# Existence Results for Fractional Differential Equations with Multistrip Riemann–Stieltjes Integral Boundary Conditions

Authors: Tongling Lv; Huihui Pang; Limei Cao
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

This paper is concerned with the existence and multiplicity of the positive solutions for a fractional boundary value problem with multistrip Riemann–Stieltjes integral boundary conditions. Our results are based on the Leggett–Williams fixed point theorem. In the end, two examples are worked out to illustrate our main work.


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## Body

## 1. Introduction

Nowadays, differential equations with fractional order have gained much attention and importance since they provided valuable tools for their applications in various sciences, such as gas dynamics, nuclear physics, electrodynamics of complex medium, and polymer rheology. With this advantage, fractional order models are regarded as more realistic and practical. For more details about fractional differential equations, we refer the readers to the monographs [1–4] and papers [5–9].Many scholars have studied the existence of nonlinear fractional differential equations with a variety boundary conditions. However, it is better to impose nonlocal conditions because they can accurately describe the actual phenomenon. Some authors studied multipoint boundary value problems; for example, [10] discussed the infinite-point boundary value problems (1)D0+αut+ft,ut=0,t∈0,1,u0=D0+βu0=0,D0+βu1=∑i=1∞ξiD0+βuηi, where 2<α⩽3,  1⩽β⩽2,  1⩽α-β,  0<ξi,ηi<1 with ∑i=1∞ξiηiα-β-1<1. Existence result of at least two positive solutions is given via fixed point theorem in a cone.Different from [10], some work focuses on the solvability of the fractional differential equations with integral boundary conditions. The details are found in [11–16] and the references therein. In [17], Sun and Zhao investigated the following fractional differential equation with integral boundary conditions: (2)D0+αut+qtft,ut=0,t∈0,1,u0=u′0=0,u1=∫01gsusds, where 2<α⩽3. By using the monotone iteration method and some inequalities technique, the existence result of positive solutions is obtained.By the same method, Zhang [18] discussed the following fractional differential equation with Riemann–Stieltjes integral boundary conditions: (3)D0+αut+ft,ut=0,t∈0,1,u0=u′0=0,D0+βu1=∫01D0+βutdAt, where 2<α⩽3,  0⩽β⩽1,  A(t) is a bounded variation, and ∫01D0+βu(t)dA(t) denotes a Riemann–Stieltjes integral with a signed measure. This includes both the multipoint and a Riemann integral in a single framework.Motivated by the wide applications of nonlocal boundary value problems and the results mentioned above, we consider the following fractional differential equation with multistrip Riemann–Stieltjes integral boundary conditions:(4)D0+αut+ft,ut,D0+βut=0,t∈0,1,(5)u0=D0+βu0=0,u1=∑i=1mαi∫IiutdAt,where 2<α⩽3,  0<β<1,  Ii⊂(0,1),  i=1,2,…,m, and D0+α is the standard Riemann-Liouville derivative; the nonlinear term f is related to the lower derivative of the function u. We emphasize that multistrip integral boundary conditions in (5) state that the value of unknown function at the right end point t=1 of the given interval is equal to the linear combination of the Riemann–Stieltjes integral values of the unknown function on the subinterval Ii, for i=1,2,…,m. The consideration of the fractional differential equation together with multistrip Riemann–Stieltjes integral boundary conditions makes problem (4) and (5) new. The proof of our main results is based on the Leggett–Williams fixed point theorem in a cone, which we present now.Theorem 1 (Leggett–Williams fixed point theorem).
LetP be a cone in a real Banach space E, Pc={x∈P∣x⩽c}, Ψ be a nonnegative continuous concave functional on P such that Ψx⩽x for all x∈P¯c, and PΨ,b,d=x∈P∣b⩽Ψx,x⩽d. Suppose that T:P¯c→P¯c is completely continuous and there exist constants 0<a<b<d⩽c such that (S1)
{x∈P(Ψ,b,d)∣Ψ(x)>b}≠⌀, and Ψ(Tx)>b for x∈P(Ψ,b,d);(S2)
Tx<a for x⩽a;(S3)
Ψ(Tx)>b for x∈P(Ψ,b,c), with Tx>d.
ThenT has at least three fixed points x1, x2, and x3, which satisfy (6)x1<a,b<Ψx2,a<x3withΨx3<b.If there holdsd=c, then condition (S1) implies condition (S3) of Theorem 1. Throughout this paper, we always make the following assumptions:(H1)
f:[0,1]×[0,+∞)×R→[0,+∞) is continuous;(H2)
2<α⩽3,  0<β<1,  1<α-β<2;(H3)
αi∈[0,+∞),  Ii⊂(0,1),  i=1,2,…,m, and A:[0,1]→R is an increasing function of bounded variation;(H4)
0<δ1<1, where δ1=∑i=1mαi∫Iitα-1dA(t).


## 2. Preliminaries

In this section, we will present several definitions and lemmas that are necessary for the proof of our main results.Definition 2 (see [1]).
The Riemann-Liouville fractional integral of orderα>0 of a function y:(0,∞)→R is given by (7)I0+αyt=1Γα∫0tt-sα-1ysds, provided the right side is pointwise defined on [0,∞).Definition 3 (see [1]).
The Riemann-Liouville fractional derivative of orderα>0 of a function y:(0,∞)→R is given by (8)D0+αyt=1Γn-αddtn∫0tt-sn-α-1ysds, where n=[α]+1,  [α] denotes the integer part of number α, provided the right side is pointwise defined on [0,∞).From the definitions of Riemann-Liouville’s derivative, we can obtain the following statement.Lemma 4.
Letα>0; if we assume u∈C(0,1)∩L1(0,1), then the fractional differential equation (9)D0+αut=0 has ut=C1tα-1+C2tα-2+⋯+CNtα-N, for some Ci∈R,i=1,2,…,N, as a unique solution, where N is the smallest integer greater than or equal to α.Lemma 5.
Letα>0; if we assume u∈C(0,1)∩L(0,1), then (10)I0+αD0+αut=ut+C1tα-1+C2tα-2+⋯+CNtα-N, for some Ci∈R,i=1,2,…,N, where N is the smallest integer greater than or equal to α.Remark 6.
The following properties are useful for our discussion:(1)
I0+αI0+βy(t)=I0+α+βy(t), for α>0,β>0,y(t)∈L1(0,1);(2)
D0+αI0+αy(t)=y(t), for α>0,y(t)∈L1(0,1).Lemma 7.
Suppose that(H4) holds. For y∈C(0,1)∩L1(0,1), the unique solution of(11)D0+αut+yt=0,t∈0,1,u0=D0+βu0=0,u1=∑i=1mαi∫IiutdAt is u(t)=∫01G(t,s)y(s)ds, in which(12)Gt,s=G0t,s+tα-11-δ1∑i=1mαi∫IiG0t,sdAt, where(13)G0t,s=1Γαtα-11-sα-1-t-sα-1,0⩽s⩽t⩽1,tα-11-sα-1,0⩽t⩽s⩽1.Proof.
In view of Lemma5, we reduce problem (11) to an equivalent integral equation (14)ut=-I0+αut+C1tα-1+C2tα-2+C3tα-3, where C1,C2,C3∈R are arbitrary constants. Consequently the general solution of the problem (11) can be written as(15)ut=-1Γα∫0tt-sα-1ysds+C1tα-1+C2tα-2+C3tα-3. By u(0)=D0+βu(0)=0, we find C2=C3=0. Set t=1 in (15), then (16)u1=-1Γα∫011-sα-1ysds+C1. Together with the boundary condition u(1)=∑i=1mαi∫Iiu(t)dA(t), we have (17)∑i=1mαi∫IiutdAt=-1Γα∫011-sα-1ysds+C1. Hence the unique solution of (11) is (18)ut=-1Γα∫0tt-sα-1ysds+tα-1Γα∫011-sα-1ysds+tα-1∑i=1mαi∫IiutdAt=∫01G0t,sysds+tα-1∑i=1mαi∫IiutdAt. Furthermore, (19)∑i=1mαi∫IiutdAt=∑i=1mαi∫Ii∫01G0t,sysdsdAt+∑i=1mαi∫Iitα-1dAt∑i=1mαi∫IiutdAt. Then (20)∑i=1mαi∫IiutdAt=11-δ1∑i=1mαi∫Ii∫01G0t,sysdsdAt. Hence, the solution of (11) is (21)ut=∫01G0t,sysds+tα-11-δ1∑i=1mαi∫Ii∫01G0t,sysdsdAt=∫01G0t,sysds+tα-11-δ1∑i=1mαi∫01∫IiG0t,sysdAtds=∫01G0t,s+tα-11-δ1∑i=1mαi∫IiG0t,sdAtysds=∫01Gt,sysds.Lemma 8.
The functionG0(t,s) defined by (13) satisfies the following inequality:(22)tα-11-sα-11-tsΓα⩽G0t,s⩽s1-sα-1Γα-1,fort,s∈0,1.Proof.
For0⩽s⩽t⩽1, we have 1-s⩾1-t, and then (23)G0t,s=1Γαtα-11-sα-1-t-sα-1=α-1Γα∫t-st-tsxα-2dx⩽t-tsα-2t-ts-t-sΓα-1=tα-21-sα-21-tsΓα-1⩽s1-sα-1Γα-1,G0t,s=1Γαtα-11-sα-1-t-sα-1=1Γαt-tsα-2t-ts-t-sα-2t-s⩾1Γαt-tsα-2t-ts-t-tsα-2t-s=1Γαtα-21-sα-21-ts⩾1Γαtα-11-sα-11-ts.
For0⩽t⩽s⩽1, since 2<α⩽3, we have (24)G0t,s=tα-11-sα-1Γα⩽α-1tα-2t1-sα-1Γα⩽tα-2s1-sα-1Γα-1⩽s1-sα-1Γα-1,G0t,s=tα-11-sα-1Γα⩾tα-11-sα-11-tsΓα.
Then the proof is completed.For convenience, denote(25)δ2=∑i=1mαi∫IidAt,δ3=∑i=1mαi∫IitαdAt,Gs=1-δ1+δ2s1-sα-11-δ1Γα-1,fors∈0,1,Λs=1-δ11-sα-β-1+δ21-sα-1Γα-β1-δ1,fors∈0,1,λ=θα1-δ1+θα-1δ1-δ31-δ1+δ2α-1,whereθ∈0,12is aconstant.δ1 is introduced (H4). It is obvious that δ2>δ1>δ3,  λ>0 and G(s)>0 for s∈(0,1).The following properties of the Green functionG(t,s) play an important role in this paper.Lemma 9.
The Green functionG(t,s) defined by (12) satisfies the following properties:(1)
G(t,s) is continuous on [0,1]×[0,1];(2)
G(t,s)⩾0 for t,s∈[0,1];(3)
Gt,s⩽Gs for t,s∈0,1;(4)
G(t,s)⩾λG(s) for s∈[0,1],t∈[θ,1-θ];(5)
|D0+βG(t,s)|⩽Λ(s) for t,s∈[0,1].Proof.
( 1 ) and (2) hold obviously; we only show that (3)–(5) are true.
( 3 ) For any t,s∈[0,1], by (12), (13), and the right inequality of (22), we get (26)Gt,s=G0t,s+tα-11-δ1∑i=1mαi∫IiG0t,sdAt⩽s1-sα-1Γα-1+tα-11-δ1∑i=1mαi∫Iis1-sα-1Γα-1dAt=s1-sα-1Γα-1+tα-1s1-sα-1δ21-δ1Γα-1⩽1-δ1+δ2s1-sα-11-δ1Γα-1=Gs.
( 4 ) For any s∈[0,1], by (12), (13), and the left inequality of (22), we get (27)mint∈θ,1-θGt,s=mint∈θ,1-θG0t,s+tα-11-δ1∑i=1mαi∫IiG0t,sdAt⩾mint∈θ,1-θtα-11-ts1-sα-1Γα+tα-11-δ1∑i=1mαi∫Iitα-11-ts1-sα-1ΓαdAt⩾θα-1θs1-sα-1Γα+θα-11-δ1∑i=1mαi∫Iitα-1-tαs1-sα-1ΓαdAt=s1-sα-1Γαθα+θα-1δ1-δ31-δ1=λGs.
( 5 ) By the definition of G0(t,s) and D0+βtμ=Γ(μ+1)/Γ(μ-β+1)tμ-β(μ>-1), we have (28)D0+βG0t,s=1Γα-βtα-β-11-sα-1-t-sα-β-1,0⩽s⩽t≤1,tα-β-11-sα-1,0⩽t⩽s≤1. Therefore,(29)D0+βG0t,s=1Γα-βtα-β-11-sα-1-t-sα-β-1,0⩽s⩽t≤1,tα-β-11-sα-1,0⩽t⩽s⩽1⩽1Γα-βmax⁡tα-β-11-sα-1,t-sα-β-1,0⩽s⩽t⩽1,1-sα-1,0⩽t⩽s⩽1⩽1Γα-βmax⁡1-sα-1,1-sα-β-1,0⩽s⩽t⩽1,1-sα-1,0⩽t⩽s⩽1⩽1Γα-β1-sα-β-1. From (13), it is evident that(30)G0t,s=G0t,s⩽1Γα1-sα-1. It follows from (29) and (30) that (31)D0+βGt,s⩽D0+βG0t,s+Γαtα-β-1Γα-β1-δ1∑i=1mαi∫IiG0t,sdAt⩽1-sα-β-1Γα-β+ΓαΓα-β1-δ1∑i=1mαi∫Ii1-sα-1ΓαdAt=1-δ11-sα-β-1+δ21-sα-1Γα-β1-δ1=Λs.
The proof of the Lemma is completed.


## 3. Existence Result

Define the spaceE={u(t)∣u(t)∈C[0,1]andD0+βu(t)∈C[0,1]} is endowed with the ordering u⩽v if u(t)⩽v(t), for all t∈[0,1], and endowed with the norm u=max{u0,D0+βu0}, where u0=maxt∈[0,1]|u(t)|.Lemma 10.
( E , · ) is a Banach space.Proof.
Let{un}n=1∞ be a Cauchy sequence in the space (E,·). Then clearly {un}n=1∞ and {D0+βun}n=1∞ are Cauchy sequences in the space C[0,1]. Therefore, {un}n=1∞ and {D0+βun}n=1∞ converge to some v and w on [0,1] uniformly and v,w∈C[0,1]. We need to proof that w=D0+βv.
Note that(32)I0+βD0+βunt-I0+βwt⩽1Γβ∫0tt-sβ-1D0+βuns-wsds⩽1Γβ+1maxs∈0,1D0+βuns-ws. By the convergence of {D0+βun}n=1∞, we have(33)limn→∞I0+βD0+βunt=I0+βwtuniformly for t∈[0,1]. On the other hand, by Lemma 5 one has I0+βD0+βunt=unt+C1tβ-1, for t∈[0,1] and some C1∈R. Further, we can obtain(34)limn→∞I0+βD0+βunt=limn→∞unt+C1tβ-1=vt+C1tβ-1. From (33) and (34), we have(35)I0+βwt=vt+C1tβ-1,fort∈0,1. Taking the β-order derivative on both sides of (35) yields (36)D0+βI0+βwt=D0+βvt+C1tβ-1,fort∈0,1. In view of Remark 6 and Lemma 4, we find that (37)wt=D0+βvt,fort∈0,1. This completes the proof.Define the coneP⊂E by (38)P=u∈E∣ut⩾0.Let the nonnegative continuous concave functionalΨ on the cone P be defined by (39)Ψu=minθ⩽t⩽1-θut.Lemma 11.
Assume conditions(H1)–(H4) hold. For any u∈E, define the operator T by (40)Tut=∫01Gt,sfs,us,D0+βusds,0⩽t⩽1,and then T:P→P is completely continuous.Proof.
First, we prove thatT:P→P. In view of the nonnegativeness and continuity of G(t,s) and f(t,u(t),D0+βu(t)),  T is continuous and (Tu)(t)⩾0 for u∈P. Hence TP⊂P.
Next, we showT is uniformly bounded. Let Ω⊂P be bounded; that is, there exists a positive constant M>0 such that u⩽M, for all u∈Ω. Let L=1+max⁡{f(t,u(t),D0+βu(t))∣0⩽t⩽1,0⩽u⩽M,-M⩽D0+βu(t)⩽M}; then for u∈Ω, from the Lemma 9, we have (41)Tut=∫01Gt,sfs,us,D0+βusds⩽L∫01Gsds,D0+βTut=∫01D0+βGt,sfs,us,D0+βusds⩽L∫01D0+βGt,sds⩽L∫01Λsds.Hence, T(Ω) is bounded.
Finally, we showT is equicontinuous. Indeed, for any u∈Ω,  t1,t2∈[0,1],  t1<t2, we have (42)Tut2-Tut1⩽∫01Gt2,s-Gt1,sfs,us,D0+βusds⩽L∫01G0t2,s-G0t1,s+t2α-1-t1α-11-δ1∑i=1mαi∫IiG0t,sdAtds=L∫011Γαt2α-1-t1α-11-sα-1-1Γαt2-sα-1-t1-sα-1+t2α-1-t1α-11-δ1∑i=1mαi∫IiG0t,sdAtds⩽LΓα∫01t2α-1-t1α-11-sα-1ds+LΓα∫01t2-sα-1-t1-sα-1ds+L1-δ1∫01t2α-1-t1α-1∑i=1mαi∫IiG0t,sdAtds. Note that, applying the mean value theorem, we arrive at t2α-1-t1α-1<α-1t2-t1 and (t2-s)α-1-(t1-s)α-1<(α-1)(t2-t1), which implies that (43)Tut2-Tut1<Lα-12Γα+11-δ1max0⩽s⩽1∑i=1mαi∫IiG0t,sdAtt2-t1⟶0,ast2⟶t1.Moreover, (44)D0+βTut2-D0+βTut1⩽∫01D0+βGt2,s-D0+βGt1,sfs,us,D0+βusds⩽L∫01D0+βG0t2,s-D0+βG0t1,s+Γαt2α-β-1-t1α-β-1Γα-β1-δ1∑i=1mαi∫IiG0t,sdAtds=L∫011Γα-βt2α-β-1-t1α-β-11-sα-1-1Γα-βt2-sα-β-1-t1-sα-β-1+Γαt2α-β-1-t1α-β-1Γα-β1-δ1∑i=1mαi∫IiG0t,sdAtds⩽LΓα-β∫01t2α-β-1-t1α-β-11-sα-1ds+LΓα-β∫01t2-sα-β-1-t1-sα-β-1ds+LΓαΓα-β1-δ1∫01t2α-β-1-t1α-β-1∑i=1mαi∫IiG0t,sdAtds<LΓα-β-12+Γα1-δ1max0⩽s⩽1∑i=1mαi∫IiG0t,sdAtt2-t1⟶0,ast2⟶t1.
Therefore, (43) and (44) imply that T:P→P is equicontinuous for all u∈Ω. By means of the Arzela-Ascoli theorem, T:P→P is completely continuous.For convenience, we denote(45)M=1max∫01Λsds,∫01Gsds,N=1λ∫θ1-θGsds.Theorem 12.
Assume that conditions(H1)–(H4) hold, there exist nonnegative numbers 0<a<b<cθ, and f(t,u,v) satisfies the following conditions: H5
f(t,u,v)⩽Mc, for (t,u,v)∈[0,1]×[0,c]×[-c,c];H6
f(t,u,v)<Ma, for (t,u,v)∈[0,1]×[0,a]×[-a,a];H7
f(t,u,v)⩾Nb, for (t,u,v)∈[θ,1-θ]×[b,c]×[-c,c].
Then BVP (4) and (5) has at least three positive solutions u1, u2, and u3 such that (46)u1<a,b<Ψu2<u2⩽c,a<u3⩽cwithΨu3<b.Proof.
We will verity that the conditions(S1)–(S3) of Theorem 1 are satisfied.
LetP¯C={u∈E∣u(t)⩾0andu⩽c}. We first prove that T:P¯c→P¯c is completely continuous. From Lemma 11, we only need to prove that TP¯c⊂P¯c. For any u∈P¯c, we have 0⩽u(t)⩽c,   -c⩽D0+βu(t)⩽c, for all t∈[0,1]. The assumption (H5) implies f(t,u(t),D0+βu(t))⩽Mc for 0⩽t⩽1. Consequently, for t∈[0,1], (47)Tut=∫01Gt,sfs,us,D0+βusds⩽Mc∫01Gsds⩽c,D0+βTut=∫01D0+βGt,sfs,us,D0+βusds⩽Mc∫01D0+βGt,sds⩽Mc∫01Λsds⩽c.
Thus,Tu⩽c and further to get TP¯c⊂P¯c. Therefore T:P¯c→P¯c is completely continuous.
Similarly, the conditions(S2) of Theorem 1 can be obtained by the assumption (H6). Here we do not do more explanation.
Finally, in order to verityu∈P(Ψ,b,c)∣Ψ(u)>b≠⌀, we make u(t)=b/θtβ,0⩽t⩽1. It is easy to find that (48)Ψu=minθ⩽t⩽1-θbθtβ⩾bθθβ>b,u=maxmaxt∈0,1ut,maxt∈0,1D0+βut=maxbθ,Γβ+1bθ=bθ<c. If u∈P(Ψ,b,c), we have b⩽u(t)⩽c,-c⩽D0+βu(t)⩽c,  for θ⩽t⩽1-θ. Then (49)ΨTu=minθ⩽t⩽1-θTut⩾∫01λGsfs,us,D0+βusds>Nb∫θ1-θλGsds=b;that is, Ψ(Tu)>b for all u∈P(Ψ,b,c). This shows that condition (S1) of Theorem 1 is also satisfied.
From the above, BVP (4) and (5) has at least three positive solutions u1, u2, and u3 such that (50)u1<a,b<Ψu2<u2⩽c,a<u3⩽cwithΨu3<b.
The proof is completed.


## 4. Example

Here we provide two cases to verify the feasibility and breadth of the conclusion, where the strip intervals in boundary condition (5) satisfy intersection relation and inclusion relation in Examples 1 and 2, respectively.Example 1.
Consider the boundary value problem of nonlinear fractional differential equations as follows:(51)D0+2.3ut+ft,ut,D0+0.5ut=0,t∈0,1,u0=D0+0.5u0=0,u1=∑i=12αi∫IiutdAt,where α1=1/2,α2=3/5,I1=1/8,5/8, and I2=3/8,7/8 satisfy I1∩I2≠⌀ and I1,  I2 do not contain each other. Let A(t)=t2+t for t∈[0,1] and (52)ft,u,v=t10+v100+5u79,u<3,t10+v100+1215,u⩾3. It is easy to see that f(t,u,v) satisfies condition (H1).
Takeθ=1/10. By a simple calculation, we obtain δ1=0.5202<1,δ2=1.1125,δ3=0.3257,  ∫01G(s)ds=0.4827,  ∫01Λ(s)ds=1.3431,λ=0.0059,  M=0.7445,  N=366.23.
Seta=1, b=3, c=2000 such that 0<a<b<cθ, and, in addition, H5
f(t,u,v)⩽1235.1<1489=Mc, for (t,u,v)∈[0,1]×[0,2000]×[-2000,2000];H6
f(t,u,v)⩽0.6656<0.7445=Ma, for (t,u,v)∈[0,1]×[0,1]×[-1,1];H7
f(t,u,v)⩾1215.01>1098.69=Nb, for (t,u,v)∈[0.1,0.9]×[3,2000]×[-2000,2000].Thus, all the conditions are satisfied. According to Theorem12, BVP (51) has at least three positive solutions u1,  u2, and u3 such that u1<1, 3<mint∈[0.1,0.9]u2(t)<u2⩽2000, and 1<u3⩽2000 with mint∈[0.1,0.9]u3(t)<3.Example 2.
Consider the boundary value problem of nonlinear fractional differential equations as follows:(53)D0+2.3ut+ft,ut,D0+0.5ut=0,t∈0,1,u0=D0+0.5u0=0,u1=∑i=12αi∫IiutdAt, where α1=1/2,  α2=3/5,  I1=1/8,7/8,  I2=3/8,5/8, and I1⊃I2. Let A(t)=t2+t for t∈[0,1], (54)ft,u,v=t20+v50+9u616,u<4,t20+v50+2304,u⩾4. It is easy to see that f(t,u,v) satisfies condition (H1).
Takeθ=1/10. By a simple calculation, we obtain δ1=0.4768<1,  δ2=1.05,  δ3=0.2929,  ∫01G(s)ds=0.4414,  ∫01Λ(s)ds=1.5333,  λ=0.0058,  M=0.6522,  N=411.18.
Seta=1,  b=4,  c=4000 such that 0<a<b<cθ, and, in addition, (H5)
f(t,u,v)⩽2384.05<2608.8=Mc, for (t,u,v)∈[0,1]×[0,4000]×[-4000,4000];(H6)
f(t,u,v)⩽0.6325<0.6522=Ma, for (t,u,v)∈[0,1]×[0,1]×[-1,1];(H7)
f(t,u,v)⩾2304.005>1644.72=Nb, for (t,u,v)∈[0.1,0.9]×[4,4000]×[-4000,4000].Thus, all the conditions are satisfied. According to Theorem12, BVP (53) has at least three positive solutions u1,  u2, and u3 such that u1<1,  4<mint∈0.1,0.9u2(t)<u2⩽4000, and 1<u3⩽4000 with mint∈[0.1,0.9]u3(t)<4.



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*Source: 2352789-2018-04-24.xml*