分数阶变分问题的欧拉–拉格朗日方程

本科毕业论文

外文文献及译文

文献、资料题目: Formulation of Euler–Lagrange equations for

fractional variational problems

文献、资料来源：爱思唯尔期刊

文献、资料发表（出版）日期：2001.7.27

外文文献：

Formulation of Euler–Lagrange equations

for fractional variational problems

1. Introduction

The field of calculus of variations is of significant importance in various disciplines such as science, engineering, and pure and applied mathematics. Reference [7] presents a Bliss-type multiplier rule for constrained variational problems with delay. Calculus of variations has been the starting point for various approximate numerical schemes such as Ritz, finite difference, and finite element methods (see [2,8]).

Functional minimization problems naturally occur in engineering and science where minimization of functionals, such as, Lagrangian, strain, potential, and total energy, etc. give the laws governing the systems behavior. In optimal control theory, minimization of certain functionals give control functions for optimum performance of the system.

Although many laws of the nature can be obtained using certain functionals and the theory of calculus of variations, not all laws can be obtained this way. For example, almost all systems contain internal damping, yet the traditional energy based approach cannot be used to obtain equations describing the behavior of a nonconservative system (see [9,10]). Recently, Refs. [9,10] presented a new approach to mechanics that allows one to obtain the equations for a nonconservative system using certain functionals. In these references, fractional derivative terms were introduced in functionals to obtain nonconservative terms in the desired differential equations.

Fractional derivatives, or more precisely derivatives of arbitrary orders, have played a significant role in engineering, science, and pure and applied mathematics in recent years. As [11] point out, there is hardly a field or science or engineering that has remained untouched by this field. Reference [12] provide an encyclopedic treatment of this subject. Additional background, survey, and application of this field in science, engineering, and mathematics can be found, among others, in [11–17].

Recent investigations have shown that many physical systems can be represented more accurately using fractional derivative formulations. Given this, one can imagine obtaining these

formulations by minimizing certain functionals. These functionals will naturally contain fractional derivative terms, and mathematical tools analogous to calculus of variations will be needed to minimize these functional. However, very little work has been done in the area of fractional calculus of variations.

This paper provides some new results in the area of fractional calculus of variations. Afractional calculus of variations problem is a problem in which either the objective functional or the constraint equations or both contain at least one fractional derivative term. In this paper we will develop necessary conditions for two problems from this field, first, minimization of a functional subjected to specified boundary conditions, and second, minimization of a functional subjected to constrains and specified boundary conditions. Both functional and the constraints will be allowed to have fractional derivative terms. 2. The simplest fractional variational problem

Several definitions of a fractional derivative have been proposed. These definitions include Riemann –Liouville, Grunwald –Letnikov, Weyl, Caputo, Marchaud, and Riesz fractional derivatives. Here, we formulate the problem in terms of the left and the right Riemann –Liouville fractional derivatives, which are defined as [16].

The left Riemann –Liouville fractional derivative

τττααα

?---??

?

??-Γ=x

a

n n x

a d f x dx d n x f D )()()(1)(1

， （1） and

The right Riemann –Liouville fractional derivative

?---??

?

??--Γ=

x

a

n n b x d f x dx d n x f D τττααα

)()()(1)(1

, （2） Where α is the order of the derivative such that n n <≤-α1.If αis an integer, these derivatives are defined in the usual sense, i.e.,

α

α

???

??=dx d x f D x

a )(， αα??? ??-=dx d x f D

b x )( ????=,2,1α （3）

These derivatives will be denoted as the LRLFD and the RRLFD, respectively. Note that in the literature the Riemann –Liouville fractional derivative generally means the LRLFD. From physical point of view, ifxis considered as a time scale, the RRLFD represents an operation

performed on the future state of the process )(x f . This derivative has generally been neglected with the assumption that the present state of a process does not depend on the results of its future development. However, the derivations to follow will show that both derivatives naturally occur in a problem of fractional calculus of variations.

Using the above definitions, the first simplest fractional calculus of variations problem can be defined as follows:Let ),,,(v u y x F be a function with continuous first and second(partial) derivatives with respect to all its arguments. Then, among all functions )(x y which have continuous LRLFD of order αand RRLFD of order βfor b x a ≤≤and satisfy the boundary conditions

b a y b y y a y ==)(,)( （4） find the function for which the functional

[]dx

y D y D y x F y J b x x a b

a ),,,(βα

?= （5）

is an extremum, where .1,0≤<βα The continuity requirement on F can be given more precisely. However, these assumptions are made for simplicity. Note that (1) we have included both the LRLFD and the RRLFD for generality. (2) We first consider 1,0≤<βα. The case of +∈R βα, will be consider shortly. (3) When 1==βα, the above problem reduces to the simplest variational problem.

To develop the necessary conditions for the extremum, assume that )(x y *is the desired function. Let R ∈ε, and define a family of curve

)()()(x x y x y εη+*= （6）

which satisfy the boundary conditions; i.e., we require that

0==b a ηη （7）

Since α

x a D and βb x D are linear operators, it follows that

)()()(x D x y D x y D x a x a x a

ηεα

αα+=*， （8a)

)()()(x D x y D x y D b x b x b x

ηεβββ+=* (8b)

Substituting Eqs. (6) and (8) into Eq. (5), we find that for each )(x η

[]?++*+*==*b

a b b x X a x a dx D y D D y D y x F J J ),,,(ηεηεεηεββαα (9)

is a function of εonly. Note that []εJ is extremum at 0=ε. Differentiating Eq. (9)with respect to ε, we obtain

dx D y D F D y D F y F d dJ b

a b x b x x a x a ???

??????+??+??=ηηηεββαα (10) Equation (10) is also called the variations of []y J at )(x y along )(x η. A necessary condition for to have an extremum is that εd dJ must be zero, and this should be true for all admissible )(x η. This leads to the condition that for []y J to have an extremum for )(x y y *=is that

0=?????

???+??+???dx D y D F D y D F y F b

a b x b x x a x a ηηηββαα (11) for all admissible η(x). Using the formula for fractional integration by parts, the second integral in Eq. (11) can be written as

dx y D F D dx D y D F b

a x a b

a b x x a x a ??????????=??ηηα

α)( (12) provided that y D F x a α

?? or ηis zero at a x = and b x =. Using Eq. (7), this condition is satisfied, and it follows that Eq. (12) is valid. Similarly, the third integral in Eq. (11) can be written as

dx y D F D dx D y D F b

a b x b

a x a

b x b x ??????????=??ηηαβ

αβ)( (13) Substituting Eqs. (12) and (13) into Eq. (11), we get

0=?????

???+??+???dx y D F D y D F D y F b

a b x x a x a b x ηββ

αα. (14) Since )(x η is arbitrary, it follows from a well established result in calculus of variations that

0=??+??+??y

D F D y D F D y F b x x a x a b x ββ

ααη. （15） Equation (15) is the Euler –Lagrange equation for the fractional calculus of variations problem. Thus, we have

Theorem 1. Let []y J be a functional of the form

dx y D y D y x F b x x a b

a

),,,(βα?

, defined on the set of functions y(x) which have continuous LRLFD of order α and RRLFD of order β in [a, b]and satisfy the boundary conditions a y a y =)(and b y b y =)(. Then a necessary condition for []y J to have an extremum for a given function )(x y is that )(x y satisfy following Euler –Lagrange equation:

0=??+??+??y

D F D y D F D y F b x x a x a b x ββααη Note that for fractional calculus of variation problems the resulting Euler –Lagrange equation contains both the LRLFD and the RRLFD. This is expected since the optimum function must

satisfy both terminal conditions. Further, for 1==βα, we have dx d D x a =αand dx d D x a =α

,and

Eq. (15) reduces to the standard Euler –Lagrange equation

0)

1(=??-??y F

dx d y F , (16) Where dx dy y =)1(.

3. The case of +∈R βα,and several functions

We now consider further generalization of the above problem. Specifically, we consider two different cases, first, in which ,......)1(,=∈+j R j j βα,i.e., one can have multiple positive

αand β, and second, in which one has more than one function. In both cases, we consider the

end points fixed.

Case 1. Fixed end points and ,......)1(,=∈+j R j j βα.

Assume that ),....,1(n j j =α and ),...,1(m k k =βare two sets of real numbers all greater than zero,

),...,,,....max(11max m n ββααα= （17）

is the maximum of all these numbers, and M is an integer such that M M <≤-max 1α. Assume that )1,...,,,(n m z z y x F +is a function with continuous first and second (partial) derivatives with respect to all its arguments, and consider a functional of the form

[]dx y D y D y D y D y x F y J m b x b x n x a b

a x a ),...,,,...,,,(11ββαα?= (18)

The problem can now be defined as follows:Among all functions )(x y satisfying the conditions

0)(a y a y =, 1)1()(a y a y =, ....., )1()1()(--=M a M y a y (19a) 0)(b y b y =, 1)1()(b y b y =, ....., )1()1()(--=M b M y b y (19b)

find the function for which Eq.(18)has an extremum. Here it is implicitly assumed that )(x y meets all the differentiability requirements.

The necessary condition for this problem can be found following the approach presented above. This leads to

Theorem 2.Let )(y J be a functional of the form given by Eq.(18)defined on the set of functions satisfying the boundary conditions given by Eq.(19).Then a necessary condition for )(y J to have an extremum for a given function )(x y is that )(x y satisfy the Euler –Lagrange equation

011=??+??+??∑∑==y D F D y D F D y F k b

x m k k

x a j x a n j j b x ββαα (20) As a special case, consider that ),....,1(n j j =α, and that F does not contain the

),...,1

(m k y D k

b x =β

terms. In this case, using Eq. (3), we have ∑==??-+??n j j j y

F

dx d y F 1)(0)( (21) Thus, for integral order derivatives, the necessary conditions obtained using fractional calculus of

variations approach reduces to that obtained using standard calculus of variations approach.

Case 2. Fixed end points and several functions.

The simplest fractional variational problem discussed in Section 2 can be generalized in a straight forward manner to problems containing several unknown functions. This problem can be defined as follows: Let ),...,,,...,,(211n n z z y y x F be a function with continuous first and second (partial) derivatives with respect to all its arguments. For 1,0≤<βα, consider the problem of finding necessary conditions for an extremum of a functional of the form

[]dx y D y D y D y D y y x F y y J n b x b x n x a b

a x a n n ),...,,,...,,,...,,(,...,1111ββα

α?= (22)

which depends on n continuously differentiable functions )(),...,(1x y x y n satisfying the boundary conditions

),...,1()(,)(n j y b y y a y jb j ja j ===. （23） Note that no relationship exists among the functions ).,...,1)((n j x y j = Therefore, the necessary condition for the functional in Eq. (22) to have an extremum can be found by considering the variations of each function one at a time.Thus we have Theorem 3. A necessary condition for the curve

),...,1)((n j x y y j j ==, (24)

which satisfies the boundary conditions given by Eq.(23) .to be an extremal of the functional given by Eq. (22)is that the functions )(x y j satisfy the following Euler –Lagrange equation:

0=??+??+??j b x x a j x a b x j y D F D y D F D y F ββ

ααη ).,...,1(n j = (25) In vector notation, the above condition can be written as

0=??+??+??y

D F D y D F D y F b x x a x a b x ββ

ααη, Where n R y ∈.

The above problem considers several functions but only one LRLFD of order 1≤αand one RRLFD of order 1≤β. The problem of finding extremum of a functional consisting of multiple

functions and multiple LRLFD and RRLFD of order greater than zero can be developed using the discussion presented in cases 1 and 2 above.

4. The problem of Lagrange and the multiplier rule

In this section we consider the following problem: Find the extremum of the functional

[]dx y D y D y x F y J b x b

a x a ),,,(βα

?= (27)

such that

0),(=Φy x (28)

and

),...,1)(()(),()()(2)(2)(1)(1m n j b y b y a y a y j s j s j s j s -===, (29)

where ,,,n m R R y m n <∈Φ∈and 1s and 2s are two sets of n numbers obtained by reordering the numbers 1 to n. It is assumed that the constrained functions ),...,1(0),(m j y x j ==Φare all independent. This problem is essentially the same as that of Lagrange except that in this case the functional contains the LRLFD and the RRLFD. For this reason, we will call this problem as the problem of Lagrange containing fractional derivatives or simply a fractional Lagrange problem. This is a special case, and in a general fractional Lagrange problem,Φmay also contain the left and the right fractional derivatives.

To develop the necessary conditions for the problem, note that y at the two ends are completely known. This follows from the fact that the constraints ),...,1(0),(m j y x j ==Φ(j =1,...,m)are all independent and the values of n?m functions ),...,1)((m j x y j =are specified at both ends. Therefore, the values of the rest of the functions at the two ends can be determined using a technique such as Newton –Raphson.

Suppose )(x y *is the solution to the above problem, and define

)()()(x x y x y εη+=*, (30)

where εis a sufficiently small number, and n R x ∈)(ηis a variation of )(x y consistent with the constraints, i.e.,)(x y satisfies Eq. (28). From the above discussion, it follows that

0)()(==b a ηη. (31)

Substituting Eq. (31) into Eq. (28), expanding the resulting vector into Taylor series, and neglecting second and higher order terms in ε,we get

.0)(=?Φ

?x y

η (32) Equation (32) clearly indicates that not all functions ),...1)((n j x j =ηcan be independent. Substituting Eq. (30) into Eq. (27), we get a function that is only dependent on . Extremum of this function requires that its derivative with respect to εmust be zero. This leads to

0=?????

???+??+???dx D y D F D y D F y F b

a b x b x x a x a ηηηββαα. (33) The left-hand side of Eq. (33) is the directional derivative of J at )(x y in the Direction )(x η. Using the formula for fractional integration by parts and Eq. (31),it follows that

0=?????

???+??+???dx y D F D y D F D y F b

a b x x a x a b x ηββ

αα. (34) Here the elements of η(x) are not all independent , and therefore its coefficients cannot be set to zero. Equation (15) motivates the following

Definition. An admissible arc )(x y *is said to satisfy the multiplier rule if there exists a vector of multipliers m R x l ∈)(continuous on [a, b], and a function

),,()(),,,(),,,,(y x x l y D y D y x F l y D y D y x F T b x x a b x x a Φ+=β

αβα (35)

such that

0=??+??+??y D F D y D F D y F b x x a x a b x ββ

αα (36) is satisfied along )(x y *.Thus:

Theorem 4. Every minimizing arc )(x y *must satisfy the multiplier rule.

Proof. To prove this, multiply Eq. (32) with )(x l T and add the results to Eq. (34)to get

0)(=?????

??Φ?+??+??+???dx y x l y D F D y D F D y F b

a T

b x x a x a b x ηββαα. (37)

It can now be shown that

0)(=?Φ?+??+??+??y

x l y D F D y D F D y F T

b x x a x a b x ββαα. (38) This follows from the fact that )(x l may be selected such that m of the n equations in Eq. (38) are zero. This is true since y ?Φ?has a full rank. Rest of theη’s can be selected as independent and therefore the other n?m equations in (38) follo ws by using Eq. (37) and applying a theorem in calculus of variations. Note that Eq. (36) can now be obtained using Eqs. (35) and (38). Equation (38) will be called the Euler –Lagrange equation for constrained fractional variational problems.

The multiplier rule is also applicable for the case when Φis also a function of the LRLFD and the RRLFD. Multiplier rule for a system containing multiple fractional derivatives can be developed in a similar manner. 5. Examples

In this section, we obtain the Euler –Lagrange equations for an unconstrained and a constrained fractional variational problems.

Example 1. As the first example, consider the following unconstrained fractional variational problem:

minimize []?=1

20)(21dx y D y J x α (39)

such that

0)0(=y and 1)1(=y . (40)

This example with 1=α, for which the solution is x x y =)(, is often considered in textbooks on variational calculus. It can be shown that for this problem, the Euler –Lagrange equation is

.0)(01=y D D x x

αα (41)

It can be shown that for 21>α, the solution is given as

[]

?

----=x

t x t dt

x y 01.))(1()12()(α

α (42) Example 2. As the second example, consider the following constrained fractional variational

problem:

minimize [][]

?+=1

2

22121dx y y y J (43)

such that

,2110y y y D x +-=α (44)

.1)0(1=y (45)

This example with integral order derivative is often considered in textbooks on optimal control. It can be shown that for this problem, the Euler –Lagrange equation is

,01=++l D l y l x α (46)

.02=-l y (47)

Note that in both examples both the LRLFD and the RRLFD occur in the resulting Euler –Lagrange equations even when the problems contain only LRLFDs. Such differential equations have not been studied much in the literature. A method to find solutions for such problems will be presented in a later work.

Remarks. In closing, we would like to make the following two remarks.

1. Here we have assumed that the terminal conditions are fixed and the functions meet all the smoothness requirements. The case of unspecified end conditions, unspecified end points, (the transversality conditions), and piecewise smoothness (the corner conditions) will be considered in a future work.

2. The theorems and their proofs presented here are very similar to those given in standard textbooks on calculus of variations. Thus, many of the concepts of classical calculus of variations can be extended with minor modifications to fractional calculus of variations. Given the fact that many systems are described more accurately using fractional derivative models and that nature attempts to minimize certain functionals, it is hoped that more research will continue in this field. 6. Conclusions

Euler –Lagrange equations have been presented for unconstrained and constrained fractional variational problems. The approach presented and the resulting equations are very similar to those for variational problems containing integral order derivatives. In special cases, when the

derivatives are of integral order only, the results of fractional calculus of variations reduce to those obtained from classical calculus of variations. Given the fact that many systems can be modeled more accurately using fractional derivative models, it is hoped that future research will continue in this area.

中文译文：

分数阶变分问题的欧拉–拉格朗日方程

1.引言

变分法领域具有重要意义在不同学科如科学、工程和纯和应用数学。文献[7]提出了一种延迟Bliss-type乘数规则约束变分问题。变分法已经成为各种近似数值方案的出发点如里兹，有限差分，有限元方法。

函数最小化问题自然发生在工程和科学最小化的泛函,如拉格朗日,应变,潜力,和总能量等。给系统行为规律。在最优控制理论,最小化的泛函,给控制功能系统的最佳性能。

虽然许多自然规律可以使用某些泛函和变分理论得到的，并不是所有的法律可以通过这种方式获得。例如,几乎所有的系统包含内部阻尼,然而,传统的基于能量的方法不能用于获得方程描述非守恒的系统的行为。最近，文献[9,10] 提出了一种新的力学方法,允许一个获得使用特定的泛函,非守恒的系统方程。在这些引用,分数导数方面介绍了泛函,得到非守恒的条件所需的微分方程。

近年来，分数阶导数，或者更准确地说是任意阶导数，在工程，科学和纯粹与应用数学中发挥了重要作用。[ 11 ]指出，几乎没有一个，一直保持不变，这一领域或科学或工程。文献[ 12 ]提供的这个问题的百科全书式的处理。额外的背景,调查,和这一领域在科学、工程和数学的应用可以在[11–17]找到。

最近的调查表明,许多物理系统可以更准确地使用分数导数公式来表示。鉴于此, 我们可以想象通过某些泛函的最小化获得这些构想。这些泛函将自然地包含分数阶导数和类似于变分法的数学工具将需要减少这些功能。然而，在分数阶微积分领域极少工作已经完成。

本文在分数微积分的变化方面提供了一些新的结果。Afractional变分法的问题是一个问题的目标函数或约束方程或者两者都包含至少一个分数导数。在本文中,我们将开发必要条件的两个问题,第一,最小化的功能被指定的边界条件,第二,最小化的功能受到约束和边界条件指定。功能和约束将被允许有分数导数的条件。

2．最简单的分数阶变分问题

一个分数阶导数的定义已经提出了几种。这些定义包括黎曼—刘维尔，格伦沃–刘维尔，外尔，卡普托, 马尔绍,和里斯分数阶导数。在这里，我们制定的左边和右边的黎曼–刘维尔分数导数项的问题，它被定义为[16]。

左黎曼–刘维尔分数阶导数

τττααα

?---??? ??-Γ=x

a

n n x a d f x dx d n x f D )()()(1)(1

（1） 右黎曼–刘维尔分数阶导数

?---??? ??--Γ=x

a

n n b x d f x dx d n x f D τττααα)()()(1)(1

（2） α是导数的次数且n

n <≤-α1，如果α是个整数，那这些导数通常定义为 α

α

??? ??=dx d x f D x

a )(，αα??

? ??-=dx d x f D b x )(

????=,2,1α （3） 这些导数分别被表示为左导数和右导数。值得注意的是，在黎曼—刘维尔分数导数一般是指左导数。从物理的角度来看，ifxis 视为一个时间尺度，右导数代表了一个操作进程)(x f 的未来状态的过程。这些导数普遍被一个进程的当前状态的假设所忽视，而不依赖于其未来发展的结果。然而，推导将表明，导数自然地出现一个分数微积分的变化问题。 使用上面的定义，第一个简单的变化分数微积分问题可以定义如下：令),,,(v u y x F 是对它所有参数有一阶和二阶导数的函数。然后，在所有的函数中，当b x a ≤≤时，)(x y 都有连续的α阶左导数和β阶右导数，并且满足边界条件

b

a y

b y y a y ==)(,)( （4） []dx y D y D y x F y J b x x a b

a ),,,(β

α?= （5）

是一个极值，在这1,0≤<βα。对F 可以给出更精确的连续性要求。然而，这些假设都很简单。注意到（1）包含一般性的左导数和右导数。（2）我们首先考虑的1

,0≤<βα。+∈R βα,的情况将会简明地考虑到。当1==βα，上述问题降低到最简单的变分问题。

为开发极值的必要条件,假设)(x y *是所需的函数。令R ∈ε，定义一组曲线 )()()(x x y x y εη+*= （6） 满足边界条件；我们要求

0=

=b a ηη （7） 因为α

x a D 和βb x D 是线性算子，它遵循

)(

)()(x D x y D x y D x a x a x

a η

εααα+=* （8a ） )(

)()(x D x y D x y D b x b x b

x η

εβββ+=* （8b ） 把公式（6）和（8）代入公式（5），我们发现，对每一个)(x η

[]?

++*+*==*b

a

b b x X a x a dx D y D D y D y x F J J ),,,(η

εηεεηε

ββαα （9） 是只有ε的函数。注意[]εJ 是0=ε的极值。我们针对ε鉴别公式（9），我们得到

dx D y D F D y D F y F d dJ b a b x b x x a x a ???

?

?????+??+??=ηηηεββαα （10） 方程（10）也被称为当)(x y 属于)(x η时[]y J 的微分，一个极限的必要条件是εd dJ 必须为0，并且这个应该对所有可采纳的)(x η为真。这导致对于[]y J 有一个极值的条件是

0=???

?????+??+???dx D y D F D y D F y F b

a

b x b x x a x a ηηηββαα （11） 对所有可采纳的)(x η。使用分级分步积分法的公式，在公式（11）的第二个积分可以写为

dx y D F D dx D y D F b

a x a b

a b x x a x a ??????????=??ηηαααα)( （12） 倘若当a x =和b x =时，y D F x

a α

??或η为0。运用公式（7），满足了这个条件，并且公式（12）是有效的。同样，公式(11)的第三个积分可写为

dx y D F D dx D y D F b

a b x b a x a b x b x ??????????=??ηηαβαβ)( （13） 把公式（12）和（13）代入公式（11）,我们得到

0=????????+??+???dx y D F D y D F D y F b

a

b x x a x a b x ηββαα （14） 因为)(x η是任意的，它遵循一个变分法已成立的结果

0=??+??+??y D F D y D F D y F b

x x a x a b x ββααη （15） 方程（15）是欧拉–拉格朗日方程的分数阶微积分的变化问题。因此，我们有

定理1.令[]y J 是函数形式dx y

D y D y x F b x x a b

a

),,,(β

α?，定义在y(x)这组函数上，有连续的α阶左导数和β阶右导数且满足边界条件a y a y =)(和b y b y =)(。那么，对于一个给定的函数

)(x y ，[]y J 有极值的必要条件是)(x y 满足欧拉—拉格朗日的方程：0=??+??+??y D F D y D F D y F b

x x a x a b x ββααη 注意,分数微积分的变分问题产生的欧拉—拉格朗日方程包含左导数和右导数。这是预期的

最优函数必须满足的两个终端条件。更进一步地，对1==βα，我们得到dx d D x a =α和dx

d D x a

=α

，并且公式（15）降低到标准的欧拉–拉格朗日方程 0)

1(=??-??y

F dx d y F ，在这dx dy y =)

1( （16） 3. +

∈R βα,和几个函数的情况

我们现在考虑上述问题的进一步推广。特别地，我们考虑两种不同的情况，第一，其

中,......1(,=∈

+j R j j βα，即：可以有多个α和β；第二，其中有一个以上的函数。在这两种情况下,我们认为端点固定。

情况1.端点固定且,......1(,=∈

+

j R j j βα。 假设),....,1(n j j =α和),...,1(m k k

=β是两组都大于0的实数， ),...,,,....max(1

1

max

m

n ββααα= (17)

是所有这些数的最大值，并且M 是一个整数且M M <≤-max

1α。假设)1,...,,,(n m z z y x F +是一个对于它所有的参数有连续的一阶和二阶导数的函数，并考虑函数的形式

[]dx y D y D y D y D y x F y J m b x b x n x a b

a x a ),...,,,...,,,(1

1ββαα?

= （18） 这个问题现在可以定义如下：在所有的函数中)(x y 满足这些情况

0)(a y a y =， 1)

1()(a y a y =，……，)1()1()(--=M a M y a y （19a ） 0)(b y b y =， 1)1()(b y b y =，…….，)1()1()(--=M b M y b y

（19b ）

可以发现，这个函数对于公式（18）有一个极值。在这里，它是隐含的假设，满足所有要求的可微性。这个问题的必要条件，可以在上面介绍的方法中发现。这导致了

定理 2. 令[]y J 是公式（18）给出的一个定义在函数集合且满足等式（19）给出的边值条件的函数形式。那么，对于一个给定的函数)(x y ，[]y J 有极值的必要条件是)(x y 满足欧拉—拉格朗日方程

011=??+??+??∑∑==y D F D y D F D y F k b x m k

k

x a j x a n j j b x ββαα （20） 作为一种特殊情况，考虑到),....,1(n j j =α，并且F 不包含),...,1(m k y

D k

b x =β

。在这种情况下，应用等式（3），我们得到

∑==??-+??n j j j y F

dx d y F 1

)

(0)( （21） 因此，对整数阶导数，使用分数阶变分法获得的必要条件归纳为使用标准的变分法得到的方法。

情况2.端点固定和多函数

第二节讨论的简单的分数变分问题可以以直接的方式推广包含多个未知函数的问题。这个问题可以定义如下:令),...,,,...,,(211n n z z y y x F 是一个对于它所有的参数有连续的一阶和二阶导数的函数。对于1

,0≤<βα，考虑寻找一个函数的极值的必要条件的形式问题 []dx y D y D y D y D y y x F y y J n b x b x n x a b

a x a n n ),...,,,...,,,...,,(,...,1111ββαα?=

（22） 这取决于N 阶连续可微函数满足的边界条件

),...,1()(,)(n j y b y y a y jb

j ja j === （23） 注意在函数).,...,1)((

n j x y j =中不存在关系。因此，对于等式（22）中的函数有一个极值的必要条件可以通过考虑每个函数的变化发现。因此我们有 定理3. 曲线的一个必要条件

),...,1)((

n j x y y j j == (24) 满足给定的方程(23)的边界条件成为给定方程(22)的极值函数即)(x y j 满足以下的欧拉—

拉格朗日方程

0=??+??+??j

b x x a j x a b x j y D F D y D F D y F ββααη，).,...,1(n j = （25） 在矢量法，以上条件可写为

0=??+??+??y D F D y D F D y F b

x x a x a b x ββααη ，n R y ∈ (26)

但是上述问题考虑的函数只有一个1≤α阶数的左导数和1≤β阶数的右导数。 函数极值问题包含多个函数和多个大于零的左右导数，这一问题可以使用例1和例2中呈现的结论进行扩展。

4.拉格朗日乘数规则的问题

在本节中,我们考虑以下问题:找到函数的极值

[]dx y D y D y x F y J b x b

a x a ),,,(β

α?

= (27) 0),(=Φy x (28) ),...,1)(()(),()()

(2)(2)(1)(1m n j b y b y a y a y j s j s j s j s -=== (29) 在这,

,,n m R R y m

n <∈Φ∈并且1s 和2s 是两组n 个数的数据，通过重新排序1至n 得到。假设约束函数),...,1(0),(m j y x j ==Φ都是独立的。这个问题本质上与拉格朗日是一样的，但在这种情况下，函数包括左导数和右导数。为此，我们将把这个问题作为拉格朗日含有分数阶导数的问题或简单的分数拉格朗日问题。这是一个特殊的情况下，在一般的分数拉格朗日问题，Φ也可能包含左、右分数阶导数。

为找出问题的必要条件，指出y 两端完全已知的。它遵循约束条件),...,1(0),(m j y x j ==Φ(j =1,...,m)是相互独立的和函数),...,1)((m j x y j =的值在两段指定，因此,其余的值函数的两端可以使用牛顿迭代等技术确定。

假设)(x y *是以上问题的，并定义

)

()()(x x y x y εη+=*

（30） ε是一个很小的数，n R x ∈)(η是)(x y 的一个变量，与约束一致满足等式（28）。从上面的讨

论中，有如下

0)()(==b a ηη （31） 把方程（31）带入（28），产生的关于ε向量展开为泰勒级数，而忽略了二阶和高阶项，我们得到

.0)(=?Φ

?x y

η （32） 方程(32)清楚地表明,并不是所有的函数),...1)((n j x j =η可以独立。把方程（30）带入（27）我们得到了一个函数,它只是依赖。这个函数极值的要求其ε的导数为零。这导致

0=???

?????+??+???dx D y D F D y D F y F b

a

b x b x x a x a ηηηββαα （33） 方程（33）左边是J 在)(x y 处)(x η方向上的方向导数。使用分级分步积分法的公式和方程（31），由此可见

0=????????+??+???dx y D F D y D F D y F b

a

b x x a x a b x ηββαα （34） 这里η(x)的元素不都独立，因此，它的系数不能设置为零。方程（15）促使以下 定义：若一个容许弧)(x y *满足乘数规则，就要看是否存在一个向量的乘数m R x l ∈)(在[a, b]上连续，并且函数

),,()(),,,(),,,,(y x x l y D y D y x F l y D y D y x F T

b

x x a b x x a Φ+=βαβα （35） 以使

0=??+??+??y D F D y D F D y F b

x x a x a b x ββαα （36） 在)(x y *上成立。 因此：

定理4.每一个最小化弧)(x y *必须满足乘数法则。

证明.为了证明这一点，方程（32）乘以)(x l T 并且把结果添加到方程（34）得到

0)(=???????Φ

?+??+??+???dx y x l y D F D y D F D y F b

a

T b x x a x a b x ηββαα （37） 现在可以证明