Download Nondegeneracy of the Lump Solution to the KP-I Equation

Nondegeneracy of the Lump Solution to the KP-I
Equation
Yong Liu
School of Mathematics and Physics,
North China Electric Power University, Beijing, China,
Email: [email protected]
Juncheng Wei
Department of Mathematics,
University of British Columbia, Vancouver, B.C., Canada, V6T 1Z2
Email: [email protected]
March 28, 2017
1
Introduction
The KP equation, introduced by Kadomtsev and Petaviashvili [18], is a classical
nonlinear dispersive equation appearing in many physical contexts, including the
motion of shallow water waves. It has the form
(
( ))
∂x ∂t u + ∂x3 u + 3∂x u2 − σ∂y2 u = 0,
(1)
where σ = ±1. In this paper, we will consider the case of σ = 1, which is
called KP-I equation. (When σ = −1, it is called KP-II equation). If u is
y independent, then equation (1) reduces to the KdV equation, a well known
integrable system. Dryuma [10] proved that the KP equation has a Lax pair.
The inverse scattering transform for KP-I equation has been carried out in
[11, 21, 31]. (See [1, 2] and the reference therein for more discussions on the
development of this topic before 1991.)
The soliton solutions of the KP equation can be obtained by various different
methods, both for the KP-I and KP-II equations. An important feature of the
KP-I equation is that it has a family of lump solutions which are also travelling
wave type. They were found in [22, 28]. Explicitly, the KP-I equation has a
lump solution of the form Q (x − t, y) , where
Q (x, y) = 4
y 2 − x2 + 3
(x2 + y 2 + 3)
1
2.
(2)
(
)
Note that the function Q decays like O r−2 at infinity. (Here r2 = x2 + y 2 .)
Because of this slow decaying property, the inverse scattering transform of KPI is quite delicate. Indeed, it is shown in [29] that a winding number could
be associated to lump type solutions. The interaction of multi-lump solutions
has been studied in [12, 20]. The lump solutions also appear in other physical
models. For instance, formal asymptotic analysis shows that in the context
of the motion in a Bose condensate, the transonic limit of the traveling wave
solutions to the GP equation approach to the lump like solutions ([3, 17]). This
is called Roberts’s Program ([16, 17]). A rigorous verification is given recently
in [4] and [6]. (For traveling waves of Gross-Pitaevskii equation, we refer to [5]
and [23] and the references therein.)
Regarding to traveling wave type solutions, it is worth mentioning that the
generalized KP-I equations
(
)
∂x2 ∂x2 u − u + up − ∂y2 u = 0
(3)
also have lump type solutions for suitable p ([7, 19]). For general p, this is not
an integrable system. Hence no explicit formula is available. These lump type
solutions are obtained via variational methods (concentration compactness). We
point out that the uniqueness of these lump type solutions is an open problem
and therefore it is not known whether Q has this variational characterization.
The stability(instability) properties of these solutions have been studied in ([9,
30]). Formal computation in [30] suggests that the lump solutions Q should be
stable.
Towards understanding the spectral and stability property of the lump solution, we shall prove in this paper that Q is nondegenerate (and hence locally
unique up to translations). Our main result in this paper is
Theorem 1 Suppose ϕ is a smooth solution to the equation
(
)
∂x2 ∂x2 ϕ − ϕ + 6Qϕ − ∂y2 ϕ = 0.
(4)
Assume
ϕ (x, y) → 0, as x2 + y 2 → +∞.
Then ϕ = c1 ∂x Q + c2 ∂y Q, for certain constants c1 , c2 .
Theorem 1 has long been conjectured to be true. See the remark after
Lemma 7 in [30] concerning the spectral property and its relation to stability.
We also expect that the linearized operator around Q has exactly one negative
eigenvalue. But we have not been able to prove this. The rigidity result, stated
in Theorem 1, may be very useful in the construction and analysis of solutions
near the KP-I equation.
Let us briefly describe the main ideas of the proof. Our main tool will be the
Bäcklund transformation. It is known that the lump solution can be obtained
from the trivial solution by performing Bäcklund transformation twice. We
consider these transformation in the linearized level. We show that a kernel
of the linearized operator around Q could be transformed to a kernel of the
2
operator ∂x2 +∂y2 −∂x4 , which is the linearized operator around the trivial solution
zero. With some information on the growth rate of the kernel function, we
are able to conclude that the only decaying solutions to (4) are corresponding
to translations in the x and y axes. The idea of using linearized Bäcklund
transformation to investigate the spectral property has already been used in
[25] in the case of Toda lattice, see also [24] for discussion in the case of Nsolitons for the KdV equation.
The organization of the paper is as follows: In Section 2, we recall some
basic facts about the bilinear derivative. In Section 3 and Section 4, we analyze
the linearized Bäcklund transformation. In Section 5, we prove Theorem 1.
Acknowledgement. The research of J. Wei is partially supported by NSERC
of Canada. Y. Liu is partially supported by the Fundamental Research Funds
for the Central Universities 13MS39.
2
Preliminaries
We will use D to denote the bilinear derivative. Explicitly,
Dxm Dtn f · g = ∂ym ∂sn f (x + y, t + s) g (x − y, t − s) |s=0,y=0.
(5)
In particular, it holds
Dx f · g = ∂x f g − f ∂x g,
Dx2 f · g = ∂x2 f g − 2∂x f ∂x g + f ∂x2 g,
Dx Dt f · g = ∂x ∂t f g − ∂x f ∂t g − ∂t f ∂x g + f ∂x ∂t g.
The KP-I equation (1) can be written in the following bilinear form [13]:
(
)
Dx Dt + Dx4 − Dy2 τ · τ = 0.
(6)
To explain this, we introduce the τ -function:
u = 2∂x2 (ln τ ) .
(7)
Then equation (1) becomes
[(
]
(
)2 )
∂x2 ∂x ∂t ln τ + ∂x4 ln τ + 6 ∂x2 ln τ
− ∂y2 ln τ = 0.
Using the identities (see Section 1.7.2 of [13])
2∂x2 ln τ =
and
2∂x4
Dx Dt τ · τ
Dx2 τ · τ
, 2∂x ∂t ln τ =
,
2
τ
τ2
D4 τ · τ
ln τ = x 2 − 3
τ
3
(
Dx2 τ · τ
τ2
)2
,
we get
(
∂x2
Dx Dt τ · τ + Dx4 τ · τ − Dy2 τ · τ
τ2
)
= 0.
Therefore, if τ satisfies the bilinear equation (6) , then u satisfies the KP-I
equation (1) .
Let i be the imaginary unit. We will investigate properties of the following
three functions:
τ0 (x, y) = 1,
τ1 (x, y) = x + iy +
√
3,
τ2 (x, y) = x2 + y 2 + 3.
Let
τ̃0 (x, y, t) = τ0 (x − t, y) ,
τ̃1 (x, y, t) = τ1 (x − t, y) ,
τ̃2 (x, y, t) = τ2 (x − t, y) .
Then τ̃0 , τ̃1 , τ̃2 are solutions of (6) .
2.1
The Bäcklund transformation
Under (7) , the function τ̃2 is corresponding to the lump solution Q (x − t, y) .
Note that the solution corresponding to τ̃1 is not real valued.
Now let µ be a constant. We first recall the following bilinear operator
identity (see [26] or P. 90 in [27] for more details, we also refer to [15] and the
references therein for the construction of more general rational solutions):
)
]
)
]
1 [(
1 [(
Dx Dt + Dx4 − Dy2 f · f gg −
Dx Dt + Dx4 − Dy2 g · g f f
2
2
[(
)
]
√
√
= Dx Dt − 3iµDy + Dx3 − 3iDx Dy f · g · (f g)
[(
)
]
1
+ 3Dx Dx2 + µDx + √ iDy f · g · (Dx g · f )
3
[(
)
]
√
1
+ 3iDy
Dx2 + µDx + √ iDy f · g · (f g) .
3
(8)
By this identity, we can consider the Bäcklund transformation from τ̃0 to τ̃1 (µ =
√1 ):
3
)
{ (
Dx2 + √13 Dx + √13 iDy τ̃0 · τ̃1 = 0,
(9)
√
(
)
Dt − iDy + Dx3 − 3iDx Dy τ̃0 · τ̃1 = 0.
(
)
The Bäcklund transformation from τ̃1 to τ̃2 is given by µ = − √13
)
{ (
Dx2 − √13 Dx + √13 iDy τ̃1 · τ̃2 = 0,
(10)
√
)
(
Dt + iDy + Dx3 − 3iDx Dy τ̃1 · τ̃2 = 0.
4
√
Throughout the paper, we set r = x2 + y 2 . We would like to relate the kernel
of the linearized KP-I equation to that of the linearized equation in bilinear
form.
Lemma 2 Suppose ϕ is a smooth function satisfying the linearized KP-I equation
(
)
∂x2 ∂x2 ϕ − ϕ + 6Qϕ − ∂y2 ϕ = 0.
Assume
ϕ (x, y) → 0, as r → +∞.
Let
∫
x
∫
η (x, y) = τ2
ϕ (s, y) dsdt.
−∞
0
Then η satisfies
t
(
)
−Dx2 + Dx4 − Dy2 η · τ2 = 0.
Moreover,
)
(
5
|η| + |∂x η| + |∂y η| + ∂x2 η + |∂x ∂y η| + ∂x3 η (1 + r) ≤ C (1 + r) 2 .
(11)
Proof. We write the equation
(
)
∂x2 ∂x2 ϕ − ϕ + 6Qϕ − ∂y2 ϕ = 0
as
∂x4 ϕ − ∂x2 ϕ − ∂y2 ϕ = −6∂x2 (Qϕ) .
Since ϕ → 0 as r → +∞, it follows from the a priori estimate of the operator
∂x2 − ∂x4 + ∂y2 (see Lemma 3.6 of [8]) that
|ϕ| + (|∂x ϕ| + |∂y ϕ|) (1 + r) ≤ C (1 + r)
Moreover,
∫ +∞
−∞
−2
.
ϕ (x, y) dx = 0, for all y. Therefore,
)
(
5
|η| + |∂x η| + |∂y η| + ∂x2 η + |∂x ∂y η| + ∂x3 η (1 + r) ≤ C (1 + r) 2 .
As a consequence,
−Dx2 η · τ2 + Dx4 η · τ2 − Dy2 η · τ2
→ 0, as r → +∞.
τ22
(12)
On the other hand, since ϕ satisfies the linearized KP-I equation, η satisfies
(
)
2
4
2
−D
η
·
τ
+
D
η
·
τ
−
D
η
·
τ
2
2
2
x
x
y
∂x2
= 0.
τ22
This together with (12) implies
−Dx2 η · τ2 + Dx4 η · τ2 − Dy2 η · τ2 = 0.
The proof is completed.
5
3
Linearized Bäcklund transformation between
τ0 and τ1
In terms of τ0 and τ1 , the Bäcklund transformation (9) can be written as
)
{ (
Dx2 + √13 Dx + √13 iDy τ0 · τ1 = 0,
(13)
√
(
)
−Dx − iDy + Dx3 − 3iDx Dy τ0 · τ1 = 0.
Linearizing this system at (τ0 , τ1 ), we obtain
{
L1 ϕ = G1 η,
M1 ϕ = N1 η.
(14)
Here for notational simplicity, we have defined
(
)
1
1
2
L1 ϕ = Dx + √ Dx + √ iDy ϕ · τ1 ,
3
3
(
)
√
3
M1 ϕ = −Dx − iDy + Dx − 3iDx Dy ϕ · τ1 ,
and
(
)
1
1
G1 η = − Dx2 + √ Dx + √ iDy τ0 · η,
3
3
(
)
√
3
N1 η = − −Dx − iDy + Dx − 3iDx Dy τ0 · η.
Proposition 3 Let η be a solution of the linearized bilinear KP-I equation at
τ1 :
−Dx2 η · τ1 + Dx4 η · τ1 − Dy2 η · τ1 = 0.
(15)
Suppose η satisfies (11) . Then the system (14) has a solution ϕ with
5
|ϕ| + |∂x ϕ| + |∂y ϕ| ≤ C (1 + r) 2 .
The rest of this section will be devoted to the proof of Proposition 3.
From the first equation in (14), we get
[
√ (
)] √
(16)
∂y ϕτ1 = i ∂x ϕτ1 + 3 ∂x2 ϕτ1 − 2∂x ϕ − 3iG1 η.
Inserting (16) into the right hand side of the second equation of (14), we get
(
)
( √
)
√
12
4∂x3 ϕτ1 + 2 3τ1 − 12 ∂x2 ϕ + −4 3 +
∂x ϕ = F 1 .
(17)
τ1
Here
√
6
3G1 η + N1 η − G1 η
τ1
√
6
= −2∂x3 η + 2 3i∂x ∂y η − G1 η.
τ1
F1 = 3∂x (G1 η) +
6
To solve the equation (17), we shall first analyze the solutions of the homogeneous equation
(√
)
(
√ )
2τ12 g ′′ +
3τ1 − 6 τ1 g ′ + 6 − 2 3τ1 g = 0.
(18)
Lemma 4 The equation (18) has two linearly independent solutions
√
3 2
τ ,
g1 = τ1 −
2 1
√
g2 = τ1 e−
3
2 τ1
Proof. This can be verified directly. Indeed, the equation (18) can be exactly
solved using software such as Mathematica. We will frequently use this software
in the rest of paper.
Let W be the Wronskian of g1 , g2 . That is,
W = g1 ∂x g2 − g2 ∂x g1 =
3 3 − √3 τ1
τ e 2 .
4 1
By the variation of parameter formula, the equation
(
)
( √
)
√
12
4τ1 g ′′ + 2 3τ1 − 12 g ′ +
− 4 3 g = F1
τ1
has a solution of the form
g ∗ (x, y) = g2 (x, y)
∫
x
−∞
g1 F1
ds − g1 (x, y)
4τ1 W
∫
x
−∞
g2 F1
ds.
4τ1 W
It follows that for each fixed y, the equation
(
)
( √
)
√
12
4τ1 ϕ′′′ + 2 3τ1 − 12 ϕ′′ +
− 4 3 ϕ′ = F
τ1
has a solution of the form
∫
x
w0 (x, y) =
g ∗ (s, y) ds.
(19)
0
( √ )
We emphasize that τ11 has a singularity at the point (x, y) = − 3, 0 . Therefore we need to be very careful about the behavior of w0 around this singular
point.
Lemma 5 Suppose η satisfies (11) . Then
5
|w0 | ≤ C (1 + r) 2 , for x ≤ 10.
7
Proof. Since η satisfies (11) , we have
3
|F1 | ≤ (1 + r) 2 .
Hence using the asymptotic behavior of W and g1 , g2 , we find that for r large,
√
g1 F1 1
≤ Ce− 23 x (1 + r)− 2 ,
W τ1 g2 F1 − 32
W τ1 ≤ C (1 + r) .
( √ )
Near the singular point − 3, 0 , using the fact that
( )
g2 − g1 = O τ13 ,
we infer
|g ∗ | ≤ C.
As a consequence, for x ≤ 10, we obtain
3
|g ∗ | ≤ C (1 + r) 2 .
It follows that
5
|w0 | ≤ C (1 + r) 2 .
(20)
( 2)
We remark that as x → +∞, since g1 = O τ1 , w0 may not satisfy (20) .
Lemma 6 The functions ξ0 (x, y) := 1,
ξ1 :=
and
solve the system
√
1 2
3 3
τ1 −
τ ,
2
6 1
(√
)
√
√
3
3
3
ξ2 :=
τ1 + 1 e− 2 x+ 4 yi
2
{
L1 ϕ = 0,
M1 ϕ = 0.
Proof. This could be checked directly. Alternatively, we can look for solutions
of L1 ϕ = 0 in the form f1 (y) ∂x−1 g1 and f2 (y) ∂x−1 g2 . This reduces to an ODE
for the unknown functions f1 and f2 .
For given function η, we have seen from (19) that the ODE (17) for the
unknown function ϕ could be solved for each fixed y. To solve the system (14) ,
we define
Φ0 (x, y) := L1 ϕ − G1 ϕ,
and
Φ1 = ∂x Φ0 , Φ2 = ∂x2 Φ0 .
8
Note that Φi depends on the function ϕ.
Consider the system of equations

 Φ0 (x, y) = 0,
Φ1 (x, y) = 0,

Φ2 (x, y) = 0,
for x = 1.
(21)
We seek a solution ϕ to (21) in the form w0 + w1 , where
w1 (x, y) = ρ0 (y) ξ0 (x, y) + ρ1 (y) ξ1 (x, y) + ρ2 (y) ξ2 (x, y) .
Lemma 7 System (21) has a solution (ρ0 , ρ1 , ρ2 ) with the initial condition
ρi (0) = 0, i = 0, 1, 2.
Proof. The equation Φ0 = 0 could be written as
L1 w1 = −L1 w0 + G1 η := H0 .
Similarly, we write Φ1 = 0 as
∂x [L1 w1 ] = −∂x [L1 w0 ] + ∂x [G1 η] := H1 .
Φ2 = 0 could be written as
∂x2 [L1 w1 ] = −∂x2 [L1 w0 ] + ∂x2 [G1 η] := H2 .
Consider the system

 L1 w1 = 0,
∂x [L1 w1 ] = 0,
 2
∂x [L1 w1 ] = 0.
(22)
In view of the definition of w1 , we know that (22) is a homogeneous system of
first order differential equations for the functions ρ0 , ρ1 , ρ2 . Explicitly, (22) has
the form
 ′ 
ρ0
A  ρ′1  = 0,
ρ′2
where


A=
iτ√
1 ξ0
3
iτ1√
∂x ξ0
3
2
iτ1 ∂x
ξ
√ 0
3
iτ√
1 ξ1
3
iτ1√
∂x ξ1
3
2
iτ1 ∂x
ξ
√ 1
3
iτ√
1 ξ2
3
iτ1√
∂x ξ2
3
2
iτ1 ∂x
ξ
√ 2
3


.
Hence (21) has a solution
∫
0

y

H0 (1, s)
A−1 (1, s)  H1 (1, s)  ds.
H2 (1, s)
This completes the proof.
9
Proposition 8 Suppose η satisfies (15) . Then
)
( √
(
)
√
3
6
15
1
3
2
∂x Φ 0 = −
+
∂x Φ 0 +
2 3−
∂x Φ0
2
τ1
τ1
τ1
(
)
√
1 15
+ 2
− 2 3 Φ0 .
τ1 τ1
Proof. This follows from quite tedious computation using Mathematica.
Intuitively, we expect that this identity follows from the compatibility properties of suitable Lax pair of the KP-I equation. But up to now we have not
been able to rigorously show this.
Lemma 9 The function ϕ = w0 + w1 satisfies (14) for all (x, y) ∈ R2 .
Proof. By the definition of Φ0 , Φ1 , Φ2 and Proposition
 

0
1
0
Φ0

0
0
1


Φ1
=  15 √
∂x
√
√
2 3− τ15
τ1 −2 3
3
6
1
Φ2
2
τ
τ − 2
τ1
1
1
8, we have


Φ0

Φ1  .

Φ2
For each fixed y ̸= 0, by Lemma 7, Φ0 (1, y) = Φ1 (1, y) = Φ2 (1, y) = 0. By the
uniqueness of solution to ODE, we obtain Φ0 (x, y) = Φ1 (x, y) = Φ2 (x, y) = 0,
for all x ∈ R. On the other hand, by the definition of w0 and w1 , the second
equation of (14) is also satisfied for all (x, y) ∈ R2 with y ̸= 0. By continuity of
ϕ, we know that ϕ = w0 + w1 satisfies (14) for all (x, y) ∈ R2 . This finishes the
proof.
Lemma 10 Suppose η satisfies (11). Let ρi , i = 0, 1, 2, be functions given by
Lemma 7. Then
ρ1 (y) = ρ2 (y) = 0, for all y ∈ R.
Proof. Dividing the equation Φ0 = 0 by ξ2 , we get
1
i
√
(ρ′0 ξ0 + ρ′1 ξ1 + ρ′2 ξ2 ) τ1 = H0 .
ξ2
3ξ2
(23)
For each fixed y ∈ R, sending x → −∞ in (23) and using the estimate (11) , we
infer
ρ′2 = 0.
This together with the initial condition ρ2 (0) = 0 tell us that ρ2 = 0.
Now Φ0 = 0 becomes
i
√ (ρ′0 ξ0 + ρ′1 ξ1 ) τ1 = H0 .
3
Dividing both sides of this equation by ξ1 and letting x → −∞, we get ρ′1 = 0.
Hence ρ1 = 0. The proof is completed.
10
Proof of Proposition 3. We have proved that ρ1 and ρ2 are both zero. Now
let us define
∫ +∞
g2 F1
k (y) =
ds, for y ̸= 0.
4τ
1W
−∞
Then
5
|w0 + k (y) ξ1 (x, y)| ≤ C (1 + r) 2 , for x large.
Using similar arguments as Lemma 10, we see that k ′ = 0. But k (y) → 0 as
|y| → +∞. Hence k (y) = 0. Then the function w0 + ρ0 ξ0 is the desired solution.
4
Linearized Bäcklund transformation between
τ1 and τ2
In terms of τ1 and τ2 , the Bäcklund transformation (10) can be written as
)
{ (
Dx2 − √13 Dx + √13 iDy τ1 · τ2 = 0,
√
(
)
−Dx + iDy + Dx3 − 3iDx Dy τ1 · τ2 = 0.
The linearization of this system is
{
Here
L2 ϕ = G2 η,
M2 ϕ = N2 η.
(24)
(
)
1
1
Dx2 − √ Dx + √ iDy ϕ · τ2 ,
3
3
(
)
√
M2 ϕ = −Dx + iDy + Dx3 − 3iDx Dy ϕ · τ2 ,
L2 ϕ =
and
)
(
1
1
2
G2 η = − Dx − √ Dx + √ iDy τ1 · η,
3
3
(
)
√
3
N2 η = − −Dx + iDy + Dx − 3iDx Dy τ1 · η.
Proposition 11 Let η = η (x, y) be a function solving the linearized bilinear
KP-I equation at τ2 :
−Dx2 ητ2 + Dx4 ητ2 = Dy2 ητ2 .
(25)
Suppose η satisfies (11) . Then the system (14) has a solution ϕ with
(
)
5
|ϕ| + |∂x ϕ| + ∂x2 ϕ + |∂x ∂y ϕ| (1 + r) ≤ C (1 + r) 2 .
11
From the first equation in (24), we get
[√ (
]
√
)
∂y ϕτ2 = i 3 ∂x2 ϕτ2 − 2∂x ϕ∂x τ2 − ∂x ϕτ2 + 2iϕτ̄1 − 3iGη.
(26)
Here we have used τ̄1 to denote the complex conjugate of τ1 . Inserting (26) into
the second equation of (24), we obtain
∂x3 ϕτ2
( √
)
√
(
)
√
3
12x2
F2
2 3x
+ −
τ2 − 6x ∂x2 ϕ + 2 3x +
τ̄1 ϕ =
. (27)
∂x ϕ −
2
τ2
τ2
4
(
Here
F2 = N 2 η −
)
√
6∂x τ2
G2 η − 3∂x (G2 η) + 3G2 η .
τ2
To solve equation (27) , we set
ϕ = τ1 κ and h = κ′ .
Equation (27) is transformed into the equation
[
( √
) ]
3
′′
T (h) := τ1 τ2 h + 3τ2 + −
τ2 − 6x τ1 h′
2
)
[ ( √
(
)]
√
3
12x2
τ2 − 6x + τ1 2 3x +
h
+ 2 −
2
τ2
=
F2
.
4
Lemma 12 The homogenous equation
T (h) = 0
has two solutions h1 , h2 , given by
2
2
h1 (x, y) = (x − yi) + √ (x − yi) + 3 +
3
)
√
τ2 (
= 2 3τ2 + 4 3 (x + τ̄1 ) ,
3τ1
and
h2 (x, y) =
√ )(
(
12 y − 3i y +
τ12
√ ) √3 x
τ2 (
x
+
yi
−
3 e2 .
τ12
Proof. This equation can be solved using Mathematica.
Lemma 13 The system
)
{ (
Dx2 − √13 Dx + √13 iDy ϕ · τ2 = 0,
√
)
(
−Dx + iDy + Dx3 − 3iDx Dy ϕ · τ2 = 0
12
√i
3
)
has three solutions ζ0 , ζ1 , ζ2 , given by ζ0 = τ1 ,
ζ1 = τ1 ∂x−1 h1
( 3
)
(
)
(
√ )
z̄
z̄ 2
i
= τ1
+ √ + 3x − 11yi − 12 y − 3i
y+ √
,
3
3
3
and
√
ζ2 = e
3yi
τ1 ∂x−1 g2
) √
(
√
3
4yi
8
2
2
√
√
x+y +
+ 7 e 2 x+ 3yi .
= x −
3
3
Proof. This is similar to Lemma 6 and can be checked by direct computation.
Let W̃ be the Wronskian of h1 , h2 . That is
W̃ = h1 ∂x h2 − h2 ∂x h1 = e
√
3
2 x
τ23
.
τ13
Variation of parameter formula gives us a solution τ1 ∂x−1 h∗ to the equation (27) ,
where
∫ x
∫ x
h1 F2
h2 F2
∗
h = h2
ds − h1
ds.
4τ
τ
1 2
+∞ W̃
+∞ W̃ 4τ1 τ2
Lemma 14 Suppose η satisfies (11) . Let w̃0 = τ1 ∂x−1 h∗ . Then
5
|w̃0 | ≤ C (1 + r) 2 , for x ≥ −10.
Proof. Since η satisfies (11) , we have
5
|F2 | ≤ C (1 + r) 2 .
Hence
Hence
√
h1 F2 3
≤ Ce− 23x (1 + r)− 2 ,
W̃ τ τ 1 2
h2 F2 − 52
W̃ τ τ ≤ C (1 + r) .
1 2
1
|h∗ (x, y)| ≤ (1 + r) 2 , for x ≥ −10.
It follows that
5
|w̃0 | ≤ C (1 + r) 2 , for x ≥ −10.
Slightly abusing the notation, we define
Φ0 = L2 ϕ − G2 η.
Then let Φ1 = ∂x Φ0 and Φ2 = ∂x2 Φ0 .
13
We would like to solve the system

 Φ0 = 0,
Φ1 = 0, , for x = 0.

Φ2 = 0,
(28)
Similarly as before, we seek a solution of this problem with the form w̃0 + w̃1 ,
with
w̃1 = β0 (y) ζ0 + β1 (y) ζ1 + β2 (y) ζ2 ,
where β0 , β1 , β2 are functions of y to be determined.
The problem (28) can be written as

 L2 w̃1 = H0 ,
∂x (L2 w̃1 ) = H1 ,
 2
∂x (L2 w̃1 ) = H2 .
(29)
Here
H0 = G2 η − L2 w̃0 ,
H1 = ∂x (G2 η − L2 w̃0 ) ,
H2 = ∂x2 (G2 η − L2 w̃0 ) .
Lemma 15 The equation (29) has a solution (β0 , β1 , β2 ) satisfying the initial
condition
βi (0) = 0, i = 0, 1, 2.
Proof. The proof is similar to that of Lemma 7.
Proposition 16 Suppose η satisfies (25) . Then
(
(
)
√ )
√
12x
6
3
4 3x 60x2
3
2
∂x Φ 0 =
+
∂x Φ0 +
−
− 2
∂x Φ 0
τ2
2
τ2
τ2
τ2
)
(
√
√
√ xτ̄1
3 36x 8 3x2
120x3
+ 2 3 2 −
− 2 +
+
Φ0 .
τ2
τ2
τ2
τ22
τ23
Proof. We can check this by Mathematica. We can also prove this directly by
tedious calculation by hand.
Lemma 17 The function ϕ = w̃0 + w̃1 solves the system (24) for all (x, y) ∈ R2 .
Proof. By Proposition 16,

 
Φ0
0
∂x  Φ1  =  0
Φ2
a31
1
0
a32
14


0
Φ0
1   Φ1  ,
a33
Φ2
where
a31
a32
√
√
√ xτ̄1
3 36x 8 3x2
120x3
− 2 +
=2 3 2 −
+
,
2
τ2
τ2
τ2
τ2
τ23
√
6
4 3x 60x2
−
− 2 ,
=
τ2
τ2
τ2
and
a33
√
12x
3
=
+
.
τ2
2
For each fixed y, since Φi (0, y) = 0, i = 0, 1, 2, we deduce from the uniqueness
of solutions to ODE that Φi (x, y) = 0, for all x ∈ R.
Lemma 18 Let βi , i = 0, 1, 2 be the functions given by Lemma 15. Then β1 =
β2 = 0.
Proof. The proof is similar to that of Lemma 10.
With Lemma 18 at hand, we can prove Proposition 11 similarly as before.
5
Proof of the main theorem
In this section, we will prove Theorem 1.
Lemma 19 Suppose η satisfies
{
Then
L1 ϕ = G1 η,
M1 ϕ = N1 η,
√
−4∂x3 η + 2 3∂x2 η = Θ1 ϕ,
where
Θ1 ϕ := −M1 ϕ −
√
3L1 ϕ + 3∂x (L1 ϕ)
In particular, if
G1 η = N1 η = 0,
and
5
|η| ≤ C (1 + r) 2 .
Then η = c1 + c2 τ1 , for some constants c1 , c2 .
Proof. The equation G1 η = L1 ϕ is
(
)
1
i
2
Dx + √ Dx + √ Dy τ0 · η = −L1 ϕ.
3
3
That is,
i
1
∂x2 η + √ (−∂x η) + √ (−∂y η) = −L1 ϕ.
3
3
15
(30)
Inserting this identity into the equation
(
)
√
−Dx − iDy + Dx3 − 3iDx Dy τ0 · η = −M1 ϕ,
we get
(
))
(
√ 2
√
√
√
1
3∂x η − ∂x3 η − 3i∂x − 3i ∂x2 η − √ ∂x η + L1 ϕ
= −M1 ϕ − 3L1 ϕ.
3
Hence
√
√
−4∂x3 η + 2 3∂x2 η = −M1 ϕ − 3L1 ϕ + 3∂x (L1 ϕ) .
If L1 ϕ = M1 ϕ = 0, then
√
−4∂x3 η + 2 3∂x2 η = 0.
The solutions of this equation are given by
c1 + c2 x + c3 e
√
3
2 x
,
where c1 , c2 , c3 are constants which may depend on y. Due to the growth estimate of η, we find that
η = c1 + c2 x.
Inserting this into the equation G1 η = 0, we find that η is a linear combination
of 1 and τ1 .
Lemma 20 We have
√
Θ1 (x) = 4 3, Θ1 (y) = 0.
and
√
√
(
)
Θ1 x2 − y 2 = 8 3x, Θ1 (xy) = 2 3iτ̄ .
Proof. This follows from direct computation. For instances, we compute
(
)
(
)
(
)
1
1
L1 x2 − y 2 = Dx2 + √ Dx + √ iDy x2 − y 2 · τ1
3
3
(
))
(
))
1 (
i (
= 2τ1 − 4x + √ 2xτ1 − x2 − y 2 + √ −2yτ1 − i x2 − y 2
3
3
2
= √ τ2 .
3
)(
√
(
) (
)
M1 x2 − y 2 = −Dx − iDy + Dx3 − 3iDx Dy x2 − y 2 · τ1
(
(
))
(
(
) )
= − 2xτ1 − x2 − y 2 − i −2yτ1 − x2 − y 2 i
√
+ (−3) 2 − 3i (−2xi − (−2y))
√
= −2τ2 − 4 3x.
16
Then
√
√ 2
(
)
4x
Θ1 x2 − y 2 = 2τ2 + 4 3x − 3 √ τ2 + 3 √
3
3
√
= 8 3x.
Lemma 21 Define z (ϕ) := (L1 (ϕ) , M1 (ϕ)) , J (ϕ) := (G1 (ϕ) , N1 (ϕ)) . Then
(
√ )
z (x) = J xτ1 − 3z ,
z (y) = J (yτ1 ) ,
and
(
)
z x2 − y 2 = J (ρ1 ) ,
(31)
z (xy) = J (ρ2 ) ,
where
√
2 3 4√ 2 4 3
2i
2√ 2
ρ1 = x +
yix − y 3 −
3x −
3y − 14yi,
3
3
3
3 )
(3
√
√
5 √ 2 1 3
y2
y3
1 2
3
3 2
i+
y x+
+
y i + 5y.
ρ2 = x y + i 3x + ix +
2
6
6
2
3
6
6
Proof. We only prove (31) . The proof of other cases are similar. Consider the
equation
√
√
(
)
−4∂x3 η + 2 3∂x2 η = Θ1 x2 − y 2 = 8 3x.
This equation has a solution
ρ1 =
2 3 4√ 2
3x + a (y) x + b (y) ,
x +
3
3
where a (y) and b (y) are functions to be determined. Since
(
)
1
i
x2 + y 2 + 3
√
∂x2 ρ1 + √ (−∂x ρ1 ) + √ (−∂y ρ1 ) = L1 x2 − y 2 = −2
,
3
3
3
we get
√
(
)
8 3
1
8√
i
x2 + y 2 + 3
2
√
4x +
−√
2x +
3x − a − √ (a′ x + b′ ) = −2
.
3
3
3
3
3
Hence
√
2i
2√ 2
4 3
yi, b (y) = − y 3 −
3y − 14yi.
3
3
3
√
2
4√ 2 4 3
2i
2√ 2
ρ1 = x3 +
3x −
yix − y 3 −
3y − 14yi.
3
3
3
3
3
a (y) = −
Thus
17
Lemma 22 Suppose η satisfies
{
Then
∂x3 ητ1
L2 ϕ = G2 η,
M2 ϕ = N2 η.
(√
)
√
(
)
√
3
3
3
2
+
τ 1 − 3 ∂x η +
− 3 ∂x η +
η = Θ2 (ϕ) .
2
τ1
τ1
Here
Θ2 (ϕ) :=
√
6
L2 ϕ − 3∂x (L2 ϕ) + M2 ϕ − 3L2 ϕ.
τ1
Proof. Explicitly, η satisfies
)
{ (
Dx2 − √13 Dx + √13 iDy τ1 · η = −L2 ϕ,
√
(
)
−Dx + iDy + Dx3 − 3iDx Dy τ1 · η = −M2 ϕ.
(32)
Let us write the first equation in this system as
(√
)
√
iDy τ1 · η = −
3Dx2 − Dx τ1 · η − 3L2 ϕ.
That is,
] √
[
√ (
)
∂y ητ1 = −i ∂x ητ1 + 3 ∂x2 ητ1 − 2∂x η − 2η − 3iL2 ϕ.
Inserting this identity into the right hand side of the second equation the system
(32), we get the following equation for η :
(
)
√
−Dx + iDy + Dx3 − 3iDx Dy τ1 · η
√
√
√
= − 3Dx2 τ1 · η + Dx3 τ1 · η − 3i [∂x ∂y ητ1 + η∂x ∂y τ1 − ∂x η∂y τ1 − ∂y η∂x τ1 ] − 3L2 ϕ
√
]
√ (
√ (
) √
) (
)
2 3ii [
= − 3 ∂x2 ητ1 − 2∂x η + −∂x3 ητ1 + 3∂x2 η − 3∂x η −
∂x ητ1 + 3 ∂x2 ητ1 − 2∂x η − 2η
τ1
√
)
(
√
√
√
√
)
(
3iL2 ϕ
− 2 3i
− 3∂x ∂x ητ1 + 3 ∂x2 ητ1 − 2∂x η − 2η − 3∂x (L2 ϕ) − 3L2 ϕ
τ1
( √
)
√
3
= −4∂x ητ1 + − 3τ1 + 3 + 6 − 3τ1 − 3 + 6 ∂x2 η
(
)
√
√
√
√
√ ) √
√
√
2 3(
4 3
6
+ 2 3− 3+
τ1 − 2 3 − 3 + 2 3 ∂x η −
η + L2 ϕ − 3∂x (L2 ϕ) − 3L2 ϕ
τ1
τ1
τ1
= −M2 ϕ.
Thus η satisfies the following third order ODE:
)
(√
√
(
)
√
3
3
3
2
3
τ 1 − 3 ∂x η +
− 3 ∂x η +
η = Θ2 .
∂x ητ1 +
2
τ1
τ1
18
Note that η = τ2 satisfies the homogeneous equation
(√
)
√
(
)
√
3
3
3
3
2
∂x ητ1 +
τ 1 − 3 ∂x η +
− 3 ∂x η +
η = 0.
2
τ1
τ1
(33)
Letting η = τ2 κ and p = κ′ , equation (33) becomes
) )
(
(√
(
) )
(√
) (3
√
3
′′
τ1 − 3 τ2 p′ + 6τ1 + 2x
− 3 τ2 p = 0.
3τ1 − 6 +
τ1 τ2 p + 6xτ1 +
2
τ1
(34)
Lemma 23 The equation (34) has two solutions given by
2
p1 =
(x + yi) − 3
τ22
( √
) 1√ τ
√
1
p2 := 8 3x − 4i 3y + 3x2 + 3y 2 + 21 e− 2 3x 2 .
τ2
and
In particular, if η satisfies (30) and
G2 η = N2 η = 0,
Then η = c1 z + c2 τ2 .
Note that ∂x−1 p1 = − τz2 . Hence we get a solution η = −z for the equation
(33) . This solution is corresponding to the translation of τ2 along the x and y
axes.
Lemma 24 We have
√
√
√
√
Θ2 (yτ1 ) = iτ1 τ2 + 2i 3x2 − 2 3xy + 12ix − 2i 3y 2 − 6i 3.
√
√
√
√
( )
Θ2 z 2 = 2τ1 τ2 − 2 3x2 − 4i 3xy − 18x + 2 3y 2 − 18iy − 12 3
)
√
√
6 (
−
−2x2 − 4ixy − 2 3x + 2y 2 − 2i 3y + 12 .
τ1
L2 ρ1 = −
√
2√ 4 4 √ 3
8
8
16 √ 2
2√ 4 4 3
3x − i 3x y+ x3 − ix2 y+
3x +8x+
3y + iy +22 3.
3
9
3
3
3
9
3
√
4√ 3
3x + 4i 3x2 y − 20x2
3
√
√
8√
2
3xy + 8ixy − 8 3x + 2y 4 + 28y 2 + 8i 3y + 42.
+
3
M2 ρ1 = −2x4 −
19
5√ 3
1
1 √
10 √
1√
3x y − ix3 − i 3x2 y 2 + 2x2 y + i 3x2 −
3xy 3
9
3
3
3
3
√
√
1 √
2
− ixy 2 − 2 3xy − ix − i 3y 4 + y 3 + y + 10i 3.
9
3
L2 ρ2 = −
8√ 2
2 √
4 √
3x y − 2ix2 − 2xy 3 + i 3xy 2
M2 ρ2 = −2x3 y − i 3x3 −
3
3
3
√
√
2√ 3
− 30xy − 17i 3x −
3y + 4iy 2 − 3y + 15i.
3
Proof. This follows from Direct computation. For instances,
(
)
1
1
L2 (ρ1 ) = Dx2 − √ Dx + √ iDy ρ1 · τ2
3
3
1
i
= ∂x2 ρ1 τ2 − 4x∂x ρ1 + 2ρ1 − √ (∂x ρ1 τ2 − 2xρ1 ) + √ (∂y ρ1 τ2 − 2yρ1 )
3
3
(
)
(
)
τ
i
2x
2yi
2
2
= ∂x ρ1 τ2 + ∂x ρ1 −4x − √
+ √ τ2 ∂y ρ1 + ρ1 2 + √ − √
3
3
3
3
√
8 3 8 2
2√ 4 4 3
2√ 4 4 √ 3
16 √ 2
3x − i 3x y + x − ix y +
3x + 8x +
3y + iy + 22 3.
=−
3
9
3
3
3
9
3
(
)
√
M2 ρ1 = −Dx + iDy + Dx3 − 3iDx Dy ρ1 · τ2
(
(
√ )
√ )
= ∂x3 ρ1 τ2 + ∂x2 ρ1 (−6x) + ∂x ρ1 −τ2 + 6 + 2 3yi + ∂y ρ1 iτ2 + 2 3xi
√
− ∂x ∂y ρ1 3iτ2 + ρ1 (2x − 2yi)
√
4√ 3
= −2x4 −
3x + 4i 3x2 y − 20x2
3
√
√
8√
2
+
3xy + 8ixy − 8 3x + 2y 4 + 28y 2 + 8i 3y + 42.
3
The other cases are similar.
We are now in a position to prove Theorem 1.
Proof of Theorem 1. Let ϕ be a solution of (4) satisfying the assumption of
Theorem 1. Then using Lemma 2, we can find η2 , a solution of (25) , satisfying
(11) . In view of Lemma 22, if G2 η2 = N2 η2 = 0. Then η2 = c1 z+c2 τ2 . Therefore,
to prove the theorem, from now on, we can assume G2 η2 ̸= 0 or N2 η2 ̸= 0.
Using Proposition 11 and the linearization of the identity (8) , there exists a
solution η1 of the equation
( 2
)
Dx − Dx4 + Dy2 η1 · τ1 = 0,
satisfying the estimate (11) .
Case 1. G1 η1 = N1 η1 = 0.
In this case, by Lemma 19, η1 = a1 + a2 τ1 . Accordingly,
η2 = c1 ∂x τ2 + c2 ∂y τ2 + c3 τ2 .
20
Case 2. G1 η1 ̸= 0 or N1 η1 ̸= 0.
In this case, by Proposition 3, there exists a solution η0 of
( 2
)
Dx − Dx4 + Dy2 η0 · τ0 = 0,
(35)
satisfying
5
|η0 | + |∂x η0 | + |∂y η0 | ≤ C (1 + r) 2 .
(36)
From (35) and (36) , we infer that
∂x2 η0 + ∂y2 η0 = 0.
Therefore,
)
(
η0 = c1 + c2 x + c3 y + c4 x2 − y 2 + c5 xy.
We claim that c2 = c3 = c4 = c5 = 0. Indeed, otherwise, using Lemma 24,
Θ2 (η1 ) would have a term like x2 y or x2 . Then using variation of parameter
formula and the asymptotic behavior of the solutions p1 , p2 , we could show that
η2 could not satisfy the estimate (11) . This is a contradiction. Hence η0 = c0 ,
a constant. This also implies
η2 = c1 ∂x τ2 + c2 ∂y τ2 + c3 τ2 .
The proof is thus finished.
References
[1] M. J. Ablowitz, H. Segur, Solitons and the inverse scattering transform.
SIAM Studies in Applied Mathematics, 4. Society for Industrial and Applied Mathematics (SIAM), Philadelphia, Pa., 1981.
[2] M. J. Ablowitz, P. A. Clarkson, Solitons, nonlinear evolution equations and
inverse scattering. London Mathematical Society Lecture Note Series, 149.
Cambridge University Press, Cambridge, 1991.
[3] N. G. Berloff, P. H. Roberts, Motions in a Bose condensate. X. New results on the stability of axisymmetric solitary waves of the Gross-Pitaevskii
equation. J. Phys. A 37 (2004), no. 47, 11333–11351.
[4] F. Béthuel, P. Gravejat, J. C. Saut, On the KP-I transonic limit of
two-dimensional Gross-Pitaevskii travelling waves. Dyn. PDE 5(3) (2008),
241?80.
[5] F. Béthuel, P. Gravejat, J. C. Saut, Travelling waves for the GrossPitaevskii equation. II. Commun. Math. Phys. 285(2) (2009), 567?51.
[6] D. Chiron and M. Maris, Rarefaction pulses for the nonlinear Schrödinger
equation in the transonic limit. Comm. Math. Phys. 326(2014), 329-392.
21
[7] A. de Bouard and J. C. Saut, Solitary waves of generalized KadomtsevPetviashvili equations. Ann. Inst. H. Poincare-Anal. Non Lineaire 14 (1997),
no. 2, 211–236.
[8] A. de Bouard and J. C. Saut, Symmetries and decay of the generalized
Kadomtsev-Petviashvili solitary waves. SIAM J. Math. Anal. 28 (1997),
no. 5, 1064–1085.
[9] A. de Bouard, Saut, J. C. Remarks on the stability of generalized KP solitary waves. in Mathematical problems in the theory of water waves (Luminy, 1995), 75–84, Contemp. Math., 200, Amer. Math. Soc., Providence,
RI, 1996.
[10] V. S. Dryuma, On the analytical solution of the two-dimensional Kortewegde Vries equation, Sov. Phys. JETP Lett. 19, 753-757 (1974).
[11] A. S. Fokas, M. J. Ablowitz, On the inverse scattering and direct linearizing transforms for the Kadomtsev-Petviashvili equation. Phys. Lett. A 94
(1983), no. 2, 67–70.
[12] K.A. Gorshkov, D. E. Pelinovsky, Y. A. Stepanyants, Normal and anormal
scattering, formation and decay of bound states of two-dimensional solitons
described by the Kadomtsev- Petviashvili equation. JETP. 77(2), 1993, 237–
245.
[13] R. Hirota, The direct method in soliton theory. Translated from the 1992
Japanese original and edited by Atsushi Nagai, Jon Nimmo and Claire
Gilson, Cambridge Tracts in Mathematics, 155. Cambridge University
Press, Cambridge, 2004.
[14] M. Hărăguş-Courcelle, R. L. Pego, Travelling waves of the KP equations
with transverse modulations. C. R. Acad. Sci. Paris. I Math. 328 (1999),
no. 3, 227–232.
[15] X. B. Hu, Rational solutions of integrable equations via nonlinear superposition formulae. J. Phys. A 30 (1997), no. 23, 8225–8240.
[16] C. Jones, S. Putterman and P. H. Roberts, Motions in a Bose condensate
V. Stability of wave solutions of nonlinear Schrödinger equations in two
and three dimensions. J. Phys. A Math. Gen. 19 (1986), 2991?011.
[17] C. Jones, P.H. Roberts, Motion in a Bose condensate IV. Axisymmetric
solitary waves. J. Phys. AMath. Gen. 15 (1982), 2599?619 (1982)
[18] Kadomtsev, Petviashvili, On the stability of solitary waves in weakly dispersive media, Soviet Physics-Doklady, 15, 539-541, 1970.
[19] Y. Liu, X. P. Wang, Nonlinear stability of solitary waves of a generalized
Kadomtsev-Petviashvili equation. Comm. Math. Phys. 183 (1997), no. 2,
253–266.
22
[20] Z. Lu, E. M. Tian, R. Grimshaw, Interaction of two lump solitons described
by the Kadomtsev-Petviashvili I equation. Wave Motion 40 (2004), no. 2,
123–135.
[21] S. V. Manakov, The inverse scattering transform for the time dependent
Schrödinger equation and Kadomtsev-Petviashvili equation, Physica, D3,
1981, 420–427.
[22] S. V. Manakov, V. E. Zakharov, L. A. Bordag, A. R. Its, and V. B. Matveev,
Two dimensional solitons of the Kadomtsev-Petviashvili equation and their
interaction, Physics letter A, 63, 205–206.
[23] M. Maris, Traveling waves for nonlinear Schrdinger equations with nonzero
conditions at infinity. Ann. Math. 178 (2013), 107–182.
[24] T. Mizumachi, Asymptotic stability of N -solitary waves of the FPU lattices.
Arch. Ration. Mech. Anal. 207 (2013), no. 2, 393–457.
[25] T. Mizumachi, R. L. Pego, Asymptotic stability of Toda lattice solitons.
Nonlinearity 21 (2008), no. 9, 2099–2111.
[26] A. Nakamura, Decay mode solution of the two-dimensional KdV equation
and the generalized Bäcklund transformation. J. Math. Phys. 22 (1981), no.
11, 2456–2462.
[27] C. Rogers, W. F. Shadwick, Bäcklund transformations and their applications. Mathematics in Science and Engineering, 161. Academic Press, Inc.
[Harcourt Brace Jovanovich, Publishers], New York-London, 1982.
[28] J. Satsuma, M. J. Ablowitz, Two-dimensional lumps in nonlinear dispersive
systems. J. Math. Phys. 20 (1979), no. 7, 1496–1503.
[29] J. Villarroel, M. J. Ablowitz, On the discrete spectrum of the nonstationary
Schrodinger equation and multipole lumps of the Kadomtsev-Petviashvili I
equation. Comm. Math. Phys. 207 (1999), no. 1, 1–42.
[30] X. P. Wang, M. J. Ablowitz, H. Segur, Wave collapse and instability of
solitary waves of a generalized Kadomtsev-Petviashvili equation. Phys. D
78 (1994), no. 3-4, 241–265.
[31] X. Zhou, Inverse scattering transform for the time dependent Schrödinger
equation with applications to the KPI equation. Comm. Math. Phys. 128
(1990), no. 3, 551–564.
23