Admissible controls and controllable sets for a linear time-varying ordinary differential equation

In this paper, for a time optimal control problem governed by a linear time-varying ordinary differential equation, we give a description to check whether the set of admissible controls is nonempty or not by finite times.


1.
Introduction. The existence of admissible controls is an important subject in the studies of time optimal control problems. This subject seems to be independent of other research issues, such as the Pontryagin Maximum Principle, the bangbang property, and so on. However, it is a prerequisite to study time optimal control problems. A control is called an admissible one if it takes value in a set of constrained controls and the corresponding solution of the controlled equation reaches a given target at some time from a starting set at the initial time.
Generally, people always connect the existence of admissible controls with controllability. In the study of the existence of admissible controls, there are two misunderstandings. The first misunderstanding is: controllability should imply the existence of admissible controls. Here is a counterexample. Consider the controlled equation:ẏ (t) = y(t) + u(t), t > 0.
(1) It is obvious that this equation is controllable. We introduce a time optimal control problem: inf{T > 0 : y(T ; 0, 2, u) = 0 and u(·) ∈ V}, where V {u : (0, +∞) → R is measurable | |u(t)| ≤ 1 for a.e. t > 0}, and y(·; 0, 2, u) denotes the solution of (1) with the control u(·) ∈ V and the initial condition y(0) = 2. We can check that there is no admissible control for the problem (2). The second misunderstanding is: if a controlled system is uncontrollable, then its corresponding time optimal control problem has no admissible control. Here is a counterexample. Consider the controlled system: It is clear that this system is uncontrollable. We introduce a time optimal control problem: inf{T > 0 : y 1 (T ; 0, (1, 0) , u) = y 2 (T ; 0, (1, 0) , u) = 0 and u(·) ∈ V}, (4) where (y 1 (·; 0, (1, 0) , u), y 2 (·; 0, (1, 0) , u)) denotes the solution of (3) with the control u(·) ∈ V and the initial condition (y 1 (0), y 2 (0)) = (1, 0) . We can check that the control defined by: is an admissible control for the problem (4). Generally, differences between an admissible control problem and a controllability problem are as follows. First, an admissible control problem has some constraints on the control set, the starting set and the target set, while a controllability problem has no such constraint. Second, controllability requires that for any two elements in state space, there is a control which drives the state of the system from one element to another element, while for the existence of admissible controls, there are two elements (which are in the starting set and the target set, respectively) in state space, there is a control which drives the state of the system from the element (in the starting set) to another element (in the target set). From these viewpoints, the existence of admissible controls is totally different from the controllability.
Though the existence of admissible controls is a base for the study of time optimal control problems, there is little work on this topic (see [1], [2], [3], [5], [8]- [11], [14] and [16]). In general, there are two angles to study the existence of admissible controls governed by differential equations. The first is: under the assumption that a set of constrained controls, a starting set and a target set are given, what conditions should a differential equation satisfy in order to guarantee the existence of admissible controls? (see [4] and [13].) The second is: under the assumption that the controlled system, the set of constrained controls and the starting set (or the target set, respectively) are given, what conditions should the target set (or the starting set, respectively) satisfy in order to guarantee the existence of admissible controls? For the second case, it turns out to be an attainability problem (or a constrained controllability problem). There are some existing works about it (see [15], [6] and [12]). In the study of the second case, a very interesting question arises: for a given linear differential equation with the null target set and a fixed set of constrained controls (closed ball with radius r), can we give a criterion for initial data, by which we can judge whether admissible control exists or not? For this question, there have been some sufficient and necessary conditions for the timeinvariant ordinary differential equations (see [4] and [13]). When the ending time is fixed, some results are derived for the controlled heat equations (see [15], [6] and [12]). However, to the best of our knowledge, for the time-varying controlled ordinary differential equations and free ending time, there is no result. In this paper, we will study this problem.
In this paper, we consider the following controlled system: where R + (0, +∞), A(·) ∈ L ∞ loc (R + ; R n×n ) and B(·) ∈ L ∞ loc (R + ; R n×m ). Throughout this paper, for a given positive constant r, we always use B r (0) to denote the closed ball in R m (or R n ) with center at the origin and of radius r if there is no confusion. And we use ∂B r (0) to denote the boundary of B r (0) in R m (or R n ). The set of constrained control is: The time optimal control problem studied in this paper is as: inf{T : y(T ; 0, y 0 , u) = 0 and u ∈ U}.
It is called the admissible control set of (P y0 ); every element in U y0 is called an admissible control of (P y0 ). We now give the definition about the solvability of the problem (P y0 ).
The problem (P y0 ) is said to be solvable if we can check whether (P y0 ) has an admissible control or not by finite times.
The main result of this paper is as follows.
In order to prove Theorem 1.2, we need some notations. Let A(·) ∈ L ∞ loc (R + ; R n×n ) and B(·) ∈ L ∞ loc (R + ; R n×m ) be given. Denote by W A, B and W r A, B controllable subspace without control constraint and controllable set with constrained controls for the controlled system: more precisely, W A, B { y 0 ∈ R n ∃ T ∈ R + and u ∈ L 2 (0, T ; R m ) so that y(T ; 0, y 0 , u) = 0} (7) and where y(·; 0, y 0 , u) is the unique solution of (6) with the initial condition y(0) = y 0 , and Further, we set and Here, Φ(·, ·) is the evolution operator generated by A(·). From (8) and (9), one can directly verify the following proposition.
Proposition 1. For arbitrarily fixed y 0 ∈ R n \ {0}, the problem (P y0 ) has an admissible control if and only if y 0 ∈ W r A,B .
Proposition 1 is indeed a judgement for the existence or non-existence of admissible control. However, it is difficult to give an exact description of W r A,B . Hence, in general, by this proposition, we hardly judge whether admissible control exists or not. In order to overcome this difficulty, we introduce candidate controllable set CW r A, B (see (18)), which plays an important role in the proof of Theorem 1.2. This paper is organized as follows. In Section 2, we present some preliminaries, which will be used in the proof of Theorem 1.2. In Section 3, based on Proposition 1 and preliminaries in Section 2, we give the proof of Theorem 1.2.
2. Preliminaries. We start this section with the following result.
Lemma 2.1. Let N A, B and W A, B be defined by (10) and (7), respectively. Then To this end, we let y 0 ∈ W A, B . According to (7), there exists a constant T ∈ R + and u ∈ L 2 (0, T ; R m ) so that y(T ; 0, y 0 , u) = 0, which indicates that From the above and (10), it follows that By the convexity and closedness of W A, B (in R n ), we use the separation theorem of convex sets to find a vector ϕ ∈ R n so that Since 0 ∈ W A, B , it follows from (12) that ϕ, z R n > 0.
Noting that z ∈ N ⊥ A, B , by the above inequality, we see that ϕ / ∈ N A, B , i.e., Hence, there is a constant T ∈ R + so that For any k > 0, we set We can easily check that y( T ; 0, −w k , −u k ) = 0. This, together with (7), implies that −w k ∈ W A, B . Thus by (12), we have that It follows from (13) and (14) that ϕ, z R n = +∞, which leads to a contradiction. In summary, we finish the proof of Lemma 2.1.
Now, for any ψ ∈ R n , we denote and an open set Furthermore, we define a candidate controllable set Here, when the supreme in (15) is achieved, Then we have the following result.
This completes the proof. Now two questions arise: are these two sets the same? If not, what conditions can we need to ensure that they are the same? For the first question, we consider the following example, which shows that generally, these two sets are not the same.
Let r = 1. We will show that (1, 1) ∈ CW 1 A, B \ W 1 A, B . Its proof will be carried out by the following three steps.
Step 1. We claim that Indeed, on one hand, by (8) and (9), we see that where and On the other hand, by (24) and (25), we can directly check that These, along with (23), imply (22).
The result follows from Steps 1 and 2 immediately.
Then, by Lemma 2.4, (45) and (44), we get that This, together with (43), implies that z ∈ W r A, B (T ). In summary, we finish the proof of Theorem 2.5.
The difficulty lies in dealing with the following case: In order to deal with this case, we need some preliminary results. First we present the following result.
is uniformly bounded.
Based on Lemma 2.6 and (33), we introduce some notations. For any ψ ∈ D A, B , we set and Here and throughout this paper, if E A, B,ψ = R + , we denote χ E c A, B,ψ (·) 0. Moreover, we write and Remark 1. We can easily check that and Indeed, (59) follows immediately from Lemma 2.1, (10) and (56). In order to prove (60), we use a contradiction argument. By contradiction and (59), we would have that This, together with Lemma 2.1, implies that On one hand, since ψ ∈ D A, B ⊆ N ⊥ A, B ∂B 1 (0) (see (33) and (32)), by (61), we get that ψ ∈ N ⊥ A,B A, B, ψ and ψ R n = 1.
On the other hand, by (55) and (56), we have that which, combined with (10), indicates that By (62) and (63), we arrive at a contradiction and (60) follows. Now we present the following result.
3. Proof of main result. We now are on the position to prove Theorem 1.2.
Proof of Theorem 1.2. For each y 0 ∈ R n \ {0}, we fix it. We can judge whether it belongs to W r A,B or not by finite times as follows.
Step 2. Is y k ∈ CW r A,B k ? If the answer is false, then y k / ∈ W r A,B k ; If the answer is yes, then we continue Step 3.
Indeed, if y k ∈ CW r A,B k , by Lemma 2.2, we have that y k ∈ W r A,B k .
Step 3. Is there ψ k ∈ D A,B k so that y k ∈ ∂(HS) r A,B k ,ψ k ? If the answer is false, then y k ∈ W r A,B k ; If the answer is yes, then we continue Step 4. In fact, by Step 2, we see that y k ∈ CW r A,B k . If y k / ∈ ψ∈ D A,B k ∂(HS) r A,B k ,ψ , by Theorem 2.5, we get that y k ∈ W r A,B k (T ) for some T ∈ R + . This implies that y k ∈ W r A,B k .
Step 5. Is there ψ ∈ D A,B k+1 so that y k+1 , ψ R n = rC A,B k+1 ,ψ when sup can not be achieved? If the answer is yes, then y k ∈ W r A,B k ; If the answer is false, then turn back to Step 3. Indeed, by Step 2 and Step 3, we have that y k ∈ CW r A,B k and there is ψ k ∈ D A,B k so that y k ∈ ∂(HS) r A,B k ,ψ k . Then by Step 4 and Theorem 2.8, we get that Now two cases may occur: can not be achieved. In this case, by (18), (20), (17) and (15), we get that y k+1 ∈ CW r A,B k+1 . This, along with Lemma 2.2, implies that From (95), we claim that Otherwise, according to (8), there would exist a constant T ∈ R + so that y k ∈ W r A,B k (T ). By Lemma 2.7 and (8) again, we have that y k+1 ∈ W r A,B k+1 (T ) ⊆ W r A,B k+1 , which contradicts (95).
In this case, by (94) and (16), we get that Hence, it follows from the latter, (94) and (18) that We now turn to Step 3.
These, along with Theorem 2.9, imply that the above judging process ends in finite times, i.e., we can check whether (P y0 ) has an admissible control or not by finite times. Thus, the problem (P y0 ) is solvable (see Defintion 1.1).
In summary, we finish the proof of Theorem 1.2.
The following result is concerned with the case that A(·) and B(·) are timeinvariant. Then there exists a constant T ∈ R + so that Since B * e −A * t ψ, t ≥ 0, is an analytic function, it follows from (109) that B * e −A * t ψ = 0, ∀ t ≥ 0.