Multiplicative Laplace transform and its applications

Optik 127 (2016) 9984–9995

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Optik journal homepage: www.elsevier.de/ijleo

Multiplicative Laplace transform and its applications Numan Yalcin a,∗ , Ercan Celik b , Ahmet Gokdogan a a b

Department of Mathematical Engineering, Gümüs¸hane University, 29100 Gümüs¸hane, Turkey Department of Mathematics, Atatürk University, 25100 Erzurum, Turkey

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 28 May 2016 Accepted 29 July 2016

In this work, taking definitions and properties of Laplace transform in classical analysis as a basis, we give some basic definitions and properties of the multiplicative Laplace transform. In addition, solutions of some multiplicative differential equations are obtained by the help of this transform. © 2016 Elsevier GmbH. All rights reserved.

Keywords: Laplace transform Multiplicative Laplace transform Multiplicative derivative Multiplicative integral

1. Introduction Non-Newtonian calculus allowed scientists to look from a different point of view to the problems encountered in science and engineering. In the period from 1967 till 1970, Non-Newtonian calculus consisting of the branches of geometric, anageometric and biogeometric calculus is studied by Michael Grossman and Robert Katz in [1]. Subsequently, geometric analysis is called as multiplicative calculus by D.stanley [2]. Then, some study with regard to multiplicative calculus is given by D. Campell [3]. Bashirov et al. [4] presented concepts of the multiplicative calculus in detail and some applications to the properties of derivative and integral operators of this calculus. Lately, some researchers [5–10] have shown that multiplicative analysis can be used effectively in the solution of problems in some science and engineering fields. In this paper, we study multiplicative Laplace transform and its application. For this purpose, firstly, we present definition of multiplicative Laplace transform. Secondly, we give basic properties of multiplicative Laplace transform as corresponding to classical calculus. Finally, we find solutions of some multiplicative differential equations applying multiplicative Laplace transform. 2. Multiplicative derivatives and integrals Here, we will give some basic definitions and properties of the multiplicative derivative theory which can be found in [2–5]. Definition 2.1.

Let f : R → R+ be a positive function. The multiplicative derivative of the function f is given by



d∗ f f (t + h) (t) = f ∗ (t) = lim dt f (t) h→0

 1h

.

∗ Corresponding author. E-mail addresses: [email protected] (N. Yalcin), [email protected] (E. Celik), [email protected] (A. Gokdogan). http://dx.doi.org/10.1016/j.ijleo.2016.07.083 0030-4026/© 2016 Elsevier GmbH. All rights reserved.

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N. Yalcin et al. / Optik 127 (2016) 9984–9995

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Assuming that f is a positive function and using properties of the classical derivative, multiplicative derivative can be written as 

f (t)  d∗ f (t) = f ∗ (t) = e f (t) = e(ln ◦f ) (t) dt

(2)

for (ln ◦f ) (t) = ln (f (t)). Definition 2.2. If multiplicative derivative f ∗ as a function has also multiplicative derivative then multiplicative derivative of f ∗ is called second order multiplicative derivative of f and it is represented by f ∗∗ . Similarly we can define nth order multiplicative derivative of f with the notation f ∗(n) . With n times repeated multiplicative differentiation operation, a positive f function has an nth order multiplicative derivative at the point t and defined as (n)

f ∗(n) (t) = e(ln ◦f )

(t)

(3)

Theorem 2.1. If a positive function f is differentiable with the multiplicative derivative at the point t, then it is differentiable in the classical sense and the relation between these two derivatives can be shown as f  (t) = f (t) ln f ∗ (t) .

(4)

Theorem 2.2. Let f and g be differentiable with the multiplicative derivative. If c be arbitrary constant, then c.f, f.g, f + g, f/g, f g functions are differentiable with the multiplicative derivative and their multiplicative derivatives can be shown as ∗

1. (c.f ) (t) = f ∗ (t), ∗ 2. (f.g) (t) = f ∗ (t) .g ∗ (t), f (t)

∗

g(t)

3.  g) (t) = f ∗ (t) f (t)+g(t) g ∗ (t) f (t)+g(t) , (f +  ∗ 4. f/g = f ∗ (t) /g ∗ (t), 

∗

5. (f g ) (t) = f ∗ (t)g(t) f (t)g (t) . Theorem 2.3.

f ∗ (t) = 1 for ∀t ∈ (a, b) ⇔ f (t) = C > 0 is fixed function in open interval (a,b).

Theorem 2.4.

Let g be differentiable in meaning of the multiplicative derivative, f be differentiable in the classical sense. If

f (t) = (g ◦ h) (t) , then, it can be written by f ∗ (t) = [g ∗ (h (t))] Theorem 2.5.

h (t)

.

(5)

Let f be a positive function. Then,

f∗

(t) = 1 ⇔

f

(t) = 0.

Definition 2.3. A multiplicative integral is also defined in [4] for positive bounded functions and if f is Riemann integrable on [a, b], then

⎛

b

f (t) = exp ⎝

b

dt

a

⎞ (ln f (t)) dt ⎠ = e

b (ln f (t))dt a

a

This multiplicative integral has the properties:

b a



 k dt

f (t)

⎛ =⎝

dt

(f (t))

a

b (f (t) g (t))

a

b  c

f (t) g(t)

=

b (f (t))

b dt (f (t)) =  ab , dt

c dt

f (t)

a

f (t)dt , a ≤ c ≤ b

f (t) a

(g (t))dt ,

b dt

=

dt

(g(t))

a



a

dt

a

dt

a b

d

⎠ , k ∈ R,

a

b b

⎞k

b

c

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where f and g are multiplicative integrable on [a, b]. 3. The multiplicative Laplace transform In this section, we will give the features of this new transform by defining multiplicative Laplace transform with the help of Laplace transform of classical analysis. Definition 3.1.. Let f (t) be a positive definite function given on the closed interval [0, ∞). Then, multiplicative laplace transform of f (t) is defined as

∞

Lm f (t)

e−st ln f (t)dt

∞ = Fm (s) =

dt e−st

=e

f (t)

= eL{ln f (t)}

0

(7)

0

∞ ∞

ln f (t)dt

f (t)dt = e

Here, multiplicative integral, which is defined as

0

, is used.

0

According to the definition, multiplicative Laplace transform of some basic functions can be given as the following expressions.

∞

Lm 1

ln 1e−st dt

= Fm (s) = e

= 1.

0

∞ t

Lm e



Lm et Lm



n

ee

at







sinhat



coshat

Lm ecosat

Theorem 3.1.

L ln e

=e



Lm e



 tn

at

L e

te−st dt

=e

0

=e

Lm esinat

Lm e

∞ lnet e−st dt

=e

(8)

1

= eL(t) = e s2 .

0

(9)

n!

n = eL(t ) = e sn+1 . 1

= e s−a .

(10) (11)

a

= eL{sinat } = e s2 +a2 .

(12)

s

= eL{cosat} = e s2 +a2 . = eL{sinhat } = e

(13)

a s2 −a2

.

(14)

s s2 −a2

.

(15)

= eL{coshat } = e

(Multiplicative Linearity Property)

Multiplicative laplace transform is multiplicatively linear. In other words, if c1 , c2 are arbitrary exponents and f1 , f2 are two given functions, which have multiplicative laplace transforms, then



Lm f1c1 f2c2 Proof.

c1

= Lm f1

c2

Lm f2

From the definition of multiplicative laplace transform, we have

∞

Lm f1c1 f2c2





ln f c1 f c2 e−st dt 1

=e

0

∞ (c1 ln f1 +c2 ln f2 )e−st dt

=e

0

2

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⎛ ∞ ⎞c1 ⎛ ∞ ⎞c2 −st −st ⎜ ln f1 e dt ⎟ ⎜ ln f2 e dt ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ = ⎜e 0 ⎟ ⎜e ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ c1

= Lm f1

c2

Lm f2

Thus, we see that



Lm f1c1 f2c2 Theorem 3.2.

c1

= Lm f1

c2

Lm f2

(Multiplicative First Shifting Property)



Let multiplicative laplace transform of function f (t) be Lm f (t) Lm Proof.



f (t)e

 at

= Fm (s). Then

= Fm (s − a)

(17)

Using the definition of multiplicative laplace transform, we have

∞ at

Lm



f (t)e

 at

ln f (t)e e−st dt

=e

0

∞ e−(s−a)t ln f (t)dt

=e

0

and thus we get Lm



f (t)e

Theorem 3.3.



Let Lm f (t)

at



(Multiplicative Second Shifting Property)





= Fm (s) and G (t) =

Then Lm G (t) Proof.

= Fm (s − a)

= Fm (s)e

−as

1, 0 < t < a . f (t − a) , t > a

.

By the definition of multiplicative laplace transform, we write

∞

Lm G (t)

ln(G(t))e−st dt

=e

a

0

∞ ln 1e−st dt+

=e

ln(t−a)e−st dt a

0

In the last expression, it is obvious that the first integral is equal to zero. And, if we use substitution in the second integral, laplace transform will be as follows

∞

Lm G (t)

ln f (u)e−s(u+a) du

=e

a

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N. Yalcin et al. / Optik 127 (2016) 9984–9995

⎛ ∞ ⎞e−as ⎜ ln f (u)e−su du ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜e a ⎟ ⎜ ⎟ ⎝ ⎠

= Fm (s)e

−as

Theorem 3.4.

(Multiplicative Change of Scale Property)





Let Lm f (t)

Lm f (at) Proof.

= Fm (s). Then we have = Fm

 s  1a

(18)

a

We can write

∞

Lm f (at)

ln f (at)e−st dt

=e

0

by the definition. If we use substitution in the integral above, we get

∞

Lm f (at)

1 a

−su

ln f (u)e a du

=e

0

⎛ ∞ ⎞ 1a −su ⎜ ln f (u)e a du ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜e 0 ⎟ ⎜ ⎟ ⎝ ⎠



Lm f (at)

 s  1a a



Lm f (t)t Proof.

= Fm

Let Lm f (t)

Theorem 3.5.



n

=



∗(n)

Fm

.

= Fm (s). Then

(−1)n (19)

(s)

We will proof this by induction. First, let us take n = 1. Using the definition, we have

∞

Lm f (t)

ln f (t)e−st dt

∞ f (t)e

= Fm (s) = 0

−st dt

=e

0

N. Yalcin et al. / Optik 127 (2016) 9984–9995

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Consequently, taking multiplicative derivative of this expression we have

⎛ ∞ ⎛ ∞ ⎞ −st ⎜ ln f (t)e dt ⎟ ⎜ ⎟ d∗ ⎜ 0 ⎟ ∗ Fm (s) = ∗ ⎜e ⎟=e ds ⎜ ⎟ ⎝ ⎠ ⎛∞ ⎞  d ⎝ ln f (t)e−st dt ⎠ ds ∗ Fm (s) = e

∞ t ln f (t)e−st dt

− ∗ Fm (s) = e

0

⎜ ⎜ ⎜e ⎜ ⎜ ⎝

0

⎟ ⎟ ⎟ ⎟ ⎟ ⎠

∞ t ln f (t)e−st dt

−

=e

0

d ds

⎞ ln f (t)e−st dt

0

⎛ ∞ ⎞−1 ⎜ ln f (t)t e−st dt ⎟ ⎜ ⎟

−1 ⎜ ⎟ = ⎜e 0 ⎟ = Lm f (t)t ⎜ ⎟ ⎝ ⎠

and get the following equality



Lm f (t)t

∗ = Fm (s)−1

(20)

Let us assume the hypothesis holds for the case n = k. Then, we have

∞ ∗(k)

Fm



tk

(s) = Lm f (t)

(−1)k

t k ln f (t)e−st dt

k

(−1)

=e

0

Now, if the multiplicative derivative is taken again for the last equation, we have

∞ t k+1 ln f (t)e−st dt

(−1)k+1 ∗(k+1)

Fm

(s) = e

0

⎛ ∞ ⎞(−1)k+1 ⎜ ln f (t)tk+1 e−st dt ⎟ ⎜ ⎟  (−1)k+1 k+1 ⎜ ⎟ = ⎜e 0 = Lm f (t)t ⎟ ⎜ ⎟ ⎝ ⎠ The proof is finished. In addition to this, we can write



tn

Lm f (t) Theorem 3.6. If

L−1 m



L−1 m

F (s)





=

n ∗(n) Fm (s)(−1)

=

∗ 1/Fm (s) , n is odd ∗ Fm (s) , n is even

.

(21)

(Multiplicative Convolution Property)



= f (t) and L−1 G (s) m

Fm (s)

G(s)

t

dx

f (x)g(t−x) .

= 0

= g (t), then

(22)

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N. Yalcin et al. / Optik 127 (2016) 9984–9995

t Proof.

dx

f (x)g(t−x) , we get

Applying multiplicative Laplace transform to integral 0

Lm

⎧t ⎨

dx

f (x)g(t−x)

⎩

⎫ ⎬ ⎭

= Lm

0

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

=e

⎛

⎜ ⎜ ⎜ L ln⎜e ⎜ ⎪ ⎜ ⎪ ⎪ ⎪ ⎝ ⎪ ⎪ ⎩

t 0

0

⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎫ ⎬

L

=e

e

⎪ ⎪ ⎪ ⎪ ⎪ ⎩

⎫ ⎪ ⎪ g(t−x) ln f (x)dx ⎪ ⎪ ⎪ ⎬

⎞⎫ ⎪ ⎪ ⎪ ⎟⎪ g(t−x) ln f (x)dx ⎪ ⎟⎪ ⎟⎬ ⎟ ⎟⎪ ⎟⎪ ⎪ ⎠⎪ ⎪ ⎪ ⎭

⎧t ⎨ g(t−x) ln f (x)dx

⎩

⎧ t ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

0

⎭ .

From the convolution property of classical analysis, we have

Lm

⎧t ⎨ ⎩

dx

f (x)g(t−x)

⎫ ⎬ ⎭

= eL{ln f (x)}L{g(x)}

0



= eL{ln f (x)}L{g(x)} = eL{ln f (x)}





= Lm f (x)

L{g(x)}

L{g(x)}

= Fm (s)G(s)

Definition 3.2. Let f (t) be a positive definite function given on the closed interval [0, ∞), and ln f (t) be a function given on the interval [0, ∞). If there exist positive constants t0 , K and ˛ such that |f (t) | ≤ Kee

˛t

(23)

for t > t0 , then f function is said to be of ˛− double exponential order. Theorem 3.7.

(Existence of Multiplicative Laplace Transform)

Let f (t) be a positive definite function, which is of ˛− double exponential order for t > t0 , given on the closed interval [0, ∞), and ln f (t) be a piecewise continuous function given on the interval [0, ∞). Then for s > ˛, Lm f (t) exists.

∞ e−st ln f (t) dt is convergent. To do this, we’ll divide integral into two integrals

Proof. We will show that for s > ˛, integral 0

as below

∞

t0 e

−st

ln f (t) dt =

0

∞ e

0

−st

e−st ln f (t) dt

ln f (t) dt +

(24)

t0

The first integral in equation (24) is convergent for any s because of ln f (t) and thus e−st ln f (t) is piecewise continuous in the interval [0, t0 ]. On the other hand, as f (t) is of ˛− double exponential order, |f (t) | ≤ Kee

˛t

N. Yalcin et al. / Optik 127 (2016) 9984–9995

for > t0 . Hence, for all t > t0



9991



|e−st ln f (t) | = e−st | ln f (t) | ≤ lnK + e˛t e−st

∞

∞ e

−st

ln f (t) ≤

ln Ke t0

0



= ln K

∞

e−st0 s

  +

−st



e−(s−˛)t0 s−˛

e−(s−˛)t dt

dt + t0

< ∞, for s > ˛

So, second integral in the right hand side of the expression (28) for t > t0 is convergent. We see that two of the integrals in the right hand side of the expression (28) exist, thus for s > ˛ Lm f (t) exists. Theorem 3.8. Let f(t) be a positive definite function, which is of ˛− double exponential order for t>t0 , given on the closed interval [0,∞), and lnf(t) be a piecewise continuous function defined on the interval [0,∞). Then the following holds



lim Lm f (t)

=1

s→∞

Proof.

(25)

From the proof of theorem 3.7, we can write

∞

Lm f (t)



ln Kee

≤e

∞

e−st

∞

0

e−(s−˛)t dt 0

 ≤e



0

ln Ke−st dt+

≤e

˛t

ln K



e−st0 s

Lm f (t)

 e

−(s−˛)t0 s−˛

 ≤e

ln K

e−st0 s



for all s > ˛. Then taking  limitfrom both sides as s → ∞, we get



lim Lm f (t)

s→∞

≤

e−st0 s

lim K

es→∞

= 1.

Theorem 3.9. Let f be a function of ˛− double exponential order defined on the interval [0,A], and let f ∗ be a piecewise continuous function defined on the interval [0,A]. Then Laplace transform of multiplicative derivative is



Lm f ∗ (t)

=

1 F(s)s f0

(26)

for s > ˛. Proof.

From the definition, we have

∞

Lm f ∗ (t)

 = Lm

e

f  (t) f (t)

 =e

A

f  (t) −st e dt f (t)

lim

A→∞

=e

0

f  (t) −st e dt f (t)

0

In the interval [0, A], denote the points of discontinuities of function f ∗ by t0 , t1 , . . ., tn . Using these points as endpoints of domain of integration, we write the integral as

A 0

f  (t) −st e dt = f (t)

t1 0

f  (t) −st e dt + f (t)

t2 t1

f  (t) −st e dt + . . . + f (t)

A tn

f  (t) −st e dt. f (t)

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Using the integration by parts method separately to each term on the right-hand side of this expression, we get

A

f  (t) −st t t A e dt = e−st ln f (t) dt|01 + e−st ln f (t) dt|t12 + . . . + e−st ln f (t) dt|tn f (t)

0

⎛t ⎞ 1 A t2 +s ⎝ ln f (t) e−st dt + ln f (t) e−st dt + . . . + ln f (t) e−st dt ⎠ t1

0

tn

As f (t) is continuous, domains of integration of the above expression can be combined in one domain. Thus, we write

⎛

A

f  (t) f (t)

e

e−sA ln f (A)−ln f (0)+s

e−st dt

⎝

=e

0

⎞

A e−st ln f (t)dt 0

As A → ∞, e−sA ln f (A) → 0 and e

e−sA

⎠ .

ln f (A)

→ 1 for s > ˛. Then for s > ˛, we obtain

⎛ ∞ ⎞s ⎜ ln f (t)e−st dt ⎟ ⎟ ∗ − ln f (0) ⎜ 1 ⎜ 0 ⎟ Lm f (t) = e ⎜e ⎟ = f F(s)s ⎜ ⎟ 0 ⎝ ⎠ This proves the theorem. If we replace f and f ∗ in this theorem by f ∗ ve f ∗∗ , respectively, multiplicative Laplace transformation of f ∗∗ is equal to



Lm f ∗∗ (t)

=

1

2

f (0)s f ∗

(0)

F(s)s .

(27)

for s > ˛. If method of induction is used to this theorem, we can generalize multiplicative Laplace transformation of function f ∗(n) as in the following result. Result. Let f, f ∗ , f ∗∗ , . . ., f ∗(n−1) be continuous functions, f ∗(n) be piecewise continuous function on the interval 0 ≤ t ≤ A. Also suppose that there exists positive real numbers K, ˛ and t0 such that ˛t

˛t

|f (t) | ≤ Kee , |f ∗ (t) | ≤ Kee , ..., |f ∗(n−1) (t) | ≤ Kee

˛t



(28)

for t ≥ t0 . Then for s > ˛ multiplicative Laplace transform Lm f ∗(n) (t) exists and can be calculated by the formula



Lm f ∗(n) (t) Theorem 3.10.

=

1 f (0)s

n−1

f ∗ (0)s



n

f ∗∗ (0)s

n−3

. . .f ∗(n−1) (0)

F(s)s .

(29)

Let f1 and f2 be positive definite continuous functions.



f1 = f2 if and only if Lm f1 Proof.

n−2



= Lm f2 .

f1 = f2 ⇔ ln f1 = ln f2 . From the Laplace transform of classical analysis, we get

L ln f1



Definition 3.3.



Lm f



L−1 F m



= L ln f2 L ln f1



= L ln f2



⇔ eL{ln f1 } = eL{ln f2 } , so Lm f1



= Lm f2 .

If Fm (s) is the multiplicative Laplace transform of a continuous function f , i.e.

=F

is called as the inverse multiplicative Laplace transform of F.

Theorem 3.11. Inverse multiplicative Laplace transform is multiplicatively linear. In other words, if c1 , c2 are arbitrary exponents and f1 , f2 are two given continuous functions, which have multiplicative Laplace transforms F1 , F2 respectively. Then,



L−1 F1c1 F2c2 m Proof.

c1

= L−1 F1 m

c2

L−1 F2 m

.

(30)



Suppose f1 and f2 are continuous functions such that F1 = Lm f1



and F2 = Lm f2 .

N. Yalcin et al. / Optik 127 (2016) 9984–9995

−1

From the definition, we know that Lm tiplicative Laplace transform, we have



Lm f1c1 f2c2

c1

= Lm f1

c2

F1

−1

= f1 and Lm

F2

= f2 . Using the multiplicative linearity property of mul-

= F1c1 F2c2

Lm f2

9993

(31)

From the definition of inverse multiplicative Laplace transform we obtain



F1c1 F2c2 L−1 m

c1

= f1c1 f2c2 = L−1 F1 m

c2

L−1 F2 m

4. Applications to multiplicative linear differential equations We have the following formula for multiplicative Laplace transforms of multiplicative derivatives



Lm f ∗(n) (t)

1

=

n

n−1 n−2 n−3 f (0)s f ∗ (0)s f ∗∗ (0)s . . .f ∗(n−1)

F(s)s .

(32)

(0)

This formula includes multiplicative Laplace transform of f and f , f ∗ , f ∗∗ , . . ., f ∗(n−1) functions, which the values of these functions at x = 0 is used. So, it can be used to obtain solutions of the initial value problems, particularly of multiplicative linear differential equations with constant exponentials. We get solutions by applying multiplicative Laplace transform to both sides of equations of these problems. For example, consider the second order initial value problem a a (y∗∗ ) 0 (y∗ ) 1 ya2 = 1,

(33)

y (0) = b0 , y∗ (0) = b1 .

(34)

Here, a0 , a1 , a2 and b0 , b1 are constants. Applying multiplicative Laplace transform to both sides of (33) and using the multiplicative linearity property of multiplicative Laplace transform, we get



Lm y∗∗

a0

a1

Lm y∗



Lm {y}a2 = Lm 1

Now, we use formula (32) and initial conditions in (34) to obtain



Y (s)s y(0)s y∗

a0 

2

(0)

Y (s)s y (0)

a1



Y (s)

a2

=1

If this equation is rearranged, we can write



Y (s)a0 s s

b0 b1

2 +a s+a 1 2

a0 a1 = 1. b0

In this case, we have Y (s)a0 s

2 +a s+a 1 2

Y (s) = (b0

= b0

a0 s+a1

a0 s+a1

a

b1 0 ) a0 s

a0 s+a1 2 +a s+a 1 2

Y (s) = (b0 ) a0 s

b1

a0

1 2 +a s+a 1 2

(b1 ) a0 s

a0 2 +a s+a 1 2

Consequently, applying inverse multiplicative Laplace transform, we obtain



Y (s) y (t) = L−1 m

This method can be applied to any order of linear differential equation with constant exponential. Example 4.1.

Consider the following multiplicative differential equation

y∗ = y with the initial condition y (0) = e Taking the multiplicative Laplace transform of both sides of the multiplicative differential equation and using the given conditions, we have



Lm y∗

= Lm {y}

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N. Yalcin et al. / Optik 127 (2016) 9984–9995

Y (s)s = Y (s) y (0) Y (s)s−1 = e 1

Y (s) = e s−1 Then, taking the inverse multiplicative Laplace transform, we obtain



Y (s) L−1 m y (t) = ee

= L−1 m



1



e s−1

t

Example 4.2.

Consider the following multiplicative differential equation

y∗∗ y = 1 with the initial condition y (0) = e, y∗ (0) = 1. Taking the multiplicative Laplace transform of both sides of the multiplicative differential equation and using the given conditions, we have



Lm y∗∗ y



= Lm (1)

Lm y∗∗ Lm {y} = Lm (1) Y (s)s

2

y(0)s y∗ (0)

Y (s) = 1.

We can rewrite last equation and obtain Y (s)s

2 +1

= es s

Y (s) = e s2 +1 Finally taking the inverse multiplicative Laplace transform, we get



Y (s) L−1 m

= L−1 m



s



e s2 +1

y (t) = ecost 5. Conclusion In this study, multiplicative Laplace transform and its applications is investigated for positive definite functions. It has shown that this transform have basic properties as linearity, convolution. Further, existence of multiplicative Laplace transform is proved. Besides, it has appeared that this transform is alternative to find solutions of some multiplicative differential equations. References [1] [2] [3] [4] [5]

M. Grossman, R. Katz, Non-Newtonian Calculus, Lee Press, Pigeon Cove, MA, 1972. D. Stanley, A multiplicative calculus, Primus IX (4) (1999) 310–326. D. Campbell, Multiplicative calculus and student projects, Primus 9 (4) (1999). A.E. Bashirov, E. Mısırlı, A. Özyapıcı, Multiplicative calculus and its applications, J. Math. Anal. Appl. 337 (2008) 36–48. A.E. Bashirov, E. Mısırlı, Y. Tando˘gdu, A. Özyapıcı, On modeling with multiplicative differential equations, Appl. Math. J. Chin. Univ. 26 (4) (2011) 425–438. [6] L. Florack, H. Assen, Multiplicative calculus in biomedical image analysis, J. Math. Imaging Vis. 42 (2012) 64–75. [7] D.A. Filip, C. Piatecki, A non-Newtonian examination of the theory of exogenous economic growth, Math. Aeterna (2014) (To appear).

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[8] M. Mora, F. Córdova-Lepe, R. Del-Valle, A non-Newtonian gradient for contour detection in images with multiplicative noise, Pattern Recognit. Lett. 33 (2012) 1245–1256. [9] D. Filip, C. Piatecki, An overview on the non-Newtonian calculus and its potential applications to economics, Appl. Math. Comput. 187 (1) (2007) 68–78. [10] D. Aniszewska, Multiplicative Runge–Kutta method, Nonlinear Dyn. 50 (2007) 265–272.