Control of Partial Differential Equations

Evolution Equations 254 Let F be another Hilbert space (with scalar product and 255 norm denoted by the same symbols as for H), the so256 called control space, and let B : F ! H be a linear opera257 tor, that we will assume to be bounded for the time being. 258 Then, given T > 0 and u0 2 H, for all f 2 L2(0; T ; F) the 259 Cauchy problem 260 ( u0(t) D Au(t) C B f (t) t 0 u(0) D u0 (12) 261 has a uniquemild solution u f 2 C([0; T];H) given by 262 u f (t) D eu0 C Z t 0 e(t s)AB f (s) 8t 0 (13) 263 Note 1 Boundary control problems can be reduced to the 264 same abstract form as above. In this case, however, B in 265 (12) turns out to be an unbounded operator related to suit266 able fractional powers of A, see, e. g., [22]. 267 For any t 0 let us denote by t : L2(0; t; F) ! H the 268 bounded linear operator 269 t f D Z t 0 e(t s)AB f (s) ds 8 f 2 L2(0; t; F) : (14) 270 The attainable (or reachable) set from u0 at time t, 271 A(u0; t) is the set of all points in H of the form u f (t) for 272 some control function f , that is 273 A(u0; t) : D eu0 C tL(0; t; F) : 274 We introduce below themain notions of controllability for 275 (7). Let T > 0. 276 Definition 1 System (7) is said to be: 277 exactly controllable in time T ifA(u0; T) D H for all 278 u0 2 H, that is, if for all u0; u1 2 H there is a control 279 function f 2 L2(0; T ; F) such that u f (T) D u1; 280 null controllable in time T if 0 2 A(u0; T) for all 281 u0 2 H, that is, if for all u0 2 H there is a control func282 tion f 2 L2(0; T ; F) such that u f (T) D 0; 283 approximately controllable in time T if A(u0; T) 284 is dense in H for all u0 2 H, that is, if for all 285 u0; u1 2 H and for any " > 0 there is a control func286 tion f 2 L2(0; T ; F) such that ku f (T) u1k < ". 287 Clearly, if a system is exactly controllable in time T, 288 then it is also null and approximately controllable in 289 time T. Although these last two notions of controllability 290 are strictly weaker than strong controllability, for specific 291 problems—like when A generates a strongly continuous 292 group—some of them may coincide. 293 Since controllability properties concern, ultimately, 294 the range of the linear operator T defined in (14), it is 295 not surprising that they can be characterized in terms of 296 the adjoint operator T : H ! L2(0; T ; F), which is de297 fined by 298 299 Z T 0 ̋ Tu(s); f (s)ids D hu0; T f i 300 8u 2 H ; 8 f 2 L2(0; T ; F) : 301 302 Such a characterization is the object of the following theo303 rem. Notice that the above identity and (14) yield 304 Tu(s) D B e(T s)A u 8s 2 [0; T] : 305 Theorem 1 System (7) is: 306 exactly controllable in time T if and only if there is 307 a constant C > 0 such that 308 Z T 0 B etA u 2 dt Ckuk2 8u 2 H ; (15) 309 null controllable in time T if and only if there is a con310 stant C > 0 such that 311 Z T 0 B etA u 2 dt C eTA u 2 8u 2 H ; (16) 312 approximately controllable in time T if and only if, for 313 every u 2 H, 314 B etA u D 0 t 2 [0; T] a.e. H) u D 0 : (17) 315 To benefit the reader who is more familiar with optimiza316 tion theory than abstract functional analysis, let us explain, 317 by a variational argument, why estimate (16) implies null 318 controllability. Consider, for every " > 0, the penalized 319 problem 320 min ̊ J"( f ) : f 2 L2(0; T ;H) ; 321 where 322 323 J"( f ) D 1 2 Z T 0 k f (t)k2 dt C 1 2" ku f (T)k2 324 8 f 2 L2(0; T ;H) : 325 326 Since J" is strictly convex, it admits a unique minimum 327 point f". Set u" D u f" . Recalling (13) we have, By Fermat’s 328 rule, 329 330 0 D J0 "( f")g D Z T 0 h f"(t); g(t)i dt 331 C 1 " hu"(T); T gi 8g 2 L2(0; T ;H) : (18) 332 333 Un co rre cte d Pr oo f 20 08 -0 731 Meyers: Encyclopedia of Complexity and Systems Science — Entry 583 — 2008/7/31 — 16:14 — page 5 — le-tex Control of Partial Differential Equations 5 Therefore, passing to the adjoint of T , 334 335 Z T 0 D f"(t) C 1 " Tu"(T) (t); g(t) E dt D 0 336 8g 2 L2(0; T ;H) ; 337 338 whence, owing to (14), 339 340 f"(t) D 1 " Tu"(T) (t) D B v"(t) 341 8t 2 [0; T] ; (19) 342 343 where v"(t) : D 1 " e(T t)A u"(T) is the solution of the dual 344 problem 345 ( v0 C A v D 0 t 2 [0; T] v(T) D 1 " u"(T) : 346 It turns out that 347 348 1 2 Z T 0 k f"(t)k dt C 1 " ku"(T)k Cku0k 349