We investigate the existence of nonnegative solutions for a Riemann-Liouville fractional differential equation with integral terms, subject to boundary conditions which contain fractional derivatives and Riemann-Stieltjes integrals. In the proof of the main results , we use the Banach contraction mapping principle and the Krasnosel ’ skii fixed point theorem for the sum of two operators.
We consider the nonlinear fractional differential equation
(E) D 0 + α x ( t ) + f ( t , x ( t ) , A x ( t ) , B x ( t ) ) = 0, t ∈ ( 0,1 ) ,
with the nonlocal boundary conditions
(BC) x ( 0 ) = x ′ ( 0 ) = ⋯ = x ( n − 2 ) ( 0 ) = 0 , D 0 + β 0 x ( 1 ) = ∑ i = 1 m ∫ 0 1 D 0 + β i x ( t ) d H i ( t ) ,
where α ∈ ℝ , α ∈ ( n − 1, n ] , n , m ∈ ℕ , n ≥ 3 , β i ∈ ℝ for all i = 0 , ⋯ , m , 0 ≤ β 1 < β 2 < ⋯ < β m ≤ β 0 < α − 1 , β 0 ≥ 1 , D 0 + k denotes the Riemann-Liouville derivative of order k (for k = α , β 0 , β 1 , ⋯ , β m ), the integrals from the boundary condition (BC) are Riemann-Stieltjes integrals with H i : [ 0 , 1 ] → ℝ , i = 1 , ⋯ , m
non-decreasing functions, A x ( t ) = ∫ 0 t K ( t , s ) x ( s ) d s , and B x ( t ) = ∫ 0 1 H ( t , s ) x ( s ) d s for all t ∈ [ 0,1 ] .
We study the existence of nonnegative solutions for problem (E)-(BC) by using the Banach contraction mapping principle and the Krasnosel’skii fixed point theorem for the sum of two operators. Equation (E) supplemented with the multi-point boundary conditions
(BC1) x ( 0 ) = x ′ ( 0 ) = ⋯ = x ( n − 2 ) ( 0 ) = 0 , D 0 + p x ( 1 ) = ∑ i = 1 m a i D 0 + q x ( ξ i ) ,
where a i , ξ i ∈ ℝ for all i = 1 , ⋯ , m , ( m ∈ ℕ ), 0 < ξ 1 < ⋯ < ξ m ≤ 1 , p , q ∈ ℝ , p ∈ [ 1, n − 2 ] , q ∈ [ 0, p ] was investigated in the paper [
(BC1) can be written as D 0 + p x ( 1 ) = ∫ 0 1 D 0 + q x ( t ) d H 0 ( t ) , where H 0 is the step function defined by
H 0 ( t ) = { 0, for t ∈ [ 0, ξ 1 ) ; a 1 , for t ∈ [ ξ 1 , ξ 2 ) ; a 1 + a 2 , for t ∈ [ ξ 2 , ξ 3 ) ; ⋯ ; ∑ i = 1 m a i , for t ∈ [ ξ m ,1 ] }
So (BC) is a generalization of (BC1), because in (BC) we have a sum of Riemann-Stieltjes integrals and various orders for the fractional derivatives. In the paper [
D c α x ( t ) + f ( t , x ( t ) , A x ( t ) , B x ( t ) ) = 0, t ∈ ( 0,1 ) ,
with the boundary conditions
(BC2) x ( 0 ) = b 0 , x ′ ( 0 ) = b 1 , ⋯ , x ( n − 3 ) ( 0 ) = b n − 3 , x ( n − 1 ) ( 0 ) = b n − 1 , x ( 1 ) = μ ∫ 0 1 x ( s ) d s ,
where n − 1 < α ≤ n , 0 ≤ μ < n − 1 , n ≥ 3 , b i ≥ 0 ( i = 1 , 2 , ⋯ , n − 3 , n − 1 ), D c α is the Caputo fractional derivative, and the operators A and B are defined as the operators from our problem, given above. In the paper [
We present in this section some auxiliary results from [
D 0 + α x ( t ) + y ( t ) = 0 , t ∈ ( 0 , 1 ) , (1)
with the boundary conditions (BC), where y ∈ C ( 0,1 ) ∩ L 1 ( 0,1 ) . We denote by
Δ = Γ ( α ) Γ ( α − β 0 ) − ∑ i = 1 m Γ ( α ) Γ ( α − β i ) ∫ 0 1 s α − β i − 1 d H i ( s ) .
Lemma 1 If Δ ≠ 0 , then the unique solution x ∈ C [ 0,1 ] of problem (1)-(BC) is given by
x ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + t α − 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 y ( s ) d s − t α − 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 y ( τ ) d τ ) d H i ( s ) , t ∈ [ 0 , 1 ] . (2)
Lemma 2 If Δ ≠ 0 , then the solution x of problem (1)-(BC) given by (2) can be written as
x ( t ) = ∫ 0 1 G ( t , s ) y ( s ) d s , t ∈ [ 0 , 1 ] , (3)
where
G ( t , s ) = g 1 ( t , s ) + t α − 1 Δ ∑ i = 1 m ( ∫ 0 1 g 2 i ( τ , s ) d H i ( τ ) ) , (4)
and
g 1 ( t , s ) = 1 Γ ( α ) { t α − 1 ( 1 − s ) α − β 0 − 1 − ( t − s ) α − 1 , 0 ≤ s ≤ t ≤ 1 , t α − 1 ( 1 − s ) α − β 0 − 1 , 0 ≤ t ≤ s ≤ 1 , g 2 i ( t , s ) = 1 Γ ( α − β i ) { t α − β i − 1 ( 1 − s ) α − β 0 − 1 − ( t − s ) α − β i − 1 , 0 ≤ s ≤ t ≤ 1 , t α − β i − 1 ( 1 − s ) α − β 0 − 1 , 0 ≤ t ≤ s ≤ 1 , (5)
for all ( t , s ) ∈ [ 0,1 ] × [ 0,1 ] , i = 1 , ⋯ , m .
By using some properties of the functions g 1 , g 2 i , i = 1 , ⋯ , m given by (5) from [
Lemma 3 We suppose that Δ > 0 . Then the function G given by (4) is a continuous function on [ 0,1 ] × [ 0,1 ] and satisfies the inequalities:
a) G ( t , s ) ≤ J ( s ) for all t , s ∈ [ 0,1 ] , where J ( s ) = h 1 ( s ) + 1 Δ ∑ i = 1 m ∫ 0 1 g 2 i ( τ , s ) d H i ( τ ) , s ∈ [ 0,1 ] , and h 1 ( s ) = 1 Γ ( α ) ( 1 − s ) α − β 0 − 1 ( 1 − ( 1 − s ) β 0 ) , s ∈ [ 0,1 ] ;
b) G ( t , s ) ≥ t α − 1 J ( s ) for all t , s ∈ [ 0,1 ] ;
c) G ( t , s ) ≤ σ t α − 1 , for all t , s ∈ [ 0,1 ] , where σ = 1 Γ ( α ) + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 τ α − β i − 1 d H i ( τ ) .
Lemma 4 We suppose that Δ > 0 , y ∈ C ( 0,1 ) ∩ L 1 ( 0,1 ) and y ( t ) ≥ 0 for all t ∈ ( 0,1 ) . Then the solution x of problem (1)-(BC) given by (3) satisfies the inequality x ( t ) ≥ t α − 1 ‖ x ‖ for all t ∈ [ 0,1 ] , where ‖ x ‖ = sup t ∈ [ 0,1 ] | x ( t ) | , and so x ( t ) ≥ 0 for all t ∈ [ 0,1 ] .
In the proof of our main theorems, we use the Banach contraction mapping principle and the Krasnosel’skii fixed point theorem for the sum of two operators presented below.
Theorem 1 (see [
Theorem 2 ( [
a) A 1 x + A 2 y ∈ M for all x , y ∈ M ;
b) A 1 is a completely continuous operator (continuous, and compact, that is, it maps bounded sets into relatively compact sets);
c) A 2 is a contraction mapping.
Then there exists z ∈ M such that z = A 1 z + A 2 z .
In this section we study the existence of nonnegative solutions for our problem (E)-(BC). We present now the assumptions that we will use in the sequel.
(I1) α ∈ ℝ , α ∈ ( n − 1, n ] , n , m ∈ ℕ , n ≥ 3 , β i ∈ ℝ for all i = 0 , ⋯ , m , 0 ≤ β 1 < β 2 < ⋯ < β m ≤ β 0 < α − 1 , β 0 ≥ 1 , H i : [ 0,1 ] → ℝ , i = 1 , ⋯ , m are nondecreasing functions, an
Δ = Γ ( α ) Γ ( α − β 0 ) − ∑ i = 1 m Γ ( α ) Γ ( α − β i ) ∫ 0 1 s α − β i − 1 d H i ( s ) > 0 .
(I2) f : [ 0,1 ] × ℝ + 3 → ℝ + is measurable with respect to t on [ 0,1 ] , ( ℝ + = [ 0 , ∞ ) ).
(I3) There exist the functions a , b , c ∈ L 1 ( ( 0,1 ) , ℝ + ) such that
| f ( t , x , y , z ) − f ( t , x ¯ , y ¯ , z ¯ ) | ≤ a ( t ) | x − x ¯ | + b ( t ) | y − y ¯ | + c ( t ) | z − z ¯ | ,
a.e. t ∈ ( 0,1 ) and for all x , y , z , x ¯ , y ¯ , z ¯ ∈ ℝ + .
(I4) There exists the function g ∈ L 1 ( ( 0,1 ) , ℝ + ) such that
| f ( t , x , y , z ) | ≤ g ( t ) , a .e . t ∈ ( 0,1 ) , ∀ x , y , z ∈ ℝ + .
(I5) K ∈ C ( E , ℝ + ) , E = { ( t , s ) ∈ [ 0,1 ] × [ 0,1 ] , t ≥ s } , and H ∈ C ( [ 0,1 ] × [ 0,1 ] , ℝ + ) .
We denote by k 0 = sup t ∈ [ 0,1 ] ∫ 0 t K ( t , s ) d s and h 0 = sup t ∈ [ 0 , 1 ] ∫ 0 1 H ( t , s ) d s .
Theorem 3 We suppose that assumptions (I1)-(I5) hold. If σ ω 1 < 1 , where ω 1 = ‖ a + k 0 b + h 0 c ‖ L 1 , then problem (E)-(BC) has at least one nonnegative solution on [ 0 , 1 ] .
Proof. By (I4) we obtain that the function s → ( t − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) is Lebesgue integrable on [ 0, t ] for all t ∈ [ 0,1 ] and x ∈ C ( [ 0,1 ] , ℝ + ) , the function s → ( 1 − s ) α − β 0 − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) is Lebesgue integrable on [ 0 , 1 ] for all x ∈ C ( [ 0,1 ] , ℝ + ) , and the function τ → ( s − τ ) α − β i − 1 f ( τ , x ( τ ) , A x ( τ ) , B x ( τ ) ) is Lebesgue integrable on [ 0, s ] for all s ∈ [ 0,1 ] , x ∈ C ( [ 0,1 ] , ℝ + ) and i = 1 , ⋯ , m .
We consider now the integral equation
x ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s + t α − 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s − t α − 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 f ( τ , x ( τ ) , A x ( τ ) , B x ( τ ) ) d τ ) d H i ( s ) , (6)
or equivalently
x ( t ) = ∫ 0 1 G ( t , s ) f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s . (7)
By Lemma 2 we easily deduce that if x is a solution of Equation (6) (or equivalently (7)), then x is a solution of problem (E)-(BC).
Let r = σ ‖ g ‖ L 1 . We define the operator T on B ¯ r = { x ∈ C ( [ 0 , 1 ] , ℝ + ) , ‖ x ‖ ≤ r } by
T x ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s + t α − 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s − t α − 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 f ( τ , x ( τ ) , A x ( τ ) , B x ( τ ) ) d τ ) d H i ( s ) ,
or equivalently
T x ( t ) = ∫ 0 1 G ( t , s ) f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s .
If x is a fixed point of operator T , then x is a solution of Equation (6) (or (7)), and hence x is a solution of problem (E)-(BC). Therefore we will study the existence (and uniqueness) of the fixed points of operator T by using the Banach contraction mapping principle.
We firstly show that if x ∈ B ¯ r , then T x ∈ B ¯ r . Indeed, we have
| T x ( t + ε ) − T x ( t ) | = | ∫ 0 1 [ G ( t + ε , s ) − G ( t , s ) ] f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s | ≤ ∫ 0 1 | G ( t + ε , s ) − G ( t , s ) | g ( s ) d s ≤ max s ∈ [ 0,1 ] | G ( t + ε , s ) − G ( t , s ) | ⋅ ‖ g ‖ L 1 → 0 , for ε → 0.
Hence T x is a continuous function. By (I1), (I2) and Lemma 4, we obtain T x ( t ) ≥ 0 for all t ∈ [ 0,1 ] , and then T x ∈ C ( [ 0,1 ] , ℝ + ) .
In addition, for any x ∈ B ¯ r and all t ∈ [ 0,1 ] , we deduce
( T x ) ( t ) = ∫ 0 1 G ( t , s ) f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s ≤ ∫ 0 1 σ t α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s ≤ σ t α − 1 ∫ 0 1 g ( s ) d s ≤ σ ‖ g ‖ L 1 = r ,
and then ‖ T x ‖ ≤ r for all x ∈ B ¯ r , so T : B ¯ r → B ¯ r .
We show now that T is a contraction mapping on B ¯ r . For x 1 , x 2 ∈ B ¯ r , and any t ∈ [ 0,1 ] , by using (I3), we find
| ( T x 1 ) ( t ) − ( T x 2 ) ( t ) | ≤ ∫ 0 1 G ( t , s ) | f ( s , x 1 ( s ) , A x 1 ( s ) , B x 1 ( s ) ) − f ( s , x 2 ( s ) , A x 2 ( s ) , B x 2 ( s ) ) | d s ≤ ∫ 0 1 G ( t , s ) [ a ( s ) | x 1 ( s ) − x 2 ( s ) | + b ( s ) | A x 1 ( s ) − A x 2 ( s ) | + c ( s ) | B x 1 ( s ) − B x 2 ( s ) | ] d s
≤ ‖ x 1 − x 2 ‖ σ t α − 1 ∫ 0 1 [ a ( s ) + k 0 b ( s ) + h 0 c ( s ) ] d s ≤ σ ‖ x 1 − x 2 ‖ ⋅ ‖ a + k 0 b + h 0 c ‖ L 1 = σ ω 1 ‖ x 1 − x 2 ‖ ,
because
A x 1 ( s ) − A x 2 ( s ) = ∫ 0 s K ( s , τ ) ( x 1 ( τ ) − x 2 ( τ ) ) d τ ≤ sup τ ∈ [ 0,1 ] | x 1 ( τ ) − x 2 ( τ ) | ∫ 0 s K ( s , τ ) d τ ≤ k 0 ‖ x 1 − x 2 ‖ , ∀ s ∈ [ 0,1 ] ,
B x 1 ( s ) − B x 2 ( s ) = ∫ 0 1 H ( s , τ ) ( x 1 ( τ ) − x 2 ( τ ) ) d τ ≤ sup τ ∈ [ 0,1 ] | x 1 ( τ ) − x 2 ( τ ) | ∫ 0 1 H ( s , τ ) d τ ≤ h 0 ‖ x 1 − x 2 ‖ , ∀ s ∈ [ 0,1 ] ,
and a , b , c ∈ L 1 ( 0,1 ) .
Therefore we obtain the inequality
‖ T x 1 − T x 2 ‖ ≤ σ ω 1 ‖ x 1 − x 2 ‖ .
Because σ ω 1 < 1 , we deduce that T is a contraction mapping. By Theorem 1, we conclude that T has a unique fixed point, which is a nonnegative solution of problem (E)-(BC).
In what follows, we denote by
ω 2 = 1 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 ( a ( s ) + k 0 b ( s ) + h 0 c ( s ) ) d s , ω 3 = ‖ a + k 0 b + h 0 c ‖ L 1 ( 1 Δ Γ ( α − β 0 ) + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ( H i ( 1 ) − H i ( 0 ) ) ) . (8)
Theorem 4 We suppose that assumptions (I1),
(I2)' f : [ 0,1 ] × ℝ + 3 → ℝ + is a continuous function and (I3), (I4), (I5) hold. If max { ω 2 , ω 3 } < 1 , then problem (E)-(BC) has at least one nonnegative solution on [ 0,1 ] .
Proof. We define R ≥ R 0 , where
R 0 = ( 1 − ω 2 ) − 1 [ τ 0 Γ ( α + 1 ) + ‖ g ‖ L 1 ( 1 Δ Γ ( α − β 0 ) + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ( H i ( 1 ) − H i ( 0 ) ) ) ] ,
τ 0 = max { f ( t , 0 , 0 , 0 ) , t ∈ [ 0 , 1 ] } and ω 2 is given by (8). We consider the set B ¯ R = { x ∈ C ( [ 0,1 ] , ℝ + ) , ‖ x ‖ ≤ R } . Then B ¯ R is a closed, convex and nonempty set of C ( [ 0,1 ] , ℝ + ) . We define the operators Q 1 and Q 2 on B ¯ R by
( Q 1 x ) ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s , t ∈ [ 0,1 ] , ( Q 2 x ) ( t ) = t α − 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s − t α − 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 f ( τ , x ( τ ) , A x ( τ ) , B x ( τ ) ) d τ ) d H i ( s ) , t ∈ [ 0,1 ] .
By (I1), (I2)' and Lemma 4, we have ( Q 1 x ) ( t ) + ( Q 2 x ) ( t ) ≥ 0 for all t ∈ [ 0,1 ] . For any x ∈ B ¯ R , by using (I3), we find
| f ( t , x ( t ) , A x ( t ) , B x ( t ) ) | ≤ | f ( t , x ( t ) , A x ( t ) , B x ( t ) ) − f ( t ,0,0,0 ) | + | f ( t ,0,0,0 ) | ≤ a ( t ) | x ( t ) | + b ( t ) | A x ( t ) | + c ( t ) | B x ( t ) | + τ 0 , ∀ t ∈ [ 0,1 ] .
Then for any x ∈ B ¯ R and all t ∈ [ 0,1 ] , we obtain by using the above inequality
| Q 1 x ( t ) | = | − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s | ≤ 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 ( a ( s ) | x ( s ) | + b ( s ) | A x ( s ) | + c ( s ) | B x ( s ) | + τ 0 ) d s ≤ ‖ x ‖ Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 ( a ( s ) + k 0 b ( s ) + h 0 c ( s ) ) d s + τ 0 Γ ( α ) ∫ 0 t ( t − s ) α − 1 d s ≤ ω 2 R + t α τ 0 Γ ( α + 1 ) ≤ ω 2 R + τ 0 Γ ( α + 1 ) ,
because ( A x ) ( t ) ≤ sup t ∈ [ 0,1 ] x ( t ) ∫ 0 t K ( t , s ) d s ≤ k 0 ‖ x ‖ and ( B x ) ( t ) ≤ sup t ∈ [ 0,1 ] x ( t ) ∫ 0 1 H ( t , s ) d s ≤ h 0 ‖ x ‖ for all t ∈ [ 0,1 ] .
Therefore for any z ∈ B ¯ R and all t ∈ [ 0,1 ] , we deduce by using (I4) that
| Q 2 z ( t ) | ≤ t α − 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 f ( s , z ( s ) , A z ( s ) , B z ( s ) ) d s + t α − 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 f ( τ , z ( τ ) , A z ( τ ) , B z ( τ ) ) d τ ) d H i ( s ) ≤ t α − 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 g ( s ) d s + t α − 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 g ( τ ) d τ ) d H i (s)
≤ 1 Δ Γ ( α − β 0 ) ∫ 0 1 g ( s ) d s + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 1 g ( τ ) d τ ) d H i ( s ) = ‖ g ‖ L 1 ( 1 Δ Γ ( α − β 0 ) + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ( H i ( 1 ) − H i ( 0 ) ) ) .
Hence for x , y ∈ B ¯ R and t ∈ [ 0,1 ] , we find
| Q 1 x ( t ) + Q 2 y ( t ) | ≤ | Q 1 x ( t ) | + | Q 2 y ( t ) | ≤ ω 2 R + τ 0 Γ ( α + 1 ) + ‖ g ‖ L 1 ( 1 Δ Γ ( α − β 0 ) + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ( H i ( 1 ) − H i ( 0 ) ) ) ≤ R .
Therefore for y 1 , y 2 ∈ B ¯ R and t ∈ [ 0,1 ] , by using (I3) we obtain
| Q 2 y 1 ( t ) − Q 2 y 2 ( t ) | ≤ t α − 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 | f ( s , y 1 ( s ) , A y 1 ( s ) , B y 1 ( s ) ) − f ( s , y 2 ( s ) , A y 2 ( s ) , B y 2 ( s ) ) | d s + t α − 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 | f ( τ , y 1 ( τ ) , A y 1 ( τ ) , B y 1 ( τ ) ) − f ( τ , y 2 ( τ ) , A y 2 ( τ ) , B y 2 ( τ ) ) | d τ ) d H i (s)
≤ 1 Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 ( a ( s ) | y 1 ( s ) − y 2 ( s ) | + b ( s ) | A y 1 ( s ) − A y 2 ( s ) | + c ( s ) | B y 1 ( s ) − B y 2 ( s ) | ) d s + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 ( a ( τ ) | y 1 ( τ ) − y 2 ( τ ) | + b ( τ ) | A y 1 ( τ ) − A y 2 ( τ ) | + c ( τ ) | B y 1 ( τ ) − B y 2 ( τ ) | ) d τ ) d H i (s)
≤ ‖ y 1 − y 2 ‖ Δ Γ ( α − β 0 ) ∫ 0 1 ( 1 − s ) α − β 0 − 1 ( a ( s ) + k 0 b ( s ) + h 0 c ( s ) ) d s + ‖ y 1 − y 2 ‖ Δ ∑ i = 1 m 1 Γ ( α − β i ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − β i − 1 ( a ( τ ) + k 0 b ( τ ) + h 0 c ( τ ) ) d τ ) d H i ( s ) .
So, we deduce
| Q 2 y 1 ( t ) − Q 2 y 2 ( t ) | ≤ ‖ y 1 − y 2 ‖ Δ Γ ( α − β 0 ) ∫ 0 1 ( a ( s ) + k 0 b ( s ) + h 0 c ( s ) ) d s + ‖ y 1 − y 2 ‖ Δ ∑ i = 1 m 1 Γ ( α − β i ) ( ∫ 0 1 ( a ( τ ) + k 0 b ( τ ) + h 0 c ( τ ) ) d τ ) ( H i ( 1 ) − H i ( 0 ) ) = ‖ y 1 − y 2 ‖ ⋅ ‖ a + k 0 b + h 0 c ‖ L 1 ( 1 Δ Γ ( α − β 0 ) + 1 Δ ∑ i = 1 m 1 Γ ( α − β i ) ( H i ( 1 ) − H i ( 0 ) ) ) = ω 3 ‖ y 1 − y 2 ‖ ,
where ω 3 is given by (8). Because ω 3 < 1 , we conclude that Q 2 is a contraction mapping.
By using assumptions (I2)' and (I5), we deduce that Q 1 is a continuous mapping. In addition, Q 1 is uniformly bounded on B ¯ R , because for any x ∈ B ¯ R , we find
| Q 1 x ( t ) | = | 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s | ≤ 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 g ( s ) d s ≤ 1 Γ ( α ) ∫ 0 1 g ( s ) d s = 1 Γ ( α ) ‖ g ‖ L 1 , ∀ t ∈ [ 0,1 ] ,
and then ‖ Q 1 x ‖ ≤ 1 Γ ( α ) ‖ g ‖ L 1 for all x ∈ B ¯ R .
The operator Q 1 is also equicontinuous on B ¯ R . Indeed, let x ∈ B ¯ R , t 1 , t 2 ∈ [ 0,1 ] , with t 1 < t 2 . We have
| Q 1 x ( t 2 ) − Q 1 x ( t 1 ) | = | 1 Γ ( α ) ∫ 0 t 2 ( t 2 − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s − 1 Γ ( α ) ∫ 0 t 1 ( t 1 − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s | ≤ 1 Γ ( α ) ∫ 0 t 1 [ ( t 2 − s ) α − 1 − ( t 1 − s ) α − 1 ] f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s + 1 Γ ( α ) ∫ t 1 t 2 ( t 2 − s ) α − 1 f ( s , x ( s ) , A x ( s ) , B x ( s ) ) d s
≤ ϕ 0 { 1 Γ ( α ) ∫ 0 t 1 [ ( t 2 − s ) α − 1 − ( t 1 − s ) α − 1 ] d s + 1 Γ ( α ) ∫ t 1 t 2 ( t 2 − s ) α − 1 d s } = ϕ 0 Γ ( α + 1 ) ( t 2 α − t 1 α ) ≤ ϕ 0 ( t 2 − t 1 ) Γ ( α ) ,
where ϕ 0 = sup { f ( t , x , y , z ) , t ∈ [ 0,1 ] , | x | ≤ R , | y | ≤ k 0 R , | z | ≤ h 0 R } . Then we obtain that | Q 1 x ( t 2 ) − Q 1 x ( t 1 ) | → 0 as t 2 − t 1 → 0 .
By using the Arzela-Ascoli theorem, we deduce that Q 1 ( B ¯ R ) is relatively compact. By Theorem 2, we conclude that operator Q 1 + Q 2 has at least one fixed point, and so problem (E)-(BC) has at least one nonnegative solution.
Let α = 10 / 3 ( n = 4 ), β 0 = 11 / 5 , m = 2 , β 1 = 1 / 2 , β 2 = 5 / 4 , H 1 ( t ) = t 2 for all t ∈ [ 0,1 ] , H 2 ( t ) = { 0, for t ∈ [ 0, 1 / 2 ) ; 1, for t ∈ [ 1 / 2 ,1 ] } .
We consider the fractional differential equation
(E0) D 0 + 10 / 3 x ( t ) + f ( t , x ( t ) , A x ( t ) , B x ( t ) ) = 0 , t ∈ ( 0 , 1 ) ,
with the boundary conditions
(BC0) x ( 0 ) = x ′ ( 0 ) = x ″ ( 0 ) = 0 , D 0 + 11 / 5 x ( 1 ) = 2 ∫ 0 1 t D 0 + 1 / 2 x ( t ) d t + D 0 + 5 / 4 x ( 1 2 ) ,
where A x ( t ) = ∫ 0 t K ( t , s ) x ( s ) d s and B x ( t ) = ∫ 0 1 H ( t , s ) x ( s ) d s for all
t ∈ [ 0,1 ] , with K ( t , s ) = e − t s 2 for all t , s ∈ [ 0,1 ] with s ≤ t , and H ( t , s ) = e − 2 t ( s 3 + 1 ) for all t , s ∈ [ 0,1 ] . Then we obtain Δ ≈ 0.85599748 > 0 and σ ≈ 1.24445843 . So assumptions (I1) and (I5) are satisfied.
We define the function
f ( t , x , y , z ) = t + 1 2 + e − t x ( 1 + k ) ( 1 + x ) + e − 2 t y ( 1 + k 2 ) ( 1 + y ) + e − 3 t z ( 1 + k 3 ) ( 1 + z ) ,
for all t ∈ [ 0,1 ] and x , y , z ∈ ℝ + with k ≥ 1 . We deduce that k 0 = 1 3 e and h 0 = 5 4 . Besides we obtain the inequalities
| f ( t , x , y , z ) − f ( t , x ¯ , y ¯ , z ¯ ) | ≤ e − t 1 + k | x − x ¯ | + e − 2 t 1 + k 2 | y − y ¯ | + e − 3 t 1 + k 3 | z − z ¯ | ,
for all t ∈ [ 0,1 ] , x , y , z , x ¯ , y ¯ , z ¯ ∈ ℝ + , and
| f ( t , x , y , z ) | ≤ t + 1 2 + e − t 1 + k + e − 2 t 1 + k 2 + e − 3 t 1 + k 3 , ∀ t ∈ [ 0 , 1 ] , x , y , z ∈ ℝ + .
We define a ( t ) = e − t 1 + k , b ( t ) = e − 2 t 1 + k 2 , c ( t ) = e − 3 t 1 + k 3 , and g ( t ) = t + 1 2 + e − t 1 + k + e − 2 t 1 + k 2 + e − 3 t 1 + k 3 , for all t ∈ [ 0,1 ] . We have a , b , c ∈ L 1 ( 0,1 ) and g ∈ L 1 ( 0,1 ) . So assumptions (I2), (I3), (I4) are also satisfied.
In addition, we find ω 1 = e − 1 e ( 1 + k ) + e 2 − 1 6 e 3 ( 1 + k 2 ) + 5 ( e 3 − 1 ) 12 e 3 ( 1 + k 3 ) ≤ e − 1 2 e + e 2 − 1 12 e 3 + 5 ( e 3 − 1 ) 24 e 3 ≈ 0.540529 , and so ω 1 < 1 / σ ≈ 0.8035624 . Therefore, by Theorem 3, we conclude that problem (E0)-(BC0) has at least one nonnegative and nontrivial solution.
In this paper, we investigated the existence of nonnegative solutions for the Riemann-Liouville fractional differential equation with integral terms (E) supplemented with the boundary conditions (BC) which contain Riemann-Liouville fractional derivatives of different orders and Riemann-Stieltjes integrals, by using the Banach contraction mapping principle and the Krasnosel’skii fixed point theorem for the sum of two operators. For some future research directions, we have in mind the study of the existence, nonexistence and multiplicity of solutions or positive solutions for fractional differential equations subject to other boundary conditions.
The authors thank the referee for his/her valuable comments and suggestions.
The authors declare no conflicts of interest regarding the publication of this paper.
Luca, R. and Tudorache, A. (2019) Nonnegative Solutions for a Riemann-Liouville Fractional Boundary Value Problem. Open Journal of Applied Sciences, 9, 749-760. https://doi.org/10.4236/ojapps.2019.910061