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Fréchet Generalized Trajectories and Minimizers for Variational Problems of Low Coercivity

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Abstract

We address consecutively two problems. First, we introduce a class of so-called Fréchet generalized controls for a multi-input control-affine system with non-commuting controlled vector fields. For each control of the class, one is able to define a unique generalized trajectory, and the input-to-trajectory map turns out to be continuous with respect to the Fréchet metric. On the other side, the class of generalized controls is broad enough to settle the second problem, which is to prove existence of generalized minimizers of Lagrange variational problem with functionals of low (in particular linear) growth. Besides, we study the possibility of Lavrentiev-type gap between the infima of the functionals in the spaces of ordinary and generalized controls. This is an abridged (due to the journal space limitations) version of a more detailed preprint with several proofs, drawings, and examples added, published in arXiv.org > math > arXiv: 1402.0477.

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  1. 1 There is a vast bibliography dedicated to relaxation and relaxed controls, which provide generalized solutions for problems where convexity is lacking

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Acknowledgments

The research of the first coauthor has been supported by FCT–Fundação para a Ciência e Tecnologia (Portugal) via strategic project PEst-OE/EGE/UI0491/2013; he is grateful to INDAM (Italy) for supporting his visit to the University of Florence in January 2014. The research of the second coauthor has been supported by MIUR (Italy) via national project (PRIN) 200894484E of MIUR (Italy); he is also grateful to CEMAPRE (Portugal) for supporting his research stay at ISEG, University of Lisbon in May 2013.

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Correspondence to Manuel Guerra.

Appendix: Proofs of Technical Results

Appendix: Proofs of Technical Results

1.1 Proof of Lemma 6

Proof

Without loss of generality, we can assume that x has finite variation in the interval [0,T] and it is constant in [T,+[.

Consider a sequence of partitions of the interval [0,T]

$$P_{k} = \left\{ 0 =t_{k,0} < t_{k,1} < \cdots < t_{k,k} =T \right\} \qquad k \in \mathbb N, $$

such that P k P k+1 for every \(k \in \mathbb N\), and \(\bigcup \limits _{k \in \mathbb N} P_{k}\) is dense in [0,T]. Let \(x_{k}:[0,+\infty [ \mapsto \mathbb R^{n}\) be the piecewise linear function interpolating the points x(t k.i ), i=0,1,…k and x k (t)=x(T) for every t>T. Then, \(\left \{ (\theta _{k},y_{k})=\left (\ell _{(t,x_{k})}^{-1}, x_{k} \circ \ell _{(t,x_{k})}^{-1} \right ) \right \}_{k \in \mathbb N} \) is a sequence in \(\mathcal Y_{n}\).

The length of the graph of x k on the interval [0,T] is

$$\begin{array}{*{20}l} \ell_{(t,x_{k})}(T)= & \sum\limits_{i=1}^{k}\sqrt{(t_{i}-t_{i-1})^{2} + \left|x(t_{i})-x(t_{i-1})\right|^{2}} \leq T+\mathrm{V}_{[0,T]}(x) . \end{array} $$

Thus, the sequence {(𝜃 k ,y k )} is uniformly bounded and equicontinuous on the interval [0,T+V[0,T](x)], and the Ascoli-Arzelà theorem guarantees that it has a subsequence converging uniformly toward some \((\theta ,y) \in \mathcal Y_{n}\). Without loss of generality, we assume that this subsequence is {(𝜃 k ,y k )}.

Notice that \(\left \{ \theta _{k}^{-1}(t) \right \}\) may fail to converge toward 𝜃 #(t) if t is a discontinuity point of 𝜃 #. Instead, we take the sequence \(\tilde \theta _{k} = \left (\theta _{k}- \left \| \theta _{k} - \theta \right \|_{L_{\infty }[0,T+\mathrm {V}_{[0,T]}(x) ]} \right )^{+} \). Notice that \(\left \{ (\tilde \theta _{k},y_{k}) \right \}\) converges uniformly toward (𝜃,y) and \(\lim \limits _{k \rightarrow \infty } \tilde \theta _{k}^\#(t) = \theta ^\#(t)\) for every t∈[0,T]. Therefore,

$$\lim_{k \rightarrow \infty} y_{k} \circ \tilde{\theta}_{k}^\#(t) = y \circ \theta^\# (t) \qquad \forall t \in [0,T] . $$

Now, suppose that x is continuous at the point t∈[0,T]. By continuity, for every ε>0 there is some δ>0, such that |x(τ)−x(t)|<ε for every τ∈[tδ,t+δ]. This implies |x k (τ)−x(t)|<ε for every sufficiently large k and every \(\tau \in \left [ t- \frac \delta 2, t+ \frac \delta 2 \right ]\) because then x k (τ) is a convex combination of points in B ε (x(t)). Thus,

$$y \circ \theta^\# (t) = \lim_{k \rightarrow \infty} y_{k} \circ \tilde \theta_{k}^\# (t) = \lim_{k \rightarrow \infty} x_{k}\left(t+ \| \theta_{k}-\theta \|_{L_{\infty}[0,T+\mathrm{V}_{[0,T]}(x)]} \right) = x(t) . $$

1.2 Proof of Lemma 13

Proof

Notice that

$$\begin{array}{*{20}l} & L \left(y, \frac{\lambda w +(1-\lambda ) \hat w}{\lambda v +(1-\lambda ) \hat v} \right) \left(\lambda v +(1-\lambda ) \hat v \right) \\ = & L \left(y, \frac{\lambda v}{\lambda v +(1-\lambda ) \hat v} \frac w v + \frac{(1-\lambda) \hat v}{\lambda v +(1-\lambda ) \hat v} \frac{\hat w}{\hat v} \right) \left(\lambda v +(1-\lambda ) \hat v \right) \\ \leq & \left(\frac{\lambda v}{\lambda v +(1-\lambda ) \hat v} L \left(y, \frac w v \right) + \frac{(1-\lambda) \hat v}{\lambda v +(1-\lambda ) \hat v} L \left(y, \frac{\hat w}{\hat v} \right) \right) \left(\lambda v +(1-\lambda ) \hat v \right) \\ = & \lambda L \left(y, \frac w v \right) v + (1-\lambda) L \left(y, \frac{\hat w}{\hat v} \right) \hat v . \end{array} $$

Therefore, \((v,w) \mapsto L \left (y, \frac w v \right ) v\) is convex.

The inequality

$$\begin{array}{*{20}l} & \liminf_{\scriptsize \begin{array}{c} (y,v,w) \rightarrow (\hat y, \hat v, \hat w) \\ v>0 \end{array}} L\left(y, \frac w v \right) v \leq \\ \leq & \liminf_{\scriptsize \begin{array}{c} (v,w) \rightarrow (\hat v, \hat w) \\ v>0 \end{array}} L\left(\hat y, \frac w v \right) v \leq \liminf_{v \rightarrow \hat v, \ v>0} L\left(\hat y, \frac {\hat w} v \right) v \end{array} $$

holds trivially. Therefore, we only need to prove that

$$\begin{array}{*{20}l} & \limsup_{v \rightarrow \hat v, \ v>0} L\left(\hat y, \frac {\hat w} v \right) v \leq \liminf_{\scriptsize \begin{array}{c} (y,v,w) \rightarrow (\hat y, \hat v, \hat w) \\ v>0 \end{array}} L\left(y, \frac w v \right) v . \end{array} $$
(42)

Due to continuity of L, this inequality holds for every \(\hat v >0\). Suppose \(\hat v =0\) and fix \(b< \limsup \limits _{v \rightarrow 0^{+}} L \left (\hat y, \frac {\hat w}{v} \right ) v \) and ε>0, then we can pick a∈]0,ε], such that \(L \left (\hat y , \frac { \hat w}{ a} \right ) > b \frac {1}{a}\). By continuity of L, there is some δ>0, such that

$$ L\left(y, \frac{w}{a} \right) > \frac{b}{a}, \quad \text{and} \quad \left| L(y,w)- L(\hat y, \hat w) \right| < \varepsilon $$
(43)

for every (y,w), such that \(|y- \hat y| < \delta \) and \(|w-\hat w| < \delta \).

Due to convexity of wL(y,w), we have

$$\begin{array}{*{20}l} L\left(y, \frac w v \right) \geq & L(y,w) + \frac{L\left(y, \frac{w}{a}\right) - L(y,w)}{\frac{1}{a}-1 } \left(\frac 1 v -1 \right) \\ = & L(y,w) + \frac{a}{1-a} \left(L\left(y, \frac{w}{a}\right) - L(y,w)\right) \frac{1-v}{v} \qquad \forall v \in ]0, a] . \end{array} $$

Using the estimates (43), this yields

$$\begin{array}{*{20}l} L \left(y, \frac w v \right) v \geq & \left(L \left(\hat y, \hat w \right) - \varepsilon \right) v + \frac{a}{1-a} \left(\frac b a -L\left(\hat y, \hat w \right) - \varepsilon \right) (1-v) \\ = & \frac{1-v}{1-a}b + L\left(\hat y , \hat w \right) \left(v-a \frac{1-v}{1-a} \right) - \varepsilon \left(v+a \frac{1-v}{1-a} \right) , \end{array} $$

that is,

$$\liminf_{\scriptsize \begin{array}{c} (y,v,w) \rightarrow (\hat y, \hat v, \hat w) \\ v>0 \end{array}} L\left(y, \frac w v \right) v \geq \frac{1}{1-a}b - \left(L\left(\hat y , \hat w \right) +\varepsilon \right) \frac{a}{1-a} . $$

Making ε tend to zero and b tend to \(\limsup \limits _{v \rightarrow 0^{+}} L \left (\hat y, \frac {\hat w}{v} \right ) v \), this implies (42).

1.3 Proof of Proposition 19

Proof

Fix (C,𝜃,y), a trajectory of the differential inclusion (32), and let \(V_{t}= \left (\dot \theta (t), \dot y(t) \right )\) for almost every t∈[0,T].

For each compact set \(K \subset \mathbb R^{1+k}\), consider the function \(F_{K}:[0,T] \mapsto \overline {\mathbb R}\), defined almost everywhere by

$$F_{K}(t) = \inf \left\{ \lambda (y(t),v,w) : (v,w) \in B^{+} \cap K , \left(v,f(y(t))v+G(y(t))w \right) =V_{t} \right\}, $$

being understood that inf=+.

First, we show that the functions F K are measurable.

For any set \(A \subset \mathbb R^{k}\) and any ε>0, let \(B_{\varepsilon }(A) = \bigcup \limits _{x \in A} B_{\varepsilon }(x)\). Then, lower semicontinuity of λ implies that for any \(\alpha \in \mathbb R\),

$$\begin{array}{*{20}l} & F_{K}^{-1}\left([-\infty, \alpha ] \right) \\ = & \left\{ t: \exists (v,w) \in B^{+} \cap K , \left(v,f(y(t))v+G(y(t))w \right) =V_{t}, \lambda(y(t),v,w) < \alpha \right\} \\ = & \bigcap_{i \in \mathbb N} \begin{array}[t]{l} \Big\{ t: \exists (v,w) \in B_{\frac 1 i} \left(B^{+} \cap K \right), \left| \left(v,f(y(t))v+G(y(t))w \right) -V_{t} \right|< \frac 1 i, \\ \hspace{2cm} \lambda(y(t),v,w) < \alpha \Big\} . \end{array} \end{array} $$

Due to Lemma 13, this is

$$\begin{array}{*{20}l} & F_{K}^{-1}\left(]-\infty, \alpha ] \right) \\ = & \bigcap_{i \in \mathbb N} \bigcup_{v \in \mathbb Q \cap ]0,1]} \begin{array}[t]{l} \Big\{ t: \exists w \in \mathbb R^{k}, (v,w) \in B_{\frac 1 i} \left(B^{+} \cap K \right) , \\ \hspace{0.3cm} \left| \left(v,f(y(t))v+G(y(t))w \right) -V_{t} \right|< \frac 1 i, L\left(y(t),\frac w v \right)v < \alpha \Big\} . \end{array} \end{array} $$

Due to continuity of L, this further reduces to

$$\begin{array}{*{20}l} & F_{K}^{-1}\left(]-\infty, \alpha ] \right) \\ = & \bigcap_{i \in \mathbb N} \bigcup_{v \in \mathbb Q \cap ]0,1]} \bigcup_{{\scriptsize \begin{array}{c}w \in \mathbb Q^{k}: \\ (v,w) \in B_{\frac 1 i} \left(B^{+} \cap K \right)\end{array}}} \begin{array}[t]{l} \Big\{ t: \left| \left(v,f(y(t))v+G(y(t))w \right) -V_{t} \right|< \frac 1 i, \\ \hspace{1cm} L\left(y(t),\frac w v \right)v < \alpha \Big\} . \end{array} \end{array} $$

Since y, V are measurable and f,G,L are continuous, it follows that F K is measurable.

Now, we construct a sequence \(\{ \mathcal A_{i} \}_{i \in \mathbb N }\) with the following properties:

  • (a) Each \(\mathcal A_{i} = \{ A_{i,1}, A_{i,2}, \ldots , A_{i,h_{i}} \}\) is a finite ordered collection of measurable subsets of B +;

  • (b)All the members of each collection \(\mathcal A_{i}\) are pairwise disjoint, \(B^{+} = \bigcup \limits _{A \in \mathcal A_{i}} A\), and each element of \(\mathcal A_{i}\) is contained in a ball of radius \(\frac 1 i\);

  • (c) For any i<j, every element of \(\mathcal A_{j}\) is a subset of some element of \(\mathcal A_{i}\). All elements of \(\mathcal A_{j}\) that are contained in A i,h precede (in the order of \(\mathcal A_{j}\)) any element of \(\mathcal A_{j}\) contained in A i,h+1.

To see that such sequences exist, let \(\mathcal A_{0}=\left \{ B^{+} \right \}\). For each \(i \in \mathbb N\), let \(\mathcal B_{i} = \{ B_{i,1},B_{i,2}, \ldots , B_{i,j_{i}} \}\), a finite cover of B + by balls of radius \(\frac 1 i\), and let

$$C_{i,h} = B_{i,h} \setminus \bigcup_{l<h}B_{i,l}, \qquad h=1,2, \ldots , j_{i} .$$

For each \(i \in \mathbb N\), let \(\mathcal A_{i}\) be the collection of intersections

$$A \cap C_{i,h}, \qquad A \in \mathcal A_{i-1}, \quad h = 1,2, \ldots , j_{i} ,$$

ordered in any way such that any C i,h A i−1,l precedes every C i,s A i−1,l+1 (discard empty intersections). So, \(\{ \mathcal A_{i} \}_{i \in \mathbb N }\) satisfies (a)–(c).

Fix a sequence \(\{ \mathcal A_{i} \}_{i \in \mathbb N }\) as above and for each \(i \in \mathbb N\), j∈{1,2,…,j i }, fix (v,w) i,j =(v i,j ,w i,j )∈A i,j . For each \(i \in \mathbb N\), consider a function \(j(i, \cdot ):[0,T] \mapsto \mathbb N\) defined almost everywhere by

$$ j(i,t) = \min \left\{ h\in \{ 1,2, \ldots , j_{i}\}: F_{\overline{A_{i,h}}}(t)=F_{B^{+} }(t) \right\} , $$
(44)

and consider the sequence \(\{(v_{i},w_{i}): [0,T] \mapsto B^{+} \}_{i \in \mathbb N}\) defined as

$$(v_{i},w_{i})(t)=(v,w)_{i,j(i,t)}, \qquad i \in \mathbb N , \ t \in [0,T]. $$

Notice that (v i ,w i )([0,T])⊂{(v,w) i,j ,j∈{1,2,…,j i }} is a finite set and

$$\begin{array}{*{20}l} & \{ t: (v_{i},w_{i})(t)=(v,w)_{i,j} \} = \left\{ t: F_{\overline{A_{i,j}}}(t)=F_{B^{+} }(t) \right\} \setminus \bigcup_{h<j} \left\{ t: F_{\overline{A_{i,h}}}(t)=F_{B^{+} }(t) \right\} . \end{array} $$

Therefore, measurability of F K guarantees measurability of (v i ,w i ).

For almost every t∈[0,T], we have the following:

  • \(\left (v_{i}(t),f(y(t))v_{i}(t) + G(y(t))w_{i}(t) \right )= V_{t} \qquad \forall i \in \mathbb N\), and

  • \( \{ (v_{i},w_{i})(t) \}_{i \in \mathbb N} \) is a Cauchy sequence.

Thus, \((v,w)(t)= \lim \limits _{i \rightarrow \infty } (v_{i},w_{i})(t)\) is a measurable function satisfying

$$\left(v(t),f(y(t))v(t) + G(y(t))w(t) \right)= V_{t} \qquad \text{a.e. }t \in [0,T].$$

Lower semicontinuity of λ and (44) imply that

$$\begin{array}{*{20}l}& \lambda (y(t),v(t),w(t)) \\ = & \inf\left\{ \lambda (y(t),\tilde v, \tilde w): (\tilde v, \tilde w) \in B^{+} , \left(\tilde v, f(y(t))\tilde v +G(y(t)) \tilde w \right)=V_{t} \right\} \\ \leq & \dot C(t) \end{array} $$

for almost every t∈[0,T].

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Guerra, M., Sarychev, A. Fréchet Generalized Trajectories and Minimizers for Variational Problems of Low Coercivity. J Dyn Control Syst 21, 351–377 (2015). https://doi.org/10.1007/s10883-014-9231-x

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