the threshold criterion of the gross-pitaevskii equation with ......gross-pitaevskii equation for...

The threshold criterion of the Gross-Pitaevskiiequation with trapped dipolar quantum gases

Dr. Li MaZhongyuan Institute of MathematicsHenan Normal University, CHINA

April 10th, 2014, Bilbao, Spain

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Outline

Abstract and references

The problem and Results of Carles-Markowich-Sparber

new results due to Ma-Wang

more results and questions?

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ABSTRACT:

We consider the Gross-Pitaevskii equation for the trapped dipolarquantum gases. The important contribution to this problem hasbeen done by Carles-Markowich-Sparber. We obtain the sharpcriterion for the global existence and finite time blow up in theunstable regime by constructing a variational problem and theso-called invariant manifold of the evolution flow.

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references

[C-M-S] R.Carles, P.Markowich, and C.Sparber, On theGross-Pitaevskii equation for trapped dipolar quantum gases,nonlinearity 21(2008)2569-2590

[M-W] L.Ma, J.Wang, Sharp Threshold of the Gross-PitaevskiiEquation with Trapped Dipolar Quantum Gases, Canad.Math. Bull. Vol. 56 (2), 2013 pp. 378-387http://dx.doi.org/10.4153/CMB-2011-181-2

[M-C] L.Ma, P.Cao, The threshold for the focusingGross-Pitaevskii equation with Trapped Dipole quantumGases, J. Math. Anal. Appl. 381 (2011) 240-246

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the problemWe consider the time-dependent Gross-Pitaevskii equation in R3:

i∂tu +12∆u = Vu + λ1|u|2u + λ2(K ∗ |u|2)u (1)

with the regular Cauchy data

u(0, x) = u0(x), x ∈ R3, (2)

where λ1, λ2 are real constants,

V (x) =12|x |2, (3)

K (x) =x21 + x2

2 − 2x23

|x |5, x ∈ R3, (4)

andK ∗ |u|2(x) =

ˆR3

K (x − y)|u(y)|2dy .

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physical background

The evolution equation of type (1) receives a lot of attentionbecause the success of atomic Bose-Einstein condensation (BEC)has stimulated great interest in the properties of trapped quantumgases. The first kind of the equation (1) has been introduced byphysicists Yi and You to describe particles which interact viashort-range repulsive forces and long-range (partly attractive)dipolar forces. Physically, the non-local term K ∗ |u|2 in (1) appearswhen an electric field is introduced along the positive z-axis.

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conserved quantities

The following two important qualities are, at least formally,conserved by the time evolution equation(1):

Mass : M = M(u) = ‖u‖2L2 = M(u0), (5)

Energy : E = E (u) =12‖∇u‖2

L2 +12

ˆR3|x |2|u|2 +

λ1

2‖u‖4

L4

+λ2

2

ˆR3

K ∗ |u|2|u|2 (6)

= E (u0).

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working space

The mass and energy conservations lead to the introduction of anatural energy space associated with the equation(1) in the linearcase λ1 = λ2 = 0:

Σ := {u ∈ H1(R3) : ‖xu‖2L2 < ∞}. (7)

Σ is a Hilbert space with the following inner product

〈u, v〉Σ :=

ˆR3

uv +

ˆR3∇u∇v +

ˆR3|x |2uv

for u, v ∈ Σ.

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Gagliardo-Nirenberg inequality and uncertainty principle

Recall here that the Gagliardo-Nirenberg inequality on Rd ,1 ≤ d ≤ 3: there exists a dimensional constant C > 0 such that

|u|44 ≤ C |u|4−d2 |∇u|d2 .

The uncertainty principle says that

ˆR3|u|2 ≤ C

(ˆR3|x |2|u|2

) 12(ˆ

R3|∇u|2

) 12

. (8)

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compactnessThe following well-known compactness lemma plays an importantrole.

Lemma

The energy space Σ is compact in Lp(R3) for any 2 ≤ p < 6.

We denote the Fourier transform of f by

f (ξ) =

ˆR3

e−ixξf (x)dx . (9)

Then the famous Plancherel formula gives us thatˆ

R3|f |2 =

1(2π)3

ˆR3|f |2. (10)

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compactness 2

R.Carles, P.Markowich, and C.Sparber (2008) proved the below.

Lemma

The Fourier transform of K is given by

K (ξ) =4π3

(3

ξ23

|ξ|2− 1

). (11)

The result above be obtained from Theorem 5 of page 73 in Stein’sbook (1970). Moreover, K is a Lp(R3) → Lp(R3) (1 < p < ∞)multiplier by the Calderon-Zygmund theory.

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well-posedness

A nice observation made by R.Carles, P.Markowich, and C.Sparber(via Mehler’s formula) is that the group U(t) = e−itH , whereH = 1

2(−∆ + |x |2), not only bounded on L2, but also enjoysdispersive properties for small time that

|U(t)φ|L∞(Rd ) ≤C

|t|d/2 |φ|L1(Rd ),

which leads to Strichartz type estimates. Roughly speaking, TheStrichartz estimate is looking for the pair (p, q) such that theoperator U is a bounded linear operator from the space-timeBanach space Lp

t Lqx to its dual space Lp′

t Lq′x . We use the pair

(p, q) = (83 , 4).

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well-posedness

Using these ingredients, Carles-Markowich-Sparber can establish in[C-M-S] the local wellposedness of the Cauchy problem for theequation (1). Moreover, they have proved that the solution isglobal if λ1 ≥ 4π

3 λ2 ≥ 0 (the corollary 4 in section 2). However,when λ1 < 4π

3 λ2(the so-called unstable regime), there is no suchresult, though global existence of the solution for small data can beobtained in this case.

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well-posedness 2One of the main results of R.Carles, P.Markowich, and C.Sparber(C-M-S, 2008) is below.

Proposition

There exists a T∗ ∈ R+ ∪ {∞} such that (1) has a unique maximalsolution in

{u ∈ C ([0, T∗); Σ) ; u,∇u, xu ∈ C([0, T∗); L2(R3)

)∩L

83loc

([0, T∗); L4(R3)

)}

such that u(0) = u0. The solution u is maximal in the sense that ifT∗ < ∞, then

‖∇u(t)‖L2 →∞ (t → T∗).

Moreover, the quantities defined as (5) and (6) are conserved for0 ≤ t < T∗.

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a corollary

As an easy consequence of proposition 3, we have

Corollary

Under the same assumptions as in Prop 3 and in addition theassumption that λ1 ≥ 4π

3 λ2 ≥ 0, we have that ‖∇u(t)‖L2 isbounded. Thus the solution u is global in time.

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global solution 1

To understand the behavior of the solution at large time, we need

Proposition

Let u0 ∈ Σ and u be the solution of the Cauchy problem of theequation (1) in the proposition 3. Define the momentum quantityby

y(t) =

ˆR3|x |2|u|2.

Then one has:

y(t) = 2ˆ

R3|∇u|2+3λ1

ˆR3|u|4+3λ2

ˆR3

K∗|u|2|u|2−2ˆ

R3|x |2|u|2.

(12)

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global solution 2Remi Carles-Markowich-Sparber obtained very nice results whichare about both finite time blow-up solutions and global solutions.

Theorem

Let d = 3 and assume the initial data u0 such that

3E ≤ |xu0|22.

Then the solution blow up at a time T∗ before π/2.

Proposition

Let d = 3 and Λ := 4π3 λ2 − λ1 > 0, λ2 ≥ 0. Then there exists a

uniform constant C0 > 0 such that for C0 = C0MΛ2 , 0 < E < C0 and

|∇u0|2 < C0, the solution is global one.

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functionals and heights

On the space Σ, we denote the global interaction energy by

P(u) :=λ1

4

ˆR3|u|4 +

λ2

4

ˆR3

K ∗ |u|2|u|2.

Then we define the following two important functionals

S(u) =σ

2M(u) +

12E (u) := S(u, σ),

R(u) = E (u) + P(u).

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functionals and heights 2

We now consider the following constrained variational problem

m = inf{S(u) : u ∈ M} := m(σ), (13)

where M is defined as

M = {u ∈ Σ \ {0}, R(u) = 0}.

Note that m is a convex function of σ > 0.

We remark that this minimization problem has nothing to do withthe ground state of the equation (1).

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M is non-empty

At first, we shall prove that M is non-empty, which is asserted bylemma 8 below.

Lemma

M is non-empty. That is, there exists u ∈ Σ \ {0} satisfyingR(u) = 0.

Though some arguments of our proof is contained in the paper[C-M-S], we give the details for completeness.

Here is it.

Let uε(x) = εα2 v(x1, x2)w(εx3), where v and w are two Schwartz

functions for some small constant ε to be determined.

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M is non-empty 2

By direct computation, we getˆ

R3|∇uε|2 = εα−1

ˆR3|∇v(x1, x2)|2|w(x3)|2

+ εα+1ˆ

R3|v(x1, x2)|2|∇w(x3)|2, (14)

ˆR3|x |2|uε|2 = εα−3

ˆR3

x23 |v(x1, x2)|2|w(x3)|2

+ εα−1ˆ

R3(x2

1 + x22 )|v(x1, x2)|2|w(x3)|2. (15)

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proof

If we define V and W as the Fourier transform of |v |2 and |w |2respectively, we obtain by the Plancherel formula (10):

ˆR3|uε|4 = ε2α−1 1

(2π)3

ˆR3|V |2|W |2, (16)

ˆR3

K ∗ |uε|2|uε|2 = ε2α−1 1(2π)3

ˆR3

3ε2ξ2

3ξ21 + ξ2

2 + ε2ξ23|V |2|W |2

− 4π3

ε2α−1 1(2π)3

ˆR3|V |2|W |2

= o(ε2α−1)− 4π3

ε2α−1, (17)

where in the last equality we have used the Lebesgue DominatedConvergence theorem and lemma (2).

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proof continued

Therefore we have

R(uε) ≈12εα−1 +

12εα−3 +

34(λ1 −

4π3

λ2)ε2α−1 + o(ε2α−1).

Hence we get R(uε) < 0 provided we choose ε small and α < −2(so the leading order term is the third one).

For such uε, we define uµε = µuε. Then we obtain

R(uµε ) ≈ µ2 + (λ1 −

4π3

λ2)µ4.

Hence for µ < 1 small, we have R(uµε ) > 0. It is obvious that

R(µuε) is continuous with respect to the variable µ. Then for someµ0 ∈ (0, 1), R(uµ0

ε ) = 0. This implies that M is non-empty.

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LemmataLemma

S(u) is uniformly bounded from below; that is, for any u ∈ M,S(u) ≥ c > 0 for some uniform constant c > 0.

Here is the proof:

For u ∈ M, we have R(u) = 0 and thenˆR3|∇u|2 +

ˆR3|x |2|u|2

= −32λ1

ˆR3|u|4 − 3

2λ2

ˆR3

K ∗ |u|2|u|2. (18)

By the Holder inequality, we have

|ˆ

R3K ∗ |u|2|u|2| ≤ ‖K ∗ |u|2‖L2(R3)‖|u|2‖L2(R3). (19)

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proof continuedNote that

‖K ∗ |u|2‖L2(R3) ≤ ‖|u|2‖L2(R3) = ‖u‖2L4(R3). (20)

By the Holder inequality, we have

|ˆ

R3K ∗ |u|2|u|2| ≤ ‖K ∗ |u|2‖L2(R3)‖|u|2‖L2(R3). (21)

Note that

‖K ∗ |u|2‖L2(R3) ≤ ‖|u|2‖L2(R3) = ‖u‖2L4(R3). (22)

Combining the inequalities (18) ,(21) and (22), we getˆ

R3|∇u|2 +

ˆR3|x |2|u|2 ≤ C

ˆR3|u|4. (23)

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proof continued 2

Applying the Gagliardo-Nirenberg inequality to the right term leadsto ˆ

R3|∇u|2 +

ˆR3|x |2|u|2 ≤ C‖u‖L2‖∇u‖3

L2 . (24)

By uncertainty principle and Hölder inequality,

ˆR3|∇u|2 +

ˆR3|x |2|u|2 ≤ C

(ˆR3|∇u|2 +

ˆR3|x |2|u|2

)2

, (25)

which insures thatˆR3|∇u|2 +

ˆR3|x |2|u|2 ≥ c > 0 (26)

for some constant c > 0.

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proof continued 3

When R(u) = 0, S(u) is reduced to

S(u) =σ

2

ˆ|u|2 +

112

ˆR3|∇u|2 +

112

ˆR3|x |2|u|2 := S(u). (27)

Hence,S(u) ≥ c > 0 (28)

for some uniform constant c > 0. This completes the proof.

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key step: the height can be attained

The height m can be attained for some u ∈ M, that is,m = min{S(u) : u ∈ M}.

Lemma

There exists at least one u ∈ M for which m = S(u).

The proof is not long. Here is it.

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arguments

We first claim that

m = inf{S(u) : u ∈ Σ \ {0}, R(u) ≤ 0}, (29)

Recall that S(u) = S(u) when R(u) = 0.

If we denote the right hand side of (29) by m, then we have thatm ≤ m.

On the other hand, for u ∈ Σ with R(u) < 0, we let uµ = µu. Then

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argument 1

R(uµ) = µ2 12

ˆR3|∇u|2 + µ2 1

2

ˆR3|x |2|u|2 + µ4 3

4λ1

ˆR3|u|4

+ µ4 34λ2

ˆR3

K ∗ |u|2|u|2, (30)

which can be written as

R(uµ) ≈ µ2 − µ4. (31)

Then R(uµ) takes values from positive to negative as µ varies from0 to 1. Hence for some 0 < µ0 < 1,

R(uµ0) = 0.

From the definition of m, we have

m ≤ S(uµ0) = S(uµ0) < S(u).

We then deduce that m ≤ m.

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argument 2

Now we choose un satisfying

R(un) ≤ 0, S(un) → m (n →∞).

By this, we know that un is bounded in Σ. Then the compactnesslemma (1) implies that up to a subsequence,

un → u (32)

in Lp(R3) for 2 ≤ p < 6.

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argument 3

Hence we have ˆR3|un|4 →

ˆR3|u|4, (33)

ˆR3

K ∗ |un|2|un|2 →ˆ

R3K ∗ |u|2|u|2. (34)

By the lower semi-continuous of the Σ norm, we haveˆ

R3|∇u|2 ≤ lim inf

n→∞

ˆR3|∇un|2, (35)

ˆR3|x |2|u|2 ≤ lim inf

n→∞

ˆR3|x |2|un|2. (36)

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argument 4From (33)(34)(35)(36), we deduce that

R(u) ≤ 0

and

S(u) ≤ lim infn→∞

S(un) = m. (37)

We claim that u 6= 0. Assume we have proved this claim. Byconsidering (29), we get that

S(u) = m; (38)

moreover,R(u) = 0, (39)

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argument 5

since otherwise a similar argument as before shows there exists0 < µ0 < 1 such that R(uµ0) = 0,m ≤ S(uµ0) = S(uµ0) < S(u) = m which leads to a contradiction.Thus u is a minimizer of the variational problem 13.

We now prove the claim. If not, we then have, by (32), un → 0 inLp(R3), 2 ≤ p < 6. Then (23) implies

ˆR3|∇un|2 +

ˆR3|x |2|un|2 → 0, (40)

which contradicts to (26). Thus the proof is complete.

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new invariants

We now introduce two sets

K+ = {u ∈ Σ; S(u) < m, R(u) > 0}, (41)K− = {u ∈ Σ; S(u) < m, R(u) < 0}. (42)

It is not hard to see, by considering S(µu) and R(µu) (for fixedu ∈ Σ) as smooth functions of the variable µ, that the set{u ∈ Σ; S(u) < m} ( and K−) is non-empty.

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new result

Theorem

K+ and K− are two invariant sets of the flow generated by theequation (1). More precisely, for any u0 ∈ K+, if u is the solutionto (1) with the initial data u0, then u(t) ∈ K+ for any t ∈ I , whereI is the maximal existence time interval of the solution u. So is K−.

Proof: By Proposition (3), we have

S(u(t)) = S(u0). (43)

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arguments

Thus from S(u0) < m, it follows that

S(u(t)) < m. (44)

To check u(t) ∈ K+, we need to show that

R(u(t)) > 0. (45)

If this is not true, then by the continuity of u and the factR(u(0)) = R(u0) > 0, we have for some t1

R(u(t1)) = 0. (46)

It follows that u(t1) ∈ M.

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argument done

At the same time, we have

S(u(t1) < m. (47)

This is impossible from the definition of m. Similarly we can provethat K− is invariant under the flow generated by the equation (1).The proof is complete.

With the help of theorem 11, we obtain the following sharp globalexistence and blow up criterion for (1).

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threshold criterion and the proof

Theorem

Assume that λ1 < 4π3 λ2. If u(t) ∈ Σ is the maximal solution to (1)

with the initial data u0 ∈ Σ satisfying S(u0) < m, then we havethat

(i) u is global for the initial data u0 ∈ K+;

(ii) u blows up at finite time for the initial data u0 ∈ K−.

(i). If u0 ∈ K+, then by Theorem 11, we conclude

u(t) ∈ K+.

That is,R(u(t)) > 0,

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proof (i)which implies that

λ2

4

ˆR3

K ∗ |u|2|u|2 > −16

ˆR3|∇u|2

− 16

ˆ|x |2|u|2 − 1

4λ1

ˆR3|u|4. (48)

By Theorem 11, we have

S(u(t)) < m. (49)

From (48) and (49), we conclude that

σ

2

ˆ|u|2 +

112

ˆR3|∇u|2 +

112

ˆR3|x |2|u|2 < m. (50)

This, by Proposition (3), implies the global result of the solutionflow u(t).

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proof (ii)

(ii). If u0 ∈ K−, then by Theorem 11, we conclude

u(t) ∈ K−,

that isR(u(t)) < 0.

Then for any fixed t, there exists 0 < µ < 1 such that

R(µu(t)) = 0, (51)

which implies via the definition of m that

S(µu) ≥ m. (52)

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proof continued

We compute by the inequality (51) that

S(u)− S(µu) =12R(u) +

σ

2(1− µ2)

ˆR3|u|2. (53)

From (53) we know that

S(u)− S(µu) ≥ 12R(u)

By (52), we conclude

S(u)−m ≥ 12R(u).

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proof being completeBy the viral identity (proposition (5)), we have

d2

dt2

ˆR3|x |2|u|2 = 4R(u)− 4

ˆR3|x |2|u|2. (54)

Then we have that

d2

dt2

ˆR3|x |2|u|2 < 4R(u). (55)

Then,

d2

dt2

ˆR3|x |2|u|2 < 8(S(u)−m) (56)

= 8(S(u0)−m) < 0. (57)

This implies the blow-up of the solution flow u(t) at some finitetime. This completes the proof of Theorem 12.

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more to say

In a joint work with Cao Pei [M-C], we can introduce other classesof threshold criterion for the Gross-Pitaevskii equation with trappeddipolar quantum gases. One interesting question is can we findmore functionals to get more invariant sets of the flow?

Note that E ′(u) · u = 2E (u) + 4P(u) and P ′(u) · u = 4P(u).Hence, when R(u) = 0, we have E = −P and then E ′(u) · u = 2Pand R(u) = E (u) + 1

2E ′(u) · u.

The geometric pictures of the functionals above and the associatedinvariants are not clear at moment. It is interesting to study themfurther.

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thanks!

I want to express my deep appreciation to Prof. Luis Vega for theinvitation.

Thank you all very much for your attending to my talk.

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