AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 10.1155/2014/958420 958420 Research Article Ground State Solutions for a Class of Fractional Differential Equations with Dirichlet Boundary Value Condition Hu Zhigang Liu Wenbin Liu Jiaying Osilike Micah Department of Mathematics, China University of Mining and Technology Xuzhou 221008 China cumt.edu.cn 2014 3062014 2014 06 03 2014 16 06 2014 1 7 2014 2014 Copyright © 2014 Zhigang Hu 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.

In this paper, we apply the method of the Nehari manifold to study the fractional differential equation (d/dt)((1/2)0Dt-β(u(t))+(1/2)tDT-β(u(t)))=  f(t,u(t)), a.e. t[0,T], and u0=uT=0, where 0Dt-β,tDT-β are the left and right Riemann-Liouville fractional integrals of order 0β<1, respectively. We prove the existence of a ground state solution of the boundary value problem.

1. Introduction

Fractional differential equations have played an important role in many fields such as engineering, science, electrical circuits, diffusion, and applied mathematics (see ). In recent years, some authors have studied the fractional differential equation by using different methods, such as fixed point theorem, coincidence degree theory, and critical point theory (see ).

By using the Mountain Pass theorem, Jiao and Zhou  studied the existence of solutions for the following boundary value problem: (1)ddt(120Dt-β(u(t))+12tDT-β(u(t)))+F(t,u(t))=0,a.e.t[0,T],  u(0)=u(T)=0, where 0<β<1,   0Dt-β and   tDT-β are the left and right Riemann-Liouville fractional integrals of order β, respectively, F:[0,T]×RNR, and F(t,x) is the gradient of F with respect to x.

By using a critical-points theorem established by G. Bonanno, Bai  investigated the following fractional boundary value problem: (2)ddt(120Dtα-1(0CDtαu(t))-12tDTα-1(tCDTαu(t)))+λa(t)f(u(t))=0,  a.e.t[0,T],u(0)=u(T)=0, where α(1/2,1],   0Dtα-1 and   tDTα-1 are the left and right Riemann-Liouville fractional integrals of order 1-α, and   0CDtαu(t) and   tCDTαu(t) are the left and right Caputo fractional derivatives of order α.

The authors in [18, 2022] further studied the existence and multiplicity of solutions for the related problems by critical point theory.

We find that the method of Nehari manifold is seldom used in the above boundary value problem. Inspired by the results in , we would like to investigate the ground state solution for the following fractional boundary value problem: (3)ddt(120Dt-β(u(t))+12tDT-β(u(t)))=f(t,u(t)),hhhhhhhhhhhhhhhhhhhhhhhhhhhhhha.e.t[0,T],u(0)=u(T)=0, where 0<β<1,   0Dt-β and   tDT-β are the left and right Riemann-Liouville fractional integrals of order β, respectively. The technical tool is the method of Nehari manifold. (see [23, 24]).

This paper is organized as follows. In Section 2, some preliminaries on the fractional calculus are presented. In Section 3, we set up the variational framework of problem (3) and give some necessary lemmas. Finally, Section 4 presents the main result and its proof.

2. Preliminaries on the Fractional Calculus

In this section, we will introduce some notations, definitions, and preliminary facts on fractional calculus which are used throughout this paper.

Definition 1 (left and right Riemann-Liouville fractional integrals).

Let f be a function defined on [a,b]. The left and right Riemann-Liouville fractional integrals of order α for function f denoted by   aDt-αf(t) and   tDb-αf(t) function, respectively, are defined by (4)  aDt-αf(t)=1Γ(α)0t(t-s)α-1f(s)ds,t[a,b],α>0,  tDb-αf(t)=1Γ(α)tb(t-s)α-1f(s)ds,t[a,b],α>0, provided that the right-hand side integral is pointwise defined on [a,b].

Definition 2 (left and right Riemann-Liouville fractional derivatives).

Let f be a function defined by [a,b]. The left and right Riemann-Liouville fractional derivatives of order α for function f denoted by   aDtαf(t) and   tDbαf(t) function, respectively, are defined by (5)  aDtαf(t)=dndtnaDtα-nf(t)=1Γ(α)dndtn0t(t-s)n-α-1f(s)ds,00000000000i00t[a,b],α>0,  tDbαf(t)=(-1)ndndtntDbα-nf(t)=(-1)nΓ(α)dndtntb(s-t)n-α-1f(s)ds,00000000000i000t[a,b],α>0, provided that the right-hand side integral is pointwise defined on [a,b].

Definition 3 (left and right Caputo fractional derivatives).

If α(n-1,n) and fACn([a,b],R), then the left and right Caputo fractional derivatives of order α for function f denoted by   aCDtαf(t) and   tCDbαf(t) function, respectively, are defined by (6)  aCDtαf(t)=aDtα-ndndtnf(t)=1Γ(α)0t(t-s)n-α-1f(n)(s)ds,00000000i0000t[a,b],α>0,  tCDbαf(t)=(-1)ntDbα-ndndtnf(t)=(-1)nΓ(α)tb(s-t)n-α-1f(n)(s)ds,000000000i0000t[a,b],α>0, respectively, where t[a,b].

Lemma 4 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

The left and right Riemann-Liouville fractional integral operators have the property of a semigroup; that is, (7)ab[aDt-αf(t)]g(t)dt=ab[tDb-αg(t)]f(t)dt,α>0, provided that fLp([a,b],R), gLq([a,b],R) and pq, q1, 1/p+1/q1+α or p1, q1, 1/p+1/q=1+α.

Lemma 5 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

Assume that n-1<α<n and fCn[a,b]. Then, (8)  aDt-α(aCDtαf(t))=f(t)-j=0n-1f(j)(a)j!(t-a)j,  tDb-α(tCDbαf(t))=f(t)-j=0n-1(-1)jf(j)(b)j!(b-t)j, for t[a,b].

Lemma 6 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

Assume that n-1<α<n. Then, (9)  aCDtαf(t)=aDtαf(t)-j=0n-1f(j)(a)Γ(j-α+1)(t-a)j-α,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhht[a,b],  tCDbαf(t)=tDbαf(t)-j=0n-1f(j)(b)Γ(j-α+1)(b-t)j-α,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhht[a,b].

3. A Variational Setting

To apply critical point theory for the existence of solutions for boundary value problem (3), we shall state some basic notations and results , which will be used in the proof of our main results.

Throughout this paper, we denote α=1-β/2 and assume that the following conditions are satisfied.

fC1(R×R).

f(t,0)=0=(f/s)(t,0) for every tR.

There are constants a,b>0 and 2<p such that (10)|fs(t,s)|a+b|s|p-2,

for every tR and sR.

There are constants μ>2,M>0 such that (11)0<μF(t,s)sf(t,s)

for all tR and |s|M. Here, (12)F(t,s)=0sf(t,x)dx.

The map tt-1sf(x,ts) is increasing on (0,+), for every xR and sR.

Now we construct appropriate function spaces. Denote by C0([0,T],R) the set of all functions uC0([0,T],R) with u(0)=u(T)=0. The fractional derivative space E0α,p is defined by the closure of C0([0,T],R) with respect to the norm (13)uα,p=(0T|u(t)|pdt+0T|0CDtαu(t)|pdt)1/p, where   0CDtα is the α-order left Caputo fractional derivative.

Remark 7.

If p=2, we define Eα=E0α,2, with respect to the norm (14)u=(0T|u(t)|2dt+0T|0CDtαu(t)|2dt)1/2. The set Eα is a reflexive and separable Hilbert space.

Remark 8.

For any uEα, noting the fact u(0)=0, we have   0Dtαu(t)=0CDtαu(t),t[0,T].

Lemma 9 (see [<xref ref-type="bibr" rid="B17">17</xref>]).

Let 0<α1 and 1<p<. For all uE0α,p, one has (15)uLpTαΓ(α+1)0CDtαuLp. Moreover, if α>1/p and 1/p+1/q=1, then (16)uTα-1/pΓ(α)[(α-1)q+1]1/q0CDtαuLp.

According to (15), we can consider Eα with respect to the equivalent norm (17)uα,p=0CDtαuLp,u=  0CDtαuL2.

Lemma 10 (see [<xref ref-type="bibr" rid="B17">17</xref>]).

Let 0<α1 and 1<p<. Assume that α>1/p and the sequence {uk} converges weakly to u in E0α,p; that is, uku. Then uku in C([0,T],R); that is, u-uk0 as k.

Similar to the proof of [17, Proposition 4.1], we have the following property.

Lemma 11.

If 1/2<α1, for any uEα, one has (18)|cos(πα)|u2-0T(0CDtαu(t),tCDTαu(t))dt1|cos(πα)|u2.

To obtain a weak solution of boundary value problem (3), we assume that u is a sufficiently smooth solution of (3). Multiplying (3) by an arbitrary vC0(0,T), we have (19)-0T(ddt(120Dt-β(u(t))+12tDT-β(u(t))),v(t))dt=0T(f(t,u(t)),v(t))dt.

Observe that (20)-120T(ddt(0Dt-βu(t)+tDT-βu(t)),v(t))dt=120T((0Dt-βu(t),v(t))+(tDT-βu(t),v(t)))dt=120T((0Dt-β/2u(t),tDT-β/2v(t))+(tDT-β/2u(t),0Dt-β/2v(t)))dt. As u(0)=u(T)=v(0)=v(T)=0, we have (21)  0Dt-β/2u(t)=0Dt1-β/2u(t),  tDT-β/2u(t)=-tDT1-β/2u(t),  0Dt-β/2v(t)=0Dt1-β/2v(t),  tDT-β/2v(t)=-tDT1-β/2v(t). Then (19) is equivalent to (22)0T-12[(0Dtαu(t),tDTαv(t))+(tDTαu(t),0Dtαv(t))]dt=0T(f(t,u(t)),v(t))dt. Since (22) is well defined for u,vEα, the weak solution of (3) can be defined as follows.

Definition 12.

A weak solution of (3) is a function uEα such that (23)0T-12[(0Dtαu(t),tDTαv(t))+(tDTαu(t),0Dtαv(t))]dt=0T(f(t,u(t)),v(t))dt for every vEα.

We consider the functional I:EαR, defined by (24)I(u)=0T[-12(0Dtαu(t),tDTαu(t))-F(t,u(t))]dt. From Theorem 4.1 of , we can get that if 1/2<α1, then the functional I is continuously differentiable on Eα. Since I is continuously differentiable on Eα, then (25)I(u),v=-0T12[(0Dtαu(t),tDTαv(t))+(tDTαu(t),0Dtαv(t))]dt-0T(f(t,u(t)),v(t))dt for u,vEα. Hence, a critical point of I is a weak solution of problem (3).

4. Main Result

In order to study the solvability of boundary value problem (3), we use the so-called Nehari method. Define (26)I(u)=-120T(0CDtαu(t),tCDTαu(t))dt-0TF(t,u(t))dt,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhuEα, where F(t,u)=0uf(t,s)ds.

There is one-to-one correspondence between the critical points of I and weak solutions of boundary value problem (3). Now, we define (27)N={uEα{0}I(u),u=0}. Then we know any nonzero critical point of I must be on N. Define (28)G(u)=I(u),u=-0T(0Dtαu(t),tDTαu(t))dt-0T(f(t,u(t)),u(t))dt.

Lemma 13.

Assume the hypotheses (H1)–(H5) hold. If uN is a critical point of I|N, then I(u)=0.

Proof.

For uN, together with (H5), (29)G(u),u=-0T2(0Dtαu(t),tDTαu(t))+uf(t,u(t))·u2(t)+f(t,u(t))u(t)dt=0T2f(t,u(t))u(t)-uf(t,u(t))·u2(t)-f(t,u(t))u(t)dt=0Tf(t,u(t))u(t)-uf(t,u(t))·u2(t)dt<0. If uN is a critical point of I|N, there exists a Lagrange multiplier λR, such that I(u)=λG(u). Then we have (30)I(u),u=λG(u),u=0. By (29), G(u),u0, and we have λ=0. So we can get that I(u)=0. The proof is complete.

Lemma 14.

Assume the hypotheses (H1)–(H5) hold. For any uEα{0}, there is a unique y=y(u) such that y(u)uN and one has I(yu)=maxy0I(yu)>0.

Proof.

First, we claim that there exist constants δ>0,   ρ>0 such that I(u)>0 for all uBρ(0){0} and I(u)δ for all uBρ(0). That is, 0 is a strict local minimizer of I. In fact, by (H3) we can get that (31)ε>0,Cε>0,  |F(t,u)|ε2|u|2+Cε|u|p. Then together with Lemmas 9 and 11, we have (32)I(u)=-120T(0CDtαu(t),tCDTαu(t))dt-0TF(t,u(t))dt-120T(0CDtαu(t),tCDTαu(t))dt-ε20T|u|2dt-Cε0T|u|pdt12|cos(πα)|u2-ε20T|u|2dt-Cε0T|u|pdt(12|cos(πα)|-ε2T2αΓ2(α+1))u2-Cε(Tp+α-1/2Γ(α)[(α-1)2+1]1/2)pup. Choose ε such that (ε/2)(T2α/Γ2(α+1))=(1/4)|cos(πα)|; then (33)I(u)14|cos(πα)|u2-Cε(Tp+α-1/2Γ(α)[(α-1)2+1]1/2)pup=u2((Tp+α-2Γ(α)[(α-1)2+1]2)p14|cos(πα)|-Cε(Tp+α-1/2Γ(α)[(α-1)2+1]1/2)pup-2). Choose ρ>0, such that Cε(Tp+α-1/2/Γ(α)[(α-1)2+1]1/2)pρp-2=(1/8)|cos(πα)|. Then we have I(u)(1/8)|cos(πα)|u2. Let δ=(1/8)|cos(πα)|u2; then we get that there exist constants δ>0, ρ>0 such that I(u)>0 for all uBρ(0){0} and I(u)δ for all uBρ(0).

Next, we claim that I(yu)-, as y. In fact, by (H4), there exists a constant A>0 such that F(t,u)A|u|μ for |u|M. On the other hand, we can easily get that there exists a constant B such that F(t,u)B for |u|M. Then together with Lemma 11, we have (34)I(yu)y22|cos(πα)|u2-Ayμ0T|u|μdt-B. Since μ>2, we can get that I(yu)-, as y.

Let g(y):=I(yu) for y>0. From what we have proved, there has at least one yu=y(u)>0 such that (35)g(yu)=maxy0g(y)=maxy0I(yu)=I(yuu). We prove next g(y) has a unique critical point for y>0. Consider a critical point (36)g(y)=I(yu),u=-0Ty(0Dtαu,tDTαu)dt-0Tf(t,yu)udt=0. Then, together with (H5), we have (37)g′′(y)=-0T(0Dtαu,tDTαu)dt-0Tf(t,yu)(yu)u2dt=0Tf(t,yu)uydt-0Tf(t,yu)(yu)u2dt=1y20Tf(t,yu)yudt-0Tf(t,yu)(yu)u2dt<0. So we know that if y is a critical point of g, then it must be a strict local maximum. This implies the uniqueness.

Finally, from (38)g(y)=I(yu),u=1yI(yu),yu, we see y is a critical point if yuN. Define m=infNI. Then we can get that minfBρ(0)Iδ>0. The proof is complete.

Lemma 15.

Assume the hypotheses (H1)–(H5) hold and m=infNI. Then there exists uN such that I(u)=m.

Proof.

We claim that both I and G are weakly lower semicontinuous. In fact, according to Lemma 10, if uku in Eα, then uku in C([0,T],R). Therefore, F(t,uk(t))F(t,u(t)) a.e. t[0,T]. By the Lebesgue dominated convergence theorem, we have 0TF(t,uk(t))dt0TF(t,u(t))dt, which means that the functional u0TF(t,u(t))dt is weakly continuous on Eα. Similarly u0Tf(t,u(t))u(t)dt is weakly continuous on Eα. Since Eα is Hilbet space, together with (17) and Lemma 11, we can easily get that -0T(0CDtαu(t),tCDTαu(t))dt is weakly lower semicontinuous on Eα. Then we get that both I and G are weakly lower semicontinuous.

Since μF(t,u)-uf(t,u) is continuous for t[0,T] and |x|M, there exists B>0, such that (39)F(t,u)1μuf(t,u)+B,t[0,T],|x|M. Together with (H4), we get (40)F(t,u)1μuf(t,u)+B,t[0,T],xR. Let {uk}N be a minimizing sequence; that is, I(uk)m, I(uk)0 as k. Then, (41)m+o(1)=I(uk)=-120T(0CDtαuk(t),tCDTαuk(t))dt-0TF(t,uk(t))dt-120T(0CDtαuk(t),tCDTαuk(t))dt-1μ0Tukf(t,uk)dt-BT=(1μ-12)0T(0CDtαuk(t),tCDTαuk(t))dt+1μI(uk),uk-BT(12-1μ)|cos(πα)|uk2-1μI(uk)uk-BT. By μ>2 and I(uk)0, we get that uk is bounded in E. Since Eα is a reflexive space, going to a subsequence if necessary, we may assume that uku in Eα. Then from Lemma 10, uku in C([0,T],R). Since G is weakly lower semicontinuous and {uk}N, we first have (42)G(u)lim_kG(uk)=0. Then we have u0. In fact, if u=0, then uk0 in C([0,T],R). By G(uk)=0, we get uk0. This is a contradiction with {uk}N.

Then from Lemma 14, there exists a unique y>0 such that yuN. Together with I which is weakly lower semicontinuous, we have (43)mI(yu)lim_kI(yuk)limkI(yuk)limkI(uk)=m. Then we get that m is achieved at yuN. The proof is complete.

Theorem 16.

Assuming the hypotheses (H1)–(H5) hold, boundary value problem (3) has a weak solution such that I(u)=m; that is, boundary value problem (3) has a ground state solution.

Proof.

By the Lemmas 14 and 15, we can get that there exists uN such that I(u)=m=infNI. Then the u is a critical point of I|N. From Lemma 13 we have I(u)=0. So boundary value problem (3) has a weak solution such that I(u)=m. The proof is complete.

5. Example

In this section, we give an example to illustrate our results.

Example 1.

Consider the following BVP: (44)-ddt(120Dt-1/2+12tDT-1/2)u(t)=u3,t[0,T],hhhhhhhhhhhhhhhhhhhhhhu(0)=0,u(T)=0. It is easy to verify all the conditions in Theorem 16, so BVP (44) has a ground state solution.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This research was supported by the Fundamental Research Funds for the Central Universities (2013XK03).

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