Bulletin T.CXXXIV de l’Académie serbe des sciences et des arts − 2007 Classe des Sciences mathématiques et naturelles Sciences mathématiques, N\underline{o} 32

A METHOD TO SOLVE AN EQUATION WITH LEFT AND RIGHT FRACTIONAL DERIVATIVES

B. STANKOVIĆ

(Presented at the 4th Meeting, held on May 25, 2007)

A b s t r a c t. We give a procedure to find a solution to the equation D_{b^{−}}^{α}D_{0^{+}}^{α}y(x) + mD_{0^{+}}^{α}D_{b^{−}}^{α}y = C_{0}(x), 0 < x < b, 0 < α < 1, solving the corresponding Fredholm integral equation of the second kind.

AMS Mathematics Subject Classification (2000): 34K06, 45E05 Key Words: left and right fractional derivatives, singular integral equa-

tions

1. Introduction

Differential equations with the left and right fractional derivatives (denoted by D_{0^{+}}^{α} and D_{b^{−}}^{α}, respectively) arise in mathematical models in many branches of sciences as: mechanics, rheology, fluid flows, electrical networks, viscoelasticity, chemical physics, polymer science, biophyscis,.. See for example the books [4] and [5]. The theory of such equations with the left fractional derivatives has been elaborated in many papers and books (see for example [5]). But equations in which arises the left fractional derivative combined with the right fractional derivative have been left unexamined. One of the reason lies in the fact that the connection between these two aspects of fractional derivatives is very complex.

We treat of the equation

D_{b^{−}}^{α}D_{0^{+}}^{α}y(x) + mD_{0^{+}}^{α}D_{b^{−}}^{α}y(x) = C_{0}(x), 0 < x < b, 0 < α < 1, (1)

where m is a constant, bringing down to a Fredholm integral equation of the second kind. This equation is interesting not only as a mathematical model in mechanics, but also it gives the connection between two operators D_{b^{−}}^{α}D_{0^{+}}^{α} and D_{0^{+}}^{α}D_{b^{−}}^{α}.

To find a solution to (1) we first construct a singular integral equation, then a quasi-Fredholm integral equation which can be brought by integration to a Fredholm integral equation with the continuous kernel.

In this paper we use the notation and defnitions as they are given in [9]. Let us repeat definitions of three classes of functions we use (see [9]):

The function f defined on the interval [a, b] belongs to the class H(λ) if and only if |f(t_{2}) − f(t_{1})| ≤ A|t_{2} − t_{1}|^{λ}, 0 < λ ≤ 1, for every t_{1}, t_{2} ∈ [a, b].

The function f ∈ H if f ∈ H(λ), for a fixed λ ∈ (0, 1].

The function f defined on [0, b] belongs to h^{λ}[0, b], 0 < λ ≤ 1, if

\frac{f(x_{2}) − f(x_{1})}{|x_{2} − x_{1}|^{λ}} → 0, x_{2} → x_{1} .

Let ρ(x) = x^{α}. We denote by L^{p}(ρ, (0, 1)) the space of functions f, such that ρ(x)f(x) ∈ L^{p}(0, 1).

2. A method to solve equation (1)

2.1. Construction of a singular integral equation related to (1)

Let C_{0}(x) = C^{(1)}(x). It is well-known that D_{0^{+}}^{α} = \frac{d}{dx}I_{0^{+}}^{1−α} and D_{b^{−}}^{α} =

−\frac{d}{dx}I_{b^{−}}^{1−α}, where I_{0^{+}}^{1−α} an I_{b^{−}}^{1−α} are the left and right fractional integral operators, respectively. Then (1) is equivalent to

−I_{b^{−}}^{1−α}D_{0^{+}}^{α}y(x) + mI_{0^{+}}^{1−α}D_{b^{−}}^{α}y(x) = C(x) + k, 0 < x < b .  (2)

We use the connection between I_{0^{+}}^{α} and I_{b^{−}}^{α} (see [9], p. 206):

I_{0^{+}}^{1−α}φ(x) = cos(1−α)πI_{b^{−}}^{1−α}φ(x)−sin(1−α)π(I_{b^{−}}^{1−α}(b−t)^{α−1}S(b−τ)^{1−α}φ)(x), which holds for φ ∈ L^{q}(0, b), q(1 − α) > 1 and x ∈ (0, b), where

S(φ) = \frac{1}{π} V P ∫_{0}^{b}(t − x)^{−1}φ(t)dt .

Then (2) is:

\begin{matrix}−I_{b^{−}}^{1−α}D_{0^{+}}^{α}y(x) + mI_{b^{−}}^{1−α}[ cos(1 − α)πD_{b^{−}}^{α}y(t) \\ − sin(1 − α)π(b − t)^{α−1}(S(b − τ)^{1−α}D_{b^{−}}^{α}y(τ))(t)](x) = C(x) + k , \end{matrix} (3)

where D_{b^{−}}^{α}y ∈ L^{q}(0, b), q(1 − α) > 1.

Appliyng the operator D_{b^{−}}^{1−α} to (3), we have:

\begin{matrix}−D_{0^{+}}^{α}y(x) + m[ cos(1 − α)πD_{b^{−}}^{α}y(x) − sin(1 − α)π(b − x)^{α−1} \\ ·(S(b − τ)^{1−α}D_{b^{−}}^{α}y(τ))(x)] = D_{b^{−}}^{1−α}(C(x) + k) . \end{matrix}  (4)

We introduce a new function g by the connection with y as

y(x) = I_{b^{−}}^{α}g(x), 0 < x < b ,  (5)

supposing that g ∈ L^{p}(0, b), αp > 1. Then y = I_{b^{−}}^{α}g ∈ h^{α−1/p}[0, b] (see [9], p. 69). With the new form of y, given by (5), equation (4) can be written as

\begin{matrix}−D_{0^{+}}^{α}I_{b^{−}}^{α}g(x) + m[ cos(1 − α)πg(x) − sin(1 − α)π(b − x)^{α−1} x \\ (S(b − τ)^{1−α}g(τ))(x)] = D_{b^{−}}^{1−α}(C(x) + k) . \end{matrix} (6)

Now we again use the connection of I_{b^{−}}^{α} and I_{0^{+}}^{α} given by (see [9], p. 206)

I_{b^{−}}^{α}φ(x) = cos απI_{0^{+}}^{α}φ(x) + sin απ(I_{0^{+}}^{α}t^{−α}Sτ^{α}φ)(x),

which holds for φ ∈ L^{p}(0, b), pα > 1, x ∈ (0, b). With such expression of I_{b^{−}}^{α} (6) is as

−D_{0^{+}}^{α}( cos απI_{0^{+}}^{α}g(x) + sin απ(I_{0^{+}}^{α}t^{−α}Sτ^{α}g)(x))

+m[ cos(1 − α)πg(x) − sin(1 − α)π(b − x)^{α−1}(S(b − τ)^{1−α}g)(x)] = D_{b^{−}}^{1−α}(C(x) + k) .

(7)

Since D_{0^{+}}^{α}I_{0^{+}}^{α}φ = φ for any sommable function φ on (0, b), and the operator S is bounded on L^{q}(τ^{α}, (0, 1)), α < q − 1 (see [9], p. 200), we can write (7) as

−[cos απg(x) + sin απx^{−α}(Sτ^{α}g)(x)] + m[cos(1 − α)πg(x)
− sin(1 − α)π(b − x)^{α−1}(S(b − τ)^{1−α}g(τ)(x)
= D_{b^{−}}^{α}(C(x) + k) .

(8)

Since 0 < α < 1, cos(1 − α)π = − cos απ and sin(1 − α)π = sin απ, finally, we obtain the singular integral equation

A(x)g(x) + \frac{1}{πi} ∫_{0}^{b} \frac{K(x, t)}{t − x}g(t) = F (x), 0 < x < b ,  (9)

where

\begin{matrix}A(x) = (m + 1) cos απx^{α}(b − x)^{1−α}, \\ F (x) = −(D_{b^{−}}^{α}C(t) + k)x)x^{α}(b − x)^{1−α}, \\ K(x, t) = i sin απ(t^{α}(b − x)^{1−α} + m(b − t)^{1−α}x^{α}) . \end{matrix}  (10)

The function g(τ) ∈ L^{p}(0, b) and τ^{α}g(τ) ∈ L^{q}(0, b), \frac{1}{p} < α < q − 1.

We show that A and K belong to H. It is well-known that x^{α} and (b − x)^{1−α} belong to H and are bounded on [0, b]. The product and the sum of two functions which belong to H, belong to H, as well (see [6], p. 24). Consequently A and K belong to H (K in two variables). We can now prove the following

Proposition. If g is a solution to (9) such that g ∈ L^{p}(0, 1), p =

max(\frac{1}{α}, \frac{1}{1 − α} , α + 1), then y = I_{b^{−}}^{α}g is a solution to (1) and belongs to

h^{α−1/p}[0, b], α − 1/p > 0.

P r o o f. The function g can be given in the form

g = D_{0^{+}}^{α}I_{0}^{α}g .

To prove this we introduce the function

F (x) = (D_{0^{+}}^{α}I_{0}^{α}g)(x), 0 < x < 1 .

Then (see [9], p. 45)

\begin{aligned}I_{0^{+}}^{α}F & = I_{0^{+}}^{α}g − (I_{0^{+}}^{1−α}I_{0^{+}}^{α}g)(0) \\ & = I_{0^{+}}^{α}g − (I_{0^{+}}^{1}g)(0) = I_{0^{+}}^{α}h . \end{aligned}

Consequently,

I_{0^{+}}^{α}(F − g) = 0, or (F − g)(x) ∗ x^{−α} = 0 .

By the Tichmarsh theorem F = φ.

We know (see [9], p. 69) that the operator I_{0^{+}}^{α} has the property

I_{0^{+}}^{α} : L^{p}(a, b) → h^{α−1/p}[0, b].

Since I_{b^{−}}^{α} = QI_{0^{+}}^{α}Q, where the operator Q maps a function f defined on (0, b) as Q : f → f(b − x), the operator I_{b^{−}}^{α} has the same cited property of I_{0^{+}}^{α}. This gives that y ∈ h^{α−1/p}[0, b].

Suppose that we find a solution g ∈ L^{p}(0, 1) to (9), p = max (\frac{1}{α} , \frac{1}{1 − α} , 1+

α). Now we can return from (9) to (1) taking in consideration that L^{p}(0, b) ⊂

L^{q}(0, b), for q = 1, q = \frac{1}{α} , q = \frac{1}{1 − α} and q = α + 1.

2.2. The characteristic equation of (9)

Our aim is to construct a Fredholm integral equation instead of singular integral equation (9). There exist different methods to do it (see for example [M] and [G]). We propose the method which uses the characteristic equation. Then the first step is to solve the characteristic equation.

Equation (9) can be written as

A(x)g(x) + \frac{B(x)}{πi} ∫_{0}^{b} \frac{g(t)}{t − x}dt + \frac{1}{πi} ∫_{0}^{b} k(x, t)g(t)dt = F (x) , (11)

where

\begin{matrix}B(x) = K(x, x) = i(1 + m) sin απx^{α}(b − x)^{1−α} and \\ k(x, t) = \frac{K(x, t) − B(x)}{t − x} . \end{matrix}  (12)

For the function k(x, t) we know (see [6], p. 90) that

k(x, t) = \frac{k^{∗}(x, t)}{|t − x|^{λ}} , 0 ≤ λ < 1, k^{∗} ∈ H .  (13)

The equation

A(x)g_{0}(x) = \frac{B(x)}{πi} ∫_{0}^{b} \frac{g_{0}(t)}{t − x}dt = F (x), 0 < x < b,  (14)

64 B.Stanković

is called the characteristic equation to (9). Because of (10) and (12) is can be given as

ag_{0}(x) + \frac{b}{πi} ∫_{0}^{b} \frac{g_{0}(t)}{t − x}dt = d(x), 0 < x < b,  (15)

where a = (m + 1) cos απ, b = i(m + 1) sin απ, d(x) = ( − D_{b^{−}}^{α}(C(t) + k))(x), To solve equation (15) we follow the procedure given in [6], §97.

Let G(t) be the function

G(t) = \frac{a − b}{a + b} = \frac{(m + 1) cos απ − i sin απ}{(m + 1) cos απ + i sin απ} = (cos απ − i sin απ)^{2} = e^{−2απi}.

Then ℓnG(t) = (k − α)2πi, k = 0, ±1, ±2, ... We fix

ℓnG(t) = (−α)2πi .  (16)

Let us analyze the function

γ(x) = \frac{1}{2πi} ∫_{0}^{b} \frac{ℓnG(t)}{t − z} dt .

With G given by (16) we have

γ(z) = \frac{1}{2πi}ℓnG(t)ℓn(\frac{b − z}{−z} ) = (−α)ℓn(\frac{b − z}{−z} )  (17)

(see [6], p. 47). Then the function

e^{γ(z)} = (\frac{b − z}{−z} )^{(−α)}

satisfies the equation

ϕ^{+}(t) = G(t)ϕ^{−}(t), 0 < t < b .  (18)

In the neighborhood of c = 0 and c = b we have

γ(z) = A m(z − c), A = ∓(−α) ,  (19)

where we take the sign ”-” if c = 0, and the sigh ”+” if c = b. We determine integers λ_{0} and λ_{b} such that −1 < α + λ_{0} < 0 and −1 < −α + λ_{b} < 0. This gives λ_{0} = −1 and λ_{b} = 0. Then the canonical solution X(z) to (18) is

X(z) = e^{γ(z)}z^{−1} = (\frac{b − z}{−z} )^{−α}z^{−1} = (b − z)^{−α}z^{−(1−α)} .  (20)

This solution is of the class h(0, b) and the index of the class h(0, b) is κ = 1 (for the definition of the class h(0, b) and the index κ see [6], p. 315). The function X^{+}(x) and X^{−}(x) are

X^{+}(x) = e^{−iαπ}(b − x)^{−α}x^{−(1−α)}, X^{−}(x) = e^{iαπ}(b − x)^{−α}x^{−(1−α)} .

With the canonical solution X(z) to (18) we construct the general solution ϕ_{0}(z) = X(z)P (z) to (18) of the given class h(0, b); P (z) is any polynomial.

The function

ϕ(z) = \frac{X(z)}{2πi} ∫_{0}^{b} \frac{d(t) dt}{(a + b)X^{+}(t)(t − z)} + ϕ_{0}(z)  (21)

is the general solution of the class h(0, b) to equation

ϕ^{+}(x) = G(x)ϕ^{−}(x) + \frac{d(x)}{a + b} .  (22)

Since we need that ϕ(∞) = 0, P (z) has to be only a constant, denoted by P.

Finally, the solution to (16) is

g_{0}(x) = ϕ^{+}(t_{0}) − ϕ^{−}(t_{0}) .  (23)

If d(x) belong to the class H, then g_{0}(x) ∈ H and is the general solution in the class H to (15).

To give the explicit form of g_{0}(x) we introduce the function Z(x) = (a + b)X^{+}(x) and the constants

a^{∗} = \frac{a}{a^{2} − b^{2}} = \frac{(m + 1) cos απ}{(m + 1)^{2} cos^{2} απ + (m + 1)^{2} sin^{2} απ} = \frac{a}{(m + 1)^{2}}

b^{∗} = \frac{b}{a^{2} − b^{2}} = \frac{b}{(m + 1)^{2}} .

By the formula of Sohocki-Plemelj (see [6], p. 66) we have

g_{0}(x) = a^{∗}d(x) − \frac{b^{∗}Z(x)}{πi} ∫_{0}^{b} \frac{d(t)dt}{Z(t)(t − x)} + b^{∗}Z(x)P^{∗}, 0 < x < b. (24)

2.3. The quasi Fredholm integral equation

We use the solution to the characteristic equation of (9) to construct a quasi Fredholm integral equation (cf. [3], §49.1). Equation (11) can be written as

\begin{aligned}ag(x) & +\frac{b}{πi} ∫_{0}^{b} \frac{g(t)}{t − x}dy = d(x) \\ & −x^{−α}(b − x)^{α−1} \frac{1}{πi} ∫_{0}^{b} k(x, t)g(t)dt, 0 < x < b . \end{aligned}  (25)

By the result of 2.2 the formal solution to (25) is

g(x) = a^{∗}(d(x) − x^{−α}(b − x)^{α−1} \frac{1}{πi} ∫_{0}^{b} k(x, t)g(t)dt)

−\frac{b^{∗}Z(x)}{πi} ∫_{0}^{b} \frac{1}{Z(t)(t − x)}[d(t)

−t^{−α}(b − t)^{α−1} \frac{1}{πi} ∫_{0}^{b} k(t, τ)g(τ)dτ]dt

+b^{∗}Z(x)P^{∗}, 0 < x < b ,

or

g(x) + ∫_{0}^{b} M(x, t)g(t)dt = E(x), 0 < x < b ,  (26)

where

\begin{aligned}M(x, t) & = a^{∗}x^{−α}(b − x)^{α−1} \frac{1}{πi}k(x, t) \\ & −\frac{b^{∗}Z(x)}{πi} ∫_{0}^{b} \frac{e^{απi}}{(a + b)t^{−1+2α}(b − t)^{1−2α}} \frac{1}{π} k(t, τ)dt, \end{aligned}  (27)

E(x) = a^{∗}d(x) − \frac{b^{∗}Z(x)}{πi} ∫_{0}^{b} \frac{d(t)}{Z(t)(t − x)}dt + b^{∗}Z(x)P^{∗} .  (28)

Equation (26) is a quasi Fredholm integral equation. It can be transformed to a Fredholm integral equation (see for example [7]) by using the process of iteration. The quasi Fredholm equations have the same properties as Fredholm’s integral equations (see [G], p. 152).

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Department of Mathematics University of Novi Sad Trg Dositeja Obradovića 4 21000 Novi Sad Serbia