AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation82658010.1155/2012/826580826580Research ArticlePositive and Nondecreasing Solutions to an m-Point Boundary Value Problem for Nonlinear Fractional Differential EquationCabreraI. J.1HarjaniJ.1SadaranganiK. B.1MomaniShaher M.1Departamento de MatemáticasUniversidad de Las Palmas de Gran CanariaCampus de Tafira Baja, 35017 Las Palmas de Gran CanariaSpainulpgc.es2012312012201230092011151120112012Copyright © 2012 I. J. Cabrera et al.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 are concerned with the existence and uniqueness of a positive and nondecreasing solution for the following nonlinear fractional m-point boundary value problem: D0+αu(t)+f(t,u(t))=0,  0<t<1,  2<α3,  u(0)=u(0)=0,  u(1)=i=1m-2aiu(ξi), where D0+α denotes the standard Riemann-Liouville fractional derivative, f:[0,1]×[0,)[0,) is a continuous function, ai0 for i=1,2,,m-2, and 0<ξ1<ξ2<<ξm-2<1. Our analysis relies on a fixed point theorem in partially ordered sets. Some examples are also presented to illustrate the main results.

1. Introduction

Many papers and books on fractional differential equations have appeared recently. Most of them are devoted to the solvability of the linear fractional equation in terms of a special function (see, e.g., [1, 2]) and to problems of analyticity in the complex domain . Moreover, Delbosco and Rodino  considered the existence of a solution for the nonlinear fractional differential equationD0+αu=f(t,u),  0<α<1, where f:[0,a]×, 0<a+, is a given continuous function in (0,a)×. They obtained existence results by using the Schauder fixed point theorem and the Banach contraction principle.

Recently, El-Shahed  considered the following nonlinear fractional boundary value problem:D0+αu+λa(t)f(t,u(t))=0,0<t<1,2<α3,u(0)=u(0)=u(1)=0. They used the Krasnoselskii’s fixed point theorem on cone expansion and compression to show the existence and nonexistence of positive solutions for the above fractional boundary value problem.

In , Liang and Zhang considered the following nonlinear fractional boundary value problem:D0+αu(t)+f(t,u(t))=0,0<t<1,3<α4,u(0)=u(0)=u(0)=u(1)=0, and by means of lower and upper solution method and fixed point theorems, they obtained some results on the existence of positive solutions to the above boundary value problem.

The question of uniqueness of the solution is not treated in .

Recently, in  Caballero et al. studied the fractional boundary value problem appearing in  by using a fixed point theorem in partially ordered sets, and the authors obtained uniqueness of the solution.

In this paper we discuss the existence and uniqueness of a positive and nondecreasing solution for the following m-point nonlinear boundary value problem of fractional order:D0+αu(t)+f(t,u(t))=0,  0<t<1,  2<α3,u(0)=u(0)=0,  u(1)=i=1m-2aiu(ξi), where ai0 for i=1,2,,m-2, 0<ξ1<ξ2<<ξm-2<1, and, moreover, 0<i=1m-2aiξiα-2<1.

Recently, this problem has been studied in [8, 9]. In  the author studies the existence and multiplicity of positive solutions for problem (1.4), and he uses Krasnoselskii and Leggett-Williams fixed point theorems. The question of uniqueness and monotonicity of the solution is not treated. In , the authors investigate the existence and uniqueness of positive and nondecreasing solutions for problem (1.4), and the main tools in this paper are a fixed point theorem in partially ordered sets and the lower and upper solution method.

Our study is based on a different fixed point theorem in partially ordered sets than the one used in .

Our main interest in this paper is to give an alternative answer to the main results of [8, 9].

Existence of fixed points in partially ordered sets has been considered recently in , among others.

For existence theorems for fractional differential equations and applications, we refer to the survey . Concerning the definitions and basic properties, we refer the reader to .

2. Preliminaries and Previous Results

For the convenience of the reader, we present here some definitions, lemmas, and results that will be used in the proofs of our main results.

Definition 2.1.

The Riemann-Liouville fractional integral of order α>0 of a function f:(0,) is given by I0+αf(t)=1Γ(α)0t(t-s)α-1f(s)ds, provided that the right-hand side is pointwise defined on (0,) and where Γ(α) denotes the Euler gamma function given by Γ(α)=0+tα-1e-tdt,  α>0.

Definition 2.2.

The Riemann-Liouville fractional derivative of order α>0 of a function f:(0,) is defined by D0+αf(t)=1Γ(n-α)(ddt)n0tf(t)(t-s)α-n+1ds, where n=[α]+1 and [α] denote the integer part of α.

The following two lemmas can be found in [16, 17].

Lemma 2.3.

Let α>0 and uC(0,1)L1(0,1). Then the fractional differential equation D0+αu(t)=0 has u(t)=c1tα-1+c2tα-2++cntα-n,ciR(i=1,2,,n),n=[α]+1, as unique solutions.

Lemma 2.4.

Assume that uC(0,1)L1(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)L1(0,1). Then I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2++cntα-n, for some ci  (i=1,2,,n) and n=[α]+1.

Using Lemma 2.4, in  the following result is proved.

Lemma 2.5.

Given fC[0,1], then the unique solution of D0+αu(t)+f(t)=0,0<t<1,2<α3,u(0)=u(0)=0,u(1)=i=1m-2aiu(ξi) is u(t)=01G(t,s)f(s)ds, where Green’s function G(t,s) is given by G(t,s)=G1(t,s)+G2(t,s),takingG1(t,s)={(1-s)α-2tα-1-(t-s)α-1Γ(α),0st1,(1-s)α-2tα-1Γ(α),0ts1,G2(t,s)=1(1-ϱ)Γ(α)(ϱ(1-s)α-2-i=1m-2ai(ξi-s)α-2χEi(s))tα-1, where ϱ=i=1m-2aiξiα-2 and χEi denotes the characteristic function of the set Ei=[0,ξi] for i=1,2,,m-2.

Remark 2.6.

Notice that Lemma 3 appears in  under assumption 2<α<3. In  the authors prove this result in the same way for 2<α3.

The following result is proved in [8, 9].

Lemma 2.7.

Under the assumption ai0 for i=1,2,,m-2,0<ξ1<ξ2<<ξm-2<1 and 1-ϱ>0 where ϱ=i=1m-2aiξiα-2, the Green’s function appearing in Lemma 2.5 satisfies G(t,s)0.

Remark 2.8.

It is easily checked that G(t,s) is a continuous function on [0,1]×[0,1].

The following lemmas appear in .

Lemma 2.9.

The function G1(t,s) appearing in Lemma 2.5 is strictly increasing in the first variable.

Lemma 2.10.

The function G(t,s) appearing in Lemma 2.5 satisfies max0t101G(t,s)ds=1(α-1)Γ(α)[1α+11-ϱ(i=1m-2aiξiα-2(1-ξi))].

For convenience, we will denote by L the constantL=1(α-1)Γ(α)[1α+11-ϱ(i=1m-2aiξiα-2(1-ξi))]. In the sequel we present the fixed point theorem we will use later and which appears in .

Firstly, we need to introduce the following class of functions. By 𝒮 we denote the class of those functions β:[0,)[0,1) satisfying β(tn)1 implies tn0.

Theorem 2.11 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

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:XX be a nondecreasing mapping such that there exists an element x0X with x0Tx0.

Suppose that there exists β𝒮 such thatd(Tx,Ty)β(d(x,y))d(x,y), for x,yX with yx.

Assume that either T is continuous or X is such thatif  (xn)  is  a  nondecreasing  sequence  with  xnx  in  X,  then  xnx,nN. Besides, if for  x,yX  there  exists  zX  which  is  comparable  to  x  and  y, then T has a unique fixed point.

In our considerations, we will work in the space C[0,1]={x:[0,1],  continuous}, with the standard distance given by d(x,y)=max0t1{|x(t)-y(t)|} for x,yC[0,1].

Notice that this space can be equipped with a partial order given byxyx(t)y(t), for x,yC[0,1] and t[0,1].

In  it is proved that (C[0,1],) with the above-mentioned distance satisfies condition (2.14) of Theorem 2.11. Moreover, for x,yC[0,1], as the function max{x,y} is continuous in [0,1], (C[0,1],) satisfies condition (2.15) of Theorem 2.11.

3. Main Result

Our starting point in this section is to present the class of functions 𝒜 which we use later. By 𝒜 we will denote the class of those functions ϕ:[0,)[0,) satisfying the following conditions:

ϕ is nondecreasing;

For any x>0, ϕ(x)<x;

ϕ(x)/x𝒮, where 𝒮 is the class of functions appearing in Section 2.

Examples of functions in 𝒜 are ϕ(x)=μx with 0μ<1,ϕ(x)=x/1+x and ϕ(x)=ln(1+x).

In what follows, we formulate our main result.

Theorem 3.1.

Suppose that the following assumptions are satisfied:

f:[0,1]×[0,)[0,) is continuous;

f(t,x) is nondecreasing with respect to the second variable for each t[0,1];

there exists 0<λ1/L and ϕ𝒜 such that

f(t,y)-f(t,x)λϕ(y-x), for x,y[0,) with yx and t[0,1].

Then problem (1.4) has a unique nonnegative and nondecreasing solution.

Proof.

Consider the cone P={uC[0,1]:u0}.

Notice that, as P is a closed set of C[0,1], P is a complete metric space with the distance given by d(x,y)=max0t1{|x(t)-y(t)|} satisfying conditions (2.14) and (2.15) of Theorem 2.11.

Now, for uP we define the operator T by(Tu)(t)=01G(t,s)f(s,u(s))ds, where G(t,s) is Green’s function defined in Section 2.

By Lemma 2.7, Remark 2.8, and (a), it is clear that T applies the cone P into itself.

Now, we will check that assumptions in Theorem 2.11 are satisfied.

Firstly, the operator T is nondecreasing.

In fact, by assumption (b), for uv we have(Tu)(t)=01G(t,s)f(s,u(s))ds01G(t,s)f(s,v(s))ds=(Tv)(t). On the other hand, for uv and uv and taking into account our hypotheses, we can obtain d(Tu,Tv)=max0t1{|(Tu)(t)-(Tv)(t)|}=max0t1[(Tu)(t)-(Tv)(t)]=max0t1[01G(t,s)(f(s,u(s))-f(s,v(s)))ds]max0t1[01G(t,s)λϕ(u(s)-v(s))ds]. Since ϕ is not nondecreasing and taking into account Lemma 2.10 and assumption (c), we get d(Tu,Tv)λϕ(u-v)max0t101G(t,s)ds=λϕ(d(u,v))Lϕ(d(u,v))=ϕ(d(u,v))d(u,v)d(u,v). Thus, for uv and uv, d(Tu,Tv)β(d(u,v))d(u,v), where β(x)=ϕ(x)/x𝒮.

Obviously, the last inequality is satisfied for u=v.

Thus condition (2.13) in Theorem 2.11 holds with β(x)=ϕ(x)/x. Moreover, since f and G are nonnegative functions,(T0)(t)=01G(t,s)f(s,0)ds0. Finally, Theorem 2.11 tells us that problem (1.4) has a unique nonnegative solution u(t).

In what follows, we will prove that the unique nonnegative solution u(t) for problem (1.4) is nondecreasing.

In fact, by Lemmas 2.5 and 2.9 it is easily seen that G(t,s) is strictly increasing and, since u(t) is a fixed point of the operator T, we haveu(t)=01G(t,s)f(s,u(s))ds. These facts and the nonnegative character of f give us that u(t) is nondecreasing.

Now, we present a sufficient condition for the existence and uniqueness of a positive and strictly increasing solution for problem (1.4) (positive solution means a solution satisfying x(t)>0 for t(0,1)). The proof of this fact is similar to the proof of Theorem 3.6 of . We present it for completeness.

Theorem 3.2.

Under assumptions of Theorem 3.1 and adding the following assumption

f(t0,0)0 for certain t0[0,1], one obtains existence and uniqueness of a positive and strictly increasing solution for problem (1.4).

Proof.

Consider the nonnegative solution x(t) for Problem (1.4) whose existence is guaranteed by Theorem 3.1.

Notice that x(t) satisfiesx(t)=01G(t,s)f(s,x(s))ds. Firstly, we will prove that x(t)>0 for t(0,1).

In fact, in contrary case we can find 0<t*<1 such that x(t*)=0. Consequently,x(t*)=01G(t*,s)f(s,x(s))ds=0. As x0,G(t,s)0, and f is nondecreasing with respect to the second variable (assumption (b)), from the last expression we can get 0=x(t*)=01G(t*,s)f(s,x(s))ds01G(t*,s)f(s,0)ds0 and, thus, 01G(t*,s)f(s,0)ds=0. This fact and the nonnegative character of the functions G(t*,s) and f(t,0) imply G(t*,s)f(s,0)=0a.e.(s). By Lemma 2.9, and since G(0,s)=0 and G(t*,s)>0 for any s[0,1], we have f(s,0)=0a.e.(s). On the other hand, assumption (d) gives us that f(t0,0)0 for certain t0[0,1] and, thus, f(t0,0)>0.

This fact and the continuity of f imply the existence of a set A[0,1] with t0A and μ(A)>0, where μ is the Lebesgue measure, such that f(t,0)>0 for any tA.

Therefore, x(t)>0 for t(0,1).

In the sequel, we will show that x(t) is strictly increasing.

In fact, since G(0,s)=0, we havex(0)=01G(0,s)f(s,x(s))ds=0. Now, we take t1,t2[0,1] with t1<t2.

We can consider two cases.

Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M193"><mml:mi>t</mml:mi><mml:mo>=</mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

Suppose that x(t2)=0.

Using a similar argument similar o the one used in the proof of the positive character of x(t), we obtain a contradiction.

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M196"><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo>></mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

Suppose that x(t1)=x(t2).

In this case, we have01(G(t1,s)-G(t2,s))f(s,x(s))ds=0.

Since G(t1,s)-G(t2,s)>0 (Lemma 2.3), we getf(s,x(s))=0a.e.(s).

Again, the same reasoning that we use earlier gives us a contradiction.

Therefore, x(t1)<x(t2).

This finished the proof.

Remark 3.3.

In Theorem 3.2, 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 (1.4), but when the solution is unique, we will see that this condition is very adjusted one. In fact, suppose that problem (1.4) has a unique nonnegative solution x(t), then we have f(t,0)=0for  any  t[0,1]  iff  x(t)0. Indeed, if f(t,0)=0 for any t[0,1], then it is easily seen that the zero function is a solution for problem (1.4) and the uniqueness of solution gives us x(t)0.

The reverse implication is obvious.

Remark 3.4.

Notice that assumptions in Theorem 3.1 are invariant by nonnegative and continuous perturbations. More precisely, if f(t,0)=0 for any t[0,1] and f satisfies conditions (a), (b), and (c) of Theorem 3.1, then g(t,x)=a(t)+f(t,x), where a:[0,1][0,) continuously and a0, satisfies assumptions of Theorem 3.2, and this means that the boundary value problem D0+αu(t)+g(t,u(t))=0,0<t<1,2<α3,u(0)=u(0)=0,u(1)=i=1m-2aiu(ξi), with ai0 for i=1,2,,m-2,0<ξ1<ξ2<<ξm-2<1, and 0<i=1m-2aiξiα-2<,1 has a unique positive and strictly increasing solution.

In the sequel, we present an example which illustrates our results.

Example 3.5.

Consider the following boundary value problem: D0+5/2u(t)+t+λu(t)1+u(t)=0,0<t<1,λ>0,u(0)=u(0)=0,u(1)=(12)u(14)+(14)u(12). In this case, α=5/2, f(t,u)=t+(λu/(1+u)), 0<ξ1=1/4<ξ2=1/2<1, a1=1/2, and a2=1/4. Besides, ϱ=(1/2)(1/2)+(1/4)(1/2)0.4267767, and L=1(3/2)Γ(5/2)[25+11-ϱ(12(14)1/234+14(12)1/212)]0.44197, and 1/L=2.2626.

It is easily seen that f(t,u) satisfies condition (a) of Theorem 3.1.

Since f/u=λ/(1+u)2>0 for u[0,), f(t,u) satisfies (b) of Theorem 3.1.

Moreover, for uv and t[0,1], we havef(t,u)-f(t,v)=t+λu1+u-λ-tv1+v=λ(u1+u-v1+v)=λ(u-v(1+u)(1+v))λ  u-v1+u-v=λϕ(u-v), where ϕ(x)=x/1+x. It is easily proved that ϕ belongs to the class 𝒜. Since f(t,0)=t0 for t0, Theorem 3.2 states that problem (3.20) has a unique positive and strictly increasing solution for 0<λ1/L2.2626.

4. Some Remark

In  the authors consider our problem (1.4) and they prove the following result.

Theorem 4.1.

Problem (1.4) has a unique positive and strictly increasing with solution u(t) if the following conditions are satisfied:

f:[0,1]×[0,)[0,) is continuous and nondecreasing with respect to the second variable and f(t,u(t))0 for tZ[0,1] with μ(Z)>0 (μ denotes the Lebesgue measure),

there exists 0<λ<1/L such that for u,v[0,) with uv and t[0,1]

f(t,u)-f(t,v)λln(u-v+1).

The main tool used by the authors in  for the proof of Theorem 4.1 is a fixed point theorem in partially ordered sets which appears in . This fixed point theorem uses the following class of functions .

By we denote the class of functions ϕ:[0,)[0,) continuously such that if φ(x)=x-ϕ(x), then the following conditions are satisfied:

φ:[0,)[0,), and it is nondecreasing;

φ(0)=0;

φ is positive in (0,).

The same proof used by the authors in  gives us that the conclusion of Theorem 4.1 is true if we replace condition 2 of Theorem 4.1 by the following:

(2) there exists 0<λ<1/L such that for u,v[0,) with uv and t[0,1]

f(t,u)-f(t,v)λϕ(u-v), where ϕ.

In what follows we will prove that the classes of functions 𝒜 (see Section 3) and are not comparable.

Example 4.2.

This example appears in .

Consider the function ϕ:[0,)[0,) defined byϕ(t)={0,0t2,2t-4,2<t3,(23)t,3<t. It is easily proved that ϕ𝒜.

On the other hand, it is easily seen that φ(t)=t-ϕ(t) is not increasing, and, consequently, ϕ.

Example 4.3.

Consider the function ϕ:[0,)[0,) given by ϕ(x)=x-arctanx.

It is easily proved that ϕ.

On the other hand, since β(x)=ϕ(x)/x=1-(arctanx)/x and β(tn)1 when tn, we have that ϕ𝒜.

Examples 4.2 and 4.3 tell us that the results of this paper cover cases which cannot be treated by Theorem 4 which appears in  and vice versa.

In  the following result is also proved ([9, Theorem 5.1]).

Theorem 4.4.

Problem (1.4) has a positive solution u(t) if the following conditions are satisfied:

f(t,u)C([0,1]×[0,),[0,)) is nondecreasing relative to u, f(t,γ(t))0 for t(0,1), and there exists a positive constant μ<1 such that

Kμf(t,u)f(t,Ku), for K[0,1], where γ(t)=01G(t,s)  ds+1(1-ϱ)Γ(α)[01ϱ(1-s)α-2ds-i=1m-2ai01(ξi-s)α-2χEi(s)  ds]tα-1.

In the sequel, we present an example which can be treated by Theorem 3.2, and it cannot be studied by Theorem 4.4.

Example 4.5.

Consider the fractional boundary value problem D0+5/2u(t)+(t2+1)(λu(t)+c)=0,0<t<1,c>0,0<λ<1,u(0)=u(0)=0,u(1)=(12)u(14)+(14)u(12). In this case, α=5/2,f(t,u)=(t2+1)(λu+c),0<ξ1=1/4<ξ2=1/2<1, a1=1/2 and a2=1/4. Besides, ϱ=(1/2)(1/4)1/2+(1/4)(1/2)1/20.4267767.

It is easily seen that f(t,u) satisfies condition (a) of Theorem 3.1. Since f/u=λ(t2+1)>0, f(t,u) satisfies (b) of Theorem 3.1.

Moreover, for uv and t[0,1], we havef(t,u)-f(t,v)=λ(t2+1)(u-v)2λ(u-v)=2ϕ(u-v), where ϕ(x)=λx. It is easily checked that ϕ𝒜.

Since f(t,0)=c(t2+1)0 for t0 and, as 2(1/L)=2.2626 (see Example 3.5), Theorem 3.2 gives us existence and uniqueness of a positive and strictly increasing solution for Problem (4.6).

On the other hand, we will show that condition (Hf) appearing in Theorem 4.4 is not satisfied.

In fact, suppose that there exists 0<μ<1 such thatKμf(t,u)f(t,Ku), for any K[0,1].

Since f(t,u)=(t2+1)(λu+c) with 0<λ<1 and c>0, we haveKμf(t,Ku)f(t,u)=(t2+1)(ϱKu+c)(t2+1)(ϱu+c)=ϱKu+cϱu+c. Taking limit when u in the last expression, we get KμK, which is false, since 0<μ<1 and the function h(α)=Kα is decreasing when 0<K<1.

Therefore, Problem (4.6) can be covered by Theorem 3.2, and it cannot be treated using Theorem 4.4.

In , the author proves the following result.

Theorem 4.6.

Assume that

f:[0,1]×[0,)[0,) is continuous,

ai0 for i=1,2,,m-2, 0<ξ1<ξ2<<ξm-2<1 and ϱ=i=1m-2aiξiα-2 with ϱ<1.

In addition, suppose that one of the following two conditions holds:

limu0mint[0,1]f(t,u)/u=,limumaxt[0,1]f(t,u)/u=0,

limu0mint[0,1]f(t,u)/u=0,limumaxt[0,1]f(t,u)/u=.

Then Problem (1.4) has at least one positive solution.

In what follows we present an example which can be treated by our results and it cannot be studied by Theorem 4.6.

Example 4.7.

Consider Problem (3.20) which appears in Example 3.5.

We prove that this example can be treated by Theorem 3.2 and we obtained a unique positive and strictly increasing solution for this problem when 0<λ<1/L2.2626.

On the other hand, in this case, since f(t,u)=t+(λu/(1+u)), we havemint[0,1](t+(λu/(1+u))u)=λu/(1+u)u=λ1+u,limu0mint[0,1](t+(λu/(1+u))u)=limu0λ1+u=λ. Since 0<λ<, Theorem 4.6 cannot be used in this case.

Example 4.8.

Consider the following boundary value problem D0+5/2u(t)+c+λarctanu(t)=0,0<t<1,c>0,λ>0,u(0)=u(0)=0,  u(1)=(12)u(14)+(14)u(12). In this case, α=5/2,f(t,u)=c+λarctanu,0<ξ1=1/4<ξ2=1/2<1, a1=1/2, and a2=1/4. Besides, ϱ=(1/2)(1/2)+(1/4)(1/2)0.4267767.

It is easily seen that f(t,u) satisfies conditions (a) and (b) of Theorem 3.1.

Moreover, in  it is proved that if uv0,arctanu-arctanvarctan(u-v). Using this fact, for uv and t[0,1], we havef(t,u)-f(t,v)=λ(arctanu-arctanv)λarctan(u-v)=λϕ(u-v), where ϕ(x)=arctanx.

In  it is proved that ϕ𝒜.

Since f(t,0)=c0, Theorem 3.2 gives us the existence and uniqueness of a positive and strictly increasing solution for Problem (4.11) when 0<λ1/L2.2626.

On the other hand, sincelimu0mint[0,1]f(t,u)u=limu0mint[0,1]c+λarctanuu=limu0c+λarctanuu=,limumaxt[0,1]f(t,u)u=limuc+λarctanuu=0, Problem (4.11) can be treated by Theorem 4.6, and we obtain the existence of at least one positive solution.

Our main contribution is that for 0<λ1/L2.2626 we obtain uniqueness and strictly increasing character for the solution of Problem (4.11).

Acknowledgment

This paper was partially supported by the Ministerio de Educación y Ciencia, Project MTM 2007/65706.

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