# Variational Methods for a Fractional Dirichlet Problem Involving Jumarie’s Derivative

Authors: Rafał Kamocki
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

We investigate a fractional Dirichlet problem involving Jumarie’s derivative. Using some variational methods a theorem on the existence and uniqueness of a solution to such problem is proved. In the proof of the main result we use a fractional counterpart of the du Bois-Reymond fundamental lemma.


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## Body

## 1. Introduction

In the last time, fractional calculus plays an essential role in the fields of mathematics, physics, electronics, mechanics, engineering, and so forth (cf. [1–5]). Many processes in physics and engineering can be described accurately by using systems of differential equations containing different type of fractional derivatives. Among definitions of derivatives of fractional order we can pick the Riemann-Liouville and the Caputo derivatives out. Unfortunately, each of them has different unusual properties. For instance, the Riemann-Liouville derivative of a constant is not zero and the Caputo derivative is defined only for differentiable functions (alternatively, for such functions that have no first order derivative but then they might have fractional derivatives of all orders less than one, see [6]).Recently, Jumarie proposed a new definition of the fractional derivative being a little modification of the Riemann-Liouville derivative (cf. [7–10]). His definition eliminates disadvantages of mentioned earlier derivatives, because the Jumarie derivative of a constant is equal to zero and it is defined for any continuous (nondifferentiable) functions.In the paper we consider the following fractional boundary problem:(1)xααt=Fxt,xt,for  a.e.t∈a,b,(2)xa=xb=0,where α∈1/2,1, F:a,b×Rn→R, and x(α) denotes Jumarie’s derivative of a function x. The above problem is a generalization of the classical Dirichlet problem of the form (3)x′′t=Fxt,xt,xa=xb=0.We discuss the problem of the existence of solutions to above problem. In our investigations we use some variational method given in [11]. First, we consider some integral functional depending on the Jumarie derivative, for which (1) is the Euler-Lagrange equation. Next, we prove existence of a critical point of mentioned functional in an appropriate space of functions and under suitable assumptions of regularity, coercivity, and convexity. In order to do it, we use the following.Proposition 1 (see [11]).
IfX is a reflexive Banach space and the functional L:X→R is coercive and sequentially weakly lower semicontinuos, then it possesses at least one minimum at x0∈X.Let us remind that a functionalL defined on a Banach space X is coercive if Lx→∞ whenever x→∞, and L is sequentially weakly lower semicontinuous at x0∈X if liminfn→∞Lxn≥Lx0 for any sequence xn⊂X such that xn⇀x0 weakly in X.Of course such critical point is a minimizer of this functional and it generates the solution to problem (1)-(2). In order to prove that a minimum point gives a solution it is sufficient to apply a fractional version of the du Bois-Reymond lemma obtained in Section 3. Results of a such type for the Dirichlet problem involving the Riemann-Liouville derivative have been obtained in [12].The paper is organized as follows. In Section2, we review some basic definitions and facts concerning fractional calculus that we need in the sequel. Moreover, we introduce the space of solutions to considered problem and give some useful properties of this space. In Section 3, we formulate and prove some fractional version of the du Bois-Reymond Lemma, which we use in the proof of the main result. Mentioned main results of the work, namely, a theorem on the existence and uniqueness of a solution to problem (1)-(2), are formulated and proved in Section 4.


## 2. Preliminaries

In the first part of this paper we recall some basic facts concerning fractional calculus (cf. [3, 7–10, 13]). Next, we introduce some function spaces, which will be used later.


### 2.1. Fractional Calculus

We will assume that[a,b]⊂R is a bounded interval.Letα>0 and f∈L1([a,b],Rn). The left-sided Riemann-Liouville integral of the function f of order α is defined by (4)Ia+αft≔1Γα∫atfτt-τ1-αdτ,t∈a,ba.e.In the rest of this paper we will assume that α∈(0,1).The left-sided Riemann-Liouville derivativeDa+αf of the function f of order α is defined in the following way:(5)Da+αft≔ddtIa+1-αft,t∈a,ba.e.,provided that Ia+1-αf has an absolutely continuous representant on [a,b] (i.e., there exists an absolutely continuous function on [a,b] which is equal a.e. on [a,b] to Ia+1-αf).Now, let us assume thatf∈C([a,b],Rn).Jumarie’s modified Riemann-Liouville derivative of the functionf of order α is defined by(6)fαt≔Da+αf·-fat=1Γ1-αddt∫atfτ-fat-ταdτ,t∈a,ba.e.,provided that Ia+1-α(f(·)-f(a)) has an absolutely continuous representant on [a,b].Remark 2.
It is easy to see that iff(a)=0, then defined above derivatives coincide. Moreover, Jumarie’s modified Riemann-Liouville derivative of a constant equals zero.Remark 3.
The definition of fractional derivative given by (6) is a consequence of the following fractional derivative via difference reads defined by Jumarie: (7)fαt≔limh→0h-α∑k=1∞-1kαkfx+α-kh.The(dt)α integral of f is given by (8)∫axftdtα≔α∫axftx-tα-1dt,x∈a,b.We have the following theorem on the integration by parts.Theorem 4.
Letf,g∈C([a,b],Rn) and there exist derivatives f(α) and g(α). Then(9)∫abfαtgtdtα=Γα+1fbgb-faga-∫abftgαtdtα.


### 2.2. SpaceLp,α([a,b],Rn)

Let1⩽p<∞.Let us define the set(10)
L
p
,
α
a
,
b
,
R
n
≔
f
:
a
,
b
⟶
R
n
;
f
-measurable
,
b
-
·
α
-
1
/
p
f
∈
L
p
a
,
b
,
R
nwith the norm (11)fLp,α≔b-·α-1/pfLp=∫abftpb-tα-1dt1/p.We will identify two functions belonging to Lp,α([a,b],Rn) that coincide a.e. on [a,b].It is easy to show thatLp,α([a,b],Rn) is a Banach space. In particular, the space L2,α([a,b],Rn), equipped with the inner product (12)f,gL2,α≔∫abftgtb-tα-1dt,is a Hilbert space.Now, we give some properties of the spaceLp,α([a,b],Rn).Proposition 5.
The injection(13)Lp,αa,b,Rn⊂Lpa,b,Rnis continuous.Proof.
Letf∈Lp,α([a,b],Rn). Then (14)∫abftpdt=∫abb-t1-αb-tα-1ftpdt⩽b-a1-α∫abb-tα-1/pftpdt.Consequently,(15)fLp⩽b-a1-α/pfLp,α.The proof is completed.Proposition 6 (Hölder inequality).
Letf∈Lp,α([a,b],Rn), g∈Lq,α([a,b],Rn), and 1/p+1/q=1. Then fg∈L1,α([a,b],Rn) and(16)∫abftgtb-tα-1dt⩽∫abftpb-tα-1dt1/p·∫abgtqb-tα-1dt1/q.Proof.
From the Hölder inequality for the spaceLp([a,b],Rn) we obtain (17)∫abftgtb-tα-1dt=∫abb-tα-1/pftb-tα-1/qgtdt⩽∫abftpb-tα-1dt1/p·∫abgtqb-tα-1dt1/q.Lemma 7.
The operatorIa+α:Lp,α([a,b],Rn)→Lp,α([a,b],Rn) is bounded, and it means there exists a constant Cp>0 such that(18)Ia+αfLp,α⩽CpfLp,α,f∈Lp,αa,b,Rn.Proof.
Using Fubini’s Theorem and [14, Lemma 1], we obtain (19)
I
a
+
α
f
L
p
,
α
p
=
∫
a
b
b
-
t
α
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1
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a
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p
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t
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c
0
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α
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1
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+
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t
p
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t
=
c
0
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α
∫
a
b
∫
a
t
b
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α
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1
·
t
-
τ
α
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1
f
τ
p
d
τ
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t
=
c
0
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α
∫
a
b
f
τ
p
·
∫
τ
b
b
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t
α
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1
t
-
τ
α
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1
d
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τ
=
t
=
b
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0
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τ
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0
1
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1
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α
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1
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α
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1
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b
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τ
d
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0
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α
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b
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-
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2
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-
1
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τ
p
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0
1
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1
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s
α
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1
d
s
d
τ
=
β
α
,
α
c
0
Γ
α
∫
a
b
b
-
τ
α
-
1
b
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τ
α
f
τ
p
d
τ
⩽
β
α
,
α
c
0
b
-
a
α
Γ
α
·
∫
a
b
b
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τ
α
-
1
f
τ
p
d
τ
=
β
α
,
α
c
0
b
-
a
α
Γ
α
f
L
p
,
α
p
,where c0=((b-a)α/Γ(α+1))p-1, β(z,u)≔∫01rz-1(1-r)u-1dr, z,u>0.
Sinceβ(z,u)=Γ(z)Γ(u)/Γ(z+u), it suffices to put Cp=(b-a)α(Γ(α)/Γ(2α)Γ(α+1)p-1)1/p.
The proof is completed.


### 2.3. SpaceIa+α(Lp,α)

Let us define the setIa+α(Lp,α([a,b],Rn)) (shortly Ia+α(Lp,α)) as follows: (20)
I
a
+
α
L
p
,
α
≔
f
:
a
,
b
⟶
R
n
;
f
=
I
a
+
α
φ
,
a.e.  on
a
,
b
,
φ
∈
L
p
,
α
a
,
b
,
R
n
.Functions belonging to Ia+α(Lp,α) and equal a.e. on [a,b] are identified.From Proposition5, we immediately obtain the following.Proposition 8.
Consider(21)Ia+αLp,α⊂Ia+αLp,where Ia+α(Lp)≔{f:[a,b]→Rn;f=Ia+αh,a.e.on[a,b],h∈Lp([a,b],Rn)}.From the above fact and [14, Proposition 2] it follows that if f∈Ia+α(Lp,α) then there exists the Riemann-Liouville derivative Da+αf almost everywhere on [a,b].Moreover, one can show thatIa+α(Lp,α) with the norm ·Ia+α(Lp,α):Ia+α(Lp,α)→[0,∞) given by(22)fIa+αLp,α≔Da+αfLp,αis complete and, consequently, is a Banach space. In particular, the space Ia+α(L2,α), equipped with the inner product(23)f,gIa+αL2,α≔∫abDa+αftDa+αgtb-tα-1dt,is a Hilbert space.Remark 9.
Let us note that in the casep>1/α, from Proposition 8 and [15, Property 4], it follows that (24)Ia+αLp,α⊂Ia+αLp⊂C0a,b,Rn,where C0([a,b],Rn):={u∈C([a,b],Rn):u(a)=0}. Consequently, if f∈Ia+α(Lp,α) then f possesses Jumarie’s modified Riemann-Liouville derivative f(α) and f(α)=Da+αf. Of course, then (25)fIa+αLp,α≔fαLp,α.Remark 10.
From Proposition8, monography [3, Lemma 2.5(a)] and Remarks 2 and 9 it follows that if f∈Ia+α(Lp,α) with p>1/α then (26)ft=Ia+αfαt,t∈a,ba.e.From the above remark and Lemma7, we immediately obtain the following.Lemma 11 (fractional Poincaré Inequality).
Letp>1/α. Then(27)∫abftpb-tα-1dt1/p≤Cp∫abfαtpb-tα-1dt1/pfor f∈Ia+αLp,α, where Cp>0 is the constant from Lemma 7.


## 2.1. Fractional Calculus

We will assume that[a,b]⊂R is a bounded interval.Letα>0 and f∈L1([a,b],Rn). The left-sided Riemann-Liouville integral of the function f of order α is defined by (4)Ia+αft≔1Γα∫atfτt-τ1-αdτ,t∈a,ba.e.In the rest of this paper we will assume that α∈(0,1).The left-sided Riemann-Liouville derivativeDa+αf of the function f of order α is defined in the following way:(5)Da+αft≔ddtIa+1-αft,t∈a,ba.e.,provided that Ia+1-αf has an absolutely continuous representant on [a,b] (i.e., there exists an absolutely continuous function on [a,b] which is equal a.e. on [a,b] to Ia+1-αf).Now, let us assume thatf∈C([a,b],Rn).Jumarie’s modified Riemann-Liouville derivative of the functionf of order α is defined by(6)fαt≔Da+αf·-fat=1Γ1-αddt∫atfτ-fat-ταdτ,t∈a,ba.e.,provided that Ia+1-α(f(·)-f(a)) has an absolutely continuous representant on [a,b].Remark 2.
It is easy to see that iff(a)=0, then defined above derivatives coincide. Moreover, Jumarie’s modified Riemann-Liouville derivative of a constant equals zero.Remark 3.
The definition of fractional derivative given by (6) is a consequence of the following fractional derivative via difference reads defined by Jumarie: (7)fαt≔limh→0h-α∑k=1∞-1kαkfx+α-kh.The(dt)α integral of f is given by (8)∫axftdtα≔α∫axftx-tα-1dt,x∈a,b.We have the following theorem on the integration by parts.Theorem 4.
Letf,g∈C([a,b],Rn) and there exist derivatives f(α) and g(α). Then(9)∫abfαtgtdtα=Γα+1fbgb-faga-∫abftgαtdtα.


## 2.2. SpaceLp,α([a,b],Rn)

Let1⩽p<∞.Let us define the set(10)
L
p
,
α
a
,
b
,
R
n
≔
f
:
a
,
b
⟶
R
n
;
f
-measurable
,
b
-
·
α
-
1
/
p
f
∈
L
p
a
,
b
,
R
nwith the norm (11)fLp,α≔b-·α-1/pfLp=∫abftpb-tα-1dt1/p.We will identify two functions belonging to Lp,α([a,b],Rn) that coincide a.e. on [a,b].It is easy to show thatLp,α([a,b],Rn) is a Banach space. In particular, the space L2,α([a,b],Rn), equipped with the inner product (12)f,gL2,α≔∫abftgtb-tα-1dt,is a Hilbert space.Now, we give some properties of the spaceLp,α([a,b],Rn).Proposition 5.
The injection(13)Lp,αa,b,Rn⊂Lpa,b,Rnis continuous.Proof.
Letf∈Lp,α([a,b],Rn). Then (14)∫abftpdt=∫abb-t1-αb-tα-1ftpdt⩽b-a1-α∫abb-tα-1/pftpdt.Consequently,(15)fLp⩽b-a1-α/pfLp,α.The proof is completed.Proposition 6 (Hölder inequality).
Letf∈Lp,α([a,b],Rn), g∈Lq,α([a,b],Rn), and 1/p+1/q=1. Then fg∈L1,α([a,b],Rn) and(16)∫abftgtb-tα-1dt⩽∫abftpb-tα-1dt1/p·∫abgtqb-tα-1dt1/q.Proof.
From the Hölder inequality for the spaceLp([a,b],Rn) we obtain (17)∫abftgtb-tα-1dt=∫abb-tα-1/pftb-tα-1/qgtdt⩽∫abftpb-tα-1dt1/p·∫abgtqb-tα-1dt1/q.Lemma 7.
The operatorIa+α:Lp,α([a,b],Rn)→Lp,α([a,b],Rn) is bounded, and it means there exists a constant Cp>0 such that(18)Ia+αfLp,α⩽CpfLp,α,f∈Lp,αa,b,Rn.Proof.
Using Fubini’s Theorem and [14, Lemma 1], we obtain (19)
I
a
+
α
f
L
p
,
α
p
=
∫
a
b
b
-
t
α
-
1
I
a
+
α
f
t
p
d
t
⩽
c
0
∫
a
b
b
-
t
α
-
1
I
a
+
α
f
t
p
d
t
=
c
0
Γ
α
∫
a
b
∫
a
t
b
-
t
α
-
1
·
t
-
τ
α
-
1
f
τ
p
d
τ
d
t
=
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0
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∫
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b
f
τ
p
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τ
b
b
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t
α
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1
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-
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d
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d
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=
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=
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-
d
s
b
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0
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∫
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b
f
τ
p
·
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0
1
s
α
-
1
b
-
τ
α
-
1
1
-
s
α
-
1
b
-
τ
α
-
1
·
b
-
τ
d
s
d
τ
=
c
0
Γ
α
∫
a
b
b
-
τ
2
α
-
1
f
τ
p
·
∫
0
1
s
α
-
1
1
-
s
α
-
1
d
s
d
τ
=
β
α
,
α
c
0
Γ
α
∫
a
b
b
-
τ
α
-
1
b
-
τ
α
f
τ
p
d
τ
⩽
β
α
,
α
c
0
b
-
a
α
Γ
α
·
∫
a
b
b
-
τ
α
-
1
f
τ
p
d
τ
=
β
α
,
α
c
0
b
-
a
α
Γ
α
f
L
p
,
α
p
,where c0=((b-a)α/Γ(α+1))p-1, β(z,u)≔∫01rz-1(1-r)u-1dr, z,u>0.
Sinceβ(z,u)=Γ(z)Γ(u)/Γ(z+u), it suffices to put Cp=(b-a)α(Γ(α)/Γ(2α)Γ(α+1)p-1)1/p.
The proof is completed.


## 2.3. SpaceIa+α(Lp,α)

Let us define the setIa+α(Lp,α([a,b],Rn)) (shortly Ia+α(Lp,α)) as follows: (20)
I
a
+
α
L
p
,
α
≔
f
:
a
,
b
⟶
R
n
;
f
=
I
a
+
α
φ
,
a.e.  on
a
,
b
,
φ
∈
L
p
,
α
a
,
b
,
R
n
.Functions belonging to Ia+α(Lp,α) and equal a.e. on [a,b] are identified.From Proposition5, we immediately obtain the following.Proposition 8.
Consider(21)Ia+αLp,α⊂Ia+αLp,where Ia+α(Lp)≔{f:[a,b]→Rn;f=Ia+αh,a.e.on[a,b],h∈Lp([a,b],Rn)}.From the above fact and [14, Proposition 2] it follows that if f∈Ia+α(Lp,α) then there exists the Riemann-Liouville derivative Da+αf almost everywhere on [a,b].Moreover, one can show thatIa+α(Lp,α) with the norm ·Ia+α(Lp,α):Ia+α(Lp,α)→[0,∞) given by(22)fIa+αLp,α≔Da+αfLp,αis complete and, consequently, is a Banach space. In particular, the space Ia+α(L2,α), equipped with the inner product(23)f,gIa+αL2,α≔∫abDa+αftDa+αgtb-tα-1dt,is a Hilbert space.Remark 9.
Let us note that in the casep>1/α, from Proposition 8 and [15, Property 4], it follows that (24)Ia+αLp,α⊂Ia+αLp⊂C0a,b,Rn,where C0([a,b],Rn):={u∈C([a,b],Rn):u(a)=0}. Consequently, if f∈Ia+α(Lp,α) then f possesses Jumarie’s modified Riemann-Liouville derivative f(α) and f(α)=Da+αf. Of course, then (25)fIa+αLp,α≔fαLp,α.Remark 10.
From Proposition8, monography [3, Lemma 2.5(a)] and Remarks 2 and 9 it follows that if f∈Ia+α(Lp,α) with p>1/α then (26)ft=Ia+αfαt,t∈a,ba.e.From the above remark and Lemma7, we immediately obtain the following.Lemma 11 (fractional Poincaré Inequality).
Letp>1/α. Then(27)∫abftpb-tα-1dt1/p≤Cp∫abfαtpb-tα-1dt1/pfor f∈Ia+αLp,α, where Cp>0 is the constant from Lemma 7.


## 3. Du Bois-Reymond Lemma

In this section, we will prove the du Bois-Reymond lemma for nondifferentiable functions.We have the following.Lemma 12 (du Bois-Reymond lemma).
Letα∈(1/2,1), f∈L2,α([a,b],Rn), and(28)∫abfthαtb-tα-1dt=0for any function h∈Ia+α(L2,α) such that h(b)=0. Then there exists a constant c∈Rn such that f(t)=c for a.e. t∈[a,b] and, consequently, f(α)(t)=0 for a.e. t∈[a,b].Proof.
First, let us note that from the Hölder inequality (cf. Proposition6) it follows that the integral (28) is well-defined. Let w(t)=c for t∈[a,b], where c=α/(b-a)α∫abf(t)(b-t)α-1dt. Then, for any function h∈Ia+α(L2,α) such that h(b)=0 (in view of Remark 9 the condition h(a)=0 is satisfied also), from assumption (28) and Theorem 4, we obtain (29)0=∫abfthαtb-tα-1dt-0=∫abfthαtb-tα-1dt-1α∫abwαthtdtα=∫abfthαtb-tα-1dt-1α∫abwthαtdtα+Γα+1αwbhb-waha=∫abft-wthαtb-tα-1dt.Thus,(30)∫abft-wthαtb-tα-1dt=0,h∈Ia+αL2,α,hb=0.Let us consider the function h^(t)=(Ia+α(f-w))(t) for a.e. t∈[a,b]. It is easy to see that h^∈Ia+α(L2,α) and, in view of Remark 9, h^(a)=0. We will show that h^(b)=0. Indeed, we have (31)h^b=1Γα∫abft-wtb-tα-1dt=1Γα∫abftb-tα-1dt-cΓα∫abb-tα-1dt=1Γα∫abftb-tα-1dt-cΓαb-aαα=0.Consequently, the function h^ satisfies equality (30). So, (32)∫abft-wt2b-tα-1dt=0.It means that (33)ft=wt=c,t∈a,ba.e.The proof is completed.Remark 13.
In [16] result of such a type, but for Caputo derivative (for differentiable functions h), had been proved.Using Lemma12, we will prove the next lemma, which will play a key role in the next section. We have the following.Lemma 14.
Letα∈(1/2,1), f1∈L2,α([a,b],Rn), f2∈L2([a,b],Rn), and(34)∫abf1thαt+f2thtb-tα-1dt=0for any function h∈Ia+α(L2,α) such that h(b)=0. Then, (35)f1t=Ia+αf2t+c,t∈a,ba.e.,where c∈Rn is the constant from Lemma 12, and consequently (36)f1αt=f2t,t∈a,ba.e.Proof.
Using the Hölder inequality for spacesLp and Lp,α, we check that integral (34) exists. Let us put v=Ia+αf2. Then v∈Ia+α(L2) and v(α)=f2. From Theorem 4 and assumption (34), we obtain (37)0=α∫abf1thαt+f2thtb-tα-1dt=α∫abf1thαtb-tα-1dt+∫abvαthtdtα=α∫abf1thαtb-tα-1dt-∫abvthαtdtα+Γα+1vbhb-vaha=α∫abf1t-vthαtb-tα-1dt,h∈Ia+αL2,α,hb=0.Thus (38)∫abf1t-vthαtb-tα-1dt=0,h∈Ia+αL2,α,hb=0.From Lemma 12 it follows that there exists a constant c∈Rn such that (39)f1t-vt=c,t∈a,ba.e.Consequently, (40)f1t=vt+c=Ia+αf2t+c,t∈a,ba.e.It is well known that the function Ia+αf2 possesses the left-sided Riemann-Liouville derivative and (41)Da+αIa+αf2t=f2t,t∈a,ba.e.Since α∈(1/2,1), from [15, Property 4] it follows that the function Ia+αf2 is continuous and (Ia+αf2)(a)=0. Consequently, it possesses also the Jumarie modified Riemann-Liouville derivative, wchich equals f2. It means that the function f1 has the Jumarie modified Riemann-Liouville derivative and (using the second part of Remark 2) (42)f1αt=f2t,t∈a,ba.e.The proof is completed.


## 4. Main Result

Let us consider Dirichlet problem (1)-(2). By a solution to such problem we will mean a function x∈Ia+α(L2,α), satysfying condition x(b)=0, such that x(α)∈Ia+α(L2)+{c;c∈R}.Let us notice that sinceα∈(1/2,1), from Remark 9 it follows that x and x(α) are continuous and x satisfies the initial condition x(a)=0.In order to prove the existence of solutions to problem (1)-(2), we use variational methods.Let us consider a functionalL of the form(43)Lx≔∫ab12xαt2+Ft,xtb-tα-1dtdefined on the following space(44)K02,αa,b,Rn=K02,α=x∈Ia+αL2,α:xb=0.We impose the following assumption on the functionF:(A1)
The function(45)a,b∋t⟼Ft,x∈Ris measurable on[a,b] for any x∈Rn and the function(46)Rn∋x⟼Ft,x∈Ris of classC1 on R for a.e. t∈[a,b].(A2)
There exist constantsc1,c2≥0 and functions γ1∈L1,αa,b,R+ and γ2∈L2a,b,R+ such that (47)Ft,x≤c1x+γ1t,Fxt,x≤c2x+γ2tfor almost everyt∈a,b and all x∈Rn.We say thatL possesses the first variation δLx,h at the point x∈K02,α in the direction h∈K02,α (cf. [17]) if there exists a finite limit (48)δLx,h=limλ→0⁡Lx+λh-Lxλ.We will prove that, under assumptions (A1) and (A2),L possesses its minimum at a point x0 which is a solution to (1).To begin with, we will prove the following.Theorem 15.
Let us assume that conditions (A1)-(A2) are satisfied. Then the functionalL is well-defined on K02,α and possesses the first variation δLx,h at any point x∈K02,α and in any direction h∈K02,α given by(49)δLx,h=∫abxαthαt+Fxt,xtht·b-tα-1dt.Proof.
The fact thatL and δL are well-defined follows directly from (A1)-(A2) and the Hölder inequality (cf. Proposition 6). Let us fix x∈K02,α and  h∈K02,α and write the functional L as(50)L=L2∘L1,where (51)L1:K02,α∋x·⟶12xα·2+F·,x·∈L1,αa,b,R,L2:L1,αa,b,R∋y·⟶∫abytb-tα-1dt∈R.It is clear that L1 and L2 are well-defined and L2 is linear. Moreover, (52)L2y⩽∫abytb-tα-1dt=yL1,α,y∈L1,αa,b,R,so L2 is continuous. Consequently, it is differentiable in the sense of Frechet on L1,α([a,b],R) and the differential at any point y∈L1,α([a,b],R) is equal to L2. Using the Lebesque dominated convergence theorem and the mean value theorem, we assert that the mapping L1 has the first variation δL1(x,h) at any point x∈K02,α and in any direction h∈K02,α given by (53)δL1x·,h·=xα·hα·+Fx·,x·h·.This means (cf. [17, Section 2.2.2]) that there exists the first variation of the mapping L given by equality (49).
The proof is completed.Using the same arguments as in [15, Proposition 1(3.3)], we can obtain the following.Theorem 16.
Letα∈(1/2,1) and (fk)k∈N⊂K02,α. If fk⇀f weakly in K02,α (with topology induced from Ia+α(L2,α)), then fk⇉f uniformly on [a,b].Now, we will prove the main result of this paper, namely, a theorem on the existence of a unique solution to problem (1)-(2). We have the following.Theorem 17.
Letα∈(1/2,1) and assume that assumptions (A1)-(A2) are satisfied. If there are constants a1<αΓ(2α)/(b-a)2α, b1,b2≥0, such that(54)Ft,x≥-a12x2-b1x-b2,t∈a,ba.e.,x∈Rn,then problem (1)-(2) possesses at least one solution which minimizes functional L. Moreover, if the function(55)Rn∋x⟼Ft,x+a1x22∈Ris convex for a.e. t∈a,b, then the solution is unique.Proof.
Letx∈K02,α. Then from condition (54), Proposition 6, and Lemma 11 it follows that (56)Lx=12xIa+αL2,α2+∫abFt,xtb-tα-1dt≥12xIa+αL2,α2-a12xL2,α2-b1b-aααxL2,α-b2b-aαα≥1-a1C222xIa+αL2,α2-b1C2b-aααxIa+αL2,α-b2b-aαα,where C2=(b-a)α/αΓ(2α) is the constant from Lemma 7.
Consequently, sincea1<αΓ(2α)/(b-a)2α, 1-a1C22>0; so L is coercive.
Now, Letxk⇀x0 weakly in K02,α. From Theorem 16 it follows that xk⇉x0, on [a,b]. Thus and from assumption (A2), using the dominated convergence theorem, we get that(57)limk→∞⁡∫abFt,xktdt=∫abFt,x0tdt.
Moreover,K02,α is a Hilbert space as the closed subspace of Ia+α(L2,α). Consequently, the mapping x↦xIa+α(L2,α) is weakly lower semicontinuous, so(58)liminfk→∞⁡Lxk=liminfk→∞⁡12xkIa+αL2,α2+limk→∞⁡⁡∫abFt,xktdt≥Lx0.This means that the functional L is sequentially weakly lower semicontinuous and, by the virtue of Proposition 1, we conclude that it possesses minimum at the point x0∈K02,α.
From Theorem15 and Fermat lemma it follows that(59)∫abx0αthαt+Fxt,x0thtb-tα-1dt=0for any h∈K02,α. Applying Lemma 14, we get(60)xααt=Fxt,xtfor a.e.t∈a,b.Since x0∈K02,α, boundary conditions xa=xb=0 are satisfied.
The proof of the existence part is completed.
Now, we will show that, under assumption (55), the solution to problem (1)-(2) is unique. First, let us note that for x,y∈K02,α,  x≠y, and λ∈0,1 we have (61)
1
-
λ
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λ
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1
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λ
x
t
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y
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λ
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1
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·
λ
∫
a
b
x
α
t
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y
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t
2
b
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α
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1
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1
∫
a
b
x
t
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y
t
2
b
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t
α
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1
d
t
.If a1≤0 then (62)
1
-
λ
λ
∫
a
b
x
α
t
-
y
α
t
2
b
-
t
α
-
1
d
t
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a
1
∫
a
b
x
t
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·
λ
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a
b
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t
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t
>
0
,otherwise, from Lemma 11 and assumption a1<αΓ(2α)/(b-a)2α, it follows that (63)
1
-
λ
λ
∫
a
b
x
α
t
-
y
α
t
2
b
-
t
α
-
1
d
t
-
a
1
∫
a
b
x
t
-
y
t
2
b
-
t
α
-
1
d
t
≥
1
-
λ
·
λ
∫
a
b
x
α
t
-
y
α
t
2
b
-
t
α
-
1
d
t
-
a
1
b
-
a
2
α
α
Γ
2
α
∫
a
b
x
α
t
-
y
α
t
2
b
-
t
α
-
1
d
t
=
1
-
λ
λ
1
-
a
1
b
-
a
2
α
α
Γ
2
α
·
∫
a
b
x
α
t
-
y
α
t
2
b
-
t
α
-
1
d
t
>
0
.It means that the functional(64)K02,α∋x⟼12∫abxαt2-a1xt2b-tα-1dt∈Ris strictly convex. Consequently, using assumption (55), we assert that the functional(65)Lx=∫abFt,xt+a12xt2b-tα-1dt+12∫abxαt2-a1xt2b-tα-1dtis strictly convex, so its minimum point x0∈K02,α is unique. On the other hand if x∈K02,α is a solution to (1)-(2) then(66)x0ααt=Fxt,x0t,t∈a,ba.e.Thus,(67)∫abx0ααthtb-tα-1dt-∫abFxt,x0thtb-tα-1dt=0for any h∈K02,α and from Theorem 4 we get(68)∫abx0αthαtb-tα-1dt+∫abFxt,x0thtb-tα-1dt=δLx0,h=0.This means that the solution x0 to problem (1)-(2) is a minimum point of L, so it is unique. The proof is completed.



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*Source: 248517-2015-06-21.xml*