We are concerned with the following superlinear fourth-order equation u4t+utφt,−ut=0, t∈0,1; −u0=u1=0, −u′0=a, − u′1=-b, where a,−b are nonnegative constants such that a+b>0 and φt,−s is a nonnegative continuous function that is required to satisfy some appropriate conditions related to a class K satisfying suitable integrability condition. Our purpose is to prove the existence, uniqueness, and global behavior of a classical positive solution to the above problem by using a method based on estimates on the Green function and perturbation arguments.
1. Introduction
A natural motivation for studying higher order BVPs lies in their applications. It is well-known (see for instance [1]) that the deformation of an elastic beam in equilibrium state, whose both ends clamped, can be described by fourth-order BVP
(1)u(4)(t)=f(t,u(t)),t∈(0,1),
subject to the boundary conditions
(2)u(0)=u(1)=u′(0)=u′(1)=0.
When the nonlinearity f is nonnegative such problems have been extensively investigated by many researchers, and various forms of the equation and boundary condition have been discussed; see, for example, [1–15] and references therein.
In particular, in 2005, Ma and Tisdel [1] studied the existence of positive solutions to the sublinear BVP
(3)u(4)(t)=p(t)uσ(t),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
where σ∈(0,1) and p:(0,1)→[0,∞) is continuous which may be singular at both ends t=0 and t=1 and satisfying some integrable conditions. Using the method of lower and upper solutions for fourth-order boundary value problems, they have given some necessary and sufficient conditions for the existence of regular positive solutions to the boundary value problem (3).
In [8], the authors used fixed-point index results, to prove the existence of a positive solution for the boundary value problem
(4)u(4)(t)=p(t)f(u(t)),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
where f:[0,∞)→[0,∞) is nondecreasing continuous function allowed to be superlinear and p:(0,1)→[0,∞) is continuous which may be singular at both ends t=0 and t=1 satisfying some adequate conditions.
Recently, in [4], the author considered problem (3), with σ∈(-1,1) and p is a nonnegative continuous function on (0,1) satisfying some hypotheses related to Karamata regular variation theory. Using the Schauder fixed-point theorem, he established the existence and uniqueness of a positive solution to (3).
Here, by using a method based on estimates on the Green function and perturbation arguments, we show that, for negative nonlinearity f, problem (1) has a unique positive classical solution subject to some boundary conditions. More precisely, we are concerned with the following superlinear fourth order problem
(5)u(4)(t)+u(t)φ(t,u(t))=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b,
where a,b are nonnegative constants such that a+b>0. Our goal is to answer the questions of existence, uniqueness, and global behavior of a classical positive solution to problem (5), where the nonnegative nonlinear term φ(t,s) is required to satisfy some appropriate conditions related to the following class K.
Definition 1.
A Borel measurable function q in (0,1) belongs to the class K if q satisfies the following condition:
(6)∫01r3(1-r)3|q(r)|dr<∞.
We will often refer in this paper to
(7)ω(t)∶=at(1-t)2+bt2(1-t),
the unique solution of the problem
(8)u(4)(t)=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b.
Observe that, for t∈[0,1],
(9)min(a,b)ρ(t)≤ω(t)≤max(a,b)ρ(t),
where ρ(t)∶=t(1-t).
Also we denote by G(t,s) the Green function of the operator u→u(4), with the conditions u(0)=u(1)=u′(0)=u′(1)=0, which can be explicitly given by
(10)G(t,s)=16{t2(1-s)2[3s-t(1+2s)],0≤t≤s≤1,s2(1-t)2[3t-s(1+2t)],0≤s≤t≤1,=16(t∧s)2(1-t∨s)2[3(t∨s)-(t∧s)(1+2(t∨s))],
where t∧s=min(t,s) and t∨s=max(t,s) (see [2]).
The outline of the paper is as follows. In Section 2, we give some sharp estimates on Green’s function G(t,s), including the following 3G-inequality: for each t,s,r∈(0,1),
(11)G(t,r)G(r,s)G(t,s)≤32[(ρ(r)ρ(t))2G(t,r)+(ρ(r)ρ(s))2G(r,s)],(ρ(r)ρ(t))2G(t,r)≤12(ρ(r))3,
where ρ(r)=r(1-r).
In particular, we derive from this 3G-inequality that, for each q∈K, one has
(12)αq∶=supt,s∈(0,1)∫01G(t,r)G(r,s)G(t,s)|q(r)|dr<∞.
In Section 3, our purpose is to study the superlinear fourth-order problem (5). The nonlinearity φ is required to satisfy a combination of the following assumptions.
φ is a nonnegative continuous function in (0,1)×[0,∞).
There exists a nonnegative function q∈K∩C(0,1) with αq≤1/2 such that for each t∈(0,1), the map s→s(q(t)-φ(t,sω(t))) is nondecreasing on [0,1].
For each t∈(0,1), the function s→sφ(t,s) is nondecreasing on [0,∞).
We will first exploit the 3G-inequality to prove that the inverse of fourth-order operators that are perturbed by a zero-order term are positivity preserving. That is, if ψ is a positive measurable function and q is a nonnegative function belonging to the class K∩(C(0,1)) with αq≤1/2, then the following problem
(13)u(4)(t)+q(t)u(t)=ψ(t),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
has a positive solution. It turns out to prove that problem (13) admits a positive Green function G(t,s).
Based on the construction of this Green function and by using perturbation arguments, we prove the following.
Theorem 2.
Assume (H1)-(H2), then problem (5) has a positive solution u in C4((0,1))∩C([0,1]) satisfying
(14)c0ω(t)≤u(t)≤ω(t),t∈[0,1],
where c0 is a constant in (0,1).
Moreover, if hypothesis (H3) is also satisfied, then the solution u to problem (5) satisfying (14) is unique.
Corollary 3.
Let f be a nonnegative function in C1([0,∞)) such that the map s→θ(s)=sf(s) is nondecreasing on [0,∞). Let p be a nonnegative continuous function on (0,1) such that the function t→p~(t)∶=p(t)max0≤ξ≤ω(t)θ′(ξ) belongs to the class K. Then for λ∈[0,1/2αp~), the following problem
(15)u(4)(t)+λp(t)u(t)f(u(t))=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b,
has a unique positive solution u in C4((0,1))∩C([0,1]) satisfying
(16)(1-λαp~)ω(t)≤u(t)≤ω(t),t∈[0,1].
Observe that in Theorem 2, we obtain a positive classical solution u to problem (5) whose behavior is not affected by the perturbed term. That is, it behaves like the solution ω of the homogeneous problem (8). As typical example of nonlinearity satisfying (H1)–(H3), we quote φ(t,s)=λp(t)sσ, where σ≥0, p is a positive continuous function on (0,1) such that
(17)∫01r3+σ(1-r)3+σp(r)dr<∞,
and q(t)=λp~(t)∶=λ(σ+1)p(t)(ω(t))σ, with λ∈[0,1/2αp~). Note that by using (9) and (17) the function p~ belongs to the class K.
As usual, let B((0,1)) be the set of Borel measurable functions in (0,1) and let B+((0,1)) be the set of nonnegative ones.
We define the kernel V:B+((0,1))→B+((0,1)) by
(18)Vψ(t)∶=∫01G(t,r)ψ(r)dr,t∈[0,1].
Let q∈B+((0,1)). We define the kernel V(q·):B+((0,1))→B+((0,1)) by
(19)V(q·)(ψ)(t)∶=∫01G(t,r)q(r)ψ(r)dr,t∈[0,1].
Remark 4 (see [1]).
Let ψ∈B+((0,1)) such that the function r→(ρ(r))2ψ(r) is continuous and integrable on (0,1), then Vψ is the unique solution in C4((0,1))∩C([0,1]) of
(20)u(4)(t)=ψ(t),t∈(0,1);u(0)=u(1)=u′(0)=u′(1)=0.
2. Estimates on the Green’s Function
From the explicit expression of Green’s function (10) we derive the following.
Proposition 5 (see [4, 7]).
For t,s∈[0,1], one has
(21)13(t∧s)2(1-t∨s)2(t∨s)(1-t∧s)≤G(t,s)≤12(t∧s)2(1-t∨s)2(t∨s)(1-t∧s),(22)13(ρ(t))2(ρ(s))2≤G(t,s)≤12ρ(t)ρ(s)(ρ(t)∧ρ(s)).
Using (22), we deduce the following.
Corollary 6.
For s,t∈(0,1), one has
(23)(ρ(s)ρ(t))2G(t,s)≤12(ρ(s))3.
For ψ∈B+((0,1)), the function t→Vψ(t) is continuous on [0,1] if and only if the integral ∫01(ρ(s))2ψ(s)ds converges.
Theorem 7 (3G inequality).
For each t,s,r∈(0,1), one has
(24)G(t,r)G(r,s)G(t,s)≤32[(ρ(r)ρ(t))2G(t,r)+(ρ(r)ρ(s))2G(r,s)],
where ρ(t)=t(1-t).
Proof.
To prove the inequality, we denote by A(t,s)=(ρ(t)ρ(s))2/G(t,s) and we claim that A is a quasi-metric, that is, for each t,s,r∈(0,1),
(25)A(t,s)≤32[A(t,r)+A(r,s)].
By symmetry, we may assume that t≤s.
Using (21), we deduce that
(26)2s(1-t)≤A(t,s)≤3s(1-t).
To show (25), we separate the proof into three cases.
Case 1 (r≤t≤s). In this case, one has
(27)A(t,r)+A(r,s)≥2(t+s)(1-r)≥2s(1-t)≥23A(t,s).
Case 2 (t≤r≤s). We obtain
(28)A(t,r)+A(r,s)≥2r(1-t)+2s(1-r)≥2s(1-t)≥23A(t,s).
Case 3 (t≤s≤r). One has
(29)A(t,r)+A(r,s)≥2r(1-t)+2r(1-s)≥2s(1-t)≥23A(t,s).
This completes the proof.
In the sequel, for any q∈B((0,1)), we recall that
(30)αq∶=supt,s∈(0,1)∫01G(t,r)G(r,s)G(t,s)|q(r)|dr,
and we denote by
(31)h1(t)=t(1-t)2,h2(t)=t2(1-t),fort∈[0,1].
Proposition 8.
Let q be a function in K; then
(32)αq≤32∫01(ρ(r))3|q(r)|dr<∞.
For t∈[0,1], one has
(33)∫01G(t,s)h1(s)|q(s)|ds≤αqh1(t).
For t∈[0,1], one has
(34)∫01G(t,s)h2(s)|q(s)|ds≤αqh2(t).
In particular, for t∈[0,1], one has
(35)∫01G(t,s)ρ(s)|q(s)|ds≤αqρ(t).
Proof.
Let q be a function in K.
(i) Using (24) and (23), one has, for t,s∈(0,1),
(36)∫01G(t,r)G(r,s)G(t,s)|q(r)|dr≤32∫01(ρ(r)ρ(t))2G(t,r)+(ρ(r)ρ(s))2G(r,s)|q(r)|dr≤32∫01(ρ(r))3|q(r)|dr.
Hence
(37)αq≤32∫01(ρ(r))3|q(r)|dr<∞.
(ii) Since, for each t,s∈(0,1), one has limr→0(G(s,r)/G(t,r))=h1(s)/h1(t), then we deduce by Fatou’s lemma and (12), that
(38)∫01G(t,s)h1(s)h1(t)|q(s)|ds≤liminfr→0∫01G(t,s)G(s,r)G(t,r)|q(s)|ds≤αq;
which implies that, for t∈[0,1],
(39)∫01G(t,s)h1(s)|q(s)|ds≤αqh1(t).
(iii) Similarly, we prove inequality (34) by observing that
(40)limr→1G(s,r)G(t,r)=h2(s)h2(t).
Inequality (35) follows from Proposition 8 (ii)-(iii) and the fact that ρ(t)=h1(t)+h2(t). This completes the proof.
3. Proofs of Main Results
In this section, we aim at proving Theorem 2 and Corollary 3. So, we need the following preliminaries results.
For a nonnegative function q in K such that αq<1, we define the function G(t,s) on [0,1]×[0,1], by
(41)G(t,s)=∑n=0∞(-1)nGn(t,s),
where G0(t,s)=G(t,s) and
(42)Gn(t,s)=∫01G(t,r)Gn-1(r,s)q(r)dr,n≥1.
Next, we establish some inequalities on Gn(t,s). In particular, we deduce that G(t,s) is well defined.
Lemma 9.
Let q be a nonnegative function in K such that αq<1; then, for each n≥0 and t,s∈[0,1], one has
Gn(t,s)≤αqnG(t,s).
In particular, G(t,s) is well defined in [0,1]×[0,1].
where Ln=(1/3n+1)(∫01(ρ(r))4q(r)dr)n and Rn=(1/2n+1)(∫01(ρ(r))3q(r)dr)n.
Gn+1(t,s)=∫01Gn(t,r)G(r,s)q(r)dr and Gn(t,s)=Gn(s,t).
∫01G(t,r)G(r,s)q(r)dr=∫01G(t,r)G(r,s)q(r)dr.
Proof.
(i) The assertion is clear for n=0.
Assume that inequality in (i) holds for some n≥0; then by using (42) and (12), we obtain
(44)Gn+1(t,s)≤αqn∫01G(t,r)G(r,s)q(r)dr≤αqn+1G(t,s).
Now, since Gn(t,s)≤αqnG(t,s), it follows that G(t,s) is well defined in [0,1]×[0,1].
(ii) Using (22) and (42), we obtain (43) by induction.
(iii) Since G(t,s)=G(s,t), then equalities in (iii) are clear for n=0.
Assume that for a given integer n≥1 and t,s∈[0,1], one has
(45)Gn(t,s)=∫01Gn-1(t,r)G(r,s)q(r)dr,=Gn-1(s,t).
Using (42) and Fubini-Tonelli theorem, we obtain
(46)Gn+1(t,s)=∫01G(t,r)(∫01Gn-1(r,ξ)G(ξ,s)q(ξ)dξ)q(r)dr=∫01(∫01G(t,r)Gn-1(r,ξ)q(r)dr)G(ξ,s)q(ξ)dξ=∫01Gn(t,ξ)G(ξ,s)q(ξ)dξ.
On the other hand, by (42) and (45), we deduce that
(47)Gn(t,s)=∫01G(t,r)Gn-1(r,s)q(r)dr=∫01Gn-1(s,r)G(r,t)q(r)dr=Gn(s,t).
(iv) Let n≥0 and t,r,s∈[0,1]. By Lemma 9 (i) and (22), one has
(48)0≤Gn(t,r)G(r,s)q(r)≤αqnG(t,r)G(r,s)q(r).
Hence the series ∑n≥0∫01Gn(t,r)G(r,s)q(r)dr converges.
So we deduce by the dominated convergence theorem and Lemma 9 (iii) that
(49)∫01G(t,r)G(r,s)q(r)dr=∑n=0∞∫01(-1)nGn(t,r)G(r,s)q(r)dr=∑n=0∞∫01(-1)nG(t,r)Gn(r,s)q(r)dr=∫01G(t,r)G(r,s)q(r)dr.
Proposition 10.
Let q be a nonnegative function in K such that αq<1. Then the function (t,s)→G(t,s) is continuous on [0,1]×[0,1].
Proof.
We claim that for n≥0, the function Gn(t,s) is continuous on [0,1]×[0,1].
The assertion is clear for n=0.
Now assume that for a given integer n≥1, the function (t,s)→Gn-1(t,s) is continuous on [0,1]×[0,1]. By (42), one has
(50)Gn(t,s)=∫01G(t,r)Gn-1(r,s)q(r)dr.
Since the function (t,s,r)→G(t,r)Gn-1(r,s) is continuous on [0,1]×[0,1]×[0,1] and by Lemma 9 (i) and (22)
(51)G(t,r)Gn-1(r,s)q(r)≤αqn-1G(t,r)G(r,s)q(r)≤(ρ(r))3q(r),
we deduce by the dominated convergence theorem that (t,s)→Gn(t,s) is continuous on [0,1]×[0,1].
This proves our claim.
Now by Lemma 9 (i) and (22), one has, for each t,s∈[0,1],
(52)Gn(t,s)≤αqnG(t,s)≤12ρ(t)(ρ(s))2αqn≤αqn.
This implies that the series ∑n≥0(-1)nGn(t,s) is uniformly convergent on [0,1]×[0,1] and therefore the function (t,s)→G(t,s) is continuous on [0,1]×[0,1].
The proof is completed.
Lemma 11.
Let q be a nonnegative function in K such that αq≤1/2. Then for t,s in [0,1], one has
(53)(1-αq)G(t,s)≤G(t,s)≤G(t,s).
Proof.
Since αq≤1/2, we deduce from Lemma 9 (i) that
(54)|G(t,s)|≤∑n=0∞(αq)nG(t,s)=11-αqG(t,s).
On the other hand, from the expression of G, one has
(55)G(t,s)=G(t,s)-∑n=0∞(-1)nGn+1(t,s).
Since the series ∑n≥0∫01G(t,r)Gn(r,s)q(r)dr is convergent, we deduce by (55) and (42) that
(56)G(t,s)=G(t,s)-∑n=0∞(-1)n∫01G(t,r)Gn(r,s)q(r)dr=G(t,s)-∫01G(t,r)(∑n=0∞(-1)nGn(r,s))q(r)dr.
That is,
(57)G(t,s)=G(t,s)-V(qG(·,s))(t).
Now from (54) and Lemma 9 (i) (with n=1), we obtain
(58)V(qG(·,s))(t)≤11-αqV(qG(·,s))(t)=11-αqG1(t,s)≤αq1-αqG(t,s).
This implies that
(59)G(t,s)≥G(t,s)-αq1-αqG(t,s)=1-2αq1-αqG(t,s)≥0.
So it follows that G(t,s)≤G(t,s) and by (57) and Lemma 9 (i) (with n=1), one has
(60)G(t,s)≥G(t,s)-V(qG(·,s))(t)≥(1-αq)G(t,s).
In the sequel, for a given nonnegative function q∈K such that αq≤1/2, we define the operator Vq:B+((0,1))→B+((0,1)) by
(61)Vqψ(t)=∫01G(t,s)ψ(s)ds,t∈[0,1].
Using (53) and (22), we obtain the following.
Corollary 12.
Let q be a nonnegative function in K such that αq≤1/2 and ψ∈B+((0,1)); then the following statements are equivalent.
The function t→Vqψ(t) is continuous on [0,1].
The integral ∫01(ρ(s))2ψ(s)ds converges.
Next, we will prove that the kernel Vq satisfies the following resolvent equation.
Lemma 13.
Let q be a nonnegative function in K such that αq≤1/2 and ψ∈B+((0,1)). Then Vqψ satisfies the following resolvent equation:
(62)Vψ=Vqψ+Vq(qVψ)=Vqψ+V(qVqψ).
In particular, if V(qψ)<∞, one has
(63)(I-Vq(q·))(I+V(q·))ψ=(I+V(q·))(I-Vq(q·))ψ=ψ.
Proof.
Let t,s∈[0,1], then by (57) one has
(64)G(t,s)=G(t,s)+V(qG(·,s))(t),
which implies by Fubini-Tonelli theorem that, for ψ∈B+((0,1)),
(65)Vψ(t)=∫01(G(t,s)+V(qG(·,s))(t))ψ(s)ds=Vqψ(t)+V(qVqψ)(t).
On the other hand, by Lemma 9 (iii) and Fubini-Tonelli theorem, we obtain for ψ∈B+((0,1)) and t∈[0,1](66)∫01∫01G(t,r)G(r,s)ψ(s)q(r)drds=∫01∫01G(t,r)G(r,s)ψ(s)q(r)drds;
that is,
(67)Vq(qVψ)(t)=V(qVqψ)(t).
So we obtain
(68)Vψ=Vqψ+V(qVqψ)=Vqψ+Vq(qVψ)(t).
This completes the proof.
Proposition 14.
Let q be a nonnegative function in K∩C(0,1) such that αq≤1/2 and let ψ∈B+((0,1)) such that t→(ρ(t))2ψ(t) is continuous and integrable on (0,1). Then Vqψ is the unique nonnegative solution in C4((0,1))∩C([0,1]) of the perturbed fourth-order equation (13) satisfying
(69)(1-αq)Vψ≤Vqψ≤Vψ.
Proof.
It is clear by Corollary 12 that the function t→q(t)Vqψ(t) is continuous on (0,1). Using (62) and (22), there exists a nonnegative constant c such that
(70)Vqψ(t)≤Vψ(t)≤12∫01ρ(t)(ρ(s))2ψ(s)ds≤cρ(t).
So we deduce that
(71)∫01(ρ(s))2q(s)Vqψ(s)ds≤c∫01(ρ(s))3q(s)ds<∞.
Hence by using Remark 4, the function u=Vqψ=Vψ-V(qVqψ) satisfies the equation
(72)u(4)(t)=ψ(t)-q(t)u(t),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
and by integration inequalities (53), we obtain (69).
It remains to prove the uniqueness. Assume that v is another nonnegative solution in C4((0,1))∩C([0,1]) of problem (13) satisfying (69).
Since the function t→q(t)v(t) is continuous on (0,1) and by (69) and (70), the function t→(ρ(t))2q(t)v(t) is integrable on (0,1); then it follows by Remark 4 that the function v~∶=v+V(qv) satisfies
(73)v~(4)(t)=ψ(t),t∈(0,1),v~(0)=v~(1)=v~′(0)=v~′(1)=0.
From the uniqueness in Remark 4, we deduce that
(74)v~∶=v+V(qv)=Vψ.
Hence
(75)(I+V(q·))(v-u)=0.
Now since by (69), (70), and (35), one has
(76)V(q|v-u|)≤2cV(qρ)≤2cαqρ<∞,
then by (63), we deduce that u=v. This completes the proof.
Proof of Theorem 2.
Let a≥0 and b≥0 with a+b>0 and recall that
(77)ω(t)=at(1-t)2+bt2(1-t)=ah1(t)+bh2(t).
Since φ satisfies (H2), then there exists a positive function q in K∩C(0,1) such that αq≤1/2 and for each t∈(0,1), the map s→s(q(t)-φ(t,sω(t))) is nondecreasing on [0,1].
Let
(78)Λ∶={u∈B+((0,1)):(1-αq)ω≤u≤ω},
and define the operator T on Λ by
(79)Tu=ω-Vq(qω)+Vq((q-φ(·,u))u).
By (62) and Proposition 8, one has
(80)Vq(qω)≤V(qω)≤αqω≤ω,
and by (H2), we obtain
(81)0≤φ(·,u)≤q,∀u∈Λ.
So we claim that Λ is invariant under T. Indeed, using (81) and (80), one has for u∈Λ(82)Tu≤ω-Vq(qω)+Vq(qu)≤ω,Tu≥ω-Vq(qω)≥(1-αq)ω.
Next, we will prove that the operator T is nondecreasing on Λ. Indeed, let u,v∈Λ be such that u≤v. Since the map s→s(q(t)-φ(t,sω(t))) is nondecreasing on [0,1], for t∈(0,1), we obtain
(83)Tv-Tu=Vq([v(q-φ(·,v))-u(q-φ(·,u))])≥0.
Now, we consider the sequence (un) defined by u0=(1-αq)ω and un+1=Tun, for n∈N. Since Λ is invariant under T, one has u1=Tu0≥u0 and by the monotonicity of T, we deduce that
(84)(1-αq)ω=u0≤u1≤⋯≤un≤un+1≤ω.
Hence by dominated convergence theorem and hypotheses (H1)-(H2), we conclude that the sequence (un) converges to a function u∈Λ satisfying
(85)u=(I-Vq(q·))ω+Vq((q-φ(·,u))u).
That is,
(86)(I-Vq(q·))u=(I-Vq(q·))ω-Vq(uφ(·,u)).
On the other hand, since by (80), one has V(qu)≤V(qω)≤ω<∞, then by applying the operator (I+V(q·)) on both sides of the above equality and using (62) and (63), we conclude that u satisfies
(87)u=ω-V(uφ(·,u)).
Next we aim at proving that u is a solution of problem (5). To this end, we remark by (81) and (9) that
(88)uφ(·,u)≤qω≤max(a,b)qρ.
This implies by Corollary 6 (ii) that the function t→V(uφ(.,u))(t) is continuous on [0,1] and so by (87), u is continuous on [0,1].
Now, since by (H1) and (88), the function t→(ρ(t))2u(t)φ(t,u(t)) is continuous and integrable on (0,1), we conclude by Remark 4 that u is the required solution.
It remains to prove that under condition (H3),u is the unique solution to problem (5) satisfying (14). Assume that v is another nonnegative solution in C4((0,1))∩C([0,1]) to problem (5) satisfying (14). Since v≤ω, we deduce by (88) that
(89)0≤vφ(·,v)≤qω≤max(a,b)qρ.
So the function t→(ρ(t))2v(t)φ(t,v(t)) is continuous and integrable on (0,1) and by Remark 4, we conclude that the function v~∶=v+V(vφ(.,v)) satisfies
(90)v~(4)(t)=0,t∈(0,1),v~(0)=v~(1)=0,v~′(0)=a,v~′(1)=-b.
From the uniqueness in problem (8), we deduce that
(91)v=ω-V(vφ(·,v)).
Now let h be the function defined on (0,1) by
(92)h(t)={v(t)φ(t,v(t))-u(t)φ(t,u(t))v(t)-u(t)ifv(t)≠u(t),0ifv(t)=u(t).
Then by (H3), h∈B+((0,1)) and by (87) and (91), one has
(93)(I+V(h·))(v-u)=0.
On the other hand, by (H2), we remark that h≤q and by (80) we deduce that
(94)V(h|v-u|)≤2V(qω)≤2αqω<∞.
Hence by (63), we conclude that u=v. This completes the proof.
Proof of Corollary 3.
Let φ(t,s)=λp(t)f(s), θ(s)=sf(s), and p~(t)=p(t)max0≤ξ≤ω(t)θ′(ξ). It is clear that hypotheses (H1) and (H3) are satisfied. Since the function q(t)∶=λp~(t) belongs to the class K, one has αq≤1/2 for λ∈[0,1/2αp~). Moreover, by a simple computation, for s∈[0,1] and t∈(0,1), we obtain
(95)dds[s(q(t)-φ(t,sω(t)))]=q(t)-λp(t)θ′(sω(t))≥0.
This implies that the function φ satisfies hypothesis (H2). So the result follows by Theorem 2.
Example 15.
Let a≥0 and b≥0 with a+b>0. Let σ≥0, and p be a positive continuous function on (0,1) such that
(96)∫01r3+σ(1-r)3+σp(r)dr<∞.
Since the function p~(t)∶=(σ+1)p(t)(ω(t))σ belongs to the class K, then, for λ∈[0,1/2αp~), the problem
(97)u(4)(t)+λp(t)uσ+1(t)=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b
has a unique positive solution u in C4((0,1))∩C([0,1]) satisfying
(98)(1-λαp~)ω(t)≤u(t)≤ω(t),t∈[0,1].
Example 16.
Let a≥0 and b≥0 with a+b>0. Let σ≥0,γ>0, and p be a positive continuous function on (0,1) such that
(99)∫01r3+σ+γ(1-r)3+σ+γp(r)dr<∞.
Let θ(s)=sσ+1log(1+sγ). Since the function p~(t)∶=p(t)max0≤ξ≤ω(t)θ′(ξ)∈K, then, for λ∈[0,1/2αp~), the problem
(100)u(4)(t)+λp(t)uσ+1(t)log(1+uγ(t))=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b
has a unique positive solution u in C4((0,1))∩C([0,1]) satisfying
(101)(1-λαp~)ω(t)≤u(t)≤ω(t),t∈[0,1].
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors thank the referees for their careful reading of the paper. This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.
MaR.TisdellC. C.Positive solutions of singular sublinear fourth-order boundary value problems200584121199122110.1080/00036810500047634MR2178767AgarwalR. P.ChowY. M.Iterative methods for a fourth order boundary value problem198410220321710.1016/0377-0427(84)90058-XMR739776ZBL0541.650552-s2.0-5844393936AgarwalR. P.O'ReganD.WongP. J. Y.1999Boston, Mass, USAKluwer Academic10.1007/978-94-015-9171-3MR1680024AlsaediR. S.Existence and global behavior of positive solutions for some fourth order boundary value problems20142014565792610.1155/2014/657926MR3193533BaiZ.WangH.On positive solutions of some nonlinear fourth-order beam equations2002270235736810.1016/S0022-247X(02)00071-9MR1915704ZBL1006.340232-s2.0-0037098593CabadaA.EnguiçaR. R.Positive solutions of fourth order problems with clamped beam boundary conditions201174103112312210.1016/j.na.2011.01.027MR27935502-s2.0-79953731706CuiY. J.SunJ. X.ZouY. M.Positive solutions of singular boundary value problems of fourth-order differential equations2009292376380MR2504612ZBL1212.34054CuiY.ZouY.Positive solutions of singular fourth-order boundary-value problems200639110MR2213583DalmassoR.Uniqueness of positive solutions for some nonlinear fourth-order equations1996201115216810.1006/jmaa.1996.0247MR1397892ZBL0856.340242-s2.0-0030186866do ÓJ. M.LorcaS.UbillaP.Multiplicity of solutions for a class of non-homogeneous fourth-order boundary value problems200821327928610.1016/j.aml.2007.02.025MR24337422-s2.0-37849045573GraefJ. R.KongL.YangB.Existence, nonexistence, and uniqueness of positive solutions to a three point fourth order boundary value problem2011184565575MR28955302-s2.0-84872781409KongL.WongJ. S. W.Positive solutions for higher order multi-point boundary value problems with nonhomogeneous boundary conditions2010367258861110.1016/j.jmaa.2010.01.063MR2607284ZBL1197.340352-s2.0-77949568164KormanP.Uniqueness and exact multiplicity of solutions for a class of fourth-order semilinear problems2004134117919010.1017/S0308210500003140MR20399102-s2.0-1942476075MaR. Y.WuH. P.Positive solutions of a fourth-order two-point boundary value problem2002222244249MR1903078YaoQ.Positive solutions for eigenvalue problems of fourth-order elastic beam equations200417223724310.1016/S0893-9659(04)90037-7MR2034772ZBL1072.340222-s2.0-1242285075