# Uniqueness of Positive Solutions for a Class of Fourth-Order Boundary Value Problems

Authors: J. Caballero; J. Harjani; K. Sadarangani
Journal: Abstract and Applied Analysis
 (2011)

---

## Abstract

The purpose of this paper is to investigate the existence and uniqueness of
positive solutions for the following fourth-order boundary value problem:y(4)(t)=f(t,y(t)), t∈[0,1], y(0)=y(1)=y′(0)=y′(1)=0. Moreover, under certain assumptions, we will prove that the above boundary
value problem has a unique symmetric positive solution.
Finally, we present some examples and we compare our results with the ones
obtained in recent papers.
Our analysis relies on a fixed point theorem in partially ordered metric spaces.


---

## Body

## 1. Introduction

The purpose of this paper is to consider the existence and uniqueness of positive solutions for the following fourth-order two-point boundary value problem:(1.1)y(4)(t)=f(t,y(t)),t∈[0,1],y(0)=y(1)=y′(0)=y′(1)=0,
which describes the bending of an elastic beam clamped at both endpoints.There have been extensive studies on fourth-order boundary value problems with diverse boundary conditions. Some of the main tools of nonlinear analysis devoted to the study of this type of problems are, among others, lower and upper solutions [1–4], monotone iterative technique [5–7], Krasnoselskii fixed point theorem [8], fixed point index [9–11], Leray-Schauder degree [12, 13], and bifurcation theory [14–16].


## 2. Background

In this section, we present some basic facts which are necessary for our results.In our study, we will use a fixed point theorem in partially ordered metric spaces which appears in [17].Letℳ denote the class of those functions β:[0,∞)→[0,1) satisfying the condition (2.1)β(tn)⟶1impliestn→0.Now, we recall the above mentioned fixed point theorem.Theorem 2.1 (see1, Theorem 2.1).
Let(X,≤) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a complete metric space. Let T:X→X be a nondecreasing mapping such that there exists an element x0∈X with x0≤Tx0. Suppose that there exists β∈ℳ such that
(2.2)d(Tx,Ty)≤β(d(x,y))⋅d(x,y),foranyx,y∈Xwithx≥y.
Assume that eitherT is continuous or X is such that
(2.3)if(xn)isanondecreasingsequenceinXsuchthatxn⟶x,thenxn≤xforalln∈N.
Besides, suppose that
(2.4)foreachx,y∈X,thereexistsz∈Xwhichiscomparabletoxandy.
Then T has a unique fixed point.In our considerations, we will work with a subset of the classical Banach space𝒞[0,1]. This space will be considered with the standard metric (2.5)d(x,y)=sup⁡0≤t≤1|x(t)-y(t)|.
This space can be equipped with a partial order given by (2.6)x,y∈C[0,1],x≤y⟺x(t)≤y(t),fort∈[0,1].
In [18], it is proved that (𝒞[0,1],≤) with the above mentioned metric satisfies condition (2.3) of Theorem 2.1. Moreover, for x,y∈𝒞[0,1], as the function max(x,y)∈𝒞[0,1], (𝒞[0,1],≤) satisfies condition (2.4).On the other hand, the boundary value problem (1.1) can be rewritten as the integral equation (see, e.g., [19]) (2.7)y(t)=∫01G(t,s)f(s,u(s))ds,fort∈[0,1],
where G(t,s) is the Green's function given by (2.8)G(t,s)=16{t2(1-s)2[(s-t)+2(1-t)s],0≤t≤s≤1,s2(1-t)2[(t-s)+2(1-s)t],0≤s≤t≤1.
Note that G(t,s) satisfies the following properties:  (i)
G(t,s) is a continuous function on [0,1]×[0,1],(ii)
G(0,s)=G(1,s)=0, for s∈[0,1],(iii)
G(t,s)≥0, for t,s∈[0,1].


## 3. Main Results

Our starting point in this section is to present the class of functions𝒜 which we use later. By 𝒜 we denote the class of functions ϕ:[0,∞)→[0,∞) satisfying the following conditions:  (i)
ϕ is nondecreasing,(ii)
for anyx>0, ϕ(x)<x,(iii)
β(x)=ϕ(x)/x∈ℳ.Examples of functions in𝒜 are ϕ(x)=μx with 0≤μ<1, ϕ(x)=x/(1+x) and ϕ(x)=ln(1+x). In the sequel, we formulate our main result.Theorem 3.1.
Consider problem (1.1) assuming the following hypotheses:  (a)
f:[0,1]×[0,∞)→[0,∞) is continuous,(b)
f(t,y) is nondecreasing with respect to the second variable, for each t∈[0,1],(c)
suppose that there exists0<α≤384, such that, for x,y∈[0,∞) with y≥x,(3.1)f(t,y)-f(t,x)≤αϕ(y-x),withϕ∈A.
Then, problem (1.1) has a unique nonnegative solution.Proof.
Consider the cone(3.2)P={x∈C[0,1]:x≥0}.
Obviously, (P,d) with d(x,y)=sup{|x(t)-y(t)|:t∈[0,1]} is a complete metric space satisfying condition (2.3) and condition (2.4) of Theorem 2.1.
Consider the operator defined by(3.3)(Tx)(t)=∫01G(t,s)f(s,x(s))ds,forx∈P,
where G(t,s) is the Green's function defined in Section 2.
It is clear thatT applies the cone P into itself since f(t,x) and G(t,s) are nonnegative continuous functions.
Now, we check that assumptions in Theorems2.1 are satisfied.
Firstly, the operatorT is nondecreasing.
Indeed, sincef is nondecreasing with respect to the second variable, for u,v∈P, u≥v and t∈[0,1], we have(3.4)(Tu)(t)=∫01G(t,s)f(s,u(s))ds≥∫01G(t,s)f(s,v(s))ds=(Tv)(t).
On the other hand, a straightforward calculation gives us
(3.5)∫01G(t,s)ds=∫0tG(t,s)ds+∫t1G(t,s)ds=t224-t312+t424,max⁡0≤t≤1∫01G(t,s)ds=max⁡0≤t≤1(t224-t312+t424)=1384.
Taking into account this fact and our hypotheses, for u,v∈P and u>v, we can obtain the following estimate:
(3.6)d(Tu,Tv)=sup⁡0≤t≤1∫01G(t,s)(f(s,u(s))-f(s,v(s)))ds|(Tu)(t)-(Tv)(t)∫01G(t,s)(f(s,u(s))-f(s,v(s)))ds|=sup⁡0≤t≤1((Tu)(t)-(Tv)(t))=sup⁡0≤t≤1∫01G(t,s)(f(s,u(s))-f(s,v(s)))ds≤sup⁡0≤t≤1∫01G(t,s)αϕ(u(s)-v(s))ds≤sup⁡0≤t≤1∫01G(t,s)αϕ(d(u,v))ds=αϕ(d(u,v))sup⁡0≤t≤1∫01G(t,s)ds=αϕ(d(u,v))⋅1384≤ϕ(d(u,v))=ϕ(d(u,v))d(u,v)⋅d(u,v).
This gives us, for u,v∈P and u>v,
(3.7)d(Tu,Tv)≤β(d(u,v))⋅d(u,v),
where β(x)=ϕ(x)/x∈ℳ.
Obviously, the last inequality is satisfied foru=v.
Therefore, the contractive condition appearing in Theorem2.1 is satisfied for u≥v. Besides, as f and G are nonnegative functions,
(3.8)T0=∫01G(t,s)f(s,0)ds≥0.
Finally, Theorem2.1 tells us that T has a unique fixed point in P, and this means that problem (1.1) has a unique nonnegative solution.
This finishes the proof.Now, we present a sufficient condition for the existence and uniqueness of positive solutions for our problem (1.1) (positive solution means x(t)>0, for t∈(0,1)). The proof of the following theorem is similar to the proof of Theorem 3.6 of [8]. We present a proof for completeness.Theorem 3.2.
Under assumptions of Theorem3.1 and suppose that f(t0,0)≠0 for certain t0∈[0,1], problem (1.1) has a unique positive solution.Proof.
Consider the nonnegative solutionx(t) given by Theorem 3.1 of problem (1.1).
Notice that this solution satisfies(3.9)x(t)=∫01G(t,s)f(s,x(s))ds.
Now, we will prove thatx is a positive solution.
In contrary case, suppose that there exists0<t*<1 such that x(t*)=0 and, consequently,(3.10)x(t*)=∫01G(t*,s)f(s,x(s))ds=0.
Since x≥0, f is nondecreasing with respect to the second variable and G(t,s)≥0, we have
(3.11)0=x(t*)=∫01G(t*,s)f(s,x(s))ds≥∫01G(t*,s)f(s,0)ds≥0,
and this gives us
(3.12)∫01G(t*,s)f(s,0)ds=0.This fact and the nonnegative character of G(t,s) and f(t,x) imply
(3.13)G(t*,s)⋅f(s,0)=0a⋅e(s).
As G(t*,s)≠0a·e (s), because G(t*,s) is given by a polynomial, we obtain
(3.14)f(s,0)=0a⋅e(s).
On the other hand, as f(t0,0)≠0 for certain t0∈[0,1] and f(t0,x)≥0, we have that f(t0,0)>0.
The continuity off gives us the existence of a set A⊂[0,1] with t0∈A and μ(A)>0, where μ is the Lebesgue measure, satisfying that f(t,0)>0 for any t∈A. This contradicts (3.14).
Therefore,x(t)>0 for t∈(0,1).
This finishes the proof.Now, we present an example which illustrates our results.Example 3.3.
Consider the nonlinear fourth-order two-point boundary value problem(3.15)y(4)(t)=c+λarctan(y(t)),t∈(0,1),c,λ>0,y(0)=y(1)=y′(0)=y′(1)=0.
In this case,f(t,y)=c+λarctany. It is easily seen that f(t,y) satisfies (a) and (b) of Theorem 3.1.
In order to prove thatf(t,y) satisfies (c) of Theorem 3.1, previously, we will prove that the function ϕ:[0,∞)→[0,∞), defined by ϕ(x)=arctanx, satisfies
(3.16)ϕ(u)-ϕ(v)≤ϕ(u-v)foru≥v.
In fact, putϕ(u)=arctanu=α and ϕ(v)=arctanv=β (notice that, as u≥v and ϕ is nondecreasing, α≥β). Then, from
(3.17)tan(α-β)=tanα-tanβ1+tanα⋅tanβ,
as α,β∈[0,π/2), then tanα,tanβ∈[0,∞), we obtain
(3.18)tan(α-β)≤tanα-tanβ.
Applying ϕ to this inequality and taking into account the nondecreasing character of ϕ, we have
(3.19)α-β≤arctan(tanα-tanβ)
or, equivalently,
(3.20)ϕ(u)-ϕ(v)=arctanu-arctanv≤arctan(u-v)=ϕ(u-v).
This proves our claim.
In the sequel, we prove thatf(t,y) satisfies assumption (c) of Theorem 3.1.
In fact, fory≥x and t∈[0,1], we can obtain
(3.21)f(t,y)-f(t,x)=λ(arctany-arctanx)≤λarctan(y-x).
Now, we will prove thatϕ(x)=arctanx belongs to 𝒜. In fact, obviously ϕ takes [0,∞) into itself and, as ϕ′(x)=1/(1+x2), ϕ is nondecreasing. Besides, as the derivative of ψ(x)=x-ϕ(x) is ψ′(x)=1-1/(1+x2)>0 for x>0, ψ is strictly increasing, and, consequently, ϕ(x)<x for x>0 (notice that ψ(0)=0). Notice that if β(x)=ϕ(x)/x=arctanx/x and β(tn)→1, then (tn) is a bounded sequence because, in contrary case, tn→∞ and, thus, β(tn)→0. Suppose that tn↛0. Then, we can find ϵ>0 such that, for each n∈ℕ, there exists pn≥n with tpn≥ϵ. The bounded character of (tn) gives us the existence of a subsequence (tkn) of (tpn) with (tkn) convergent. Suppose that tkn→a. From β(tn)→1, we obtain arctantkn/tkn→arctana/a=1 and, as the unique solution of arctanx=x is x0=0, we obtain a=0. Thus, tkn→0, and this contradicts the fact that tkn≥ϵ for any n∈ℕ. Therefore, tn→0. This proves that f(t,y) satisfies assumption (c) of Theorem 3.1. Finally, as f(t,0)=c>0, Problem (3.15) has a unique positive solution for 0<λ≤384 by Theorems 3.1 and 3.2.Remark 3.4.
In Theorem3.2, the condition f(t0,0)≠0 for certain t0∈[0,1] seems to be a strong condition in order to obtain a positive solution for Problem (1.1), but when the solution is unique, we will see that this condition is very adjusted one. More precisely, under assumption that Problem (1.1) has a unique nonnegative solution x(t), then
(3.22)f(t,0)=0fort∈[0,1]iffx(t)≡0.
In fact, iff(t,0)=0 for t∈[0,1], then it is easily seen that the zero function satisfies Problem (1.1) and the uniqueness of solution gives us x(t)≡0.
The other implication is obvious since if the zero function is solution of Problem (1.1), then 0=f(t,0) for any t∈[0,1].Remark 3.5.
Notice that assumptions in Theorem3.1 are invariant by continuous perturbations. More precisely, if f(t,0)=0 for any t∈[0,1] and f satisfies (a), (b), and (c) of Theorem 3.1, then g(t,x)=a(t)+f(t,x), with a:[0,1]→[0,∞) continuous and a≠0, satisfies assumptions of Theorem 3.2, and this means that the following boundary value problem
(3.23)y(4)(t)=g(t,y(t)),t∈[0,1],y(0)=y(1)=y′(0)=y′(1)=0,
has a unique positive solution.


## 4. Some Remarks

In this section, we compare our results with the ones obtained in recent papers. Recently, in [19], the authors present as main result the following theorem.Theorem 4.1 (Theorem 3.1 of [19]).
Suppose that(H1)
f:[0,1]×[0,∞)→[0,∞) is continuous,(H2)
f(t,y) is nondecreasing in y, for each t∈[0,1],(H3)
f(t,y)=f(1-t,y) for each (t,y)∈[0,1]×[0,∞). Moreover, suppose that there exist positive numbers a>b such that
(4.1)max⁡0≤t≤1f(t,a)≤a⋅A,min⁡1/4≤t≤3/4f(t,b16)≥b⋅B,
where
(4.2)A=(max⁡0≤t≤1∫01G(t,s)ds)-1,B=(max⁡0≤t≤1∫1/43/4G(t,s)ds)-1,
with G(t,s) being the Green's function defined in Section 2.Then, Problem (1.1) has at least one symmetric positive solution y*∈𝒞[0,1] such that b≤∥y*∥≤a and, moreover, y*=limk→∞Tky0 in the uniform norm, where T is the operator defined by
(4.3)(Tx)(t)=∫01G(t,s)f(s,x(s))ds,forx∈C[0,1]
and y0 is the function given by y0(t)=b·q(t), for t∈[0,1], with q(t)=min(t2,(1-t)2), for t∈[0,1] (symmetric solution means a solution y(t) satisfying y(t)=y(1-t), for t∈[0,1]).In what follows, we present a parallel result to Theorem3.2 where we obtain uniqueness of a symmetric positive solution of Problem (1.1).Theorem 4.2.
Adding assumption(H3) of Theorem 4.1 to the hypotheses of Theorem 3.2, one obtains a unique symmetric positive solution of Problem (1.1).Proof.
As in the proof of Theorem3.1, instead of P, we consider the following setK(4.4)K={x∈C[0,1]:x≥0andxissymmetric}.
It is easily seen thatK is a closed subset of 𝒞[0,1]. Thus, (K,d), where d is the induced metric given by
(4.5)d(x,y)=sup⁡0≤t≤1|x(t)-y(t)|,forx,y∈K,
is a complete metric space.
Moreover,K with the induced order by (𝒞[0,1],≤) satisfies condition (2.3) of Theorem 2.1, and it is easily proved that the function max(x,y)∈K, for x,y∈K and, consequently, (K,≤), satisfies condition (2.4) of Theorem 2.1.
Now, as in Theorem2.1, we consider the operator defined by(4.6)(Tx)(t)=∫01G(t,s)f(s,x(s))ds,forx∈K.
In the sequel, we prove that, under our assumptions, T applies K into itself.
In fact, suppose thatx is symmetric, then for t∈[0,1], we have(4.7)(Tx)(1-t)=∫01G(1-t,s)f(s,x(s))ds.
Making the change of variables s=1-u, we obtain
(4.8)(Tx)(1-t)=-∫10G(1-t,1-u)f(1-u,x(1-u))du=∫01G(1-t,1-u)f(1-u,x(1-u))du.
Now, it is easily seen that G(t,s)=G(1-t,1-s) for t,s∈[0,1] and taking into account assumption (H3) of Theorem 4.1 and the symmetric character of x, we have
(4.9)(Tx)(1-t)=∫01G(t,u)f(u,x(1-u))du=∫01G(t,u)f(u,x(u))du=(Tx)(t).
The rest of the proof follows the lines of Theorems 3.1 and 3.2.
This finishes the proof.Now, we present an example which illustrates Theorem4.2.Example 4.3.
Consider the following problem(4.10)y(4)(t)=c+λsin(πt)arctan(y(t)),t∈(0,1),c,λ>0,y(0)=y(1)=y′(0)=y′(1)=0.
In this case,f(t,y)=c+λsin(πt)arctany. It is easily checked that f(t,y) satisfies (a) and (b) of Theorem 3.1 and f(t,y)=f(1-t,y), for (t,y)∈[0,1]×[0,∞).
On the other hand, taking into account Example3.3, we can obtain, for y≥x and t∈[0,1],
(4.11)f(t,y)-f(t,x)=λsinπt[arctany-arctanx]≤λsinπt[arctan(y-x)]≤λarctan(y-x).
Finally, as it is proved in Example3.3, ϕ(x)=arctanx belongs to 𝒜. Therefore, Theorem 4.2 tells us that Problem (4.10) has a unique symmetric positive solution for 0<λ≤384. In what follows, we prove that Problem (4.10) can be treated using Theorem 4.1. In fact, in this case, f(t,y)=c+λsin(πt)arctany. Moreover, A=384 (see proof of Theorem 3.1); it can be proved that B=531.61. As we have seen in Example 4.3, f(t,y) satisfies assumptions (H1), (H2), and (H3) of Theorem 4.1. Moreover,(4.12)max⁡0≤t≤1f(t,a)=f(12,a)=c+λarctana,min⁡1/4≤t≤3/4f(t,b16)=f(14,b16)=c+λsinπ4arctan(b16)=c+λ22arctan(b16).Consider the function φ(a)=384·a-(c+λarctana), with 0<λ≤384 and a∈[0,∞). Obviously, φ(0)=-c<0 and, as lima→∞φ(a)=∞, we can find a0>0 such that φ(a0)>0. This means that
(4.13)c+λarctana0≤384a0.
On the other hand, we consider the function ψ(b)=c+λ(2/2)arctan(b/16)-531.61·b, with 0<λ≤384 and b∈[0,∞).
Then, asψ(0)=c>0 and ψ is a continuous function, we can find b0 such that(4.14)min⁡1/4≤t≤3/4f(t,b016)=c+λ22arctan(b016)≥b0⋅531.61.
Therefore, Problem (4.10) can be treated using Theorem 4.1, and we obtain the existence of a symmetric positive solution.
Our main contribution is the uniqueness of the solution.
In what follows, we present the following example which can be treated by Theorem4.2 and Theorem 4.1 cannot be used.Example 4.4.
Consider the following problem which is a variant of Example4.3:
(4.15)y(4)(t)=c(t)+λsin(πt)arctan(y(t)),t∈(0,1),λ>0y(0)=y(1)=y′(0)=y′(1)=0,
where c(t) is a symmetric positive function satisfying c(1/4)=0, for example,
(4.16)c(t)={1-4t,0≤t≤14,0,14≤t≤344t-3,34≤t≤1.
In this case,f(t,y)=c(t)+λsin(πt)arctan(y(t)). Taking into account Example 4.3, it is easily proved that f(t,y) satisfies assumptions of Theorem 4.2, and, consequently, Problem (4.15) has a unique symmetric positive solution for 0<λ≤384.
Now, we prove thatf(t,y) does not satisfy assumptions of Theorem 4.1 and, consequently, Problem (4.15) cannot be treated using this theorem. In fact, in this case (notice that c(1/4)=0),(4.17)min⁡1/4≤t≤3/4f(t,b16)=f(14,b16)=λsinπ4arctan(b16)=λ22arctan(b16),
and we cannot find a positive number b0 such that
(4.18)λ22arctan(b016)≥b0⋅531.61,for0<λ≤384.
This proves that Problem (4.15) cannot be treated by Theorem 4.1.
Now, we compare our results with the ones obtained in [14]. In [14], the author studies positive solutions of the problem(4.19)u(iv)(x)=λf(u(x)),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
using theory of bifurcation.
His main result works with functionsf(u) satisfying  (i)
f(u)>0, for u≤0,(ii)
limu→∞f(u)/u=∞,(iii)
f′(0)≥0,(iv)
f′′(u)>0, for u>0,
and the author proves that there exists a criticalλ0 such that Problem (4.19) has exactly two, exactly one, or no symmetric positive solution depending on whether 0<λ<λ0,λ=λ0 or λ>λ0.
Our Example3.3 cannot be treated by the results of [14], because, in this case, f(u)=c+λarctanu and f does not satisfy assumptions (ii) and (iv) above mentioned.



---

*Source: 543035-2011-05-29.xml*