?F?F?F???xDb?D??0. （15） ax??y?aDxy?xDb?yEquation (15) is the Euler–Lagrange equation for the fractional calculus of variations problem. Thus, we have
Theorem 1. Let J?y? be a functional of the form
b??F(x,y,Dy,Dy)dx, axxb?adefined on the set of functions y(x) which have continuous LRLFD of order ? and RRLFD of order? in [a, b]and satisfy the boundary conditionsy(a)?yaandy(b)?yb. Then a necessary condition for J?y?to have an extremum for a given functiony(x)is thaty(x)satisfy following Euler–Lagrange equation:
?F?F?F???xDb?D??0 ax??y?aDxy?xDb?yNote 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, we haveaDx?ddxandaDx?ddx,and
Eq. (15) reduces to the standard Euler–Lagrange equation
?Fd?F??0, (16) ?ydx?y(1)Where y(1)?dydx.
3. The case of ?,??R?and several functions
We now consider further generalization of the above problem. Specifically, we consider two different cases, first, in which ?j,?j?R?(j?1,......),i.e., one can have multiple positive
?and?, 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 ?j,?j?R?(j?1,......).
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Assume that ?j(j?1,....,n) and ?k(k?1,...,m)are two sets of real numbers all greater than zero,
?max?max(?1,....?n,?1,...,?m) （17）
is the maximum of all these numbers, and M is an integer such that M?1??max?M. Assume that F(x,y,z1,...,zm?n)is a function with continuous first and second (partial) derivatives with respect to all its arguments, and consider a functional of the form
?n?1J?y???F(x,y,aDxy,...,aDxy,xDb?1y,...,xDb?my)dx (18)
abThe problem can now be defined as follows:Among all functionsy(x)satisfying the conditions
y(a)?ya0, y(1)(a)?ya1, ....., y(M?1)(a)?ya(M?1) (19a) y(b)?yb0, y(1)(b)?yb1, ....., y(M?1)(b)?yb(M?1) (19b)
find the function for which Eq.(18)has an extremum. Here it is implicitly assumed thaty(x)meets all the differentiability requirements.
The necessary condition for this problem can be found following the approach presented above. This leads to
Theorem 2.LetJ(y)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 forJ(y)to have an extremum for a given function y(x)is thaty(x)satisfy the Euler–Lagrange equation
m?Fn?F?F?j?k??xDb?D?0 (20) ?ax?j?k?yj?1?Dy?aDxyk?1xbAs a special case, consider that?j(j?1,....,n), and that F does not contain the
k,...,m)terms. In this case, using Eq. (3), we have xDby(k?1??Fnd?F??(?)j(j)?0 (21) ?yj?1dx?yThus, for integral order derivatives, the necessary conditions obtained using fractional calculus of
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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 F(x,y1,...,yn,z1,...,z2n)be a function with continuous first and second (partial) derivatives with respect to all its arguments. For 0??,??1, consider the problem of finding necessary conditions for an extremum of a functional of the form
??J?y1,...,yn???F(x,y1,...,yn,aDxy1,...,aDxyn,xDb?y1,...,xDb?yn)dx (22)
abwhich depends on n continuously differentiable functionsy1(x),...,yn(x)satisfying the boundary conditions
yj(a)?yja,yj(b)?yjb(j?1,...,n). （23） Note that no relationship exists among the functionsyj(x)(j?1,...,n). 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
yj?yj(x)(j?1,...,n), (24)
which satisfies the boundary conditions given by Eq.(23) .to be an extremal of the functional given by Eq. (22)is that the functionsyj(x)satisfy the following Euler–Lagrange equation:
?F?F?F???xDb?D??0 (j?1,...,n). (25) ax??yj?aDxyj?xDb?yjIn vector notation, the above condition can be written as
?F?F?F???xDb?D??0, ax???y?aDxy?xDbyWhere y?Rn.
The above problem considers several functions but only one LRLFD of order??1and one RRLFD of order??1. The problem of finding extremum of a functional consisting of multiple
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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
?J?y???F(x,y,aDxy,xDb?y)dx (27)
absuch that
?(x,y)?0 (28)
and
ys1(j)(a)?ys1(j)(a),ys2(j)(b)?ys2(j)(b)(j?1,...,n?m), (29)
where y?Rn,??Rm,m?n,ands1ands2are two sets of n numbers obtained by reordering the numbers 1 to n. It is assumed that the constrained functions?j(x,y)?0(j?1,...,m)are 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,?may 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?j(x,y)?0(j?1,...,m)(j =1,...,m)are all independent and the values of n?m functions yj(x)(j?1,...,m)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.
Supposey?(x)is the solution to the above problem, and define
y(x)?y?(x)???(x), (30)
where?is a sufficiently small number, and?(x)?Rnis a variation of y(x)consistent with the constraints, i.e.,y(x)satisfies Eq. (28). From the above discussion, it follows that
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?(a)??(b)?0. (31)
Substituting Eq. (31) into Eq. (28), expanding the resulting vector into Taylor series, and neglecting second and higher order terms in?,we get
???(x)?0. (32) ?yEquation (32) clearly indicates that not all functions?j(x)(j?1,...n)can 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?must be zero. This leads to
??F?F?F?????D??D?dx?0. (33) ??ax?xb???y?aDxy?xDby?a?The left-hand side of Eq. (33) is the directional derivative ofJaty(x)in the Direction?(x). Using the formula for fractional integration by parts and Eq. (31),it follows that
b??F?F?F????D?D?dx?0. (34) xbax??????y?aDxy?xDby?a?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 arcy?(x)is said to satisfy the multiplier rule if there exists a vector of multipliersl(x)?Rmcontinuous on [a, b], and a function
b????F(x,y,aDxy,xDby,l)?F(x,y,aDxy,xDby)?lT(x)?(x,y), (35)
such that
?F?F?F???xDb?D?0 (36) ax??y?aDxy?xDb?yis satisfied alongy?(x).Thus:
Theorem 4. Every minimizing arcy?(x)must satisfy the multiplier rule.
Proof. To prove this, multiply Eq. (32) withlT(x)and add the results to Eq. (34)to get
??F?F?F?????T?D?D?l(x)xbax???dx?0. (37) ????y?Dy?Dy?yaxxb?a?- 9 -
b