Home Mathematics Analysis of solutions for the fractional differential equation with Hadamard-type
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Analysis of solutions for the fractional differential equation with Hadamard-type

  • Huijuan Zhu , Yuanfang Ru EMAIL logo and Fanglei Wang
Published/Copyright: October 20, 2023

Abstract

We mainly consider the existence and stability results of the positive solutions for the fractional differential equation with Hadamard-type by applying fixed point theorems, if the nonlinearity may be continuous or singular. We also construct some examples to show the applicability of the results.

MSC 2010: 34C25; 34B15

1 Introduction

During the past two decades, many real-world processes and phenomena in signal processing, fluid flow, electrical systems, control theory, etc., are characterized by fractional differential equations (FDEs), which are considered much better and improved over ones that are only dependent on integer-order operators (see [13]). Therefore, many researchers focused on the study of the FDEs using various methods, such as fixed points [413], variational methods [1416], and upper and lower solution methods [17,18]. Especially, in [1922], combining some fixed point theorems with the monotonic iteration technique, the authors studied the existence of solutions of fractional equations.

In 1892, the well-known Hadamard-type fractional derivative was defined in [23] with salient features, which contains a logarithmic function of any exponent in the integral kernel. Since then, it is important to study the property of the Hadamard-type fractional derivative and applications. For example, Ahmad and Ntouyas [24] applied the fixed point theorem to study the solutions of the FDEs:

D 1 + α z ( t ) = g ( z ( t ) ) , 1 < t < e , z ( 1 ) = 0 , z ( e ) = α z ( ρ ) ,

where D 1 + α denotes the Hadamard-type fractional derivative, 1 < γ < 2 , and ρ ( 1 , e ) . Butzer et al. [25,26] studied semigroups and bounded properties of the Hadamard-type fractional operators that are used in [27]. Also, many researchers applied the fixed point theorem with the monotonic iteration technique or numerical techniques to study the solutions of the Hadamard-type FDEs (see [2831]). In [32], the authors studied the general Hadamard-type singular FDEs. Using the double monotone iterative technique, the uniqueness of positive solutions is obtained. Especially, the authors [31] have studied the existence of positive solutions for a class of weakly singular Hadamard-type FDE with various kinds of boundary conditions.

Inspired by these references, we are interested in studying the problem:

(1.1) ( D 1 + α x ( t ) ) + λ 2 D 1 + α x ( t ) + f ( t , x ( t ) , D 1 + α x ( t ) ) = 0 , t ( 1 , e ) , x ( 1 ) = x ( 1 ) = D 1 + α + 1 x ( 1 ) = 0 , x ( e ) = D 1 + α + 1 x ( e ) = 0 ,

where 2 < α 3 and 0 < λ 1 e ( e 1 ) . When the function f may be continuous or singular, we will investigate the properties of positive solutions of this equation, such as existence, uniqueness, and stability. The discussions are based on Banach’s contraction mapping and some fixed point theorems. Compared to the results in the references, the main results obtained by us have some new features. First, the problem satisfies the Neumann boundary condition, which is obviously different to the periodic boundary value condition. Second, we give some sufficient conditions for the uniqueness result, which are based on the iterative method. Finally, we further study the stability.

The organization of this article is as follows. In Section 2, we present some definitions and preliminary lemmas. In Section 3, we study the regular or singular Problem (1.1) and give the proof of the related theorems. In Section 4, we further show the stability results. In Section 5, some examples are given to illustrate the main results.

2 Preliminaries

For convenience, the basic definition for the Hadamard-type fractional derivatives is given as follows.

Definition 2.1

[10] The Hadamard derivative of fractional order α > 0 for a function x : [ a , + ) R is defined by:

D a + α x ( t ) = 1 Γ ( n α ) t d d t n a t ln t s n α 1 x ( s ) d s s ,

where n = [ α ] + 1 , [ α ] stands for the largest integer, which is less than or equal to α .

Now, let D 1 + α x ( t ) = u ( t ) . Then, combining with the boundary condition, it follows from [31] that

x ( t ) = 1 e G 1 ( t , s ) u ( s ) d s s ,

where

G 1 ( t , s ) = 1 Γ ( α ) ( ln t ) α 1 ( 1 ln s ) α 2 ( ln t ln s ) α 1 , 1 s t e , ( ln t ) α 1 ( 1 ln s ) α 2 , 1 t s e .

Thus, Problem (1.1) can be reduced to the Neumann boundary value problem:

(2.1) u ( t ) + λ 2 u ( t ) = f t , 1 e G 1 ( t , s ) u ( s ) d s s , u ( t ) , t ( 1 , e ) , u ( 1 ) = u ( e ) = 0 .

Meanwhile, the solution of the nonhomogeneous problem

(2.2) u ( t ) + λ 2 u ( t ) = h ( t ) , u ( 1 ) = u ( e ) = 0 ,

can be expressed by u ( t ) = 1 e G 2 ( t , s ) h ( s ) d s , where

G 2 ( t , s ) = 1 λ sin λ ( e 1 ) cos λ ( 1 t ) cos λ ( s e ) + sin λ ( e 1 ) sin λ ( t s ) , 1 s t e , cos λ ( 1 t ) cos λ ( s e ) , 1 t s e .

Next, we give the positive estimates of G i ( t , s ) ( i = 1 , 2 ) .

Lemma 2.2

G i ( t , s ) ( i = 1 , 2 ) satisfy the properties:

  1. G i ( t , s ) 0 , i = 1 , 2 ;

  2. for all t , s [ 1 , e ] ,

    1. ( ln t ) α 1 G 1 ( e , s ) G 1 ( t , s ) ( ln t ) α 1 Γ ( α ) ;

    2. G 2 ( s , s ) G 2 ( t , s ) cos λ ( e t ) λ sin λ ( e 1 ) .

Proof

It is obvious that Property (1) is satisfied. Now, we assert Property (2).

(i) For 1 s t e ,

G 1 ( t , s ) = ( ln t ) α 1 Γ ( α ) ( 1 ln s ) α 2 1 ln s ln t α 1 ( ln t ) α 1 Γ ( α ) ln s ( 1 ln s ) α 2 = ( ln t ) α 1 G 1 ( e , s ) .

Similarly, we have G 1 ( t , s ) ( ln t ) α 1 G 1 ( e , s ) , with 1 t s e for t , s [ 1 , e ] .

Moreover, for 1 s t e ,

G 1 ( t , s ) 1 Γ ( α ) ( ln t ) α 1 ( 1 ln s ) α 2 ( ln t ) α 1 Γ ( α ) .

Hence, for 1 t s e , G 1 ( t , s ) ( ln t ) α 1 Γ ( α ) is easily visible. Then, (i) is demonstrated.

(ii) Since G 2 ( t , s ) is continuous, taking the derivative of G 2 ( t , s ) with respect to t , we obtain the following:

For 1 s t e ,

G 2 ( t , s ) t = 1 sin λ ( e 1 ) [ cos λ ( e s ) sin λ ( 1 t ) + sin λ ( e 1 ) cos λ ( t s ) ]

and

2 G 2 ( t , s ) t 2 = λ sin λ ( e 1 ) [ cos λ ( e s ) cos λ ( 1 t ) sin λ ( e 1 ) sin λ ( t s ) ] 0 ,

so G 2 ( t , s ) t G 2 ( e , s ) t = sin λ ( s 1 ) sin λ ( e 1 ) 0 .

For 1 t s e ,

G 2 ( t , s ) t = 1 sin λ ( e 1 ) cos λ ( e s ) sin λ ( 1 t ) 0 .

Therefore, G 2 ( t , s ) is decreasing with respect to t on [ 1 , s ] and increasing with respect to t on [ s , e ] , which means that G 2 ( t , s ) G 2 ( s , s ) for t , s [ 1 , e ] .

Furthermore, for 1 t s e , from 0 < λ 1 e ( e 1 ) < π 2 ( e 1 ) , we have 0 < cos λ ( s e ) 1 and

G 2 ( t , s ) = 1 λ sin λ ( e 1 ) cos λ ( 1 t ) cos λ ( s e ) cos λ ( 1 t ) λ sin λ ( e 1 ) .

For 1 s t e , taking the derivative of G 2 ( t , s ) with respect to s , using the similar discussion earlier, G 2 ( t , s ) s is monotonically decreasing with respect to s on [ 1 , t ] . Since G 2 ( t , 1 ) s = 0 , G 2 ( t , s ) is decreasing with respect to s on [ 1 , t ] , namely,

G 2 ( t , s ) G 2 ( t , 1 ) = cos λ ( e t ) λ sin λ ( e 1 ) .

Hence,

G 2 ( s , s ) G 2 ( t , s ) cos λ ( e t ) λ sin λ ( e 1 ) ,

which completes the proof.□

3 Existence

For convenience, in this part, let E = C [ 1 , e ] be a Banach space equipped with the norm x max t [ 1 , e ] x ( t ) . Let B r = { u E : u < r } , where r > sup t [ 1 , e ] f ( t , 0 , 0 ) Γ ( α ) ( e 1 ) λ Γ ( α ) sin λ ( e 1 ) ( 1 + 2 Γ ( α ) ) ( e 1 ) .

Theorem 3.1

Let f : [ 1 , e ] × R × R R + be continuous. If the following assumption holds:

(H1) there exist 1 , 2 > 0 such that

f ( t , x , u ) f ( t , y , v ) 1 x y + 2 u v , x , y , u , v E , and t [ 1 , e ] .

Then, (1.1) has a unique solution if 1 Γ ( α ) + 2 ( e 1 ) λ sin λ ( e 1 ) < 1 .

Proof

Define the operator T by:

T u ( t ) = 1 e G 2 ( t , s ) f s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ , u ( s ) d s .

Then, for any u B r , by virtue of Lemma 2.2 and (H1), we have

T u ( t ) < 1 e G 2 ( t , s ) f s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ , u ( s ) f ( s , 0 , 0 ) + sup t [ 1 , e ] f ( t , 0 , 0 ) d s < 1 Γ ( α ) + 2 u + sup t [ 1 , e ] f ( t , 0 , 0 ) 1 e G 2 ( t , s ) d s < 1 Γ ( α ) + 2 u + sup t [ 1 , e ] f ( t , 0 , 0 ) e 1 λ sin λ ( e 1 ) < r ,

which follows that T ( B r ) B r .

In addition, for any u , v E , we obtain

T u ( t ) T v ( t ) < 1 e G 2 ( t , s ) 1 1 e G 1 ( s , τ ) u ( τ ) v ( τ ) d τ τ + 2 u ( s ) v ( s ) d s < 1 Γ ( α ) + 2 u v 1 e G 2 ( t , s ) d s < 1 Γ ( α ) + 2 e 1 λ sin λ ( e 1 ) u v ,

namely, T is contractional.

At last, we conclude that T has a unique fixed point by Banach’s contraction mapping principle, which means that (2.1) has a unique solution.□

Theorem 3.2

Let f : [ 1 , e ] × R + × R + R + be continuous, and the following assumption holds:

(H2) There exist a constant γ ( 0 , 1 ) and two functions h 1 , h 2 P with h 1 , h 2 0 on any subinterval of ( 1 , e ) , such that

h 1 ( t ) ( u + v ) γ f ( t , u , v ) h 2 ( t ) ( u + v ) γ , ( u , v ) [ 0 , ) × [ 0 , ) , and t [ 1 , e ] .

Then, Problem (1.1) has at least one positive solution.

Proof

Let

P = { u C [ 1 , e ] : u ( t ) 0 }

and

P 0 = { u P : l u ( t ) L } .

Obviously, P 0 is a subset of P and P E . Now, we show that T has a fixed point in P 0 .

By Lemma 2.2, for any u P 0 , we obtain

(3.1) l ( ln s ) α 1 α ( α 1 ) Γ ( α ) 1 e G 1 ( s , τ ) u ( τ ) d τ τ L ( ln s ) α 1 Γ ( α ) L Γ ( α ) .

From (3.1) and (H2), we have

h 1 ( s ) l γ f s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ , u ( s ) h 2 ( s ) L γ 1 + 1 Γ ( α ) γ .

Let

M 1 = 1 e G 2 ( s , s ) h 1 ( s ) d s , M 2 = ( 1 + Γ ( α ) ) γ ( Γ ( α ) ) γ λ sin λ ( e 1 ) 1 e h 2 ( s ) d s .

Then, we have

T u ( t ) l γ 1 e G 2 ( s , s ) h 1 ( s ) d s = M 1 l γ

and

T u ( t ) L γ 1 + 1 Γ ( α ) γ λ sin λ ( e 1 ) 1 e h 2 ( s ) d s = M 2 L γ .

Take l = 1 L . Since M i > 0 and 0 < γ < 1 , actually there exists a L > 1 such that M 1 L 1 γ 1 and M 2 L γ 1 1 , which implies that T : P 0 P 0 . Thus, T has a fixed point u P 0 by Schauder’s fixed point theorem. Therefore, Theorem 3.2 is proved.□

If f ( t , u , v ) is nonnegative and continuous on [ 1 , e ] × ( 0 , ) 2 and may be singular at u = 0 , v = 0 . Now, we study the problem:

(3.2) u ( t ) + λ 2 u ( t ) = f t , 1 e G 1 ( t , s ) u ( s ) d s s + 1 n , u ( t ) + 1 n , u ( 1 ) = u ( e ) = 0 ,

where n N + with n 2 . Similar to the previous discussion, u is a solution of equation of (3.2) if and if only u satisfies the following equation:

u ( t ) = 1 e G 2 ( t , s ) f s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ + 1 n , u ( s ) + 1 n d s .

Theorem 3.3

Suppose that the following conditions hold:

(H3) f ( t , u , v ) = φ ( t , u , v ) + ψ ( t , u , v ) , where φ : [ 1 , e ] × [ 0 , + ) × [ 0 , + ) [ 0 , + ) , ψ : [ 1 , e ] × ( 0 , + ) × ( 0 , + ) [ 0 , + ) are continuous, φ ( t , u , v ) is nondecreasing, and ψ ( t , u , v ) is nonincreasing in u , v > 0 .

(H4) For any u , v > 0 , c ( 0 , 1 ) , there exists 0 < σ < 1 2 such that

c σ φ ( t , u , v ) φ ( t , c u , c v ) , c σ ψ ( t , u , v ) ψ ( t , c 1 u , c 1 v ) .

Then, there exists a unique positive solution x * for Problem (1.1) satisfying

M cos λ ( e t ) x * ( t ) 1 M cos λ ( e t ) ,

where 0 < M < 1 is a constant.

Furthermore, for any given initial u 0 , v 0 P 0 ˜ , construct two sequences as follows:

u n = 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) u n 1 ( τ ) d τ τ , u n 1 ( s ) + ψ s , 1 e G 1 ( s , τ ) v n 1 ( τ ) d τ τ , v n 1 ( s ) d s , n N + , v n = 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) v n 1 ( τ ) d τ τ , v n 1 ( s ) + ψ s , 1 e G 1 ( s , τ ) u n 1 ( τ ) d τ τ , u n 1 ( s ) d s , n N + ,

which converge uniformly to D 1 + α x * ( t ) on [ 1 , e ] as n , i.e.,

u n D 1 + α x * a n d v n D 1 + α x * as n .

Remark 3.4

By (H4), for c 1 and u , v > 0 , one has

c σ φ ( t , u , v ) φ ( t , c u , c v ) , c σ ψ ( t , u , v ) ψ ( t , c 1 u , c 1 v ) .

We first give the following necessary definition and lemma, before proving Theorem 3.3.

Definition 3.5

[22] Assume P to be a normal cone of a Banach space E , T : P × P P is said to be mixed monotone if T ( y , z ) is non-decreasing in y and non-increasing in z , i.e., y 1 y 2 ( y 1 , y 2 P ) implies T ( y 1 , z ) T ( y 2 , z ) for any z P , and z 1 z 2 ( z 1 , z 2 P ) implies T ( y , z 1 ) T ( y , z 2 ) for any y P . The element y * P is called a fixed point of T if T ( y * , y * ) = y * .

Lemma 3.6

[33] Let P be a normal, solid cone of Banach space E and T : P 0 ˜ P 0 ˜ be a mixed monotone operator. Assume that there exists 0 < σ < 1 such that

(3.3) T ( c y , c 1 z ) c σ T ( y , z ) , y , z P 0 ˜ , 0 < c < 1 .

Then, the operator T has a unique fixed point y * P 0 ˜ . Moreover, for any initial y 0 , z 0 P 0 ˜ , by constructing successively the sequences

y n = T ( y n 1 , z n 1 ) , z n = T ( y n 1 , z n 1 ) , n = 1 , 2 , ,

we have y n y * 0 , z n z * 0 as n + .

Proof of Theorem 3.3

Let P 0 ˜ be a normal and solid cone in E , which is defined by:

P 0 ˜ = u E : M cos λ ( e t ) u ( t ) 1 M cos λ ( e t ) ,

where M ( 0 , 1 ) with M < min { 1 , L 1 , L 2 } ,

L 1 = 1 e ( [ a ( cos λ ( e s ) + 1 ) ] σ M σ φ ( s , 1 , 1 ) + ( M b ) σ ( ln s ) 2 σ ψ ( s , 1 , 1 ) ) d s λ sin λ ( e 1 ) 1 , L 2 = 1 e M σ [ a ( cos λ ( e s ) + 1 ) ] σ G 2 ( s , s ) ψ ( s , 1 , 1 ) d s ,

in which

a = max N 2 Γ ( α ) , 1 , b = min N 1 Γ ( α ) , 1

and

N 1 = 1 e ln τ ( 1 ln τ ) α 2 cos λ ( e τ ) d τ τ , N 2 = 1 e cos λ ( e τ ) d τ τ .

Now, we define an operator T by:

T ( u , v ) ( t ) = 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ + 1 n , u ( s ) + 1 n + ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n d s .

First, for u P 0 ˜ , we have

(3.4) M N 1 ( ln s ) α 1 Γ ( α ) 1 e G 1 ( s , τ ) u ( τ ) d τ τ N 2 M Γ ( α ) .

Let f ( s ) = ln s cos λ ( e s ) , s [ 1 , e ] , and then,

f ( s ) = 1 s λ sin λ ( e s ) = 1 s g ( s ) ,

which g ( s ) = 1 λ s sin λ ( e s ) . Since 0 < λ 1 e ( e 1 ) , we have g ( s ) 0 , which implies that f ( s ) 0 . Therefore, f ( s ) is increasing on [ 1 , e ] , which follows that f ( s ) f ( e ) = 0 . This means that ln s cos λ ( e s ) , i.e., ( ln s ) 2 cos 2 λ ( e s ) for s [ 1 , e ] .

For any u , v P 0 ˜ , from (H3)–(H4) and Remark 3.4, it follows that

φ s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ + 1 n , u ( s ) + 1 n a σ M σ φ ( s , ( ln s ) α 1 + 1 , cos λ ( e s ) + 1 ) [ a ( cos λ ( e s ) + 1 ) ] σ M σ φ ( s , 1 , 1 )

and

ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n ψ s , M N 1 ( ln s ) α 1 Γ ( α ) + 1 n , M cos λ ( e s ) + 1 n M σ ψ s , N 1 ( ln s ) α 1 Γ ( α ) , cos λ ( e s ) ( M b ) σ ψ ( s , ( ln s ) α 1 , cos λ ( e s ) ) ( M b ) σ ψ ( s , ( ln s ) 2 , cos 2 λ ( e s ) ) ( M b ) σ ( ln s ) 2 σ ψ ( s , 1 , 1 ) .

Consequently, applying Lemma 2.2, we have

T ( u , v ) = 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ + 1 n , u ( s ) + 1 n + ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n d s cos λ ( e t ) λ sin λ ( e 1 ) 1 e ( [ a ( cos λ ( e s ) + 1 ) ] σ M σ φ ( s , 1 , 1 ) + ( M b ) σ ( ln s ) 2 σ ψ ( s , 1 , 1 ) ) d s 1 M cos λ ( e t ) .

Moreover,

ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n ψ s , N 2 ( ln s ) α 1 M Γ ( α ) + 1 n , 1 M cos λ ( e s ) + 1 n M σ a σ ψ ( s , ( ln s ) α 1 + 1 , cos λ ( e s ) + 1 ) M σ a σ ψ ( s , ln s + 1 , cos λ ( e s ) + 1 ) M σ [ a ( cos λ ( e s ) + 1 ) ] σ ψ ( s , 1 , 1 ) .

Meanwhile, Lemma 2.2 yields that

T ( u , v ) = 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ , u ( s ) + ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ , v ( s ) d s 1 e G 2 ( t , s ) ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ , v ( s ) d s cos λ ( e t ) 1 e M σ [ a ( cos λ ( e s ) + 1 ) ] σ G 2 ( s , s ) ψ ( s , 1 , 1 ) d s M cos λ ( e t ) .

From the aforementioned discussions, it concludes that T : P 0 ˜ × P 0 ˜ P 0 ˜ .

Next, for any u 1 , u 2 P 0 ˜ with u 1 u 2 , from the monotonicity of 1 e G 1 ( s , τ ) u ( τ ) d τ τ and φ , we have

T ( u 1 , v ) = 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) u 1 ( τ ) d τ τ + 1 n , u 1 ( s ) + 1 n + ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n d s 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) u 2 ( τ ) d τ τ + 1 n , u 2 ( s ) + 1 n + ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n d s = T ( u 2 , v ) ,

i.e., T ( u , v ) is non-decreasing with respect to u , for any v P 0 ˜ . In an analogous manner, if v 1 , v 2 P 0 ˜ with v 1 v 2 , we obtain

T ( u , v 1 ) ( t ) T ( u , v 2 ) ( t ) , u P 0 ˜ .

Hence, T is a mixed monotone operator on P 0 ˜ × P 0 ˜ P 0 ˜ .

Now, for every u , v P 0 ˜ , t [ 1 , e ] and c ( 0 , 1 ) , by (H4), we have

T ( c u , c 1 v ) ( t ) 1 e G 2 ( t , s ) c σ φ s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ + 1 n , u ( s ) + 1 n + ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n d s = c σ 1 e G 2 ( t , s ) φ s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ + 1 n , u ( s ) + 1 n + ψ s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ + 1 n , v ( s ) + 1 n d s = c σ T ( u , v ) ( t ) ,

which shows that (3.3) of Lemma 3.6 holds.

By virtue of Lemma 3.6, one can show that Problem (2.1) has a unique positive solution u * satisfying T ( u * , u * ) = u * . Hence, Problem (1.1) has a unique positive solution x * P 0 ˜ , where x * ( t ) = 1 e G 1 ( t , s ) u ( s ) d s s and

M cos λ ( e t ) x * ( t ) 1 M cos λ ( e t ) .

Finally, for any given u 0 , v 0 P 0 ˜ , construct two iterative sequences { u n } and { v n } by:

u n = T ( u n 1 , v n 1 ) , v n = T ( v n 1 , u n 1 ) , n = 1 , 2 , ,

which converges uniformly to u * on [ 1 , e ] , as n , i.e.,

u n D 1 + α x * ( t ) , v n D 1 + α x * ( t ) as n .

Therefore, Theorem 3.3 is proved.□

4 Stability

Theorem 4.1

Suppose that (H1) holds and f is nonnegative and continuous on [ 1 , e ] × R × R , then Problem (1.1) is Ulam-Hyers-Rassias stable if 1 Γ ( α ) + 2 e 1 λ sin λ ( e 1 ) < 1 .

Proof

For Ψ ( t ) C [ 1 , e ] , let y C [ 1 , e ] , which satisfies the following inequality:

( D 1 + α y ( t ) ) + λ 2 D 1 + α y ( t ) + f ( t , y ( t ) , D 1 + α y ( t ) ) ε Ψ ( t ) .

Let D 1 + α y ( t ) = v ( t ) , then we obtain that, for ε > 0 ,

v ( t ) 1 e G 2 ( t , s ) f s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ , v ( s ) d s ε Ψ ( t ) , t [ 1 , e ] .

Since Problem (1.1) can be equivalently rewritten to Problem (2.1), it is obvious that we only need to discuss the stability of Problem (2.1).

By Theorem 3.1, the solution u ( t ) of Problem (2.1) can be rewritten as:

u ( t ) = 1 e G 2 ( t , s ) f s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ , u ( s ) d s .

Then, for t [ 1 , e ] , according to Lemma 2.2, we have

u ( t ) v ( t ) v ( t ) 1 e G 2 ( t , s ) f s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ , v ( s ) d s + 1 e G 2 ( t , s ) f s , 1 e G 1 ( s , τ ) u ( τ ) d τ τ , u ( s ) f s , 1 e G 1 ( s , τ ) v ( τ ) d τ τ , v ( s ) d s ε Ψ ( t ) + 1 Γ ( α ) + 2 u v 1 e G 2 ( t , s ) d s < ε Ψ ( t ) + 1 Γ ( α ) + 2 e 1 λ sin λ ( e 1 ) u v ,

which yields

v ( t ) u ( t ) ε 1 1 Γ ( α ) + 2 e 1 λ sin λ ( e 1 ) Ψ ( t ) = Λ ε Ψ ( t ) , t [ 1 , e ] .

Consequently, the solution of Problem (2.1) is Ulam-Hyers-Rassias stable with the Ψ ( t ) .□

5 Examples

Example 5.1

Considering the following equation:

D 1 + 5 2 y ( t ) π 2 4 e 2 D 1 + 5 2 y ( t ) = t y + D 5 2 y ( t ) 1 2 , t [ 1 , e ] , y ( 1 ) = y ( 1 ) = y ( e ) = 0 , D 1 + 7 2 y ( 1 ) = D 1 + 7 2 y ( e ) = 0 .

Denote f ( t , y ( t ) , D α y ( t ) ) = t y + D 5 2 y ( t ) 1 2 and γ = 1 2 . Taking h 1 ( t ) 1 , h 2 ( t ) e , then

( u + v ) 1 2 f ( t , u , v ) e ( u + v ) 1 2 ,

which means (H2) holds. Let α = 5 2 and λ = π 2 e , then

G 1 ( t , s ) = 1 Γ 5 2 ( ln t ) 3 2 ( 1 ln s ) 1 2 ( ln t ln s ) 3 2 , 1 s t e , ( ln t ) 3 2 ( 1 ln s ) 1 2 , 1 t s e ,

and

G 2 ( t , s ) = 2 e π sin π 2 e ( e 1 ) cos π 2 e ( 1 t ) cos π 2 e ( s e ) + sin π 2 e ( e 1 ) sin π 2 e ( t s ) , 1 s t e , cos π 2 e ( 1 t ) cos π 2 e ( s e ) , 1 t s e .

Moreover,

M 1 = 2 e π sin π 2 e ( e 1 ) 1 e cos π 2 e ( 1 s ) cos π 2 e ( s e ) d s 2.46679963

and

M 2 = 1 + 1 Γ 5 2 1 2 e ( e 1 ) π 2 e sin π 2 e ( e 1 ) 1 10.5179295 .

Let L = 38,000, then M 1 L 1 2 > 1 and M 2 L 1 2 > 1 hold. Therefore, this equation satisfies the conditions of Theorem 3.2; then, the equation has at least one positive solution.

Example 5.2

Considering the equation:

D 1 + 5 2 y ( t ) π 2 4 e 2 D 1 + 5 2 y ( t ) = t y 1 5 ( t ) D 1 + 5 2 y ( t ) 1 4 + t 1 2 y 1 8 ( t ) D 1 + 5 2 y ( t ) 1 6 , t [ 1 , e ] , y ( 1 ) = y ( 1 ) = y ( e ) = 0 , D 1 + 7 2 y ( 1 ) = D 1 + 7 2 y ( e ) = 0 .

Let φ ( t , u , v ) = t ( u 1 5 + v 1 4 ) , ψ ( t , u , v ) = t 1 2 ( u 1 8 + v 1 6 ) and σ = 1 4 , then

φ ( t , c u , c v ) = t ( c 1 5 u 1 5 + c 1 4 v 1 4 ) c 1 4 t ( u 1 5 + v 1 4 ) = c 1 4 φ ( t , u , v )

and

ψ ( t , c 1 u , c 1 v ) = t 1 2 c 1 8 u 1 8 + c 1 6 v 1 6 c 1 4 t 1 2 u 1 8 + v 1 6 = c 1 4 ψ ( t , u , v ) ,

for every u , v > 0 and c ( 0 , 1 ) . Obviously, ψ : [ 1 , e ] × ( 0 , + ) 2 [ 0 , + ) and φ : [ 1 , e ] × [ 0 , + ) 2 [ 0 , + ) are continuous. Moreover, φ ( t , u , v ) is non-decreasing and ψ ( t , u , v ) is nonincreasing in u , v > 0 . Then, from Theorem 3.3, the equation has a unique positive solution x * satisfying

M cos π 2 e ( e t ) x * ( t ) 1 M cos π 2 e ( e t ) ,

where 0 < M < 1 .

6 Conclusion

In this article, we first consider the existence and stability of positive solutions to a class of FDEs with Hadamard-type under continuous nonlinearities using the immovable point theorem. Then, we consider the existence of positive solutions to this model under singular nonlinearities using the mixed immovable point theorem. Finally, we construct some examples to illustrate the applicability of the results.

Acknowledgement

The authors are grateful to the anonymous referee whose careful reading of this manuscript and valuable comments enhanced the presentation of this manuscript.

  1. Funding information: The work was supported by the Scientific Research Plan of Jiangsu Drug Administration (No. 202102) and the Fundamental Research Funds for the Central Universities (2013/B220202082).

  2. Author contributions: H. Zhu was mainly responsible for the calculation of this article, F. Wang was responsible for the typesetting of this article, and Y. Ru was responsible for the revisions and touch-ups.

  3. Conflict of interest: The authors declare that they have no conflict of interest.

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Received: 2023-02-20
Revised: 2023-09-19
Accepted: 2023-09-20
Published Online: 2023-10-20

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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