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Existence results for optimal control problems with some special non-linear dependence on state and control

We present a general approach to prove existence of solutions for optimal control problems not based on typical convexity conditions which quite often are very hard, if not impossible, to check. By taking advantage of several relaxations of the problem, we isolate an assumption which guarantees the existence of solutions of the original optimal control problem. To show the validity of this crucial hypothesis through various means and in various contexts is the main body of this contribution. In each such situation, we end up with some existence result. In particular, we would like to stress a general result that takes advantage of the particular structure of both the cost functional and the state equation. One main motivation for our work here comes from a model for guidance and control of ocean vehicles. Some explicit existence results and comparison examples are given.


Introduction
This paper focuses on the analysis of optimal control problems of the general form (P 1 ) Minimize in u : subject to and u ∈ L ∞ (0, T ), u(t) ∈ K, where K ⊂ R m is compact. The state x : (0, T ) → R N takes values in R N . The mappings as well as the restriction set K ⊂ R m will play a fundamental role. We assume, at this initial stage, that c i are continuous, φ i are of class C 1 , and each Q i is Lipschitz so that the state system is well-posed. In such a general form, we cannot apply results for non-necessarily convex problems like the ones in [2], [5], [24] or [29]. Besides, techniques based on Bauer' Maximum Principle ( [3]) are quite difficult to extend to our general setting because it is hard to analyze the concavity of the cost functional when the dependence on both state and control comes in product form. Also the Rockafellar's variational reformulation introduced in [26], and well-described in [6], [11] or recently in [22] or [23], looks as if it cannot avoid assuming a separated dependence on the state and control variables, since this is the structure of the variational problem for which the existence of solution has been so far ensured ( [7]).
Concerning the classical Filippov-Roxin theory introduced in [12] and [30], it is not easy at all to know if typical convexity assumptions hold, or when they may hold, as we can see from the examples and counter-examples in [6]. When analyzing explicit examples, one realizes such difficulties coming from the need of a deep understanding of typical orientor fields. The same troubles would arise when applying refinements of this result as the ones in [19] and [20].
Recently ([8]), an existence result has been shown for minimum time problems where the typical convexity assumptions over the set valued function on the differential inclusion has been replaced by more general conditions. In fact, the intersection of this result with the ones we present here is not empty although, as we will comment, our frame extends to situations not covered by this result. Such analysis can be done by writing problem (P 1 ) as a minimum time problem as suggested in [6].
Our aim is to provide hypotheses on the different ingredients of the problem so that existence of solutions can be achieved through an independent road. Actually, it is not easy to claim whether our results improve on classical or more recent general results. They provide an alternative tool which can be more easily used in practice than such results when one faces an optimal control problem under the special structure we consider here. As a matter of fact, convexity will also occur in our statements but in an unexpected and non-standard way.
Before stating our main general result, a bit of notation is convenient. We will write with components c i , φ i , and Q i , respectively. Consider also a new ingredient of the problem related to φ. Suppose that there is a C 1 mapping Ψ : R s → R s−n , Ψ = (ψ 1 , ..., ψ s−n ), (s > n), so that φ(K) ⊂ {Ψ = 0}. This is simply saying, in a rough way, that the embedded (parametrized) manifold φ(K) of R s is part of the manifold defined implicitly by Ψ = 0. In practical terms, it suffices to check that the composition Ψ(φ(u)) = 0 for u ∈ K. For a pair (c, Q), put Similarly, set Our main general result is the following.
Theorem 1.1. Assume that the mapping Ψ as above is strictly convex (componentwise) and then the corresponding optimal control problem (P 1 ) admits at least one solution.
As it stands, this result looks rather abstract, and it is hard to grasp to what extent may be applied in more specific situations.
A particular, yet still under some generality, situation where this result can be implemented is the case of polynomial dependence where the φ i 's are polynomials of various degrees. The main structural assumption, in addition to the one coming from the set K, is concerned with the convexity of the corresponding mapping Ψ.
Suppose we take φ i (u) = u i , for i = 1, 2, . . . , n, and φ n+i (u), i = 1, 2, . . . , s − n, convex polynomials of whatever degree, or simply polynomials whose restriction to K is convex. In particular, K itself is supposed to be convex. Then we can take In this case, it is clear that Ψ(φ(u)) = 0 for u ∈ K, by construction, and, in addition, Ψ is smooth and convex. The important constraint (8) can also be analyzed in more concrete terms, if we specify in a better way the structure of the problem.
As an illustration, though more general results are possible, we will concentrate on an optimal control problem of the type subject to We are taking here N = n. Q 1 and Q 2 are n × n matrices that, together with the vector Q 0 , comply with appropriate technical hypotheses so that the state law is a well-posed problem.
where Q 1 is a non-singular n × n matrix, and c 1 ∈ R n . In addition, we put where the e i 's stand for the vectors of the canonical basis of R n , and id is the identity matrix of size n × n.
Theorem 1.2. Suppose that for the ingredients (c, Q, K) of (P ), we have 1. the matrix U is always non-singular for u ∈ K, and x ∈ R n ; 2. for such pairs (u, x), we always have U −T E < 0, componentwise.
Then the optimal control problem admits solutions.
As a more specific example of the kind of existence result that can be obtained through this approach, we state the following corollary whose proof amounts to going carefully through the arithmetic involved after Theorem 1.2. Corollary 1.1. Consider the optimal control problem Minimize in u : and an initial condition and Then there is, at least, one optimal solution of the problem.
Our strategy to prove these results is not new as it is based on the well-established philosophy of relying on relaxed versions of the original problem, and then, under suitable assumptions, prove that there are solutions of the relaxed problem which are indeed solutions of the original one ( [9], [14], [16], [17], [32] and [33]). From this perspective, it is a very good example of the power of relaxed versions in optimization problems.
The relaxed version of the problem that we will be using is formulated in terms of Young measures associated with sequences of admissible controls. These so-called parametrized measures where introduced by L. C. Young ([33], [34] and [35]), and have been extensively used in Calculus of Variations and Optimal Control Theory ( see for example [20], [21], [27] and [28]). Because of the special structure of the dependence on u, we will be concerned with (generalized) "moments" of such probability measures. Namely, the set and the space of moments will play a fundamental role. Here P (K) is the convex set of all probability measures supported in K. Since the mapping is linear, we easily conclude that Λ is a convex set of vectors, and, in addition, that the set of its extreme points is contained in L. In fact, for some particular φ i 's of polynomial type, the set of the extreme points of Λ is precisely L. We examine and comment on the set Λ in Section 3. This is closely related to the classical moment problem ([1], [31] or more recently [10], [18]). A crucial fact in our strategy is the following. It is interesting to realize the meaning of this assumption. If we drop the linear constraint ξ = Qm on the above minimum, then the minimum is always attained in a certain point in L simply because a linear function on a convex set will always take its extreme values on extreme points of such convex set. However, precisely the presence of the linear constraint ξ = Qm makes the hypothesis meaningful as the extreme points of the section of Λ by such set of linear constraints may not (indeed most of the time they do not) belong to L, so that the extreme points of the linear function c · m over such convex section may not attain its minimum on L.
Our main hypothesis establishes that this should be so, and fundamentally, that the minimum is only attained in L.
Under this assumption, and the other technical requirements indicated at the beginning, one can show a general existence theorem of optimal solutions for our problem. Notice that we are not assuming any convexity on the set K in this statement. The proof of this theorem can be found in Section 2. As remarked before, the proof is more-or-less standard, and it involves the use of an appropriate relaxed formulation of the problem in terms of moments of Young measures ( [20], [28]).
Condition (8) in Theorem 1.1 is nothing but a sufficient condition to ensure Assumption 1.1 in a more explicit way. As a matter of fact, all of our efforts are directed towards finding in various ways more explicit conditions for the validity of this assumption. In this vein, the rest of the paper focuses on exploring more fully our Assumption 1.1 either through duality, geometric arguments, or in order to prove Theorem 1.1. Ideally, one would like to provide explicit results saying that for a certain set M, In fact, by looking at Assumption 1.1 from the point of view of duality, one can write a general statement whose proof is a standard exercise. We briefly comment on this in Section 3. One then says that (c, Q) ∈ M if this pair verifies the condition on this proposition. A full analysis of this set M turns out to depend dramatically on the ingredients of the problem. In particular, we will treat the cases n = N = 1, and the typical situation of algebraic moments of degree 2 and 3 in Section 4, Section 5, and Section 6. In Section 7 we apply our results to a few explicit examples and compare it with the application of the classical Filippov-Roxin theory.
Situations where either N > 1 or n > 1 are much harder to deal with, specially because existence results are more demanding on the structure of the underlying problem. In particular, we need a convexity assumption on how the non-linear dependence on controls occurs. We found that (8) turns out to be a general sufficient condition for the validity of Assumption 1.1, thus permitting to prove Theorem 1.1 based on Theorem 1.3. Both Theorem 1.2, and Corollary 1.1 follow then directly from Theorem 1.1 after some algebra. This can be found in Section 8.
Finally, we would like to point out that one particular interesting example, from the point of view of applications, that adapts to our results comes from the control of underwater vehicles (submarines). See [4], [13], and [15]. This served as a clear motivation for our work. We plan to go back to this problem in the near future.

Proof of Theorem 1.3
Consider the following four formulations of the same underlying optimal control problem.
(P 2 ) The relaxed formulation in terms of Young measures ( [20], [21], [27], [28]) associated with sequences of admissible controls: subject to (P 3 ) The above relaxed formulation (P 2 ) rewritten by taking advantage of the moment structure of the cost density and the state equation. If we put c = (c 1 , ..., c s ) ∈ R s , Q ∈ M N ×s and m such that then we pretend to (P 4 ) Variational reformulation of formulation (P 3 ) ( [6], [22], [23], [26]). This amounts to defining an appropriate density by setting Then we would like to Minimize in x(t) : We know that the three versions of the problem (P 2 ), (P 3 ), and (P 4 ) admit solutions because they are relaxations of the original problem (P 1 ). In fact, since K is compact, (P 2 ) is a particular case of the relaxed problems studied in [20] and [28]. The existence of solution for the linear optimal control problem (P 3 ) is part of the classical theory ( [6]). Indeed, (P 3 ) is nothing but (P 2 ) rewritten in terms of moments, so that the equivalence is immediate. (P 4 ) is the reformulated problem introduced in [26] whose equivalence to (P 3 ) was largely explored in [6] and [22], [23].

The set Λ and duality
The moment set Λ deserves some comments before proceeding further. Consider the mapping φ as in (4) and L as in (15). We can regard L as part of an embedded n-manifold in R s , s > n, and φ its standard or canonical parametrization. The moment set Λ defined in (16) is contained in the convex hull of this manifold.
The most important fact about Λ that one may need in our analysis is stated in the next proposition.
Proposition 3.1. The set of extreme points of Λ is contained in L.
Proof. First notice that, as it was shown in [18] in a context similar to ours, the compactness of K implies co(L) = co(L) =Λ = Λ.
The fact of K being bounded plays an important role because otherwise Λ can be shown to be not necessarily closed ( [10]). Since Λ = co(L) then it is known from convex analysis ( [25]) that where ext(Λ) represents the extreme points of Λ.
Remark 3.1. For some φ's it is possible to conclude that ext(Λ) = L. This is the case, for example when φ contains all the linear and quadratic terms of a n-variable polynomial. However this is not essential in what follows.
Due to this result the proof of Proposition 1.1 is standard (see [25]), so that we shall only make a few remarks. Since which is a compact set, the minimum of (c + ηQ)m in Λ is always attained at least in one point of L (it can be attained also in points of Λ \ L). However, if this point happens to be unique, because of Proposition 3.1, it is also immediate to check that it must be the unique minimizer in Λ.
The condition (17)  for ξ ∈ Q(x)Λ, that is, such that Assumption 1.1 is non empty.
In particular the associated Karush-Kuhn-Tucker vectorη verifies (see [25]) for a single a ∈ K complying with Qφ(a) = ξ. As a consequence, for all admissible m ∈ Λ different from φ(a), we have c · m > c · φ(a).
4 Polynomial Dependence. The case N = n = 1, p = 2 Until Section 8, we concentrate in the situation where φ : R n → R s is such that φ i (u) = u i , for i = 1, 2, . . . , n, and φ n+i (u), i = 1, 2, . . . , s−n are convex polynomials of some degree p, or simply polynomials whose restriction to K is convex. We will consider K itself to be convex. Our goal is to explore different possibilities to apply directly Theorem 1.3 by ensuring Assumption 1.1. In other words, we will search for functions During the following three sections we will focus on the one dimensional case N = n = 1 and use some ideas based in duality (Proposition 1.1) and in geometric interpretations.
In Sections 4, 5, and 6, we explore various scenarios where Assumption 1.1 can be derived, and defer explicit examples until Section 7. In particular, we consider in this section the situation where φ is given by φ(a) = (a, a 2 ). We are talking about polynomial components of degree less or equal than p = 2.
Let K = [a 1 , a 2 ], L, and Λ as in (15)- (16). Here, we have s = 2 and can be identified with vectors in R 2 , or more precisely, with plane curves parametrized by x.
To emphasize that function Q is not a matrix-valued but vector-valued, we will call it q. Next we describe sufficient conditions for (c(x), q(x)) ∈ M.
Lemma 4.1. Let K, L and φ be as above. For every x ∈ R, let q = q(x) and c = c(x) be vectors such that one of the following conditions is verified 1. q 1 + q 2 (a 1 + a 2 ) = 0 and det c 1 c 2 q 1 q 2 = 0; 2. q 1 + q 2 (a 1 + a 2 ) = 0 and Then (c, q) ∈ M, and consequently Assumption 1.1 is verified.
Proof. Suppose there is η such that the minimum of (c + ηq) · m is attained in more than one point of L = φ(K). This means that the real function has more than one minimum point over K. For that to happen, either g is constant on t, i. e., which contradicts our hypothesis; or else we must have This condition can be written as c 1 + (a 1 + a 2 )c 2 + η[q 1 + (a 1 + a 2 )q 2 ] = 0.
If q 1 + q 2 (a 1 + a 2 ) = 0, but c 1 + (a 1 + a 2 )c 2 = 0 (condition 1. in statement of lemma), then this equation can never be fulfilled. Otherwise, there is a unique value for η, by solving this equation, which should also verify the condition on the sign of c 2 + ηq 2 . It is elementary, after going through the algebra, that the condition on this sign cannot be true under the second condition on the statement of the lemma.
5 The case N = n = 1, p = 3 We study the case where φ(a) = (a, a 2 , a 3 ), s = 3, and c and q can be identified as vectors in R 3 . The understanding of the set Λ and its sections by planes in R 3 is much more subtle however.
To repeat the procedure used for p = 2, and apply Proposition 1.1, we would like to give sufficient conditions for the function to have a single minimum over K = [a 1 , a 2 ] for every η. As indicated, and after some reflection, a complete analysis of the situation is rather confusing and the conditions on the vectors c and q much more involved. To illustrate this, we give a sufficient condition in the following form.
Lemma 5.1. For all x ∈ R, let c = c(x) and q = q(x) be vectors in R 3 such that then (c, q) ∈ M, and Assumption 1.1 is verified. Proof. The proof consists in the realization that the conditions on the vectors c and q ensure that the cubic polynomial (19) is monotone in all of R (avoiding degenerate situations), and thus it can only attain the minimum in a single point of any finite interval. Notice that this condition is independent of the interval. In fact, we have to discard the possibility for the derivative of the polynomial g(t) to have roots. This amounts to the negativity of the corresponding discriminant. And this, in turn, is a quadratic expression in η that ought to be always negative. This occurs when that parabola has a negative discriminant, and the leading coefficient is also negative. These two conditions are exactly the ones in the statement of this lemma.
A more general condition would focus on considering the local maximizer and the local minimizer of g(t), M + and M − , respectively, and demanding that the interval [a 1 , a 2 ] have an empty intersection with the interval determined by M + and M − . But this would lead to rather complicated expressions. Even so, some times under more specific hypotheses on the form of the vectors c and q, these conditions can be exploited. is not true for a general K if it has positive and negative values. However, it is true if we consider a 1 > 0 or a 2 < 0.
Then the assumptions of Proposition 1.1 are valid, and consequently so is Assumption 1.1.
Proof. In this situation, the maximizer M + referred to above is given by If q 2 > 0, then q 3 > 0, and we have − In any case M + (η) ≤ 0, thus a 1 > M + .
which has a unique minimum in K since we have assumed a 1 > 0. We conclude that the condition (17) in Proposition 1.1 is verified.
In a very similar way we can prove the following.
Then (c, q) ∈ M and consequently Assumption 1.1 is valid.
6 A geometric approach to the case N = n = 1, p = 3.
As we have seen, the use of Proposition 1.1 is simpler only when restricted to some particular classes of examples. Thus we propose a general criteria for obtaining Assumption 1.1, based on a geometric approach. We first give a result that generalizes the strictly convexity of a φ-parametrized plane curve for a 3-dimensional one. Lemma 6.1. Let K = [a 1 , a 2 ] with a 1 > 0, φ(t) = (t, t 2 , t 3 ), and L the curve parametrized by φ for t in K.
1. Given t in K, then for all s ∈ K such that s = t we have where N (t) is the normal vector to φ at t.

For every
Proof. To check the first part of the statement notice that since φ (t) = (1, 2t, 3t 2 ) and φ (t) = (0, 2, 6t) we have that the normal vector, colinear to φ (t) × φ (t), is given by where c t > 0 is a normalizing constant. Setting we find that the solution s of (φ(s) − φ(t)) · N (t) = 0 also verifies This means thatŝ is the only solution different from t, but also that it is negative for all t > 0, and consequently that it should be excluded. Once we assumed K ⊂ R + and s = t the conclusion is immediate.
By using the previous discussion, proving the second part of the statement is trivial once we notice that both m and φ(t) can be rewritten as Another useful lemma.
Lemma 6.2. If q and c are such that does not change sign for t, s ∈ K, s = t, then if v ∈ Λ, v = φ(t), and q · (v − φ(t)) = 0 we have This means that the linear function c cannot take the same value over φ(t) and any v = φ(t) in the plane section {v ∈ Λ : q · v = q · φ(t)}.
Proof. Notice that for v ∈ Λ, so that the condition stated is also verified for any v ∈ Λ. Suppose now that v ∈ Λ verifies q · (v − φ(t)) = 0 for given t ∈ K with v = φ(t) and is such that c · (v − φ(t)) = 0, then a contradiction concerning the argument above.
We now define the set M 1 of pairs (c, q) ∈ R 3 × R 3 through the following requirements: • the quantity in (20) does not change sign over the pairs t, s ∈ K, s = t; • whenever there is a unique a ∈ K = [a 1 , a 2 ] such that then (φ(a 1 ) + φ(a 2 ) − 2φ(a)) · c > 0.
Once more we can establish the following result.
Proof. 1. Suppose first that there is a ∈ K such that we have (21). Let v a = φ(a 1 ) + φ(a 2 ) 2 . Consider and consider the continuous function over the bounded path connecting (v a , φ(a)) and (v, φ(a) given by It is easy to check that every component of a vector of S is contained in the section {v ∈ Λ : q · v = q · φ(a)}.
Then there exists α such that which by Lemma 6.2 means that necessarily Consequently and this is in contradiction with Lemma 6.1. Hence Lett be such that q · φ(a 1 ) = q · φ(t) and t = a, t ≥t, such that Considering once more the continuous function G(v, u) over the path connecting (v t , φ(t)) and (v a , φ(a)), as we can, as we did above, conclude that if for any t ≥t. The same type of arguments show that If t <t, there exists s ∈ K such that q · φ(s) = q · φ(t).
If there ist = a 2 such that q · φ(a 2 ) = q · φ(t) we shall proceed in an analogous way. 2. Suppose now that there are a, b ∈ K such that Then it is not difficult to conclude that a = a 1 and b = a 2 .
Hence assuming (without loss of generality) that (φ(a 1 ) − φ(a 2 )) · c > 0 we can, once again, use the continuity of G to conclude where φ(s) and φ(t), verify (φ(s) − φ(t)) · q = 0 and after that, for a general v such that Remark 6.1. This type of argument can be also deduced for the case N = n = 1, p = 2 where it can be seen to be equivalent to the conditions in Lemma 4.1. However when the parameters N , n and p increase their values, it becomes very hard to give geometrically-based sufficient conditions in such an exhaustive manner as we have done here. Even so, in Section 8 we show how to give more restrictive yet more general sufficient conditions (Theorem 1.2 and Theorem 1.1) for interesting high dimensional particular situations, where some geometrical ideas can be used as a way to verify Assumption 1.1.

Examples
Before going further to higher dimensional situations we gather in this section some typical, academic examples for which either Lemma 4.1, Lemma 5.2, or Proposition 6.1 can be applied.

Example 1
Let us consider the optimal control problem Minimize in u : where |u(t)| ≤ 1.
We have the following remarkable existence result. then the optimal control problem admits solution.
The proof reduces to performing some elementary algebra to check the conditions of Lemma 4.1.
Instead of applying that lemma, as both our cost and state-equation functions have cross dependence on x and on u so that we can't apply results in [3], [24], one can try the classical existence result based on the classical Filippov-Roxin theory. For that we need to check if the orientor field is a convex set. Notice that K is bounded so coercivity is not an issue here. Proceeding in that direction, we can see that ξ = q(x)u + u 2 is equivalent to which are possible solutions when ξ is such that ξ ≥ − q 2 4 , and at least one of them belongs to for all ξ as above. Consequently refers to the pre-image of the solutions u i as functions of ξ. Because of the assumption on (c, q) it is easy to see that A 2 x = ∅, and consequently that the convexity of A x reduces to the convexity of the function over a certain convex set This can be checked by elementary calculus. We now turn over the possibility of applying the result in [8] to this example. First, in order to write our problem as a minimum time problem, we need that c(x)u + u 2 never changes sign in R × K ( [6]). So a first restriction must be imposed. For example, consider c(.) and q(.) such that q(x) > c(x) > 1.
The right member of the differential equation of the minimum time problem is given by The result in [8] doesn't ask for the convexity of the set-valued map F, but it requires a linear boundedness in the sense that ∃α, β s. t. ∀x ∈ R, ∀ξ ∈ F(x, K) then It is easy to see that this condition places a real constraint on the relative growth of pairs (c, q), even before verifying the remaining assumptions in [8].

Example 2
Look at the problem Minimize in u :  Like we have seen in the previous example we need to check the convexity of the orientor field In this case, accordingly to the discriminant we will have from one to three possible real solutions. Consider for each ξ where u i = u x i (ξ), i = 1, 2, 3 are the three, possible equal, real solutions. Then } Checking the convexity of this set, or alternatively, of the function is not an easy task at all, specially when compared to the almost immediate exercise of verifying the conditions of Lemma 5.2. It is also plausible that the inherent difficulties to apply classical theory will increase until a practically impossible scenario when we let N , n and p grow.

Example 3
In order to give an heuristic for using the criteria given in Proposition 6.1 let us consider the previous problem, just by rewriting q as c − β and for a specific K. then the optimal control problem admits solutions.

Minimize in u :
Proof. First notice that for a ∈ K = [a 1 , a 2 ], we can find α such that the vector Moreover, it is not difficult to see that and in the projection plane yz, (B 2 , B 3 ) belongs to the line of slope m passing through (a 2 , a 3 ), where B 2 − a 2 = (a − a 1 )[a 2 (a 2 + a 1 ) − a 2 (a + a 1 )] a 2 2 + a 1 a 2 + a 2 1 − m(a 2 + a 1 ) and m = − q 2 q 3 . In our case K = [1,2], so, because of what we have just seen, taking a 1 = 1 and a 2 = 2 we see that for a ∈ K, we can find Furthermore, it is easy to see that the equation α = 1 2 has a unique solution in K. Consequently, if we consider q = (0, c − β, 1) andc = (0, c, 1), there exists a unique a ∈ K such that Also, because of what we have seen above In addition, given t, s ∈ K, s = t, which is impossible since s ∈ K = [1, 2] and s = t. The result follows then by applying Proposition 6.1.

The case N , n > 1
The previous analysis makes it very clear that checking Assumption 1.1 may be a very hard task as soon as n and/or N become greater than 1. Yet in this section we would like to show that there are chances to prove some non-trivial results. The three main ingredients in Assumption 1.1 are: • the vector c ∈ R s in the cost functional; • the matrix Q ∈ M N ×s occurring in the state equation; • the convexification Λ of the set of moments L.
For (c, Q) given, consider the set N (c, Q) as it was defined in (6). Let Ψ be as in (5) and such that ∇Ψ(m) is a rank s − n matrix and L can be seen as the embedded (parametrized) manifold of R s in the manifold defined implicitly by Ψ = 0. This means that Ψ(φ(u)) = 0 for all u ∈ K.
Consider also the set of vectors N (K, φ) described in (7), that is, the set of "ascent" directions for Ψ at points of L.
We are now in conditions to prove Theorem 1.1.
Proof. The proof is rather straightforward. Firstly, note that due to the convexity assumption on Ψ, and the fact that L ⊂ {Ψ = 0}, we have Λ ⊂ {Ψ ≤ 0}. Suppose that m 0 ∈ L and m 1 ∈ Λ, so that Though Theorem 1.1 can be applied to more general cases, we will focus on a particular situation motivated by the control of underwater vehicles ( [4]). We will briefly describe the structure of the state equation. Indeed, it is just where the state x ∈ R 12 incorporates the position and orientation in body and world coordinates, and the control u ∈ R 10 accounts for guidance and propulsion. Under suitable simplifying assumptions ( [4], [13], [15]), the components of the control vector u only occur as either linear or pure squares, in such a way that φ(u) = (u, u 2 ) ∈ R 20 , and u 2 = (u 2 i ) i , componentwise. Q 1 and Q 0 are matrices which may have essentially any kind of dependence on the state x.
To cover this sort of situations just described, we will concentrate on the optimal control problem (P ) already stated in (10)- (12), and set D, E and U as in (13)-(14).
We can now prove Theorem 1.2.
Proof. Notice that accordingly to (9) A more explicit way of writing this is So v 1 = Dv 2 , Ev 2 ≤ 0.
We have to check that such a vector v is not a direction of descent for every function ψ j , or it is an ascent direction for at least one of them. Note that It is an elementary Linear Algebra exercise to check that if U −T E < 0, then condition (8) is verified so that Theorem 1.1 can be applied. Corollary 1.1 is a specific example of the kind of existence result that can be obtained through this approach. Its proof amounts to going carefully through the arithmetic while checking that matrix U and vector E defined from such given class of (c(.), Q(.)) verify the assumptions of Theorem 1.2.
By using the same ideas, more general situations can be treated, for example the number of controls could be greater than the components of the state. This is in fact the situation in the model that has served as an inspiration for us. We will pursue a closer analysis of such a particular situation, even stressing the more practical issues, in a forthcoming work.