On the Continuous-Time Complementarity Problem

ABSTRACT

This work deals with solving continuous-time nonlinear complementarity problems defined on two types of nonempty closed convex cones: a polyhedral cone (positive octant) and a second-order cone. Theoretical results that establish a relationship between such problems and the variational inequalities problem are presented. We show that global minimizers of an unconstrained continuous-time programming problem are solutions to the continuous-time nonlinear complementarity problem. Moreover, a relation is set up so that a stationary point of an unconstrained continuous-time programming problem, in which the objective function involves the Fischer-Burmeister function, is a solution for the continuous-time complementarity problem. To guarantee the validity of the K.K.T. conditions for some auxiliary continuous-time problems which appear during the theoretical development, we use the linear independence constraint qualification. These constraint qualification are posed in the continuous-time context and appeared in the literature recently. In order to exemplify the developed theory, some simple examples are presented throughout the text.

Keywords:
continuous-time complementarity problem; variational inequalities problem; continuous-time programming problems

1 INTRODUCTION

Complementarity problems were firstly proposed as the question of finding an n−vector x which satisfies the system of inequalities

$x\ge 0,Mx+b\ge 0\text{and}{x}^{\top }\left(Mx+b\right)=0,$ (1.1)

where M is an n ×n matrix, b is an n−vector of real numbers and “⊤” denotes the transposition of vectors and matrices. Such problems are elegant generalizations of certain linear programming, quadratic programming, and game theory problems.

The importance of problem (1.1) lies in the fact that its form includes several problems by appropriate choices of the vector b and the matrix M. As examples of applications, we can cite the problem of the existence of solutions to linear programs (Cottle ⁵, Dorn ⁷) that can be reduced to a problem in the formulation (1.1), the equilibrium point problem of bimatrix games (Lemke ¹⁴) and the unloading problem for plane curves (Du Val ⁸). For other examples of applications, see Isac ¹¹.

The formulation (1.1) was expanded to include a broader class of problems such as nonlinear programming and was rewritten as the problem of finding an n−vector x which satisfies the system of inequalities

$x\ge 0,f\left(x\right)\ge 0\text{and}{x}^{\top }f\left(x\right)=0,$ (1.2)

where f is a mapping of ℝⁿ into itself. Among other authors who studied the formulation (1.2), Cottle ⁵ gave sufficient conditions for the existence of x, and Karamardian ¹² established sufficient conditions for the existence of an unique solution.

Bodo and Hanson ³ extended the results of Karamardian ¹² to the case where x is an essentially bounded measurable function which maps some finite interval into ℝⁿ . Sufficient conditions for the existence of a unique solution are given and applications to continuous linear and nonlinear programming are presented. Such a problem will be called here as Continuous-Time Complementarity Problem (CCP).

In ⁴, Giannessi first introduced Variational Inequalities Problem (VIP) in a finite-dimensional vector space. Since then, the (VIP) has been extensively studied in a general setting by many authors (see, for example, ^{9), (13), (21} ). Zalmai ²² generalized the VIP for the continuous-time case, presented a generalized sufficiency criteria in continuous-time programming under the concept of invexity and used it to study the existence of a solution for the VIP. Differently of Zalmai, in this paper we use the VIP in order to find solutions to the continuous-time complementarity problem, without assuming convexity.

The main contributions of this work are:

• ⋆ To show that CCP and VIP have the same solutions sets;

• ⋆ Reformulation of the CCP as an Auxiliary Optimization Problem (AOP) and apply optimality conditions to solve it;

• ⋆ Analysis of the conditions under which an AOP global optimal solution is a CCP solution;

• ⋆ Analysis of the conditions under which an AOP stationary point is a CCP solution;

• ⋆ For the analysis mentioned above, we considered two types of cones: polyhedral cone $\left({\mathrm{ℝ}}_{+}^n\right)$ and second-order cone.

We note that Bodo and Hanson defined the variables of the CCP problem in L _∞([0, T]; ℝⁿ ) space (Banach space of all Lebesgue-measurable essentially bounded n-dimensional vector functions), while Zalmai worked in W ([0, T]; ℝⁿ ) space (Hilbert space of all absolutely continuous n-dimensional vector functions). For standardization purposes, we will work in the L _∞([0, T]; ℝⁿ ) space.

The paper is organized as follows. Some preliminaries and results from literature related to continuous-time programming problems with equality and inequality constraints are given in Section 2. In Section 3, we define the continuous-time complementarity problem and the variational-type inequalities problem, and we establish the relationship between these problems.

In Section 4, we consider the particular case in which the cone is equal to ${\mathrm{ℝ}}_{+}^n$ and, using the Fischer-Burmeister function ¹⁰, we derive an unconstrained equivalent problem in the sense that a stationary point of the unconstrained equivalent problem is a solution of the continuoustime complementarity problem. Second-order cones are considered in Section 5. An auxiliary problem with inequality constraints is formulated. Assumptions about the constraint set are made in order to guarantee optimality conditions at optimal solutions. Examples are presented. It is worth mentioning that the approach we followed in Sections 4 and 5 is original even in finite dimension: to the best of our knowledge, there is no work in the literature where results from Sections 4 and 5 were obtained through the approach employed here. Final comments are given in Section 6.

2 PRELIMINARIES

We denote by ∥ · ∥ the Euclidean norm and given f: Rⁿ ×[0, T ] → ℝ, we denote by ∇f(x, t) the gradient of f(·, t) at x. Consider the continuous-time programming problem with equality and inequality constraints

$$\begin{gathered} \text{maximize}P\left(x\right)={\int }_0^T\varphi \left(x\left(t\right),t\right)dt \\ \text{subject to}{h}_j\left(x\left(t\right),t\right)=0\text{a.e. in}\left[0,T\right],j\in J=\left\{1,\dots ,p\right\}, \\ {g}_i\left(x\left(t\right),t\right)\ge 0\text{a.e. in}\left[0,T\right],i\in I=\left\{1,\dots ,m\right\}, \\ x\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}^n\right), \end{gathered}$$ (2.1)

where ϕ: ℝⁿ × [0, T] → ℝ, h _j: ℝⁿ × [0, T] → ℝ, j ∈ J, and g _i: ℝⁿ × [0, T] → ℝ, i ∈ I, are given functions. Here for each t ∈ [0, T], x(t) ∈ ℝⁿ and all integrals are given in the Lebesgue sense. L _∞([0, T ]; ℝⁿ ) denotes the Banach space of all Lebesgue-measurable essentially-bounded n-dimensional vector functions defined on the compact interval [0, T ] ⊂ ℝ, with the norm ∥ · ∥_∞ defined by

${||x||}_{\infty }=\underset{1\le i\le n}{max}\underset{t\in \left[0,T\right]}{essup}\left|x_i\left(t\right)\right|.$

The set of feasible solutions for (2.1) is denoted by

${\Omega }_0=\left\{x\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}^n\right)|h_j\left(x\left(t\right),t\right)=0,g_i\left(x\left(t\right),t\right)\ge 0\text{a.e. in}\left[0,T\right],i\in I,j\in J\right\}.$

Definition 2.1.x^∗ ∈ Ω₀ is said to be a local optimal solution for (2.1) if there exists ε > 0 such that P(x) ≤ P(x ^∗) for all x ∈ Ω₀ with x(t) ∈ x ^∗(t) + εB a.e. t ∈ [0, T], where B denotes the open unit ball with center at the origin in ℝⁿ . x ^∗ ∈ Ω₀ is said to be a global optimal solution if P(x) ≤ P(x ^∗) for all x ∈ Ω₀ .

Let ε > 0 and x ^∗ a local optimal solution for (2.1). We will assume that

• (H1) the functions ϕ(·, t), h _j (·, t), j ∈ J, and g _i (·, t), i ∈ I, are twice continuously differentiable on $x*\left(t\right)+\epsilon \overline{B}$ a.e. in [0, T], where $\overline{B}$ denotes the closed unit ball with center at the origin in ℝⁿ ;

• (H2) the functions ϕ(x, ·), h _j (x, ·), j ∈ J, and g _i (x, ·), i ∈ I, are Lebesgue measurable for each x, ϕ(x(·), ·), h _j (·, ·), j ∈ J, and g _i (x(·), ·), i ∈ I, are essentially bounded in [0, T] for all x ∈ L _∞([0, T]; ℝⁿ );

• (H3) there exist K _ϕ > 0 and K ₀ > 0 such that, for a.e. t ∈ [0, T], we have that

$||\nabla \varphi \left(x*\left(t\right),t\right)||\le K_{\varphi }\text{and}||\nabla \left[h,g\right]\left(x*,\left(t\right),t\right)]||\le K_0.$

Karush-Kuhn-Tucker type optimality conditions can be obtained under additional assumptions, for example, the linear independence constraint qualification (LICQ). The hypothesis below is necessary for the application of the Uniform Implicit Function Theorem (see ¹⁵ and ⁶).

• (H4) There exists an increasing function $\overline{\theta }:\left(0,\infty \right)\to \left(0,\infty \right),\overline{\theta }\left(s\right)\downarrow 0$ when s ↓ 0, such that, for all $\tilde{x},x\in x*\left(t\right)+\epsilon \overline{B}$,

$||\nabla \left[h,g\right]\left(\tilde{x},t\right)-\nabla \left[h,g\right]\left(x,t\right)||\le \overline{\theta }\left(||\tilde{x}-x||\right)\text{a.e.}t\in \left[0,T\right].$

The following definition refers to the continuous-time case of the linear independence constraint qualification. Note that we define LICQ requiring a “uniform invertibility” for almost every t ∈ [0, T]. This definition is different from the definition used in the finite dimensional case. For more details, see ¹⁵.

Definition 2.2.We say that the constraint qualification (LICQ) is satisfied at x^∗ ∈ Ω₀ if there exists K > 0 such that

$det\left\{{\rm Y}\left(t\right){\rm Y}{\left(t\right)}^{\top }\right\}\ge K\text{a.e.}t\in \left[0,T\right],$

where

${\rm Y}\left(t\right)=\left(\begin{array}{cc}\nabla h\left(x*\left(t\right),t\right)& 0\\ \nabla g\left(x*\left(t\right),t\right)& diag{\left\{-2w_j^{*}\left(t\right)\right\}}_{j\in I}\end{array}\right)$

and$w_j^{*}\left(t\right)=\sqrt{g_j\left(x*\left(t\right),t\right)}$a.e. t ∈ [0, T], j ∈ I.

Proposition 2.1.(Theorem 4.2 in do Monte and de Oliveira¹⁵) Assume that (H1)-(H4) and (LICQ) hold at x^∗ ∈ Ω₀ and x ^∗ is a local optimal solution for (2.1). Then there exists$\left(\tilde{u},\tilde{v}\right)\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}_p\times {\mathrm{ℝ}}_{+}^m\right)$such that, for almost every$t\in \left[0,T\right],{\tilde{v}}_i\left(t\right)g_i\left(x*\left(t\right),t\right)=0$, i ∈ I, and

$\nabla \varphi \left(x*\left(t\right),t\right)+\sum _{j=1}^p{\tilde{u}}_j\left(t\right)\nabla h_j\left(x*\left(t\right),t\right)+\sum _{i=1}^m{\tilde{v}}_i\left(t\right)\nabla g_i\left(x*\left(t\right),t\right)=0.$

We will use the Mangasarian-Fromovitz constraint qualification presented in (⁶). For this purpose, let b > 0 real number. We will denote by I _b (t) the index set of b−active constraints at x ^∗ ∈ Ω₀, that is, I _b (t) = {i ∈ I | 0 ≤ g _i (x ^∗(t), t) ≤ b} for almost every t ∈ [0, T]. For all i ∈ I, let us define the function ${\delta }_i^b:\left[0,T\right]\to \mathrm{ℝ}$ as

${\delta }_i^b\left(t\right)=\left\{\begin{array}{l}1,i\in I_b\left(t\right),\\ 0,\text{otherwise}.\end{array}\right.$

The next definition refers to continuous-time case of the Mangasarian-Fromovitz constraint qualification. Note that we define MFCQ requiring a “uniform positivity” for almost every t ∈ [0, T]. This definition is different from the definition used in the finite dimensional case. For more details, see ⁶.

Definition 2.3.We say that the constraint qualification (MFCQ) is satisfied at x^∗ ∈ Ω₀ if

• (i) There exist$\hat{\gamma }\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}^n\right)$and$\hat{b}>0$such that, for almost every t ∈ [0, T],

$\nabla h{\left(x*\left(t\right),t\right)}^{\top }\hat{\gamma }\left(t\right)=0\text{and}\nabla g_j{\left(x*\left(t\right),t\right)}^{\top }\hat{\gamma }\left(t\right)\ge \beta ,j\in I_{\hat{b}}\left(t\right),$

for some β > 0;

• (ii) There exists K > 0 such that, for almost every t ∈ [0, T],

$det\left\{\nabla h\left(x*\left(t\right),t\right)\nabla h{\left(x*\left(t\right),t\right)}^{\top }\right\}\ge K.$

For x ∈ ℝⁿ , let

$\left\{\begin{array}{l}{f}_0\left(x,t\right):=-f_0^T\nabla \varphi {\left(x*\left(t\right),t\right)}^{\top }\overline{H}\left(t\right)xdt<0,\\ f_j\left(x,t\right):=-g_j\left(x*\left(t\right),t\right)-{\delta }_j^{\hat{b}}\left(t\right)\nabla g_j{\left(x*\left(t\right),t\right)}^{\top }\overline{H}\left(t\right)\gamma \le 0,j\in J,\end{array}\right.$ (2.2)

where $\hat{b}$ is given in (H4) and

$\overline{H}\left(t\right)=I_n-\nabla h{\left(x*\left(t\right),t\right)}^{\top }{\left[\nabla h\left(x*\left(t\right),t\right)\nabla h{\left(x*\left(t\right),t\right)}^{\top }\right]}^{-1}\nabla h\left(x*\left(t\right),t\right),\text{a.e. in}\left[0,T\right].$

Above, I _n denotes the identity matrix of order n.

$$\begin{gathered} F_0\left(x,t\right)=-{\int }_0^T\nabla \varphi {\left(x*\left(t\right),t\right)}^{\top }xdt<0, \\ {F}_i\left(x,t\right)=-g_i\left(x*\left(t\right),t\right)-{\delta }_i^b\left(t\right)\nabla g_i{\left(x*\left(t\right),t\right)}^{\top }x\le 0,i\in I, \end{gathered}$$ (2.3)

be a system corresponding to Problem (2.1) and

$\mathcal{I}\left(x,t\right)=\left\{j|F_j\left(x,t\right)=max\left\{F_0\left(x,t\right),F_1\left(x,t\right),\dots ,F_m\left(x,t\right)\right\}\right\},t\in \left[0,T\right],x\in {\mathrm{ℝ}}^n.$

Definition 2.4. (Arutyunov et al.²) System (2.3) is said to be regular when there exist a function$\overline{x}\left(·\right)\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}^n\right)$, real numbers R ≥ 0 and α > 0 such that for a. e. t ∈ [0, T] and for all x ∈ ℝⁿ with$||x-\overline{x}\left(t\right)||\ge R$, there exists a vector e = e(x, t) ∈ ℝⁿ with ∥e∥ = 1, satisfying

where ∂_xF_jdenotes the partial subdifferential of F_jat (x, t) with respect to x in the sense of convex analysis. For more information on convex analysis, the reader is referred to Rockafellar¹⁷.

Proposition 2.2.(Theorem 3.8 in do Monte and de Oliveira⁶) Assume that (H1)-(H4) hold and (MFCQ) is satisfied at x^∗ ∈ Ω₀ which is a local optimal solution for (2.1). If the system (2.3) is regular, then there exists$\left(\tilde{u},\tilde{v}\right)\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}_p\times {\mathrm{ℝ}}_{+}^m\right)$such that, for almost every t ∈ [0, T],${\tilde{v}}_i\left(t\right)\ge 0,i\in I$, and

$\nabla \varphi \left(x*\left(t\right),t\right)+\sum _{j=1}^p{\tilde{u}}_j\left(t\right)\nabla h_j\left(x*\left(t\right),t\right)+\sum _{i=1}^m{\tilde{v}}_i\left(t\right)\nabla g_i\left(x*\left(t\right),t\right)=0.$

3 VARIATIONAL-TYPE INEQUALITIES PROBLEM

The continuous-time complementarity problem CTCP(f , K) is posed as to find x in L _∞([0, T]; ℝⁿ ) such that, for a.e. t ∈ [0, T], we have that

$x\left(t\right)\in K,f\left(x\left(t\right),t\right)\in K^{\circ }\text{and}x{\left(t\right)}^{\top }f\left(x\left(t\right),t\right)=0,$

where K ⊂ ℝⁿ is a nonempty closed convex cone with vertex at 0, namely, if x ∈ K, αx ∈ K for all α > 0. The polar cone K ^◦ of K is given by

$K^{\circ }=\left\{y\in {\mathrm{ℝ}}^n|y^{\top }x\ge 0,\forall x\in K\right\}.$

f: ℝⁿ ×[0, T] → ℝⁿ is a nonlinear function with f _j (·, t), j = 1, . . . , n, twice continuously differentiable throughout [0, T] and f _j (x, ·), ∇f _j (x, ·) and ∇² f _j (x, ·) measurable and essentially bounded for all x ∈ ℝⁿ , j = 1, . . . , n. Let

$\Omega =\left\{x\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}^n\right)|x\left(t\right)\in K\text{a.e.}t\in \left[0,T\right]\right\}.$

Definition 3.5.The Variational-type Inequalities Problem VIP(f , Ω) consists in finding x ^∗ ∈ Ω such that

${\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }\left(x\left(t\right)-x*\left(t\right)\right)dt\ge 0,\mathit{\text{for all}}x\in \Omega .$

Lemma 3.1.x^∗is a solution of the CTCP( f , K) if, and only if, x ^∗ is a solution of VIP( f , Ω).

Proof. If x ^∗ ∈ L _∞([0, T ], ℝⁿ ) is a solution of CTCP(f , K) then f(x ^∗(t), t) ∈ K ^◦ a.e. t ∈ [0, T], and we can conclude that

$x{\left(t\right)}^{\top }f\left(x*\left(t\right),t\right)\ge 0,\text{a.e.}t\in \left[0,T\right],\text{for all}x\in \Omega .$ (3.1)

Using (3.1) and the hypothesis, for a.e. t ∈ [0, T] and for all x ∈ Ω, we have that

$$\begin{gathered} f{\left(x*\left(t\right),t\right)}^{\top }\left(x\left(t\right)-x*\left(t\right)\right)=f{\left(x*\left(t\right),t\right)}^{\top }x\left(t\right)-f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right) \\ =f{\left(x*\left(t\right),t\right)}^{\top }x\left(t\right) \\ \ge 0, \end{gathered}$$

resulting that x ^∗ ∈ Ω is a solution of VIP(f , Ω). Conversely, if x ^∗ ∈ Ω is a solution of VIP(f , Ω), then x ^∗(t) ∈ K a.e. t ∈ [0, T]. The inequality in Definition 3.5 holds for all x ∈ Ω. Particularly, for x = 0 ∈ Ω and x = 2x ^∗ ∈ Ω we have that

${\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)dt\le 0\text{and}{\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)dt\ge 0,$

respectively, resulting in

${\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)dt=0.$ (3.2)

Statement: For all x ∈ Ω, f(x ^∗(t), t)^⊤ x(t) ≥ 0 a.e. t ∈ [0, T]. Indeed, suppose that there exists $\tilde{x}\in \Omega$ and a subset D ⊂ [0, T], with positive measure, such that $f{\left(x*\left(t\right),t\right)}^{\top }\tilde{x}\left(t\right)<0$ for all t ∈ D. Define $\overline{x}\in \Omega$ given by

$\overline{x}\left(t\right)=\left\{\begin{array}{l}\tilde{x}\left(t\right)\text{if}t\in D,\\ 0\text{if}t\in \left[0,T\right]\backslash D.\end{array}\right.$

Then, using (3.2) and the definition of $\overline{x}$, we have that

$$\begin{gathered} {\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }\left(\overline{x}\left(t\right)-x*\left(t\right)\right)dt \\ ={\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }\overline{x}\left(t\right)dt-{\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)dt \\ ={\int }_{D}f{\left(x*\left(t\right),t\right)}^{\top }\tilde{x}\left(t\right)dt \\ <0, \end{gathered}$$

contradicting the fact that x ^∗ is a solution of VIP(f , Ω). Therefore, by the above statement, f(x ^∗(t), t) ∈ K ^◦ a.e. t ∈ [0, T]. Besides that, as f(x ^∗(t), t)^⊤ x ^∗(t) ≥ 0 a.e. t ∈ [0, T], it results from (3.2) that f(x ^∗(t), t)^⊤ x ^∗(t) = 0 a.e. t ∈ [0, T], concluding the proof. □

Now, consider the auxiliary continuous-time problem

$$\begin{gathered} \text{maximize}P\left(x\right)=-{\int }_0^{T}f{\left(x\left(t\right),t\right)}^{\top }x\left(t\right)dt \\ \text{subject to}x\left(t\right)\in K\text{a.e.}t\in \left[0,T\right], \\ f\left(x\left(t\right),t\right)\in K^{\circ }\text{a.e.}t\in \left[0,T\right], \\ x\in L_{\infty }\left(\left[0,T\right],{\mathrm{ℝ}}^n\right). \end{gathered}$$ (3.3)

Proposition 3.3.x^∗ ∈ Ω is a solution of Problem CTCP(f , K) if, and only if, x ^∗ is a global maximum point of Problem (3.3) with P(x ^∗) = 0.

Proof. If x ^∗ is a solution of CTCP(f , K), then x ^∗(t) ∈ K, f(x ^∗(t), t) ∈ K ^◦ a.e. t ∈ [0, T] and

$P\left(x*\right)=-{\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)dt=0.$

Let x ∈ L _∞([0, T], ℝⁿ ) a feasible point of (3.3). Then, x(t) ∈ K a.e. t ∈ [0, T] and f(x(t), t) ∈ K ^◦ a.e. t ∈ [0, T], imply that

$P\left(x\right)=-{\int }_0^{T}f{\left(x\left(t\right),t\right)}^{\top }x\left(t\right)dt\le 0.$

Therefore, P(x) ≤ P(x ^∗) for all x ∈ Ω and x ^∗ is a global maximum point of (3.3) with P(x ^∗) = 0. Conversely, if x ^∗ is a global maximum point of (3.3), then x ^∗(t) ∈ K a.e. t ∈ [0, T], f(x ^∗(t), t) ∈ K ^◦ a.e. t ∈ [0, T] and P(x ^∗) = 0. If x ∈ Ω, from definition of polar cone K ^◦ we obtain

$-{\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x\left(t\right)dt\le 0=P\left(x*\right).$

Therefore,

$$\begin{gathered} -{\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x\left(t\right)dt\le -{\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)dt \\ \iff {\int }_0^{T}f{\left(x*\left(t\right),t\right)}^{\top }\left(x\left(t\right)-x*\left(t\right)\right)dt\ge 0, \end{gathered}$$

that is, x ^∗ is a solution of VIP(f , Ω) and, by Lemma 3.1, x ^∗ is a solution of CTCP(f , K). □

Remark 1.If x^∗is a global maximum point of (3.3) and a regularity condition is verified on the constraint set Ω, then optimality conditions which are found in the literature can be applied to solve (3.3). In our case, we will use the Proposition 2.1 and Proposition 2.2. The next example illustrates the Proposition 3.3.

Example 3.1.Consider Problem CT CP(f , K) with x: [0, 1] → ℝ, f(x, t) = [x]² − tx and K = ℝ₊ = K ^◦. Then, (3.3) is given as

$$\begin{gathered} \text{maximize}P\left(x\right)=-{\int }_0^1\left\{{\left[x\left(t\right)\right]}^2-tx\left(t\right)\right\}x\left(t\right)dt \\ \text{subject to}x\left(t\right)\ge 0\text{a.e.}t\in \left[0,1\right], \\ {\left[x\left(t\right)\right]}^2-tx\left(t\right)\ge 0\text{a.e.}t\in \left[0,1\right], \\ x\in L_{\infty }\left(\left[0,T\right],{\mathrm{ℝ}}^n\right). \end{gathered}$$

Note that the two constraints are satisfied when x(t) − t ≥ 0 a.e. t ∈ [0, T], and the problem can be written in the form

$$\begin{gathered} \text{maximize}P\left(x\right)=-{\int }_0^1\left\{{\left[x\left(t\right)\right]}^2-tx\left(t\right)\right\}x\left(t\right)dt \\ \text{subject to}x\left(t\right)-t\ge 0\text{a.e.}t\in \left[0,1\right], \\ x\in L_{\infty }\left(\left[0,1\right],{\mathrm{ℝ}}^n\right). \end{gathered}$$

If x^∗is a global maximum point of (3.4), then the Proposition 2.1 guarantees us that there exists u^∗∈ L_∞ ([0, 1]; ℝ) such that, for a.e. t ∈ [0, 1],

• (i) −3[x ^∗ (t)]² + 2tx ^∗ (t) + u ^∗ (t) = 0 ⇒ u ^∗ (t) = 3[x ^∗ (t)]² − 2tx ^∗ (t),

• (ii)u^∗ (t) ≥ 0 and u ^∗ (t)[x ^∗ (t) − t] = 0,

resulting that x^∗ (t) = t and u ^∗ (t) = t ² for a.e. t ∈ [0, 1], with P(x ^∗ ) = 0. Observe that x ^∗ (t) = t a.e. t ∈ [0, T ] is a candidate solution to the problem (3.4), but it is a solution of CT CP(f , K). Then, by Proposition 3.3, we can guarantee that x ^∗ is a global minimizer of (3.4).

4 THE CASE $K={\mathrm{ℝ}}_{+}^n$

Let us consider K to be the positive octant of ℝⁿ . In this case, K = K ^◦. The Fischer-Burmeister function (see ¹⁰) φ: ℝ² → ℝ is given by

$\phi \left(a,b\right)=\sqrt{a^2+b^2}-a-b.$ (4.1)

This function has the property that φ(a, b) = 0 ⇔ a ≥ 0, b ≥ 0, ab = 0. The partial derivates of φ will be denoted by $\frac{\partial \phi }{\partial a}\left(a,b\right)$ and $\frac{\partial \phi }{\partial b}\left(a,b\right)$. Suppose that Problem (3.3) satisfy the assumptions (H1)-(H4) and (LICQ) at the local optimal solution x ^∗. Then, the Proposition 2.1 guarantees that there exist u ^∗, v ^∗ ∈ L _∞([0, T ]; ℝⁿ ), with u ^∗(t) ≥ 0 and v ^∗(t) ≥ 0 a.e. t ∈ [0, T], such that

$$\begin{gathered} ||f\left(x*\left(t\right),t\right)+\sum _{i=1}^n\nabla f_i\left(x*\left(t\right),t\right)\left[x_i^{*}\left(t\right)-v_i^{*}\left(t\right)\right]-u*\left(t\right)||=0, \\ \phi \left(x_i^{*}\left(t\right),u_i^{*}\left(t\right)\right)=0\text{and}\phi \left(f_i\left(x_i^{*}\left(t\right),t\right),v_i^{*}\left(t\right)\right)=0 \end{gathered}$$

for a.e. t ∈ [0, T] and i = 1, . . . , n. Simplifying the notation, define

$$\begin{gathered} \tilde{\phi }\left(x*\left(t\right),u*\left(t\right)\right)=2\left(\begin{array}{c}\phi \left(x_1^{*}\left(t\right),u_1^{*}\left(t\right)\right)\\ ⋮\\ \phi \left(x_n^{*}\left(t\right),u_n^{*}\left(t\right)\right)\end{array}\right), \\ \tilde{\phi }\left(f\left(x*\left(t\right),t\right)v*\left(t\right)\right)=2\left(\begin{array}{c}\phi \left(f_1\left(x*\left(t\right),t\right),v_1^{*}\left(t\right)\right)\\ ⋮\\ \phi \left(f_n\left(x*\right)\left(t\right),t\right),v_n^{*}\left(t\right)\end{array}\right), \\ {\Phi }_x\left(t\right)=diag{\left(\frac{\partial \phi }{\partial a}\left(x_i^{*}\left(t\right),u_i^{*}\left(t\right)\right)\right)}_{i=1}^n,{\Phi }_u\left(t\right)=diag{\left(\frac{\partial \phi }{\partial b}\left(x_i^{*}\left(t\right),u_i^{*}\left(t\right)\right)\right)}_{i=1}^n, \\ {\Phi }_v\left(t\right)diag{\left(\frac{\partial \phi }{\partial b}\left(f_i\left(x*\left(t\right),t\right),v_i^{*}\left(t\right)\right)\right)}_{i=1}^n \end{gathered}$$

and the n × n matrix Φ_f (t) = (c _ij (t)) where

$c_{ij}\left(t\right)=\frac{\partial \phi }{\partial a}\left(f_j\left(x*\left(t\right),t\right),v_j^{*}\left(t\right)\right)\frac{\partial f_j}{\partial x_i}\left(x*\left(t\right),t\right),i,j=1,2,\dots ,n.$

Also, we write

$\nabla f{\left(x\left(t\right),t\right)}^{\top }=\left(\nabla f_1\left(x\left(t\right),t\right)\nabla f_2\left(x\left(t\right),t\right)\dots \nabla f_n\left(x\left(t\right),t\right)\right)\text{a.e.}t\in \left[0,T\right].$

Consider the following unconstrained continuous-time problem:

$$\begin{gathered} \text{maximize}Q\left(x,u,v\right)=-{\int }_0^{T}F\left(x\left(t\right),u\left(t\right),v\left(t\right),t\right)dt \\ \text{subject to}x,u,v\in L_{\infty }\left(\left[0,T\right],{\mathrm{ℝ}}^n\right), \end{gathered}$$ (4.2)

where F: ℝⁿ × ℝⁿ × ℝⁿ × [0, T] → ℝ is defined by

$F\left(x,u,v,t\right)={||f\left(x,t\right)+\nabla f{\left(x,t\right)}^{\top }\left[x-v\right]-u||}^2+\sigma \left(x,u,v\right),$

with $\sigma \left(x,u,v\right)=\sum _{i=1}^n{\left[\phi \left(x_i,u_i\right)\right]}^2+{\left[\phi \left(f_i\left(x,t\right),v_i\right)\right]}^2\text{a.e.}t\in \left[0,T\right]$.

Theorem 4.1.If$\left(x*,u*,v*\right)\in L_{\infty }\left(\left[0,T\right],{\mathrm{ℝ}}^n\times {\mathrm{ℝ}}_{+}^n\times {\mathrm{ℝ}}_{+}^n\right)$is a global optimal solution for Problem (4.2) with Q(x^∗, u^∗, v^∗ ) = 0 and [x ^∗ (t) − v ^∗ (t)] ∈ Ker(∇f(x(t), t)^⊤) a.e. t ∈ [0, T], then x ^∗ is a solution of Problem CT CP(f, K).

Proof. By definition of F, if Q(x ^∗ , u ^∗ , v ^∗ ) = 0 then F(x ^∗ (t), u ^∗ (t), v ^∗ (t), t) = 0 a.e. t ∈ [0, T]. Then, σ(x ^∗ (t), u ^∗ (t), v ^∗ (t)) = 0 a.e. t ∈ [0, T] implies that

$$\begin{gathered} \sum _{i=1}^n{\left[\phi \left(x_i^{*}\left(t\right),u_i^{*}\left(t\right)\right)\right]}^2=0\text{a.e.}t\in \left[0,T\right] \\ \iff \phi \left(x_i^{*}\left(t\right),u_i^{*}\left(t\right)\right)=0\text{a.e.}t\in \left[0,T\right],i=1,\dots ,n \\ \iff x_i^{*}\left(t\right)\ge 0,u_i^{*}\left(t\right)\ge 0,x_i^{*}\left(t\right)u_i^{*}\left(t\right)=0\text{a.e.}t\in \left[0,T\right],i=1,\dots ,n. \end{gathered}$$

Similarly, we conclude that $f_i\left(x*\left(t\right),t\right)\ge 0,v_i^{*}\left(t\right)\ge 0,f_i\left(x*\left(t\right),t\right)v_i^{*}\left(t\right)=0\text{a.e.}t\in \left[0,T\right],i=1,\dots ,n$. Thus, we have that x ^∗(t) ∈ K and f(x ^∗ (t),t) ∈ K ^◦ a.e. t ∈ [0, T]. Moreover, since

$f\left(x*\left(t\right),t\right)+\nabla f{\left(x*\left(t\right),t\right)}^{\top }\left[x*\left(t\right)-v*\left(t\right)\right]-u*\left(t\right)=0\text{a.e.}t\in \left[0,T\right],$

and using the fact that [x ^∗ (t) − v ^∗ (t)] ∈ Ker(∇f(x ^∗ (t),t)^⊤) a.e. t ∈ [0, T], it results that

$x*{\left(t\right)}^{\top }f\left(x*\left(t\right),t\right)=-x*{\left(t\right)}^{\top }\nabla f{\left(x*\left(t\right),t\right)}^{\top }\left[x*\left(t\right)-v*\left(t\right)\right]+x*{\left(t\right)}^{\top }u*\left(t\right)=0.$

Therefore, x ^∗ is a solution for Problem CT CP(f, K). □

Theorem 4.2.Assume that the Problem (3.3) satisfy the assumptions (H1)-(H4) and (LICQ) at the local optimal solution x^∗. If x^∗∈ L_∞ ([0, T]; ℝⁿ ) is a solution of Problem CT CP(f, K), then there exist u ^∗ , v ^∗ in$L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}_{+}^n\right)$such that (x ^∗ , u ^∗ , v ^∗ ) is a global optimal solution for Problem (4.2) with Q(x ^∗ , u ^∗ , v ^∗ ) = 0.

Proof. If x^∗ is a solution of Problem CT CP(f, K), by Proposition 3.3, x ^∗ is a global maximum point of Problem (3.3). Using the Proposition 2.1, we conclude that F(x ^∗ (t), u ^∗ (t), v ^∗ (t),t) = 0 a.e. t ∈ [0, T], in other words, (x ^∗ , u ^∗ , v ^∗ ) is a global maximum point of (4.2) with Q(x ^∗ , u ^∗ , v ^∗ ) = 0. □

Most of the algorithms used in the resolution of Problem (4.2) guarantee convergence only to stationary points. A global minimum point is very hard to find (see ¹). For this purpose, we will derive conditions for stationary points of (4.2). We say that p = (x, u, v) ∈ L _∞([0, T ], ℝ³ⁿ ) is a stationary point of (4.2) if, and only if, ∇F(p(t), t) = 0 a.e. t ∈ [0, T].

Definition 4.6.The n ×n matrix M(x, t), with elements m _{i j} (x, t), x ∈ ℝⁿ , t ∈ [0, T], i, j = 1, . . . , n, is positive definite at x ^∗ ∈ L _∞([0, T]; ℝⁿ ) if, for all y ∈ L _∞([0, T]; ℝⁿ ) and a.e. t ∈ [0, T],

$y{\left(t\right)}^{\top }M\left(x*\left(t\right),t\right)y\left(t\right)>0\text{whenever}y\left(t\right)\ne 0.$

The next theorem relates stationary points of (4.2) to solutions of Problem CTCP(f, K).

Theorem 4.3.Let (x ^∗, u ^∗, v ^∗) is a stationary point of (4.2). Regarding the Problem (4.2), assume that

• (i) D(t) is definite positive a.e. t ∈ [0, T], where

$D\left(t\right)=2\nabla f\left(x*\left(t\right),t\right)+\sum _{i=1}^n\left[x_i^{*}\left(t\right)-v_i^{*}\left(t\right)\right]{\nabla }^2{f}_i\left(x*\left(t\right),t\right)e.e.t\in \left[0,T\right];$

• (ii) $\left[x*\left(t\right)-v*\left(t\right)\right]\in Ker\left(\nabla f{\left(x*\left(t\right),t\right)}^{\top }\right)$;

• (iv) $\left[{\Phi }_u\left(t\right)\tilde{\phi }\left(x*\left(t\right),u*\left(t\right)\right)\right]\in K\text{and}\left[{\Phi }_f\left(t\right)\tilde{\phi }\left(f\left(x*\left(t\right),t\right),v*\left(t\right)\right)\right]\in K^{\circ }\text{a.e.}t\in \left[0,T\right]$.

Then x^∗is a solution of CT CP(f, K).

Proof. Note that, if w(t) = f(x(t), t) +∇f(x(t), t)^⊤[x(t) − v(t)] − u(t) a.e. t ∈ [0, T], we have that

$$\begin{gathered} {\nabla }_x\left\{{||f\left(x\left(t\right),t\right)+\nabla f{\left(x\left(t\right),t\right)}^{\top }\left[x\left(t\right)-v\left(t\right)\right]-u\left(t\right)||}^2\right\} \\ =2{\left\{\nabla f\left(x\left(t\right),t\right)+\sum _{i=1}^n\left[x_i\left(t\right)-v_i\left(t\right)\right]{\nabla }^2{f}_i\left(x\left(t\right),t\right)+\nabla f\left(x\left(t\right),t\right)\right\}}^{\top }w\left(t\right) \\ =2{\left\{2\nabla f\left(x\left(t\right),t\right)+\sum _{i=1}^n\left[x_i\left(t\right)-v_i\left(t\right)\right]{\nabla }^2{f}_i\left(x\left(t\right),t\right)\right\}}^{\top }w\left(t\right). \end{gathered}$$

For a.e. t ∈ [0, T], let us denote

$w*\left(t\right)=f\left(x*\left(t\right),t\right)+\nabla f{\left(x*\left(t\right),t\right)}^{\top }\left[x*\left(t\right)-v*\left(t\right)\right]-u*\left(t\right).$

If (x ^∗, u ^∗, v ^∗) is a stationary point of (4.2), then ∇F(x ^∗(t), u ^∗(t), v ^∗(t), t) = 0 a.e. t ∈ [0, T], that is, for almost every t ∈ [0, T],

$D\left(t\right)w*\left(t\right)+{\Phi }_x\left(t\right)\tilde{\phi }\left(x*\left(t\right),u*\left(t\right)\right)+{\Phi }_f\left(t\right)\tilde{\phi }\left(f\left(x*\left(t\right),t\right),v*\left(t\right)\right)=0,$ (4.3)

$-w*\left(t\right)+{\Phi }_u\left(t\right)\tilde{\phi }\left(x*\left(t\right),u*\left(t\right)\right)=0,$ (4.4)

$-\nabla f\left(x*\left(t\right),t\right)w*\left(t\right)+{\Phi }_v\left(t\right)\tilde{\phi }\left(f\left(x*\left(t\right),t\right),v*\left(t\right)\right)=0.$ (4.5)

From (4.3) and (4.4) we obtain, for almost every t ∈ [0, T],

$$\begin{gathered} w*{\left(t\right)}^{\top }D\left(t\right)w*\left(t\right)+w*{\left(t\right)}^{\top }{\Phi }_x\left(t\right)\tilde{\phi }\left(x*\left(t\right),u*\left(t\right)\right) \\ +w*{\left(t\right)}^{\top }{\Phi }_f\left(t\right)\tilde{\phi }\left(f\left(x*\left(t\right),t\right),v*\left(t\right)\right)=0 \end{gathered}$$ (4.6)

and

$w*{\left(t\right)}^{\top }=\tilde{\phi }{\left(x*\left(t\right),u*\left(t\right)\right)}^{\top }{\Phi }_u\left(t\right),$ (4.7)

respectively. Using (4.6) and (4.7) we have, for almost every t ∈ [0, T], that

$$\begin{gathered} w*{\left(t\right)}^{\top }D\left(t\right)w*\left(t\right) \\ =-\tilde{\phi }{\left(x*\left(t\right),u*\left(t\right)\right)}^{\top }\left\{{\Phi }_u\left(t\right){\Phi }_x\left(t\right)\right\}\tilde{\phi }\left(x*\left(t\right),u*\left(t\right)\right) \\ -\tilde{\phi }{\left(x*\left(t\right),u*\left(t\right)\right)}^{\top }\left\{{\Phi }_u\left(t\right){\Phi }_f\left(t\right)\right\}\tilde{\phi }\left(f\left(x*\left(t\right),t\right),v*\left(t\right)\right). \end{gathered}$$ (4.8)

Noting that

$\frac{\partial \phi }{\partial a}\left(x_i^{*}\left(t\right),u_i^{*}\left(t\right)\right)\frac{\partial \phi }{\partial b}\left(x_i^{*}\left(t\right),u_i^{*}\left(t\right)\right)\ge 0\text{a.e.}t\in \left[0,T\right],i=1,\dots ,n,$

we have that Φ_u (t)Φ_x (t) is positive semi-definite for a.e. t ∈ [0, T] and using assumption (iii) we conclude that w ^∗(t)^⊤ D(t)w ^∗(t) ≤ 0 a.e. t ∈ [0, T]. But, from (i), it results that w(t)^⊤ D(t)w(t) > 0 for all w ∈ L _∞([0, T ]; ℝⁿ ), whenever w(t) ≠ 0. Then

$w*\left(t\right)=0\text{a.e.}t\in \left[0,T\right].$ (4.9)

From (4.9), (ii) and the definition of w ^∗(t), it follows that u ^∗(t) = f (x ^∗(t), t) a.e. t ∈ [0, T]. Replacing (4.9) in (4.5), we obtain

${\Phi }_v\left(t\right)\tilde{\phi }\left(f\left(x*\left(t\right),t\right),v*\left(t\right)\right)=0\iff \phi \left(f_i\left(x*\left(t\right),t\right),v_i^{*}\left(t\right)\right)\frac{\partial \phi }{\partial b}\left(f_i\left(x*\left(t\right),t\right),v_i^{*}\left(t\right)\right)=0,$

for i = 1, . . . , n. So, for almost every t ∈ [0, T] and for each i = 1, . . . , n, we have that $\phi \left(f_i\left(x*\left(t\right),t\right),v_i^{*}\left(t\right)\right)=0$ or $\frac{\partial \phi }{\partial b}\left(f_i\left(x*\left(t\right),t\right),v_i^{*}\left(t\right)\right)=0$. Observe that

$\frac{\partial \phi }{\partial b}\left(f_i\left(x*\left(t\right),t\right),v_i^{*}\left(t\right)\right)=0\Rightarrow \frac{v_i^{*}\left(t\right)}{\sqrt{{\left[f_i\left(x*\left(t\right),t\right)\right]}^2+{\left[v_i^{*}\left(t\right)\right]}^2}}-1=0$

implying that f _i (x ^∗(t), t) = 0 and $v_i^{*}\left(t\right)>0$, that is, $\phi \left(f_i\left(x*\left(t\right),t\right),v_i^{*}\left(t\right)\right)=0$. Therefore,

$\tilde{\phi }\left(f\left(x*\left(t\right),t\right),v*\left(t\right)\right)=0\text{a.e.}t\in \left[0,T\right].$ (4.10)

It results from (4.10) that f (x ^∗(t), t) ≥ 0 a.e. t ∈ [0, T], that is, f(x ^∗(t), t) ∈ K ^◦ a.e. t ∈ [0, T]. With a similar argument, replacing (4.9) in (4.4), we obtain x ^∗(t) ≥ 0, u ^∗(t) ≥ 0, and x ^∗(t)^⊤ u ^∗(t) = 0 a. e. t ∈ [0, T], namely x ^∗(t) ∈ K a.e. t ∈ [0, T]. Remembering that u ^∗(t) = f (x ^∗(t), t) a.e. t ∈ [0, T], we conclude that x ^∗(t)^⊤ f(x ^∗(t), t) = 0 for a.e. t ∈ [0, T]. □

Example 4.1.Considering Example 3.1, note that the (t, 0, t) a.e. t ∈ [0, 1] is a stationary point for

$$\begin{gathered} \text{maximize}Q\left(x,u,v\right)=-{\int }_0^{1}F\left(x\left(t\right),u\left(t\right),v\left(t\right),t\right)dt \\ \text{subject to}x,u,v\in L_{\infty }\left(\left[0,T\right],\mathrm{ℝ}\right), \end{gathered}$$

where

$F\left(x\left(t\right),u\left(t\right),v\left(t\right),t\right)={\left|{\left[x\left(t\right)\right]}^2-tx\left(t\right)+\left(2x\left(t\right)-t\right)\left[x\left(t\right)-v\left(t\right)\right]-u\left(t\right)\right|}^2+\sigma \left(x\left(t\right),u\left(t\right),v\left(t\right)\right),$

satisfying all assumptions of Theorem 4.3. Then, x^∗(t) = t a.e. t ∈ [0, 1] is a solution of CT CP(f, K).

5 THE SECOND-ORDER CONE CASE

In this section, we will consider the second-order cone defined as

$K=\left\{x\in {\mathrm{ℝ}}^n|x_1\ge {\left(\sum _{i=2}^n{a}_i^2{x}_i^2\right)}^{\frac{1}{2}}\right\},$ (5.1)

where a _i ∈ ℝ, i = 2, . . . , n, and its polar cone given by

$$\begin{gathered} K^{\circ }=\left\{x\in {\mathrm{ℝ}}^n|x_1\ge {\left(\sum _{i=2 \\ i\notin J}^n\frac{1}{a_i^2}{x}_i^2\right)}^{\frac{1}{2}},x_i=0\text{for}i\in J\right\}, \end{gathered}$$ (5.2)

where J = { j ∈ {2,..., n} | a _j = 0}. Our goal is to write K.K.T. optimality conditions for the problem (3.3) with

$\Omega =\left\{x\in L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}^n\right)|x\left(t\right)\in K\text{a.e.}t\in \left[0,T\right]\right\}$

and K is the second-order cone. But, note that the constraint of this problem is non-differentiable. To work around this problem, let us rewrite (5.1) as

$K=\left\{x\in {\mathrm{ℝ}}^n|x_1^2\ge \sum _{i=2}^n{a}_i^2{x}_i^2,x_1\ge 0\right\}$ (5.3)

and (5.2) as

$$\begin{gathered} K^{\circ }=\left\{x\in {\mathrm{ℝ}}^n|x_1^2\ge \sum _{i=2 \\ i\notin J}^n\frac{1}{a_i^2}{x}_i^2,x_1\ge 0,x_i=0\text{for}i\in J\right\}. \end{gathered}$$ (5.4)

Remark 1.In definition of K^◦, the condition “x _i = 0 for i ∈ J” is essential. For example, consider

$K=\left\{x\in {\mathrm{ℝ}}^3|x_1\ge 4x_2^2,x_1\ge 0\right\}\text{and}{K}^{\circ }=\left\{x\in {\mathrm{ℝ}}^3|x_1^2\ge \frac{1}{4}{x}_2^2,x_1\ge 0\right\}.$

Note that x = (0, 0, 1) ∈ K, y = (0, 0, −1) ∈ K ^◦ , but x ^⊤ y = −1 contradicting the definition of polar cone. With the purpose of applying Mangasarian-Fromovitz constraint qualification, we will assume that a _i ≠ 0, i = 1, 2, . . . , n, in the definition of K.

Define the diagonal n × n matrices

$A=diag\left(1,-a_2^2,\dots ,-a_n^2\right)\text{and}\overline{A}=diag\left(1,\frac{1}{a_2^2},\dots ,-\frac{1}{a_n^2}\right).$

In the matrix form, with e ₁ = (1, 0, . . . , 0)^⊤ ∈ ℝⁿ , (5.3) and (5.4) become

$K=\left\{x\in {\mathrm{ℝ}}^n|\frac{1}{2}{x}^{\top }Ax\ge 0,e_1^{\top }x\ge 0\right\}$ (5.5)

and

$K^{\circ }=\left\{x\in {\mathrm{ℝ}}^n|\frac{1}{2}{x}^{\top }\overline{A}x\ge 0,e_1^{\top }x\ge 0\right\}.$ (5.6)

Then, for fixed x ^∗ ∈ Ω, Problem (3.3) is given by

$$\begin{gathered} \text{maximize}P\left(x\right)=-{\int }_0^{T}f{\left(x\left(t\right),t\right)}^{\top }x\left(t\right)dt \\ \text{subject to}\frac{1}{2}x{\left(t\right)}^{\top }Ax\left(t\right)\ge 0\text{a.e. in}\left[0,T\right], \\ {e}_1^{\top }x\left(t\right)\ge 0\text{a.e. in}\left[0,T\right], \\ \frac{1}{2}f{\left(x\left(t\right),t\right)}^{\top }\overline{A}f\left(x\left(t\right),t\right)\ge 0\text{a.e. in}\left[0,T\right], \\ {e}_1^{\top }f\left(x\left(t\right),t\right)\ge 0\text{a.e. in}\left[0,T\right]. \end{gathered}$$ (5.7)

Note that Problem (5.7) is a particular case of (2.1), with

$$\begin{gathered} \varphi \left(x,t\right)=-f{\left(x,t\right)}^{\top }x,g_1\left(x,t\right)=\frac{1}{2}{x}^{\top }Ax,g_2\left(x,t\right)=e_1^{\top }x, \\ {g}_3\left(x,t\right)=\frac{1}{2}f{\left(x,t\right)}^{\top }\overline{A}f\left(x,t\right),g_4\left(x,t\right)=e_1^{\top }f\left(x,t\right), \end{gathered}$$

and satisfy (H1)-(H4). To ensure regularity of the constraints, we will assume that Problem (5.7) satisfies MFCQ.

Now, we will define a unconstrained problem related to CT CP(f, K), where K is the secondorder cone. To this end, we will use the Fischer-Burmeister function (4.1) again. For almost every t ∈ [0, T], define

$g\left(x\left(t\right),t\right)=\left(\begin{array}{c}{g}_1\left(x\left(t\right),t\right)\\ g_2\left(x\left(t\right),t\right)\\ g_3\left(x\left(t\right),t\right)\\ g_4\left(x\left(t\right),t\right)\end{array}\right),\nabla g\left(x\left(t\right),t\right)=\left(\begin{array}{c}\nabla g_1{\left(x\left(t\right),t\right)}^{\top }\\ \nabla g_2{\left(x\left(t\right),t\right)}^{\top }\\ \nabla g_3{\left(x\left(t\right),t\right)}^{\top }\\ \nabla g_4{\left(x\left(t\right),t\right)}^{\top }\end{array}\right),$

and consider u, z functions in L _∞([0, T]; ℝ⁴) with

$u\left(t\right)=\left(\begin{array}{c}{u}_1\left(t\right)\\ u_2\left(t\right)\\ u_3\left(t\right)\\ u_4\left(t\right)\end{array}\right)andz\left(t\right)=\left(\begin{array}{c}{z}_1\left(t\right)\\ z_2\left(t\right)\\ z_3\left(t\right)\\ z_4\left(t\right)\end{array}\right),$

and v ∈ L _∞([0, T], ℝⁿ ). Define the following unconstrained continuous-time problem:

$$\begin{gathered} \text{maximize}R\left(x,u,z\right)=-{\int }_0^{T}G\left(x\left(t\right),u\left(t\right),z\left(t\right),t\right)dt, \\ \text{subject to}u,z\in L_{\infty }\left(\left[0,T\right],{\mathrm{ℝ}}^4\right), \\ x\in L_{\infty }\left(\left[0,T\right],{\mathrm{ℝ}}^n\right), \end{gathered}$$ (5.8)

where

$G\left(x,u,z,t\right)={||f\left(x,t\right)+\nabla f{\left(x,t\right)}^{\top }x-\nabla g{\left(x,t\right)}^{\top }u||}^2+{||g\left(x,t\right)-z||}^2+\sigma \left(u,z\right)$

and

$\sigma \left(u\left(t\right),z\left(t\right)\right)=\sum _{i=1}^4{\left[{\phi }_i\left(u_i\left(t\right),z_i\left(t\right)\right)\right]}^2\text{a.e.}t\in \left[0,T\right],$

with φ _i representing Fischer-Burmeister functions at (u _i (t), z _i (t)) a.e. t ∈ [0, T] and i ∈ {1, 2, 3, 4}.

Theorem 5.4.If (x ^∗, u ^∗, z ^∗) ∈ L _∞([0, T]; ℝⁿ × ℝ⁴ × ℝ⁴) is a global optimal solution of (5.8) with R(x ^∗, u ^∗, z ^∗) = 0 and$\left[x*\left(t\right)-\overline{A}f\left(x*\left(t\right),t\right)u_3^{*}\left(t\right)-e_1{u}_4^{*}\left(t\right)\right]\in Ker\left(\nabla f{\left(x*\left(t\right),t\right)}^{\top }\right)$, then x ^∗ is a solution of CT CP(f, K).

Proof. By hypothesis, for almost every t ∈ [0, T], we have that

• (a) $f\left(x*\left(t\right),t\right)+\nabla f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)-\nabla g{\left(x*\left(t\right),t\right)}^{\top }u*\left(t\right)=0;$

• (b) $\frac{1}{2}x*{\left(t\right)}^{\top }Ax*\left(t\right)-z_1^{*}\left(t\right)=0;$

• (c) $e_1^{\top }x*{\left(t\right)}^{\top }-z_2^{*}\left(t\right)=0;$

• (d) $\frac{1}{2}f{\left(x*\left(t\right),t\right)}^{\top }\overline{A}f\left(x*\left(t\right),t\right)-z_3^{*}\left(t\right)=0;$

• (e) $e_1^{\top }f\left(x*\left(t\right),t\right)-z_4^{*}\left(t\right)=0;$

• (f) $\sigma \left(u*\left(t\right),z*\left(t\right)\right)=0.$

Using the property of the Fischer-Burmeister function, for almost every t ∈ [0, T], from (f) it results

${\phi }_i\left(u_i^{*}\left(t\right),z_i^{*}\left(t\right)\right)=0\Rightarrow u_i^{*}\left(t\right)\ge 0,z_i^{*}\left(t\right)\ge 0\text{and}{u}_i^{*}\left(t\right)z_i^{*}\left(t\right)=0,i=1,2,3,4.$

As $z_1^{*}\left(t\right)\ge 0$ and $z_2^{*}\left(t\right)\ge 0$ a.e. t ∈ [0, T], from (b) and (c) it results that

$\frac{1}{2}x*{\left(t\right)}^{\top }Ax*\left(t\right)=z_1^{*}\left(t\right)\ge 0\text{a.e. in}\left[0,T\right],$ (5.9)

$x_1^{*}\left(t\right)=z_2^{*}\left(t\right)\ge 0\text{a.e. in}\left[0,T\right],$ (5.10)

so that x ^∗(t) ∈ K a.e. t ∈ [0, T]. Analogously, as $z_3^{*}\left(t\right)\ge 0$ and $z_4^{*}\left(t\right)\ge 0$ a.e. t ∈ [0, T], from (d) and (e) it results that f(x ^∗(t), t) ∈ K ^◦ a.e. t ∈ [0, T]. Now, for each t ∈ [0, T],

• (i) if $x_1^{*}\left(t\right)=0$, it results from (5.9) that $x_i^{*}\left(t\right)=0,2\le i\le n$. Therefore, we have that $x*{\left(t\right)}^{\top }f\left(x*\left(t\right),t\right)=0$.

• (ii) if $x_1^{*}\left(t\right)>0$, from (c) we have that $z_2^{*}\left(t\right)>0$; from (f), we see that this implies in $u_2^{*}\left(t\right)=0$.

Therefore, using (a), we conclude that

$$\begin{gathered} f\left(x*\left(t\right),t\right)=-\nabla f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)+Ax*\left(t\right)u_1^{*}\left(t\right) \\ +\nabla f{\left(x*\left(t\right),t\right)}^{\top }\overline{A}f\left(x*\left(t\right),t\right)u_3^{*}\left(t\right)+\nabla f_1\left(x*\left(t\right),t\right)u_4^{*}\left(t\right) \end{gathered}$$

or

$$\begin{gathered} f\left(x*\left(t\right),t\right)=-\nabla f{\left(x*\left(t\right),t\right)}^{\top }\left[x*\left(t\right)-\overline{A}f\left(x*\left(t\right),t\right)u_3^{*}\left(t\right)-e_1{u}_4^{*}\left(t\right)\right] \\ +Ax*\left(t\right)u_1^{*}\left(t\right). \end{gathered}$$

Multiplying the left side by x ^∗(t)^⊤, it results that

$$\begin{gathered} x*{\left(t\right)}^{\top }f\left(x*\left(t\right),t\right) \\ =-*x*{\left(t\right)}^{\top }\nabla f{\left(x*\left(t\right),t\right)}^{\top }\left[x*\left(t\right)-\overline{A}f\left(x*\left(t\right),t\right)u_3^{*}\left(t\right)-e_1{u}_4^{*}\left(t\right)\right] \\ +x*{\left(t\right)}^{\top }Ax*\left(t\right)u_1^{*}\left(t\right) \\ =0 \end{gathered}$$

because $x*{\left(t\right)}^{\top }Ax*\left(t\right)u_1^{*}\left(t\right)=2z_1^{*}\left(t\right)u_1^{*}=0$ and $\left[x*\left(t\right)-\overline{A}f\left(x*\left(t\right),t\right)u_3^{*}\left(t\right)-e_1{u}_4^{*}\left(t\right)\right]\in Ker\left(\nabla f{\left(x*\left(t\right),t\right)}^{\top }\right)$ a.e. t ∈ [0, T].

From (i) and (ii) we can conclude that x ^∗(t)^⊤ f(x ^∗(t), t) = 0 for almost every t ∈ [0, T]. □

In the next example, we illustrate the use of Theorem 5.4.

Example 5.1.We want to solve CT CP(f, K) with

$$\begin{gathered} f\left(x,t\right)=\left(x_3^2+t,-x_2,x_1-x_2\right)\mathit{\text{and}} \\ K=K^{\circ }=\left\{x\in {\mathrm{ℝ}}^3|x_1^2\ge x_2^2+x_3^2,x_1\ge 0\right\}. \end{gathered}$$

If (x ^∗, u ^∗, z ^∗) ∈ L _∞([0, T]; ℝ³ × ℝ⁴ × ℝ⁴) is a solution of (5.8) with R(x ^∗, u ^∗, z ^∗) = 0, where

$$\begin{gathered} g_1\left(x,t\right)=\frac{1}{2}\left(x_1^2-x_2^2-x_3^2\right), \\ {g}_2\left(x,t\right)=x_1, \\ {g}_3\left(x,t\right)=\frac{1}{2}\left[{\left(x_3^2+t\right)}^2-x_2^2-{\left(x_1-x_2\right)}^2\right], \\ {g}_4\left(x,t\right)=x_3^2+t, \end{gathered}$$

then for almost every t ∈ [0, T], we have that

• (a) $x_3^{*}{\left(t\right)}^2+t+x_3^{*}\left(t\right)-x_1^{*}\left(t\right)u_1^{*}\left(t\right)-u_2^{*}\left(t\right)+\left(x_1^{*}\left(t\right)-x_2^{*}\left(t\right)\right)u_3^{*}\left(t\right)=0;$

• (b) $-x_2^{*}\left(t\right)-x_2^{*}\left(t\right)-x_3^{*}\left(t\right)+x_2^{*}\left(t\right)u_1^{*}\left(t\right)-\left(x_1^{*}\left(t\right)-2x_2^{*}\left(t\right)\right)u_3^{*}\left(t\right)=0;$

• (c) $x_1^{*}\left(t\right)-x_2^{*}\left(t\right)+2x_1^{*}\left(t\right)x_3^{*}\left(t\right)+x_3^{*}\left(t\right)u_1^{*}\left(t\right)-2\left(x_3^{*}{\left(t\right)}^2+t\right)x_3^{*}\left(t\right)u_3^{*}\left(t\right)-2x_3^{*}\left(t\right)u_4^{*}\left(t\right)=0;$

• (d) $u_1^{*}\left(t\right)\ge 0,g_i\left(x*\left(t\right),t\right)\ge 0,u_i^{*}\left(t\right)g_i\left(x*\left(t\right),t\right)=0,i\in \left\{1,2,3,4\right\}.$

Observe that (x ^∗, u ^∗, z ^∗) satisfies (a)-(d) with x ^∗(t) = (t, t, 0), u ^∗(t) = (1, 0, 1, 0) and z ^∗(t) = g(x ^∗(t), t) a.e. t ∈ [0, T] and is a global optimal solution of (5.8). Moreover,

$\left[x*\left(t\right)-\overline{A}f\left(x*\left(t\right),t\right)u_3^{*}\left(t\right)-e_1{u}_4^{*}\left(t\right)\right]\in Ker\left(\nabla f{\left(x*\left(t\right),t\right)}^{\top }\right)$

is satisfied for a.e. t ∈ [0, T]. Then, by Theorem 5.4, x ^∗ is a solution for CT CP(f, K), that is, x ^∗(t) ∈ K, f(x ^∗(t), t) ∈ K ^◦ and x ^∗(t)^⊤ f(x ^∗(t), t) = 0 a.e. t ∈ [0, 1].

Theorem 5.5.Let x^∗be solution of CT CP(f, K) and suppose that the data of Problem (5.7) satisfies (H1)-(H4) and Definition 2.3 hold and that the system (2.3) is regular. Then there exist u^∗and z^∗in$L_{\infty }\left(\left[0,T\right];{\mathrm{ℝ}}_{+}^4\right)$such that (x ^∗, u ^∗, z ^∗) is a global optimal solution of (5.8), with R(x ^∗, u ^∗, z ^∗) = 0.

Proof. The assumptions guarantee that Proposition 2.2 can be applied to Problem (5.7), that is, that there exists u ^∗ in L _∞([0, T ]; ℝ⁴) such that, for almost every t ∈ [0, T], we have

• (a) $f\left(x*\left(t\right),t\right)+\nabla f{\left(x*\left(t\right),t\right)}^{\top }x*\left(t\right)-\nabla g{\left(x*\left(t\right),t\right)}^{\top }u*\left(t\right)=0;$

• (b) $g_1\left(x*\left(t\right),t\right)=\frac{1}{2}x*{\left(t\right)}^{\top }Ax*\left(t\right)\ge 0;$

• (c) $g_2\left(x*\left(t\right),t\right)=e_1^{\top }x*\left(t\right)\ge 0;$

• (d) $g_3\left(x*\left(t\right),t\right)=\frac{1}{2}f{\left(x*\left(t\right),t\right)}^{\top }\overline{A}f\left(x*\left(t\right),t\right)\ge 0;$

• (e) $g_4\left(x*\left(t\right),t\right)=e_1^{\top }f\left(x*\left(t\right),t\right)\ge 0;$

• (f) $u_i^{*}\left(t\right)\ge 0,g_i\left(x*\left(t\right),t\right)\ge 0,u_i^{*}\left(t\right)g_i\left(x*\left(t\right),t\right)=0,i\in \left\{1,2,3,4\right\};$

From (a) it results

$||f\left(x*\left(t\right),t\right)+\nabla f\left(x*\left(t\right),t\right)x*\left(t\right)-\nabla g\left(x*\left(t\right),t\right)u*\left(t\right)||=0$

a.e. in [0, T]. Defining $z_i^{*}\left(t\right)g_i\left(x*\left(t\right),t\right)$, i = 1, 2, 3, 4, it follows that ∥g(x ^∗(t), t) − z ^∗(t)∥ = 0 a.e. in [0, T]. From (f) we obtain ${\phi }_i\left(u_i^{*}\left(t\right),z_i^{*}\left(t\right)\right)=0$ a.e. in [0, T]. Therefore, we can conclude that (x ^∗, u ^∗, z ^∗) is a global optimal solution of (5.8), with R(x ^∗, u ^∗, z ^∗) = 0. □

6 FINAL COMMENTS

In this work, we presented an approach to the resolution of the continuous-time complementarity problem by reformulating it as an equivalent unconstrained optimization problem. The definition and properties of the VIP( f ,Ω) given by Zalmai in ²² are applied on the resolution of the continuous-time nonlinear complementarity problem presented by Bodo an Hanson in ³.

We proved that a solution of Problem CT CP(f, K) is a global minimizer of Problem (3.3) with zero objective function value and vice versa. A discretization approach can be used to solve linear Problem (3.3) (for example, see ^{16), (18), (19), (20}).

For the polyhedral cone $K={\mathrm{ℝ}}_{+}^n$, we showed that a stationary point of the unconstrained Problem (4.2) with zero objective function value is a solution of Problem CT CP(f, K). Moreover, we verified that stationary points of Problem (4.2) also are solutions of CT CP(f, K).

The Fischer-Burmeister function was used to write an unconstrained auxiliary problem and to obtain solutions of CT CP(f, K) when K is the second order cone. In this case, to ensure that some constraint qualification holds, assumptions over the constraints should be made (see Remark 1).

This article opens new perspectives for research in the area, such as the possibility of studying the generalized continuous-time nonlinear complementarity problem that consists of find x ∈ L _∞([0, T], ℝⁿ ) such that

$F\left(x,t\right)\in K,G\left(x,t\right)\in K^{\circ }\text{and}F{\left(x,t\right)}^{\top }G\left(x,t\right)=0,$

where F: ℝⁿ × [0, T] → ℝⁿ , G: ℝⁿ × [0, T] → ℝⁿ , K ⊂ ℝⁿ is a nonempty closed convex cone with vertex at 0 and K ^◦ is the polar cone of K.

Acknowledgments

The authors are grateful for the support of the São Paulo Research Foundation (FAPESP), grant number 2013/07375-0, and of the Minas Gerais Research Foundation (FAPEMIG), grant number APQ-00453-21.

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Publication Dates

• Publication in this collection
  11 Nov 2024
• Date of issue
  2024

History

• Received
  06 July 2023
• Accepted
  04 June 2024