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Article

A Generalized Lyapunov Inequality for a Pantograph Boundary Value Problem Involving a Variable Order Hadamard Fractional Derivative

by
John R. Graef
1,*,
Kadda Maazouz
2 and
Moussa Daif Allah Zaak
2
1
Department of Mathematics, University of Tennessee at Chattanooga, Chattanooga, TN 37401, USA
2
Department of Mathematics, University of Ibn Khaldoun, Tiaret P.O. Box 78, Algeria
*
Author to whom correspondence should be addressed.
Submission received: 8 June 2023 / Revised: 28 June 2023 / Accepted: 30 June 2023 / Published: 4 July 2023
(This article belongs to the Special Issue Fractional Differential Equations and Control Problems)

Abstract

:
The authors obtain existence and uniqueness results for a nonlinear fractional pantograph boundary value problem containing a variable order Hadamard fractional derivative. This type of model is appropriate for applications involving processes that occur in strongly anomalous media. They also derive a generalized Lyapunov-type inequality for the problem considered. Their results are obtained by the fractional calculus and Krasnosel’skii’s fixed point theorem. An example is given to illustrate their approach.

1. Introduction

Pantograph equations are usually referred to as differential equations having proportional delays, and have became a primary example for delay differential equations over the last several decades. They arise in many different areas of pure and applied mathematics including number theory, dynamical systems, probability, quantum mechanics, and electro dynamics. They have been well studied by many researchers (for more details see [1,2,3,4,5]). In [6], Ockendon and Taylor studied how an electric current is collected by the pantograph of an electric locomotive; this is in fact how the problem got its name.
Fractional calculus has undergone a remarkable evolution in recent decades (see [7,8,9,10,11]) thanks to its usefulness in modeling complex real-life phenomena. This paved the way for the definition of variable order differential and integral operators such as those of the Grunwald-Letnikov, Erdlyi-Kober, Riesz, Riemann-Liouville, Caputo, Hadamard, and Hilfer types [12,13]. Their extensive applications has led to considerable systematic studies of the existence and uniqueness of solutions to problems involving these operators.
In [14], Harikrishnan et al. discussed the nonlocal initial value problems for pantograph equations with a ψ -Hilfer fractional derivative of the form
D a + α , β ; ψ φ ( t ) = g ( t , φ ( t ) , φ ( λ t ) ) , 0 < λ < 1 , t J : = ( a , T ] , I a + 1 γ ; ψ φ ( t ) t = a = k = 1 n C k φ ( t k ) , t k ( a , T ] ,
where D a + α , β ; ψ is the ψ -Hilfer fractional derivative of order α , 0 < α < 1 , and type β , 0 β 1 , with γ = α + β α β . Here, E is a Banach space, g : J × E × E R is a given continuous function, t k are fixed points satisfying a < t 1 t k < T , and C k , k = 1 , 2 , , n are real numbers.
Lyapunov inequalities are important in many different applications of mathematics. The result, as proven by Lyapunov [15], states that if f : [ a , b ] R is a continuous function, then the boundary value problem
φ ( t ) + f ( t ) φ ( t ) = 0 , t ( a , b ) , φ ( a ) = φ ( b ) = 0 ,
has a non-trivial solution if
a b | f ( t ) | d t > 4 b a , a < b < + .
Lyapunov’s inequality has had applications even in fractional calculus, and takes different forms depending on the type of fractional derivative involved [16]. For example, if we have
D a + ψ φ ( t ) + f ( t ) φ ( t ) = 0 , t ( a , b ) , φ ( a ) = φ ( b ) = 0 ,
where D a + ψ is either the Riemann-Liouville or Caputo fractional derivative of order ψ ( 1 , 2 ] and f : [ a , b ] R is a continuous function, then the Lyapunov inequality (1) takes the fractional form
a b | f ( t ) | d t > Γ ( ψ ) 4 b a ψ 1 , a < b < + .
Recently, in [17], the authors obtained a generalized Lyapunov type inequality for the Hadamard fractional boundary value problem
H D 1 + ψ φ ( t ) + f ( t ) φ ( t ) = 0 , t ( 1 , e ) , φ ( 1 ) = φ ( e ) = 0 ,
where H D a + ψ is the Hadamard fractional derivative of order ψ ( 1 , 2 ] and f : [ 1 , e ] R is again a real continuous function. Under some further assumptions on the nonlinear term f, condition (1) becomes
1 e | f ( t ) | d t > Γ ( ψ ) μ 1 ψ ( 1 μ ) 1 ψ exp ( μ ) ,
where μ = 2 ψ 1 ( 2 ψ 2 ) 2 + 1 2 .
Inspired by the work mentioned above, here we study the existence and uniqueness of a solution to the nonlinear pantograph boundary value problem containing a Hadamard fractional derivative of variable order
H D 1 + ψ ( t ) φ ( t ) = g ( t , φ ( t ) , φ ( λ t ) ) , 1 t T < + , φ ( 1 ) = φ ( T ) = 0 ,
where 1 < ψ ( t ) < 2 , 0 < λ < 1 , g : [ 1 , T ] × R × R R is a continuous function, and H D 1 + ψ ( t ) is the left hand variable order Hadamard fractional derivative.
Under additional assumptions on the nonlinear term, we will obtain a generalized Lyapunov inequality for this problem.
This paper is organized as follows. In Section 2, we present some definitions and necessary lemmas associated with the boundary value problem. In Section 3, we establish the existence and uniqueness of solutions for the problem (2) by using the Krasnosel’skii fixed point theorem. In Section 4, we discuss generalized Lyapunov-type inequalities for (2), and we end the paper with an example to illustrate our results.

2. Preliminaries

In this section we introduce some fundamental definitions and lemmas that will be essential in obtaining our results in the following sections.
Definition 1
([18]). Let 1 a < b < + and ψ : [ a , b ] ( 0 , + ) . The left hand Hadamard fractional integral of variable order ψ ( t ) for a function φ is given by
H I a + ψ ( t ) φ ( t ) = 1 Γ ( ψ ( t ) ) a t ln t s ψ ( t ) 1 φ ( s ) s d s , t > a .
Definition 2
([18]). Let n N and ψ : [ a , b ] ( n 1 , n ) . The left hand Hadamard derivative of variable order ψ ( t ) for a function φ is given by
H D a + ψ ( t ) φ ( t ) = t n Γ ( n ψ ( t ) ) d n d t n a t ln t s n 1 ψ ( t ) φ ( s ) s d s , t > a .
It is clear that when the order ψ ( t ) is just a constant ψ , then the variable order Hadamard fractional operators coincide with its constant order counterparts, and so the semi-group property holds:
H I 1 + ψ 1 H I 1 + ψ 2 = H I 1 + ψ 2 H I 1 + ψ 1 = H I 1 + ψ 1 + ψ 2 .
With these properties, a fractional order differential equation can be transformed into an equivalent integral equation, thus paving the way for the application of fixed point theorems to prove the existence and uniqueness of the solutions. However, such properties do not hold for variable order fractional operators (see the example below) making it difficult to transform such a fractional differential equation into an equivalent integral equation.
Example 1.
To show that in general
H I 1 + ψ 1 ( t ) H I 0 + ψ 2 ( t ) φ ( t ) H I 1 + ψ 1 ( t ) + ψ 2 ( t ) φ ( t ) ,
let
ψ 1 ( t ) = t + 1 , t [ 1 , 3 ] , 1 t , t ( 3 , 4 ] , ψ 2 ( t ) = 1 , t [ 1 , 3 ] , 2 , t ( 3 , 4 ] , a n d φ ( t ) = 1 f o r 1 t 4 .
Then,
H I 1 + ψ 1 ( t ) H I 1 + ψ 2 ( t ) φ ( t ) = 1 Γ ( ψ 1 ( t ) ) 1 t 1 s ln t s ψ 1 ( t ) 1 1 Γ ( ψ 2 ( s ) ) 1 s 1 h ln s h ψ 2 ( s ) 1 φ ( h ) d h d s = 1 Γ ( ψ 1 ( t ) ) 1 3 1 s ln t s ψ 1 ( t ) 1 1 Γ ( 1 ) 1 s 1 h ln s h 1 1 d h d s + 1 Γ ( ψ 1 ( t ) ) 3 t 1 s ln t s ψ 1 ( t ) 1 1 Γ ( 2 ) 1 s 1 h ln s h 2 1 d h d s .
Therefore,
H I 1 + ψ 1 ( t ) H I 1 + ψ 2 ( t ) φ ( t ) | t = 3 = 1 Γ ( 4 ) 1 3 ln 3 s 3 ln s d s t = 3 0.06069 .
On the other hand,
H I 1 + ψ 1 ( t ) + ψ 2 ( t ) φ ( t ) t = 3 = 1 Γ ( ψ ( t ) + ψ 2 ( t ) ) 1 t 1 s ln t s ψ 1 ( t ) + ψ 2 ( t ) 1 d s t = 3 = 1 Γ ( 4 + 1 ) 1 t 1 s ln 3 s 4 + 1 1 d s = 1 Γ ( 6 ) ( ln 3 ) 5 0.01333 .
In a similar way, we can show that
H D 1 + ψ 1 ( t ) H D 1 + ψ 2 ( t ) φ ( t ) H D 1 + ψ 2 ( t ) H D 1 + ψ 1 ( t ) φ ( t ) H D 1 + ψ 1 ( t ) + ψ 2 ( t ) φ ( t )
by taking 0 < ψ 1 ( t ) < 1 and 0 < ψ 2 ( t ) < 1 with ψ 1 ( t ) ψ 2 ( t ) , and letting φ ( t ) = 1 for t [ 1 , T ] .
We now let E be a Banach space and Ω be a subset of E . By C ( Ω , E ) , we mean the Banach space of continuous functions φ : Ω E , with the usual supremum norm
| | φ | | = sup { | φ ( t ) | , t Ω } .
Definition 3
([19]). Let S be a subset of R .
(i
S is called a generalized interval if it is either a standard interval, a point, or ∅.
(ii
If S is a generalized interval, a finite set P consisting of generalized intervals contained in S is called a partition of S, provided that every x S belongs to exactly one of the generalized intervals in the finite set P .
(iii
The function ψ : t R is a piecewise constant with respect to the partition P of S, if for any W S , ψ is constant on W.
The following two propositions will be needed in the proof of our existence results.
Proposition 1
([20]). Let ψ C ( [ 1 , T ] , ( 1 , 2 ] ) . If φ C α ( [ 1 , T ] , R ) = { φ ( t ) C ( [ 1 , T ] , R ) : ( ln t ) α φ ( t ) C ( [ 1 , T ] , R ) , 0 < α < 1 } , then the left hand Hadamard variable order integral H I 1 + ψ ( t ) φ ( t ) exists for each t [ 1 , T ] .
Proposition 2
([20]). Let ψ C ( [ 1 , T ] , ( 1 , 2 ] ) . Then H I 1 + ψ ( t ) φ ( t ) C ( [ 1 , T ] , R ) for every φ C ( [ 1 , T ] , R ) .
Lemma 1
([21]). Let α > 0 , 1 < a < b , φ L 1 ( a , b ) , and H D a + α φ L 1 ( a , b ) . Then, the differential equation
H D a + α φ = 0
has a solution
φ ( t ) = C 1 ln t a α 1 + C 2 ln t a α 2 + + C n ln t a α n = i = 1 n C i ln t a α i ,
where n = [ α ] + 1 and the C i R , i = 1 , 2 , , n are arbitrary constants. Moreover,
H I a + α ( H D a + α ) φ ( t ) = φ ( t ) + i = 1 n C i ln t a α i
and
H D a + α ( H I a + α ) φ ( t ) = φ ( t ) .
Theorem 1
([22]). Let Ω be a non-empty bounded closed convex subset of a real Banach space E , and let F 1 and F 2 be operators on Ω satisfying the following conditions:
( a )
F 1 ( Ω ) + F 2 ( Ω ) Ω ;
( b )
F 1 is continuous on Ω and F 1 ( Ω ) is a relatively compact subset of E;
( c )
F 2 is a strict contraction on Ω, i.e., there exists k [ 0 , 1 ) such that
| | F 2 ( x ) F 2 ( y ) | | k | | x y | | , for every x , y Ω .
Then, the equation  F 1 ( x ) + F 2 ( x ) = x  has a solution in Ω.

3. Existence of Solution

In this section, we present our main results.
Let P = [ 1 , t 1 ] , ( t 1 , t 2 ] , ( t 2 , t 3 ] , , ( t n , T ] be a partition of the interval [ 1 , T ] , and let ψ ( t ) : [ 1 , T ] ( 1 , 2 ) be the piecewise constant function with respect to P given by
ψ ( t ) = i = 1 n ψ i I i ( t ) , t [ 1 , T ] ,
where 1 < ψ i < 2 , i = 1 , 2 , , n , are constants, and I i is the characteristic function for the interval [ t i 1 , t i ] , i = 1 , 2 , , n , i.e.,
I i ( t ) = 1 , t [ t i 1 , t i ] , 0 , otherwise .
Hence, from Definition 2
H D 1 + ψ ( t ) φ ( t ) = t 2 Γ 2 i = 1 n ψ i I i ( t ) d 2 d t 2 1 t ln t s 1 i = 1 n ψ i I i ( t ) φ ( s ) s d s = t 2 Γ ( 2 ψ ( t ) ) d 2 d t 2 1 t 1 ln t s 1 ψ 1 φ ( s ) s d s + t 2 Γ ( 2 ψ ( t ) ) d 2 d t 2 t 1 t 2 ln t s 1 ψ 2 φ ( s ) s d s + + t 2 Γ ( 2 ψ ( t ) ) d 2 d t 2 t n t ln t s 1 ψ n φ ( s ) s d s = t 2 Γ ( 2 ψ ( t ) ) d 2 d t 2 i = 1 n t i 1 t i ln t s 1 ψ i φ ( s ) s d s + t n t ln t s 1 ψ n φ ( s ) s d s = t 2 Γ ( 2 ψ ( t ) ) i = 1 n d 2 d t 2 t i 1 t i ln t s 1 ψ i φ ( s ) s d s + d 2 d t 2 t n t ln t s 1 ψ n φ ( s ) s d s .
Thus, the equation in (2) can be written as
H D 1 + ψ ( t ) φ ( t ) = t 2 Γ ( 2 ψ ( t ) ) i = 1 n d 2 d t 2 t i 1 t i ln t s 1 ψ i φ ( s ) s d s + d 2 d t 2 t n t ln t s 1 ψ n φ ( s ) s d s = g ( t , φ ( t ) , φ ( λ t ) ) .
Therefore, in the interval [ 1 , t 1 ] , (3) can be written as
H D 1 + ψ 1 φ ^ ( t ) = t 2 Γ ( 2 ψ 1 ) d 2 d t 2 1 t ln t s 1 ψ 1 φ ^ ( s ) s d s ,
in the interval ( t 1 , t 2 ] , it can be written as
H D 1 + ψ 2 φ ^ ( t ) = t 2 Γ ( 2 ψ 1 ) d 2 d t 2 1 t 1 ln t s 1 ψ 1 φ ^ ( t ) s d s + d 2 d t 2 t 1 t ln t s 1 ψ 2 φ ^ ( t ) s d s ,
and in general, in the interval ( t i 1 , t i ] , it can be written as
H D 1 + ψ i φ ^ ( t ) = t 2 Γ ( 2 ψ i ) k = 1 i 1 d 2 d t 2 1 t k ln t s 1 ψ k φ ^ ( t ) s d s + d 2 d t 2 t i 1 t ln t s 1 ψ i φ ^ ( s ) s d s .
We denote by E i = C ( [ λ , t i ] , R ) the class of functions that form a Banach space with the norm
| | φ | | E i = sup t [ λ , t i ] | φ ( t ) | , i { 1 , 2 , , n }
Let φ ^ E i with φ ^ ( t ) = 0 for all t [ λ , t i 1 ] , i { 2 , , n } be the solutions to the above equations for any i 1 , , n , and consider the auxiliary boundary value problems for Hadamard fractional equations of constant order
H D t i 1 + ψ i φ ^ ( t ) = t 2 Γ ( 2 ψ i ) d 2 d t 2 t i 1 t ln t s 1 ψ i φ ^ ( s ) s d s = g ( t , φ ^ ( t ) , φ ^ ( λ t ) ) , ( t , λ t ) [ t i 1 , t i ] , φ ^ ( t i 1 ) = φ ^ ( t i ) = 0 .
Definition 4.
We say that the problem (2) has a solution φ, if there exist functions φ i , such that φ 1 E 1 satisfies Equation (4) and φ 1 ( 1 ) = φ 1 ( t 1 ) = 0 ; φ 2 E 2 satisfies Equation (5) and φ 2 ( t 1 ) = φ 2 ( t 2 ) = 0 ; φ i E i satisfies equation (6) and φ i ( t i 1 ) = φ i ( t i ) = 0 for i = 3 , , n .
Remark 1.
We say that problem (2) has a unique solution if the functions φ i are unique.
Based on the previous discussion, we have the following results.
Lemma 2.
Let i { 1 , , n } . Then, the function φ ^ is a solution of (7) if and only if φ ^ is a solution of the integral equation
φ ^ ( t ) = ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 H I t i 1 + ψ i g ( t i , φ ^ ( t i ) , φ ^ ( λ t i ) ) + H I t i 1 + ψ i g ( t , φ ^ ( t ) , φ ^ ( λ t ) )
for t ( t i 1 , t i ] for each i = 1 , 2 , , n .
Proof. 
Assume φ ^ satisfies (7); then, we transform (7) into an equivalent integral equation as follows. Let t i 1 < t t i ; then Lemma 1 implies
H I t i 1 + ψ i H D t i 1 + ψ i φ ^ ( t ) = H I t i 1 + ψ i g ( t , φ ^ ( t ) , φ ^ ( λ t ) )
so
φ ^ ( t ) = k = 1 2 C k ln t t i 1 ψ i k + H I t i 1 + ψ g ( t , φ ^ ( t ) , φ ^ ( λ t ) ) .
Using the boundary conditions φ ^ ( t i ) = φ ^ ( t i 1 ) = 0 , we obtain
C 1 = ln t i t i 1 1 ψ i C 2 ln t t i 1 ψ i 2 + H I t i 1 + ψ i g ( t i , φ ^ ( t i ) , φ ^ ( λ t i ) ) 0 = C 2 ln t i t i 1 ψ i 2 H I t i 1 + ψ i g ( t i , φ ^ ( t i ) , φ ^ ( λ t i ) ) .
Hence,
C 2 = 0 , C 1 = ln t i t i 1 1 ψ i H I t i 1 + ψ i g ( t i , φ ^ ( t i ) , φ ^ ( λ t i ) ) .
Therefore, the solution of the auxiliary boundary value problem (7) is given by
φ ^ ( t ) = ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 H I t i 1 + ψ i g ( t i , φ ^ ( t i ) , φ ^ ( λ t i ) ) + H I t i 1 + ψ i g ( t , φ ^ ( t ) , φ ^ ( λ t ) ) , t i 1 < t t i .
A straight forward calculation shows that if φ ^ is given by (8), then it is a solution of (7). □
Before presenting our main results, we first state the following hypotheses that will be needed:
Hypothesis 1 (H1).
Let g : [ 1 , T ] × R × R R be a continuous function such that for all x 1 , x 2 , y 1 , y 2 R , there exists a positive constant L > 0 so that
| g ( t , x 1 , y 1 ) g ( t , x 2 , y 2 ) | L ( | x 1 x 2 | + | y 1 y 2 | ) ;
Hypothesis 2 (H2).
2 L Γ ( ψ i ) ln t i t i 1 ψ i 1 < 1 for each i = 1 , , n .
Theorem 2.
Under conditions (H1) and (H2), the boundary value problem (2) has a unique solution in C ( [ 1 , T ] , R ) .
Proof. 
Consider the operator F : E i E i given by
( F φ ^ ) ( t ) = ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 H I t i 1 + ψ i g ( t i , φ ^ ( t i ) , φ ^ ( λ t i ) ) + H I t i 1 + ψ i g ( t , φ ^ ( t ) , φ ^ ( λ t ) ) .
Clearly, F is well defined. For each i = 1 , , n , let B R i = φ E i : | | φ | | E i R i be a non-empty, closed, bounded, convex subset of E i , where
R i 2 Γ ( ψ i + 1 ) sup t [ 1 , T ] | g ( t , 0 , 0 ) | ln t i t i 1 ψ i 1 4 L Γ ( ψ i ) ln t i t i 1 ψ i 1 .
We split the mapping F into two maps F 1 and F 2 on B R i , as follows:
( F 1 φ ) ( t ) = ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 H I t i 1 + ψ i g ( t i , φ ( t i ) , φ ( λ t i ) ) , ( F 2 φ ) ( t ) = H I t i 1 + ψ i g ( t , φ ( t ) , φ ( λ t ) ) .
We will now show that the conditions of Theorem 1 are satisfied.
Step 1:
F 1 ( B R i ) + F 2 ( B R i ) B R i . For each i = 1 , , n , we have
| F 1 ( φ ) ( t ) + F ( φ ) ( t ) | = | ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 H I t i 1 + ψ i g ( t i , φ ( t i ) , φ ( λ t i ) ) + H I t i 1 + ψ i g ( t , φ ( t ) , φ ( λ t ) ) | H I t i 1 + ψ i g ( t i , φ ( t i ) , φ ( λ t i ) ) + H I t i 1 + ψ i g ( t , φ ( t ) , φ ( λ t ) ) H I t i 1 + ψ i g ( t i , φ ( t i ) , φ ( λ t i ) ) + H I t i 1 + ψ i g ( t , φ ( t ) , φ ( λ t ) ) 1 Γ ( ψ i ) t i 1 t i ln t i s ψ i 1 | g ( s , φ ( s ) , φ ( λ s ) ) | s d s + 1 Γ ( ψ i ) t i 1 t ln t s ψ i 1 | g ( s , φ ( s ) , φ ( λ s ) ) | s d s 2 Γ ( ψ i ) t i 1 t i ln t i s ψ i 1 | g ( s , φ ( s ) , φ ( λ s ) ) g ( s , 0 , 0 ) | s d s + 2 Γ ( ψ i ) t i 1 t i ln t i s ψ i 1 | g ( s , 0 , 0 ) | s d s 2 Γ ( ψ i ) ln t i t i 1 ψ i 1 t i 1 t i ( L ( | φ ( s ) | + | φ ( λ s ) | ) d s + 2 Γ ( ψ i ) sup t [ 1 , T ] | g ( t , 0 , 0 ) | ) t i 1 t i ln t i s ψ i 1 d s 4 L | | φ | | E i Γ ( ψ i ) ln t i t i 1 ψ i 1 + 2 Γ ( ψ i + 1 ) sup t [ 1 , T ] | g ( t , 0 , 0 ) | ln t i t i 1 ψ i 4 L | | φ | | E i Γ ( ψ i ) ln t i t i 1 ψ i 1 + 2 Γ ( ψ i + 1 ) sup t [ 1 , T ] | g ( t , 0 , 0 ) | ln t i t i 1 ψ i R i ,
which is what we wanted to show.
Step 2:
F 1 is a contraction. For each i = 1 , , n , we have
| F 1 ( φ 1 ) ( t ) F 1 ( φ 2 ) ( t ) | ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 Γ ( ψ i ) × t i 1 t i ln t i s ψ i 1 | g ( s , φ 1 ( s ) , φ 1 ( λ s ) ) g ( s , φ 2 ( s ) , φ 2 ( λ s ) ) | s d s 1 Γ ( ψ i ) ln t i t i 1 ψ i 1 t i 1 t i L ( | φ 1 ( s ) φ 2 ( s ) | + | φ 1 ( λ s ) φ 2 ( λ s ) | ) d s 2 L Γ ( ψ i ) ln t i t i 1 ψ i 1 | | φ 1 ( t ) φ 2 ( t ) | | E i .
Hence, F 1 is a contraction by condition (H2).
Step 3:
F 2 is continuous and F 2 ( B R i ) is relatively compact. To show that F 2 is continuous, let { φ n } be a sequence such that φ n φ in B R i . Then, for t ( t i , t i + 1 ] , i = 1 , , n ,
| F 2 ( φ n ) ( t ) F 2 ( φ ) ( t ) | 1 Γ ( ψ i ) t i 1 t 1 s ln t s ψ i 1 | g ( s , φ n ( s ) , φ n ( λ s ) ) g ( s , φ ( s ) , φ ( λ s ) ) | d s 1 Γ ( ψ i ) t i 1 t i 1 s ln t i s ψ i 1 | g ( s , φ n ( s ) , φ n ( λ s ) ) g ( s , φ ( s ) , φ ( λ s ) ) | d s L Γ ( ψ i ) ln t i t i 1 ψ i 1 t i 1 t i ( | φ n ( s ) φ ( s ) | + | φ n ( λ s ) φ ( λ s ) | ) d s 2 L Γ ( ψ i ) ln t i t i 1 ψ i 1 | | φ n φ | | E i .
From the Lebesgue Dominated Convergence Theorem, we have
| | F 2 ( φ n ) ( t ) F 2 ( φ ) ( t ) | | E i 0 as n .
In order to show that F 2 is equicontinuous, we need to show that for every ϵ > 0 there exists α > 0 such that for every φ B R i and t 1 , t 2 Ω , | t 2 t 1 | < α implies | | F 2 ( φ ) ( t 2 ) F 2 ( φ ) ( t 1 ) | | E i < ϵ . Now we have
| F 2 ( φ ) ( t 2 ) F 2 ( φ ) ( t 1 ) | = 1 γ ( ψ i ) t i 1 t 2 1 s ln t 2 s ψ i 1 g ( s , φ ( s ) , φ ( λ s ) ) d s 1 Γ ( ψ i ) t i 1 t 1 1 s ln t 1 s ψ i 1 g ( s , φ ( s ) , φ ( λ s ) ) d s = 1 Γ ( ψ i ) t i 1 t 1 1 s ln t 2 s ψ i 1 ln t 1 s ψ i 1 g ( s , φ ( s , φ ( λ s ) ) ) d s + 1 Γ ( ψ i ) t 1 t 2 1 s ln t 2 s ψ i 1 g ( s , φ ( s , φ ( λ s ) ) ) d s .
As t 1 t 2 , the right-hand side of the above inequality tends to zero, so the mapping F 2 is equicontinuous. Since it is uniformly bounded by Step 1, by the Ascoli-Arzelà Theorem, F 2 is relatively compact on B R i .
It then follows from Theorem 1 that the auxiliary boundary value problem (7) has at least one solution in B R i for each i { 1 , 2 , , n } . Hence, the boundary value problem (HFPE) has a solution in C ( [ λ , T ] , R ) , given by
φ ( t ) = φ 1 ( t ) = 0 , t [ λ , 1 ] φ ^ 1 ( t ) , t [ 1 , t 1 ] φ 2 ( t ) = 0 , t [ λ , t 1 ] φ ^ 2 ( t ) , t [ t 1 , t 2 ] φ n ( t ) = 0 , t [ λ , t n 1 ] φ ^ n ( t ) , t [ t n 1 , T ] .
The uniqueness of the solution obtained above is easy to show by using Gronwall’s inequality, as follows. Let i { 1 , 2 , , n } and let φ i and φ i * be two solutions of (7); then,
| φ i ( t ) φ i * ( t ) | = H I t i 1 + ψ i ( t ) g ( t i , φ ( t i ) , φ ( λ t i ) ) + H I t i 1 + ψ i ( t ) g ( t , φ ( t ) , φ ( λ t ) ) + H I t i 1 + ψ i ( t ) g ( t i , φ * ( t i ) , φ * ( λ t i ) ) H I t i 1 + ψ i ( t ) g ( t i , φ * ( t ) , φ * ( λ t ) ) 2 Γ ( ψ i ) t i 1 t i 1 s ln t i s ψ i 1 | g ( s , φ i ( s ) , φ i ( λ s ) ) g ( s , φ i * ( s ) , φ i * ( λ s ) ) | d s 2 Γ ( ψ i ) ln t i t i 1 ψ i 1 t i 1 t i L ( | φ i ( s ) φ i * ( s ) | + | φ i ( λ s ) φ i * ( λ s ) | ) d s 4 L Γ ( ψ i ) ln t i t i 1 ψ i 1 t i 1 t i | φ ( s ) φ * ( s ) | d s .
from which, uniqueness of φ i ( t ) follows. In view of Remark 1, we have uniqueness of solutions to (2). □

4. Generalized Lyapunov-Type Inequalities

In this section, we discuss generalized Lyapunov-type inequalities for the boundary value problem (2).
Proposition 3.
The Green function for the auxiliary boundary value problem (7) for each i { 1 , 2 , , n } is given by
Φ i ( s , t ) = Φ 1 , i ( s , t ) , t i 1 s t t i , Φ 2 , i ( s , t ) , t i 1 t s t i ,
where
Φ 1 , i ( s , t ) = 1 s Γ ( ψ i ) ln t s ψ i 1 ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 ln t i s ψ i 1
and
Φ 2 , i ( s , t ) = 1 s Γ ( ψ i ) ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 ln t i s ψ i 1 .
Proof. 
From the proof of Theorem 2, we have
φ i ^ ( t ) = 1 Γ ( ψ i ) ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 t i 1 t i 1 s ln t i s ψ i 1 | g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) | d s + 1 Γ ( ψ i ) t i 1 t 1 s ln t s ψ i 1 | g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) | d s = 1 Γ ( ψ i ) ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 t i 1 t 1 s ln t i s ψ i 1 | g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) | d s + t t i 1 s ln t i s ψ i 1 | g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) | d s + 1 Γ ( ψ i ) t i 1 t 1 s ln t s ψ i 1 | g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) | d s = 1 Γ ( ψ i ) t i 1 t 1 s ln t s ψ i 1 ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 ln t i s ψ i 1 | g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) | d s 1 Γ ( ψ i ) ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 t t i 1 s ln t i s ψ i 1 | g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) | d s = t i 1 t i Φ i ( s , t ) g ( s , φ i ^ ( s ) , φ i ^ ( λ s ) ) d s ,
which proves the proposition. □
Lemma 3.
Let the Green function Φ i be defined as in Proposition 3. Then, for i { 1 , 2 , , n } ,
max t [ t i 1 , t i ] | Φ i ( s , t ) | 1 Γ ( ψ i ) ln t i t i 1 1 ψ i μ i ln t i 1 ln t i μ i ψ i 1 exp ( μ i ) ,
where μ i = 2 ψ i 2 + ln t i t i 1 ( 2 ψ i 2 + ln t i t i 1 ) 2 4 ( ψ i 1 ) ln t i t i 1 + ln t i ln t i 1 2 .
Proof. 
It is easy to see that Φ i ( s , t ) 0 for all t i 1 s , t t i . Thus, Φ 1 , i ( t i , t i ) = 0 and Φ 2 , i ( t i 1 , s ) = 0 , which are the maximum values of Φ 1 , i and Φ 2 , i , respectively.
For t s , we see that
Φ 2 , i ( s , t ) = 1 s Γ ( ψ i ) ln t i t i 1 1 ψ i ln t t i 1 ψ i 1 ln t i s ψ i 1 1 s Γ ( ψ i ) ln t i t i 1 1 ψ i ln s t i 1 ψ i 1 ln t i s ψ i 1 1 s Γ ( ψ i ) ln t i t i 1 1 ψ i ln t i ln s ln 2 s ln t i 1 ln t i + ln t i 1 ln s ψ i 1 1 s Γ ( ψ i ) ln t i t i 1 1 ψ i ( ln t i t i 1 ) ln s ln 2 s ln t i 1 ln t i ψ i 1 .
It follows that we need to determine where the maximum value of the function
y ( s ) = 1 s ( ln t i t t 1 ) ln s ln 2 s ln t i 1 ln t i ψ i 1 .
occurs. Now,
y ( s ) = 1 s 2 ( ln t i t t 1 ) ln s ln 2 s ln t i 1 ln t i ψ i 1 + 1 s 2 ( ln t i t i 1 ) 2 ln s ( ψ i 1 ) ( ln t i t t 1 ) ln s ln 2 s ln t i 1 ln t i ψ i 2 = 1 s 2 ( ln t i t t 1 ) ln s ln 2 s ln t i 1 ln t i ψ i 2 × ( ψ i 1 ) ( ln t i t i 1 2 ln s ) ( ln t i t i 1 ) ln s ln 2 s ln t i ln t i 1 = 1 s 2 ( ln t i t t 1 ) ln s ln 2 s ln t i 1 ln t i ψ i 2 × ( 2 2 ψ i ln t i t i 1 ) ln s + ln 2 s + ( ψ i 1 ) ln t i t i 1 + ln t i ln t i 1 ,
and y ( s ) = 0 if and only if
ln 2 s ( 2 ψ i 2 + ln t i t i 1 ) ln s + ( ψ i 1 ) ln t i t i 1 + ln t i ln t i 1 = 0 ,
so
ln s = 2 ψ i 2 + ln t i t i 1 ± ( 2 ψ i 2 + ln t i t i 1 ) 2 4 ( ψ i 1 ) ln t i t i 1 + ln t i ln t i 1 2 .
However, since ψ i 1 , we have
ln s = 2 ψ i 2 + ln t i t i 1 + ( 2 ψ i 2 + ln t i t i 1 ) 2 4 ( ψ i 1 ) ln t i t i 1 + ln t i ln t i 1 2 ln t i t i 1 + ( ln t i t i 1 ) 2 4 ln t i ln t i 1 2 ln t i + ln t i 1 + ln t i ln t i 1 2 = ln t i ,
which is a contradiction, since s t i . Therefore, min s [ t , t i ] Φ 2 , i ( s , t ) = Φ 2 , i ( s * , s * ) , where
s * = exp 2 ψ i 2 + ln t i t i 1 ( 2 ψ i 2 + ln t i t i 1 ) 2 4 ( ψ i 1 ) ln t i t i 1 + ln t i ln t i 1 2 .
For the function Φ 1 , i , we already know that Φ 1 , i 0 , and that the maximum value of Φ 1 , i is 0 for s = t = t i . To obtain the minimum value of Φ 1 , i , for fixed s, a computation gives
t Φ 1 , i ( s , t ) = ( ψ i 1 ) s t Γ ( ψ i ) ln t s ψ i 2 ln t i t i 1 1 ψ i ln t t i 1 ψ i 2 ln t i s ψ i 1 .
Observing that ln t s ln t t i 1 and ψ i 2 0 , we see that
ln t s ψ i 2 ln t t i 1 ψ i 2 ln t i t i 1 1 ψ i ln t t i 1 ψ i 2 ln t i s ψ i 1 .
So, t Φ 1 , i ( s , t ) is increasing with respect to t, which means
min t [ s , t i ] Φ 1 , i ( s , t ) = Φ 1 , i ( s , s ) = 1 s Γ ( ψ i ) ln t i t i 1 1 ψ i ln s t i 1 ψ i 1 ln t i s ψ i 1 = min s [ t , t i ] Φ 2 , i ( s , t ) = Φ 2 , i ( s * , s * ) = 1 Γ ( ψ i ) ln t i t i 1 1 ψ i ( μ i ln t i 1 ) ( ln t i μ i ) ψ i 1 exp ( μ i ) .
This implies that for i { 1 , 2 , , n } ,
max t [ t i 1 , t i ] | Φ i ( s , t ) | 1 Γ ( ψ i ) ln t i t i 1 1 ψ i μ i ln t i 1 ln t i μ i ψ i 1 exp ( μ i ) ,
and completes the proof of the lemma. □
We are now ready to present our Lyapunov inequality for our problem.
Theorem 3.
Assume there exists a non-negative continuous function h : [ 1 , T ] R such that
| g ( t , φ ( t ) , φ ( λ t ) ) | h ( t ) | φ ( t ) | + | φ ( λ t ) | , 1 t T .
If the boundary value problem (2) has a non-trivial solution φ, then
1 T h ( s ) d s > i = 1 n Γ ( ψ i ) 2 ln t i t i 1 ψ i 1 ( μ i ln t i 1 ) ( ln t i μ i ) 1 ψ i exp ( μ i ) ,
where μ i , i { 1 , 2 , , n } are given in Lemma 3.
Proof. 
Let φ be a non-trivial solution of (2). We know that
φ ( t ) = φ 1 ( t ) = φ 1 ^ ( t ) , t [ 1 , t 1 ] , φ 2 ( t ) = 0 , t [ 1 , t 1 ] , φ 2 ^ ( t ) , t ( t 1 , t 2 ] , φ n ( t ) = 0 , t [ 1 , t n 1 ] , φ n ^ ( t ) , t ( t n 1 , T ] .
Therefore,
| | φ 1 | | C ( [ 1 , t 1 ] , R ) = max t [ 1 , t 1 ] | φ 1 ( t ) | max t [ 1 , t 1 ] 1 t 1 Φ 1 ( t , s ) | g ( s , φ 1 ( s ) , φ 1 ( λ s ) ) | d s 1 t 1 | Φ 1 ( t , s ) [ h ( s ) | φ 1 ( s ) | + φ ( λ s ) | ] d s < 2 | | φ 1 | | C ( [ 1 , t 1 ] , R ) Γ ( ψ 1 ) ln t 1 1 ψ 1 μ 1 ln t 1 μ 1 ψ 1 1 exp ( μ 1 ) 1 t 1 h ( s ) d s ,
which implies that
1 t 1 h ( s ) d s > Γ ( ψ 1 ) 2 ln t 1 ψ 1 1 μ 1 ln t 1 μ 1 1 ψ 1 exp ( μ 1 ) .
Similarly, for i { 2 , 3 , , n } , we have
| | φ i | | C ( [ 1 , t i ] , R ) = max t [ t i 1 , t i ] | φ i ^ ( t ) | max t [ t i 1 , t i ] t i 1 t i Φ i ( t , s ) | g ( s , φ i ^ ( s ) φ i ^ ( λ s ) ) | d s t i 1 t i | Φ i ( t , s ) | h ( s ) | φ i ^ ( s ) + φ i ^ ( λ s ) | d s < 2 | | φ i | | C ( [ 1 , t i ] , R ) Γ ( ψ i ) ln t i t i 1 1 ψ i μ i ln t i 1 ln t i μ i ψ i 1 exp ( μ i ) t i 1 t i h ( s ) d s ,
so
t i 1 t i h ( s ) d s > Γ ( ψ i ) 2 ln t i t i 1 ψ i 1 μ i ln t i 1 ln t i μ i 1 ψ i exp ( μ i ) .
Hence,
1 T h ( s ) d s > i = 1 n Γ ( ψ i ) 2 ln t i t i 1 ψ i 1 ( μ i ln t i 1 ) ( ln t i μ i ) 1 ψ i exp ( μ i ) ,
which proves the theorem. □

5. Example

In this section, we illustrate the usefulness of the results obtained in this paper. Consider the boundary value problem
H D 1 + ψ ( t ) φ ( t ) = 1 t 2 + 4 φ ( t ) + φ ( λ t ) , 1 t e , φ ( 1 ) = φ ( e ) = 0 .
where α = 1 9 and t 0 = 1 , t 1 = 2 , t 2 = e , so that our partition of [ 1 , e ] becomes [ 1 , 2 ] , ( 2 , e ] . Also, we take
ψ ( t ) = 12 10 , if t 1 , 2 , 17 10 , if t 2 , e ,
and see that g ( t , x , y ) = 1 t 2 + 4 [ x + y ] . Since
| g ( t , φ 1 ( t ) , φ 1 ( λ t ) ) g ( t , φ 2 ( t ) , φ 2 ( λ t ) ) | = 1 t 2 + 4 φ 1 ( t ) + φ 1 ( λ t ) 1 t 2 + 4 φ 2 ( t ) + φ 2 ( λ t ) 1 t 2 + 4 | φ 1 ( t ) φ 2 ( t ) | + | φ 1 ( λ t ) φ 2 ( λ t ) | 1 5 | φ 1 ( t ) φ 2 ( t ) | + | φ 1 ( λ t ) φ 2 ( λ t ) | ,
so condition (H1) is satisfied with L = 1 5 .
Consider the auxiliary boundary value problems corresponding to (7)
H D 1 + 12 10 φ 1 ( t ) = 1 t 2 + 4 φ 1 ( t ) + φ 1 ( λ t ) , 1 t 2 , φ ( 1 ) = 0 , φ ( 2 ) = 0 ,
and
H D 1 + 17 10 φ 2 ( t ) = 1 t 2 + 4 φ 2 ( t ) + φ 2 ( λ t ) , 2 < t e , φ ( 2 ) = 0 , φ ( e ) = 0 .
Now,
4 L Γ ( ψ i ) ln t i t i 1 ψ i 1 = 4 5 Γ ( 12 2 ) ln 2 1 12 10 1 0.80968 < 1 , 4 L Γ ( ψ i ) ln t i t i 1 ψ i 1 = 4 5 Γ ( 17 10 ) ln e 2 17 10 1 0.38512 < 1 ,
so (H2) is satisfied. Therefore, by Theorem 2, the problem (13) has a solution given by
φ ( t ) = φ 1 ( t ) = φ 1 ^ ( t ) , t [ 1 , 2 ] , φ 2 ( t ) = 0 , t [ 1 , 2 ] , φ 2 ^ ( t ) , t ( 2 , e ] .
If we take h ( t ) = 1 t on [ 1 , e ] , then condition (11) holds for h ( t ) . We need to check conditions (10); we have
μ 1 = 4 10 + ln 2 ( 4 10 + ln 2 ) 2 4 ( 2 10 ) ln 2 2 0.14639 , μ 2 = 14 10 + ln 2 e ( 14 10 + ln 2 e ) 2 4 ( 7 10 ) ln 2 e + ln 2 2 0.82991 ,
Thus,
1 e 1 s d s = 1 > i = 1 2 Γ ( ψ i ) 2 ln t i t i 1 ψ i 1 ( μ i ln t i 1 ) ( ln t i μ i ) 1 ψ i exp ( μ i ) 0.30254 + 0.45641 0.75895 .
Therefore, the Lyapunov inequality is established, i.e., (12) holds.

Author Contributions

Conceptualization, K.M. and M.D.A.Z.; Methodology, J.R.G.; Formal analysis, J.R.G., KM. and M.D.A.Z.; Investigation, J.R.G., K.M., M.D.A.Z.; Writing—original draft, J.R.G. and K.M.; Writing—review & editing, J.R.G., K.M. and M.D.A.Z. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Graef, J.R.; Maazouz, K.; Zaak, M.D.A. A Generalized Lyapunov Inequality for a Pantograph Boundary Value Problem Involving a Variable Order Hadamard Fractional Derivative. Mathematics 2023, 11, 2984. https://0-doi-org.brum.beds.ac.uk/10.3390/math11132984

AMA Style

Graef JR, Maazouz K, Zaak MDA. A Generalized Lyapunov Inequality for a Pantograph Boundary Value Problem Involving a Variable Order Hadamard Fractional Derivative. Mathematics. 2023; 11(13):2984. https://0-doi-org.brum.beds.ac.uk/10.3390/math11132984

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Graef, John R., Kadda Maazouz, and Moussa Daif Allah Zaak. 2023. "A Generalized Lyapunov Inequality for a Pantograph Boundary Value Problem Involving a Variable Order Hadamard Fractional Derivative" Mathematics 11, no. 13: 2984. https://0-doi-org.brum.beds.ac.uk/10.3390/math11132984

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