Citation: Djamila Chergui, Taki Eddine Oussaeif, Merad Ahcene. Existence and uniqueness of solutions for nonlinear fractional differential equations depending on lower-order derivative with non-separated type integral boundary conditions[J]. AIMS Mathematics, 2019, 4(1): 112-133. doi: 10.3934/Math.2019.1.112
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Boundary value problems for nonlinear fractional differential equations have recently been investigated by several researchers. The study of fractional equations ranges from the theoretical aspects of existence and uniqueness of solutions to the analytic and numerical methods for finding solutions. A strong motivation for studying fractional differential equations comes from the fact that they have been proved to be valuable tools in the modeling of many phenomena in engineering and sciences such as physics, mechanics, chemistry, economics and biology, etc. [4,7,10,14]
Ahmad and Ntouyas [6] investigated the existence of solutions for a fractional boundary value problem with fractional separated boundary conditions given by
{cDqx(t)=f(t,x(t)),t∈[0,T],1≤q≤2,α1x(0)+β1(cDpx(0))=γ1,α2x(1)+β2(cDpx(1))=γ2,0<p<1. |
Where
Also Xiaoyou Liu and Zhenhai Liu [17] investigated the existence and uniqueness of solutions for the nonlinear fractional boundary value problem with fractional separated boundary conditions given by :
{cDαx(t)=f(t,x(t),cDβx(t)),t∈[0,T],1≤q≤2,1<β≤1a1x(0)+b1(cDγx(0))=c1,a2x(T)+b2(cDγx(T))=c2,0<γ<1. |
Where
Bashir Ahmad, Juan J, Nieto and Ahmed Alsaedi [5] investigated the existence and uniqueness of the solutions for a new class of boundary value problems of nonlinear fractional differential equations with non-separated type integral boundary conditions. Precisely, they consider the following problem
{cDqx(t)=f(t,x(t)),t∈[0,T],T>0,1<q≤2,x(0)−λ1x(T)=μ1∫T0g(s,x(s))ds,x′(0)−λ2x′(T)=μ2∫T0h(s,x(s))ds. |
Where
In this paper, we discuss the existence and uniqueness of solutions for a new class of boundary value problems of nonlinear fractional differential equations depending with non-separated type integral boundary conditions. Precisely, we consider the following problem
{cDqx(t)=f(t,x(t),cDrx(t)),t∈[0,T],T>0,1<q≤2,0<r≤1x(0)−λ1x(T)=μ1∫T0g(s,x(s))ds,x′(0)−λ2x′(T)=μ2∫T0h(s,x(s))ds. | (1.1) |
Where
The rest of the paper is arranged as follows. In Section 2, we establish a basic result that lays the foundation for defining a fixed point problem equivalent to the given problem (1.1). The main results, based on Banach's contraction mapping principal, Schauder fixed point theorem and nonlinear alternative of Leray-Schauder type, are obtained in Section 3. Illustrating examples are discussed in Section 4.
For convenience of the reader, we present here some necessary definitions about fractional calculus theory, which can be found in [1,9,12,13].
Definition 2.1. The Riemann-Liouville fractional integral of order
Iqf(t)=1Γ(q)∫t0(t−s)q−1f(s)ds,q>0, |
provided the right-hand side is point-wise defined on
Definition 2.2. For a at least n-times continuously differentiable function
cDqf(t)=1Γ(n−q)∫t0(t−s)n−q−1f(n)(s)ds,n−1<q<n,n=[q]+1, |
where
Lemma 2.3. Let
cDαh(t)=0 |
has solutions
IαcDαh(t)=h(t)+c0+c1t+c2t2+…+cn−1tn−1, |
here
Theorem 2.4. (Schauder fixed Point theorem)(see [2]) Let
Theorem 2.5. (Nonlineair alternative of Leray-Schauder type)(see [2]) Let
Let
Now we present the Green's function for boundary value problem of fractional differential equation.
Lemma 3.1. For a given
{cDqx(t)=y(t),t∈[0,T],T>0,1<q≤2,x(0)−λ1x(T)=μ1∫T0g(s,x(s))ds,x′(0)−λ2x′(T)=μ2∫T0h(s,x(s))ds, | (3.1) |
is given by :
x(t)=∫T0G(t,s)f(s,x(s),cDrx(s))ds+μ2[λ1T+(1−λ1)t](λ2−1)(λ1−1)∫T0h(s,x(s))ds−μ1λ1−1∫T0g(s,x(s))ds, |
where
G(t,s)={(t−s)q−1Γ(q)−λ1(t−s)q−1(λ1−1)Γ(q)+λ2[λ1T+(1−λ1)t](t−s)q−1(λ2−1)(λ1−1)Γ(q−1)s≤t,−λ1(t−s)q−1(λ1−1)Γ(q)+λ2[λ1T+(1−λ1)t](t−s)q−1(λ2−1)(λ1−1)Γ(q−1)t≤s, | (3.2) |
Proof. We omit the proof as it employs the standard arguments for instance, see [3].
In this section, we given some existence results for the problem (1.1) In view of Lemma 3.1 we define an operator
(Fx)(t)=∫t0(t−s)q−1Γ(q)f(s,x(s),cDrx(s))ds−ξ1λ1∫T0(T−s)q−1Γ(q)f(s,x(s),cDrx(s))ds+ξ2λ2[λ1(T−t)+t]∫T0(T−s)q−2Γ(q−1)f(s,x(s),cDrx(s))ds+ξ2μ2[λ1(T−t)+t]∫T0h(s,x(s))ds−μ1ξ1∫T0g(s,x(s))ds,t∈[0,T], | (3.3) |
ξ1=1λ1−1,ξ2=1(λ2−1)(λ1−1). |
It is clear that the problem (1.1) has solutions if and only if the operator equation
(Nx)(t)=f(t,x(t),cDrx(t)),t∈[0,T]. |
Since the function
(cDrFx)(t)=(Iq−rNx)(t)−kt1−rΓ(2−r). | (3.4) |
We know that the operator
k=λ2(λ2−1)Γ(q−1)∫T0(T−s)q−2f(s,x(s),cDrx(s))ds+μ2(λ2−1)∫T0h(s,x(s))ds. |
We put
(F1x)(t)=∫t0(t−s)q−1Γ(q)f(s,x(s),cDrx(s))ds, |
(F2x)(t)=−ξ1λ1∫T0(T−s)q−1Γ(q)f(s,x(s),cDrx(s))ds+ξ2λ2[λ1(T−t)+t]∫T0(T−s)q−2Γ(q−1)f(s,x(s),cDrx(s))ds+ξ2μ2[λ1(T−t)+t]∫T0h(s,x(s))ds−μ1ξ1∫T0g(s,x(s))ds. |
Observe that problem (1.1) has solution if the operator Eq. (3.3) has fixed points, our first result is based on the Banach fixed point theorem (see [11]).
Theorem 3.2. We suppose that
|g(t,x)−g(t,y)|≤L1|x−y|,|h(t,x)−h(t,y)|≤L2|x−y|,fort∈[0,T],x,y∈R, |
0<c2<Γ(2−r)[Lc1−c1|ξ2μ2||1+λ1|T2L2−c1|μ1ξ1|TL1]T2−r|μ2ξ2||λ1−1|L2, |
L>|ξ2μ2||1+λ1|T2L2+|μ1ξ1|TL1 |
and
θ1,θ2≥0 |
with
θ1≤M′1N′1,θ2≤M′2N′2, |
such that
|f(t,x1,y1)−f(t,x2,y2)|≤θ1|x1−x2|+θ2|y1−y2|,fort∈[0,T],x1,x2,y1,y2∈R. |
where :
M′1=Γ(q+1)Γ(2−r)Γ(q−r+1){Γ(2−r)[Lc1−c1|ξ2μ2||1+λ1|T2L2−c1|μ1ξ1|TL1]−T2−r|μ2ξ2||λ1−1|L2c2}, |
N′1=Γ(2−r)2Γ(q−r+1)c1Tq+Γ(2−r)2Γ(q−r+1)c1|ξ1λ1|Tq+Γ(2−r)2Γ(q−r+1)c1|ξ2λ2||1+λ1|Tqq+Γ(2−r)2Γ(q+1)Tq−rc2+Γ(2−r)Γ(q−r+1)Tq−r|λ2ξ2||λ1−1|c2q |
and
M′2=Lc2Γ(q+1)Γ(2−r)Γ(q−r+1), |
N′2=Γ(2−r)Γ(q−r+1)Tqc1+Γ(2−r)Γ(q−r+1)|ξ1λ1|Tqc1+Γ(2−r)Γ(q−r+1)|ξ2λ2||1+λ1|Tqc1q+Γ(q+1)Γ(2−r)Tq−rc2+Γ(q−r+1)Tq−r|λ2ξ2||λ1−1|c2q. |
Then the boundary value problem (1.1) has a unique solution.
Proof. Let us set
γ=MTqc1Γ(q+1)+|ξ1λ1|TqMc1Γ(q+1)+|ξ2λ2||1+λ1|TqMc1Γ(q)+|ξ2μ2||1+λ1|T2M2c1+|μ1ξ1|TM1c1+Tq−rMc2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|Mc2Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|M2c2Γ(2−r). |
Now we show that
|(Fx)(t)|≤(Tqθ1Γ(q+1)+|ξ1λ1|Tqθ1Γ(q+1)+|ξ2λ2||1+λ1|Tqθ1Γ(q)+|ξ2μ2||1+λ1|T2L2+|μ1ξ1|TL1)|x|+≤(Tqθ2Γ(q+1)+|ξ1λ1|Tqθ2Γ(q+1)+|ξ2λ2||1+λ1|Tqθ2Γ(q))|cDrx|+TqMΓ(q+1)+|ξ1λ1|TqMΓ(q+1)+∣ξ2λ2||1+λ1|TqMΓ(q)+|ξ2μ2||1+λ1|T2M2+|μ1ξ1|TM1. |
Similarly, we have
|(cDrfx)(t)|≤1Γ(q−r)∫t0(t−s)q−r−1(θ1|x|+θ2|cDrx|+M)ds+T1−rΓ(2−r)|λ2||λ2−1|Γ(q−1)∫T0(T−s)q−2(θ1|x|+θ2|cDrx|+M)ds+T1−rΓ(2−r)|μ2||λ2−1|∫T0(L2|x|+M2)ds≤(Tq−rθ1Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ1Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|L2Γ(2−r))|x|+(Tq−rθ2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ2Γ(2−r)Γ(q))|cDrx|+Tq−rMΓ(q−r+1)+Tq−r|λ2ξ2||λ1−1|MΓ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|M2Γ(2−r). |
From the above inequalities, we obtain :
‖Fx‖≤c1((Tqθ1Γ(q+1)+|ξ1λ1|Tqθ1Γ(q+1)+|ξ2λ2||1+λ1|Tqθ1Γ(q)+|ξ2μ2||1+λ1|T2L2+|μ1ξ1|TL1)|x|+(Tqθ2Γ(q+1)+|ξ1λ1|Tqθ2Γ(q+1)+|ξ2λ2||1+λ1|Tqθ2Γ(q))|cDrx|+TqMΓ(q+1)+|ξ1λ1|TqMΓ(q+1)+∣ξ2λ2||1+λ1|TqMΓ(q)+|ξ2μ2||1+λ1|T2M2+|μ1ξ1|TM1)+c2((Tq−rθ1Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ1Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|L2Γ(2−r))|x|+(Tq−rθ2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ2Γ(2−r)Γ(q))|cDrx|+Tq−rMΓ(q−r+1)+Tq−r|λ2ξ2||λ1−1|MΓ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|M2Γ(2−r))≤(c1Tqθ1Γ(q+1)+c1|ξ1λ1|Tqθ1Γ(q+1)+c1|ξ2λ2||1+λ1|Tqθ1Γ(q)+c1|ξ2μ2||1+λ1|T2L2+c1|μ1ξ1|TL1+Tq−rθ1c2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ1c2Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|L2c2Γ(2−r))|x|+(Tqθ2c1Γ(q+1)+|ξ1λ1|Tqθ2c1Γ(q+1)+|ξ2λ2||1+λ1|Tqθ2c1Γ(q)+Tq−rθ2c2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ2c2Γ(2−r)Γ(q))|cDrx|+MTqc1Γ(q+1)+|ξ1λ1|TqMc1Γ(q+1)+|ξ2λ2||1+λ1|TqMc1Γ(q)+|ξ2μ2||1+λ1|T2M2c1+|ξ1μ1|TM1c1+Tq−rMc2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|Mc2Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|M2c2Γ(2−r)≤L(c1|x|+c2|cDrx|)+γ≤LR+γ≤R. |
Now, for any
|(F1x)(t)−(F1y)(t)|≤TqΓ(q+1)(θ1|x−y|+θ2|cDrx−cDry|), |
|(F2x)(t)−(F2y)(t)|≤(Tq−rθ1Γ(q−r−1)+Tq−r|λ2ξ2||λ1−1|θ1Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|L2Γ(2−r))|x−y|+(Tq−rθ2Γ(q−r−1)+Tq−r|λ2ξ2||λ1−1|θ2Γ(2−r)Γ(q))|cDrx−cDry|. |
We obtain :
|(Fx)(t)−(Fy)(t)|≤(Tqθ1Γ(q+1)+|ξ1λ1|Tqθ1Γ(q+1)+|ξ2λ2||1+λ1|Tqθ1Γ(q)+|ξ2μ2||1+λ1|T2L2+|μ1ξ1|TL1)|x−y|+(Tqθ2Γ(q+1)+|ξ1λ1|Tqθ2Γ(q+1)+|ξ2λ2||1+λ1|Tqθ2Γ(q))|cDrx−cDry|. |
Similary, we have :
|(cDrFx)(t)−(cDrFy)(t)|=|(Iq−rNx)(t)−t1−rΓ(2−r)λ2(λ2−1)Γ(q−1)∫T0(T−s)q−2N(x)(s)ds−t1−rΓ(2−r)μ2(λ2−1)∫T0h(s,x(s))ds−(Iq−rNy)(t)+t1−rΓ(2−r)λ2(λ2−1)Γ(q−1)∫T0(T−s)q−2N(y)(s)ds+t1−rΓ(2−r)μ2(λ2−1)∫T0h(s,y(s))ds|≤(Tq−rθ1Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ1Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|L2Γ(2−r))|x−y|+(Tq−rθ2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ2Γ(2−r)Γ(q))|cDrx−cDry|. |
From the above inequalities, we obtain
‖(Fx)(t)−(Fy)(t)‖≤(Tqθ1c1Γ(q+1)+|ξ1λ1|Tqθ1c1Γ(q+1)+|ξ2λ2||1+λ1|Tqθ1c1Γ(q)+|ξ2μ2||1+λ1|T2L2c1+|μ1ξ1|TL1c1+Tq−rθ1c2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ1c2Γ(2−r)Γ(q)+T2−r|μ2ξ2||λ1−1|L2c2Γ(2−r))|x−y|+(Tqθ2c1Γ(q+1)+|ξ1λ1|Tqθ2c1Γ(q+1)+|ξ2λ2||1+λ1|Tqθ2c1Γ(q)+Tq−rθ2c2Γ(q−r+1)+Tq−r|λ2ξ2||λ1−1|θ2c2Γ(2−r)Γ(q))|cDrx−cDry|≤L(c1|x−y|+c2|cDrx−cDrx|)≤L‖x−y‖. |
Which implies that
Now, we state a known result due to Schauder which is needed to prove the existence of at least one solution of (1.1).
Theorem 3.3. Let
∣f(t,x,y)∣≤m1(t)+d1|x|ρ1+d′1|y|ρ′1, |
∣g(t,x)∣≤m2(t)+d2|x|ρ2,∣h(t,x)∣≤m3(t)+d3|x|ρ3, |
for each
Proof. Schauder's Fixed point theorem is used to prove that
Step 1:
Denote
|(F1x)(t)|≤‖m1‖TqΓ(q+1)+(d1rρ1+d′1rρ′1)TqΓ(q+1), |
|(F2x)(t)|≤(|ξ1λ1|TqΓ(q+1)+|ξ2λ2||1+λ1|TqΓ(q))‖m1‖+|ξ2μ2||1+λ1|T‖m3‖+|ξ1μ1|‖m2‖+(|ξ1λ1|TqΓ(q+1)+|ξ2μ2||1+λ1|TqΓ(q))(d1rρ1+d′1rρ′1)+|ξ1μ1|Td2rρ2+|ξ2μ2||1+λ1|T2d3rρ3. |
So, we have
|(Fx)(t)|≤(TqΓ(q+1)+|ξ1λ1|TqΓ(q+1)+ξ2λ2||1+λ1|TqΓ(q))‖m1‖+|ξ1μ1|‖m2‖+|ξ2μ2||1+λ1|T‖m3‖+(TqΓ(q+1)+|ξ1λ1|TqΓ(q+1)+|ξ2λ2||1+λ1|TqΓ(q))(d1rρ1+d′1r′ρ1)+|ξ1μ1|Td2rρ2+|ξ2μ2||1+λ1|T2d3rρ3. |
Then from Eq. (3.4), we have
|(cDrFx)(t)|≤Tq−r‖m1‖Γ(q−r+1)+Tq−rΓ(q−r+1)(d1rρ1+d′1r′ρ′1)+|k|T1−rΓ(2−r), |
where
|k|≤|λ2ξ2||λ1−1|Tq−1Γ(q)‖m1‖+|μ2ξ2||λ2−1|‖m3‖+|ξ2λ2||λ1−1|Tq−1Γ(q)(d1rρ1+d′1rρ′1)+|ξ2μ2||λ2−1|Td3|r|ρ3. |
So, we have:
|(cDrFx)(t)|≤(Tq−rΓ(q−r+1)+|λ2ξ2||λ1−1|Tq−rΓ(2−r)Γ(q))‖m1‖+|μ2ξ2||λ1−1|T1−rΓ(2−r)‖m3‖+(Tq−rΓ(q−r+1)+|λ2ξ2||λ1−1|Tq−rΓ(q)Γ(2−r))(d1rρ1+d′1rρ′1)+T2−rΓ(2−r)|μ2ξ2||λ1−1|d3rρ3. |
From above inequalities, we obtain
‖(Fx)(t)‖≤(c1TqΓ(q+1)+|ξ1λ1|c1TqΓ(q+1)+ξ2λ2||1+λ1|c1TqΓ(q))‖m1‖+c1|ξ1μ1|‖m2‖+c1|ξ2μ2||1+λ1|T‖m3‖+(c1TqΓ(q+1)+|ξ1λ1|c1TqΓ(q+1)+|ξ2λ2||1+λ1|c1TqΓ(q))(d1rρ1+d′1r′ρ1)+c1|ξ1μ1|Td2rρ2+c1|ξ2μ2||1+λ1|T2d3rρ3+(c2Tq−rΓ(q−r+1)+|λ2ξ2||λ1−1|c2Tq−rΓ(2−r)Γ(q))‖m1‖+|μ2ξ2||λ1−1|c2T1−rΓ(2−r)‖m3‖+(c2Tq−rΓ(q−r+1)+|λ2ξ2||λ1−1|c2Tq−rΓ(q)Γ(2−r))(d1rρ1+d′1rρ′1)+c2T2−rΓ(2−r)|μ2ξ2||λ1−1|d3rρ3. |
Denote:
L=(c1TqΓ(q+1)+|ξ1λ1|c1TqΓ(q+1)+ξ2λ2||1+λ1|c1TqΓ(q))‖m1‖+c1|ξ2μ2||1+λ1|T‖m3‖+c1|ξ1μ1|‖m2‖+(c2Tq−rΓ(q−r+1)+|λ2ξ2||λ1−1|c2Tq−rΓ(2−r)Γ(q))‖m1‖+|μ2ξ2||λ1−1|c2T1−rΓ(2−r)‖m3‖, |
M1=(c1TqΓ(q+1)+|ξ1λ1|c1TqΓ(q+1)+|ξ2λ2||1+λ1|c1TqΓ(q)+c2Tq−rΓ(q−r+1)+|λ2ξ2||λ1−1|c2Tq−rΓ(q)Γ(2−r))(d1rρ1+d′1r′ρ1), |
M2=c1|ξ1μ1|T,M3=c1|ξ2μ2||1+λ1|T2+c2|μ2ξ2||λ1−1|T2−rΓ(2−r). |
Now let
R≥max(5L,(5M1d1)11−ρ1,(5M1d′1)11−ρ′1,(5M2d2)11−ρ2,(5M3d3)11−ρ3). |
Then it is obvious that for any
‖Fx‖≤L+M1(d1rρ1+d′1rρ′1)+M2d2rρ2+M3d3rρ3≤R5+R5+R5+R5+R5=R. |
This implies that
Step 2:
Suppose that
So we have
limn→∞‖xn−x‖∞=0andlimn→∞‖cDrxn−cDrx‖∞=0, |
which implies that
limn→∞xn(t)=x(t)andlimn→∞cDrxn(t)=cDrx(t),t∈[0,T], |
therefore
limn→∞f(t,xn(t),cDrxn(t))=f(t,x(t),cDrx(t)) |
limn→∞g(t,xn(t))=g(t,x(t)),limn→∞h(t,xn(t))=h(t,x(t)),t∈[0,T], |
which gives
|(Fxn)(t)−(Fx)(t)|≤∫t0(t−s)q−1Γ(q)|f(s,xn,cDrxn)−f(s,x,cDrx)|ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)|f(s,xn,cDrxn)−f(s,x,cDrx)|ds+|ξ2λ2||1+λ1|T∫T0(T−s)q−2Γ(q−1)|f(s,xn,cDrxn)−f(s,x,cDrx)|ds+|ξ2μ2||1+λ2|T∫T0|h(s,xn)−h(s,x)|ds+|μ1ξ1|∫T0|g(s,xn)−g(s,x)|ds |
and
|(cDrFxn)(t)−(cDrFx)(t)|≤∫t0(t−s)q−r−1Γ(q−r)(|f(s,xn,cDrxn)−f(s,x,cDrx)|)ds+T1−rΓ(2−r)|ξ2λ2||λ1−1|∫T0(T−s)q−2Γ(q−1)×(|f(s,xn,cDrxn)−f(s,x,cDrx)|)ds+T1−rΓ(2−r)|ξ2μ2||λ1−1|∫T0|h(s,xn)−h(s,x)|ds. |
Finely, we have
‖(Fxn)(t)−(Fx)(t)‖=c1‖(Fxn)(t)−(Fx)(t)‖∞+c2‖(cDrFxn)(t)−(cDrFx)(t)‖∞⟶n→∞0, |
which means that
Step 3:
Since
Now let,
|(F1x)(t2)−(F1x)(t1)|=|∫t20(t2−s)q−1Γ(q)f(s,x(s),cDrx(s))ds−∫t10(t1−s)q−1Γ(q)f(s,x(s),cDrx(s))ds|≤∫t10|(t2−s)q−1−(t1−s)q−1|Γ(q)|f(s,x(s),cDrx(s))|ds+∫t2t1(t2−s)q−1Γ(q)|f(s,x(s),cDrx(s))|ds≤|tq2−(t2−t1)q−tq1|Γ(q+1)N1+(t2−t1)qΓ(q+1)N1≤2N1(t2−t1)qΓ(q+1)+N1|tq2−tq1|Γ(q+1), |
|(F2x)(t2)−(F2x)(t1)|≤(|ξ2λ2||1−λ1|Tq−1Γ(q)N1+|ξ2μ2||1−λ2|TN2)|t2−t1|. |
So, we have:
|(Fx)(t2)−(Fx)(t1)|≤2N1(t2−t1)qΓ(q+1)+N1|tq2−tq1|Γ(q+1)+(|ξ2λ2||1−λ1|Tq−1Γ(q)N1+|ξ2μ2||1−λ2|TN2)|t2−t1|, |
we find that
|(cDrFx)(t2)−(cDrFx)(t1)|=|1Γ(q−r)∫t10((t2−s)q−r−1−(t1−s)q−r−1)f(s,x(s),cDrx(s))ds+1Γ(q−r)∫t2t1(t2−s)q−r−1f(s,x(s),cDrx(s))ds−(λ2(λ2−1)Γ(q−1)∫T0(T−s)q−2f(s,x(s),cDrx(s))ds+μ2(λ2−1)∫T0h(s,x(s))ds)(t1−r2−t1−r1)Γ(2−r)|≤N1|tq−r2−tq−r1|Γ(q−r+1)+2N1(t2−t1)q−rΓ(q−r+1)+(|λ2ξ2||λ1−1|Tq−1Γ(q)N1+|μ1ξ2||λ1−1|TN2)|t1−r2−t1−r1|Γ(2−r). |
Hence we have (since
‖(Fx)(t2)−(Fx)(t1)‖⟶0ast2⟶t1 |
and the limit is independent of
Now, we prove the existence of solution of (1.1) by applying Alternative of Leray-Schauder fixed point theorem.
Theorem 3.4. Let
|f(t,x,y)|≤a1(t)+a2(t)|x|+a3(t)|y|, |
|g(t,x)|≤b1(t)+b2(t)|x|,|h(t,x)|≤d1(t)+d2(t)|x|,∀t∈[0,T]. |
Then the problem (1.1) has at least one solution.
Proof. It is trivially that
We have shown in Theorem 3.3 that
Firstly, Let
|(Fx)(t)|≤{∫t0(t−s)q−1Γ(q)ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)ds+|ξ2λ2||1+λ1|T∫T0(T−s)q−2Γ(q−1)ds}|f(s,x(s),cDrx(s))|+|ξ2μ2||1+λ1|T∫T0|h(s,x(s))|ds+|μ1ξ1|∫T0|g(s,x(s))|ds≤{TqΓ(q+1)+|ξ1λ1|TqΓ(q+1)+|ξ1λ1||1+λ1|TqΓ(q)}M1+|ξ2μ2||1+λ1|TM2+|μ1ξ1|M3. |
So, we have
|(cDrFx)(t)|≤{1Γ(q−r)∫t0(t−s)q−r−1ds+T1−rΓ(2−r)|ξ2λ2||λ1−1|∫T0(T−s)q−2Γ(q−1)ds}|f(s,x(s),cDrx(s))|+|ξ2μ2||λ1−1|∫T0|h(s,x(s))|ds≤{Tq−rΓ(q−r+1)+|ξ2λ2||λ1−1|Tq−rΓ(q)}M1+|μ2ξ2||λ1−1|T1−rΓ(2−r)M2. |
Finely, we have
|(Fx)(t)|≤{c1TqΓ(q+1)+c1|ξ1λ1|TqΓ(q+1)+c1|ξ1λ1||1+λ1|TqΓ(q)+c2Tq−rΓ(q−r+1)+c2|ξ2λ2||λ1−1|Tq−rΓ(q)}M1+{c1|ξ2μ2||1+λ1|T+c2|μ2ξ2||λ1−1|T1−rΓ(2−r)}M2+c1|μ1ξ1|M3≤K1M1+K2M2+K3M3. |
Where:
M1=max(s,z1,z2)∈[0,T]×R2|f(s,z1,z2)|,M2=max(s,z1)∈[0,T]×R∫T0|h(s,z1)|, |
M3=max(s,z1)∈[0,T]×R∫T0|g(s,z1)| |
K1=c1TqΓ(q+1)+c1|ξ1λ1|TqΓ(q+1)+c1|ξ1λ1||1+λ1|TqΓ(q)+c2Tq−rΓ(q−r+1)+c2|ξ2λ2||λ1−1|Tq−rΓ(q) |
K2=c1|ξ2μ2||1+λ1|T+c2|μ2ξ2||λ1−1|T1−rΓ(2−r),K3=c1|μ1ξ1|. |
Hence
Secondly, we prove the compactness of the operator
|(Fx)(t2)−(Fx)(t1)|≤2(t2−t1)q+|tq2−tq1|Γ(q+1)N1+(|ξ2λ2||λ1−1|Tq−1Γ(q)N1+|ξ2μ2||λ1−1|TN2)|t2−t1|. |
So, we have
|(cDrFx)(t2)−(cDrFx)(t1)|≤2|t2−t1|q−r+(tq−r2−tq−r1)Γ(q−r+1)N1+(|ξ2λ2||λ1−1|Tq−1Γ(q)N1+|ξ2μ2||λ1−1|TN2)|t1−r2−t1−r1|Γ(2−r). |
Hence
‖(Fx)(t2)−(Fx)(t1)‖≤2c1(t2−t1)q+c1|tq2−tq1|Γ(q+1)N1+(|ξ2λ2||λ1−1|Tq−1Γ(q)c1N1+|ξ2μ2||λ1−1|Tc1N2)|t2−t1|+2c2|t2−t1|q−r+c2(tq−r2−tq−r1)Γ(q−r+1)N1+(|ξ2λ2||λ1−1|Tq−1Γ(q)c2N1+|ξ2μ2||λ1−1|Tc2N2)|t1−r2−t1−r1|Γ(2−r)⟶t2→t10 |
and the limit is independent of
Thirdly, the result will follow from the Leray-Schauder nonlinear alternative (Theorem 2.5) once we have proved the boundeness of the set of all solutions to equations
Let
|(Fx)(t)|≤∫t0(t−s)q−1Γ(q)a1(s)ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)a1(s)ds+|ξ2λ2[λ1(T−t)+T]|∫T0(T−s)q−2Γ(q)a1(s)ds+|ξ2μ2[λ2(T−t)+T]|∫T0d1(s)ds+|μ1ξ1|∫T0b1(s)ds+{∫t0(t−s)q−1Γ(q)a2(s)ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)a2(s)ds+|ξ2λ2[λ1(T−t)+T]|∫T0(T−s)q−2Γ(q)a2(s)ds+|ξ2μ2[λ2(T−t)+T]|∫T0d2(s)|x|ds+|μ1ξ1|∫T0b2(s)ds}|x|+{∫t0(t−s)q−1Γ(q)a3(s)|cDrx|ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)a3(s)ds+|ξ2λ2[λ1(T−t)+T]|∫T0(T−s)q−2Γ(q)a3(s)ds}|cDrx|≤A1+A2|x|+A3|cDrx|. |
Where
A1=∫t0(t−s)q−1Γ(q)a1(s)ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)a1(s)ds+|ξ2λ2[λ1(T−t)+T]|∫T0(T−s)q−2Γ(q)a1(s)ds+|ξ2μ2[λ2(T−t)+T]|∫T0d1(s)ds+|μ1ξ1|∫T0b1(s)ds |
A2=∫t0(t−s)q−1Γ(q)a2(s)ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)a2(s)ds+∣ξ2λ2[λ1(T−t)+T]|∫T0(T−s)q−2Γ(q)a2(s)ds+∣ξ2μ2[λ2(T−t)+T]|∫T0d2(s)|x|ds+|μ1ξ1|∫T0b2(s)ds |
A3=∫t0(t−s)q−1Γ(q)a3(s)|cDrx|ds+|ξ1λ1|∫T0(T−s)q−1Γ(q)a3(s)ds+∣ξ2λ2[λ1(T−t)+T]|∫T0(T−s)q−2Γ(q)a3(s)ds |
By the definition of the Caputo fractional derivative with
|cDr(Fx)(t)|≤1Γ(q−r)∫t0(t−s)q−r−1a1(s)ds+T1−r|λ2|Γ(2−r)Γ(q−1)|λ2−1|∫T0(T−s)q−2a1(s)ds+T1−r|μ2|Γ(2−r)|λ2−1|∫T0d1(s)ds +{1Γ(q−r)∫t0(t−s)q−r−1a2(s)ds+T1−r|λ2|Γ(2−r)Γ(q−1)|λ2−1|∫T0(T−s)q−2a2(s)ds+T1−r|μ2|Γ(2−r)|λ2−1|∫T0d2(s)ds}|x|+{1Γ(q−r)∫t0(t−s)q−r−1a3(s)ds+T1−r|λ2|Γ(2−r)Γ(q−1)|λ2−1|∫T0(T−s)q−2a3(s)ds}|cDrx|≤A′1+A′2|x|+A′3|cDrx|. |
Where
A′1=1Γ(q−r)∫t0(t−s)q−r−1a1(s)ds+T1−r|λ2|Γ(2−r)Γ(q−1)|λ2−1|∫T0(T−s)q−2a1(s)ds+T1−r|μ2|Γ(2−r)|λ2−1|∫T0d1(s)ds |
A′2=1Γ(q−r)∫t0(t−s)q−r−1a2(s)ds+T1−r|λ2|Γ(2−r)Γ(q−1)|λ2−1|∫T0(T−s)q−2a2(s)ds+T1−r|μ2|Γ(2−r)|λ2−1|∫T0d2(s)ds |
A′3=1Γ(q−r)∫t0(t−s)q−r−1a3(s)ds+T1−r|λ2|Γ(2−r)Γ(q−1)|λ2−1|∫T0(T−s)q−2a3(s)ds |
Therefore, we can obtain that
‖Fx(t)‖≤c1(A1+A2|x(t)|+A3|cDrx(t)|)+c2(A′1+A′2|x(t)|+A′3|cDrx(t)|)=(c1A1+c2A′1)⏟=A+(c1A2+c2A′2)|x(t)|+(c1A3+c2A′3)|cDrx(t)|≤A+ρ(c1maxt∈I|x(t)|+c2maxt∈I|cDrx(t)|)≤A+ρ‖x‖. |
Suppose there exists a
R=‖x‖=λ‖Fx‖<A+ρ‖x‖=R, |
which is a contradiction. By Theorem 2.5, there exists a fixed point
Example 4.1. Consider the following boundary value problem :
{cD32x(t)=1(t+4)2tan−1(x)+|cD34x(t)|(t+3)2(1+|cD34x(t)|),t∈[0,1]x(0)+12x(1)=13∫103|x|4+|x|dx,x′(0)+13x(1)=32∫102|x|3+|x|dx. | (4.1) |
Here,
q=32,r=34,λ1=−12,λ2=−13,μ1=13,μ2=32,c1=12,c2=13,ξ1=−23, |
ξ2=12,T=1and|f(t,x1,y1)−f(t,x2,y2)|≤116|x1−x2|+19|y1−y2|, |
∣g(t,x)−g(t,y)∣≤34|x−y|,∣h(t,x)−h(t,y)∣≤23|x−y|,θ1=116,θ2=19, |
L1=34,L2=23. |
Furthermore,
θ1≤0.14971,θ2≤0.54173,1>L>0,6666. |
Thus, by Theorem 3.2 the boundary value problem (4.1) has a unique solution on
Example 4.2. Consider the following boundary value problem :
{cD53x(t)=(4t2−9t)e−x3(t)+14|x(t)|12+12(|cD34x(t)|1+cos2x(t))13,t∈[0,1]x(0)+12x(1)=15∫10[(s−1)e−x2(s)+13|x(s)|12]ds,x′(0)+15x′(1)=13∫10[(s3−2s)e−x2(s)+19|x(s)|12]ds. | (4.2) |
In this case, we have
f(t,x,y)=(4t2−9t)e−x3+14|x|12+12(|y|1+cos2x)13 |
and
g(t,x)=(t−1)e−x2+13|x(s)|12,h(t,x)=(t3−2t)e−x2+19|x|12, |
since
|f(t,x,y)|≤|4t2−9t|+14|x|12+12|y|13, |
|g(t,x)|≤|t−1|+13|x|12,|h(t,x)|≤|t3−2t|+19|x|12. |
Let
d1=14,d2=12,ρ1=12,ρ2=13,d3=13,d4=19,ρ3=ρ4=12 |
and
m(t)=|4t2−9t|∈L∞(0,1),m1(t)=|t−1|∈L1(0,1),m2(t)=|t3−2t|∈L1(0,1) |
Now it is easy to verify that all conditions of Theorem 3.3 are satisfied. Therfore, the fractional boundary value problem (4.2) has at least one solution on
The authors declare no conflict of interest.
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