Singular solutions of the subcritical nonlinear Schrodinger equation

0.16 MB·English

by  Gadi Fibich

Singular solutions of the subcritical nonlinear Schro¨dinger equation
Gadi Fibich
School of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel
1 Abstract
1
0 We show that the subcritical d-dimensional nonlinear Schro¨dinger equation iψ +∆ψ+ ψ 2σψ = 0,
t
2 | |
where 1 < σd < 2, admits smooth solutions that become singular in Lp for p∗ < p , where

n p∗ := σd . Sincelim p∗ = 2, thesesolutions cancollapse atany 2< p , and≤inp∞articular
a σd−1 σd→2− ≤ ∞
J for p = 2σ+2.

0
3

1. Introduction
]
P
A The focusing nonlinear Schro¨dinger equation (NLS)
.
h iψ (t,x)+∆ψ+ ψ 2σψ = 0, ψ(0,x) = ψ (x), (1)
t t | | 0
a
m where x = (x ,...,x ) Rd and ∆ = d ∂ is the Laplacian, has been the subject of
1 d ∈ j=1 xjxj
[ intense study, due to its role in various areas of physics, such as nonlinear optics and Bose-Einstein
P
2 Condensates (BEC). The NLS is called subcritical, critical, and supercritical if σd < 2, σd = 2,
v and σd > 2, respectively. It is well-known that in the critical and supercritical cases, the NLS (1)
7
possesses solutions that become singular in a finite time [1]. In this study we show that, contrary
2
8 to common belief, the subcritical NLS also admits solutions that become singular in a finite time.
1 Most of NLS theory has been developed for solutions that are in H1(Rd). In this case, the
.
4 initial condition ψ H1, and the NLS solution is said to become singular at t = T , if ψ(t) H1
0 c
0 ∈ ∈
for 0 t < T , and lim ψ(t) = . In 1983, Weinstein proved that all H1 solutions of the
0 ≤ c t→Tc|| ||H1 ∞
1 subcritical NLS exist globally:
:
v
Theorem 1 ([2]). Let ψ be a solution of the NLS (1), let 0 < σd < 2, and let ψ H1. Then,
i 0
X ψ exists globally in H1. ∈
r
a
Until now, this result has been interpreted as implying that the subcritical NLS does not admit
singular solutions. In this study we show that if we do not restrict ourselves to H1 solutions, then
the subcritical NLS also admits singular solutions. Here, by singular we mean that there exists
some 2< p < , such that ψ becomes infinite in a finite time.1 Our main result is as follows:
p
∞ || ||
Theorem 2. Let
σd
p∗ < p , p∗ := . (2)
≤ ∞ σd 1
−
Email address: [email protected] (Gadi Fibich)
1InthecaseofH1solutionsoftheNLS,blowupoftheH1normimpliesblowupoftheLpnormsfor2σ+2≤p≤∞,
see Appendix A.
Preprint submitted to Elsevier February 1, 2011
Then, the subcritical NLS with 1 < σd < 2 admits classical solutions that becomes singular at a
finite time T in Lp, i.e.,
c
ψ(t) < , 0 t < T ,
p c
|| || ∞ ≤
and
lim ψ(t) = .
p
t→Tc|| || ∞
Theorem 2 follows from the following Theorem:
Theorem 3. Let p be in the range (2), let 1 < σd < 2, let a > 0 be a positive constant, and let
Q(ρ) be the solution of
1
∆Q(ρ) Q+ia Q+ρQ′ + Q 2σQ = 0, 0< ρ < , (3a)
− σ | | ∞
(cid:18) (cid:19)
subject to
0= Q(0) C, Q′(0) = 0. (3b)
6 ∈
Let
1
ψexplicit(t,r) = Q(ρ)eiτ(t), (4a)
Q L1/σ(t)
where
r = x, L(t) = 2a(T t), (4b)
c
| | −
and p
r t 1 1 T
c
ρ= , τ = ds = log . (4c)
L(t) L2(s) 2a T t
Z0 c−
Then, ψexplicit is an explicit solution of the subcritical NLS that becomes singular in Lp as t T .
Q −→ c
Remark. Although Q, hence also ψexplicit, is not in H1, it is smooth, and it decays to zero as
Q
x , see Lemma 2.
| | −→ ∞
Since
lim p∗ = 2+,
σd→2−
then for any 2< p < , there exists a singular solution of a subcritical NLS that becomes singular
in Lp. In particular,∞if σd is sufficiently close to 2 from below, then ψexplicit becomes singular in
Q
L2σ+2.
Remark. The linear Schro¨dinger equation iψ + ∆ψ = 0 admits the fundamental solution
t
ψ = 1 ei|x|2/4t, which becomes singular in finite time in L∞ [3]. Unlike ψexplicit, however, this
(4πit)d/2 Q
solution does not become singular in Lp for any finite p.
2. Proof of Theorem 3
We begin with the following result.
Lemma 1. let ψexplicit be defined as in Theorem 3. Then, ψexplicit is an explicit solution of the
Q Q
NLS (1).
2
Proof. Substituting ψexplicit in the NLS (1) and carrying out the differentiation proves the
Q
result. (cid:3)
The result of Lemma 1 was used by Zakharov [4], and subsequently by others (see [1] and the
references therein), in the study of singular H1 solutions of the supercritical NLS. These solutions
undergo a quasi self-similar collapse, in which ψexplicit is the asymptotic blowup profile of the
Q
explicit
collapsing core of the solution. Here, in contrast, ψ is an explicit, “truly” self-similar solution
Q
of the subcritical NLS.
We now establish the decay at infinity of all solutions of equation (3):
Lemma 2. Let a > 0 and let 1 < σd < 2. Then, for any Q(0) C, the solution of equation (3)
∈
exists, is unique, and decays to zero as ρ , so that
−→ ∞
Q = O(ρ−d+1/σ), ρ .
| | −→ ∞
Therefore, Q is in Lp for any p∗ <p .
≤ ∞
Proof. Theproofisnearlyidentical totheproofofJohnsonandPaninthesupercriticalcase[5],
see Appendix B. (cid:3)
Lemma 3. For any p∗ < p , ψexplicit becomes singular in Lp as t T .
≤ ∞ Q −→ c
Proof. Since
Q
explicit p
ψ (t) = || || ,
|| Q ||p L1/σ(t)
and Q < , the result follows. (cid:3)
p
|| || ∞
This concludes the proof of Theorem 3.
3. The Q equation in the subcritical case
As in [6], the far-field asymptotics of Q can be calculated using the WKB method:
Lemma 4. Let Q(ρ) be a solution of equation (3), where 1 < σd < 2. Then,
Q(ρ) c Q (ρ)+c Q (ρ), ρ , (5)
1 1 2 2
∼ −→ ∞
where c and c are complex numbers, and
1 2
Q ρ−i/a−1/σ, Q e−iaρ2/2ρi/a−d+1/σ.
1 2
∼ ∼
Proof. See Appendix C. (cid:3)
Corollary 1. If 1 <σd < 2, then
Q L2(Rd), Q L2(Rd),
1 1
∈ ∇ ∈
3
and
Q L2(Rd), Q L2(Rd).
2 2
6∈ ∇ 6∈
In addition, Q Lp(Rd) for any 2 p , and
1
∈ ≤ ≤ ∞
σd
Q Lp(Rd), < p .
2
∈ σd 1 ≤ ∞
−
Proof. This follows from Lemma 4. (cid:3)
Inthesupercriticalcase,akeyroleisplayedbythezero-Hamiltonian solutionsoftheQequation,
whichbehave asc Q atlarge ρ[1]. We nowshow thattherearenosuch solutions inthesubcritical
1 1
case:
Lemma 5. When 1 < σd < 2, there are no nontrivial solutions of the Q equation (3), such that
c = 0, i.e., that
2
Q(ρ) c Q (ρ), ρ .
1 1
∼ −→ ∞
Proof. By negation. Assume that there is such a Q. In this case, it follows from Corollary 1
that Q H1. Hence, ψexplicit is a solution of the subcritical NLS that becomes singular in H1,
∈ Q
which is in contradiction with Theorem 1. (cid:3)
4. Simulations
1.5 2
1.5
1
|Q| |Q|
1
0.5
0.5
0 0
0 20 40 0 20 40
ρ ρ
Figure 1: Solutions of equation (3) with d = 1, σ = 1.9, a = 0.5145, and with A: Q(0) = 1.2953,
and B: Q(0) = 3.
Figure 1 shows two numerical solutions of equation (3). As expected, see Lemma 4,
Q c ρ−i/a−1/σ +c e−iaρ2/2ρi/a−d+1/σ
1 2
| | ∼
(cid:12) (cid:12)
decreases to zero as ρ , while(cid:12)undergoing faster and faster oscilla(cid:12)tions. The “cleaner picture”
−→ ∞
in Figure 1A has to do with the fact that the values of a and Q(0) were chosen so as to minimize
the value of c .23
2
2These values of a and Q(0) were calculated using the shooting algorithm of Budd, Chen and Russel [7] for
calculating the zero-Hamiltonian solutions in thesupercritical case.
3The value of c cannot be equalto zero, see Lemma 5.
2
4
L=1 L=0.7
4 4
(a) (b)
3 3
|ψ| |ψ|
2 2
1 1
0 0
0 2 4 6 8 0 2 4 6 8
x x
L=0.4 L=0.2
4 4
(c) (d)
3 3
|ψ| |ψ|
2 2
1 1
0 0
0 2 4 6 8 0 2 4 6 8
x x
Figure2: NumericalsolutionofthesubcriticalNLSwiththeinitialconditionequationψ (x) =Q(x)
0
explicit
(solid). Dotted line is the analytic solution ψ . a) t = 0, L = 1; b) t = 0.4956, L = 0.7;
Q
c) t = 0.8163, L = 0.4; d) t = 0.9329, L = 0.2.
In Figure 2 we solve numerically the subcritical NLS with d = 1, σ = 1.9, and the initial
condition ψ (x) = Q(x), where Q is taken from Figure 1A. By Lemma 1, the analytic solution of
0
this equation is given by ψexplicit with L = √1 2at. As expected, the numerical solution agrees
Q −
with the analytic solution, thus providing the first ever simulation of a singular solution of the
subcritical NLS.
In thesesimulations, we usedastandard fourth-orderfinite-differenceimplicit scheme with dx =
0.05 and dt = 0.001 over the spatial domain 70 x 70. Nevertheless, becauseof the slow decay
− ≤ ≤
and the ever faster oscillations as x , the agreement between the analytic and numerical
−→ ∞
solutions breaks down after focusing by less than 3 (see Figure 2d). We could, of course, take an
even larger domain with a finer mesh. In that case, the numerical solution would simply bifurcate
from the analytic one at a higher focusing level. The point of this simulation, however, is not to
establish numerically the existence of a singular subcritical solution (which we prove rigorously),
but rather to illustrate the numerical difficulties in computing this solution, by showing that even
with a relatively large domain and a fine grid, the numerical solution breaks down after focusing
by less than 3. This suggests that numerical simulations may be useless in studying the stability
of these solutions.
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5. Final remarks
Until now, it was believed that only the critical and supercritical NLS admits singular solutions.
In this study we showed that if we do not limit ourselves to H1 solutions, then the subcritical NLS
also admits solutions that become singular at a finite time. This finding raises several questions,
which are currently open. One question is whether the explicit singular solutions are stable. As
note, this question is hard to study numerically, because of the slow decay, coupled with the ever
faster oscillations, of the solution at infinity. Another open question is whether the subcritical NLS
admits singular solutions that are not self-similar. The answers to these questions will determine
whether singularity formation in the subcritical NLS will remain as an anecdote, or lead to a new
line of research.
Acknowledgments.
This research was partially supported by the Israel Science Foundation (ISF grant No. 123/08).
We would like to thank Moran Klein for the NLS simulations of Figure 2.
References
[1] C. Sulem and P.L. Sulem. The Nonlinear Schro¨dinger Equation. Springer, New-York, 1999.
[2] M.I. Weinstein. Nonlinear Schro¨dinger equations and sharp interpolation estimates. Comm.
Math. Phys., 87:567–576, 1983.
[3] R. Cordero-Soto, R. M. Lopez, E. Suaz, and S. K Suslov . Propagator of a charged particle
with aspinin uniformmagnetic and perpendicularelectric fields. Lett. Math. Phys, 84:159–178,
2008.
[4] V.E. Zakharov. Handbook of plasma physics. volume 2. Elsevier, New York, 1984.
[5] R. Johnson and X. Pan. On an elliptic equation related to the blow-up phenomenon in the
nonlinear Schro¨dinger equation. Proc. Roy. Soc. Edinburgh Sect. A, 123:763–782, 1993.
[6] B.J. LeMesurier, G. Papanicolaou, C. Sulem, and P.L. Sulem. Focusing and multi-focusing
solutions of the nonlinear Schro¨dinger equation. Physica D, 31:78–102, 1988.
[7] C. J. Budd, S. Chen, and R.D. Russel. New self-similar solutions of the nonlinear Schro¨dinger
equation with moving mesh computations. J. Comput. Phys., 152:756–789, 1999.
Appendix A. Blowup of ||ψ||
p
The NLS conserves the power (mass, L2 norm) and the Hamiltonian, i.e.,
1
ψ 2 ψ 2, H(t) := ψ 2 ψ 2σ+2 H(0).
|| ||2 ≡|| 0||2 ||∇ ||2 − σ+1|| ||2σ+2 ≡
Therefore, when ψ becomes infinite, then so does ψ , hence ψ . Therefore, since
H1 2 2σ+2
|| || ||∇ || || ||
ψ is conserved, it follows from theinterpolation inequality for Lp normsthat ψ also becomes
2 p
|| || || ||
infinite for 2σ+2 p .
≤ ≤ ∞
6
Appendix B. Proof of Lemma 2
As in the proof of Johnson and Pan in the supercritical case [5], let
Q(ρ) = u(ρ)e−iaρ2/4, u(ρ) = u (ρ)+iu (ρ),
1 2
where u and u are real functions, let
1 2
v (ρ) = ρ(d−1)/2u (ρ), j = 1,2,
j j
let
dx
t = ρ2, x (t)= v (ρ), y (t) = j,
j j j
dt
let
df
f (t) = t1/4x (t), g (t) = j,
j j j
dt
and let
1 1 1 e 1
H(t)= (g2+g2)+ λ (f2+f2)+ t−β(f2+f2)σ+1,
2 1 2 8 − t − t2 1 2 4(2σ+2) 1 2
(cid:18) (cid:19)
where
a2 σd 1
λ = , β = 1+ , e = d(d 4).
4 2 4 −
Then,
B 1 2e β
H′(t) = (f g f g )+ 1+ (f2+f2) t−β−1(f2+f2)σ+1,
4t 1 2− 2 1 8t2 t 1 2 − 4(2σ +2) 1 2
(cid:18) (cid:19)
where
d 1
B = a < 0.
2 − σ
(cid:18) (cid:19)
In addition, from the Cauchy-Schwartz inequality [5],
1 √λ 2 1 √λ 2
f g f2+ g2 , f g f2+ g2 .
| 1 2|≤ 2 " 2 1 √λ 2# | 2 1| ≤ 2" 2 2 √λ 1#
Since β > 0 and B <0,
B 1 2e
H′(t) | |(f g + f g )+ 1+ (f2+f2).
≤ 4t | 1 2| | 2 1| 8t2 t 1 2
(cid:18) (cid:19)
B √λ 2 1 2e
| | (f2+f2)+ (g2+g2) + 1+ (f2+f2)
≤ 8t 2 1 2 √λ 1 2 ! 8t2 t 1 2
(cid:18) (cid:19)
B λ 1 1 2e
= | | (f2+f2)+ (g2 +g2) + 1+ (f2+f2)
2√λt 8 1 2 2 1 2 8t2 t 1 2
(cid:18) (cid:19) (cid:18) (cid:19)
B 1
| | 1+O H(t).
≤ 2√λt t
(cid:18) (cid:18) (cid:19)(cid:19)
7
Since H(t) > 0 for large t,
H′ B 1
| | +O .
H ≤ 2√λt t2
(cid:18) (cid:19)
Therefore, as in [5], there exists a constant c> 0, such that
H(t) c(1+t2|α|), 0 t < ,
≤ ≤ ∞
where
B 1 d
2α = | | = .
| | 2√λ σ − 2
Hence,
f (t) c(1+t|α|),
j
| | ≤
x (t) ct−1/4(1+t|α|),
j
| | ≤
v (ρ) cρ−1/2(1+ρ2|α|),
j
| | ≤
and
u (ρ) cρ−d/2(1+ρ2|α|).
j
| | ≤
Therefore,
u = O(ρ−d+1/σ), ρ .
| | −→ ∞
Remark. The only difference from the original proof of Johnson and Pan is that in in the
supercritical case B > 0. Therefore, we take the absolute value of B, instead of B, in the bounds
for H′.
Appendix C. Proof of Lemma 4
Let
Q(ρ) = e−12R(cid:16)d−ρ1+iaρ(cid:17)Z(ρ)= e−iaρ2/4ρ−(d−1)/2Z(ρ).
Therefore, the equation for Z is given by
a2 dσ 2 (d 1)(d 3)
Z′′(ρ)+ ρ2 1 ia − − − + Q 2σ Z = 0. (C.1)
4 − − 2σ − 4ρ2 | |
(cid:18) (cid:19)
Since by Lemma 2, lim Q = 0, let us look for an asymptotic solution of the form
ρ→∞
Z = ew(ρ), w(ρ) w (ρ)+w (ρ)+ .
0 1
∼ ···
The equation for w (t) is given by
i
{ }
a2 dσ 2 (d 1)(d 3)
(w′′+w′′+ )+(w′ +w′ + )2+ ρ2 1 ia − − − + Q 2σ = 0. (C.2)
0 1 ··· 0 1 ··· 4 − − 2σ − 4ρ2 | |
(cid:18) (cid:19)
A-priori, the equation for the leading-order terms is
a2
w′′+(w′)2+ ρ2 = 0.
0 0 4
8
The substitution w = cρn shows that the order of the terms in this equation is ρn−2, ρ2n−2, and
0
ρ2, respectively. Since the only consistent way to balance the leading-order terms is if n = 2, the
equation for the leading-order terms is given by
a2
(w′)2+ ρ2 = 0.
0 4
Therefore,
ia ia
w′ = ρ, w = ρ2.
0 ± 2 0 ± 4
The balance of the next-order terms is given by
dσ 2
w′′+2w′w′ 1 ia − = 0.
0 0 1− − 2σ
Substituting w′ = iaρ/2 and rearranging gives,
0 ±
i dσ 21 1 i dσ 2 1
w′ = − , w = − logρ.
1 ∓aρ ± 2σ ρ − 2ρ 1 ∓a ± 2σ − 2
(cid:18) (cid:19)
We will now show that w = o(1). Therefore, we obtained the two solutions
2
ρ2 i 1 d 1
w(1)(ρ) = ia + − logρ+o(1),
4 −a − 2 − σ
(cid:18) (cid:19)
ρ2 i 1 d 1
w(2)(ρ) = ia + + − − + logρ+o(1).
− 4 a 2 σ
(cid:18) (cid:19)
Substituting Q (ρ) = e−iaρ2/4ρ−(d−1)/2ew(i)(ρ) leads to the result.
i
In order to confirm that w = o(1), we note that the equation for w is given by
2 2
(d 1)(d 3)
w′′ +(w′)2+2w′w′ − − + Q 2σ = 0. (C.3)
1 1 0 2− 4ρ2 | |
In the case of Q , since Q 2σ ρ−2, substituting the expressions for w and w gives
1 1 0 1
| | ∼
1 1
w′ = O , w = O .
2 ρ3 2 ρ2
(cid:18) (cid:19) (cid:18) (cid:19)
In the case of Q , since Q 2σ ρ−2σd+2 ρ−2, the leading-order equation for w becomes
2 2 2
| | ∼ ≫
2w′w′ + Q 2σ = 0.
0 2 | 2|
Since w′ ρ, then w′ ρ−2σd+1 and w ρ−2σd+2 = o(1).
0 ∼ 2 ∼ 2 ∼
Finally, we note that this this proof is rigorous, since solutions of linear ODEs always have their
asymptotics obtained by WKB calculations, and the ODE (C.1) for Z is “linear”, since it can be
written as
a2 dσ 2 (d 1)(d 3)
Z′′(ρ)+ ρ2 1 ia − − − +O(ρ−d+1/σ) Z = 0.
4 − − 2σ − 4ρ2
(cid:18) (cid:19)
9