Fractional Leibniz Integral Rules for Riemann-Liouville and Caputo

Home , Leibniz integral rule

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t t RL α RL α−1 RL α t Dt K(t, τ)dτ = lim τ Dt K(t, τ)+ τ Dt K(t, τ)dτ, α (0, 1), (1.3) 0 τ→t−0 ∈ Z0 Z0 where lower terminal t0 = 0. The following important particular case for convolution operator whenever we have K(t τ)f(τ) instead of K(t, τ), the relationship (1.3) takes the form: −

t t RL α RL α−1 RL α 0Dt K(t τ)f(τ)dτ = lim f(t τ) τ Dt K(τ)+ 0Dτ K(τ)f(t τ)dτ, α (0, 1). (1.4) − τ→t−0 − − ∈ Z0 Z0 It is important to note that the above tools are necessary for checking by substitution method for fractional differential equations with variable and constant coefficients. Fractional differential equations (FDEs) are differential equations involving derivatives of arbitrary (fractional) order. FDEs provide one of the most accurate tools to describe hereditary properties of natural phenomenon. Using fractional derivatives instead of integer-order derivatives allows us for the modeling of a wider variety of behaviours. However, sometimes, FDEs involving one fractional order of differentiation are not sufficient to demonstrate physical processes. Therefore, recently, several authors have studied more general types of fractional-order models, such as multi-term equations [10, 11, 12, 13, 14] and multi-dimensional systems [4, 6, 7, 15, 16, 17, 18]. Multi-term differential equations with fractional-order have been studied and solved using various mathematical methods, of which we mention a few as follows. Luchko and several collaborates [10, 11] have used the method of operational calculus to solve multi-term FDEs with constant coefficients with regard to various types of fractional derivatives. Bazhlekova [12] has considered multi-term fractional relaxation equations with Caputo fractional derivatives by using a Laplace transform technique, and studied the fundamental and impulse-response solutions of the initial value problem (IVP). Kaczorek and Idczak [17] have considered existence and uniqueness results and a Cauchy formula for the analytical solution of the time-varying linear system with Caputo fractional derivative. Pak et al. [13] has recently investigated multi-term FDEs with variable coefficients using a new method to construct analytical solutions. As one of the important special cases of multi-term FDEs, Bagley-Torvik equations have been discussed 1 in terms of analytical [8, 19, 20, 21] and numerical methods [22, 23]. Bagley-Torvik equations with 2 -order 3 or 2 -order derivative describe the motion of real physical systems in a Newtonian fluid [19]. In 1984, Bagley and Torvik [19] have considered the following Cauchy problem under the homogeneous initial conditions:

′′ 2S√µρ my (r)+ CDα y (r)+ ky(r)= g(r), r> 0, (1.5) m 0+ ′ y(0) = y (0) = 0, where C Dα y ( ) is Caputo fractional differential operator of order α = 1 or α = 3 , S- an area of the rigid 0+ · 2 2 plate, µ- viscosity, ρ-fluid density, m-mass, k- spring of stiffness and g( )-an external force. An analytical solution of (1.5) has introduced by Podlubny [8] in the form: ·

r y(r)= G(r τ)g(τ)dτ, r> 0, (1.6) − Z0 with ∞ l 1 ( 1)l k 2S√µρ G r r2l+1 (l) − √r , ( )= − 1 ,2+ 3l m l! m E 2 2 m l X=0

2 (l) where α,β( ) is the lth-derivative of two-parameter Mittag-Leﬄer function. In [20], Mahmudov et al. have studiedE explicit· analytical solutions for several families of generalized multidimensional Bagley-Torvik equations with permutable matrices. In [21], Wang et al. have modiﬁed the following Bagley-Torvik equation

′′ C α y (r)+ µ D0+ y (r)+ y(r)=0, µ,r> 0, (1.7)

1 3 where α = 2 or α = 2 , to the sequential FDEs and introduced a general solution of (1.7) by using the technique related to characteristic roots. The numerical point of view Diethelm and Ford in [22] have used linear multi-steps , Srivastava et al. in [23] have applied wavelet approach to obtain approximate solutions of the Bagley-Torvik equations. Therefore, the plan of this paper is systematized as below. Section 2 is a mathematical preliminary section where we recall main definitions and results from fractional calculus, special functions and necessary lemmas from fractional differential equations. Section 3 is devoted to formulating the Leibniz integral rule for higher order derivatives of Lebesgue integration which depends on parameter in classical sense. In Section 4, we have introduced fractional differentiation under the integral sign in Riemann-Liouville and Caputo sense. Moreover, we have considered the derivative of convolution operator which has more importance for differential equations with classical or fractional order. In Section 5, we have acquired explicit analytical solutions of Bagley-Torvik equations with Riemann-Liouville and Caputo type fractional derivatives in terms of recently defined bivariate Mittag-Leffler type functions in accordance with fractional Green’s function method and tested the candidate solutions by using our newly defined tools which are natural generalization of well-known Leibniz integral rule. At the end, in Section 6 we give the conclusions and future directions.

2. Mathematical preliminaries

We embark on this section by briefly introducing the essential structure of fractional calculus, special functions and fractional differential operators (for the more salient details on the matter, see the textbooks [8, 25, 26, 28, 42, 44]). We begin by defining some notations, Riemann-Liouville and Caputo fractional differentiation operators which are fundamental for fractional calculus and fractional differential equations. Let Rn be Euclidean space and J be some interval of the real line, i.e. J R. We suppose that J = [t ,T ] ⊂ 0 for some t Jˆ and denote Jˆ = (t ,T ). Assume that f : J R is an absolutely continuous function. ∈ 0 → Definition 2.1 ([28, 42, 44]). The Riemann-Liouville derivative operator of fractional order n 1 < α n for n N is defined by − ≤ ∈ t n n d − 1 d − − RLDαg (t)= In αg (t)= (t s)n α 1g(s)ds, t Jˆ, (2.1) t0 t dtn t0 t Γ(n α) dtn − ∈ − tZ0 α where t0 It is the Riemann-Liouville integral operator of order α> 0 which is defined by

t 1 − Iαg (t)= (t s)α 1g(s)ds, t Jˆ. (2.2) t0 t Γ(α) − ∈ Z t0 Furthermore, the following equality holds true: RLDα Iαg (t)= g(t), α> 0, t Jˆ. (2.3) t0 t t0 t ∈ Deﬁnition 2.2 ([8, 25, 27]). The Caputo derivative operator of fractional order n 1 < α n for n N is deﬁned by − ≤ ∈

t n n − d 1 − − d C Dαg (t)= In α g (t)= (t s)n α 1 g(s)ds, t Jˆ. (2.4) t0 t t0 t dtn Γ(n α) − dsn ∈ − tZ0 Moreover, the next relation holds true: CDα Iαg (t)= g(t), α> 0, t Jˆ. (2.5) t0 t t0 t ∈ 3 The relationship between Riemann-Liouville and Caputo fractional derivatives are as follows [8]:

n− 1 (t t )k−αf (k)(t ) C Dαg (t)= RLDαg (t) − 0 0 , n 1 < α n, n N. (2.6) t0 t t0 t − Γ(k α + 1) − ≤ ∈ k=0 X − The following results are useful in solving fractional differential equations. Definition 2.3 ([47, 48]). A function g is said to be exponentially bounded on [0, ) if it satisfies an inequality of the form: ∞

g(t) Meσt, t T, | |≤ ≥ for some real constants M > 0, T > 0 and σ R. ∈ Deﬁnition 2.4 ([47, 48]). If g : [0, ) R is exponentially bounded for t 0, then the Laplace integral transform L g(t) (s) deﬁned by ∞ → ≥ { }

∞ − G(s)= L g(t) (s)= e stg(t)dt, { } Z0 exists for s C and is an analytic function of s for (s) > 0 and Laplace inversion formula is deﬁned as ∈ ℜ − 1 L 1 G(s) (t) := estG(s)ds, { } 2πi LZ where g(t)= L −1 G(s) (t), t 0 and L is a closed contour which enclosing the poles (singularities) of g. { } ≥ Deﬁnition 2.5 ([8]). The Laplace integral transform of Riemann-Liouville fractional derivative of order α (n 1,n], n N is given by [8]: ∈ − ∈ n − − L RLDαy (t) (s)= sαY (s) sk 1 RLDα ky (0), (2.7) 0 t − 0 t k=1 X where Y (s) represents the Laplace transform of the function y(t).

Remark 2.1 ([8]). In the special cases, the Laplace integral transform of the Riemann-Liouville fractional diﬀerentiation is:

• If α (0, 1], then − ∈ L RLDαy (t) (s)= sαY (s) RLDα 1y (0). 0 t − 0 t • If α (1, 2], then ∈ − − L RLDαy (t) (s)= sαY (s) RLDα 1y (0) s RLDα 2y (0). 0 t − 0 t − 0 t Deﬁnition 2.6. The Laplace integral transform of Caputo fractional derivative of order α (n 1,n], n N is given by [8]: ∈ − ∈ n−1 − − L C Dαy (t) (s)= sαY (s) sα k 1 y(k)(0), (2.8) 0 t − k=0 X where Y (s) represents the Laplace transform of the function y(t). Remark 2.2. In the special cases, the Laplace integral transform of the Caputo fractional diﬀerentiation is:

4 • If α (0, 1], then ∈ − L C Dαy (t) (s)= sαY (s) sα 1y , where y = y(0). 0 t − 0 0 • If α (1, 2], then ∈ − − ′ ′ ′ L C Dαy (t) (s)= sαY (s) sα 1y sα 2y , where y = y(0) and y = y (0). 0 t − 0 − 0 0 0 Definition 2.7([47, 48]) . Let f and g be both piece-wise continuous functions on [0, ). Then the integral in ∞ t f g := (f g)(t)= f(t s)g(s)ds, ∗ ∗ − Z0 is called the convolution operator of two functions f and g which is well-defined and finite for any t 0 and it has the commutativity property: ≥ f g = g f. ∗ ∗ Theorem 2.1 ([47, 48]). Suppose that f and g are piece-wise continuous and exponentially bounded functions on [0, ). Then the Laplace transform of convolution operator of two functions f and g, given on [0, ), has the∞ following property : ∞ L (f g) (t) (s)= L f(t) (s)L g(t) (s), s C. { ∗ } { } { } ∈ The Mittag-Leffler function is a generalization of the exponential function, first proposed in 1903 [29] as a single-parameter function of one variable, defined using a convergent infinite series. Extensions to two, three and multi-parameters are well known and thoroughly studied in textbooks such as [26, 31] which are involving single power series in one variable [32, 33, 34]. Extensions to two, three, or more variables, involving correspondingly double, triple, or multiple power series, have been studied more recently [18, 35, 36, 37]. Definition 2.8 ([29]). The classical Mittag-Leffler function is defined by ∞ ti Eα(t)= , α> 0,t R. Γ(iα + 1) ∈ i=0 X Remark 2.3 ([25]). The Mittag-Leffler functions are often used in a form where the variable inside the brackets is not t but a fractional power tα, or even a constant multiple λtα, as follows: ∞ i iα α λ t Eα(λt )= , α> 0,t,λ R. Γ(iα + 1) ∈ i=0 X The two-parameter Mittag-Leffler function [26] is given by ∞ ti Eα,β(t)= , α> 0,β R,t R. Γ(iα + β) ∈ ∈ i=0 X The l-th derivative of two-parameter Mittag-Leffler function [26] is defined by ∞ l i d (l) (i + l)! t Eα,β(t)= E (t)= , l N,α> 0,β R,t R. dtl α,β i! Γ(iα + lα + β) ∈ ∈ ∈ i=0 X The three-parameter Mittag-Leffler function [30] is determined by ∞ (γ) ti Eγ (t)= i , α> 0,β,γ R,t R, α,β Γ(iα + β) i! ∈ ∈ i=0 X Γ(γ+i) where (γ)i is the Pochhammer symbol denoting Γ(γ) . These series are convergent, locally uniformly in τ , provided the α> 0 condition is satisfied. Note that

1 Eα,β(t)= Eα,β(t), Eα,1(t)= Eα(t), E1(t) = exp(t).

5 The next lemma includes Laplace integral transform of three-parameter Mittag-Leﬄer function which will be used throughout the proof of Lemma 2.2.

Lemma 2.1. For α>β> 0, λ R, l N0 = 0, 1, 2,... and (s) > 0, we have: ∈ ∈ { ∞ } ℜ k k(α−β) − 1 − l + k λ t L 1 (τ)= t(l+1)α 1 (sα λsβ )l+1 k Γ(k(α β) + (l + 1)α) k n − o X=0 − (l+1)α−1 l+1 α−β := t Eα−β,(l+1)α(λt ). 1 Proof. By using the Taylor series representation of +1 ,l N of the form (1−t)l ∈ 0 ∞ 1 l + k = tk, t < 1, (1 t)l+1 k | | k − X=0 we achieve that ∞ 1 1 1 1 l + k λ k = = α β l+1 α l+1 λ l+1 (l+1)α α−β (s λs ) (s ) (1 − ) s k s − − sα β k=0 ∞ X l + k λk = . k sk(α−β)+(l+1)α k X=0 Taking inverse Laplace transform of the above function, we get the desired result: ∞ − 1 l + k − 1 L 1 (t)= λk L 1 (t) (sα λsβ)l+1 k sk(α−β)+(l+1)α − k=0 n o X∞ n o l + k tk(α−β)+(l+1)α−1 = λk k Γ(k(α β) + (l + 1)α) k X=0 − (l+1)α−1 l+1 α−β = t Eα−β,(l+1)α(λt ), which is the required result. We have required an extra condition on s for convergence of the binomial type series in the Laplace domain, namely that sα−β > λ . | | However, this condition can be removed at the end, by analytic continuation of both sides of the identity, to give the desired result for all s C satisfying (s) > 0. The proof is complete. ∈ ℜ Definition 2.9 ([35]). We consider the bivariate Mittag-Leffler function defined by ∞ ∞ (δ) ulvk Eδ (u, v)= l+k , α,β> 0,γ,δ R,u,v R. (2.9) α,β,γ Γ(lα + kβ + γ) l!k! ∈ ∈ l k X=0 X=0 If we write u = λtα and v = µtβ for a single variable t, and multiply by a power function tγ−1, we derive the following univariate version: ∞ ∞ l k − (δ) λ µ − tγ 1Eδ (λtα,µtβ)= l+k tlα+kβ+γ 1. (2.10) α,β,γ Γ(lα + kβ + γ) l!k! l k X=0 X=0 Note that when δ = 1, ∞ ∞ l k (1) λ µ − E1 (λtα,µtβ)= l+k tlα+kβ+γ 1 α,β,γ Γ(lα + kβ + γ) l!k! l k X=0 X=0 ∞ ∞ l k (l + k)! λ µ − = tlα+kβ+γ 1 l!k! Γ(lα + kβ + γ) l=0 k=0 X∞ X∞ l k l + k λ µ − = tlα+kβ+γ 1. k Γ(lα + kβ + γ) l k X=0 X=0

6 1 α β α β For simplicity, we denote Eα,β,γ(λt ,µt ) := Eα,β,γ(λt ,µt ) in our results for this paper. Lemma 2.2. For α>β, α>γ, λ, µ R and (s) > 0, the following result holds true: ∈ ℜ ∞ ∞ γ l k lα+k(α−β) − s − − l + k λ µ t L 1 (t)= tα γ 1 sα µsβ λ k Γ(lα + k(α β)+ α γ) l k n − − o X=0 X=0 − − α−γ−1 α α−β = t Eα,α−β,α−γ(λt ,µt ).

sγ Proof. sα−µsβ −λ can be written via a series expansion as follows:

∞ sγ sγ 1 λlsγ = = . sα µsβ λ sα µsβ λ (sα µsβ)l+1 1 sα−µsβ l − − − − X=0 − Then applying Lemma 2.1 to the last expression, we acquire that

∞ sγ λlsγ 1 α β = (l+1)α µ l+1 s µs λ s (1 − ) − − l=0 − sα β X∞ ∞ λlsγ l + k µ k = s(l+1)α k sα−β l=0 k=0 X∞ ∞ X l + k λlµk = . k s(l+1)α+k(α−β)−γ l k X=0 X=0 Taking inverse Laplace transform of the aforementioned function, we attain:

∞ ∞ γ − s l + k − 1 L 1 (t)= µlλkL 1 (t) sα µsβ λ k s(l+1)α+k(α−β)−γ − − l=0 k=0 n o X X ∞ ∞ n o l k lα+k(α−β) − − l + k λ µ t = tα γ 1 k Γ(lα + k(α β)+ α γ) l k X=0 X=0 − − α−γ−1 α α−β = t Eα,α−β,α−γ(λt ,µt ), which is the desired result. We have required extra conditions on s for convergence of the binomial type series in the Laplace domain, namely that

sα−β > µ , | | sα µsβ > λ . | − | | | However, these conditions can be removed at the end, by analytic continuation of both sides of the identity, to give the desired result for all s C satisfying (s) > 0. The proof is complete. ∈ ℜ Lemma 2.3. For any parameters α,β,γ,λ,µ R satisfying α,β > 0 and γ 1 > α , we have ∈ − ⌊ ⌋ C α γ−1 α β γ−α−1 α β 0Dt t Eα,β,γ(λt ,µt ) = t Eα,β,γ−α(λt ,µt ), t> 0. (2.11) h i Proof. We have the following formula for Caputo derivatives of power functions [8, 27]:

− tη ν η Γ(η−ν+1) , η> ν , C ν t ⌊ ⌋ Dt = 0, η =0, 1, 2,..., ν , (2.12) 0 Γ(η + 1)  ⌊ ⌋ undeﬁned, otherwise.



7 Therefore, the given condition γ 1 > α , from (2.12) we can attain − ⌊ ⌋ ∞ ∞ l k lα+kβ+γ−1 − l + k λ µ t C Dα tγ 1E (λtα,µtβ) = C Dα 0 t α,β,γ 0 t k Γ(lα + kβ + γ) "l=0 k=0 # ∞ ∞ X X l + k tlα+kβ+γ−1 = λlµk CDα k 0 t Γ(lα + kβ + γ) l=0 k=0 X∞ X∞ l + k λlµktlα+kβ+γ−α−1 = k Γ(lα + kβ + γ α) l k X=0 X=0 − γ−α−1 α β = t Eα,β,γ−α(λt ,µt ), t> 0.

The proof is complete. Lemma 2.4. For any parameters α,β,γ,λ,µ R satisfying α,β,γ > 0, we have ∈ RL α γ−1 α β γ−α−1 α β 0Dt t Eα,β,γ(λt ,µt ) = t Eα,β,γ−α(λt ,µt ), t> 0. (2.13) h i Proof. We have the following formula for Riemann-Liouville derivatives of power functions [8, 27]:

tη tη−ν RLDν = , ν,η R, η> 1. (2.14) 0 t Γ(η + 1) Γ(η ν + 1) ∈ − − Therefore, given the condition γ > 0, in accordance with (2.14) we will get the same result with (2.11). The proof is complete.

Deﬁnition 2.10. Let λi,µj R, αi,βj R, i =1, 2,...,p, j =1, 2,...,q. Generalized Wright function or ∈ ∈ Fox-Wright function Ψq( ): R R is deﬁned by p · → p ∞ Γ(λi + αik) k (λi, αi)1,p i=1 t pΨq(t)= pΨq t = q . (2.15) (µj ,βj )1,q Q k! k=0 µ β k X Γ( j + j ) j=1 Q The Fox-Wright function was established by Fox [45] and Wright [46]. If the following condition holds

q p

βj αi > 1, − − j=1 i=1 X X then the series in (2.15) is convergent for arbitrary t R. ∈

3. Leibniz integral rule

In this section, we formulate Leibniz integral rule for higher order derivatives on Lebesgue integration. It is known that according to the suitable conditions, we can differentiate under the integral sign for Lebesgue integrals [38]. We begin with the first derivative of a Lebesgue integral on X R. ⊆ Theorem 3.1 ([38]). Assume that X, Y R are intervals. Suppose also that the function f : X Y R satisfies the following assumptions: ⊆ × → (a) For every fixed y Y , the function f( ,y) is measurable on X; (b) The partial derivative∈ ∂ f(x, y) exists· for every interior point (x, y) X Y ; ∂y ∈ × (c) There exists a non-negative integrable function g such that ∂ f(x, y) g(x) exists for every interior ∂y ≤ point (x, y) X Y ; ∈ × (d) There exists y Y such that f(x, y ) is integrable on X. 0 ∈ 0

8 Then for every y Y , the Lebesgue integral ∈ f(x, y)dx

XZ exists. Furthermore, the function F : Y R, deﬁned by → F (y)= f(x, y)dx

for every y Y, is differentiable at every interior point of Y , and the derivative of F (y) satisfies ∈ ′ ∂ F (y)= f(x, y)dx. ∂y XZ The well-known rule for the differentiation of an integral depending on a parameter with the upper limit also depends on the same parameter, namely:

Corollary 3.1 ([39]). If X = (y0,y) and assumptions of Theorem 3.1 are fulﬁlled, then the following relation holds true for all y Y : ∈ y y d ∂ f(x, y)dx = f(x, y)dx + lim f(x, y), y X. (3.1) dy ∂y x→y−0 ∈ yZ0 yZ0

So, the formula of diﬀerentiation under the integral sign for K(t,s) with respect to t is

t t d ∂ K(t,s)ds = K(t,s)ds + lim K(t,s), t Jˆ. (3.2) dt ∂t s→t−0 ∈ tZ0 tZ0

Using the formula (3.2), we deﬁne the second-order derivative of the integral depending on t:

t t t d2 d d d ∂ K(t,s)ds = K(t,s)ds = lim K(t,s)+ K(t,s)ds dt2 dt dt  dt s→t−0 ∂t  tZ0 tZ0 tZ0   t   d d ∂ = lim K(t,s)+ K(t,s)ds dt s→t−0 dt ∂t tZ0 t d ∂ ∂2 = lim K(t,s) + lim K(t,s)+ K(t,s)ds dt s→t−0 s→t−0 ∂t ∂t2 tZ0 t 2 l−1 2−l 2 d ∂ ∂ ˆ = − lim − K(t,s)+ K(t,s)ds, t J. dtl 1 s→t−0 ∂t2 l ∂t2 ∈ l=1 X tZ0 Then, the third-order diﬀerentiation of the integral will be:

t t d3 d d2 K(t,s)ds = K(t,s)ds dt3 dt dt2  tZ0 tZ0   t d d ∂ ∂2 = lim K(t,s) + lim K(t,s)+ K(t,s)ds dt dt s→t−0 s→t−0 ∂t ∂t2  tZ0   9 t d2 d ∂ d ∂2 = lim K(t,s)+ lim K(t,s)+ K(t,s)ds dt2 s→t−0 dt s→t−0 ∂t dt ∂t2 tZ0 d2 d ∂ = lim K(t,s)+ lim K(t,s) dt2 s→t−0 dt s→t−0 ∂t t ∂2 ∂3 + lim K(t,s)+ K(t,s)ds s→t−0 ∂t2 ∂t3 tZ0 t 3 l−1 3−l 3 d ∂ ∂ ˆ = − lim − K(t,s)+ K(t,s)ds, t J. dtl 1 s→t−0 ∂t3 l ∂t3 ∈ l=1 X tZ0 Thus, we establish nth derivative of the integral for n N which depends on t by recursively as follows: ∈

t t dn d dn−1 K(t,s)ds = K(t,s)ds dtn dt dtn−1  tZ0 tZ0 dn−1 dn−2 ∂  d ∂n−2 = − lim K(t,s)+ − lim K(t,s)+ + lim − K(t,s) dtn 1 s→t−0 dtn 2 s→t−0 ∂t ··· dt s→t−0 ∂tn 2 t ∂n−1 ∂n + lim − K(t,s)+ K(t,s)ds s→t−0 ∂tn 1 ∂tn tZ0 t n l−1 n−l n d ∂ ∂ ˆ = − lim − K(t,s)+ K(t,s)ds, t J. dtl 1 s→t−0 ∂tn l ∂tn ∈ l=1 X tZ0 In the next theorem, we state and prove the Leibniz rule for higher order derivatives. Theorem 3.2. Let the function K : J J R be such that the following assumptions are fulﬁlled: − J × →∂n 1 J J (a) For every ﬁxed t , the function ∂tn−1 K(t,s) is measurable on and integrable on with respect to for some t∗ J; ∈ ∈ ∂n Jˆ Jˆ (b) The partial derivative ∂tn K(t,s) exists for every interior point (t,s) ; n ∈ × (c) There exists a non-negative integrable function g such that ∂ K(t,s) g(s) for every interior point ∂tn ≤ (t,s) Jˆ Jˆ; − − ∈ × dl 1 ∂n l ˆ ˆ (d) The derivative l−1 lim n−l K(t,s), l =1, 2,...,n exists for every interior point (t,s) J J. dt s→t−0 ∂t ∈ × Then, the following relation holds true for nth derivative under Lebesgue integration for n N: ∈ t t n n l−1 n−l n d d ∂ ∂ ˆ K(t,s)ds = − lim − K(t,s)+ K(t,s)ds, t J. (3.3) dtn dtl 1 s→t−0 ∂tn l ∂tn ∈ l=1 tZ0 X tZ0 Proof. Using mathematical induction principle, we prove above theorem. It is obvious that the equation (3.3) is true for n = 1 [8]:

t t d ∂ K(t,s)ds = lim K(t,s)+ K(t,s)ds. dt s→t−0 ∂t tZ0 tZ0 We assume that (3.3) holds true for n = k:

t t k k l−1 k−l k d d ∂ ∂ ˆ K(t,s)ds = − lim − K(t,s)+ K(t,s)ds, t J. dtk dtl 1 s→t−0 ∂tk l ∂tk ∈ l=1 tZ0 X tZ0

10 We prove that (3.3) is true for n = k + 1:

t t dk+1 d dk K(t,s)ds = K(t,s)ds dtk+1 dt dtk  tZ0 tZ0   t d k dl−1 ∂k−l ∂k = − lim − K(t,s)+ K(t,s)ds dt  dtl 1 s→t−0 ∂tk l ∂tk  l=1 X tZ0 d  ∂k−1 d ∂k−2  dk−1 = lim − K(t,s)+ lim − K(t,s)+ + − lim K(t,s) dt s→t−0 ∂tk 1 dt s→t−0 ∂tk 2 ··· dtk 1 s→t−0 t d ∂k + K(t,s)ds dt ∂tk tZ0 d ∂k−1 d2 ∂k−2 dk = lim − K(t,s)+ lim − K(t,s)+ + lim K(t,s) dt s→t−0 ∂tk 1 dt2 s→t−0 ∂tk 2 ··· dtk s→t−0 t ∂k ∂k+1 + lim K(t,s)+ K(t,s)ds s→t−0 ∂tk ∂tk+1 tZ0 t k+1 dl−1 ∂k−l+1 ∂k ∂k+1 = − lim − K(t,s) + lim K(t,s)+ K(t,s)ds dtl 1 s→t−0 ∂tk l+1 s→t−0 ∂tk ∂tk+1 l=2 X tZ0 t k+1 l−1 k−l+1 k+1 d ∂ ∂ ˆ = − lim − K(t,s)+ K(t,s)ds, t J. dtl 1 s→t−0 ∂tk l+1 ∂tk+1 ∈ l=1 X tZ0 Therefore, the formula (3.3) holds true for all n N and t Jˆ. ∈ ∈ The following important particular case must be deﬁned for convolution operator of the functions f and g.

Corollary 3.2. If K(t,s)= f(t s)g(s) and t0 =0, and assumptions of Theorem 3.2 are satisﬁed, then the following relation is true for any−n N: ∈ t dn n ∂n−l dl−1 f(t s)g(s)ds = lim − f(t s) − lim g(s) dtn − s→t−0 ∂tn l − dtl 1 s→t−0 l Z0 X=1 t ∂n + f(t s)g(s)ds, t> 0. (3.4) ∂tn − Z0

Proof. If we write f(t s)g(s) instead of K(t,s) in (3.3), then we obtain − t dn n dl−1 ∂n−l f(t s)g(s)ds = − lim − f(t s)g(s) dtn − dtl 1 s→t−0 ∂tn l − l Z0 X=1 t ∂n + f(t s)g(s)ds ∂tn − Z0 n dl−1 ∂n−l = − lim − f(t s) lim g(s) dtl 1 s→t−0 ∂tn l − s→t−0 l X=1

11 t ∂n + f(t s)g(s)ds ∂tn − Z0 n ∂n−l dl−1 = lim − f(t s) − lim g(s) s→t−0 ∂tn l − dtl 1 s→t−0 l X=1 t ∂n + f(t s)g(s)ds, t> 0. ∂tn − Z0 Thus, the proof is complete.

4. Fractional Leibniz integral rules

Now, we are starting to prove fractional Leibniz integral rule for Riemann-Liouville fractional derivative of order α (n 1,n],n N. For this, firstly, let us consider partial Riemann-Liouville fractional differentiation operator∈ of order− n ∈1 < α n, n N with respect to t of a function K(t,s) of two variables (t,s) J J, K : J J R, defined− by [28,≤ 42, 44]:∈ ∈ × × → t n n ∂ − 1 ∂ − − RL,tDαK(t,s)= tIn αK(t,s)= (t s)n α 1K(s, τ)ds, t Jˆ, (4.1) t0 t ∂tn t0 t Γ(n α) ∂tn − ∈ − tZ0 t α where t0 It is the partial Riemann-Liouville integral operator of order α> 0 which is given by:

t 1 − tIαK(t,s)= (t s)α 1K(s, τ)ds, for t Jˆ. t0 t Γ(α) − ∈ tZ0

The following important result in the theory of fractional calculus was ﬁrst proposed by Podlubny [8] for α (0, 1] in Riemann-Liouville sense as follows: ∈ t t RL α t 1−α RL,t α Dt K(t,s)ds = lim It K(t,s)+ Dt K(t,s)ds, t>t0. t0 s→t−0 s s tZ0 tZ0

Now, we are going to state and prove the following theorem for more general case where α (n 1,n],n N which is more useful tool for the testing particular solution of inhomogeneous linear multi-order∈ − differential∈ equations with variable coefficients. Note that Matychyn has proposed [41] Leibniz integral rule for Riemann- Liouville derivative of order 0 < α 1 on Lebesgue integration. ≤ Theorem 4.1. Let the function K : J J R be such that the following assumptions are fulfilled: × → RL,t J ˆ α−1 J J (a) For every fixed t , the function K(t,s)= sDt K(t,s) is measurable on and integrable on with respect to some t∗ ∈J; (b) The partial derivative∈ RL,tDαK(t,s) exists for every interior point (t,s) Jˆ Jˆ; s t ∈ × (c) There exists a non-negative integrable function g such that RL,tDαK(t,s) g(s) for every interior s t ≤ point (t,s) Jˆ Jˆ; ∈ × l−1 d RL,t α−l Jˆ Jˆ (d) The derivative l−1 lim Dt K(t,s), l =1, 2,...,n exists for every interior point (t,s) ; dt s→t−0 s ∈ × Then, the following relation holds true for fractional derivative in Riemann-Liouville sense under Lebesgue integration:

t t n l−1 RL α d RL,t α−l RL,t α ˆ Dt K(t,s)ds = − lim Dt K(t,s)+ Dt K(t,s)ds, t J. (4.2) t0 dtl 1 s→t−0 s s ∈ l=1 tZ0 X tZ0

12 Proof. Using the Deﬁnition 2.1 and Fubini’s theorem [38], we have

t t τ n 1 d − − RLDα K(t,s)ds = (t τ)n α 1dτ K(τ,s)ds t0 t Γ(n α) dtn − tZ0 − tZ0 tZ0 t τ n 1 d − − = (t τ)n α 1K(τ,s)dsdτ Γ(n α) dtn − − tZ0 tZ0 t t n 1 d − − = (t τ)n α 1K(τ,s)dτds Γ(n α) dtn − − tZ0 Zs t t n 1 d − − = ds (t τ)n α 1K(τ,s)dτ Γ(n α) dtn − − tZ0 Zs t t n d 1 − − = (t τ)n α 1K(τ,s)dτ ds dtn Γ(n α) −  tZ0 − Zs t   n d t − = In αK(t,s)ds. dtn s t tZ0 Using the formula (3.3) for the last part of above expression, we get a desired result:

t t n l−1 n−l n RL α d ∂ t n−α ∂ t n−α Dt K(t,s)ds = − lim − It K(t,s)+ It K(t,s)ds t0 dtl 1 s→t−0 ∂tn l s ∂tn s l=1 tZ0 X tZ0 t n l−1 d RL,t α−l RL,t α ˆ = − lim Dt K(t,s)+ Dt K(t,s)ds, t J. dtl 1 s→t−0 s s ∈ l=1 X tZ0

Corollary 4.1. If we have K(t,s)= f(t s)g(s), t0 =0 and assumptions of Theorem 4.1 are fulﬁlled, then following equality holds true for convolution− operator in Riemann-Liouville sense for any n N: ∈

t n l−1 RL α RL,t α−l d Dt f(t s)g(s)ds = lim Dt f(t s) − lim g(s) 0 − s→t−0 s − dtl 1 s→t−0 l Z0 X=1 t + RL,tDαf(t s)g(s)ds, t> 0. (4.3) s t − Z0 Proof. If we write f(t s)g(s) instead of K(t,s) in (4.2), then we obtain − t n l−1 n−l RL α d ∂ t n−α Dt f(t s)g(s)ds = − lim − It f(t s)g(s) 0 − dtl 1 s→t−0 ∂tn l s − l Z0 X=1 t + RL,tDαf(t s)g(s)ds s t − Z0 n n−l ∂ t n−α = lim − It f(t s) lim g(s) s→t−0 ∂tn l s − s→t−0 l X=1

13 t + RL,tDαf(t s)g(s)ds s t − Z0 n l−1 RL,t α−l d = lim Dt f(t s) − lim g(s) s→t−0 s − dtl 1 s→t−0 l X=1 t + RL,tDαf(t s)g(s)ds, t> 0. s t − Z0 Thus, the proof is complete. Then, we are going to introduce fractional differentiation under the integral sign in Caputo sense which will be useful for checking the candidate solutions of fractional differential equations with multi-orders. For this, firstly, let us consider partial Caputo fractional differentiation operator of order n 1 < α n, n N with respect to t of a function K(t,s) of two variables (t,s) J J, K : J J R, defined− by [28,≤ 42, 44]:∈ ∈ × × → t n n − ∂ 1 − − ∂ C,tDαK(t,s)= tIn α K(t,s)= (t s)n α 1 K(s, τ)ds, t Jˆ, (4.4) t0 t t0 t ∂tn Γ(n α) − ∂sn ∈ − tZ0 Matychyn and Onyshchenko [40] showed that the fractional Leibniz integral rule for Caputo fractional derivative coincide with Riemann-Liouville one when α (0, 1]: ∈ t t C α t 1−α RL,t α ˆ Dt K(t,s)ds = lim It K(t,s)+ Dt K(t,s)ds, t J. t0 s→t−0 s s ∈ tZ0 tZ0

More generally, the fractional Leibniz integral rule for fractional derivative of order α (n 1,n], n 2 in Caputo sense is stated and proved in the following theorem. ∈ − ≥ Theorem 4.2. Let the function K : J J R be such that the following assumptions are fulﬁlled. × → C,t J ˆ α−1 J (a) For every ﬁxed t , the function K(t,s) = sDt K(t,s) is measurable and integrable on with respect to some t∗ J; ∈ (b) The partial∈ derivative C,tDαK(t,s) exists for every interior point (t,s) Jˆ Jˆ; s t ∈ × (c) There exists a non-negative integrable function g such that C,tDαK(t,s) g(s) for every interior s t ≤ point (t,s) Jˆ Jˆ; ∈ × l−1 n−l t n−α d ∂ N (d) The integral It l−1 lim n−l K(t,s) , l =1, 2,...,n, n exists for every interior point t0 dt s→t−0 ∂t ∈ (t,s) Jˆ Jˆ; Then,∈ × the following relation holds true for fractional derivative in Caputo sense under Lebesgue integration:

t t n l−1 n−l C α t n−α d ∂ C,t α ˆ Dt K(t,s)ds = It − lim − K(t,s) + Dt K(t,s)ds, t J. (4.5) t0 t0 dtl 1 s→t−0 ∂tn l s ∈ (l=1 ) tZ0 X tZ0 Proof. Using the Deﬁnition 2.2, Fubini’s theorem [38], and the formula (3.3) we have

t t τ n 1 − − d C Dα K(t,s)ds = (t τ)n α 1dτ K(τ,s)ds t0 t Γ(n α) − dτ n tZ0 − tZ0 tZ0 t n l−1 n−l 1 n−α−1 d ∂ = (t τ) − lim − K(τ,s)dτ Γ(n α) − dτ l 1 s→τ−0 ∂τ n l l=1 − tZ0 X

14 t τ n 1 − − ∂ + (t τ)n α 1dτ K(τ,s)ds Γ(n α) − ∂τ n − tZ0 tZ0 n l−1 n−l t n−α d ∂ = It − lim − K(t,s) t0 dtl 1 s→t−0 ∂tn l ( l ) X=1 t τ n 1 − − ∂ + (t τ)n α 1 K(τ,s)dsdτ Γ(n α) − ∂τ n − tZ0 tZ0 n l−1 n−l t n−α d ∂ = It − lim − K(t,s) t0 dtl 1 s→t−0 ∂tn l ( l ) X=1 t t n 1 − − ∂ + (t τ)n α 1 K(τ,s)dτds Γ(n α) − ∂τ n − tZ0 Zs n l−1 n−l t n−α d ∂ = It − lim − K(t,s) t0 dtl 1 s→t−0 ∂tn l ( l ) X=1 t t n 1 − − ∂ + ds (t τ)n α 1 K(τ,s)dτ Γ(n α) − ∂τ n − tZ0 Zs n l−1 n−l t n−α d ∂ = It − lim − K(t,s) t0 dtl 1 s→t−0 ∂tn l ( l ) X=1 t t n 1 − − ∂ + (t τ)n α 1 K(τ,s)dτ ds Γ(n α) − ∂τ n  tZ0 − Zs  n l−1 n−l  t n−α d ∂ = It − lim − K(t,s) t0 dtl 1 s→t−0 ∂tn l ( l ) X=1 t n t − ∂ + In α K(t,s)ds s t ∂tn tZ0 n l−1 n−l t n−α d ∂ = It − lim − K(t,s) t0 dtl 1 s→t−0 ∂tn l ( l ) X=1 t + C,tDαK(t,s)ds, t Jˆ. s t ∈ tZ0 Therefore, the proof is compete. However, the fractional Leibniz integral rule for Caputo derivative of order 0 < α 1 is different from the general case which is given in the relation (4.5). ≤ Theorem 4.3. Let the function K : J J R be such that the following assumptions are fulfilled. J × ˆ→ t 1−α J J (a) For every fixed t , the function K(t,s)= sIt K(t,s) is measurable on and integrable on with respect to some t∗ J; ∈ (b) The partial∈ derivative RL,tDαK(t,s) exists for every interior point (t,s) Jˆ Jˆ; s t ∈ × (c) There exists a non-negative integrable function g such that RL,tDαK(t,s) g(s) for every interior s t ≤ Jˆ Jˆ point (t,s) ; Then, the∈ Caputo× fractional derivative under Lebesgue integration coincides with the fractional differentiation of an integral in Riemann-Liouville sense for 0 < α 1: ≤

15 t t C α t 1−α RL,t α ˆ Dt K(t,s)ds = lim It K(t,s)+ Dt K(t,s)ds, t J. (4.6) t0 s→t−0 s s ∈ tZ0 tZ0 Proof. In accordance the formula (2.6), it is obvious for α (0, 1]: ∈ −α C α RL α (t t0) D g (t)= D g (t) − g(t0), t Jˆ. (4.7) t0 t t0 t − Γ(1 α) ∈ − Since the formula (4.7), we can attain that

t t t (t t )−α CDα K(t,s)ds = RLDα K(t,s)ds K(t,s)ds − 0 t0 t t0 t −   × Γ(1 α) tZ0 tZ0 tZ0 t=t0 − t   = RLDα K(t,s)ds, t Jˆ. t0 t ∈ tZ0

Therefore, fractional Leibniz integral rule for Caputo derivative is identical with the Riemann-Liouville one whenever 0 < α 1: ≤ t t C α t 1−α RL,t α ˆ Dt K(t,s)ds = lim It K(t,s)+ Dt K(t,s)ds, t J. t0 s→t−0 s s ∈ tZ0 tZ0

It is important to introduce Caputo fractional derivative of convolution operator in general sense which is so accurate tool for testing particular solution of Caputo type multi-term FDEs.

Corollary 4.2. If we have K(t,s)= f(t s)g(s), t0 =0, and assumptions of Theorem 4.3 are fulﬁlled, then following equality holds true for convolution− operator in Caputo’s sense of order α (n 1,n], where n 2: ∈ − ≥ t t n n−l l−1 C α t n−α ∂ d C,t α Dt f(t s)g(s)ds = It lim − f(t s) − lim g(s) + Dt f(τ s)g(s)ds, t> 0. 0 − 0 s→t−0 ∂tn l − dtl 1 s→t−0 s − (l ) Z0 X=1 Z0 (4.8) Proof. If we write f(t s)g(s) instead of K(t,s) in (4.8), then we acquire − t n l−1 n−l C α t n−α d ∂ Dt f(t s)g(s)ds = It − lim − f(t s)g(s) 0 − 0 dtl 1 s→t−0 ∂tn l − (l ) Z0 X=1 t + C,tDαf(t s)g(s)ds s t − Z0 n l−1 n−l t n−α d ∂ = It − lim − f(t s) lim g(s) 0 dtl 1 s→t−0 ∂tn l − s→t−0 (l ) X=1 t + C,tDαf(t s)g(s)ds s t − Z0 n n−l l−1 t n−α ∂ d = It lim − f(t s) − lim g(s) 0 s→t−0 ∂tn l − dtl 1 s→t−0 (l ) X=1

16 t + C,tDαf(t s)g(s)ds, t> 0. s t − Z0 Thus, the proof is complete.

Corollary 4.3. If we have K(t,s)= f(t s)g(s), t0 =0, and assumptions of Theorem 4.3 are fulﬁlled, then following equality holds true for convolution− operator in Caputo’s sense for α (0, 1]: ∈ t t C α t 1−α RL,t α Dt f(t s)g(s)ds = lim It f(t s) lim g(s)+ Dt f(t s)g(s)ds, t> 0. (4.9) 0 − s→t−0 s − s→t−0 s − Z0 Z0 Proof. If we make use of the substitution K(t,s) = f(t s)g(s) in the relation (4.6), the proof is straight- forward. So, we omit it here. − Theorem 4.4. The relationship between Leibniz integral rule for Riemann-Liouville and Caputo fractional diﬀerentiation operators of order n 1 < α n, n 2 holds true: − ≤ ≥ − t t n 1 i l−1 i−l i−α RL α C α d ∂ (t t0) ˆ Dt K(t,s)ds = Dt K(t,s)ds + − lim − K(t,s) − ,t J. t0 t0 dtl 1 s→t−0 ∂ti l × Γ(i α + 1) ∈ Zt0 Zt0 i=1 l=1 t=t0 − X X (4.10) Proof. Using the relationship between Riemann-Liouville and Caputo fractional derivatives (2.6), we get

n−1 t t (t t )i−α di t RLDα K(t,s)ds = C Dα K(t,s)ds + 0 K(t,s)ds t0 t t0 t − i t0 t0 Γ(i α + 1) dt t0 t=t0 Z Z i=0 − h Z i X− t n 1 i−α i l−1 i−l C α (t t0) d ∂ = Dt K(t,s)ds + − − lim − K(t,s) t0 l 1 s→t−0 i l t0 Γ(i α + 1) dt ∂t t=t0 Z i=1 − h l=1 i X − X −α t n 1 i−α t i (t t0) (t t0) ∂ + − K(t,s)ds + − i K(t,s)ds Γ(1 α) t0 Γ(i α + 1) t0 ∂t t=t0 − Z t=t0 i=0 − h Z i − X t n 1 i l−1 i−l i−α C α d ∂ (t t0) ˆ = Dt K(t,s)ds + − lim − K(t,s) − ,t J. t0 dtl 1 s→t−0 ∂ti l × Γ(i α + 1) ∈ Zt0 i=1 l=1 t=t0 − X X (4.11)

Corollary 4.4. If we replace K(t,s) with f(t s)g(s) and consider t0 = 0 for lower bound of the integral in Theorem 4.4, the relationship between Riemann-Liou− ville and Caputo type Leibniz integral rules for convolution operator of the functions f and g holds true for n 1 < α n,n 2: − ≤ ≥

t t RL α C α 0Dt f(t s)g(s)ds = 0Dt f(t s)g(s)ds 0 − 0 − Z − Z n 1 i ∂i−l dl−1 t−α + lim − f(t s) − lim g(s) , t> 0. (4.12) s→t−0 ∂ti l − dtl 1 s→t−0 Γ(1 α) i=1 l t=0 X X=1 − Corollary 4.5. The fractional Leibniz rule for Riemann-Liouville and Caputo type fractional diﬀerential operators coincides for 0 < α 1: ≤ t t RLDα K(t,s)ds = C Dα K(t,s)ds, t Jˆ, (4.13) t0 t t0 t ∈ Zt0 Zt0 t t RLDα f(t s)g(s)ds = C Dα f(t s)g(s)ds, t> 0. (4.14) 0 t − 0 t − Z0 Z0

17 5. Fractional Green’s function method

The Laplace transform is a convenient technique for solving the Cauchy problem associated with multi- term FDEs with constant coeﬃcients. For instance, let us consider linear in-homogeneous FDE with multi- orders in Caputo’s sense and constant coeﬃcients:

C αn C αn−1 C αn−2 C α1 0Dt + λ1 0Dt + λ2 0Dt + ... + λn−1 0Dt + λn y(t)= g(t), t> 0, (5.1) under the homogeneous initial conditions:

y(k)(0) = 0, k =0, 1,...,n 1, (5.2) − C αi where D y( ),i =1, 2,...,n, are the Caputo fractional differentiation operators of orders i 1 αi i, 0 (·) · − ≤ ≤ λi R for i =1, 2,...,n denote constants and g C([0, ), R) is the continuous force or input function. ∈The classical analogue of the same problem is∈ considered∞ by Miller in [24]. Let us consider IVP for n-th order linear differential equation with constant coefficients:

n n−1 n−2 D + λ1D + λ2D + ... + λn−1D + λn y(t)=0, t> 0, (5.3) with zero initial conditions Dky(0) = 0, 0 k n 1. (5.4) ≤ ≤ − The fractional Green’s function is a very useful and applicable practical as well as theoretical tool for solving the IVP (5.1)-(5.2) for multi-order FDE. If we let

n n−1 n−2 P (x)= x + λ1x + λ2x + ...λn−1x + λn (5.5) be a polynomial which is related to the equation (5.3), then according to the Laplace transform method, the unique solution of the following diﬀerential system:

P (D)y(t)= g(t), t> 0, Dky(0) = 0, k =0, 1,...,n 1, ( − can be represented in terms of a convolution integral

t y(t)= H(t s)g(s)ds, t> 0, (5.6) − Z0 where H( ) is the Green or weight function associated with the diﬀerential operator P (D) that is evaluated by taking· inverse Laplace transform of the transfer function.

5.1. Applications of fractional Leibniz rules In this subsection, we study applications of the fractional Leibniz integral rule in Riemann-Liouville and Caputo sense using the generalized Bagley-Torvik equations. Moreover, we have used Leibniz integral rule for checking candidate solution of the oscillator equation in classical sense. In the following cases, to obtain analytical representation of solutions for the Cauchy problem we will apply fractional Green’s function method as we mentioned in Section 5. Case 1: We consider the IVP for generalized Bagley-Torvik equations with Riemann-Liouville fractional derivatives of order 1 < α 2 and 0 <β 1 in the form of: ≤ ≤ RLDαy (t) µ RLDβy (t) λy(t)= g(t), t> 0, 0 t − 0 t − (5.7) 2−α 1−α ( I y(t) t = I y(t)t =0, λ,µ R. 0 t | =0 0 t | =0 ∈

18 Theorem 5.1. A unique solution y C2([0, ), R) of the Cauchy problem (5.7) has the following formula: ∈ ∞ t α−1 α α−β y(t)= (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds. (5.8) − − − Z0 Proof. We assume that (5.7) has a unique solution y(t) and g(t) is continuous on [0, ) and exponentially RL α RL β ∞ bounded, then y(t), 0Dt y (t), and 0Dt y (t) are exponentially bounded, thus their Laplace transform exist. Applying Laplace integral transform for Riemann-Liouville fractional derivative using the formula (2.7) to the both sides of (5.7) yields: sα µsβ λ Y (s)= G(s). (5.9) − − Then we solve (5.9) with respect to Y (s), G(s) Y (s)= . (5.10) sα µsβ λ − − Taking inverse Laplace transform of (5.10) and applying Lemma 2.2, we ﬁnd an explicit representation of solution to (5.7):

t α−1 α α−β y(t)= (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds. (5.11) − − − Z0

Verification by substitution. Having found explicit form for y(t), it remains to confirm that y(t) is an analytical solution of (5.7) indeed. Firstly, for make the use of checking by substitution, we apply fractional Leibniz integral rule in Riemann-Liouville sense for the first and second terms of (5.7). Then the first term will be as follows:

t RL α RL α α−1 α α−β D y (t)= D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t 0 t − − − Z 0 2 l−1 RL,t α−l α−1 α α−β d = lim Dt (t s) Eα,α−β,α(λ(t s) ,µ(t s) ) − lim g(s) s→t−0 0 − − − dtl 1 s→t−0 l X=1 t RL,t α α−1 α α−β + D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 RL,t α−1 α−1 α α−β = lim Dt (t s) Eα,α−β,α(λ(t s) ,µ(t s) ) lim g(s) s→t−0 0 − − − s→t−0 RL,t α−2 α−1 α α−β d + lim 0Dt (t s) Eα,α−β,α(λ(t s) ,µ(t s) ) lim g(s) s→t−0 − − − dt s→t−0 t RL,t α α−1 α α−β + D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 α α−β = lim Eα,α−β,1(λ(t s) ,µ(t s) ) lim g(s) s→t−0 − − s→t−0 α α−β d + lim (t s)Eα,α−β,2(λ(t s) ,µ(t s) ) lim g(s) s→t−0 − − − dt s→t−0 t RL,t α α−1 α α−β + D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds. 0 t − − − Z0

19 From now on, we apply Pascal’s rule for binomial coeﬃcients to the ﬁrst term of above expression and the limit of the second term is equal to zero as s t 0, we obtain → −

t RL α RL α α−1 α α−β D y (t)= D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t 0 t − − − Z 0 α α α−β = g(t)+ λ lim (t s) Eα,α−β,α+1(λ(t s) ,µ(t s) ) lim g(s) s→t−0 − − − s→t−0 α−β α α−β + µ lim (t s) Eα,α−β,α−β+1(λ(t s) ,µ(t s) ) lim g(s) s→t−0 − − − s→t−0 t RL,t α α−1 α α−β + D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 t RL,t α α−1 α α−β = g(t)+ D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds. 0 t − − − Z0 Now, using Lemma 2.4 and again applying Pascal’s rule for the expression under above integral, we attain

t RL,t α α−1 α α−β D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 t −1 α α−β = (t s) Eα,α−β, (λ(t s) ,µ(t s) )g(s)ds − 0 − − Z0 t ∞ ∞ l + k λlµk(t s)lα+k(α−β)−1 = − g(s)ds k Γ(lα + k(α β)) l k Z0 X=0 X=0 − t t ∞ ∞ (t s)−1 l + k 1 λlµk(t s)lα+k(α−β)−1 = − g(s)ds + − − g(s)ds Γ(0) k Γ(lα + k(α β)) l k Z0 Z0 X=1 X=0 − t ∞ ∞ l + k 1 λlµk(t s)lα+k(α−β)−1 + − − g(s)ds k 1 Γ(lα + k(α β)) l k Z0 X=0 X=1 − − t ∞ ∞ l + k λl+1µk(t s)(l+1)α+k(α−β)−1 = − g(s)ds k Γ((l + 1)α + k(α β)) l k Z0 X=0 X=0 − t ∞ ∞ l + k λlµk+1(t s)lα+(k+1)(α−β)−1 + − g(s)ds k Γ(lα + (k + 1)(α β)) l k Z0 X=0 X=0 − t α−1 α α−β =λ (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds − − − Z0 t α−β−1 α α−β +µ (t s) Eα,α−β,α−β(λ(t s) ,µ(t s) )g(s)ds. − − − Z0 Therefore, we have

t RL α α−1 α α−β D y (t)= g(t)+ λ (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z 0

20 t α−β−1 α α−β + µ (t s) Eα,α−β,α−β(λ(t s) ,µ(t s) )g(s)ds. (5.12) − − − Z0

Similarly, the second term of (5.7) will be

t RL β RL β α−1 α α−β D y (t)= D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t 0 t − − − Z0 t α−β−1 α α−β = (t s) Eα,α−β,α−β(λ(t s) ,µ(t s) )g(s)ds. (5.13) − − − Z0

Taking linear combination of and (5.12) and (5.13) together with (5.8), we get the desired result. Case 2: We consider the Cauchy problem for generalized Bagley-Torvik equations which is the special case of Caputo type fractional multi-term diﬀerential equations with constant coeﬃcients of order 1 < α 2 and 0 <β 1 in the form of: ≤ ≤ C α C β 0Dt y (t) µ 0Dt y (t) λy(t)= g(t), t> 0, ′ − − (5.14) (y(0) = y (0) = 0, λ,µ R. ∈ Theorem 5.2. A unique solution y C2([0, ), R) of the Cauchy problem (5.14) has the following formula: ∈ ∞ t α−1 α α−β y(t)= (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds. (5.15) − − − Z0

Proof. We assume that (5.14) has a unique solution y(t) and g(t) is continuous on [0, ) and exponentially C α C β ∞ bounded, then y(t), 0Dt y (t), and 0Dt y (t) are exponentially bounded, thus their Laplace transform exist. Applying the formula of Laplace transform for Caputo fractional derivative (2.8) to the both sides of (5.14) yields:

sα µsβ λ Y (s)= G(s). (5.16) − − Then we solve (5.16) with respect to Y (s),

G(s) Y (s)= . (5.17) sα µsβ λ − − Taking inverse Laplace transform of (5.17) and applying Lemma 2.2, we ﬁnd an explicit representation of solution to (5.14):

t α−1 α α−β y(t)= (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds. (5.18) − − − Z0

Remark 5.1. Since initial conditions equal to zero, in accordance with the formula (2.6), analytical solutions should be coincide with each other for the Cauchy problems in Riemann-Liouville (5.7) and Caputo (5.14) senses. Veriﬁcation by substitution. Having found explicit form for y(t), it remains to conﬁrm that y(t) is an analytical solution of (5.14) indeed.

21 Now, we again make use of checking by substitution via Caputo fractional Leibniz integral rule. In this α−1 C α (t−s) case, we apply ﬁrst Pascal’s rule before applying fractional Leibniz rule since sDt Γ(α) is undeﬁned in accordance with (2.12). Then according to the formula (2.3), we obtain

t C α C α α−1 α α−β D y (t)= D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t 0 t − − − Z 0 t C α 2α−1 α α−β = g(t)+ λ D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0 t C α 2α−β−1 α α−β + µ D (t s) Eα,α−β, α−β(λ(t s) ,µ(t s) )g(s)ds. (5.19) 0 t − 2 − − Z0 Then using the formula for fractional Leibniz integral rule in Caputo sense (4.8) of order 1 < α 2, we have ≤

t C α 2α−1 α α−β D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0 2 2−l l−1 t 2−α ∂ 2α−1 α α−β d = It lim − (t s) Eα,α−β,2α(λ(t s) ,µ(t s) ) − lim g(s) 0 s→t−0 ∂t2 l − − − dtl 1 s→t−0 l h X=1 i t C,t α 2α−1 α α−β + D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0

t 2−α ∂ 2α−1 α α−β = It lim (t s) Eα,α−β,2α(λ(t s) ,µ(t s) ) lim g(s) 0 s→t−0 ∂t − − − s→t−0 h i t 2−α 2α−1 α α−β d + It lim (t s) Eα,α−β,2α(λ(t s) ,µ(t s) ) lim g(s) 0 s→t−0 − − − dt s→t−0 t h i C,t α 2α−1 α α−β + D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0 t 2−α 2α−2 α α−β = It lim (t s) Eα,α−β,2α−1(λ(t s) ,µ(t s) ) lim g(s) 0 s→t−0 − − − s→t−0 h i t 2−α 2α−1 α α−β d + It lim (t s) Eα,α−β,2α(λ(t s) ,µ(t s) ) lim g(s) 0 s→t−0 − − − dt s→t−0 t h i C,t α 2α−1 α α−β + D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0 t C,t α 2α−1 α α−β = D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds. 0 t − 2 − − Z0 Thus, by Lemma 2.3, we get

t C α 2α−1 α α−β D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0 t C,t α 2α−1 α α−β = D (t s) Eα,α−β, α(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0

22 t α−1 α α−β = (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds. (5.20) − − − Z0 Similarly, by applying the formula (4.8), we also have

t C α 2α−β−1 α α−β D (t s) Eα,α−β, α−β(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0 t C,t α 2α−β−1 α α−β = D (t s) Eα,α−β, α−β(λ(t s) ,µ(t s) )g(s)ds 0 t − 2 − − Z0 t α−β−1 α α−β = (t s) Eα,α−β,α−β(λ(t s) ,µ(t s) )g(s)ds. (5.21) − − − Z0 On the other hand, since 0 < β 1, we will apply the formula (4.9) for second term of the equation (5.14): ≤

t C β α−1 α α−β D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 t 1−β α−1 α α−β = lim 0It (t s) Eα,α−β,α(λ(t s) ,µ(t s) ) lim g(s) s→t−0 − − − s→t−0 t RL,t β α−1 α α−β + D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 α−β α α−β = lim (t s) Eα,α−β,α−β+1(λ(t s) ,µ(t s) ) lim g(s) s→t−0 − − − s→t−0 t RL,t β α−1 α α−β + D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 t RL,t β α−1 α α−β = D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 t α−β−1 α α−β = (t s) Eα,α−β,α−β(λ(t s) ,µ(t s) )g(s)ds. (5.22) − − − Z0 where α−1 α−β − (t t ) (t t ) tI1 β − 0 = − 0 , t> 0. 0 t Γ(α) Γ(α β + 1) − Next we plug (5.20) and (5.21) into (5.19), we therefore get

t C α α−1 α α−β D (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds 0 t − − − Z0 t α−1 α α−β = g(t)+ λ (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds − − − Z0 t α−β−1 α α−β + µ (t s) Eα,α−β,α−β(λ(t s) ,µ(t s) )g(s)ds. − − − Z0

23 Taking linear combination of above equation together with (5.15) and (5.22), we arrive at

C Dαy (t) µ C Dβy (t) λy(t)= g(t), t> 0. 0 t − 0 t − Remark 5.2. Kilbas et al. [27] have obtained the analytical solutions of the Cauchy problems (5.7) and (5.14) in terms of Fox-Wright functions below:

t α−1 y(t)= (t s) Hα,β λ,µ(t s)g(s)ds, t> 0, (5.23) − ; − Z0 where ∞ l lα λ t (l +1, 1) − H (t) := Ψ µtα β . α,β;λ,µ l! 1 1 (lα + α, α β) l=0 − X Proof. Using the deﬁnition of Fox-Wright function [45, 46], we arrive at

t ∞ l lα+α−1 λ (t s) (l +1, 1) − y(t)= − Ψ µ(t s)α β g(s)ds l! 1 1 (lα + α, α β) − Z0 l=0 − X t ∞ ∞ l + k λlµk(t s)lα+k(α−β)+α−1 = − g(s)ds k Γ(lα + k(α β)+ α) l k Z0 X=0 X=0 − t α−1 α α−β = (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds, t> 0. − − − Z0 Therefore, our solution in terms of univariate version of bivariate Mittag-Leﬄer type functions coincide with the solution by means of Fox-Wright type functions shown in [27]. Remark 5.3. Podlubny [8] have attained the analytical solutions of the Cauchy problems (5.7) and (5.14) in terms of l-th derivative of two-parameter Mittag-Leﬄer functions below:

t α−1 y(t)= (t s) Hα,β λ,µ(t s)g(s)ds, t> 0, (5.24) − ; − Z0 where ∞ l lα λ t (l) α−β Hα,β λ,µ(t) := (µt ). ; l! Eα−β,α+lβ l X=0 Proof. Using the deﬁnition of l-th derivative of two-parameter Mittag-Leﬄer function, we arrive at

t ∞ l lα+α−1 λ (t s) l − y(t)= − ( ) (µ(t s)α β)g(s)ds l! Eα−β,α+lβ − l Z0 X=0 t ∞ ∞ l k k(α−β) λ − (l + k)! µ (t s) = (t s)(l+1)α 1 − g(s)ds l! − k! Γ(k(α β)+ l(α β)+ lβ + α) l k Z0 X=0 X=0 − − t ∞ ∞ l + k λlµk(t s)lα+k(α−β)+α−1 = − g(s)ds k Γ(lα + k(α β)+ α) l k Z0 X=0 X=0 − t α−1 α α−β = (t s) Eα,α−β,α(λ(t s) ,µ(t s) )g(s)ds, t> 0. − − − Z0

24 Therefore, our solution in terms of univariate version of bivariate Mittag-Leffler type functions coincide with the solution by means of l-th derivative of two-parameter Mittag-Leffler type functions shown in [8]. Case 3: In special case, we substitute α = 2 and β = 1 in (5.7) and (5.14), then we get the following Cauchy problem for the classical second order linear differential equation - the oscillator equation with constant coefficients: y′′(t) µy′(t) λy(t)= g(t), t> 0, − − (5.25) y′(0) = y(0) = 0, λ,µ R, ( ∈ Theorem 5.3. A unique solution y C2([0, ), R) of the Cauchy problem (5.25) has the following formula: ∈ ∞ t 2 y(t)= (t s)E , , (λ(t s) ,µ(t s))g(s)ds, t> 0. (5.26) − 2 1 2 − − Z0 Proof. By using verification by substitution, we have

t 2 d 2 (t s)E , , (λ(t s) ,µ(t s))g(s)ds dt2 − 2 1 2 − − Z0 2 2−l l−1 ∂ 2 d = lim − (t s)E2,1,2(λ(t s) ,µ(t s)) − lim g(s) s→t−0 ∂t2 l − − − dtl 1 s→t−0 l X=1 t 2 ∂ 2 + (t s)E , , (λ(t s) ,µ(t s))g(s)ds ∂t2 − 2 1 2 − − Z0

∂ 2 = lim (t s)E2,1,2(λ(t s) ,µ(t s)) lim g(s) s→t−0 ∂t − − − s→t−0 2 d + lim (t s)E2,1,2(λ(t s) ,µ(t s)) lim g(s) s→t−0 − − − dt s→t−0 t 2 ∂ 2 + (t s)E , , (λ(t s) ,µ(t s))g(s)ds ∂t2 − 2 1 2 − − Z0 2 = lim E2,1,1(λ(t s) ,µ(t s)) lim g(s) s→t−0 − − s→t−0 t −1 2 + (t s) E , , (λ(t s) ,µ(t s))g(s)ds − 2 1 0 − − Z0 2 2 = g(t)+ λ lim (t s) E2,1,3(λ(t s) ,µ(t s))g(s)ds s→t−0 − − − 2 + µ lim (t s)E2,1,2(λ(t s) ,µ(t s))g(s)ds s→t−0 − − − t −1 2 + (t s) E , , (λ(t s) ,µ(t s))g(s)ds − 2 1 0 − − Z0 t −1 2 = g(t)+ (t s) E , , (λ(t s) ,µ(t s))g(s)ds. (5.27) − 2 1 0 − − Z0 Applying Pascal’s rule for binomial coeﬃcients for the last term of above equality, we get

t −1 2 (t s) E , , (λ(t s) ,µ(t s))g(s)ds − 2 1 0 − − Z0

25 t t −1 (t s) 2 = − g(s)ds + λ (t s)E , , (λ(t s) ,µ(t s))g(s)ds Γ(0) − 2 1 2 − − Z0 Z0 t 2 + µ E , , (λ(t s) ,µ(t s))g(s)ds 2 1 1 − − Z0 t 2 = λ (t s)E , , (λ(t s) ,µ(t s))g(s)ds − 2 1 2 − − Z0 t 2 + µ E , , (λ(t s) ,µ(t s))g(s)ds. (5.28) 2 1 1 − − Z0 Next, we have

t d 2 (t s)E , , (λ(t s) ,µ(t s))g(s)ds dt − 2 1 2 − − Z0 t ∂ 2 = (t s)E , , (λ(t s) ,µ(t s))g(s)ds ∂t − 2 1 2 − − Z0 t 2 = E , , (λ(t s) ,µ(t s))g(s)ds. (5.29) 2 1 1 − − Z0

Again taking linear combination of above equations (5.28) and (5.29) together with (5.26), we prove the desired result.

6. Conclusions and future work

The theory of Leibniz integral rule allows us to study particular solutions of classical and fractional multi-term differential equations. To the best of our knowledge, we derive explicit analytical solutions of well-known Bagley-Torvik and oscillator equations in terms of bivariate Mittag-Leffler functions via the technique of fractional Green’s function, since this theory has not been presented in recent literature. The major contributions of our research work are as below: • we have proposed a Leibniz rule for higher order derivatives in classical sense which is more productive tool for testing solutions of multi-order differential equation;

• we have introduced fractional Leibniz rule for Riemann-Liouville and Caputo type fractional differentiation operators; • we have investigated differentiation of convolution operator which is more crucial in theory of differential equations with constant coefficients of classical and fractional-order derivatives; • analytical explicit solutions of the generalized Bagley-Torvik and oscillator equations are derived in terms of univariate version of bivariate Mittag-Leffler type functions in accordance with the method of Laplace integral transform; • We have showed that our analytical solutions are coincide with Fox-Wright type and l-th derivative of two-parameter Mittag-Leffler type functions; • we tested the candidate solutions of Cauchy problems for Bagley-Torvik equations with fractional-order sense and oscillator equation with classical-order one via our new fractional Leibniz integral rules.

26 There are a number of potential directions in which the results acquired here can be extended. Our future work will proceed to study the Leibniz integral rule results for ψ-Hilfer and Hadamard type fractional derivatives and the analytical explicit solutions of multi-term fractional diﬀerential equations in terms of natural extensions of Mittag-Leﬄer type functions.

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