# Ground State Solutions for a Class of Fractional Differential Equations with Dirichlet Boundary Value Condition

Authors: Zhigang Hu; Wenbin Liu; Jiaying Liu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

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 0≤β<1, respectively. We prove the existence of a ground state solution of the boundary value problem.


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## Body

## 1. Introduction

Fractional differential equations have played an important role in many fields such as engineering, science, electrical circuits, diffusion, and applied mathematics (see [1–4]). 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 [5–22]).By using the Mountain Pass theorem, Jiao and Zhou [17] 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]×RN→R, and ∇F(t,x) is the gradient of F with respect to x.By using a critical-points theorem established by G. Bonanno, Bai [19] 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, 20–22] 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 [16–22], 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 Section2, 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).
Letf 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).
Letf 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)dndtn∫0t‍(t-s)n-α-1f(s)ds,00000000000i00t∈[a,b],α>0,tDbαf(t)=(-1)ndndtntDbα-nf(t)=(-1)nΓ(α)dndtn∫tb‍(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).
Ifα∈(n-1,n) and f∈ACn([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 [18]).
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 f∈Lp([a,b],R), g∈Lq([a,b],R) and p≥q, q≥1, 1/p+1/q≤1+α or p≠1, q≠1, 1/p+1/q=1+α.Lemma 5 (see [18]).
Assume thatn-1<α<n and f∈Cn[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 [18]).
Assume thatn-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 [18], which will be used in the proof of our main results.Throughout this paper, we denoteα=1-β/2 and assume that the following conditions are satisfied.(H1)
f∈C1(R×R).(H2)
f(t,0)=0=(∂f/∂s)(t,0) for every t∈R.(H3)
There are constantsa,b>0 and 2<p such that
(10)|∂f∂s(t,s)|≤a+b|s|p-2,for everyt∈R and s∈R.(H4)
There are constantsμ>2,M>0 such that
(11)0<μF(t,s)≤sf(t,s)for allt∈R and |s|≥M. Here,
(12)F(t,s)=∫0s‍f(t,x)dx.(H5)
The mapt→t-1sf(x,ts) is increasing on (0,+∞), for every x∈R and s∈R.Now we construct appropriate function spaces. Denote byC0∞([0,T],R) the set of all functions u∈C0∞([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)∥u∥α,p=(∫0T‍|u(t)|pdt+∫0T‍|0CDtαu(t)|pdt)1/p,
where 0CDtα is the α-order left Caputo fractional derivative.Remark 7.
Ifp=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 anyu∈Eα, noting the fact u(0)=0, we have 0Dtαu(t)=0CDtαu(t),t∈[0,T].Lemma 9 (see [17]).
Let0<α≤1 and 1<p<∞. For all u∈E0α,p, one has
(15)∥u∥Lp≤TαΓ(α+1)∥0CDtαu∥Lp.
Moreover, if α>1/p and 1/p+1/q=1, then
(16)∥u∥∞≤Tα-1/pΓ(α)[(α-1)q+1]1/q∥0CDtαu∥Lp.According to (15), we can consider Eα with respect to the equivalent norm
(17)∥u∥α,p=∥0CDtαu∥Lp,∥u∥=∥0CDtαu∥L2.Lemma 10 (see [17]).
Let0<α≤1 and 1<p<∞. Assume that α>1/p and the sequence {uk} converges weakly to u in E0α,p; that is, uk⇀u. Then uk→u in C([0,T],R); that is, ∥u-uk∥∞→0 as k→∞.Similar to the proof of [17, Proposition 4.1], we have the following property.Lemma 11.
If1/2<α≤1, for any u∈Eα, one has
(18)|cos⁡(πα)|∥u∥2≤-∫0T‍(0CDtαu(t),tCDTαu(t))dt≤1|cos⁡(πα)|∥u∥2.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 v∈C0∞(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)-12∫0T‍(ddt(0Dt-βu′(t)+tDT-βu′(t)),v(t))dt=12∫0T‍((0Dt-βu′(t),v′(t))+(tDT-βu′(t),v′(t)))dt=12∫0T‍((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,v∈Eα, the weak solution of (3) can be defined as follows.Definition 12.
A weak solution of (3) is a function u∈Eα such that
(23)∫0T‍-12[(0Dtαu(t),tDTαv(t))+(tDTαu(t),0Dtαv(t))]dt=∫0T‍(f(t,u(t)),v(t))dt
for every v∈Eα.We consider the functionalI:Eα→R, defined by
(24)I(u)=∫0T‍[-12(0Dtαu(t),tDTαu(t))-F(t,u(t))]dt.
From Theorem 4.1 of [17], we can get that if 1/2<α≤1, 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,v∈Eα. 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)=-12∫0T‍(0CDtαu(t),tCDTαu(t))dt-∫0T‍F(t,u(t))dt,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh∀u∈Eα,
where F(t,u)=∫0u‍f(t,s)ds.There is one-to-one correspondence between the critical points ofI and weak solutions of boundary value problem (3). Now, we define
(27)N={u∈Eα∖{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 u∈N is a critical point of I|N, then I′(u)=0.Proof.
Foru∈N, 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 u∈N is a critical point of I|N, there exists a Lagrange multiplier λ∈R, such that I′(u)=λG′(u). Then we have
(30)〈I′(u),u〉=λ〈G′(u),u〉=0.
By (29), 〈G′(u),u〉≠0, 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 u∈Eα∖{0}, there is a unique y=y(u) such that y(u)u∈N and one has I(yu)=max⁡y≥0I(yu)>0.Proof.
First, we claim that there exist constantsδ>0, ρ>0 such that I(u)>0 for all u∈Bρ(0)∖{0} and I(u)≥δ for all u∈∂Bρ(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)=-12∫0T‍(0CDtαu(t),tCDTαu(t))dt-∫0T‍F(t,u(t))dt≥-12∫0T‍(0CDtαu(t),tCDTαu(t))dt-ε2∫0T‍|u|2dt-Cε∫0T‍|u|pdt≥12|cos⁡(πα)|∥u∥2-ε2∫0T‍|u|2dt-Cε∫0T‍|u|pdt≥(12|cos⁡⁡(πα)|-ε2T2αΓ2(α+1))∥u∥2-Cε(Tp+α-1/2Γ(α)[(α-1)2+1]1/2)p∥u∥p.
Choose ε such that (ε/2)(T2α/Γ2(α+1))=(1/4)|cos⁡(πα)|; then
(33)I(u)≥14|cos⁡(πα)|∥u∥2-Cε(Tp+α-1/2Γ(α)[(α-1)2+1]1/2)p∥u∥p=∥u∥2((Tp+α-2Γ(α)[(α-1)2+1]2)p14|cos⁡⁡(πα)|-Cε(Tp+α-1/2Γ(α)[(α-1)2+1]1/2)p∥u∥p-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)|∥u∥2. Let δ=(1/8)|cos⁡(πα)|∥u∥2; then we get that there exist constants δ>0, ρ>0 such that I(u)>0 for all u∈Bρ(0)∖{0} and I(u)≥δ for all u∈∂Bρ(0).
Next, we claim thatI(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)|∥u∥2-Ayμ∫0T‍|u|μdt-B.
Since μ>2, we can get that I(yu)→-∞, as y→∞.
Letg(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)=max⁡y≥0g(y)=max⁡y≥0I(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-∫0T‍∂f(t,yu)∂(yu)u2dt=∫0T‍f(t,yu)uydt-∫0T‍∂f(t,yu)∂(yu)u2dt=1y2∫0T‍f(t,yu)yudt-∫0T‍∂f(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〉=1y〈I′(yu),yu〉,
we see y is a critical point if yu∈N. Define m=inf⁡NI. Then we can get that m≥inf⁡∂Bρ(0)I≥δ>0. The proof is complete.Lemma 15.
Assume the hypotheses (H1)–(H5) hold and m=inf⁡NI. Then there exists u∈N such that I(u)=m.Proof.
We claim that bothI and G are weakly lower semicontinuous. In fact, according to Lemma 10, if uk⇀u in Eα, then uk→u 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))dt→∫0T‍F(t,u(t))dt, which means that the functional u→∫0T‍F(t,u(t))dt is weakly continuous on Eα. Similarly u→∫0T‍f(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.
SinceμF(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],x∈R.
Let {uk}∈N be a minimizing sequence; that is, I(uk)→m, I′(uk)→0 as k→∞. Then,
(41)m+o(1)=I(uk)=-12∫0T‍(0CDtαuk(t),tCDTαuk(t))dt-∫0T‍F(t,uk(t))dt≥-12∫0T‍(0CDtαuk(t),tCDTαuk(t))dt-1μ∫0T‍ukf(t,uk)dt-BT=(1μ-12)∫0T‍(0CDtαuk(t),tCDTαuk(t))dt+1μ〈I′(uk),uk〉-BT≥(12-1μ)|cos⁡(πα)|∥uk∥2-1μ∥I′(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 uk⇀u in Eα. Then from Lemma 10, uk→u in C([0,T],R). Since G is weakly lower semicontinuous and {uk}∈N, we first have
(42)G(u)≤lim⁡_k→∞⁡G(uk)=0.
Then we have u≠0. In fact, if u=0, then uk→0 in C([0,T],R). By G(uk)=0, we get ∥uk∥→0. This is a contradiction with {uk}∈N.
Then from Lemma14, there exists a unique y>0 such that yu∈N. Together with I which is weakly lower semicontinuous, we have
(43)m≤I(yu)≤lim⁡_k→∞I(yuk)≤lim⁡k→∞I(yuk)≤lim⁡k→∞I(uk)=m.
Then we get that m is achieved at yu∈N. 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 Lemmas14 and 15, we can get that there exists u∈N such that I(u)=m=inf⁡NI. 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.



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