We establish the existence and uniqueness of a positive solution to the following fourth-order value problem: u(4)(x)=a(x)uσ(x), x∈(0,1) with the boundary conditions u(0)=u(1)=u'(0)=u'(1)=0, where σ∈(-1,1) and a is a nonnegative continuous function on (0, 1) that may be singular at x=0 or x=1. We also give the global behavior of such a solution.

1. Introduction

The purpose of this paper is to study the existence and uniqueness with a precise global behavior of a positive solution u∈C4((0,1))∩C([0,1]) for the following fourth-order two-point boundary value problem:
(1)u(4)(x)=a(x)uσ(x),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
where -1<σ<1 and a is a nonnegative continuous function on (0,1) that may be singular at x=0 or x=1 and satisfies some hypotheses related to the class of Karamata regularly varying functions.

There have been extensive studies on fourth-order boundary value problems with diverse boundary conditions via many methods; see, for example, [1–9] and the references therein.

A naturel motivation for studying higher order boundary value problems lies in their applications. For example, it is well known that the deformation of an elastic beam in equilibrium state, whose both ends clamped, can be described by fourth-order boundary value problem
(2)u(4)(x)=g(x,u(x)),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0.

Our aim in this paper is to give a contribution to the study of these problems by exploiting the properties of the Karamata class of functions.

To state our result, we need some notations. We denote by C([0,1]) the set of all continuous functions f on [0,1], and we will use 𝒦 to denote the set of Karamata functions L defined on (0,η] by
(3)L(t):=cexp(∫tηz(s)sds),
for some η>1, where c>0 and z∈C([0,η]) such that z(0)=0. It is clear that a function L is in 𝒦 if and only if L is a positive function in C1((0,η]) such that
(4)limt→0+tL′(t)L(t)=0.

For two nonnegative functions f and g defined on a set S, the notation f(x)≈g(x), x∈S, means that there exists c>0 such that (1/c)f(x)≤g(x)≤cf(x), for all x∈S. We denote by x+=max(x,0), x∧t=min(x,t), x∨t=max(x,t), for x, t∈ℝ, and B+((0,1)) the set of all measurable functions on (0,1).

Throughout this paper, we assume that a is nonnegative on (0,1) and satisfies the following condition:

a∈C((0,1)) such that for t∈(0,1)(5)a(t)≈t-λL1(t)(1-t)-μL2(1-t),

where λ≤3+σ, μ≤3+σ, L1,L2∈𝒦 satisfying
(6)∫0ηt2+σ-λL1(t)dt<∞,∫0ηt2+σ-μL2(t)dt<∞.

In the sequel, we introduce the function θλ,μ defined on (0,1) by
(7)θλ,μ(x)=xmin(2,(4-λ)/(1-σ))(L~1(x))1/(1-σ)×(1-x)min(2,(4-λ)/(1-σ))(L~2(1-x))1/(1-σ),
where
(8)L~1(x)={1ifλ<2(1+σ),∫xηL1(s)sdsifλ=2(1+σ),L1(x)if2(1+σ)<λ<3+σ,∫0xL1(s)sdsifλ=3+σ,L~2(x)={1ifμ<2(1+σ),∫xηL2(s)sdsifμ=2(1+σ),L2(x)if2(1+σ)<μ<3+σ,∫0xL2(s)sdsifμ=3+σ.

Our main result is the following.

Theorem 1.

Let σ∈(-1,1) and assume that a satisfies (H0). Then, problem (1) has a unique positive solution u∈C4((0,1))∩C([0,1]) satisfying for x∈(0,1)(9)u(x)≈θλ,μ(x).

This paper is organized as follows. Some preliminary lemmas are stated and proved in the next section, involving some already known results on Karamata functions. In Section 3, we give the proof of Theorem 1.

2. Technical Lemmas

To let the paper be self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [10, 11].

Lemma 2.

The following assertions hold.

Let L∈𝒦 and ɛ>0; then, one has
(10)limt→0+tɛL(t)=0.

Let L1,L2∈𝒦 and let p∈ℝ. Then, one has L1+L2∈𝒦, L1L2∈𝒦, and L1p∈𝒦.

Example 3.

Let m be a positive integer. Let c>0, let (μ1,μ2,…,μm)∈ℝm, and let d be a sufficiently large positive real number such that the function
(11)L(t)=c∏k=1m(logk(dt))μk
is defined and positive on (0,η], for some η>1, where logkx=log∘log∘⋯∘logx (k times). Then, L∈𝒦.

Applying Karamata’s theorem (see [10, 11]), we get the following.

Lemma 4.

Let μ∈ℝ and let L be a function in 𝒦 defined on (0,η]. One has the following:

if μ<-1, then ∫0ηsμL(s)ds diverges and ∫tηsμL(s)ds~t→0+-(t1+μL(t)/(μ+1));

if μ>-1, then ∫0ηsμL(s)ds converges and ∫0tsμL(s)ds~t→0+(t1+μL(t)/(μ+1)).

Lemma 5 (see [<xref ref-type="bibr" rid="B2">12</xref>] or [<xref ref-type="bibr" rid="B14">13</xref>]).

Let L∈𝒦 be defined on (0,η]. Then, one has
(12)limt→0+L(t)∫tη(L(s)/s)ds=0.

If further ∫0η(L(s)/s)ds converges, then one has
(13)limt→0+L(t)∫0t(L(s)/s)ds=0.

Remark 6.

Let L∈𝒦 be defined on (0,η]; then, using (4) and (12), we deduce that
(14)t⟶∫tηL(s)sds∈𝒦.

If further ∫0η(L(s)/s)ds converges, we have by (12) that
(15)t⟶∫0tL(s)sds∈𝒦.

Lemma 7.

Given that f∈C([0,1]), then the unique continuous solution of
(16)u(4)(x)=f(x),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0
is given by
(17)u(x)=Gf(x):=∫01G(x,t)f(t)dt,
where
(18)G(x,t)=16(x∧t)2(1-x∨t)2[3(x∨t)-(x∧t)(1+2(x∨t))]
is Green’s function for the boundary value problem (16).

Remark 8.

For x,t∈(0,1), we have G(1-x,1-t)=G(x,t).

In the following, we give some estimates on the Green function G(x,t) that will be used later.

Let f be a measurable function such that the function t→t2(1-t)2f(t) is continuous and integrable on (0,1). Then, Gf is the unique solution in C4((0,1))∩C([0,1]) of the problem
(21)u(4)(x)=f(x),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0.

Proof.

From Corollary 10, the function Gf is defined on (0,1) and, by Proposition 9, we have
(22)G(f)(x)≤12x(1-x)∫01t2(1-t)2|f(t)|dt.

Now, since t→t2f(t) is integrable near 0 and t→(1-t)2f(t) is integrable near 1, then, for x∈(0,1), we have
(23)Gf(x)=12x(1-x)2∫0xt2f(t)dt+12x2∫x1t(1-t)2f(t)dt-16(1+2x)(1-x)2∫0xt3f(t)dt-16x3∫x1(1+2t)(1-t)2f(t)dt.
This gives
(24)(Gf)′(x)=12(1-3x)(1-x)∫0xt2f(t)dt+x∫x1t(1-t)2f(t)dt+x(1-x)∫0xt3f(t)dt-12x2∫x1(1+2t)(1-t)2f(t)dt,(Gf)′′(x)=(3x-2)∫0xt2f(t)dt+∫x1t(1-t)2f(t)dt+(1-2x)∫0xt3f(t)dt-x∫x1(1+2t)(1-t)2f(t)dt,(Gf)′′′(x)=∫0x(3t2-2t3)f(t)dt-∫x1(1+2t)(1-t)2f(t)dt,(Gf)(4)(x)=f(x).

Moreover, we have Gf(0)=Gf(1)=(Gf)′(0)=(Gf)′(1)=0.

Finally, we prove the uniqueness. Let u,v∈C4((0,1))∩C([0,1]) be two solutions of (21) and put w=v-u. Then, w∈C4((0,1))∩C([0,1]) and w(4)=0. Hence, it follows that w(x)=ax3+bx2+cx+d. Using the fact that w(0)=w(1)=w′(0)=w′(1)=0, we conclude that w=0 and so u=v.

In the sequel, we assume that β≤3 and γ≤3 and we put
(25)b(t)=t-βL3(t)(1-t)-γL4(1-t),
where L3,L4∈𝒦 satisfy
(26)∫0ηt2-βL3(t)dt<∞,∫0ηt2-γL4(t)dt<∞.

So, we aim to give some estimates on the potential function Gb(x).

We define the Karamata functions ψβ, ϕγ by
(27)ψβ(x)={∫0xL3(t)tdtifβ=3,L3(x)if2<β≤3,∫xηL3(t)tdtifβ=2,1ifβ<2,ϕγ(x)={∫0xL4(t)tdtifγ=3,L4(x)if2<γ<3,∫xηL4(t)tdtifγ=2,1ifγ<2.

Then, we have the following.

Proposition 12.

For x∈(0,1),
(28)Gb(x)≈xmin(2,4-β)(1-x)min(2,4-γ)ψβ(x)ϕγ(1-x).

Proof.

Using Proposition 9, we have
(29)Gb(x)≈∫01(1-t)-γt-β(x∧t)2×(1-x∨t)2(x∨t)(1-x∧t)L3(t)L4(1-t)dt≈x(1-x)2∫0x(1-t)1-γt2-βL3(t)L4(1-t)dt+x2(1-x)∫x1(1-t)2-γt1-βL3(t)L4(1-t)dt=x(1-x)2I(x)+x2(1-x)J(x).

For 0<x≤1/2, we have I(x)≈∫0xt2-βL3(t)dt. So, using Lemma 4 and hypothesis (26), we deduce that
(30)I(x)≈{∫0xL3(t)tdtifβ=3,x3-βL3(x)ifβ<3.

Now, we have
(31)J(x)≈∫x1/2t1-βL3(t)dt+∫1/21(1-t)2-γL4(1-t)dt≈1+∫x1/2t1-βL3(t)dt.

This implies by Lemma 4 that
(32)J(x)≈{x2-βL3(x)if2<β≤3,∫xηL3(t)tdtifβ=2,1ifβ<2.

Hence, it follows by Lemma 5 and hypothesis (26) that, for 0<x≤1/2, we get
(33)Gb(x)≈{x∫0xL3(t)tdtifβ=3,x4-βL3(x)if2<β<3,x2∫xηL3(t)tdtifβ=2x2ifβ<2,

That is, for 0<x≤1/2,
(34)Gb(x)≈xmin(2,4-β)ψβ(x).
Now, since G(1-x,1-t)=G(x,t), we use similar arguments as above applied to L4 instead of L3 to obtain
(35)Gb(x)≈(1-x)min(2,4-γ)ϕγ(1-x)for12≤x≤1.

This together with (34) implies that, for x∈(0,1), we have
(36)Gb(x)≈xmin(2,4-β)(1-x)min(2,4-γ)ψβ(x)ϕγ(1-x).

3. Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>

In order to prove Theorem 1, we need the following lemma.

Lemma 13.

Assume that the function a satisfies (H0) and put ω(t)=a(t)(θλ,μ(t))σ for t∈(0,1). Then, one has, for x∈(0,1),
(37)Gω(x)≈θλ,μ(x).

Proof.

Put r=min(2,(4-λ)/(1-σ)) and s=min(2,(4-μ)/(1-σ)). Then, for t∈(0,1), we have
(38)ω(t)=t-λ+rσL1(t)(L~1(t))σ/(1-σ)(1-t)-μ+sσ×L2(1-t)(L~2(1-t))σ/(1-σ).

Let β=λ-rσ, γ=μ-sσ, L3(t)=L1(t)(L~1(t))σ/(1-σ), and L4(t)=L2(t)(L~2(t))σ/(1-σ). Then, using Proposition 12, we obtain by a simple computation that
(39)G(ω)(x)≈θλ,μ(x).

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

From Lemma 13, there exists M>1 such that for each x∈(0,1)(40)1Mθλ,μ(x)≤Gω(x)≤Mθλ,μ(x),
where ω(t)=a(t)(θλ,μ(t))σ.

Put c0=M1/(1-|σ|) and let
(41)Λ={u∈C([0,1]):1c0θλ,μ≤u≤c0θλ,μ}.

In order to use a fixed point theorem, we define the operator T on Λ by
(42)Tu(x)=G(auσ)(x)=∫01G(x,t)a(t)uσ(t)dt.

For this choice of c0, we can easily prove that, for u∈Λ, we have Tu≤c0θλ,μ and Tu≥(1/c0)θλ,μ.

Now, since the function (x,t)→G(x,t) is continuous on [0,1]×[0,1] and, by Proposition 9, Corollary 10, and Lemma 13, the function t→t2(1-t)2a(t)θλ,μσ(t) is integrable on (0,1), we deduce that the operator T is compact from Λ to itself. It follows by the Schauder fixed point theorem that there exists u∈Λ such that Tu=u. Then, u∈C([0,1]) and u satisfies the equation
(43)u(x)=G(auσ)(x).

Since the function t→t2(1-t)2a(t)uσ(t) is continuous and integrable on (0,1), then by Proposition 11, the function u is a positive solution in C4((0,1))∩C([0,1]) of problem (1).

Finally, let us prove that u is the unique positive continuous solution satisfying (9). To this aim, we assume that (1) has two positive solutions u,v∈C4((0,1))∩C([0,1]) satisfying (9) and consider the nonempty set J={m≥1:1/m≤u/v≤m} and put c=infJ. Then, c≥1 and we have (1/c)v≤u≤cv. It follows that uσ≤c|σ|vσ and consequently
(44)(c|σ|v-u)(4)=a(c|σ|vσ-uσ):=f≥0,(c|σ|v-u)(0)=(c|σ|v-u)(1)=(c|σ|v-u)′(0)=(c|σ|v-u)′(1)=0.

Since the function t→t2(1-t)2f(t) is continuous and integrable on (0,1), it follows by Proposition 11 that c|σ|v-u=G(a(c|σ|vσ-uσ))≥0. By symmetry, we obtain also that v≤c|σ|u. Hence, c|σ|∈J and c≤c|σ|. Since |σ|<1, then c=1 and consequently u=v.

Example 14.

Let σ∈(-1,1) and let a be a positive continuous function on (0,1) such that
(45)a(t)≈t-λ(1-t)-μlog(21-t),
where λ<3+σ and μ<3+σ. Then, using Theorem 1, problem (1) has a unique positive continuous solution u satisfying the following estimates:
(46)u(x)≈xmin(2,(4-λ)/(1-σ))(L~1(x))1/(1-σ)×(1-x)min(2,(4-μ)/(1-σ))(L~2(1-x))1/(1-σ),
where
(47)L~1(x)={1ifλ≠2(1+σ),log(2x)ifλ=2(1+σ),L~2(x)={1ifμ<2(1+σ),(log(2x))2ifμ=2(1+σ),log(2x)if2(1+σ)<μ<3+σ.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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