fractional calculus

CHAPTER 3 FRACTIONAL CALCULUS ANDFRACTIONAL DIFFERENTIALEQUATIONS [This chapter is based on the lectures of Professor R.K. Saxen a of Jai Narain Vyas University, Jodhpur, Rajasthan, India . ] 3.0. Introduction This section deals with certain properties of fractional ca lculus associated with Laplace and Mellin transforms. Composition relations between Riemann- Liouville fractional calculus and generalized Mittag-Le ffler functions are pre- sented. Applications of fractional calculus in the solutio n of differential and integralequationsoffractionalorderaredemonstrated. T hisstudywillbringthe reader totheresearch level. 3.1. LaplaceTransformoftheFractionalIntegral 3.1.1. Laplace transform Notation3.1.1. F(s)=L{f(t);s}=(Lf)(s) : Laplace transform of f(t) withparam- eters. Notation3.1.2. L−1{f(s);t}: Inverse Laplace transform Definition3.1.1. TheLaplacetransformofafunction f(t),denotedby F(s), isdefined by theequation F(s)=(Lf)(s)=L{f(t);s}=/integraldisplay∞ 0e−stf(t)dt, (3.1.1) 175 176 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS whereℜ(s)>0, which maybesymbolicallywrittenas F(s)=L{f(t);s}orf(t)=L−1{F(s);t}, provided that the function f(t) is continuous for t≥0, it being tacitly assumed thattheintegralin(3.1.1)exists. Example 3.1.1. Provethat L−1{s−ρ}=tρ−1 Γ(ρ),ℜ(s)>0,ℜ(ρ)>0. (3.1.2) It follows from the Laplace integral /integraldisplay∞ 0e−sttρ−1dt=Γ(s) sρ,ℜ(s)>0,ℜ(ρ)>0. (3.1.3) Example 3.1.2. Find the inverse Laplace transform ofF(s) a+sα;a,α >0; where ℜ(s)>0,F(s)=L{f(t);s}. Solution3.1.1. Let G(s)=1 a+sα=∞/summationdisplay r=0(−a)rs−α−αr,|a sα|<1. Therefore, L−1{G(s)}=g(t)=L−1∞/summationdisplay r=0(−a)rs−α−αr =tα−1Eα,α(−atα). (3.1.4) Applicationofconvolutiontheorem ofLaplace transformyi eldstheresult L−1/braceleftBiggF(s) a+sα;t/bracerightBigg =/integraldisplayx 0(x−t)α−1Eα,α(−a(x−t)α)f(t)dt (3.1.5) whereℜ(α)>0. 3.1. LAPLACE TRANSFORM OF THE FRACTIONAL INTEGRAL 177 3.1.2. Laplace transformof the fractionalintegral Wehave 0I−ν xf(x)=1 Γ(ν)/integraldisplayx 0(x−t)ν−1f(t)dt, (3.1.6) whereℜ(ν)>0. ApplicationofconvolutiontheoremoftheLaplacetransfor mgives L{0I−ν xf(x);s}=L/braceleftBiggtν−1 Γ(ν)/bracerightBigg L{f(t);s} =s−νF(s), (3.1.7) whereℜ(s)>0,ℜ(ν)>0. 3.1.3. Laplace transformof the fractionalderivative Ifn∈N, thenby thetheory oftheLaplace transform,weknowthat L/braceleftBiggdn dxnf;s/bracerightBigg =snF(s)−n−1/summationdisplay k=0sn−k−1f(k)(0+) (3.1.8) =snF(s)−n−1/summationdisplay k=0skf(n−k−1)(0+),(n−1≤α<n) (3.1.9) whereℜ(s)>0 andF(s)istheLaplacetransformof f(t). By virtueofthedefinitionofthederivative,wefind that L{0Dα xf;s}=L/braceleftBiggdn dxn0In−α xf;s/bracerightBigg 178 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS =snL/braceleftbig 0In−α xf;s/bracerightbig−n−1/summationdisplay k=0skdn−k−1 dxn−k−10In−α xf(0+) =sαF(s)−n−1/summationdisplay k=0skDα−k−1f(0+),/parenleftBigg D=d dx/parenrightBigg (3.1.10) =sαF(s)−n/summationdisplay k=1sk−1Dα−kf(0+) (3.1.11) whereℜ(s)>0. 3.1.4. Laplace transformofCaputo derivative Notation3.1.3. C 0aDα x Definition 3.1.2. The Caputo derivative of a casual function f(t) ( that is f(t)=0fort<0)withα>0 was defined by Caputo(1969)intheform C 0aDα xf(x)=aIn−α xdn dxnf(x)=aD−(n−α) tfn(t) (3.1.12) =1 Γ(n−α)/integraldisplayx a(x−t)n−α−1fn(t)dt,(n−1<α<n) (3.1.13) wheren∈N. From (3.1.7)and(3.1.13), itfollowsthat L{C 00Dα tf(t);s}=s−(n−α)L{f(n)(t)}. (3.1.14) On using(3.1.8), weseethat L{C 00Dα tf(t);s}=s−(n−α)snF(s)−n−1/summationdisplay k=0sn−k−1f(k)(0+) =sαF(s)−n−1/summationdisplay k=0sα−k−1f(k)(0+),(n−1<α≤n),(3.1.15) whereℜ(s)>0 andℜ(α)>0. 3.1. LAPLACE TRANSFORM OF THE FRACTIONAL INTEGRAL 179 Note3.1.1. From (3.1.12),it can beseen that C 00Dα tA=0,whereAisaconstant , whereas theRiemann-Liouvillederivative 0Dα tA=At−α Γ(1−α),(α/nequal1,2,···), (3.1.16) whichis aremarkableresult. Exercises 3.1. 3.1.1.Provethat (0I−ν xf)(x)=L−1x−νL{f(x)}, (3.1.17) whereℜ(ν)>0. 3.1.2.Provethat (xWν ∞Lf)(x)=Lx−νL−1f(x), (3.1.18) whereℜ(ν)>0. 3.1.3.ProvethatthesolutionofAbelintegralequationoftheseco ndkind φ(x)−λ Γ(α)/integraldisplayx 0φ(t)dt (x−t)1−α=f(x),0<x<1 α>0,is givenby φ(x)=d dx/integraldisplayx 0Eα[λ(x−t)α]f(t) dt, (3.1.19) whereEα(x) is theMittag-Lefflerfunctiondefined by equation(3.5.1). 3.1.4.Show that λ Γ(α)/integraldisplayx 0Eα(λtα) (x−t)1−αdt=Eα(λxα)−1, α>0. (3.1.20) 180 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS 3.2. MellinTransformoftheFractionalIntegralsand theFractionalDerivatives 3.2.1. Mellin transform Notation3.2.1. m{f(x);s},f∗(s),the Mellin transform Notation3.2.2. m−1{f∗(s);x},Inverse Mellin transform Definition3.2.1. TheMellintransformofafunction f(x),denotedby f∗(s), isdefined by f∗(s)=m{f(x);s}=/integraldisplay∞ 0xs−1f(x)dx,x>0. (3.2.1) TheinverseMellintransformis givenbythecontourintegra l f(x)=m−1{f∗(s);x}=1 2πi/integraldisplayγ+i∞ γ−i∞f∗(s)x−sds,i=√ −1 (3.2.2) whereγisreal. 3.2.2. Mellin transformofthe fractionalintegral Theorem 3.2.1. Thefollowingresultholdstrue. m(0Iα xf)(s)=Γ(1−α−s) Γ(1−α)f∗(s+α), (3.2.3) whereℜ(α)>0andℜ(α+s)<1. Proof3.2.1. Wehave m(0Iα xf)(s)=/integraldisplay∞ 0zs−11 Γ(α)/integraldisplayz 0(z−t)α−1f(t) dtdz =1 Γ(α)/integraldisplay∞ 0f(t) dt/integraldisplay∞ tzs−1(z−t)α−1dz(3.2.4) onsetting z=t u, thez-integralbecomes 3.2. MELLIN TRANSFORM OF THE FRACTIONAL INTEGRALS 181 tα+s−1/integraldisplay1 0u−α−s(1−u)α−1du=tα+s−1B(α,1−α−s),(3.2.5) whereℜ(α)>0,ℜ(α+s)<1. Puttingtheabovevalueof z-integral,theresult follows. Similarlywecan establish Theorem 3.2.2. Thefollowingresult holdstrue. m(xIα ∞f)(s)=Γ(s) Γ(s+α)m{tαf(t);s} =Γ(s) Γ(s+α)f∗(s+α), (3.2.6) whereℜ(α)>0,ℜ(s)>0. Note 3.2.1. If we set f(x)=x−αφ(x),then using the property of the Mellin transform xαφ(x)↔φ∗(s+α), (3.2.7) theresults(3.2.3)and (3.2.6)become (0Iα xx−αf(x))(s)=Γ(1−α−s) Γ(1−s)f∗(s), (3.2.8) whereℜ(α)>0,ℜ(α+s)<1 and (xIα ∞x−αf(x))(s)=Γ(s) Γ(s+α)f∗(s), (3.2.9) whereℜ(α)>0,andℜ(s)>0,respectively. 3.2.3. Mellin transformof the fractionalderivative Theorem 3.2.3. Ifn∈N, then 182 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS m{f(n)(t); (s)}=(−1)nΓ(s) Γ(s−n)m{f(t);s−n}, (3.2.10) whereℜ(s)>0,ℜ(s−n)>0. Proof3.2.2. Integratebyparts andusingthedefinitionoftheMellintran sform, theresultfollows. Example 3.2.1. Find the Mellin transform of the fractional derivative. Solution3.2.1. Wehave 0Dα xf=0Dn x0Dα−n xf=0Dn x0In−α xf. (3.2.11) Therefore, m(0Dα xf)(s)=(−1)nΓ(s) Γ(s−n)m/braceleftbig 0In−α xf/bracerightbig(s−n),(n−1≤ℜ(α)<n) (3.2.12) =(−1)nΓ(s)Γ(1−(s−α)) Γ(s−n)Γ(1−s+n)m{f(t);s−α}, (3.2.13) whereℜ(s)>0,ℜ(s)<1+ℜ(α). Remark3.2.1. Analternativeformof(3.2.13) isgivenin Exercise3.2.2. Exercises3.2. 3.2.1ProveTheorem 3.2.2. 3.2.2Provethat theMellintransformoffractional derivativeis givenby m(0Dα xf)(s)=(−1)nΓ(s)sin[π(s−n)] Γ(s−α)sin[π(s−α)]m{f(t);s−α}, (3.2.14) whereℜ(s)>0,ℜ(α−s)>−1. 3.2.3Find theMellintransformof(1 +xa)−b;a,b>0. 3.3. KOBER OPERATORS 183 3.3. KoberOperators Kober operators are the generalization of Riemann - Liouvil leand Weyl op- erators. Theseoperatorshavebeenusedbymanyauthorsinde rivingthesolution ofsingle,dualandtripleintegralequationspossessingsp ecialfunctionsofmath- ematicalphysics,as theirkernels. Notation3.3.1. Kober operator of thefirst kind I[f(x)],I[α,η:f(x)],I(α,η)f(x),Eα,η 0,xf,In,α xf. Notation3.3.2. Kober operator of thesecond kind R[f(x)],R[α,ζ:f(x)],R(α,ζ)f(x),Kα,ζ x,∞f,Kζ,α xf. Definition 3.3.1. I[f(x)]=I[α,η:f(x)]=I(α,η)f(x)=Eα,η 0,xf =Iη,α xf=x−η−α Γ(α)/integraldisplayx 0(x−t)α−1tηf(t)dt, (3.3.1) whereℜ(α)>0. Definition 3.3.2. R[f(x)]=R[α,ζ:f(x)]=R(α,ζ)f(x)=Kα,ζ x,∞f =Kζ,α xf=xζ Γ(α)/integraldisplay∞ x(t−x)α−1t−ζ−αf(t)dt,(3.3.2) whereℜ(α)>0. (3.3.1)and (3.3.2)holdtrueunderthefollowingcondition s: f∈Lp(0,∞),ℜ(α)>0,ℜ(η)>−1 q,ℜ(ζ)>−1 p,1 p+1 q=1,p≥1. Whenη=0,(3.3.1)reduces toRiemann -Liouvilleoperator. Thatis, I0,α xf=x−α 0Iα xf. (3.3.3) Forζ=0,(3.3.2)yieldstheWeyl operatorof t−αf(t). Thatis, 184 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS K0,α xf=xWα ∞t−αf(t). (3.3.4) Theorem 3.3.1. [Kober (1940)]. Ifℜ(α)>0,ℜ(η−s)>−1,f∈Lp(o,∞),1≤p≤2( or f∈Mp(o,∞), a subspace of L p(o,∞)and p>2),ℜ(η)>−1 q,1 p+1 q=1,then there holds the formula m{I(α,η)f}(s)=Γ(1+η−s) Γ(α+η+1−s)m{f(x);s}. (3.3.5) Proof3.3.1. It issimilarto theproofofTheorem3.2.1. In asimilarmanner, wecan establish Theorem 3.3.2. [Kober (1940)]. Ifℜ(α)>0,ℜ(s+ζ)>0,f∈Lp(o,∞),1≤p≤2( or f∈Mp(o,∞),a subspaceof L p(o,∞)and p>2) ℜ(ζ)>−1 p,1 p+1 q=1, then, m{ℜ(α,ζ)f}(s)=Γ(ζ+s) Γ(α+ζ+s)m{f(x);s}. (3.3.6) Semigroup propertyoftheKoberoperators hasbeen givenint heformof Theorem3.3.3. If f∈Lp(o,∞),g∈Lq(o,∞),1 p+1 q=1,ℜ(η)>−1 q,ℜ(ζ)> −1 p,1≤p≤2,(or f∈Mp(o,∞),a subspaceof L p(o,∞)and p>2), then /integraldisplay∞ 0g(x)(I(α,η:f))(x)dx=/integraldisplay∞ 0f(x)(R(α,ζ:g))(x)dx. (3.3.7) Proof3.3.2. Interchangetheorderofintegration. 3.3. KOBER OPERATORS 185 Remark3.3.1. Operatorsdefinedby(3.3.1.) and(3.3.2)arealsocalledErd ´ elyi- Koberoperators. Exercises 3.3. 3.3.1Provetheorem 3.3.1. 3.3.2For the modified Erd´ elyi-Kober operators, defined by the fol lowing equa- tionsform>0: I(α,η:m)f(x)=I(f(x) :α,η,m) =m Γ(α)x−η−mα+m−1/integraldisplayx otη(xm−tm)α−1f(t)dt,(3.3.8) and R(α,ζ:m)f(x)=R(f(x) :α,ζ,m) =mxζ Γ(α)/integraldisplay∞ xt−ζ−mα+m−1(tm−xm)α−1f(t)dt, (3.3.9) wheref∈Lp(0,∞),ℜ(α)>0,ℜ(η)>−1 q,ℜ(ζ)>−1 p,1 p+1 q=1,find the Mellin transforms of (i) I(α,η:m)f(x) and (ii)R(α,ζ:m)f(x) , giving theconditionsofvalidity. 3.3.3Fortheoperators defined by (3.3.8)and (3.3.9.),showthat /integraldisplay∞ 0R(f(x) :α,η,m)g(x)dx=/integraldisplay∞ 0f(x)I(g(x) :α,η,m)dx,(3.3.10) where the parameters α,η,mare the same in both the operators IandR. Give conditionsofvalidityof(3.3.10). 3.3.4FortheErd´ elyi-Koberoperator, defined by Iη,αf(x)=2x−2α−2η Γ(α)/integraldisplayx 0(x2−t2)α−1t2η+1f(t)dt, (3.3.11) 186 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS whereℜ(α)>0,establishthefollowingresults(Sneddon(1975)): (i)Iη,αx2βf(x)=x2βIη+β,αf(x) (3.3.12) (ii)Iη,αIη+α,β=Iη,α+β=Iη+α,β,Iη,α (3.3.13) (iii) I−1 η,α=Iη+α,−α. (3.3.14) Remark 3.3.2. The results of Exercise 3.3.4 also hold for the operator, defi ned by Kη,αf(x)=2x2η Γ(α)/integraldisplay∞ x(t2−x2)α−1t−2α−2η+1f(t)dt,(3.3.15) whereℜ(α)>0. Remark 3.3.3. Operators more general than the operators defined by (3.3.11 ) and (3.3.15) are recently defined by Galu´ e et al [Integral Tr ansform & Spec. Funct. Vol. 9 (2000),No. 3,pp. 185-196]intheform aIη,α xf(x)=x−η−α Γ(α)/integraldisplayx a(x−t)α−1tηf(t)dt, (3.3.16) whereℜ(α)>0. 3.4. GeneralizedKoberOperators Notation3.4.1. I[α,β,γ:m,µ,η,a:f(x)],I[f(x)] Notation3.4.2. I[α,β,γ:m,µ,δ,a:f(x)],I[f(x)] Notation3.4.3. R[f(x)],R/bracketleftBigα,β,γ; σ,ρ,a;:f(x)/bracketrightBig Notation3.4.4. K[f(x)],K/bracketleftBigα,β,γ; δ,ρ,a;:f(x)/bracketrightBig Notation3.4.5. Iα,β,η; 0,xf(x) (Saigo, 1978) Notation3.4.6. Jα,β,η; x,αf(x) (Saigo, 1978) 3.4. GENERALIZED KOBER OPERATORS 187 Definition 3.4.1. I[f(x)]=I[α,β,γ:m,µ,η,a:f(x)] =µx−η−1 Γ(1−α)/integraldisplayx 02F1(α,β+m,γ;atµ xµ)tηf(t)dt,(3.4.1) where 2F1(·)istheGausshypergeometricfunction. Definition 3.4.2. I[f(x)]=I[α,β,γ:m,µ,δ,a:f(x)] =µxδ Γ(1−α)/integraldisplay∞ x2F1(α,β+m;γ;axµ tµ)t−δ−1f(t)dt.(3.4.2) Operatorsdefinedby(3.4.1)and(3.4.2)existunderthefoll owingconditions: (i) 1≤p,q<∞,p−1+q−1=1,|arg(1−a)|<π (ii)ℜ(1−α)>m,ℜ(η)>−1 q,ℜ(δ)>−1 p,ℜ(γ−α−β−m)>−1,m∈N0;γ/nequal 0,−1,−2,··· (iii)f∈Lp(0,∞) Equations(3.4.1)and (3.4.2)are introducedby Kallaand Sa xena(1969). Forγ=β, (3.4.1) and (3.4.2) reduce to generalized Kober operators , given bySaxena(1967). Definition 3.4.3. R[f(x)]=R/bracketleftBigα,β,γ; σ,ρ,a;f(x)/bracketrightBig =x−σ−ρ Γ(ρ)/integraldisplayx 0tσ(x−t)ρ−1 2F1[α,β;γ;a(1−t x)]f(t)dt.(3.4.3) Definition 3.4.4. K[f(x)]=K/bracketleftBigα,β,γ; δ,ρ,a;f(x)/bracketrightBig =xδ Γ(ρ)/integraldisplay∞ xt−δ−ρ(t−x)ρ−1 2F1[α,β;γ;a(1−x t)]f(t)dt.(3.4.4) 188 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS The conditions of validity of the operators (3.4.3) and (3.4 .4) are given be- low: (i)p≥1,q<∞,p−1+q−1=1,|arg(1−a)|<π. (ii)ℜ(σ)>−1 q,ℜ(δ)>−1 p,ℜ(ρ)>0. (iii)γ/nequal0,−1,−2,···;ℜ(γ−α−β)>0. (iv)f∈Lp(0,∞). Theoperatorsdefined by (3.4.3)and (3.4.4)aregivenby Saxe naandKumb- hat(1973). When aisreplacedbya αandαtendstoinfinity,theoperatorsdefined by(3.4.3)and(3.4.4)reducetothefollowingoperatorsass ociatedwithconfluent hypergeometricfunctions. Definition 3.4.5. R/bracketleftBigβ,γ; σ,ρ,a;f(x)/bracketrightBig =lim α→∞R/bracketleftbigg α,β,γ; σ,ρ,a α;f(x)/bracketrightbigg =x−σ−ρ Γ(ρ)/integraldisplayx 0Φ[β,γ;a(1−t x)]tσ(x−t)ρ−1f(t)dt.(3.4.5) Definition 3.4.6. K/bracketleftBigβ,γ; σ,ρ,a;f(x)/bracketrightBig =lim α→∞K/bracketleftbigg α,β,γ; δ,ρ,a α;f(x)/bracketrightbigg =xδ Γ(ρ)/integraldisplay∞ xΦ[β,γ;a(1−x t)]t−δ−ρ(t−x)ρ−1f(t)dt,(3.4.6) whereℜ(ρ)>0,ℜ(δ)>0. Remark 3.4.1. Many interesting and useful properties of the operators defi ned by (3.4.3) and (3.4.4) are investigated by Saxena and Kumbha t (1975), which deal withrelations of theseoperators with well-knowninte graltransforms, such as Laplace,MellinandHankel transforms. Equation(3.4.3) was firstconsidered byLove(1967). Remark 3.4.2. In the special case, when αis replaced by α+β,γbyα,σby zero,ρbyαandβby−η, then (3.4.3)reduces to theoperator(3.4.7)considered bySaigo(1978). Similarly,(3.4.4)reducestoanotheroper ator(3.4.9)introduced bySaigo (1978). 3.4. GENERALIZED KOBER OPERATORS 189 Definition 3.4.7. Letα,β,η∈C, and let x∈R+the fractional integral (ℜ(α)>0)andthefractionalderivative( ℜ(α)<0)ofthefirstkindofafunction f(x)onR+are defined bySaigo (1978)intheform Iα,β,η 0,xf(x)=x−α−β Γ(α)/integraldisplayx 0(x−t)α−1 ×2F1(α+β,−η;α;1−t x)f(t)dt,ℜ(α)>0 (3.4.7) =dn dxnIα+n,β−n,η−n 0,xf(x),0<ℜ(α)+n≤1,(n∈N0).(3.4.8) Definition3.4.8. Thefractionalintegral( ℜ(α)>0)andfractionalderivative (ℜ(α)<0)ofthesecondkindofafunction f(x)onR+aregivenbySaigo(1978) intheform Jα,β,η x,∞f(x)=1 Γ(α)/integraldisplay∞ x(t−x)α−1t−α−β ×2F1(α+β,−η;α;1−x t)f(t)dt,ℜ(α)>0 (3.4.9) =(−1)ndn dxnJα+n,β−n,η x,∞f(x),0<ℜ(α)+n≤1,(n∈N0).(3.4.10) Example3.4.1. Find the value of Iα,β,η 0,x/braceleftBig xσ−12F1(a,b;c;−a′x)/bracerightBig . Solution3.4.1. Wehave K=Iα,β,η 0,x/braceleftBig xσ−1 2F1(a,b;c;−ax)/bracerightBig =∞/summationdisplay r=0(a)r(b)r(−1)r(a′)r (c)rr!Iα,β,η 0,xxr+σ−1. ApplyingtheresultofExercise3.4.1,weobtain 190 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS K=xσ−β−1∞/summationdisplay r=0(−1)r(a)r(b)rΓ(σ+r)Γ(σ−β+η+r)(a′)r (c)rr!Γ(σ−β+r)Γ(α+η+σ+r)xr =xσ−β−1Γ(σ)Γ(σ+η−β) Γ(σ−β)Γ(σ+α+η) ×4F3(a,b,σ,σ+η−β;c,σ−β,σ+α+η;−a′x), whereℜ(α)>0,ℜ(σ)>0,ℜ(σ+η−β)>0,c/nequal0,−1,−2,···;|a′x|<1. Example 3.4.2. Find the value of Jα,β,η x,∞(xλ2F1(a,b;c;a′ x)). Solution 3.4.2. Following a similar procedure and using the result of Exerci se 3.4.3,itgives Jα,β,η x,∞(xλ 2F1(a,b;c;a′ x))=Γ(β−λ)Γ(η−λ) Γ(−λ)Γ(α+β+η−λ)xλ−β ×4F3(a,b,β−λ,η−λ;c,−λ,α+β+η−λ;a′ x), whereℜ(α)>0,ℜ(β−λ)>0,ℜ(η−λ)>0,x>0,c/nequal0,−1,−2,···;|x|>|a′|. Remark3.4.3. Specialcasesoftheoperators Iα,β,η 0,xandJα,β,η x,∞aretheoperatorsof Riemann -Liouville: Iα,−α,η 0,xf(x)=0D−α xf(x)=1 Γ(α)/integraldisplayx 0(x−t)α−1f(t)dt,(ℜ(α)>0) (3.4.11) theWeyl: Jα,−α,η x,∞f(x)=xWα ∞f(x)=1 Γ(α)/integraldisplay∞ x(t−x)α−1f(t)dt,(ℜ(α)>0) (3.4.12) 3.4. GENERALIZED KOBER OPERATORS 191 and theErd´ elyi- Koberoperators: Iα,0,η 0,xf(x)=Eα,η 0,xf(x)=x−α−η Γ(α)/integraldisplayx 0(x−t)α−1tηf(t)dt,(ℜ(α)>0) (3.4.13) and Jα,0,η x,∞f(x)=Kα,η x,∞f(x)=xη Γ(α)/integraldisplay∞ x(t−x)α−1t−α−ηf(t)dt,(ℜ(α)>0) (3.4.14) Example3.4.3. Provethe following theorem. Ifℜ(α)>0andℜ(s)<1+min[0,ℜ(η−β)],then the following formula holds for f(x)∈Lp(0,∞) with 1≤p≤2orf(x)∈Mp(0,∞) withp>2: m/braceleftBig xβIα,β,η 0,xf/bracerightBig =Γ(1−s)Γ(η−β+1−s) Γ(1−s−β)Γ(α+η+1−s)m{f(x)}. (3.4.15) Solution3.4.3. Usetheintegral /integraldisplay∞ xu−σ−γ(u−x)γ−1 2F1(α,β;γ;1−x u)du=Γ(γ)Γ(σ)Γ(γ+σ−α−β) Γ(γ+σ−α)Γ(γ+σ−β), (3.4.16) whereℜ(γ)>0,ℜ(σ)>0,ℜ(γ+σ−α−β)>0. Exercises3.4. 3.4.1.Provethat Iα,β,η 0,xxλ=Γ(1+λ)Γ(1+λ+η−β) Γ(1+λ−β)Γ(1+λ+α+η)xλ−β, (3.4.17) and givetheconditionsofvalidity. 3.4.2.FindtheMellintransformof xβJα,β,η x,∞f(x),givingconditionsofitsvalidity. 3.4.3.Provethat Jα,β,η x,∞xλ=Γ(β−λ)Γ(η−λ) Γ(−λ)Γ(α+β+η−λ)xλ−β(3.4.18) 192 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS and givetheconditionsofvalidity. 3.4.4.Provethat Iα,β,η 0,x(xke−λx)=Γ(k+1)Γ(η+k−β+1) Γ(k−β+1)Γ(α+η+k+1)xk−β ×2F2(k+1,η+k−β+1;k−β+1,α+η+k+1;−λx),(3.4.19) and givetheconditionsofvalidity. 3.4.5.Provethat Jα,β,η x,∞e−sx=sηxη−βΓ(β−η) Γ(α+β)Φ(1−α−β,1+η−β;−sx) +sβΓ(η−β) Γ(α+η)Φ(1−α−η,1+β−η;−sx), (3.4.20) and give the conditions of its validity. Deduce the results f orL[xWα ∞f](s) and L[Kα,η x,∞f](s). 3.4.6.Provethat [Saxenaand Nishimoto(2002)] Iα,β,η 0,x[xσ−1(a+bx)c]=acΓ(σ)Γ(σ+η−β) Γ(σ−β)Γ(σ+α+η)xσ−β−1 ×3F2(σ,σ+η−β,−c;σ−β,σ+α+η;−bx a), (3.4.21) whereℜ(σ)>max[0,ℜ(β−η)],|bx a|<1. 3.4.7.Evaluate Iα,β,η 0,x/braceleftBig xσ−1Hm,n p,q/bracketleftBig axλ|(ap,Ap) (bq,Bq)/bracketrightBig/bracerightBig , λ>0, (3.4.22) and givetheconditionsofitsvalidity. 3.4.8.Evaluate Jα,β,η x,∞/braceleftBig xσ−1Hm,n p,q/bracketleftBig ax−λ|(ap,Ap) (bq,Bq)/bracketrightBig/bracerightBig , λ>0, (3.4.23) and givetheconditionsofitsvalidity. 3.4. GENERALIZED KOBER OPERATORS 193 3.4.9.EstablishthefollowingpropertyofSaigo operators called “Integrationby parts”. /integraldisplay∞ 0f(x)/parenleftBig Iα,β,η 0,xg/parenrightBig (x)dx=/integraldisplay∞ 0g(x)/parenleftBig Jα,β,η x,∞f/parenrightBig (x)dx. 3.4.10.FromExercise3.4.6,deducetheformulafor Iα,−α,η 0,x(a+bx)c, (3.4.24) givenby B. Ross (1993). 3.4.11.Provethat Rα 0,xxk=Γ(k+1) Γ(α+k+1)xk+α, (3.4.25) whereℜ(α)>0,ℜ(k)>−1, 3.4.12.Provethat Wα x,∞xk=Γ(−α−k) Γ(−k)xk+α, (3.4.26) whereℜ(α)>0,ℜ(k)<−ℜ(α). 3.4.13.Showthat Jα,β,η x,∞(xλe−px)=xλ−βG3,0 2,3/bracketleftBig px|−λ,α+β+η−λ 0,β−λ,η−λ/bracketrightBig , (3.4.27) whereG3,0 2,3(·)is theMeijer’sG-function, ℜ(px)>0,ℜ(α)>0. Hint:Usetheintegral e−px=1 2πi/integraldisplay LΓ(−s)(px)sds. (3.4.28) 3.4.14.Evaluate Iα,β,η 0,xxσ−1Hm,n p,q/bracketleftbigg ax−λ|(ap,Ap) (bq,Bq)/bracketrightbigg , λ>0, (3.4.29) givingtheconditionsofits validity. 194 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS 3.4.15.Evaluate Jα,β,η x,∞xσ−1Hm,n p,q/bracketleftbigg axλ|(ap,Ap) (bq,Bq)/bracketrightbigg , λ>0 (3.4.30) and givetheconditionsofvalidityoftheresult. 3.4.16.With the help of the following chain rules for Saigo operator s ( Saigo, 1985) Iα,β,η 0,xIγ,δ,α+η 0,xf=Iα+γ,β+δ,η 0,xf, (3.4.31) and Jα,β,η x,∞Jγ,δ,α+η x,∞f=Jα+γ,β+δ,η x,∞f, (3.4.32) derivetheinverses (Iα,β,η 0,x)−1=I−α,−β,α+η 0,x. (3.4.33) and (Jα,β,η x,∞)−1=J−α,−β,α+η x,∞. (3.4.34) 3.5. Compositions of Riemann-Liouville Fractional CalculusOperatorsandGeneralizedMittag-Le ffler Functions In this section, composition relations between Riemann-Li ouville fractional calculus operators and generalized Mittag-Le ffler functions are derived. These relations may be useful in the solutionof fractional di fferintegralequations. For details,onecan refer totheworkofSaxenaand Saigo (2005). 3.5. COMPOSITIONS OF RIEMANN-LIOUVILLE FRACTIONAL CALCUL US 195 3.5.1. CompositionrelationsbetweenR-Loperatorsand Eδ β,γ(z) Notation3.5.1. Eα(x) : Mittag-Leffler function. Notation3.5.2. Eα,β(x) : Generalized Mittag-Le ffler function. Notation3.5.3. Iα 0+f: Riemann-Liouville left-sided integral. Notation3.5.4. Iα −f: Riemann-Liouville right-sided integral. Notation3.5.5. Dα 0+f: Riemann-Liouville left-sided derivative. Notation3.5.6. Dα −f: Riemann-Liouville right-sided derivative. Notation3.5.7. Eδ β,γ(z) : Generalized Mittag-Le ffler function (Prabhakar, 1971). Definition 3.5.1. Eα(z) :=∞/summationdisplay k=0zk Γ(αk+1),(α∈C,ℜ(α)>0). (3.5.1) Definition 3.5.2. Eα,β(z) :=∞/summationdisplay k=0zk Γ(αk+β),(α,β∈C,ℜ(α)>0,ℜ(β)>0).(3.5.2) Definition 3.5.3. (Iα 0+f)(x) :=1 Γ(α)/integraldisplayx 0f(t) (x−t)1−αdt,ℜ(α)>0. (3.5.3) Definition 3.5.4. (Iα −f)(x) :=1 Γ(α)/integraldisplay∞ xf(t) (t−x)1−αdt,ℜ(α)>0. (3.5.4) Definition 3.5.5. (Dα 0+f)(x) :=/parenleftbiggd dx/parenrightbigg[α]+1/parenleftbigg I1−{α} 0+/parenrightbigg (x);ℜ(α)>0 (3.5.5) =1 Γ(1−{α})/parenleftbiggd dx/parenrightbigg[α]+1/integraldisplayx 0f(t) (x−t){α}dt,ℜ(α)>0.(3.5.6) 196 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS Definition 3.5.6. (Dα −f)(x) :=/parenleftBiggd dx/parenrightBigg[α]+1 (I1−{α} −f)(x),ℜ(α)>0 (3.5.7) =1 Γ(1−{α})/parenleftBigg −d dx/parenrightBigg[α]+1/integraldisplay∞ xf(t) (t−x){α}dt,ℜ(α)>0.(3.5.8) Remark 3.5.1. Here [α] means the maximal integer not exceeding αand{α}is thefractionalpart of α. Definition 3.5.7. Eδ β,γ(z) :=∞/summationdisplay k=0(δ)kzk Γ(βk+γ)k!,(β,γ,δ∈C;ℜ(γ)>0,ℜ(β)>0).(3.5.9) Forδ=1,(3.5.9)reduces to(3.5.2). Theorem 3.5.1. Letα >0, β >0, γ >0andα∈R. Let Iα 0+be the left-sided operator of Riemann-Liouville fractional inte gral (3.5.3). Then there holdstheformula (Iα 0+[tγ−1Eδ β,γ(atβ)])(x)=xα+γ−1Eδ β,α+γ(axβ). (3.5.10) Proof3.5.1. By virtueof(3.5.3)and (3.5.9),wehave K≡(Iα 0+[tγ−1Eδ β,γ(atβ)])(x)=1 Γ(α)/integraldisplayx 0(x−t)α−1∞/summationdisplay n=0(δ)nantnβ+γ−1 Γ(βn+γ)n!dt. Interchanging the order of integration and summation and ev aluating the inner integralbymeans ofbeta-functionformula,itgives K≡xα+γ−1∞/summationdisplay n=0(δ)n(axβ)n Γ(α+βn+γ)(n)!=xα+γ−1Eδ β,α+γ(axβ). ThiscompletestheproofofTheorem 3.5.1. Corollary3.5.1. Forα>0,β>0,γ>0 andα∈R, thereholdstheformula (Iα 0+[tγ−1Eβ,γ(atβ)])(x)=xα+γ−1Eβ,α+γ(axβ). (3.5.11) 3.5. COMPOSITIONS OF RIEMANN-LIOUVILLE FRACTIONAL CALCUL US 197 Remark3.5.2. Forβ=α, (3.5.11)reduces to (Iα 0+[tγ−1Eα,γ(atβ)])(x)=xγ−1 a[Eα,γ(axα)−1 Γ(γ)],(a/nequal0) (3.5.12) byvirtueoftheidentity Eα,γ(x)=1 Γ(γ)+xEα,α+γ(x),(a/nequal0). (3.5.13) Theorem 3.5.2. Letα >0,β >0,γ >0andα∈R,(a/nequal0)and let Iα 0+ be the left-sided operator of Riemann-Liouvillefractiona lintegral(3.5.3). Then thereholdstheformula (Iα 0+[tγ−1Eδ β,γ(atβ)])(x)=1 axα+γ−β−1[Eδ β,α+γ−β(axβ)−Eδ−1 β,α+γ−β(axβ)].(3.5.14) Proof. UseTheorem3.5.1. Thefollowingtwotheoremscan beestablishedin thesameway . Theorem 3.5.3. Letα >0,β >0,γ >0andα∈Rand let Iα −be the right- sided operator of Riemann-Liouville fractional integral ( 3.5.4). Then we arrive atthefollowingresult: (Iα −[t−α−γEδ β,γ(at−β)])(x)=x−γ[Eδ β,α+γ(ax−β)] (3.5.15) Corollary3.5.2. Forα>0,β>0,γ>0 andα∈R, thereholdstheformulas: (Iα −[t−α−γEβ,γ(at−β)])(x)=x−γ[Eβ,α+γ(ax−β)] (3.5.16) and (Iα −t−α−1Eβ(at−β))(x)=x−1[Eβ,α+1(ax−β)]. (3.5.17) Theorem 3.5.4. Letα >0,β>0,γ >0,α∈R,(a/nequal0),α+γ >βand let Iα −be the right-sided operator of Riemann-Liouville fraction al integral (3.5.4). Then thereholdstheformula 198 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS (Iα −[t−α−γEδ β,γ(at−β)])(x)=1 axβ−γ[Eδ β,α+γ−β(ax−β)−Eδ−1 β,α+γ−β(ax−β)].(3.5.18) Corollary3.5.3. Forα>0,β>0,γ>0withα+γ>βandforα∈R,(a/nequal0), thereholdstheformula (Iα −[t−α−γEβ,γ(at−β)])(x)=1 axβ−γ/bracketleftBigg Eβ,α+γ−β(ax−β)−1 Γ(α+γ−β)/bracketrightBigg .(3.5.19) Remark3.5.3. (Kilbasand Saigo, (1998)) (Iα −[t−α−γEα,γ(at−α)])(x)=xα−γ a[Eα,γ(ax−α)−1 Γ(γ)],(a/nequal0) (3.5.20) (Iα −[t−α−1Eα(at−α)])(x)=xα−1 a/bracketleftbigg Eα(ax−α)−1/bracketrightbigg ,(a/nequal0). (3.5.21) Theorem 3.5.5. Letα >0,β >0,γ >0,γ > α,α∈Rand let Dα 0+be the left-sided operator of Riemann -Liouville fractional d erivative (3.5.6). Then thereholdstheformula. (Dα 0+[tγ−1Eδ β,γ(atβ)])(x)=xγ−α−1Eδ β,γ−α(axβ). (3.5.22) Proof3.5.2. By virtueof(3.5.9)and (3.5.6),wehave K≡(Dα 0+[tγ−1Eδ β,γ(atβ)])(x)=/parenleftbiggd dx/parenrightbigg[α]+1/parenleftbigg I1−{α} 0+/bracketleftBig tγ−1Eδ β,γ(atβ)/bracketrightBig/parenrightbigg (x) =∞/summationdisplay n=0an(δ)n Γ(γ+nβ)Γ(1−{a})n!/parenleftbiggd dx/parenrightbigg[α]+1/integraldisplayx 0tnβ+γ−1(x−t)−{α}dt =∞/summationdisplay n=0an(δ)n Γ(γ+nβ+1−{α})n!/parenleftbiggd dx/parenrightbigg[α]+1 xnβ+γ−{α} =∞/summationdisplay n=0an(δ)nxγ+nβ−α−1 Γ(nβ+γ−α)n!=xγ−α−1Eδ β,γ−α(axβ), whichprovesthetheorem. 3.5. COMPOSITIONS OF RIEMANN-LIOUVILLE FRACTIONAL CALCUL US 199 By usingasimilarprocedure, wearriveat thefollowingtheo rem. Theorem 3.5.6. Letα >0,γ > β > 0,α∈R,(a/nequal0), γ > α+βand let Dα 0+betheleft-sidedoperatorofRiemann-Liouvillefractiona lderivative(3.5.6). Then thereholdstheformula /parenleftBig Dα 0+[tγ−1Eδ β,γ(atβ)]/parenrightBig (x)=1 axγ−α−β−1/bracketleftBig Eδ β,γ−α−β(axβ)−Eδ−1 β,γ−α−β(axβ)/bracketrightBig .(3.5.23) Corollary 3.5.4. Letα >0,γ > β > 0,α∈R,(a/nequal0), γ > α+β, then there holdstheformula. /parenleftBig Dα 0+[tγ−1Eβ,γ(atβ)]/parenrightBig (x)=1 axγ−α−β−1/bracketleftBigg Eβ,γ−α−β(axβ)−1 Γ(γ−α−β)/bracketrightBigg .(3.5.24) Theorem 3.5.7. Letα >0,γ >0,γ−α >0withγ−α+{α}>1,α∈R, andletDα −betheright-sidedoperatorofRiemann-Liouvillefraction alderivative (3.5.8). Then thereholdstheformula. /parenleftBig Dα −[tα−γEδ β,γ(at−β)]/parenrightBig (x)=x−γEδ β,γ−α(ax−β). (3.5.25) Theorem3.5.8. Letα>0,β>0withγ−{α}>1, α∈R, γ>α+β,(a/nequal0) andletDα −betheright-sidedoperatorofRiemann-Liouvillefraction alderivative (3.5.8). Then thereholdstheformula /parenleftBig Dα −[tα−γEδ β,γ(at−β)]/parenrightBig (x)=xβ−γ a/bracketleftBig Eδ β,γ−α−β(ax−β)−Eδ−1 β,γ−α−β(ax−β)/bracketrightBig .(3.5.26) Exercises3.5. 3.5.1.Showthat axβEδ β,γ(axβ)=Eδ β,γ−β(axβ)−Eδ−1 β,γ−β(axβ),(a/nequal0) (3.5.27) 3.5.2.Showthat 200 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS /parenleftBig Iα 0+[tγ−1Eα,γ(atα)]/parenrightBig (x)=xγ−1 a/bracketleftBigg Eα,γ(axα)−1 Γ(γ)/bracketrightBigg ,(a/nequal0).(3.5.28) 3.5.3.ProveTheorem 3.5.3. 3.5.4.ProveTheorem 3.5.4. 3.5.5.ProveTheorem 3.5.6. 3.5.6.ProveTheorem 3.5.7. 3.5.7.ProveTheorem 3.5.8. 3.5.8.Provethat /parenleftBig Iα 0+tωHm,n p,q/bracketleftBig tσ|(ap,Ap) (bq,Bq)/bracketrightBig/parenrightBig (x)=xω+αHm,n+1 p+1,q+1/bracketleftBig xσ|(−ω,σ),(ap,Ap) (bq,Bq),(−ω−α,σ)/bracketrightBig ,(3.5.29) givingconditionsofvalidity. 3.5.9.Evaluate /parenleftBig Iα −tωHm,n p,q/bracketleftBig tσ|(ap,Ap) (bq,Bq)/bracketrightBig/parenrightBig (x), (3.5.30) and givetheconditionsofvalidity. 3.6. FractionalDi fferentialEquations Differential equations contain integer order derivatives, whe reas fractional differential equations involve fractional derivatives, likedα dxα, which are de- fined forα >0. Hereαis not necessarily an integer and can be rational, irrational or even complex-valued. Today, fractional calc ulus models find ap- plications in physical, biological, engineering, biomedi cal and earth sciences. Most of the problems discussed involve relaxation and di ffusion models in the so called complex or disordered systems. Thus, it gives rise to the generaliza- tion of initial value problems involving ordinary di fferential equations to gen- eralized fractional-order di fferential equations and Cauchy problems involving 3.6. FRACTIONAL DIFFERENTIAL EQUATIONS 201 partial differential equations to fractional reaction, fractional di ffusion and frac- tional reaction-diffusion equations. Fractional calculus plays a dominant role in thesolutionofall thesephysicalproblems. 3.6.1. Fractionalrelaxation In order toformulatearelaxationprocess, werequirea phys icallaw,say the relaxationequation d dtf(t)+1 cf(t)=0,t>0,c>0, (3.6.1) to be solved for the initial value f(t=0)=f0. The unique solution of (3.6.1) is givenby f(t)=f0e−t c,t≥0,c>0. (3.6.2) Now the problem is as to how we can generalize the initial-val ue problem (3.6.1)intoafractional valueproblemwithphysicalmotiv ation. If we incorporate the initial value f0into the integrated relaxation equation (3.6.1),wefind that f(t)−f0=−1 c0D−1 tf(t), (3.6.3) where 0D−1 tis thestandard Riemann integralof f(t). On replacing1 c0D−1 tf(t) by1 cα0D−α tf(t), it yields the fractional integral equation f(t)−f0=−/parenleftBigg1 cα/parenrightBigg 0D−α tf(t),α>0 (3.6.4) withinitialvalue f0=f(t=0). ApplyingtheRiemann-Liouvilledi fferentialoperator 0Dα tfromtheleftandmak- inguseoftheformula(3.1.16),wearriveat 0Dα tf(t)=f0t−α Γ(1−α)=−c−αf(t), α>0,c>0, (3.6.5) withinitialcondition f0=f(t=0). 202 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS Theorem3.6.1. Thesolutionofthefractionaldi fferentialequation(3.6.4)is given by f(t)=f0H1,1 1,2/bracketleftbigg/parenleftbiggt c/parenrightbiggα/vextendsingle/vextendsingle/vextendsingle(0,1) (0,1),(0,α)/bracketrightbigg , (3.6.6) whereα>0,c>0. Proof3.6.1. If weapplytheLaplacetransformtoequation(3.6.4), itgiv es F(s)−f0s−1=−1 cαs−αF(s), (3.6.7) where we have used the result (3.1.7) and F(s) is the Laplace transform of f(t). Solvingfor F(s), wehave F(s)=L{f(t)}=f0/bracketleftBiggs−1 1+(cs)−α/bracketrightBigg . (3.6.8) TakinginverseLaplacetransform, (3.6.8)gives f(t)=L−1{F(s)}=f0L−1/bracketleftBiggs−1 1+(cs)−α/bracketrightBigg =f0L−1∞/summationdisplay k=0(−1)kc−αks−αk−1 =f0∞/summationdisplay k=0(−1)k(t c)αk Γ(αk+1) =f0Eα/bracketleftbigg −(t c)α/bracketrightbigg , (3.6.9) whereEα(·) is the Mittag-Leffler function. (3.6.9) can be written in terms of the H-functionas f(t)=f0H1,1 1,2/bracketleftbigg/parenleftbiggt c/parenrightbiggα/vextendsingle/vextendsingle/vextendsingle(0,1) (0,1),(0,α)/bracketrightbigg , (3.6.10) wherec>0,α>0. ThiscompletestheproofoftheTheorem3.6.1. Alternativeformofthe solution. By virtueoftheidentity Hm,n p,q/bracketleftbigg xµ/vextendsingle/vextendsingle/vextendsingle(ap,Ap) (bq,Bq)/bracketrightbigg =1 µHm,n p,q/bracketleftBigg x/vextendsingle/vextendsingle/vextendsingle(ap,Ap µ) (bq,Bq µ)/bracketrightBigg ,(µ>0) (3.6.11) 3.6. FRACTIONAL DIFFERENTIAL EQUATIONS 203 thesolution(3.6.10)can bewrittenas f(t)=f0 αH1,1 1,2/bracketleftbiggt c/vextendsingle/vextendsingle/vextendsingle(0,1 α) (0,1 α),(0,1)/bracketrightbigg , (3.6.12) whereα>0,c>0. Remark3.6.1. In thelimitas α→1,onerecovers theresult(3.6.2) f(t)=f0exp(−t c)=f0E1(t c). (3.6.13) Remark 3.6.2. In terms of Wright’s function, the solution (3.6.10) can be e x- pressed intheform f(t)=f0 1ψ1/bracketleftbigg (1,1) (1,α)/vextendsingle/vextendsingle/vextendsingle−(t c)α/bracketrightbigg , (3.6.14) whereα>0,c>0. In asimilarmanner,wecan establishTheorems3.6.2and 3.6. 3givenbelow. Theorem 3.6.2. Thesolutionofthefractionalintegralequation N(t)−N0tµ−1=−cν 0D−ν tN(t), (3.6.15) isgiven by N(t)=N0Γ(µ)tµ−1Eν,µ(−cνtµ), (3.6.16) where E ν,µ(·)isthegeneralizedMittag-Le ffler function(3.5.2), ν>0,µ>0. Remark 3.6.3. Whenµ=1, we obtain theresult given by Haubold and Mathai (2000). Theorem 3.6.3. If c>0,ν >0,µ >0, then for the solution of the integral equation N(t)−N0tµ−1Eγ ν,µ[−(ct)ν]=−cν 0D−ν tN(t), (3.6.17) thereholdstheformula N(t)=N0tµ−1Eγ+1 ν,µ[−(ct)ν]. (3.6.18) 204 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS Hint: Usetheformula L−1/braceleftBig s−β(1−as−α)−γ/bracerightBig =tβ−1Eγ α,β(atα), (3.6.19) whereℜ(α)>0,ℜ(β)>0,ℜ(s)>|a|1 ℜ(α),ℜ(s)>0. Corollary3.6.1. Ifc>0,µ>0,ν>0, thenfor thesolutionof N(t)−N0tµ−1Eν,µ[−cνtν]=−cν 0D−ν tN(t), (3.6.20) thereholdstherelation N(t)=N0 νtµ−1/bracketleftBig Eν,µ−1(−cνtν)+(1+ν−µ)Eν,µ(−cνtν)/bracketrightBig . (3.6.21) Theorem 3.6.4. TheCauchyproblemfor theintegro-di fferentialequation 0Dµ xf(x)+λ0D−ν xf(x)=h(x),(λ,µ,ν∈C) (3.6.22) with theinitialcondition Dµ−k−1 xf(0)=ak,k=0,1,···,[µ], (3.6.23) whereℜ(ν)>0,ℜ(µ)>0and h(x)is any integrable function on the finite interval[0,b]hastheuniquesolution,givenby f(x)=/integraldisplayx 0(x−t)µ−1Eµ+ν,µ[−λ(x−t)µ+ν]h(t)dt +n−1/summationdisplay k=0akxα−k−1Eµ+ν,µ−k(−λxµ+ν) (3.6.24) Proof3.6.2. Exercise. Theorem 3.6.5. Thesolutionof theequation 0D1 2 tf(t)+bf(t)=t>0;/bracketleftbigg 0D−1 2 tf(t)/bracketrightbigg t=0=C, (3.6.25) whereC is aconstantisgiven by 3.6. FRACTIONAL DIFFERENTIAL EQUATIONS 205 f(t)=C t−1 2E1 2,1 2(−bt1 2), (3.6.26) where E 1 2,1 2(·)istheMittag-Leffler function. Proof3.6.3. Exercise. Remark3.6.4. Theorem3.6.5givesthegeneralizedformoftheequationsol ved byOldhamand Spanier(1974). Exercises3.6. 3.6.1.Provethat if c>0,ν>0,µ>0,then thesolutionof N(t)−N0tµ−1E2 ν,µ(cνtν)=−cν 0D−ν tN(t), (3.6.27) isgivenby N(t)=N0tµ−1E3 ν,µ(−cνtν)=N0tµ−1 2ν2/bracketleftbigg Eν,µ−2(−cνtν) +{3(ν+1)−2µ}Eν,µ−1(−cνtν) +/braceleftBig 2ν2+µ2+3ν−2µ−3νµ+1/bracerightBig Eν,µ(−cνtν)/bracketrightbigg , (3.6.28) whereℜ(ν)>0,ℜ(µ)>2. 3.6.2.Prove that if ν >0,c>0,d>0,µ>0,c/nequald, then for the solution of the equation N(t)−N0tµ−1Eν,µ(−dνtν)=−cν 0D−ν tN(t), (3.6.29) thereholdstheformula. N(t)=N0tµ−ν−1 cν−dν/bracketleftBig Eν,µ−ν(−dνtν)−Eν,µ−ν(−cνtν)/bracketrightBig .(3.6.30) 3.6.3.Provethat if c>0,ν>0,µ>0,then forthesolutionoftheequation N(t)−N0tµ−1Eν,µ(−cνtν)=−cν 0D−ν tN(t), (3.6.31) thefollowingresult holds: 206 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS N(t)=N0 νtµ−1/bracketleftBig Eν,µ−1(−cνtν)+(1+ν−µ)Eν,µ(−cνtν)/bracketrightBig . (3.6.32) 3.6.4.Solvetheequation 0DQ tf(t)+0Dq tf(t)=g(t), whereq−Qisnot an integerorahalfintegerandtheinitialconditioni s /bracketleftbigg 0Dq−1 tf(t)+0DQ−1 tf(t)/bracketrightbigg t=0=C (3.6.33) whereCisaconstant. 3.6.5.Solvetheequation 0Dα tx(t)−λx(t)=h(t),(t>0), (3.6.34) subjectto theinitialconditions /bracketleftBig 0Dα−k th(t)/bracketrightBig t=0=bk,(k=1,···,n) (3.6.35) wheren−1<α<n. 3.6.6ProveTheorem 3.6.4 3.6.6ProveTheorem 3.6.5. 3.6.2. Fractionaldi ffusion Theorem 3.6.6. The solution of the following initial value problem for the fractionaldiffusionequationin onedimension 0Dα tU(x,t)=λ2∂2U(x,t) ∂x2,(t>0,−∞<x<∞) (3.6.36) 3.6. FRACTIONAL DIFFERENTIAL EQUATIONS 207 withinitialconditions: lim x→±∞U(x,t)=0;/bracketleftBig 0Dα−1 tU(x,t)/bracketrightBig t=0=φ(x) (3.6.37) isgiven by U(x,t)=/integraldisplay∞ −∞G(x−ζ,t)φ(ζ)dζ, (3.6.38) where G(x,t)=1 π/integraldisplay∞ 0tα−1Eα,α(−k2λ2tα)coskxdk. (3.6.39) Solution 3.6.1. Let 0< α <1. Using the boundary conditions (3.6.37), the Fouriertransformof(3.6.36)withrespect to variable xgives 0Dα x¯U(k,t)+λ2k2¯U(k,t)=0 (3.6.40) /bracketleftBig 0Dα−1 t¯U(k,t)/bracketrightBig t=0=¯φ(k), (3.6.41) wherekis a Fourier transform parameter and ‘ −’ indicates Fourier transform. ApplyingtheLaplacetransformto (3.6.40)and using(3.6.4 1),itgives ≃ U(k,s)=¯φ(k) sα+k2λ2, (3.6.42) where ‘∼’ indicates Laplace transform. The inverse Laplace transfo rm of (3.6.42)yields ¯U(k,t)=tα−1¯φ(k)Eα,α(−λ2k2t2), (3.6.43) and then the solutionis obtained by taking inverseFourier t ransform. By taking inverseFouriertransformof(3.6.43)and usingtheformula 1 2π/integraldisplay∞ −∞e−ikxf(k)dk=1 π/integraldisplay∞ 0f(k)cos(kx)dk (3.6.44) wehave 208 3. FRACTIONAL CALCULUS AND FRACTIONAL DIFFERENTIAL EQU ATIONS U(x,t)=/integraldisplay∞ −∞G(x−ζ,t)φ(ζ)dζ, (3.6.45) where G(x,t)=1 π/integraldisplay∞ 0tα−1Eα,α(−k2λ2tα)cos(kx)dk (3.6.46) withℜ(α)>0,k>0. 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