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. This produces a gap between necessary conditions and sufficient conditions, which was already pointed out in finite
dimensions [7]. Finally, we stress that the exact controllability assumption cannot be weakened to null con-
trollability, as pointed out in [6, Ex. 4.5].
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AMS Subject Classification: 49N10, 49J20.
Francesca BUCCI Universit`a degli Studi di Firenze
Dipartimento di Matematica Applicata “G. Sansone” Via S. Marta 3, I–50139 Firenze, Italy
e-mail:
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