AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 657926 10.1155/2014/657926 657926 Research Article Existence and Global Behavior of Positive Solutions for Some Fourth-Order Boundary Value Problems Alsaedi Ramzi S. Saker Samir Department of Mathematics College of Sciences and Arts King Abdulaziz University Rabigh Campus P.O. Box 344 Rabigh 21911 Saudi Arabia kau.edu.sa 2014 3132014 2014 10 11 2013 27 02 2014 31 3 2014 2014 Copyright © 2014 Ramzi S. Alsaedi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We establish the existence and uniqueness of a positive solution to the following fourth-order value problem: u ( 4 ) ( x ) = a ( x ) u σ ( 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 C 4 ( ( 0,1 ) ) C ( [ 0,1 ] ) for the following fourth-order two-point boundary value problem: (1) u ( 4 ) ( x ) = a ( x ) u σ ( 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,  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 ) : = c exp ( t η z ( s ) s d s ) , 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 C 1 ( ( 0 , η ] ) such that (4) lim t 0 + t L ( 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 ) c f ( 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 , 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 - λ L 1 ( t ) ( 1 - t ) - μ L 2 ( 1 - t ) ,

where λ 3 + σ , μ 3 + σ , L 1 , L 2 𝒦 satisfying (6) 0 η t 2 + σ - λ L 1 ( t ) d t < , 0 η t 2 + σ - μ L 2 ( t ) d t < .

In the sequel, we introduce the function θ λ , μ defined on ( 0,1 ) by (7) θ λ , μ ( x ) = x min ( 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 ) = { 1 if λ < 2 ( 1 + σ ) , x η L 1 ( s ) s d s if λ = 2 ( 1 + σ ) , L 1 ( x ) if 2 ( 1 + σ ) < λ < 3 + σ , 0 x L 1 ( s ) s d s if λ = 3 + σ , L ~ 2 ( x ) = { 1 if μ < 2 ( 1 + σ ) , x η L 2 ( s ) s d s if μ = 2 ( 1 + σ ) , L 2 ( x ) if 2 ( 1 + σ ) < μ < 3 + σ , 0 x L 2 ( s ) s d s if μ = 3 + σ .

Our main result is the following.

Theorem 1.

Let σ ( - 1,1 ) and assume that a satisfies ( H 0 ) . Then, problem (1) has a unique positive solution u C 4 ( ( 0,1 ) ) C ( [ 0,1 ] ) satisfying for x ( 0,1 ) (9) u ( x ) θ λ , 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) lim t 0 + t ɛ L ( t ) = 0 .

Let L 1 , L 2 𝒦 and let p . Then, one has L 1 + L 2 𝒦 , L 1 L 2 𝒦 , and L 1 p 𝒦 .

Example 3.

Let m be a positive integer. Let c > 0 , let ( μ 1 , μ 2 , , μ m ) m , and let d be a sufficiently large positive real number such that the function (11) L ( t ) = c k = 1 m ( log k ( d t ) ) μ k is defined and positive on ( 0 , η ] , for some η > 1 , where log k x = log log log x ( k times). Then, L 𝒦 .

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

Lemma 4.

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

if μ < - 1 , then 0 η s μ L ( s ) d s diverges and t η s μ L ( s ) d s ~ t 0 + - ( t 1 + μ L ( t ) / ( μ + 1 ) ) ;

if μ > - 1 , then 0 η s μ L ( s ) d s converges and 0 t s μ L ( s ) d s ~ t 0 + ( t 1 + μ L ( 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) lim t 0 + L ( t ) t η ( L ( s ) / s ) d s = 0 .

If further 0 η ( L ( s ) / s ) d s converges, then one has (13) lim t 0 + L ( t ) 0 t ( L ( s ) / s ) d s = 0 .

Remark 6.

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

If further 0 η ( L ( s ) / s ) d s converges, we have by (12) that (15) t 0 t L ( s ) s d s 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 ) = G f ( x ) : = 0 1 G ( x , t ) f ( t ) d t , where (18) G ( x , t ) = 1 6 ( 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.

Proposition 9.

On ( 0,1 ) × ( 0,1 ) , one has the following:

( 1 / 3 ) ( x t ) 2 ( 1 - x t ) 2 ( x t ) ( 1 - x t ) G ( x , t ) ( 1 / 2 ) ( x t ) 2 ( 1 - x t ) 2 ( x t ) ( 1 - x t ) ;

( 1 / 3 ) x 2 ( 1 - x ) 2 t 2 ( 1 - t ) 2 G ( x , t ) ( 1 / 2 ) x ( 1 - x ) t 2 ( 1 - t ) 2 .

Proof.

(i) It follows from the fact that for x , t ( 0,1 ) × ( 0,1 ) we have (19) 2 ( x t ) ( 1 - x t ) 3 ( x t ) - ( x t ) ( 1 + 2 x t ) 3 ( x t ) ( 1 - x t ) .

(ii) Since for x , t ( 0,1 ) we have x 2 ( 1 - x ) 2 t 2 ( 1 - t ) 2 ( x t ) 2 ( 1 - x t ) 2 ( x t ) ( 1 - x t ) , the result follows from (i).

As a consequence of the assertion (ii) of Proposition 9, we obtain the following.

Corollary 10.

Let f B + ( ( 0,1 ) ) and put G f ( x ) : = 0 1 G ( x , t ) f ( t ) d t , for x ( 0,1 ] .

Then, (20) G f ( x ) < f o r x ( 0,1 ) i f f 0 1 t 2 ( 1 - t ) 2 f ( t ) d t < .

Proposition 11.

Let f be a measurable function such that the function t t 2 ( 1 - t ) 2 f ( t ) is continuous and integrable on ( 0,1 ) . Then, G f is the unique solution in C 4 ( ( 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 G f is defined on ( 0,1 ) and, by Proposition 9, we have (22) G ( f ) ( x ) 1 2    x ( 1 - x ) 0 1 t 2 ( 1 - t ) 2 | f ( t ) | d t .

Now, since t t 2 f ( t ) is integrable near 0 and t ( 1 - t ) 2 f ( t ) is integrable near 1, then, for x ( 0,1 ) , we have (23) G f ( x ) = 1 2 x ( 1 - x ) 2 0 x t 2 f ( t ) d t + 1 2 x 2 x 1 t ( 1 - t ) 2 f ( t ) d t - 1 6 ( 1 + 2 x ) ( 1 - x ) 2 0 x t 3 f ( t ) d t - 1 6 x 3 x 1 ( 1 + 2 t ) ( 1 - t ) 2 f ( t ) d t . This gives (24) ( G f ) ( x ) = 1 2 ( 1 - 3 x ) ( 1 - x ) 0 x t 2 f ( t ) d t + x x 1 t ( 1 - t ) 2 f ( t ) d t + x ( 1 - x ) 0 x t 3 f ( t ) d t - 1 2 x 2 x 1 ( 1 + 2 t ) ( 1 - t ) 2 f ( t ) d t , ( G f ) ′′ ( x ) = ( 3 x - 2 ) 0 x t 2 f ( t ) d t + x 1 t ( 1 - t ) 2 f ( t ) d t + ( 1 - 2 x ) 0 x t 3 f ( t ) d t - x x 1 ( 1 + 2 t ) ( 1 - t ) 2 f ( t ) d t , ( G f ) ′′′ ( x ) = 0 x ( 3 t 2 - 2 t 3 ) f ( t ) d t - x 1 ( 1 + 2 t ) ( 1 - t ) 2 f ( t ) d t , ( G f ) ( 4 ) ( x ) = f ( x ) .

Moreover, we have G f ( 0 ) = G f ( 1 ) = ( G f ) ( 0 ) = ( G f ) ( 1 ) = 0 .

Finally, we prove the uniqueness. Let u , v C 4 ( ( 0,1 ) ) C ( [ 0,1 ] ) be two solutions of (21) and put w = v - u . Then, w C 4 ( ( 0,1 ) ) C ( [ 0,1 ] ) and w ( 4 ) = 0 . Hence, it follows that w ( x ) = a x 3 + b x 2 + c x + 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 - β L 3 ( t ) ( 1 - t ) - γ L 4 ( 1 - t ) , where L 3 , L 4 𝒦 satisfy (26) 0 η t 2 - β L 3 ( t ) d t < , 0 η t 2 - γ L 4 ( t ) d t < .

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

We define the Karamata functions ψ β , ϕ γ by (27) ψ β ( x ) = { 0 x L 3 ( t ) t d t if β = 3 , L 3 ( x ) if 2 < β 3 , x η L 3 ( t ) t d t if β = 2 , 1 if β < 2 , ϕ γ ( x ) = { 0 x L 4 ( t ) t    d t if γ = 3 , L 4 ( x ) if 2 < γ < 3 , x η L 4 ( t ) t    d t if γ = 2 , 1 if γ < 2 .

Then, we have the following.

Proposition 12.

For x ( 0,1 ) , (28) G b ( x ) x min ( 2,4 - β ) ( 1 - x ) min ( 2,4 - γ ) ψ β ( x ) ϕ γ ( 1 - x ) .

Proof.

Using Proposition 9, we have (29) G b ( x ) 0 1 ( 1 - t ) - γ t - β ( x t ) 2 × ( 1 - x t ) 2 ( x t ) ( 1 - x t ) L 3 ( t ) L 4 ( 1 - t ) d t x ( 1 - x ) 2 0 x ( 1 - t ) 1 - γ t 2 - β L 3 ( t ) L 4 ( 1 - t ) d t + x 2 ( 1 - x ) x 1 ( 1 - t ) 2 - γ t 1 - β L 3 ( t ) L 4 ( 1 - t ) d t = x ( 1 - x ) 2 I ( x ) + x 2 ( 1 - x ) J ( x ) .

For 0 < x 1 / 2 , we have I ( x ) 0 x t 2 - β L 3 ( t ) d t . So, using Lemma 4 and hypothesis (26), we deduce that (30) I ( x ) { 0 x L 3 ( t ) t d t if β = 3 , x 3 - β L 3 ( x ) if β < 3 .

Now, we have (31) J ( x ) x 1 / 2 t 1 - β L 3 ( t ) d t + 1 / 2 1 ( 1 - t ) 2 - γ L 4 ( 1 - t ) d t 1 + x 1 / 2 t 1 - β L 3 ( t ) d t .

This implies by Lemma 4 that (32) J ( x ) { x 2 - β L 3 ( x ) if 2 < β 3 , x η L 3 ( t ) t d t if β = 2 , 1 if β < 2 .

Hence, it follows by Lemma 5 and hypothesis (26) that, for 0 < x 1 / 2 , we get (33) G b ( x ) { x 0 x L 3 ( t ) t d t if β = 3 , x 4 - β L 3 ( x ) if 2 < β < 3 , x 2 x η L 3 ( t ) t d t if β = 2 x 2 if β < 2 ,

That is, for 0 < x 1 / 2 , (34) G b ( x ) x min ( 2,4 - β ) ψ β ( x ) . Now, since G ( 1 - x , 1 - t ) = G ( x , t ) , we use similar arguments as above applied to L 4 instead of L 3 to obtain (35) G b ( x ) ( 1 - x ) min ( 2,4 - γ ) ϕ γ ( 1 - x ) for 1 2 x 1 .

This together with (34) implies that, for x ( 0,1 ) , we have (36) G b ( x ) x min ( 2,4 - β )    ( 1 - x ) min ( 2,4 - γ ) ψ β ( x ) ϕ γ ( 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 ( H 0 ) and put ω ( t ) = a ( t ) ( θ λ , μ ( t ) ) σ for t ( 0,1 ) . Then, one has, for x ( 0,1 ) , (37) G ω ( x ) θ λ , 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 + r σ    L 1 ( t ) ( L ~ 1 ( t ) ) σ / ( 1 - σ ) ( 1 - t ) - μ + s σ × L 2 ( 1 - t ) ( L ~ 2 ( 1 - t ) ) σ / ( 1 - σ ) .

Let β = λ - r σ , γ = μ - s σ , L 3 ( t ) = L 1 ( t ) ( L ~ 1 ( t ) ) σ / ( 1 - σ ) , and L 4 ( t ) = L 2 ( t ) ( L ~ 2 ( t ) ) σ / ( 1 - σ ) . Then, using Proposition 12, we obtain by a simple computation that (39) G ( ω ) ( x ) θ λ , 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) 1 M    θ λ , μ ( x ) G ω ( x ) M θ λ , μ ( x ) , where ω ( t ) = a ( t ) ( θ λ , μ ( t ) ) σ .

Put c 0 = M 1 / ( 1 - | σ | ) and let (41) Λ = { u C ( [ 0,1 ] ) : 1 c 0    θ λ , μ u c 0 θ λ , μ } .

In order to use a fixed point theorem, we define the operator T on Λ by (42) T u ( x ) = G ( a u σ ) ( x ) = 0 1 G ( x , t ) a ( t ) u σ ( t ) d t .

For this choice of c 0 , we can easily prove that, for u Λ , we have T u c 0 θ λ , μ and T u ( 1 / c 0 ) θ λ , 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 t 2 ( 1 - t ) 2    a ( t ) θ λ , μ σ ( 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 T u = u . Then, u C ( [ 0,1 ] ) and u satisfies the equation (43) u ( x ) = G ( a u σ ) ( x ) .

Since the function t t 2 ( 1 - t ) 2 a ( t ) u σ ( t ) is continuous and integrable on ( 0,1 ) , then by Proposition 11, the function u is a positive solution in C 4 ( ( 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 C 4 ( ( 0,1 ) ) C ( [ 0,1 ] ) satisfying (9) and consider the nonempty set J = { m 1 : 1 / m u / v m } and put c = inf J . Then, c 1 and we have ( 1 / c ) v u c v . It follows that u σ c | σ | v σ and consequently (44) ( c | σ | v - u ) ( 4 ) = a ( c | σ | v σ - u σ ) : = 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 t 2 ( 1 - t ) 2 f ( t ) is continuous and integrable on ( 0,1 ) , it follows by Proposition 11 that c | σ | v - u = G ( a ( c | σ | v σ - u σ ) ) 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 ) - μ log ( 2 1 - 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 ) x min ( 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 ) = { 1 if λ 2 ( 1 + σ ) , log ( 2 x ) if λ = 2 ( 1 + σ ) , L ~ 2 ( x ) = { 1 if μ < 2 ( 1 + σ ) , ( log ( 2 x ) ) 2 if μ = 2 ( 1 + σ ) , log ( 2 x ) if 2 ( 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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