# Lump type solutions: Backlund transformation and spectral properties

Yong Liu  Department of Mathematics, University of Science and Technology of
China, Hefei, P.R. China<EMAIL_ADDRESS>,  Juncheng Wei  Department of
Mathematics, University of British Columbia, Vancouver, B.C. Canada, V6T 1Z2.
<EMAIL_ADDRESS>and  Wen Yang  Department of Mathematics, Faculty of Science
and Technology, University of Macau, Macau, P.R. China<EMAIL_ADDRESS>
###### Abstract.

There are various different ways to obtain traveling waves of lump type for
the KP equation. We propose a general and simple approach to derive them via a
Backlund transformation. This enables us to establish an explicit connection
between those low energy solutions and high energy ones. Based on this
construction, spectral analysis of the degree $6$ solutions is then carried
out in details. The analysis of higher energy ones can be done in an inductive
way.

##  1\.
Introduction and statement of main results

KP equation is a two dimensional analogy of the classical KdV equation. It
naturally appears in the theory of shallow water waves. As an important $2+1$
dimensional integrable system, it has been extensively studied for more than
forty years, and many other integrable systems can be regarded as its suitable
reduction. However, there are still some interesting questions remained to be
answered for this equation. As a matter of fact, even the properties of its
traveling wave solutions are not fully understood yet. For traveling waves,
the KP equation reduces to the so called Boussinesq equation:

(1)
$\partial_{x}^{2}\left(\partial_{x}^{2}u-u+3u^{2}\right)-\partial_{y}^{2}u=0.$

In principle, solutions to this equation should play important role in the
long time dynamics of the KP equation.

The Boussinesq equation in the form of $\left(\ref{Bou}\right)$ is of elliptic
type and closely related to other PDEs such as GP equation. While there
already exist a lot of works concerning elliptic equations of second order,
the study of fourth order equations with both mathematical and physical
significance is relatively few. In this paper, we would like to study the
spectral properties of “lump type” solutions to $\left(\ref{Bou}\right)$. By
lump type solutions, we mean solutions of $\left(\ref{Bou}\right)$ which decay
to zero at infinity. This is a natural class of physically meaningful
solutions. The name “lump type” comes from the fact that the following
“classical” lump solution solves $\left(\ref{Bou}\right)$:

$U\left(x,y\right)=4\frac{y^{2}-x^{2}+3}{\left(x^{2}+y^{2}+3\right)^{2}}.$

The analysis of $U$ has a long history. It is first obtained in [16, 22] by
parameter degeneration. The spectral property of $U$ is now well understood.
Indeed, we have proved in [14] using Backlund transformation that $U$ is
nondegenerated, in the sense that the linearized operator at $U$ does not have
any nontrivial kernels. A direct consequence of this property is that the lump
is orbitally stable under the KP-I flow, which is globally well posed [11, 17,
18]. The asymptotical stability of $U$ remains to be an unsolved open problem
in this direction.

If we introduce the tau function $\tau$ by $u=2\partial_{x}^{2}\ln\tau,$ then
the equation $\left(\ref{Bou}\right)$ turns into the following bilinear
equation:

(2) $\left(D_{x}^{4}-D_{x}^{2}-D_{y}^{2}\right)\tau\cdot\tau=0.$

Here $D$ is the bilinear derivative operator introduced by Hirota [10]. One
easily checks that the lump solution $U$ corresponds to $\tau=x^{2}+y^{2}+3.$

Our recent result [15] shows that real valued solution(with a mild decaying
assumption) of $\left(\ref{Bou}\right)$ has to be rational, and the
corresponding tau function, which solves $\left(\ref{bilinear}\right),$ will
be a polynomial of degree $k\left(k+1\right)$ with $k\in\mathbb{N}$. In the
case of $k=1,$ it is not difficult to show that up to translation in the $x,y$
variables and multiplication by a constant, real valued solution to
$\left(\ref{bilinear}\right)$ has to be $x^{2}+y^{2}+3.$

According to our classification result mentioned above, the next family of tau
functions of $\left(\ref{bilinear}\right)$ are polynomials of degree $6,$
corresponding to $k=2.$ In Section 2, we show, using Backlund transformation,
that the following family of polynomials $h_{A,B}$ solves
$\left(\ref{bilinear}\right),$ where

$\displaystyle h_{A,B}\left(x,y\right)=~{}$ $\displaystyle
x^{6}+3x^{4}y^{2}+3x^{2}y^{4}+y^{6}+25x^{4}+90x^{2}y^{2}+17y^{4}$
$\displaystyle+Bx^{3}+3Ax^{2}y-3Bxy^{2}-Ay^{3}-125x^{2}+475y^{2}$
$\displaystyle-Bx+5Ay+1875+\frac{A^{2}}{4}+\frac{B^{2}}{4}.$

Here $A,B\in\mathbb{R}$ are parameters. The solutions $2\partial_{x}^{2}\ln
h_{A,B}$ of $\left(\ref{Bou}\right)$ will be denoted by $u_{A,B}.$ Presumably,
any degree $6$ solution should belong to this family. Note that this family of
solutions are first obtained in [9], using a limiting procedure on the
involved parameters.

As we know, solutions of most elliptic equations do not have explicit
expression. Therefore the existence of such a family of non-radially-
symmetric, rational, solutions $u_{A,B}$ to the Boussinesq equation is by
itself an interesting phenomenon and provides an important scenario for us to
analyze various properties of the solutions relevant in the subject of
elliptic equations.

Our first result is the following

######  Theorem 1.

For any $A,B,$ the solution $u=u_{A,B}$ is nondegenerated in the following
sense: If $\phi$ is a solution of the linearized operator

$\partial_{x}^{2}\left(\partial_{x}^{2}\phi-\phi+6u\phi\right)-\partial_{y}^{2}\phi=0,$

with

$\phi\left(x,y\right)\rightarrow 0\text{, as }x^{2}+y^{2}\rightarrow+\infty,$

then there exist constants $c_{1},...,c_{4},$ such that

$\phi=c_{1}\partial_{x}u+c_{2}\partial_{y}u+c_{3}\partial_{A}u+c_{4}\partial_{B}u.$

Moreover, the Morse index of $u$ is equal to $3.$

The Morse index of $u$ is, by definition, the number of negative
eigenvalues(counted by multiplicity) of the operator

$\eta\rightarrow\partial_{x}^{2}\phi-\phi+6u\phi-\partial_{x}^{-2}\partial_{y}^{2}\phi.$

In the special case that $A=B=0,$ the solution $u_{0,0}$ will be even in both
variables. This is precisely the solution obtained for the first time in [19].
If we consider kernel which is also even in both variables, then from Theorem
1, we know that it has to be $0.$ As a direct application of this fact, one
can then construct new subsonic traveling wave solutions to the GP equation
from $u_{0,0}$, using the same perturbation method developed in [13]. More
precisely, we have

######  Corollary 2.

For each $\varepsilon>0$ sufficiently small, there exists a traveling wave
solution

$\Phi\left(t,x,y\right)=\phi\left(x-\left(\sqrt{2}-\varepsilon^{2}\right)t,y\right),$

to the following GP equation

$i\partial_{t}\Phi+\Delta\Phi+\left(1-\left|\Phi\right|^{2}\right)\Phi=0,$

which has the asymptotic expansion

$\phi\left(x,y\right)=1+i\varepsilon\partial_{x}^{-1}u^{\ast}+O\left(\varepsilon^{2}\right),$

where

$u^{\ast}\left(x,y\right)=-\frac{3}{2}u_{0,0}\left(2^{\frac{3}{4}}x,2^{\frac{1}{2}}y\right).$

To the best of our knowledge, this is the first result in the construction of
high energy traveling wave solution for the GP equation in the subsonic
regime, although there are quite a few existence results on the least energy
solutions.

From the even solution $u_{0,0}$ and our nondegeneracy result, one can
actually also construct new nontrivial solutions to following generalized KP-I
equation:

$\partial_{x}^{2}u-u+\left|u\right|^{\alpha}-\partial_{x}^{-2}\partial_{y}^{2}u=0,$

provided that the exponent $\alpha$ is sufficiently close to $2.$ Note that
for this type of generalized KP-I equation, one can use variational method
(See [4, 5, 12]) to construct the ground state solution, whose Morse index is
presumably equal to $1.$ On the other hand, the new solutions close to
$u_{0,0}$ will have Morse index $3,$ and it seems to be hard to construct them
by variational method. It is also interesting to see whether this type of
solutions exist for $\alpha$ not close to $2.$

The method developed in this paper to prove Theorem 1 can actually be used to
construct and analyze the spectral properties of higher energy lump type
solutions. We emphasize that general lump type solutions of KP equation with
degree $k\left(k+1\right),$ and $k$ free parameters have already been found in
[8, 19, 20, 21], using the Wronskian representation. It is also worth
mentioning that there are also other methods to construct these solutions, see
[1, 2, 6, 7, 23, 24]. Our method has the advantage that it establishes
explicit connection between low energy solutions and high energy ones. The
complete classification of lump type solutions remains open.

The paper is organized in the following way. In Section 2, we explain how to
use the Backlund transformation to create higher energy solutions from the low
energy solutions. We point out that these solutions are in general complex
valued. In Section 3, we analyze the precise asymptotic behavior of the
eigenfunctions of the linearized operator and show that the Morse index of
$h_{A,B}$ is equal to $3.$ This is based on a “reverse” Lyapunov-Schmidt
reduction type argument. The very delicate point here is that the reduced
problem is actually degenerated.

Acknowledgement Y. Liu was supported by the National Key R&D Program of China
2022YFA1005400 and NSFC 11971026, NSFC 12141105. J.Wei was supported by NSERC
of Canada. W. Yang was supported by National Key R&D Program of China
2022YFA1006800, NSFC No.12171456 and NSFC No.12271369.

##  2\. Backlund transformation from low energy solutions to high energy ones

In this section, we propose a general scheme to create high energy lump type
solutions from low energy ones. Although there are other methods to construct
these solutions, our method has the advantage that it establishes explicit
link between solutions with different energy, which in turn enables us to
study their spectral properties using an inductive argument.

The Boussinesq equation has the following Backlund transformation(See [14]):

(3) $\begin{cases}\left(D_{x}^{2}+\mu
D_{x}+\frac{i}{\sqrt{3}}D_{y}-\lambda\right)f\cdot g=0,\\\ \\\
\left(\left(3\lambda-1\right)D_{x}-\sqrt{3}i\mu
D_{y}+D_{x}^{3}-\sqrt{3}iD_{x}D_{y}+v\right)f\cdot g=0.\end{cases}$

Here $\mu,\lambda,v$ are arbitrary parameters, and throughout the paper, $i$
will be the imaginary unit. The Backlund transformation
$\left(\ref{Back}\right)$ has the following property: If $f$ satisfies the
bilinear equation $\left(\ref{bilinear}\right),$ and $f,g$ satisfy the system
$\left(\ref{Back}\right),$ then $g$ will automatically be a solution of
$\left(\ref{bilinear}\right).$ In this way, to get a new solution, we only
need to solve a linear problem involving third order derivatives, instead of
the original nonlinear problem with fourth order derivatives.

If $f,g$ are polynomials, then in view of the highest degree terms,
necessarily $\lambda=v=0.$ Then inspecting the highest degree terms, we find
that $\mu=\pm\frac{1}{\sqrt{3}}$. We are thus lead to consider the following

(4) $\begin{cases}\left(D_{x}^{2}+\mu
D_{x}+\frac{i}{\sqrt{3}}D_{y}\right)f\cdot g=0,\\\ \\\
\left(-D_{x}-\sqrt{3}i\mu D_{y}+D_{x}^{3}-\sqrt{3}iD_{x}D_{y}\right)f\cdot
g=0.\end{cases}$

Let $f$ be a real valued polynomial solution of $\left(\ref{bilinear}\right)$
of degree $2n.$ The classification result in [15] tells us that

$n=\frac{k\left(k+1\right)}{2},\text{ for some }k\in\mathbb{N}.$

Those degree $j$ terms in $f$ will be denoted by $f_{j}.$ Our first key
observation is that it will be more convenient to consider the problem in the
$z$-$\bar{z}$ coordinate, rather than the usual $x$-$y$ coordinate, where

$z=x+yi\text{ and }\bar{z}=x-yi.$

It follows from Lemma 13 of [15] that $f_{2n}=z^{n}\bar{z}^{n}$, and by
suitable translation in the $x$ and $y$ variables, we can assume that
$f_{2n-1}=0.$ We also observe that

$D_{x}+iD_{y}=2D_{\bar{z}}\text{ \ and }D_{x}-iD_{y}=2D_{z}.$

This formula, although very simple, will be frequently used in this section.

The following lemma tells us that if $f,g$ are connected through Backlund
transformation, then the degree of $g$ will be a square number and essentially
determined by that of $f,$ and more importantly, $g$ will be complex valued.

######  Lemma 3.

Let $f$ be a real valued polynomial solution of $\left(\ref{bilinear}\right)$
of degree $2n$ with $f_{2n-1}=0.$ Suppose $f,g$ satisfies
$\left(\ref{Back2}\right).$ Then the highest degree term $g_{m}$ of $g$ has
the form

$g_{m}=z^{j}\bar{z}^{n},$

where $j$ satisfies

(5) $n\left(n-1\right)-2nj+j\left(j-1\right)=0.$

In particular, if $n=\frac{k\left(k+1\right)}{2}$ for some integer $k,$ then
the degree $j+n$ of $g$ is equal to

$k^{2}\text{ or }\left(k+1\right)^{2}.$

Moreover, $g_{m-1}=\sqrt{3}nz^{j}\bar{z}^{n-1}+cz^{j-1}\bar{z}^{n},$ where

$c=-\frac{\left(j-n\right)\left(j+n-1\right)}{2\sqrt{3}\left(n-j+1\right)},$

and $g_{m-2}$ solves

$2D_{\bar{z}}f_{2n}\cdot g_{m-2}+2D_{\bar{z}}f_{2n-2}\cdot
g_{m}+\sqrt{3}D_{x}^{2}f_{2n}\cdot g_{m-1}=0.$

###### Proof.

Balancing highest degree terms in $\left(\ref{Back2}\right)$ requires

$D_{\bar{z}}f_{2n}\cdot g_{m}=0.$

This readily implies

$g_{m}=cz^{j}\bar{z}^{n},$

for some constant $c\ $and non-negative integer $j.$ The constant $c$ can be
normalized to be $1.$

Since $f_{2n-1}=0,$ the degree $m-1$ terms in $g$ should satisfy

(6) $\begin{cases}2D_{\bar{z}}f_{2n}\cdot g_{m-1}+\sqrt{3}D_{x}^{2}f_{2n}\cdot
g_{m}=0,\\\ \\\ 2D_{\bar{z}}f_{2n}\cdot g_{m-1}+\sqrt{3}iD_{x}D_{y}f_{2n}\cdot
g_{m}=0.\end{cases}$

We compute

$\displaystyle
D_{x}^{2}\left(z^{n}\bar{z}^{n}\right)\cdot\left(z^{j}\bar{z}^{k}\right)=~{}$
$\displaystyle\left[n\left(n-1\right)-2nj+j\left(j-1\right)\right]z^{j+n-2}\bar{z}^{n+k}$
$\displaystyle+\left[2n^{2}-2nk-2nj+2jk\right]z^{j+n-1}\bar{z}^{n+k-1}$
$\displaystyle+\left[n\left(n-1\right)-2nk+k\left(k-1\right)\right]z^{j+n}\bar{z}^{n+k-2}.$

Similarly, we have

$\displaystyle
D_{x}D_{y}\left(z^{n}\bar{z}^{n}\right)\cdot\left(z^{j}\bar{z}^{k}\right)=~{}$
$\displaystyle\left[n\left(n-1\right)-2nj+j\left(j-1\right)\right]iz^{j+n-2}\bar{z}^{k+n}$
$\displaystyle+\left[-n\left(n-1\right)+2nk-k\left(k-1\right)\right]iz^{j+n}\bar{z}^{k+n-2}.$

From the system $\left(\ref{gm-1}\right),$ we deduce

$D_{x}^{2}f_{2n}\cdot g_{m}=iD_{x}D_{y}f_{2n}\cdot g_{m},$

which implies that when $k=n,$ there holds

$\displaystyle\left[n\left(n-1\right)-2nj+j\left(j-1\right)\right]z^{j+n-2}\bar{z}^{n+k}+\left[n\left(n-1\right)-2nk+k\left(k-1\right)\right]z^{j+n}\bar{z}^{n+k-2}$
$\displaystyle=-\left[n\left(n-1\right)-2nj+j\left(j-1\right)\right]z^{j+n-2}\bar{z}^{k+n}$
$\displaystyle\quad-\left[-n\left(n-1\right)+2nk-k\left(k-1\right)\right]z^{j+n}\bar{z}^{k+n-2}.$

From this we get the relation $\left(\ref{jn}\right)$ between $j$ and $n.$

Now $g_{m-1}$ satisfies

$D_{\bar{z}}f_{2n}\cdot g_{m-1}=-\frac{\sqrt{3}}{2}D_{x}^{2}f_{2n}\cdot
g_{m}=\sqrt{3}nz^{j+n}\bar{z}^{2n-2}.$

Therefore we obtain

$g_{m-1}=\sqrt{3}nz^{j}\bar{z}^{n-1}+cz^{j-1}\bar{z}^{n},$

where $c$ is a constant to be determined.

To find $c,$ we consider the equations to be satisfied by $g_{m-2}:$

$\begin{cases}2D_{\bar{z}}f_{2n}\cdot g_{m-2}+2D_{\bar{z}}f_{2n-2}\cdot
g_{m}+\sqrt{3}D_{x}^{2}f_{2n}\cdot g_{m-1}=0,\\\ \\\ 2D_{\bar{z}}f_{2n}\cdot
g_{m-2}+2D_{\bar{z}}f_{2n-2}\cdot g_{m}+\sqrt{3}iD_{x}D_{y}f_{2n}\cdot
g_{m-1}-D_{x}^{3}f_{2n}\cdot g_{m}=0.\end{cases}$

We compute

$\displaystyle
D_{x}^{3}\left(z^{n}\bar{z}^{n}\right)\cdot\left(z^{j}\bar{z}^{k}\right)$
$\displaystyle=\left[n\left(n-1\right)\left(n-2\right)-3kn\left(n-1\right)+3nk\left(k-1\right)-k\left(k-1\right)\left(k-2\right)\right]z^{j+n}\bar{z}^{k+n-3}$
$\displaystyle\quad+\left[3n^{2}\left(n-1\right)-3jn\left(n-1\right)-6n^{2}k+6njk+3nk\left(k-1\right)-3jk\left(k-1\right)\right]z^{j+n-1}\bar{z}^{k+n-2}$
$\displaystyle\quad+\left[3n^{2}\left(n-1\right)-6n^{2}j-3nk\left(n-1\right)+6njk+3nj\left(j-1\right)-3j\left(j-1\right)k\right]z^{j+n-2}\bar{z}^{k+n-1}$
$\displaystyle\quad+\left[n\left(n-1\right)\left(n-2\right)-3nj\left(n-1\right)+3nj\left(j-1\right)-j\left(j-1\right)\left(j-2\right)\right]z^{j+n-3}\bar{z}^{k+n}.$

We find that if $j$ and $n$ satisfy $\left(\ref{jn}\right),$ then

$D_{x}^{3}\left(z^{n}\bar{z}^{n}\right)\cdot\left(z^{j}\bar{z}^{n}\right)=\left(6jn-6n^{2}\right)z^{j+n-1}\bar{z}^{2n-2}+2\left(j-n\right)\left(j+n-1\right)z^{j+n-3}\bar{z}^{2n},$

and moreover,

$\displaystyle\sqrt{3}D_{x}^{2}f_{2n}\cdot
g_{m-1}-\sqrt{3}iD_{x}D_{y}f_{2n}\cdot g_{m-1}+D_{x}^{3}f_{2n}\cdot g_{m}$
$\displaystyle=2\sqrt{3}D_{x}D_{z}\left(z^{n}\bar{z}^{n}\right)\left(\sqrt{3}nz^{j}\bar{z}^{n-1}+cz^{j-1}\bar{z}^{n}\right)+D_{x}^{3}\left(z^{n}\bar{z}^{n}\right)\left(z^{j}\bar{z}^{n}\right)$
$\displaystyle=2\sqrt{3}c\left(2n-2j+2\right)z^{j+n-3}\bar{z}^{2n}+2\left(j-n\right)\left(j+n-1\right)z^{j+n-3}\bar{z}^{k+n}.$

Compatibility of the two equations in $\left(\ref{Back2}\right)$ requires the
right hand side to be $0.$ It follows that

$c=-\frac{\left(j-n\right)\left(j+n-1\right)}{2\sqrt{3}\left(n-j+1\right)}.$

This proves the assertion. ∎

This lemma tells us that if $n=1,$ then the degree of $g$ has to be $1$ or
$4.$ We now proceed to construct explicit Backlund transformations from

(7) $f=x^{2}+y^{2}+3,$

the first nontrivial solution, to a family of degree $4$ polynomial. We will
see that there will be a free complex parameter appearing in the process.

######  Proposition 4.

Let $f\ $be given by $\left(\ref{deg2}\right)$ and

$g=z^{3}\bar{z}+\sqrt{3}z^{3}+\sqrt{3}z^{2}\bar{z}+12z^{2}-3\bar{z}^{2}+3z\bar{z}+9\sqrt{3}z+\alpha\bar{z}-36+\sqrt{3}\alpha,$

where $\alpha\in\mathbb{C}$ is a parameter. Then $f,g$ satisfies

$\begin{cases}\left(D_{x}^{2}+\frac{1}{\sqrt{3}}D_{x}+\frac{i}{\sqrt{3}}D_{y}\right)f\cdot
g=0,\\\ \\\ \left(D_{x}^{3}-\sqrt{3}iD_{x}D_{y}-D_{x}-iD_{y}\right)f\cdot
g=0.\end{cases}$

As a consequence, $g$ is a solution to the bilinear equation
$\left(\ref{bilinear}\right)$.

###### Proof.

Note that $f_{2}=z\bar{z}$ and $f_{0}=3.$ Applying Lemma 3, we find
$g_{4}=z^{3}\bar{z}$ and $g_{3}=\sqrt{3}z^{3}+\sqrt{3}z^{2}\bar{z}.$

To get $g_{2},$ we observe that it satisfies

$2D_{\bar{z}}f_{2}\cdot g_{2}+2D_{\bar{z}}f_{0}\cdot
g_{4}+\sqrt{3}D_{x}^{2}f_{2}\cdot g_{3}=0.$

Solving this equation gives

$g_{2}=12z^{2}-3\bar{z}^{2}+az\bar{z},$

where $a$ is a parameter still to be determined in the next step.

To find $a$ and $g_{1},$ we use the fact that $g_{1}$ satisfies

$\begin{cases}2D_{\bar{z}}f_{2}\cdot g_{1}+2D_{\bar{z}}f_{0}\cdot
g_{3}+\sqrt{3}D_{x}^{2}f_{2}\cdot g_{2}+\sqrt{3}D_{x}^{2}f_{0}\cdot
g_{4}=0,\\\ \\\ 2D_{\bar{z}}f_{2}\cdot g_{1}+2D_{\bar{z}}f_{0}\cdot
g_{3}+\sqrt{3}iD_{x}D_{y}f_{2}\cdot g_{2}+\sqrt{3}iD_{x}D_{y}f_{0}\cdot
g_{4}-D_{x}^{3}f_{2}\cdot g_{3}=0.\end{cases}$

Compatibility of these two equations implies that $a=3$ and

$g_{2}=12z^{2}-3\bar{z}^{2}+3z\bar{z}.$

Now solving the following equation for $g_{1}:$

$2D_{\bar{z}}f_{2}\cdot g_{1}+2D_{\bar{z}}f_{0}\cdot
g_{3}+\sqrt{3}D_{x}^{2}f_{2}\cdot g_{2}+\sqrt{3}D_{x}^{2}f_{0}\cdot g_{4}=0,$

we get

$g_{1}=9\sqrt{3}z+\alpha\bar{z},$

where $\alpha$ is a parameter.

To see whether or not $\alpha$ can be arbitrary, we consider the equations
satisfied by $g_{0}:$

$\begin{cases}2D_{\bar{z}}f_{2}\cdot g_{0}+2D_{\bar{z}}f_{0}\cdot
g_{2}+\sqrt{3}D_{x}^{2}f_{2}\cdot g_{1}+\sqrt{3}D_{x}^{2}f_{0}\cdot
g_{3}=0,\\\ \\\ 2D_{\bar{z}}f_{2}\cdot g_{0}+2D_{\bar{z}}f_{0}\cdot
g_{2}+\sqrt{3}iD_{x}D_{y}f_{2}\cdot g_{1}+\sqrt{3}iD_{x}D_{y}f_{0}\cdot
g_{3}-D_{x}^{3}f_{2}\cdot g_{2}-D_{x}^{3}f_{0}\cdot g_{4}=0.\end{cases}$

Direct computation tells us that this system is compatible for any $\alpha,$
and we are lead to

$2D_{\bar{z}}f_{2}\cdot g_{0}+\left(72-2\sqrt{3}\alpha\right)z=0,$

which implies that $g_{0}=-36+\sqrt{3}\alpha.$ One then checks that the
function $g$ obtained in this way indeed solves $\left(\ref{Back2}\right).$
This finishes the proof. ∎

###  2.1. A family of degree $6$ tau functions and their Backlund
transformation

We have obtained a family of degree $4$ polynomials:

$g=z^{3}\bar{z}+\sqrt{3}z^{3}+\sqrt{3}z^{2}\bar{z}+12z^{2}-3\bar{z}^{2}+3z\bar{z}+9\sqrt{3}z+\alpha\bar{z}-36+\sqrt{3}\alpha.$

We would like to find all the degree $6$ polynomial $h$ such that $g,h$ are
connected through the Backlund transformation. They are supposed to satisfy
the following system

$\begin{cases}\left(D_{x}-iD_{y}-\sqrt{3}D_{x}^{2}\right)g\cdot h=0,\\\ \\\
\left(D_{x}-iD_{y}+\sqrt{3}iD_{x}D_{y}-D_{x}^{3}\right)g\cdot h=0.\end{cases}$

In the $z$-$\bar{z}$ coordinate, it takes the form:

(8) $\begin{cases}\left(2D_{z}-\sqrt{3}D_{x}^{2}\right)g\cdot h=0,\\\ \\\
\left(2D_{z}+\sqrt{3}iD_{x}D_{y}-D_{x}^{3}\right)g\cdot h=0.\end{cases}$

The following result provides the explicit formula of a family of degree 6
polynomial solutions.

######  Proposition 5.

Let $\alpha,\beta$ be parameters and

$\displaystyle h=~{}$ $\displaystyle
z^{3}\bar{z}^{3}-2\sqrt{3}z^{2}\bar{z}^{3}+2\sqrt{3}z^{3}\bar{z}^{2}-3z^{4}+15z^{2}\bar{z}^{2}+6z\bar{z}^{3}-3\bar{z}^{4}+6z^{3}\bar{z}$
$\displaystyle+\beta
z^{3}+24\sqrt{3}z^{2}\bar{z}-24\sqrt{3}z\bar{z}^{2}+\left(3\sqrt{3}+\alpha\right)\bar{z}^{3}$
$\displaystyle-\left(90+2\sqrt{3}\right)z^{2}+63z\bar{z}-\left(72-2\sqrt{3}\alpha\right)\bar{z}^{2}$
$\displaystyle+\left(189\sqrt{3}-3\alpha+6\beta\right)z+\left(-180\sqrt{3}+6\alpha-3\beta\right)\bar{z}$
$\displaystyle+1161-6\sqrt{3}\alpha+9\sqrt{3}\beta+\alpha\beta.$

Then $g,h$ satisfy $\left(\ref{gh}\right).$

###### Proof.

The highest degree terms of $h$ has to be $h_{6}=z^{3}\bar{z}^{3}.$ The
$h_{5}$ term can be obtained by solving the equation:

$2D_{z}g_{4}\cdot h_{5}=\sqrt{3}D_{x}^{2}g_{4}\cdot h_{6}-2D_{z}g_{3}\cdot
h_{6},$

which gives

$h_{5}=-2\sqrt{3}z^{2}\bar{z}^{3}+cz^{3}\bar{z}^{2},$

where $c$ is a parameter to be determined using information of $h_{4},$ which
satisfies

$2D_{z}g_{4}\cdot h_{4}+2D_{z}g_{3}\cdot h_{5}+2D_{z}g_{2}\cdot
h_{6}-\sqrt{3}D_{x}^{2}g_{4}\cdot h_{5}-\sqrt{3}D_{x}^{2}g_{3}\cdot h_{6}=0,$

and

$\displaystyle 2D_{z}g_{4}\cdot h_{4}+2D_{z}g_{3}\cdot h_{5}+2D_{z}g_{2}\cdot
h_{6}+\sqrt{3}iD_{x}D_{y}g_{4}\cdot h_{5}$
$\displaystyle+\sqrt{3}iD_{x}D_{y}g_{3}\cdot h_{6}-D_{x}^{3}g_{4}\cdot
h_{6}=0.$

The compatibility of these two equations gives $c=2\sqrt{3}\ $and hence

$h_{5}=-2\sqrt{3}z^{2}\bar{z}^{3}+2\sqrt{3}z^{3}\bar{z}^{2}.$

With $h_{5}$ at hand, $h_{4}$ can be found by solving the equation

$2D_{z}g_{4}\cdot h_{4}+2D_{z}g_{3}\cdot h_{5}+2D_{z}g_{2}\cdot
h_{6}-\sqrt{3}D_{x}^{2}g_{4}\cdot h_{5}-\sqrt{3}D_{x}^{2}g_{3}\cdot h_{6}=0.$

This gives

$h_{4}=-3z^{4}+15z^{2}\bar{z}^{2}+6z\bar{z}^{3}-3\bar{z}^{4}+cz^{3}\bar{z}.$

Again, $c$ is a parameter to be determined, using the equation of $h_{3}.$

Now $h_{3}$ satisfies

$\displaystyle 2D_{z}g_{4}\cdot h_{3}+2D_{z}g_{3}\cdot h_{4}+2D_{z}g_{2}\cdot
h_{5}+2D_{z}g_{1}\cdot h_{6}$ $\displaystyle-\sqrt{3}D_{x}^{2}g_{4}\cdot
h_{4}-\sqrt{3}D_{x}^{2}g_{3}\cdot h_{5}-\sqrt{3}D_{x}^{2}g_{2}\cdot h_{6}=0,$

and

$\displaystyle 2D_{z}g_{4}\cdot h_{3}+2D_{z}g_{3}\cdot h_{4}+2D_{z}g_{2}\cdot
h_{5}+2D_{z}g_{1}\cdot h_{6}+\sqrt{3}iD_{x}D_{y}g_{4}\cdot h_{4}$
$\displaystyle+\sqrt{3}iD_{x}D_{y}g_{3}\cdot
h_{5}+\sqrt{3}iD_{x}D_{y}g_{2}\cdot h_{6}-D_{x}^{3}g_{4}\cdot
h_{5}-D_{x}^{3}g_{3}\cdot h_{6}=0.$

The compatibility implies that $c=6,$ and we deduce that

$h_{4}=-3z^{4}+15z^{2}\bar{z}^{2}+6z\bar{z}^{3}-3\bar{z}^{4}+6z^{3}\bar{z}.$

Having obtained $h_{4},$ we then proceed to solve the equation for $h_{3}$ and
find that for some parameter $\beta,$

$h_{3}=24\sqrt{3}z^{2}\bar{z}-24\sqrt{3}z\bar{z}^{2}+\left(3+\frac{\alpha}{\sqrt{3}}\right)\sqrt{3}\bar{z}^{3}+\beta
z^{3}.$

The rest of terms $h_{2},h_{1},h_{0}$ follow from routine computation, and it
turns out that $\beta$ is a free parameter. We omit the details. ∎

Observe that the function $h$ given by Proposition 5 is not real valued. But
we are mainly interested in real valued solutions. In view of this, we replace
$y$ by $y-\frac{2\sqrt{3}i}{3},$ and choose the complex parameters
$\alpha,\beta$ such that

$\displaystyle\alpha-\beta$ $\displaystyle=Ai-\frac{211}{3\sqrt{3}},$
$\displaystyle\alpha+\beta$ $\displaystyle=B-3\sqrt{3},$

where $A,B$ are real numbers. Then the function $h$ becomes

$\displaystyle h_{A,B}\left(x,y\right):=~{}$ $\displaystyle
x^{6}+3x^{4}y^{2}+3x^{2}y^{4}+y^{6}+25x^{4}+90x^{2}y^{2}+17y^{4}$
$\displaystyle+Bx^{3}+3Ax^{2}y-3Bxy^{2}-Ay^{3}-125x^{2}+475y^{2}$
$\displaystyle-Bx+5Ay+1875+\frac{A^{2}}{4}+\frac{B^{2}}{4}.$

This is a family of real valued solution. Note that translation along the $x$
or $y$ direction still yields a solution(the rotation will not), hence there
are all together $4$ free real parameters in the whole family of solutions.
Now if

$A=0\text{ and }B=0,$

then $h$ will be an even solution, which equals

$\displaystyle
x^{6}+3x^{4}y^{2}+3x^{2}y^{4}+y^{6}+25x^{4}+90x^{2}y^{2}+17y^{4}$
$\displaystyle-125x^{2}+475y^{2}+1875.$

Note that this function is not radially symmetry. For general parameters
$A,B$, the solution does not have any symmetry.

We should emphasize that although $h_{A,B}$ is real valued, it is connected by
Backlund transformation, via a degree $4$ polynomial, to the following degree
$2$ polynomial:

$x^{2}+\left(y-\frac{2\sqrt{3}i}{3}\right)^{2}+3,$

which is not real valued.

###  2.2. The general case

Our construction can then be iterated to created higher energy solutions(Note
that energy is completely determined by the degree, [9]). One can indeed
directly write down an algorithm to do this computation. At each stage, one
find the solutions from their highest degree terms to lower degree terms. More
precisely, suppose we have already found a polynomial solution $f$ to the
bilinear equation, with highest degree term equals $z^{n}\bar{z}^{n},$ and we
want to find its Backlund transformation $g.$ Then once we have found $g_{j}$
for $j>m,$ then from the Backlund transformation, we see that $g_{m}$ will
satisfy a system of equations of the form

$\displaystyle D_{\bar{z}}\left(z^{n}\bar{z}^{n}\right)\cdot g_{m}$
$\displaystyle=\text{RHS}_{1}\text{,}$ $\displaystyle
D_{\bar{z}}\left(z^{n}\bar{z}^{n}\right)\cdot g_{m}$
$\displaystyle=\text{RHS}_{2}\text{,}$

where RHS1 and RHS2 contain terms explicitly polynomial terms from $g_{j}$
with $j>m.$

An important question is, in which form, should a free parameter appear.
Assume that the term $g_{m+1}$ has a parameter term $\sigma z^{j}\bar{z}^{n}$
to be determined, where $j+n=m+1.$ The compatibility of these two equations is
RHS${}_{1}-$RHS${}_{2}=0.$ In this equation, the parameter $\sigma$ appears as

$\sigma\left(D_{x}^{2}-iD_{x}D_{y}\right)\left(z^{n}\bar{z}^{n}\right)\cdot\left(z^{j}\bar{z}^{n}\right).$

Direct computation tells us that this equals

$\sigma\left[n\left(n-1\right)-2nj+j\left(j-1\right)\right]z^{j+n-2}\bar{z}^{2n}.$

Hence we conclude that free parameter(that is, no restriction on $\sigma$) can
occur only if $j$ satisfies

$n\left(n-1\right)-2nj+j\left(j-1\right)=0.$

In the case of $n=3,$ the free parameter term is $\sigma z\bar{z}^{3}.$

Next, let us assume that the polynomial solution $g,$ with highest degree term
$z^{p}\bar{z}^{n}$ has been found. We would like to find function $h$ by
another Backlund transformation. Similar as above, once we have found $h_{j}$
for $j>m,$ then $h_{m}$ will satisfy a system of equations of the form

$\displaystyle D_{z}\left(z^{p}\bar{z}^{n}\right)\cdot h_{m}$
$\displaystyle=\text{RHS}_{1}\text{,}$ $\displaystyle
D_{z}\left(z^{p}\bar{z}^{n}\right)\cdot h_{m}$
$\displaystyle=\text{RHS}_{2}\text{.}$

Assume that $h_{m+1}$ has a parameter term $\alpha z^{p}\bar{z}^{j}$ to be
determined. Then $\alpha$ enters into the compatibility condition as

$\alpha\left(D_{x}^{2}+iD_{x}D_{y}\right)\left(z^{p}\bar{z}^{n}\right)\cdot\left(z^{p}\bar{z}^{j}\right).$

This equals

$\alpha\left[n\left(n-1\right)-2nj+j\left(j-1\right)\right]z^{2p}\bar{z}^{j+n-2}.$

Therefore, again, free parameter can occur only if

$n\left(n-1\right)-2nj+j\left(j-1\right)=0.$

In the case of $n=3,$ the free parameter term is $\alpha z^{6}\bar{z}.$

######  Remark 6.

Our algorithm tells us that, at least formally, if we consider those
solutions(complex valued) whose leading term is $z^{n}\bar{z}^{n},$ where
$n=\frac{1}{2}k\left(k+1\right),$ then the space of these solutions should
have complex dimension $2k.$ Moreover, the space of real valued solutions with
degree $k\left(k+1\right)$ is expected to have real dimension $2k.$

##  3\. Nondegeneracy and Morse index of degree $6$ solutions

In this section, we will show that the family of real valued solutions
$u_{A,B}$ to the Boussinesq equation corresponding to $h_{A,B}$ have Morse
index $3.$

Our starting point in the computation of Morse index is to analyze the
asymptotic behavior of $u=u_{0,B}$ for $B$ large.

In view of the fact that

$u=2\partial_{x}^{2}\ln
h_{0,B}=2\frac{h_{0,B}\partial_{x}^{2}h_{0,B}-\left(\partial_{x}h_{0,B}\right)^{2}}{h_{0,B}^{2}},$

for $B$ large, the maximum of $u$ should take place around the points
$\left(x,y\right)$ which solve the system of algebraic equations:

(9) $\begin{cases}\phi=0,\\\ \partial_{x}\phi=0,\end{cases}$

where $\phi$ designates the main order term of $h_{0,B}$(away from the maximum
of $u$) and is defined by

$\phi\left(x,y\right)=x^{6}+3x^{4}y^{2}+3x^{2}y^{4}+y^{6}+Bx^{3}-3Bxy^{2}+\frac{B^{2}}{4}.$

Solving $\left(\ref{peak}\right)$ and setting
$\gamma=\left(\frac{B}{2}\right)^{\frac{1}{3}},$ we obtain the following three
points $P_{j}$ on the $\left(x,y\right)$ plane:

(10) $P_{1}=\left(-\gamma,0\right),\text{
}P_{2}=\frac{1}{2}\left(\gamma,-\sqrt{3}\gamma\right),\text{
}P_{3}=\frac{1}{2}\left(\gamma,\sqrt{3}\gamma\right).$

These three points are the vertices of an equilateral triangle. This is in
agreement with the formal computation for the dynamics of peaks of KP-I
equation, carried out in [9]. The reason that $P_{j}$ are in this position
will be clear later on.

Let us set

$L\left(x,y\right)=x^{2}+y^{2}+3.$

Recall that we use $U$ to denote the classical lump solution. That is,

(11) $U\left(x,y\right)=2\partial_{x}^{2}\ln
L=4\frac{y^{2}-x^{2}+3}{\left(x^{2}+y^{2}+3\right)^{2}}.$

Then we define

$L_{j}\left(\cdot\right)=L\left(\cdot-P_{j}\right),\text{ and
}U_{j}=2\partial_{x}^{2}L_{j},\text{ \ }j=1,2,3.$

The following result describes the asymptotic behavior of $u$ as $B$(or
$\gamma$) tends to $+\infty.$

######  Lemma 7.

The error between $u$ and $U_{1}+U_{2}+U_{3}$ satisfies

$\left\|u-\left(U_{1}+U_{2}+U_{3}\right)\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\rightarrow
0,\text{ as }B\rightarrow+\infty.$

###### Proof.

We have

$\displaystyle u-\left(U_{1}+U_{2}+U_{3}\right)$
$\displaystyle=2\partial_{x}^{2}\left(\ln h-\ln L_{1}-\ln L_{2}-\ln
L_{3}\right)$
$\displaystyle=-2\partial_{x}^{2}\left[\ln\left(1-\frac{h-L_{1}L_{2}L_{3}}{h}\right)\right].$

Direct computation tells us that

$\displaystyle\eta:=~{}$ $\displaystyle h-L_{1}L_{2}L_{3}$ $\displaystyle=~{}$
$\displaystyle 16x^{4}+72x^{2}y^{2}+8y^{4}-\left(152+9\gamma^{2}\right)x^{2}$
$\displaystyle+\left(448-9\gamma^{2}\right)y^{2}-Bx+1848-27\gamma^{2}-9\gamma^{4}.$

Observe that at $P_{j},$ the main contribution to $h$ comes from those degree
$4$ terms. Therefore, to estimate the error around $P_{j},$ we need to have a
better control of $\eta$ at $P_{j}.$ It turns out that
$\eta\left(P_{j}\right)$ is of the order $O\left(\gamma^{2}\right).$ More
precisely,

$\displaystyle\eta\left(P_{1}\right)=-179\gamma^{2}+1848,\quad\eta\left(P_{2}\right)=\eta\left(P_{3}\right)=271\gamma^{2}+1848.$

From these, we then conclude with little work that

$\left\|\frac{h-L_{1}L_{2}L_{3}}{h}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\rightarrow
0,$

which readily implies

$\left\|u-\left(U_{1}+U_{2}+U_{3}\right)\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\rightarrow
0,\text{ as }B\rightarrow+\infty.$

This finishes the proof. ∎

Lemma 7 provides a rough picture of the solution $u.$ However, to analyze its
Morse index, we need more precise expansion of $u.$ To obtain the required
expansion, it turns out that the explicit form of the function $h$ does not
help too much. Therefore we use the mapping property of the linearized
Boussinesq operator. This will be done in sequel.

For a function $q,$ we use $\mathcal{L}_{q}$ to denote the linearized
Boussinesq operator at $q,$ with the following form:

$\mathcal{L}_{q}\eta=\partial_{x}^{2}\eta-\eta+6q\eta-\partial_{x}^{-2}\partial_{y}^{2}\eta.$

Here $\partial_{x}^{-1}=\int_{-\infty}^{x}.$ One of the reason that we
integrate twice in the original form of the Boussinesq equation is that the
operator $\mathcal{L}_{q}$ is self adjoint. We emphasize that in view of the
definition of $\partial_{x}^{-1},$ one should be very careful about the
integrability of the function.

We let $\left(x_{j}^{\ast},y_{j}^{\ast}\right)$ be the point close to
$P_{j}=\left(\tilde{x}_{j},\tilde{y}_{j}\right)$, which is introduced in
$\left(\ref{Pj}\right).$ Setting

$U_{j}^{\ast}\left(x,y\right):=U\left(x-x_{j}^{\ast},y-y_{j}^{\ast}\right)$

and writing

$u=U^{\ast}+\xi,$

where $U^{\ast}$ is the “approximate” solution defined by

$U^{\ast}=U_{1}^{\ast}+U_{2}^{\ast}+U_{3}^{\ast},$

and $\xi$ is a perturbation term satisfying the following orthogonality
condition: For $k=1,2,3,$

$\int_{\mathbb{R}^{2}}\xi\partial_{x}U_{k}^{\ast}dxdy=0\text{ \ \ and \
}\int_{\mathbb{R}^{2}}\xi\partial_{y}U_{k}^{\ast}dxdy=0.$

This can always be achieved by perturbing
$\left(\tilde{x}_{k},\tilde{y}_{k}\right)$ into
$\left(x_{k}^{\ast},y_{k}^{\ast}\right).$ Note that
$\partial_{x}U,\partial_{y}U$ are kernels of the operator $\mathcal{L}_{U}.$
Hence $\partial_{x}U_{k}^{\ast},\partial_{y}U_{k}^{\ast}$ are “approximate”
kernels of $\mathcal{L}_{U^{\ast}}.$

By Lemma 7, if the distance between $\left(x_{j}^{\ast},y_{j}^{\ast}\right)$
and $P_{j}$ is close enough, then
$\left\|\xi\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}$ will be small,
provided that $B$ is large. Since $u$ satisfies the Boussinesq equation, the
perturbation $\xi$ should satisfy the following nonlinear equation

(12) $\mathcal{L}_{U^{\ast}}\xi=-E\left(U^{\ast}\right)-3\xi^{2},$

where $E\left(U^{\ast}\right)$ is the “error” of the approximate solution
$U^{\ast}:$

$\displaystyle E\left(U^{\ast}\right)$
$\displaystyle=\partial_{x}^{2}U^{\ast}-U^{\ast}+3U^{\ast
2}-\partial_{x}^{-2}\partial_{y}^{2}U^{\ast}$ (13)
$\displaystyle=6\left(U_{1}^{\ast}U_{2}^{\ast}+U_{2}^{\ast}U_{3}^{\ast}+U_{1}^{\ast}U_{3}^{\ast}\right).$

We see that essentially $E\left(U^{\ast}\right)$ gives the interaction between
$U_{j}^{\ast},$ and therefore the presence of $\xi$ is due to this
interaction. $E\left(U^{\ast}\right)$ is of the order
$O\left(\gamma^{-2}\right).$

We introduce the complex numbers
$z_{j}^{\ast}=x_{j}^{\ast}+iy_{j}^{\ast},j=1,2,3,$ and define

(14) $d^{\ast}=24\int_{\mathbb{R}^{2}}U^{2}dxdy.$

An important ingredient of the analysis is to understand the projection of the
error $E=E\left(U^{\ast}\right)$ onto the kernels. This is the content of the
following

######  Lemma 8.

There holds

(15)
$\int_{\mathbb{R}^{2}}E\partial_{x}U_{j}^{\ast}dxdy=-d^{\ast}\sum\limits_{k\neq
j}\operatorname{Re}\frac{1}{\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{3}}+O\left(\gamma^{-4}\right),\quad
j=1,2,3,$

and

(16)
$\int_{\mathbb{R}^{2}}E\partial_{y}U_{j}^{\ast}dxdy=d^{\ast}\sum\limits_{k\neq
j}\operatorname{Im}\frac{1}{\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{3}}+O\left(\gamma^{-4}\right),\quad
j=1,2,3.$

###### Proof.

We shall prove (15) and (16) for the case $j=1$, the other ones can be treated
similarly. For the former one, using (13) we get

$\int_{\mathbb{R}^{2}}E\partial_{x}U_{1}^{\ast}dxdy=6\int_{\mathbb{R}^{2}}\left(U_{1}^{\ast}U_{2}^{\ast}+U_{1}^{\ast}U_{3}^{\ast}+U_{2}^{\ast}U_{3}^{\ast}\right)\partial_{x}U_{1}^{\ast}dxdy.$

Let us compute each integral appeared in the right hand side. Integrating by
parts, we get

(17)
$\int_{\mathbb{R}^{2}}U_{2}^{\ast}U_{1}^{\ast}\partial_{x}U_{1}^{\ast}dxdy=-\frac{1}{2}\int_{\mathbb{R}^{2}}U_{1}^{\ast
2}\partial_{x}U_{2}^{\ast}dxdy.$

To estimate this integral, we need the identity

(18)
$\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}=-\frac{1}{2}\left(\frac{1}{z^{2}}+\frac{1}{\bar{z}^{2}}\right),$

which implies

(19)
$\partial_{x}\left(\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right)=2\operatorname{Re}\frac{1}{z^{3}},\quad\partial_{y}\left(\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right)=-2\operatorname{Im}\frac{1}{z^{3}},$
$\partial_{x}^{2}\left(\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right)=-6\operatorname{Re}\frac{1}{z^{4}},\quad\partial_{y}^{2}\left(\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right)=6\operatorname{Re}\frac{1}{z^{4}},$

and

$\partial_{x}\partial_{y}\left(\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right)=6\operatorname{Im}\frac{1}{z^{4}}.$

We then use the explicit form of the lump solution $U$ to deduce that within a
ball of radius $\frac{\gamma}{2}$ centered at $P_{1},$

$\left|\partial_{x}U_{2}^{\ast}-8\operatorname{Re}\frac{1}{z^{3}}\right|=O\left(\gamma^{-4}\right).$

Inserting this estimate into $\left(\ref{I1}\right),$ we then get

$6\int_{\mathbb{R}^{2}}U_{2}^{\ast}U_{1}^{\ast}\partial_{x}U_{1}^{\ast}dxdy=-d^{\ast}\operatorname{Re}\frac{1}{\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{3}}+O\left(\gamma^{-4}\right),$

where $d^{\ast}$ is defined in $\left(\ref{cstar}\right).$

Similarly,

$6\int_{\mathbb{R}^{2}}U_{3}^{\ast}U_{1}^{\ast}\partial_{x}U_{1}^{\ast}dxdy=-d^{\ast}\operatorname{Re}\frac{1}{\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{3}}+O\left(\gamma^{-4}\right).$

Moreover, we use the decay of $U_{2}^{\ast},U_{3}^{\ast}$ to conclude directly
that

$\int_{\mathbb{R}^{2}}U_{2}^{\ast}U_{3}^{\ast}\partial_{x}U_{1}^{\ast}dxdy=O\left(\gamma^{-4}\right).$

Combining all these estimates, we then get

$\int_{\mathbb{R}^{2}}E\partial_{x}U_{1}^{\ast}dxdy=-d^{\ast}\operatorname{Re}\frac{1}{\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{3}}-d^{\ast}\operatorname{Re}\frac{1}{\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{3}}+O\left(\gamma^{-4}\right).$

The equation $\left(\ref{proony}\right)$ can be obtained in a very similar
way, using the second identity of $\left(\ref{dy}\right).$ ∎

In view of Lemma 8, the position $z_{j}^{\ast}$ of the single lumps should
approximately satisfy the following balancing condition: For each fixed $j,$

(20) $\sum\limits_{k\neq
j}\frac{1}{\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{3}}=0.$

As we have mentioned above, this condition has already been observed in
Section 3 of [9] for the KP-I equation, from a more physically inspired point
of view. On the other hand, the space $M$ of points $z_{j}^{\ast}$ satisfying
these balancing equations $\left(\ref{balan}\right)$ has been investigated in
[3], where rational solutions of the KdV equation has been studied.

Taking into account of the previous computation, we shall define the map

$\mathcal{F}:\left(z_{1},z_{2},z_{3}\right)^{T}\rightarrow\left(F_{1},F_{2},F_{3}\right)^{T},\text{
}z_{j}\in\mathbb{C},$

where

$F_{j}={\displaystyle\sum\limits_{k\neq
j}}\frac{1}{\left(z_{j}-z_{k}\right)^{3}},\text{ }j=1,2,3.$

Note that for
$\tilde{z}_{1}:=-1,\tilde{z}_{2}:=\frac{1+\sqrt{3}i}{2},\tilde{z}_{3}:=\frac{1-\sqrt{3}i}{2},$
we have

$\mathcal{F}\left(\tilde{z}_{1},\tilde{z}_{2},\tilde{z}_{3}\right)=0.$

The linearization of $\mathcal{F}$ will play important role in our later
analysis. The derivative $D\mathcal{F}$ of $\mathcal{F}$ at
$\left(z_{1},z_{2},z_{3}\right)$ is a matrix of the form
$\left[F_{j,k}\right],$ where $F_{j,k}=\partial_{z_{k}}F_{j}$.

The next lemma follows from direct computation of eigenvectors.

######  Lemma 9.

The kernels of $M:=$
$D\mathcal{F}|_{\left(\tilde{z}_{1},\tilde{z}_{2},\tilde{z}_{3}\right)}$ are
given by

$c_{1}\mathbf{b}_{1}+c_{2}\mathbf{b}_{2},$

where $c_{1},c_{2}$ are complex numbers and

$\mathbf{b}_{1}=\left(1,1,1\right)^{T},\text{
}\mathbf{b}_{2}=\left(\tilde{z}_{1},\tilde{z}_{2},\tilde{z}_{3}\right)^{T}.$

We remark that the vector $\mathbf{b}_{1}$ reflects the translation invariance
of the system, and $\mathbf{b}_{2}$ is corresponding to scaling and
rotation(multiplication by a complex number $c$).

To proceed, we use $d_{k}\left(x,y\right)$ to denote the distance between
$\left(x,y\right)$ and $\left(x_{k}^{\ast},y_{k}^{\ast}\right).$ Let

$\theta_{\alpha}\left(x,y\right)=\left(\sum\limits_{k=1}^{3}\left(1+d_{k}\right)^{-1}\right)^{-\alpha}.$

We need some apriori estimates for the linearized operator.

######  Lemma 10.

Let $\varepsilon>0$ be a fixed small constant. Suppose $\eta$ satisfies
$\left\|\eta\theta_{\varepsilon}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}<+\infty$
and

(21) $\mathcal{L}_{U^{\ast}}\eta=f,$

where
$\left\|f\theta_{2}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}<+\infty.$
Assume for $k=1,2,3,$

$\int_{\mathbb{R}^{2}}\eta\partial_{x}U_{k}^{\ast}dxdy=0,\text{ \ and
}\int_{\mathbb{R}^{2}}\eta\partial_{y}U_{k}^{\ast}dxdy=0.$

Then for any $\sigma\in\left(0,2-\varepsilon\right),$ there holds

$\left\|\eta\theta_{\sigma}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq
C\left\|f\theta_{2}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)},$

where $C$ is independent of $\sigma$ and $f.$

###### Proof.

Consider the cone $\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3}$ with
vertex at the origin containing those points whose angle coordinate are in the
range
$\left(\frac{2\pi}{3},\frac{4\pi}{3}\right),\left(\frac{2\pi}{3},2\pi\right),\left(0,\frac{2\pi}{3}\right),$
respectively. Then $P_{k}\in\mathcal{C}_{k}.$ Let $\rho_{1},\rho_{2},\rho_{3}$
be a partition of unity such that $\rho_{k}$ equals $1$ in most part of
$\mathcal{C}_{k}$ and $\nabla\rho_{k}$ is supported in a radius $1$ tubular
neighborhood of $\partial\mathcal{C}_{k}.$

We rewrite the equation $\left(\ref{etaf}\right)$ into the form

$\mathcal{L}_{0}\eta:=\partial_{x}^{2}\eta-\eta-\partial_{x}^{-2}\partial_{y}^{2}\eta=f-6U^{\ast}\eta.$

Hence $\eta=\eta_{1}+\eta_{2}+\eta_{3},$ where $f_{k}=\rho_{k}f$ and
$\eta_{k}$ is determined by the equation

$\mathcal{L}_{0}\eta_{k}=f_{k}-6U_{k}^{\ast}\eta.$

Observe that $\mathcal{L}_{0}$ is a hypo-elliptic operator with constant
coefficient, and the decay properties of its Green function $K$ have been
established in [5]. In particular,

$\left\|r^{2}K\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq C.$

Using this decay estimate, we obtain

$\left\|\left(1+d_{k}\right)^{\sigma}\eta_{k}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq
C\left\|\left(1+d_{k}\right)^{2}f_{k}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}+C\left\|\left(1+d_{k}\right)^{\sigma}U_{k}^{\ast}\eta\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}.$

It follows that for some $r_{0}$ sufficiently large,

$\left\|\left(1+d_{k}\right)^{\sigma}\eta_{k}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq
C\left\|\left(1+d_{k}\right)^{2}f_{k}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}+C\left\|\eta\right\|_{L^{\infty}\left(B_{r_{0}}\left(P_{k}\right)\right)}.$

Now we claim

$\left\|\eta\right\|_{L^{\infty}\left(B_{r_{0}}\left(P_{k}\right)\right)}\leq
C\left\|\left(1+d_{k}\right)^{2}f_{k}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}.$

Otherwise, there would exist a sequence of
$\eta\left(x-x_{k}^{\ast},y-y_{k}^{\ast}\right)$ which converges to solution
of the equation

$\mathcal{L}_{U}\eta_{\infty}=0.$

However, this contradicts with the nondegeneracy of lump solution [14] and the
assumption that $\eta$ is orthogonal to the
$\partial_{x}U_{k}^{\ast},\partial_{y}U_{k}^{\ast}.$ ∎

The next lemma deals with the explicit expression of the function related to
the main order correction of the approximate solution.

######  Lemma 11.

Let $U$ be the classical lump solution defined by $\left(\ref{lumpc}\right)$
and

$\omega=\partial_{x}\left[\frac{24x\left(y^{2}-3\right)}{\left(x^{2}+y^{2}+3\right)^{2}}\right].\
$

Then

(22) $\mathcal{L}_{U}\omega=-6U,\text{ and
}\mathcal{L}_{U}\left[\partial_{x}\omega\right]=-6\partial_{x}U-6\partial_{x}U\omega.$

###### Proof.

This follows from direct computation. Note that $\omega$ decays at the rate
$O\left(r^{-2}\right)$ at infinity. ∎

With Lemma 11 being understood, we introduce the notation

$\mathbf{p}_{k}=-2\sum\limits_{j\neq
k}\operatorname{Re}\frac{1}{\left(z_{k}^{\ast}-z_{j}^{\ast}\right)^{2}},\text{
\ \ for \ \ }k=1,2,3.$

From the explicit formula of $U$ and $\left(\ref{ident}\right),$ we deduce

$U_{2}^{\ast}\left(x_{1}^{\ast},y_{1}^{\ast}\right)+U_{3}^{\ast}\left(x_{1}^{\ast},y_{1}^{\ast}\right)=\mathbf{p}_{1}+O\left(\gamma^{-4}\right),$

and $\left|\mathbf{p}_{k}\right|=O\left(\gamma^{-2}\right).$ We then define,
for $k=1,2,3,$

$\omega_{k}\left(x,y\right):=\mathbf{p}_{k}\omega\left(x-x_{k}^{\ast},y-y_{k}^{\ast}\right).$

With the help of $\omega_{k}$, we will prove the following result, which gives
us a more precise description of the solution $u.$

######  Proposition 12.

There exists $\varepsilon>0$ such that the function $\xi=u-U^{\ast}$ has the
following expansion:

$\left\|\left(\xi-\sum\limits_{k=1}^{3}\omega_{k}\right)\theta_{\varepsilon}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq
C\gamma^{-3}.$

###### Proof.

We write $\xi=\sum\limits_{k=1}^{3}\omega_{k}+\eta,$ then $\eta$ satisfies

$\mathcal{L}_{U^{\ast}}\eta=-\sum\limits_{k=1}^{3}\mathcal{L}_{U^{\ast}}\omega_{k}-E\left(U^{\ast}\right)-3\left(\sum\limits_{k=1}^{3}\omega_{k}+\eta\right)^{2}.$

We have

$\left\|\left(-\sum\limits_{k=1}^{3}\mathcal{L}_{U^{\ast}}\omega_{k}-E\left(U^{\ast}\right)\right)\theta_{2}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq
C\gamma^{-3}.$

Applying Lemma 10, we conclude that for some $\varepsilon>0,$

$\left\|\eta\theta_{\varepsilon}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq
C\gamma^{-3}.$

This finishes the proof. ∎

In the next result, we need to use the following constants:

$\displaystyle a^{\ast}$
$\displaystyle=\int_{\mathbb{R}^{2}}\left(3\omega^{2}+6\omega\right)\partial_{x}^{2}Udxdy,$
$\displaystyle b^{\ast}$
$\displaystyle=\int_{\mathbb{R}^{2}}\left(3\omega^{2}+6\omega\right)\partial_{x}\partial_{y}Udxdy,$
$\displaystyle c^{\ast}$
$\displaystyle=\int_{\mathbb{R}^{2}}\left(3\omega^{2}+6\omega\right)\partial_{y}^{2}Udxdy.$

######  Lemma 13.

For any index $j\in\\{1,2,3\\}$ it holds that

(23)
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{x}U_{j}^{\ast}\right]dxdy=-3d^{\ast}\sum_{k\neq
j}\operatorname{Re}\left[\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{-4}\right]+a^{\ast}\mathbf{p}_{j}^{2}+O\left(\gamma^{-5}\right),$
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{j}^{\ast}\right]dxdy=3d^{\ast}\sum_{k\neq
j}\operatorname{Im}\left[\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{-4}\right]+b^{\ast}\mathbf{p}_{j}^{2}+O\left(\gamma^{-5}\right),$
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{y}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{j}^{\ast}\right]dxdy=3d^{\ast}\sum_{k\neq
j}\operatorname{Re}\left[\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{-4}\right]+c^{\ast}\mathbf{p}_{j}^{2}+O\left(\gamma^{-5}\right).$

While for different indices $j,k\in\\{1,2,3\\}$, we have

(24)
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{x}U_{k}^{\ast}\right]dxdy=3d^{\ast}\operatorname{Re}\left[\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{-4}\right]+O\left(\gamma^{-5}\right),$
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{k}^{\ast}\right]dxdy=-3d^{\ast}\operatorname{Im}\left[\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{-4}\right]+O\left(\gamma^{-5}\right),$
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{y}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{k}^{\ast}\right]dxdy=-3d^{\ast}\operatorname{Re}\left[\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{-4}\right]+O\left(\gamma^{-5}\right).$

###### Proof.

We shall firstly verify the three equations in (23). Without loss of
generality we may assume that $j=1.$ Since $\partial_{x}U_{1}^{\ast}$ is a
kernel of the operator $\mathcal{L}_{U_{1}^{\ast}},$ we have

(25)
$\mathcal{L}_{u}\left[\partial_{x}U_{1}^{\ast}\right]=\mathcal{L}_{U_{1}^{\ast}}\left[\partial_{x}U_{1}^{\ast}\right]+6\left(u-U_{1}^{\ast}\right)\partial_{x}U_{1}^{\ast}=6\left(U_{2}^{\ast}+U_{3}^{\ast}+\xi\right)\partial_{x}U_{1}^{\ast}.$

As a direct consequence,

(26)
$\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{x}U_{1}^{\ast}\right]dxdy=6\int_{\mathbb{R}^{2}}\left(\partial_{x}U_{1}^{\ast}\right)^{2}\left(U_{2}^{\ast}+U_{3}^{\ast}+\xi\right)dxdy.$

To estimate the right hand side, we first differentiate the equation

$\mathcal{L}_{U^{\ast}}\xi=-E\left(U^{\ast}\right)-3\xi^{2}:=J$

with respect to $x$. This yields

$\mathcal{L}_{U^{\ast}}\left[\partial_{x}\xi\right]+6\partial_{x}U^{\ast}\xi=\partial_{x}J.$

Multiplying both sides with $\partial_{x}U_{1}^{\ast}$ and integrating by
parts, we obtain

$\int_{\mathbb{R}^{2}}\left(-\partial_{x}\left(\mathcal{L}_{U^{\ast}}\left[\partial_{x}U_{1}^{\ast}\right]\right)+6\partial_{x}U_{1}^{\ast}\partial_{x}U^{\ast}\right)\xi
dxdy=\int_{\mathbb{R}^{2}}\partial_{x}J\partial_{x}U_{1}^{\ast}dxdy.$

Reorganizing terms, we have

$\displaystyle
6\int_{\mathbb{R}^{2}}\left(\partial_{x}U_{1}^{\ast}\right)^{2}\xi dxdy=$
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}J\partial_{x}U_{1}^{\ast}dxdy-6\int_{\mathbb{R}^{2}}\left(\partial_{x}U_{2}^{\ast}+\partial_{x}U_{3}^{\ast}\right)\partial_{x}U_{1}^{\ast}\xi
dxdy$
$\displaystyle+6\int_{\mathbb{R}^{2}}\partial_{x}\left[\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{x}U_{1}^{\ast}\right]\xi
dxdy$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}J\partial_{x}U_{1}^{\ast}dxdy+6\int_{\mathbb{R}^{2}}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{x}^{2}U_{1}^{\ast}\xi
dxdy.$

Inserting this into $\left(\ref{u1u1}\right),$ we get

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{x}U_{1}^{\ast}\right]dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\left(\partial_{x}U_{1}^{\ast}\right)^{2}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy+\int_{\mathbb{R}^{2}}\partial_{x}J\partial_{x}U_{1}^{\ast}dxdy+6\int_{\mathbb{R}^{2}}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{x}^{2}U_{1}^{\ast}\xi
dxdy$ $\displaystyle=3\int_{\mathbb{R}^{2}}U_{1}^{\ast
2}\partial_{x}^{2}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy-3\int_{\mathbb{R}^{2}}\partial_{x}\left(\xi^{2}\right)\partial_{x}U_{1}^{\ast}dxdy$
$\displaystyle\quad+6\int_{\mathbb{R}^{2}}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{x}^{2}U_{1}^{\ast}\xi
dxdy+O\left(\gamma^{-5}\right).$

Note that

$\int_{\mathbb{R}^{2}}U_{1}^{\ast
2}\partial_{x}^{2}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy=-d^{\ast}\operatorname{Re}\left[\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{-4}+\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{-4}\right]+O\left(\gamma^{-5}\right).$

We then get

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{x}U_{1}^{\ast}\right]dxdy=$
$\displaystyle-3d^{\ast}\operatorname{Re}\left[\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{-4}+\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{-4}\right]$
$\displaystyle+\mathbf{p}_{1}^{2}\int_{\mathbb{R}^{2}}\left(3\omega_{1}^{2}+6\omega_{1}\right)\partial_{x}^{2}U_{1}^{\ast}dxdy+O\left(\gamma^{-5}\right).$

This is the required identity.

Next we compute

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{1}^{\ast}\right]dxdy$
$\displaystyle=\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\mathcal{L}_{U_{1}^{\ast}}\left[\partial_{y}U_{1}^{\ast}\right]dxdy+6\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\left(u-U_{1}^{\ast}\right)\partial_{y}U_{1}^{\ast}dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\partial_{y}U_{1}^{\ast}\left(U_{2}^{\ast}+U_{3}^{\ast}+\xi\right)dxdy.$

On the other hand,

$\mathcal{L}_{U^{\ast}}\left[\partial_{y}\xi\right]+6\partial_{y}U^{\ast}\xi=\partial_{y}J,$

which implies

$\int_{\mathbb{R}^{2}}\left(-\partial_{y}\left(\mathcal{L}_{U^{\ast}}\left[\partial_{x}U_{1}^{\ast}\right]\right)+6\partial_{x}U_{1}^{\ast}\partial_{y}U^{\ast}\right)\xi
dxdy=\int_{\mathbb{R}^{2}}\partial_{y}J\partial_{x}U_{1}^{\ast}dxdy.$

From this identity, we get, using similar computation as before,

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{1}^{\ast}\right]dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\partial_{y}U_{1}^{\ast}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy+\int_{\mathbb{R}^{2}}\partial_{y}J\partial_{x}U_{1}^{\ast}dxdy+6\int_{\mathbb{R}^{2}}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{x}\partial_{y}U_{1}^{\ast}\xi
dxdy$
$\displaystyle=3\int_{\mathbb{R}^{2}}\partial_{x}\partial_{y}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)U_{1}^{\ast
2}dxdy-3\int_{\mathbb{R}^{2}}\partial_{y}\left(\xi^{2}\right)\partial_{x}U_{1}^{\ast}dxdy$
$\displaystyle\quad+6\int_{\mathbb{R}^{2}}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{x}\partial_{y}U_{1}^{\ast}\xi
dxdy+O\left(\gamma^{-5}\right).$

In view of the estimate

$\int_{\mathbb{R}^{2}}U_{1}^{\ast
2}\partial_{x}\partial_{y}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy=-3d^{\ast}\operatorname{Im}\left[\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{-4}+\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{-4}\right]+O\left(\gamma^{-5}\right),$

we then arrive at

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{1}^{\ast}\right]dxdy=~{}$
$\displaystyle
3d^{\ast}\operatorname{Im}\left[\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{-4}+\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{-4}\right]$
$\displaystyle+\mathbf{p}_{1}^{2}\int_{\mathbb{R}^{2}}\left(3\omega_{1}^{2}+6\omega_{1}\right)\partial_{x}\partial_{y}U_{1}^{\ast}dxdy+O\left(\gamma^{-5}\right).$

Regarding the last equation in (23), as what we have done for the first one we
get that

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{y}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{1}^{\ast}\right]dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\left(\partial_{y}U_{1}^{\ast}\right)^{2}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy+\int_{\mathbb{R}^{2}}\partial_{y}J\partial_{y}U_{1}^{\ast}dxdy+6\int_{\mathbb{R}^{2}}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{y}^{2}U_{1}^{\ast}\xi
dxdy$ $\displaystyle=3\int_{\mathbb{R}^{2}}U_{1}^{\ast
2}\partial_{y}^{2}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy-3\int_{\mathbb{R}^{2}}\partial_{y}\left(\xi^{2}\right)\partial_{y}U_{1}^{\ast}dxdy$
$\displaystyle\quad+6\int_{\mathbb{R}^{2}}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)\partial_{y}^{2}U_{1}^{\ast}\xi
dxdy+O\left(\gamma^{-5}\right).$

Using the fact that

$\int_{\mathbb{R}^{2}}U_{1}^{\ast
2}\partial_{y}^{2}\left(U_{2}^{\ast}+U_{3}^{\ast}\right)dxdy=d^{\ast}\operatorname{Re}\left[\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{-4}+\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{-4}\right]+O\left(\gamma^{-5}\right).$

We obtain that

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{y}U_{1}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{1}^{\ast}\right]dxdy=~{}$
$\displaystyle
3d^{\ast}\operatorname{Re}\left[\left(z_{1}^{\ast}-z_{2}^{\ast}\right)^{-4}+\left(z_{1}^{\ast}-z_{3}^{\ast}\right)^{-4}\right]$
$\displaystyle+\mathbf{p}_{1}^{2}\int_{\mathbb{R}^{2}}\left(3\omega_{1}^{2}+6\omega_{1}\right)\partial_{y}^{2}U_{1}^{\ast}dxdy+O\left(\gamma^{-5}\right).$

Now we verify the equations in (24).

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{x}U_{k}^{\ast}\right]dxdy$
$\displaystyle=\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{U_{k}^{\ast}}\left[\partial_{x}U_{k}^{\ast}\right]dxdy+6\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\left(u-U_{k}^{\ast}\right)\partial_{x}U_{k}^{\ast}dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\partial_{x}U_{k}^{\ast}\left(\sum_{\ell\neq
k}U_{\ell}^{\ast}+\xi\right)dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\partial_{x}U_{k}^{\ast}U_{j}^{\ast}dxdy+O\left(\gamma^{-5}\right)$
$\displaystyle=3d^{\ast}\operatorname{Re}\left[\left(z_{j}^{\ast}-z_{k}^{\ast}\right)^{-4}\right]+O\left(\gamma^{-5}\right).$

Since the computation of other integrals like
$\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}U_{k}^{\ast}\right]dxdy$
is quite similar, the details of these computation will be omitted. ∎

The key result of this section is the following

######  Proposition 14.

The Morse index of $u$ is equal to $3,$ provided that $B$ is sufficiently
large.

###### Proof.

Let $\lambda_{B}$ be a negative eigenvalue of the linearized operator, with
$\phi_{B}$ being an eigenfunction normalized such that
$\left\|\phi_{B}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}=1.$ Then

$\partial_{x}^{2}\phi_{B}-\phi_{B}+6u\phi_{B}-\partial_{x}^{-2}\partial_{y}^{2}\phi_{B}=\lambda_{B}\phi_{B}.$

Taking $x$-derivative twice, we get

$\partial_{x}^{2}\left(\partial_{x}^{2}\phi_{B}-\phi_{B}+6u\phi_{B}\right)-\partial_{y}^{2}\phi_{B}=\lambda_{B}\partial_{x}^{2}\phi_{B}.$

Our first step is to show that there exists $c_{0}$ independent of $B$ such
that

(27) $\lambda_{B}\leq c_{0}<0.$

Assume to the contrary that $\left(\ref{bound}\right)$ was not true. Then
there was a sequence $\lambda_{j}\rightarrow 0$ with corresponding normalized
eigenfunctions $\phi_{j}.$

Consider the translated sequence
$\tilde{\phi}_{j}\left(x,y\right):=\phi_{j}\left(x-x_{1}^{\ast},y-y_{1}^{\ast}\right).$
Since we have assumed that
$\left\|\phi_{j}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}=1,$ the new
functions $\tilde{\phi}_{j}$ will converge to a function $\Phi,$ solution of
the equation

$\partial_{x}^{2}\left(\partial_{x}^{2}\Phi-\Phi+6U\Phi\right)-\partial_{y}^{2}\Phi=0.$

By the nondegeneracy of lump, there exist constants $c_{1},c_{2}$(could be
zero) such that

$\Phi=c_{1}\partial_{x}U+c_{2}\partial_{y}U.$

To simplify the notation, we omit the subscript $j$, if no confusion will
arise. Then we can write $\phi$ as $\phi^{\ast}+\zeta,$ where

$\phi^{\ast}={\displaystyle\sum\limits_{k=1}^{3}}\left(\sigma_{k}\partial_{x}U_{k}^{\ast}+\tau_{k}\partial_{y}U_{k}^{\ast}\right),$

and $\zeta$ is small and satisfies the orthogonality condition: For $k=1,2,3,$

$\int_{\mathbb{R}^{2}}\zeta\partial_{x}U_{k}^{\ast}dxdy=0,\text{ \ and
}\int_{\mathbb{R}^{2}}\zeta\partial_{y}U_{k}^{\ast}dxdy=0.$

We would like to estimate $\zeta,$ using the equation

$\mathcal{L}_{u}\zeta=-\mathcal{L}_{u}\phi^{\ast}+\lambda\phi^{\ast}+\lambda\zeta.$

To do this, we first observe that $\mathcal{L}_{u}\phi^{\ast}$ is of the order
$O\left(\gamma^{-2}\right).$ Hence to obtain a precise expansion for $\zeta,$
we write

$\zeta=\sum\limits_{k=1}^{3}\left(\sigma_{k}\partial_{x}\omega_{k}+\tau_{k}\partial_{y}\omega_{k}\right)+\zeta^{\ast}.$

Using $\left(\ref{Lw}\right),\left(\ref{LU1}\right),$ for any $k=1,2,3,$ we
can estimate

(28)
$\mathcal{L}_{u}\left[\partial_{x}U_{k}^{\ast}\right]+\mathcal{L}_{u}\left[\partial_{x}\omega_{k}\right]=O\left(\gamma^{-3}\right),$
(29)
$\mathcal{L}_{u}\left[\partial_{y}U_{k}^{\ast}\right]+\mathcal{L}_{u}\left[\partial_{y}\omega_{k}\right]=O\left(\gamma^{-3}\right).$

It follows that

$\displaystyle\mathcal{L}_{u}\zeta^{\ast}-\lambda\xi^{\ast}=$
$\displaystyle-\sum\limits_{k=1}^{3}\left(\sigma_{k}\mathcal{L}_{u}\left[\partial_{x}U_{k}^{\ast}+\partial_{x}\omega_{k}\right]+\tau_{k}\mathcal{L}_{u}\left[\partial_{y}U_{k}^{\ast}+\partial_{y}\omega_{k}\right]\right)$
$\displaystyle+\lambda\left(\phi^{\ast}+\sum\limits_{k=1}^{3}\left(\sigma_{k}\partial_{x}\omega_{k}+\tau_{k}\partial_{y}\omega_{k}\right)\right).$

Let us denote the right hand side by $Q.$

For a function $\eta,$ we define

$\displaystyle\eta_{k,x}^{\parallel}=\int_{\mathbb{R}^{2}}\eta\partial_{x}U_{k}^{\ast}dxdy,\quad\eta_{k,y}^{\parallel}=\int_{\mathbb{R}^{2}}\eta\partial_{y}U_{k}^{\ast}dxdy,$

and

$\displaystyle\eta^{\|}=\sum\limits_{k=1}^{3}\left(\eta_{k,x}^{\parallel}\partial_{x}U_{k}^{\ast}+\eta_{k,y}^{\parallel}\partial_{y}U_{k}^{\ast}\right),$

and $\eta^{\perp}=\eta-\eta^{\|}.$ The function $\eta^{\perp}$ can be
understood as the projection orthogonal to the kernels of the linearized
operator. Then

(30) $\mathcal{L}_{u}\zeta^{\ast}-\lambda\xi^{\ast}=Q^{\perp}+Q^{\|}.$

Estimates $\left(\ref{e1}\right)$ and $\left(\ref{e11}\right)$ imply that
$Q^{\perp}=O\left(\gamma^{-3}\right).$ On the other hand, multiplying
$\left(\ref{kistar}\right)$ by
$\partial_{x}U_{k}^{\ast},\partial_{y}U_{k}^{\ast}$ and integrating tell us
that $Q^{\|}=o\left(\zeta^{\ast}\right).$ Following the proof of Lemma 10, we
find that there exists $\varepsilon>0$, such that

$\left\|\zeta^{\ast}\theta_{\varepsilon}\right\|_{L^{\infty}\left(\mathbb{R}^{2}\right)}\leq
C\gamma^{-3}.$

With the estimate of $\zeta$ at hand, now we project the equation
$\mathcal{L}_{u}\phi=\lambda\phi$ onto the kernels $\partial_{x}U_{k}^{\ast}$
and $\partial_{y}U_{k}^{\ast}.$ More precisely, we consider the following two
identities

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{k}^{\ast}\left(\mathcal{L}_{u}\phi-\lambda\phi\right)dxdy=0,\quad\int_{\mathbb{R}^{2}}\partial_{y}U_{k}^{\ast}\left(\mathcal{L}_{u}\phi-\lambda\phi\right)dxdy=0.$

Now we compute

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{x}\omega_{j}\right]dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\left(\sum_{k\neq
j}U_{k}^{\ast}+\xi\right)\partial_{x}\omega_{j}dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\left(-\omega_{j}\mathbf{p}_{j}-\frac{1}{2}\omega_{j}^{2}\right)\partial_{x}^{2}U_{j}^{\ast}dxdy+O\left(\gamma^{-5}\right)$
$\displaystyle=-a^{\ast}\mathbf{p}_{j}^{2}+O\left(\gamma^{-5}\right).$

Similarly, we have

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}\omega_{j}\right]dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\left(\sum_{k\neq
j}U_{k}^{\ast}+\xi\right)\partial_{y}\omega_{j}dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\left(-\omega_{j}\mathbf{p}_{j}-\frac{1}{2}\omega_{j}^{2}\right)\partial_{x}\partial_{y}U_{j}^{\ast}dxdy$
$\displaystyle=-b^{\ast}\mathbf{p}_{j}^{2}+O\left(\gamma^{-5}\right),$

and

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{y}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}\omega_{j}\right]dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\partial_{y}U_{j}^{\ast}\left(\sum_{k\neq
j}U_{k}^{\ast}+\xi\right)\partial_{y}\omega_{j}dxdy$
$\displaystyle=6\int_{\mathbb{R}^{2}}\left(-\omega_{j}\mathbf{p}_{j}-\frac{1}{2}\omega_{j}^{2}\right)\partial_{y}^{2}U_{j}^{\ast}dxdy$
$\displaystyle=-c^{\ast}\mathbf{p}_{j}^{2}+O\left(\gamma^{-5}\right).$

While for $j\neq k$, we have

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{x}\omega_{k}\right]dxdy=O\left(\gamma^{-5}\right),\quad\int_{\mathbb{R}^{2}}\partial_{x}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}\omega_{k}\right]dxdy=O\left(\gamma^{-5}\right),$

and

$\displaystyle\int_{\mathbb{R}^{2}}\partial_{y}U_{j}^{\ast}\mathcal{L}_{u}\left[\partial_{y}\omega_{k}\right]dxdy=O\left(\gamma^{-5}\right).$

From the above estimates and Lemma 13, we deduce that for $k=1,2,3,$

$\int_{\mathbb{R}^{2}}\left(\partial_{x}U_{k}^{\ast}-i\partial_{y}U_{k}^{\ast}\right)\left(\mathcal{L}_{u}\phi-\lambda\phi\right)dxdy=\sum\limits_{j=1}^{3}\left(F_{k,j}\mathbf{e}_{j}\right)+O\left(\gamma^{-5}\right),$

where $\mathbf{e}_{j}=\sigma_{j}+i\tau_{j}$ is complex number. Moreover, there
exists universal positive constants $\delta_{1},\delta_{2},$ such that

$\delta_{1}\leq\sum\limits_{j}\left|\mathbf{e}_{j}\right|^{2}\leq\delta_{2}.$

Hence, after a scaling, in terms of the matrix $M$ defined in Lemma 9, we
obtain

$M\left(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\right)^{T}=O\left(\gamma^{-1}\right)+O\left(\left|\lambda\right|\right).$

We then deduce that for some constants $c_{1},c_{2},$ with
$c_{1}^{2}+c_{2}^{2}$ uniformly bounded away from zero, such that

$\left(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\right)=c_{1}\mathbf{b}_{1}+c_{2}\mathbf{b}_{2}+O\left(\gamma^{-1}\right)+O\left(\left|\lambda\right|\right).$

On the other hand, using the explicit four-parameter family of solutions
$u_{A,B}$ for the Boussinesq equation, we deduce that there exists a kernel
$\varphi$ of $\mathcal{L}_{u}$ whose projection on
$\partial_{x}U_{k}^{\ast},\partial_{y}U_{k}^{\ast}$ is close to
$c_{1}\mathbf{b}_{1}+c_{2}\mathbf{b}_{2}.$ We then conclude that $\phi_{j}$
will not be $L^{2}$-orthogonal to the $\varphi$ for $j$ large enough. This
contradicts with the fact that $\phi_{j}$, as an eigenfunction with respect to
a negative eigenvalue, has to be $L^{2}$-orthogonal to $\varphi.$

Now we have proved that negative eigenvalue $\lambda_{j}$ will be uniformly
bounded away from $0.$ Using this information, we then deduce that
$\lambda_{j}$ has to converge to the unique negative eigenvalue of the
operator $\mathcal{L}_{U}.$ By result in [14], the Morse index of the standard
lump $U$ is equal to $1,$ we then find that the Morse index of $u$ is at most
$3.$ On the other hand, by constructing explicit test functions using the
eigenfunctions(of negative eigenvalue) of $\mathcal{L}_{U}$, we know that the
Morse index of $u$ is at least $3.$ Hence we conclude that the Morse index of
$u$ is equal to $3.$ This finishes the proof. ∎

Having analyzed the solutions $u_{0,B}$ for $B$ large, we proceed to show that
all the solutions $u_{A,B}=2\partial_{x}^{2}\ln h_{A,B}$ has Morse index $3.$

###### Proof of Theorem 1.

Straightforward application of the results in the previous section tells us
that there is a Backlund transformation from the translated degree $2$
polynomial

$f\left(x,y\right)=x^{2}+\left(y-\frac{2\sqrt{3}i}{3}\right)^{2}+3$

to the degree $4$ polynomial $g,$ which is defined explicitly by

$\displaystyle g\left(x,y\right)=~{}$ $\displaystyle
x^{4}+2ix^{3}y+2ixy^{3}-y^{4}+\frac{10x^{3}}{\sqrt{3}}+4\sqrt{3}ix^{2}y+2\sqrt{3}xy^{2}+\frac{\sqrt{3}iy^{3}}{8}$
$\displaystyle+20x^{2}+30ixy+2y^{2}+\left(\frac{50}{\sqrt{3}}+\frac{Ai+B}{2}\right)x$
$\displaystyle+\left(\frac{80i}{\sqrt{3}}+\frac{A-Bi}{2}\right)y-25+\frac{Ai+B}{2\sqrt{3}}.$

Then the function $g$ is Backlund transformed to $h_{A,B}.$ Observe that both
$f$ and $g$ have finitely many simple zeros. Using this fact, the
nondegeneracy of the linearized operator $\mathcal{L}_{u_{A,B}}$ then follows
directly from the same argument as that of [14], that is, by analyzing the
associated linearized Backlund transformations.

To show that the Morse index of $u_{A,B}$ is equal to $3,$ let $m$ be
sufficiently large and consider the family of solutions
$u_{At,Bt+\left(1-t\right)m}.$ When $t=1,$ it is $u_{A,B},$ and for $t=0,$ the
solution is $u_{0,m}$. We now know that for any $t\in\left[0,1\right],$ the
solution is nondegenerated in the sense that the linearized operator has no
nontrivial kernels. This together with the continuous dependence of the
negative eigenvalues with respect to $t$, implies that as $t$ decreases from
$1$ to $0,$ a negative eigenvalue can not diminish to the zero eigenvalue. We
then see that the Morse index of $u_{A,B}$ should be the same as that of
$u_{0,m}.$ By Proposition 14, $u_{0,m}$ has Morse index $3$, provided that $m$
is large. We then conclude that for any $A,B,$ the Morse index of $u_{A,B}$
equals $3.$ This completes the proof. ∎

As a final remark, we point out that for solutions with higher degrees, the
above arguments also work. The only delicate part is, we need to show, as one
of the parameter tends to infinity, the solution splits into a number of
classical lumps which are far away from each other. This is to ensure that
procedure of reverse Lyapunov-Schmidt reduction can be started. For polynomial
tau functions of degree $k\left(k+1\right),$ the corresponding solution should
have Morse index $k\left(k+1\right)/2.$ Once this is proved, it will yield the
existence of infinitely many solutions for the GP and generalized KP equation.
Rigorous justification of this fact would require a complete classification of
the moduli space of lump type solutions. This is an ongoing project.

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