Approximate controllability for linear degenerate parabolic problems with bilinear control

Appro ximate con trollabilit y for linear degenerate parab olic problems with bilinear con trol Piermarco Cannarsa, Dipartimen to di Matematica , Univ ersit` a di Roma “T or V ergata”, I-00161 Roma, Italy Giusepp e Floridia, Dipartimen to di Matematica e Informatica , Univ ersit` a di Catania, I-95125 Catania, Italy Abstract In this w ork we study the global approximate multiplicativ e controllabil- it y for the linear degenerate parab olic Cauc hy-Neumann problem          v t − ( a ( x ) v x ) x = α ( t, x ) v in Q T = (0 , T ) × ( − 1 , 1) a ( x ) v x ( t, x ) | x = ± 1 = 0 t ∈ (0 , T ) v (0 , x ) = v 0 ( x ) x ∈ ( − 1 , 1) , with the bilinear control α ( t, x ) ∈ L ∞ ( Q T ) . The problem is strongly degen- erate in the sense that a ∈ C 1 ([ − 1 , 1]) , p ositiv e on ( − 1 , 1) , is allow ed to v anish at ± 1 provided that a certain integrabilit y condition is fulfilled. W e will show that the ab o ve system can b e steered in L 2 (Ω) from any nonzero, nonnegativ e initial state in to any neigh b orhoo d of any desirable nonnegativ e target-state b y bilinear static controls. Moreo ver, w e extend the ab ov e result relaxing the sign constrain t on v 0 . Key w ords: approximate con trollability , degenerate parabolic equations, bilinear con trol AMS sub ject classifications: 35K65, 93B05, 34B24 1 In tro duction Motiv ation Climate dep ends on v arious parameters such as temp erature, humidit y , wind in- tensit y , the effect of greenhouse gases, and so on. It is also affected by a complex set of in teractions in the atmosphere, o ceans and con tinents, that inv olv e physical, c hemical, geological and biological pro cesses. 1 One of the first attempts to mo del the effects of interaction b et ween large ice masses and solar radiation on climate is the one due, indep endently , b y Budyko [5, 6] and Sellers [25] (see also [12, 13, 17] and the references therein). Such a mo del studies ho w extensive the climate resp onse is to an even t such as a sharp increase in greenhouse gases; in this case we talk about climate sensitivit y . A pro cess that c hanges climate sensitivity is called fe e db ack . If the pro cess increases the intensit y of resp onse we say that it has p ositive fe e db ack , whereas it has ne gative fe e db ack if it reduces the intensit y of resp onse. The Budyko-Sellers mo del studies the role pla yed by con tinental and o ceanic areas of ice on climate c hange. In suc h a mo del, the sea lev el mean zonally a veraged temp erature u ( t, x ) on the Earth, where t denotes time and x the sine of latitude, satisfies the following degenerate Cauchy-Neumann problem (1.1) in the b ounded domain ( − 1 , 1). The effect of solar radiation on climate can b e summarized in the follo wing figure Figure 1: www.edu-design-principles.org (copyrigh ted by DPD) W e hav e the following ener gy b alanc e : Heat v ariation = R a − R e + D • R a = absorb ed energy • R e = emitted energy • D = diffusion The general formulation of the Budyk o-Sellers mo del on a compact surface M without b oundary is as follows u t − ∆ M u = R a ( t, x, u ) − R e ( t, x, u ) 2 where u ( t, x ) is the distribution of temp erature and ∆ M is the classical Laplace- Beltrami op erator. Moreov er, • R a ( t, x, u ) = Q ( t, x ) β ( x, u ) • R e ( t, x, u ) = A ( t, x ) + B ( t, x ) u In the ab o ve, Q is the insolation function, and β is the c o alb e do function (that is, 1- alb e do function). Alb edo is the reflecting p ow er of a surface. It is defined as the ratio of reflected radiation from the surface to incident radiation up on it. It may also be expressed as a p ercentage, and is measured on a scale from zero for no reflecting p o w er of a p erfectly blac k surface, to 1 for p erfect reflection of a white surface. Figure 2: www.esr.org (copyrigh ted by ESR) The main difference b et w een Budyk o’s mo del and the one by Sellers, is that in the former the coalb edo function is discontin uous, while in the latter it is a con tinuous function. In fact we ha ve 3 • Budyk o β ( u ) =          β 0 u < − 10 [ β 0 , β 1 ] u = − 10 β 1 u > − 10 , • Sellers β ( u ) =          β 0 u < u − line u − ≤ u ≤ u + β 1 u > u + , where u ± = − 10 ± δ , δ > 0 . On M = Σ 2 the Laplace-Beltrami op erator is ∆ M = 1 sin φ n ∂ ∂ φ  sin φ ∂ u ∂ φ  + 1 sin φ ∂ 2 u ∂ λ 2 o where φ is the c olatitude and λ is the longitude . Figure 3: www.globalwarmingart.com (copyrigh ted by Global W arming Art) In the one-dimensional Budyko-Sellers we take the av erage of the temp erature at x = cos φ and the Budyk o-Sellers mo del reduces to    u t −  (1 − x 2 ) u x  x = g ( t, x ) h ( x, u ) + f ( t, x ) , x ∈ ( − 1 , 1) (1 − x 2 ) u x | x = ± 1 = 0 . (1.1) 4 Problem formulation Let us consider the follo wing Cauchy-Neumann strongly degenerate b oundary lin- ear problem in div ergence form, go v erned in the bounded domain ( − 1 , 1) b y means of the biline ar c ontr ol α ( t, x )          v t − ( a ( x ) v x ) x = α ( t, x ) v in Q T = (0 , T ) × ( − 1 , 1) a ( x ) v x ( t, x ) | x = ± 1 = 0 t ∈ (0 , T ) v (0 , x ) = v 0 ( x ) x ∈ ( − 1 , 1) . (1.2) W e assume that 1. v 0 ∈ L 2 ( − 1 , 1) 2. α ∈ L ∞ ( Q T ) 3. a ∈ C 1 ([ − 1 , 1]) satisfies (a) a ( x ) > 0 ∀ x ∈ ( − 1 , 1) , a ( − 1) = a (1) = 0 (b) A ∈ L 1 ( − 1 , 1) , where A ( x ) = R x 0 ds a ( s ) . Remark W e observe that 1. 1 a 6∈ L 1 ( − 1 , 1) , so a ( x ) is strongly degenerate 2. the principal part of the op erator in (1.2) coincides with that of the Budyko- Sellers mo del for a ( x ) = 1 − x 2 . In this case A ( x ) = 1 2 ln  1+ x 1 − x  ∈ L 1 ( − 1 , 1) 3. a sufficient condition for 3.b) is that a 0 ( ± 1) 6 = 0 (if a ∈ C 2 ([ − 1 , 1]) the ab ov e condition is also necessary). W e are in terested in studying the m ultiplicative con trollability of problem (1.2) b y the biline ar c ontr ol α ( t, x ). In particular, for the ab o ve linear problem, we will discuss results guaran teeing global nonnegativ e appro ximate controllabilit y in large time (for m ultiplicative controllabilit y see [20, 23, 8]). No w w e recall one definition from con trol theory . Definition 1.2 We say that the system (1.2) is nonne gatively glob al ly appr oximately c ontr ol- lable in L 2 ( − 1 , 1) , if for every ε > 0 and for every nonne gative v 0 ( x ) , v d ( x ) ∈ L 2 ( − 1 , 1) with v 0 6≡ 0 ther e ar e a T = T ( ε, v 0 , v d ) and a biline ar c ontr ol α ( t, x ) ∈ L ∞ ( Q T ) such that for the c orr esp onding solution v ( t, x ) of (1.2) we obtain k v ( T , · ) − v d k L 2 ( − 1 , 1) ≤ ε . 5 In the follo wing, w e will sometimes use k · k instead of k · k L 2 ( − 1 , 1) . Main results In this work at first the nonne gative glob al appr oximate c ontr ol lability result is obtained for the linear system (1.2) in the follo wing theorem. Theorem 1.3 The line ar system (1.2) is nonne gatively appr oximately c ontr ol lable in L 2 ( − 1 , 1) by me ans of static c ontr ols in L ∞ ( − 1 , 1) . Mor e over, the c orr esp onding solution to (1.2) r emains nonne gative at al l times. Then the results present in Theorem 1.3 can b e extended to a larger class of initial states. Theorem 1.4 F or any v d ∈ L 2 ( − 1 , 1) , v d ≥ 0 and any v 0 ∈ L 2 ( − 1 , 1) such that Z 1 − 1 v 0 v d dx > 0 , (1.3) for every ε > 0 , ther e ar e T = T ( ε, v 0 , v d ) ≥ 0 and a static biline ar c ontr ol, α = α ( x ) , α ∈ L ∞ ( − 1 , 1) such that k v ( T , · ) − v d k L 2 ( − 1 , 1) ≤ ε . Remark The solution v ( t, x ) of the problem (1.2) in the assumptions of Theorem 1.4 does not remain nonnegativ e in Q T , lik e in Theorem 1.3, but it can also assume negativ e v alues. Mathematical motiv ation This note is inspired by [20, 8]. In [20] A.Y. Khapalov studied the global nonnega- tiv e approximate controllabilit y of the one dimensional non-de gener ate semilinear con vection-diffusion-reaction equation go verned in a b ounded domain via the bi- linear con trol α ∈ L ∞ ( Q T ) . In [8], the same appro ximate controllabilit y prop erty is derived in suitable classes of functions that change sign. In this note we extend some of the results of [20] to degenerate linear equations. 6 General references for multiplic ative c ontr ol lability are, e.g., [18, 19, 21, 22, 23, 3]. In control theory , b oundary and in terior lo cally distributed controls are usually emplo yed (see, e.g., [9, 10, 11, 14, 15, 16]). These controls are additiv e terms in the equation and ha ve lo calized support. How ever, such mo dels are unfit to study sev eral interesting applied problems such as chemical reactions controlled b y catalysts, and also smart materials, which are able to change their principal parameters under certain conditions. This explains the growing interest in multi- plic ative c ontr ol lability . 2 Preliminaries P ositiv e and negativ e part Giv en Ω ⊆ R n , v : Ω − → R we consider the p ositive-part function v + ( x ) = max ( v ( x ) , 0) , ∀ x ∈ Ω , and the negativ e-part function v − ( x ) = max (0 , − v ( x )) , ∀ x ∈ Ω . Then we hav e the follo wing equalit y v = v + − v − in Ω F or the functions v + and v − the following result of regularity in Sob olev’s spaces will b e useful (see [24], App endix A ). Theorem 2.1 L et Ω ⊂ R n , u : Ω − → R , u ∈ H 1 ,s (Ω) , 1 ≤ s ≤ ∞ . Then u + , u − ∈ H 1 ,s (Ω) and for 1 ≤ i ≤ n ( u + ) x i =    u x i in { x ∈ Ω : u ( x ) > 0 } 0 in { x ∈ Ω : u ( x ) ≤ 0 } , (2.4) and ( u − ) x i =    − u x i in { x ∈ Ω : u ( x ) < 0 } 0 in { x ∈ Ω : u ( x ) ≥ 0 } . (2.5) 7 Gron w all’s Lemma Lemma 2.2 Gron wall’s inequality (differential form). L et η ( t ) b e a nonne gative, absolutely c ontinuous function on [0 , T ] , which sat- isfies for a.e. t ∈ [0 , T ] the differ ential ine quality η 0 ( t ) ≤ φ ( t ) η ( t ) + ψ ( t ) , (2.6) wher e φ ( t ) and ψ ( t ) ar e nonne gative, summable functions on [0 , T ] . Then η ( t ) ≤ e R t 0 φ ( s ) ds  η (0) + Z t 0 ψ ( s ) ds  (2.7) for al l 0 ≤ t ≤ T . In p articular, if ψ ( t ) ≡ 0 in (2.6), i.e. η 0 ≤ φ η for a.e. t ∈ [0 , T ] , and η (0) = 0 , then η ≡ 0 in [0 , T ] . W ell-posedness in w eighted Sob olev spaces In order to deal with the well-posedness of problem (1.2), it is necessary to in tro- duce the follo wing Sob olev w eighted spaces H 1 a ( − 1 , 1) := := { u ∈ L 2 ( − 1 , 1) : u locally absolutely contin uous in ( − 1 , 1) , √ au x ∈ L 2 ( − 1 , 1) } and H 2 a ( − 1 , 1) := { u ∈ H 1 a ( − 1 , 1) | au x ∈ H 1 ( − 1 , 1) } = = { u ∈ L 2 ( − 1 , 1) | u locally absolutely contin uous in ( − 1 , 1) , au ∈ H 1 0 ( − 1 , 1) , au x ∈ H 1 ( − 1 , 1) and ( a u x )( ± 1) = 0 } resp ectiv ely with the following norms k u k 2 H 1 a := k u k 2 L 2 ( − 1 , 1) + | u | 2 1 ,a and k u k 2 H 2 a := k u k 2 H 1 a + k ( au x ) x k 2 L 2 ( − 1 , 1) ; where | u | 1 ,a = k √ au x k L 2 ( − 1 , 1) is a seminorm. In this note we obtain the following result. Lemma 2.3 H 1 a ( − 1 , 1)  → L 2 ( − 1 , 1) with c omp act emb e dding. (2.8) 8 Pro of: Giv en u ∈ H 1 a ( − 1 , 1), let ¯ u ( x ) =    u if x ∈ [ − 1 , 1] 0 elsew ere . It is sufficien t to prov e that, for every R > 0 , sup k u k 1 ,a ≤ R Z R | ¯ u ( x + h ) − ¯ u ( x ) | 2 dx − → 0 , as h → 0 (2.9) Let h > 0( 1 ) and let u ∈ H 1 a ( − 1 , 1) b e suc h that k u k 1 ,a ≤ R , we ha v e the following equalit y Z R | ¯ u ( x + h ) − ¯ u ( x ) | 2 dx = = Z − 1 − 1 − h | u ( x + h ) | 2 dx + Z 1 − h − 1 | u ( x + h ) − u ( x ) | 2 dx + Z 1 1 − h | u ( x ) | 2 dx = = Z − 1+ h − 1 | u ( x ) | 2 dx + Z 1 − h − 1 | u ( x + h ) − u ( x ) | 2 dx + Z 1 1 − h | u ( x ) | 2 dx First, let us prov e that sup k u k 1 ,a ≤ R Z 1 − h − 1 | u ( x + h ) − u ( x ) | 2 dx − → 0 , as h → 0 + . (2.10) Recalling that A ( x ) = R x 0 ds a ( s ) , we hav e | u ( x + h ) − u ( x ) | ≤ Z x + h x p a ( s ) | u 0 ( s ) | 1 p a ( s ) ds ≤ ≤  Z 1 − 1 a ( s ) | u 0 ( s ) | 2 ds  1 2 Z x + h x ds a ( s ) ! 1 2 = | u | 1 ,a [ A ( x + h ) − A ( x )] 1 2 . By integrating on [ − 1 , 1 − h ], since A ∈ L 1 ( − 1 , 1) (by assumption 3.b)), w e obtain Z 1 − h − 1 | u ( x + h ) − u ( x ) | 2 dx ≤ | u | 2 1 ,a Z 1 − h − 1 ( A ( x + h ) − A ( x )) dx ≤ ≤ R 2 " Z 1 − 1+ h A ( x ) dx − Z 1 − h − 1 A ( x ) dx # = 1 In the case h < 0 we pro ceed similarly . 9 = R 2 " Z 1 1 − h A ( x ) dx − Z − 1+ h − 1 A ( x ) dx # − → 0 , as h → 0 + . No w, let us pro ve that sup k u k 1 ,a ≤ R Z 1 1 − h | u ( x ) | 2 dx − → 0 , as h → 0 + . (2.11) W e hav e | u (0) | ≤ | u ( x ) | + Z x 0 p a ( s ) | u 0 ( s ) | 1 p a ( s ) ds ≤ ≤ | u ( x ) | +  Z 1 − 1 a ( s ) | u 0 ( s ) | 2 ds  1 2  Z x 0 ds a ( s )  1 2 ≤ | u ( x ) | + | u | 1 ,a p A ( x ) . By integrating on [0 , 1] , we obtain | u (0) | ≤ Z 1 0 | u ( x ) | dx + | u | 1 ,a Z 1 0 p A ( x ) dx ≤ ≤ k u k L 2 ( − 1 , 1) + | u | 1 ,a Z 1 0 p A ( x ) dx ≤ C k u k 1 ,a . Then, | u (0) | ≤ C R . (2.12) No w, it follows that | u ( x ) | 2 ≤ 2 | u (0) | 2 + 2 A ( x ) | u | 2 1 ,a ≤ C R 2 + 2 A ( x ) R 2 . Finally , since A ∈ L 1 ( − 1 , 1), b y in tegrating on [1 − h, 1] we obtain Z 1 1 − h | u ( x ) | 2 dx ≤ C hR 2 + 2 R 2 Z 1 1 − h A ( x ) dx − → 0 , as h → 0 + . Similarly , we can pro ve that sup k u k 1 ,a ≤ R Z − 1+ h − 1 | u ( x ) | 2 dx − → 0 , as h → 0 + . (2.13) By (2.10), (2.11) and (2.13) we obtain (2.9). W e now recall the existence and uniqueness result for system (1.2) obtained in [7] (see also [1]). Let us consider, first, the op erator ( A 0 , D ( A 0 )) defined b y    D ( A 0 ) = H 2 a ( − 1 , 1) A 0 u = ( au x ) x , ∀ u ∈ D ( A 0 ) . (2.14) 10 Observ e that A 0 is a closed, self-adjoint, dissipative op erator with dense domain in L 2 ( − 1 , 1). Therefore, A 0 is the infinitesimal generator of a C 0 − semigroup of con tractions in L 2 ( − 1 , 1). Next, given α ∈ L ∞ ( − 1 , 1) , let us in tro duce the operator    D ( A ) = D ( A 0 ) A = A 0 + αI . (2.15) F or such an operator we hav e the following prop osition. Prop osition 2.4 • D ( A ) is c omp actly emb e dde d and dense in L 2 ( − 1 , 1) . • A : D ( A ) − → L 2 ( − 1 , 1) is the infinitesimal gener ator of a str ongly c ontinu- ous semigr oup, e tA , of b ounde d line ar op er ators on L 2 ( − 1 , 1) . Observ e that problem (1.2) can b e recast in the Hilb ert space L 2 ( − 1 , 1) as    u 0 ( t ) = A u ( t ) , t > 0 u (0) = u 0 . (2.16) where A is the op erator in (2.15). W e recall that a we ak solution of (2.16) is a function u ∈ C 0 ([0 , T ]; L 2 ( − 1 , 1)) suc h that for every v ∈ D ( A ∗ ) the function h u ( t ) , v i is absolutely contin uous on [0 , T ] and d dt h u ( t ) , v i = h u ( t ) , A ∗ v i , for almost t ∈ [0 , T ] (see [2]). Theorem 2.5 F or every α ∈ L ∞ ((0 , T ) × ( − 1 , 1)) and every u 0 ∈ L 2 ( − 1 , 1) , ther e exists a unique u ∈ C 0 ([0 , T ]; L 2 ( − 1 , 1)) ∩ L 2 (0 , T ; H 1 a ( − 1 , 1)) we ak solution to (1.2), which c oincides with e tA u 0 . In the space B (0 , T ) = C 0 ([0 , T ]; L 2 ( − 1 , 1)) ∩ L 2 (0 , T ; H 1 a ( − 1 , 1)) 11 let us define the follo wing norm k u k 2 B (0 ,T ) = sup t ∈ [0 ,T ] k u ( t, · ) k 2 L 2 ( − 1 , 1) + 2 Z T 0 Z 1 − 1 a ( x ) u 2 x dx , ∀ u ∈ B (0 , T ) . (2.17) 3 Some auxiliary lemmas and the pro ofs of main results Let A = A 0 + α I , where the op erator A 0 is defined in (2.14) and α ∈ L ∞ ( − 1 , 1) . Since A is self-adjoint and D ( A )  → L 2 ( − 1 , 1) is compact (see Prop osition 2.4), w e ha ve the following (see also [4]). Lemma 3.1 Ther e exists an incr e asing se quenc e { λ k } k ∈ N , with λ k − → + ∞ as k → ∞ , such that the eigenvalues of A ar e given by {− λ k } k ∈ N , and the c orr esp onding eigenfunctions { ω k } k ∈ N form a c omplete orthonormal system in L 2 ( − 1 , 1) . In this note we obtain the following result Lemma 3.2 L et v ∈ C ∞ ([ − 1 , 1]) , v > 0 on [ − 1 , 1] , let α ∗ ( x ) = − ( a ( x ) v x ( x )) x v ( x ) , x ∈ ( − 1 , 1) . L et A b e the op er ator define d in (2.15) with α = α ∗    D ( A ) = H 2 a ( − 1 , 1) A = A 0 + α ∗ I , (3.18) and let { λ k } , { ω k } b e the eigenvalues and eigenfunctions of A, r esp e ctively, given by L emma 3.1. Then λ 1 = 0 and | ω 1 | = v k v k . Mor e over, v k v k and − v k v k ar e the only eigenfunctions of A with norm 1 that do not change sign in ( − 1 , 1) . Remark Problem (3.18) is equiv alent to the following differential problem    ( a ( x ) ω x ) x + α ∗ ( x ) ω + λ ω = 0 in ( − 1 , 1) a ( x ) ω x ( x ) | x = ± 1 = 0 . (3.19) 12 Pro of: (of Lemma 3.2) STEP .1 W e denote by {− λ k } k ∈ N and { ω k } k ∈ N , resp ectiv ely , the eigen v alues and orthonormal eigenfunctions of the operator (3.18) (see Lemma 3.1). Therefore, h ω k , ω h i L 2 ( − 1 , 1) = Z 1 − 1 ω k ( x ) ω h ( x ) dx = 0 , if h 6 = k . (3.20) W e can see, by easy calculations, that an eigenfunction of the operator defined in (3.18) is the function v ( x ) k v k , asso ciated with the eigenv alue λ = 0. T aking into account the abov e and consid- ering that v ( x ) > 0 , ∀ x ∈ ( − 1 , 1) ∃ k ∗ ∈ N : ω k ∗ ( x ) = v ( x ) k v k > 0 or ω k ∗ ( x ) = − v ( x ) k v k < 0 , ∀ x ∈ ( − 1 , 1) . (3.21) W riting (3.20) with k = k ∗ w e obtain h ω k ∗ , ω h i L 2 ( − 1 , 1) = Z 1 − 1 ω k ∗ ( x ) ω h ( x ) dx = 0 , ∀ h 6 = k ∗ . (3.22) Therefore, considering (3.22) and keeping in mind that ω k ∗ > 0 or ω k ∗ < 0 in ( − 1 , 1), w e observe that ω k ∗ is the only eigenfunction of the op erator defined in (3.18) that do esn’t c hange sign in ( − 1 , 1). STEP .2 Let us now prov e that k ∗ = 1 , (3.23) that is, λ 1 = 0. By a well-kno wn v ariational characterization of the first eigenv alue, w e ha ve λ 1 = inf u ∈ H 1 a ( − 1 , 1) R 1 − 1  a u 2 x − α ∗ u 2  dx R 1 − 1 u 2 dx . By Lemma 3.1, since λ k ∗ = 0 , it is sufficient to prov e that λ 1 ≥ 0, or Z 1 − 1 α ∗ u 2 dx ≤ Z 1 − 1 a u 2 x dx, ∀ u ∈ H 1 a ( − 1 , 1) (3.24) In tegrating b y parts, we hav e Z 1 − 1 α ∗ u 2 dx = − Z 1 − 1 ( a v x ) x v u 2 dx = Z 1 − 1 a v x  u 2 v  x dx = 13 = Z 1 − 1 a v x 2 uu x v dx − Z 1 − 1 a v 2 x  u 2 v 2  dx = = 2 Z 1 − 1 √ a v x v u √ au x dx − Z 1 − 1 a v 2 x  u 2 v 2  dx ≤ ≤ Z 1 − 1 a  v x u v  2 dx + Z 1 − 1 au 2 x dx − Z 1 − 1 a v 2 x  u 2 v 2  dx = Z 1 − 1 au 2 x dx , from which (3.24). F or the pro of of Theorem 1.3 the following Lemma is necessary . Lemma 3.4 L et T > 0 , α ∈ L ∞ ( Q T ) , let v 0 ∈ L 2 ( − 1 , 1) , v 0 ( x ) ≥ 0 a.e. x ∈ ( − 1 , 1) and let v ∈ B (0 , T ) b e the solution to the line ar system          v t − ( a ( x ) v x ) x = α ( t, x ) v in Q T = (0 , T ) × ( − 1 , 1) a ( x ) v x ( t, x ) | x = ± 1 = 0 t ∈ (0 , T ) v (0 , x ) = v 0 ( x ) x ∈ ( − 1 , 1) . Then v ( t, x ) ≥ 0 , ∀ ( t, x ) ∈ Q T . Pro of: Let v ∈ B (0 , T ) b e the solution to the system (1.2), and we consider the p ositiv e-part and the negative-part. It is sufficient to prov e that v − ( t, x ) ≡ 0 in Q T . Multiplying b oth members equation of the problem (1.2) by v − and integrating it on ( − 1 , 1) we obtain Z 1 − 1  v t v − − ( a ( x ) v x ) x v − − αv v −  dx = 0 . (3.25) Recalling the definition v + and v − , we obtain Z 1 − 1 v t v − dx = Z 1 − 1 ( v + − v − ) t v − dx = − Z 1 − 1 ( v − ) t v − dx = − 1 2 d dt Z ( v − ) 2 dx . In tegrating by parts and applying Theorem 2.1, we obtain v − ∈ H 1 a ( − 1 , 1) and the following equality Z 1 − 1 ( a ( x ) v x ) x v − dx = [ a ( x ) v x v − ] 1 − 1 − Z 1 − 1 a ( x ) v x ( − v ) x dx = Z 1 − 1 a ( x ) v 2 x dx . 14 W e also hav e Z 1 − 1 αv v − dx = − Z 1 − 1 α ( v − ) 2 dx and therefore (3.25) b ecomes − 1 2 d dt Z 1 − 1 ( v − ) 2 dx + Z 1 − 1 α ( v − ) 2 dx = Z 1 − 1 a ( x ) v 2 x ≥ 0 , from which d dt Z 1 − 1 ( v − ) 2 dx ≤ 2 Z 1 − 1 α ( v − ) 2 dx ≤ 2 k α k ∞ Z 1 − 1 ( v − ) 2 dx. F rom the ab o ve inequality , applying Gron wall’s lemma we obtain Z 1 − 1 ( v − ( t, x )) 2 dx ≤ e 2 t k α k ∞ Z 1 − 1 ( v − (0 , x )) 2 dx . Since v (0 , x ) = v 0 ( x ) ≥ 0 , w e ha ve v − (0 , x ) = 0 . Therefore, v − ( t, x ) = 0 , ∀ ( t, x ) ∈ Q T . F rom this, as we mentioned initially , it follows that v ( t, x ) = v + ( t, x ) ≥ 0 ∀ ( t, x ) ∈ Q T . W e are now ready to prov e our main result. Pro of: (of Theorem 1.3) STEP .1 T o pro ve Theorem 1.3 it is sufficient to consider the set of target states v d ∈ C ∞ ([ − 1 , 1]) , v d > 0 on [ − 1 , 1] . (3.26) Indeed, regularizing by conv olution, every function v d ∈ L 2 ( − 1 , 1) , v d ≥ 0 can b e appro ximated b y a sequence of strictly p ositiv e C ∞ ([ − 1 , 1]) − functions. STEP .2 T aking any nonzero, nonnegativ e initial state v 0 ∈ L 2 ( − 1 , 1) and an y target state v d as describ ed in (3.26) in STEP .1, let us set α ∗ ( x ) = − ( a ( x ) v dx ( x )) x v d ( x ) , x ∈ ( − 1 , 1) . (3.27) 15 Then, by (3.26), α ∗ ( x ) ∈ L ∞ ( − 1 , 1) . W e denote by {− λ k } k ∈ N and { ω k } k ∈ N , resp ectiv ely , the eigen v alues and orthonormal eigenfunctions 2 of the spectral prob- lem Aω + λω = 0 , with A = A 0 + α ∗ I (see Lemma 3.1). W e can see, by Lemma 3.2, that λ 1 = 0 and ω 1 ( x ) = v d ( x ) k v d k > 0 , ∀ x ∈ ( − 1 , 1) . (3.28) STEP .3 Let us now choose the following s tatic bilinear control α ( x ) = α ∗ ( x ) + β , ∀ x ∈ ( − 1 , 1) , with β ∈ R ( β to b e determined b elow). The corresp onding solution of (1.2), for this particular bilinear co efficien t α, has the following F ourier series representation ( 3 ) v ( t, x ) = ∞ X k =1 e ( − λ k + β ) t  Z 1 − 1 v 0 ( s ) ω k ( s ) ds  ω k ( x ) = = e β t  Z 1 − 1 v 0 ( s ) ω 1 ( s ) ds  ω 1 ( x ) + X k> 1 e ( − λ k + β ) t  Z 1 − 1 v 0 ( s ) ω k ( s ) ds  ω k ( x ) Let r ( t, x ) = X k> 1 e ( − λ k + β ) t  Z 1 − 1 v 0 ( s ) ω k ( s ) ds  ω k ( x ) where, recalling that λ k < λ k +1 , we obtain − λ k < − λ 1 = 0 for ever k ∈ N , k > 1 . Owing to (3.28), k v ( t, · ) − v d k ≤     e β t  Z 1 − 1 v 0 ( s ) ω 1 ( s ) ds  ω 1 − k v d k ω 1     + k r ( t, x ) k = =     e β t  Z 1 − 1 v 0 ( x ) ω 1 ( x ) dx  − k v d k     + k r ( t, x ) k 2 As first eigenfunction we take the one which is positive in ( − 1 , 1). 3 Observe that adding β ∈ R in the co efficien t α ∗ there is a shift of the eigenv alues corre- sponding to α ∗ from {− λ k } k ∈ N to {− λ k + β } k ∈ N , but the eigenfunctions remain the same for α ∗ and α ∗ + β . 16 Since − λ k < − λ 2 , ∀ k > 2, applying P arsev al’s equality we hav e k r ( t, x ) k 2 ≤ e 2( − λ 2 + β ) t X k> 1     Z 1 − 1 v 0 ω k ds     2 k ω k ( x ) k 2 = = e 2( − λ 2 + β ) t X k> 1 h v 0 , ω k i 2 = e 2( − λ 2 + β ) t k v 0 k 2 . Fixed ε > 0, we choose T ε > 0 such that e − λ 2 T ε = ε R 1 − 1 v 0 v d dx k v 0 kk v d k 2 . (3.29) Since v 0 ∈ L 2 ( − 1 , 1) , v 0 ≥ 0 and v 0 6≡ 0 in ( − 1 , 1) and by (3.28), we obtain h v 0 , ω 1 i = Z 1 − 1 v 0 ( x ) ω 1 ( x ) dx > 0 . (3.30) Then, it is p ossible choose β ε so that e β ε T ε Z 1 − 1 v 0 ω 1 dx = k v d k , that is, since ω 1 = v d k v d k , β ε = 1 T ε ln  k v d k 2 R 1 − 1 v 0 v d dx  . (3.31) So, b y (3.29), (3.31) and the ab o v e estimates for k v ( T ε , · ) − v d ( · ) k and k r ( T ε , · ) k w e conclude that k v ( T ε , · ) − v d ( · ) k ≤ e ( − λ 2 + β ε ) T ε k v 0 k = e − λ 2 T ε k v d k 2 R 1 − 1 v 0 v d dx k v 0 k = ε . F rom which we ha ve the conclusion. Pro of: (of Theorem 1.4) The pro of of Theorem 1.3 can be adapted to Theorem 1.4, keeping in mind that, in STEP .3, inequality (3.30) contin ues to hold in this new setting. In fact we hav e Z 1 − 1 v 0 ( x ) ω 1 ( x ) dx = Z 1 − 1 v 0 ( x ) v d ( x ) k v d k dx = = 1 k v d k Z 1 − 1 v 0 v d dx > 0 , by assumptions (1.3). 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