis, we

deduce

S
[
(v(k)(t))′

]
=

s

u
S
[
v(k)(t)

]
−v(k)(0)

=
s

u

[
sk

uk
S [v(t)] −

k−1∑
i=0

(s
u

)k−(i+1)
v(i)(0)

]
−v(k)(0)

=
(s
u

)k+1
S [v(t)] −

k∑
i=0

(s
u

)k−i
v(i)(0),

which implies that Equ. (2.8) holds for n = k+ 1. By induction hypothesis the proof is complete. �

The following important properties are obtain using the Leibniz’s rule

S
[
∂v(x,t)

∂x

]
=

∫ ∞
0

exp

(
−st
u

)
∂v(x,t)

∂x
dt =

∂

∂x

∫ ∞
0

exp

(
−st
u

)
v(x,t) dt

=
∂

∂x
[V (x,s,u)] ⇒ S

[
∂v(x,t)

∂x

]
=

d

dx
[V (x,s,u)] ,

S
[
∂2v(x,t)

∂x2

]
=

∫ ∞
0

exp

(
−st
u

)
∂2v(x,t)

∂x2
dt =

∂2

∂x2

∫ ∞
0

exp

(
−st
u

)
v(x,t) dt

=
∂2

∂x2
[V (x,s,u)] ⇒ S

[
∂2v(x,t)

∂x2

]
=

d2

dx2
[V (x,s,u)] ,

and

S
[
∂nv(x,t)

∂xn

]
=

∫ ∞
0

exp

(
−st
u

)
∂nv(x,t)

∂xn
dt =

∂n

∂xn

∫ ∞
0

exp

(
−st
u

)
v(x,t)dt

=
∂n

∂xn
[V (x,s,u)] ⇒ S

[
∂nv(x,t)

∂xn

]
=

dn

dxn
[V (x,s,u)] .

3. Some useful results of Shehu transform

Property 3. Let the function v(t) = 1 be in set A. Then its Shehu transform is given by

S [1] =
u

s
. (3.1)

Proof. Using Equ.(2.1), we deduce

S [1] =
∫ ∞

0

exp

(
−st
u

)
dt = −

u

s
lim
γ→∞

[
exp

(
−st
u

)]γ
0

=
u

s
.

This ends the proof. �

Property 4. Let the function v(t) = t be in set A. Then its Shehu transform is given by

S [t] =
u2

s2
. (3.2)



Int. J. Anal. Appl. 17 (2) (2019) 174

Proof. Using the Definition 1 of the Shehu transform and integration by parts, we get

S [t] =
∫ ∞

0

t exp

(
−st
u

)
dt =

u

s
lim
γ→∞

[
t exp

(
−st
u

)]γ
0

+
u

s

∫ ∞
0

exp

(
−st
u

)
dt

= −
u2

s2
lim
γ→∞

[
exp

(
−st
u

)]γ
0

=
u2

s2
.

Thus the proof ends. �

Property 5. Let the function v(t) = t
n

n!
n = 0, 1, 2.. be in set A. Then its Shehu transform is given by

S
[
tn

n!

]
=
(u
s

)n+1
. (3.3)

Proof. From the Definition 1 of the Shehu transform and integration by parts, we deduce

S [tn] =
∫ ∞

0

tn exp

(
−st
u

)
dt =

u

s
n

∫ ∞
0

tn−1 exp

(
−st
u

)
dt

=
u2

s2
n(n− 1)

∫ ∞
0

tn−2 exp

(
−st
u

)
dt

=
u3

s3
n(n− 1)(n− 2)

∫ ∞
0

tn−3 exp

(
−st
u

)
dt

=
u4

s4
n(n− 1)(n− 2)(n− 3)

∫ ∞
0

tn−4 exp

(
−st
u

)
dt

=
u5

s5
n(n− 1)(n− 2)(n− 3)(n− 4)

∫ ∞
0

tn−5 exp

(
−st
u

)
dt = · · · = n!

(u
s

)n+1
.

The proof is completed. �

Property 6. Let the function v(t) = t
n

Γ(n+1)
n = 0, 1, 2, · · · be in set A. Then its Shehu transform is given

by

S
[

tn

Γ(n + 1)

]
=
(u
s

)n+1
. (3.4)

The proof of property 6 follows immediately from the previous property 5. �

Property 7. Let the function v(t) = exp(αt) be in A. Then its Shehu transform is given by

S [exp(αt)] =
u

s−αu
. (3.5)

Proof. Using Equ.(2.1), we get

S [exp(αt)] =
∫ ∞

0

exp

(
−

(s−αu)t
u

)
dt

= −
u

s−αu
lim
γ→∞

[
exp

(
−

(s−αu)t
u

)]γ
0

=
u

s−αu
.

This ends the proof. �



Int. J. Anal. Appl. 17 (2) (2019) 175

Property 8. Let the function v(t) = t exp(αt) be in set A. Then its Shehu transform is given by

S [t exp(αt)] =
u2

(s−αu)2
. (3.6)

Proof. Using the Definition 1 of the Shehu transform and integration by parts, we get

S [t exp(αt)] =
∫ ∞

0

t exp

(
−

(s−αu)t
u

)
dt

= −
u

s−αu
lim
γ→∞

[
t exp

(
−

(s−αu)t
u

)]γ
0

+
u

s−αu

∫ ∞
0

exp

(
−

(s−αu)t
u

)
dt

= −
u2

(s−αu)2
lim
γ→∞

[
exp

(
−

(s−αu)t
u

)]γ
0

=
u2

(s−αu)2
.

The proof is complete. �

Property 9. Let the function v(t) =
tn exp(αt)

n!
n = 0, 1, 2, ... be in set A. Then its Shehu transform is given

by

S
[
tn exp(αt)

n!

]
=

un+1

(s−αu)n+1
. (3.7)

Proof. Using the Definition 1 of the Shehu transform and integration by parts, we deduce

S [tn exp(αt)] =
∫ ∞

0

tn exp

(
−

(s−αu)t
u

)
dt

=
un

(s−αu)

∫ ∞
0

tn−1 exp

(
−

(s−αu)t
u

)
dt

=
u2n(n− 1)
(s−αu)2

∫ ∞
0

tn−2 exp

(
−

(s−αu)t
u

)
dt = · · · =

n!

(s−αu)n+1
.

Thus the proof is complete. �

Property 10. Let the function v(t) = t
n

Γ(n+1)
exp(αt) n = 0, 1, 2, ... be in set A. Then its Shehu transform

is given by

S
[
tn exp(αt)

Γ(n + 1)

]
=

un+1

(s−αu)n+1
. (3.8)

The proof of Property 10 follows as a direct consequence of Property 9. �

Property 11. Let the function v(t) = sin(αt) be in set A. Then its Shehu transform is given by

S [sin(αt)] =
αu2

s2 + α2u2
. (3.9)



Int. J. Anal. Appl. 17 (2) (2019) 176

Proof. Using the Definition 1 of the Shehu transform and integration by parts, we get

S [sin(αt)] =
∫ ∞

0

exp

(
−
st

u

)
sin(αt)dt

= −
u

s
lim
γ→∞

[
exp

(
−
st

u

)
sin(αt)

]γ
0

+
uα

s

∫ ∞
0

exp

(
−
st

u

)
cos(αt)dt

= −
αu2

s2
lim
γ→∞

[
exp

(
−
st

u

)
cos(αt)

]γ
0

−
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
sin(αt)dt

=
αu2

s2
−
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
sin(αt)dt.

Simplifying the required integrals complete the proof of Property 11. �

Property 12. Let the function v(t) = cos(αt) be in set A. Then its Shehu transform is given by

S [cos(αt)] =
us

s2 + α2u2
. (3.10)

Proof. Using the Definition 1 of the Shehu transform and integration by parts, we deduce

S [cos(αt)] =
∫ ∞

0

exp

(
−
st

u

)
cos(αt)dt

= −
u

s
lim
γ→∞

[
exp

(
−
st

u

)
cos(αt)

]γ
0

−
αu

s

∫ ∞
0

exp

(
−
st

u

)
sin(αt)dt

=
u

s
−
αu2

s2
lim
γ→∞

[
exp

(
−
st

u

)
sin(αt)

]γ
0

−
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
cos(αt)dt

=
u

s
−
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
cos(αt)dt.

Simplifying the required integrals complete the proof of Property 12. �

Property 13. Let the function v(t) =
sinh(αt)

α
be in set A. Then its Shehu transform is given by

S
[

sinh(αt)

α

]
=

u2

s2 −α2u2
. (3.11)

Proof. From the Definition 1 of the Shehu transform and integration by parts, we get

S [sinh(αt)] =
∫ ∞

0

exp

(
−
st

u

)
sinh(αt)dt

= −
u

s
lim
γ→∞

[
exp

(
−
st

u

)
sinh(αt)

]γ
0

+
uα

s

∫ ∞
0

exp

(
−
st

u

)
cosh(αt)dt

= −
αu2

s2
lim
γ→∞

[
exp

(
−
st

u

)
cos(αt)

]γ
0

+
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
sinh(αt)dt

=
αu2

s2
+
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
sinh(αt)dt.

Simplifying the required integrals complete the proof of Property 13. �



Int. J. Anal. Appl. 17 (2) (2019) 177

Property 14. Let the function v(t) = cosh(αt) be in set A. Then its Shehu transform is given by

S [cosh(αt)] =
us

s2 −α2u2
. (3.12)

Proof. Applying the Definition 1 of the Shehu transform and integration by parts, we get

S [cosh(αt)] =
∫ ∞

0

exp

(
−
st

u

)
cosh(αt)dt

= −
u

s
lim
γ→∞

[
exp

(
−
st

u

)
cos(αt)

]γ
0

+
αu

s

∫ ∞
0

exp

(
−
st

u

)
sinh(αt)dt

=
u

s
−
αu2

s
lim
γ→∞

[
exp

(
−
st

u

)
sinh(αt)

]γ
0

+
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
cos(αt)dt

=
u

s
+
α2u2

s2

∫ ∞
0

exp

(
−
st

u

)
cosh(αt)dt.

Collecting the required integrals complete the proof of Property 14. �

Property 15. Let the function
exp(βt) sin(αt)

α
be in set A. Then its Shehu transform is given by

S
[

exp (βt) sin(αt)

α

]
=

u2

(s−βu)2 + α2u2
. (3.13)

Proof. Using the Definition 1 of the Shehu transform and integration by parts, we deduce

S [exp (βt) sin(αt)] =
∫ ∞

0

exp

(
−

(s−βu)
u

t

)
sin(αt)dt

=
−u

(s−βu)
lim
γ→∞

[
exp

(
−

(s−βu)
u

t

)
sin(αt)dt

]γ
0

+
uα

s−βu

∫ ∞
0

exp

(
−

(s−βu)
u

t

)
cos(αt)dt

= −
u2α

(s−βu)2
lim
γ→∞

[
exp

(
−

(s−βu)
u

t

)
cos(αt)dt

]γ
0

−
αu2α2

(s−βu)2

∫ ∞
0

exp

(
−

(s−βu)
u

t

)
sin(αt)dt

=
u2α

(s−βu)2
−

u2α2

(s−βu)2

∫ ∞
0

exp

(
−

(s−βu)
u

t

)
sin(αt)dt.

Simplifying the required integrals complete the proof of property 15. This ends the proof. �

Property 16. Let the function exp (βt) cos(αt) be in set A. Then its Shehu transform is given by

S [exp (βt) cos(αt)] =
u(s−αu)

(s−βu)2 + α2u2
. (3.14)



Int. J. Anal. Appl. 17 (2) (2019) 178

Proof. Applying the Definition 1 of the Shehu transform and integration by parts, we get

S [exp (βt) cos(αt)] =
∫ ∞

0

exp

(
−

(s−βu)
u

t

)
cos(αt)dt

= −
u

s−βu
lim
γ→∞

[
exp

(
−

(s−βu)
u

t

)
cos(αt)

]γ
0

+
αu

s−βu

∫ ∞
0

exp

(
−

(s−βu)
u

t

)
sin(αt)dt

=
u

s−βu
+

αu

s−βu

∫ ∞
0

exp

(
−

(s−βu)
u

t

)
sin(αt)dt

=
u

s−βu
+

αu2

(s−αu)2
lim
γ→∞

[
exp

(
−

(s−βu)
u

t

)
sin(αt)

]γ
0

−
α2u2

(s−βu)2

∫ ∞
0

exp

(
−

(s−βu)
u

t

)
cos(αt)dt

=
u

s−βu
−

α2u2

(s−βu)2

∫ ∞
0

exp

(
−

(s−βu)
u

t

)
cos(αt)dt.

Simplifying the required integrals complete the proof of property 16. �

Property 17. Let the function
exp(βt)−exp(αt)

β−α be in set A. Then its Shehu transform is given by

S
[

exp (αt)

β −α

]
=

u2

(s−αu)(s−βu)
. (3.15)

Proof. Using the definition of Shehu transform, we get

S
[

exp (αt)

β −α

]
=

u

β −α

∫ ∞
0

exp

(
−st
u

)
(exp(βt) − exp (αt)) dt

=
1

β −α

∫ ∞
0

e
(βu−s)
u

tdt−
1

β −α

∫ ∞
0

exp

(
(αu−s)t

u

)
dt

=
u

(β −α)(βu−s)
lim
γ→∞

[
exp

(
−

(s−βu)t
u

)]γ
0

−
u

(β −α)(αu−s)
lim
γ→∞

[
exp

(
−

(s−βu)t
u

)]γ
0

= −
u

(β −α)(βu−s)
+

u

(β −α)(αu−s)

=
−u(αu−s) + u(βu−s)
(β −α)(αu−s)(βu−s)

=
u2

(s−αu)(s−βu)
.

The proof is complete. �

Property 18. Let the function
β exp(βt)−α exp(αt)

β−α be in set A. Then its Shehu transform is given by:

S
[
β exp (βt) −α exp (αt)

β −α

]
=

us

(s−αu)(s−βu)
. (3.16)



Int. J. Anal. Appl. 17 (2) (2019) 179

Proof:

Using the definition of Shehu transform, we get

S
[
β exp (βt) −α exp (αt)

β −α

]
=

1

β −α

∫ ∞
0

exp

(
−st
u

)
(β exp (βt) −α exp (αt)) dt

=
β

β −α

∫ ∞
0

exp

(
(βu−s)

u
t

)
dt−

α

β −α

∫ ∞
0

exp

(
(αu−s)t

u

)
dt

=
uβ

(β −α)(βu−s)
lim
γ→∞

[
exp

(
−

(s−βu)t
u

)]γ
0

−
uα

(β −α)(αu−s)
lim
γ→∞

[
exp

(
−

(s−αu)t
u

)]γ
0

= −
uβ

(β −α)(βu−s)
+

uα

(β −α)(αu−s)

=
−uβ(αu−s) + uα(βu−s)

(β −α)(αu−s)(βu−s)
=

us

(s−αu)(s−βu)
.

This ends the proof. �

More properties of the Shehu transform and their converges to the natural transform, the Sumudu trans-

form, and the Laplace transform are presented in table 1. The comprehensive summary of Shehu transform

properties are presented in table 2.

4. Applications

In this section, the applications of the proposed transform are presented. The simplicity, efficiency and

high accuracy of the Shehu transform are clearly illustrated.

Example 1. Consider the following first order ordinary differential equation

dv(t)

dt
+ v(t) = 0, (4.1)

subject to the initial condition

v(0) = 1. (4.2)

Applying the Shehu transform on both sides of Equ. (4.1), we get

s

u
V (s,u) −v(0) + V (s,u) = 0. (4.3)

Substituting the given initial condition and simplifying, we deduce

V (s,u) =
u

s + u
(4.4)

Taking the inverse Shehu transform of Equ. (4.4), yields

v(t) = exp(−t). (4.5)



Int. J. Anal. Appl. 17 (2) (2019) 180

Example 2. Consider the following second order ordinary differential equation

d2v(t)

dt2
+
dv(t)

dt
= 1 (4.6)

subject to the initial conditions

v(0) = 0,
dv(0)

dt
= 0. (4.7)

Applying the Shehu transform on both sides of Equ. (4.6), we obtain

s2

u2
V (s,u) −

s

u
v(0) −v′(0) +

s

u
V (s,u) −v(0) =

u

s
. (4.8)

Substituting the given initial conditions and simplifying, we deduce

V (s,u) = −
u

s
+
u2

s2
+

u

s + u
. (4.9)

Taking the inverse Shehu transform of Equ. (4.9), we get

v(t) = −1 + t + exp(−t). (4.10)

Example 3. Consider the following second nonhomogeneous order ordinary differential equation

d2v(t)

dt2
− 3

dv(t)

dt
+ 2v(t) = exp(3t). (4.11)

subject to the initial conditions

v(0) = 1,
dv(0)

dt
= 0. (4.12)

Applying the Shehu transform on both sides of Equ. (4.11), yields

s2

u2
V (s,u) −

s

u
v(0) −v′(0) − 3

(s
u
V (s,u) −v(0)

)
+ 2V (s,u) =

u

s− 3u
. (4.13)

Substituting the given initial conditions and simplifying, we obtain

V (s,u) =
5

2

u

(s−u)
− 2

u

s− 2u
+

1

2

u

(s− 3u)
. (4.14)

Taking the inverse Shehu transform of Equ. (4.14), we get

v(t) =
5

2
exp(t) − 2 exp(2t) +

1

2
exp(3t). (4.15)

Example 4. Consider the following ordinary differential equation

d2v(t)

dt2
+ 2

dv(t)

dt
+ 5v(t) = exp(−t) sin(t). (4.16)

subject to the initial conditions

v(0) = 0,
dv(0)

dt
= 1. (4.17)

Applying the Shehu transform on both sides of Equ. (4.16), we get

s2

u2
V (s,u) −

s

u
v(0) −v′(0) + 2

(s
u
V (s,u) −

s

u
v(0)

)
+ 5V (s,u) =

u2

(s + u)2 + u2
(4.18)



Int. J. Anal. Appl. 17 (2) (2019) 181

Substituting the given initial conditions and simplifying, we get

V (s,u) =
1

3

u2

((s + u)2 + u2)
+

2

3

u2

((s + u)2 + (2u)2)
(4.19)

Taking the inverse Shehu transform of Equ. (4.19), we get

v(t) =
1

3
exp(−t) sin(t) +

2

3
exp(−t) sin(2t). (4.20)

Example 5. Consider the following homogeneous partial differential equation

∂v(x,t)

∂t
=
∂2v(x,t)

∂x2
(4.21)

subject to the boundary and initial conditions

v(0, t) = 0, v(1, t) = 0, v(x, 0) = 3sin(2πx). (4.22)

Applying the Shehu transform on both sides of Equ. (4.21), we get

s

u
V (x,s,u) −v(x, 0) =

d2V (x,s,u)

dx2
. (4.23)

Substituting the given initial condition and simplifying, we get

d2V (x,s,u)

dx2
−
s

u
V (x,s,u) = −3sin(2πx). (4.24)

The general solution of Equ. (4.24) can be written as

V (x,s,u) = Vh(x,s,u) + Vp(x,s,u), (4.25)

where Vh(x,s,u) is the solution of the homogeneous part which is given by

Vh(x,s,u) = α1 exp

(√
s

u
x

)
+ α2 exp

(
−
√
s

u
x

)
, (4.26)

and Vp(x,s,u) is the solution of the nonhomogeneous part which is given by

Vp(x,s,u) = β1sin(2πx) + β2cos(2πx). (4.27)

Applying the boundary conditions on Equ. (4.26), we get

α1 + α2 = 0 and α1 exp

(√
s

u

)
+ α2 exp

(
−
√
s

u

)
= 0 ⇒ Vh(x,s,u) = 0,

since α1 = α2 = 0.

Using the method of undetermined coefficients on the nonhomogeneous part, we get

Vp(x,s,u) =
3u

s + 4π2u
sin(2πx), (4.28)

Since, β1 =
3u

s+4π2u
, and β2 = 0.

Then Equ. (4.25) will become

V (x,s,u) =
3u

s + 4π2u
sin(2πx), (4.29)



Int. J. Anal. Appl. 17 (2) (2019) 182

Taking the inverse Shehu transform of Equ. (4.29), we get

v(x,t) = 3 exp(−4π2t) sin(2πx). (4.30)

Figure 1. 3D and 2D surfaces of the analytical solution of Equ. (4.21) in the ranges −1 < x < 1, and

−1 < t < 1, when t = 1.

Example 6. Consider the following nonhomogeneous partial differential equation

∂2v(x,t)

∂t2
= β2

∂2v(x,t)

∂x2
+ sin(πx) (4.31)

subject to the boundary and initial conditions

v(0, t) = 0, v(1, t) = 0, v(x, 0) = 0,
∂v(x, 0)

∂t
= 0, β2 = 1. (4.32)

Applying the Shehu transform on both sides of Equ. (4.31), we get

s2

u2
V (x,s,u) −

s

u
v(x, 0) −v′(x, 0) =

d2V (x,s,u)

dx2
+
u

s
sin(πx). (4.33)

Substituting the given initial condition and simplifying, we get

d2V (x,s,u)

dx2
−
s2

u2
V (x,s,u) = −

u

s
sin(πx). (4.34)

The general solution of Equ. (4.34) can be written as

V (x,s,u) = Vh(x,s,u) + Vp(x,s,u), (4.35)

where Vh(x,s,u) is the solution of the homogeneous part which is given by

Vh(x,s,u) = λ1 exp
(s
u
x
)

+ λ2 exp
(
−
s

u
x
)
, (4.36)

and Vp(x,s,u) is the solution of the nonhomogeneous part which is given by

Vp(x,s,u) = η1 sin(πx) + η2 cos(πx). (4.37)

Applying the boundary conditions on Equ. (4.36), we deduce

λ1 + λ2 = 0 and λ1 exp
(s
u

)
+ λ2 exp

(
−
s

u

)
= 0 ⇒ Vh(x,s,u) = 0,



Int. J. Anal. Appl. 17 (2) (2019) 183

since λ1 = λ2 = 0.

Using the method of undetermined coefficients on the nonhomogeneous part, we get

Vp(x,s,u) =
1

π2

(
u

s
−

us

s2 + u2π2

)
sin(πx), (4.38)

since,

η1 =
u3

s(s2 + u2π2)
=

1

π2

(
u

s
−

us

s2 + u2π2

)
, and η2 = 0.

Then Equ. (4.35) will becomes

V (x,s,u) =
1

π2

(
u

s
−

us

s2 + u2π2

)
sin(πx). (4.39)

Taking the inverse Shehu transform of Equ. (4.39), we get

v(x,t) =
1

π2
(1 − cos(πt)) sin(πx). (4.40)

Figure 2. 3D and 2D surfaces of the analytical solution of Equ. (4.31) in the ranges −1 < x < 1, and

−1 < t < 1.

5. Conclusion

We introduced an efficient Laplace-type integral transform called the Shehu transform for solving both

ordinary and partial differential equations. We presented its existence and inverse transform. We presented

some useful properties and their applications for solving ordinary and partial differential equations. We

provide a comprehensive list of the Laplace transform, Sumudu transform, and the natural transform in

table 1 to show their mutual relationship with the Shehu transform. Finally, based on the mathematical

formulations, simplicity and the findings of the proposed integral transform, we conclude that it is highly

efficient because of the following advantages:

• It is a generalization of the Laplace and the Sumudu integral transforms.

• Its visualization is easier then the Sumudu transform, the natural transform and the Elzaki transform.



Int. J. Anal. Appl. 17 (2) (2019) 184

• The Laplace-type integral transform become Laplace transform when the variable u = 1 and the

Yang integral transform when the variable s = 1.

• It can easily be applied directly to some class of ordinary and the partial differential equations as

demonstrated in the application section.

• For advanced research in physical science and engineering, the proposed integral transform can be

considered a stepping-stone to the Sumudu transform, the natural transform, the Elzaki transform,

and the Laplace transform.

6. ACKNOWLEDGEMENTS

The authors are highly grateful to the editor’s and the anonymous referees’ for their useful comments and

suggestions in this paper. This research is partially supported by the National Natural Science Foundations

of China under Grants No. 11571206. The first author also acknowledges the financial support of Chinese

Scholarship Council (CSC) in Shandong University with grand (CSC No: 2017GXZ025381).

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Appendix

Table 1: Here we present a comprehensive list of the Shehu transform of some special functions and their

relationship with the natural transform N [v(t)], the Sumudu transform S [v(t)], and the Laplace transform.

S.NO. v(t) S [v(t)] N [v(t)] S [v(t)] £ [v(t)]

1 1 u
s

1
s

1 1
s

2 t u
2

s2
u
s2

u 1
s2

3 exp(α(t)) u
s−αu

1
s−αu

1
1−αu

1
s−α

4
sin(αt)
α

u2

s2+α2u2
u

s2+α2u2
u

1+α2u2
1

s2+α2

5 cos(αt) us
s2+α2u2

s
s2+α2u2

1
1+α2u2

s
s2+α2

6 cosh αt us
s2−u2

s
s2−u2

1
1−u2

s
s2−1

7 t
n

n!
n = 0, 1, 2...

(
u
s

)n+1 un
sn+1

un 1
sn+1

8 t
n

Γ(n+1)
n = 0, 1, 2, ...

(
u
s

)n+1 un
sn+1

un 1
sn+1

9 cos(t) us
s2+u2

s
s2+u2

1
1+u2

1
s2+1

10 sin(t) u
2

s2+u2
u

s2+u2
u

1+u2
1

s2+1

11
sinh(αt)

α
u2

s2−α2u2
αu

s2−α2u2
αu2

1−α2u2
α

s2−α2

12 cosh(αt) us
s2−α2u2

s
s2−α2u2

1
1−α2u2

s
s2−α2

13 exp(βt)cosh(αt)
u(s−βu)

(s−βu)2−α2u2
s−βu

(s−βu)2−α2u2
1−βu

(s−βu)2−α2u2
s−β

(s−βu)2−α2

14
exp(βt) sinh(αt)

α
u2

(s−βu)2−α2u2
u

(s−βu)2−α2u2
u

(1−βu)2−α2u2
1

(s−β)2−α2

15
t sin(αt)

2α
u3s

(s2+α2u2)2
u2s

(s2+α2u2)2
u3

(1+α2u2)2
s

(s2+α2)2

16 t cos(αt)
u2(s2−α2u2)2
(s2+α2u2)2

u(s2−α2u2)2
(s2+α2u2)2

u(1−α2u2)2
(1+α2u2)2

(s2−α2)2
(s2+α2)2

17
sin(αt)+αt cos(αt)

2α
u2s2

(s2+α2u2)2
us2

(s2+α2u2)2
u

(1+α2u2)2
s2

(s2+α2)2

18 cos αt− αt sin(αt)
2

us3

(s2+α2u2)2
s3

(s2+α2u2)2
1

(1+α2u2)2
s3

(s2+α2)2

19
sin(αt)−αt cos(αt)

2α3
u4

(s2+α2u2)2
u3

(s2+α2u2)2
u3

(1+α2u2)2
1

(s2+α2)2



Int. J. Anal. Appl. 17 (2) (2019) 188

20 t sinh(αt) + t cosh(αt) u
2

(s−αu)2
u

(s−αu)2
u2

(1−αu)2
1

(s−α)2

21
t sinh(αt)

2α
u3s

(s2−α2u2)2
u2s

(s2−α2u2)2
u2

(1−α2u2)2
s

(s2−α2)2

22 Si(αt) (Sine integral) u
s
tan−1

(
αu
s

)
1
s
tan−1

(
αu
s

)
tan−1

(
u
√
α2
)

1
s
tan−1

(
α
s

)
23 Ci(αt) (Cosine integral) − u

2s
log
(
s2+α2

α2

)
− 1

2s
log
(
s2+α2u2

α2u2

)
−1

2
log
(
α2u2+1
α2u2

)
− 1

2s
log
(
s2+α2

α2

)
24 Ei(αt) (Exp. integral) −u

s
log
(
αu−s
αu

)
−1
s

log
(
αu−s
αu

)
log
(
αu−1
αu

)
−1
s

log
(
α−s
α

)
25

(3−α2t2) sin(αt)−3αt cos(αt)
8α5

u6

(s2+α2u2)3
u5

(s2+α2u2)3
u5

(1+α2u2)3
1

(s2+α2)3

26
(3−α2t2) sin(αt)+5αt cos(αt)

8α
u2s4

(s2+α2u2)3
us4

(s2+α2u2)3
u

(1+α2u2)3
s4

(s2+α2)3

27
(8−α2t2) cos(αt)−7αt sin(αt)

8
us5

(s2+α2u2)3
s5

(s2+α2u2)3
1

(1+α2u2)3
s5

(s2+α2)3

28
t2 sin(αt)

2α
u4(3s2−α2u2)
(s2+α2u2)3

u3(3s2−α2u3)
(s2+α2u2)3

u3(−3+α2u2)
(1+α2u2)3

(3s2−α2)
(s2+α2)3

29
t2 cos(αt)

2
u3(s3−3α2u2s)

(s2+α2u2)3
u2(s3−3α2u2s)

(s2+α2u2)3
u2(1−3α2u2)
(1+α2u2)3

(s3−3α2s)
(s2+α2)3

30
t3 sin(αt)

24α
su5(s−αu)2
(s2+α2u2)4

su4(s−αu)2
(s2+α2u2)4

u4(1−αu)2
(1+α2u2)4

s(s−α)2
(s2+α2)4

31
exp(αt)−exp(βt)

α−β α 6= β
u2

(s−αu)(s−βu)
u

(s−αu)(s−βu)
u

(1−βu)(1−αu)
1

(s−β)(s−α)

32
α exp(αt)−β exp(βt)

α−β α 6= β
us

(s−βu)(s−αu)
s

(s−βu)(s−αu)
1

(1−βu)(1−αu)
s

(s−β)(s−α)

33 I0(αt)
u√

s2−α2u2
1√

s2−α2u2
1√

1−α2u2
1√

s2−α2

34 δ(t−α) u exp
(−αs

u

)
1
u

exp
(−αs

u

)
1
u

exp
(−α
u

)
exp(−αs)

35 J0(αt)
u√

s2+α2u2
1√

s2+α2u2
1√

1+α2u2
1√

s2+α2



Int. J. Anal. Appl. 17 (2) (2019) 189

Table 2: General properties of Shehu transform

S.NO. Definition/Property S [v(t)] = V (s,u) Transforms

1 Definition S [v(t)]
∫∞

0
exp

(−st
u

)
v(t) dt; s > 0, u > 0

2 Inverse v(t) = S−1 [V (s,u)] 1
2πi

∫α+i∞
α−i∞

1
u

exp
(
st
u

)
V (s,u) ds

3 Linearity S [αv(t) + βw(t)] αS [v(t)] + βS [w(t)]

4 Change of scale S [v(αt)] u
α
V
(
s
α
,u
)

5 Derivatives S [v′(t)] s
u
V (s,u) −v(0)

S [v′′(t)] s
2

u2
V (s,u) − s

u
v(0) −v′(0)

S [v′′′(t)] s
3

u3
V (s,u) − s

2

u2
v(0) − s

u
v′(0) −v′′(0)

...

S
[
v(n)(t)

]
sn

un
V (s,u) −

∑n−1
k=0

(
s
u

)n−(k+1)



Int. J. Anal. Appl. 17 (2) (2019) 190

Table 3: Summary of some integral transform and their definitions

S.No. Integral transform Definition

1 Laplace transform £[f(t)] = F(s) =
∫∞

0
exp (−st) f(t)dt

2 Fourier transform z[f(t)] = f(ω) = 1√
2π

∫∞
−∞ exp (−iωt) f(t)dt

3 Melling transform M[f(s); s] = f∗(s) =
∫∞

0
xs−1f(x)dx

4 Hankel’s transform Fv(s) = Hv[f(r)] =
∫∞

0
rf(r)Jv(sr)dr, r ≥ 0

5 Sumudu transform S[f(t)] = G(u) = 1
u

∫∞
0

exp
(−t
u

)
f(t)dt

6 Laplace-Carson transform f̂C(p) = p
∫∞

0
exp (−pt) f(t)dt, t ≥ 0

7 Atangana-Kilicman transform Mn(s) = Mn[f(x)](s) =
∫∞

0
xn exp (−xs) f(x)dx

8 El-zaki transform E[f(t)] = T(u) = u
∫∞

0
exp

(−t
u

)
f(t)dt

9 Yang transform Y [φ(τ)] = φ(ω) =
∫∞

0
exp

(−τ
ω

)
φ(τ)dτ

10 natural transform N+[f(t)] = R(s,u) = 1
u

∫∞
0

exp
(−st
u

)
f(t)dt, s > 0, u > 0

11 z-transform X(z) = Z{x[n]} =
∑∞
n=−∞x[n]z

−n, n ∈ Z, z ∈ C

12 M-transform Mρ,m[f(t)](u,v) =
∫∞

0
exp(−ut)f(vt)

(tm+vm)ρ
dt, ρ ∈ C, <(ρ) ≥ 0, m ∈ Z+


	1. Introduction
	2. Main result
	3. Some useful results of Shehu transform
	4. Applications
	5. Conclusion
	6. ACKNOWLEDGEMENTS
	References
	Appendix