Doktorov E.V., Leble S.B. A Dressing Method in Mathematical Physics
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10
Generating solutions via
¯
∂ problem
This chapter is devoted to a brief exposition of the
¯
∂ formalism, as applied
to nonlinear equations. The first three sections deal with the so-called non-
linear equations with self-consistent sources (or with nonanalytic dispersion
relations ). This class of nonlinear equations is physically interesting because
nonanalytic dispersi on relations are directly associated with the resonant in-
teraction of radiation with matter. In Sects. 10.1 and 10.2 we consider the
(1+1)-dimensional nonlinear Schr¨odinger (NLS) and modified NLS equations
with self-consistent sources, resp ectively, along with their gauge equivalents,
while Sect. 10.3 is devoted to the Davey–Stewartson I equation with a non-
analytic dispersion relation. We analyze these equations by means of the
¯
∂
approach. It should be noted that the Riemann–Hilbert (RH) problem could
be applied as well for this aim but, in our opinion, the
¯
∂ approach is frequently
the most transparent and leads directly to the final results. In the first three
sections, the
¯
∂ formalism is outlined in a rather unusual setting, but we prove
its usefulness for practical calculations.
The last two sections comprise examples of nonlinear equations where
using the
¯
∂ problem is necessary. Namely, in Sect. 10.4 we consider the
Kadomtsev–Petviashvili II (KP II) equation and Sect. 10.5 is concerned with
the Davey–Stewartson II (DS II) equations. The exposition in Sect. 10.4 is
fairly standard. Section 10.5 contains some recent results pertaining to mul-
tiple poles of discrete eigenfunctions.
10.1 Nonlinear equations with singular dispersion
relations: 1+1 dimensions
One of the ways to generalize nonlinear equations integrable by the IST
is to add a source (the so-called self-consistent source) to a given equa-
tion. Needless to say, this operation has to preserve the integr ability of the
equation. Mel’nikov [329] represented the source as a Fourier transform of
eigenfunctions of the recursion operator associated with the spectral problem.
319
320 10 Generating solutions via
¯
∂ problem
Claude et al. [90, 295, 293, 291] related the source to the singular (nonana-
lytic) component of the disper sion law which follows from the evolution part
of the Lax pair. Besides, (1+1)-dimensional nonlinear equations with a source
can arise as a result of reductions via the symmetry constraints of (2+1)-
dimensional equations [247, 479]. The importance of the singular dispersion
relations (SDR) stems from the fact that adding a source transforms, as a
rule, the initial-value problem to the initial boundary value problem. As is
well known, boundaries play a vital part in many physical applications.
We start the study of nonlinear equations with SDR from the NLS equation
with a source. A general approach to the solution of the Cauchy problem for
nonlinear equations with SDR a ssociated with the Zakharov–Shabat spectral
problem was discussed by Leon [292] in terms of the RH problem. Our aim
in this section is mainly to demonstrate the basic rules of working within the
framework of the
¯
∂ formalism. We will closely follow the formalism used by
Beals and Coifman [42] in their review article. A different approach to the
¯
∂
problem can be found in [220].
10.1.1 Spectral transform and Lax pair
We start from the matrix
¯
∂ problem in the complex k-plane,
¯
∂ψ = ψR, (10.1)
where R(x, t, k) is a spectral transform matrix which will be associated with
a nonlinear equation. For s i mpl icity we omit
¯
k in arguments of R(x, t, k)and
ψ(x, t, k), so the quantities like ψ and R are, in general, nonanalytic in some
domains in the k-plane (this may be everywhere in the k-plane). It is the
operator
¯
∂ that measures the “departure from analyticity,” when
¯
∂ψ =0.As
shown in Sect. 1.11, a solution of the
¯
∂ problem (10.1) with the canonical
normalization is written as
ψ(k)=
11+
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ψ(ℓ)R(ℓ) ≡ 11+ψRC
k
. (10.2)
Here C
k
is the Cauchy–Green integral operator acting on the left. It transforms
the argument k to ℓ in the function in front of it and integrates the result with
the weight (2πi)
−1
(ℓ−k)
−1
over the whole complex plane. The representation
(10.2) enables us to write formally a solution of the
¯
∂ problem (10.1) in terms
of the matrix R:
ψ(k)=11 · (11 − RC
k
)
−1
. (10.3)
We will see later that though (10.3) lo oks rather symbolic, we can do with it
all the manipulations we need.
Define a pairing
f,g =
1
2πi
dk ∧ d
¯
kf(k)g
T
(k), f,g
T
= g, f, (10.4)
10.1 Nonlinear equations with singular dispersion relations: 1+1 dimensions 321
where the superscript T means transposition. The pairing (10.4) possesses
easily verified properties:
fR,g = f, gR
T
, fC
k
,g = −f, gC
k
. (10.5)
It is important that the space–time dependence of the matrix R(x, t, k)
dictates completely the form of the Lax pair of a given equation. Let the
x-dependence be given by a simple linear equation
R
x
=ik[R, σ
3
]. (10.6)
Then we can perform with account of (10.3) the following calculation:
ψ
x
(k)=11·(11 − RC
k
)
−1
R
x
C
k
(11 − RC
k
)
−1
=ikψ(Rσ
3
− σ
3
R)C
k
(11 − RC
k
)
−1
=ikψRσ
3
C
k
(11 − RC
k
)
−1
− ikψσ
3
RC
k
(11 − RC
k
)
−1
.
The first term on the right-hand side is transformed in this way:
ikψRC
k
=
i
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ℓψ(ℓ)R(ℓ)=
i
2πi
dℓ ∧d
¯
ℓ
1+
k
ℓ − k
ψ(ℓ)R(ℓ)
=iψR,
11 +ik(ψRC
k
)=iψR, 11 +ik(ψ − 11) ≡ iψR +ik(ψ − 11) (10.7)
(sometimes we will omit 11in·, 11 unless its presence is important). As regards
the second term, we write RC
k
(11 − RC
k
)
−1
=(11 − RC
k
)
−1
− 11; hence,
ψ
x
(k)=iψR−ikσ
3
(11 − RC
k
)
−1
+ikψσ
3
.
Now the only problem is concerned with the term k(11 − RC
k
)
−1
.Wehave
from (10.7) kψRC
k
= ψR + kψ − k; hence, k = ψR + kψ(11 − RC
k
)and
finally
k(11 − RC
k
)
−1
= ψR·(11 − RC
k
)
−1
+ kψ =(ψR+ k)ψ. (10.8)
As a result, ψ
x
= −i[σ
3
, ψR]ψ − ik[σ
3
,ψ]. Introducing a potential
Q =
0 q
r 0
= −i[σ
3
, ψR], (10.9)
we arrive at the Zakharov–Shabat spectral problem
ψ
x
+ik[σ
3
,ψ] − Qψ =0. (10.10)
Hence, the x-equation for R (10.6) with linear dependence on k leads to the
ZS spectral problem.
For the time dependence of R we choose a linear equation as well:
R
t
=[R, Ω], (10.11)
322 10 Generating solutions via
¯
∂ problem
where Ω(k) is a dispersion relation. Suppose Ω(k) comprises b oth a polyno-
mial part Ω
p
(k) and a singular (nonanalytic) part Ω
s
(k). We put
Ω = Ω
p
+ Ω
s
= α
n
k
n
σ
3
+
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ω(ℓ)σ
3
. (10.12)
Here α
n
is a constant and ω(k) is some scalar function. Note that
¯
∂Ω
s
(k)
= ω(k)σ
3
, in accordance with (1.100). To derive the evolution Lax equation,
we proceed once again from the representation (10.3). First we consider the
polynomial dispersion relation only, Ω =2ik
2
σ
3
. Hence,
ψ
t
= ψR
t
C
k
(11 − RC
k
)
−1
= ψRΩC
k
(11 − RC
k
)
−1
− ψΩRC
k
(11 − RC
k
)
−1
= ψRΩC
k
(11 − RC
k
)
−1
− ψΩ(11 − RC
k
)
−1
+ ψΩ
=2i
k
2
ψRC
k
σ
3
(11 − RC
k
)
−1
− k
2
ψσ
3
(11 − RC
k
)
−1
+ ψΩ.
Note that we cannot factor out k
2
from the brackets because different op er-
ators C
k
σ
3
(11 − RC
k
)
−1
and (11 − RC
k
)
−1
act on k
2
.Thetermk
2
ψRC
k
is
transformed as follows:
k
2
ψRC
k
=
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ℓ
2
ψ(ℓ)R(ℓ)=
1
2πi
dℓ∧d
¯
ℓ
ℓ + k +
k
2
ℓ − k
ψ(ℓ)R(ℓ)
= kψR+ kψR + k
2
ψ − k
2
. (10.13)
This yields
ψ
t
=2i
kψRσ
3
ψ + ψRσ
3
k(11 − RC
k
)
−1
− k
2
σ
3
(11 − RC
k
)
−1
+ ψΩ.
(10.14)
The expression for k(11 − RC
k
)
−1
was o btained in (10.8). As regards k
2
(11 − RC
k
)
−1
, it follows from (10.13) that
k
2
= kψR + kψR+ k
2
ψ(11 − RC
k
)
and hence
k
2
(11 − RC
k
)
−1
=
kψR + ψR
2
+ kψR+ k
2
ψ.
Then we obtain from (10.14)
ψ
t
= −2i
[σ
3
, kψR]+iQψR +ikQ + k
2
σ
3
ψ + ψΩ. (10.15)
We can further simplify this equation owing to the relation between kψR
and Q
x
. Indeed,
Q
x
= −i[σ
3
, ψR
x
]=−2iψR
a
x
.
Here the superscript a means the off-diagonal part of a matrix (the superscript
d will denote the diagonal part). Further,
(ψR)
x
= −i[σ
3
, ψR,11]ψR − i[σ
3
,kψR].
10.1 Nonlinear equations with singular dispersion relations: 1+1 dimensions 323
Hence, ψR
x
= QψR−i[σ
3
, kψR]andψR
a
x
= QψR
d
− 2iσ
3
kψR
a
.
Therefore, we derive the following expression for Q
x
:
Q
x
= −2iσ
3
QψR
d
− 2iσ
3
kψR
a
,
which enables us to express kψR
a
as
kψR
a
= −
1
4
Q
x
−
i
2
σ
3
QψR
d
.
Substituting this formula into (10.15), we obtain the well -known second Lax
equation for the NLS equation:
ψ
t
=2i(−k
2
σ
3
− ikQ +
1
2
σ
3
Q
x
−
1
2
σ
3
Q
2
)ψ + ψΩ
p
.
Now we account for the SDR Ω
s
(10.12). Again we calculate ψ
t
:
ψ
t
=(ψRΩ
s
C
k
− ψΩ
s
)(11 − RC
k
)
−1
+ ψΩ
s
. The first term gives
ψRΩ
s
C
k
=
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ψ(ℓ)R(ℓ)
1
2πi
dm ∧ d
m
m − ℓ
ω(m)σ
3
.
The denominator is transformed as
1
ℓ − k
1
m − ℓ
=
1
m − k
1
ℓ − k
−
1
ℓ − m
.
Hence,
ψRΩ
s
C
k
=
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ψ(ℓ)R(ℓ)
1
2πi
dm ∧ d
m
m − k
ω(m)σ
3
−
1
2πi
dm ∧ d
m
m − k
ω(m)
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − m
ψ(ℓ)R(ℓ)σ
3
.
The first term of this formula gives [ψ(k) − 11]Ω(k), while the second one
produces
−
dm ∧ dm
m − k
ψ(m)ω(m)σ
3
+ Ω(k),
because the integral over ℓ gives ψRC
m
= ψ(m) − 11. As a result,
ψ
t
=
1
2πi
dm ∧ d
m
k − m
ψ(m)ω(m)σ
3
(11 − RC
k
)
−1
+ ψΩ
s
.
How does one calculate (k −m)
−1
(11 −RC
k
)
−1
in this integral? Following our
above experience, we consider first the term (k − m)
−1
ψRC
k
:
1
k − m
ψRC
k
=
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
1
ℓ − m
ψ(ℓ)R(ℓ)
324 10 Generating solutions via
¯
∂ problem
=
1
k − m
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ψ(ℓ)R(ℓ) −
1
k − m
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − m
ψ(ℓ)R(ℓ)
=
1
k − m
[(ψ(k) − 11) − (ψ(m) − 11)] =
1
k − m
[ψ(k) − ψ(m)] .
Hence,
1
k − m
ψ(k)(11 − RC
k
)=
1
k − m
ψ(m).
Taking the inverse of the last formula yields
1
k − m
(11 − RC
k
)
−1
=
1
k − m
ψ
−1
(m)ψ(k).
Therefore,
ψ
t
= −
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
ω(ℓ)ψ(ℓ)σ
3
ψ
−1
(ℓ)ψ(k)+ψΩ
s
= V
s
ψ + ψΩ
s
. (10.16)
Hence, we derived the Lax operator V
s
with distinctive squared eigenfunction
structure ψσ
3
ψ
−1
.
In fact, working with the
¯
∂ method, we need neither the spectral problem
(10.10) nor the evolution part (10.15) and (10.16) of the Lax pair. All we need
is the
¯
∂ problem (10.1) together with the linear equations (10.6) and (10.11)
governing the space–time dependence of the spectral transform R(x, t, k). In-
deed, we will establish later a gauge equiva lence of the NLS and Heisenberg
spin chain equations with SDR, derive the recursion op erators for them, and
find their soliton solutions without any resort to the Lax pair. It should be
also noted that when searching for the time evolution of spectral data, we use
only the asymptotic of the second Lax operator, i.e., the dispersion relation.
Just this necessary information is contained in the evolution equation (10.11)
for the spectral transform matrix.
10.1.2 Recursion operator
By means of (10.9) and (10.11) we will find the time evolution of the p otential
Q:
Q
t
= −i[σ
3
, ψR
t
].
Calculation gives
(ψR)
t
=
¯
∂ψ
t
=
¯
∂
&
11 · (11 − RC
k
)
−1
R
i
C
k
(11 − RC
k
)
−1
'
=
¯
∂
&
ψR
t
C
k
(11 − RC
k
)
−1
'
=
¯
∂
&
ψR
t
(11 − RC
k
)
−1
'
C
k
= ψR
t
(11 − RC
k
)
−1
.
In p erforming the last step we made use of the evident relation
¯
∂f(k)C
k
= f (k). Therefore, in virtue of the properties (10.5), we can write
Q
t
= −i
σ
3
ψR
t
(11 − RC
k
)
−1
, 11
= −i
σ
3
, ψR
t
, 11 · (11+R
T
C
k
)
−1
.
(10.17)
10.1 Nonlinear equations with singular dispersion relations: 1+1 dimensions 325
It can be shown, owing to
¯
∂(ψ
−1
)
T
= −(ψ
−1
)
T
R
T
,that
11 · (11+R
T
C
k
)
−1
=(ψ
−1
)
T
. (10.18)
Hence,
Q
t
= −i
σ
3
, ψ(RΩ − ΩR), (ψ
−1
)
T
= −i
σ
3
, (
¯
∂ψ)Ωψ
−1
, 11
+i
σ
3
, ψΩ,
¯
∂(ψ
−1
)
T
= −i
σ
3
, (
¯
∂ψ)Ωψ
−1
+ ψΩ
¯
∂ψ
−1
= −i
σ
3
,
¯
∂(ψΩ
p
ψ
−1
)−ψ(
¯
∂Ω
s
)ψ
−1
,
wherewehaveusedΩ
s
→ 0atk →∞. Taking into account (10.12), we obtain
Q
t
= −iα
n
[σ
3
,
¯
∂(k
n
ψσ
3
ψ
−1
)]+i[σ
3
, ω(k)ψσ
3
ψ
−1
]. (10.19)
Denote M(k)=ψσ
3
ψ
−1
. This function satisfies the equation
M
x
+ik[σ
3
,M] − [Q, M ]=0. (10.20)
The next steps in deriving the recursion operator are rather standard [148].
Let us write the 2 ×2matrixM as a sum of diagonal and off-diagonal parts,
M = M
d
+ M
a
=
1
2
σ
3
tr(Mσ
3
)+
1
4
[σ
3
, [σ
3
,M]] . (10.21)
Then (10.20) gives two equations
M
d
x
=[Q, M
a
],M
a
x
+2ikσ
3
M
a
=[Q, M
d
]. (10.22)
According to the asymptotic condition ψ → 11atx →∞, we obtain from the
first equation in (10.22) M
d
= σ
3
+ ∂
−1
[Q, M
a
], which enables us to write the
second equation as
M
a
x
+2ikσ
3
M
a
=
Q, σ
3
+ ∂
−1
[Q, M
a
]
. (10.23)
Hence, it is natural to introduce a recursion op erator in the form
Λ· =
i
2
σ
3
∂
x
−
Q, ∂
−1
[Q, ·]
, (10.24)
which evidently does not depend on k. Then (10.23) gives M
a
= −i(Λ−k)
−1
Q
and
Q
t
= −α
n
σ
3
,
¯
∂
k
n
(Λ − k)
−1
Q
+i[σ
3
, ω(k)M (k)].
Now we expand (Λ−k)
−1
in a series, (Λ−k)
−1
= −
∞
j=1
k
−j
Λ
j−1
,andmake
use of
¯
∂k
n−j
= πδ(k)δ
j,n+1
.Then
⎡
⎣
σ
3
,
∞
j=1
¯
∂k
n−j
Λ
j−1
Q
⎤
⎦
= −[σ
3
,Λ
n
Q]=−2σ
3
Λ
n
Q.
326 10 Generating solutions via
¯
∂ problem
As a result, we obtain the hierarchy of equations with SDR associated with the
particular x-dep endence (10.6) of the spectral transform (or, in other words,
with the ZS spectral problem):
Q
t
+2α
n
σ
3
Λ
n
Q =i[σ
3
, ω(k)M (k)], (10.25)
M
x
=[−ikσ
3
+ Q, M].
Let us consider two examples of the initial boundary value problem de-
scribed by the system (10.25).
1. n =1,α
1
= −i, ω(k)=πg(k
R
)δ(k
I
), k = k
R
+ik
I
,
Q =
1
2
0 E
−E
∗
0
,M=
i
2
−np
p
∗
n
.
In this case ΛQ =(i/2)σ
3
Q
x
. Therefore,
E
t
+ E
x
= p,p
x
+2ik
R
p = En, n
x
= −
1
2
(E ¯p +
¯
Ep). (10.26)
Here the double brackets p =
∞
−∞
dk
R
g(k
R
)p(k
R
) stand for the av-
erage over the inhomogeneous broadening with the distribution function
g(k
R
), when referring to the model of the radiation–matter interaction.
Equations (10.26) are solved with the initial condition for E at t =0and
boundary conditions for p and N at x →−∞or x → +∞.Equations
(10.26) are o f the Maxwell–Bloch equation type and describe a number
of p h enomena like self-induced transparency [259, 325] and stimulated
Raman scattering [91, 230, 410]. Leon [294] has shown that the system
(10.26) is integrable for arbitrary boundary values.
2. n =2,Λ
2
Q = −(1/4)Q
xx
+(1/2)Q
3
,
Q =
0 E
−E
∗
0
,M=i
−np
p
∗
n
.
Equations (10.25) give
iE
t
+ E
xx
+2|E|
2
E =2ip,p
x
+2ik
R
p = −2En, n
x
= −(E ¯p +
¯
Ep).
(10.27)
These equations (without inhomogeneous broadening) were derived from
physical motivations by Doktorov and Vlasov [127, 441] to describe the dy-
namics of picosecond optical pulses in a co mbined re sonant-cubic medium.
This system can also be applied to nonlinear interac tion of the electro-
static high-frequency wave with an ion-acoustic wave in two-component
homogeneous plasma [90].
10.1.3 NLS–Maxwell–Bloch soliton
Here we will obtain the soliton solution of equations (10.27) within the
¯
∂
method. The soliton corresponds to the spectral transform matrix located at
10.1 Nonlinear equations with singular dispersion relations: 1+1 dimensions 327
the points k
1
and
¯
k
1
of the complex plane where a solution ψ of the
¯
∂ problem
has simple poles. Namely,
R =2πie
−ikσ
3
x
0 cδ(k − k
1
)
¯cδ(k −
¯
k
1
)0
e
ikσ
3
x
, (10.28)
where c = c(t) and this time dependence should be found from (10.11). As
follows from (10.9), the soliton solution is given by
q
s
= −2iψR
12
= −
1
π
dk ∧ d
¯
kψ
11
(k)R
12
(k). (10.29)
Substituting the explicit form of R (10.28) into (10.29), we derive a linear
algebraic system
ψ
11
(k)=1+
2i¯c
k − k
1
ψ
12
(
¯
k
1
)e
2i
¯
k
1
x
,ψ
12
=
2ic
k − k
1
ψ
11
(k
1
)e
−2ik
1
x
.
Solving this system with respect to ψ
11
(k) and substituting the result into
(10.29) yields
q
s
=2ηe
−2iξx+iφ
sech2η(x − x
0
). (10.30)
Here k
1
= ξ +iη, c = −η exp(−2ηx
0
+iφ), and x
0
and φ are time-d ependent.
The time dependence is found from (10.11). Let us denote
Ω
s
(k
1
)=
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
1
ω(ℓ)σ
3
≡ (ω
1
− iω
2
)σ
3
.
Then c
t
= −2c(2ik
2
1
+ ω
1
− iω
2
). On the other hand, c
t
= c(−2ηx
0t
+iφ
t
).
Comparing these two equation s, we find
x
0
=
−4ξ +
ω
1
η
t + ξ
0
,φ= −4(ξ
2
− η
2
)t +2ω
2
t + φ
0
, (10.31)
where ξ
0
and φ
0
are constants. As a result, the soliton solution of the NLS–
Maxwell–Bloch system has the form of the standard NLS soliton (8.32) but
with modulated velocity and phase. This modulation is caused by the res-
onant comp onent of the combined medium and manifests itself through the
frequencies ω
1
and ω
2
. Other quantities p and n are easily calculated with the
known soliton solution (10.30) by means of (10.26).
10.1.4 Gauge equivalence
Up to now we considered the
¯
∂ problem wit h the canonical normalization
ψ(k) → 11+O(1/k). To what extent is the demand of the canonical normal-
ization critical ? To answer this question, let us multiply (10.2) by a nondegen-
erate matrix g
−1
(x, t) from the left [matrix function g(x, t) does not depend
on k] and put ϕ(k)=g
−1
ψ(k). Then evidently we get
ϕ(k)=g
−1
(11 − RC
k
)
−1
. (10.32)
328 10 Generating solutions via
¯
∂ problem
Calculation of the x-derivative of ϕ yields
ϕ
x
=
−g
−1
g
x
+iϕRσ
3
g − ig
−1
σ
3
gϕRg
ϕ−ikg
−1
σ
3
gϕ+ikϕσ
3
. (10.33)
Now we choose g(x, t) in such a way that the expression in brackets in (10.33)
vanishes. This gives g
x
+i[σ
3
, ψR]g =0org
x
= Qg. It means that we can
put g = ψ(k =0), with ψ being a solution of the
¯
∂ problem. Denoting
S(x, t)=g
−1
σ
3
g, (10.34)
we obtain from (10.32)
ϕ
x
+ikSϕ − ikϕσ
3
=0. (10.35)
This equation is nothing more than the spectral problem for the Heisenberg
spin chain model [424] which is described by the equation
S
t
=
1
2i
[S, S
xx
],S
2
= 11. (10.36)
In the same way we could derive the evolution part of the Lax pair for
(10.36) and r eproduce the well-known gauge equivalence between the Lax
operators for the NLS and Heisenberg equations [476]. However, because we
work with the
¯
∂ problem, we do not need the Lax operators. Within the
¯
∂
method, the gauge equivalence is realized as a change of the normalization
condition by means of the function ψ(k =0).
10.1.5 Recursion operator for Heisenberg spin chain equation
with SDR
To derive the recursion operator for the Heisenberg spin chain equation with
SDR, we start, as in Sect. 10.1.2, from the calculation of the time derivative
of S: S
t
=[S, g
−1
g
t
]. Because
g = ψ(k =0)=11+
1
2πi
dℓ ∧ d
¯
ℓ
ℓ
ψ(ℓ)R(ℓ)=11+
1
k
ψR,11,
we obtain
g
t
=
1
k
ψR
t
(11 − C
k
R)
−1
, 11 = ψR
t
,
1
k
(11+R
T
C
k
)
−1
. (10.37)
We have shown already [see (10.18)] that the operator 11+R
T
C
k
is intimately
related to the function (ψ
−1
)
T
. Therefore, to calculate (1/k)(11+R
T
C
k
)
−1
in
(10.37), we begin with
1
k
(ψ
−1
)
T
R
T
C
k
=
1
2πi
dℓ ∧ d
¯
ℓ
ℓ − k
1
ℓ
(ψ
−1
)
T
(ℓ)R
T
(ℓ)