Sufficient conditions Bucci THE NON-STANDARD LQR PROBLEM

The non-standard LQR problem 111 A better result can be provided in the case of holomorphic semigroup systems, by using the smoothing properties of the operator L τ . Somehow the ‘analytic case’ parallels the case when the input operator is bounded. See [14, Ch. 9, Theorem 3.1] for the proof. T HEOREM 3. Assume that H 1 holds, with γ 12. Then 14 is a necessary condition in order that 4 is satisfied, even when P T 6= 0.

3. Sufficient conditions

In this section we provide sufficient conditions in order that 4 is satisfied. Let us go back to the non-negativity conditions of Theorem 1. It can be easily shown that neither 14, nor 13, are, by themselves, sufficient to guarantee that the cost functional is bounded from below. E XAMPLE 1. Let X = U = , and set τ = 0, A = −1, B = 0 in 1; moreover, let Fx, u = xu. Note that here R = 0. For any x , the solution to 1 is given by xt = x e −t , so that J 0,T x ; u = x Z T e −t ut dt for any admissible control u. Therefore J 0,T ; u ≡ 0, and 13 holds true, whereas it is readily verified that when x 6= 0, inf u J 0,T x ; u = −∞ if x 0 take, for instance, the sequence u k t = −k on [0, T ]. If T = +∞, the same example shows that Theorem 1 cannot be reversed without further assumptions. However it turns out that, over an infinite horizon, the necessary non-negativity condition 13 is also sufficient in order that 4 is satisfied, if system 1 is completely control- lable. This property is well known in the finite dimensional case, since the early work [10]. Recently, the aforementioned result has been extended to boundary control systems, under the following assumptions: i ′ A : D A ⊂ X → X is the generator of a s.c. group e At on X , t ∈ ; H 2 ′ there exists a T 0 and a constant k T 0 such that Z T |B ∗ e A ∗ t x | 2 dt ≤ k T |x| 2 ∀x ∈ DA ∗ ; 18 H 3 ′ system 1 is completely controllable, namely for each pair x , x 1 ∈ X there is a T and an admissibile control v · such that xT ; 0, x , v = x 1 . For simplicity of exposition, we state the theorem below under the additional condition that e At is exponentially stable. T HEOREM 4 [22]. Assume i ′ –H 2 ′ –H 3 ′ . If J ∞ ; u ≥ 0 ∀u ∈ L 2 0, ∞; U , then for each x ∈ X there exists a constant c ∞ x such that inf L 2 0,∞;U J ∞ x ; u ≥ c ∞ x ∀u ∈ L 2 0, ∞; U . 112 F. Bucci We now return to the finite time interval [0, T ] and introduce the assumption that the system is exactly controllable on a certain interval [0, r ] see, e.g., [3]: H 3 there is an r 0 such that, for each pair x , x 1 ∈ X, there exists an admissibile control v · ∈ L 2 0, r ; U yielding xr; 0, x , v = x 1 . Equivalently, ∃ r, c r 0 : Z r |B ∗ e A ∗ t x | 2 dt ≥ c r |x| 2 ∀x ∈ DA ∗ . 19 On the basis of Theorem 4, one would be tempted to formulate the following claim. C LAIM 5. Assume i ′ –H 2–H 3. If J 0,T ; u ≥ 0 ∀u ∈ L 2 0, T ; U , then 4 is satisfied for 0 ≤ τ ≤ T . It turns out that this claim is false, as it can been shown by means of examples: see [7, 6]. A correct counterpart of Theorem 4 over a finite time interval has been given in [6]. T HEOREM 6 [6]. Assume i ′ –H 2–H 3. If J 0,T +r ; u ≥ 0 ∀u ∈ L 2 0, T + r; U , 20 then 4 is satisfied for 0 ≤ τ ≤ T . We point out that in fact a proof of Theorem 6 can be provided which does not make explicit use of assumption i ′ , see Theorem 7 below. Let us recall however that, when the input oper- ator B is bounded, controllability of the pair A, B on [0, r ], namely assumption H 3 above, implies that the semigroup e At is right invertible, [16]. Therefore, the actual need of some kind of ‘group property’ in Theorem 6 is an issue which is left for further investigation. T HEOREM 7. Assume H 2–H 3. If 20 holds, then 4 is satisfied for 0 ≤ τ ≤ T . Proof. Let x 1 ∈ X be given. By H3 there exists a control v· ∈ L 2 0, r ; U steering the solution of 1 from x = 0 to x 1 in time r , namely xr ; 0, 0, v = x 1 . Obviously, v depends on x 1 : more precisely, it can be shown that, as a consequence of assumptions H 2 and H 3, a constant K exists such that |v| L 2 0,r;U ≤ K |x 1 | , see [6]. For arbitrary u ∈ L 2 r, T + r; U, set now u v t = v t ≤ t ≤ r ut r t ≤ T + r . Readily u v · ∈ L 2 0, T + r; U, and J 0,T +r ; u v ≥ 0 due to 20. On the other hand, J 0,T +r ; u v = Z r Fxs ; 0, 0, v, vs ds + J r,T +r x 1 ; u , The non-standard LQR problem 113 where the first summand is a constant which depends only on x 1 . A straightforward computa- tion shows that the second summand equals J 0,T x 1 ; u r , with u r t = ut + r an arbitrary admissible control on [0, T ]. In conclusion, J 0,T x 1 ; u r ≥ − Z r Fxs ; 0, 0, v, vs ds =: cx 1 , and 4 holds for τ = 0. The case τ 0 can be treated by using similar arguments. R EMARK 4. In conclusion, we have provided the sufficiency counterpart of item i of Theorem 2, under the additional condition that system 1 is exactly controllable in time r 0. Apparently, in order that 4 is satisfied, the non-negativity condition need to be required on a larger interval than [0, T ], precisely on an interval of lenght T + r. 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