Generalized fractional Dirac type operators
aa r X i v : . [ m a t h . C A ] J a n GENERALIZED FRACTIONAL DIRAC TYPE OPERATORS
JOEL E. RESTREPO, MICHAEL RUZHANSKY, AND DURVUDKHAN SURAGAN
Abstract.
We introduce a class of fractional Dirac type operators with time-variable coefficients by means of a Witt basis and the Riemann-Liouville fractionalderivative with respect to another function. Direct and inverse fractional Cauchytype problems are studied for the introduced operators. We give explicit solutionsof the considered fractional Cauchy type problems. We also use a recent method[17] to recover a variable coefficient solution of some inverse fractional wave andheat type equations. Illustrative examples are provided.
Contents
1. Introduction 12. Preliminaries 22.1. Fractional Laplacian 22.2. Fractional integro-differential operators 32.3. Fractional Cauchy type problem 42.4. Clifford Analysis 73. Main results 83.1. Generalized fractional Dirac type operators with time variable coefficients 83.2. Explicit solution of fractional Cauchy type problems 93.3. Generalized fractional Dirac type operators with constant coefficients 134. Special cases of Dirac type operators 174.1. Wave Dirac type operator 184.2. Fractional telegraph Dirac operator 215. Inverse problems 225.1. Fractional wave type equations 225.2. Fractional heat type equations 245.3. Examples 266. Conflict of interest statement 287. Acknowledgements 28References 281.
Introduction
The current paper gives an extension of some direct and inverse fractional Cauchytype problems to the fractional Clifford analysis. In fact, we use the recent results
Mathematics Subject Classification.
Key words and phrases.
Fractional integro-differential operator, Cauchy problem, time-fractionalDirac operators, inverse problem. from [35] to define a large class of fractional Dirac type operators, which involvestime-variable coefficients, Witt basis and the Riemann-Liouville fractional derivativewith respect to another function. These operators lead us to study some generalCauchy problems of similar type of those in [2, 32, 34].Here we generalize some of the ideas given in [13, 14] where fundamental solu-tions of time-fractional telegraph, diffusion-wave and parabolic Dirac operators wereobtained. We also extend some recent results given in [2]. We mainly introduce aclass of fractional Dirac type operators that factorize a general fractional Laplace-type operator which involves Riemann-Liouville fractional derivatives with respect toanother function and time variable functions. These type of Dirac operators can bevery useful to analyze the solvability of the in-stationary Navier–Stokes equations [6],as well as Maxwell equations, Lame equations, among others [22, 23].Notice that fractional direct and inverse Cauchy type problems have been studiedby many authors since their applications and the intrinsic development of the frac-tional calculus theory. We refer, for instance, the sources [10, 18, 28, 29, 30, 37, 38,39, 41, 42, 47] and references therein. The following books [9, 19, 27, 40, 46] as well.With respect to the Dirac type operators, its great impact and applications inClifford analysis and PDE’s are well-known, see e.g. the books [3, 4, 7, 8, 20], and alsothe papers [5, 6, 11, 13, 14, 43]. For some works related to more general presentationsand applications of Dirac type operators, see e.g. [1, 15, 31, 36].In some theoretical frames, our results and the generalized fractional Dirac typeoperators will allow one in the future to explore different questions between fractionalcalculus and some topics like Clifford analysis, quantum mechanics, physics, etc [24,33, 44, 45].The paper is organized as follows: In Section 2, we recall some facts and defi-nitions on fractional integro-differential operators, fractional Cauchy type equationsand Clifford analysis. Section 3 is devoted to the main results of the paper. Indeed, byusing a class of generalized time-fractional Dirac type operators, we study fractionalCauchy type problems and give their explicit solutions. In Section 4 we discuss somespecial cases of the introduced Dirac type operators. While, in Section 5, we studysome inverse fractional wave and heat type equations. We also give some examples.2.
Preliminaries
In this section we recall some definitions and auxiliary results on fractional integro-differential operators, fractional Cauchy type equations and Clifford analysis, whichwill be used throughout the whole paper.2.1.
Fractional Laplacian.
We first recall the Fourier transform of a function f : f ( ξ ) = ( F ϕ )( ξ ) = b ϕ ( ξ ) = Z R n e iξ · x ϕ ( x ) dx, while the inverse Fourier transform is defined by ϕ ( ξ ) = (cid:0) F − f (cid:1) ( ξ ) = 1(2 π ) n Z R n e − iξ · τ f ( τ ) dτ, where “ · ” is the usual inner product of vectors in R n . ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 3
The fractional Laplacian ( − ∆) λ is defined by [40, Chapter 5]:( F ( − ∆) λ f )( ξ ) = | ξ | λ ( F f )( ξ ) , ξ ∈ R n , (2.1)where 0 < λ < m , m ∈ N and F is the Fourier transform. The above operator canbe also given by ( − ∆) λ f ( x ) = 1 d n,m ( λ ) Z R n (∆ my f )( x ) | y | n +2 λ dy, where (∆ my f )( x ) is the difference operator defined in [40, formulas (25.57) and (25.58)]and d n,m ( β ) is a normalization constant. Note that for λ = 1 we get the classicalLaplacian in R n , i.e. ∆ x = P nk =1 ∂ x k .2.2. Fractional integro-differential operators.
Now we recall some definitionsand properties of the fractional integro-differential operators with respect to anotherfunction, see e.g. [40, Chapter 4], also [26].
Definition 2.1.
Let α ∈ C , Re( α ) > −∞ a < b ∞ , let f be an integrablefunction on [ a, b ], and let φ ∈ C [ a, b ] be such that φ ′ ( t ) > t ∈ [ a, b ]. Theleft-sided Riemann-Liouville fractional integral of f with respect to another function φ is defined by [26, formula (2.5.1)]: I α,φa + f ( t ) = 1Γ( α ) Z ta φ ′ ( s )( φ ( t ) − φ ( s )) α − f ( s ) ds. (2.2) Definition 2.2.
Let α ∈ C , Re( α ) > −∞ a < b ∞ , let f be an integrablefunction on [ a, b ], and let φ ∈ C [ a, b ] be such that φ ′ ( t ) > t ∈ [ a, b ]. The left-sided Riemann-Liouville fractional derivative of a function f with respect to anotherfunction φ is defined by [26, Formula 2.5.17]: D α,φa + f ( t ) = (cid:18) φ ′ ( t ) ddt (cid:19) n (cid:0) I n − α,φa + f (cid:1) ( t ) , (2.3)where n = ⌊ Re( α ) ⌋ + 1 (or n = −⌊− Re( α ) ⌋ ) and ⌊·⌋ is the floor function ( n − < Re( α ) n ).Below we always assume that φ ∈ C [ a, b ] is such that φ ′ ( t ) > t ∈ [ a, b ]when we use the operators I α,φa + or D α,φa + .Let us recall a result which will be useful in some examples in the next sections.Taking into account [40, Theorem 2.4] it can be proved similarly that the followingstatement holds. Theorem 2.3. If α ∈ C (Re( α ) > and f ∈ L ( a, b ) , then D α,φa + I α,φa + f ( t ) = f ( t ) holds almost everywhere on [ a, b ] . In this paper we will use the following modified fractional derivative with respectto another function: C D α,φ f ( t ) = D α,φ (cid:20) f ( t ) − n − X j =0 f [ j ] φ (0) j ! (cid:0) φ ( t ) − φ (0) (cid:1) j (cid:21) , α ∈ C , Re( α ) > , (2.4) J. E. RESTREPO, M. RUZHANSKY, AND D. SURAGAN where n = − [ − Re( α )] for α / ∈ N , n = α for α ∈ N and f [ j ] φ ( t ) = (cid:18) φ ′ ( t ) ddt (cid:19) j f ( t ) . Note that for φ ( t ) = t , C D α,φ f ( t ) becomes the modified fractional derivative used in[25, formula (1.3)]. We also have: If α > n − < α < n and f ∈ C n [ a, b ], then C D α,φ of (2.4) becomes the so-called Caputo fractional derivative: C D α,φa + f ( t ) = I n − α,φa + (cid:18) φ ′ ( t ) ddt (cid:19) n f ( t ) , (2.5)where n = ⌊ α ⌋ + 1 for α / ∈ N and n = α for α ∈ N .We must mention that the existence of the fractional derivate (2.5) is guaranteedby f ( n ) ∈ L [ a, b ]. And, the stronger condition f ∈ C n [ a, b ] gives the continuity ofthe derivative. Furthermore, if α = n , we have C D α,φ f ( t ) = (cid:18) φ ′ ( t ) ddt (cid:19) n f ( t ) . For α = n and φ ( t ) = t , it follows that C D α,φ f ( t ) = D n f ( t ) = f ( n ) ( t ) . Fractional Cauchy type problem.
Here we recall some useful results from[34] that will help us to prove our main results in the next sections.We first introduce some necessary notation. We denote by C ∂ α,φt w ( x, t ) := C D α,φ w ( x, t ) = D α,φ " w ( x, t ) − n − X j =0 w [ j ] φ ( x, j ! (cid:0) φ ( t ) − φ (0) (cid:1) j , where α ∈ C , Re( α ) > x ∈ R n , t ∈ (0 , T ], n = − [ − Re( α )] for α / ∈ N , n = α for α ∈ N and w [ j ] φ ( x, t ) = (cid:18) φ ′ ( t ) ddt (cid:19) j w ( x, t ) . Let K j := { i : 0 Re( β i ) j , i = 1 , . . . , m } , j = 0 , , . . . , n − , and κ j = min { K j } , if K j = ∅ . Note that the inclusion s ∈ K j implies Re( β s ) j ,while K j ⊂ K j for j < j . Besides, if β m = 0, then K j = ∅ , j = 0 , , . . . , n − j = 0 , . . . , n − K κ j j ( t, | s | λ , Θ , . . . , Θ m ) := + ∞ X k =0 ( − k +1 I β ,φ (cid:18) m X i =1 d i ( t ) I β − β i ,φ (cid:19) k m X i = κ j d i ( t ) D β i ,φ Ψ j ( t ) , and K j ( t, | s | λ , Θ , . . . , Θ m ) := + ∞ X k =0 ( − k +1 I β ,φ m X i =1 d i ( t ) I β − β i ,φ ! k m X i =1 d i ( t ) D β i ,φ Ψ j ( t ) , where d m ( t ) = | s | λ Θ m ( t ), d i ( t ) = Θ i ( t ), i = 1 , . . . , m − κ j = min { K j } andΨ j ( t ) = ( φ ( t ) − φ (0)) j Γ( j + 1) , j ∈ N ∪ { } . (2.6) ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 5
Besides, we set G ( b h ( s, t )) := + ∞ X k =1 ( − k I β ,φ m X i =1 d i ( t ) I β − β i ,φ ! k b h ( s, t ) . As in the classical case without fractional operators, it is a natural to require thatthe kernels K κ j j , K j ( j = 0 , . . . , n −
1) and G ( b h ( s, t )) are some functions of L ( R n ).We also recall the function space C n − ,β [0 , T ] := { u ( t ) ∈ C n − [0 , T ] , C D β ,φ u ( t ) ∈ C [0 , T ] } endowed with the norm k u k C n − ,β [0 ,T ] = n − X k =0 (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:18) φ ′ ( t ) ddt (cid:19) k u (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) C [0 ,T ] + (cid:13)(cid:13) C D β ,φ u (cid:13)(cid:13) C [0 ,T ] . Now we recall the space-time fractional Cauchy problem studied in [34]: C ∂ β ,φt w ( x, t ) + m − X i =1 Θ i ( t ) C ∂ β i ,φt w ( x, t )+Θ m ( t )( − ∆) λ w ( x, t )= h ( x, t ) , t ∈ (0 , T ] , x ∈ R n ,w ( x, t ) | t =0 = w ( x ) ,∂ t w ( x, t ) | t =0 = w ( x ) , ... ∂ n t w ( x, t ) | t =0 = w n − ( x ) , (2.7)where 0 < λ β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > n i = ⌊ Re β i ⌋ +1(or n i = −⌊− Re( β i ) ⌋ ), i = 0 , , . . . , m − n i − < Re( β i ) n i ). Also, it is assumedthat h ( t, · ) ∈ C [0 , T ] and Θ i ( t ) ∈ C [0 , T ], i = 1 , . . . , m .The explicit solution of equation 2.7 was given in [34, Theorems 3.1, 3.2, 3.3].Below we recall the last two ones since it will be used frequently in this paper. Theorem 2.4.
Let n > n , β m = 0 and h ( · , t ) , Θ i ∈ C [0 , T ] ( i = 1 , . . . , m ) . Assumealso that P mi =1 k Θ i k max I β − β i ,φ e νt Ce νt for some ν > and some constant Theorem 2.5. Let n = n , β m = 0 , let h ( · , t ) , Θ i ∈ C [0 , T ] ( i = 1 , . . . , m ) . Assumealso that P mi =1 k Θ i k max I β − β i ,φ e νt Ce νt for some ν > and a constant < C < independent of t . Then the problem (2.7) has a unique solution given by: w ( x, t ) − n − X j =0 w j ( x )Ψ j ( t ) = n − X j =0 Z R n F − s (cid:0) K κ j j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) w j ( y ) dy − I β ,φ h ( x, t ) + F − s (cid:0) G ( b h ( s, t )) (cid:1) ( x ) , j = 0 , . . . , n − . For the case of constant coefficients Θ i ( t ) = λ i ∈ C in equation (2.7), we havethe following explicit representations for the solution. For more details, see [34,Theorems 4.6, 4.7]. First, we need to recall the multivariate Mittag-Leffler func-tion E ( a ,...,a n ) ,b ( z , . . . , z n ), where the variables z , . . . , z n ∈ C and any parameters a , . . . , a n , b ∈ C with positive real parts, which is defined by E ( a ,...,a n ) ,b ( z , . . . , z n ) = + ∞ X k =0 X l + ··· + l n = k, l ,...,l n ≥ (cid:18) kl , . . . , l n (cid:19) Q ni =1 z l i i Γ ( b + P ni =1 a i l i ) , (2.8)where the multinomial coefficients are (cid:18) kl , . . . , l n (cid:19) = k ! l ! × · · · × l n ! . Theorem 2.6. Let n > n , β m = 0 and h ( · , t ) ∈ C [0 , T ] . Suppose that in equation(2.7) we have Θ i ( t ) = λ i ∈ C , i = 1 , . . . , m and P mi =1 k Θ i k max I β − β i ,φ e νt Ce νt forsome ν > and a constant < C < independent of t . Then the initial valueproblem (2.7) has a unique solution given by: w ( x, t ) = n − X j =0 w j ( x )Ψ j ( t ) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:19) ( x − y ) w j ( y ) dy + n − X j = n Z R n F − r (cid:18) m X i =0 λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:19) ( x − y ) w j ( y ) dy + Z t φ ′ ( s )( φ ( t ) − φ ( s )) β − Z R n F − r (cid:0) E ( β − β ,...,β − β m ) ,β (cid:0) − λ ( φ ( t ) − φ ( s )) β − β , . . .. . . , −| r | λ λ m ( φ ( t ) − φ ( s )) β − β m (cid:1)(cid:1) ( x − y ) h ( y, s ) dyds, where λ ⋆i = λ i ( i = 0 , , . . . , m − , λ ⋆m = | r | λ λ m and Ψ j ( t ) is that of (2.6). Theorem 2.7. Let n = n , β m = 0 and h ( · , t ) ∈ C [0 , T ] . Suppose that in equation(2.7) we have Θ i ( t ) = λ i ∈ C , i = 1 , . . . , m and P mi =1 k Θ i k max I β − β i ,φ e νt Ce νt for ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 7 some ν > and a constant < C < independent of t . Then the initial valueproblem (2.7) has a unique solution given by: w ( x, t ) = n − X j =0 w j ( x )Ψ j ( t ) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:19) ( x − y ) w j ( y ) dy + Z t φ ′ ( s )( φ ( t ) − φ ( s )) β − Z R n F − r (cid:16) E ( β − β ,...,β − β m ) ,β (cid:0) − λ ( φ ( t ) − φ ( s )) β − β , . . .. . . , −| r | λ λ m ( φ ( t ) − φ ( s )) β − β m (cid:1)(cid:17) ( x − y ) h ( y, s ) dyds, where λ ⋆i = λ i ( i = 0 , , . . . , m − , λ ⋆m = | r | λ λ m and Ψ j ( t ) is that of (2.6). Clifford Analysis. Below we recall some necessary facts and notions on Cliffordanalysis. Nevertheless, for more details on this topic, see e.g. [16]. Let us start byrecalling the universal real Clifford algebra. We then take the n -dimensional vectorspace R n endowed with an orthonormal basis { e , . . . , e n } . The universal real Cliffordalgebra Cl ,n is defined as the 2 n -dimensional associative algebra which satisfies thefollowing multiplication rule e i e j + e j e i = − δ ij , i, j = 1 , . . . , n. A vector space basis for Cl ,n is generated by the elements e = 1 and e B = e r ,...,r k ,where B = { r , . . . , r k } ⊂ N = { , . . . , n } for 1 ≤ r < · · · < r k ≤ n . Hence, for any y ∈ Cl ,n we have that y = P B x B e B with x B ∈ R . Now we recall the complexifiedClifford algebra C n : C n = C ⊗ Cl ,n = ( v = X B v B e B , v B ∈ C , B ⊂ N ) , where the imaginary unit i of C commutes with the basis elements ( ie j = e j i forany j = 1 , . . . , n ). A C n -valued function defined on an open subset V ⊂ R n can berepresented by f = P B f B e B with C -valued components f B . As usual, the continu-ity, differentiability and other properties are normally assumed component-wisely bymeans of the classical notions on C .For the next definition we need to recall the Euclidean Dirac operator D x = P nk =1 e k ∂ x k . Note also that D x = − ∆ = − P nk =1 ∂ x k . Definition 2.8. [16, Chapter 2] A Clifford valued C function f is left-monogenic if D x f = 0 on V , respectively right-monogenic if f D x = 0 on V .The above definition will be used implicitly in Section 4 to illustrate some particularcases of the main results of the present paper.In the next section we will introduce a new class of generalized fractional Dirac typeoperators. Hence, we need to use and describe a Witt basis. Let us embed R n into R n +2 by considering two new elements e + and e − which satisfy e = 1, e − = − e + e − + e − e + = 0. We also suppose that e − , e + anti-commute with each element from J. E. RESTREPO, M. RUZHANSKY, AND D. SURAGAN { e , . . . , e n } . Then { e , . . . , e n , e + , e − } spans R n +1 , . By using the elements e + , e − wecompose two nilpotent elements usually denoted by f and f + . They are defined by: f = e + − e − f + = e + + e − . Some useful properties:(1) ( f ) = ( f + ) = 0 , (2) ff + + f + f = 1 , (3) f e i + e i f = f + e i + e i f + = 0 , i = 1 , . . . , n. Main results In this section, we study some general fractional Cauchy type problems by usingsome generalized fractional Dirac type operators. We show in all cases the explicitsolutions.3.1. Generalized fractional Dirac type operators with time variable coeffi-cients. By using the Witt basis { e , . . . , e n , f , f + } we formally introduce a new classof generalized fractional Dirac type operators with time variable coefficients and withrespect to a given function φ by x,t D λ,β ,...,β m − Θ ,..., Θ m ; φ := Θ / m ( t )( − ∆) λ/ x + f C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt ! + f + , (3.1)where x ∈ R n , t > 0, 0 < λ β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > n i = ⌊ Re β i ⌋ + 1 (or n i = −⌊− Re( β i ) ⌋ ), i = 0 , , . . . , m − n i − < Re( β i ) n i ).We also assume that Θ i ( t ) ∈ C [0 , T ], i = 1 , . . . , m . Remark 3.1. Notice that the generalized fractional Dirac type operator of (3.1)becomes the one introduced in [2, Formula (3.1)] when φ ( t ) = t . Proposition 3.2. Let x ∈ R n , t > , < λ , β i ∈ C , Re( β ) > Re( β ) >. . . > Re( β m − ) > and n i = ⌊ Re β i ⌋ + 1 , i = 0 , , . . . , m − n i − < Re( β i ) n i ) . We also suppose that Θ i ( t ) ∈ C [0 , T ] , i = 1 , . . . , m . If f ( · , t ) ∈ L ( R n ) and | y | β ( F f ( · , t ))( y ) ∈ L ( R n ) then the following factorization holds: (cid:0) x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ (cid:1) = Θ m ( t )( − ∆) λx + C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt . (3.2) Proof. We know that( x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ ) = Θ / m ( t )( − ∆) λ/ x + f C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt ! + f + ! . Notice that for E = Θ / m ( t )( − ∆) λ/ x , F = C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt , it follows that ( x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ ) = (cid:0) E + f F + f + (cid:1)(cid:0) E + f F + f + (cid:1) . ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 9 By the properties (1), (2), (3) and ( − ∆) λ/ x ( − ∆) λ/ x = ( − ∆) λx we obtain: (cid:0) E + f F + f + (cid:1)(cid:0) E + f F + f + (cid:1) = EE − f EF − f + f EF + ( f ) F F + ff + F + f + E + f + f F + ( f ) = EE + ff + F + f + f F = EE + ( ff + + f + f ) F = EE + F, which complete the proof. (cid:3) Explicit solution of fractional Cauchy type problems. Now we give themain results of the paper. Theorem 3.3. Let P mi =1 k Θ i k max I β − β i ,φ e νt Ce νt for some ν > and some con-stant < C < which does not depend on t . Let n > n , < λ , β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > and n i = −⌊− Re( β i ) ⌋ , i = 0 , , . . . , m − n i − < Re( β i ) n i ) . We also assume that Θ i ( t ) ∈ C [0 , T ] , i = 1 , . . . , m . Thefollowing fractional Cauchy type problem Θ / m ( t )( − ∆) λ/ x + f C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt ! + f + ! w ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,w ( x, t ) | t =0 = r ( x ) ,∂ t w ( x, t ) | t =0 = r ( x ) , ... ∂ n t w ( x, t ) | t =0 = r n − ( x ) , (3.3) is soluble, and the solution is given by w ( x, t ) = n − X j =0 Θ / m ( t )( − ∆) λ/ x (cid:0) r j ( x ) (cid:1) Ψ j ( t ) (3.4)+ n − X j =0 Θ / m ( t )(2 π ) n Z R n e − ix · τ | τ | λ H j ( t, | s | λ , Θ , . . . , Θ m )( τ ) b r j ( τ ) dτ + f m − X i =1 n − X j =0 Θ i ( t ) r j ( x ) C ∂ β i ,φt Ψ j ( t )+ n − X j =0 Z R n F − s (cid:0) C ∂ β ,φt H j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy + m − X i =1 n − X j =0 Θ i ( t ) Z R n F − s (cid:0) C ∂ β i ,φt H j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy ! + f + n − X j =0 r j ( x )Ψ j ( t ) + n − X j =0 Z R n F − s (cid:0) H j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy ! , (3.5) where H j ( t, | s | λ , Θ , . . . , Θ m ) = (cid:26) K κ j j ( t, | s | λ , Θ , . . . , Θ m ) if j = 0 , . . . , n − ,K j ( t, | s | λ , Θ , . . . , Θ m ) if j = n , . . . , n − ,K κ j j ( t, | s | λ , Θ , . . . , Θ m ) := + ∞ X k =0 ( − k +1 I β ,φ (cid:18) m X i =1 d i ( t ) I β − β i ,φ (cid:19) k m X i = κ j d i ( t ) D β i ,φ Ψ j ( t ) , and K j ( t, | s | λ , Θ , . . . , Θ m ) := + ∞ X k =0 ( − k +1 I β ,φ m X i =1 d i ( t ) I β − β i ,φ ! k m X i =1 d i ( t ) D β i ,φ Ψ j ( t ) , with d m ( t ) = | s | λ Θ m ( t ) , d i ( t ) = Θ i ( t ) , i = 1 , . . . , m − , κ j = min { K j } and Ψ j ( t ) isthat of (2.6). Proof. Notice first that equation (3.3) is equivalent to x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ w ( x, t ) = 0 , x ∈ R n , t > . (3.6)Applying the operator x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ to (3.6) implies thatΘ m ( t )( − ∆) λx w ( x, t ) + C ∂ β ,φt w ( x, t ) + m − X i =1 Θ i ( t ) C ∂ β i ,φt w ( x, t ) = 0 , due to the factorization (3.2). We then obtain the equation (2.7) with h ≡ 0. Thus,by Theorem 2.4, the solution of equation (3.3) is given by x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ w ( x, t ), where w ( x, t ) = n − X j =0 r j ( x )Ψ j ( t ) + n − X j =0 Z R n F − s (cid:0) H j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy, and H j ( t, | s | λ , Θ , . . . , Θ m ) = (cid:26) K κ j j ( t, | s | λ , Θ , . . . , Θ m ) if j = 0 , . . . , n − ,K j ( t, | s | λ , Θ , . . . , Θ m ) if j = n , . . . , n − . The explicit representation of the solution follows by calculating each of the compo-nents of x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ w ( x, t ), separately. In fact, we have x,t D λ,β ,...,β m − Θ , ··· , Θ m ; φ w ( x, t ) =Θ / m ( t )( − ∆) λ/ x w ( x, t ) + f C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt ! w ( x, t ) + f + w ( x, t ) . ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 11 Clearly by (2.1) and changing the order of integration we getΘ / m ( t )( − ∆) λ/ x w ( x, t ) = n − X j =0 Θ / m ( t )( − ∆) λ/ x (cid:0) r j ( x ) (cid:1) Ψ j ( t )+ n − X j =0 Θ / m ( t )(2 π ) n Z R n e − ix · τ | τ | λ H j ( t, | s | λ , Θ , . . . , Θ m )( τ ) b r j ( τ ) dτ. We also have C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt ! w ( x, t ) = m − X i =1 n − X j =0 Θ i ( t ) r j ( x ) C ∂ β i ,φt Ψ j ( t )+ n − X j =0 Z R n F − s (cid:0) C ∂ β ,φt H j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy + m − X i =1 n − X j =0 Θ i ( t ) Z R n F − s (cid:0) C ∂ β i ,φt H j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy, since C ∂ β ,φt Ψ j ( t ) = 0 for any j = 0 , , . . . , n − (cid:3) The proof of the next result follows the same steps of the proof of Theorem 3.3.Moreover, instead of using Theorem 2.4 in the proof we need now to apply Theorem2.5. We then omit the proof and leave it to the reader. Theorem 3.4. Let P mi =1 k Θ i k max I β − β i ,φ e νt Ce νt for some ν > and some con-stant < C < which does not depend on t . Let n = n , < λ , β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > and n i = −⌊− Re( β i ) ⌋ , i = 0 , , . . . , m − n i − < Re( β i ) n i ) . We also assume that Θ i ( t ) ∈ C [0 , T ] , i = 1 , . . . , m . Thefollowing fractional Cauchy type problem Θ / m ( t )( − ∆) λ/ x + f C ∂ β ,φt + m − X i =1 Θ i ( t ) C ∂ β i ,φt ! + f + ! w ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,w ( x, t ) | t =0 = r ( x ) ,∂ t w ( x, t ) | t =0 = r ( x ) , ... ∂ n t w ( x, t ) | t =0 = r n − ( x ) , (3.7) is soluble, and the solution is given by w ( x, t ) = n − X j =0 Θ / m ( t )( − ∆) λ/ x (cid:0) r j ( x ) (cid:1) Ψ j ( t ) (3.8)+ n − X j =0 Θ / m ( t )(2 π ) n Z R n e − xi · τ | τ | λ K κ j j ( t, | s | λ , Θ , . . . , Θ m )( τ ) b r j ( τ ) dτ + f m − X i =1 n − X j =0 Θ i ( t ) r j ( x ) C ∂ β i ,φt Ψ j ( t )+ n − X j =0 Z R n F − s (cid:0) C ∂ β ,φt K κ j j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy + m − X i =1 n − X j =0 Θ i ( t ) Z R n F − s (cid:0) C ∂ β i ,φt K κ j j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy ! + f + n − X j =0 r j ( x )Ψ j ( t ) + n − X j =0 Z R n F − s (cid:0) K κ j j ( t, | s | λ , Θ , . . . , Θ m ) (cid:1) ( x − y ) r j ( y ) dy ! , (3.9) where K κ j j ( t, | s | λ , Θ , . . . , Θ m ) = + ∞ X k =0 ( − k +1 I β ,φ (cid:18) m X i =1 d i ( t ) I β − β i ,φ (cid:19) k m X i = κ j d i ( t ) D β i ,φ Ψ j ( t ) , with d m ( t ) = | s | λ Θ m ( t ) , d i ( t ) = Θ i ( t ) , i = 1 , . . . , m − , κ j = min { K j } and Ψ j ( t ) isthat of (2.6). ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 13 Generalized fractional Dirac type operators with constant coefficients. Let us now consider a class of generalized fractional Dirac type operators with con-stant coefficients (related to those ones introduced in formula (3.1)) and with respectto a given function φ by x,t D λ,β ,...,β m − λ ,...,λ m ; φ : λ / m ( − ∆) λ/ x + f C ∂ β ,φt + m − X i =1 λ i C ∂ β i ,φt ! + f + , (3.10)where x ∈ R n , t > 0, 0 < λ β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > n i = ⌊ Re β i ⌋ + 1 (or n i = −⌊− Re( β i ) ⌋ ), i = 0 , , . . . , m − n i − < Re( β i ) n i ).We also assume that λ i ∈ C , i = 1 , . . . , m .By formula (3.2) it is clear that (cid:0) x,t D λ,β ,...,β m − λ ,...,λ m ; φ (cid:1) = λ m ( − ∆) λx + C ∂ β ,φt + m − X i =1 λ i C ∂ β i ,φt . (3.11)Now we establish the next results following the proof of Theorem 3.3. In this case,we just apply Theorems 2.6 and 2.7 respectively. We leave the proofs to the reader. Theorem 3.5. Let P mi =1 | λ i | I β − β i ,φ e νt Ce νt for some ν > and some constant < C < which does not depend on t . Let n > n , < λ , β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > and n i = −⌊− Re( β i ) ⌋ , i = 0 , , . . . , m − n i − < Re( β i ) n i ) . We also assume that λ i ∈ C , i = 1 , . . . , m . The following fractionalCauchy type problem λ / m ( − ∆) λ/ x + f C ∂ β ,φt + m − X i =1 λ i C ∂ β i ,φt ! + f + ! w ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,w ( x, t ) | t =0 = r ( x ) ,∂ t w ( x, t ) | t =0 = r ( x ) , ... ∂ n t w ( x, t ) | t =0 = r n − ( x ) , (3.12) is soluble, and the solution is given by w ( x, t ) = λ / m n − X j =0 ( − ∆) λ/ x (cid:0) r j ( x ) (cid:1) Ψ j ( t )+ n − X j =0 λ / m (2 π ) n Z R n e − ix · τ | τ | λ m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | s | λ ( φ ( t ) − φ (0)) β − β m )( τ ) b r j ( τ ) dτ + n − X j = n λ / m (2 π ) n Z R n e − ix · τ | τ | λ m X i =0 λ ⋆i (cid:0) φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:1) ( τ ) b r j ( τ ) dτ + f C ∂ β ,φt + m − X i =1 λ i C ∂ β i ,φt ! n − X j =0 r j ( x )Ψ j ( t )+ n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:1) ( x − y ) r j ( y ) dy + n − X j = n Z R n F − r (cid:18) m X i =0 λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) + f + n − X j =0 r j ( x )Ψ j ( t ) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:1) ( x − y ) r j ( y ) dy + n − X j = n Z R n F − r (cid:18) m X i =0 λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) where λ ⋆i = λ i ( i = 0 , , . . . , m − , λ ⋆m = | r | λ λ m and Ψ j ( t ) is that of (2.6). ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 15 Theorem 3.6. Let P mi =1 | λ i | I β − β i ,φ e νt Ce νt for some ν > and some constant < C < which does not depend on t . Let n = n , < λ , β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > and n i = −⌊− Re( β i ) ⌋ , i = 0 , , . . . , m − n i − < Re( β i ) n i ) . We also assume that λ i ∈ C , i = 1 , . . . , m . The following fractionalCauchy type problem λ / m ( − ∆) λ/ x + f C ∂ β ,φt + m − X i =1 λ i C ∂ β i ,φt ! + f + ! w ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,w ( x, t ) | t =0 = r ( x ) ,∂ t w ( x, t ) | t =0 = r ( x ) , ... ∂ n t w ( x, t ) | t =0 = r n − ( x ) , (3.13) is soluble, and the solution is given by w ( x, t ) = n − X j =0 λ / m ( − ∆) λ/ x (cid:0) r j ( x ) (cid:1) Ψ j ( t )+ n − X j =0 λ / m (2 π ) n Z R n e − ix · τ | τ | λ m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:1) ( τ ) b r j ( τ ) dτ + f C ∂ β ,φt + m − X i =1 λ i C ∂ β i ,φt ! n − X j =0 r j ( x )Ψ j ( t )+ n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) + f + n − X j =0 r j ( x )Ψ j ( t ) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i ( φ ( t ) − φ (0)) j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ ( φ ( t ) − φ (0)) β − β , · · · . . . , λ m | r | λ ( φ ( t ) − φ (0)) β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) where λ ⋆i = λ i ( i = 0 , , . . . , m − , λ ⋆m = | r | λ λ m and Ψ j ( t ) is that of (2.6). We mention some special results when φ ( t ) = t in Theorems 3.5 and 3.6. Wedenote I β , C D β instead of I β,φ , C D β,φ when φ ( t ) ≡ t . Corollary 3.7. Let n = n , < λ , β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > and n i = −⌊− Re( β i ) ⌋ , i = 0 , , . . . , m − n i − < Re( β i ) n i ) . Wealso assume that λ i ∈ C , i = 1 , . . . , m . The following fractional Cauchy type problem λ / m ( − ∆) λ/ x + f C ∂ β t + m − X i =1 λ i C ∂ β i t ! + f + ! w ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,w ( x, t ) | t =0 = r ( x ) ,∂ t w ( x, t ) | t =0 = r ( x ) , ... ∂ n t w ( x, t ) | t =0 = r n − ( x ) , (3.14) is soluble, and the solution is given by w ( x, t ) = n − X j =0 λ / m ( − ∆) λ/ x (cid:0) r j ( x ) (cid:1) t j Γ( j + 1) + n − X j =0 λ / m (2 π ) n Z R n e − ix · τ | τ | λ m X i = κ j λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:1) ( τ ) b r j ( τ ) dτ + f C ∂ β t + m − X i =1 λ i C ∂ β i t ! n − X j =0 r j ( x ) t j Γ( j + 1) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) + f + n − X j =0 r j ( x ) t j Γ( j + 1) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) , where λ ⋆i = λ i ( i = 0 , , . . . , m − and λ ⋆m = | r | λ λ m . Corollary 3.8. Let n > n , < λ , λ j ∈ C , j = 1 , . . . , m , β i ∈ C , Re( β ) > Re( β ) > . . . > Re( β m − ) > and n i = −⌊− Re( β i ) ⌋ , i = 0 , , . . . , m − n i − < Re( β i ) n i ) . The fractional Cauchy type problem (3.14) is soluble, and the solution ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 17 is given by w ( x, t ) = λ / m n − X j =0 ( − ∆) λ/ x (cid:0) r j ( x ) (cid:1) t j Γ( j + 1) + n − X j =0 λ / m (2 π ) n Z R n e − ix · τ | τ | λ m X i = κ j λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:1) ( τ ) b r j ( τ ) dτ + n − X j = n λ / m (2 π ) n Z R n e − ix · τ | τ | λ m X i =0 λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:1) ( τ ) b r j ( τ ) dτ + f C ∂ β ,φt + m − X i =1 λ i C ∂ β i ,φt ! n − X j =0 r j ( x ) t j Γ( j + 1) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ t β − β , · · · . . . , λ m | r | λ t β − β m ) (cid:19) ( x − y ) r j ( y ) dy + n − X j = n Z R n F − r (cid:18) m X i =0 λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i ( λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) + f + n − X j =0 r j ( x ) t j Γ( j + 1) + n − X j =0 Z R n F − r (cid:18) m X i = κ j λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i ( λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:19) ( x − y ) r j ( y ) dy + n − X j = n Z R n F − r (cid:18) m X i =0 λ ⋆i t j + β − β i × E ( β − β ,...,β − β m ) ,j +1+ β − β i (cid:0) λ t β − β , . . . , λ m | r | λ t β − β m ) (cid:19) ( x − y ) r j ( y ) dy (cid:19) where λ ⋆i = λ i ( i = 0 , , . . . , m − and λ ⋆m = | r | λ λ m . Special cases of Dirac type operators In this section, we show some relevant Dirac type operators as special cases of thoseones introduced in formulas (3.1) and (3.10). We also denote I β , C D β instead of I β,φ , C D β,φ when φ ( t ) ≡ t . Some of the following examples can be found in [2] butwe include here for the sake of completeness. Wave Dirac type operator. We begin with the following wave Dirac typeoperator: x,t D ,α t α ; t := t α / D x + f (cid:0) C ∂ α t (cid:1) + f + , where 1 < α D x = P nk =1 e k ∂ x k is the Dirac operator, which factorizes theLaplacian as D x = − ∆ = − P nk =1 ∂ x k . . We have that( x,t D ,α t α ; t ) = − t α ∆ x + C ∂ α t . Let us now recall the following fractional initial value problem C ∂ α t w ( x, t ) − t α ∆ x w ( x, t ) = 0 ,w ( x, t ) | t =0+ = w ( x ) ,∂ t w ( x, t ) | t =0+ = w ( x ) , (4.1)where 1 < α 2. It was shown in [34, Section 3.2] that the solution is given by w ( x, t ) = w ( x ) + w ( x ) t − Z R n F − s (cid:0) I α (cid:0) | s | t α E α , α ,α ( −| s | t α ) (cid:1)(cid:1) ( x − y ) w ( y ) dy − Z R n F − s (cid:0) I α (cid:0) | s | t α +1 E β , α ,α +1 ( −| s | t α ) (cid:1) ( x − y ) w ( y ) dy, (4.2)where E λα,β,γ = + ∞ X k =0 c k z k , z ∈ C , with c = 1 , c k = k − Y j =0 Γ( α [ jβ + γ ] + 1)Γ( α [ jβ + γ ] + λ + 1) , k = 1 , , . . . , α, β, λ ∈ R , γ ∈ C . Note that in the case λ = α the function E αα,β,γ becomes the generalized (Kilbas-Saigo) Mittag–Leffler type function [21, Chapter 5].By using the above solution we can establish the following result. Corollary 4.1. Let < α . Then the fractional Cauchy problem of wave type (cid:0) t α / D x + f (cid:0) C ∂ α t (cid:1) + f + (cid:1) v ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,v ( x, t ) | t =0 = h ( x ) ,∂ t v ( x, t ) | t =0 = h ( x ) , (4.3) ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 19 can be solved, and the solution is given by v ( x, t ) = t α / D x ( h ( x )) + t α / D x ( h ( x ))+ t α / Γ( α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α ( x − y )(2 π | x − y | ) n/ ×× Z + ∞ r E α , α ,α ( − r u α ) r n/ rJ n ( r | x − y | ) drdu (cid:19) dy + t α / Γ( α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α +1 ( x − y )(2 π | x − y | ) n/ ×× Z + ∞ r E α , α ,α +1 ( − r u α ) r n/ rJ n ( r | x − y | ) drdu (cid:19) dy + f (cid:18) − t α Z R n h ( y ) (cid:18) | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α ( − r t α ) r n/ J n − ( r | x − y | ) dr (cid:19) dy − Z R n h ( y ) (cid:18) | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α +1 ( − r t α ) r n/ J n − ( r | x − y | ) dr (cid:19) dy (cid:19) f + (cid:18) h ( x ) + h ( x ) t − α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α ( − r u α ) r n/ J n − ( r | x − y | ) drdu (cid:19) dy − α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α +1 | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α +1 ( − r u α ) r n/ J n − ( r | x − y | ) drdu (cid:19) dy (cid:19) . Proof. By Theorem 3.3 we have that the solution of equation (4.3) is given bythe application of (cid:0) t β / D x + f (cid:16) C ∂ β t (cid:17) + f + (cid:1) to the representation (4.2). Let usthen calculate each component of the solution. First we need to recall some usefulestimates. Note that formula (25.11) in [40, Lemma 25.1] implies that1(2 π ) n Z R n e − is · x ϕ ( | s | ) ds = | x | − n/ (2 π ) n/ Z + ∞ ϕ ( r ) r n/ J n − ( r | x | ) dr, (4.4)where J ν denotes the Bessel function with index ν (for more details see e.g. [12]) and ϕ is a radial function such that Z + ∞ τ n − (1 + τ ) (1 − n ) / | ϕ ( τ ) | dτ < + ∞ , provided that the integral on the left-hand side of (4.4) is interpreted as conventionallyconvergent. It converges absolutely if Z + ∞ τ n − | ϕ ( τ ) | dτ < + ∞ . And, we also have that [13]: D x (cid:0) | x | − n J n − ( r | x | ) (cid:1) = − rx | x | n J n ( r | x | ) , x ∈ R n , r ≥ . (4.5)By (4.2) and (4.4) we get w ( x, t ) = h ( x ) + h ( x ) t − α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α ( − r u α ) r n/ J n − ( r | x − y | ) drdu (cid:19) dy − α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α +1 | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α +1 ( − r u α ) r n/ J n − ( r | x − y | ) drdu (cid:19) dy. (4.6)Let us calculate each component of x,t D ,α t α ; t w ( x, t ) where w ( x, t ) is given in (4.6).By (4.5), we get the first component: t α / D x w ( x, t ) = t α / D x ( h ( x )) + t α / D x ( h ( x ))+ t α / Γ( α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α ( x − y )(2 π | x − y | ) n/ ×× Z + ∞ r E α , α ,α ( − r u α ) r n/ rJ n ( r | x − y | ) drdu (cid:19) dy + t α / Γ( α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α +1 ( x − y )(2 π | x − y | ) n/ ×× Z + ∞ r E α , α ,α +1 ( − r u α ) r n/ rJ n ( r | x − y | ) drdu (cid:19) dy. ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 21 By (4.6) and Theorem 2.3 we obtain the second component as follows: C ∂ α t w ( x, t )= − t α Z R n h ( y ) (cid:18) | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α ( − r t α ) r n/ J n − ( r | x − y | ) dr (cid:19) dy − t α +1 Z R n h ( y ) (cid:18) | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α +1 ( − r t α ) r n/ J n − ( r | x − y | ) dr (cid:19) dy. (cid:3) Similarly as the above statement the following assertion can be obtained. Corollary 4.2. Let < α . Then the fractional Cauchy problem of heat type ((cid:0) t α / D x + f (cid:0) C ∂ α t (cid:1) + f + (cid:1) v ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,v ( x, t ) | t =0 = h ( x ) , (4.7) can be solved, and the solution is given by v ( x, t ) = t α / D x ( h ( x )) + t α / Γ( α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α ( x − y )(2 π | x − y | ) n/ ×× Z + ∞ r E α , α ,α ( − r u α ) r n/ rJ n ( r | x − y | ) drdu (cid:19) dy + f (cid:18) − t α Z R n h ( y ) (cid:18) | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α ( − r t α ) r n/ J n − ( r | x − y | ) dr (cid:19) dy (cid:19) f + (cid:18) h ( x ) − α ) Z R n h ( y ) (cid:18)Z t ( t − u ) α − u α | x − y | − n/ (2 π ) n/ ×× Z + ∞ r E α , α ,α ( − r u α ) r n/ J n − ( r | x − y | ) drdu (cid:19) dy (cid:19) . Fractional telegraph Dirac operator. Now we consider the following frac-tional telegraph Dirac operator x,t D ,α ,α a,c ; t := c D x + f (cid:0) C ∂ α t + a C ∂ α t (cid:1) + f + , where D x is the Dirac operator, a > c > 0, 0 < α < α 2. We have( x,t D ,α ,α a,c ; t ) = − c ∆ x + C ∂ α t + a C ∂ α t . Here we consider the case of constant coefficients. By using the obtained resultswe can directly prove the following statement. Nevertheless, we omit all calculationssince it is an analogue of [15, Theorem 4.1]. Corollary 4.3. Let < α , < α , a > and c > . Then the fractionalCauchy type problem (cid:0) cD x + f (cid:0) C ∂ α t + a C ∂ α t (cid:1) + f + (cid:1) v ( x, t ) = 0 , x ∈ R n , t ∈ (0 , T ] ,v ( x, t ) | t =0 = h ( x ) ,∂ t v ( x, t ) | t =0 = h ( x ) , (4.8) can be solved, and the solution is given by v ( x, t ) = Z R n H α ,α ( x − y, t ) h ( y ) dy + Z R n H α ,α ( x − y, t ) h ( y ) dy, where H α ,α and H α ,α are the first and second fundamental solutions given in for-mulas (4.3) and (4.4) of [15] . For the particular case a = 0, the above result coincides with the result in [14] forthe time-fractional parabolic Dirac operator.5. Inverse problems Now we combine some results given in Section 3 with a new method to findingthe variable coefficient of an inverse Cauchy type problem by the consideration oftwo (direct) fractional Cauchy type problems. The method was introduced recentlyin [17], and extended to some fractional differential equations in [32, 34] as well. Inthis section, we extend some recent results from [2] by using the Riemann-Liouvillefractional derivative of complex order, with respect to another function. Some recentresults of [34, 35] are used to establish the newer statements. We mainly focus insolving some inverse fractional Cauchy problems of wave and heat type.5.1. Fractional wave type equations. We will recover the variable coefficient Θ( t )for the following fractional Cauchy problem of wave type: (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ α,φt (cid:1) + f + (cid:1) w ( x, t ) = 0 , x ∈ R n , < t T < ∞ ,w ( x, t ) | t =0 = w ( x ) ,∂ t w ( x, t ) | t =0 = w ( x ) , (5.1)where 1 < α t ) > C ∂ α,φt w ( x, t ) := C D α,φ w ( x, t ) = D α,φ (cid:0) w ( x, t ) − w ( x, t ) | t =0 − ∂ t w ( x, t ) | t =0 ( φ ( t ) − φ (0)) (cid:1) . Notice that the formal passage α → α = 2. Now applying (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ αt (cid:1) + f + (cid:1) to equation (5.1), it follows that (cid:0) − Θ( t )∆ x + C ∂ α,φt (cid:1) w ( x, t ) = 0 , x ∈ R n , < t T < ∞ ,w ( x, t ) | t =0 = w ( x ) ,∂ t w ( x, t ) | t =0 = w ( x ) , (5.2) ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 23 whose solution is given by (Theorem 2.4) w ( x, t ) − w ( x ) − w ( x )( φ ( t ) − φ (0)) = Z R n (cid:0) F − H ( t, | s | , Θ (cid:1) ( x − y ) (cid:0) χ Ω w (cid:1) ( y ) dy + Z R n (cid:0) F − H ( t, | s | , Θ (cid:1) ( x − y ) (cid:0) χ Ω w (cid:1) ( y ) dy. (5.3)To recover the continuous variable coefficient Θ in the inverse problem (5.1), westudy the following two direct fractional Cauchy type problems: (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ α,φt (cid:1) + f + (cid:1) w ( x, t ) = 0 , x ∈ R n , < t T < + ∞ ,w ( x, t ) | t =0 = 0 ,∂ t w ( x, t ) | t =0 = w ( x ) , (5.4)and (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ α,φt (cid:1) + f + (cid:1) v ( x, t ) = 0 , x ∈ R n , < t T < + ∞ ,v ( x, t ) | t =0 = 0 ,∂ t v ( x, t ) | t =0 = ∆ x w ( x ) , (5.5)where Ω ⊂ R n is an open, bounded set with piecewise smooth boundary ∂ Ω and w ∈ C (Ω) ∩ C (Ω) is a given function which satisfy supp( w ) ⊂ Ω. To apply themethod showed in [34, Section 5], we need an additional initial data at a fixed point q ∈ Ω. Indeed, let us fix an arbitrary observation point q ∈ Ω for two time dependentvalues: h ( t ) := w ( x, t ) | x = q − w ( x ) | x = q ( φ ( t ) − φ (0)) , h ( t ) := v ( x, t ) | x = q , < t T, where w ( x, t ) and v ( x, t ) are the solutions (which form is given by formula (5.3))of the transformed equations (5.4) and (5.5) after the application of (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ αt (cid:1) + f + (cid:1) respectively. In the proof of the next result will be very clear thosequantities.Let us now establish one of the main results on this section. Theorem 5.1. Let the following conditions be satisfied: (1) k Θ k max I α,φ e νt Ce νt for some ν > and a constant < C < independentof t , (2) w ∈ C (Ω) ∩ C (Ω) and supp ( w ) ⊂ Ω , (3) h ∈ C [0 , T ] and h ( t ) = 0 for any t ∈ (0 , T ) , (4) D α,φ h ( t ) h ( t ) > K > for any t ∈ (0 , T ] .Then, the fractional inverse Cauchy problem (5.4), (5.5) has a solution given by Θ( t ) = D α,φ h ( t ) h ( t ) , t ∈ (0 , T ] . Proof. By Theorem 2.4, the solutions of equations (5.4) and (5.5) are given by theapplication of (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ αt (cid:1) + f + (cid:1) to the following representations respectively: w ( x, t ) − w ( x )( φ ( t ) − φ (0)) = Z R n F − (cid:0) H ( t, | s | , Θ (cid:1) ( x − y ) (cid:0) χ Ω w (cid:1) ( y ) dy, and v ( x, t ) − ∆ x w ( x )( φ ( t ) − φ (0)) = Z R n F − (cid:0) H ( t, | s | , Θ (cid:1) ( x − y ) (cid:0) χ Ω ∆ y w (cid:1) ( y ) dy, where χ Ω is the characteristic function of Ω. Due to the additional data at q ∈ Ω, h ( t ) = Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q w ( y ) dy, and h ( t ) = Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q ∆ y w ( y ) dy + ∆ x w ( x ) | x = q ( φ ( t ) − φ (0)) . By (5.2), (5.3), the definition of h and h at the point x = q , and by the Greensecond formula we arrive at D α,φ h ( t ) = C ∂ α,φt w ( x, t ) | x = q = Θ( t )∆ x (cid:18)Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q w ( y ) dy + w ( x ) | x = q ( φ ( t ) − φ (0)) (cid:19) = Θ( t ) (cid:18)Z Ω ∆ y (cid:0) F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q (cid:1) w ( y ) dy + ∆ x w ( x ) | x = q ( φ ( t ) − φ (0)) (cid:19) = Θ( t ) (cid:18)Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q ∆ y w ( y ) dy + ∆ x w ( x ) | x = q ( φ ( t ) − φ (0)) (cid:19) = Θ( t ) (cid:0) w ( x, t ) | x = q − ∆ x w ( x ) | x = q ( φ ( t ) − φ (0)) + ∆ x w ( x ) | x = q ( φ ( t ) − φ (0)) (cid:1) = Θ( t ) h ( t ) . Since the function Θ is assumed to be continuous, we are to require that h ( t ) = 0for any t ∈ (0 , T ]. Note also that we have considered Θ( t ) > K > 0, hence we haveto request at the beginning D α h ( t ) h ( t ) > K > t ∈ (0 , T ]. (cid:3) Fractional heat type equations. Let us study the inverse problem in recov-ering the thermal diffusivity Θ in the following fractional Cauchy problem ((cid:0) Θ / ( t ) D x + f (cid:0) C ∂ α,φt (cid:1) + f + (cid:1) w ( x, t ) = 0 , x ∈ R n , < t T < ∞ ,w ( x, t ) | t =0 = w ( x ) , (5.6)where 0 < α t ) > C ∂ α,φt w ( x, t ) = C D α,φ w ( x, t ) = D α,φ (cid:0) w ( x, t ) − w ( x, t ) | t =0 (cid:1) , x ∈ R n , t > , and we get the Schr¨odinger equation in the particular case α = 1 of (5.6).To reconstruct the variable coefficient in (5.6), we use a similar procedure to thecase of the fractional wave equation. As before, we suppose that Ω ⊂ R n is an open,bounded set with a piecewise smooth boundary ∂ Ω and q ∈ Ω is a fixed point. Thus,we solve the inverse problem (5.6) by studying two fractional Cauchy problems withadditional data at the point q ∈ Ω: (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ α,φt (cid:1) + f + (cid:1) w ( x, t ) = 0 , x ∈ R n , < t T < + ∞ ,w ( x, t ) | t =0 = w ( x ) ,w ( x, t ) | x = q − w ( x ) | x = q = h ( t ) , (5.7) ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 25 and (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ α,φt (cid:1) + f + (cid:1) v ( x, t ) = 0 , x ∈ R n , < t ≤ T < + ∞ ,v ( x, t ) | t =0 = ∆ x w ( x ) ,v ( x, t ) | x = q = h ( t ) , (5.8)where w ∈ C (Ω) ∩ C (Ω) and supp( w ) ⊂ Ω ⊂ R n . Theorem 5.2. Let the following conditions be satisfied: (1) k Θ k max I α,φ e νt Ce νt for some ν > and a constant < C < independentof t , (2) w ∈ C (Ω) ∩ C (Ω) and supp ( w ) ⊂ Ω , (3) h ∈ C [0 , T ] such that h ( t ) = 0 for any t ∈ (0 , T ) , (4) D α,φ h ( t ) h ( t ) > K > for any t ∈ (0 , T ] .Then, the fractional inverse Cauchy problem (5.7), (5.8) has a solution given by Θ( t ) = D α,φ h ( t ) h ( t ) , t ∈ (0 , T ] . Proof. The solutions of the fractional Cauchy problems (5.7) and (5.8) are givenby the application of (cid:0) Θ / ( t ) D x + f (cid:0) C ∂ α,φt (cid:1) + f + (cid:1) to the following representationsrespectively: w ( x, t ) − w ( x ) = Z R n F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) (cid:0) χ Ω w (cid:1) ( y ) dy, and v ( x, t ) − ∆ x w ( x ) = Z R n F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) (cid:0) χ Ω ∆ y w (cid:1) ( y ) dy. By the additional data at the point q ∈ Ω, we also have h ( t ) = Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q w ( y ) dy and h ( t ) = Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q ∆ y w ( y ) dy + ∆ x w ( x ) | x = q . By (5.2), (5.3), the definition of h and h at the point x = q , and by the Greensecond formula we get the following equivalences: D α,φ h ( t ) = C ∂ α,φt w ( x, t ) | x = q = Θ( t )∆ x (cid:18)Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q w ( y ) dy + w ( x ) | x = q (cid:19) = Θ( t ) (cid:18)Z Ω ∆ y (cid:0) F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q (cid:1) w ( y ) dy + ∆ x w ( x ) | x = q (cid:19) = Θ( t ) (cid:18)Z Ω F − (cid:0) H ( t, | s | , Θ) (cid:1) ( x − y ) | x = q ∆ y w ( y ) dy + ∆ x w ( x ) | x = q (cid:19) = Θ( t ) (cid:0) v ( x, t ) | x = q − ∆ x w ( x ) | x = q + ∆ x w ( x ) | x = q (cid:1) = Θ( t ) h ( t ) , which finishes the proof. (cid:3) Examples. Let us consider the particular case 1 < β φ ( t ) = t ofTheorem 5.1. We have (cid:0) t β / D x + f (cid:0) C ∂ β t (cid:1) + f + (cid:1) w ( x, t ) = 0 , x ∈ R n , < t T < ∞ ,w ( x, t ) | t =0 = 0 ,∂ t w ( x, t ) | t =0 = w ( x ) ,w ( x, t ) | x = q − w ( x ) | x = q t = h ( t ) , q ∈ Ω , (5.9)and (cid:0) t β / D x + f (cid:0) C ∂ β t (cid:1) + f + (cid:1) v ( x, t ) = 0 , x ∈ R n , < t T < ∞ ,v ( x, t ) | t =0 = 0 ,∂ t v ( x, t ) | t =0 = ∆ x w ( x ) ,v ( x, t ) | x = q = h ( t ) , q ∈ Ω , (5.10)where supp( w ) ⊂ Ω ⊂ R n . The solutions of (5.9) and (5.10) are given by theapplication of (cid:0) t β / D x + f (cid:0) C ∂ β t (cid:1) + f + (cid:1) to the following representations respectively(see formula (4.2)): w ( x, t ) = w ( t ) t − Z R n F − (cid:0) I β (cid:0) | s | t β +1 E β , β ,β +1 ( −| s | t β ) (cid:1) ( x − y ) w ( y ) dy, (5.11)and v ( x, t ) = ∆ x w ( t ) t − Z R n F − (cid:0) I β (cid:0) | s | t β +1 E β , β ,β +1 ( −| s | t β ) (cid:1) ( x − y )∆ y w ( y ) dy. (5.12)Further, by Theorem 2.3 and (5.11) we obtain D β +0 h ( t ) = Z Ω w ( y ) (cid:16) t β +1 F − (cid:0) | s | E β , β ,β +1 ( −| s | t β ) (cid:1) ( x − y ) (cid:17) dy. Since (5.11) is the solution of (4.1), we get D β +0 h ( t ) = C ∂ β t w ( x, t ) | x = q = t β ∆ x w ( x, t ) | x = q . On the other hand, we get − h ( t ) = Z Ω ∆ y w ( y ) I β (cid:16) t β +1 F − (cid:0) | s | E β , β ,β +1 ( −| s | t β ) (cid:1) ( x − y ) | x = q (cid:17) dy − ∆ x w ( x ) | x = q t. ENERALIZED FRACTIONAL DIRAC TYPE OPERATORS 27 Now, applying Green’s second formula and (5.11) we get − h ( t ) = Z Ω ∆ y w ( y ) I β (cid:16) t β +1 F − (cid:0) | s | E β , β ,β +1 ( −| s | t β ) (cid:1) ( x − y ) | x = q (cid:17) dy − ∆ x w ( x ) | x = q t = Z Ω w ( y ) I β (cid:16) t β +1 ∆ y F − (cid:0) | s | E β , β ,β +1 ( −| s | t β ) (cid:1) ( x − y ) | x = q (cid:17) dy − ∆ x w ( x ) | x = q t = ∆ x (cid:18)Z Ω w ( y ) I β (cid:16) t β +1 F − (cid:0) | s | E β , β ,β +1 ( −| s | t β ) (cid:1) ( x − y ) | x = q (cid:17) dy − w ( x ) | x = q t (cid:19) = − ∆ x w ( x, t ) | x = q . Hence Θ( t ) = t β = D β +0 h ( t ) h ( t ) = t β ∆ x w ( x, t ) | x = q ∆ x w ( x, t ) | x = q . We finish this section with the following example. We consider the one-dimensionalcase of the equations (5.4) and (5.5) with φ ( t ) = t , α = 2: (cid:0) Θ / ( t ) ∂ x + f (cid:0) ∂ t (cid:1) + f + (cid:1) u ( x, t ) = 0 , x ∈ R , < t T < + ∞ ,u ( x, t ) | t =0 = 0 ,∂ t u ( x, t ) | t =0 = u ( x ) ,u ( x, t ) | x = q − u | x = q t = h ( t ) , q ∈ Ω , (5.13)and (cid:0) Θ / ( t ) ∂ x + f (cid:0) ∂ t (cid:1) + f + (cid:1) w ( x, t ) = 0 , x ∈ R , < t T < + ∞ ,w ( x, t ) | t =0 = 0 ,∂ t w ( x, t ) | t =0 = ∂ x u ( x ) ,w ( x, t ) | x = q = h ( t ) , q ∈ Ω , (5.14)where u ( x ) = sin x in Ω and supp( u ) ⊂ Ω ⊂ R . For simplicity, we also assume thatΘ( t ) = c for some nonzero constant c ∈ R . Equations (5.13) and (5.14) are of wavetype in the one-dimensional space, where their solutions are given by the applicationof (cid:0) Θ / ( t ) ∂ x + f (cid:0) ∂ t (cid:1) + f + (cid:1) to the following formulas respectively (for more details,see [34, Section 5.3]): u ( x, t ) = − c (cid:0) cos( x + ct ) − cos( x − ct )) , and w ( x, t ) = 12 c (cid:0) cos( x + ct ) − cos( x − ct )) . Hence, D α h ( t ) = D t h ( t ) = ∂ t u ( t, q ) = c c (cid:0) cos( q + ct ) − cos( q − ct )) , and we get Θ( t ) = c = D α h ( t ) h ( t ) . Conflict of interest statement This work does not have any conflicts of interest.7. Acknowledgements First and third authors were supported by the Nazarbayev University Program091019CRP2120. The second author was supported by the FWO Odysseus 1 grantG.0H94.18N: Analysis and Partial Differential Equations and by the EPSRC GrantEP/R003025/1. Joel E. 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Restrepo:Department of MathematicsNazarbayev UniversityKazakhstan&Department of MathematicsUniversity of AntioquiaColombia E-mail address [email protected]; [email protected] Michael Ruzhansky:Department of MathematicsGhent University, BelgiumandSchool of Mathematical SciencesQueen Mary University of LondonUnited Kingdom E-mail address Durvudkhan Suragan:Department of MathematicsNazarbayev UniversityKazakhstan