Abstract
In this paper we consider an impulsive extension of an optimal control problem with unbounded controls, subject to endpoint and state constraints. We show that the existence of an extended-sense minimizer that is a normal extremal for a constrained Maximum Principle ensures that there is no gap between the infima of the original problem and of its extension. Furthermore, we translate such relation into verifiable sufficient conditions for normality in the form of constraint and endpoint qualifications. Links between existence of an infimum gap and normality in impulsive control have previously been explored for problems without state constraints. This paper establishes such links in the presence of state constraints and of an additional ordinary control, for locally Lipschitz continuous data.
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This research is partially supported by the INdAM-GNAMPA Project 2020 “Extended control problems: gap, higher order conditions and Lyapunov functions” and by the Padua University grant SID 2018 “Controllability, stabilizability and infimum gaps for control systems”, prot. BIRD 187147.
Appendix
Appendix
Proof Proof of Theorem 4.2
Thanks to Remark 4.6, when (CQn)b is in force the proof of Theorem 4.2 is analogous to the proof of [25, Th. 4.2], while under assumption (CQn)f it requires some adaptation. For this reason, we limit ourselves to give the proof in the last case.
By standard truncation and mollification arguments, we can assume h Lipschitz continuous, with Lipschitz constant L > 0, and f, \(g_{1},\dots ,g_{m}\), and their limiting subdifferentials in (t,x), \(L^{\infty }\)-bounded by some constant \(\tilde {M} >0\). Set \(M:=(1+m)\tilde {M}\).
By assumption, the local minimizer \((\bar S, \bar \omega ^{0} ,\bar \omega ,\bar \alpha , \bar y^{0} ,\bar y , \bar \nu )\) has a set of multipliers (p0,p,π,λ,μ) and some functions (m0,m) such that the conditions (i)-(vi) of Theorem 2.1 hold true, and verifying the strengthened non-triviality condition (??). Let us first assume that \(\bar y^{0}(\bar S)>\bar y^{0}(0)\) and suppose by contradiction that
Set
Observe that \(\bar s<\bar S\). Indeed, if not, \(\mu (]0,\bar S])=0\). But in this case q(s) = p(s) + μ({0})m(0), so that it is absolutely continuous and by the adjoint equation with initial condition q(0) = 0 it follows that q ≡ 0. Precisely, by known properties of the convex hull of the limiting subdifferential of locally Lipschitz continuous functions (see e.g. [35, Ch. 6]), we have
which implies that q ≡ 0 by Gronwall’s Lemma. Since λ = 0 by Eq. 1, this is in contradiction with the first relation in Eq. ??. When \(\bar y^{0}(\bar S)=\bar y^{0}(0)\) and we assume by contradiction that
the value \(\bar s\) defined as above is still strictly smaller than \(\bar S\), since otherwise \(\mu (]0,\bar S])=0\), so that (q0,q) ≡ 0, again by the adjoint equation. In view of Eq. 2, this yields contradiction with the second relation in Eq. ??. Obviously, \((\bar y^{0}(\bar s),\bar y(\bar s))\in \partial {{\varOmega }}\).
From now on, the proof is the same for both cases. Introduce
so that, for any \(s\in [0,\bar S[\),
By the adjoint equation, (z0,z) verifies
Since the integral on the right hand side is identically zero in \(]0,\bar s[\), arguing as above we derive that z(s) = 0 and therefore q(s) = 0 for all \(s\in [0,\bar s[\), by continuity. Moreover, Gronwall’s Lemma implies that \(|z(s)|\leq C \mu ([\bar s,s[)\) for all \(s\in [\bar s,\bar S[\), for some C > 0, so that
As a consequence of Eq. 3, for every \(s\in [\bar s,\bar S]\) one gets \(q(s) =z(s) + {\int \limits }_{[\bar s,s[}m(\sigma ) \mu (d\sigma )\), and Eqs. 4 and 5 imply
where \(\bar C:=M(L+C)\). In view of (CQn)\(^{\prime }_{f}\) in Remark 4.4, there exist ε, δ > 0 and a measurable control \((\tilde \omega ,\hat \alpha ): [0,\bar S]\to (\mathcal {C}\cap \partial \mathbb {B})\times A\), verifying for all \((\xi _{0},\xi )\in \partial ^{*}h(\bar y^{0}(s),\bar y(s))\) with \(s\in ]\bar s,\bar s+\varepsilon [\cap [0,\bar S]\):
where \( (\hat \omega ^{0}(s),\hat \omega (s),\hat \alpha (s)):=\left (\bar \omega ^{0}(s), (1-\bar \omega ^{0}(s))\tilde \omega (s), \hat \alpha (s)\right )\) for a.e. \(s\in [0,\bar S]\). Observe that, being \(\hat \omega ^{0}\equiv \bar \omega ^{0}\), one has \(|\hat \omega |=1-\bar \omega ^{0}=|\bar \omega |\) a.e. As observed in Remark 4.6, \(\ell ({{\varGamma }}(\bar s,\varepsilon ))\) is > 0 for any ε > 0 sufficiently small. Moreover, Eq. 7 is valid for any \((\xi _{0},\xi )\in \partial ^{C} h(\bar y^{0}(s),\bar y(s))\), so that it is true, in particular, for \((\xi _{0},\xi )\in \partial ^{>}h(\bar y^{0}(s),\bar y(s))\). On the other hand, by the maximization condition (??) of Theorem 2.1, it follows that, for a.e. \(s\in ]\bar s,\bar s+\varepsilon [\cap [0,\bar S]\),
Putting together Eqs. 6, 7, and 8 we get the desired contradiction. Indeed, for ε > 0 small enough, for any \(s^{\prime }\in {{\varGamma }}(\bar s, \varepsilon )\), one has
for ε > 0 sufficiently small. This concludes the proof. □
Proof Proof of Proposition 4.3
The proof follows the same lines of the proof of [25, Prop. 4.1], where however only condition (TQn)b for an implicit state constraint \({{\varOmega }}\subseteq \mathbb {R}^{n}\) is considered. Let us prove (i). Assume by contradiction λ = 0. Then the transversality condition (??) implies that
where \((\zeta _{t_{2}},\zeta _{x_{2}})\ne (0,0)\) and, in particular, \(\zeta _{x_{2}}\ne 0\) if \(\bar y^{0}(\bar S)>\bar y^{0}(0)\) by Eq. ??. By hypothesis (TQn)b, there is some ε > 0 such that \((\bar y^{0}(s),\bar y(s))\in \)Int(Ω) for all \(s \in [\bar S-\varepsilon , \bar S[\). Hence \(\mu ([\bar S-\varepsilon ,\bar S[)=0\), so that, for any \(s\in ]\bar S-\varepsilon ,\bar S[\), (q0,q) is continuous at s and
Set \( (q_{0}(\bar S^{-}),q(\bar S^{-})):=\lim _{s\to \bar S^{-}}(q_{0}(s),q(s))=(p_{0},p)(\bar S)+{\int \limits }_{[0,\bar S-\varepsilon ]} (m_{0},m)(r)\mu (dr). \) We get
where \((\tilde \zeta _{t_{2}},\tilde \zeta _{x_{2}}):=\Big (\zeta _{t_{2}}+\mu (\{\bar S\})m_{0}(\bar S),\zeta _{x_{2}}+\mu (\{\bar S\})m(\bar S)\Big )\). Thus, in particular, the pair \((\tilde \zeta _{t_{2}},\tilde \zeta _{x_{2}})\) verifies
The continuity of (q0,q) on \(]\bar S-\varepsilon ,\bar S[\) also implies that the equality (??) in the Maximum Principle is verified for all \(s\in ]\bar S-\varepsilon ,\bar S[\). Hence, passing to the limit in it as s tends to \(\bar S^{-}\), we obtain
Suppose first that condition (a) in (TQn)b is satisfied. Then, from Eq. ?? we deduce that \((\tilde \zeta _{t_{2}},\tilde \zeta _{x_{2}}) \neq (0,0)\) and choosing ω = 0 in Eq. 10 we obtain a contradiction to Eq. ??.
If instead condition (b) in (TQn)b is valid, π = 0 and Eq. ?? implies that \(\zeta _{x_{2}}\ne 0\). In view of Eq. 9 and hypothesis (??), this yields \(\tilde \zeta _{x_{2}} \neq 0\). At this point, we get a contradiction to Eq. ?? by choosing ω0 = 0 in Eq. 10. The proof of (ii) is very similar, hence we omit it. □
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Fusco, G., Motta, M. No Infimum Gap and Normality in Optimal Impulsive Control Under State Constraints. Set-Valued Var. Anal 29, 519–550 (2021). https://doi.org/10.1007/s11228-021-00576-2
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DOI: https://doi.org/10.1007/s11228-021-00576-2
Keywords
- Impulsive optimal control problems
- Maximum principle
- State constraints
- Gap phenomena
- Normality
- Degeneracy