Exact Penalty Functions for Optimal Control Problems I: Main Theorem and Free-Endpoint Problems

TL;DR

This paper introduces a comprehensive framework for designing exact penalty functions in optimal control, enabling the transformation of constrained problems into unconstrained variational problems for easier analysis and computation.

Contribution

It extends existing results on penalty functions to infinite-dimensional spaces and applies them specifically to free-endpoint optimal control problems, providing new theoretical insights.

Findings

Strengthened results on penalty functions in infinite-dimensional spaces

Reduced free-endpoint control problems to variational problems

Provided auxiliary results on integral functionals and Nemytskii operators

Abstract

In this two-part study we develop a general approach to the design and analysis of exact penalty functions for various optimal control problems, including problems with terminal and state constraints, problems involving differential inclusions, as well as optimal control problems for linear evolution equations. This approach allows one to simplify an optimal control problem by removing some (or all) constraints of this problem with the use of an exact penalty function, thus allowing one to reduce optimal control problems to equivalent variational problems, apply numerical methods for solving, e.g. problems without state constraints, to problems including such constraints, etc. In the first part of our study we strengthen some existing results on exact penalty functions for optimisation problems in infinite dimensional spaces and utilise them to study exact penalty functions for free-endpoint optimal control problems, which reduce these problems to equivalent variational ones. We also prove several auxiliary results on integral functionals and Nemytskii operators that are helpful for verifying the assumptions under which the proposed penalty functions are exact.

Equations

$$\underset{x\in {\mathbb{R}}^d}{\min }f(x)\text{subject to}{g}_i(x)\le 0,i\in \{1,2,\dots ,m\},$$

$$\underset{x\in {\mathbb{R}}^d}{\min }{\Phi }_{\lambda }(x)=f(x)+\lambda \sum _{i=1}^m\max \{0,g_i(x)\}.$$

$$\min \mathcal{I}(x)\text{subject to}x\in M\cap A.$$

$$\min {\Phi }_{\lambda }(x)\text{subject to}x\in A.$$

$$g_K^{\downarrow }(x)=\underset{y\in K,y\to x}{\mathrm{liminf}}\frac{g(y)-g(x)}{d(y,x)}$$

$$g^{\downarrow }(x)\le \underset{\Vert v\Vert =1}{\inf }{g}^{\prime }(x,v),\text{ where }{g}^{\prime }(x,v)=\underset{\alpha \to +0}{\lim }\frac{g(x+\alpha v)-g(x)}{\alpha }.$$

$${\Psi }_{\lambda }(x)=\{\begin{array}{ll}\mathcal{I}(x)+\lambda \frac{\phi (x)}{\delta -\phi (x)},& \text{if }\phi (x)<\delta ,\\ +\infty ,& \text{otherwise.}\end{array}$$

$$g^{\prime }(x)[y-x]\ge 0\forall y\in K.$$

$$\begin{array}{rl}& \min \mathcal{I}(x,u)={\int }_0^T\theta (x(t),u(t),t)dt+\zeta (x(T)),\\ & \text{subject to }\dot{x}(t)=f(x(t),u(t),t),t\in [0,T],x(0)=x_0,u\in U.\end{array}$$

$$\Vert u{\Vert }_q={\left({\int }_0^T|u(t){|}^{q}dt\right)}^{\frac{1}{q}}\forall u\in L_q^m(0,T),$$

$$\|x\|_{1,p}\leq\big{(}1+\max\{T,T^{1/p}\}\big{)}\|x\|_{0}\quad\forall x\in W^{d}_{1,p}(0,T),$$

$$M=\big{\{}(x,u)\in X\bigm{|}F(x,u)=0\big{\}},\quad A=\big{\{}(x,u)\in X\bigm{|}x(0)=x_{0},\>u\in U\big{\}},$$

$$\underset{(x,u)\in X}{\min }\mathcal{I}(x,u)\text{subject to}(x,u)\in M\cap A.$$

$$\varphi(x,u)=\|F(x,u)\|_{p}=\left(\int_{0}^{T}\big{|}\dot{x}(t)-f(x(t),u(t),t)\big{|}^{p}\,dt\right)^{\frac{1}{p}}.$$

$$\underset{(x,u)\in X}{\min }{\Phi }_{\lambda }(x,u)=\mathcal{I}(x,u)+\lambda \phi (x,u)\text{subject to}(x,u)\in A.$$

$$\begin{split}{}&\min\>\int_{0}^{T}\theta(x(t),u(t),t)\,dt+\lambda\left(\int_{0}^{T}\big{|}\dot{x}(t)-f(x(t),u(t),t)\big{|}^{p}\,dt\right)^{\frac{1}{p}}+\zeta(x(T))\\ {}&\text{subject to }x(0)=x_{0},\quad u\in U,\end{split}$$

$$\dot{x}(t)=f(x(t),u(t),t)+w(t),t\in [0,T],x(0)=x_0$$

$$0>({\gamma }_u{)}^{\downarrow }(\dot{x})=\underset{z\to \dot{x}}{\mathrm{liminf}}\frac{{\gamma }_u(z)-{\gamma }_u(\dot{x})}{\Vert z-\dot{x}{\Vert }_p}$$

$$\ge (1+T)\underset{y\in A_s,y\to x}{\mathrm{liminf}}\frac{{\phi }_u(y)-{\phi }_u(x)}{\Vert y-x{\Vert }_{1,p}}=(1+T)({\phi }_u{)}_{A_s}^{\downarrow }(x).$$

$$H(x)=\{\begin{array}{ll}|x{|}^{p-2}x,& \text{if }x\ne 0,\\ 0,& \text{if }x=0,\end{array}$$

$$\frac{1}{\gamma_{u}(z)^{p-1}}\int_{0}^{T}\big{\langle}G(z,u)(t),F(Jz,u)(t)\big{\rangle}\,dt=\|F(Jz,u)\|_{p}=\gamma_{u}(z),$$

$$F_u^{\prime }(x)[h]=\dot{h}(\cdot )-{\nabla }_{x}f(x(\cdot ),u(\cdot ),\cdot )h(\cdot )\forall h\in W_{1,p}^d(0,T).$$

$$\gamma_{u}^{\prime}(z)[h]=\frac{1}{\gamma_{u}(z)^{p-1}}\int_{0}^{T}\Big{\langle}G(z,u)(t),h(t)-\nabla_{x}f(x(t),u(t),t)\int_{0}^{t}h(\tau)\,d\tau\Big{\rangle}\,dt\quad\forall h\in L_{p}^{d}(0,T)$$

$${\gamma }_u^{\prime }(z)[h]={\int }_0^T⟨w(x,u)(t)-{\int }_t^T{\nabla }_{x}f(x(\tau ),u(\tau ),\tau {)}^{T}w(x,u)(\tau )d\tau ,h(t)⟩dt,$$

$$({\gamma }_u{)}^{\downarrow }(z)\le -\Vert {\gamma }_u^{\prime }(z)\Vert =-\Vert (I-{\mathcal{K}}_y)w(x,u){\Vert }_{p^{\prime }}\le -\frac{1}{\Vert (I-{\mathcal{K}}_y{)}^{-1}\Vert },$$

$$\big{(}\mathcal{K}_{y}h\big{)}(t)=\int_{t}^{T}y(s)h(s)\,ds\quad\forall h\in L_{p^{\prime}}^{d}(0,T),\quad y(s)=\nabla_{x}f(x(s),u(s),s)^{T}.$$

$$|f(x,u,t)|\le C_R|x|+{\omega }_R(t),\theta (x,u,t)\ge -C_R|x|-{\omega }_R(t)$$

$$|f(x,u,t)|\leq C_{f}\big{(}|x|+|u|^{\frac{q}{p}}\big{)}+\omega_{f}(t)$$

$$|f(x,u,t)|\leq C\big{(}|x|^{r}+|u|^{\frac{q}{p}}\big{)}+\omega(t),\quad\theta(x,u,t)\geq C\big{(}|x|^{s}+|u|^{q}\big{)}-\omega(t)$$

$$\dot{x}(t)=f(x(t),u(t),t)+w(t)\text{for a.e. }t\in [0,T]$$

$$x(t)=x_0+{\int }_0^{t}f(x(\tau ),u(\tau ),\tau )d\tau +{\int }_0^{t}w(\tau )d\tau \forall t\in [0,T].$$

Full text

In the first part of our study we strengthen some existing results on exact penalty functions for optimisation problems in infinite dimensional spaces and utilise them to study exact penalty functions for free-endpoint optimal control problems, which reduce these problems to equivalent variational ones. We also prove several auxiliary results on integral functionals and Nemytskii operators that are helpful for verifying the assumptions under which the proposed penalty functions are exact.

1 Introduction

The idea of using so-called exact penalty functions for solving constrained optimisation problems was suggested practically simultaneously by Eremin [32] and Zangwill [71] in the 1960s. Since then, exact penalty functions have been extensively studied and applied to various optimisation problems by many researchers (see, e.g. [33, 43, 24, 25, 23, 28] and the references therein).

The main idea behind the exact penalty approach consists in replacing a constrained optimisation problem, say

[TABLE]

by the unconstrained problem of minimising the nonsmooth penalty function:

[TABLE]

Under some natural assumptions this penalised problem is equivalent to the original one in the sense that these problems have the same optimal value and the same locally/globally optimal solutions, provided the penalty parameter $\lambda$ is sufficiently large (but finite). Thus, the exact penalty approach allows one to reduce constrained optimisation problems to equivalent unconstrained ones and apply numerical methods of unconstrained optimisation to constrained problems. However, most papers on the theory and applications of exact penalty functions deal only with the finite dimensional case or a local analysis of an exact penalty function.

In the infinite dimensional case, globally exact penalty functions were probably first studied by Demyanov et al. for a problem of finding optimal parameters in a system described by ordinary differential equations [18], free-endpoint optimal control problems [19, 17, 49], the simplest problem of the calculus of variations [13, 14], and variational problems with nonholonomic inequality constraints [12, 16]. The main results of these papers were further extended to isoperimetric problems of the calculus of variations [22], variational problems involving higher order derivatives [68], parametric moving boundary variational problems [21], control problems involving differential inclusions [36], and certain optimal control problems for implicit control systems with strict inequality constraints [20]. Numerical methods for solving optimal control problems based on the use of exact penalty functions in the infinite dimensional setting were probably first considered by Outrata [62] (see also [64, 63]), and later on were also studied in [37]. However, in [62] only the local exactness of a penalty function was considered under the assumption that an abstract constraint qualification holds true, and it is unclear how to verify this assumption for any particular problem. In [17, 37], the global exactness of penalty functions was stated without proof. The main results on exact penalty functions for various variational problems from [12, 16, 13, 14, 22, 68, 21] are based on the assumptions that the objective function is Lipschitz continuous on a rather complicated and possibly unbounded set, and a penalty function attains a global minimum in the space of piecewise continuous functions for any sufficiently large value of the penalty parameter, and it is, once again, unclear how to verify these assumptions in any particular case. The same remark is true for the main results of the papers [18, 19, 17, 49, 20, 36] devoted to exact penalty functions for optimal control problems. To the best of authors’ knowledge, the only verifiable sufficient conditions for the global exactness of an exact penalty function in the infinite dimensional setting were obtained by Gugat and Zuazua in [41], where the exact penalisation of the terminal constraint for optimal control problems involving linear evolution equations was considered.

The main goal of our study is to develop a general theory of exact penalty functions for optimal control problems that contains verifiable sufficient conditions for the global/complete exactness of penalty functions. To this end, in the first paper we strengthen some existing results on exact penalty functions for optimisation problems in infinite dimensional spaces and apply them to free-endpoint problems. We also obtain a number of auxiliary results that are helpful for verifying the exactness of penalty functions for optimal control problems in particular cases. For instance, we provide simple sufficient conditions for the Lipschitz continuity of integral functionals, the boundedness of sublevel sets of penalty functions, the existence of global minimisers, etc. Thus, in this paper we obtain first simple and verifiable sufficient conditions for the global exactness of penalty functions for nonlinear optimal control problems, which allow one to reduce free-endpoint optimal control problems to equivalent variational problems. In the second paper we apply our general results on exact penalty functions to optimal control problems with terminal and pointwise state constraints, including such problems for linear evolution equations in Hilbert spaces.

Let us point out that in our study we consider only so-called simple linear penalty functions, i.e. such penalty functions that depend linearly on the objective function and do not depend on derivatives of the objective function or constraints. Such exact penalty functions are inherently nonsmooth (see, e.g. Remark 3 in [28] and Sect. 2.3 in [41]), and one has to utilise a well-developed apparatus of nonsmooth optimisation to minimise them. In particular, one can apply such popular and efficient modern methods of nonsmooth optimisation as bundle methods [61, 42, 44, 39], gradient sampling methods [9, 52, 10], nonsmooth quasi-Newton methods [55, 51], discrete gradient method [7] (see also [50, 6]), etc. Alternatively, one can utilise smoothing approximations of nonsmooth penalty functions as in [65, 59, 60, 57, 29] or the smooth penalty function proposed by Huyer and Neumaier [45]. This penalty function was analysed in detail in [70, 29] and applied to discretised optimal control problems in [56, 48, 58]. In [27] it was shown that Huyer and Neumaier’s penalty function is exact if and only if a corresponding standard nonsmooth penalty function is exact. With the use of this result and the main results of our two-part study one can easily verify the global exactness of Huyer and Neumaier’s penalty function for various optimal control problems without discretisation.

The paper is organised as follows. Some general results on exact penalty functions for optimisation problems in infinite dimensional spaces are presented in Section 2. In particular, in this section we formulate “the Main Theorem” (Theorem 2), which is the main tool for proving the global/complete exactness of penalty functions for optimal control problems. We extensively utilise this theorem throughout both parts of our study. In Section 3, we study an exact penalty function for free-endpoint optimal control problems, while in Section 4 these results are extended to the case of free-endpoint variational problems involving differential inclusions. Finally, a proof of the main theorem, as well as a number of auxiliary results on integral functionals and Nemytskii operators that are helpful for verifying the assumptions of the main theorem in the case of optimal control problems, are given in Appendices A and B respectively.

2 Exact Penalty Functions in Metric Spaces

In this section we present some general results on exact penalty functions for optimisation problems in metric spaces that are utilised throughout the paper. Let $(X,d)$ be a metric space, $M,A\subset X$ be nonempty sets such that $M\cap A\neq\emptyset$, and $\mathcal{I}\colon X\to\mathbb{R}\cup\{+\infty\}$ be a given function. Denote by $\operatorname{dom}\mathcal{I}=\{x\in X\mid\mathcal{I}(x)<+\infty\}$ the effective domain of $\mathcal{I}$.

Consider the following optimisation problem:

[TABLE]

Here the sets $M$ and $A$ correspond to two different types of constraints of the optimisation problem. In particular, it can be equality/inequality constraints or linear/nonlinear constraints or terminal/pointwise constraints. Denote by $\Omega=M\cap A$ the feasible region of $(\mathcal{P})$. Hereinafter, we suppose that there exists a globally optimal solution $x^{*}\in\operatorname{dom}\mathcal{I}$ of the problem $(\mathcal{P})$, i.e. $\mathcal{I}$ attains a global minimum on $\Omega$, and the optimal value is finite.

Let a function $\varphi\colon X\to[0,+\infty]$ be such that $\varphi(x)=0$ iff $x\in M$. For any $\lambda\geq 0$ introduce the function $\Phi_{\lambda}(x)=\mathcal{I}(x)+\lambda\varphi(x)$. This function is called a penalty function for the problem $(\mathcal{P})$, $\lambda$ is called a penalty parameter, and $\varphi$ is called a penalty term for the constraint $x\in M$. Note that the function $\Phi_{\lambda}(x)$ is non-decreasing in $\lambda$, $\Phi_{\lambda}(x)\geq\mathcal{I}(x)$ for all $x\in X$, and $\Phi_{\lambda}(x)=\mathcal{I}(x)$ for any $x$ satisfying the constraint $x\in M$. Therefore, it is natural to consider the penalised problem

[TABLE]

Observe that only the constraint $x\in M$ is penalised, i.e. only this constraint is incorporated into the penalty function $\Phi_{\lambda}(x)$. This approach allows one to choose which constraints of an optimisation problem are to be “removed” via the exact penalty function technique in order to simplify the problem under consideration.

We would like to know when the penalised problem (1) is, in some sense, equivalent to the original problem $(\mathcal{P})$, i.e. when the penalisation does not distort information about minimisers of the problem $(\mathcal{P})$.

Definition 1**.**

The penalty function $\Phi_{\lambda}$ is called (globally) exact, if there exists $\lambda^{*}\geq 0$ such that for any $\lambda\geq\lambda^{*}$ the set of globally optimal solutions of the penalised problem (1) coincides with the set of globally optimal solutions of the problem $(\mathcal{P})$. The greatest lower bound of all such $\lambda^{*}$ is denoted by $\lambda^{*}(\mathcal{I},\varphi,A)$ and is called the least exact penalty parameter of the penalty function $\Phi_{\lambda}$.

One can easily verify (see [28, Corollary 3.3]) that the penalty function $\Phi_{\lambda}$ is exact iff there exists $\lambda\geq 0$ such that $\inf_{x\in A}\Phi_{\lambda}(x)=\inf_{x\in\Omega}\mathcal{I}(x)$, i.e. iff the optimal values of the problems $(\mathcal{P})$ and (1) coincide. Furthermore, the greatest lower bound of all such $\lambda$ coincides with the least exact penalty parameter.

Thus, if the penalty function $\Phi_{\lambda}$ is globally exact, then for any $\lambda\geq 0$ large enough the penalised problem (1) is equivalent to the original problem $(\mathcal{P})$ in the sense that it has the same optimal value and the same globally optimal solutions.

Let us provide simple sufficient conditions for the global exactness of the penalty function $\Phi_{\lambda}$. To this end, we need to recall the definition of the rate of steepest descent of a function defined on a metric space [11, 15, 69]. Let $g\colon X\to\mathbb{R}\cup\{+\infty\}$ be a given function, $K\subset X$ be a nonempty set, and $x\in K$ be such that $g(x)<+\infty$. The quantity

[TABLE]

is called the rate of steepest descent of the function $g$ with respect to the set $K$ at the point $x$ (if $x$ is an isolated point of $K$, then $g^{\downarrow}_{K}(x)=+\infty$ by definition). In the case $K=X$ we denote $g^{\downarrow}(x)=g^{\downarrow}_{X}(x)$. Let us note that the rate of steepest descent of the function $g$ at $x$ is closely connected to the so-called strong slope $|\nabla g|(x)$ of $g$ at $x$ [5, 53]. See [5, 53, 29] for some calculus rules for strong slope/rate of steepest descent, and the ways one can estimate them in various particular cases. Here we only note that if $X$ is a normed space, and $g$ is Fréchet differentiable at a point $x\in X$, then $g^{\downarrow}(x)=-\|g^{\prime}(x)\|_{X^{*}}$, where $g^{\prime}(x)$ is the Fréchet derivative of $g$ at $x$, and $\|\cdot\|_{X^{*}}$ is the standard norm in the topological dual space $X^{*}$. If $g$ is Gâteaux differentiable at $x$, then $g^{\downarrow}(x)\leq-\|g^{\prime}(x)\|_{X^{*}}$, where $g^{\prime}(x)$ is the Gâteaux derivative of $g$ at $x$. Finally, if $g$ is merely directionally differentiable at $x$, then

[TABLE]

The following theorem, which is a particular case of [29, Theorem 3.6], contains simple sufficient conditions for the global exactness of the penalty function $\Phi_{\lambda}(x)$. For any $\delta>0$ define $\Omega_{\delta}=\{x\in A\mid\varphi(x)<\delta\}$.

Theorem 1**.**

Let $X$ be a complete metric space, $A$ be closed, $\mathcal{I}$ and $\varphi$ be lower semi-continuous (l.s.c.) on $X$. Suppose also that there exist a feasible point $x_{0}\in X$, $\lambda_{0}>0$ and $\delta>0$ such that

the function $\mathcal{I}$ is Lipschitz continuous on an open set containing the set $C(\delta,\lambda_{0})=\{x\in\Omega_{\delta}\mid\Phi_{\lambda_{0}}(x)<\mathcal{I}(x_{0})\}$; 2. 2.

there exists $a>0$ such that $\varphi^{\downarrow}_{A}(x)\leq-a$ for all $x\in C(\delta,\lambda_{0})\setminus\Omega$.

Then the penalty function $\Phi_{\lambda}$ is globally exact if and only if it is bounded below on $A$ for some $\lambda\geq 0$.

Remark 1*.*

If the assumptions of the theorem above are satisfied, but the penalty function $\Phi_{\lambda}$ is not bounded below, one can consider the penalty function

[TABLE]

One can check that under the assumptions of Theorem 1 the penalty function $\Psi_{\lambda}$ is exact iff it is bounded below. In particular, $\Psi_{\lambda}$ is exact, provided the function $\mathcal{I}$ is bounded below on $C(\delta,\lambda_{0})$.

As was noted above, if the penalty function $\Phi_{\lambda}$ is globally exact, then the penalised problem (1) is equivalent to the problem $(\mathcal{P})$ in the sense that it has the same optimal value and the same globally optimal solutions. However, optimisation methods often can find only local minimisers or even only stationary (critical) points of an optimisation problem. That is why it is desirable to ensure that local minimisers/stationary points of the penalty function $\Phi_{\lambda}$ coincide with locally optimal solutions/stationary points of the problem $(\mathcal{P})$. Our aim is to show that this “complete” equivalence can be achieved under assumptions that are very similar to the ones in Theorem 1. To this end, let us recall a natural extension of the definition of stationary point to the case of functions defined on metric spaces (see [11, 15]).

Let $g\colon X\to\mathbb{R}\cup\{+\infty\}$ be a given function, and $K$ be a nonempty set. A point $x\in K\cap\operatorname{dom}g$ is called an inf-stationary point of the function $g$ on the set $K$, if $g^{\downarrow}_{K}(x)\geq 0$. In the case when $X$ is a normed space, $K$ is convex, and $g$ is Fréchet differentiable at $x$ this condition is reduced to the standard necessary optimality condition

[TABLE]

Let us also note that if $(\mathcal{P})$ is a mathematical programming problem with equality and inequality constraints, and $\Phi_{\lambda}$ is the $\ell_{1}$ penalty function for this problem, then condition $\Phi_{\lambda}^{\downarrow}(x)\geq 0$ for some $\lambda>0$ and a feasible point $x$ is satisfied iff KKT optimality conditions hold true at $x$.

For any $\lambda\geq 0$ and $c\in\mathbb{R}$ denote $S_{\lambda}(c)=\{x\in A\mid\Phi_{\lambda}(x)<c\}$.

Theorem 2**.**

Let $X$ be a complete metric space, $A$ be closed, $\mathcal{I}$ and $\varphi$ be l.s.c. on $A$, and $\varphi$ be continuous at every point of the set $\Omega$. Suppose also that there exist $c>\mathcal{I}^{*}=\inf_{x\in\Omega}\mathcal{I}(x)$, $\lambda_{0}>0$, and $\delta>0$ such that

$\mathcal{I}$* is Lipschitz continuous on an open set containing the set $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$;* 2. 2.

there exists $a>0$ such that $\varphi^{\downarrow}_{A}(x)\leq-a$ for all $x\in S_{\lambda_{0}}(c)\cap(\Omega_{\delta}\setminus\Omega)$; 3. 3.

$\Phi_{\lambda_{0}}$* is bounded below on $A$.*

Then there exists $\lambda^{*}\geq 0$ such that for any $\lambda\geq\lambda^{*}$ the following statements hold true:

the optimal values of the problems $(\mathcal{P})$ and (1) coincide; 2. 2.

globally optimal solutions of the problems $(\mathcal{P})$ and (1) coincide; 3. 3.

$x^{*}\in S_{\lambda}(c)$* is a locally optimal solution of the penalised problem (1) iff $x^{*}$ is a locally optimal solution of the problem $(\mathcal{P})$;* 4. 4.

$x^{*}\in S_{\lambda}(c)$* is an inf-stationary point of $\Phi_{\lambda}$ on $A$ iff $x^{*}\in\Omega$, and it is an inf-stationary point of $\mathcal{I}$ on $\Omega$.*

A proof of Theorem 2 is given in Appendix A. If the penalty function $\Phi_{\lambda}$ satisfies the four statements of this theorem, then it is said to be completely exact on the set $S_{\lambda}(c)$.

Remark 2*.*

In the general case, under the assumptions of Theorem 2 nothing can be said about locally optimal solutions of the penalised problem (1)/inf-stationary points of $\Phi_{\lambda}$ on $A$ that do not belong to the set $S_{\lambda}(c)$. In order to ensure that the penalty function $\Phi_{\lambda}$ is completely exact on $A$ (i.e. when $c=+\infty$) one must suppose that the objective function $\mathcal{I}$ is globally Lipschitz continuous, and there exists $a>0$ such that $\varphi^{\downarrow}_{A}(x)\leq-a$ for all $x\in A\setminus\Omega$ (see [29, Section 3.3]).

Remark 3*.*

Let us note that the assumptions of Theorem 2 cannot be improved (see [29, Theorem 3.13]). On the other hand, the global exactness of the penalty function $\Phi_{\lambda}$ can be proved under weaker assumptions on the penalty term $\varphi$. See [28] for more details.

In the following section we utilise Theorem 2 and several auxiliary results on integral functionals and Nemytskii operators given in Appendix B to design exact penalty functions for free-endpoint optimal control problems.

3 Exact Penalty Functions for Free-Endpoint Optimal Control Problems

Consider the following optimal control problem:

[TABLE]

Here $x(t)\in\mathbb{R}^{d}$ is the system state at time $t$, $t\to u(t)\in\mathbb{R}^{m}$ is a control input, $f\colon\mathbb{R}^{d}\times\mathbb{R}^{m}\times[0,T]\to\mathbb{R}^{d}$, $\theta\colon\mathbb{R}^{d}\times\mathbb{R}^{m}\times[0,T]\to\mathbb{R}$, and $\zeta\colon\mathbb{R}^{d}\to\mathbb{R}$ are given functions, while $T>0$ and $x_{0}\in\mathbb{R}^{d}$ are fixed. We suppose that $x(\cdot)$ belongs to the space $W_{1,p}^{d}(0,T)$, which is the Cartesian product of $d$ copies of the Sobolev space $W^{1,p}(0,T)$, while the control inputs $u(\cdot)$ belong to a closed subset $U$ of the Cartesian product $L_{q}^{m}(0,T)$ of $m$ copies of $L^{q}(0,T)$. Here $1<p<+\infty$ and $1\leq q\leq+\infty$ (the cases $p=1$ and $p=+\infty$ are excluded to avoid differentiability issues and the use of subdifferentials). For any $r\in[1,+\infty]$ denote by $r^{\prime}\in[1,+\infty]$ the conjugate exponent of $r$, i.e. $1/r+1/r^{\prime}=1$. Also, for any differentiable function $g(x,u,t)$ the gradient of the function $x\mapsto g(x,u,t)$ is denoted by $\nabla_{x}g(x,u,t)$, and a similar notation is used for the gradient of the function $u\mapsto g(x,u,t)$.

As usual (see, e.g. [54]), we identify the Sobolev space $W^{1,p}(0,T)$ with the space consisting of all those absolutely continuous functions $x\colon[0,T]\to\mathbb{R}$ for which $\dot{x}\in L^{p}(0,T)$. The space $L_{q}^{m}(0,T)$ with $1\leq q<+\infty$ is equipped with the norm

[TABLE]

where $|\cdot|$ is the Euclidean norm, while the space $L_{\infty}^{m}(0,T)$ is equipped with the norm $\|u\|_{\infty}=\operatorname*{ess\,sup}_{t\in[0,T]}|u(t)|$. The Sobolev space $W_{1,p}^{d}(0,T)$ is endowed with the norm $\|x\|_{1,p}=\|x\|_{p}+\|\dot{x}\|_{p}$. Also, below we suppose that the Cartesian product $X\times Y$ of normed spaces $X$ and $Y$ is endowed with the norm $\|(x,y)\|=\|x\|_{X}+\|y\|_{Y}$.

Remark 4*.*

For the sake of completeness let us recall two basic facts about norms in Sobolev spaces (see [54]) that will be utilised below. From the equality $x(t)=x(0)+\int_{0}^{t}\dot{x}(\tau)\,d\tau$ it follows that

[TABLE]

where $\|x\|_{0}=|x(0)|+\|\dot{x}\|_{p}$. Hence with the use of the bounded inverse theorem one gets that the norms $\|\cdot\|_{1,p}$ and $\|\cdot\|_{0}$ are equivalent. Moreover, by applying Hölder’s inequality and the equality $x(t)=x(0)+\int_{0}^{t}\dot{x}(\tau)\,d\tau$ again one obtains that there exists $C>0$ such that $\|x\|_{\infty}\leq\max\{1,T^{1/p^{\prime}}\}\|x\|_{0}\leq C\|x\|_{1,p}$, which implies that any bounded set in $W^{d}_{1,p}(0,T)$ is also bounded in $L_{\infty}^{d}(0,T)$. Let us finally note that from the fact that the operator $A\colon L^{q}(0,T)\to C[0,T]$, $(Ax)(t)=\int_{0}^{t}x(\tau)\,d\tau$ is compact (which can be easily verified with the use of the Arzelà-Ascoli theorem) it follows that for any weakly converging sequence $\{x_{n}\}\subset W^{d}_{1,p}(0,T)$ there exists a subsequence $\{x_{n_{k}}\}$ strongly converging in $C[0,T]$. This result is a simple particular case of the Rellich-Kondrachov theorem (see [1, Theorem 6.2]).

Our aim is to reduce optimal control problem (4) to a variational one. To this end we consider the differential equation $\dot{x}(t)=f(x(t),u(t),t)$ as a constraint that we want to incorporate into a penalty function. Define $X=W_{1,p}^{d}(0,T)\times L_{q}^{m}(0,T)$ and

[TABLE]

where $F(x,u)=\dot{x}(\cdot)-f(x(\cdot),u(\cdot),\cdot)$. Note that the set $A$ is obviously closed. Problem (4) can be rewritten as follows:

[TABLE]

Formally introduce the penalty term

[TABLE]

Clearly, $M=\{(x,u)\in X\mid\varphi(x,u)=0\}$. Therefore one can consider the penalised problem

[TABLE]

Observe that this is a variational problem of the form:

[TABLE]

With the use of Theorem 2 we can prove that under some natural assumptions on the functions $\theta$, $f$ and $\zeta$ this variational problem is equivalent to problem (4), provided $\lambda>0$ large enough. This result allows one to apply methods for solving variational problems to find optimal solutions of free-endpoint optimal control problems.

Remark 5*.*

In most (if not all) optimal control problems appearing in applications, control inputs $u(\cdot)$ are bounded in $L_{\infty}^{m}(0,T)$, and the bounds are known in advance, which raises the question of why to consider the case $1\leq q<+\infty$. The reason behind this is as follows. Firstly, some authors consider optimal control problems with only $L^{2}$-bounded control inputs (see, e.g. [2]), and to apply our results to such problems one must consider the case $q=2$. Secondly, the case $q<+\infty$ does not exclude known bounds on control inputs, since one can define, e.g. $U=\{u\in L_{q}^{m}(0,T)\mid\|u\|_{\infty}\leq C\}$ for some $C>0$, even in the case $1\leq q<+\infty$. The reason to suppose $q<+\infty$ in this case is related to the analysis of numerical methods. Although “discretise-then-optimise”-type methods are prevalent, there exist some continuous methods for solving optimal control problems (see, e.g. [62, 63, 37]), some of which are based on the minimisation of the penalty function $\Phi_{\lambda}(x,u)$. These methods are usually formulated and analysed in the case $p=q=2$, i.e. in the Hilbert space setting, when one can utilise inner products. Therefore, to make the theory of exact penalty functions consistent with these methods one must consider the case $q=2$, even in the presence of known bounds on control inputs. Finally, from the mathematical standpoint it is important to analyse the general case $1\leq q\leq+\infty$ to understand the limitations of the general theory of exact penalty functions and, in particular, Theorem 2.

Recall that for any $c,\delta>0$ we define $S_{\lambda}(c)=\{(x,u)\in A\mid\Phi_{\lambda}(x,u)<c\}$ and $\Omega_{\delta}=\{(x,u)\in A\mid\varphi(x,u)<\delta\}$. Note that $\Omega_{\delta}$ consists of all those $(x,u)\in W^{d}_{1,p}(0,T)\times U$ that satisfy the perturbed equation

[TABLE]

for some $w\in L_{p}^{d}(0,T)$ with $\|w\|_{p}<\delta$. Let $\mathcal{I}^{*}$ be the optimal value of problem (4). We also need the following definition to conveniently formulate assumptions on the functions $\theta$ and $f$.

Definition 2**.**

Let $g\colon\mathbb{R}^{d}\times\mathbb{R}^{m}\times[0,T]\to\mathbb{R}^{k}$ be a given function. We say that $g$ satisfies the growth condition of order $(l,s)$ with $0\leq l<+\infty$ and $1\leq s\leq+\infty$, if for any $R>0$ there exist $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{s}(0,T)$ such that $|g(x,u,t)|\leq C_{R}|u|^{l}+\omega_{R}(t)$ for a.e. $t\in[0,T]$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|\leq R$.

With the use of several auxiliary results on integral functionals and Nemytskii operators from Appendix B we can prove the following theorem, which provides conditions under which free-endpoint optimal control problem (4) and variational problem (6) are equivalent.

Theorem 3**.**

Let the following assumptions be valid:

$\zeta$* is locally Lipschitz continuous, $\theta$ is continuous and differentiable in $x$ and $u$, $f$ is continuous and differentiable in $x$, and the functions $\nabla_{x}\theta$, $\nabla_{u}\theta$ and $\nabla_{x}f$ are continuous;* 2. 2.

either $q=+\infty$ or the functions $\theta$ and $\nabla_{x}\theta$ satisfy the growth condition of order $(q,1)$, the function $\nabla_{u}\theta$ satisfies the growth condition of order $(q-1,q^{\prime})$, and the functions $f$ and $\nabla_{x}f$ satisfy the growth condition of order $(q/p,p)$; 3. 3.

there exists a globally optimal solution of problem (4); 4. 4.

there exist $\lambda_{0}>0$, $c>\mathcal{I}^{*}$, and $\delta>0$ such that the set $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ is bounded in $W^{d}_{1,p}(0,T)\times L_{q}^{m}(0,T)$, and the function $\Phi_{\lambda_{0}}(x,u)$ is bounded below on $A$.

Then there exists $\lambda^{*}\geq 0$ such that for any $\lambda\geq\lambda^{*}$ the penalty function $\Phi_{\lambda}$ for problem (4) is completely exact on $S_{\lambda}(c)$.

Proof.

Our aim is to apply Theorem 2. By Proposition 3 below the growth condition on $\theta$ ensures that the functional $\mathcal{I}(x,u)$ is correctly defined and finite for any $(x,u)\in X$. In turn, the growth conditions on $\nabla_{x}\theta$ and $\nabla_{u}\theta$ guarantee that the mapping $(x,u)\mapsto\int_{0}^{T}\theta(x(t),u(t),t)\,dt$ is Lipschitz continuous on any bounded subset of $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ by Proposition 4. From Remark 4 it follows that any bounded subset of $X$ is bounded in $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$. Therefore by applying the fact that $\zeta$ is locally Lipschitz continuous one obtains that the functional $\mathcal{I}(x,u)$ is correctly defined and Lipschitz continuous on any bounded subset of $X$ (in particular, on any bounded open set containing $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$; recall that $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ is bounded by our assumption). Finally, by applying the growth condition on the function $f$ and Proposition 5 one gets that the penalty term $\varphi(x,u)$ is correctly defined and continuous on $X$. Let us check that for any bounded set $K\subset A$ there exists $a>0$ such that $\varphi^{\downarrow}_{A}(x,u)\leq-a$ for any $(x,u)\in K\setminus\Omega$. Then by applying Theorem 2 one gets the desired result.

Define $A_{s}=\{x\in W^{d}_{1,p}(0,T)\mid x(0)=x_{0}\}$, and for any $u\in U$ introduce the function $\varphi_{u}(x)=\varphi(x,u)$. Observe that $\varphi^{\downarrow}_{A}(x,u)\leq(\varphi_{u})^{\downarrow}_{A_{s}}(x)$ for any $(x,u)\in A$ due to the fact that $A=A_{s}\times U$. Therefore, it is sufficient to check that for any bounded set $K\subset A$ there exists $a>0$ such that $(\varphi_{u})^{\downarrow}_{A_{s}}(x)\leq-a$ for all $(x,u)\in K\setminus\Omega$, i.e. for all $(x,u)\in K$ such that $\varphi_{u}(x)>0$. To simplify the computation of $(\varphi_{u})^{\downarrow}_{A_{s}}(x)$ we apply a change of variables called “transition into the space of derivatives” that was widely utilised by Demyanov in his works on exact penalty functions (see, e.g. [13, 14, 12]).

For any $z\in L_{p}^{d}(0,T)$ define $(Jz)(t)=x_{0}+\int_{0}^{t}z(\tau)d\tau$ for all $t\in[0,T]$ and $\gamma_{u}(z)=\varphi_{u}(Jz)$. From the Lebesgue differentiation theorem it follows that the operator $J$ is a one-to-one correspondence between $L_{p}^{d}(0,T)$ and $A_{s}$ (see [54]). Furthermore, by Hölder’s inequality one has $\|Jz-Jw\|_{1,p}\leq(1+T)\|z-w\|_{p}$ for any $z,w\in L_{p}^{d}(0,T)$. Consequently, if for some $x\in A_{s}$ and $u\in U$ one has $(\gamma_{u})^{\downarrow}(\dot{x})<0$, then

[TABLE]

Therefore, it is sufficient to check that for any bounded sets $Z\subset L_{p}^{d}(0,T)$ and $V\subseteq U$ there exists $a>0$ such that $(\gamma_{u})^{\downarrow}(z)\leq-a$ for all $(z,u)\in Z\times V$ such that $\gamma_{u}(z)>0$ (note that the transition into the space of derivatives allowed us to “remove” the constraint $x\in A_{s}$).

Fix any bounded sets $Z\subset L_{p}^{d}(0,T)$ and $V\subseteq U$. Introduce the function $H\colon\mathbb{R}^{d}\to\mathbb{R}^{d}$,

[TABLE]

and for any $(z,u)\in Z\times V$ define $G(z,u)(\cdot)=H(F(Jz,u)(\cdot))$, where, as above, $F(x,u)=\dot{x}(\cdot)-f(x(\cdot),u(\cdot),\cdot)$. It is easy to verify that $H$ is a continuous function (recall that $p>1$), which implies that that the function $G(z,u)(\cdot)$ is measurable. Furthermore, for any $x\in L_{p}^{d}(0,T)$ one has $H(x(\cdot))\in L_{p^{\prime}}^{d}(0,T)$, which by the growth condition on $f$ and Proposition 5 implies that $G(z,u)\in L_{p^{\prime}}^{d}(0,T)$ for any $(z,u)\in Z\times V$. Note also that

[TABLE]

provided $\gamma_{u}(z)>0$. Here $\langle\cdot,\cdot\rangle$ is the inner product in $\mathbb{R}^{d}$.

Taking into account the growth condition on $\nabla_{x}f$ and Proposition 6 (see Appendix B) one gets that the mapping $F_{u}(x)=F(x,u)$ is Gâteaux differentiable for any $(x,u)\in X$, and its Gâteaux derivative has the form

[TABLE]

Hence by applying the fact that the norm $\|\cdot\|_{p}$ is Fréchet differentiable (this fact follows, e.g. from [26, Theorem 2.2.1]) and the chain rule one obtains that for any $(z,u)\in Z\times V$ such that $\gamma_{u}(z)>0$ the function $\gamma_{u}$ is Gâteaux differentiable at $z$, and

[TABLE]

(here $x=Jz$). Integrating by parts one obtains that

[TABLE]

where $w(x,u)=\gamma_{u}(z)^{1-p}G(x,u)$. Consequently, taking into account the fact that $\|w(x,u)\|_{p^{\prime}}=1$ one gets that

[TABLE]

where $I$ is the identity operator and

[TABLE]

From the facts that $\nabla_{x}f$ satisfies the growth condition of order $(q/p,p)$, and the sets $Z$ and $V$ are bounded it follows that the set $\{\nabla_{x}f(Jz(\cdot),u(\cdot),\cdot)\mid(z,u)\in Z\times V\}$ of kernels of the integral operators $\mathcal{K}_{y}$ is bounded in $L_{p}^{d\times d}(0,T)$. Hence by applying Lemma 3 on the resolvent of a Volterra-type integral equation one obtains that there exists $a>0$ such that $\|(I-\mathcal{K}_{y})^{-1}\|<1/a$, i.e. $(\gamma_{u})^{\downarrow}(z)\leq-a$, for any $(z,u)\in Z\times V$, and the proof is complete. ∎

Remark 6*.*

Let $1<q<+\infty$, the set $U$ be convex (or, more generally, weakly closed), and the following assumptions be valid:

1. (i)

$f(x,u,t)=f_{0}(x,t)+g(x,t)u$, where the functions $f_{0}$ and $g$ are continuous; 2. (ii)

$\theta(x,u,t)$ is convex in $u$ for all $x\in\mathbb{R}^{d}$ and $t\in[0,T]$.

Then under assumptions 1, 2, and 4 of Theorem 3 a globally optimal solution of problem (4) exists iff there exists a feasible point of this problem, i.e. iff there exists $u\in U$ such that there exists an absolutely continuous solution of the differential equation $\dot{x}=f(x,u,t)$ with $x(0)=x_{0}$ defined on $[0,T]$.

Indeed, if a feasible point exists, then the sublevel set $\{(x,u)\in\Omega\mid\mathcal{I}(x,u)<c\}\subset S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ is nonempty and bounded. Let $\{(x_{n},u_{n})\}\subset\Omega$ be a sequence such that $\mathcal{I}(x_{n},u_{n})\to\mathcal{I}^{*}$ as $n\to\infty$. Since $c>\mathcal{I}^{*}$, the sequence $\{(x_{n},u_{n})\}$ is bounded, which due to the reflexivity of $L_{q}^{m}(0,T)$ and $W^{d}_{1,p}(0,T)$ for $1<q,p<+\infty$ implies that one can extract a subsequence $\{(x_{n_{k}},u_{n_{k}})\}$ weakly converging to some $(x^{*},u^{*})$. Note that $u^{*}\in U$, since $U$ is weakly closed. Furthermore, by Remark 4 one can suppose that $x_{n_{k}}$ converges to $x^{*}$ uniformly on $[0,T]$. Hence by applying assumption (i) one can easily check that $(x^{*},u^{*})$ is a feasible point of problem (4), while assumption (ii) ensures that $\mathcal{I}(x^{*},u^{*})\leq\liminf_{k\to\infty}\mathcal{I}(x_{n_{k}},u_{n_{k}})=\mathcal{I}^{*}$ (see [38, Section 7.3.2] and [46]), which implies that $(x^{*},u^{*})$ is a globally optimal solution of problem (4).

Note finally that the existence of a feasible point of problem (4) can be proved with the use of various standard results from the theory of differential equations. For example, it exists, if there exist $u\in U$, $C>0$, and a.e. nonnegative function $\omega\in L^{1}(0,T)$ such that $|f(x,u(t),t)|\leq C|x|+\omega(t)$ for a.e. $t\in(0,T)$ and for all $x\in\mathbb{R}^{d}$ (cf. Proposition 1 and Remark 7 below).

Let us also point out several simple ways to verify the boundedness of the set $S_{\lambda}(c)\cap\Omega_{\delta}$ and the boundedness below of $\Phi_{\lambda}(x,u)$ on the set $A$. One can utilise a combination of these ways and a structure of the problem in order to verify the boundedness conditions for particular optimal control problems.

Proposition 1**.**

Let $\theta$ and $f$ be continuous, and one of the following assumptions be valid:

the set $U$ is bounded in $L_{\infty}^{m}(0,T)$, and for any $R>0$ there exist $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{1}(0,T)$ such that

[TABLE]

for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|u|\leq R$ and for a.e. $t\in(0,T)$, and there exist $K_{1},K_{2}\geq 0$ such that $\zeta(x)\geq-K_{1}|x|-K_{2}$ for all $x\in\mathbb{R}^{d}$; 2. 2.

$1\leq q<+\infty$, there exist $C_{f}>0$ and a.e. nonnegative function $\omega_{f}\in L^{p}(0,T)$ such that

[TABLE]

for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ and a.e. $t\in(0,T)$, and one of the two following assumptions is valid:

1. (a)

$U$* is bounded in $L_{q}^{m}(0,T)$, and there exist $C,K_{1},K_{2}>0$, and an a.e. nonnegative function $\omega\in L^{1}(0,T)$ such that for all $(x,u)$ and a.e. $t\in(0,T)$ one has $\theta(x,u,t)\geq-C(|x|+|u|^{q})-\omega(t)$ and $\zeta(x)\geq-K_{1}|x|-K_{2}$;* 2. (b)

$\zeta$* is bounded below, and there exist $C>0$ and $\omega\in L^{1}(0,T)$ such that $\theta(x,u,t)\geq C|u|^{q}+\omega(t)$ for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ and a.e. $t\in(0,T)$.* 3. 3.

$1\leq q<+\infty$, $\zeta$ is bounded below, there exist $s\geq r\geq 1$, $C>0$, and an a.e. nonnegative function $\omega\in L^{p}(0,T)$ such that

[TABLE]

for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ and a.e. $t\in(0,T)$.

Then there exists $\lambda_{0}\geq 0$ such that for any $c\in\mathbb{R}$, $\delta>0$, and $\lambda\geq\lambda_{0}$ the set $S_{\lambda}(c)\cap\Omega_{\delta}$ is bounded, and the function $\Phi_{\lambda}$ is bounded below on $A$. Furthermore, if either assumption 7 or assumption 2 is satisfied, then there exists a feasible point of problem (4).

Proof.

Part 7. Fix $\delta>0$. By definition for any $(x,u)\in\Omega_{\delta}$ one has $\|F(x,u)\|<\delta$. Hence there exists $w\in L^{d}_{p}(0,T)$ with $\|w\|_{p}<\delta$ such that

[TABLE]

or, equivalently,

[TABLE]

Since $U$ is bounded in $L_{\infty}^{m}(0,T)$, there exists $R>0$ such that for all $u\in U$ one has $|u(t)|\leq R$ for a.e. $t\in(0,T)$. Consequently, there exists $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{1}(0,T)$ such that for any $(x,u)\in\Omega_{\delta}$ one has

[TABLE]

for some $w\in L^{d}_{p}(0,T)$ with $\|w\|_{p}<\delta$. By applying the Grönwall-Bellman inequality one obtains that

[TABLE]

where $\alpha(t)=|x_{0}|+\int_{0}^{t}(\omega_{R}(\tau)+|w(\tau)|)\,d\tau$. With the use of Hölder’s inequality one gets that

[TABLE]

Therefore, the set $\Omega_{\delta}$ is bounded in $L_{\infty}^{d}(0,T)\times L_{\infty}^{m}(0,T)$, which implies that there exists $C>0$ such that $|f(x(t),u(t),t)|\leq C$ for a.e. $t\in[0,T]$ and for all $(x,u)\in\Omega_{\delta}$. Hence and from (10) it follows that for all $(x,u)\in\Omega_{\delta}$ one has

[TABLE]

Integrating this inequality from [math] to $T$ and taking into account the fact that $\|w\|_{p}<\delta$ one obtains that $\Omega_{\delta}$ is bounded in $X$. Thus, $S_{\lambda}(c)\cap\Omega_{\delta}$ is bounded in $X$ for any $\lambda\geq 0$, $c\in\mathbb{R}$ and $\delta>0$.

Fix $(x,u)\in A$, and let $\delta=\varphi(x,u)$. From (13) and (14) it follows that there exist $C_{1},C_{2}>0$ depending only on $C_{R}$, $\omega_{R}$ and $T$ such that $\|x\|_{\infty}\leq C_{1}+C_{2}\delta$. By applying the second inequality in (7) one obtains that

[TABLE]

for any $\lambda\geq C_{2}(TC_{R}+K_{1})$. Consequently, the penalty function $\Phi_{\lambda}$ is bounded below on $A$ for any $\lambda\geq C_{2}(TC_{R}+K_{1})$.

Let us now prove the existence of a feasible point. Fix $u\in U$. From the fact that $f$ is continuous it follows that a solution $x(\cdot)$ of (10) with $w(\cdot)\equiv 0$ and $x(0)=x_{0}$ is defined at least on some subinterval $[0,T_{0})$ of $[0,T]$. By applying (12) and the Grönwall-Bellman inequality one can easily check that $T_{0}=T$, and $x(\cdot)$ is bounded on $[0,T)$. Furthermore, from the continuity of $f$ it obviously follows that $x\in W^{d}_{1,\infty}(0,T)$, which implies that $(x,u)$ is a feasible point of problem (4).

Part 2. Fix $c\in\mathbb{R}$ and $\delta>0$. By applying either the boundedness of the set $U$ in $L_{q}^{m}(0,T)$ or the boundedness below of $\zeta$ and the inequalities $\theta(x,u,t)\geq C|u|^{q}+\omega(t)$ and $\Phi_{\lambda}(\cdot)\geq\mathcal{I}(\cdot)$ one obtains that there exists $K>0$ such that $\|u\|_{q}\leq K$ for all $(x,u)\in S_{\lambda}(c)$ and any $\lambda\geq 0$.

As was pointed out above, for any $(x,u)\in\Omega_{\delta}$ there exists $w\in L^{d}_{p}(0,T)$ with $\|w\|_{p}<\delta$ such that (11) holds true. By applying (8) one gets that

[TABLE]

for any $t\in[0,T]$. Hence with the use of the Grönwall-Bellman and Hölder’s inequalities one obtains that

[TABLE]

for any $(x,u)\in S_{\lambda}(c)\cap\Omega_{\delta}$, which implies that the set $S_{\lambda}(c)\cap\Omega_{\delta}$ is bounded in $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ for any $\lambda\geq 0$. Hence by applying (8), (10), and Hölder’s inequality one can easily check that this set is bounded in $X$ for any $\lambda\geq 0$.

If $\zeta$ is bounded below and $\theta(x,u,t)\geq C|u|^{q}+\omega(t)$, then the boundedness below of the penalty function $\Phi_{\lambda}$ follows from the inequality $\Phi_{\lambda}(\cdot)\geq\mathcal{I}(\cdot)$. On the other hand, if $U$ is bounded, and the inequalities $\theta(x,u,t)\geq-C(|x|+|u|^{q})-\omega(t)$ and $\zeta(x)\geq-K_{1}|x|-K_{2}$ are satisfied, then the boundedness below of the penalty function $\Phi_{\lambda}$ can be proved in the same way as in part 7 of the proposition.

Finally, let us prove the existence of a feasible point. Fix any $u\in U$. From the continuity of $f$ and (8) it follows that the function $(t,x)\mapsto f(x,u(t),t)$ satisfies the Carathéodory condition, and $|f(x,u(t),t)|\leq C_{f}|x|+\eta(t)$ for all $x\in\mathbb{R}^{d}$ and a.e. $t\in(0,T)$, where $\eta(\cdot)=C_{f}|u(\cdot)|^{q/p}+\omega_{f}(\cdot)\in L^{p}(0,T)$. Consequently, there exists an absolutely continuous solution $x(\cdot)$ of (10) with $w(\cdot)\equiv 0$ and $x(0)=x_{0}$ defined at least on some subinterval $[0,T_{0})$ of $[0,T]$. By applying (15) and the Grönwall-Bellman inequality one obtains that $T_{0}=T$, and $x(\cdot)$ is bounded on $[0,T)$. Hence and from (8) and (10) with $w(\cdot)\equiv 0$ it follows that $x\in W^{d}_{1,p}(0,T)$, which implies that $(x,u)$ is a feasible point of problem (4).

Part 9. Fix $c\in\mathbb{R}$ and $\delta>0$. From the second inequality in (9) and the fact that $\zeta$ is bounded below it obviously follows that for any $\lambda\geq 0$ the penalty function $\Phi_{\lambda}$ is bounded below on $A$, and the set $S_{\lambda}(c)$ is bounded in $L_{s}^{d}(0,T)\times L_{q}^{m}(0,T)$.

By applying (11) and the first inequality in (9) one obtains that for any $(x,u)\in\Omega_{\delta}$ there exists $w\in L^{d}_{p}(0,T)$ with $\|w\|_{p}<\delta$ such that

[TABLE]

for any $t\in(0,T)$. Hence with the use of Hölder’s inequality and the fact that $s\geq r$ one gets that there exists $K>0$ such that $\|x\|_{\infty}\leq K$ for any $(x,u)\in S_{\lambda}(c)\cap\Omega_{\delta}$. Therefore, by applying (10) and the first inequality in (9) one obtains that

[TABLE]

for a.e. $t\in(0,T)$, where $(x,u)\in S_{\lambda}(c)\cap\Omega_{\delta}$ and $\|w\|_{p}<\delta$. Consequently, taking into account the facts that $\omega\in L^{p}(0,T)$ and the set $(x,u)\in S_{\lambda}(c)\cap\Omega_{\delta}$ is bounded in $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$, one obtains that there exists $R>0$ such that $\|\dot{x}\|_{p}\leq R$ for any $S_{\lambda}(c)\cap\Omega_{\delta}$, i.e. this set is bounded in $X$. ∎

Remark 7*.*

Let us note that the assumptions of the first two parts of the proposition above can be relaxed. For example, let $R=\sup\{\|u\|_{\infty}\mid u\in U\}$, and suppose that instead of the first inequality in (7) the inequality $|f(x,u,t)|\leq\eta_{R}(|x|)+\omega_{R}(t)$ holds true for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|u|\leq R$ and for a.e. $t\in(0,T)$, where $\omega_{R}\in L^{1}(0,T)$, and $\eta_{R}\colon[0,+\infty)\to[0,+\infty)$ is a continuous non-decreasing function such that $\eta_{R}(s)>0$ for any $s>0$. Then arguing in the same way as in the proof of Proposition 1, but utilising the Bihari-LaSalle inequality instead of the Grönwall-Bellman inequality one can easily verify that the set $\Omega_{\delta}$ is bounded, provided $T>0$ satisfies the assumptions of the Bihari-LaSalle inequality. However, to ensure the boundedness below of the penalty function $\Phi_{\lambda}$ in this case one must suppose that both functions $\theta$ and $\zeta$ are bounded below.

Remark 8*.*

Proposition 1 demonstrates how one can prove the boundedness of the set $S_{\lambda}(c)\cap\Omega_{\delta}$ and the boundedness below of the function $\Phi_{\lambda}$ using various information about the functions $f$, $\theta$, $\zeta$, and the set of admissible control inputs $U$. In the first part of this proposition we suppose that the set $U$ is bounded in $L_{\infty}^{m}(0,T)$, which, in essence, allows us not to impose any assumptions on the behaviour of the functions $u\mapsto f(x,u,t)$ and $u\mapsto\theta(x,u,t)$. If $U$ is bounded only in $L_{q}^{m}(0,T)$, then appropriate growth conditions on these functions must be imposed to prove the required result (see part two of Proposition 1). Finally, if no information about the set $U$ is available, one must impose some assumptions that ensure the coercivity of the functional $\mathcal{I}(x,u)$, as it is done in the last part of Proposition 1.

4 Exact Penalty Functions for Free-Endpoint Differential Inclusions

Let us extend the main results of the previous section to the case of variational problems involving differential inclusions. Consider the following problem:

[TABLE]

Here $\theta\colon\mathbb{R}^{d}\times[0,T]\to\mathbb{R}$ and $\zeta\colon\mathbb{R}^{d}\to\mathbb{R}$ are given functions, $F\colon\mathbb{R}^{d}\times[0,T]\rightrightarrows\mathbb{R}^{d}$ is a set-valued mapping with nonempty compact convex values, $T>0$ and $x_{0}\in\mathbb{R}^{d}$ are fixed, and $x\in W^{d}_{1,p}(0,T)$.

Remark 9*.*

Although problem (16) does not include control inputs, it encompasses many optimal control problems, including some optimal feedback control problems, which cannot be tackled with the use of the approach presented in the previous section. For example, the system

[TABLE]

where $U\colon\mathbb{R}^{d}\times[0,T]\rightrightarrows\mathbb{R}^{m}$ is a multifunction, can be rewritten as the differential inclusion $\dot{x}(t)\in F(x(t),t)$ with $F(x,t)=f(x,U(x,t),t)$, which allows one to reduce optimal control problems involving such systems to problem (16). Thus, the results of this section have many direct applications to various optimal control problems.

Let us introduce a penalty function for problem (16). Define $X=W^{d}_{1,p}(0,T)$, and put

[TABLE]

and $A=\{x\in X|x(0)=x_{0}\}$ (note that this set is obviously closed). Then problem (16) can be rewritten as follows:

[TABLE]

In order to introduce a penalty term $\varphi(x)$, denote $S=\{\psi\in\mathbb{R}^{d}\mid|\psi|=1\}$, and for any convex set $Y\subset\mathbb{R}^{d}$ and $\psi\in\mathbb{R}^{d}$ denote by $s(Y,\psi)=\sup_{y\in Y}\langle y,\psi\rangle$ the support function of $Y$. By [67, Theorem 13.1] a function $x\in X$ satisfies the differential inclusion $\dot{x}(t)\in F(x(t),t)$ iff

[TABLE]

or equivalently iff $h(x(t),\dot{x}(t),t)=0$ for a.e. $t\in[0,T]$, where

[TABLE]

Note that the maximum over all $\psi\in S$ in the definition of $h(x,z,t)$ is achieved, since the mapping $\psi\to s(F(x,t),\psi)$ is continuous, which, in turn, follows from the fact that $F(x,t)$ is a compact set.

Observe that $h(x,z,t)=\operatorname{dist}(z,F(x,t))$ for all $z,x\in\mathbb{R}^{d}$ and $t\in[0,T]$. Indeed, for any $\psi\in S$ and $y\in F(x,t)$ one has

[TABLE]

Taking the minimum over all $y\in F(x,t)$ one gets that

[TABLE]

Consequently, $\max\{0,\langle z,\psi\rangle-s(F(x,t),\psi)\}\leq\operatorname{dist}(z,F(x,t))$ by virtue of the fact that $\operatorname{dist}(z,F(x,t))\geq 0$. Hence taking the maximum over all $\psi\in S$ one obtains that $h(x,z,t)\leq\operatorname{dist}(z,F(x,t))$. Clearly, this inequality turns into an equality when $z\in F(x,t)$. Moreover, in the case $z\notin F(x,t)$ from the necessary conditions for a minimum (3) with $g(x)=|x|$ and $K=\{z-y\mid y\in F(x,t)\}$ it follows that for $\psi^{*}=(z-y^{*})/|z-y^{*}|$, where $y^{*}\in F(x,t)$ is such that $\operatorname{dist}(z,F(x,t))=|z-y^{*}|$, one has

[TABLE]

Taking the maximum over all $y\in F(x,t)$ one gets that $s(F(x,t),\psi^{*})-\langle z,\psi^{*}\rangle\leq-\operatorname{dist}(z,F(x,t))$, which obviously implies that $h(x,z,t)\geq\operatorname{dist}(z,F(x,t))$. Thus, one has $h(x,z,t)=\operatorname{dist}(z,F(x,t))$ for all $z,x\in\mathbb{R}^{d}$ and $t\in[0,T]$.

Now one can formally define

[TABLE]

Clearly, $M=\{x\in X\mid\varphi(x)=0\}$, which implies that one can consider the penalised problem

[TABLE]

Our aim is to provide sufficient conditions for the penalty function $\Phi_{\lambda}$ to be completely exact. For the sake of simplicity, below we analyse only the simplest case when the support function $s(F(x,t),\psi)$ is differentiable in $x$ for all $x\in\mathbb{R}^{d}$, $\psi\in S$, and $t\in[0,T]$.

Before we can proceed to the theorem on the exactness of $\Phi_{\lambda}$, we need to obtain an auxiliary result on the differentiability of the penalty term $\varphi$. Denote by $\psi^{*}(x,z,t)$ a vector $\psi\in S$ at which the maximum in the definition of $h(x,z,t)$ is attained in the case $h(x,z,t)>0$, and define $\psi^{*}(x,z,t)=\psi_{0}$ otherwise, where $\psi_{0}\in S$ is a fixed vector. Note that in the case $h(x,z,t)>0$ such $\psi^{*}(x,z,t)$ is unique. Indeed, if the maximum is attained for $\psi=\psi_{1}\in S$ and $\psi=\psi_{2}\in S$ with $\psi_{1}\neq\psi_{2}$, then by applying the fact that the function $h_{0}(\psi)=\langle z,\psi\rangle-s(F(x,t),\psi)$ is concave one obtains that $h_{0}(\xi)\geq 0.5h_{0}(\psi_{1})+0.5h_{0}(\psi_{2})$, where $\xi=0.5\psi_{1}+0.5\psi_{2}$. Note that $h_{0}(\psi_{1})=h_{0}(\psi_{2})=h(x,z,t)$ by the fact that $h(x,z,t)>0$; furthermore, $|\xi|<1$, since the space $\mathbb{R}^{d}$ endowed with the Euclidean norm is strictly convex. Hence taking into account the fact that the function $h_{0}$ is positively homogeneous of degree one we obtain that

[TABLE]

which is impossible. Thus, $\psi^{*}(x,z,t)$ is well-defined in the case $h(x,z,t)>0$.

Proposition 2**.**

Let the multifunction $F(x,t)$ be continuous on $\mathbb{R}^{d}\times[0,T]$, its support function $s(F(x,t),\psi)$ be differentiable in $x$, and the function $(x,t,\psi)\mapsto\nabla_{x}s(F(x,t),\psi)$ be continuous on $\mathbb{R}^{d}\times[0,T]\times S$. Then the penalty term $\varphi(x)$ is correctly defined and finite on $X$, Gâteaux differentiable at every point $x\in W^{d}_{1,p}(0,T)$ such that $\varphi(x)>0$, and

[TABLE]

where $\psi^{*}(x,t)=\psi^{*}(x(t),\dot{x}(t),t)$ for a.e. $t\in[0,T]$.

Proof.

From the continuity of $F(x,t)$ it obviously follows that the function $h(x,z,t)$ is continuous (see (18)), which implies that the function $h(x(\cdot),\dot{x}(\cdot),\cdot)$ is measurable for any $x\in W^{d}_{1,p}(0,T)$. Hence by applying the inequality

[TABLE]

one gets that the penalty term $\varphi(x)$ is correctly defined and finite on $X$. Let us check that the functional

[TABLE]

is Gâteaux differentiable and compute its derivative. Then by applying the chain rule one obtains the required result.

Bearing in mind the facts that the support function $s(F(x,t),\psi)$ is differentiable in $x$, and the function $(x,t,\psi)\mapsto\nabla_{x}s(F(x,t),\psi)$ is continuous, and utilising a generalisation of the Danskin-Demyanov theorem (see [8, Theorem 4.13]) one obtains that the function $h^{p}(x,z,t)$ is Gâteaux differentiable in $x$ and $z$ at any point $(x,z,t)$ such that $h(x,z,t)>0$, and

[TABLE]

Let us consider the case $h(x,z,t)=0$. Note that the function $x\mapsto s(F(x,t),\psi)$ is locally Lipschitz continuous with the same Lipschitz constant for all $t\in[0,T]$ and $\psi\in S$, since its derivative in $x$ is continuous. Hence, as it is easy to check, the function $(x,z)\to h(x,z,t)$ is locally Lipschitz continuous with the same Lipschitz constant for any $t\in[0,T]$. Utilising this fact and the inequality $p>1$ one obtains that if $(x,z,t)$ is such that $h(x,z,t)=0$, then the function $h^{p}(x,z,t)$ is Gâteaux differentiable in $x$ and $z$ as well, and $\nabla_{x}h^{p}(x,z,t)=0$ and $\nabla_{z}h^{p}(x,z,t)=0$, since

[TABLE]

for any $\Delta x$ and $\Delta z$ in a neighbourhood of zero, and for some $L>0$. Moreover, with the use of the facts that the mapping $\psi^{*}(\cdot)$ is continuous on the open set $\{(x,z,t)\mid h(x,z,t)>0\}$ by [8, Proposition 4.4] and $|\psi^{*}(\cdot)|\equiv 1$ one can easily check that the functions $\nabla_{x}h^{p}(\cdot)$ and $\nabla_{z}h^{p}(\cdot)$ are continuous on $\mathbb{R}^{d}\times\mathbb{R}^{d}\times[0,T]$.

Fix $x,v\in W^{d}_{1,p}(0,T)$. By the mean value theorem for any $\alpha\in(0,1]$ and for a.e. $t\in(0,T)$ there exists $\alpha(t)\in(0,\alpha)$ such that

[TABLE]

for a.e. $t\in[0,T]$. The right hand side of this equality converges to

[TABLE]

as $\alpha\to+0$ for a.e. $t\in[0,T]$ due to the continuity of $\nabla_{x}h^{p}(\cdot)$ and $\nabla_{z}h^{p}(\cdot)$. Taking into account the obvious inequalities

[TABLE]

and the fact that $\|x\|_{\infty}\leq C\|x\|_{1,d}$ for some $C>0$ (see Remark 4) one obtains that there exist $C_{1},C_{2}>0$ such that

[TABLE]

and

[TABLE]

for a.e. $t\in[0,T]$ and for any $\alpha\in(0,1]$. Note that the right-hand sides of these inequalities belong to $L^{1}(0,T)$. Therefore, by applying Lebesgue’s dominated convergence theorem one gets that

[TABLE]

which along with (19) completes the proof. ∎

Let $\mathcal{I}^{*}$ be the optimal value of problem (16).

Theorem 4**.**

Let the following assumptions be valid:

$\zeta$* is locally Lipschitz continuous, $\theta(x,t)$ is continuous, differentiable in $x$, and its derivative in $x$ is continuous;* 2. 2.

the multifunction $F(x,t)$ is continuous, its support function $s(F(x,t),\psi)$ is differentiable in $x$, and the function $(x,t,\psi)\mapsto\nabla_{x}s(F(x,t),\psi)$ is continuous; 3. 3.

there exists a globally optimal solution of problem (16); 4. 4.

there exist $\lambda_{0}>0$, $\delta>0$, and $c>\mathcal{I}^{*}$ such that the set $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ is bounded in $W^{d}_{1,p}(0,T)$, and the function $\Phi_{\lambda_{0}}(x)$ is bounded below on $A$.

Then there exists $\lambda^{*}\geq 0$ such that for any $\lambda\geq\lambda^{*}$ the penalty function $\Phi_{\lambda}$ for problem (16) is completely exact on $S_{\lambda}(c)$.

Proof.

Note that the functional $\mathcal{I}(x)$ is Lipschitz continuous on any bounded subset of $W^{d}_{1,p}(0,T)$ by Proposition 4, Remark 4, and the fact that $\zeta$ is locally Lipschitz continuous. Furthermore, taking into account the obvious estimate

[TABLE]

and utilising Vitali’s convergence theorem (see [30, Theorem III.6.15]) one can easily check that the operator $x\to h(x(\cdot),\dot{x}(\cdot),\cdot)$ continuously maps $W^{d}_{1,p}(0,T)$ to $L^{p}(0,T)$ (cf. [66, Theorem 5.1]), i.e. the penalty term $\varphi$ is continuous. Thus, by Theorem 2 it remains to check that for any bounded set $K\subset A$ there exists $a>0$ such that $\varphi^{\downarrow}_{A}(x)\leq-a$ for any $x\in K\setminus\Omega$.

By applying the technique of “transition into the space of derivatives” as in the proof of Theorem 3 one obtains that it is sufficient to check that for any bounded set $Z\subset L^{d}_{p}(0,T)$ there exists $a>0$ such that $\gamma^{\downarrow}(z)\leq-a$ for all $z\in Z$ with $\gamma(z)>0$, where $\gamma(z)=\varphi(Jz)$ and $(Jz)(t)=x_{0}+\int_{0}^{t}z(\tau)\,d\tau$. Let $Z\subset L^{d}_{p}(0,T)$ be a bounded set. Utilising Proposition 2 and integrating by parts one gets that for any $z\in Z$ such that $\gamma(z)>0$ the functional $\gamma(\cdot)$ is Gâteaux differentiable at $z$, and

[TABLE]

where $x=Jz$ and $w(z)(t)=\gamma(z)^{1-p}h(x(t),z(t),t)^{p-1}$. Therefore, for any such $z$ one has $\gamma^{\downarrow}(z)\leq-\|\gamma^{\prime}(z)\|=-\|H(z)\|_{p^{\prime}}\leq 0$, where

[TABLE]

Arguing by reductio ad absurdum, suppose that there exists a sequence $\{z_{n}\}\subset Z$ such that $\gamma(z_{n})>0$ for all $n\in\mathbb{N}$, and $\gamma^{\downarrow}(z_{n})\to 0$ as $n\to\infty$. Then $\|H(z_{n})\|_{p^{\prime}}\to 0$ as $n\to\infty$ as well.

For any $n\in\mathbb{N}$ define $\psi_{n}(\cdot)=\psi^{*}(x_{n},\cdot)=\psi^{*}(x_{n}(\cdot),z_{n}(\cdot),\cdot)$, where $x_{n}=Jz_{n}$. Let us check that these functions are measurable. From the continuity of the multifunction $F(x,t)$ it follows that the function $h(x,z,t)$ is continuous as well. Therefore, the function $h(x_{n}(\cdot),z_{n}(\cdot),\cdot)$ is measurable, which implies that the set $E_{n}=\{t\in[0,T]\mid h(x_{n}(t),z_{n}(t),t)>0\}$ is measurable. As was pointed out above, the function $\psi^{*}(x,z,t)$ is continuous on the open set $V=\{(x,z,t)\mid h(x,z,t)>0\}$ by [8, Proposition 4.4]. Consequently, the function $E_{n}\ni t\mapsto\psi^{*}(x_{n}(t),z_{n}(t),t)$ is measurable as the composition of the restriction of $\psi^{*}(x,z,t)$ to $V$ and the measurable mapping $E_{n}\ni t\to(x_{n}(t),z_{n}(t),t)$. Hence one obtains that

[TABLE]

is measurable (recall that $\psi_{0}\in S$ is a fixed vector; see the discussion before Proposition 2).

Recall that $\|H(z_{n})\|_{p^{\prime}}\to 0$ as $n\to\infty$. Hence and from the fact that by definition $|\psi_{n}(\cdot)|\equiv 1$ it follows that $\|\langle\psi_{n},H(z_{n})\rangle\|_{p^{\prime}}\to 0$ as $n\to\infty$ as well. On the other hand, from the equalities $\|w(z_{n})\|_{p^{\prime}}=1$ and $|\psi_{n}(\cdot)|\equiv 1$ it follows that

[TABLE]

(see (20)), where $I$ is the identity operator and

[TABLE]

Observe that by Lemma 3 one has $\sup_{n\in\mathbb{N}}\|(I-\mathcal{K}_{y_{n}})^{-1}\|<+\infty$ due to the inequality

[TABLE]

and the fact that the sequence $\{x_{n}\}$ is bounded in $L_{\infty}^{d}(0,T)$, which in turn follows from the boundedness of the set $Z$ and Remark 4. Thus, $\inf_{n\in\mathbb{N}}\|\langle\psi_{n},H(z_{n})\rangle\|_{p^{\prime}}>0$, which contradicts the fact that $\|\langle\psi_{n},H(z_{n})\rangle\|_{p^{\prime}}\to 0$ as $n\to\infty$. Therefore, there exists $a>0$ such that $\gamma^{\downarrow}(z)\leq-a$ for any $z\in Z$ with $\gamma(z)>0$, and the proof is complete. ∎

Remark 10*.*

Note that under the assumptions 1, 2, and 4 of the theorem above a globally optimal solution of problem (16) exists iff there exists a feasible point of this problem, i.e. there exists an absolutely continuous solution of the differential inclusion $\dot{x}\in F(x,t)$ starting at $x_{0}$ and defined on $[0,T]$. Indeed, from the existence of a feasible point of problem (16) and the inequality $c>\mathcal{I}^{*}$ it follows that the sublevel set $\{x\in\Omega\mid\mathcal{I}(x)\leq c\}\subseteq S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ is nonempty and bounded. Therefore, there exists a bounded sequence $\{x_{n}\}\subset\Omega$ such that $\mathcal{I}(x_{n})\to\mathcal{I}^{*}$ as $n\to\infty$. Taking into account the reflexivity of the space $W^{d}_{1,p}(0,T)$ one obtains that there exists a subsequence $\{x_{n_{k}}\}$ weakly converging to some $x^{*}\in W^{d}_{1,p}(0,T)$. By Remark 4 one can suppose that $x_{n_{k}}$ converges to $x^{*}$ uniformly on $[0,T]$, which, as it is easily seen, implies that $\mathcal{I}(x^{*})=\lim_{k\to\infty}\mathcal{I}(x_{n_{k}})=\mathcal{I}^{*}$ and $x^{*}\in A$. Thus, it remains to check that $x^{*}$ is a solution of the differential inclusion $\dot{x}\in F(x,t)$.

From (17) and the fact that $x_{n}\in\Omega$ it follows that

[TABLE]

Passing to the limit as $k\to\infty$ with the use of Lebesgue’s dominated convergence theorem, and the facts that $x_{n_{k}}$ converges to $x^{*}$ uniformly in $[0,T]$ and the compact-valued multifunction $F(x,t)$ is continuous one obtains that

[TABLE]

By Filippov’s theorem (see, e.g. [34] or [4, Theorem 8.2.10]) there exists a measurable selection $y(t)$ of $F(x^{*}(t),t)$ such that for almost every $t\in[0,T]$ one has $s(F(x^{*}(t),t)-\dot{x}^{*}(t),\psi(t))=\langle y(t)-\dot{x}^{*}(t),\psi(t)\rangle$. Consequently, one has

[TABLE]

where $\mathcal{F}(x^{*})$ is the set of all measurable selections of the multifunction $F(x^{*}(\cdot),\cdot)$. With the use of the facts that the set-valued map $F$ is continuous and its values are compact and convex one can check that the set $\mathcal{F}(x^{*})$ is convex, closed and bounded in $L_{p}^{d}(0,T)$, i.e. it is weakly compact in $L_{p}^{d}(0,T)$. Hence by applying the separation theorem and the inequality above one gets that $0\in\mathcal{F}(x^{*})-\dot{x}^{*}$ or, equivalently, $\dot{x}^{*}(t)\in F(x^{*}(t),t)$ for a.e. $t\in[0,T]$.

Let us finally note that a feasible point of problem (16) exists, for example, if there exist $C>0$ and a.e. nonnegative function $\omega\in L^{p}(0,T)$ such that $|v|\leq C|x|+\omega(t)$ for all $v\in F(x,t)$, $x\in\mathbb{R}^{d}$ and a.e. $t\in(0,T)$ (this result follows directly from the Grönwall-Bellman inequality).

5 Conclusions

In the first paper of our two-part study we strengthened some existing results on exact penalty functions for optimisation problems in infinite dimensional spaces and applied them to free-endpoint optimal control problems. We obtained simple and verifiable sufficient conditions for the complete exactness of penalty functions for such problems with the use of a number of auxiliary results on integral functionals and Nemytskii operators proved in this paper. Our results allow one to reduce free-endpoint point optimal control problems (including such problems involving differential inclusions) to equivalent variational problems and, thus, apply numerical methods of nonsmooth optimisation to such optimal control problems.

The equivalent variational problems obtained with the use of exact penalty functions can be discretised directly and then solved with the use of many modern methods of nonsmooth optimisation (see [50, 6] for a comparative analysis of existing nonsmooth optimisation software). Alternatively, they can be solved via continuous methods, such as the method of hypodifferential descent (see [37] for more details and numerical examples) and an infinite dimensional version of bundle methods [62, 64, 63]. Furthermore, one can solve the equivalent variational problems via smooth optimisation methods (both continuous and based on discretisation) by applying smoothing approximations of nonsmooth penalty functions [65, 59, 60, 57, 29] or by replacing these problems with equivalent problems of minimising the smooth penalty function proposed by Huyer and Neumaier [29, 45, 70, 27].

Thus, our results pave the way for a comparative analysis of various nonsmooth optimisation methods for solving optimal control problems, as well as for a comparative analysis of smooth and nonsmooth approaches to the solution of optimal control problems. Moreover, our results can be extended to the case of nonsmooth optimal control problems and utilised to develop new numerical methods for solving such problems.

The second paper of our two-part study will be devoted to the analysis of exact penalty functions for optimal control problems with terminal and pointwise state constraints, including optimal control problems for linear evolution equations in Hilbert spaces.

Appendix A. The Proof of Theorem 2

For the sake of completeness, let us present an almost self-contained proof of Theorem 2, although some parts of this theorem can be found in [28, 29]. The only result we will use is Ekeland’s variational principle [31]. Let us recall that to apply this principle one must suppose that the function under consideration is defined on a complete metric space, l.s.c., and bounded below. That is why we must suppose that $X$ is complete, $A$ is closed, and both functions $\mathcal{I}$ and $\varphi$ are lower semi-continuous.

Proof of part 1. Arguing by reductio ad absurdum, suppose that the optimal values of the problems $(\mathcal{P})$ and (1) do not coincide for any $\lambda\geq 0$ (note that if they coincide for some $\lambda^{*}\geq 0$, then they coincide for all $\lambda\geq\lambda^{*}$ due to the fact that $\Phi_{\lambda}(x)$ is nondecreasing in $\lambda$). Recall that the penalty term $\varphi$ is nonnegative and $\varphi(x)=0$ iff $x\in M$. Consequently,

[TABLE]

which implies that $\inf_{x\in A}\Phi_{\lambda}(x)<\mathcal{I}^{*}=\inf_{x\in\Omega}\mathcal{I}(x)$ for any $\lambda\geq 0$ (note also that $\inf_{x\in A}\Phi_{\lambda}(x)>-\infty$ for any $\lambda\geq\lambda_{0}$ due to the fact that $\Phi_{\lambda}$ is nondecreasing in $\lambda$). Hence, in particular, for any $n\in\mathbb{N}$ there exists $x_{n}\in A$ such that $\Phi_{n}(x_{n})<\mathcal{I}^{*}$. Define $\varepsilon_{n}=\Phi_{n}(x_{n})-\inf_{x\in A}\Phi_{n}(x)+1/n$. By applying Ekeland’s variational principle one obtains that for any $n\in\mathbb{N}$, $n\geq\lambda_{0}$, and $t>0$ there exists $y_{n}\in A$ such that $\Phi_{n}(y_{n})\leq\Phi_{n}(x_{n})$, and the following inequalities hold true:

[TABLE]

Setting $t=\varepsilon_{n}$, dividing the last inequality by $d(y,y_{n})$, and passing to the limit inferior as $y\to y_{n}$, $y\in A$ one gets that

[TABLE]

(note that if $y_{n}$ is an isolated point of $A$, then by definition $(\Phi_{n})^{\downarrow}_{A}(y_{n})=+\infty$). From the facts that $\Phi_{n}(y_{n})\leq\Phi_{n}(x_{n})<\mathcal{I}^{*}<c$ and $\Phi_{n}(x)=\mathcal{I}(x)$ for any $x\in\Omega$ it follows that $y_{n}\in S_{n}(c)$ and $y_{n}\notin\Omega$ for any $n\in\mathbb{N}$. Observe also that for any $n\geq\lambda_{0}$, $m\in\mathbb{N}$, and $x\in A$ such that $x\notin\Omega_{\delta}=\{x\in A\mid\varphi(x)<\delta\}$ one has

[TABLE]

Consequently, for any sufficiently large $n$ one has $\Phi_{n}(x)\geq\mathcal{I}^{*}$, provided $x\in A\setminus\Omega_{\delta}$, which implies that there exists $n_{0}\geq\lambda_{0}$ such that $y_{n}\in\Omega_{\delta}$ for all $n\geq n_{0}$.

Thus, $y_{n}\in S_{\lambda_{0}}(c)\cap(\Omega_{\delta}\setminus\Omega)$ for any $n\geq n_{0}$ (here we used the fact that $S_{n}(c)\subseteq S_{\lambda_{0}}(c)$ for any $n\geq\lambda_{0}$, since $\Phi_{\lambda}$ is nondecreasing in $\lambda$). Therefore, $\varphi^{\downarrow}_{A}(y_{n})\leq-a$ for all $n\geq n_{0}$. By the definition of the rate of steepest descent for any $n\geq n_{0}$ there exists a sequence $\{y_{n}^{k}\}\subset A$, $k\in\mathbb{N}$, converging to $y_{n}$ such that $\varphi(y_{n}^{k})-\varphi(y_{n})\leq-0.5ad(y_{n}^{k},y_{n})$ for all $k\in\mathbb{N}$. Hence taking into account the fact that the function $\mathcal{I}$ is Lipschitz continuous on an open set containing the set $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ with a Lipschitz constant $L\geq 0$ one obtains that for any $n\geq n_{0}$ there exists $k(n)\in\mathbb{N}$ such that for all $k\geq k(n)$ one has

[TABLE]

Dividing this inequality by $d(y_{n}^{k},y_{n})$, and passing to the limit inferior as $k\to+\infty$ one obtains that $(\Phi_{n})^{\downarrow}_{A}(y_{n})\leq L-0.5na<-1$ for any sufficiently large $n$, which contradicts (22). ∎

Proof of part 2. By the first part of the theorem there exists $\lambda^{*}\geq 0$ such that for any $\lambda\geq\lambda^{*}$ one has $\inf_{x\in A}\Phi_{\lambda}(x)=\mathcal{I}^{*}=\inf_{x\in\Omega}\mathcal{I}(x)$. Hence by applying (21) one gets that $\operatorname*{arg\,min}_{x\in\Omega}\mathcal{I}(x)\subseteq\operatorname*{arg\,min}_{x\in A}\Phi_{\lambda}(x)$ for all $\lambda\geq\lambda^{*}$. On ther other hand, if $x\in A\setminus\Omega$, then $\varphi(x)>0$, and for any $\lambda>\lambda^{*}$ one has $\Phi_{\lambda}(x)>\Phi_{\lambda^{*}}(x)\geq\mathcal{I}^{*}$. Therefore, for any $\lambda>\lambda^{*}$ one has $\operatorname*{arg\,min}_{x\in A}\Phi_{\lambda}(x)\subset\Omega$, which with the use of (21) implies that $\operatorname*{arg\,min}_{x\in\Omega}\mathcal{I}(x)=\operatorname*{arg\,min}_{x\in A}\Phi_{\lambda}(x)$. ∎

Before we proceed to the proof of the last two statements of Theorem 2, let us first prove two auxiliary lemmas. The first one is a modification of the main lemma from [47], while the second one is a generalisation of [28, Proposition 2.7].

Lemma 1**.**

Let the assumptions of Theorem 2 be valid. Then for any $x_{0}\in S_{\lambda_{0}}(c)\cap\Omega$ there exists $r>0$ such that $\varphi(x)\geq a\operatorname{dist}(x,\Omega)$ for all $x\in B(x_{0},r)\cap A$, where $B(x_{0},r)=\{y\in X\mid d(y,x_{0})\leq r\}$.

Proof.

Fix $x_{0}\in S_{\lambda_{0}}(c)\cap\Omega$. Note that $\varphi(x_{0})=0$, since $x_{0}\in\Omega$. Due to the continuity of $\varphi$ there exists $\eta>0$ such that $\varphi(x)<\delta$ for any $x\in B(x_{0},\eta)$, i.e. $B(x_{0},\eta)\subset\Omega_{\delta}$. Furthermore, one can choose $\eta>0$ so small that $\mathcal{I}$ is Lipschitz continuous on $B(x_{0},\eta)$. Consequently, decreasing $\eta$ further if necessary, one can suppose that $B(x_{0},\eta)\subset S_{\lambda_{0}}(c)\cap\Omega_{\delta}$, which implies that $\varphi^{\downarrow}(x)\leq-a<0$ for all $x\in B(x_{0},\eta)\setminus\Omega$.

The continuity of $\varphi$ implies that there exists $r>0$ such that

[TABLE]

Moreover, one can obviously suppose that $r<\eta/2$. Let $x\in B(x_{0},r)\cap A$ be arbitrary. If $x\in\Omega$, then $\varphi(x)=0$, and the inequality $\varphi(x)\geq a\operatorname{dist}(x,\Omega)$ is satisfied. Suppose that $x\notin\Omega$. Denote $\varepsilon=\varphi(x)>0$, and choose an arbitrary $t\in(a/2,a)$. By applying Ekeland’s variational principle one obtains that there exists $y\in A$ such that $\varphi(y)\leq\varphi(x)$, and

[TABLE]

Let us check that $\varphi(y)=0$. Indeed, if $\varphi(y)>0$, then taking into account (23) one obtains that $y\in B(x_{0},\eta)\setminus\Omega$, since

[TABLE]

Hence $\varphi^{\downarrow}(y)\leq-a$, which implies that there exists $z\in A$, $z\neq y$, such that $\varphi(z)-\varphi(y)\leq-td(z,y)$. Therefore $\varphi(z)+td(z,y)\leq\varphi(y)$, which contradicts the second inequality in (24). Thus, $\varphi(y)=0$, i.e. $y\in\Omega$. Now, by applying the first inequality in (24) one obtains that

[TABLE]

or equivalently $\varphi(x)\geq t\operatorname{dist}(x,\Omega)$. Hence taking into account the fact that $t\in(a/2,a)$ was chosen arbitrarily one gets the required result. ∎

Lemma 2**.**

Let the assumptions of Theorem 2 be valid, and $L$ be a Lipschitz constant of $\mathcal{I}$ on an open set containing the set $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$. Suppose also that $x^{*}\in S_{\lambda_{0}}(c)\cap\Omega$ is an inf-stationary point of $\mathcal{I}$ on $\Omega$. Then for any $L^{\prime}>L$ there exists $r>0$ such that $\mathcal{I}(x)-\mathcal{I}(x^{*})\geq-L^{\prime}\operatorname{dist}(x,\Omega)-(L^{\prime}-L)d(x,x^{*})$ for all $x\in B(x^{*},r)$.

Proof.

Choose $L^{\prime}>L$. By the definition of inf-stationary point there exists $r_{0}>0$ such that $\mathcal{I}(x)-\mathcal{I}(x^{*})\geq-(L^{\prime}-L)d(x,x^{*})$ for all $x\in B(x^{*},r_{0})\cap\Omega$. Decreasing $r_{0}$, if necessary, one can suppose that $\mathcal{I}$ is Lipschitz continuous on $B(x^{*},r_{0})$ with Lipschitz constant $L$.

Put $r=r_{0}/2$, and fix an arbitrary $x\in B(x^{*},r)$. By definition there exists a sequence $\{x_{n}\}\subset\Omega$ such that $d(x,x_{n})\to\operatorname{dist}(x,\Omega)$ as $n\to\infty$ and $d(x,x_{n})\leq d(x,x^{*})\leq r$. Hence, in particular, one has

[TABLE]

i.e. $\{x_{n}\}\subset B(x^{*},r_{0})\cap\Omega$ for all $n\in\mathbb{N}$. Therefore,

[TABLE]

for any $n\in\mathbb{N}$, where the last inequality follows from (25). Passing to the limit as $n\to\infty$ we arrive at the required result. ∎

Remark 11*.*

Note that if $x^{*}$ is a point of local minimum of $\mathcal{I}$ on $\Omega$, then in the lemma above one can obviously set $L^{\prime}=L$, and check that for any $x\in B(x^{*},r)$ one has $\mathcal{I}(x)-\mathcal{I}(x^{*})\geq-L\operatorname{dist}(x,\Omega)$ (see [28, Proposition 2.7]).

Now we are ready to prove the last two statements of Theorem 2.

Proof of part 3. At first, note that without loss of generality one can suppose that $\delta=+\infty$. Indeed, denote $\eta=\inf_{x\in A}\Phi_{\lambda_{0}}(x)>-\infty$. Then for any $x\in A\setminus\Omega_{\delta}$ and $\lambda>\widehat{\lambda}:=\lambda_{0}+(c-\eta)/\delta$ one has

[TABLE]

which implies that $S_{\lambda}(c)\subseteq S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ for any $\lambda>\widehat{\lambda}$. Thus, increasing if necessary $\lambda_{0}$ one can suppose that $\delta=+\infty$, i.e. one can replace $S_{\lambda_{0}}(c)\cap\Omega_{\delta}$ with $S_{\lambda_{0}}(c)$. Note also that

[TABLE]

by virtue of the fact that $\Phi_{\lambda}$ is non-decreasing in $\lambda$.

Let $L>0$ be a Lipschitz constant of $\mathcal{I}$ on an open set $V$ containing the set $S_{\lambda_{0}}(c)$. By our assumption for any $x\in S_{\lambda_{0}}(c)\setminus\Omega$ one has $\varphi^{\downarrow}_{A}(x)\leq-a$. Hence by the definition of the rate of steepest descent there exists a sequence $\{x_{n}\}\subset A$ converging to $x$ and such that

[TABLE]

One can obviously suppose that $\{x_{n}\}\subset V$. Therefore for any $\lambda>0$ one has

[TABLE]

which along with (26) implies that

[TABLE]

Fix $\lambda>\max\{\lambda_{0},L/a\}$. Let $x^{*}\in S_{\lambda}(c)$ be a point of local minimum of the penalised problem (1). Then, as it is easy to check, $(\Phi_{\lambda})^{\downarrow}_{A}(x)\geq 0$, which with the use of (27) implies that $x^{*}\in\Omega$. Hence taking into account the fact that $\Phi_{\lambda}(x)=\mathcal{I}(x)$ for any $x\in\Omega$ one obtains that $x^{*}$ is a point of local minimum of the problem $(\mathcal{P})$.

Let now $x^{*}\in S_{\lambda}(c)$ be a point of local minimum of the problem $(\mathcal{P})$. Then by applying Lemma 1 and Remark 11 one gets that there exists $r>0$ such that for any $\lambda\geq L/a$ and $x\in B(x^{*},r)\cap A$ one has

[TABLE]

i.e. $x^{*}$ is a point of local minimum of the penalised problem (1). ∎

Proof of part 4. Fix $\lambda>\max\{L/a,\lambda_{0}\}$. Let $x^{*}\in S_{\lambda}(c)$ be an inf-stationary point of $\Phi_{\lambda}$ on $A$. Then by (27) one has $x^{*}\in\Omega$. Hence taking into account the fact that $\Phi_{\lambda}(x)=\mathcal{I}(x)$ for any $x\in\Omega$ one can easily check that $x^{*}$ is an inf-stationary point of $\mathcal{I}$ on $\Omega$.

Let now $x^{*}\in S_{\lambda}(c)\cap\Omega$ be an inf-stationary point of the function $\mathcal{I}$ on $\Omega$. Note that one can suppose that $x^{*}$ is not an isolated point of the set $A$, since otherwise $(\Phi_{\lambda})^{\downarrow}_{A}(x^{*})=+\infty$, i.e. $x^{*}$ is obviously an inf-stationary point of $\Phi_{\lambda}$ on $A$.

By the definition of the rate of steepest descent there exists a sequence $\{x_{n}\}\subset A$ converging to $x^{*}$ such that

[TABLE]

If there exists a subsequence $\{x_{n_{k}}\}\subset\Omega$, then taking into account the fact that $\varphi(x)=0$ for all $x\in\Omega$ one gets that

[TABLE]

Thus, one can suppose that $\{x_{n}\}\subset A\setminus\Omega$ and, moreover, $\Phi_{\lambda}(x_{n})<c$ for all $n\in\mathbb{N}$, since otherwise there exists a subsequence $\{x_{n_{k}}\}$ such that

[TABLE]

which obviously implies that $(\Phi_{\lambda})^{\downarrow}_{A}(x^{*})\geq 0$. Thus, $\{x_{n}\}\subset S_{\lambda_{0}}(c)\setminus\Omega$.

Choose $L^{\prime}\in(L,\lambda a)$. By applying Lemmas 1 and 2 one obtains that

[TABLE]

for any sufficiently large $n$. Dividing this inequality by $d(x_{n},x^{*})$, and passing to the limit as $n\to\infty$ one gets that $(\Phi_{\lambda})^{\downarrow}_{A}(x^{*})\geq-(L^{\prime}-L)$, which implies that $(\Phi_{\lambda})^{\downarrow}_{A}(x^{*})\geq 0$ due to the fact that $L^{\prime}\in(L,\lambda a)$ was chosen arbitrarily. Thus, $x^{*}$ is an inf-stationary point of $\Phi_{\lambda}$ on $A$, and the proof is complete. ∎

Appendix B. Integral Functionals and Nemytskii Operators

In this section we obtain several results on the functional

[TABLE]

and the nonlinear operator $F(x,u)=\dot{x}(\cdot)-f(x(\cdot),u(\cdot),\cdot)$ defined in Section 3 that significantly simplify the verification of the assumptions of general theorems on exact penalty functions in the case of optimal control problems. Namely, we obtain conditions under which the functional $\mathcal{I}(x,u)$ is correctly defined, finite valued, and Lipschitz continuous on bounded sets, and the nonlinear operator $F(x,u)=\dot{x}(\cdot)-f(x(\cdot),u(\cdot),\cdot)$ defining the constraint $F(x,u)=0$ maps $W_{1,p}^{d}(0,T)\times L_{q}^{m}(0,T)$ to $L_{p}^{d}(0,T)$ (note that the codomain of $F(x,u)$ is $L_{p}^{d}(0,T)$, since $\dot{x}$ belongs to this space) and is differentiable in $x$. In order to obtain necessary and sufficient conditions, we shall suppose that $x\in L_{\infty}^{d}[0,T]$ and $\zeta(\cdot)\equiv 0$. In the case when $x\in W^{d}_{1,p}(0,T)$ these conditions become only sufficient.

The following result is a particular case of [38, Theorem 7.3].

Proposition 3**.**

Let the function $\theta$ be continuous and $\zeta(\cdot)\equiv 0$. Then $\mathcal{I}(x,u)$ is correctly defined and finite valued for every $x\in L^{d}_{\infty}(0,T)$ and $u\in L_{q}^{m}(0,T)$ if and only if one of the two following conditions is satisfied:

$q=+\infty$; 2. 2.

$1\leq q<+\infty$, and for every $R>0$ there exist $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{1}(0,T)$ such that

[TABLE]

for a.e. $t\in[0,T]$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|\leq R$ (i.e. $\theta$ satisfies the growth condition of order $(q,1)$).

Next we obtain necessary and sufficient conditions for $\mathcal{I}(x,u)$ to be Lipschitz continuous on bounded sets, which are needed for the verification of the assumptions of Theorem 2.

Proposition 4**.**

Let $\zeta(\cdot)\equiv 0$, $\theta=\theta(x,u,t)$ be continuous, differentiable in $x$ and $u$, and let the functions $\nabla_{x}\theta$ and $\nabla_{u}\theta$ be continuous as well. Suppose also that either $q=+\infty$ or $\theta$ satisfies (28). Then the functional $\mathcal{I}(x,u)$ is Lipschitz continuous on any bounded subset of $L^{d}_{\infty}(0,T)\times L^{m}_{q}(0,T)$ if and only if one of the following conditions is satisfied:

$q=+\infty$; 2. 2.

$1\leq q<+\infty$, and for every $R>0$ there exist $C_{R}>0$, and a.e. nonnegative functions $\omega_{R}\in L^{1}(0,T)$ and $\eta_{R}\in L^{q^{\prime}}(0,T)$ such that

[TABLE]

for a.e. $t\in[0,T]$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|\leq R$ (i.e. $\nabla_{x}\theta$ satisfies the growth condition of order $(q,1)$, while $\nabla_{u}\theta$ satisfies the growth condition of order $(q-1,q^{\prime})$).

Proof.

Let us prove the “if” part of the theorem first. For any $r>0$ denote

[TABLE]

Choose $x,h\in L^{d}_{\infty}(0,T)$, $u,v\in L^{m}_{q}(0,T)$, and $\alpha>0$. By the mean value theorem for a.e. $t\in(0,T)$ there exists $\alpha(t)\in(0,\alpha)$ such that

[TABLE]

for a.e. $t\in(0,T)$. From the continuity of $\nabla_{x}\theta$ and $\nabla_{u}\theta$ it follows that the right-hand side of the above equality converges to

[TABLE]

as $\alpha\to+0$ for a.e. $t\in(0,T)$, and this function is measurable. Hence by applying one of the two conditions of the proposition and Lebesgue’s dominated convergence theorem one can easily verify that

[TABLE]

i.e. the functional $\mathcal{I}(x,u)$ is Gâteaux differentiable, and its Gâteaux derivative $\mathcal{I}^{\prime}(x,u)[h,v]$ is equal to the expression above. Consequently, one has

[TABLE]

Let $C\subset L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ be a bounded set. Then $C\subseteq B_{r}$ for some $r>0$. As is well-known and easy to check, the functional $\mathcal{I}$ is Lipschitz continuous on $B_{r}$ with Lipschitz constant $L>0$ iff $L=\sup_{(x,u)\in B_{r}}\|\mathcal{I}^{\prime}(x,u)\|<+\infty$. It remains to note that from (31) and the assumptions of the proposition it follows that this supremum is indeed finite.

Let us now prove the converse statement. We derive only the necessary growth conditions for the function $\nabla_{u}\theta$, since the derivation of the growth conditions for the function $\nabla_{x}\theta$ is essentially the same (even slightly simpler) as in the case of $\nabla_{u}\theta$.

Let $1<q<+\infty$, and fix some $x\in L^{d}_{\infty}(0,T)$ and $r>\|x\|_{\infty}$. Choose $u\in L_{q}^{m}(0,T)\cap L_{\infty}^{m}(0,T)$ and $v\in(C[0,T])^{m}$ with $\|u\|_{q}<r$ and $\|v\|_{q}<r$, where $C[0,T]$ is the space of of continuous functions defined on $[0,T]$.

Choose $\alpha\in(0,1]$. By the mean value theorem for a.e. $t\in(0,T)$ there exists $\alpha(t)\in(0,\alpha)$ such that

[TABLE]

The right-hand side of the above equality converges to $\nabla_{u}\theta(x(t),u(t),t)$ for a.e. $t\in(0,T)$ as $\alpha\to+0$ due to the continuity of $\nabla_{u}\theta$. Hence integrating the left-hand side of (32) from [math] to $T$, and passing to the limit as $\alpha\to+0$ with the use of Lebesgue’s dominated convergence theorem and the fact that all functions $x$, $u$ and $v$ are essentially bounded one gets that

[TABLE]

By our assumption the functional $\mathcal{I}(x,u)$ is Lipschitz continuous on $B_{2r}$ with Lipschitz constant $L_{2r}>0$. Therefore

[TABLE]

Choose now arbitrary $v\in L_{q}^{m}(0,T)$ with $\|v\|<r$. Since $C[0,T]$ is dense in $L^{q}(0,T)$ (see, e.g. [38, Theorem 2.78]), there exists a sequence $\{v_{n}\}\subset(C[0,T])^{m}$ converging to $v$ in $L_{q}^{m}(0,T)$ and such that $\|v_{n}\|<r$ for all $n\in\mathbb{N}$. By applying inequality (33) with $v=v_{n}$ and passing to the limit as $n\to\infty$ one gets that

[TABLE]

(here we used the fact that the mapping $v\mapsto\int_{0}^{T}\langle\nabla_{u}\theta(x(t),u(t),t),v(t)\rangle\,dt$ is a continuous linear functional on $L_{q}^{m}(0,T)$ due to the essential boundedness of $x$ and $u$ and the continuity of $\nabla_{u}\theta$). Taking the supremum over all $v\in L_{q}^{m}(0,T)$ with $\|v\|_{q}<r$ one gets that

[TABLE]

for all $u\in L_{q}^{m}(0,T)\cap L_{\infty}^{m}(0,T)$ with $\|u\|_{q}<r$. Let us check that this inequality holds true for any $u\in L_{q}^{m}(0,T)$ with $\|u\|_{q}<r$. Then taking into account the fact that $x$ and $r$ were chosen arbitrarily one obtains that

[TABLE]

Consequently, by applying [38, Theorem 7.3] one gets that for any $R>0$ there exist $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{1}(0,T)$ such that $|\nabla_{u}\theta(x,u,t)|^{q^{\prime}}\leq C_{R}|u|^{q}+\omega_{R}(t)$ for a.e. $t\in(0,T)$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|<R$. Therefore for any such $x$, $u$ and $t$ one has

[TABLE]

i.e. the desired growth condition (see (29)) holds true (note that $q/q^{\prime}=q-1$).

Thus, it remains to prove that (34) is valid for all $u\in L_{q}^{m}(0,T)$ with $\|u\|_{q}<r$. Fix any such $u$. For any $n\in\mathbb{N}$ define

[TABLE]

Clearly, $u_{n}\in L_{q}^{m}(0,T)\cap L_{\infty}^{m}(0,T)$ and $\|u_{n}\|_{q}<r$ for any $n\in\mathbb{N}$. Furthermore, $|\nabla_{u}\theta(x(t),u_{n}(t),t)|^{q^{\prime}}$ converges to $|\nabla_{u}\theta(x(t),u(t),t)|^{q^{\prime}}$ as $n\to\infty$ for a.e. $t\in(0,T)$. Hence by applying Fatou’s Lemma (see, e.g. [35, Lemma 2.18]) and (34) one obtains that

[TABLE]

and the proof of the case $1<q<+\infty$ is complete.

To obtain the necessary growth condition in the case $q=1$ note that in this case inequality (33) holds true for any $v\in L_{1}^{m}(0,T)$ with $\|v\|_{1}<r$. Taking the supremum over all such $v$ one gets that $|\nabla_{u}\theta(x(t),u(t),t)|\leq L_{2r}$ for a.e. $t\in(0,T)$. Let $u\in L_{1}^{m}(0,T)$ with $\|u\|_{1}<r$ be arbitrary. By definition $u_{n}\in L_{\infty}^{m}(0,T)$ for any $n\in\mathbb{N}$. Therefore $|\nabla_{u}\theta(x(t),u_{n}(t),t)|\leq L_{2r}$ for a.e. $t\in(0,T)$ and for all $n\in\mathbb{N}$, which obviously implies that $|\nabla_{u}\theta(x(t),u(t),t)|\leq L_{2r}$ for a.e. $t\in(0,T)$. Hence taking into account the fact that $x$ and $r$ were chosen arbitrarily one obtains that $\nabla_{u}\theta(x(\cdot),u(\cdot),\cdot)\in L_{\infty}^{m}(0,T)$ for any $x\in L_{\infty}^{d}(0,T)$ and $u\in L_{1}^{m}(0,T)$. Utilising this result one can easily obtain the required growth condition on $\nabla_{u}\theta$ in the case $q=1$. ∎

Remark 12*.*

Note that in the case $q=1$ the second inequality in (29) simply means that for any $R>0$ there exists $C_{R}>0$ such that $|\nabla_{u}\theta(x,u,t)|\leq C_{R}$ for a.e. $t\in(0,T)$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|\leq R$.

Let us also point conditions under which the nonlinear operator $F(x,u)=\dot{x}(\cdot)-f(x(\cdot),u(\cdot),\cdot)$ is correctly defined and differentiable in $x$. The following result is a simple generalisation of the standard theorem on Nemytskii operators (see, e.g. [3]).

Proposition 5**.**

Let $f$ be continuous. Then the Nemytskii operator $(x(\cdot),u(\cdot))\mapsto f(x(\cdot),u(\cdot),\cdot)$ maps $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ to $L_{p}^{d}(0,T)$ if and only if one of the two following conditions is satisfied:

$q=+\infty$; 2. 2.

$1\leq q<+\infty$, and for every $R>0$ there exist $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{p}(0,T)$ such that

[TABLE]

for a.e. $t\in[0,T]$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|\leq R$ (i.e. $f$ satisfies the growth conditions of order $(q/p,p)$).

Moreover, if one of these conditions is satisfied, then $F$ is a continuous nonlinear operator from $W_{1,p}^{d}(0,T)\times L_{q}^{m}(0,T)$ to $L_{p}^{d}(0,T)$.

Proof.

Let $1\leq q<+\infty$ (the validity of the statement in the case $q=+\infty$ follows directly from the continuity of the function $f$). By definition the operator $(x(\cdot),u(\cdot))\mapsto f(x(\cdot),u(\cdot),\cdot)$ maps $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ to $L_{p}^{d}(0,T)$ iff

[TABLE]

Hence by applying [38, Theorem 7.3] one obtains that this operator maps $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ to $L_{p}^{d}(0,T)$ iff for every $R>0$ there exist $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{1}(0,T)$ such that

[TABLE]

for a.e. $t\in[0,T]$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|\leq R$. If this inequality is satisfied, then

[TABLE]

which implies the validity of the “only if” part of the proposition (note that $\omega_{R}^{1/p}\in L^{p}(0,T)$). Conversely, if

[TABLE]

for some $C_{R}>0$ and $\omega_{R}\in L^{p}(0,T)$, then

[TABLE]

which implies the validity of the “if” part of the proposition.

Let one of the conditions be satisfied. From the fact that every $x\in W^{d}_{1,p}(0,T)$ belongs to $L_{\infty}^{d}(0,T)$ it follows that the operator $(x(\cdot),u(\cdot))\mapsto f(x(\cdot),u(\cdot),\cdot)$ maps $W_{1,p}^{d}(0,T)\times L_{q}^{m}(0,T)$ to $L_{p}^{d}(0,T)$, and, therefore, so does the operator $F$. The continuity of this operator can be easily verified in the case $1\leq q<+\infty$ with the use of Vitali’s theorem characterising convergence in $L^{p}$ spaces (cf. the proof of [66, Theorem 5.1]), and it can be proved via a simple $\varepsilon$-$\delta$ argument in the case $q=+\infty$. ∎

Proposition 6**.**

Let $f=f(x,u,t)$ be continuous, differentiable in $x$, and let the function $\nabla_{x}f$ be continuous. Suppose also that either $q=+\infty$ or inequality (35) holds true. Then the Nemytskii operator $G_{u}\colon L_{\infty}^{d}(0,T)\to L_{p}^{d}(0,T)$ defined as $G_{u}(x)=f(x(\cdot),u(\cdot),\cdot)$ is Gâteaux differentiable at every point $x\in L_{\infty}^{d}(0,T)$, and its Gâteaux derivative has the form

[TABLE]

for any $u\in L_{q}^{m}(0,T)$ if and only if one of the following conditions is satisfied

$q=+\infty$; 2. 2.

$1\leq q<+\infty$, and for any $R>0$ there exist $C_{R}>0$ and an a.e. nonnegative function $\omega_{R}\in L^{p}(0,T)$ such that

[TABLE]

for a.e. $t\in(0,T)$ and for all $(x,u)\in\mathbb{R}^{d}\times\mathbb{R}^{m}$ with $|x|\leq R$ (i.e. $\nabla_{x}f$ satisfies the growth condition of order $(q/p,p)$).

Proof.

If one the conditions is satisfied, then by applying Lebesgue’s dominated convergence theorem one can easily check that $G_{u}(x)$ is Gâteaux differentiable, and (36) holds true. Conversely, if $G_{u}(x)$ is Gâteaux differentiable, and (36) holds true, then, as it is easily seen, the operator $(x,u)\mapsto\nabla_{x}f(x(\cdot),u(\cdot),\cdot)$ maps $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ to $L_{p}^{d\times d}(0,T)$. Hence arguing in the same way as in the proof of Proposition 5 one obtains that inequality (37) is valid. ∎

Remark 13*.*

Arguing in a similar way to the proof of Proposition 6 one can check that the functional $\mathcal{I}(x,u)$ is Gâteaux differentiable at every point $(x,u)\in L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ and its Gâteaux derivative has the natural form (30) iff one of the two conditions of Proposition 4 are satisfied. Therefore, if $\theta(x,u,t)$ is continuous, differentiable in $x$ and $u$, and the functions $\nabla_{x}\theta$ and $\nabla_{u}\theta$ are continuous, then the functional $\mathcal{I}(x,u)$ is Lipschitz continuous on bounded subsets of the space $L_{\infty}^{d}(0,T)\times L_{q}^{m}(0,T)$ iff it is Gâteaux differentiable at every point of this space, and its Gâteaux derivative has the natural form (30).

Let us also prove a simple auxiliary result on the resolvent of a Volterra-type integral equation. This result is well-known. Nevertheless, we briefly outline its proof for the sake of completeness, and because of the fact that the equation that we analyse slightly differs from the classical one [40] (instead of integrating from [math] to $t$ we integrate from $t$ to $T$).

Denote by $I$ the identity operator, and define $(\mathcal{K}_{y}x)(t)=\int_{t}^{T}y(t,s)x(s)\,ds$ for a.e. $t\in(0,T)$, where $x\colon(0,T)\to\mathbb{R}^{d}$ and $y\colon(0,T)^{2}\to\mathbb{R}^{d\times d}$ are measurable functions such that $y(t,\cdot)x(\cdot)\in L_{1}^{d}(0,T)$ for a.e. $t\in(0,T)$.

Lemma 3**.**

Let a function $y\colon(0,T)^{2}\to\mathbb{R}^{d\times d}$ be measurable and satisfy the inequality $|y(t,s)|\leq y_{0}(s)$ for a.e. $t,s\in(0,T)$ and for some a.e. nonnegative function $y_{0}\in L^{p^{\prime}}(0,T)$. Then the operator $I-\mathcal{K}_{y}$ maps $L_{p}^{d}(0,T)$ to $L_{p}^{d}(0,T)$ and is invertible. Furthermore, there exists a continuous function $\omega\colon[0,+\infty)\times[0,+\infty)\to[0,+\infty)$ such that $\|(I-\mathcal{K}_{y})^{-1}\|\leq\omega(T,\|y_{0}\|_{p^{\prime}})$.

Proof.

Fix a measurable function $y(t,s)$ satisfying the assumptions of the lemma for some $y_{0}\in L^{p^{\prime}}(0,T)$. By applying Hölder’s inequality one can easily check that the operator $\mathcal{K}_{y}$ maps $L_{p}^{d}(0,T)$ to $L_{p}^{d}(0,T)$, and $|(\mathcal{K}_{y}x)(t)|\leq\|y_{0}\|_{p^{\prime}}\|x\|_{p}$ for a.e. $t\in(0,T)$.

It is well-known and easy to check that if the Neumann series $\sum_{n=0}^{\infty}\mathcal{K}_{y}^{n}$ converges in the operator norm, then its limit is the inverse of $I-\mathcal{K}_{y}$. Let us check the convergence of this series. Indeed, with the use of Hölder’s inequality one gets that

[TABLE]

for a.e. $t\in(0,T)$. Similarly, one has

[TABLE]

for a.e. $t\in(0,T)$. By induction one can easily check that

[TABLE]

for a.e. $t\in(0,T)$ and for any $n\in\mathbb{N}$. Therefore

[TABLE]

for all $n\in\mathbb{N}$. Consequently, the Neumann series $\sum_{n=0}^{\infty}\mathcal{K}^{n}_{y}$ converges, and the norm of its limit does not exceed $\omega(T,\|y_{0}\|_{p^{\prime}})$, where

[TABLE]

It remains to note that the series in the definition of $\omega$ converges to a continuous function uniformly on bounded sets by the Weierstrass M-test. ∎