The variational problem of fractional-order control systems

Abstract

This article discusses the control system of fractional endpoint variable variational problems. For this problem, we prove the Euler-Lagrange type necessary conditions which must be satisfied for the given functional to be extremum. Finally, one example is provided to show the application of our results.

Liu et al. Advances in Dierence Equations (2015) 2015:110 DOI 10.1186/s13662-015-0450-6

*Correspondence: [email protected] of Mathematical Sciences, Anhui University, Hefei, 230039, China

R E S E A R C H Open Access

The variational problem of fractional-order control systems

Biao Liu*, Yanhua Wen and Xian-feng Zhou

Abstract

This article discusses the control system of fractional endpoint variable variational problems. For this problem, we prove the Euler-Lagrange type necessary conditions which must be satised for the given functional to be extremum. Finally, one example is provided to show the application of our results.

Keywords: variational problem; necessary condition; fractional-order control systems

1 Introduction

Fractional calculus is an old mathematical topic since the seventeenth century, yet it has only received much attention and interest in the past years. For more details on the basic theory of fractional calculus, one can see the monographs [] and the references therein.

As one of the important topics in control theory, the variational method plays an important role in the analysis control systems and is an important branch of mathematics study functional extremum. In recent years, the variational method is widely used in physics, economics, electrical engineering, image processing elds, etc.

Moreover, the numerical methods for solving fractional dierential equations, optimal control and variational problems have a good development. In [], Bhrawy et al. investigated a new spectral collocation scheme, which obtained a numerical solution of this equation with variable coecients on a semi-innite domain. In [], Doha et al. introduced a numerical technique for solving a general form of the fractional optimal control problem. And in [], Bhrawy et al. used the Rayleigh-Ritz method for the necessary conditions of optimization and the operational matrix of fractional derivatives together with the help of the properties of the shifted Legendre orthonormal polynomials to reduce the fractional optimal control problem to solving a system of algebraic equations that greatly simplies the problem. In [], Bhrawy and Zaky proposed and analyzed an ecient operational formulation of spectral tau method for a multi-term time-space fractional dierential equation with Dirichlet boundary conditions. In [], based on the shifted Legendre orthonormal polynomials, Ezz-Eldien et al. employed the operational matrix of fractional derivatives, the Legendre-Gauss quadrature formula and the Lagrange multiplier method for reducing such a problem into a problem consisting of solving a system of algebraic equations.

2015 Liu et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 2 of 10

However, at present, very little work has been done in the area of fractional calculus of variations []. In [], Jiao and Zhou used the critical point theory to solve the existence of solutions for a class of fractional boundary value problems. In [, ], the author presented a new approach to mechanics that allows one to obtain the equations for a nonconservative system using certain functionals. In [], the author proposed the fractional-order variational problem and gave solution for a class of variational problems with xed boundary value

J[x] =

tf

[integraldisplay] t F[parenleftbig]t, x, Dt,tx, Dt,tf x[parenrightbig] dtand the boundary conditions x(t) = x, x(tf ) = xf are xed. But [] did not discuss the problem of variable boundary conditions and transversal conditions. In other words, the boundary condition x(t) = x is xed, but x(tf ) = C(tf ) is variable, or vice versa. Inspired by the above-mentioned works, in this paper, we follow the ideas to investigate the optimality of control systems.

This paper is organized as follows. In Section , we briey review the denitions of Riemann-Liouville fractional integrals and derivatives and some lemmas. In Section , we give the necessary conditions for the fractional-order functional variational problem with xed and variable boundary. In Section , we give an example to show the application of our results, and Section briey summarizes the results of this paper and future work.

2 Preliminaries and lemmas

In this section, we give some basic denitions and results that are used throughout this paper. For more details, please see [].

Denition . [] Let [a, b] be a nite interval on the real axis R. The Riemann-Liouville fractional integrals Ia,tf and It,bf of order > are dened by

Ia,tf (t) =

()

[integraldisplay]

f ()(t ) d

, t > a (.)

t

a

and

It,bf (t) =

()

[integraldisplay]

f ()(t ) d

b

t

, t < b, (.)

respectively. Here, () denotes the gamma function. These integrals are called the left-

sided and the right-sided fractional integrals.

Denition . [] The Riemann-Liouville fraction derivatives Da,ty and Dt,by of order

> are dened by

Da,ty(t) = [parenleftbigg]

d dt

n

Ina,ty(t)[parenrightbig] (.)

=

(n )

[parenleftbigg]

d dt

n

[integraldisplay]

f ()(t )n+ d

n = [] + , t > a[parenrightbig] (.)

t

a

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 3 of 10

and

Dt,by(t) = [parenleftbigg]

d dt

n

Int,by(t)[parenrightbig] (.)

=

(n )

[parenleftbigg]

d dt

n

[integraldisplay]

f ()(t )n+ d

b

t

n = [] + , x < b

, (.)

respectively, where [] means the integral part of .

Remark . [] Obviously, if < < , then

Da,ty(t) =

( )

d dt

[integraldisplay]

t

f ()(t ) d

( < < , x > a) (.)

and

Dt,by(t) =

( )

a

d dt

[integraldisplay]

f ()(t ) d

b

t

( < < , x < b). (.)

Lemma . (Fractional integration by parts []) Assume that Da,tf (x) and Dt,bg(x) are existent and continuous on [a, b]. Let f (t) and g(t) be two continuous functions dened on t [a, b], > , and f (t) or g(t) and their until m derivatives are zero at t = a, b, and m is

less than the largest integer of . Then(i)

[integraldisplay]

b

a

(t)

Ia,t(t)[parenrightbig] dt = [integraldisplay]ba(t)

It,b(t)[parenrightbig] dt, (.)

(ii)

[integraldisplay] Da,tf (t)[parenrightbig]g(t) dt = [integraldisplay]ba f (t)[parenleftbig]Dt,bg(t)[parenrightbig] dt. (.)

3 The fractional variational problems-variable endpoints

Denition . Assume that F = F(t, x, Dt,tx, Dt,tf x) is a function with continuous rst and second (partial) derivatives with respect to all its arguments. Then, among all functions x(t) which have continuous LRLFD of order and RRLFD of order for t t tf and satisfy the boundary conditions

x(t) = x, x(tf ) = C(tf ). (.)

Looking for a function x(t) such that the functional

J(x) =

tf

b

a

[integraldisplay] t F[parenleftbig]t, x, Dt,tx, Dt,tf x[parenrightbig] dt (.) has extreme value, where < , and the endpoint C(tf ) is variable.

Theorem . Let J(x) be a functional of the form

J(x) =

tf

[integraldisplay] t F[parenleftbig]t, x, Dt,tx, Dt,tf x[parenrightbig] dt (.)

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 4 of 10

dened on the set of functions x(t) which have continuous LRLFD of order and RRLFD of order in [t, tf ] and satisfy the boundary conditions x(t) = x and x(tf ) = C(tf ). Then a necessary condition for J[x] to have an extremum for a given function x(t) is that x(t) satises the following Euler-Lagrange equation and terminal transversality condition:

F

x Dt,tf

F

Dt,tx Dt,t

F Dt,tf x

= , (.)

[parenleftbigg]

F Dt,tf x

Dt,tf (C x) + F[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

Dt,txDt,t(C x) +

= . (.)

Proof To prove the necessary conditions for the extremum, assume that x(t) is the desired function. Let R, and dene a family of curves

x(t) = x(t) + (t), (.)

where (t) is a continuous dierentiable function for all given, which satisfy the boundary conditions, i.e.,

(t) = . (.)

Due to the changing terminal time tf , each has its own trajectory terminal point tf . Therefore we must dene a terminal times set corresponding to x(t),

tf = tf + (tf ). (.)

Since Da,t and Dt,tf are linear operations, it follows that

Dt,tx(t) = Dt,tx(t) + Dt,t(t), (.)

Dt,tf x(t) = Dt,tf x(t) + Dt,tf (t). (.)

Substituting Eqs. (.) and (.)-(.) into Eq. (.),

J() =

t

f +(tf )

t=tf

[integraldisplay] t F[parenleftbig]t, x +

, Dt,tx + Dt,t, Dt,tf x + Dt,t[parenrightbig] dt

, Dt,tx + Dt,t, Dt,tf x + Dt,t[parenrightbig] dt

+

t

f

= [integraldisplay] t F[parenleftbig]t, x +

[integraldisplay]

t

f +(tf )

t

f

F

t, x + , Dt,tx + Dt,t, Dt,tf x + Dt,t[parenrightbig] dt

[integraldisplay]

, Da,tx + Dt,t, Dt,tf x + Dt,t[parenrightbig] dt

+ (tf )F

t

f

t F[parenleftbig]t, x +

, Dt,tf x[parenleftbig]tf[parenrightbig][parenrightbig] (.)

we nd that for each (t) there is only just a function of . Note that J() is extremum at

= , and the dierential of Eq. (.) with respect to , we obtain

J

[vextendsingle][vextendsingle][vextendsingle][vextendsingle]=

=

tf, x[parenleftbig]t f

, Dt,tx[parenleftbig]t f

[integraldisplay]

t

f

t

F

x

+

F

Dt,txDt,t

+

F Dt,tf x

Dt,tf [parenrightbigg] dt + (tf )F|t

f . (.)

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 5 of 10

Equation (.) is also called the variations of J(x) at x(t) along (t). A necessary condition for J() to have an extremum is that J |= = , and it is true for any admissible (t).

This leads to the condition that for J(x) to have an extremum for x = x(t),

[integraldisplay]

t

f

t

F

x

+

F

Dt,txDt,t

+

F Dt,tf x

Dt,tf [parenrightbigg] dt + (tf )F|t

f = (.) for all admissible (t). Using the denition of fractional derivatives and the formula for fractional integration by parts, the second and third integral in Eq. (.) can be written as

[integraldisplay]

t

f

t

F

Dt,tx(t)Dt,t

(t) dt

[vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

= Dt,tx(t)It,t

tf

(t)

t [integraldisplay]

tf

t

d dt

[parenleftbigg] Dt,tx(t)[parenrightbigg]It,t

(t) dt

F

[vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

= Dt,tx(t)It,t

tf

(t)

t [integraldisplay]

tf

t

(t)Dt,tf [parenleftbigg]

Dt,tx(t)[parenrightbigg] dt, (.)

F

[integraldisplay]

t

f

t

F Dt,tf x(t)

Dt,tf (t) dt

It,tf(t)[vextendsingle][vextendsingle][vextendsingle][vextendsingle]

tf

=

F Dt,tf x(t)

t [integraldisplay]

tf

t

d dt

[parenleftbigg]

F Dt,tx(t)[parenrightbigg]

It,t(t) dt

=

F Dt,tf x(t)

It,tf(t)[vextendsingle][vextendsingle][vextendsingle][vextendsingle]

tf

t

[integraldisplay]

tf

t

(t)Dt,t[parenleftbigg]

F Dt,tf x(t)[parenrightbigg]

dt. (.)

Substituting Eqs. (.) and (.) into Eq. (.) yields

[integraldisplay]

t

f

t

F

x Dt,tf [parenleftbigg]

Dt,tx[parenrightbigg] Dt,t[parenleftbigg]

F

F Dt,tf x[parenrightbigg][bracerightbigg]

dt

+

F Dt,tf x

It,tf[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

tf

t

F

+ Dt,txIt,t

+ (tf )F|t

[parenleftbigg] f = . (.) In virtue of (t) = , Eq. (.) can be written in the form

[integraldisplay]

t

f

t

F

x Dt,tf [parenleftbigg]

Dt,tx[parenrightbigg] Dt,t[parenleftbigg]

F

F Dt,tf x[parenrightbigg][bracerightbigg]

dt

[parenleftbigg]

+

F Dt,tf x

It,tf + (tf )F[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

+ Dt,txIt,t

= . (.)

Note that (tf) and (tf ) are not independent of each other, they are aected by the terminal constraint x(t)|t=tf = C(tf ), namely

x tf + (tf )[parenrightbig] +

tf + (tf )[parenrightbig] = C[parenleftbig]tf +(tf )

. (.)

t

f

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 6 of 10

Dierentiating Eq. (.) with respect to and letting , we have

(tf )x[parenleftbig]t

f

[parenrightbig] +

tf[parenrightbig] =(tf ) C[parenleftbig]t f

. (.)

That is,

tf[parenrightbig] =(tf )

[bracketleftbig] C[parenleftbig]t

[parenrightbig][bracketrightbig]. (.)

Substituting Eq. (.) into Eq. (.), we get

[integraldisplay]

t

f

t

f

[parenrightbig] x[parenleftbig]t

f

F

x Dt,tf

F

Dt,tx Dt,t

F Dt,tf x[bracerightbigg]

dt

+ (tf )

[parenleftbigg]

F Dt,tf x

Dt,tf (C x) + F[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

Dt,txDt,t(C x) +

= . (.)

Since (t) and (tf ) are arbitrary, it follows from a well-established result in the calculus of variations that

F

x Dt,tf

F

Dt,tx Dt,t

t

f

F Dt,tf x

= , (.)

[parenleftbigg]

F Dt,tf x

Dt,tf (C x) + F[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

Dt,txDt,t(C x) +

= , (.)

where Eq. (.) is called fractional-order variational problem of Euler-Lagrange equation, and Eq. (.) is called terminal transversality condition. The proof is completed.

Remark . Note that for the fractional calculus of variation problems, the resulting Euler-Lagrange equation contains both LRLFD and RRLFD. This is expected since the optimum function must satisfy both terminal conditions. Further, for = = , we have Dt,t = d/dt and Dt,tf = d/dt. Then Eqs. (.)-(.) can be turned into the standard

Euler-Lagrange equation and the terminal transversality condition

L

x

d dt

tf

L

x

= , (.)

( C x) F

x

+ F

[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

t=tf

= , (.)

where x = dx/dt.

Remark . In control engineering, the majority target line is parallel to the t axis, since C(t) = , hence

F Dt,tf xF

Dt,tx Dt,tx

F Dt,tf x[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

t=tf

= . (.)

If C(t) is perpendicular to the t axis, since C(t) = , we get

[parenleftbigg]

F

Dt,tx +

FDt,tf x[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]t=t

f

= . (.)

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 7 of 10

Remark . In a similar way, it is not dicult to infer that when the terminal is xed, and the initial c(t) is variable, the terminal transversality condition will change to

F F

Dt,txDt,t(x c)

F Dt,tf x

Dt,tf (x c)[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

t=t = . (.)

If F(t) is parallel to the t axis, we have

F F

Dt,txDt,tx

F Dt,tf x

Dt,tf x[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

t=t = . (.)

If F(t) is perpendicular to the t axis, then

[parenleftbigg]

F

Dt,tx +

F Dt,tf x[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

t=t = . (.)

Corollary . Let J[x] =

tft F(t, x, Dt,tx, . . . , Dnt,tx, Dt,tf x, . . . , Dmt,tfx) dt be a functional function satisfying the boundary conditions. Then a necessary condition for J[x] to have an extremum for a given function x(t) is that x(t) satises the Euler-Lagrange equation and the terminal transversality condition:

F

x

[integraltext]

n

[summationdisplay]

j=

Djt,tf

F Djt,tx

m

[summationdisplay]

k=

Dkt,t

F Dkt,tf x

= , (.)

M

[summationdisplay]

i=

[parenleftBigg]

n

[summationdisplay]

j=

F Djt,tx

Djt,t(Ci x) +

m

[summationdisplay]

k=

F Dkt,tf x

Dkt,tf (Ci x) + F[parenrightBigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle][vextendsingle]t=t

= , (.)

where j R+ (j = , . . . , n) and k R+ (k = , . . . , m) are two sets of real numbers all greater

than zero,

max = max(, . . . , n, , . . . , m) (.)

is the maximum of all these numbers, and M is the integer such that M max < M. And F is a function with continuous rst and second (partial) derivatives with respect to all its arguments. Meanwhile, all functions x(t) satisfy the following conditions:

x(a) = xa, x()(a) = xa, . . . , x(M)(a) = xa(M), (.)

x(tf ) = C(tf ), x()(tf ) = C(tf ), . . . , x(M)(tf ) = CM(tf ). (.)

Corollary . Let F(t, x, . . . , xn, y, . . . , yn) be a function with continuous rst and second (partial) derivatives with respect to all its arguments. For < , , consider the problem

of nding necessary conditions for an extremum of a functional of the form

J[x, . . . , xn] =

tf

f

[integraldisplay] t F[parenleftbig]t, x, . . . , xn, Dt,tx, . . . , Dt,txn, Dt,tf x, . . . , Dt,tf xn[parenrightbig] dt, (.)

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 8 of 10

which depends on n continuously dierentiable functions x(t), . . . , xn(t) satisfying the boundary conditions

xj(t) = xj, xj(tf ) = Cj(tf ) (j = , . . . , n). (.)

Then, a necessary condition for the curve is

xj = xj(t) (j = , . . . , n), (.)

which satises the boundary conditions given by Eq. (.) to be an extremal of the functional given by Eq. (.), i.e., the functions xj(t) satisfy the following Euler-Lagrange equation and terminal transversality condition:

F

xj Dt,tf

F

Dt,txj Dt,t

F Dt,tf xj

= (j = , . . . , n), (.)

[parenleftbigg]

F Dt,tf xj

Dt,tf (Cj xj) + F[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

Dt,txj Dt,t(Cj xj) +

= (j = , . . . , n). (.)

tf

In vector notation, the above conditions can be written as

F

X Dt,tf

F

Dt,tX Dt,t

F Dt,tf X

= , (.)

[parenleftbigg]

F Dt,tf X

Dt,tf (C X) + F[parenrightbigg][vextendsingle][vextendsingle][vextendsingle][vextendsingle]

F

Dt,tX Dt,t(C X) +

= , (.)

tf

where X, C Rn.

4 Application

In this section, we illustrate the importance of our results through one example.

Example . Consider the terminal variable fractional variational problem

J(x) =

[integraldisplay] D,tx(t)[parenrightbig] dt, (.)

where x() = , x(tf ) = tf .

Proof We know that

F =

D,tx(t)[parenrightbig] (.)

and the fractional Euler-Lagrange equation

Dt,tf D,tx = . (.)

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

x(t) = ( )

[integraldisplay]

tf

tf

(tf )(t )

d. (.)

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 9 of 10

Figure 1 Example 4.1.

Now, we consider the transversality condition, by Eq. (.), we obtain

C(t) x(t)

[parenrightbig][bracketrightbig]D,tx[bracerightbigg]t=tf= . (.)

By simplication, we get

D,tC(t) =

D,tx(t). (.)

By Eqs. (.)-(.), C(t) = t and the initial value, we obtain the optimal trajectory ( > )

x(t) = (

)

[braceleftbigg] D,tx[parenrightbig] + [bracketleftbig]D ,t

[integraldisplay]

tf

d. (.)

Thus when t = tf , we can get the optimal terminal times

tf = tf . (.)

By numerical simulation Eq. (.), we know that the equation has some solutions (see Figure ).

This example with = , for which the optimal trajectory is x(t) = t, and then we have obtained the same result.

5 Conclusions

Necessary conditions for the optimality control of those systems are established. The case of piecewise continuous conditions and researching the minimum value of fractional differential equations principles will be considered in a future work.

(tf )(t )

Competing interests

The authors declare that they have no competing interests.

Authors contributions

The authors have equal contributions to each part of this paper. All authors read and approved the nal manuscript.

Liu et al. Advances in Dierence Equations (2015) 2015:110 Page 10 of 10

Acknowledgements

This article was supported by the Natural Science Foundation of China (Grant Nos. 11471015, 11371027), the natural Science Foundation of Anhui Province (No. 1508085MA01) and the Youth Superior Talent Key Foundation of Anhui Province (No. 2013SQRL142ZD). The authors would like to thank the referees for their valuable suggestions and comments.

Received: 16 January 2015 Accepted: 23 March 2015

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