In this paper, an augmented memoryless BFGS quasi-Newton method was proposed for solving unconstrained optimization problems. Based on a new modified secant equation, an augmented memoryless BFGS update formula and an efficient optimization algorithm were established. To improve the stability of the numerical experiment, we obtained the scaling parameter by minimizing the upper bound of the condition number. The global convergence of the algorithm was proved, and numerical experiments showed that the algorithm was efficient.
Citation: Yulin Cheng, Jing Gao. An efficient augmented memoryless quasi-Newton method for solving large-scale unconstrained optimization problems[J]. AIMS Mathematics, 2024, 9(9): 25232-25252. doi: 10.3934/math.20241231
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In this paper, an augmented memoryless BFGS quasi-Newton method was proposed for solving unconstrained optimization problems. Based on a new modified secant equation, an augmented memoryless BFGS update formula and an efficient optimization algorithm were established. To improve the stability of the numerical experiment, we obtained the scaling parameter by minimizing the upper bound of the condition number. The global convergence of the algorithm was proved, and numerical experiments showed that the algorithm was efficient.
The models of the differential and integral equations have been appeared in different applications (see [1,2,3,4,6,7,9,12,13,14,15,16,17,18,19,20]).
Boundary value problems involving fractional differential equations arise in physical sciences and applied mathematics. In some of these problems, subsidiary conditions are imposed locally. In some other cases, nonlocal conditions are imposed. It is sometimes better to impose nonlocal conditions since the measurements needed by a nonlocal condition may be more precise than the measurement given by a local condition. Consequently, a variety of excellent results on fractional boundary value problems (abbreviated BVPs) with resonant conditions have been achieved. For instance, Bai [4] studied a type of fractional differential equations with m-points boundary conditions. The existence of nontrivial solutions was established by using coincidence degree theory. Applying the same method, Kosmatov [17] investigated the fractional order three points BVP with resonant case.
Although the study of fractional BVPs at resonance has acquired fruitful achievements, it should be noted that such problems with Riemann-Stieltjes integrals are very scarce, so it is worthy of further explorations. Riemann-Stieltjes integral has been considered as both multipoint and integral in a single framework, which is more common, see the relevant works due to Ahmad et al. [1].
The boundary value problems with nonlocal, integral and infinite points boundary conditions have been studied by some authors (see, for example [8,10,11,12]).
Here, we discuss the boundary value problem of the nonlinear differential inclusions of arbitrary (fractional) orders
dxdt∈F1(t,x(t),Iγf2(t,Dαx(t))),α,γ∈(0,1]t∈(0,1) | (1.1) |
with the nonlocal boundary condition
m∑k=1akx(τk)=x0,ak>0τk∈[0,1], | (1.2) |
the integral condition
∫10x(s)dg(s)=x0 | (1.3) |
and the infinite point boundary condition
∞∑k=1akx(τk)=x0,ak>0andτk∈[0,1]. | (1.4) |
We study the existence of solutions x∈C[0,1] of the problems (1.1) and (1.2), and deduce the existence of solutions of the problem of (1.1) with the conditions (1.3) and (1.4). Then the existence of the maximal and minimal solutions will be proved. The sufficient condition for the uniqueness and continuous dependence of the solution will be studied.
This paper is organised as: In Section 2, we prove the existence of continuous solutions of the problems (1.1) and (1.2), and deduce the existence of solutions of the problem of (1.1) with the conditions (1.3) and (1.4). In Section 3, the existence of the maximal and minimal solutions is proved. In Section 4, the sufficient condition for the uniqueness and continuous dependence of the solution are studied. Next, in Section 5, we extend our results to the nonlocal problems (1.3) and (2.1). Finally, some existence results is proved for the nonlocal problems (1.4) and (2.1) in Section 6.
Consider the following assumptions:
(I) (i) The set F1(t,x,y) is nonempty, closed and convex for all (t,x,y)∈[0,1]×R×R.
(ii) F1(t,x,y) is measurable in t∈[0,1] for every x,y∈R.
(iii) F1(t,x,y) is upper semicontinuous in x and y for every t∈[0,1].
(iv) There exist a bounded measurable function a1:[0,1]⟶R and a positive constant K1, such that
‖F1(t,x,y)‖=sup{|f1|:f1∈F1(t,x,y)}≤|a1(t)|+K1(|x|+|y|). |
Remark 2.1. From the assumptions (i)–(iv) we can deduce that (see [3,6,7,13]) there exists f1∈F1(t,x,y), such that
(v) f1:[0,1]×R×R⟶R is measurable in t for every x,y∈R and continuous in x,y for t∈[0,1], and there exist a bounded measurable function a1:[0,1]→R and a positive constant K1>0 such that
|f1(t,x,y)|≤|a1(t)|+K1(|x|+|y|), |
and the functional f1 satisfies the differential equation
dxdt=f1(t,x(t),Iγf2(t,Dαx(t))),α,γ∈(0,1]andt∈(0,1]. | (2.1) |
(II) f2:[0,1]×R⟶R is measurable in t for any x∈R and continuous in x for t∈[0,1], and there exist a bounded measurable function a2:[0,1]→R and a positive constant K2>0 such that
|f2(t,x)|≤|a2(t)|+K2|x|,∀t∈[0,1]andx∈R |
and
supt∈[0,1]|ai(t)|≤ai,i=1,2. |
(III) 2K1γ+K1K2α<αγΓ(2−α),α,γ∈(0,1].
Remark 2.2. From (I) and (v) we can deduce that every solution of (1.1) is a solution of (2.1). Now, we shall prove the following lemma.
Lemma 2.1. If the solution of the problems (1.2)–(2.1) exists then it can be expressed by the integral equation
y(t)=∫t0(t−s)−αΓ(1−α)f1(s,1m∑k=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(ϕ,y(θ))dθ)ds. | (2.2) |
Proof. Consider the boundary value problems (1.2)–(2.1) be satisfied. Operating by I1−α on both sides of (2.1) we can obtain
Dαx(t)=I1−αdxdt=I1−αf1(t,x(t),Iγf2(t,Dαx(t))). | (2.3) |
Taking
Dαx(t)=y(t), | (2.4) |
then we obtain
x(t)=x(0)+Iαy(t). | (2.5) |
Putting t=τ and multiplying (2.5) by Σmk=1ak, then we get
Σmk=1akx(τk)=Σmk=1akx(0)+Σmk=1akIαy(τk), | (2.6) |
x0=Σmk=1akx(0)+Σmk=1akIαy(τk) | (2.7) |
and
x(0)=1Σmk=1ak[x0−Σmk=1akIαy(τk)]. | (2.8) |
Then
x(t)=1Σmk=1ak[x0−Σmk=1akIαy(τk)]+Iαy(t). | (2.9) |
Substituting (2.8) and (2.9) in (2.5), which completes the proof.
Theorem 2.1. Let assumptions (I)–(III) be satisfied. Then the integral equation (2.2) has at least one continuous solution.
Proof. Define a set Qr as
Qr={y∈C[0,1]:‖y‖≤r}, |
r=Γ(γ+1)[a1+K1A|x0|]+k1a2αγΓ(2−α)−[2K1α+K1K2γ],(m∑k=1ak)−1=A, |
and the operator F by
Fy(t)=∫t0(t−s)−αΓ(1−α)f1(s,1m∑k=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y(θ))dθ)ds. |
For y∈Qr, then
|Fy(t)|=|∫t0(t−s)−αΓ(1−α)f1(s,1m∑k=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y(θ))dθ)ds|≤K1∫t0(t−s)−αΓ(1−α)[|x0|m∑k=1ak+m∑k=1ak∫τk0(τk−θ)α−1Γ(α)|y(θ)|dθm∑k=1ak+∫s0(s−θ)α−1Γ(α)|y(θ)|dθ+∫s0(s−ϕ)γ−1Γ(γ)(K2|y(θ)|+|a2(t)|)dθ]+|a1(t)|)ds≤1Γ(2−α)[K1A|x0|+K1‖y‖Γ(α+1)+K1‖y‖Γ(α+1)+K1K2‖y‖Γ(γ+1)+K1a2Γ(γ+1)+a1]≤1Γ(2−α)[K1A|x0|+K1rΓ(α+1)+K1rΓ(α+1)+K1K2rΓ(γ+1)+K1a2Γ(γ+1)+a1]≤1Γ(2−α)[K1A|x0|+2K1rΓ(α+1)+K1K2rΓ(γ+1)+K1a2Γ(γ+1)+a1]≤r. |
Thus, the class of functions {Fy} is uniformly bounded on Qr and F:Qr→Qr. Let y∈Qr and t1,t2∈[0,1] such that |t2−t1|<δ, then
|Fy(t2)−Fy(t1)|=|∫t20(t2−s)−αΓ(1−α)(f1(s,x(s),Iγf2(s,y(s)))ds−∫t10(t1−s)−αΓ(1−α)(f1(s,x(s),Iγf2(s,y(s)))ds|≤|∫t20(t2−s)−αΓ(1−α)(f1(s,x(s),Iγf2(s,y(s)))ds−∫t10(t2−s)−αΓ(1−α)(f1(s,x(s),Iγf2(s,y(s)))ds+∫t10(t2−s)−αΓ(1−α)(f1(s,x(s),Iγf2(s,y(s)))ds−∫t10(t1−s)−αΓ(1−α)(f1(s,x(s),Iγf2(s,y(s)))ds|≤∫t2t1(t2−s)−αΓ(1−α)|(f1(s,x(s),Iγf2(s,y(s)))|ds+∫t10(t2−s)−α−(t1−s)−αΓ(1−α)|(f1(s,x(s),Iγf2(s,y(s)))ds|≤∫t2t1(t2−s)−αΓ(1−α)|(f1(s,x(s),Iγf2(s,y(s)))|ds+∫t10(t2−s)α−(t1−s)αΓ(1−α)(t2−s)α(t1−s)α.|(f1(s,x(s),Iγf2(s,y(s)))|ds. |
Thus, the class of functions {Fy} is equicontinuous on Qr and {Fy} is compact operator by the Arzela-Ascoli Theorem [5].
Now we prove that F is continuous operator. Let yn⊂Qr be convergent sequence such that yn→y, then
Fyn(t)=∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yn(θ)dθ]+∫s0(s−θ)α−1Γ(α)yn(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,yn(θ)))ds. |
Using Lebesgue dominated convergence Theorem [5] and assumptions (iv)–(II) we have
limn→∞Fyn(t)=limn→∞∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yn(θ)dθ]+∫s0(s−θ)α−1Γ(α)yn(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,yn(θ)))ds=∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)limn→∞yn(θ)dθ]+∫s0(s−θ)α−1Γ(α)limn→∞yn(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,limn→∞yn(θ)))ds=∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y(θ)))ds=Fy(t). |
Then F:Qr→Qr is continuous, and by Schauder Fixed Point Theorem[5] there exists at least one solution y∈C[0,1] of (2.2). Now
dxdt=ddt[x(0)+Iαy(t)]=ddtIαI1−αf1(t,x(t),Iγf2(t,y(t)))=ddtIf1(t,x(t),Iγf2(t,y(t)))=f1(t,x(t),Iγf2(t,y(t))). |
Putting t=τ and using (2.9), we obtain
x(t)=1Σmk=1ak[x0−Σmk=1akIαy(τk)]+Iαy(t),Σmk=1akx(τk)=Σmk=1ak1Σmk=1ak[x0−Σmk=1akIαy(τk)]+Σmk=1akIαy(τk), |
then
Σmk=1akx(τk)=x0,ak>0andτk∈[0,1]. |
This proves the equivalence between the problems (1.2)–(2.1) and the integral equation (2.2). Then there exists at least one solution y∈C[0,1] of the problems (1.2)–(2.1).
Here, we shall study the maximal and minimal solutions for the problems (1.2) and (2.1). Let y(t) be any solution of (2.2), let u(t) be a solution of (2.2), then u(t) is said to be a maximal solution of (2.2) if it satisfies the inequality
y(t)≤u(t),t∈[0,1]. |
A minimal solution can be defined by similar way by reversing the above inequality.
Lemma 3.1. Let the assumptions of Theorem 2.1 be satisfied. Assume that x(t) and y(t) are two continuous functions on [0,1] satisfying
y(t)≤∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y(θ)))dst∈[0,1],x(t)≥∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)x(θ)dθ]+∫s0(s−θ)α−1Γ(α)x(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,x(θ)))dst∈[0,1], |
where one of them is strict.
Let functions f1 and f2 be monotonic nondecreasing in y, then
y(t)<x(t),t>0. | (3.1) |
Proof. Let the conclusion (3.1) be not true, then there exists t1 with
y(t1)<x(t1),t1>0andy(t)<x(t),0<t<t1. |
Since f1 and f2 are monotonic functions in y, then we have
y(t1)≤∫t10(t1−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y(θ)))ds<∫t10(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)x(θ)dθ]+∫s0(s−θ)α−1Γ(α)x(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,x(θ)))ds<x(t1),t1∈[0,1]. |
This contradicts the fact that y(t1)=x(t1), then y(t)<x(t). This completes the proof.
For the existence of the continuous maximal and minimal solutions for (2.1), we have the following theorem.
Theorem 3.1. Let the assumptions of Theorem 2.1 be hold. Moreover, if f1 and f2 are monotonic nondecreasing functions in y for each t∈[0,1], then Eq (2.1) has maximal and minimal solutions.
Proof. First, we should demonstrate the existence of the maximal solution of (2.1). Let ϵ>0 be given. Now consider the integral equation
yϵ(t)=∫t0(t−s)−αΓ(1−α)f1,ϵ(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵ(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2,ϵ(θ,yϵ(θ))dθ)ds,t∈[0,1], |
where
f1,ϵ(s,xϵ(s),yϵ(s))=f1(s,xϵ(s),yϵ(s))+ϵ, |
f2,ϵ(s,xϵ(s))=f2(s,xϵ(s))+ϵ. |
For ϵ2>ϵ1, we have
yϵ2(t)=∫t0(t−s)−αΓ(1−α)f1,ϵ2(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵ2(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ2(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2,ϵ2(θ,yϵ2(θ))dθ)ds,t∈[0,1], |
yϵ2(t)=∫t0(t−s)−αΓ(1−α)[f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵ2(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ2(θ)dθ,∫s0(s−θ)γ−1Γ(γ)(f2(θ,yϵ2(θ))+ϵ2)dθ)+ϵ2]ds,t∈[0,1]. |
Also
yϵ1(t)=∫t0(t−s)−αΓ(1−α)f1,ϵ1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵ1(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ1(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2,ϵ1(θ,yϵ1(θ))dθ)ds,yϵ1(t)=∫t0(t−s)−αΓ(1−α)[f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−ϕ)α−1Γ(α)yϵ1(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ1(θ)dθ,∫s0(s−θ)γ−1Γ(γ)(f2(θ,yϵ1(θ))+ϵ1)dθ)+ϵ1]ds>∫t0(t−s)−αΓ(1−α)[f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵ2(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ2(θ)dθ,∫s0(s−θ)γ−1Γ(γ)(f2(θ,yϵ2(τ))+ϵ2)dθ)+ϵ2]ds. |
Applying Lemma 3.1, we obtain
yϵ2<yϵ1,t∈[0,1]. |
As shown before, the family of function yϵ(t) is equi-continuous and uniformly bounded, then by Arzela Theorem, there exist decreasing sequence ϵn, such that ϵn→0 as n→∞, and u(t)=limn→∞yϵn(t) exists uniformly in [0,1] and denote this limit by u(t). From the continuity of the functions, f2,ϵn(t,yϵn(t)), we get f2,ϵn(t,yϵn(t))⟶f2(t,y(t)) as n→∞ and
u(t)=limn→∞yϵn(t)=∫t0(t−s)−αΓ(1−α)[f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵn(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵn(θ)dθ,∫s0(s−θ)γ−1Γ(γ)(f2(θ,yϵn(θ))+ϵn))+ϵn]ds,t∈[0,1]. |
Now we prove that u(t) is the maximal solution of (2.1). To do this, let y(t) be any solution of (2.1), then
y(t)=∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y(θ))dθ)ds,yϵ(t)=∫t0(t−s)−αΓ(1−α)[f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵ(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ(θ)dθ,∫s0(s−θ)γ−1Γ(γ)(f2(θ,yϵ(θ))+ϵ)dθ)+ϵ]ds |
and
yϵ(t)>∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)yϵ(θ)dθ]+∫s0(s−θ)α−1Γ(α)yϵ(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,yϵ(θ)))ds. |
Applying Lemma 3.1, we obtain
y(t)<yϵ(t),t∈[0,1]. |
From the uniqueness of the maximal solution it clear that yϵ(t) tends to u(t) uniformly in [0,1] as ϵ→0.
By a similar way as done above, we can prove the existence of the minimal solution.
Here, we study the sufficient condition for the uniqueness of the solution y∈C[0,1] of problems (1.2) and (2.1). Consider the following assumptions:
(I∗) (i) The set F1(t,x,y) is nonempty, closed and convex for all (t,x,y)∈[0,1]×R×R.
(ii) F1(t,x,y) is measurable in t∈[0,1] for every x,y∈R.
(iii) F1 satisfies the Lipschitz condition with a positive constant K1 such that
H(F1(t,x1,y1),F1(t,x2,y2)|≤K1(|x1−x2|+|y1−y2|), |
where H(A,B) is the Hausdorff metric between the two subsets A,B∈[0,1]×E.
Remark 4.1. From this assumptions we can deduce that there exists a function f1∈F1(t,x,y), such that
(iv) f1:[0,1]×R×R→R is measurable in t∈[0,1] for every x,y∈R and satisfies Lipschitz condition with a positive constant K1 such that (see [3,7])
|f1(t,x1,y1)−f1(t,x2,y2)|≤K1(|x1−x2|+|y1−y2|). |
(II∗)f2:[0,1]×R⟶R is measurable in t∈[0,T] and satisfies Lipschitz condition with positive constant K2, such that
|f2(t,x)−f2(t,y)|≤K2|x−y|. |
From the assumption (I∗), we have
|f1(t,x,y)|−|f1(t,0,0)|≤|f1(t,x,y)−f1(t,0,0)|≤K1(|x|+|y|). |
Then
|f1(t,x,y)|≤K1(|x|+|y|)+|f1(t,0,0)|≤K1(|x|+|y|)+∣a1(t)∣, |
where |a1(t)|=supt∈I|f1(t,0,0)|.
From the assumption (II∗), we have
|f2(t,y)|−|f2(t,0)|≤|f2(t,y)−f2(t,0)|≤K2|y|. |
Then
|f2(t,x)|≤K2|x|+|f2(t,0)|≤K2|x|+∣a2(t)∣, |
where |a2(t)|=supt∈I|f1(t,0)|.
Theorem 4.1. Let the assumptions (I∗) and (II∗) be satisfied. Then the solution of the problems (1.2) and (2.1) is unique.
Proof. Let y1(t) and y2(t) be solutions of the problems (1.2) and (2.1), then
![]() |
Then
‖y1−y2‖[αγΓ(2−α)−(2K1α+K1K2γ)]<0. |
Since (αγΓ(2−α)−(2K1α+K1K2γ))<1, then y1(t)=y2(t) and the solution of (1.2) and (2.1) is unique.
Definition 4.1. The unique solution of the problems (1.2) and (2.1) depends continuously on initial data x0, if ϵ>0, ∃δ>0, such that
|x0−x∗0|≤δ⇒‖y−y∗‖≤ϵ, |
where y∗ is the unique solution of the integral equation
y∗(t)=∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x∗0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y∗(θ)dθ]+∫s0(s−θ)α−1Γ(α)y∗(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y∗(θ))dθ)ds|. |
Theorem 4.2. Let the assumptions (I∗) and (II∗) be satisfied, then the unique solution of (1.2) and (2.1) depends continuously on x0.
Proof. Let y(t) and y∗(t) be the solutions of problems (1.2) and (2.1), then
![]() |
then we obtain
|y(t)−y∗(t)|≤(AK1δ)(αγΓ(2−α)−(2K1α+K1K2γ))−1≤ϵ |
and
‖y(t)−y∗(t)‖≤ϵ. |
Definition 4.2. The unique solution of the problems (1.2) and (2.1) depends continuously on initial data ak, if ϵ>0, ∃δ>0, such that
m∑k=1|ak−a∗k|≤δ⇒‖y−y∗‖≤ϵ, |
where y∗ is the unique solution of the integral equation
y∗(t)=∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1a∗k∫τk0(τk−θ)α−1Γ(α)y∗(θ)dθ]+∫s0(s−θ)α−1Γ(α)y∗(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y∗(θ))dθ)ds|. |
Theorem 4.3. Let the assumptions (I∗) and (II∗) be satisfied, then the unique solution of problems (1.2) and (2.1) depends continuously on ak.
Proof. Let y(t) and y∗(t) be the solutions of problems (1.2) and (2.1) and (∑mk=1a∗k)−1=A∗,
|y(t)−y∗(t)|=|∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1ak[x0−m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y(θ))dθ)ds−∫t0(t−s)−αΓ(1−α)f1(s,1∑mk=1a∗k[x0−m∑k=1a∗k∫τk0(τk−θ)α−1Γ(α)y∗(θ)dθ]+∫s0(s−θ)α−1Γ(α)y∗(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(θ,y∗(θ))dθ)ds|≤∫t0(t−s)−αΓ(1−α)[|x0|(∑mk=1a∗k−∑mk=1ak)∑mk=1ak∑mk=1a∗k+∑mk=1ak∫τk0(τk−θ)α−1Γ(α)|y(θ)|dθ∑mk=1ak−∑mk=1a∗k∫τk0(τk−θ)α−1Γ(α)|y∗(θ)|dθ∑mk=1a∗k+K1∫s0(s−θ)α−1Γ(α)|y(θ)−y∗(θ)|dθ+K1K2∫s0(s−θ)γ−1Γ(γ)|y(ϕ)−y∗(θ)|dθ]ds≤1Γ(2−α)[K1|x0|∑mk=1|a∗k−ak|∑mk=1ak∑mk=1a∗k+K1∑mk=1ak(∑mk=1a∗k∫τk0(τk−θ)α−1Γ(α)|y(θ)|dθ−∑mk=1ak∫τk0(τk−θ)α−1Γ(α)|y(θ)|dθ)∑mk=1ak∑mk=1a∗k+K1∑mk=1a∗k(∑mk=1ak∫τk0(τk−θ)α−1Γ(α)|y(θ)|dθ−∑mk=1a∗k∫τk0(τk−θ)α−1Γ(α)|y(θ)|dθ)∑mk=1ak∑mk=1a∗k+K1∫s0(s−θ)α−1Γ(α)|y(θ)−y∗(θ)|dθ+K1K2∫s0(s−θ)γ−1Γ(γ)|y(ϕ)−y∗(θ)|dθ]ds≤1Γ(2−α)[K1|x0|δ∑mk=1ak∑mk=1a∗k+K1(∑mk=1akδrΓ(α+1)+∑mk=1a∗kδrΓ(α+1))∑mk=1ak∑mk=1a∗k+K1‖y−y∗‖Γ(α+1)+K1K2‖y−y∗‖Γ(γ+1)≤K1δ|x0|Γ(2−α)∑mk=1ak∑mk=1a∗k+K1(∑mk=1akδrΓ(α+1)+∑mk=1a∗kδrΓ(α+1))Γ(2−α)Γ(α+1)∑mk=1ak∑mk=1a∗k+K1‖y−y∗‖Γ(2−α)Γ(α+1)+K1K2‖y−y∗‖Γ(γ+1)Γ(2−α), |
then we obtain
|y(t)−y∗(t)|≤(AA∗K1δ[|x0|+r(m∑k=1ak+m∑k=1a∗k)]).(αγΓ(2−α)−(2K1α+K1K2γ))−1≤ϵ |
and
‖y(t)−y∗(t)‖≤ϵ. |
Let y∈C[0,1] be the solution of the nonlocal boundary value problems (1.2) and (2.1). Let ak=(g(tk)−g(tk−1), g is increasing function, τk∈(tk−1−tk), 0=t0<t1<t2,...<tm=1, then, as m⟶∞ the nonlocal condition (1.2) will be
m∑k=1(g(tk)−g(tk−1)y(τk)=x0. |
As the limit m⟶∞, we obtain
limm→∞m∑k=1(g(tk)−g(tk−1)y(τk)=∫10y(s)dg(s)=x0. |
Theorem 5.1. Let the assumptions (I)–(III) be satisfied. If ∑mk=1ak be convergent, then the nonlocal boundary value problems of (1.3) and (2.1) have at least one solution given by
y(t)=∫t0(t−s)−αΓ(1−α)f1(s,1g(1)−g(0)[x0−∫10∫s0(s−θ)α−1Γ(α)y(θ)dθdg(s)]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(ϕ,y(θ))dθ)ds. |
Proof. As m→∞, the solution of the nonlocal boundary value problem (2.1) will be
limm→∞y(t)=∫t0(t−s)−αΓ(1−α)f1(s,(g(1)−g(0))−1x0−(g(1)−g(0))−1.limm→∞m∑k=1[g(tk)−g(tk−1]∫τk0(τk−θ)α−1Γ(α)y(θ)dθ+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(ϕ,y(θ))dθ)ds=∫t0(t−s)−αΓ(1−α)f1(s,1g(1)−g(0)[x0−∫10∫s0(s−θ)α−1Γ(α)y(θ)dθdg(s)]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(ϕ,y(θ))dθ)ds. |
Theorem 6.1. Let the assumptions (I)–(III) be satisfied, then the nonlocal boundary value problems of (1.4) and (2.1) have at least one solution given by
y(t)=∫t0(t−s)−αΓ(1−α)f1(s,1∑∞k=1ak[x0−∞∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(ϕ,y(θ))dθ)ds. |
Proof. Let the assumptions of Theorem 2.1 be satisfied. Let ∑mk=1ak be convergent, then take the limit to (1.4), we have
limm→∞y(t)=∫t0(t−s)−αΓ(1−α)f1(s,limm→m1∑mk=1ak[x0−limm→∞m∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(ϕ,y(θ))dθ)ds. |
Now
|ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ|≤|ak|∫τk0(τk−θ)α−1Γ(α)|y(θ)|dθ≤|ak|‖y‖Γ(α+1)≤|ak|rΓ(α+1) |
and by the comparison test (∑mk=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ) is convergent,
y(t)=∫t0(t−s)−αΓ(1−α)f1(s,1∑∞k=1ak[x0−∞∑k=1ak∫τk0(τk−θ)α−1Γ(α)y(θ)dθ]+∫s0(s−θ)α−1Γ(α)y(θ)dθ,∫s0(s−θ)γ−1Γ(γ)f2(ϕ,y(θ))dθ)ds. |
Furthermore, from (2.9) we have
∞∑k=1akx(τk)=∞∑k=1ak[1∑∞k=1ak(x0−∞∑k=1ak∫τk0(τk−s)α−1Γ(α)y(s)ds)]+∫τk0(τk−s)α−1Γ(α)y(s)ds)]=x0. |
Example 6.1. Consider the following nonlinear integro-differential equation
dxdt=t4e−t+x(t)√t+2+13Iγ(cos(5t+1)+19[t5sinD13x(t)+e−3tx(t)]) | (6.1) |
with boundary condition
m∑k=1[1k−1k+1]x(τk)=x0,ak>0τk∈[0,1]. | (6.2) |
Let
f1(t,x(t),Iγf2(t,Dαx(t)))=t4e−t+x(t)√t+2+13Iγ(cos(5t+1)+19[t5sinD13x(t)+e−3tx(t)]), |
then
|f1(t,x(t),Iγf2(t,Dαx(t)))=|t4e−t+x(t)√2t+4+14(43Iγ(cos(5t+1)+19[t5sinD13x(t)+e−3tx(t)]))| |
and also
|Iγf2(t,Dαx(t))|≤14(43Iγ|cos(5t+1)+13[t5sinD13x(t)+e−3tx(t)]|. |
It is clear that the assumptions (I) and (II) of Theorem 2.1 are satisfied with a1(t)=t4e−t∈L1[0,1], a2(t)=43Iγ|(cos(5t+1)|∈L1[0,1], and let α=13, γ=23, then 2K1γ+K1K2α=12<αγΓ(2−α)=12Γ(2−α). Therefore, by applying Theorem 2.1, the nonlocal problems (6.1) and (6.2) has a continuous solution.
In this paper, we have studied a boundary value problem of fractional order differential inclusion with nonlocal, integral and infinite points boundary conditions. We have prove some existence results for that a single nonlocal boundary value problem, in of proving some existence results for a boundary value problem of fractional order differential inclusion with nonlocal, integral and infinite points boundary conditions. Next, we have proved the existence of maximal and minimal solutions. Then we have established the sufficient conditions for the uniqueness of solutions and continuous dependence of solution on some initial data and on the coefficients ak are studied. Finally, we have proved the existence of a nonlocal boundary value problem with Riemann-Stieltjes integral condition and with infinite-point boundary condition. An example is given to illustrate our results.
The authors declare no conflict of interest.
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