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Article

A Cotangent Fractional Derivative with the Application

Department of Mathematics, Faculty of Sciences, Chouaib Doukkali University, El Jadida 24000, Morocco
Fractal Fract. 2023, 7(6), 444; https://doi.org/10.3390/fractalfract7060444
Submission received: 4 April 2023 / Revised: 2 May 2023 / Accepted: 29 May 2023 / Published: 30 May 2023

Abstract

:
In this work, we present a new type of fractional derivatives (FD) involving exponential cotangent function in their kernels called Riemann–Liouville D σ , γ and Caputo cotangent fractional derivatives C D σ , γ , respectively, and their corresponding integral I σ , γ . The advantage of the new fractional derivatives is that they achieve a semi-group property, and we have special cases; if γ = 1 we obtain the Riemann–Liouville FD (RL-FD), Caputo FD (C-FD), and Riemann–Liouville fractional integral (RL-FI). We give some theorems and lemmas, and we give solutions to linear cotangent fractional differential equations using the Laplace transform of the D σ , γ , C D σ , γ and I σ , γ . Finally, we give the application of this new type on the SIR model. This new type of fractional calculus can help other researchers who still work on the actual subject.

1. Introduction

Fractional calculus (FC) is used in modeling in chemistry, physics, electricity, and mechanics; see [1,2,3,4]. There are many works on FC, see [5,6,7,8,9,10,11,12,13,14,15,16].
In [17], the authors presented the conformable derivative (CD) of x of order γ [ 0 , 1 ] is:
D γ x ( t ) = lim θ 0 x t + θ t 1 γ x ( t ) θ ,
where the drawback is that lim γ 0 + D γ x ( t ) x ( t ) . The author [18] presented some concepts of CD and raised an open problem about how to use CD to produce a more general FD. The general FD and fractional integrals (FI) proposed and studied [19,20] provided a response to this problem.
In [21,22], Anderson hence improved the CD, i.e., lim γ 0 + D γ x ( t ) = x ( t ) . In [23,24,25,26,27], the authors presented new types of FD that allow the appearance of the kernel (exponential function or the Mittag–Leffler (ML) function). Nevertheless, the new nonsingular kernel does not possess a semi-group property which makes it difficult to solve certain complicated fractional systems. Concurrently, remarkable efforts have been made to define different types of FD and integrals involving ML functions in their representations; see the papers [28,29,30]. Motivated by the above-mentioned background, we introduce a new type of FC.
Cotangent FD has three features that make it different and special:
1.
The kernel operator is the exponential of the cotangent function,
2.
The I σ , γ , C D σ , γ and D σ , γ achieve a semi-group property,
3.
If order γ = 1 , we obtain the RL-FD, C-FD, and RL-FI.
Contained in this paper, in Section 2, we provide preliminaries of FC. In Section 3, we present the Riemann–Liouville cotangent fractional derivatives and their corresponding integrals, giving the main results and studying their properties. In Section 4, we present the Laplace transforms for the cotangent Riemann–Liouville fractional derivative and use them to solve linear cotangent fractional differential equations of the Riemann–Liouville type. In Section 5, we present the Caputo cotangent fractional derivatives, the Laplace transforms for the cotangent, and the Caputo fractional derivative and use them to solve linear cotangent fractional differential equations of the Caputo type. Finally, in Section 6, we present the application.

2. Preliminaries of FC

In this section, we give some definitions for FD and FI that will be for the sake of comparison. Let δ C ,   Re ( δ ) > 0 and a function x : [ t i , t f ] R , we state the following definitions:
1.
The left RL-FI of x of order δ is:
t i I δ x ( t ) = 1 Γ ( δ ) t i t ( t s ) δ 1 x ( s ) d s .
2.
The right RL-FI of x of order δ is:
I t f δ x ( t ) = 1 Γ ( δ ) t t f ( s t ) δ 1 x ( s ) d s .
3.
The left RL-FD of x of order δ is:
t i D δ x ( t ) = d d t n δ t i I n δ δ x ( t ) , n δ = [ δ ] + 1 .
4.
The right RL-FD of x of order δ is:
D t f δ x ( t ) = d d t n δ I t f n δ δ x ( t ) , n δ = [ δ ] + 1 .
5.
The left C-FD of x of order δ is:
t i C D δ x ( t ) = t i I n δ δ x ( n δ ) ( t ) , n δ = [ δ ] + 1 .
6.
The right C-FD of x of order δ is:
C D t f δ x ( t ) = I t f n δ δ ( 1 ) n δ x ( n δ ) ( t ) , n δ = [ δ ] + 1 .
For comparison with cotangent integrals and cotangent derivatives, we give the main definitions presented [19,20,21]. In [19], Katugampola gives the left fractional integral (K-FI) by
t i I δ , γ x ( t ) = 1 Γ ( δ ) t i t t γ s γ γ δ 1 x ( s ) d s s 1 γ ,
and right K-FI by
I t f δ , γ x ( t ) = 1 Γ ( δ ) t t f s γ t γ γ δ 1 x ( s ) d s s 1 γ .
The left and right Katugampola FD (K-FD) [20] are defined, respectively, as:
t i D δ , γ x ( t ) = κ n δ t i I n δ δ , γ x ( t ) ,
and
D t f δ , γ x ( t ) = ( κ ) n δ I t f n δ δ , γ x ( t ) ,
where γ [ 0 , 1 ] and κ = t 1 γ d d t .
In [30], Jarad et al. give the left Caputo FDs (GC-FD) by
t i C D δ , γ x ( t ) = t i I n δ δ , γ κ n δ x ( t ) ,
and right GC-FD by
C D t f δ , γ x ( t ) = I t f n δ δ , γ ( κ ) n δ x ( t ) ,
where n δ = [ δ ] + 1 . We give some reference for ML functions see [29,31,32]. It is worth mentioning here that once γ = 1 , the K-FI (8) and (9), we obtain RL-FI (2) and (3), the K-FD (10) and (11) become the RL-FD (4) and (5) and the GC-FD (12) and (13) have the forms of the C-FD (6) and (7).

3. The Riemann–Liouville Cotangent Fractional Derivatives

Now, we present the Riemann–Liouville cotangent fractional derivatives and their corresponding integrals, giving the main results and studying their properties. The first time has been presented CD by Khalil et al. [17] as x of order γ is Equation (1).
Notice that lim γ 0 + D γ x ( t ) x ( t ) and lim γ 1 D γ x ( t ) = x ( t ) . From the article [21], Anderson et al. gave the following definition.
Definition 1.
([21,22]). Let γ [ 0 , 1 ] and f , g : [ 0 , 1 ] × R [ 0 , + ] be continuous such that
lim γ 1 g ( γ , t ) = 0 , lim γ 1 f ( γ , t ) = 1 , t R , lim γ 0 + g ( γ , t ) = 1 , lim γ 0 + f ( γ , t ) = 0 , t R , g ( γ , t ) 0 , γ [ 0 , 1 [ , f ( γ , t ) 0 , γ [ 0 , 1 ] , t R .
Then, the proportional derivatives (PD) of x of order γ is:
D γ x ( t ) = g ( γ , t ) x ( t ) + f ( γ , t ) x ( t ) .
We will confine ourselves to an important special case when f ( γ , t ) = sin ( π 2 γ ) and g ( γ , t ) = cos ( π 2 γ ) . Therefore, (14) becomes
D γ x ( t ) = cos ( π 2 γ ) x ( t ) + sin ( π 2 γ ) x ( t ) .
Notice that lim γ 0 + D γ x ( t ) = x ( t ) and lim γ 1 D γ x ( t ) = x ( t ) .
We want to search for the integral associated with PD in (15). Let us use the following equation:
D γ y ( t ) = cos ( π 2 γ ) y ( t ) + sin ( π 2 γ ) y ( t ) = x ( t ) , t t i .
the solution of (16) is:
y ( t ) = 1 sin ( π 2 γ ) t i t e cot ( π 2 γ ) ( t s ) x ( s ) d s ,
where cot is the cotangent function, defined by cot ( t ) = cos ( t ) sin ( t ) . The proportional integral (cotangent fractional integral) associated with D γ is defined by:
t i I 1 , γ x ( t ) = 1 sin ( π 2 γ ) t i t e cot ( π 2 γ ) ( t s ) x ( s ) d s ,
where we accept that t i I 0 , γ x ( t ) = x ( t ) .
Remark 1.
Let γ [ 0 , 1 ] . The x ( t ) = e cot ( π 2 γ ) t is a nonconstant function. However, D γ x ( t ) = 0 .
D γ x ( t ) = cos ( π 2 γ ) e cos ( π 2 γ ) sin ( π 2 γ ) t + sin ( π 2 γ ) ( e cos ( π 2 γ ) sin ( π 2 γ ) t ) = cos ( π 2 γ ) e cos ( π 2 γ ) sin ( π 2 γ ) t + sin ( π 2 γ ) ( cos ( π 2 γ ) sin ( π 2 γ ) e cos ( π 2 γ ) sin ( π 2 γ ) t ) = cos ( π 2 γ ) e cos ( π 2 γ ) sin ( π 2 γ ) t cos ( π 2 γ ) e cos ( π 2 γ ) sin ( π 2 γ ) t = 0 .
Proposition 1.
Let x be defined on [ t i , + ] and differentiable on [ t i , + ] and γ [ 0 , 1 ] . Then, we obtain
t i I 1 , γ D γ x ( t ) = x ( t ) e cot ( π 2 γ ) ( t t i ) x ( t i ) .
Proof. 
We have
t i I 1 , γ D γ x ( t ) = 1 sin ( π 2 γ ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) D γ x ( s ) d s = 1 sin ( π 2 γ ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) cos ( π 2 γ ) f ( s ) + sin ( π 2 γ ) x ( s ) d s = 1 sin ( π 2 γ ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) cos ( π 2 γ ) x ( s ) d s + t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) x ( s ) d s = cos ( π 2 γ ) sin ( π 2 γ ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) x ( s ) d s + e cos ( π 2 γ ) sin ( π 2 γ ) ( t τ ) x ( τ ) τ = t i τ = t cos ( π 2 γ ) sin ( π 2 γ ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t r ) x ( r ) d r = e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) x ( s ) s = t i s = t = x ( t ) e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) x ( t i ) .
For producing a general type of FI depending on the cotangent fractional integral Equation (17), we have
t i I n , γ x ( t ) = 1 sin ( π 2 γ ) t i t e cot ( π 2 γ ) ( t τ 1 ) d τ 1 1 sin ( π 2 γ ) t i τ 1 e cot ( π 2 γ ) ( τ 1 τ 2 ) d τ 2 1 sin ( π 2 γ ) t i τ n 1 e cot ( π 2 γ ) ( τ n 1 τ n ) x ( τ n ) d τ n = 1 sin ( π 2 γ ) n Γ ( n ) t i t e cot ( π 2 γ ) ( t s ) ( t s ) n 1 x ( s ) d s .
From (18), we have the following definition.
Definition 2.
Let γ [ 0 , 1 ] and σ C such that R e ( σ ) > 0 , the left Riemann–Liouville cotangent fractional integral of x is:
( t i I σ , γ x ) ( t ) = 1 sin ( π 2 γ ) σ Γ ( σ ) t i t e cot ( π 2 γ ) ( t s ) ( t s ) σ 1 x ( s ) d s ,
and right Riemann–Liouville cotangent fractional integral of x is:
( I t f σ , γ x ) ( t ) = 1 sin ( π 2 γ ) σ Γ ( σ ) t t f e cot ( π 2 γ ) ( s t ) ( s t ) σ 1 x ( s ) d s .
Let n N , we use the notation
D n , γ x ( t ) = ( D γ D γ D γ n times x ) ( t ) .
Definition 3.
Let γ [ 0 , 1 ] and R e ( σ ) 0 . The left Riemann–Liouville cotangent derivative of x is given as:
( t i D σ , γ x ) ( t ) = ( D n σ , γ t i I n σ σ , γ x ) ( t ) = D n σ , γ sin ( π 2 γ ) n σ σ Γ ( n σ σ ) t i t e cot ( π 2 γ ) ( t s ) × ( t s ) n σ σ 1 x ( s ) d s ,
and right Riemann–Liouville cotangent derivative of x is given as:
( D t f σ , γ x ) ( t ) = ( D n σ , γ I t f n σ σ , γ x ) ( t ) = D n σ , γ sin ( π 2 γ ) n σ σ Γ ( n σ σ ) t t f e cot ( π 2 γ ) ( s t ) × ( s t ) n σ σ 1 x ( s ) d s ,
where n σ = [ R e ( σ ) ] + 1 .
Remark 2.
We have
  • If γ = 1 in Definition 3, then we have the RL-FD (4) and (5).
  • lim σ 1 t i D σ , γ x ( t ) = D γ x ( t ) and lim σ 0 t i D σ , γ x ( t ) = x ( t ) .
Lemma 1.
Let function y ( t ) , we have
D γ y ( t ) e cot ( π 2 γ ) t = sin ( π 2 γ ) y ( t ) e cot ( π 2 γ ) t .
Proposition 2.
Let γ [ 0 , 1 ] and Re ( σ 1 ) > 0 , Re ( σ 2 ) > 0 . We have
1. 
t i I σ 1 , γ e cot ( π 2 γ ) ( t t i ) ( t t i ) σ 2 1 = Γ ( σ 2 ) sin ( π 2 γ ) σ 1 Γ ( σ 1 + σ 2 ) e cot ( π 2 γ ) ( t t i ) ( t t i ) σ 1 + σ 2 1 .
2. 
I t f σ 1 , γ e cot ( π 2 γ ) ( t f t ) ( t f t ) σ 2 1 = Γ ( σ 2 ) sin ( π 2 γ ) σ 1 Γ ( σ 1 + σ 2 ) e cot ( π 2 γ ) ( t f t ) ( t f t ) σ 1 + σ 2 1 .
3. 
t i D σ 1 , γ e cot ( π 2 γ ) ( t t i ) ( t t i ) σ 2 1 = sin ( π 2 γ ) σ 1 Γ ( σ 2 ) Γ ( σ 2 σ 1 ) e cot ( π 2 γ ) ( t t i ) ( t t i ) σ 2 1 σ 1 .
4. 
D t f σ 1 , γ e cot ( π 2 γ ) ( t f t ) ( t f t ) σ 2 1 = sin ( π 2 γ ) σ 1 Γ ( σ 2 ) Γ ( σ 2 σ 1 ) e cot ( π 2 γ ) ( t f t ) ( t f t ) σ 2 1 σ 1 .
Proof. 1.     
We have
t i I σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 = 1 sin ( π 2 γ ) σ 1 Γ ( σ 1 ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) × ( t s ) σ 1 1 e cos ( π 2 γ ) sin ( π 2 γ ) s ( s t i ) σ 2 1 d s = 1 sin ( π 2 γ ) σ 1 Γ ( σ 1 ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) t × ( t s ) σ 1 1 ( s t i ) σ 2 1 d s = e cos ( π 2 γ ) sin ( π 2 γ ) t sin ( π 2 γ ) σ 1 Γ ( σ 1 ) t i t ( t s ) σ 1 1 ( s t i ) σ 2 1 d s ,
making the change of variable y = t s t t i , we obtain
t i I σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 = e cos ( π 2 γ ) sin ( π 2 γ ) t sin ( π 2 γ ) σ 1 Γ ( σ 1 ) t i t ( t s ) σ 1 1 ( s t i ) σ 2 1 d s = e cos ( π 2 γ ) sin ( π 2 γ ) t sin ( π 2 γ ) σ 1 Γ ( σ 1 ) ( t t i ) σ 1 + σ 2 1 × 0 1 y σ 1 1 ( 1 y ) σ 2 1 d y ,
and using the Beta function defined by, B ( σ 1 , σ 2 ) = 0 1 u σ 1 1 ( 1 u ) σ 2 1 d u and the fact that B ( σ 1 , σ 2 ) = Γ ( σ 1 ) Γ ( σ 2 ) Γ ( σ 1 + σ 2 ) , so
t i I σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 = e cos ( π 2 γ ) sin ( π 2 γ ) t sin ( π 2 γ ) σ 1 Γ ( σ 1 ) ( t t i ) σ 1 + σ 2 1 × Γ ( σ 1 ) Γ ( σ 2 ) Γ ( σ 1 + σ 2 ) = Γ ( σ 2 ) e cos ( π 2 γ ) sin ( π 2 γ ) t sin ( π 2 γ ) σ 1 Γ ( σ 1 + σ 2 ) ( t t i ) σ 1 + σ 2 1 .
2.
Similar to 1.
3.
Let x ( t ) = e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 , using the Lemma 1 and we have
t i D σ 1 , γ x ( t ) = D t n σ 1 , γ t i I n σ 1 σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 = D t n σ 1 , γ Γ ( σ 2 ) e cos ( π 2 γ ) sin ( π 2 γ ) t sin ( π 2 γ ) n σ 1 σ 1 Γ ( n σ 1 σ 1 + σ 2 ) ( t t i ) n σ 1 σ 1 + σ 2 1 = sin ( π 2 γ ) n σ 1 Γ ( σ 2 ) ( n σ 1 σ 1 + σ 2 1 ) ( n σ 1 σ 1 + σ 2 1 ) ( σ 2 σ 1 ) sin ( π 2 γ ) n σ 1 σ 1 Γ ( n σ 1 σ 1 + σ 2 ) × e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 σ 1 = sin ( π 2 γ ) σ 1 Γ ( σ 2 ) Γ ( σ 2 σ 1 ) e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 σ 1 .
4.
Similar to 3.
Lemma 2.
Let λ R and cot σ 1 , σ 2 γ ( t 1 , t 2 ) = e cot ( π 2 γ ) t 1 E σ 1 , σ 2 ( t 2 ) , where E σ 1 , σ 2 is the Mittag–Lefler function [3]. Then
t i D σ 1 , γ cot σ 1 , 1 γ ( t t i , λ ( t t i ) σ 1 ) = λ sin ( π 2 γ ) σ 1 cot σ 1 , 1 γ ( t t i , λ ( t t i ) σ 1 ) .
Proof. 
We have
t i D σ 1 , γ cot σ 1 , 1 γ ( t t i , λ ( t t i ) σ 1 ) = t i D σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ 1 , 1 ( λ ( t t i ) σ 1 ) = t i D σ 1 , γ k = 0 + λ k Γ ( σ 1 k + 1 ) e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) k σ 1 = k = 0 + λ k Γ ( σ 1 k + 1 ) t i D σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) k σ 1 = k = 1 + λ k Γ ( σ 1 k + 1 ) sin ( π 2 γ ) σ 1 Γ ( σ 1 k + 1 ) Γ ( σ 1 k + 1 σ 1 ) e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) σ 1 k σ 1 = sin ( π 2 γ ) σ 1 e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) k = 1 + λ k Γ ( σ 1 ( k 1 ) + 1 ) ( t t i ) σ 1 ( k 1 ) = λ sin ( π 2 γ ) σ 1 e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) k = 1 + λ k 1 Γ ( σ 1 ( k 1 ) + 1 ) ( t t i ) σ 1 ( k 1 ) = λ sin ( π 2 γ ) σ 1 e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ 1 , 1 ( λ ( t t i ) σ 1 ) .
In the Theorem 1, we present the semi-group property for the Riemann–Liouville cotangent integral.
Theorem 1.
Let γ [ 0 , 1 ] , Re ( σ 1 ) > 0 , Re ( σ 2 ) > 0 and x be continuous and defined for t t i . Then,
( t i I σ 1 , γ t i I σ 2 , γ x ) ( t ) = ( t i I σ 2 , γ t i I σ 1 , γ x ) ( t ) = t i I σ 1 + σ 2 , γ x ( t ) .
Proof. 
We have
( t i I σ 1 , γ t i I σ 2 , γ x ) ( t ) = 1 sin ( π 2 γ ) σ 1 + σ 2 Γ ( σ 1 ) Γ ( σ 2 ) × t i t t i u e cos ( π 2 γ ) sin ( π 2 γ ) ( t u ) e cos ( π 2 γ ) sin ( π 2 γ ) ( u s ) × ( t u ) σ 1 1 ( u s ) σ 2 1 x ( s ) d s d u = 1 sin ( π 2 γ ) σ 1 + σ 2 Γ ( σ 1 ) Γ ( σ 2 ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) x ( s ) × s t ( t u ) σ 1 1 ( u s ) σ 2 1 d u d s ,
making the change of variable y = u s t s , we obtain
( t i I σ 1 , γ t i I σ 2 , γ x ) ( t ) = 1 sin ( π 2 γ ) σ 1 + σ 2 Γ ( σ 1 ) Γ ( σ 2 ) × t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) ( t s ) σ 1 + σ 2 1 x ( s ) d s × 0 1 ( 1 y ) σ 1 1 y σ 2 1 d y = 1 sin ( π 2 γ ) σ 1 + β Γ ( σ 1 + σ 2 ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) × ( t s ) σ 1 + σ 2 1 x ( s ) d s = t i I σ 1 + σ 2 , γ x ( t ) .
Theorem 2.
Let x be integrable in each interval [ t i , t ] , t > t i and 0 l < [ Re ( σ ) ] + 1 . Then
D l , γ t i I σ , γ x ( t ) = t i I σ l , γ x ( t ) .
Proof. 
Using the definition and the Lemma 1.7 in [21] (or D 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) = 0 ), we have
D l , γ t i I σ , γ x ( t ) = D l 1 , γ D 1 , γ t i I σ , γ x ( t ) = D l 1 , γ 1 sin ( π 2 γ ) σ 1 Γ ( σ 1 ) t i t e cot ( π 2 γ ) ( t s ) × ( t s ) σ 2 x ( s ) d s .
we continue l-times in this method until we reach (24). □
Corollary 1.
Let 0 < Re ( σ 2 ) < Re ( σ 1 ) , l 1 < Re ( σ 2 ) l . We obtain
( t i D σ 2 , γ t i I σ 1 , γ x ) ( t ) = t i I σ 1 σ 2 , γ x ( t ) .
Proof. 
From the help of the Theorems 1 and 2, we obtain
( t i D σ 2 , γ t i I σ 1 , γ x ) ( t ) = D l , γ t i I l σ 2 , γ t i I σ 1 , γ x ( t ) = D l , γ t i I l σ 2 + σ 1 , γ x ( t ) = ( t i I σ 1 σ 2 , γ x ) ( t ) .
Theorem 3.
Let Re [ σ ] > 0 , γ [ 0 , 1 ] , n σ = [ Re ( σ ) ] + 1 and x be integrable on t t i . Then, we obtain
( t i D σ , γ t i I σ , γ x ) ( t ) = x ( t ) .
Proof. 
From the definition and Theorem 1, we obtain
( t i D σ , γ t i I σ , γ x ) ( t ) = D n σ , γ t i I n σ σ , γ t i I σ , γ x ( t ) = D n σ , γ t i I n σ , γ x ( t ) = x ( t ) .
Lemma 3.
Let σ > 0 , γ [ 0 , 1 ] and l N , then
t i I σ , γ D l , γ x ( t ) = D l , γ t i I σ , γ x ( t ) k = 0 l 1 ( t t i ) σ l + k e cot ( π 2 γ ) ( t t i ) Γ ( σ + k l + 1 ) sin ( π 2 γ ) σ l + k D k , γ x ( t i ) .
In particular, if l = 1 , then
t i I σ , γ D γ x ( t ) = D γ t i I σ , γ x ( t ) ( t t i ) σ 1 e cot ( π 2 γ ) ( t t i ) sin ( π 2 γ ) σ 1 Γ ( σ ) x ( t i ) .
Proof. 
Observing that
D γ x y ( t ) = y ( t ) D γ x ( t ) + x ( t ) D γ y ( t ) cos ( π 2 γ ) x ( t ) y ( t ) ,
we obtain
D γ ( t t i ) n σ σ 1 e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) x ( t i ) = sin ( π 2 γ ) ( n σ σ 1 ) ( t t i ) n σ σ 2 × e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) x ( t i ) ,
so we have
L t i D γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) X t i ( p ) sin ( π 2 γ ) x ( t i ) ,
where the Laplace transform starting from t i is:
L t i { x ( t ) } ( p ) = t i e p ( τ t i ) x ( τ ) d τ ,
and L t i x ( t ) ( p ) = X t i ( p ) . We use Equation (29) in Theorem 5, we obtain
L t i t i I σ , γ D γ x ( t ) ( p ) = L t i D γ x ( t ) ( p ) ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) X t i ( p ) sin ( π 2 γ ) x ( t i ) ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ L t i D γ t i I σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) X t i ( p ) ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ ,
and
L t i ( t t i ) σ 1 e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) sin ( π 2 γ ) σ 1 Γ ( σ ) x ( t i ) ( p ) = x ( t i ) sin ( π 2 γ ) ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) .
Alternatively, since
( D γ t i I σ , γ x ) ( t ) = ( σ 1 ) sin ( π 2 γ ) sin ( π 2 γ ) σ Γ ( σ ) t i t e cos ( π 2 γ ) sin ( π 2 γ ) ( t s ) ( t s ) σ 2 x ( s ) d s ,
the Equation (26) can be proved by integration by parts.
Relation (25) follows by applying (26) inductively through making use of (27). □
Remark 3.
We have
  • t i D σ , γ x ( t ) = t i I σ , γ x ( t ) .
  • Using Lemma 3 we obtain
    D γ t i D σ , γ x ( t ) = D n σ , γ D γ t i I n σ σ , γ x ( t ) = D n σ , γ t i I n σ σ , γ D γ ( t t i ) n σ σ 1 e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) Γ ( n σ σ ) sin ( π 2 γ ) n σ σ 1 x ( t i ) = t i I σ , γ D γ x ( t ) ( t t i ) σ 1 e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) sin ( π 2 γ ) σ 1 Γ ( σ ) x ( t i ) .
  • Lemma 3 is valid for any real σ.
Theorem 4.
Let Re ( σ ) > 0 , n σ = [ Re ( σ ) ] , x L 1 [ t i , t f ] and t i I σ , γ x ( t ) A C n σ [ t i , t f ] . Then
t i I σ , γ t i D σ , γ x ( t ) = x ( t ) e cot ( π 2 γ ) ( t t i ) m = 1 n σ t i I m σ , γ x t i + × ( t t i ) σ m sin ( π 2 γ ) σ m Γ ( σ + 1 m ) .
Proof. 
By the Definition 3, we have
( t i I σ , γ t i D σ , γ x ) ( t ) = ( t i I σ , γ D n σ , γ t i I n σ σ , γ x ) ( t ) ,
from applying (25) in Lemma 3, we obtain
( t i I σ , γ t i D σ , γ x ) ( t ) = ( D n σ , γ t i I σ , γ t i I n σ σ , γ x ) ( t ) k = 0 n σ 1 ( t t i ) σ n σ + k e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) Γ ( σ + k n σ + 1 ) sin ( π 2 γ ) σ n σ + k × ( D k , γ t i I n σ σ , γ x ) ( t i ) ,
and use of the first point in Remark 3, we have
( t i I σ , γ t i D σ , γ x ) ( t ) = ( D n σ , γ t i I σ , γ t i I n σ σ , γ x ) ( t ) k = 0 n σ 1 ( t t i ) σ n σ + k e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) Γ ( σ + k n σ + 1 ) sin ( π 2 γ ) σ n σ + k × ( t i I k , γ t i I n σ σ , γ x ) ( t i ) ,
and the Theorem 1, we have
( t i I σ , γ t i D σ , γ x ) ( t ) = x ( t ) k = 0 n σ 1 ( t t i ) σ n σ + k e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) Γ ( σ + k n σ + 1 ) sin ( π 2 γ ) σ n σ + k t i I n σ σ k , γ x t i + = x ( t ) e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) m = 1 n σ t i I m σ , γ x t i + × ( t t i ) σ m sin ( π 2 γ ) σ m Γ ( σ + 1 m ) ,
with the change of variable m = n σ k has been used.   □

4. The Laplace Transforms for Cotangent Fractional Integrals

Theorem 5.
Let x to be exponential order, γ [ 0 , 1 ] and σ C where Re ( σ ) > 0 . We have
L t i t i I σ , γ x ( t ) ( p ) = 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ L t i { x ( t ) } ( p ) , p > d .
Proof. 
We have
L t i t i I σ , γ x ( t ) ( p ) = 1 sin ( π 2 γ ) σ Γ ( σ ) L t i e cos ( π 2 γ ) sin ( π 2 γ ) t t σ 1 x ( t ) ( p ) = 1 sin ( π 2 γ ) σ Γ ( σ ) Γ ( σ ) p cos ( π 2 γ ) sin ( π 2 γ ) σ L t i { x ( t ) } ( p ) = 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ L t i { x ( t ) } ( p ) .
Theorem 6.
Let σ sash that n σ = [ Re ( σ ) ] + 1 and x C n σ 1 [ t i , + ] be such that x ( k ) , k = 1 , 2 , , n σ 1 are of exponential order on each subinterval [ t i , t f ] . We have,
L t i D n σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ L t i { x ( t ) } ( p ) sin ( π 2 γ ) k = 0 n σ 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ 1 k × D k · γ x ( t i ) .
Proof. 
By using that L t i x ( t ) ( p ) = p L t i { x ( t ) } ( p ) x ( t i ) , we have for n σ = 1
L t i D 1 , γ x ( t ) ( p ) = L t i cos ( π 2 γ ) x ( t ) + sin ( π 2 γ ) x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) L t i { x ( t ) } ( p ) sin ( π 2 γ ) x ( t i ) .
From applying (31) we obtain the relation (30). □
Theorem 7.
Let σ C where Re ( σ ) > 0 and n σ = [ Re ( σ ) ] + 1 and γ [ 0 , 1 ] , then
L t i t i D σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ X t i ( p ) sin ( π 2 γ ) k = 0 n σ 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ k 1 I n σ σ k , γ x t i + ,
with X t i ( p ) = L t i { x ( t ) } ( p ) . In particular, if x is continuous at t i then
L t i t i D σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) s ) σ X t i ( p ) .
Proof. 
By applying Theorems 5 and 6 we have
L t i t i D σ , γ x ( t ) ( p ) = L t i t i D t i n σ , γ I n σ σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ L t i t i I n σ σ , γ x ( t ) ( p ) sin ( π 2 γ ) k = 0 n σ 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ 1 k × D t i k , γ I n σ σ , γ x t i + = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ n σ × X t i ( p ) sin ( π 2 γ ) k = 0 n σ 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ 1 k × t i I n σ σ k , γ x t i + .
Theorem 8.
Let the linear cotangent fractional differential equation:
t i D σ , γ x ( t ) sin ( π 2 γ ) σ λ x ( t ) = y ( t ) , 0 < σ , γ 1 , t i I 1 σ , γ x t i + = x t i .
Then, the solution of Equation (32) is:
x ( t ) = x t i sin ( π 2 γ ) 1 σ cot σ , 1 γ ( t t i , λ ( t t i ) σ ) + sin ( π 2 γ ) σ t i t cot σ , σ γ ( t s , λ ( t s ) σ ) ( t s ) σ 1 y ( s ) d s .
Proof. 
We apply L t i to (32) and make use of Theorem 7 with n σ = 1 , then we obtain
( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ λ sin ( π 2 γ ) σ X t i ( p ) = sin ( π 2 γ ) x t i + Y t i ( p ) ,
where X t i ( p ) = L t i { x ( t ) } ( p ) and Y t i ( p ) = L t i { y ( t ) } ( p ) . Hence,
X t i ( p ) = sin ( π 2 γ ) 1 σ x t i p cos ( π 2 γ ) sin ( π 2 γ ) σ λ + sin ( π 2 γ ) σ x t i ( p ) p cos ( π 2 γ ) sin ( π 2 γ ) σ λ .
Using the inverse of L t i and the fact that (see Theorem 1.9.13 in [3])
L t i ( t t i ) σ 1 E σ , σ λ ( t t i ) σ ( p ) = 1 p σ λ ,
Using the convolution formula we obtain (33). □

5. The Caputo Cotangent Fractional Derivative

Now, we present the Caputo cotangent fractional derivatives with a solution of their linear cotangent fractional equations.
Definition 4.
Let γ [ 0 , 1 ] and R e ( σ ) 0 . The left Caputo cotangent fractional derivative is:
( t i C D σ , γ x ) ( t ) = ( t i I n σ σ , γ D n σ , γ x ) ( t ) = 1 sin ( π 2 γ ) n σ σ Γ ( n σ σ ) t i t e cot ( π 2 γ ) ( t s ) ( t s ) n σ σ 1 × ( D n , γ x ) ( s ) d s ,
and the right Caputo cotangent fractional derivative is:
( C D t f σ , γ x ) ( t ) = ( I t f n σ σ , γ D n σ , γ x ) ( t ) = 1 sin ( π 2 γ ) n σ σ Γ ( n σ σ ) t t f e cot ( π 2 γ ) ( s t ) ( s t ) n σ σ 1 × ( D n σ , γ x ) ( s ) d s ,
where n σ = [ R e ( σ ) ] + 1 .
Proposition 3.
Let γ [ 0 , 1 ] , Re ( σ 1 ) > 0 and Re ( σ 2 ) > 0 . We have
1. 
t i C D σ 1 , γ e cot ( π 2 γ ) ( t t i ) ( t t i ) σ 2 1 = sin ( π 2 γ ) σ 1 Γ ( σ 2 ) Γ ( σ 2 σ 1 ) e cot ( π 2 γ ) ( t a ) ( t t i ) σ 2 1 σ 1 .
2. 
C D t f σ 1 , γ e cot ( π 2 γ ) ( t f t ) ( t f t ) σ 2 1 = sin ( π 2 γ ) σ 1 Γ ( σ 2 ) Γ ( σ 2 σ 1 ) e cot ( π 2 γ ) ( t f t ) ( t f t ) σ 2 1 σ 1 .
Let n σ 1 = [ R e ( σ 1 ) ] + 1 , for k = 0 , 1 , , n σ 1 1 , we have
t i C D σ 1 , γ e cot ( π 2 γ ) t ( t t i ) k = 0 and C D t f σ 1 , γ e cot ( π 2 γ ) ( t f t ) ( t f t ) k = 0 .
In particular, t i C D σ 1 , γ e cot ( π 2 γ ) t = 0 and C D t f σ 1 , γ e cot ( π 2 γ ) ( t f t ) = 0 .
Proof. 
Let n σ 1 = [ R e ( σ 1 ) ] + 1 , using Proposition 2, we have
t i C D σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 = t i I n σ 1 σ 1 , γ D n σ 1 , γ e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 = t i I n σ 1 σ 1 , γ × sin ( π 2 γ ) n σ 1 ( σ 2 1 ) ( σ 2 2 ) ( σ 2 1 n σ 1 ) ( t t i ) σ 2 n σ 1 1 e cos ( π 2 γ ) sin ( π 2 γ ) t = sin ( π 2 γ ) n σ 1 ( σ 2 1 ) ( σ 2 2 ) ( σ 2 1 n σ 1 ) Γ ( σ 2 n ) Γ ( σ 2 σ 1 ) sin ( π 2 γ ) n σ 1 × ( t t i ) σ 2 σ 1 1 e cos ( π 2 γ ) sin ( π 2 γ ) t = sin ( π 2 γ ) σ 1 Γ ( σ 2 ) Γ ( σ 2 σ 1 ) e cos ( π 2 γ ) sin ( π 2 γ ) t ( t t i ) σ 2 1 σ 1 .
For the relation 2 is similar.  □
Lemma 4.
Let γ [ 0 , 1 ] , then
t i C D σ , γ cot σ , 1 γ ( t t i , λ ( t t i ) α ) = λ sin ( π 2 γ ) σ cot σ , 1 γ ( t t i , λ ( t t i ) σ ) .
Proof. 
t i C D σ , γ cot σ , 1 γ ( t t i , λ ( t t i ) σ ) = t i C D σ , γ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ , 1 ( λ ( t t i ) σ ) = t i C D σ , γ k = 0 + λ k Γ ( σ k + 1 ) e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) k σ = k = 0 + λ k Γ ( σ k + 1 ) t i C D σ , γ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) k σ = k = 1 + λ k Γ ( σ k + 1 ) sin ( π 2 γ ) σ Γ ( σ k + 1 ) Γ ( σ k + 1 σ ) e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) σ k σ = sin ( π 2 γ ) σ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) k = 1 + λ k Γ ( σ ( k 1 ) + 1 ) ( t a ) σ ( k 1 ) = λ sin ( π 2 γ ) σ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) k = 1 + λ k 1 Γ ( σ ( k 1 ) + 1 ) ( t t i ) σ ( k 1 ) = λ sin ( π 2 γ ) σ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ , 1 ( λ ( t t i ) σ ) .
Theorem 9.
Let γ [ 0 , 1 ] and n σ = [ Re ( σ ) ] + 1 , then
t i I σ , γ t i C D σ , γ x ( t ) = x ( t ) k = 0 n 1 D k , γ x ( t i ) sin ( π 2 γ ) k k ! ( t t i ) k e cot ( π 2 γ ) ( t t i ) .
Proof. 
From the Theorem 4 where σ = n σ , we obtain
t i I σ , γ t i C D σ , γ x ( t ) = t i I σ , γ t i I n σ σ , γ D n σ , γ x ( t ) = t i I n σ , γ D n σ , γ x ( t ) = x ( t ) e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) j = 1 n σ t i I j n σ , γ x t i + ( t t i ) n σ j sin ( π 2 γ ) n σ j Γ ( n σ j + 1 ) = x ( t ) k = 0 n σ 1 D k , γ x ( t i ) sin ( π 2 γ ) k k ! ( t t i ) k e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) .
Theorem 10.
Let γ [ 0 , 1 ] and Re ( σ ) > 0 , n σ = [ Re ( σ ) ] + 1 . Let X t i ( p ) = L t i { x ( t ) } ( p ) , then
L t i t i C D σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ X t i ( p ) sin ( π 2 γ ) k = 0 n σ 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ 1 k D k , γ x ( t i ) .
Proof. 
By using Theorems 5 and 6, we obtain
L t i t i C D σ , γ x ( t ) ( p ) = L t i t i I n σ σ , γ D n σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ n σ L t i D σ , γ x ( t ) ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ n σ × [ ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ X t i ( p ) sin ( π 2 γ ) k = 0 n σ 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n σ 1 k D k , γ x ( t i ) ] = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ X t i ( p ) sin ( π 2 γ ) k = 0 n σ 1 ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ 1 k D k , γ x ( t i ) .
By using Theorems 7 and 10, we obtain the following proposition.
Proposition 4.
Let γ [ 0 , 1 ] and σ C where Re ( σ ) > 0 n σ = [ Re ( σ ) ] + 1 , then
t i C D σ , γ x ( t ) = t i D σ , γ x ( t ) k = 0 n σ 1 sin ( π 2 γ ) σ k Γ ( k + 1 σ ) ( t t i ) k σ × e cot ( π 2 γ ) ( t t i ) D k , γ x ( t i ) ,
and
C D t f σ , γ x ( t ) = D t f σ , γ x ( t ) k = 0 n σ 1 sin ( π 2 γ ) σ k Γ ( k + 1 σ ) ( t f t ) k σ × e cot ( π 2 γ ) ( t f t ) D k , γ x ( t f ) .
Remark 4.
We have
  • t i C D σ , γ c ( t ) 0 , γ [ 0 , 1 ] .
  • Let n σ = [ Re ( σ ) ] + 1 then L t i t i C D σ , γ 1 ( t ) ( p ) = sin ( π 2 γ ) n σ ( cos ( π 2 γ ) ) n ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) n 1 p , γ [ 0 , 1 ] .
  • D γ e cos ( π 2 γ ) sin ( π 2 γ ) t = 0 implies that t i C D σ , γ e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) = 0 .
Theorem 11.
Let the linear Caputo cotangent fractional differential equation:
t i C D σ , γ x ( t ) sin ( π 2 γ ) σ λ x ( t ) = y ( t ) , 0 < σ , γ 1 , x ( t i ) = x t i ,
Then the solution of (40) is:
x ( t ) = x t i cot σ , 1 γ ( t t i , λ ( t t i ) σ ) + sin ( π 2 γ ) σ t i t cot σ , α γ ( t s , λ ( t s ) σ ) ( t s ) σ 1 y ( s ) d s .
Proof. 
Applying L t i to (40) and use the Theorem 10 where n σ = 1 , we obtain
( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ λ sin ( π 2 γ ) σ X t i ( p ) = sin ( π 2 γ ) x t i ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ 1 + Y t i ( p ) .
where X t i ( p ) = L t i { x ( t ) } ( p ) and Y t i ( p ) = L t i { y ( t ) } ( p ) . Hence,
X t i ( p ) = p cos ( π 2 γ ) sin ( π 2 γ ) σ 1 x t i p cos ( π 2 γ ) sin ( π 2 γ ) σ λ + sin ( π 2 γ ) σ Y t i ( p ) cos ( π 2 γ ) sin ( π 2 γ ) p σ λ .
We applying the inverse of L t i and using the Theorem (see Theorem 1.9.13 in [3])
L t i ( t t i ) σ 1 E σ , σ λ ( t t i ) σ ( p ) = 1 p σ λ ,
and
L t i E σ λ ( t t i ) σ ( p ) = p σ 1 p σ λ ,
Using the convolution formula we obtain (41).   □
Remark 5.
Let γ [ 0 , 1 ] and K : [ 0 , 1 ] [ 0 , + ] be continuous such that
lim γ 1 K ( γ ) = π 2 + 2 k π , k Z , lim γ 0 + K ( γ ) = 2 k π , K ( γ ) 0 , γ [ 0 , 1 ] .
Then, the left Riemann–Liouville cotangent fractional integral of x is:
( t i I σ , γ x ) ( t ) = 1 sin ( K ( γ ) ) σ Γ ( σ ) t i t e cot ( K ( γ ) ) ( t s ) ( t s ) σ 1 x ( s ) d s ,
the right Riemann–Liouville cotangent fractional integral of x is:
( I t f σ , γ x ) ( t ) = 1 sin ( K ( γ ) ) σ Γ ( σ ) t t f e cot ( K ( γ ) ) ( s t ) ( s t ) σ 1 x ( s ) d s .
the left Riemann–Liouville cotangent derivative of x is:
( t i D σ , γ x ) ( t ) = D n σ , γ sin ( K ( γ ) ) n σ σ Γ ( n σ σ ) t i t e cot ( K ( γ ) ) ( t s ) × ( t s ) n σ σ 1 x ( s ) d s ,
the right Riemann–Liouville cotangent derivative of x is:
( D t f σ , γ x ) ( t ) = D n σ , γ sin ( K ( γ ) ) n σ σ Γ ( n σ σ ) t t f e cot ( K ( γ ) ) ( s t ) × ( s t ) n σ σ 1 x ( s ) d s ,
the left Caputo cotangent fractional derivative of x is:
( t i C D σ , γ x ) ( t ) = 1 sin ( K ( γ ) ) n σ σ Γ ( n σ σ ) t i t e cot ( K ( γ ) ) ( t s ) × ( t s ) n σ σ 1 ( D n , γ x ) ( s ) d s ,
and the right Caputo cotangent fractional derivative of x is:
( C D t f σ , γ x ) ( t ) = 1 sin ( K ( γ ) ) n σ σ Γ ( n σ σ ) t t f e cot ( K ( γ ) ) ( s t ) × ( s t ) n σ σ 1 ( D n σ , γ x ) ( s ) d s ,
where n σ = [ R e ( σ ) ] + 1 ,
D n σ , γ x ( t ) = ( D γ D γ D γ n σ times x ) ( t ) .
and D γ x ( t ) = cos ( K ( γ ) ) x ( t ) + sin ( K ( γ ) ) x ( t ) .
This new type of fractional calculus can help other researchers who still work on the actual subject, for example [33,34,35,36,37,38,39,40,41].

6. Application

Now, we present the application of the SIR model (see for example [42,43,44,45,46,47,48,49,50]). Let the following model:
x ( t ) = Λ μ x ( t ) a x ( t ) y ( t ) , y ( t ) = a x ( t ) y ( t ) ( b + μ ) y ( t ) , z ( t ) = b y ( t ) μ z ( t ) ,
where x ( t ) the number of susceptible, y ( t ) the number of infected and z ( t ) the number of removed individuals at time t. The parameters μ , a and b represent the recruitment rate, the natural death rate, the infection rate, and the removal rate, respectively. Let T ( t ) be the total population. Then
T ( t ) = x ( t ) + y ( t ) + z ( t ) ,
so
T ( t ) = Λ μ T ( t ) .
The exact solution of Equation (51) is:
T ( t ) = ( T ( t i ) Λ μ ) e μ ( t t i ) + Λ μ .
We replace the classical derivative by t i C D σ , γ , so from Equation (51), we obtain
t i C D σ , γ T ( t ) = Λ μ T ( t ) .
We are interested in solving Equation (53), which plays a significant role in virology as well as in epidemiology.
We apply L t i to Equation (40) and use the Theorem 10 where n σ = 1 , we obtain
( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ L t i { T ( t ) } ( p ) sin ( π 2 γ ) ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ 1 T ( t i ) = μ L t i { T ( t ) } ( p ) + Λ L t i { 1 } ( p ) .
Which is equivalent to
( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ + μ L t i { T ( t ) } ( p ) = sin ( π 2 γ ) ( sin ( π 2 γ ) p + cos ( π 2 γ ) ) σ 1 T ( t i ) + Λ L t i { 1 } ( p ) .
Thus,
L t i { T ( t ) } ( p ) = ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ + μ 1 sin ( π 2 γ ) ( sin ( π 2 γ ) p + cos ( π 2 γ ) ) σ 1 T ( t i ) + ( cos ( π 2 γ ) + sin ( π 2 γ ) p ) σ + μ 1 Λ L t i { 1 } ( p ) .
Equation (54) can be written as:
L t i { T ( t ) } ( p ) = p + cos ( π 2 γ ) sin ( π 2 γ ) σ + sin ( π 2 γ ) σ μ 1 p + cos ( π 2 γ ) sin ( π 2 γ ) σ 1 T ( t i ) + sin ( π 2 γ ) σ p + cos ( π 2 γ ) sin ( π 2 γ ) σ + sin ( π 2 γ ) σ μ 1 × Λ L t i { 1 } ( p ) ,
using
L t i { e c ( t t i ) x ( t ) } ( p ) = L t i { x ( t ) } ( p c ) ,
as follows:
L t i ( t t i ) σ 1 E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ p + cos ( π 2 γ ) sin ( π 2 γ ) = L t i e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) σ 1 E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ ( p ) .
Additionally,
L t i E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ p + cos ( π 2 γ ) sin ( π 2 γ ) = L t i e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ ( p ) .
In our case, using Equations (42) and (43) as follows:
L t i { ( t t i ) σ 1 E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ p + cos ( π 2 γ ) sin ( π 2 γ ) = ( p + cos ( π 2 γ ) sin ( π 2 γ ) ) σ + sin ( π 2 γ ) σ μ 1 .
Additionally,
L t i E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ p + cos ( π 2 γ ) sin ( π 2 γ ) = p + cos ( π 2 γ ) sin ( π 2 γ ) σ 1 p + cos ( π 2 γ ) sin ( π 2 γ ) σ + sin ( π 2 γ ) σ μ 1 .
Hence, by using Equations (57) and (59), we have:
( p + cos ( π 2 γ ) sin ( π 2 γ ) σ 1 p + cos ( π 2 γ ) sin ( π 2 γ ) σ sin ( π 2 γ ) σ Λ 1 = L t i E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ p + cos ( π 2 γ ) sin ( π 2 γ ) = L t i e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ ( p ) .
Further, by using Equations (56) and (58), we obtain:
p + cos ( π 2 γ ) sin ( π 2 γ ) σ + sin ( π 2 γ ) σ μ 1 = L t i ( t t i ) σ 1 E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ p + cos ( π 2 γ ) sin ( π 2 γ ) = L t i e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) σ 1 E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ ( p ) .
We may obtain the following result by combining Equations (60) and (61) in Equation (55), we have:
L t i [ T ( t ) ] ( p ) = L t i e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ ( p ) T ( t i ) + sin ( π 2 γ ) σ L t i { e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) σ 1 × E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ } ( p ) L t i [ Λ ] ( p ) .
Hence, by using convolution formula in Equation (62), we obtain:
L t i [ T ( t ) ] ( p ) = L t i e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ ( p ) T ( t i ) + sin ( π 2 γ ) σ L t i e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) σ 1 × E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ Λ ( p ) .
Applying the Laplace inverse of Equation (63), we obtain:
T ( t ) = e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) E σ , 1 sin ( π 2 γ ) σ μ ( t t i ) σ T ( t i ) + sin ( π 2 γ ) σ { e cos ( π 2 γ ) sin ( π 2 γ ) ( t t i ) ( t t i ) σ 1 × E σ , σ sin ( π 2 γ ) σ μ ( t t i ) σ Λ } .
Therefore,
T ( t ) = cot σ , 1 γ ( t t i , ( μ sin ( π 2 γ ) σ ( t t i ) σ ) T ( t i ) + sin ( π 2 γ ) σ t i t cot σ , σ γ ( t s , μ sin ( π 2 γ ) σ × ( t s ) σ ) ( t s ) σ 1 Λ d s .
Remark 6.
We have,
  • For γ = 1 , Equation (64) becomes
    T ( t ) = E σ , 1 ( μ t σ ) T ( t i ) + t i t E σ , σ ( μ ( t s ) σ ) ( t s ) σ 1 Λ d s .
  • For γ = 1 and σ = 1 , Equation (64) coincide with Equation (52).
Now, we trace the impact of the order of the new type of FD on the dynamics behavior of the solution given by (64). We choose Λ = 10 cells μ L 1 day 1 , μ = 0.0139 day 1 , and T ( 0 ) = 600 cells μ L 1 .
In Figure 1, left T ( t ) in (64) for different values of σ with γ = 1 ; Right T ( t ) in (64) for different values of γ with σ = 0.5 . When σ = γ = 1 , the graph of (64) coincides with that of the ordinary differential equation given by (52).

7. Conclusions

We have presented the cotangent fractional derivatives D σ , γ (Riemann–Liouville type) and C D σ , γ (Caputo type) whose kernel contains exponential cotangent function. The advantage of the new type of FD is that they achieve a semi-group property, and we have special cases; if γ = 1 we obtain the RL-FD, C-FD, and RL-FI. We noticed that the function x ( t ) = e cot ( π 2 γ ) t is a nonconstant function, however, C D σ , γ of x ( t ) is zero. Using the Laplace transform of cotangent derivatives and integrals and we give the exact solution for linear cotangent fractional differential equations. Finally, we give the application of this new type on the SIR model. This new type of fractional calculus can help other researchers who still work on the actual subject.

Funding

This research received no external funding.

Data Availability Statement

Data sharing is not applicable to this article, as no data sets were generated or analyzed during the current study.

Acknowledgments

The author would like to thank to the referees for their useful comments and remarks.

Conflicts of Interest

The author declares no conflict of interest.

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Figure 1. Left T ( t ) in (64) for different values of σ with γ = 1 ; Right T ( t ) in (64) for different values of γ with σ = 0.5 .
Figure 1. Left T ( t ) in (64) for different values of σ with γ = 1 ; Right T ( t ) in (64) for different values of γ with σ = 0.5 .
Fractalfract 07 00444 g001
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Sadek, L. A Cotangent Fractional Derivative with the Application. Fractal Fract. 2023, 7, 444. https://doi.org/10.3390/fractalfract7060444

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Sadek L. A Cotangent Fractional Derivative with the Application. Fractal and Fractional. 2023; 7(6):444. https://doi.org/10.3390/fractalfract7060444

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Sadek, Lakhlifa. 2023. "A Cotangent Fractional Derivative with the Application" Fractal and Fractional 7, no. 6: 444. https://doi.org/10.3390/fractalfract7060444

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