Abstract
This paper deals with optimal control problems of integral equations, with initial–final and running state constraints. The order of a running state constraint is defined in the setting of integral dynamics, and we work here with constraints of arbitrary high orders. First-order necessary conditions of optimality are given by the description of the set of Lagrange multipliers. Second-order necessary conditions are expressed by the nonnegativity of the supremum of some quadratic forms. Second-order sufficient conditions are also obtained in the case where these quadratic forms are of Legendre type.
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Acknowledgements
We thank the two anonymous referees for their useful remarks. The research leading to these results has received funding from the EU 7th Framework Programme (FP7-PEOPLE-2010-ITN), under GA No. 264735-SADCO. The first and third authors also thank the MMSN (Modélisation Mathématique et Simulation Numérique) Chair (EADS, Inria, and Ecole Polytechnique) for its support.
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Communicated by Hans Joseph Pesch.
Appendix
Appendix
1.1 A.1 Functions of Bounded Variation
The main reference here is [26], Sect. 3.2. Recall that with the definition of \(\mathit{BV} ([0,T];\mathbb {R}^{n*} )\) given at the beginning of Sect. 2.2, for \(h \in\mathit{BV} ([0,T];\mathbb{R}^{n*} )\), there exist \(h_{0_{-}}, h_{T_{+}} \in\mathbb{R}^{n*}\) such that (6) holds.
Lemma A.1
Let \(h \in\mathit{BV} ([0,T];\mathbb{R}^{n*} )\). Let h l, h r be defined for all t∈[0,T] by
Then they are both in the same equivalence class of h, h l is left continuous, h r is right continuous, and, for all t∈[0,T],
Proof
Theorem 3.28 in [26]. □
The identification between measures and functions of bounded variation that we mention at the beginning of Sect. 2.2 relies on the following:
Lemma A.2
The linear map
is an isomorphism between \(\mathbb{R}^{r*} \times\mathit{M} ( [0,T];\mathbb{R}^{r*} )\) and \(\mathit{BV} ( [0,T];\mathbb{R}^{r*} )\), whose inverse is
Proof
Theorem 3.30 in [26]. □
Let us now prove Lemma 2.1:
Proof of Lemma 2.1
By (149), a solution in P of (11) is any \(p \in L^{1}([0,T];\mathbb{R}^{n*})\) such that, for a.e. t∈[0,T],
We define \(\varTheta\colon L^{1}([0,T];\mathbb{R}^{n*}) \rightarrow L^{1}([0,T];\mathbb{R}^{n*})\) by
for a.e. t∈[0,T], and we show that Θ has a unique fixed point. Let C>0 be such that ∥D y f∥∞, \(\| D^{2}_{y,\tau} f \| _{\infty}\le C\) along (u,y). Then
We consider the family of equivalent norms on \(L^{1}([0,T];\mathbb{R}^{n*})\)
Then
For K big enough, Θ is a contraction on \(L^{1}([0,T];\mathbb {R}^{n*})\) for ∥⋅∥1,K ; its unique fixed point is the unique solution of (11). □
Another useful result is the following integration by parts formula:
Lemma A.3
Let h,k∈BV([0,T]). Then h l∈L 1(dk), k r∈L 1(dh), and
Proof
Let Ω:={0≤y≤x≤T}. Since χ Ω ∈L 1(dh⊗dk), we have by Fubini’s theorem (Theorem 7.27 in [27]) and Lemma A.1 that h l∈L 1(dk), k r∈L 1(dh), and we can compute dh⊗dk(Ω) in two different ways:
□
1.2 A.2 The Hidden Use of Assumption (A3)
We use assumption (A3) to prove Lemma 4.3 (and then Lemma 4.1, …) through the following:
Lemma A.4
Recall that \(M_{t}:= D_{\tilde{u}}G^{(q)}_{I^{\varepsilon_{0}}(t)}(t,\bar {u}_{t},\bar{y}_{t},\bar {u},\bar{y}) \in M_{|I^{\varepsilon_{0}}_{t}|,m} (\mathbb{R} )\), t∈[0,T]. Then \(M_{t}M_{t}^{*}\) is invertible and \(| ( M_{t}M_{t}^{*} ) ^{-1}| \le\gamma^{-2}\) for all t∈[0,T].
Proof
For any \(x \in\mathbb{R}^{|I^{\varepsilon_{0}}(t)|}\),
Then \(M_{t} M_{t}^{*} x =0 \) implies x=0, and the invertibility follows.
Let \(y \in\mathbb{R}^{|I^{\varepsilon_{0}}(t)|}\) and \(x:= ( M_{t}M_{t}^{*} ) ^{-1}y\).
For y≠0, we have x≠0; dividing the previous inequality by |x|, we get
The result follows. □
Before we prove Lemma 4.3, we define the truncation of an integrable function:
Definition A.1
Given any ϕ∈L s(J) (s∈[1,∞[ and J interval), we will call the truncation of ϕ the sequence ϕ k∈L ∞(J) defined for \(k \in\mathbb{N}\) and a.a. t∈J by
Observe that \(\phi^{k} \xrightarrow[k \to \infty]{L^{s}} \phi\).
Proof of Lemma 4.3
In the sequel we omit z 0 in the notations.
(i) Let v∈V s . We claim that v satisfies
iff there exists \(w \in L^{s}(J_{l};\mathbb{R}^{m})\) such that (v,w) satisfies
Clearly, if (v,w) satisfies (158), then v satisfies (157). Conversely, suppose that v satisfies (157). With Lemma A.4 in mind, we define \(\alpha\in L^{s}(J_{l};\mathbb{R}^{|I_{l}|})\) and \(w\in L^{s}(J_{l};\mathbb {R}^{m})\) by
Then
We derive from (157) and (159) that
Using again Lemma A.4 and (159), we get (158).
(ii) Given \((v,h,w) \in\mathit{V}_{s} \times L^{s}(J_{l};\mathbb{R}^{|I_{l}|}) \times L^{s}(J_{l};\mathbb{R}^{m})\), there exists a unique \(\tilde{v}\in\mathit{V}_{s}\) such that
Indeed, one can define a mapping from V s to V s , using the right-hand side of (160). Then it can be shown, as in the proof of Lemma 2.1, that this mapping is a contraction for a well-suited norm, using Lemmas 2.2, 2.3, and A.4. The existence and uniqueness follow. Moreover, a version of the contraction mapping theorem with parameter (see, e.g., Théorème 21-5 in [28]) shows that \(\tilde{v}\) depends continuously on (v,h,w).
(iii) Let us prove (a): let \((\bar{h},v) \in L^{s}(J_{l};\mathbb{R}^{|I_{l}|}) \times\mathit {V}_{s}\), and let w:=0. Let \(\tilde{v}\in\mathit{V}_{s}\) be the unique solution of (160) for \((v,\bar{h},w)\). Then \(\tilde{v}\) is a solution of (112) by (i).
(iv) Let us prove (b): let \((\bar{h},\bar{v}) \in L^{s}(J_{l};\mathbb {R}^{|I_{l}|}) \times \mathit{V}_{s}\) as in the statement, and let \(\bar{w}\) be given by (i). Then \(\bar{v}\) is the unique solution of (160) for \((\bar{v},\bar {h},\bar{w})\).
Let \((h^{k},v^{k}) \in L^{\infty}(J_{l}; \mathbb{R}^{|I_{l}|}) \times\mathit {U}\), \(k \in \mathbb{N}\), be such that \((h^{k},v^{k})\xrightarrow[]{L^{s}\times L^{s}} (\bar{h},\bar{v})\), and let \(w^{k} \in L^{\infty}(J_{l};\mathbb{R}^{m})\), \(k \in\mathbb{N}\), be the truncation of \(\bar{w}\). It is obvious from Definition A.1 that
Let \(\tilde{v}^{k} \in\mathit{U}\) be the unique solution of (160) for (v k,h k,w k), \(k \in\mathbb{N}\). Then, by the uniqueness and continuity in (ii),
and \(\tilde{v}^{k}\) is a solution of (114) by (i). □
We finish this section with an example where assumption (A3) can be satisfied or not.
Example A.1
We consider the scalar Example 2.1.2 with q=1 and f(t,s)=f(2t−s):
where f is a continuous function and is not a polynomial, and the trajectory \((\bar {u},\bar{y})=(0,0)\). Then
and (A3) is satisfied iff
1.3 A.3 Approximations in W q,2
We will prove in this section Lemmas 4.2 and 4.6. First, we give the statement and the proof of a general result:
Lemma A.5
Let \(\hat{x}\in W^{q,2}([0,1])\). For j=0,…,q−1, we denote
and we consider \(\alpha_{j}^{k}, \beta_{j}^{k} \in\mathbb{R}^{q}\), \(k\in \mathbb{N}\), such that \((\alpha_{j}^{k} ,\beta_{j}^{k}) \longrightarrow(\hat{\alpha}_{j},\hat {\beta}_{j})\). Then there exists x k∈W q,∞([0,1]), \(k \in\mathbb{N}\), such that \(x^{k} \xrightarrow[]{W^{q,2}} \hat{x}\) and, for j=0,…,q−1,
Proof
Given u∈L 2([0,1]), we define x u ∈W q,2([0,1]) by
Then \(x_{u}^{(q)} = u\) and, for j=0,…,q−1,
where a j ∈C([0,1]) is defined by
Indeed, a straightforward induction shows that
Then integrations by parts give the expression of the a j . Note that the a j (j=0,…,q−1) are linearly independent in L 2([0,1]). Then
is such that A ∗ A is invertible (here A ∗ is the adjoint operator), and
Going back to the lemma, let \(\hat{u}:= \hat{x}^{(q)} \in L^{2}([0,1])\). Observe that
and that \(A^{*} \hat{u}= (\hat{\gamma}_{0},\ldots,\hat{\gamma }_{q-1})^{T}\), where
Then we consider, for \(k \in\mathbb{N}\), the truncation (Definition A.1) \(\hat{u}^{k} \in L^{\infty}([0,1])\) of \(\hat{u}\) and
It is clear that u k∈L ∞([0,1]) (by the definition of A); then x k∈W q,∞([0,T]). Since A ∗ u k=γ k and in view of (165), (166), and (167), (164) is satisfied. Finally, \(\gamma^{k}_{j} \longrightarrow\hat{\gamma}_{j}\), for j=1 to q−1; then \(\gamma^{k} \longrightarrow A^{*} \hat{u}\) and \(u^{k} \longrightarrow\hat{u}\). □
We can also prove the following:
Lemma A.6
Let \(\hat{x}\in W^{q,2}([0,1])\) be such that \(\hat{x}^{(j)}(0)=0\) for j=0,…,q−1. Then for δ>0, there exists x δ∈W q,∞([0,1]) such that \(x^{\delta}\xrightarrow[\delta\to0]{W^{q,2}} \hat{x}\) and
Proof
We consider u δ∈L ∞([0,1]), δ>0, such that u δ=0 on [0,δ] and \(u^{\delta}\xrightarrow[\delta\to0]{L^{2}}\hat{u}:= \hat{x}^{(q)}\). Then we define \(x^{\delta}:=x_{u^{\delta}}\) (see the previous proof). □
Now the proof of Lemma 4.6 is straightforward.
Proof of Lemma 4.6
We observe that \(\bar{b}_{i}= 0\) on I i implies that \(\bar {b}_{i}^{(j)}=0\) at the end points of I i for j=0,…,q i −1 (note that with the definition (68), if one component of I i is a singleton, then q i =1). Then the conclusion follows with Lemma A.6 applied on each component of \(\mathit{I}_{i}^{\varepsilon}\setminus\mathit{I}_{i}\). □
Finally, we use Lemma A.5 to prove Lemma 4.2.
Proof of Lemma 4.2
In the sequel we omit z 0 in the notations. We define a connection in W q,∞ between ψ 1 at t 1 and ψ 2 at t 2 as any ψ∈W q,∞([t 1,t 2]) such that
(a) We define \(\tilde{b}_{i}\) on [0,t 0] by \(\tilde{b}_{i} := g_{i}'(\bar {y})z[v]\), i=1,…,r. We need to explain how we define \(\tilde{b}_{i}\) on ]t 0,T], using \(\bar{b}_{i}\) and connections, to have \(\tilde{b}_{i} \in W^{q_{i},s}([0,T])\) and \(\tilde{b}_{i}=\bar {b}_{i}\) on each component of \(\mathit{I}_{i}^{\varepsilon}\cap\mathopen {]}t_{0},T]\). The construction is slightly different whether \(t_{0} \in\mathit{I}_{i}^{\varepsilon}\) or not, i.e., whether \(i \in I^{\varepsilon}_{t_{0}}\) or not. Note that by the definition of ε 0 and of t 0, \(I^{\varepsilon}_{t}\) is constant for t in a neighborhood of t 0. We now distinguish the two cases just mentioned:
-
1.
\(i \in I^{\varepsilon}_{t_{0}}\): We denote by [t 1,t 2] the connected component of \(\mathit {I}_{i}^{\varepsilon}\) such that t 0∈]t 1,t 2[. We derive from (105) that \(\tilde{b}_{i}= \bar{b}_{i}\) on [t 1,t 0]. Then we define \(\tilde{b}_{i} : = \bar{b}_{i} \) on ]t 0,t 2].
If \(\mathit{I}_{i}^{\varepsilon}\) has another component in ]t 2,T], we denote the first one by \([t_{1}',t_{2}']\). Let ψ be a connection in \(W^{q_{i},\infty}\) between \(\tilde{b}_{i}\) at t 2 to \(\bar{b}_{i}\) at \(t_{1}'\). We define \(\tilde{b}_{i}:= \psi\) on \(\mathopen{]}t_{2},t_{1}'\mathclose{[}\), \(\tilde{b}_{i}:=\bar{b}_{i}\) on \([t_{1}',t_{2}']\), and so forth on \(\mathopen {]}t_{2}',T\mathopen{]}\).
If \(\mathit{I}_{i}^{\varepsilon}\) has no more component, we define \(\tilde{b}_{i}\) on what is left as a connection in \(W^{q_{i},\infty}\) between \(\bar{b}_{i}\) and \(g_{i}'(\bar {y})z[v]\) at T.
-
2.
\(i \not\in I^{\varepsilon}_{t_{0}}\): If \(\mathit{I}_{i}^{\varepsilon}\) has a component in [t 0,T], we denote the first one by [t 1,t 2]. Note that t 1−t 0≥ε 0−ε>0. We consider a connection in \(W^{q_{i},\infty}\) between \(\tilde{b}_{i}\) at t 0 and \(\bar{b}_{i}\) at t 1 and continue as in 1.
If \(\mathit{I}_{i}^{\varepsilon}\) has no component in [t 0,T], we do as in 1.
(b) For all \(k \in\mathbb{N}\), we apply (a) to (b k,v k), and we get \(\tilde{b}^{k}\). We just need to explain how we can get, for i=1,…,r,
By construction we have
Then it is enough to show that every connection which appears when we apply (a) to (b k,v k), for example, \(\psi_{i}^{k} \in W^{q_{i},\infty}([t_{1},t_{2}])\), can be chosen in such a way that
This is possible by Lemma A.5. □
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Bonnans, J.F., de la Vega, C. & Dupuis, X. First- and Second-Order Optimality Conditions for Optimal Control Problems of State Constrained Integral Equations. J Optim Theory Appl 159, 1–40 (2013). https://doi.org/10.1007/s10957-013-0299-3
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DOI: https://doi.org/10.1007/s10957-013-0299-3