1 0 {|x|≤1} L (R), then the equation Assume that u ∈ 1 operators in a Banach space X and u : R → R is a realis in ψ). dynamics play a role in r>1 valued control, which involves a control control term uBψ that loc u {|x|≤1} and ψ0 ∈ H (R(1) r loc ). Theorem 1. Let K(x) = ln(|x|)1 r>1 valued control, involves a uBψ that r Φ(R), is bilinear bilinear in (u, (u,which ψ). Such Such dynamics play term a major major role in admits (t)(ψ ). a global flow ψ(t) = L then the equation (1) Assume that u ∈ u 0 L (R), then the equation (1) Assume that u ∈ u (t)(ψ loc r>1 valued control, which involves a control term uBψ that ). admits a global flow ψ(t) = Φ loc r>1 physics and are the subject of a vast literature Khapalov r is bilinear in (u, ψ). Such dynamics play a major role in 0 0 u is bilinear in (u, ψ). Such dynamics play a major role in L (R), then the equation (1) Assume that u ∈ physics and are the subject of a vast literature Khapalov admits u (t)(ψ ). aa global flow r>1 = 1Φ 3 0 global flow ψψ(t) ψ(t) =loc Φ is bilinear inare (u,etthe ψ). Such dynamics play a major role in admits 0 ). attainable set (2010). al. (1982), Marsden, and Slemrod H (R Moreover, for every physics and a literature u (t)(ψ 3 ), the 0 ∈ physics andBall are subject ofBall, a vast vast literature Khapalov 3 (2010). In In Ball etthe al.subject (1982),of Ball, Marsden, andKhapalov Slemrod admits (t)(ψ a global flow ψψ(t) =111Φ ∈ H (R ), the set Moreover, for every 0 ). attainable 0 0 3 physics andBall areet ofBall, a vast literature 3 ), (2010). In al. (1982), Marsden, and Slemrod ∈ attainable set Moreover, ψ u the (2010). In Ball etthe al.subject (1982), Marsden, andKhapalov Slemrod Φ ∈H H11 (R (R ), the attainable set (3) Moreover, for for every every ψ00 T. Chambrion is supported byBall, the grant ”QUACO” ANR-17(t)(ψ ) 3u 0 ) u (t)(ψ T. Chambrion is supported byBall, the grant ”QUACO” ANR-17(2010). In Ball et al. (1982), Marsden, and”BEKAM” Slemrod ∈ H (R ), the attainable set (3) Moreover, for every ψ Φ 0 0 CE40-0007-01. L. Thomann is supported by the grants 0 u r T. Chambrion is supported by the grant ”QUACO” ANR-17u t∈R Φ (t)(ψ ) (3) T. Chambrion supported by the grant ”QUACO” ANR-17u∈Lrloc (R), Φ (t)(ψ0 CE40-0007-01. L. is Thomann is supported by the grants ”BEKAM” 0 ) (3) r (R), t∈R u ANR-15-CE40-0001 and ”ISDEEC” ANR-16-CE40-0013. t∈R u∈L u∈L (R), T. Chambrion is supported by the grant ”QUACO” ANR-17loc CE40-0007-01. L. Thomann is supported by the grants ”BEKAM” r>1 loc Φ (t)(ψ ) (3) r CE40-0007-01. L. Thomann is supported by the grants ”BEKAM” ANR-15-CE40-0001 and ”ISDEEC” ANR-16-CE40-0013. 0 r t∈R t∈R u∈Lr>1(R),
loc u∈Lr>1 loc (R), CE40-0007-01. L. Thomann is supported by the grants ”BEKAM” ANR-15-CE40-0001 and ”ISDEEC” ”ISDEEC” ANR-16-CE40-0013. r ANR-15-CE40-0001 and ANR-16-CE40-0013. r>1 t∈R u∈Lr>1 loc (R), ANR-15-CE40-0001 and ”ISDEEC” ANR-16-CE40-0013. 2405-8963 © © 2019 2019, IFAC IFAC (International Federation of Automatic Control) rights reserved. Copyright 446 Hosting by Elsevier Ltd. All r>1 Copyright © 2019 IFAC 446 Peer review under responsibility of International Federation of Automatic Control. Copyright © 446 Copyright © 2019 2019 IFAC IFAC 446 10.1016/j.ifacol.2019.11.796 Copyright © 2019 IFAC 446
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is a countable union of compact subsets of H1 (R3 ). In this paper, the solutions to (1) are understood in the mild sense t itH u(τ )ei(t−τ )H (Kψ(τ ))dτ ψ(t) = e ψ0 − i 0 t ei(t−τ )H (|ψ|2 ψ)dτ. + iσ 0
305
In the sequel c, C > 0 denote constants the value of which may change from line to line. These constants will always be universal, or uniformly bounded. For x ∈ R3 , we write x = (1 + |x|2 )1/2 . We will sometimes use the notations LpT = Lp ([0, T ]) and LpT X = Lp ([0, T ]; X) for T > 0. 3. PROOF OF THEOREM 1 3.1 Global existence theory for dynamics (1)
1.4 Content of the paper
Notice that using the reversibility of the equation (1), it is enough to consider non-negative times in the proofs.
The rest of this note provides a proof of Theorem 1. The proof crucially relies on classical Strichartz estimates, which we recall in Section 2. The proof itself is split in two parts. The global well-posedness of the problem (1) is established in Section 3.1, using among other some energy estimates. The proof of the obstruction result follows the strategy used in the paper Ball et al. (1982) and is given in Section 3.2.
We first state two technical lemmas which will be useful to control the bilinear term in (1). Lemma 2. Let K(x) = ln(|x|)1{|x|≤1} . Then for all 1 ≤ p < 2 there exists C > 0 such that KψW 1,p (R3 )) ≤ CψH1 (R3 ) .
2. STRICHARTZ ESTIMATES As in Chambrion and Thomann (2018a), the Strichartz estimates play a major role in the argument, let us recall them in the three-dimensional case. A couple (q, r) ∈ [2, +∞]2 is called admissible if 3 2 3 + = , q r 2 and if one defines XT1 := Lq [−T, T ] ; W 1,r (R3 ) , (q,r) admissible
then for all T > 0 there exists CT > 0 so that for all ψ0 ∈ H1 (R3 ) we have (4) eitH ψ0 XT1 ≤ CT ψ0 H1 (R3 ).
Using interpolation theory one can prove that XT1 = L∞ [−T, T ] ; H1 (R3 ) ∩ L2 [−T, T ] ; W 1,6 (R3 ) , so that one can define ψXT1 = ψL∞ ([−T,T ];H1 (R3 )) + ψL2 ([−T,T ];W 1,6 (R3 )) .
T
(5) where q and r are the H¨ older conjugate of q and r. In particular, they are related by 2 3 7 (6) + = . q r 2 We refer to Poiret (2012) for a proof. Let us point out that (5) implies that
t i(t−τ )H
F (τ )dτ
≤ CT F1 L1 ([−T,T ],H1 (R3 )) +F2 L2 ([−T,T ],W 1,6/5 (R3 )) ,(7) for any F1 , F2 such that F1 + F2 = F , which will prove useful. e
0
1 XT
with 1/q1 = 1/p − 1/2 and 1/q2 = 1/p − 1/6. Now we use the expression of K to get KLq1 < ∞ (since q1 < ∞). Then we write, by (2) KW 1,q2 ≤ CxKLq2 + C∇KLq2 , and since |∇K| ≤ C|x|−1 , we check that KW 1,q2 < ∞ because q2 < 3.
The case p = 2 does not hold true in the previous statement, but we have the following bound after averaging in time. Lemma 3. Let K(x) = ln(|x|)1{|x|≤1} and T > 0. Then for all 2 ≤ q < ∞, there exists CT > 0 such that for all ψ ∈ XT1 KψLq ([−T,T ];H1 (R3 )) ≤ CT ψXT1 . Proof. Firstly by (2) we have Kψ 1 ≤ c∇Kψ 2 + cK∇ψ H
We will also need the inhomogeneous version of the Strichartz estimates: for all T > 0, there exists CT > 0 so that for all admissible couple (q, r) and function F ∈ Lq ([−T, T ]; W 1,r (R3 )), t ei(t−τ )H F (τ )dτ X 1 ≤ CT F Lq ([−T,T ],W 1,r (R3 )) , 0
Proof. Let 1 ≤ p < 2, then by (Chambrion and Thomann, 2018a, Lemma A.1), Kψ 1,p ≤ CKLq1 ψH1 + CKW 1,q2 ψL6 , (8) W
447
L
L2
+ cKxψ L2 . (9)
• Let us study the first term in (9). Since |∇K| ≤ C|x|−1 , we can use the Hardy inequality −1 |x| ψ 2 3 ≤ C ψ 1 3 (10) L (R )
H (R )
(we refer to (Tao, 2006, Lemma A.2) for the general statement and proof of this inequality), and therefore the contribution of the first term reads ∇Kψ Lq L2 ≤ 1 . CT 1/q ψL∞ H1 ≤ CT ψXT T
T
• To bound the contribution of the two last terms in (9), we will use that K ∈ Lp (R3 ) for any 1 ≤ p < ∞. Given 2 ≤ q < ∞, we choose r > 2 such that the couple (q, r) is (Strichartz) admissible and write, using H¨older K∇ψ 2 ≤ KLp ∇ψLr ≤ cKLp ψW 1,r , L with 1/p + 1/r = 1/2. Thus K∇ψ q 2 ≤ cKLp ψLq W 1,r ≤ cKLp ψX 1 . L L T T T
Similarly, Kxψ q 2 ≤ cKLp ψLq W 1,r ≤ cKLp ψX 1 , L L T T T
which achieves the proof.
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We now state a global existence result for (1) adapted to our control problem. Proposition 4. Let r > 1 and u ∈ Lrloc (R). Set K(x) = ln(|x|)1{|x|≤1} . Let ψ0 ∈ H1 (R3 ), then the equation (1) admits a unique global solution ψ ∈ C(R; H1 (R3 )) ∩ L2loc (R; W 1,6 (R3 )) which moreover satisfies the bound ψL∞ ([−T,T ];H1 (R3 )) + ψL2 ([−T,T ];W 1,6 (R3 )) ≤ C T, ψ0 H1 (R3 ) , uLr (−T,T ) .(11)
Proof. The proof is in the spirit of the proof of (Chambrion and Thomann, 2018a, Proposition 1.5), but here we use moreover the Hardy inequality (10) to control the bilinear term in (1). Energy bound: Assume for a moment that the solution exists on a time interval [0, T ]. For 0 ≤ t ≤ T , we define σ E(t) = ψHψ + |ψ|2 + |ψ|4 dx 2 3 R σ = |∇ψ|2 + |x|2 |ψ|2 + |ψ|2 + |ψ|4 dx. 2 R3
Then, using that ∂t ψ = −i(Hψ + σ|ψ|2 ψ) + iu(t)K(x)ψ, we get E (t) = 2 ∂t ψ ψ + Hψ + σ|ψ|2 ψ dx R3 = −2u(t) KψHψdx 3 R = 2u(t) ψ∇K · ∇ψdx.
Using that |∇K| ≤ C|x| we get
R3 −1
, by the Hardy inequality (10)
As a consequence, 2 Φ(ψ)XT1 ≤ cψ0 H1 + cψL∞ H1 ψL2 L∞ T T
+CT uLrT ψL∞ H1 . T By the Gagliardo-Nirenberg and Sobolev inequalities on R3 , 1/2
1/2
1/2
1/2
ψL∞ ≤ CψL6 ψW 1,6 ≤ CψH1 ψW 1,6 , 1/2
1/2
thus ψL2T L∞ ≤ cT 1/4 ψL∞ H1 ψL2 W 1,6 , and for ψ ∈ T T BT,R we get Φ(ψ)XT1 ≤ cψ0 H1 + cT 1/2 R3 + CT RuLrT .
We now choose R = 2cψ0 H1 . Then for T > 0 small enough, Φ maps BT,R into itself. With similar estimates we can show that Φ is a contraction in BT,R , namely Φ(ψ1 ) − Φ(ψ2 )XT1 ≤ cT 1/2 R2 + CT uLrT ψ1 − ψ2 XT1 . As a conclusion there exists a unique fixed point to Φ, which is a local solution to (1). The local time of existence only depends on u and on the H1 -norm. Therefore one can use the energy bound to show the global existence. Proof of the bound (11): It remains to prove that ψL2 ([−T,T ];W 1,6 (R3 )) ≤ C T, ψ0 H1 (R3 ) , uLr (−T,T ) . (12) The argument follows the main lines of (Chambrion and Thomann, 2018a, Bound (1.18)), so we only give the key steps. Similarly to Chambrion and Thomann (2018a), for any δ > 0 small enough we can prove the bound ψL2 ([τ,τ +δ];W 1,6 ) ≤ cψL∞ H1 T 2 +cδ 1/2 ψ L∞ H1 ψL2 ([τ,τ +δ];W 1,6 ) T
+CT ψL∞ H1 uLr (−T,T ) . T
E (t) ≤ 2|u(t)|ψ∇K L2 ∇ψ L2 ≤ C|u(t)||x|−1 ψ L2 ∇ψ L2 2 ≤ C|u(t)|ψ H1
≤ C|u(t)|E(t). Thus, using that σ ≥ 0, we deduce the first part of the bound (11). Local existence and global existence: We consider the map t ei(t−τ )H (|ψ|2 ψ)dτ Φ(ψ)(t) = eitH ψ0 + iσ 0 t u(τ )ei(t−τ )H (Kψ)dτ, −i 0
and we will show that it is a contraction in the space BT,R := ψXT1 ≤ R ,
with R > 0 and T > 0 to be fixed.
Then we choose δ = δ(T ) > 0 such that 2 1 cδ 1/2 ψ L∞ H1 = , T 2 thus the previous estimate gives ψL2 ([τ,τ +δ];W 1,6 ) ≤ 2ψL∞ H1 c + CT uLr (−T,T ) , T
and by summing up in δ, we get (12). 3.2 Meagerness of the attainable set
Let > 0 and let u, un ∈ L1+ ([0, T ]; R) such that un u weakly in L1+ ([0, T ]; R). This implies a bound un L1+ ≤ C(T ) for some C(T ) > 0, uniformly in n ≥ 1. T We have t ψ(t) = eitH ψ0 − i u(τ )ei(t−τ )H (Kψ(τ ))dτ 0 t ei(t−τ )H (|ψ|2 ψ)dτ, +iσ 0
and
By the Strichartz inequalities (4) and (5) Φ(ψ)XT1 ≤ cψ0 H1 + c|ψ|2 ψ L1 H1 + cuKψL1T H1 . T
Then, for all r > 1, there exists q < ∞ sur that 1/r + 1/q = 1, and we have by Lemma 3 uKψL1T H1 ≤ uLrT KψLqT H1 ≤ CT uLrT ψXT1 . 448
ψn (t) = e
itH
ψ0 − i
+iσ
t
un (τ )ei(t−τ )H (Kψn (τ ))dτ 0
t 0
ei(t−τ )H (|ψn |2 ψn )dτ.
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We set zn = ψ − ψn , then zn satisfies
zn = L(ψ, ψn ) + N (ψ, ψn ),
with
L(ψ, ψn ) = −i
t
(13)
u(τ ) − un (τ ) ei(t−τ )H (Kψ)dτ 0 t un (τ )ei(t−τ )H K(ψ − ψn ) dτ −i
By the Minkowski inequality, the unitarity of eiτ H and the H¨older inequality t un (τ ) − u(τ ) ei(t−τ )H (Kψ(τ ))dτ H1 (R3 ) tn t un (τ ) − u(τ )Kψ(τ ) 1 3 dτ ≤ H (R ) tn
≤ un − uL1+ KψLq
N (ψ, ψn ) = iσ
t
e 0
i(t−τ )H
+iσ
0
T
ei(t−τ )H (ψ − ψn )ψn2 dτ.
Let us prove that zn tends to 0 in L∞ ([0, T ]; H1 (R3 )). Lemma 5. Denote by t un (τ ) − u(τ ) ei(t−τ )H (Kψ(τ ))dτ ∞ . n := Then n tends to 0, when n tends to +∞.
Proof. We proceed by contradiction. Assume that there exists > 0, a subsequence of un (still denoted by un ) and a sequence tn −→ t ∈ [0, T ] such that tn ≥ . un (τ ) − u(τ ) ei(tn −τ )H (Kψ(τ ))dτ H1 (R3 )
(14) Up to a subsequence, we can assume that for all n ≥ 1, tn ≤ t or tn ≥ t. We only consider the first case, since the second is similar. By the Minkowski inequality and the unitarity of eiτ H
=
T 0
+∞
+∞
(2k + 1)|αk (τ )|2 and
k=0
(2k + 1)|αk (τ )|
k=0
2 q /2
dτ
1/q
.
(19)
k=0
Denote by ρ = supn≥0 un − uL1+ . We claim that there T exists M > 0 large enough such that the function g(τ, x) = M k=0 αk (τ )hk (x) satisfies v−gLq ([0,T ];H1 (R3 )) ≤ /(4ρ). Actually, T 1/q +∞ q /2 v − gLqT H1 = (2k + 1)|αk (τ )|2 dτ
tn un (τ ) − u(τ ) ei(tn −τ )H− ei(t−τ )H (Kψ(τ ))dτ 1 3 H (R ) 0 itn H itH (Kψ(τ )) q ≤ un − uL1+ e −e , 1 3
0
k=M +1
tends to zero when M −→ +∞, by the Lebesgue theorem and (19), hence the claim.
Lτ ∈[0,T ] H (R )
where 1 < q < ∞ is such that 1/(1 + ) + 1/q = 1. Now, by Lemma 3, we have (15)
Now we apply (Chambrion and Thomann, 2018a, Lemma 3.2) (with d = 3 and s = 1) together with the previous lines, and we get that tn un (τ )−u(τ ) ei(tn −τ )H −ei(t−τ )H (Kψ(τ ))dτ 1 3 0
so that we have v(τ, ·)2H1 =
This implies in particular that +∞ q /2 (2k + 1)|αk (τ )|2 ∈ LqT .
Then by H¨ older
KψLqT H1 (R3 ) ≤ CT, ψXT1 < ∞.
k=0
v
H (R )
T
t Let us now prove that 0 un (τ )−u(τ ) ei(t−τ )H (Kψ(τ ))dτ tends to 0 in H1 (R3 ), to reach a contradiction with (14). We set v(τ ) = ei(t−τ )H (Kψ(τ )). Then by the unitarity of H, we have v(τ )H1 = Kψ(τ )H1 , thus by (15), v ∈ Lq ([0, T ]; H1 (R3 )). We expand v on the Hermite functions (hk )k≥0 (which are the eigenfunctions of H) which form a Hilbertian basis of L2 (R3 ) +∞ αk (τ )hk (x), v(τ, x) =
LqT H1
tn un (τ )− u(τ ) ei(tn −τ )H− ei(t−τ )H (Kψ(τ ))dτ 1 3 H (R ) 0 tn ≤ un (τ )− u(τ ) ei(tn −τ )H − ei(t−τ )H (Kψ(τ )) dτ H1 (R3 ) 0 tn it H itH un (τ ) − u(τ ) (Kψ(τ )) 1 3 dτ. = e n −e 0
H (R )
0
LT H1 (R3 )
0
(17)
T
Using Lemma 3 and the fact that un − uL1+ ≤ C, we T deduce that the term (17) tends to 0. We combine this with (16) to deduce tn un (τ ) − u(τ ) ei(tn −τ )H (Kψ(τ ))dτ − 0 t un (τ ) − u(τ ) ei(t−τ )H (Kψ(τ ))dτ −→ 0.(18) 1 3
0
H1 (R3 )
≤ |t − tn |1/q un − uL1+ KψL2q H1 (R3 ) .
(ψ − ψn )(ψ + ψn )ψ dτ
t
τ ∈[tn ,t]
T
0
and
307
H (R )
(16)
tends to zero when n tends to +∞.
449
We have t t M un (τ )−u(τ ) αk (τ )dτ. un (τ )−u(τ ) g(τ )dτ = hk 0
k=0
0
Then, by (19), for all k ≥ 0, αk ∈ LqT , which implies
M
t 0
2 un (τ ) − u(τ ) g(τ )dτ H1 (R3 ) =
t 2 s (2k + 1) un (τ ) − u(τ ) αk (τ )dτ −→ 0,
k=0
0
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by the weak convergence of (un ). Finally, for n large enough, t un (τ ) − u(τ ) v(τ )dτ H1 (R3 ) 0 t ≤ un − uL1+ + un (τ ) − u(τ ) g(τ )dτ T 4ρ H1 (R3 ) 0 ≤ , 2 which together with (14) and (18) gives the contradiction. Thanks to (Chambrion and Thomann, 2018a, Lemma A.3) we get N (ψ, ψn )(t)H1 (R3 ) t ≤ (ψ − ψn )(ψ + ψn )ψH1 (R3 ) dτ 0 t (ψ − ψn )ψn2 H1 (R3 ) dτ + 0 t zn H1 (R3 ) ψ2W 1,6 + ψn 2W 1,6 dτ. ≤c
(20)
1/2
1/2
H zn (t) = −iH 0 t i(t−τ )H 1/2 Kzn dτ + H 1/2 N (ψ, ψn )(t). un (τ )e H −i 0
where p < 2 is given by
=
−
2 1+ ),
(21)
see (6).
≤ C3 (T )n , and this latter quantity tends to 0 when n tends to +∞. Lemma 6. (Gr¨onwall inequality). Let f, A, B : [0, T ] −→ R+ be non-negative functions with f (0) = 0, and assume that A is an increasing function. Assume that t f (t) ≤ A(t) + B(τ )f (τ )dτ, ∀ 0 ≤ t ≤ T. (23)
t
0
B(τ )dτ .
(24)
Then by the classical Gr¨onwall inequality (A is constant in s) we get s f (s) ≤ A(t) exp B(τ )dτ ,
for all 0 ≤ s ≤ t, which implies (24) for s = t.
4. CONCLUSION
un H 1/2 Kzn L1+ Lp (R3 ) t t 1/(1+) ≤C |un (τ )|1+ zn (τ )1+ . (22) H1 (R3 ) dτ 0
As a conclusion, from (20), (21) and (22) we infer
This note provides an example of a Ball-Marsden-Slemrod like obstruction to the controllability of a nonlinear partial differential equation with a bilinear control term. The novelty of the result lies in the unboundedness of the biinear control term. The possible relations of this obstruction result and the concepts introduced in Boussa¨ıd et al. (2013) will be the subject of further investigations in future works.
zn (t)H1 (R3 ) ≤ t 1/(1+) n + C |un (τ )|1+ zn (τ )1+ + H1 (R3 ) dτ 0 t zn (τ )H1 (R3 ) ψ(τ )2W 1,6 + ψn (τ )2W 1,6 dτ. +c
REFERENCES Ball, J.M., Marsden, J.E., and Slemrod, M. (1982). Controllability for distributed bilinear systems. SIAM J. Control Optim., 20(4), 575–597. doi:10.1137/0320042. URL https://doi.org/10.1137/0320042. Boussa¨ıd, N., Caponigro, M., and Chambrion, T. (2013). Weakly coupled systems in quantum control. IEEE Trans. Automat. Control, 58(9), 2205–2216. doi:10.1109/TAC.2013.2255948. URL https://doi.org/10.1109/TAC.2013.2255948.
0
We now apply Lemma 6 with t 1/(1+) A(t) = n + C |un (τ )|1+ zn (τ )1+ dτ 1 3 H (R ) 0
thus
0
|un (τ )|1+ dτ
0
By Lemma 2 H 1/2 Kzn Lp (R3 ) = Kzn W 1,p (R3 ) ≤ Czn H1 (R3 ) , which in turn implies
and
T
0
t
1 7 3(2
Proof. In the case A constant, the result is classical (see e.g. (Tao, 2006, Theorem 1.10)). Now assume that A is an increasing function and let 0 ≤ s ≤ t ≤ T , then by (23) s B(τ )f (τ )dτ, ∀ 0 ≤ s ≤ t. f (s) ≤ A(t) +
Thus from the Strichartz inequality (7) we deduce
1 p
≤ C2 (T )n exp C2 (T )
f (t) ≤ A(t) exp
u(τ ) − un (τ ) ei(t−τ )H (Kψ)dτ
+N (ψ, ψn )(t)H1 (R3 ) ,
0
zn (t)H1 (R3 ) ≤ n + un H 1/2 Kzn L1+ Lp (R3 )
0
Finally, by the classical Gr¨onwall inequality we deduce
Then for all 0 ≤ t ≤ T ,
By (13) we have
t
0
where in the last line, we used (11). We raise the previous inequality to the power (1 + ), and get t 1+ dτ . zn (t)H1 (R3 ) ≤ C2 (T ) n+ |un (τ )|1+ zn (τ )1+ 1 3 H (R ) zn 1+ H1 (R3 ) L∞ T
0
zn (t)H1 (R3 ) t 1/(1+) ≤ n + C |un (τ )|1+ zn (τ )1+ dτ 1 3 H (R ) 0 t c ψ(τ )2W 1,6 +ψn (τ )2W 1,6 dτ e 0 t 1/(1+) , ≤ C1 (T ) n + |un (τ )|1+ zn (τ )1+ H1 (R3 ) dτ
B(τ ) = c ψ(τ )2W 1,6 + ψn (τ )2W 1,6 ,
450
2019 IFAC NOLCOS Vienna, Austria, Sept. 4-6, 2019
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