We consider a Cauchy-type problem associated with the equation

D a α ( x - a ) r D a β y ( x ) = f ( x , y ) , x > a , 0 < α < 1 , 0 ≤ β ≤ 1 , r < α ,

where D a α and D a β are the Riemann-Liouville fractional derivatives.

In recent years there has been a considerable interest in the theory and applications of fractional differential equations. As for the theory, the investigations include the existence and uniqueness of solutions, asymptotic behavior, stability, etc. See for example the books 123 and the articles 45678910 and the references therein.

As for the applications, fractional models provide a tool for capturing and understanding complex phenomena in many fields. See for example the surveys in 111 and the collection of applications in 12.

Some recent applications include control systems 1314, viscoelasticity 15161718, and nanotechnology 19. Also fractional models are used to model a vibrating string 20, and anomalous transport 21, anomalous diffusion 222324.

Another field of applications is in random walk and stochastic processes 252627 and its applications in financial modeling 282930. Other physical and engineering processes are given in 3132

In a series of articles, 333435, Glushak studied the uniform well-posedness of a Cauchy-type problem with two fractional derivatives and bounded operator. He also proposed a criterion for the uniform correctness of unbounded operator.

In this article we prove an existence and uniqueness result for a nonlinear Cauchy-type problem associated with the Equation (1) in the space of weighted continuous functions.

We start with some preliminaries in Section 2. In Section 3 we define the sequential derivative and develop some properties and composition identities. In Section 4 we set up the Cauchy-type problem and establish the equivalence with the associated integral equation. Finally, in Section 5 we prove the existence and uniqueness of the solution.

In this section we present some definitions, lemmas, properties and notation which we use later. For more details please see 1.

Let -∞ < a < b < ∞. Let C[a, b] denote the spaces of continuous functions on [a, b]. We denote by L(a, b) the spaces of Lebesgue integrable functions on (a, b). Let CL(a, b) = L(a, b) ⋂ C(a, b].

We introduce the weighted spaces of continuous functions

C γ [ a , b ] = f : ( a , b ] → ℝ : ( x - a ) γ f ( x ) ∈ C [ a , b ] , γ ∈ ℝ ,

with the norm

f C γ [ a , b ] = ( x - a ) γ f ( x ) C [ a , b ] ,

where

f C [ a , b ] = max x ∈ [ a , b ] f ( x ) .

In the case f is not defined at x = a or γ < 0 we let ( x - a ) γ f ( x ) x = a = lim x - a + ( x - a ) γ f ( x ) . The spaces C_γ[a, b] satisfy the following properties.

• C₀[a,b] = C[a,b].

• C γ 1 [ a , b ] ⊂ C γ 2 [ a , b ] , γ 1 < γ 2 .

• C_γ[a,b] ⊂CL(a,b),γ < 1.

• f ∈ C_γ [a, b] if and only if f ∈ C(a, b) and lim x → a + ( x - a ) γ f ( x ) exists and is finite.

The left-sided Riemann-Liouville fractional integrals and derivatives are defined as follows.

Definition 1 Let f ∈ L(a,b). The integral

I a α f ( x ) : = 1 Γ ( α ) ∫ a x f ( s ) ( x - s ) 1 - α d s , x > a , α > 0 ,

is called the left-sided Riemann-Liouville fractional integral of order α of the function f.

Definition 2 The expression

D a α f ( x ) : = D I a α - 1 f ( x ) , x > a , 0 < α < 1 , D = d d x ,

is called the left-sided Riemann-Liouville fractional derivative of order α of f provided the right-hand side exists.

For power functions we have the following formulas.

Lemma 3 For x > a we have

I a α ( t - a ) β - 1 ( x ) = Γ ( β ) Γ ( β + α ) ( x - a ) β + α - 1 , α ≥ 0 , β > 0 .

D a α ( t - a ) α - 1 ( x ) = 0 , 0 < α < 1 .

Next we present some mapping properties of the operator I a α .

Lemma 4 For α > 0, Iaα maps L(a, b) into L(a, b).

The proof of Lemma 4 is given in 36. The following lemma is proved in 37.

Lemma 5 For α > 0, I a α maps C[a, b] into C[a, b].

The following lemma is proved in 38.

Lemma 6 Let α ≥ 0. If f ∈ CL(a, b) then I a α f ∈ C L ( a , b ) .

The mapping properties of Iaα in the spaces C_γ[a, b], 0 ≤ α ≤ γ < 1, are given in 1, Lemma 2.8 which is proved in 39 in Russian. For completeness we present here a more general result for α > 0 and γ < 1. First we prove the necessity condition at the left end.

Lemma 7 Let α ≥ 0 and γ < 1. If f ∈ C_γ[a, b] then

lim x → a + ( x - a ) γ - α I a α f ( x ) = c Γ ( 1 - γ ) Γ ( 1 + α - γ ) ,

where c = lim x → a + ( x - a ) γ f ( x ) .

Proof. Note that from Lemma 3 we have

I a α ( x - a ) - γ = Γ ( 1 - γ ) Γ ( 1 + α - γ ) ( x - a ) α - γ .

Thus

( x - a ) γ - α I a α f ( x ) - c Γ ( 1 - γ ) Γ ( 1 + α - γ ) = ( x - a ) γ - α I a α f ( x ) - c ( x - a ) γ - α I a α ( x - a ) - γ = ( x - a ) γ - α I a α f ( x ) - c I a α ( x - a ) - γ = ( x - a ) γ - α I a α ( x - a ) - γ { ( x - a ) γ f ( x ) - c } .

Now, given ϵ > 0 there exists δ > 0 such that x - a < δ implies that

( x - a ) γ f ( x ) - c < ε Γ ( 1 + α - γ ) Γ ( 1 - γ ) .

Thus

( x - a ) γ - α I a α f ( x ) - c Γ ( 1 - γ ) Γ ( 1 + α - γ ) = ( x - a ) γ - α I a α ( x - a ) - γ { ( x - a ) γ f ( x ) - c } < ε Γ ( 1 + α - γ ) Γ ( 1 - γ ) ( x - a ) γ - α I a α ( x - a ) - γ = ε Γ ( 1 + α - γ ) Γ ( 1 - γ ) ( x - a ) γ - α Γ ( 1 - γ ) Γ ( 1 + α - γ ) ( x - a ) α - γ = ε .

This yields the limit (9).

Next we present the mapping properties of Iaα in the spaces C_γ[a, b].

Lemma 8 Let α > 0 and γ < 1. If f ∈ C_γ[a, b] then I a α f ∈ C γ - α [ a , b ] and for x ∈ (a, b] we have

I a α f ( x ) ≤ Γ ( 1 - γ ) Γ ( 1 + α - γ ) ( x - a ) α - γ f C γ [ a , b ] .

Proof. From Lemmas 6 and 7 we have I^αf ∈ C(a, b) and lim x → a + ( x - a ) γ - α I a α f ( x ) exists and is finite. Thus I a α f ∈ C γ - α [ a , b ] . Now for x ∈ (a, b] we have

I a α f ( x ) = 1 Γ ( α ) ∫ a x ( x - t ) α - 1 ( t - a ) - γ ( t - a ) γ f ( t ) d t ≤ 1 Γ ( α ) f C γ [ a , b ] ∫ a x ( x - t ) α - 1 ( t - a ) - γ d t .

The relation (10) follows by applying Lemma 3.

Consequently, from Lemma 8 we have the following property.

Lemma 9 Let α > 0, γ < 1, and r ∈ ℝ. If f ∈ C_γ[a, b] then ( x - a ) - r I a α f ∈ C γ + r - α [ a , b ] . In particular, if γ + r - α < 1 then I^αf ∈ CL(a, b).

Later, the following observation is important.

Lemma 10 Let α > 0 and r < α. If f ∈ CL(a, b) then ( x - a ) - r I a α f ∈ C L ( a , b ) .

Proof. When r ≤ 0 the result follows clearly from Lemma 6. When r > 0 it follows from Lemma 6 that ( x - a ) - r I a α f ∈ C ( a , b ) and we only need to show that ( x - a ) - r I a α f ∈ L ( a , b ) .

For any x ∈ (a, b] we have the following inequality.

I a α f ( x ) = 1 Γ ( α ) ∫ a x ( x - t ) α - 1 f ( t ) d t = 1 Γ ( α ) ∫ a x ( x - t ) α - r - 1 ( x - t ) r f ( t ) d t ≤ ( x - a ) r Γ ( α ) ∫ a x ( x - t ) α - r - 1 f ( t ) d t = Γ ( α - r ) Γ ( α ) ( x - a ) r I a α - r f ( x ) .

Or,

( x - a ) - r I a α f ( x ) ≤ Γ ( α - r ) Γ ( α ) I a α - r f ( x ) .

From Lemma 4 the right-hand side is in L(a, b) and thus ( x - a ) - r I a α f ∈ L ( a , b ) . This completes the proof.

The following lemma follows by direct calculations using Dirichlet formula, 36.

Lemma 11 Let α ≥ 0, β ≥ 0, and f ∈ CL(a, b). Then

I a α I a β f ( x ) = I a α + β f ( x ) ,

for all x ∈ (a, b].

Lemma 11 leads to the left inverse operator.

Lemma 12 Let α > 0 and f ∈ CL(a, b). Then

D a α I a α f ( x ) = f ( x ) ,

for all x ∈ (a, b].

Now we present a version of the fundamental theorem of fractional calculus.

Lemma 13 Let 0 < α < 1. If f ∈ C(a, b) and D a α f ∈ C L ( a , b ) , then f ∈ C L ( a , b ) , I a α f ( a + ) exists and is finite, and

I a α D a α f ( x ) = f ( x ) - I a 1 - α f ( a + ) Γ ( α ) ( x - a ) α - 1 ,

for all x ∈ (a, b].

Proof. From Lemma 12 we have for all x ∈ (a, b] the relation

D a α I a α D a α f ( x ) = D a α f ( x ) ,

which we can write as

D a α f - I a α D a α f ( x ) = 0 .

This implies that

f ( x ) - I a α D a α f ( x ) = c ( x - a ) α - 1 ,

for some constant c. Since Lemma 6 implies that I a α D a α f ∈ C L ( a , b ) , we also have f ∈ CL(a, b). Also, if we apply I a 1 - α to both sides of (14) we obtain

I a 1 - α f ( x ) = I a D a α f ( x ) = c Γ ( α ) .

Taking the limit yields I a 1 - α f ( a + ) = c Γ ( α ) and (13) is obtained.

In the proof of our existence and uniqueness result we will use the following results.

Lemma 14 Let γ ∈ ℝ, a < c < b, g ∈ C_γ[a, c], g ∈ C[c, b] and g is continuous at c. Then g ∈ C_γ[a, b].

Theorem 15 (1, Banach Fixed Point Theorem) Let (U, d) be a nonempty complete metric space. Let T : U → U be a map such that for every u, v ∈ U, the relation

d ( T u , T v ) ≤ w d ( u , v ) , 0 ≤ w < 1 ,

holds. Then the operator T has a unique fixed point u* ∈ U.

In this section we define the sequential derivative and integral that we consider and develop some of their properties. In particular, we derive the composition identities.

Definition 16 Let α > 0, β > 0, r ∈ ℝ. Let f ∈ CL(a, b). Define the sequential integral J r , a α , β f and the sequential derivative D r , a α , β f by

J r , a α , β f ( x ) = I a α ( x - a ) - r I a β f ( x ) ,

and

D r , a α , β f ( x ) = D a α ( x - a ) r D a β f ( x ) ,

if the right-hand sides exist.

From Lemma 3 we have the following formula for the power function.

Lemma 17 Let α > 0, β > 0, r ∈ ℝ. If

ρ > max { - 1 , β - r - 1 } ,

then for x > a,

D r , a α , β ( x - a ) ρ = Γ ( ρ + 1 ) Γ ( ρ - β + 1 ) Γ ( ρ + r - β + 1 ) Γ ( ρ + r - β - α + 1 ) ( x - a ) ρ + r - β - α .

Moreover, from Lemmas 3 and 17 we have the following vanishing derivatives.

Lemma 18

(a) Let α > 0, 0 < β < 1, r ∈ ℝ. Then for x > a,

D r , a α , β ( x - a ) β - 1 = 0 .

(b) Let 0 < α < 1 and β > 0. Let r ∈ ℝ be such that r < α + β. Then for x > a,

D r , a α , β ( x - a ) α + β - r - 1 = 0 .

Lemma 19 (Left inverse) Let α > 0, β > 0, and r ∈ ℝ. If f ∈ CL(a, b) such that ( x - a ) - r I a α f ∈ C L ( a , b ) then

D r , a α , β J r , a β , α f ( x ) = f ( x ) ,

for all x ∈ (a, b].

Proof. Relation (20) follows directly by applying Lemma 12 twice.

From Lemmas 8 and 9 we have the following mapping property of the operator J r , a β , α .

Lemma 20 Let α > 0, β > 0, and r < 1 + α. Let 0 ≤ γ < min{1, 1 + α - r}. If f ∈ C_γ[a, b] then J r , a β , α f ∈ C γ + r - α - β [ a , b ] and for x ∈ (a, b] we have

J r , a β , α f ( x ) ≤ k f C γ [ a , b ] ( x - a ) α + β - r - γ ,

where

k = Γ ( 1 - γ ) Γ ( 1 + α - r - γ ) Γ ( 1 - γ + α ) Γ ( 1 + α + β - r - γ ) .

Lemma 20 implies the following.

Lemma 21 Let α > 0, β > 0, and r < 1 + α. Let 0 ≤ γ < min{1, 1 + α-r}. If r ≤ α + β, then Jr,aβ,α is bounded in C_γ[a, b] and

J r , a β , α f C γ [ a , b ] ≤ k ( b - a ) α + β - r f C γ [ a , b ] ,

where k is given by (22).

Proof. Since γ + r - α - β ≤ γ, then from Lemma 20 we have

J r , a β , α f ∈ C γ + r - α - β [ a , b ] ⊂ C γ [ a , b ] .

The bound in (23) follows by multiply (21) by (x - a)^γ and taking the maximum.

As a special case of Lemma 21, we have

Lemma 22 Let α > 0, β > 0, and r < min{α + 1, α + β}. Then Jr,aβ,α , maps C[a, b] into C[a, b] and

J r , a β , α f C [ a , b ] ≤ L ( b - a ) α + β - r f C [ a , b ] ,

where

L = Γ ( 1 + α - r ) Γ ( 1 + α ) Γ ( 1 + α + β - r ) .

The following is an analogous result to the result for the Riemann-Liouville integral proved in 10.

Lemma 23 Let α > 0, β > 0, and r < α. Let f ∈ CL(a, c). Let

g ( x ) = 1 Γ ( β ) ∫ a c ( x - t ) β - 1 ( t - a ) - r I a α f ( t ) d t .

Then

lim x → c + g ( x ) = J r , a β , α f ( c ) .

Proof. Since r < α, Lemma 10 implies that ( x - a ) - r I a α f ( x ) ∈ C L ( a , c ) . Thus J r , a β , α f ( c ) is finite and

g ( x ) - J r , a β , c f ( c ) ≤ 1 Γ ( β ) ∫ a c k ( x , t ) ( t - a ) - r I a α f ( t ) d t ,

where

k ( x , t ) = ( c - t ) β - 1 - ( x - t ) β - 1 .

Since lim x → c + k ( x , t ) = 0 , the limit of the right-hand side vanishes and the proof is complete.

The following lemma relates the fractional derivative D r , a α , β to the Riemann-Liouville derivative D a β .

Lemma 24 Let 0 < α < 1, β ≥ 0, and r ∈ ℝ. If D a β y ( x ) ∈ C ( a , b ) and D r , a α , β y ∈ C L ( a , b ) then ( x - a ) r D a β y ∈ C L ( a , b ) , I a 1 - α ( x - a ) r D a β y ( a + ) exists and finite, and

D a β y ( x ) = ( x - a ) - r I a α D r , a α , β y ( x ) + I 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) ( x - a ) α - r - 1 ,

for all x ∈ (a, b]. If in addition, r < α then D a β y ∈ C L ( a , b ) .

Proof. Clearly D a β y ∈ C ( a , b ) implies that ( x - a ) r D a β y ∈ C ( a , b ) . Thus we can apply Lemma 13 to ( x - a ) r D a β y and obtain

I a α D r , a α , β y ( x ) = I a α D a α ( x - a ) r D a β y ( x ) = ( x - a ) r D a β y ( x ) - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) ( x - a ) α - 1 .

By multiplying both sides by (x - a)^r we obtain (26). If r < α then Lemma 10 implies that ( x - a ) - r I a α D r , a α , β y ( x ) ∈ C L ( a , b ) and thus from (26) we have D a β y ∈ C L ( a , b ) . This proves the result.

The Next lemma gives an analogous result to the fundamental theorem of calculus in terms of the operators Dr,aα,β and J r , a α , β .

Lemma 25 Let 0 < α < 1 and 0 < β < 1. Let y ∈ C(a, b) be such that D r , a α , β y ∈ C L ( a , b ) and Daβy∈CL(a,b) . Then both I a 1 - α ( x - a ) r D a β y ( a + ) and I a 1 - β y ( a + ) exist, y ∈ CL(a, b), and

J r , a β , α D r , a α , β y ( x ) = y ( x ) - I a 1 - β y ( a + ) Γ ( β ) ( x - a ) β - 1 - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) Γ ( α + r ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 ,

for all x ∈ (a, b].

Proof. By applying Lemma 13 twice we obtain

J r , a β , α D r , a α , β y ( x ) = I a β ( x - a ) - r I a α D a α ( x - a ) r D a β y ( x ) = I β ( x - a ) - r ( x - a ) r D a β y ( x ) - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) ( x - a ) α - 1 = I a β D a β y - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) I a β ( x - a ) α - 1 - r = y ( x ) - I a 1 - β y ( a + ) Γ β ( x - a ) β - 1 - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) Γ ( α + r ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 .

Consider the Cauchy-type problem

D r , a α , β y ( x ) = f ( x , y ( x ) ) , a < x ≤ b , 0 < α < 1 , 0 < β < 1 , r ∈ ℝ ,

lim x → a + I a 1 - α ( x - a ) r D a β y ( x ) = c 1 ,

lim x → a + I a 1 - β y ( x ) = c 0 ,

where c₀ and c₁ are real numbers.

In this problem there are two conditions even when 0 < α + β < 1. The two initial conditions are based on the composition (27). The condition (29) is of one order less than that in the differential Equation (28) while the condition (30) is one order less than the equation for D a β y .

In addition, from [1, Lemma 3.2], the condition (30) follows from the condition

lim x → a + ( x - a ) 1 - β y ( x ) = c 0 Γ ( β ) ,

and if 0 < α - r < 1 then (29) follows from the condition

lim x → a + ( x - a ) 1 - α + r D a β y ( x ) = c 1 Γ ( α ) .

Consequently, the results below hold under conditions of the type (31) and (32).

Now, Based on the composition in Lemma 24, in the next theorem we establish an equivalence with the following fractional integro-differential equation:

D a β y = ( x - a ) - r I a α f [ x , y ( x ) ] + c 1 Γ ( α ) ( x - a ) α - r - 1 ,

lim x → a + I a 1 - β y ( x ) = c 0 .

Theorem 26 Let 0 < α < 1, β > 0 and r ∈ ℝ. Let f : (a, b] × ℝ → ℝ be a function such that f(.,y(.)) ∈ C_1-α [a, b] for any y ∈ C_1-α [a, b]. Then we have the following.

(a) If y ∈ C_1-α[a, b] satisfies (33) and (34) then y(x) satisfies (28-30).

(b) If y ∈ C_1-α[a, b] with Daβy∈C(a,b) satisfy (28-30), then y(x) satisfies (33-34).

Proof.

For assertion (a), let y ∈ C_1-α[a, b] satisfy (33-34). We multiply (33) by (x - a)^r to obtain

( x - a ) r D a β y = I a α f ( x , y ( x ) ) + c 1 Γ ( α ) ( x - a ) α - 1 .

Next we apply Daα to both sides of (35) to obtain (28). As for the initial condition, apply Ia1-α to both sides of (35) and then take the limit to obtain (29).

For assertion (b), let y ∈ C_1-α[a, b] satisfy (28-30). Since f(x, y(x)) ∈ C_1-α[a, b], then from (28) we have D r , a α , β y ∈ C 1 - α [ a , b ] . Since also by hypothesis Daβy∈C(a,b) , we can apply Lemma 24 and the formula (26) holds. By substituting the initial condition we obtain (33). This completes the proof.

The composition in Lemma 25 leads to the nonlinear integral equation,

y ( x ) = J r , a β , α f ( x , y ( x ) ) + c 0 Γ ( β ) ( x - a ) β - 1 + c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 .

The following theorem establishes an equivalence with this equation.

Theorem 27 Let 0 < α < 1, 0 < β < 1 and r < α. Let f : (a, b] × ℝ → ℝ be a function such that f(.,y(.)) ∈ C_1-β[a, b] for any y ∈ C_1-β[a, b]. Then the following statements hold.

(a) If y ∈ C_1-β[a, b] satisfies the integral Equation (36) then y(x) satisfies the Cauchy-type problem (28-30).

(b) If y ∈ C_1-β[a, b] with Daβy∈C(a,b) satisfies the Cauchy-type problem (28-30), then y(x) satisfies the integral Equation (36).

Proof. (a). Let y ∈ C_1-β[a, b] satisfy the integral Equation (36). By hypothesis we have f ∈ C_1-β[a, b]. Moreover, from Lemma 9 and the hypothesis r < α, we have

( x - a ) - r I a α f ∈ C 1 + r - α - β [ a , b ] ⊂ C L ( a , b ) .

Thus the hypothesis of Lemmas 18 and 19 are satisfied. Applying the operator Dr,aα,β to both sides of (36) and using Lemmas 18 and 19 yields (28) as follows.

D r , a α , β y ( x ) = D r , a α , β J r , a β , α f ( x , y ( x ) ) + D r , a α , β c 0 Γ ( β ) ( x - a ) β - 1 + c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 = f ( x , y ( x ) ) .

Next, applying I a 1 - β to both sides of (36) yields

I a 1 - β y ( x ) = J r , a 1 , α f ( x , y ( x ) ) + c 0 + c 1 Γ ( α ) Γ ( α - r ) Γ ( α + β - r ) Γ ( α + β - r ) Γ ( α - r + 1 ) ( x - a ) α - r .

Since r < α, taking the limit we obtain the initial condition (30).

Applying I a 1 - α ( x - a ) r D a β to both sides of (36) and using Lemmas 3, 11, and 12 yields

I a 1 - α ( x - a ) r D a β y ( x ) = I a 1 - α ( x - a ) r D a β J r , a β , α f ( x , y ( x ) ) + I a 1 - α ( x - a ) r D a β c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 = I a 1 f ( x , y ( x ) ) + c 1 .

Again, taking the limit we obtain the initial condition (29).

(b). Let y ∈ C_1-β[a, b] satisfy (28-30). Since f(x, y(x)) ∈ CL(a, b) then from (28), Dr,aα,βy∈CL(a,b) . Since r < α then from Lemma 24 we have Daβy∈CL(a,b) . Thus we can apply Lemma 25 and the formula (27) holds. By using the initial conditions we obtain (36). This completes the proof.

In the next section we use this equivalence to prove the existence and uniqueness of solutions.

In this section we prove an existence and uniqueness result for the Cauchy-type problem (28-30) using the integral Equation (36). For this purpose we introduce the following lemma.

Lemma 28 Let 0 < r < α < 1, 0 < β < 1, then the fractional differentiation operator Jr,aβ,α is bounded in C_1-β[a, b] and

J r , a β , α f C 1 - β [ a , b ] ≤ K ( b - a ) α + β - r f C 1 - β [ a , b ] ,

where

K = Γ ( β ) Γ ( α + β - r ) Γ ( α + β ) Γ ( α + 2 β - r ) .

Proof. Clearly from the hypothesis we have r < α + β and 0 < 1 - β < min{1, 1 + α - r}. Thus the result follows by taking γ = 1 - β in Lemma 21.

Theorem 29 Let 0 < r < α < 1, 0 ≤ β < 1. Let f : (a, b] × ℝ → ℝ be a function such that f(.,y(.)) ∈ C_1-β[a, b] for any y ∈ C_1-β[a, b] and the condition:

f ( x , y 1 ) - f ( x , y 2 ) ≤ A y 1 - y 2 , A > 0 ,

is satisfied for all x ∈ (a, b] and for all y₁, y₂ ∈ ℝ.

Then the Cauchy-type problem (28-30) has a solution y ∈ C_1-β[a, b]. Furthermore, if for this solution Daβy∈C(a,b) , then this solution is unique.

Proof.

According to Theorem 27(a), we can consider the existence of an C_1-β[a, b] solution for the integral Equation (36). This equation holds in any interval (a, x₁] ⊂ (a, b], a < x₁ < b. Choose x₁ such that

w 1 : = A K ( x 1 - a ) α + β - r < 1 ,

where K is given by (39). We rewrite the integral equation in the form y(x) = Ty(x), where

T y ( x ) = v 0 ( x ) + J r , a β , α f ( x , y ( x ) ) ,

and

v 0 ( x ) = c 0 Γ ( β ) ( x - a ) β - 1 + c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 .

Since r < α then v₀ ∈ C_1-β[a, b]. Thus, it follows from Lemma 28 that if y ∈ C_1-β[a, x₁] then Ty ∈ C_1-β[a, x₁]. Also, for any y₁, y₂ in C_1-β[a, x₁], we have

T y 1 - T y 2 C 1 - β [ a , x 1 ] ≤ J r , a β , α f ( x , y 1 ( x ) ) - f ( x , y 2 ( x ) ) C 1 - β [ a , x 1 ] ≤ A J r , a β , α y 1 ( x ) - y 2 ( x ) C 1 - β [ a , x 1 ] ≤ A K ( x 1 - a ) α + β - r y 1 - y 2 C 1 - β [ a , x 1 ] ≤ w 1 y 1 - y 2 C 1 - β [ a , x 1 ] , 0 < w 1 < 1 .