Doktorov E.V., Leble S.B. A Dressing Method in Mathematical Physics
Подождите немного. Документ загружается.
10.3 Nonlinear equations with singular dispersion relation: 2+1 dimensions 339
10.3.2 Dual function
Let us calculate the time evolution of Q given explicitly by (10.67):
∂
t
Q = −i[σ
3
, ∂
t
(ψR
k
F )].
The right-hand side can be transformed as follows:
∂
t
(ψR
k
F )=VψR
k
F + ψR
k
FΩ.
Further calculation, owing to (10.70), yields
∂
t
(ψR
k
F )=ψR
k
FΩC
k
(11 − R
k
FC
k
)
−1
R
k
F −ψΩ(11 − R
k
FC
k
)
−1
R
k
F
+ψR
k
FΩ = ψR
k
FΩ(11 − C
k
R
k
F )
−1
− ψΩR
k
F (11 − C
k
R
k
F )
−1
.
Hence,
∂
t
Q = −i[σ
3
, ψR
k
FΩ(11 − C
k
R
k
F )
−1
, 11−ψΩR
k
F (11 − C
k
R
k
F )
−1
, 11].
Taking into account the pairing properties, we get
∂
t
Q = −i[σ
3
, ψR
k
FΩ,11 · (11+
ˆ
R
k
FC
k
)
−1
−ψΩ, 11 ·(11+
ˆ
R
k
FC
k
)
−1
ˆ
R
k
F ].
(10.72)
Now we introduce a dual function
˜
ψ(k) by means of the relation [120]
˜
ψ
T
= 11 · (11+
ˆ
R
k
FC
k
)
−1
. (10.73)
The
¯
∂ problem for the dual function has the form
¯
∂
˜
ψ(k)=−
dℓ ∧ d
¯
ℓR(ℓ, k)
˜
ψ(ℓ), or
¯
∂
˜
ψ
T
= −
˜
ψ
T
(k)R
k
F (10.74)
and
˜
ψ(k) satisfies the dual spectral problem
∂
x
˜
ψ + ∂
y
˜
ψσ
3
−
˜
ψQ − ik[σ
3
,
˜
ψ]=0. (10.75)
In order to derive (10.74), we proceed from
˜
ψ
T
= 11 −
˜
ψ
T
ˆ
R
k
FC
k
and take into
account the identity
¯
∂f(k)C
k
= f(k). Then
¯
∂
˜
ψ
T
= −
˜
ψ
T
ˆ
R
k
F = −
dℓ ∧ d
¯
ℓ
˜
ψ
T
(ℓ)
ˆ
R(k, ℓ)
= −
dℓ ∧ d
¯
ℓ
˜
ψ
T
(ℓ)R
T
(ℓ, k)=−
dℓ ∧ d
¯
ℓ[R(ℓ, k)
˜
ψ(ℓ)]
T
and (10.74) follows. The derivation of (10.75) is slightly more cumbersome.
Differentiating (10.73) in x, we find ∂
x
˜
ψ
T
= −
˜
ψ
T
∂
x
ˆ
R
k
FC
k
(11+
ˆ
R
k
FC
k
)
−1
.
Accounting for
ˆ
R(k, ℓ)=R
T
(ℓ, k), we obtain from (10.63)
∂
x
ˆ
R(k, ℓ)=ik
ˆ
R(k, ℓ)σ
3
− iℓσ
3
ˆ
R(k, ℓ),∂
y
ˆ
R(k, ℓ)=−i(k − ℓ)
ˆ
R(k, ℓ).
340 10 Generating solutions via
¯
∂ problem
Then following the derivation of the Zakharov–Shabat spectral problem, we
obtain
∂
x
˜
ψ
T
=ik
˜
ψ
T
σ
3
− i
˜
ψ
T
ˆ
R
k
F, 11σ
3
˜
ψ
T
− ikσ
3
(11+
ˆ
R
k
FC
k
)
−1
and
∂
x
˜
ψ =ikσ
3
˜
ψ − i
˜
ψσ
3
11,
˜
ψ
T
ˆ
R
k
F −ik
!
(11+
ˆ
R
k
FC
k
)
−1
"
T
σ
3
.
Similarly,
∂
y
˜
ψ = −ik
˜
ψ +i
˜
ψ11,
˜
ψ
T
ˆ
R
k
F +ik
!
(11+
ˆ
R
k
FC
k
)
−1
"
T
.
Hence,
∂
x
˜
ψ + ∂
y
˜
ψσ
3
− ik[σ
3
,
˜
ψ]+i
˜
ψ[σ
3
, 11 ,
˜
ψ
T
ˆ
R
k
F ]=0.
Now we need a connection between 11,
˜
ψ
T
ˆ
R
k
F and Q. It can be found as
follows:
ψR
k
F, 11 = 11 · (11 − R
k
FC
k
)
−1
R
k
F, 11 = 11 · (11 − R
k
FC
k
)
−1
,
ˆ
R
k
F
= 11,
ˆ
R
k
F (11+
ˆ
R
k
FC
k
)
−1
= 11, 11 · (11+
ˆ
R
k
FC
k
)
−1
ˆ
R
k
F = 11,
˜
ψ
T
ˆ
R
k
F .
Hence,
Q = −i[σ
3
, ψR
k
F, 11]=−i[σ
3
, 11 ,
˜
ψ
T
ˆ
R
k
F ]
and we arrive at the spectral equation (10.75).
Taking into account the above relations concerning the dual function, we
write the evolution (10.72) in the form
∂
t
Q = −i[σ
3
, ψR
k
FΩ,
˜
ψ−ψΩ,
˜
ψ
T
ˆ
R
k
F ]=−i[σ
3
, (
¯
∂ψ)Ω
˜
ψ + ψΩ
¯
∂
˜
ψ].
Finally, dividing the dispersion relation into the regular and singular parts,
we obtain under the condition Ω
s
(k) → 0fork →∞
∂
t
Q = −i[σ
3
,
¯
∂(ψΩ
p
˜
ψ−ω
s
ψσ
3
˜
ψ]. (10.76)
As a result, it is the dual function
˜
ψ that is a true (2+1)-dimensional gener-
alization of inverse functions. It should be stressed that the definition (10.73)
of the dual function arises naturally within the framework of the formalism
based on the representation (10.62).
10.3.3 Recursion operator
To derive the recursion operator, we introduce a bilo cal object [243]
M
12
(x, y
1
,y
2
,k)=ψ(x, y
1
,k)σ
3
˜
ψ(x, y
2
,k) ≡ ψ
1
σ
3
˜
ψ
2
.
10.3 Nonlinear equations with singular dispersion relation: 2+1 dimensions 341
The function M
12
satisfies the equation
∂
x
M
12
+ σ
3
∂
y
1
M
12
+ ∂
y
2
M
12
σ
3
−ik[σ
3
,M
12
]+Q
1
M
12
−M
12
Q
2
=0, (10.77)
where Q
i
≡ Q(x, y
i
); hence, (10.76) takes the form
δ
12
∂
t
Q
2
= −iα
n
δ
12
[σ
3
,
¯
∂(k
n
M
12
)]+iδ
12
[σ
3
, ω(k)M
12
], (10.78)
where δ
12
= δ(y
1
− y
2
). Following [384], we introduce the notations
P
12
M
12
= ∂
x
M
12
+ σ
3
∂
y
1
M
12
+ ∂
y
2
M
12
σ
3
,Q
±
12
M
12
= Q
1
M
12
± M
12
Q
2
.
(10.79)
Let M
d
12
and M
a
12
be the diagonal and off-diagonal parts of the matrix M .
Then (10.77) and (10.79) yield
P
12
M
d
12
+ Q
−
12
M
a
12
=0,P
12
M
a
12
− 2ikσ
3
M
a
12
+ Q
−
12
M
d
12
=0. (10.80)
Because M
d
12
= σ
3
−P
−1
12
Q
−
12
M
a
12
, the second equation in (10.80) is written in
the form (Λ − k)M
a
12
=(2i)
−1
Q
+
12
· 11[Q
+
12
· 11=Q
1
+ Q
2
in accordance with
the definition (10.79)], where the op erator Λ is defined as
Λ =
1
2i
σ
3
(P
12
− Q
−
12
P
−1
12
Q
−
12
).
Then M
a
12
=(2i)
−1
(Λ − k)
−1
Q
+
12
· 11 and after the expansion (Λ − k)
−1
= −
∞
m=1
k
−m
Λ
m−1
we can write the polynomial contribution to ∂
t
Q in
(10.78) as
−iα
n
δ
12
[σ
3
,
¯
∂(k
n
M
12
)]=α
n
σ
3
δ
12
∞
m=1
¯
∂k
n−m
Λ
m−1
Q
+
12
· 11
= −
i
2
α
n
σ
3
δ
12
Λ
n
Q
+
12
·11.
Now we have all we need to formulate a closed system of equations de-
scribing the evolution of the potential Q with account for both parts of the
dispersion relations:
δ
12
∂
t
Q
2
= −
i
2
α
n
σ
3
δ
12
Λ
n
Q
+
12
·11+iδ
12
[σ
3
, ω
s
(k)M
12
],
P
12
M
12
− ik[σ
3
,M
12
]+Q
−
12
M
12
=0. (10.81)
Here the operator Λ plays the role of a recursion operator (more precisely, Λ
is related to the true recursion operator by means of σ
3
). If M
12
= σ
3
and
ω
s
(k) = 0, we get from (10.81) the hierarchy including the Davey–Stewartson I
equation derived by Santini and Fokas [384] on the basis of an integral repre-
sentation for the Lax evolution op erator V .
For Ω
p
= 0 the system (10.81) takes the form
δ
12
∂
t
Q
2
=iδ
12
[σ
3
, ω
s
(k)M
12
], (10.82)
P
12
M
12
− ik[σ
3
,M
12
]+Q
−
12
M
12
=0.
It is seen that the structure of this system is similar to that (10.25) with
α
n
= 0; hence, we can treat (10.82) as the Maxwell–Bloch equation in 2+1
dimensions. Its soliton solution can be found in [61].
342 10 Generating solutions via
¯
∂ problem
10.4 Kadomtsev–Petviashvili II equation
In the previous sections we applied the
¯
∂ formalism to solve some problems
for nonlinear equations in 1+1 and 2+1 dimensions. Though this approach
has proved its efficiency, the
¯
∂ formalism was not absolutely necessary to solve
these problems. In par ticular, the RH problem could be applied equally well
for this aim.
The present section is devoted to analysis of the KP II equation. The KP II
equation plays a distinctive role in the theory of nonlinear equ ations. It is the
KP II equation that demonstrated for the first time the nonuniversality of the
nonlocal RH problem for solving nonlinear equations.
Ablowitz et al. [1] showed in a beautiful paper that the inverse problem
for the KP II equation can be successfully solved by means of the
¯
∂ problem.
Following this paper, we demonstrate in this section the main steps in realizing
the program for solution of the KP II equation in the framework of the
¯
∂
method.
10.4.1 Eigenfunctions and scattering equation
The KP II equation
(u
t
+6uu
x
+ u
xxx
)
x
+3u
yy
= 0 (10.83)
describes the evolution of weakly nonlinear, weakly dispersive, and weakly
two-dimensional water waves (all these effects are of the same order) when
gravity dominates surface tension. The physical derivation of the KP II equa-
tion can be found in [3]. The KP II equation represents the compatibility
condition of two linear Lax equations
− ψ
y
+ ψ
xx
+ uψ =0, (10.84)
ψ
t
+4ψ
xxx
+6uψ
x
+3u
x
ψ +3(∂
−1
x
u
y
)ψ =0,
where ∂
−1
x
f =(1/2)
!
x
−∞
−
∞
x
"
dx
′
f(x
′
). In order to introduce a spec-
tral parameter, we transform the Lax pair (10.84) to the function m(x, y, k)
= ψ(x, y)exp(−ikx + k
2
y). As a result, we will work with the Lax pair of the
form
− m
y
+ m
xx
+2ikm
x
+ um =0, (10.85)
m
t
+4m
xxx
+ 12ikm
xx
− 12k
2
m
x
+6um
x
+6ikum +3u
x
m
+3(∂
−1
x
u
y
)m +(α(k) − 4ik
3
)m =0, (10.86)
where α(k) is an arbitrary function. As usual, we choose two linearly inde-
pendent solutions of (10.85) with zero potential:
M
0
=1,N
0
=exp
−i(k +
¯
k)x +(k
2
−
¯
k
2
)y
.
10.4 Kadomtsev–Petviash vili II equation 343
One of them, M
0
, provides the canonical normalization of the solution M of
the full spectral equation (10.85):
M(x, y, k)=1+
∞
−∞
dx
′
dy
′
G(x −x
′
,y− y
′
,k)u(x
′
,y
′
)M(x
′
,y
′
,k), (10.87)
which is bounded for all complex k. The other solution reads
N(x, y, k)=exp[−i(k +
¯
k)x +(k
2
−
¯
k
2
)y] (10.88)
+
∞
−∞
dx
′
dy
′
G(x − x
′
,y− y
′
,k)u(x
′
,y
′
)N(x
′
,y
′
,k).
Here G(x, y, k) is the Green function which obeys the equation
−G
y
+ G
xx
+2ikG
x
= −δ(x)δ(y)
and can be written as
G(x, y, k)=
1
4π
2
∞
−∞
dξdη
e
i(ξx+ηy)
ξ
2
+2kξ +iη
. (10.89)
As distinct from the KP I equation, the Green function (10.89) has no jump
across the real axis. Moreover, this function is analytic nowhere in the k-plane,
as can be explicitly seen after integration in η. Indeed, calculating by residues
the integral (10.89) with respect to η, we obtain the following formula for the
Green function [1]:
G(x, y, k)=
1
2π
1
θ(k
R
)
#
−θ(−y)
0
−2k
R
dξ + θ(y)
∞
0
dξ +
−2k
R
−∞
dξ
$
+θ(−k
R
)
#
−θ(−y)
−2k
R
0
dξ + θ(y)
0
−∞
dξ +
∞
−2k
R
dξ
$2
e
iξx−(ξ
2
+2kξ)y
.
(10.90)
Formula (10.90) contains explicitly k
R
and hence the Green function is non-
analytic. As a result, the eigenfunctions M and N are analytic nowhere in the
k-plane. We will stress this fact denoting the eigenfunctions as M(x, y, k,
¯
k)
and N(x, y, k,
¯
k).
Nonanalyticity of eigenfunctions prevents us from making use of the
RH problem. It was found by Ablowitz et al. [1] that it is the
¯
∂ prob-
lem that should be employed to formulate scattering equations and to solve
the inverse scatterin g problem. Hence, we cal culate first the
¯
∂ derivative
¯
∂M =(1/2)(∂
k
R
+i∂
k
I
)M(x, y, k,
¯
k):
¯
∂M(x, y, k,
¯
k)=
∞
−∞
dx
′
dy
′
¯
∂G(x − x
′
,y−y
′
,k,
¯
k)
u(x
′
,y
′
)M(x
′
,y
′
,k,
¯
k)
+
∞
−∞
dx
′
dy
′
G(x − x
′
,y− y
′
,k,
¯
k)u(x
′
,y
′
)
¯
∂M(x
′
,y
′
,k,
¯
k).
344 10 Generating solutions via
¯
∂ problem
It is easily obtained from (10.90) that
¯
∂G(x, y, k,
¯
k)=
1
2π
sign(−k
R
)exp
−i(k +
¯
k)x +(k
2
−
¯
k
2
)y
.
Therefore,
¯
∂M(x, y, k,
¯
k)=F (k,
¯
k)e
−i(k+
¯
k)x+(k
2
−
¯
k
2
)y
(10.91)
+
∞
−∞
dx
′
dy
′
G(x − x
′
,y− y
′
,k,
¯
k)u(x
′
,y
′
)
¯
∂M(x
′
,y
′
,k,
¯
k),
where the spectral data are given by the function
F (k,
¯
k)=
1
2π
sign(−k
R
)
∞
−∞
dxdyu(x, y)M(x, y, k,
¯
k)e
i(k+
¯
k)x−(k
2
−
¯
k
2
)y
.
(10.92)
Comparing (10.91) with the integral equation (10.88), we obtain the scattering
equation in the form of the linear
¯
∂ problem:
¯
∂M(x, y, k,
¯
k)=F (k,
¯
k)N (x, y, k,
¯
k). (10.93)
The next step consists in finding a symmetry (closure) relation, in order to
express N in terms of M . The Green function obeys the symmetry property
G(x, y, −
¯
k, −k)=G(x, y, k,
¯
k)exp
i(k +
¯
k)x − (k
2
−
¯
k
2
)y
.
Now from comparison of (10.87), where substitutions k →−
¯
k and
¯
k →−k
have b een performed, with (10.88) the discrete closure relation follows:
N(x, y, k,
¯
k)=M (x, y, −
¯
k, −k)exp
−i(k +
¯
k)x +(k
2
−
¯
k
2
)y
.
Thereby, the
¯
∂ problem for the eigenfunction M is written in terms of the
scattering data:
¯
∂M(x, y, k,
¯
k)=F (k,
¯
k)M(x, y, −
¯
k, −k)exp
−i(k +
¯
k)x +(k
2
−
¯
k
2
)y
.
(10.94)
It should b e stressed that the above calculation is valid under the assumption
that there are no nontrivial solutions of the homogeneous integral equation
obtained from (10.88).
10.4.2 Inverse spectral problem
The inverse spectral problem is solved by means of the Cauchy–Green formula
[see (1.99)]
M(x, y, k,
¯
k)=1+
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
¯
∂M(x, y, ℓ,
¯
ℓ)
=1+
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
F (ℓ,
¯
ℓ)M(x, y −
¯
ℓ, −ℓ)e
−i(ℓ+
¯
ℓ)x+(ℓ
2
−
¯
ℓ
2
)y
.
10.5 Davey–Stewartson II equation 345
To reconstruct the potential u(x, y), note that we have two representation for
M − 1:
M(x, y, k,
¯
k)−1=
⎧
⎪
⎨
⎪
⎩
dx
′
dy
′
G(x − x
′
,y−y
′
,k,
¯
k)u(x
′
,y
′
)M(x
′
,y
′
,k,
¯
k)
1
2πi
dℓ ∧d
¯
ℓ
ℓ − k
F (ℓ,
¯
ℓ)M(x, y, −
¯
ℓ, −ℓ)e
−i(ℓ+
¯
ℓ)x+(ℓ
2
−
¯
ℓ
2
)y
.
Now we compare them in the order of O(k
−1
). From (10.89) for |k|→∞we
obtain
G =
1
8π
2
k
∞
−∞
dη v.p.
∞
−∞
dξ
ξ
e
i(ξx+ηy)
+ O (k
−2
)=
i
4k
sign(x)δ(y)+O(k
−2
).
Besides, M =1+O(k
−1
). Hence, the Green function representation of M −1
yields
M − 1=
i
4k
⎛
⎝
x
−∞
dx
′
u(x
′
,y) −
∞
x
dx
′
u(x
′
,y)
⎞
⎠
+ O (k
−2
). (10.95)
From the
¯
∂ representation we obtain
M − 1=
i
2πk
dℓ ∧ d
¯
ℓF (ℓ,
¯
ℓ)M(x, y, −
¯
ℓ, −ℓ)e
−i(ℓ+
¯
ℓ)x+(ℓ
2
−
¯
ℓ
2
)y
+ O (k
−2
).
(10.96)
Comparing (10.95) and (10.96), we get the reconstruction formula
u(x, y)=
1
π
∂
x
dk ∧ d
¯
kF(k,
¯
k)M(x, y, −
¯
k, −k)e
−i(k+
¯
k)x+(k
2
−
¯
k
2
)y
.
We will be able to completely solve the KP II equation if the time evolution of
the spectral data is determined. By the standard manipulations with (10.86)
considered at the asymptotic x
2
+ y
2
→∞we obtain α(k)=4ik
3
and
F (k,
¯
k, t)=F (k,
¯
k, 0)e
−4i(k
3
+
¯
k
3
)t
.
The uniqueness of a solution of the KP II equation for small initial data
u(x, y, 0) was proved by Wickerhauser [451].
10.5 Davey–Stewartson II equation
TheDSIIequationinthefocusedcase
iu
t
=
1
2
(u
xx
− u
yy
)+(φ + |u|
2
)u, φ
xx
+ φ
yy
=2|u|
2
xx
(10.97)
describes an evolution of quasimonochromatic wave packets with slowly vary-
ing amplitude u(x, y, t) on a two-dimensional water surface under gravity
346 10 Generating solutions via
¯
∂ problem
[12, 105], where φ(x, y, t) is the velocity potential. Besides, the DS II equation
found use in plasma physics [352]. The IST method for (10.97) was realized
in terms of the
¯
∂ problem by Fokas and Ablowitz [159]. Rational nonsingu-
lar lo c alized solutions (lumps) of the DS II equation decaying at infinity as
(x
2
+ y
2
)
−1
have been derived by Arkadiev et al. [31]. Various aspects of the
IST approach for solving the DS II equation have been discussed by Beals
and Coifman [41, 42] and K onop elchenko and Matkarimov [246]. The Dar-
boux method was used in [308] to obtain soliton solutions which demonstrate
nontrivial dynamics under interaction. The completeness of the eigenf unction
system of the elliptic spectral problem associated with the DS II equation was
established in [36 4].
In the pap ers cited above, solitons of the DS II equation correspond to sim-
ple poles of the solutions of the spectral equation. As we know from the exam-
ple of the KP I equation, a novel class of solutions with more diverse properties
arises if the eigenfunctions allow multiple poles. The same situation exists for
the DS II equation. Villarroel and Ablowitz [440] found a variety of rationally
decaying, regular, localized solutions of the DS II equation which stem from
meromorphic eigenfunctions with multiple poles in the spectral parameter.
10.5.1 Eigenfunctions and scattering equation
The DS II equation (10.97) allows the matrix-valued Lax representation
ψ
x
+iσ
3
ψ
y
− Qψ =0,Q=
0 u
−¯u 0
(10.98)
ψ
t
= Aψ − Qψ
y
+iσ
3
ψ
yy
. (10.99)
Here ψ and A are 2 ×2 matrices. The compatibility condition for (10.98) and
(10.99) gives the DS II equation (10.97) provided the entries of A are given
by
(∂
x
+i∂
y
)A
11
=
1
2i
(∂
x
− i∂
y
)|u|
2
,A
12
=
1
2i
(∂
x
− i∂
y
)u, (10.100)
(∂
x
− i∂
y
)A
21
= −
1
2i
(∂
x
+i∂
y
)|u|
2
,A
21
=
1
2i
(∂
x
+i∂
y
)¯u,
while the potential φ is expressed in terms of A
ij
as φ =i(A
11
− A
22
) −|u|
2
.
Because the spectral problem (10.98) is elliptic, it is reasonable to introduce
complex coor dinates z = x +iy and ¯z = x −iy. Then the DS II equation takes
the form
iu
t
= u
zz
+ u
¯z¯z
+(g +¯g)u, g
¯z
=
1
2
|u|
2
z
, (10.101)
where g =iA
11
. The Lax representation for (10.101) is written as
Dψ =
1
2
Qψ, ψ
t
= Aψ − iQ(∂
z
− ∂
¯z
)ψ − iσ
3
(∂
z
− ∂
¯z
)
2
ψ. (10.102)
10.5 Davey–Stewartson II equation 347
Here
D =
∂
¯z
0
0 ∂
z
, and now A = −i
gu
z
u
¯z
−¯g
.
Let us cho ose a solution of the free equation Dψ =0intheform
E =
e
ikz
0
0e
−ik¯z
(10.103)
with a spectral parameter k and transform ψ as ψ = ME. Then the Lax pair
(10.102) takes the form
DM −
i
2
k[σ
3
,M]=
1
2
QM, (10.104)
M
t
= AM − iQ(∂
z
− ∂
¯z
)M + kQM − iσ
3
(∂
z
− ∂
¯z
+ik)
2
M.(10.105)
Taking the asymptotic expansion
M = 11+
1
ik
m + O (k
−2
)
in (10.104), we obtain the p otential reconstruction formula
Q = −[σ
3
,m], or u(x, y, t)=−2
m
21
(x, y, t). (10.106)
In virtue of the specific structure of the potential matrix Q,thereisa
symmetry relation between the entries of the matrix M .Namely,
M =
M
11
(k) −
M
21
(
¯
k)
M
21
(k)
M
11
(
¯
k)
. (10.107)
Hence, it is sufficient to study only the first column M
1
of the matrix M .The
components of the column M
1
=(M
1
,M
2
)
T
obey the following (spectral)
equations
∂
¯z
M
1
=
1
2
uM
2
,∂
z
M
2
= −ikM
2
−
1
2
¯uM
1
, (10.108)
with the boundary condition
lim
|k|→∞
M
1
(z, ¯z,k,
¯
k) ≡ M
0
=
1
0
. (10.109)
As the second linearly independent solution of the free equations (10.108) with
u = 0 we can choose [159]
N
0
=
0
1
e
−i(kz+
¯
k¯z)
. (10.110)
348 10 Generating solutions via
¯
∂ problem
Note that equations (10.108) can b e treated as
¯
∂ (∂) problems in the
coordinate space. Therefore, in accordance with the Cauchy–Green formula
(1.98) and boundary condition (10.109) we can write a solution of (10.108) in
the integral form:
M
1
(z, ¯z,k,
¯
k)=1+
1
2πi
dz
′
∧d¯z
′
z
′
− z
1
2
u(z
′
, ¯z
′
)M
2
(z
′
, ¯z
′
,k,
¯
k), (10.111)
M
2
(z, ¯z,k,
¯
k)=−
1
2πi
dz
′
∧ d¯z
′
¯z
′
− ¯z
1
2
¯u(z
′
, ¯z
′
)M
1
(z
′
, ¯z
′
,k,
¯
k)e
−ik(z−z
′
)−i
¯
k(¯z−¯z
′
)
.
Similarly,
N
1
(z, ¯z,k,
¯
k)=
1
2πi
dz
′
∧ d¯z
′
z
′
− z
1
2
u(z
′
, ¯z
′
)N
2
(z
′
, ¯z
′
,k,
¯
k), (10.112)
N
2
(z, ¯z,k,
¯
k)=e
−i(kz+
¯
k¯z)
−
1
2πi
dz
′
∧ d¯z
′
¯z
′
− ¯z
1
2
¯u(z
′
, ¯z
′
)N
1
(z
′
, ¯z
′
,k,
¯
k)e
−ik(z−z
′
)−i
¯
k(¯z−¯z
′
)
.
It is easily seen that equations (10.111) can be written in the standard
form with the Green function,
(GM)(z, ¯z, k,
¯
k)=
1
0
, (10.113)
where
(GM)(z, ¯z, k,
¯
k) ≡ M −
dz
′
R
dz
′
I
G(z
R
− z
′
R
,z
I
− z
′
I
,k,
¯
k)(QM)(z
′
, ¯z
′
,k,
¯
k)
and
G(z, ¯z, k,
¯
k)=
G
1
0
0 G
2
,G
1
=
1
2πz
,G
2
=
1
2π¯z
e
−i(kz+
¯
k¯z)
. (10.114)
The presence of exp[−i(kz +
¯
k¯z)] = exp[−2i(k
R
x − k
I
y)] in (10.114)
means that the Green function is nowhere analytic. In t u rn, the eig envec-
tors M and N are nowhere analytic as well. “Departure from analyticity”
¯
∂M ≡ ∂M/∂
¯
k determines the continuous spectrum and can be calculated
directly from (10.111). Indeed,
¯
∂M =
1
2πi
dz
′
∧ d¯z
′
z
′
− z
1
2
u(z
′
, ¯z
′
)
¯
∂M
2
(z
′
, ¯z
′
,k,
¯
k),
¯
∂M
2
= b(k,
¯
k)e
−i(kz+
¯
k¯z)
(10.115)
−
1
2πi
dz
′
∧ d¯z
′
¯z
′
− ¯z
1
2
¯u(z
′
, ¯z
′
)
¯
∂M
1
(z
′
, ¯z
′
,k,
¯
k)e
−ik(z−z
′
)−i
¯
k(¯z−¯z
′
)
,