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Article

A Study of Positive Solutions for Semilinear Fractional Measure Driven Functional Differential Equations in Banach Spaces

1
College of Science, Gansu Agricultural University, Lanzhou 730070, China
2
College of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070, China
*
Author to whom correspondence should be addressed.
Mathematics 2024, 12(17), 2696; https://doi.org/10.3390/math12172696
Submission received: 19 May 2024 / Revised: 24 August 2024 / Accepted: 25 August 2024 / Published: 29 August 2024

Abstract

:
In this paper, we deal with the delayed measure differential equations with nonlocal conditions via measure of noncompactness in ordered Banach spaces. Combining ( β , γ k ) -resolvent family, regulated functions and fixed point theorem with respect to convex-power condensing operator and measure of noncompactness, we investigate the existence of positive mild solutions for the mentioned system under the situation that the nonlinear function satisfies measure conditions and order conditions. In addition, we provide an example to verify the rationality of our conclusion.

1. Introduction

Measure-driven differential equations, also known as measure differential equations, have been widely used in non-smooth mechanics, game theory, and many other fields, etc., (see [1,2,3] and their references). The theory of Measurement Differential Equations (MDE) encompasses some well-known situations. When absolutely continuous function, a step function, and the sum of an absolutely continuous function with a step function are given, these systems correspond to typical ordinary differential equations, impulse differential equations, respectively. In addition, the advantage of MDE is that we can model Zeno trajectories, since gases as bounded variation functions may exhibit infinite discontinuities over finite intervals. This class of system appears in many fields of applied mathematics, such as nonsmooth mechanics, game theory, etc., see [3,4,5]. An early investigation of MDE was conducted by [6,7,8,9,10,11]. Recently, the MDE theory in the R n space was developed, we refer to papers [12,13,14,15].
This important property makes it possible for measure differential equations to model some non-classical problems, such as quantum effects and Zeno trajectories (see [16,17]. Measure differential equations were initially studied in [7,9,18]. For a complete introduction to measure differential systems, we refer the reader to [11]. Recently, Tvrdý [10] introduced the regulated function and Kurzweil–Stieltjes integral to study the regulated solutions for general measure linear systems. The papers [12,13,19] studied nonlinear measure functional differential equations. The paper [20] discusses the solutions of nonlinear measurement driven systems using Hausdorff non-compact measures. Other than that, the authors in [21] studied the existence of mild solutions for semilinear measure driven systems.
In the past decade, there are only a few scholars that have focused on the existence of positive solutions to fractional differential equations, we refer the reader to [22,23,24,25,26]. In [23], Bai and Lü, used some fixed point theorems on cones to study the existence of positive solutions for fractional-order ordinary differential equations. In [24], by using the contraction mapping principle of regular cones, the existence and uniqueness of positive mild solutions for neutral fractional evolution equations were obtained. In addition, fractional delayed evolution equations have been extensively studied and some interesting results have been obtained, we refer to [22,27,28] and their references. However, there is little research on the positive solutions of fractional measure evolution equations.
In our work, based on some ideas and methods of previous works [12,26,29,30,31], we will survey the existence of positive mild solution of measure differential equations
D t 1 + β c u ( t ) + k = 1 n α k D t γ k c u ( t ) = A u ( t ) + F ( t , u ( t ) , u t ) d h ( t ) , t [ 0 , b ] , u ( t ) = ρ ( t ) , t [ a , 0 ] , u ( 0 ) + m ( u ) = ϕ ,
where u ( · ) take values in a Banach space E; J = [ 0 , b ] with b > 0 , D t η c stand for the Caputo fractional derivative, α k > 0 and all γ k , k = 1 , 2 , , n , n N , such that 0 < β γ n γ 1 1 , A : D ( A ) E E is a generator of a bounded and strongly continuous cosine family, nonlinear function F : J × K × K B K is Carathéodory continuous and h : J R is nondeacreasing and continuous from the left, d h is the distributional derivative of h (see [15,20]), the function m : G ( J , E ) E , where G ( J , E ) denotes the space of regulated functions on J, B : = G ( [ a , 0 ] , E ) . For t 0 , u t B with u t ( s ) = u ( t + s ) for s [ a , 0 ] , here u : [ a , b ] E is a regulated function, ρ B and ρ ( 0 ) D ( A ) , ϕ E , a > 0 is a constant.
In [20], the author studied the measure driven equations of integer order, they obtained the existence of regulated solutions for measure integral equations driven by a nondecreasing function, bu means of the generalization of Darbo’s fixed point theorem. But in this article, we investigate the existence positive mild solutions for nonlinear fractional evolution measure driven equations in ordered Banach spaces, by using an extension of Sadoveskii’s fixed point theorem. Compare the results with paper in [20], the equation we study is more general than that in [20]. The fixed point theorem we use is a generalization in [20]. In other words, the fixed point theorem used in [20] is a special case of our fixed point theorem. More importantly, the solution we get is mild solution and our workspace is ordered Banach space.
We note that in previous research on measure driven equations, many researchers focus on discussing the case of operators A 0 (see [20]), using the assumptions in [29,30,32,33]. The first innovation of this work is the assumption that the operator A is the generator of a bounded strongly continuous cosine function. Moreover, this is very interesting and of great significance for studying measure evolution equations. On the other hand, all corresponding calculations to the nonlinear function F are completed in a Henstock–Lebesgue–Stieltjes sense. The second novelty of the this paper lies in the fact that the techniques used can be extended to study the positive mild solutions of measure differential equation by suitably introducing the abstract ordered Banach space. Therefore, the purpose of this work is to fill this gap by using the ( β , γ k ) -resolvent operator to study this type of equation.
The significance of our present work compared to previous existing publications lies in:
(1)
We gain the existence result under that the nonlinearity term F satisfies non-compactness measure conditions and order conditions, none Assumptions of the Lipschitz-type as those in [12,13].
(2)
The key behind our method is to use a new estimation technique of convex-power condensing operator and Hausdorff measure of noncompactness, which enables us to obtain the existence of positive solutions without uniform continuity of the nonlinearity.
(3)
Our tool is to utilize ( β , γ k ) -resolvent family, regulated functions and fixed point theorem with respect to convex-power condensing operator and measure of noncompactness.
An outline of this paper is organized as follows. The second part of the paper presents preliminary details. The third part states the existence of positive mild solution by convex-power condensing operator and a new estimation technique of Hausdorff measure of noncompactness. Finally, an example is provided to demonstrate the application of our results.

2. Preliminaries

Throughout this paper, let ( E , · ) be an ordered Banach space with partial order “≤” and P be a cone in E. The positive cone K = { u P | u θ } is normal. Denote by L ( E ) the Banach space of all linear and bounded operators in E equipped with the norm · L ( E ) .   G ( J , E ) denotes the Banach space of regulated functions from J to E equipped with a norm u G = sup t J u ( t ) and B : = G ( [ a , 0 ] , E ) the Banach space of regulated functions from [ a , 0 ] to E with the norm ϕ B = sup s [ a , 0 ] ϕ ( s ) , a > 0 .
Define a positive cone K G by
K G = { u G ( J , E ) | u ( t ) K , t J } ,
then K G is a normal with a normal constant N. Therefore, G ( J , E ) is an ordered Banach spaces endow with cone K G , B is also an order Banach space endow with positive cone
K B = { ϕ B | ϕ ( s ) K , s [ a , 0 ] } .
Definition 1.
Let P be a cone in E, if there exists a positive real constant N such that every x 1 , x 2 P with θ x 1 x 2 satisfy the inequality x 1   N x 2 , then we say that the cone P is normal in E (θ denotes the zero vector in E). The constant N is called the normal constant of P.
Definition 2
([20,31]). The function x : [ c , d ] E is said to be regulated on [ c , d ] , if the limits
lim τ s x ( τ ) = x ( s ) , s ( c , d ] a n d lim τ s + x ( τ ) = x ( s + ) , s [ c , d )
exist and are finite.
Definition 3
([20]). The set D G ( [ c , d ] , E ) is called equiregulated, if ϵ > 0 and s [ c , d ] , there exists σ > 0 such that
(i) 
If x D , τ [ c , d ] and τ ( s σ , s ) , then x ( s ) x ( τ ) E   < ϵ .
(ii) 
If x D , τ [ c , d ] and τ ( s , s + σ ) , then x ( τ ) x ( s + ) E   < ϵ .
Lemma 1
([20]). If { x m } m = 1 is a sequence of functions, x m converge pointwisely to x 0 , m and the sequence { x m } m = 1 is equiregulated, then x m converges uniformly to x 0 .
Lemma 2
([20,32]). If D G ( [ c , d ] , E ) is bounded and equiregulated, then the set c o ¯ ( D ) is also bounded and equiregulated.
Lemma 3
([19]). If D G ( [ c , d ] , E ) is equiregulated, for τ [ c , d ] , the set { x ( τ ) : x D } is relatively compact in E. Then D is relatively compact in G ( [ c , d ] , E ) .
Definition 4
([20]). The function φ : J E is said to be Henstock–Lebesgue–Stieltjes integrable with respect to h : J R , if there exists ( H L S ) 0 · : J E , for ϵ > 0 , δ ϵ on J with
m = 1 n φ ( e m ) ( h ( t m ) h ( t m 1 ) ) ( H L S ) 0 t m φ ( τ ) d h ( τ ) ( H L S ) 0 t m 1 φ ( τ ) d h ( τ ) < ϵ ,
for every δ ϵ -fine partition { ( e m , [ t m 1 , t m ] ) : m = 1 , 2 , , n } of J.
Denote by HLS h p ( J , R ) ( p > 1 ) the space of all p-ordered Henstock–Lebesgue–Stieltjes integral regulated from J to R with norm · HLS h p defined by
ρ HLS h p = ( H L S ) 0 b ρ ( τ ) p d h ( τ ) 1 p .
Lemma 4
([29]). Let φ HLS h p ( J , R + ) and h : J R be regulated. Then L ( τ ) = ( H L S ) 0 τ ( τ t ) β φ ( t ) d h ( t ) is regulated and
L ( τ ) L ( τ ) τ τ ( τ t ) q β d h ( t ) 1 q φ ( τ ) ( Δ h ( τ ) ) 1 p , τ ( 0 , b ] ,
L ( τ + ) L ( τ ) τ τ + ( τ + t ) q β d h ( t ) 1 q φ ( τ ) ( Δ + h ( τ ) ) 1 p , τ [ 0 , b ) ,
where Δ + h ( τ ) = h ( τ + ) h ( τ ) , Δ h ( τ ) = h ( τ ) h ( τ ) and p , q > 1 , 1 p + 1 q = 1 .
Definition 5
([34]). An ( β , γ k ) -resolvent family { R β , γ k ( τ ) } τ 0 is said to be positive, if for x θ , x E , τ 0 , R β , γ k ( τ ) x θ .
Definition 6
([34]). An ( β , γ k ) -resolvent family { R β , γ k ( τ ) } t 0 is said to be equicontinuous, if τ R β , γ k ( τ ) is continuous on the operator norm · L ( E ) .
Now, let us recall the following basic definitions and properties related to fractional calculus.
Let Γ ( · ) denote the gamma function and define φ k for k > 0 by
φ k ( t ) = t k 1 Γ ( k ) , t > 0 0 , t 0 .
Definition 7.
The Riemann–Liouville fractional integral of x L l o c 1 ( R + , E ) of order α > 0 as follows
I α x ( s ) = 0 s ( s τ ) α 1 Γ ( α ) x ( τ ) d τ
and I 0 x ( s ) = x ( s ) .
Definition 8.
Let η > 0 be given and denote m = [ η ] . The Caputo fractional derivative of order α > 0 of x C m ( R + , E ) is given by
D α c x ( s ) = I m α D m x ( s ) = 0 s ( s τ ) m α 1 Γ ( m α ) D m x ( τ ) d τ ,
and D 0 c x ( s ) = x ( s ) , where D m = d m / d t m .
Definition 9
([35,36]). Let β > 0 , γ k , α k > 0 , k = 1 , 2 , n . Then A is called the generator of ( β , γ k ) -resolvent family if κ 0 and a strongly continuous function R β , γ k : R + L ( E ) such that
λ β + 1 + k = 1 n α k λ γ k : R e ( λ ) > κ ρ ( A )
and
λ β λ β + 1 + k = 1 n α k λ γ k A 1 x = 0 e λ t R β , γ k ( t ) x d t , x E ,
where ρ ( A ) is the resolvent set of A .
Definition 10
([35,36]). Let 0 < β γ n γ n 1 γ 1 1 and α k 0 ,   k = 1 , 2 , , n be given and let A be a generator of a bounded and strongly continuous cosine family { C ( t ) } t R . Then, A generates a bounded ( β , γ k ) -resolvent family { R β , γ k ( t ) } t 0 .
Definition 11
([31]). A regulated function u ( · ) : J E is mild solution of problem (1) if u ( t ) = ρ ( t ) , u ( 0 ) + m ( u ) = ϕ and satifies
u ( t ) = R β , γ k ( t ) ρ ( 0 ) + ( φ 1 R β , γ k ) ( t ) [ ϕ m ( u ) ] + k = 1 n α k ( φ 1 + β γ k R β , γ k ) ( t ) ρ ( 0 ) + 0 t Q β , γ k ( t s ) F ( s , u ( s ) , u s ) d h ( s ) , t J .
where Q β , γ k ( t ) = ( φ β R β , γ k ) ( t ) .
Next, we denote sup t J ( R β , γ k ( t ) ) = M > 0 , then we have
Q β , γ k ( t ) L ( E ) M Γ ( 1 + β ) t β .
Definition 12
([37]). The Hausdorff measure of noncompactness α ( · ) defined on bounded set D of Banach space E is
α ( D ) : = inf { ε > 0 : D k = 1 n B ( ξ k , d k ) , ξ k E , d k < ε ( k = 1 , , n ) , n N } ,
where B ( ξ k , d k ) is ball centered at ξ k and of radius d k .
Now, we give the following useful lemmas.
Lemma 5
([37,38,39]). Let C , D be bounded subsets of E and δ R . Then
(1) 
α ( D ) = 0 if and only if D is relatively compact;
(2) 
C D implies α ( C ) α ( D ) ;
(3) 
α ( D ¯ ) = α ( D ) ;
(4) 
α ( C D ) = max { α ( C ) , α ( D ) } ;
(5) 
α ( λ C ) = | λ | α ( C ) , where λ C = { x = λ z : z C } ;
(6) 
α ( C + D ) α ( C ) + α ( D ) , where C + D = { x = y + z : y C , z D } ;
(7) 
α ( c o ( D ) ) = α ( D ) ;
(8) 
If the map Q : D ( Q ) E Z is Lipschitz continuous with a constant k, then α z ( Q Ω ) k α ( Ω ) for any bounded subset Ω D ( Ω ) , where Z is a Banach space.
In this article, we denote by α ( · ) and α G ( · ) the Hausdorff measure of noncompactness on the bounded set of E and G ( J , E ) , respectively. For any W G ( [ c , d ] , E ) and t J , set W ( t ) = { x ( t ) | x W } then W ( t ) E . If W G ( [ c , d ] , E ) is bounded, then W ( t ) is bounded in E and α ( D ( t ) ) α G ( D ) .
Lemma 6
([32]). Let W G ( [ c , d ] , E ) be bounded and equiregulated on [ c , d ] . Then α ( W ( t ) ) is regulated on [ c , d ] .
Lemma 7
([32]). Let W G ( [ c , d ] , E ) be bounded and equiregulated on [ c , d ] . Then α ( W ) = sup { α ( W ( t ) ) : t [ c , d ] } .
Lemma 8
([40]). Let E be a Banach space and W E be bounded. Then there exists a countable subset W 0 W , such that α ( W ) α ( W 0 ) .
Denote by LS h ( [ c , d ] , E ) the space of all functions f : [ c , d ] E that are Lebesgue–Stieltjes integrable with respect to h. Let μ h be the Lebesgue–Stieltjes measure on [ c , d ] induced by h.
Lemma 9
([41]). Let W 0 LS h ( [ c , d ] , E ) be a countable set. Assume that there exists a positive function k LS h ( [ c , d ] , R + ) such that w ( t ) k ( t ) μ h -a.e. holds for all w W 0 .
α c d W 0 ( t ) d h ( t ) 2 c d α ( W 0 ( t ) ) d h ( t ) .
The following fixed point theorem is an extension of Sadoveskii’s fixed point theorem. It was introduced by Sun and Zhang [42] in the sense of Kuratowski measure of noncompactness. Given the relationship between two types of noncompact measures (see Theorem 5.13 in [38]), by repeating the same process as Lemma 2.4 in [43], we can obtain the same conclusion in the setting of Hausdorff noncompact measures.
Definition 13.
Let E be a real Banach space. If O : E E is a continuous and bounded operator, u 0 E and n 0 > 0 , for any bounded and nonprecompact subset W E ,
α ( O ( n 0 , u 0 ) ( W ) ) < α ( W ) ,
where
O ( 1 , u 0 ) ( W ) O ( W ) , O ( n , u 0 ) ( W ) = O ( c o ¯ O ( n 1 , u 0 ) ( W ) , u 0 ) , n = 2 , 3 , .
Then we call O a convex-power condensing operator about u 0 and n 0 .
Lemma 10
([42]). Let E be a real Banach space, and D E is a bounded, closed and convex set in E. If u 0 D and integer n 0 > 0 such that O : D D is a convex-power condensing operator with u 0 and n 0 , then O has at least one fixed point in D.
Lemma 11
([37]). Assuming E is a Banach space, and D E is a bounded closed convex sets, O : D D is a condensing operator. Then O has at least a fixed point in D.
The following lemma is Bellman integral inequality for Henstock–Lebesgue–Stieltjes integrals, see [44] (Theorem 4.4).
Lemma 12
([44]). Let T > 0 . Assume that a , m G ( [ 0 , T ] , R + ) . If y G ( [ 0 , T ] , R + ) with
y ( t ) m ( t ) + 0 t a ( s ) y ( s ) d g ( s )
for t [ 0 , T ] , then
y ( t ) m ( t ) + 0 t a ( s ) m ( s ) e s t a ( τ ) d g ( τ ) d g ( s ) .
Lemma 13
([20]). Let h : J R be nondecreasing, left-continuous and γ 1 ( t ) = 0 t d h ( s ) and γ k ( t ) = 0 t γ k 1 ( s ) d h ( s ) , for each k N . Then, for t J ,
γ k ( t ) h k ( t ) k ! , k N .

3. Main Result

Define function u ρ ( t ) : [ a , b ] K by
u ρ ( t ) = u ( t ) , t J ρ ( t ) , t [ a , 0 ] ,
where ρ K B , u G ( J , K ) .
A closed subspace of G ( J , K ) given by
G ρ ( K ) : = { u G ( J , K ) | u ( 0 ) = ρ ( 0 ) }
with the norm · ρ .
Now we will show the main result in this section.
Theorem 1.
Let E be an ordered Banach space, and its positive cone K E be normal. If ρ K B , ρ ( 0 ) K D ( A ) , ϕ K and the following conditions hold:
(H1)
A is the generator of a positive and bounded strongly continuous cosine family { R β , γ k ( t ) } t 0 .
(H2)
Function F ( · , x , ϕ ) is measurable for u G ( J , E ) ϕ B , and F ( t , · , · ) is continuous for a.e t J .
(H3)
There exist constant l 1 , l 2 , L g > 0 , for D K , D K B , t J ,
α ( F ( t , D , D ) ) l 1 α ( D ) + l 2 sup τ [ r , 0 ] α ( D ( τ ) ) ,
α ( { m ( D ) } ) L g t β 1 α ( D ) .
(H4)
There exist constants c 1 , c 2 > 0 and function g G ( J , K ) ,
F ( t , x , ϕ ) c 1 x + c 2 ϕ ( · ) + g ( t ) ,
for t J , x K , ϕ K B .
(H5)
Q 0 : G ( J , K ) K is continuous mapping, and there are constants c 0 , d 0 > 0 ,
m ( u ) c 0 u + d 0 .
(H6)
The condition
Θ = b M c 0 + M b β Γ ( 1 + β ) ( c 1 + c 2 ) ( 0 , 1 ) .
The nonlocal problem (1) has at least one positive mild solution u ρ G ( J , K ) .
Proof. 
Give an operator O on G ρ ( K ) by
( O u ) ( t ) = R β , γ k ( t ) ρ ( 0 ) + ( φ 1 R β , γ k ) ( t ) [ ϕ m ( u ) ] + k = 1 n α k ( φ 1 + β γ k R β , γ k ) ( t ) ρ ( 0 ) + 0 t Q β , γ k ( t s ) F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) , t J .
By (H2), 0 t Q β , γ k ( t s ) F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) is well defined. Then O : G ρ ( K ) G ρ ( K ) is well defined. And by Definition 11, the fixed point u of O in G ρ ( K ) , i.e., u ρ is a mild solution to problem (1).
In the following, we will prove that O has at least one fixed point by using Lemma 10.
Let
Ω R = { u G ρ ( K ) | u G R } .
Obviously, Ω R is a closed ball in G ρ ( K ) with a center of θ . Then u ρ ( t ) R for u Ω R , t J .
We complete the proof via four steps.
Step 1.
We check that R 0 > 0 , O ( Ω R 0 ) Ω R 0 .
Indeed, if there were not so, it would prove that for R > 0 , u Ω R with O u G > R . By the definition of O , for t J ,
( O u ) ( t ) R β , γ k ( t ) ρ ( 0 ) + ( φ 1 R β , γ k ) ( t ) [ ϕ m ( u ) ] + k = 1 n α k 0 t ( t s ) β γ k Γ ( 1 + β γ k ) R β , γ k ( s ) ρ ( 0 ) d s + 0 t Q β , γ k ( t s ) F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) R β , γ k ( t ) ρ ( 0 ) + ( φ 1 R β , γ k ) ( t ) [ ϕ m ( u ) ] + k = 1 n α k 0 t ( t s ) β γ k Γ ( 1 + β γ k ) R β , γ k ( s ) ρ ( 0 ) d s + 0 t Q β , γ k ( t s ) · F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) M ρ B + b M ( ϕ + c 0 u + d 0 ) + k = 1 n α k M b 1 + β + γ k Γ ( 1 + β γ k ) ρ B + M b β Γ ( 1 + β ) ( c 1 + c 2 ) R + M Γ ( 1 + β ) 0 t ( t s ) q β d h ( s ) 1 q 0 t [ h ( s ) ] p d h ( s ) 1 p M ρ B + b M ( ϕ + c 0 u + d 0 ) + k = 1 n α k M b 1 + β + γ k Γ ( 1 + β γ k ) ρ B + M b β Γ ( 1 + β ) ( c 1 + c 2 ) R + M Γ ( 1 + β ) sup t J 0 t ( t s ) q β d h ( s ) 1 q h HLS h p .
Hence, we have
R < M + k = 1 n α k M b 1 + β + γ k Γ ( 1 + β γ k ) ρ B + b M ( ϕ + c 0 R + d 0 ) + M b β Γ ( 1 + β ) ( c 1 + c 2 ) R + M Γ ( 1 + β ) sup t J 0 t ( t s ) q β d h ( s ) 1 q h HLS h p .
Taking the lower limit as R , we get
b M c 0 + M b β Γ ( 1 + β ) ( c 1 + c 2 ) 1 ,
which contradiction Θ < 1 . Thus, R 0 > 0 such that O ( Ω R 0 ) Ω R 0 .
Step 2.
Set { O u : u ( · ) Ω } is equiregulated.
For t 0 [ 0 , b ) , we obtain
( O u ) ( t ) ( O u ) ( t 0 + ) ( R β , γ k ( t ) R β , γ k ( t 0 + ) ) ρ ( 0 ) + [ ( φ 1 S β , γ k ) ( t ) ( φ 1 R β , γ k ) ( t 0 + ) ] [ ϕ m ( u ) ] + α k M k = 1 n 1 Γ ( 1 + β γ k ) | 0 t ( t s ) β γ k d s 0 t 0 + ( t 0 + s ) β γ k d s | ρ ( 0 ) + 0 t 0 + [ Q β , γ k ( t s ) Q β , γ k ( t 0 + s ) ] F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) + t 0 + t Q β , γ k ( t s ) ( t s ) F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) R β , γ k ( t ) R β , γ k ( t 0 + ) L ( E ) · ρ ( 0 ) + M | t t 0 + | · ( ϕ + R c 0 + d 0 ) + α k M k = 1 n | t 1 + β γ k ( t 0 + ) 1 + β γ k Γ ( 2 + β γ k ) | ρ ( 0 ) + 0 t 0 + Q β , γ k ( t s ) Q β , γ k ( t 0 + s ) L ( E ) · F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) + M Γ ( 1 + β ) t 0 + t ( t s ) β F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) = I 1 ( t ) + I 2 ( t ) + I 3 ( t ) + I 4 ( t ) + I 5 ( t ) ,
where
I 1 ( t ) = R β , γ k ( t ) R β , γ k ( t 0 + ) L ( E ) · ρ ( 0 ) ,
I 2 ( t ) = M | t t 0 + | · ( ϕ + R c 0 + d 0 ) ,
I 3 ( t ) = α k M k = 1 n | t 1 + β γ k ( t 0 + ) 1 + β γ k Γ ( 2 + β γ k ) | · ρ ( 0 ) ,
I 4 ( t ) = 0 t 0 + Q β , γ k ( t s ) Q β , γ k ( t 0 + s ) L ( E ) · F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) ,
I 5 ( t ) = M Γ ( 1 + β ) t 0 + t ( t s ) β F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) .
By I 2 ( t ) and I 3 ( t ) , we infer that I 2 ( t ) 0 , I 3 ( t ) 0 as t t 0 + . Since R β , γ k ( t ) and Q β , γ k ( t ) for t > 0 are continuous. We infer that I 1 ( t ) 0 , by applying dominated convergence theorem, we get I 4 ( t ) 0 as t t 0 + .
Set L ( t ) = 0 t ( t s ) β F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) . In view of Lemma 4, then L ( t ) is a regulated function on J. Thus
I 5 ( t ) = M Γ ( 1 + β ) t 0 + t ( t s ) β F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) M Γ ( 1 + β ) ( L ( t ) L ( t 0 + ) + 0 t 0 + ( ( t s ) β ( t 0 + s ) β ) F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) ) 0 a s t t 0 + .
Hence, ( O u ) ( t ) ( O u ) ( t 0 + ) Ω 0 , t t 0 + .
Similarly, we prove that for t 0 ( 0 , b ] , ( O u ) ( t ) ( O u ) ( t 0 + ) G 0 , t t 0 + . Thus, claim that { O u : u ( · ) Ω } is equiregulated.
Step 3.
O : Ω Ω is continuous. Let { u ( n ) } Ω , u ( n ) u ( t ) , when n , then, u ( n ) ( t ) u ( t ) , t J as n . Moreover, from (3), u ρ ( n ) u ρ as n . For t J ,
lim n F ( t , u ( n ) ( t ) , u ρ ( n ) ( t ) ) = F ( t , u ( t ) , u ρ ( t ) ) , lim n m ( u ( n ) ) = m ( u ) .
By (H1), we get
F ( t , u ( n ) ( t ) , u ρ ( n ) ( t ) ) F ( t , u ( t ) , u ρ ( t ) ) 2 N ( 2 R 0 + g G ) .
By the inequalities (9), we have
O ( u n ) ( t ) O ( u ) ( t ) ( φ 1 R β , γ k ) ( t ) m ( u ( n ) ) m ( u ) + M Γ ( 1 + β ) 0 t ( t s ) β F ( s , u ( n ) ( s ) , u ρ ( n ) ( s ) ) F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) , t J .
Thus,
O ( u ( n ) ) O ( u ) G = sup t J O ( u ( n ) ) ( t ) O ( u ) ( t ) 0 , n .
Furthermore, in view of { O ( u ( n ) ) } n = 1 is equiregulated. Therefore, according to Lemma 1, O ( u ( n ) ) O ( u ) . Therefore, O is a continuous operator.
Step 4.
O : Θ = c o ¯ O ( Ω R ) G φ ( K ) Θ is a convex power condensing operator.
Based on the above proof process, it is easy to infer O and O ( Θ ) Ω R is bounded and equiregulated. Let u 0 Θ . Next, we prove that m 0 > 0 , for any bounded subset B Θ ,
α G O ( m 0 , u 0 ) ( B ) < α G ( B ) .
For B Θ and u 0 Θ , by O ( m 0 , u 0 ) and the equiregulated of Θ , we know that O ( m 0 , u 0 ) ( B ) Ω R is also equiregulated. Thus, by Lemma 10, we obtain
α G O ( m , u 0 ) ( B ) = sup t J α O ( m , u 0 ) ( B ) ( t ) , m = 1 , 2 , .
By Lemma 11, B 1 = { u n 1 } B ,
α ( O ( B ) ( t ) ) α ( O ( B 1 ) ( t ) ) .
Moreover, for t J ,
α B ( { u 1 ( n ) } ρ ) = sup τ [ a , 0 ] α ( { u 1 ( n ) ( t + τ ) } ) α ( { u 1 ( n ) ( t ) } ρ ) .
By (5), (13), (14), (H3), and Lemmas 6–8 and 12, we get
α O ( 1 , u 0 ) ( B ) ( t ) α O ( B 1 ) ( t ) 2 α ( { R β , γ k ( t ) ρ ( 0 ) } ) + 2 α 0 t R β , γ k ( t ) [ ϕ m ( u 1 ( n ) ) ] d s + 2 α k = 1 n α k ( φ 1 + β γ k R β , γ k ) ( t ) ρ ( 0 ) + 2 α 0 t Q β , γ k ( t s ) F ( s , u 1 ( n ) ( s ) , u 1 ( n ) ( s ) ρ ) d h ( s ) 4 M L g Γ ( β ) t β Γ ( 1 + β ) α G ( B ) + 4 M ( l 1 + l 2 ) Γ ( 1 + β ) 0 t ( t s ) β α ( B 1 ( s ) ) d h ( s ) 4 M L g Γ ( β ) t β Γ ( 1 + β ) α G ( B ) + 4 M ( l 1 + l 2 ) t β · g ( t ) Γ ( 1 + β ) α G ( B ) .
And by Lemma 10, we know that B 2 = { u 2 ( n ) } c o ¯ { O ( 1 , u 0 ) ( B ) , u 0 } ,
α O ( c o ¯ { O ( 1 , u 0 ) ( O ) , u 0 } ) ( t ) = α ( O ( B 2 ) ( t ) ) .
For t J ,
α B ( { u 2 ( n ) ( t ) ρ } ) = sup t [ a , 0 ] α ( { u 2 ( n ) ( t + τ ) ρ } ) α ( { u 2 ( n ) ( t ) } ) .
Moreover, by (5), (15)–(17) and (H3), we have
α O ( 2 , u 0 ) ( B ) ( t ) = α O ( c o ¯ { O ( 1 , u 0 ) ( B ) , u 0 } ) ( t ) = α ( O ( B 2 ) ( t ) ) 2 α ( { R β , γ k ( t ) ρ ( 0 ) } ) + 2 α 0 t R β , γ k ( t ) [ ϕ m ( u 2 ( n ) ) ] d s + 2 α k = 1 n α k ( φ 1 + β γ k R β , γ k ) ( t ) ρ ( 0 ) + 2 α 0 t Q β , γ k ( t s ) F ( s , u 2 ( n ) ( s ) , u 2 ( n ) ( s ) ρ ) d h ( s ) 4 M L g 0 t α ( B 2 ( s ) ) d s + 4 M ( l 1 + l 2 ) Γ ( 1 + β ) 0 t ( t s ) β α ( B 2 ( s ) ) d h ( s ) 4 M L g 0 t α O ( c o ¯ { O ( 1 , u 0 ) ( B ) , u 0 } ) ( s ) d s + 4 M ( l 1 + l 2 ) Γ ( 1 + β ) 0 t ( t s ) β α O ( c o ¯ { O ( 1 , u 0 ) ( B ) , u 0 } ) ( s ) d h ( s ) 4 M L g 0 t α 4 M L g Γ ( β ) t β Γ ( 1 + β ) + 4 M ( l 1 + l 2 ) t β · h ( t ) Γ ( 1 + β ) d s · α G ( B ) + 4 M ( l 1 + l 2 ) Γ ( 1 + β ) 0 t ( t s ) β α 4 M L g Γ ( β ) t β Γ ( 1 + β ) + 4 M ( l 1 + l 2 ) t β · h ( t ) Γ ( 1 + β ) d h ( s ) · α G ( B ) 4 M L g · 4 M L g Γ 2 ( β ) t 2 β Γ ( 1 + 2 β ) α G ( B ) + 2 · 4 M L g · 4 M ( l 1 + l 2 ) Γ ( β ) t 2 β · h 2 ( t ) Γ ( 1 + 2 β ) α G ( B ) + 4 M ( l 1 + l 2 ) · 4 M ( l 1 + l 2 ) t 2 β · h 2 ( t ) Γ ( 1 + 2 β ) α G ( B ) .
For t J ,
α O ( k , u 0 ) ( D B ) ( t ) C k 0 ( 4 M L g Γ ( β ) t β ) k · ( 4 M ( l 1 + l 2 ) t β ) 0 Γ ( 1 + k β ) α G ( B ) + C k 1 ( 4 M L g Γ ( β ) t β ) k 1 · ( 4 M ( l 1 + l 2 ) t β ) 1 h k ( t ) Γ ( 1 + k β ) α G ( B ) + + C k k 1 ( 4 M L g Γ ( β ) t β ) 1 · ( 4 M ( l 1 + l 2 ) t β ) k 1 h k ( t ) Γ ( 1 + k β ) α G ( B ) + C k k ( 4 M L g Γ ( β ) t β ) 0 · ( 4 M ( l 1 + l 2 ) t β ) k h k ( t ) Γ ( 1 + k β ) α G ( B ) .
By Lemma 8, B k + 1 = { u k + 1 ( n ) } c o ¯ { O ( k , u 0 ) ( B ) , u 0 } ,
α O ( c o ¯ { O ( k , u 0 ) ( B ) , u 0 } ) ( t ) = α ( O ( B k + 1 ) ( t ) ) .
For t J ,
α B ( { u k + 1 ( n ) ( t ) ρ } ) = sup t [ a , 0 ] α ( { u k + 1 ( n ) ( t + τ ) ρ } ) α ( { u k + 1 ( n ) ( t ) } ) .
By (5), (19)–(21) and (H3), we get
α O ( k + 1 , u 0 ) ( B ) ( t ) = α Q ( c o ¯ { O ( k , u 0 ) ( B ) , u 0 } ) ( t ) 4 M L g 0 t ( t s ) β 1 α ( B k + 1 ( s ) ) d s + 4 M ( l 1 + l 2 ) Γ ( 1 + β ) 0 t ( t s ) β α ( B k + 1 ( s ) ) d h ( s ) 4 M L g 0 t ( t s ) β 1 α Q ( c o ¯ { O ( k , u 0 ) ( B ) , u 0 } ) ( s ) d s + 4 M ( l 1 + l 2 ) Γ ( 1 + β ) 0 t ( t s ) β α Q ( c o ¯ { O ( k , u 0 ) ( B ) , u 0 } ) ( s ) d h ( s ) C k + 1 0 ( 4 M L g Γ ( β ) t ) k + 1 · ( 4 M ( l 1 + l 2 ) t β ) 0 Γ ( 1 + k β ) α G ( B ) + C k + 1 1 ( 4 M L g Γ ( β ) t β ) k · ( 4 M ( l 1 + l 2 ) t β ) 1 h k ( t ) Γ ( 1 + k β ) α G ( B ) + + C k + 1 k ( 4 M L g Γ ( β ) t β ) 1 · ( 4 M ( l 1 + l 2 ) t β ) k 1 h k ( t ) Γ ( 1 + k β ) α G ( B ) + C k + 1 k + 1 ( 4 M L g Γ ( β ) t β ) 0 · ( 4 M ( l 1 + l 2 ) t β ) k + 1 h k ( t ) Γ ( 1 + ( k + 1 ) β ) α G ( B ) .
Thus, by the method of mathematical induction, for m > 0 , t J ,
α O ( m , u 0 ) ( B ) ( t ) C m 0 ( 4 M L g Γ ( β ) t β ) m · ( 4 M ( l 1 + l 2 ) t β ) 0 Γ ( 1 + m β ) α G ( B ) + C m 1 ( 4 M L g Γ ( β ) t β ) m 1 · ( 4 M ( l 1 + l 2 ) t β ) 1 h m ( t ) Γ ( 1 + m β ) α G ( B ) + + C m m 1 ( 4 M L g Γ ( β ) t β ) 1 · ( 4 M ( l 1 + l 2 ) t β ) m 1 h m ( t ) Γ ( 1 + m β ) α G ( B ) + C m m ( 4 M L g Γ ( β ) t β ) 0 · ( 4 M ( l 1 + l 2 ) t β ) m h m ( t ) Γ ( 1 + m β ) α G ( B ) .
which implies that
α G O ( m , u 0 ) ( B ) = sup t J α O ( m , u 0 ) ( B ) ( t ) [ ( 4 M L g Γ ( β ) ) m · b m β Γ ( 1 + m β ) + C m 1 ( 4 M L g Γ ( β ) ) m 1 · ( 4 M ( l 1 + l 2 ) ) 1 · b m β · h ( b ) m Γ ( 1 + m β ) + + C m m 1 ( 4 M L g Γ ( β ) ) 1 · ( 4 M ( l 1 + l 2 ) ) m 1 · b m β · h ( b ) m Γ ( 1 + m β ) + ( 4 M L g ( l 1 + l 2 ) ) m · b m β · h ( b ) m Γ ( 1 + m β ) ] α G ( B ) .
In view of the Stirling formula
Γ ( 1 + m β ) = 2 π m β m β e m β · e v 12 m β , v ( 0 , 1 ) .
Thus, we get that m , then
( 4 M L g Γ ( β ) ) m · a m β Γ ( 1 + m β ) = ( 4 M L g Γ ( β ) ) m · a m β 2 π m β m β e m β · e v 12 m β 0 ,
C m 1 ( 4 M L g Γ ( β ) ) m 1 · ( 4 M ( l 1 + l 2 ) ) 1 · b m β · h ( b ) m Γ ( 1 + m β ) = C m 1 ( 4 M L g Γ ( β ) ) m 1 · ( 4 M ( l 1 + l 2 ) ) 1 · b m β · h ( b ) m 2 π m β m β e m β · e v 12 m β 0 , C m m 1 ( 4 M L g Γ ( β ) ) 1 · ( 4 M ( l 1 + l 2 ) ) m 1 · b m β · h ( b ) m Γ ( 1 + m β ) = C m m 1 ( 4 M L g Γ ( β ) ) 1 · ( 4 M ( l 1 + l 2 ) ) m 1 · b m β · h ( b ) m 2 π m β m β e m β · e v 12 m β 0 ,
( 4 M ( l 1 + l 2 ) ) m · b m β · h m ( b ) Γ ( m β + 1 ) = ( 4 M ( l 1 + l 2 ) ) m · b m β · h m ( b ) 2 π m β m β e m β · e v 12 m β 0 ,
Thus, m 0 > 0 , which is large enough, such that
( 4 M L g Γ ( β ) ) m 0 · a m 0 Γ ( 1 + m 0 β ) + + ( 4 M ( l 1 + l 2 ) ) m 0 · b m 0 β · h m 0 ( b ) Γ ( 1 + m 0 β ) < 1 .
By (24), we have
α G O ( m , u 0 ) ( B ) < α G ( B ) .
Hence, by Lemma 10, we get that the operator O defined by (5) has at least one fixed point u Θ G φ ( K ) , which is a positive mild solution to (1). □
Theorem 2.
Let E be an ordered Banach space, whose positive cone K E is normal. If ρ K B , ρ ( 0 ) K D ( A ) , ϕ K and (H1)–(H6) satisfy, then (1) has at least one positive mild solution u ρ G ( J , K ) with
4 M L g Γ ( β ) b β Γ ( 1 + β ) + 4 M ( l 1 + L 2 ) b β · h ( b ) Γ ( 1 + β ) < 1 .
Proof. 
Let O : G φ ( K ) G φ ( K ) . By Theorem 1, we get that O : Ω R Ω R is a continuous operator and mild solution to (1) is equivalent to the fixed point of the operator O defined by (5).
Similar to Step 3 in Theorem 1, we verify { O u : u Ω R } is equiregulated in G φ ( J , K ) , for B c o ¯ O ( Ω R ) , O ( B ) Ω R is equiregulated. By Lemma 7, we have
α G ( O ( B ) ) = sup t J α ( O ( B ) ( t ) ) .
By Lemma 7, B 0 = u m B ,
α G ( O ( B ) ) α G ( O ( B 0 ) ) .
For t J ,
α B ( { u m ( t ) ρ } ) = sup τ [ a , 0 ] α ( { u m ( t + τ ) ρ } ) α ( { u m ( t ) } ) .
In view of Lemma 12 and (H3), (5), (28), we have
α O ( 1 , u 0 ) ( B ) ( t ) = α O ( B ) ( t ) 2 α O ( B 1 ) ( t ) 2 α ( { R β , γ k ( t ) ρ ( 0 ) } ) + 2 α 0 t R β , γ k ( t ) [ ϕ m ( u 1 ( n ) ) ] d s + 2 α k = 1 n α k ( φ 1 + β γ k R β , γ k ) ( t ) ρ ( 0 ) + 2 α 0 t Q β , γ k ( t s ) F ( s , u 1 ( n ) ( s ) , u 1 ( n ) ( s ) ρ ) d h ( s ) 4 M L g Γ ( β ) b β Γ ( 1 + β ) α G ( B ) + 4 M ( l 1 + l 2 ) b β · h ( b ) Γ ( 1 + β ) 0 t ( t s ) β α ( B 1 ( s ) ) d h ( s ) .
And by (25), (26) and Lemma 11, we have
α G ( O ( B ) ) 4 M L g Γ ( β ) b β Γ ( 1 + β ) + 4 M ( l 1 + L 2 ) b β · h ( b ) Γ ( 1 + β ) α G ( O ( B 0 ) ) < α G ( O ( B 0 ) ) α G ( B ) .
Thus, α G ( O ( B ) ) α G ( B ) , i.e., O : B B is a condensing operator. In view of Lemma 11, O has a fixed point u B c o ¯ O ( Ω R ) . Thus, u ρ G φ ( J , K ) is a positive mild solution to (1). □
Theorem 3.
Let E be an ordered Banach space, whose positive cone K E is normal. If ρ K B , ρ ( 0 ) K D ( A ) , ϕ K . If (H1)–(H3), (H5) and
(H7)
For t 0 , x 1 , x 2 K with θ x 1 x 2 , x i R and ϕ 1 , ϕ 2 K B with ϕ i B R ,   θ ϕ 1 ϕ 2 ,
F ( t , x 2 , ϕ 2 ) F ( t , x 1 , ϕ 1 ) θ ;
(H8)
t 0 , x E , ϕ B , p i ( · ) HLS h p ( R + , R + ) , for p > 1 , K ¯ > 0
F ( t , x , ϕ ) p 1 ( t ) x + p 2 ( t ) ϕ B + K ¯ ,
then (1) has the minimal positive mild solution u G ( J , K ) with
M b + M Γ ( 1 + β ) sup t 0 0 t ( t s ) q β d h ( s ) 1 q p 1 HLS h p < 1 , 4 M L g Γ ( β ) b β < Γ ( 1 + β ) ,
and 1 q + 1 p = 1 .
Proof. 
We verify that O : G φ ( K ) G φ ( K ) be defined by (5) is continuous by Theorem 1. By Definition 11, we know that the fixed point u of the operator O is the mild solution of problem (1). By (H3) and (5), for u G ρ ( K ) , we get
( O u ) ( t ) M ρ B + M t ( ϕ + c 0 u + d 0 ) + k = 1 n α k M b 1 + β + γ k Γ ( 1 + β γ k ) ρ B + 0 t Q β , γ k ( t s ) · F ( s , u ( s ) , u ρ ( s ) ) d h ( s ) M ρ B + M t ( ϕ + c 0 u + d 0 ) + k = 1 n α k M b 1 + β + γ k Γ ( 1 + β γ k ) ρ B + M Γ ( 1 + β ) 0 t ( t s ) β ( p 1 u ( s ) + p 2 ( s ) sup t [ a , 0 ] u ( s + τ ) ρ + K ¯ ) d h ( s ) ( M + k = 1 n α k M b 1 + β + γ k Γ ( 1 + β γ k ) ) ρ B + M b ( ϕ + c 0 u G + d 0 ) + M Γ ( 1 + β ) sup t 0 0 t ( t s ) q β d h ( s ) 1 q ( u G · p 1 HLS h p + ρ B · p 2 HLS h p + K ¯ ) ,
where
L = ( M + k = 1 n α k M b 1 + β + γ k Γ ( 1 + β γ k ) ) ρ B + M b ( ϕ + d 0 )
+ M Γ ( 1 + β ) sup t 0 0 t ( t s ) q β d h ( s ) 1 q ( ρ B · p 2 HLS h p + K ¯ ) ,
η = M b + M Γ ( 1 + β ) sup t 0 0 t ( t s ) q β d h ( s ) 1 q p 1 HLS h p < 1 .
Next, we verify that the existence of positive mild solutions.
Step 1.
For u , v s . G ρ ( K ) , t J , u v , u ρ ( t ) , v ρ ( t ) K B satisfying u ρ ( t ) v ρ ( t ) . By (H7), (5) and ρ K B , ϕ K , for t J ,
θ O u ( t ) R β , γ k ( t ) ρ ( 0 ) + ( φ 1 R β , γ k ) ( t ) [ ϕ m ( v ) ] + k = 1 n α k ( φ 1 + β γ k R β , γ k ) ( t ) ρ ( 0 ) + 0 t Q β , γ k ( t s ) F ( s , v ( s ) , v ρ ( s ) ) d h ( s ) = O v ( t ) ,
thus, for u v , θ O u O v .
Let v 0 ( t ) ρ ( 0 ) . Thus, v 0 G ρ ( J , K ) , v 0 ( t ) ρ K B .
Define a sequence { v m } with
v m = O v m 1 , m = 1 , 2 , .
By monotonicity of O , we have { v m } G ρ ( J , K ) and
v 0 v 1 v m , v m G L + η v m 1 G .
Since v 0 ρ B , 0 < β < 1 , we get
v m G L + L η + + L η m 1 + η m v 0 G L ( 1 η m ) 1 η + η m ρ B L 1 η + ρ B ,
i.e., the sequence { v m } is uniformly bounded.
Step 2.
We verify that { v m } is convergent in G ρ ( J , K ) . Let V = { v m } , V 0 = V { v 0 } , thus O ( V 0 ) = V . By Step 3 in Theorem 1, we get V G ρ ( K ) is equiregulated. By (H3), Lemma 7, Lemma 8, (5), for t J ,
α V ( t ) = α O V 0 ( t ) 2 α ( { S β , γ k ( t ) ρ ( 0 ) } ) + 2 α 0 t R β , γ k ( t ) [ ϕ m ( v m 1 ) ] d s + 2 α k = 1 n α k ( φ 1 + β γ k R β , γ k ) ( t ) ρ ( 0 ) + 2 α 0 t Q β , γ k ( t s ) F ( s , v m 1 ( s ) , v m 1 ( s ) ρ ) d h ( s ) 4 M L g Γ ( β ) b β Γ ( 1 + β ) α ( V ( t ) ) + 4 M ( l 1 + l 2 ) b β Γ ( 1 + β ) 0 t α ( V ( s ) ) d h ( s ) .
Since 4 M L g Γ ( β ) b β Γ ( 1 + β ) < 1 , it gives that
α V ( t ) 4 M ( l 1 + l 2 ) b β ( Γ ( 1 + β ) 4 b M L g Γ ( β ) b β ) 0 t α ( V ( s ) ) d h ( s ) .
By Lemma 12, we get α ( { v m ( t ) } ) 0 on J. And by the uniform boundedness and equiregulated of { v m } , { v m } is relatively compact in G φ ( J , K ) . Thus, we get that { v m } itself is convergent, i.e, there exists u G φ ( J , K ) , lim m v m = u , i.e, u = O u , which is also a positive mild solution of the problem (1). Next, we verify that u is the minimal positive mild solution. Let u ˜ G ( J , E ) be a positive mild solution to (1), then u ˜ G ρ ( K ) with u ˜ = O u ˜ . Since for t J , u ˜ ( t ) v 0 ( t ) = ρ ( 0 ) , we have
u ˜ ( t ) = O u ˜ ( t ) O v 0 ( t ) = v 1 ( t ) , t J ,
thus, u ˜ v 1 . Obviously, u ˜ v m , m = 1 , 2 , . When m , we have u ˜ u . □

4. Applications

Consider the following measure driven differential equation:
D 0 + 1 + β c u ( t , x ) + k = 1 n α k D γ k c u ( t , x ) = Δ u ( t , x ) + τ u ( t , x ) + sin ( u ( t + s ) ) 1 + e 2 t d h ( t ) , ( t , x ) [ 0 , b ] × [ 0 , π ] , s [ a , 0 ] , u ( t , 0 ) = u ( t , π ) = 0 , t [ 0 , b ] , u ( t , x ) = ρ ( t , x ) , ( t , x ) [ a , 0 ] × [ 0 , π ] , u ( t , x ) t | t = 0 = | u ( t , x ) | 6 + | u ( t , x ) | + ϕ ( x ) , x [ 0 , π ] ,
where Δ is Laplace operator, τ < 0 is constant, Assume that f : [ 0 , b ] × [ 0 , π ] × K × K B E be measurable function, ρ ( · , · ) G ( Ω ¯ × [ r , 0 ] , [ 0 , ) ) , ϕ ( · ) 0 , b , a > 0 are constants, h : [ 0 , π ] R is a nondecreasing function. A : D ( A ) E E with A u = Δ u + τ u ,
D ( A ) = { u E : u , u are   absolutely   continuous , u E , u ( 0 ) = u ( π ) = 0 } .
Then
A u = n = 1 n 2 ( u , w n ) w n , u D ,
where w n ( θ ) = 2 π sin ( n θ ) , n = 1 , 2 , . Thus, A generates a strongly continuous cosine family { C ( t ) } t R given by
C ( t ) u = n = 1 cos ( n t ) ( u , w n ) w n , u E .
Since β , γ k > 0 , k = 1 , 2 , , m be such that 0 < β γ m γ 1 1 , by Definition 9, thus A generates ( β , γ k ) -resolvent family
R β , γ k ( t ) u = 0 1 t ( 1 + β ) 2 Φ ( 1 + β ) 2 ( s t ( 1 + β ) 2 ) C ( s ) u d s , u E , t [ 0 , 1 ] ,
where
Φ ( 1 + β ) 2 ( y ) = 0 ( y ) n n ! Γ ( ( β ( n + 1 ) ) n ) , y C ,
are the Wright functions.
We choose the workspace E = L 2 ( [ 0 , π ] , [ 0 , b ] ) , which is an ordered Banach space with L 2 norm · 2 and partial-order “≤”, K = { u L 2 ( [ 0 , π ] , [ 0 , b ] ) : u ( x ) 0 , a . e . x [ 0 , π ] } is a normal regeneration cone with the normal constant N = 1 . Note B : = G ( [ a , 0 ] × [ 0 , π ] , E ) with the normal cone K B = { u B : u ( t , x ) K , t [ a , 0 ] , a . e . x [ 0 , π ] } . We define
f ( t , x , u ( t , x ) , u ( t + s , x ) ) = sin ( u ( t + s ) ) 1 + e 2 t , t [ 0 , b ] , s [ a , 0 ] .
m ( u ( t , x ) ) = | u ( t , x ) | 6 + | u ( t , x ) | .
For u [ 0 , π ] , we set φ ( t ) = φ ( t , · ) , ψ = ψ ( · ) , u ( t ) = u ( t , · ) , u t ( s ) = u ( t + s , · ) and
F ( t , u ( t ) , u t ) = f ( t , · , u ( t , · ) , u ( t + s , · ) ) , m ( u ) = | u | 6 + | u | .
Then, Equation (34) can be transformed into (1) in L 2 ( [ 0 , π ] , [ 0 , b ] ) .
By the definition of functions f and m,
F ( t , u ( t ) , u t ) 1 2 u , m ( u ( t , x ) ) 1 6 u .
We deduce that condition (H5) is satisfied with c 0 = 1 6 and d 0 = 0 . Additionally, (H8) is satisfied with p 1 ( t ) = 1 2 , p 2 ( t ) = 0 and K ¯ = 0 .
Theorem 4.
If assumption (H1), (H3) and the following condition
(A1) 
For ξ [ 0 , π ] , t [ 0 , b ] , θ x 1 x 2 , θ ϕ 1 ϕ 2 ,
f ( ξ , t , x 2 , ϕ 2 ) f ( ξ , t , x 1 , ϕ 1 ) θ ,
hold, then (34) has at least one positive mild solution.
Proof. 
From assumptions condition f : [ 0 , b ] × [ 0 , π ] × K × K B E is continuous measurable function, guarantee condition (H2) is established. By the condition (A1), one can find that the condition (H7) holds. Therefore, from Theorem 1, we can obtain that the problem (34) has at least one positive mild solution. □

Author Contributions

Methodology, J.Z. and H.G.; Validation, J.Z.; Formal analysis, H.G.; Investigation, J.Z.; Writing—original draft, H.G.; Writing—review & editing, J.Z.; Funding acquisition, J.Z. and H.G. All authors have read and agreed to the published version of the manuscript.

Funding

Supported by the National Natural Science Foundation of China (12061062), Science Research Project for Colleges and Universities of Gansu Province (No. 2022A-010), Lanzhou Youth Science and Technology Talent Innovation Project (2023-QN-106), Gansu University Youth Doctoral Support Project 2024QB-068.

Data Availability Statement

My manuscript has no associate data.

Acknowledgments

This work is supported by the National Natural Science Foundation of China (12061062), Science Research Project for Colleges and Universities of Gansu Province (No. 2022A-010), Lanzhou Youth Science and Technology Talent Innovation Project (2023-QN-106).

Conflicts of Interest

The authors declare no conflicts of interest.

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Zhang, J.; Gou, H. A Study of Positive Solutions for Semilinear Fractional Measure Driven Functional Differential Equations in Banach Spaces. Mathematics 2024, 12, 2696. https://doi.org/10.3390/math12172696

AMA Style

Zhang J, Gou H. A Study of Positive Solutions for Semilinear Fractional Measure Driven Functional Differential Equations in Banach Spaces. Mathematics. 2024; 12(17):2696. https://doi.org/10.3390/math12172696

Chicago/Turabian Style

Zhang, Jing, and Haide Gou. 2024. "A Study of Positive Solutions for Semilinear Fractional Measure Driven Functional Differential Equations in Banach Spaces" Mathematics 12, no. 17: 2696. https://doi.org/10.3390/math12172696

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