T >0. To this end, we first estimate 1
2k∇ψ(t)k2L2 ≤EW(t) + Z
W(t, x)|ψ(t, x)|2 dx
≤EW(t) +Ckxψ(t)k2L2,
under the assumption that λ > 0. Integrating equation (3.23) and having in mind that W(t, x) is sub-quadratic inx, we obtain that
k∇ψ(t)k2L2 ≤EW(0) + Z t
0
d
dsEW(s)ds+Ckxψ(t)k2L2
(3.24)
≤C0
1 +kxψ(t)k2L2 + Z t
0
kxψ(s)k2L2 ds
.
Recalling inequality (3.17), we infer d
dtkxψ(t)k2L2 +kxψ(t)k2L2 ≤C0
1 +kxψ(t)k2L2 + Z t
0
kxψ(s)k2L2 ds
,
which by Gronwall’s inequality yields an uniform bound onkxψ(t)kL2 for every time inter- val [0, T]. With this in hand, we can boundk∇ψ(t)kL2 by simply using inequality (3.24) once more.
Remark 3.3.7. In particular, for Ω = (0,0, ω)> and U(x) given by (3.2), we explicitly find
W(t, x) = 1 2
γ12cos2(ωt) +γ22sin2(ωt) x21 + γ21sin2(ωt) +γ22cos2(ωt)
x22+ sin(2ωt) γ12−γ22
x1x2+γ32x23 . Clearly,W = 12(γ12x21+γ22x22+γ32x23) in the axially symmetric case γ12 =γ22.
Note that the condition for the energy of the initial data ψ0 are not identical in both cases. The reason will become clear in the proof given below.
Proof. To simplify the arguments later on, let us first compute the conservation laws for the mass and momentum densities, i.e. ρ := |ψ|2 and J := Im(ψ∇ψ). Indeed a straightforward calculation yields
(3.25) ∂tρ+ divJ =iΩ·Lρ.
On the other hand, for the current density J we find
(3.26)
∂t Im(ψ∇ψ)
= Im
−i
2∆ψ+iU(x)ψ+iλ|ψ|2σψ+iΩ·Lψ
∇ψ
+ Im
ψ∇
i
2∆ψ−iU(x)ψ −iλ|ψ|2σψ+i(Ω·Lψ)
.
Next, we calculate
Im ψ∇(iΩ·Lψ)
= Im ψ(iΩ·L)∇ψ
−Ω∧J.
Thus we can combine the two terms in (3.26) which stem from the rotation via Im (iΩ·L)ψ∇ψ
+ Im ψ(iΩ·L)∇ψ
−Ω∧J = (iΩ·L)J−Ω∧J
where we have used that iΩ·L is real-valued. The other terms in (3.26) are usual in quantum hydrodynamics, see e.g. [3], yielding the following equation for J:
(3.27) ∂tJ+ div Re(∇ψ⊗ ∇ψ)
+ λσ
σ+ 1∇|ψ|2σ+2+ρ∇U = 1
4∆∇ρ+ (iΩ·L)J−Ω∧J.
The proof of finite time blow-up now follows by the classical argument of Glassey [32].
To this end, we consider the time evolution of I(t) := 1
2 Z
Rd
|x|2|ψ(t, x)|2dx.
Differentiating with respect to time and using (3.25), we obtain d
dtI(t) = Z
Rd
x·J(t, x)dx+ Z
Rd
|x|2
2 (iΩ·L)ρ(t, x)dx.
Integrating by parts and using (Ω ·L)|x|2 = 0 shows that the second integral in fact vanishes, i.e. we have
d
dtI(t) = Z
Rd
x·J dx.
Differentiating in time once more and using (3.27), we obtain d
dt Z
x·J dx = Z
x·
−div Re(∇ψ⊗ ∇ψ)
−λ σ
σ+ 1∇|ψ|2σ+2−ρ∇U +1
4∆∇ρ+ (iΩ·L)J−Ω∧J
dx, which we rewrite as
(3.28)
d dt
Z
x·J dx= Z
|∇ψ|2+λ dσ
σ+ 1|ψ|2σ+2−ρx· ∇U +x·(iΩ·L)J−x·Ω∧J
dx.
Now we first note that for any potential U(x) of the form (3.2) we have x· ∇U = 2U. Moreover, we compute
Z
Rd
x·(iΩ·LJ)dx=− Z
Rd
(Ω·Lx)·J dx =− Z
Rd
(Ω·(x∧ ∇)x)·J dx
=− Z
Rd
((Ω∧x)· ∇)x·J dx=− Z
Rd
(Ω∧x)·J dx,
which shows that the last two terms in (3.28) cancel each other. In summary we arrive at the following identity
(3.29) d2
dt2I(t) = Z
|∇ψ|2+λ dσ
σ+ 1|ψ|2σ+2−2U|ψ|2
dx,
which is in fact exactly the same as in the case of NLS without rotation, c.f. [19].
We can now prove assertion (i): Recall from Corollary 3.3.6 that if the potential U(x) is axially symmetric, thenE0(t) =E0(0), with E0 defined in (3.18). Hence from (3.29) and U ≥0 we can write
d2
dt2I(t)≤2E0+λdσ−2 σ+ 1
Z
Rd
|ψ|2σ+2dx.
AssumingE0 <0,λ <0, and σ ≥2/d, we consequently obtain d2
dt2I(t)<−C,
for some constantC > 0. Integrating this relation twice, we obtain I(t)<−C
2t2+c1t+c2
with some integration constants c1 and c2. Thus, if the solution ψ(t) ∈ Σ were to exist for all times, there would be a time T∗ < +∞, such that I(T∗) < 0. This however is
in contradiction with the fact that, by definition, I(t) ≥ 0 for all t ∈ R and hence the assertion is proved.
In order to prove assertion (ii) we again consider (3.29): The problem is that in the case of a non-axially symmetric potential (Ω·LU(x)6= 0), the energyE0 is no longer conserved.
Rather we only have the conservation law for EΩ(t) = EΩ(0). In order to use this piece of information, we first add and subtract to (3.29) a multiple of the angular momentum LΩ(t), i.e.
d2
dt2I(t) = Z
Rd
|∇ψ|2+ λσd
σ+ 1|ψ|2σ+2−2U|ψ|2+κψΩ·Lψ
dx− Z
κψΩ·Lψ dx, where κ > 0 is a parameter to be chosen later on. Using Cauchy-Schwarz and Young’s inequality, the last term on the right hand side can be bounded by
κ Z
Rd
ψΩ·Lψdx≤κ|Ω|k∇ψkL2kxψkL2 ≤ κθ
2 k∇ψk2L2 +κ|Ω|2
2θ kxψk2L2,
where θ >0 is another free parameter to be chosen later on. We consequently estimate d2
dt2I(t)≤ Z
Rd
1 + κθ
2
|∇ψ|2+ λσd
σ+ 1|ψ|2σ+2+
−2U +κ|Ω|2 2θ |x|2
|ψ|2dx
− Z
Rd
κψΩ·Lψ dx.
Now, we choose θ such that 2(1 + κθ2 ) =κ, that isθ = κ−2κ . In this way we have d2
dt2I(t)≤ Z
Rd
κ 1
2|∇ψ|2+λ 1
σ+ 1|ψ|2σ+2+U|ψ|2−ψΩ·Lψ
dx +
Z
Rd
λσd−κ
σ+ 1 |ψ|2σ+2dx+ Z
Rd
−(κ+ 2)U + κ2|Ω|2 2(κ−2)|x|2
|ψ|2dx.
Let γ := min(γ1, γ2, γ3), and chooseκ such that (κ+ 2)
2 γ2 =κ2 |Ω|2 2(κ−2). This yields κ=κΩ with
κΩ =
s 4γ2 γ2− |Ω|2.
By doing so, the last term in the previous inequality is seen to be non-positive and furthermore we conclude that, for λ <0 and σ≥ κdΩ, it holds:
(3.30) d2
dt2I(t)≤κΩEΩ(t)≡κΩEΩ(0).
Thus, if the initial energy EΩ(0) < 0 the second derivative of I(t) is again negative and we can argue (by contradiction) as before.
The next lemma gives a proof of assertion (2)(ii) in Theorem 3.2.3.
Lemma 3.4.2. Let λ <0, σ <2/(d−2), d = 2,3, and U(x) be a quadratic potential of the form (3.2). Denote γ = min{γj}dj=1 and suppose that Ω = (0,0, ω)> if d= 3 or Ω·L is of the form (3.5) if d= 2. If there exists a T >0 such that
T < 2γ2
|(γ12−γ22)ω|
as well as
E0(0)T2+ ˙I(0)T +I(0) <0,
then the corresponding solution to equation (3.4) necessarily blows up in finite time.
Note that if one does not care about the orientation of the rotation, one can always change the sign of Ω such thatLΩ(0)≤0 in which case EΩ <0 implies E0(0)<0.
Proof. Again, we let
I(t) = Z
x2|ψ(t, x)|2 dx
denote the second moment of |ψ|2. In equation (3.29), we already showed that d2
dt2I(t) = Z
Rd
|∇ψ|2+ λdσ
σ+ 1|ψ|2σ+2−2U|ψ|2 dx.
In the special case where Ω = (0,0, ω)>, equation (3.14) yields d
dtLΩ(t) = (γ12−γ22)ω Z
Rd
x1x2|ψ|2 dx.
Integrating twice and applying H¨older’s inequality yields Z t
0
LΩ(s)ds =tLΩ(0) + Z t
0
Z s 0
(γ12−γ22)ω Z
Rd
x1x2|ψ(τ)|2 dx dτ ds
≤tLΩ(0) + Z t
0
Z s 0
|(γ12−γ22)ω|
Z
Rd
1
2 x21+x22
|ψ(τ)|2 dx dτ ds.
The integrand on the right hand side is positive and we can estimate the integral over {0≤τ ≤s} by the integral over the larger set{0≤τ ≤t}. Hence it holds that
Z t 0
LΩ(s)ds ≤tLΩ(0) +t Z t
0
|(γ12 −γ22)ω|
Z
Rd
1
2 x21 +x22
|ψ(s)|2 dx ds
≤tLΩ(0) +t Z t
0
|(γ12−γ22)ω|
γ2
Z
Rd
U|ψ(s)|2 dx ds (3.31)
by the definition of γ. Integrating (3.29) yields I˙(t) = ˙I(0) +
Z t 0
Z
Rd
|∇ψ|2+ λdσ
σ+ 1|ψ|2σ+2−2U|ψ|2 dx ds
Adding and substracting the integral over 2LΩ yields I(t) = ˙˙ I(0) +
Z t 0
Z
Rd
|∇ψ|2+ λdσ
σ+ 1|ψ|2σ+2−2U|ψ|2 dx ds + 2
Z t 0
LΩ(s)ds−2 Z t
0
LΩ(s)ds.
The definition of the energy (3.6) thus yields I(t) = ˙˙ I(0) +
Z t 0
EΩ ds+λ(dσ−2) σ+ 1
Z t 0
Z
Rd
|ψ|2σ+2 dx ds
−4 Z t
0
Z
Rd
U|ψ|2 dx ds−2 Z t
0
LΩ(s)ds.
The energyEΩ is constant, the term involving the nonlinearity is non-positive, and hence estimate (3.31) yields
I˙(t)≤I˙(0) + 2t(EΩ−LΩ(0)) + Z t
0
2
t|(γ12−γ22)ω|
γ2 −2 Z
Rd
U|ψ(s)|2 dx
ds
Of course,E0(0) =EΩ−LΩ(0). Under the conditions of the theorem, we obtainI(T)≤0 upon another integration with respect to time, thus yielding a contradiction!