The analytic solution to an optimal control problem is investigated using the canonical dual method. By means of the Pontryagin principle and a transformation of the cost functional, the optimal control of a nonconvex problem is obtained. It turns out that the optimal control can be expressed by the costate via canonical dual variables. Some examples are illustrated.

1. Introduction

Consider the following optimal control problem (primal problem (𝒫) in short):

ξ€œ(𝒫)min𝑇0[]𝐹(π‘₯)+𝑃(𝑒)𝑑𝑑,(1)s.t.Μ‡π‘₯=𝐴(𝑑)π‘₯+𝐡(𝑑)𝑒,π‘₯(0)=π‘₯0βˆˆπ‘…π‘›,[],π‘‘βˆˆ0,𝑇‖𝑒‖≀1,(2) where 𝐹(β‹…) is continuous on 𝑅𝑛, and 𝑃(β‹…) is twice continuously differentiable in π‘…π‘š. An admissible control, taking values on the unit ball 𝐷∢={π‘’βˆˆπ‘…π‘šβˆ£β€–π‘’β€–β‰€1}, is integrable or piecewise continuous on [0,𝑇]. In (2) we assume that 𝐴(𝑑),𝐡(𝑑) are continuous matrix functions in 𝐢([0,𝑇],𝑅𝑛×𝑛) and 𝐢([0,𝑇],π‘…π‘›Γ—π‘š), respectively. This problem often comes up as a main objective in general optimal control theory [1].

By the classical control theory [2], we have the following Hamilton-Jacobi-Bellman function:

𝐻(𝑑,π‘₯,𝑒,πœ†)=πœ†βˆ—(𝐴(𝑑)π‘₯+𝐡(𝑑)𝑒)+𝐹(π‘₯)+𝑃(𝑒).(3) The state and costate systems are


In general, it is difficult to obtain an analytic form of the optimal feedback control for the problem (1)-(2). It is well known that, in the case of unconstraint, if 𝑃(𝑒) is a positive definite quadratic form and 𝐹(π‘₯) is a positive semidefinite quadratic form, then a perfect optimal feedback control is obtained by the solution of a Riccati matrix differential equation. The primal goal of this paper is to present an analytic solution to the optimal control problem (𝒫).

We know from the Pontryagin principle [1] that if the control ̂𝑒 is an optimal solution to the problem (𝒫), with Μ‚π‘₯(β‹…) and Μ‚πœ†(β‹…) denoting the state and costate corresponding to ̂𝑒(β‹…), respectively, then ̂𝑒 is an extremal control, that is, we have


By means of the Pontryagin principle and the dynamic programming theory, many numerical algorithms have been suggested to approximate the solution to the problem (𝒫) (see, [3–5]). This is due to the nonlinear integrand in the cost functional. It is even difficult for the case of 𝑃(𝑒) being nonconvex on the unit ball 𝐷 in π‘…π‘š. We know that when 𝑃(𝑒) is nonconvex on the unit ball 𝐷, sometimes the optimal control of the problem (𝒫) may exist. Let us see the following simple example for 𝑛=π‘š=1:

ξ€œ(𝒫)min101π‘₯βˆ’2𝑒2ξ‚„[]𝑑𝑑,s.t.Μ‡π‘₯=π‘₯+𝑒,π‘₯(0)=0,π‘‘βˆˆ0,𝑇,|𝑒|≀1.(8) In fact, it is easy to see that ̂𝑒(𝑑)β‰‘βˆ’1;π‘‘βˆˆ[0,𝑇] is an optimal control.

In this paper, we consider 𝑃(𝑒) to be nonconvex. If the optimal control of the problem (𝒫) exists, we solve the problem (1) to find the optimal control which is an expression of the costate. We see that, with respect to 𝑒, the minimization in (7) is equivalent to the following global nonconvex optimization over a sphere:

min‖𝑒‖≀1̂𝑃(𝑒)+πœ†(𝑑)βˆ—ξ€»[]𝐡(𝑑)𝑒,a.e.π‘‘βˆˆ0,𝑇,(9) when 𝑃(𝑒) is a nonconvex quadratic function, the problem (9) can be solved completely by the canonical dual transformation [6–8]. In [9], the global concave optimization over a sphere is solved by use of a differential system with the canonical dual function. Because the Pontryagin principle is a necessary condition for a control to be optimal, it is not sufficient for obtaining an optimal control to solve only the optimization (9). In this paper, combing the method given in [6, 9] with the Pontryagin principle, we solve problem (1)-(2) which has nonconvex integrand on the control variable in the cost functional and present the optimal control expressed by the costate via canonical dual variables.

2. Global Optimization over a Sphere

In this section we present a differential flow to deal with the global optimization, which is used to find the optimal control expressed by the costate in the next section. Here we use the method in our another paper (see [9]).

In what follows we consider the function 𝑃(π‘₯) to be twice continuously differentiable and nonconvex on the unit ball in π‘…π‘š. Define the set

ξ€½ξ€Ίβˆ‡πΊ=𝜌>0∣2𝑃(π‘₯)+𝜌𝐼>0,π‘₯βˆ—ξ€Ύπ‘₯≀1.(10) Since 𝑃(π‘₯) is nonconvex and βˆ‡2𝑃(π‘₯) is bounded on 𝐷∢={π‘₯βˆˆπ‘…π‘šπ‘₯βˆ—π‘₯≀1}, 𝐺 is an open interval (𝜌,+∞) for the nonnegative real number 𝜌 depending on 𝑃(π‘₯). Let πœŒβˆ—βˆˆπΊ and Μƒπ‘₯∈{π‘₯βˆ—π‘₯≀1} satisfy the following KKT equation:

βˆ‡π‘ƒ(Μƒπ‘₯)+πœŒβˆ—Μƒπ‘₯=0.(11) We focus on the flow Μ‚π‘₯(𝜌) defined near πœŒβˆ— by the following backward differential equation:

𝑑̂π‘₯+ξ€Ίβˆ‡π‘‘πœŒ2𝑃(Μ‚π‘₯)+πœŒπΌβˆ’1ξ€·πœŒΜ‚π‘₯=0,πœŒβˆˆβˆ—βˆ’π›Ώ,πœŒβˆ—ξ€»,ξ€·πœŒ(12)Μ‚π‘₯βˆ—ξ€Έ=Μƒπ‘₯.(13) The flow Μ‚π‘₯(𝜌) can be extended to wherever 𝜌∈𝐺∩(0,πœŒβˆ—] [10]. The dual function [6] with respect to a given flow Μ‚π‘₯(𝜌) is defined as

π‘ƒπ‘‘πœŒ(𝜌)=𝑃(Μ‚π‘₯(𝜌))+2Μ‚π‘₯βˆ—πœŒ(𝜌)Μ‚π‘₯(𝜌)βˆ’2.(14) We have

𝑑̂π‘₯(𝜌)ξ‚Άπ‘‘πœŒβˆ—ξ€Ίβˆ‡2𝑃(Μ‚π‘₯(𝜌))+πœŒπΌπ‘‘Μ‚π‘₯(𝜌)=π‘‘πœŒβˆ’12𝑑̂π‘₯βˆ—ξ€»(𝜌)Μ‚π‘₯(𝜌)π‘‘π‘‘πœŒ=βˆ’2𝑃𝑑(Μ‚πœŒ)π‘‘πœŒ2β‰₯0.(15) Consequently


It means that 𝑑𝑃𝑑(𝜌)/π‘‘πœŒ decreases when 𝜌 increases in 𝐺. If, for a Μ‚πœŒβˆˆπΊ, 𝑑𝑃𝑑(Μ‚πœŒ)/π‘‘πœŒβ‰€0, then 𝑑𝑃𝑑(𝜌)/π‘‘πœŒβ‰€0 for 𝜌∈𝐺∩[Μ‚πœŒ,∞). Therefore,

𝑃𝑑(Μ‚πœŒ)β‰₯𝑃𝑑(𝜌),(17) as long as 𝜌β‰₯Μ‚πœŒ.

Theorem 1. If the flow Μ‚π‘₯(𝜌),𝜌∈𝐺∩(0,πœŒβˆ—], defined by (11)–(13), passes through a boundary point of the ball 𝐷={π‘₯βˆˆπ‘…π‘šβ€–π‘₯‖≀1} at Μ‚πœŒβˆˆπΊ, that is, []Μ‚π‘₯(Μ‚πœŒ)βˆ—ξ€·Μ‚π‘₯(Μ‚πœŒ)=1,Μ‚πœŒβˆˆπΊβˆ©0,πœŒβˆ—ξ€»,(18) then Μ‚π‘₯ is a global minimizer of 𝑃(π‘₯) over the ball 𝐷. Further one has min𝐷𝑃(π‘₯)=𝑃(Μ‚π‘₯)=𝑃𝑑(Μ‚πœŒ)=max𝜌β‰₯π‘ƒΜ‚πœŒπ‘‘(𝜌).(19)

Proof. Since Μ‚πœŒβˆˆπΊ, Μ‚πœŒ>0. For each π‘₯∈𝐷 and whenever 𝜌β‰₯Μ‚πœŒ we have π‘ƒπœŒ(π‘₯)β‰₯𝑃(π‘₯)+2ξ€Ίπ‘₯βˆ—ξ€»π‘₯βˆ’1β‰₯infπ·ξ‚ƒπœŒπ‘ƒ(π‘₯)+2ξ€Ίπ‘₯βˆ—ξ€»ξ‚„π‘₯βˆ’1=𝑃𝑑(𝜌).(20) By (17), (18), we have 𝑃(π‘₯)β‰₯max𝜌β‰₯π‘ƒΜ‚πœŒπ‘‘(𝜌)=π‘ƒπ‘‘πœŒ(Μ‚πœŒ)=𝑃(Μ‚π‘₯(Μ‚πœŒ))+2ξ€Ί(Μ‚π‘₯(Μ‚πœŒ))βˆ—ξ€»Μ‚π‘₯(Μ‚πœŒ)βˆ’1=𝑃(Μ‚π‘₯(Μ‚πœŒ)).(21) Thus min𝐷𝑃(π‘₯)=max𝜌β‰₯π‘ƒΜ‚πœŒπ‘‘(𝜌).(22) This concludes the proof of Theorem 1.

To illustrate the canonical dual method, let us present several examples as follows.

Example 2. Let us consider the following one-dimensional concave minimization problem: π‘βˆ—=min𝑃(π‘₯)=βˆ’1π‘₯124βˆ’π‘₯2+π‘₯,s.t.π‘₯2≀1.(23) We have π‘ƒξ…ž(π‘₯)=βˆ’13π‘₯3βˆ’2π‘₯+1,π‘ƒξ…žξ…ž(π‘₯)=βˆ’π‘₯2βˆ’2<0,βˆ€π‘₯2≀1.(24) By choosing πœŒβˆ—=10, we solve the following equation in {π‘₯2<1}: βˆ’13π‘₯3βˆ’2π‘₯+1+10π‘₯=0(25) to get a solution Μƒπ‘₯=βˆ’0.1251. Next we solve the following boundary value problem of the ordinary differential equation: 𝑑π‘₯(𝜌)=π‘‘πœŒπ‘₯(𝜌)π‘₯2ξ€·πœŒ(𝜌)+2βˆ’πœŒ,π‘₯βˆ—ξ€Έ=βˆ’0.1251,πœŒβ‰€10.(26) To find a parameter such that π‘₯2(𝜌)=1,(27) we get Μ‚πœŒ=103,(28) which satisfies π‘ƒξ…žξ…ž(π‘₯)+Μ‚πœŒ=π‘ƒξ…žξ…ž(π‘₯)+103>0,βˆ€π‘₯2≀1.(29) Let π‘₯(10/3) be denoted by Μ‚π‘₯. To find the value of Μ‚π‘₯, we compute the solution of the following algebra equation: βˆ’13π‘₯3βˆ’2π‘₯+1+103π‘₯=0,π‘₯2=1(30) and get Μ‚π‘₯=βˆ’1. It follows from Theorem 1 that Μ‚π‘₯=βˆ’1 is the global minimizer of 𝑃(π‘₯) over [βˆ’1,1].

Example 3. We now consider the nonconvex minimization problem: π‘βˆ—1=min𝑃(π‘₯)=3π‘₯3+2π‘₯,s.t.π‘₯2≀1.(31) By choosing πœŒβˆ—=√72, we solve the following equation in {π‘₯2<1}: π‘₯2√+2+72π‘₯=0(32) to get a solution βˆšΜƒπ‘₯=βˆ’2/(4+32). Next we solve the following boundary value problem of the ordinary differential equation: Μ‡π‘₯=βˆ’π‘₯√2π‘₯+𝑑,𝑑≀π‘₯ξ‚€βˆš72,=72βˆ’2√4+32.(33) To find a parameter such that π‘₯2(𝜌)=1,(34) we get Μ‚πœŒ=3,(35) which satisfies π‘ƒξ…žξ…ž(π‘₯)+Μ‚πœŒ=π‘ƒξ…žξ…ž(π‘₯)+3=2π‘₯+3>0,βˆ€π‘₯2≀1.(36) Let π‘₯(3) be denoted by Μ‚π‘₯. To find the value of Μ‚π‘₯, we compute the solution of the following algebra equation: π‘₯2+2+3π‘₯=0,π‘₯2=1(37) and get Μ‚π‘₯=βˆ’1. It follows from Theorem 1 that Μ‚π‘₯=βˆ’1 is the global minimizer of 𝑃(π‘₯) over [βˆ’1,1].

Example 4. Given a symmetric matrix πΊβˆˆπ‘…π‘šΓ—π‘š and a nonzero vector π‘“βˆˆπ‘…π‘š, let 𝑃(π‘₯)=(1/2)π‘₯βˆ—πΊπ‘₯βˆ’π‘“βˆ—π‘₯ be a nonconvex quadratic function. Consider the following global optimization problem over a sphere: 1min𝑃(π‘₯)∢=2π‘₯βˆ—πΊπ‘₯βˆ’π‘“βˆ—π‘₯,s.t.π‘₯βˆ—π‘₯≀1.(38) Suppose that 𝐺 has π‘β‰€π‘š distinct eigenvalues π‘Ž1<π‘Ž2<β‹―<π‘Žπ‘. Since 𝑃(π‘₯)=(1/2)π‘₯βˆ—πΊπ‘₯βˆ’π‘“βˆ—π‘₯ is nonconvex, π‘Ž1<0. Let us choose a large πœŒβˆ—>βˆ’π‘Ž1 such that β€–β€–ξ€·0<𝐺+πœŒβˆ—πΌξ€Έβˆ’1𝑓‖‖<1.(39) By solving the boundary value problem of ordinary differential equation 𝑑π‘₯π‘‘πœŒ=βˆ’(𝐺+𝜌𝐼)βˆ’1ξ€·πœŒπ‘₯,βˆ—ξ€Έ=𝐺+πœŒβˆ—πΌξ€Έβˆ’1𝑓,πœŒβ‰€πœŒβˆ—,(40) we get the unique solution π‘₯(𝜌)=(𝐺+𝜌𝐼)βˆ’1𝑓,πœŒβ‰€πœŒβˆ—.(41) Since 𝐺 is symmetric, there exists an orthogonal matrix 𝑅 such that π‘…πΊπ‘…βˆ—=𝐷∢=(π‘Žπ‘–π›Ώπ‘–π‘—) (a diagonal matrix) and correspondingly 𝑅𝑓=π‘”βˆΆ=(𝑔𝑖) (a vector). By (41), we have π‘₯βˆ—(𝜌)π‘₯(𝜌)=π‘“βˆ—(𝐺+𝜌𝐼)βˆ’2𝑓=𝑝𝑖=1𝑔2π‘–ξ€·π‘Žπ‘–ξ€Έ+𝜌2.(42) Since π‘“βˆ—(𝐺+πœŒβˆ—πΌ)βˆ’2𝑓<1 and lim𝜌>βˆ’π‘Ž1,πœŒβ†’βˆ’π‘Ž1𝑝𝑖=1𝑔2π‘–ξ€·π‘Žπ‘–ξ€Έ+𝜌2=+∞,(43) there exists Μ‚πœŒβˆˆ(βˆ’π‘Ž1,πœŒβˆ—) uniquely such that π‘₯βˆ—(Μ‚πœŒ)π‘₯(Μ‚πœŒ)=π‘“βˆ—(𝐺+Μ‚πœŒπΌ)βˆ’2𝑓=𝑝𝑖=1𝑔2π‘–ξ€·π‘Žπ‘–ξ€Έ+Μ‚πœŒ2=1.(44) By Theorem 1, we see that π‘₯(Μ‚πœŒ)=(𝐺+Μ‚πœŒπΌ)βˆ’1𝑓 is a global minimizer of the problem.

3. Find an Analytic Solution to the OptimalControl Problem

In this section, we consider 𝐴(𝑑),𝐡(𝑑) in problem (1)-(2) to be constant matrices, 𝐹(π‘₯)=π‘βˆ—π‘₯ and


where π‘βˆˆπ‘…π‘›Γ—1,π‘βˆˆπ‘…π‘šΓ—1, and 𝐺(βˆˆπ‘…π‘šΓ—π‘š) is a symmetric matrix. Suppose that 𝐺 has π‘β‰€π‘š distinct eigenvalues π‘Ž1<π‘Ž2<β‹―<π‘Žπ‘ and π‘Ž1<0. Moreover, we need the following basic assumption:


We consider the following optimal control problem:


To solve the above problem, we define the function πœ™(𝑑,π‘₯)=πœ“βˆ—(𝑑)π‘₯, where πœ“(𝑑) is the solution to the following Cauchy boundary value problem of the ordinary differential equation:

Μ‡πœ“(𝑑)=βˆ’π΄βˆ—πœ“(𝑑)+𝑐,(48)πœ“(𝑇)=0.(49) By comparing (48)-(49) with (6) in terms of this special problem (46)-(47), we see that


Noting that πœ“(𝑇)=0 and π‘₯(0)=π‘₯0, we have

ξ€œπ½(𝑒)=𝑇0ξ‚ƒπ‘βˆ—1π‘₯+2π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’ξ‚„=ξ€œπ‘‘π‘‘π‘‡0ξ‚ƒξ€·Μ‡πœ“(𝑑)+π΄βˆ—ξ€Έπœ“(𝑑)βˆ—1π‘₯+2π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’ξ‚„=ξ€œπ‘‘π‘‘π‘‡0ξ‚ƒΜ‡πœ“βˆ—(𝑑)π‘₯+πœ“(𝑑)βˆ—1𝐴π‘₯+2π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’ξ‚„=ξ€œπ‘‘π‘‘π‘‡0ξ‚ƒΜ‡πœ“βˆ—(𝑑)π‘₯+πœ“(𝑑)βˆ—(𝐴π‘₯+𝐡𝑒)βˆ’πœ“(𝑑)βˆ—1𝐡𝑒+2π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’ξ‚„=ξ€œπ‘‘π‘‘π‘‡0ξ‚ƒΜ‡πœ“βˆ—(𝑑)π‘₯(𝑑)+πœ“(𝑑)βˆ—Μ‡π‘₯(𝑑)βˆ’πœ“(𝑑)βˆ—1𝐡𝑒+2π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’ξ‚„=ξ€œπ‘‘π‘‘π‘‡0ξ‚ƒΜ‡πœ™(𝑑,π‘₯(𝑑))βˆ’πœ“(𝑑)βˆ—1𝐡𝑒+2π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’ξ‚„+ξ€œπ‘‘π‘‘=πœ™(𝑇,π‘₯(𝑇))βˆ’πœ™(0,π‘₯(0))𝑇012π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’βˆ’πœ“(𝑑)βˆ—ξ‚„ξ€œπ΅π‘’π‘‘π‘‘=βˆ’πœ™(0,π‘₯(0))+𝑇012π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’βˆ’πœ“(𝑑)βˆ—ξ‚„π΅π‘’π‘‘π‘‘.(51) Thus,


Consequently, we deduce that, for almost every 𝑑 in [0,𝑇], the optimal control is

̂𝑒(𝑑)=argminπ‘’βˆ—π‘’β‰€112π‘’βˆ—πΊπ‘’βˆ’π‘βˆ—π‘’βˆ’πœ“(𝑑)βˆ—ξ‚„π΅π‘’.(53) By the relation between πœ“(𝑑) and the costate in (50), for given π‘‘βˆˆ[0,𝑇], we need to solve the following nonconvex optimization:

1min2π‘’βˆ—ξ€·πΊπ‘’βˆ’π‘βˆ’π΅βˆ—ξ€Έπœ†(𝑑)βˆ—π‘’,s.t.π‘’βˆ—π‘’β‰€1.(54) It follows from the basic assumption (βˆ—) that π‘βˆ’π΅π‘‡πœ†(𝑑)β‰ 0 for each π‘‘βˆˆ[0,𝑇]. By Example 4 and (53), for almost every 𝑑 in [0,𝑇], we have

̂𝑒(𝑑)=𝐺+πœŒπ‘‘πΌξ€Έβˆ’1ξ€Ίπ‘βˆ’π΅βˆ—ξ€»πœ†(𝑑),(55) with the dual variable πœŒπ‘‘>βˆ’π‘Ž1 satisfying

ξ€·π‘βˆ’π΅βˆ—ξ€Έπœ†(𝑑)βˆ—ξ€·πΊ+πœŒπ‘‘πΌξ€Έβˆ’2ξ€·π‘βˆ’π΅βˆ—ξ€Έπœ†(𝑑)=1.(56) We define the function 𝜌(πœ†) with respect to πœ† by the following equation:

ξ€·π‘βˆ’π΅βˆ—πœ†ξ€Έβˆ—(𝐺+𝜌(πœ†)𝐼)βˆ’2ξ€·π‘βˆ’π΅βˆ—πœ†ξ€Έ=1,𝜌(πœ†)>βˆ’π‘Ž1(57) and obtain an analytic solution to the optimal control problem via a costate expression


On the other hand, by the solution of the Cauchy boundary value problem of the ordinary differential equation (48)-(49), we have

πœ†(𝑑)=βˆ’πœ“(𝑑)=π‘’π΄βˆ—π‘‡ξ€œ0π‘‡βˆ’π‘‘π‘’βˆ’π΄βˆ—π‘‘π‘’βˆ’π΄βˆ—π‘ π‘‘π‘ π‘=π‘’π΄βˆ—(π‘‡βˆ’π‘‘)ξ‚Έξ€œ0π‘‡βˆ’π‘‘π‘’βˆ’π΄βˆ—π‘ ξ‚Ήπ‘‘π‘ π‘.(59)

Example 5. Consider the following optimal control problem: ξ€œ(𝒫)min101π‘₯βˆ’2𝑒2ξ‚„π‘₯[],𝑑𝑑s.t.Μ‡π‘₯=π‘₯+𝑒,(0)=0,π‘‘βˆˆ0,1|𝑒|≀1.(60) This is a simple case of (46),(47). We have 𝐺=βˆ’1,𝑐=1,𝑏=0,𝐴=1,𝐡=1,𝑇=1. By (59), we have πœ†(𝑑)=𝑒1βˆ’π‘‘ξ€œ01βˆ’π‘‘π‘’βˆ’π‘ π‘‘π‘ =𝑒1βˆ’π‘‘βˆ’1β‰ 0(𝑑≠1).(61) To find an analytic solution of the optimal control problem, we solve the equation (πœŒβˆ’1)βˆ’2πœ†2(𝑑)=1,𝜌>1(62) to get ||||ξ€Ίπ‘’πœŒ=1+πœ†(𝑑)=1+1βˆ’π‘‘ξ€»βˆ’1.(63) By (58), we obtain an analytic solution of the optimal control problem which can be expressed as ̂𝑒(𝑑)=(πœŒβˆ’1)βˆ’1[]=ξ€·π‘’βˆ’πœ†(𝑑)1βˆ’π‘‘ξ€Έβˆ’1βˆ’1ξ€Ί1βˆ’π‘’1βˆ’π‘‘ξ€»=βˆ’1,(𝑑≠1).(64)

4. Concluding Remarks

In this paper, a new approach to optimal control problems has been investigated using the canonical dual method. Some nonlinear and nonconvex problems can be solved by global optimizations, and therefore, the differential flow defined by the KKT equation (see (11)) can produce an analytic solution of the optimal control problem. Meanwhile, by means of the canonical dual function, an optimality condition is proved (see Theorem 1). The global optimization problem is solved by a backward differential equation with an equality condition (see (12), (18)). More research needs to be done for the development of applicable canonical dual theory.


This research was partly supported by the National Science Foundation of China under grants No. 10671145.