# Positive Solutions ton-Order Fractional Differential Equation with Parameter

Authors: Jing-jing Tan; Cong Tan; Xueling Zhou
Journal: Journal of Function Spaces
 (2018)

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## Abstract

By employing the properties of the sum operators, we investigate the solutions of then-order fractional differential equations multiple point boundary value problem (in short BVP) with the boundary conditions contains a parameter. We not only obtain the existence and uniqueness of solutions about this BVP but also construct an iterative scheme to approximate the solution which is important for practical application. An example is given to demonstrate the validity of our main results.


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## Body

## 1. Introduction

In this paper, we focus on then-order fractional differential equations of multiple point BVP:(1)-D0+αxt=f1t,xt,xt+f2t,xt,0<t<1,n-1<α≤n,n≥2,x0=x′0=⋯=xk0=0,0≤k≤n-2,x1=∑i=1m-2αixξi+λ,m≥3,where D0+α is the α-order fractional derivative, f1:[0,1]×[0,+∞)×[0,+∞)→[0,+∞) and f2:[0,1]×[0,+∞)→[0,+∞) are continuous, αi>0(i=1,2,…,m-2),  0<ξ1<ξ2<⋯<ξm-2<1, ∑i=1m-2αiξiα-1<1, λ is a parameter.Boundary value problem is an important branch of differential equation. In recent years, as a mathematical model, BVP can be processed in many engineering and scientific models. Fractional differential equations BVP has become a hot issue; see the monographs of Lakshmikantham and Vatsala [1], Rudin [2], Samko et al. [3], Agarwal et al. [4, 5], and Webb and Zima [6]. Many excellent results have been reported; see [7–23]. In [24], El-Sayed and Bin-Taher study the following m-point BVP:(2)-x′′t=ft,D0+αxt,0<t<1,0<α<1,x0=0,x1=∑i=1m-2αixξi,0<a<ξi<b<1.By using the technical of Kolmogorov compactness criterion, they obtained the results of existence at least of one positive solution. In [25], Zhang studied the following n-order BVP:(3)-D0+αxt=qtft,xt,…,xn-2t,0<t<1,α∈n-1,n,n≥2,x0=x′0=⋯=xn-20=0,xn-20=1.The authors obtained the existence of positive solutions. However, many aspects of the theory to the solutions of the fractional differential equations BVP need to be explored, such as the solutions of high-order multipoint BVP.Motivated by the work mentioned above, we focus on the existence and uniqueness of positive solutions for the nonlocal BVP (1) by the technical of a fixed point theorem of a sum operator. In [12], we study the fractional equations with two and three point boundary conditions, where α∈(1,2). In [20], we study the fractional equations of with integral boundary conditions. The new features of this paper are as follows. Firstly, the order of the fractional differential equation is high order with α∈(n-1,n], and the boundary condition is multipoint boundary condition. Compared with [12], the BVP is more general; secondly, the boundary conditions are dependent on a parameter λ which is different from paper [20], so the properties of Green function and the construct of the operator are differential. The Green function and the operator are the key to the two BVP.The paper is organized as follows. In Section2, we recall some definitions and lemmas. In Section 3, the existence of positive solutions to BVP (1) is obtained. Finally, in Section 4, an illustrative example is also presented.


## 2. Preliminaries

We first recall four classes operator.The basic space used in this paper is the spaceC[0,1]. Set(4)P=x∈C0,1∣xt≥0,t∈0,1.LetD=P or D=oP and r be a real number with 0≤r<1. An operator T:P→P is said to be(1)
positive homogeneous if it satisfies(5)Ttx=tTx,for  anyt∈0,1,x∈P;(2)
subhomogeneous if it satisfies(6)Ttx≥tTx,for  anyt∈0,1,x∈P;(3)
r-concave if it satisfies(7)Ttx≥trTx,for  anyt∈0,1,x∈D;(4)
T:P×P→P is said to be a mixed monotone operator if it satisfies T(x1,y1)≤T(x2,y2) for any xi,yi(i=1,2)∈P,x1≤x2,y1≥y2.Lemma 1 (see [26]).
Leth>θ and β∈(0,1).  T:P×P→P is a mixed monotone operator; S:P→P is an increasing subhomogeneous operator for any x,y∈P, and there exists a constant δ0>0 such that T(x,y)≥δ0Sx. Assume that(i)
for anyt∈(0,1),x,y∈P,(8)Ttx,t-1y≥tβTx,y,(ii)
there exists a constanth0∈Ph such that T(h0,h0)∈Ph and Sh0∈Ph;
Then, the operator equationT(x,x)+Sx=x has a unique solution x∗ in Ph and for any initial values x0,y0∈Ph, constructing successively sequences xn=T(xn-1,yn-1)+Sxn-1, yn=T(yn-1,xn-1)+Syn-1, n=1,2,…, we have xn→x∗ and yn→x∗ as n→∞.Lemma 2.
Given∈C(0,1)∩L(0,1), the unique solution of(9)-D0+αxt=ft,0<t<1,x0=x′0=⋯=xk0=0,x1=∑i=1m-2αixξi+λ,is(10)xt=∫01Gt,sfsds+λtα-11-∑i=1m-2αiξiα-1,where n-1<α≤n,n≥2,m≥3,0≤k≤n-2,(11)Gt,s=Ht,s+tα-11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sHt,s=t1-sα-1-t-sα-1Γα,0≤s≤t≤1,t1-sα-1Γα,0≤t≤s≤1.Proof.
By using the properties of the Riemann-Liouville fractional derivative, we can easily get the conclusion, so we omit.In the follow, we verify the properties ofG(t,s).Lemma 3.
For allt,s∈(0,1), G(t,s) is continuous on [0,1]×[0,1] and(12)tα-11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,s≤Gt,s≤tα-11+11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,s.Proof.
Form the definition ofG(t,s), we can get G(t,s) is continuous. Now, we need only to prove that (12) is satisfied. For all t,s∈(0,1), we have(13)Gt,s=Ht,s+tα-11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,s≤t1-sα-1+tα-11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,s=tα-11+11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,s,Gt,s=Ht,s+tα-11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,s≥tα-11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,s=11-∑i=1m-2αiξiα-1tα-1∑i=1m-2αiHξi,s.Hence (3) is held; the proof is complete.


## 3. Main Results

In this section, we establish the existence and uniqueness of positive solutions results for the problems (1).Theorem 4.
Assume thatf1∈C[[0,1]×[0,∞)×[0,∞),[0,∞)] and f2∈C[[0,1]×[0,∞),[0,∞)] are continuous. A function x∈P is a solution of the BVP (1) if and only if it is a solution of the integral equation(14)xt=∫01Gt,sf1s,xs,xs+f2s,xsds+λtα-11-∑i=1m-2αiξiα-1.Proof.
The proof of this result is quite similar to that given earlier in Lemma3 and omitted.Define operatorsT and S as follows:(15)Tx,yt=∫01Gt,sf1s,xs,ysds,Sxt=∫01Gt,sf2s,xsds+λtα-11-∑i=1m-2αiξiα-1.Theorem 5.
Assume that(H1)
f1(t,x,y)∈C[[0,1]×[0,+∞)×[0,+∞),[0,+∞)], f2(t,x)∈C[[0,1]×[0,+∞)→[0,+∞)], f1 and f2 increasing in x∈[0,+∞), f decreasing in y∈[0,+∞), and there exists a constant δ0>0 such that f1(t,x,y)≥δ0f2(t,x)+δ021/1-∑i=1m-2αiξiα-1;(H2)
for allμ∈(0,1),t∈[0,1],x,y∈[0,+∞), there exists a constant ξ∈(0,1) such that f2(t,μx)≥μf2(t,x) and f1(t,μx,μ-1y)≥μξf1(t,x,y).
Then, for any0<λ≤δ0, BVP (1) has a unique positive solution x∗ in Ph, where h(t)=tα-1, and for any x0,y0∈Ph, construct successively the sequences(16)xn+1t=∫01Gt,sf1s,xns,yns+f2s,xnsds+λtα-11-∑i=1m-2αiξiα-1,n=0,1,2,…,yn+1t=∫01Gt,sf1s,yns,xns+f2s,ynsds+λtα-11-∑i=1m-2αiξiα-1,n=0,1,2,…,we have xn-x∗→0 and yn-x∗→0 as n→∞.Proof.
According to Theorem4 and the definition of operators T and S, it is straightforward to show that x is the solution of BVP (1) if and only if x solves the operator equation x=T(x,x)+Sx. We only need to investigate the solution of the above operator equation.
Now, we will show thatT:P×P→P and S:P→P satisfy all the assumptions of Lemma 1. To begin with, we prove that A is a mixed monotones operator. In fact, for xi, yi∈P, i=1,2 with x1≥x2, y1≤y2, we know that x1(t)≥x2(t) and y1(t)≤y2(t) for all t∈[0,1]. According to (H1) and Lemma 3, we get(17)f1s,x1s,y1s≥f1s,x2s,y2s,since G(t,s)>0, that is, T(x1,y1)≥T(x2,y2). Furthermore, an argument similar to the one used in T shows that S is increasing. Next we prove that T satisfies (8). For any μ,ξ∈(0,1) and x,y∈P, from(18)f1s,μxs,μ-1ysds≥μξf1s,xs,yswe have T(μx,μ-1y)≥μξT(x,y) for μ,ξ∈(0,1),  x,y∈P. Hence (8) holds. Also, for any μ∈(0,1),x∈P, taking (H2) into consideration, we have(19)Sμxt=∫01Gt,sf2s,μxsds+λtα-11-∑i=1m-2αiξiα-1≥μ∫01Gt,sf2s,xsds+μλtα-11-∑i=1m-2αiξiα-1=μSxt.Thus, S(μx)≥μS(x) for any μ∈(0,1),  x∈P. Hence the operator S is subhomogeneous. Now, we are in the position to show that T(h,h)∈Ph and Sh∈Ph. It follows from (H1), (H2) and Lemma 3 that, for any t∈[0,1],(20)Th,ht=∫01Gt,sf1s,hs,hsds≤ht∫011+11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sf1s,0,1ds=a2ht,where(21)a2≔∫011+11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sf1s,0,1ds,Th,ht=∫01Gt,sf1s,hs,hsds≥ht1-∑i=1m-2αiξiα-1∫01∑i=1m-2αiHξi,sf1s,1,0ds=a1ht,where(22)a1≔∫0111-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sf1s,1,0ds,Thanks to (H1), (H2) and (H3), we get(23)f1s,1,0≥f1s,0,1≥δ0f2s,0+δ0211-∑i=1m-2αiξiα-1>0.So(24)∫01f1s,1,0ds≥δ0∫01f2s,0ds+δ0211-∑i=1m-2αiξiα-1>0,and, in consequence, a1>0 and a2>0. Thus, for t∈[0,1], inequality a1h(t)≤T(h,h)(t)≤a2h(t) holds, and then we get T(h,h)∈Ph. An argument similar to the one used in T shows that(25)Sht=∫01Gt,sf2s,hsds+λtα-11-∑i=1m-2αiξiα-1≤ht∫011+11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sf2s,0ds+δ01-∑i=1m-2αiξiα-1=b2ht,where(26)b2≔∫011+11-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sf2s,0ds+δ01-∑i=1m-2αiξiα-1,Sht=∫01Gt,sf2s,hsds+λtα-11-∑i=1m-2αiξiα-1≥ht∫0111-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sf2s,1ds=b1ht,where(27)b1≔∫0111-∑i=1m-2αiξiα-1∑i=1m-2αiHξi,sf2s,1ds.Obviously, b1>0 and b2>0; thus we are led to the conclusion that b1h(t)≤Sh(t)≤b2h(t) with t∈[0,1], i.e., Sh∈Ph. Hence condition (i) of Lemma 1 is proved. It remains to show that condition (ii) of Lemma 1 is satisfied. For any x,y∈P,  t∈[0,1], from (H3) we have(28)Tx,yt=∫01Gt,sf1s,xs,ysds≥δ0∫01Gt,sf2s,xs+δ01-∑i=1m-2αiξiα-1ds≥δ0Sxt.That is, for any x,y∈P, inequality T(x,y)≥δ0Sx holds. Applying Lemma 2, we can get the conclusion of Theorem 5.As a special case of Theorem5, we obtain the following results.Corollary 6.
Assume thatH1′
f(t,x,y)∈C[[0,1]×[0,+∞)×[0,+∞),[0,+∞)] is increasing in x∈[0,+∞), decreasing in y∈[0,+∞) and f(t,0,1)≢0;H2′
for allt∈[0,1],  μ∈(0,1),  x,y∈[0,+∞), there exists a constant ξ∈(0,1) such that f(t,μx,μ-1y)≥μξf(t,x,y).
Then, the problem(29)-D0+αxt=ft,xt,xt,0<t<1,n-1<α≤n,n≥2,x0=x′0=⋯=xk0=0,0≤k≤n-2,x1=∑i=1m-2αixξi+λ,m≥3has a unique positive solution x∗ in Ph, where h(t)=tα-1 and, for any x0,y0∈Ph, constructing successively the sequences(30)xn+1t=∫01Gt,sfs,xns,ynsds+λtα-11-∑i=1m-2αiξiα-1,n=0,1,2,…,yn+1t=∫01Gt,sfs,yns,xnsds+λtα-11-∑i=1m-2αiξiα-1,n=0,1,2,…,we have xn-x∗→0 and yn-x∗→0 as n→∞.Proof.
Lettingg≡0 in Theorem 5, by suitable modification to the proof of Theorem 5, we can prove the desired conclusion.IfT:P×P→P is a mixed monotone operator and satisfies(31)Ttx,t-1y≥tTx,y,∀t∈0,1,x,y∈P.S:P→P is an increasing α-concave operator. We can also obtain the following result.Theorem 7.
Assume that(H3)
f1(t,x,y)∈C[[0,1]×[0,+∞)×[0,+∞),[0,+∞)], f2(t,x)∈C[[0,1]×[0,+∞)→[0,+∞)], f1 and f2 increasing in x∈[0,+∞), f decreasing in y∈[0,+∞), and there exists a constant δ0>0 such that f1(t,x,y)≤δ0f2(t,x)+δ021/1-∑i=1m-2αiξiα-1;(H4)
for allt∈[0,1],  μ∈(0,1),  x,y∈[0,+∞), there exists a constant ξ∈(0,1) such that f2(t,μx)≥μξf2(t,x) and f1(t,μx,μ-1y)≥μf1(t,x,y).
Then, BVP (1) has a unique positive solution x∗ in Ph, where h(t)=tα-1 and, for any x0,y0∈Ph, constructing successively the sequences(32)xn+1t=∫01Gt,sf1s,xns,yns+f2s,xnsds+λtα-11-∑i=1m-2αiξiα-1,n=0,1,2,…,yn+1t=∫01Gt,sf1s,yns,xns+f2s,ynsds+λtα-11-∑i=1m-2αiξiα-1,n=0,1,2,…,we have xn-x∗→0 and yn-x∗→0 as n→∞.Proof.
In the light of [26, Theorem 2.4], a similar argument used in Theorem 5. shows that the desire result. We omit it.


## 4. Example

In this section, we give one example to illustrate our results.Example 1.
Consider the BVP (1). Letting α=5/2, α1=1/2, α2=1/4, ξ1=1/16, ξ2=1/4, f1(t,x,y)=x1/2+y-1/2+t2+8/123 and f2(t,x)=x1/2+t3. A simple calculation gives ∑i=1m-2αiξiα-1=5/128<1. It is not difficult to verify that (H1) and (H2) are satisfied. Taking δ0=1/4>0, we have(33)f1t,x,y=x1/2+y-1/2+t2+8123≥x1/2+t3+8123≥δ0x1/2+t3+δ0128123=δ0f2t,x+δ011-∑i=1m-2αiξiα-1.Thus, (H3) is satisfied. Moreover, for any μ∈(0,1), t∈[0,1], x∈[0,∞), y∈[0,∞), we have(34)f2t,μx=μ1/2x1/2+t3≥μ1/2x1/2+t3≥μf2t,x,f1t,μx,μ-1y=μ1/2x1/2+μ1/2y-1/2+t2+8123≥μ1/2x1/2+y-1/2+t2+8123=μγf1t,x,y,where γ=1/2. We conclude that condition (H4) is satisfied. Therefore, Theorem 5 ensures that the BVP (1) has a unique positive solution in Ph with h(t)=t3/2.



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