Second order analysis of control-affine problems with scalar state constraint

Abstract

In this article we establish new second order necessary and sufficient optimality conditions for a class of control-affine problems with a scalar control and a scalar state constraint. These optimality conditions extend to the constrained state framework the Goh transform, which is the classical tool for obtaining an extension of the Legendre condition.

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Notes

  1. 1.

    Actually \(H_u\) is continuous on B since p does not jump on B.

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Acknowledgments

We wish to thank the anonymous referees for their bibliographical advices. This work was partially supported by the European Union under the 7th Framework Pro-gramme FP7-PEOPLE-2010-ITN Grant Agreement Number 264735-SADCO. The last stage of this research took place while the first author was holding a postdoctoral position at IMPA, Rio de Janeiro, with CAPES-Brazil funding.

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Correspondence to J. F. Bonnans.

Appendix: On second order necessary conditions

Appendix: On second order necessary conditions

General constraints

We will study an abstract optimization problem of the form

$$\begin{aligned} \min f(x); \quad G_E(x)=0, \,\, G_I(x) \in K_I, \end{aligned}$$
(5.1)

where X, \(Y_E\), \(Y_I\) are Banach spaces, \(f:X\rightarrow {\mathbb {R}}\), \(G_E:X\rightarrow Y_E\), and \(G_I:X\rightarrow Y_I\) are functions of class \(C^2\), and \(K_I\) is a closed convex subset of \(Y_I\) with nonempty interior. The subindex E is used to refer to ‘equalities’ and I to ‘inequalities’.

Setting

$$\begin{aligned} Y:= Y_E\times Y_I, \quad G(x):=(G_E(x),G_I(x)), \quad K:= \{0\}_{K_E} \times K_I, \end{aligned}$$
(5.2)

we can rewrite problem (5.1) in the more compact form

$$\begin{aligned} \min f(x); \quad G(x) \in K. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (P_A) \end{aligned}$$

We use \(F(P_A)\) to denote the set of feasible solutions of \((P_A)\).

Remark 9

We refer to [10] for a systematic study of problem \((P_A)\). Here we will take advantage of the product structure (that one can find in essentially all practical applications) to introduce a non qualified version of second order necessary conditions specialized to the case of quasi radial directions, that extends in some sense [10, Theorem3.50]. See Kawasaki [25] for non radial directions.

The tangent cone (in the sense of convex analysis) to \(K_I\) at \(y\in K_I\) is defined as

$$\begin{aligned} T_{K_I}(y) := \{ z\in Y_I: \mathop {\mathrm{dist}}(y+ t z,K_I)=o(t), \;\; \text {with } t\ge 0\}, \end{aligned}$$
(5.3)

and the normal cone to \(K_I\) at \(y\in K_I\) is

$$\begin{aligned} N_{K_I}(y) := \{ z^* \in Y_I^*: \langle z^*, y'-y \rangle \le 0,\,\, \text {for all } y'\in K_I \}. \end{aligned}$$
(5.4)

In what follows, we shall study a nominal feasible solution \(\hat{x}\in F(P_A)\) that may satisfy or not the qualification condition

$$\begin{aligned} \left\{ \begin{array}{lll} \mathrm{(i)} &{} DG_E(\hat{x}) \text { is onto,} \\ \mathrm{(ii)} &{} \text {there exists } z \in \mathop {\mathrm{Ker}}DG_E(\hat{x}) \text { such that } G_I(\hat{x}) + DG_I(\hat{x})z \in \mathop {\mathrm{int}}(K_I). \end{array}\right. \end{aligned}$$
(5.5)

The latter condition coincides with the qualification condition in (2.21) which was introduced for the optimal control problem (P).

Remark 10

Condition (5.5) is equivalent to the Robinson qualification condition in [39]. See the discussion in [10, Section 2.3.4].

The Lagrangian function of problem \((P_A)\) is defined as

$$\begin{aligned} L(x,\lambda ) := \beta f(x) + \langle \lambda _E, G_E(x) \rangle + \langle \lambda _I, G_I(x) \rangle , \end{aligned}$$
(5.6)

where we set \(\lambda :=(\beta ,\lambda _E,\lambda _I) \in {\mathbb {R}}_+ \times Y_E^* \times Y_I^*.\) Define the set of Lagrange multipliers associated with \( x \in F(P_A)\) as

$$\begin{aligned} {\varLambda }(x) := \{ \lambda \in {\mathbb {R}}_+ \times Y_E^* \times N_{K_I}(G_I(x)): \lambda \ne 0, \,D_x L(x,\lambda ) = 0 \}. \end{aligned}$$
(5.7)

Let \(y_I \in \mathop {\mathrm{int}}(K_I),\) \(y_I \ne G_I(\hat{x}).\) We consider the following auxiliary problem, where \((x,\gamma ) \in X \times {\mathbb {R}}\):

$$\begin{aligned} \begin{array}{lll} \displaystyle \min _{x,\gamma } \gamma ; &{} f(x) - f(\hat{x}) \le \gamma , \quad G_E(x)=0, \quad \gamma \ge -1/2, \\ &{} y_I + (1+\gamma )^{-1} \big ( G_I(x) - y_I \big ) \in K_I. \end{array}\qquad \qquad \qquad (AP_A) \end{aligned}$$

Note that we recognize the idea of a gauge function (see e.g. [40]) in the last constraint.

Lemma 4

Assume that \(\hat{x}\) is a local solution of \((P_A)\). Then \((\hat{x},0)\) is a local solution of \((AP_A)\).

Proof

We easily check that \((\hat{x},0)\in F(AP_A)\). Now take \((x,\gamma )\in F(AP_A).\) Let us prove that if \(-1/2 \le \gamma <0,\) then x cannot be closed to \(\hat{x}\) (in the norm of the Banach space X). Assuming that \(-1/2 \le \gamma <0,\) we get \(G_E(x)=0,\) \(G_I(x) \in K_I + (-\gamma ) y_I \subseteq K_I\), and \(f(x) < f(\hat{x})\). Since \(\hat{x}\) is a local solution of \((P_A)\), the x cannot be too closed to \(\hat{x}.\) The conclusion follows.

The Lagrangian function of \((AP_A)\), in qualified form, is

$$\begin{aligned} \gamma + \beta ( f(x) - f(\hat{x}) - \gamma ) + \langle \lambda _E, G_E(x) \rangle + \langle \lambda _I, y_I + (1+\gamma )^{-1} (G_I(x) - y_I) \rangle . \end{aligned}$$
(5.8)

or equivalently

$$\begin{aligned} L(x,\lambda ) + (\beta _0 - \beta ) \gamma + \big ((1+\gamma )^{-1} -1 \big ) \langle \lambda _I, G_I(x) - y_I\rangle . \end{aligned}$$
(5.9)

Setting \(\hat{\lambda }=(\beta _0,\beta ,\lambda _E,\lambda _I)\), we see that the set of Lagrange multipliers of the auxiliary problem \((AP_A)\) at \((\hat{x},0)\) is

$$\begin{aligned} \hat{{\varLambda }} := \left\{ \begin{array}{lll} \hat{\lambda } \in {\mathbb {R}}_+ \times {\mathbb {R}}_+ \times Y_E^* \times N_{K_I}(G_I(\hat{x})): \lambda \ne 0,\\ D_x L(\hat{x},\lambda ) =0; \; \beta + \langle \lambda _I, G_I(\hat{x}) - y_I \rangle = 1 \end{array} \right\} . \end{aligned}$$
(5.10)

Proposition 8

Suppose that (5.5)(i) holds. Then, the mapping

$$\begin{aligned} (\beta ,\lambda _E,\lambda _I) \mapsto \frac{(\beta + \langle \lambda _I, G_I(x) - y_I \rangle , \beta ,\lambda _E,\lambda _I) }{\beta + \langle \lambda _I, G_I(x) - y_I \rangle } \end{aligned}$$
(5.11)

is a bijection between \({\varLambda }(\hat{x})\) and \(\hat{{\varLambda }}_1\) (recall the definition in (2.16)).

Proof

Since (5.5)(i) holds, then we necessarily have that \((\beta ,\lambda _I) \ne 0\) for all \(\lambda =(\beta ,\lambda _E,\lambda _I) \in {\varLambda }(\hat{x}).\) Therefore, if \(\lambda _I = 0\) then \(\beta >0\) and \(\beta + \langle \lambda _I, G_I(x) - y_I \rangle > 0.\) If by the contrary, \(\lambda _I \ne 0,\) then \(\langle \lambda _I, G_I(x) - y_I \rangle > 0\) and again, \(\beta + \langle \lambda _I, G_I(x) - y_I \rangle > 0.\) Hence, the mapping in (5.11) is well-defined and is a bijection from \({\varLambda }(\hat{x})\) to \(\hat{{\varLambda }}_1,\) as we wanted to show.

Theorem 6

Let \(\hat{x}\) be a local solution of \((P_A)\), such that \(DG_E(\hat{x})\) is surjective. Then \(\hat{{\varLambda }}_1\) is non empty and bounded.

Proof

By lemma 4, \((\hat{x},0)\) is a local solution of \((AP_A)\). In addition the qualification condition for the latter problem at the point \((\hat{x},0)\) states as follows: there exists \((z,\delta ) \in \mathop {\mathrm{Ker}}DG_E(\hat{x}) \times {\mathbb {R}}\) such that

$$\begin{aligned}&Df(\hat{x})z<\delta ,\quad \delta >0,\\&\quad G_I(\hat{x}) + D G_I(\hat{x})z -\delta (G_I(\hat{x}) - y_I) \in \mathop {\mathrm{int}}(K_I). \end{aligned}$$

These conditions trivially hold for \((z,\delta )=(0,1).\) Hence, in view of classical results by e.g. Robinson [39], the conclusion follows.

Second order necessary optimality conditions

Let us introduce the notation [ab] to refer to the segment \(\{\rho a+(1-\rho )b;\ \text {for }\rho \in [0,1]\},\) defined for any pair of points ab in an arbitrary vector space Z.

Definition 6

Let \(y\in K\). We say that \(z\in Y\) is a radial direction to K at y if \([y,y+\varepsilon z] \subset K\) for some \(\varepsilon >0\), and a quasi-radial direction if \(\mathop {\mathrm{dist}}(y+\sigma z,K) = o(\sigma ^2)\) for \(\sigma >0\).

Note that any radial direction is also quasi-radial, and both radial and quasi radial directions are tangent. With \(\hat{x}\in F(P_A)\), we associate the critical cone

$$\begin{aligned} C(\hat{x}) := \{z\in X: Df(\hat{x})z \le 0, \; DG_E(\hat{x})z = 0, \; DG_I(\hat{x})z \in T_K(G_I(\hat{x})) \}. \end{aligned}$$
(5.12)

Definition 7

We say that \(z\in C(\hat{x})\) is a radial (quasi radial) critical direction for problem \((P_A)\) if \(D G_I(\hat{x}) z\) is a radial (quasi radial) direction to \(K_I\) at \(G_I(\hat{x})\). We write \(C_{QR}(\hat{x})\) for the set of quasi radial critical directions. The critical cone \(C(\hat{x})\) is quasi radial if \(C_{QR}(\hat{x})\) is a dense subset of \(C(\hat{x})\).

It is immediate to check that \(C_{QR}(\hat{x})\) is a convex cone.

We next state primal second order necessary conditions for the problem \((P_A)\). Consider the following optimization problem, where \(z\in X\), \(w\in X\) and \(\theta \in {\mathbb {R}}\):

$$\begin{aligned} \left\{ \begin{aligned}&\min _{w,\theta } \;\; \theta , \\&Df(\hat{x}) w + D^2f(\hat{x})(z,z) \le \theta , \\&D G_E(\hat{x}) w + D^2 G_E(\hat{x})(z,z) = 0, \\&D G_I(\hat{x}) w + D^2 G_I(\hat{x})(z,z) - \theta (G(\hat{x})-y_I) \in T_K(G_I(\hat{x})). \end{aligned} \right. \qquad \qquad \qquad \qquad (Q_z) \end{aligned}$$

Theorem 7

Let \((\hat{x},0)\) be a local solution of \((AP_A)\), such that \(DG_E(\hat{x})\) is surjective, and let \(h\in C_{QR}(\hat{x})\). Then problem \((Q_z)\) is feasible, and has a nonnegative value.

Proof

We shall first show that \((Q_z)\) is feasible. Since \(DG_E(\hat{x})\) is surjective, there exists \(w \in X\) such that \(D G_E(\hat{x}) w + D^2 G_E(\hat{x})(z,z) = 0\). Since \(T_K(G_I(\hat{x}))\) is a cone, the last equation divided by \(\theta >0\) is equivalent to

$$\begin{aligned} \theta ^{-1} \big (D G_I(\hat{x}) w + D^2 G_I(\hat{x})(z,z)\big ) + y_I - G(\hat{x})\in T_K(G_I(\hat{x})). \end{aligned}$$
(5.13)

Since \(y_I\in \mathop {\mathrm{int}}(K_I)\), we have that \(y_I - G(\hat{x})\in \mathop {\mathrm{int}}T_K(G_I(\hat{x}))\), and therefore the last constraint of \((Q_z)\) holds when \(\theta \) is large enough. So it does the first constraint, and hence, \((Q_z)\) is feasible.

We next have to show that we cannot have \((w,\theta _0)\in F(Q_z)\) with \(\theta _0<0.\) Let us suppose, on the contrary, that there is such a feasible solution \((w,\theta _0).\) Set \(\theta :=\frac{1}{2}\theta _0\). Then \(Df(\hat{x}) w + D^2f(\hat{x})(z,z) < \theta .\) Using (5.13) and \(y_I\in \mathop {\mathrm{int}}(K_i)\), we can easily show that, for some \(\varepsilon >0\):

$$\begin{aligned} D G_I(\hat{x}) w + D^2 G_I(\hat{x})(z,z) - \theta (G_I(\hat{x})-y_I) + \varepsilon B \in T_K(G_I(\hat{x})). \end{aligned}$$
(5.14)

Consider, for \(\sigma >0\), the path

$$\begin{aligned} x_\sigma := \hat{x}+ \sigma z + \frac{1}{2}\sigma ^2 w. \end{aligned}$$
(5.15)

By a second order Taylor expansion we obtain that \(G_E(x_\sigma ) = o(\sigma ^2)\). Since \(DG_E(\hat{x})\) is onto, by Lyusternik’s theorem [27], there exists a path \(x'_\sigma = x_\sigma + o(\sigma ^2)\), such that \(G_E(x'_\sigma )=0\). Assuming, without loss of generality, that \(G_I(\hat{x})=0\), we get

$$\begin{aligned} G_I(x'_\sigma ) = \sigma D G_I(\hat{x})z+ \frac{1}{2}\sigma ^2 \left[ D G_I(\hat{x}) w + D^2 G_I(\hat{x})(z,z) \right] + o(\sigma ^2). \end{aligned}$$
(5.16)

Setting

$$\begin{aligned} \left\{ \begin{aligned} k_1(\sigma )&:= (1-\sigma )^{-1}\sigma D G_I(\hat{x})z, \\ k_2(\sigma )&:= \sigma \big ( D G_I(\hat{x}) w + D^2 G_I(\hat{x})(z,z) \big ), \end{aligned} \right. \end{aligned}$$
(5.17)

we can rewrite (5.16) as

$$\begin{aligned} G_I(x'_\sigma ) = (1-\sigma ) k_1(\sigma ) + \frac{1}{2}\sigma k_2(\sigma ) + o(\sigma ^2). \end{aligned}$$
(5.18)

Since z is a quasi radial critical direction, there exists \(k'_1(\sigma )\in K_I\) such that

$$\begin{aligned} \sigma D G_I(\hat{x})z = k'_1(\sigma ) + o(\sigma ^2), \end{aligned}$$
(5.19)

and so,

$$\begin{aligned} G_I(x'_\sigma ) \in (1-\sigma ) K_I + \frac{1}{2}\sigma k_2(\sigma ) + o(\sigma ^2). \end{aligned}$$
(5.20)

Using (5.14) and \(G_I(\hat{x})=0\) we obtain

$$\begin{aligned} k_2(\sigma ) + \sigma \theta y_I + \sigma \varepsilon B \in K_I. \end{aligned}$$
(5.21)

Therefore, for \(\sigma >0\) small enough

$$\begin{aligned} G_I(x'_\sigma ) \in (1-\frac{1}{2}\sigma ) K_I + \frac{1}{2}\sigma K_I - \frac{1}{2}\sigma ^2(\theta y_I +o(1)) \subset K_I, \end{aligned}$$
(5.22)

where we have used the fact that since \(0=G_I(\bar{x})\in K_I\), we have that (remember that \(\theta <0\)): \(\frac{1}{2}\sigma ^2 (-\theta ) (y_I + \varepsilon B) \subset K_I\).

We check easily that \(f(x'_\sigma ) <0\), and so, we have constructed a feasible path for \((AP_A)\), contradicting the local optimality of \((\hat{x},0)\).

We conclude that such a solution \((w,\theta _0)\) of \((Q_z)\) with \(\theta _0<0\) cannot exist and, therefore, \((Q_z)\) has nonnegative value.

We now present dual second order necessary conditions.

Theorem 8

Let \(\hat{x}\) be a local minimum of \((P_A)\), that satisfies the qualification condition (5.5). Then, for every \(z\in C_{QR}(\hat{x}),\)

$$\begin{aligned} \max _{\lambda \in {\varLambda }(\hat{x})} D^2_{xx} L(\hat{x},\lambda )(z,z) \ge 0. \end{aligned}$$
(5.23)

Proof

Since problem \((Q_z)\) is qualified with a finite nonnegative value, by the convex duality theory [14], its dual has a nonnegative value and a nonempty set of solutions. The Lagrangian of problem \((Q_z)\) in qualified form (\(\beta _0=1\)) can be written as

$$\begin{aligned} D_xL(\hat{x},\lambda )w + D^2_{xx} L(\beta ,\hat{x},\lambda )(z,z)+ \big (1-\beta + \langle \lambda _I,G(\hat{x})-y_I\rangle \big ) \theta , \end{aligned}$$
(5.24)

where \(\lambda =(\beta ,\lambda _E,\lambda _I)\) as before, and so, the dual problem of \((Q_z)\) can be written as

$$\begin{aligned} \mathop {\mathrm{Max}}_{\lambda \in {\varLambda }(\hat{x})} D^2_{xx} L(\beta ,\hat{x},\lambda )(z,z); \quad \beta + \langle \lambda _I,G(\hat{x})-y_I\rangle ) \theta =1. \end{aligned}$$

The conclusion follows.

Remark 11

Whereas the above theorem follows from Cominetti [12] or Kawasaki [25], our proof avoids the concepts of second order tangent set and its associated calculus, used in these references. This considerably simplifies the proof.

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Aronna, M.S., Bonnans, J.F. & Goh, B.S. Second order analysis of control-affine problems with scalar state constraint. Math. Program. 160, 115–147 (2016). https://doi.org/10.1007/s10107-015-0976-0

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Mathematics Subject Classification

  • 49K15
  • 49K27