# On Solution Representation of Generalized Abel Integral Equation

Authors: Arsen Pskhu
Journal: Journal of Mathematics
 (2013)

---

## Abstract

An explicit representation for solution of the generalized Abel integral
equation∑k=0mλkD0x-αku(x)=f(x), where D0x-αk is the Riemann-Liouville fractional integral, in terms of the Wright function, is
constructed.


---

## Body

## 1. Introduction

Consider the equation(1)∑k=0m‍λkD0x-αku(x)=f(x),
where αk≥0,  λk∈ℝ,  x∈(0,l), and  D0xβ is the Riemann-Liouville operator of fractional integrodifferentiation of order β, which is defined as follows:
(2)D0xβφ(x)=1Γ(-β)∫0xφ(t)(x-t)-β-1dtforβ<0,D0x0φ(x)=φ(x),D0xβφ(x)=dndxnD0xβ-nφ(x)forβ∈(n-1,n],n∈ℕ.For suitable functionf(x), (1) with m=0 and α0=α>0,  λ0=1; that is, the Abel integral equation of the first kind (see [1])
(3)D0x-αu(x)=f(x)
may be inverted by the formula
(4)u(x)=D0xαf(x)
(see [2]). For m=1,  α0=0,  α1=α>0,  λ0=1, and  λ1=λ≠0, (1) has the following form
(5)u(x)+λD0x-αu(x)=f(x),
which is the Abel equation of the second kind, and its solution may be given by the Hille-Tamarkin formula
(6)u(x)=f(x)-λ∫0x(x-t)α-1Eα,α(-λ(x-t)α)f(t)dt
(see [3]), which can be rewritten as
(7)u(x)=ddx∫0xEα,1(-λ(x-t)α)f(t)dt.
Here,
(8)Eα,β(z)=∑n=0∞‍znΓ(αn+β)
is the Mittag-Leffler function.Equations of type (1) with rational parameters αk were considered in [4, 5]. Generalized Abel integral equations of the second kind were investigated by operational method in [6]. For a more complete survey on results relating Abel type integral equations and fractional order integral and differential equations and their applications, we refer to [6–8].In this paper, we obtain a representation for solution of (1). The results cover both cases, the solution of equation of the first kind (α0>0) and that of the second kind (α0=0). The solution is constructed in terms of the Wright function. It should be noted that (1) can be reduced to the generalized Abel integral equation of the second kind, and the method developed in [6] can be applied for (1). This provides an alternative approach to the equation under study.


## 2. Preliminaries

Now, we recall several properties of the operators of fractional integration and differentiation and special functions which are required for what follows.The following three relations are valid:(9)D0xεD0xδφ(x)=D0xε+δφ(x)
for ε∈ℝ and δ≤0 or ε∈ℕ and δ∈ℝ;
(10)D0xεD0xδφ(x)=D0xε+δφ(x)-∑j=1nx-ε-jΓ(1-ε-j)[D0xδ-jφ(x)]x=0
for ε∈ℝ and δ∈(n-1,n], n∈ℕ;
(11)D0xεxδ-1Γ(δ)=xδ-ε-1Γ(δ-ε)
for ε∈ℝ and δ>0 or ε∈ℕ and δ∈ℝ (e.g., see [2] and [9, Section 2.1]).By(φ*ψ)(x), we denote the Laplace convolution of functions φ(x) and ψ(x):
(12)(φ*ψ)(x)=∫0xφ(x-t)ψ(t)dt.OperatorD0xβ may be written in the form of convolution
(13)D0xβφ(x)=φ*x-β-1Γ(-β)forβ<0,D0xβφ(x)=dndxn(φ*xn-β-1Γ(n-β))forβ∈(n-1,n],n∈ℕ.Using (11) and (13) and taking into account the associativity and commutativity of convolution, we get
(14)1∏i=1mΓ(εi)(xε1-1*⋯*xεm-1)=xε-1Γ(ε),
where ε=∑j=1m‍εj,  εj>0, and
(15)(D0x-δ1φ1)*⋯*(D0x-δmφm)=D0x-δ(φ1*⋯*φm),
with δ=∑j=1m‍δj and δj≥0. The relation
(16)D0x-δ(φ*ψ)(x)=(D0x-δφ*ψ)(x)=(φ*D0x-δψ)(x)
is a particular case of (15).Recall that the Wright function(17)ϕ(β,δ;z)=∑n=0∞znn!Γ(βn+δ)
satisfies the relations
(18)ddzϕ(β,δ;z)=ϕ(β,δ+β;z)
(see [10]) and
(19)D0xεxδ-1ϕ(β,δ;zxβ)=xδ-ε-1ϕ(β,δ-ε;zxβ)forε∈ℝ,δ>0.
For β>0, relation (19) may be proved by termwise applying formula (11) to the representation of the Wright function as a series (17). Combining (18) and (19), we obtain
(20)∂∂zxδ-1ϕ(β,δ;λzxβ)=λD0x-βxδ-1ϕ(β,δ;λzxβ)(λ∈ℝ).


## 3. Auxiliary Results

Here, we investigate properties of functions in terms of which a solution of integral equation (1) will be constructed.Consider the function(21)Smμ(x;z1,…,zm;β1,…,βm)=(h1*h2*⋯*hm)(x)
(see [11]), where
(22)hk=hk(x)=xμk-1ϕ(βk,μk;zkxβk).
In what follows, the parameters and arguments of function (21) satisfy
(23)x>0,zk∈ℝ,βk>0,μk>0,μ=∑k=1m‍μk.Remark 1.
The functionSmμ(x;z1,…,zm;β1,…,βm) is independent of the distribution of parameters μk but depends only on their sum μ. Indeed, let S and S~ be the values of functions Smμ corresponding to the sets {μ1,…,μm} and {μ~1,…,μ~m}, with same sets of x, zk, and βk, and
(24)μ=∑k=1m‍μk=∑k=1m‍μ~k.
Let
(25)ν=min{μ1,…,μm,μ~1,…,μ~m},gk=xν-1ϕ(βk,ν;zkxβk).
Then, using (15) and (19), we obtain
(26)S=(D0xν-μ1g1)*⋯*(D0xν-μmgm)=D0xmν-μ(g1*⋯*gm)=(D0xν-μ~1g1)*⋯*(D0xν-μ~mgm)=S~.Lemma 2.
If(27)Z=max{|z1|,…,|z2|},ρ=min{β1,…,βm},τ=max{β1,…,βm},
then
(28)|Smμ(x;z1,…,zm;β1,…,βm)|≤Cxμ-1exp(σZ1/(1+ρ)xτ/(1+τ))
with positive constants C and σ independent of x and zk.Proof.
Using the asymptotic formula of the Wright function (see [10]), we have
(29)|ϕ(βk,μk;zk)|≤Ckxμ-1exp(σk|zk|1/(1+βk)xkβk/(1+βk)),
where Ck and σk are positive constants independent of x and zk. If we recall the definition (21), we get
(30)|Smμ(x;z1,…,zm;β1,…,βk)|≤C(xμ1-1*⋯*xμm-1)exp(σZ1/(1+ρ)xτ/(1+τ)).
Using (14), we obtain (28).Lemma 3.
Ifμ>ν, then
(31)D0xνSmμ(x;z1,…,zm;β1,…,βm)=Smμ-ν(x;z1,…,zm;β1,…,βm).Proof.
Ifν<0, then combining (15), (19), and (21), we get
(32)D0xνSmμ(x;z1,…,zm;β1,…,βm)=(D0xνh1)*(h2*⋯*hm)(x)=Smμ-ν(x;z1,…,zm;β1,…,βm).
Now, letν∈(q-1,q],  q∈ℕ. As follows from Remark 1, in (21), we can take a parameter μ1 such that μ1>ν. Therefore, in view of the equality
(33)ddx(ϕ*ψ)(x)=(φ′*ψ)-φ(0)ψ(x),
formula (19), and the relation
(34)limx→0D0xν-jh1(x)=0,1≤j≤q,
we see that
(35)D0xνSmμ(x;z1,…,zm;β1,…,βm)=dqdxq(D0xν-qh1)*(h2*⋯*hm)(x)=(D0xνh1)*(h2*⋯*hm)(x)=Smμ-ν(x;z1,…,zm;β1,…,βm).
This completes the proof.Now, we set(36)Gmμ(x;γ1,…,γm;β1,…,βm)=∫0∞e-tSmμ(x;γ1t,…,γmt;β1,…,βm)dt.
The convergence of the integral in (36) follows from (28).Lemma 4.
The functionGmμ=Gmμ(x;γ1,…,γm;β1,…,βm) satisfies the relations
(37)Gmμ=O(xμ-1)asx→0,(38)D0xνGmμ=Gmμ-νforμ>ν.Proof.
From (28) and (36), we obtain
(39)|Gmμ|≤Cxμ-1∫0∞exp(-t+σ(γt)1/(1+ρ)xτ/(1+τ))dt≤Cxμ-1exp[supt>0(-t2+σ(γt)1/(1+ρ)xτ/(1+τ))]×∫0∞e-t/2dt,
where
(40)ρ=min{β1,…,βm},τ=max{β1,…,βm},γ=max{|γ1|,…,|γm|},
and C and σ are positive constants independent of x, γ, and t. By the relation
(41)supt>0(-t+atε)=a1/(1-ε)εε/(1-ε)(1-ε),
which holds for a>0 and ε∈(0,1), (39) yields (37).
Relation (38) follows from (28), (31), and definition (36).Lemma 5.
Letλ0≠0, 0≤α0<αk for k=1,2,…,m, and μ>α0. Denote that
(42)wμ(x)=1λ0Gmμ-α0(x;-λ1λ0,…,-λmλ0;α1-α0,…,αm-α0).
Then, the function wμ(x) is a solution of the integral equation
(43)∑k=0m‍λkD0x-αkwμ(x)=xμ-1Γ(μ).Proof.
Let(44)gk=xμk-1ϕ(αk-α0,μk;-λktλ0xαk-α0),μ=∑k=1m‍μk.
Using (20), (21), and (31) and bearing in mind Remark 1, we get
(45)∂∂tSmμ(x;-λ1tλ0,…,-λmtλ0;α1-α0,…,αm-α0)=∑k=1m‍(g1*⋯*gk-1*∂gk∂t*gk+1*⋯*gm)=-1λ0∑k=1m‍λk(D0xα0-αkg1*⋯*gk-1*D0xα0-αkgk*gk+1*⋯*gmD0xα0-αk)=-1λ0∑k=1m‍λkD0x-αkSmμ-α0(x;-λ1tλ0,…,-λmtλ0;α1-α0,…,αm-α0λmtλ0).
Hence, in view of (36) and (42),
(46)∑k=1m‍λkD0x-αkwμ(x)=-∫0∞‍e-t∂∂tSmμ(x;-λ1tλ0,…,-λmtλ0;α1-α0,…,αm-α0λmtλ0)dt=Smμ(x;-λ1tλ0,…,-λmtλ0;α1-α0,…,αm-α0)|t=0-λ0D0x-α0wμ(x).
Taking account of the equality
(47)Smμ(x;-λ1tλ0,…,-λmtλ0;α1-α0,…,αm-α0)|t=0=xμ1-1Γ(μ1)*⋯*xμm-1Γ(μm)
and (14), the relation (46) yields (43).


## 4. Main Result

Theorem 6.
Letλ0≠0, 0≤α0<αk for k=1,2,…,m, f(x)∈L(0,l), and there exists a function g(x)∈L(0,l) such that f(x)=D0x-α0g(x). Then, (1) has a unique integrable solution u(x), which for each μ>α0 can be represented as follows:
(48)u(x)=D0xμ(f*wμ)(x),
where wμ=wμ(x) is defined by (42).Proof.
Letu(x) be a solution of (1). Taking into account (16) and (43), we obtain
(49)f*wμ=(∑k=0m‍λkD0x-αku)*wμ=u*(∑k=0m‍λkD0x-αkwμ)=u*xμ-1Γ(μ),
or, in view of (13),
(50)(f*wμ)(x)=D0x-μu(x).
Applying the operator D0xμ to both sides of this equality, bearing in mind (9), we obtain (48).
Now, we prove that a functionu(x) that is defined by (48) is a solution of (1). Let μ∈(q-1,q],  q∈ℕ. Consider the function
(51)v(x)=D0xμ-q(f*wμ)(x).
Bearing in mind (16), (38), and (42), we get
(52)v(x)=D0xμ-q(D0x-α0g*wμ)(x)=(g*wq+α0)(x).
Using (33), (37), and (38), we obtain
(53)D0xμ-j(f*wμ)(x)=dq-jdxq-jv(x)=(g*wj+α0)(x)forj=1,2,…,q.
Taking this and relation (37) into account, we may conclude that the function D0xμ-j(f*wμ)(x) is absolutely continuous and
(54)limx→0D0xμ-j(f*wμ)(x)=dq-jdxq-jv(x)|x=0=0
for j=1,2,…,q, and
(55)D0xμ(f*wμ)(x)=dqdxqv(x)∈L(0,l).
Using (10) and (54), we obtain
(56)D0x-αkD0xμ(f*wμ)(x)=D0xμD0x-αk(f*wμ)(x)=D0xμ(f*D0x-αkwμ)(x)fork=0,1,…,m.
Then, it follows that
(57)∑k=0m‍λkD0x-αkD0xμ(f*wμ)(x)=D0xμ(f*∑k=0m‍λkD0x-αkwμ)(x)=D0xμ(f*xμ-1Γ(μ))=D0xμD0x-μf(x)=f(x).
This completes the proof.Remark 7.
Letα0=α, λ0=1, and λk=0 for k≥1. In this case, (1) will appear like (3), and we get
(58)wμ(x)=Gmμ-α(x;0,…,0;α1-α,…,αm-α)=xμ-α-1Γ(μ-α).
In addition, it follows from the equalities
(59)G1μ(x;λ;β)=∫0∞‍e-txμ-1ϕ(β,μ;λtxβ)dt=∑k=0∞‍λkxβk+μ-1k!Γ(βk+μ)∫0∞‍e-ttkdt=xμ-1Eβ,μ(λxβ)
that if m=1,  α0=0,  α1=α>0,  λ0=1, and λ1=λ≠0, then
(60)wμ(x)=G1μ(x;-λ;α)=xμ-1Eα,μ(-λxα).
Thus, formula (48) coincides with (4) and (7) for the cases of (3) and (5) (i.e., for m=0 and m=1), respectively.



---

*Source: 106251-2013-03-11.xml*