Solutions of a Class of Deviated-Advanced Nonlocal Problems for the Differential Inclusion 𝑥¹(𝑡)∈𝐹(𝑡,𝑥(𝑡))

Abstract

We study the existence of solutions for deviated-advanced nonlocal and integral condition problems for the differential inclusion 𝑥¹(𝑡)∈𝐹(𝑡,𝑥(𝑡)).

1. Introduction

Problems with nonlocal conditions have been extensively studied by several authors in the last two decades. The reader is referred to [1–12] and references therein. Consider the deviated-advanced nonlocal problem 𝑑𝑥(𝑡)𝑑𝑡∈𝐹(𝑡,𝑥(𝑡)),a.e.𝑡∈(0,1),(1.1)_𝑚_𝑘=1𝑎_𝑘𝑥𝜙𝜏_𝑘=𝛼^𝑛_𝑗=1𝑏_𝑗𝑥𝜓𝜂_𝑗,𝑎_𝑘,𝑏_𝑗>0,(1.2) where 𝜏_𝑘,𝜂_𝑗∈(0,1), 𝛼>0 is a parameter, and 𝜓 and 𝜙 are, respectively, deviated and advanced given functions.

Our aim here is to study the existence of at least one absolutely continuous solution 𝑥∈AC[0,1] for the problem (1.1)-(1.2) when the set-valued function 𝐹∶𝑅→𝑃(𝑅) is 𝐿¹-Carathéodory.

As an application, we deduce the existence of a solution for the nonlocal problem of the differential inclusion (1.1) with the deviated-advanced integral condition ¹⁰𝑥(𝜙(𝑠))𝑑𝑠=𝛼¹⁰𝑥(𝜓(𝑠))𝑑𝑠.(1.3)

It must be noticed that the following nonlocal and integral conditions are special cases of our nonlocal and integral conditions 𝑥(𝜙(𝜏))=𝛼𝑥(𝜓(𝜂)),𝜏,𝜂∈(0,1),_𝑚_𝑘=1𝑎_𝑘𝑥𝜙𝜏_𝑘=𝛼𝑥(𝜓(𝜂)),𝜏_𝑘,𝜂∈(0,1),_𝑚_𝑘=1𝑎_𝑘𝑥𝜙𝜏_𝑘=0,𝜏_𝑘∈(0,1),₁₀𝛼𝑥(𝜙(𝑠))𝑑𝑠=𝛼𝑥(𝜓(𝜂)),𝜂∈(0,1),₁₀𝑥(𝜓(𝑠))𝑑𝑠=𝑥(𝜙(𝜏)),𝜏∈(0,1),₁₀𝑥(𝜙(𝑠))𝑑𝑠=0,₁₀𝑥(𝜓(𝑠))𝑑𝑠=0.(1.4) As an example of the deviated function 𝜙∶(0,1)→(0,1), we have 𝜙(𝑡)=𝛽𝑡,𝛽∈(0,1). As an example of the advanced function 𝜓∶(0,1)→(0,1), we have 𝜓(𝑡)=𝑡^𝛽,𝛽∈(0,1).

2. Preliminaries

The following preliminaries are needed.

Definition 2.1. A set-valued function 𝐹∶[0,1]×𝑅→𝑃(𝑅) is called 𝐿¹-Carathéodory if(a)𝑡→𝐹(𝑡,𝑥) is measurable for each 𝑥∈𝑅,(b)𝑥→𝐹(𝑡,𝑥) is upper semicontinuous for almost all 𝑡∈[0,1],(c)there exists 𝑚∈𝐿¹([0,1],𝐷),𝐷⊂𝑅 such that ||𝐹(𝑡,𝑥)||=sup{|𝑣|∶𝑣∈𝐹(𝑡,𝑥)}≤𝑚(𝑡),foralmostall𝑡∈[0,1].(2.1)

Definition 2.2. A single-valued function 𝑓∶[0,1]×𝑅→𝑅  is called 𝐿¹-Carathéodory if(i)𝑡→𝑓(𝑡,𝑥) is measurable for each 𝑥∈𝑅,(ii)𝑥→𝑓(𝑡,𝑥) is continuous for almost all 𝑡∈[0,1],(iii)there exists 𝑚∈𝐿¹([0,1],𝐷),𝐷⊂𝑅 such that |𝑓|≤𝑚.

Definition 2.3. The set 𝑆^{1𝐹(⋅,𝑥(𝑡))}={𝑓∈([0,1],𝑅)∶𝑓(𝑡,𝑥)∈𝐹(𝑡,𝑥(𝑡))fora.e.𝑡∈[0,1]}(2.2) is called the set of selections of the set-valued function 𝐹.

Theorem 2.4. For any 𝐿¹-Carathéodory set-valued function 𝐹, the set 𝑆^{1𝐹(⋅,𝑥(𝑡))} is nonempty [1, 13].

Theorem 2.5 (Carathéodory, [14]). Let 𝑓∶[0,1]×𝑅→𝑅 be 𝐿¹-Carathéodory. Then the problem 𝑑𝑥(𝑡)𝑑𝑡=𝑓(𝑡,𝑥(𝑡)),fora.e.𝑡>0,𝑥(0)=𝑥₀,(2.3) has at least one solution 𝑥∈AC[0,𝑇].

3. Existence of Solution

Consider the following assumptions. (i)𝐹∶[0,1]×𝑅→𝑃(𝑅⁺) is  𝐿¹-Carathéodory.(ii)𝛼^𝑛_𝑗=1𝑏_𝑗≠^𝑚_𝑘=1𝑎_𝑘.(3.1)(iii)𝜙∶(0,1)→(0,1),𝜙(𝑡)≤𝑡 is a deviated continuous function.(iv)𝜓∶(0,1)→(0,1),𝜓(𝑡)≥𝑡 is an advanced continuous function.

Now we have the following lemma.

Lemma 3.1. Let assumptions (i)-(ii) be satisfied. The solution of the nonlocal problem (1.1)-(1.2) can be expressed by the integral equation 𝑥(𝑡)=𝐴^𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠+^𝑡0𝑓(𝑠,𝑥(𝑠))𝑑𝑠,(3.2) where 𝑓(𝑡,𝑥)∈𝐹(𝑡,𝑥),forall𝑥∈𝑅, and 𝐴=(𝛼∑^𝑛𝑗=1𝑏_𝑗−∑^𝑚𝑘=1𝑎_𝑘)⁻¹.

Proof. From the assumption that the set-valued function 𝐹∶[0,1]×𝑅→𝑃(𝑅⁺) is 𝐿¹-Carathéodory, then (Theorem 2.4) there exists a single-valued selection 𝑓∶[0,1]×𝑅→𝑅⁺ such that 𝑑𝑑𝑡𝑥(𝑡)=𝑓(𝑡,𝑥)∈𝐹(𝑡,𝑥),∀𝑥∈𝑅.(3.3) This selection 𝑓(𝑡,𝑥) is 𝐿¹-Carathéodory.
Integrating (3.3), we get 𝑥(𝑡)=𝑥(0)+^𝑡0𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.4) Let 𝑡=𝜙(𝜏_𝑘). Then ^𝑚_𝑘=1𝑎_𝑘𝑥𝜙𝜏_𝑘=^𝑚_𝑘=1𝑎_𝑘𝑥(0)+^𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.5) Let 𝑡=𝜓(𝜂_𝑗). Then 𝛼^𝑛_𝑗=1𝑏_𝑗𝑥𝜓𝜂_𝑗=𝛼^𝑛_𝑗=1𝑏_𝑗𝑥(0)+𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.6) From (3.5) and (3.6), we obtain 𝑥(0)=𝐴^𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠,(3.7) where 𝐴=(𝛼∑^𝑛𝑗=1𝑏_𝑗−∑^𝑚𝑘=1𝑎_𝑘)⁻¹, 𝛼∑^𝑛𝑗=1𝑏_𝑗≠∑^𝑚𝑘=1𝑎_𝑘.
Substituting (3.7) into (3.4), we obtain 𝑥(𝑡)=𝐴^𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠+^𝑡0𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.8) This proves that the solution of the nonlocal problem (1.1)-(1.2) can be expressed by the integral equation (3.2).

For the existence of the solution, we have the following theorem.

Theorem 3.2. Assume that (i)–(iv) are satisfied. Then the integral equation (3.2) has at least one continuous solution 𝑥∈𝐶[0,1].

Proof. Define a subset 𝑄_𝑟⊂𝐶[0,1] by 𝑄_𝑟=𝑥∈𝐶[0,1]∶||𝑥(𝑡)||≤𝑟,𝑟=𝐴𝑀1+^𝑚_𝑘=1𝑎_𝑘+𝛼^𝑛_𝑗=1𝑏_𝑗.(3.9) Clearly, the set 𝑄_𝑟 is nonempty, closed, and convex.
Let 𝐻 be an operator defined by (𝐻𝑥)(𝑡)=𝐴^𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠+^𝑡0𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.10) Let 𝑥∈𝑄_𝑟. Let {𝑥_𝑛(𝑡)} be a sequence in 𝑄_𝑟 converging to 𝑥(𝑡),𝑥_𝑛(𝑡)→𝑥(𝑡),forall𝑡∈𝐼. Then lim_𝑛→∞𝐻𝑥_𝑛(𝑡)=𝐴^𝑚_𝑘=1𝑎_𝑘lim_𝑛→∞^{𝜙(𝜏𝑘)0}𝑓𝑠,𝑥_𝑛(𝑠)𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗lim_𝑛→∞^{𝜓(𝜂𝑗)0}+lim_𝑛→∞𝑓𝑠,𝑥_𝑛(𝑠)𝑑𝑠^𝑡0𝑓𝑠,𝑥_𝑛(𝑠)𝑑𝑠,(3.11) By assumptions (i)-(ii) and the Lebesgue dominated convergence theorem, we deduce that lim_𝑛→∞𝐻𝑥_𝑛(𝑡)=(𝐻𝑥)(𝑡).(3.12) Then 𝐻 is continuous.
Now, letting𝑥∈𝑄_𝑟, (then𝜙(𝑡)≤𝑡 and 𝜓(𝑡)≥𝑡), we obtain (𝐻𝑥)(𝑡)≤𝐴^𝑚_𝑘=1𝑎_𝑘^𝜏𝑘₀𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^𝜂𝑗₀+𝑓(𝑠,𝑥(𝑠))𝑑𝑠_𝑡0||(𝐻𝑥)(𝑡)||≤𝐴𝑓_𝑚(𝑠,𝑥(𝑠))𝑑𝑠,_𝑘=1𝑎_𝑘^𝜏𝑘₀||𝑓(𝑠,𝑥(𝑠))||𝑑𝑠+𝛼^𝑛_𝑗=1𝑏_𝑗^𝜂𝑗₀+||𝑓(𝑠,𝑥(𝑠))||𝑑𝑠_𝑡0≤𝐴||𝑓_𝑚(𝑠,𝑥(𝑠))||𝑑𝑠_𝑘=1𝑎_𝑘^𝜏𝑘₀𝑚(𝑠)𝑑𝑠+𝛼^𝑛_𝑗=1𝑏_𝑗^𝜂𝑗₀≤𝐴_𝑚𝑚(𝑠)𝑑𝑠+^𝑡0𝑚(𝑠)𝑑𝑠_𝑘=1𝑎_𝑘𝑀+𝛼^𝑛_𝑗=1≤𝐴𝑀1+_𝑚𝑏_𝑗𝑀+𝑀_𝑘=1𝑎_𝑘+𝛼^𝑛_𝑗=1𝑏_𝑗≤𝑟.(3.13) Then {𝐻𝑥(𝑡)} is uniformly bounded in 𝑄_𝑟.
Also for 𝑡₁,𝑡₂∈(0,1),𝑡₁<𝑡₂ such that |𝑡₂−𝑡₁|<𝛿, we have (𝐻𝑥)𝑡₂||(𝐻−𝑥(𝐻)𝑥𝑡₂)𝑡₁=^𝑡2₀𝑓(𝑠,𝑥(𝑠))𝑑𝑠−^𝑡1₀𝑓(𝑠,𝑥(𝑠))𝑑𝑠,−(𝐻𝑥)𝑡₁≤||≤^𝑡2_𝑡1||𝑓(𝑠,𝑥(𝑠))||𝑑𝑠_𝑡2𝑡1||(𝐻𝑥)𝑡₂𝑚(𝑠)𝑑𝑠,−(𝐻𝑥)𝑡₁||≤𝜀.(3.14) Hence the class of functions {𝐻𝑥(𝑡)} is equicontinuous. By Arzela-Ascoli's theorem, {𝐻𝑥(𝑡)} is relatively compact. Since all conditions of Schauder's theorem hold, then 𝐻 has a fixed point in 𝑄_𝑟.
Therefore the integral equation (3.2) has at least one continuous solution 𝑥∈𝐶(0,1). Now, lim_𝑡→0𝑥(𝑡)=𝐴lim_𝑡→0^𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}+lim_𝑡→0𝑓(𝑠,𝑥(𝑠))𝑑𝑠^𝑡0=𝐴_𝑚𝑓(𝑠,𝑥(𝑠))𝑑𝑠_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠=𝑥(0).(3.15) Also 𝑥(1)=lim_𝑡→1𝑥(𝑡)=𝐴^𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼^𝑛_𝑗=1𝑏_𝑗^{𝜙(𝜂𝑗)0}+𝑓(𝑠,𝑥(𝑠))𝑑𝑠₁₀𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.16) Then the integral equation (3.2) has at least one continuous solution 𝑥∈𝐶[0,1].

The following theorem proves the existence of at least one solution for the nonlocal problem(1.1)-(1.2).

Theorem 3.3. Let (i)–(iv) be satisfied. Then the nonlocal problem (1.1)-(1.2) has at least one solution 𝑥∈AC[0,1].

Proof. From Theorem 3.2 and the integral equation (3.2), we deduce that there exists at least one solution, 𝑥∈AC[0,1], of the integral equation (3.2).
To complete the proof, we prove that the integral equation (3.2) satisfies nonlocal problem (1.1)-(1.2).
Differentiating (3.2), we get 𝑑𝑥𝑑𝑡=𝑓(𝑡,𝑥(𝑡))∈𝐹(𝑡,𝑥(𝑡)),a.e.𝑡∈(0,1).(3.17) Letting 𝑡=𝜙(𝜏_𝑘) in (3.2), we obtain ^𝑚_𝑘=1𝑎_𝑘𝑥𝜙𝜏_𝑘=^𝑚_𝑘=1𝑎_𝑘𝐴^𝑚_𝑘=1𝑎_𝑘+1^{𝜙(𝜏𝑘)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠−𝛼𝐴^𝑚_𝑘=1𝑎_𝑘𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.18) Also, letting 𝑡=𝜓(𝜂_𝑗) in (3.2), we obtain 𝛼^𝑛_𝑗=1𝑏_𝑗𝑥𝜓𝜂_𝑗=𝛼𝐴^𝑛_𝑗=1𝑏_𝑗𝑚_𝑘=1𝑎_𝑘^{𝜙(𝜏𝑘)0}+𝛼_𝑛𝑓(𝑠,𝑥(𝑠))𝑑𝑠_𝑗=1𝑏_𝑗1−𝛼𝐴^𝑛_𝑗=1𝑏_𝑗^{𝜓(𝜂𝑗)0}𝑓(𝑠,𝑥(𝑠))𝑑𝑠.(3.19) And from (3.19) from (3.18), we obtain ^𝑚_𝑘=1𝑎_𝑘𝑥𝜙𝜏_𝑘=𝛼^𝑛_𝑗=1𝑏_𝑗𝑥𝜓𝜂_𝑗.(3.20) This complete the proof of the equivalence between the nonlocal problem (1.1)-(1.2) and the integral equation (3.2).
This implies that there exists at least one absolutely continuous solution 𝑥∈AC[0,1] of the nonlocal problem (1.1)-(1.2).

4. Nonlocal Integral Condition

Let 𝑥∈[0,1] be a solution of the nonlocal problem (1.1)-(1.2). Let 𝑎_𝑘=𝑡_𝑘−𝑡_𝑘−1,𝜏_𝑘∈(𝑡_𝑘−1,𝑡_𝑘)⊂(0,1). Also, let 𝑏_𝑗=𝑡_𝑗−𝑡_𝑗−1,𝜂_𝑗∈(𝑡_𝑗−1,𝑡_𝑗)⊂(0,1). Then the nonlocal condition (1.2) will be ^𝑚_𝑘=1𝑡_𝑘−𝑡_𝑘−1𝑥𝜙𝜏_𝑘=𝛼^𝑛_𝑗=1𝑡_𝑗−𝑡_𝑗−1𝑥𝜓𝜂_𝑗.(4.1) From the continuity of the solution 𝑥 of the nonlocal condition (1.2) we obtain lim_{𝑚𝑚→∞}_𝑘=1𝑡_𝑘−𝑡_𝑘−1𝑥𝜙𝜏_𝑘=lim_𝑛→∞𝛼^𝑛_𝑗=1𝑡_𝑗−𝑡_𝑗−1𝑥𝜓𝜂_𝑗.(4.2) That is, the nonlocal condition (1.2) is transformed to the integral condition ¹⁰𝑥(𝜙(𝑠))𝑑𝑠=𝛼¹⁰𝑥(𝜓(𝑠))𝑑𝑠,(4.3) and the solution of the integral equation (3.2) will be 𝑥(𝑡)=𝐴^∗¹⁰^0𝜙(𝑠)𝑓(𝜃,𝑥(𝜃))𝑑𝜃𝑑𝑠−𝛼¹⁰^0𝜓(𝑠)+𝑓(𝜃,𝑥(𝜃))𝑑𝜃𝑑𝑠_𝑡0𝑓(𝑠,𝑥(𝑠))𝑑𝑠,𝐴^∗=(𝛼−1)⁻¹.(4.4)

Now, we have the following theorem.

Theorem 4.1. Let assumptions (i)–(iv) of Theorem 3.2 be satisfied. Then the nonlocal problem with the integral condition 𝑑𝑥(𝑡)𝑑𝑡=𝑓(𝑡,𝑥(𝑡))∈𝐹(𝑡,𝑥(𝑡)),fora.e.𝑡∈(0,1),₁₀𝑥(𝜙(𝑠))𝑑𝑠=𝛼¹⁰𝑥(𝜓(𝑠))𝑑𝑠(4.5) has at least one solution 𝑥∈𝐴𝐶[0,1] represented by (4.4).