# A Study of Nonlinear Fractional Differential Equations of Arbitrary Order with Riemann-Liouville Type Multistrip Boundary Conditions

Authors: Bashir Ahmad; Sotiris K. Ntouyas; Ahmed Alsaedi
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

We develop the existence theory for nonlinear fractional differential equations of arbitrary order with Riemann-Liouville type boundary conditions involving nonintersecting finite many strips of arbitrary length. Our results are based on some standard tools of fixed point theory. For the illustration of the results, some examples are also discussed.


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## Body

## 1. Introduction

The subject of fractional calculus has recently developed into a hot topic for the researchers in view of its numerous applications in the field of physics, mechanics, chemistry, engineering, and so forth. One can find the systematic progress of the topic in the books ([1–6]). A significant characteristic of a fractional-order differential operator distinguishing it from the integer-order differential operator is that it is nonlocal in nature, that is, the future state of a dynamical system or process involving fractional derivative depends on its current state as well its past states. In fact, this feature of fractional-order operators has contributed towards the popularity of fractional-order models, which are recognized as more realistic and practical than the classical integer-order models. In other words, we can say that the memory and hereditary properties of various materials and processes can be described by differential equations of arbitrary order. There has been a rapid development in the theoretical aspects such as periodicity, asymptotic behavior, and numerical methods for fractional equations. For some recent work on the topic, see ([7–23]) and the references therein. In particular, Ahmad et al. [22] studied nonlinear fractional differential equations and inclusions of arbitrary order with multistrip boundary conditions.In this paper, we continue the study initiated in [22] and consider a boundary value problem of fractional differential equations of arbitrary order q∈(n-1,n], n≥2 with finite many multistrip Riemann-Liouville type integral boundary conditions:
(1)cDqx(t)=f(t,x(t)),t∈[0,T],x(0)=0,x′(0)=0,…,x(n-2)(0)=0,x(T)=∑i=1m‍γi[Iβix(ηi)-Iβix(ζi)],
where cDq denotes the Caputo fractional derivative of order q, f is a given continuous function, Iβi is the Riemann-Liouville fractional integral of order βi>0, i=1,2,…,m, 0<ζ1<η1<ζ2<η2<…<ζm<ηm<T, and γi∈ℝ are suitable chosen constants.Regarding the motivation of the problem, we know that the strip conditions appear in the mathematical modeling of certain real world problems, for instance, see [24, 25]. In [22], the authors considered the nonlocal strip conditions of the form:
(2)x(1)=∑i=1n-2‍αi∫ζiηi‍x(s)ds,0<ζi<ηi,<1,i=1,2,…,(n-2).
In the problem (1), we have introduced Riemann-Liouville type multistrip integral boundary conditions which can be interpreted as the controller at the right-end of the interval under consideration is influenced by a discrete distribution of finite many nonintersecting sensors (strips) of arbitrary length expressed in terms of Riemann-Liouville type integral boundary conditions. For some engineering applications of strip conditions, see ([26–32]).The main objective of the present study is to develop some existence results for the problem (1) by using standard techniques of fixed point theory. The paper is organized as follows. In Section 2 we discuss a linear variant of the problem (1), which plays a key role in developing the main results presented in Section 3. For the illustration of the theory, we have also included some examples.


## 2. Preliminary Result

Let us begin this section with some basic definitions of fractional calculus [2–4].Definition 1.
Ifg(t)∈ACn[a,b], then the Caputo derivative of fractional order q is defined as
(3)cDa+qg(t)=1Γ(n-q)∫at‍(t-s)n-q-1g(n)(s)ds=Ia+n-qDng(x),n-1<q<n,n=[q]+1,
where [q] denotes the integer part of the real number q. For details, see Theorem 2.1 ([4, page 92]). Here ACn[a,b] denote the space of real valued functions g(t) which have continuous derivatives up to order n-1 on [a,b] such that gn-1(t)∈AC[a,b].Definition 2.
The Riemann-Liouville fractional integral of orderq is defined as
(4)Iqg(t)=1Γ(q)∫0t‍g(s)(t-s)1-qds,q>0,
provided the integral exists.The following result associated with a linear variant of problem (1) plays a pivotal role in establishing the main results.Lemma 3.
Forh∈C[0,T], the fractional boundary value problem
(5)cDqx(t)=h(t),t∈[0,T],q∈(n-1,n]x(0)=0,x′(0)=0,…,x(n-2)(0)=0,x(T)=∑i=1m‍γi[Iβix(ηi)-Iβix(ζi)],
has a unique solution x(t)∈ACn[0,T] given by
(6)x(t)=1Γ(q)∫0t‍(t-s)q-1h(s)ds-tn-1λΓ(q)∫0T‍(T-s)q-1h(s)ds+tn-1λΓ(q)∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1×(s-u)q-1h(u)duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×h(u)duds∫0ηi‍],
where
(7)λ=(Tn-1-∑i=1m‍γi(ηiβi+n-1-ζiβi+n-1)Γ(n)Γ(βi+n))≠0.Proof.
The general solution of fractional differential equations in (5) can be written as
(8)x(t)=1Γ(q)∫0t‍(t-s)q-1h(s)ds-c0-c1t-⋯-cn-1tn-1.
Using the given boundary conditions, it is found that c0=0, c1=0,…,cn-2=0. Applying the Riemann-Liouville integral operator Iβi on (8), we get
(9)Iβix(t)=1Γ(βi)∫0t‍(t-s)βi-1×(1Γ(q)∫0s‍(s-u)q-1h(u)du-cn-1sn-1)ds=1Γ(βi)Γ(q)∫0t‍∫0s‍(t-s)βi-1(s-u)q-1h(u)duds-cn-11Γ(βi)∫0t‍(t-s)βi-1sn-1ds.
Using the condition x(T)=∑i=1m‍γi[Iβix(ηi)-Iβix(ζi)], together with the fact that
(10)1Γ(βi)∫0t‍(t-s)βi-1sn-1ds=tβi+n-1Γ(n)Γ(βi+n),
we obtain
(11)1Γ(q)∫0T‍(T-s)q-1h(s)ds-cn-1Tn-1=∑i=1m‍γiΓ(q)Γ(βi)×[∫0ζi‍∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1h(u)duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1h(u)duds]-cn-1∑i=1m‍γi(ηiβi+n-1-ζiβi+n-1)Γ(n)Γ(βi+n),
which yields
(12)cn-1=1λΓ(q)∫0T‍(T-s)q-1h(s)ds-1λΓ(q)∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1×(s-u)q-1h(u)duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×h(u)duds∫0ηi‍],
where λ is given by (7). Substituting the values of c0,c1,…,cn-2,cn-1 in (8), we obtain (6). This completes the proof.


## 3. Main Results

Let𝒞:=C([0,T],ℝ) denotes the Banach space of all continuous functions defined on [0,T]×ℝ endowed with a topology of uniform convergence with the norm ∥x∥=supt∈[0,T]|x(t)|.By Lemma3, we define an operator 𝒫:𝒞→𝒞 as
(13)(𝒫x)(t)=1Γ(q)∫0t‍(t-s)q-1f(s,x(s))ds-tn-1λΓ(q)∫0T‍(T-s)q-1f(s,x(s))ds+tn-1λΓ(q)∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1×f(u,x(u))duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×f(u,x(u))duds∫0ηi‍],t∈[0,T].Observe that the problem (1) has a solution if and only if the associated fixed point problem 𝒫x=x has a fixed point.In the first result we prove an existence and uniqueness result by means of Banach’s contraction mapping principle. For the sake of convenience, we set(14)Λ=TqΓ(q+1)+Tq+n-1|λ|Γ(q+1)+Tn-1|λ|∑i=1m‍γiηiq+βi-ζiq+βiΓ(q+βi+1).Theorem 4.
Suppose thatf:[0,T]×ℝ→ℝ is a continuous function and satisfies the following assumption:(A3)
(15)|f(t,x)-f(t,y)|≤L|x-y|,∀t∈[0,1],L>0,x,y∈ℝ.Then the boundary value problem (1) has a unique solution provided
(16)L<1Λ,
where Λ is given by (14).Proof.
Withr≥MΛ/(1-LΛ), we define Br={x∈𝒞:∥x∥≤r}, where M=supt∈[0,T]|f(t,0)|<∞ and Λ is given by (14). Then we show that 𝒫Br⊂Br. For x∈Br, by means of the inequality |f(s,x(s))|≤|f(s,x(s))-f(s,0)|+|f(s,0)|≤L∥x∥+M≤Lr+M, it can easily be shown that
(17)∥𝒫x∥=(Lr+M)Λ≤r.
Now, for x,y∈𝒞 and for each t∈[0,T], we obtain
(18)∥(𝒫x)-(𝒫y)∥≤supt∈[0,T]{1Γ(q)∫0t‍(t-s)q-1|f(s,x(s))-f(s,y(s))|ds+tn-1λΓ(q)∫0T‍(T-s)q-1×|f(s,x(s))-f(s,y(s))|ds+tn-1λΓ(q)×∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1×(s-u)q-1du×|f(u,x(u))-f(u,y(u))|ds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×|f(u,x(u))-f(u,y(u))|duds1Γ(q)∫0t‍]}≤LΛ∥x-y∥.
Note that Λ depends only on the parameters involved in the problem. As LΛ<1, therefore 𝒫 is a contraction. Hence, by Banach’s contraction mapping principle, the problem (1) has a unique solution on [0,T].Example 5.
Let us consider the following4-strip nonlocal boundary value problem:
(19)cD9/2x(t)=f(t,x(t)),t∈[0,2],x(0)=0,x′(0)=0,x′′(0)=0,x′′′(0)=0,x(T)=∑i=14‍γi[Iβix(ηi)-Iβix(ζi)],
where q=9/2, n=5, ζ1=1/4, η1=1/2, ζ2=2/3, η2=1, ζ3=5/4, η3=4/3, ζ4=3/2, η4=7/4, γ1=5, γ2=10, γ3=15, γ4=25, β1=5/4, β2=7/4, β3=9/4, β4=11/4.With the given values of the parameters involved, we find that(20)λ=(Tn-1-∑i=1m‍γi(ηiβi+n-1-ζiβi+n-1)Γ(n)Γ(βi+n))≃9.334784,Λ=TqΓ(q+1)+Tq+n-1|λ|Γ(q+1)+Tn-1|λ|∑i=1m‍γiηiq+βi-ζiq+βiΓ(q+βi+1)≃1.406972.
Let us choose
(21)f(t,x(t))=1(t+8)3(tan-1x)+4+3sin2t.
Clearly L=1/2 as |f(t,x)-f(t,y)|≤(1/2)|x-y| and L<1/Λ, where Λ≃1.406972. Therefore all the conditions of Theorem 4 hold and consequently there exists a unique solution for the problem (19) with f(t,x(t)) given by (21).In case of the following unbounded nonlinear function:(22)f(t,x(t))=x7+1(t+8)3(tan-1x)+4+3sin2t,
we have L=9/14 and L<1/Λ (Λ≃1.406972). As before, the problem (19) with f(t,x(t)) given by (22) has a unique solution.In the second result we use the Leray-Schauder alternative.Theorem 6 ((Leray-Schauder alternative) [33, page 4]).
LetX be a Banach space. Assume that T:X→X is completely continuous operator and the set
(23)V={u∈X∣u=μTu,0<μ<1}
is bounded. Then T has a fixed point in X.Theorem 7.
Assume that there exists a positive constantL1 such that |f(t,x)|≤L1 for t∈[0,T], x∈ℝ. Then the problem (1) has at least one solution.Proof.
First of all, we show that the operator𝒫 is completely continuous. Note that the operator 𝒫 is continuous in view of the continuity of f. Let ℬ⊂𝒞 be a bounded set. By the assumption that |f(t,x)|≤L1, for x∈ℬ, we have
(24)|(𝒫x)(t)|≤1Γ(q)∫0t‍(t-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)∫0T‍(T-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1×|f(u,x(u))|duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×|f(u,x(u))|duds∫0ηi‍]≤L1[1Γ(q)∫0t‍(t-s)q-1ds+tn-1|λ|Γ(q)∫0t‍(T-s)q-1ds+tn-1|λ|Γ(q)∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1×(s-u)q-1duds-∫0ζi‍∫0s‍(ζi-s)βi-1×(s-u)q-1duds∫0ηi‍]1Γ(q)∫0t‍]≤L1{ηiq+βi-ζiq+βiΓ(q+βi+1)TqΓ(q+1)+Tq+n-1|λ|Γ(q+1)+Tn-1|λ|∑i=1m‍γiηiq+βi-ζiq+βiΓ(q+βi+1)}=L2,
which implies that ∥(𝒫x)∥≤L2. Further, we find that
(25)|(𝒫x)′(t)|=1Γ(q-1)∫0t‍(t-s)q-2|f(s,x(s))|ds+(n-1)tn-2|λ|Γ(q)∫0T‍(T-s)q-1|f(s,x(s))|ds+(n-1)tn-2|λ|Γ(q)×∑i=1m‍γiΓ(βi)[∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1×|f(u,x(u))|duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×|f(u,x(u))|duds∫0ηi‍]≤L1[1Γ(q-1)∫0t‍(t-s)q-2ds+(n-1)tn-2|λ|Γ(q)∫0T‍(T-s)q-1ds+(n-1)tn-2|λ|Γ(q)×∑i=1m‍γiΓ(βi)(∫0ηi‍∫0s‍(ηi-s)βi-1×(s-u)q-1×|f(u,x(u))|duds-∫0ζi‍∫0s‍(ζi-s)βi-1×(s-u)q-1×|f(u,x(u))|×duds1Γ(q-1)∫0ηi‍)]≤L1{ηiq+βi-ζiq+βiΓ(q+βi+1Tq-1Γ(q)+(n-1)Tq+n-2|λ|Γ(q+1)+(n-1)Tn-2|λ|∑i=1m‍γjηiq+βi-ζiq+βiΓ(q+βi+1)}=L3.
Hence, for t1,t2∈[0,T], we have
(26)|(𝒫x)(t2)-(𝒫x)(t1)|≤∫t1t2‍|(𝒫x)′(s)|ds≤L3(t2-t1).
This implies that 𝒫 is equicontinuous on [0,T]. Thus, by the Arzelá-Ascoli theorem, the operator 𝒫:𝒞→𝒞 is completely continuous.
Next, we consider the set(27)V={x∈𝒞∣x=μ𝒫x,0<μ<1},
and show that the set V is bounded. Let x∈V, then x=μ𝒫x,0<μ<1. For any t∈[0,T], we have
(28)|x(t)|=μ|(𝒫x)(t)|≤1Γ(q)∫0t‍(t-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)∫0T‍(T-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1×|f(u,x(u))|duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×|f(u,x(u))|duds∫0ηi‍]≤L1{ηiq+βi-ζiq+βiΓ(q+βi+1TqΓ(q+1)+Tq+n-1|λ|Γ(q+1)+Tn-1|λ|∑i=1m‍γiηiq+βi-ζiq+βiΓ(q+βi+1)}=M1.
Thus,∥x∥≤M1 for any t∈[0,T]. So, the set V is bounded. Thus, by the conclusion of Theorem 6, the operator 𝒫 has at least one fixed point, which implies that (1) has at least one solution.Example 8.
Consider the boundary value problem of Example5 with
(29)f(t,x(t))=3e(2-|x(t)|)3[cos4t+2ln(1+4sin2x(t))](10+cosx(t)).
Observe that |f(t,x)|≤L1 with L1=e22(1+ln25). Thus the hypothesis of Theorem 7 is satisfied. Hence by the conclusion of Theorem 7, the problem (19) with f(t,x(t)) given by (29) has at least one solution.In the next we prove one more existence result for problem (1), based on the following known result.Theorem 9 (see [34]).
LetX be a Banach space. Assume that Ω is an open bounded subset of X with θ∈Ω and let T:Ω¯→X be a completely continuous operator such that
(30)∥Tu∥≤∥u∥,∀u∈∂Ω.
Then T has a fixed point in Ω¯.Theorem 10.
Let there exist a small positive numberτ such that |f(t,x)|≤ν|x| for 0<|x|<τ, with 0<ν≤1/Λ, where Λ is given by (14). Then the problem (1) has at least one solution.Proof.
Let us defineℬτ={x∈𝒞∣∥x∥<τ} and take x∈𝒞 such that ∥x∥=τ, that is, x∈∂ℬτ. As before, it can be shown that 𝒫 is completely continuous and
(31)∥𝒫x∥≤supt∈[0,t]{1Γ(q)∫0t‍(t-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)×∫0T‍(T-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)×∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1×|f(u,x(u))|duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×|f(u,x(u))|duds∫0ηi‍]1Γ(q)∫0t‍(t-s)q-1}≤Λν∥x∥,
which in view of the given condition (νΛ≤1), gives ∥𝒫x∥≤∥x∥, x∈∂ℬτ. Therefore, by Theorem 9, the operator 𝒫 has at least one fixed point, which in turn implies that the problem (1) has at least one solution.Example 11.
Consider the boundary value problem of Example5 and let us consider
(32)f(t,x(t))=x(b5+x4(t))1/5+2(1+cos(t4+3))5×(1-cosx(t)),x≠0,b>0.
For sufficiently small x (ignoring x2 and higher powers of x), we have
(33)|x(b5+x4(t))1/5+2(1+cos(t4+3))5(1-cosx(t))|≤b|x|.
Choosing b≤1/Λ, all the assumptions of Theorem 10 hold. Therefore, the conclusion of Theorem 10 implies that the problem (19) with f(t,x(t)) given by (32) has at least one solution.Our final existence result is based on Leray-Schauder nonlinear alternative.Lemma 12 ((Nonlinear alternative for single valued maps) [33, page 135]).
LetE be a Banach space, C a closed, convex subset of E, U an open subset of C and 0∈U. Suppose that F:U¯→C is a continuous, compact (i.e., F(U¯) is a relatively compact subset of C) map. Then either(i)
Fhas a fixed point in U¯, or(ii)
there is au∈∂U (the boundary of U in C) and λ∈(0,1) with u=λF(u).Theorem 13.
Assume that(A1)
there exist a functionσ∈C([0,1],ℝ+), and a nondecreasing function ψ:ℝ+→ℝ+ such that |f(t,x)|≤σ(t)ψ(∥x∥), for all (t,x)∈[0,T]×ℝ;(A2)
there exists a constantM>0 such that
(34)Mψ(M)Λ∥σ∥>1.Then the boundary value problem (1) has at least one solution on
[
0
,
T
].Proof.
Consider the operator𝒫:𝒞→𝒞 defined by (13). We show that 𝒫  maps bounded sets into bounded sets inC([0,T],ℝ). For a positive number r, let Br={x∈C([0,T],ℝ):∥x∥≤r} be a bounded set in C([0,T],ℝ). Then
(35)∥𝒫x∥≤supt∈[0,T]{1Γ(q)∫0t‍(t-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)∫0T‍(T-s)q-1|f(s,x(s))|ds+tn-1|λ|Γ(q)×∑i=1m‍γiΓ(βi)×[∫0ηi‍∫0s‍(ηi-s)βi-1×(s-u)q-1×|f(u,x(u))|duds-∫0ζi‍∫0s‍(ζi-s)βi-1×(s-u)q-1×|f(u,x(u))|×duds∫0ηi‍]1Γ(q)∫0t‍}≤ψ(r){ηiq+βi-ζiq+βiΓ(q+βi+1TqΓ(q+1)+Tq+n-1|λ|Γ(q+1)+Tn-1|λ|∑i=1m‍γiηiq+βi-ζiq+βiΓ(q+βi+1)}∥σ∥.
Next we show that Fmaps bounded sets into equicontinuous sets ofC([0,1],ℝ). Let t′,t′′∈[0,1] with t′<t′′ and x∈Br, where Br is a bounded set of C([0,1],ℝ). Then we obtain
(36)|(𝒫x)(t′′)-(𝒫x)(t′)|=|1Γ(q)∫0t′′‍(t′′-s)q-1f(s,x(s))ds-1Γ(q)∫0t′‍(t′-s)q-1f(s,x(s))ds-[(t′′)n-1-(t′)n-1]λΓ(q)∫0T‍(T-s)q-1f(s,x(s))ds+[(t′′)n-1-(t′′)n-1]λΓ(q)×∑i=1m‍γiΓ(βi)[∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1×|f(u,x(u))|duds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1×|f(u,x(u))|duds∫0ηi‍]∫0t′′‍|≤1Γ(q)∫0t′‍|(t′′-s)q-1-(t′-s)q-1|ψ(r)σ(s)ds+1Γ(q)∫t′t′′‍|t′′-s|q-1ψ(r)σ(s)ds+|(t′′)n-1-(t′)n-1||λ|Γ(q)∫0T‍|T-s|q-1ψ(r)σ(s)ds+|(t′′)n-1-(t′)n-1||λ|Γ(q)×∑i=1m‍γiΓ(βi)×[∫0ζi‍∫0ηi‍∫0s‍(ηi-s)βi-1(s-u)q-1duψ(r)σ(s)ds-∫0ζi‍∫0s‍(ζi-s)βi-1(s-u)q-1ψ(r)σ(s)duds].
Obviously the right hand side of the above inequality tends to zero independently of x∈Br as t′′-t′→0. As 𝒫:C([0,T],ℝ)→C([0,T],ℝ) satisfies the above assumptions, therefore it follows by the Arzelá-Ascoli theorem that 𝒫 is completely continuous.
Letx be a solution. Then, for t∈[0,T], and following the similar computations as before, we find that
(37)|x(t)|=|μ(𝒫x)(t)|≤ψ(r){ηiq+βi-ζiq+βiΓ(q+βi+1TqΓ(q+1)+Tq+n-1|λ|Γ(q+1)+Tn-1|λ|∑i=1m‍γiηiq+βi-ζiq+βiΓ(q+βi+1)}∥σ∥.
In consequence, we have
(38)∥x∥ψ(∥x∥)Λ∥σ∥≤1.
Thus, by (A2), there exists M such that ∥x∥≠M. Let us set
(39)V={x∈C([0,T],ℝ):∥x∥<M+1}.
Note that the operator 𝒫:V¯→C([0,T],ℝ) is continuous and completely continuous. From the choice of V, there is no x∈∂V such that x=μ𝒫(x) for some μ∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 12), we deduce that 𝒫 has a fixed point x∈V¯ which is a solution of the problem (1). This completes the proof.Example 14.
Consider the boundary value problem of Example5 with
(40)f(t,x(t))=1t+4(1+|x|1+|x|)≤σ(t)ψ(∥x∥).
Then σ(t)=1/t+4 and ψ(∥x∥)=2. Using ∥σ∥=1/2,Λ≃1.406972, we find by the condition (A2) that M>Λ. Thus all the assumptions of Theorem 13 are satisfied. Hence, it follows by Theorem 13 that the problem (19) with f(t,x(t)) defined by (40) has at least one solution.
If we choose an unbounded nonlinearity as follows:(41)f(t,x(t))=1t+4(1+|x|1+|x|+|x|2).
Then f(t,x(t))≤σ(t)ψ(∥x∥) with σ(t)=1/t+4 and ψ(∥x∥)=2+∥x∥/2. Using the earlier arguments, with ∥σ∥=1/2, Λ≃1.406972, we find that M>M1, M1≈2.170392. Hence the problem (19) with f(t,x(t)) given by (41) has at least one solution.



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*Source: 320415-2013-03-05.xml*