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Journal of Mathematical Physics, Analysis, Geometry

2017, Vol. 13, No. 2, pp. 119–153

doi: 10.15407/mag13.02.119

Maxwell–Bloch Equations without Spectral

Broadening: Gauge Equivalence, Transformation

Operators and Matrix Riemann–Hilbert Problems

M.S. Filipkovska, V.P. Kotlyarov,

and E.A. Melamedova (Moskovchenko)

B. Verkin Institute for Low Temperature Physics and Engineering

of the National Academy of Sciences of Ukraine

47 Nauky Ave., Kharkiv, 61103, Ukraine

E-mail: ﬁlipkovskaya@ilt.kharkov.ua

kotlyarov@ilt.kharkov ua

melamedova@ilt.kharkov.ua

Received January 26, 2017, revised March 26, 2017

A mixed initial-boundary value problem for nonlinear Maxwell–Bloch

(MB) equations without spectral broadening is studied by using the inverse

scattering transform in the form of the matrix Riemann–Hilbert (RH) prob-

lem. We use transformation operators whose existence is closely related with

the Goursat problems with nontrivial characteristics. We also use a gauge

transformation which allows us to obtain Goursat problems of the canoni-

cal type with rectilinear characteristics, the solvability of which is known.

The transformation operators and a gauge transformation are used to obtain

the Jost type solutions of the Ablowitz–Kaup–Newel–Segur equations with

well-controlled asymptotic behavior by the spectral parameter near singular

points. A well posed regular matrix RH problem in the sense of the feasibil-

ity of the Schwartz symmetry principle is obtained. The matrix RH problem

generates the solution of the mixed problem for MB equations.

Key words: Maxwell–Bloch equations, gauge equivalence, transformation

operators, matrix Riemann–Hilbert problems.

Mathematics Subject Classiﬁcation 2010: 34L25, 34M50, 35F31, 35Q15,

35Q51.

c

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko), 2017

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

The paper is dedicated to the 95th anniversary

of Vladimir Aleksandrovich Marchenko

1. Introduction

Special integral operators that transform solutions of diﬀerential equations

with constant coeﬃcients into solutions of equations with variable coeﬃcients

are a distinctive feature of the Marchenko’s papers. In this paper, we propose

a systematic use of the transformation operators for the construction of matrix

Riemann–Hilbert problems which lead to the solution of the initial-boundary

value problem for nonlinear Maxwell–Bloch equations without spectral broaden-

ing. The transformation operators proposed are closely related to the Goursat

problems with nontrivial characteristics. The use of a gauge transformation al-

lows one to obtain the Goursat problems of the canonical type with rectilinear

characteristics as well as their solvability. We use the same transformation to

establish a gauge equivalence between two pairs of the Ablowitz–Kaup–Newel–

Segur (AKNS) equations to construct their Jost type solutions with the well-

controlled asymptotic behavior by a spectral parameter on the complex plane

near the singular points. As a result, the well-posed regular matrix RH problem,

which generates the solution of the mixed problem for MB equations, is obtained.

The Maxwell–Bloch equations in the integrable case have the following form

(sf. [16]):

∂E

∂t +∂E

∂x =hρi,(1)

∂ρ

∂t + 2iλρ =N E ,(2)

∂N

∂t =−1

2Eρ+Eρ.(3)

Here the symbol ¯ denotes a complex conjugation, E=E(t, x) is a complex valued

function of the space variable xand the time t,ρ=ρ(t, x, λ), and N(t, x, λ) are

the complex valued and real functions of t,xand a spectral parameter λ. The

angular brackets hi mean the averaging by λwith the given ”weight” function

n(λ),

hρi=

∞

Z

−∞

ρ(t, x, λ)n(λ)dλ,

∞

Z

−∞

n(λ)dλ =±1.(4)

If n(λ)>0, then an unstable medium is considered (the so-called two-level

laser ampliﬁer). If n(λ)<0, then a stable medium is considered (the so-called

attenuator).

Equations (1)–(4) have appeared in many physical models. However, ﬁrst

they were studied in [22–25]. The most important is a model of the propagation

120 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

of electromagnetic waves in a medium with distributed two-level atoms. For

example, there are models of self-induced transparency [1,2,14–17], and two-level

laser ampliﬁers [14–16, 27–30]. For these models, E(t, x) is the complex valued

envelope of electromagnetic wave of ﬁxed polarization, N(t, x, λ) and ρ(t, x, λ)

are the entries of a density matrix of the atomic subsystem

ˆρ(t, x, λ) = N(t, x, λ)ρ(t, x, λ)

ρ(t, x, λ)−N (t, x, λ).

The parameter λdenotes a deviation of the transition frequency from its mean

value. The weight function n(λ) characterizes the inhomogeneous broadening

which is the diﬀerence between the initial population of the upper and lower

levels. Short reviews on the Maxwell–Bloch equations and applying to them of

the inverse scattering transform (IST) method can be found in [1, 2, 16].

We restrict our study to the case where n(λ) = δ(λ), i.e., without spectral

broadening. Then hρi=ρand the system (1)–(4) is written as

∂E

∂t +∂E

∂x =ρ, ∂ρ

∂t =N E ,∂N

∂t =−1

2Eρ+Eρ.(5)

These equations are simpler than (1)–(4). However, applying of the IST method

to (5) is somewhat complicated. The matrix Riemann–Hilbert problem for MB

equations (1)–(4) was studied in [19]. The main goal of this paper is to study a

mixed problem for the Maxwell–Bloch equations which is deﬁned by the initial

and boundary conditions:

E(0, x) = E0(x), ρ(0, x) = ρ0(x),N(0, x) = N0(x),E(t, 0) = E1(t),(6)

where x∈(0, l) (l≤ ∞) and t∈R+. The function E1(t) is a Schwartz-type

function (smooth and fast decreasing at inﬁnity). The functions E0(x), ρ0(x), 1 −

N0(x) are smooth if x∈[0, l] or of Schwartz type if x∈R+.

The functions ρ(t, x), N(t, x) are not independent. Indeed, equations (2), (3)

give ∂

∂t |ρ(t, x)|2+N(t, x)2= 0,and we put

|ρ(t, x)|2+N(t, x)2≡1.

If we deﬁne ρ(0, x), then

N(0, x) = ∓p1− |ρ(0, x)|2.

If one chooses the sign ”minus”, then the problem is considered in a stable

medium, in the so-called attenuator (for example, a model of self-induced trans-

parency). The matrix Riemann–Hilbert problems were studied for this case

in [18, 20]. If the sign ”plus” is chosen, then the problem is considered in an

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 121

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

unstable medium (for example, a model of a two-level laser ampliﬁer), which is

the subject of our study.

We suppose that the functions E(t, x), ρ(t, x) and N(t, x) satisfy the MB

equations (5) in the domain x, t ∈(0, l)×(0,∞). We develop the IST method

in the form of the matrix Riemann–Hilbert problem in a complex z-plane. The

method is based on using the transformation operators for constructing the Jost

type solutions of the AKNS equations. The RH problem is deﬁned by spectral

functions which, in turn, are deﬁned through the given initial and boundary

conditions for the MB equations. The RH problem is meromorphic and simple in

some sense: it is deduced by using standard approaches to the inverse scattering

transform for the quarter of the xt-plane. Unfortunately, this RH problem has

an essential deﬁciency because it may have multiple eigenvalues and spectral

singularities. Therefore we deduce a new regular matrix RH problem, free from

the mentioned deﬁciency, which has the unique solution. Then we prove that

this regular RH problem generates a system of compatible diﬀerential equations,

which is the AKNS system of linear equations [2] for the MB equations without

broadening. Thus the RH problem generates a solution to the MB equations.

Our approach diﬀers from those considered for the mixed problem to the MB

equations given in [1,16,18, 27,28]. We develop an approach to the simultaneous

spectral analysis proposed in [10–13] and in [3–7, 21] for other nonlinear equations

and prove that the mixed problem for the MB equations is completely linearizable

by the appropriate matrix RH problem.

2. Gauge Transformation of the Ablowitz–Kaup–Newel–Segur

Equations

The Ablowitz–Kaup–Newel–Segur equations for the Maxwell–Bloch equations

without spectral broadening have the form:

Φt=U(t, x, λ)Φ, U(t, x, λ) = −(iλσ3+H(t, x)),(7)

Φx=V(t, x, λ)Φ, V (t, x, λ)=iλσ3+H(t, x) + iF(t, x)

4λ,(8)

where H(t, x) = 1

20E(t, x)

−E(t, x) 0 , F (t, x) = N(t, x)ρ(t, x)

ρ(t, x)−N (t, x), and

σ3=1 0

0−1is the Pauli matrix. It is well known [2] that the over-determined

system of diﬀerential equations (7), (8) is compatible if and only if the compati-

bility condition

Ux−Vt+ [U, V ] = 0 (9)

holds. Equation (9) is equivalent to the system of nonlinear equations

∂H

∂t +∂H

∂x =1

4[σ3, F ],∂F

∂t = [F, H],(10)

122 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

which are the matrix form of MB equations (5).

Two systems of the AKNS equations,

Φt=U(t, x, λ)Φ,

Φx=V(t, x, λ)Φ

and

Ψt=ˆ

U(t, x, λ)Ψ,(11)

Ψx=ˆ

V(t, x, λ)Ψ,(12)

are called gauge equivalent [9] if their compatible solutions are related by

Φ(t, x, λ) = g(t, x)Ψ(t, x, λ),

where the unitary matrix g(t, x) does not depend on λ. Obviously, the matrices

(ˆ

U,ˆ

V) and (U,V) are connected by the relations:

ˆ

U=g−1Ug −g−1gt,(13)

ˆ

V=g−1V g −g−1gx.(14)

The nonlinear equations, deﬁned by the corresponding compatibility conditions

Ux−Vt+ [U, V ] = 0,ˆ

Ux−ˆ

Vt+ [ ˆ

U, ˆ

V]=0,

are also called gauge equivalent [9].

Among all gauge transformations we are interested only in those which make

the matrix F(t, x) be diagonal, i.e.,

F(t, x) = g(t, x)σ3g−1(t, x), g−1(t, x) = g∗(t, x),(15)

where ∗means the Hermitian conjugation. This equality does not deﬁne the

unitary matrix g(t, x) uniquely. It is deﬁned up to the unitary diagonal matrix

g(t, x) = D(t, x)eχ(t,x)σ3,(16)

where χ=χ(t, x) is an imaginary scalar function. It is convenient to chose the

matrix D(t, x) in the form

D(t, x) = 1

p2(1 + N(t, x)) 1 + N(t, x)−ρ(t, x)

¯ρ(t, x) 1 + N(t, x),(17)

where the root is its arithmetic value. It is easy to verify that det D(t, x)≡1

and D−1(t, x) = D∗(t, x). Then for the matrix ˆ

U(13) we obtain

ˆ

U=−iλS(t, x) + ˆ

H(t, x),

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 123

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

where

S(t, x) :=g−1(t, x)σ3g(t, x),

ˆ

H(t, x) :=g−1(t, x)H(t, x)g(t, x) + g−1(t, x) ˙gt(t, x).

The matrix ˆ

H(t, x)=e−χ(t,x)σ3(D−1HD +D−1˙

Dt+ ˙χtσ3)eχ(t,x)σ3. It will be

identically equal to zero if the matrix ˆ

D=D−1HD +D−1˙

Dtcommutes with σ3.

Indeed, in this case, the matrix ˆ

Dis proportional to σ3,

ˆ

D=f(t, x)σ3.(18)

Then putting ˙χt=−f(t, x), one ﬁnds that ˆ

H(t, x)≡0. In order to prove (18),

we use the second equation from (10). Since F=gσ3g−1=Deχσ3σ3e−χσ3D−1=

Dσ3D−1, then ˙

Ft−[F, H ] = D[D−1˙

Dt, σ3]D−1−[Dσ3D−1, H ] =

=D[D−1˙

Dt+D−1HD, σ3]D−1=D[ˆ

D, σ3]D−1= 0

It means that [ ˆ

D, σ3] = 0 and hence ˆ

D=f(t, x)σ3+β(t, x)I, where f=f(t, x)

and β=β(t, x) are arbitrary scalars. Further, since tr ˆ

D= tr H+ tr D−1˙

Dt=

tr D−1˙

Dt= 2N˙

Nt+ ˙ρt¯ρ+ρ

¯

˙ρt≡0, one ﬁnds β= 0. Finally, in view of

˙χt=−f(t, x), f(t, x)=(D−1HD +D−1˙

Dt)11,

where (·)11 means (11) element of the matrix, we have that ˆ

H(t, x)≡0, and the

matrix ˆ

Uis equal to

ˆ

U(t, x, λ) = −iλS(t, x), S(t, x) = ν(t, x)p(t, x)

¯p(t, x)−ν(t, x),

where S(t, x) = g−1(t, x)σ3g(t, x) = S∗(t, x) and S2≡I.

For the matrix ˆ

V=ˆ

V(t, x, λ), we have

ˆ

V=g−1V g −g−1g0

x= iλS(t, x) + R(t, x) + iσ3

4λ,

where R=g−1Hg −g−1g0

x= e−χσ3(D−1HD −D−1D0

x−χ0

xσ3)eχσ3. Since tr R=

0, then the matrix Rtakes the form

R= (h(t, x)−χ0

x)σ3+σ3

2[σ3, g−1Hg −g−1g0

x],

where h(t, x)=(D−1HD −D−1D0

x)11. By putting χ0

x=h(t, x), we ﬁnd that

R(t, x) = σ3

2[σ3,e−χσ3(D−1HD −D−1D0

x)eχσ3] = 0r(t, x)

−¯r(t, x) 0 .

124 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

All written above is true if the equations

˙χt=−f(t, x), f(t, x) = ¯ρ

4E − ˙ρt

1 + N−ρ

4¯

E − ¯

˙ρt

1 + N,

˙χx=h(t, x), h(t, x) = ¯ρ

4E+ρ0

x

1 + N−ρ

4¯

E+¯

ρ0x

1 + N

are compatible. Indeed, these simple equations are compatible if and only if

∂h

∂t +∂f

∂x = 0.

By using the Maxwell–Bloch equations (5), this condition can be veriﬁed by

routine calculations. Then χ(t, x) is deﬁned as an integral

χ(t, x) =

(t,x)

Z

(0,0)

h(t, s)ds −f(τ, x)dτ +χ(0,0),

which does not depend on a path of integration. The free parameter χ(0,0) will

be used in Sec. 5.

The gauge Eqs. (11) and (12) with

ˆ

U=−iλS(t, x),ˆ

V= iλS(t, x) + R(t, x) + iσ3

4λ

are also compatible ˆ

Ux−ˆ

Vt+ [ ˆ

U, ˆ

V] = 0.

The last equation is equivalent to the system of nonlinear equation

∂S

∂t +∂S

∂x = [R, S ], Rt=1

4[S, σ3].

Thus we have proved the theorem below.

Theorem 1. The Maxwell–Bloch Eqs. (5) are gauge equivalent to the

equations

∂ν

∂t +∂ν

∂x =p¯r+ ¯pr, ∂p

∂t +∂p

∂x =−2νr, ∂r

∂t =−1

2p,

where ν=ν(t, x)is real, and p=p(t, x),r=r(t, x)are complex val-

ued functions which constitute the matrices S=ν(t, x)p(t, x)

¯p(t, x)−ν(t, x)and R=

0r(t, x)

¯r(t, x) 0 .

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 125

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

R e m a r k 1. The second matrix equation Rt=1

4[S, σ3] gives

S=νσ3+ [Rt, σ3] = ν σ3+ 2Rtσ3=ν−2rt

−2¯rt−ν.

Then the ﬁrst matrix equation ∂S

∂t +∂S

∂x = [R, S ] is read as

∂ν

∂t +∂ν

∂x + 2 ∂|r|2

∂t = 0,∂2r

∂t2+∂2r

∂x∂t =νr.

The gauge AKNS equations (11) and (12) will be used below for the trans-

formation of some Goursat problems and also for the construction of compatible

solutions of the AKNS equations (7) and (8) which have a well-controlled asymp-

totic behavior as z→0.

3. Basic Solutions of the Ablowitz–Kaup–Newel–Segur Linear

Equations

We suppose here that the solution (E(t, x), N(t, x), ρ(t, x)) of the mixed

problem (5), (6) for the Maxwell–Bloch equations in the domain t∈R+, 0 ≤x≤

l≤ ∞ exists, and it is unique and smooth. Then the AKNS linear equations (7)

and (8) are compatible. To construct their solutions we use the following lemma.

Lemma 1. Let Eqs. (7) and (8) be compatible for all t, x, λ ∈R. Let

Φ(t, x, λ)be a matrix satisfying the t-equation (7) for all x(the x-equation (8)

for all t). Assume that Φ(t0, x, λ)satisﬁes the x-equation (8) for some t=t0≤

≤ ∞ (the t-equation (7) for some x=x0≤ ∞). Then Φ(t, x, λ)satisﬁes the

x-equation (8) for all t(satisﬁes the t-equation (7) for all x).

P r o o f. The proof can be found in [3] (Lemma 2.1).

Let Y(t, x, λ) be a product of the matrices

Y(t, x, λ) = W(t, x, λ)Φ(t, λ),(19)

where W(t, x, λ) satisﬁes the x-equation (8) for all tand W(t, 0, λ) = I, and

Φ(t, λ) satisﬁes the t-equation (7) for x= 0 under the initial condition

lim

t→∞ Φ(t, λ)eiλtσ3=I.

Let Z(t, x, λ) be a product of the matrices

Z(t, x, λ) = Ψ(t, x, λ)w(x, λ),(20)

126 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

where Ψ(t, x, λ) satisﬁes the t-equation (7) for all xand Ψ(0, x, λ) = I, and

w(x, λ) satisﬁes the x-equation (8) for t= 0 under the initial condition w(l, λ) =

eilµ(λ)σ3, where µ(λ) = λ+1

4λ. If l=∞, the initial condition takes the form

lim

x→∞ w(x, λ)e−ixµ(λ)σ3=I.

It is easy to see that due to Lemma 1, the matrices Y(t, x, λ) and Z(t, x, λ)

are compatible solutions of the AKNS system of Eqs. (7), (8).

Lemma 2. Let E(t, 0) = E1(t)be smooth and fast decreasing as t→ ∞.

Then for Im λ= 0 there exists the Jost solution Φ(t, λ)of the t-equation (7) with

x= 0 represented by the transformation operator

Φ(t, λ) = e−iλtσ3+

∞

Zt

K(t, τ )e−iλτσ3dτ, Im λ= 0.(21)

The kernel K(t, τ )satisﬁes the symmetry condition K(t, τ ) = ΛK(t, τ )Λ−1with

matrix Λ = 0 1

−1 0and it is deﬁned by the Goursat problem:

σ3

∂K (t, τ )

∂t +∂K(t, τ )

∂τ σ3=H(t, 0)σ3K(t, τ ),

σ3K(t, t)−K(t, t)σ3=σ3H(t, 0),

lim

t+τ→+∞K(t, τ )=0.

The kernel K(t, τ )is smooth and fast decreasing as t+τ→ ∞.

The proof uses the Goursat problem and the corresponding integral equations

which allow one to prove their unique solvability and thus to prove the inte-

gral representation (21) (sf. [9]). Due to (21), the vector columns Φ[1](t, λ) and

Φ[2](t, λ) of the matrix Φ(t, λ) = (Φ[1](t, λ),Φ[2](t, λ)) have analytic continua-

tions Φ[1](t, z) and Φ[2](t, z) to the lower half-plane C−and the upper half-plane

C+of the complex z-plane, respectively. Thus the vector columns Φ[1](t, z),

Φ[2](t, z) are analytic in C−,C+, respectively, continuous in C−∪R,C+∪Rand

have the following asymptotics:

Φ[1](t, z)eizt =1

0+ O(z−1),Im z≤0, z → ∞;

Φ[2](t, z)e−izt =0

1+ O(z−1),Im z≥0, z → ∞.

The symbol O(.) means a matrix whose entries have the indicated order.

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 127

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

Lemma 3. Let E(t, x),N(t, x),ρ(t, x)be smooth. Then for any xand

Im λ= 0 there exists the solution Ψ(t, x, λ)of the t-equation (7) represented by

the transformation operator

Ψ(t, x, λ) = e−iλtσ3+

t

Z

−t

K0(t, τ, x)e−iλτ σ3dτ, Im λ= 0.(22)

The kernel K0(t, τ, x)is smooth, it satisﬁes the symmetry condition K0(t, τ, x) =

ΛK0(t, τ, x)Λ−1with matrix Λ = 0 1

−1 0and is deﬁned by the Goursat problem:

σ3

∂K0(t, τ , x)

∂t +∂K0(t, τ, x)

∂τ σ3=H(t, x)σ3K0(t, τ , x),

σ3K0(t, t, x)−K0(t, t, x)σ3=H(t, x)σ3,

σ3K0(t, −t, x) + K0(t, −t, x)σ3=0.

The proof of the lemma can be found in [3].

The integral representation (22) gives the analyticity of the solution Ψ(t, x, z)

for z∈Cand its asymptotic behavior as z→ ∞:

Ψ(t, x, z)eizt =1 0

0e2izt+ O(z−1) + O(e2iztz−1), z → ∞;

Ψ(t, x, z)e−izt =e−2izt 0

0 1+ O(z−1) + O(e−2iztz−1), z → ∞.

Since tis positive, these asymptotics mean that Ψ(t, x, z)eizt is bounded in C+,

and Ψ(t, x, z)e−izt is bounded in C−as z→ ∞. For any ﬁxed tand x, they are

bounded on any compact set of the complex plane.

Now we pass to the construction of the solutions of the x-equation (8) which

has two singular points ∞and 0 (t-equation (7) had the singular point at ∞

only). First, we consider the Jost solution of (8) which has a good behavior as

z→ ∞. With this purpose, we represent the x-equation (8) in the form

Wx=iµ(λ)σ3+H(t, x) + i

4λ(F(t, x)−σ3)W, µ(λ) = λ+1

4λ.(23)

If H(t, x)≡0 and F(t, x)≡σ3, the x-equation has the exact solution eixµ(λ)σ3.

Lemma 4. Let E(t, x),N(t, x),ρ(t, x)be the smooth functions for 0≤

x≤l, and

N2(t, x) + |ρ(t, x)|2≡1.

128 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

Then for any tand Im µ(λ)=0there exists the solution W(t, x, λ)of the x-

equation (23) represented by the transformation operators

W(t, x, λ) = eixµ(λ)σ3+

x

Z

−x

L(x, y, t)eiyµ(λ)σ3dy +1

iλ

x

Z

−x

M(x, y, t)eiyµ(λ)σ3dy. (24)

The kernels L(x, y, t)and M(x, y, t)are smooth, they satisfy the symmetry con-

dition L(x, y, t) = ΛL(x, y, t)Λ−1,M(x, y.t) = ΛM(x, y, t)Λ−1with matrix Λ =

0 1

−1 0, and are deﬁned by the formulas L(x, y, t) = ˆ

L(x, y, t)and M(x, y, t) =

g(t, x)ˆ

M(x, y, t), where the unitary matrix g(t, x)is the same as in Section 2. The

matrices ˆ

L(x, y, t)and ˆ

M(x, y, t)are the unique solution of the Goursat problem:

σ3

∂ˆ

L

∂x +∂ˆ

L

∂y σ3=σ3H(t, x)ˆ

L+σ3[σ3, g(t, x)] ˆ

M,

σ3

∂ˆ

M

∂x +∂ˆ

M

∂y σ3=σ3R(t, x)ˆ

M+σ3

4[g−1(t, x), σ3]ˆ

L,

σ3ˆ

L(x, x, t)−ˆ

L(x, x, t)σ3=σ3H(t, x),

σ3ˆ

M(x, x, t)−ˆ

M(x, x, t)σ3=1

4[σ3, g−1(t, x)]σ3,(25)

σ3ˆ

L(x, −x, t) + ˆ

L(x, −x, t)σ3=0,

σ3ˆ

M(x, −x, t) + ˆ

M(x, −x, t)σ3=0,

where R(t, x) = g−1(t, x)H(t, x)g(t, x)−g−1(t, x)g0

x(t, x).

P r o o f. Substituting (24) into equation (23) and integrating by parts, we

get the Goursat problem (−x<y<x):

∂L

∂x +σ3

∂L

∂y σ3=H(t, x)L+ (σ3−F(t, x))M,

∂M

∂x +F(t, x)∂M

∂y σ3=H(t, x)M+1

4(σ3−F(t, x)) M,

σ3L(x, x, t)−L(x, x, t)σ3=σ3H(t, x),

F(t, x)M(x, x, t)−M(x, x, t)σ3=−1

4(σ3−F(t, x)) σ3,

σ3L(x, −x, t) + L(x, −x, t)σ3=0,

F(t, x)M(x, −x, t) + M(x, −x, t)σ3=0.

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 129

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

To integrate the summand with the factor 1

λ2we have to use the identity 1

λ2=

4µ

λ−4.

We have obtained the Goursat problem with variable coeﬃcients at deriva-

tives. To reduce this problem to the standard form with constant matrices at

derivatives, we use the same gauge transformation as in the previous section. We

put F(t, x) = g(t, x)σ3g−1(t, x) where the unitary matrix g(t, x) is the same as in

Section 2. Further we put L(x, y, t) = ˆ

L(x, y, t) and M(x, y, t) = g(t, x)ˆ

M(x, y, t).

Taking into account that ∂M

∂x =g(t, x)∂ˆ

M

∂x +g0

x(t, x)ˆ

Mand ∂M

∂y =g(t, x)∂ˆ

M

∂y ,we

ﬁnd that the Goursat problem reduces to the form:

∂ˆ

L

∂x +σ3

∂ˆ

L

∂y σ3=H(t, x)ˆ

L+ [σ3, g(t, x)] ˆ

M,

∂ˆ

M

∂x +σ3

∂ˆ

M

∂y σ3=R(t, x)ˆ

M+1

4[g−1(t, x), σ3]ˆ

L,

σ3ˆ

L(x, x, t)−ˆ

L(x, x, t)σ3=σ3H(t, x),

σ3ˆ

M(x, x, t)−ˆ

M(x, x, t)σ3=1

4[σ3, g−1(t, x)]σ3,

σ3ˆ

L(x, −x, t) + ˆ

L(x, −x, t)σ3=0,

σ3ˆ

M(x, −x, t) + ˆ

M(x, −x, t)σ3=0.

This problem coincides with (25). Thus we obtain the classical Goursat problem

which in turn gives the existence of representation (24).

The integral representation (24) gives the analyticity of the solution W(t, x, z)

for z∈C\ {0}and its asymptotic behavior as z→ ∞ and z→0:

W(t, x, z)e−ixµ(z)=

1 0

0e−2ixµ(z)!+ O(z−1) + O(e−2ixµ(z)z−1), z → ∞;

O(1) + O(e−2ixµ(z)), z →0;

W(t, x, z)eixµ(z)=

e2ixµ(z)0

0 1!+ O(z−1) + O(e2ixµ(z)z−1), z → ∞;

O(1) + O(e2ixµ(z)), z →0.

Taking into account that Im µ(z) = (1 −1

4|z|2) Im z, and since x > 0, the above

asymptotics mean that W(t, x, z)e−ixµ(z)is bounded in the domain {z∈C:

Im µ(z)≤0}, and W(t, x, z)eixµ(z)is bounded in the domain {z∈C: Im µ(z)≥

0}.

130 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

Lemma 5. Let the initial functions E0(x),ρ0(x),N0(x)from (6) be

smooth or of Schwartz type if x∈R+, i.e., l=∞, and

N2

0(x) + |ρ0(x)|2≡1.

Then the Jost solution w(x, λ)can be represented in the form (Im µ(λ) = 0),

w(x, λ) = eixµ(λ)σ3+

2l−x

Z

x

L(x, y)eiyµ(λ)σ3dy +1

iλ

2l−x

Z

x

M(x, y)eiyµ(λ)σ3dy. (26)

The kernels L(x, y)and M(x, y)satisfy the symmetry conditions

L(x, y)=ΛL(x, y)Λ−1, M(x, y) = ΛM(x, y)Λ−1

with matrix Λ = 0 1

−1 0and they are deﬁned by the formulas L(x, y) = ˆ

L(x, y)

and M(x, y) = g(0, x)ˆ

M(x, y), where the unitary matrix g(0, x)is the same as in

Section 2 with t= 0. The matrices ˆ

L(x, y)and ˆ

M(x, y)are the unique solution

of the Goursat problem:

σ3

∂ˆ

L

∂x +∂ˆ

L

∂y σ3=σ3H(0, x)ˆ

L+σ3[σ3, g(0, x)] ˆ

M,

σ3

∂ˆ

M

∂x +∂ˆ

M

∂y σ3=σ3R(0, x)ˆ

M+σ3

4[g−1(0, x), σ3]ˆ

L,

σ3ˆ

L(x, x)−ˆ

L(x, x)σ3=H(0, x)σ3,

σ3ˆ

M(x, x)−ˆ

M(x, x)σ3=1

4[g−1(0, x), σ3]σ3,

σ3ˆ

L(x, 2l−x) + ˆ

L(x, 2l−x)σ3=0,

σ3ˆ

M(x, 2l−x) + ˆ

M(x, 2l−x)σ3=0,

or

lim

x+y→+∞

ˆ

L(x, y) = lim

x+y→+∞

ˆ

M(x, y) =0,if l=∞,

where R(0, x) = g−1(0, x)H(0, x)g(0, x)−g−1(0, x)g0

x(0, x). The kernels L(x, y)

and M(x, y)are smooth and fast decreasing as x+y→ ∞ if l=∞.

P r o o f. By substituting (26) into equation (23) and integrating by parts,

we get the Goursat problem (x < y < 2l−x):

∂L

∂x +σ3

∂L

∂y σ3=H(0, x)L+ (σ3−F(0, x))M,

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 131

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

∂M

∂x +F(0, x)∂M

∂y σ3=H(0, x)M+1

4(σ3−F(0, x)) M,

σ3L(x, x)−L(x, x)σ3=H(0, x)σ3,

F(0, x)M(x, x)−M(x, x)σ3=1

4(σ3−F(0, x)) σ3,

σ3L(x, 2l−x) + L(x, 2l−x)σ3=0,

F(0, x)M(x, 2l−x) + M(x, 2l−x)σ3=0.

If l=∞, then the last two conditions (for y= 2l−x) are changed with

lim

x+y→+∞L(x, y) = lim

x+y→+∞M(x, y) = 0,if l=∞.

Further we put L(x, y) = ˆ

L(x, y) and M(x, y) = g(0, x)ˆ

M(x, y), and we ﬁnish

the proof in the same way as in Lemma (4).

Introduce the notations: Ω±={z∈C±|z|>1

2},D±={z∈C±|z|<

1

2}, Σ = R∪Cup ∪Clow , where the semicircles Cup and Clow are: Cup ={z∈

C|z|=1

2,arg z∈(0, π)}and Clow ={z∈C|z|=1

2,arg z∈(π, 2π)}. Let Ω±

and D±be the closures of the domains Ω±and D±, respectively. The contour Σ

is the set where Im µ(λ) = 0:

Σ = {λ∈C: Im λ+1

4λ= 0}=R∪Cup ∪Clow.

The orientation on Σ is depicted in Figure (1).

Fig. 1. The domains Ω±,D±and the oriented contour Σ.

132 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

The vector columns of the matrix w(x, λ) have analytic continuations

w[1](x, z) and w[2](x, z) in the domains Ω+∪D−and Ω−∪D+, respectively.

These vectors have the asymptotics:

w[1](x, z)e−ixµ(z)=

1

0!+ O(z−1), z ∈Ω+, z → ∞,

O(1), z ∈D−\ {0}, z →0.

(27)

w[2](x, z)eixµ(z)=

0

1!+ O(z−1), z ∈Ω−, z → ∞,

O(1), z ∈D+\ {0}, z →0.

(28)

Formula (19), Lemmas 2, 4 and Eqs. (27), (28) imply the following properties

of Y(t, x, λ)=(Y[1](t, x, λ)Y[2](t, x, λ)):

1) Y(t, x, λ) (λ6= 0) satisﬁes the t−and x−equations (7), (8);

2) Y(t, x, λ) = ΛY(t, x, λ)Λ−1,λ∈R\ {0}, where Λ = 0 1

−1 0;

3) det Y(t, x, λ)≡1, λ ∈R\ {0};

4) the map (x, t)7−→ Y(t, x, λ) (λ6= 0) is smooth in tand x;

5) the vector functions Y[1](t, x, z)eizt−iµ(z)x,Y[1](t, x, z)e−izt+iµ(z)xand

Y[2](t, x, z)e−izt+iµ(z)x,Y[2](t, x, z)eizt−iµ(z)xare analytic in C−and C+, respec-

tively, continuous up to the boundary with exception of λ= 0 and have the

following asymptotic behavior:

Y[1](t, x, z)eizt−iµ(z)x=1

0+ O(z−1), z ∈Ω−, z → ∞,(29)

Y[1](t, x, z)e−izt+iµ(z)x= O(1), z ∈D−\ {0}, z →0,

Y[2](t, x, z)e−izt+iµ(z)x=0

1+ O(z−1), z ∈Ω+, z → ∞,(30)

Y[2](t, x, z)eizt−iµ(z)x= O(1), z ∈D+\ {0}, z →0.

Formula (20) and Lemmas 3, 5 imply the following properties of

Z(t, x, λ)=(Z[1](t, x, λ)Z[2](t, x, λ)):

1) Z(t, x, λ) (λ6= 0) satisﬁes the t- and x-equations (7), (8);

2) Z(t, x, λ)=ΛZ(t, x, λ)Λ−1,λ∈R\ {0};

3) det Z(t, x, λ)≡1, λ ∈R\ {0};

4) the map (x, t)7−→ Z(t, x, λ) (λ6= 0) is smooth in tand x;

5) the maps z7−→ Z[1](t, x, z) and z7−→ Z[2](t, x, z) are analytic in Ω+∪D−

and Ω−∪D+, respectively, and the asymptotic behavior of Z[1](t, x, z)eizt−ixµ(z),

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 133

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

Z[2](t, x, z)e−izt+ixµ(z)is as follows:

Z[1](t, x, z)eizt−ixµ(z)=1

0+ O(z−1), z ∈Ω+, z → ∞,(31)

Z[1](t, x, z)eizt−ixµ(z)= O(1), z ∈D−\ {0}z→0,

Z[2](t, x, z)e−izt+ixµ(z)=0

1+ O(z−1), z ∈Ω−, z → ∞,(32)

Z[2](t, x, z)e−izt+ixµ(z)= O(1), z ∈D+\ {0}, z →0.

Since the matrices Y(t, x, λ) and Z(t, x, λ) are the solutions of the t−and

x−equations (7), (8), they are linear dependent. Consequently, there exists a

transition matrix T(λ), independent of xand t, such that

Y(t, x, λ) = Z(t, x, λ)T(λ).(33)

The transition matrix is equal to

T(λ) = Z−1(0,0, λ)Y(0,0, λ) = w−1(0, λ)Φ(0, λ),

and hence T(λ)=ΛT(λ)Λ−1,λ∈R\ {0}, i.e., T(λ) has the form

T(λ) = a(λ)b(λ)

−b(λ)a(λ).(34)

The scattering relation (33) can be written in the form

Y[1](t, x, λ) =a(λ)Z[1](t, x, λ)−b(λ)Z[2](t, x, λ), λ ∈R\ {0},(35)

Y[2](t, x, λ) =a(λ)Z[2](t, x, λ) + b(λ)Z[1](t, x, λ), λ ∈R\ {0}.(36)

Relations (35), (36) give

a(λ) = det(Z[1](t, x, λ), Y [2](t, x, λ)), a(λ) = det(Y[1](t, x, λ), Z [2](t, x, λ)),

b(λ) = det(Y[2](t, x, λ), Z[2](t, x, λ)), b(λ) = −det(Z[1](t, x, λ), Y [1](t, x, λ)).

It is easy to see that the matrix

w(0, λ) = α(λ)−β(λ)

β(λ)α(λ)(37)

is the spectral function of the x-equation for t= 0, which is uniquely deﬁned by

the given initial functions E(0, x), ρ(0, x) and N(0, x), and the matrix

Φ(0, λ) = A(λ)B(λ)

−B(λ)A(λ)(38)

134 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

is the spectral function of the t-equation for x= 0, which is uniquely deﬁned by

the boundary condition E(t, 0).

The functions α(λ), β(λ) and α(λ), β(λ) have analytic continuations in Ω+∪

∪D−and Ω−∪D+, respectively, the functions A(λ), B(λ) and A(λ), B(λ) have

analytic continuations in C+and C−, respectively.

Let f?(z) := f(z) denote the Schwartz conjugate of a function f(z). Then an-

alytic continuations are denoted as (α(z), β(z), α?(z), β?(z), A(z), B(z), A?(z),

B?(z)) for zin the domains of their analyticity. They have the following asymp-

totic behavior:

α(z) =1 + O(z−1), β(z) =O(z−1), z → ∞, z ∈Ω+;

α(z) =O(1), β(z) =O(1), z →0, z ∈D−;

α?(z) =1 + O(z−1), β?(z) =O(z−1), z → ∞, z ∈Ω−;

α?(z) =O(1), β?(z) =O(1), z →0, z ∈D+;

A(z) =1 + O(z−1), B(z) =O(z−1), z → ∞, z ∈C+;

A?(z) =1 + O(z−1), B?(z) =O(z−1), z → ∞, z ∈C−;

A(z) =O(1), B(z) =O(1), z →0, z ∈C+;

A?(z) =O(1), B?(z) =O(1), z →0, z ∈C−.

The entries of the transition matrix T(λ) in the domains of their analyticity

are equal to

a(z) =α(z)A(z)−β(z)B(z), z ∈Ω+;

b(z) =α?(z)B(z) + β?(z)A(z), z ∈D+;

a?(z) =α?(z)A?(z)−β?(z)B?(z), z ∈Ω−;

b?(z) =α(z)B?(z) + β(z)A?(z), z ∈D−.

The spectral functions a(z) and b(z) are deﬁned and smooth for z∈Σ\ {0}. The

determinant of T(z)≡1 and, hence, a(z)a?(z) + b(z)b?(z)≡1 for z∈Σ\ {0}.

The spectral functions have the asymptotics:

a(z) =1 + O(z−1), z → ∞, z ∈Ω+; (39)

b(z) =O(1), z →0, z ∈D+;

a?(z) =1 + O(z−1), z → ∞, z ∈Ω−; (40)

b?(z) =O(1), z →0, z ∈D−.

If the function a(z) has zeroes zj∈Ω+,j= 1, n, then

a(zj) = det(Z[1](t, x, zj), Y [2](t, x, zj)) = 0.

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 135

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

Therefore, the vector columns are linear dependent:

Y[2](t, x, zj) = γjZ[1](t, x, zj), γj=B(zj)

α(zj)=A(zj)

β(zj), j = 1, n. (41)

At the conjugate points zj∈Ω−,j= 1, n,

a?(zj) = det(Y[1](t, x, zj), Z [2](t, x, zj)) = 0.

Consequently,

Y[1](t, x, zj) = γjZ[2](t, x, zj), γj=B?(zj)

α?(zj)=A?(zj)

β?(zj), j = 1, n. (42)

If the function b(z) has zeroes ζk∈D+,k= 1, m, then

b(ζk) = det(Y[2](t, x, ζk), Z[2](t, x, ζk)) = 0.

Therefore,

Z[2](t, x, ζk) = ηkY[2](t, x, ζk), ηk=B(ζk)

β?(ζk)=−A(ζk)

α?(ζk), k = 1, m. (43)

At the conjugate points ζk∈D−,k= 1, m,

b?(ζk) = −det(Z[1](t, x, ζk), Y [1](t, x, ζ k)) = 0.

Hence,

Z[1](t, x, ζk) = −ηkY[1](t, x, ζk), ηk=B?(ζk)

β(ζk)=−A?(ζk)

α(ζk), k = 1, m. (44)

Let us deﬁne the matrix

M(t, x, z) =

Z[1](t, x, z)eizt−ixµ(z)Y[2](t, x, z)

a(z)e−izt+ixµ(z), z ∈Ω+,

Y[1](t, x, z)

a?(z)eizt−ixµ(z)Z[2](t, x, z)e−izt+ixµ(z), z ∈Ω−,

Y[2](t, x, z)

b(z)eizt−ixµ(z)Z[2](t, x, z)e−izt+ixµ(z), z ∈D+,

Z[1](t, x, z)eizt−ixµ(z)−Y[1](t, x, z)

b?(z)e−izt+ixµ(z), z ∈D−.

(45)

The matrix Mis analytic for z∈C\Σ if a(z)6= 0 and b(z)6= 0. It is

meromorphic for z∈C\Σ and has poles at zj∈Ω+and zj∈Ω−, where a(zj) =

136 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

0 and a?(zj)=0j= 1, n. The matrix Malso has poles at ζk∈D+and ζk∈D−,

where b(ζk) = 0 and b?(ζk) = 0, k= 1, m. It has the asymptotics M(t, x, z ) =

I+ O(z−1) as z→ ∞.

Let the contour Σ have the orientation shown in Fig. 1. Then the matrix (45)

has jumps over the contour Σ,

M(t, x, λ)−=M+(t, x, λ)J(t, x, λ), λ ∈Σ\ {0},

where

J(t, x, λ) =

1 + |r(λ)|2−r(λ)e−2iθ(t,x,λ)

−r(λ)e2iθ(t,x,λ)1

, λ ∈R,|z|>1

2,

=

1−r−1(λ)e−2iθ(t,x,λ)

−r−1(λ)e2iθ(t,x,λ)1 + |r(λ)|−2

, λ ∈R,|z|<1

2, λ 6= 0;

(46)

=

0−r(λ)e−2iθ(t,x,λ)

r−1(λ)e2iθ(t,x,λ)1

, λ ∈Cup,

=

0r?(λ)e−2iθ(t,x,λ)

−r?−1(λ)e2iθ(t,x,λ)1

, λ ∈Clow,(47)

where r(λ) := b(λ)/a(λ) is deﬁned on R∪Cup,r?(λ) := b?(λ)/a?(λ) is deﬁned

on R∪Clow, and θ(t, x, λ) = λt −xµ(λ), λ∈Σ\ {0}. It is easy to see that

det J(t, x, λ)≡1, λ∈Σ\ {0}.

Suppose the number of zeroes is ﬁnite and these zeroes are simple poles of

M(t, x, z), i.e., a(zj) = 0, a0(zj) = da(z)

dz z=zj

6= 0, zj∈Ω+,j= 1, n, and b(ζk) =

0, b0(ζk) = db(z)

dz z=ζk

6= 0, ζk∈D+,k= 1, m. Then

res

z=zj

M[2](t, x, z) = γj

a0(zj)e−2iθ(t,x,zj)M[1](t, x, zj),(48)

res

z=zj

M[1](t, x, z) = γj

a?0(zj)e2iθ(t,x,zj)M[2](t, x, zj),(49)

res

z=ζk

M[2](t, x, z) = ηk

b0(ζk)e−2iθ(t,x,ζk)M[1](t, x, ζk),(50)

res

z=ζk

M[1](t, x, z) = ηk

b?0(ζk)e2iθ(t,x,ζk)M[2](t, x, ζ k),(51)

where γj,γjare deﬁned in (41), (42), and ηk,ηkare deﬁned in (43), (44).

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4. Matrix Riemann–Hilbert Problems

In this section we give a reconstruction of the solution to the MB equations in

terms of the spectral functions a(λ), b(λ), which are deﬁned through the spectral

functions A(λ), B(λ), deﬁned by input pulse E1(t), and α(λ), β(λ) deﬁned by

initial data E0(x), ρ0(x), N0(t). In the previous section we proved that the

matrices (45) are the solutions of the following matrix RH problem:

Find a 2 ×2 matrix M(t, x, z) such that

•M(t, x, z) is meromorphic in z∈C\Σ (or analytic if a(z)6= 0 and b(z)6=

0) and continuous up to the set Σ \ {−1

2,0,1

2};RH1

•If a(zj) = a?(zj)=0, j = 1,2, . . . , n and all these zeroes are simple, then

M(x, t, z) has poles at the points z=zj,z=zj,j= 1,2, . . . , n, and the

corresponding residues satisfy relations (48) and (49); RH2

•If b(ζj) = b?(ζj)=0, j = 1,2, . . . , m and all these zeroes are simple, then

M(x, t, z) has poles at the points z=ζj,z=ζj,j= 1,2, . . . , m, and the

corresponding residues satisfy relations (50) and (51); RH3

•M−(t, x, λ) = M+(t, x, λ)J(t, x, λ), λ ∈Σ\ {−1

2,0,1

2}, where J(t, x, λ)

is deﬁned in (46) and (47); RH4

•M(t, x, z) is bounded in the neighborhoods of the points {−1

2,0,1

2};

•M(t, x, z) = I+O(z−1),|z|→∞. RH 5

Theorem 2. Let the functions E(t, x),N(t, x)and ρ(t, x)be the solution

of the mixed problem (5)–(6) for the Maxwell–Bloch equations. Let the corre-

sponding spectral functions a(λ)and b(λ)have ﬁnite and simple zeroes in the

domains of their analyticity. Then there exists the matrix M(t, x, z)which is the

solution of the Riemann–Hilbert problem RH1–RH5, and the complex electric

ﬁeld envelope E(t, x)is deﬁned by the relation

E(t, x) = −lim

z→∞ 4izM12(t, x, z).(52)

P r o o f. The existence of the matrix M(t, x, z) follows from the above con-

siderations. We only need to prove equation (52). The matrix M(t, x, z) deﬁnes

the solution Φ(t, x, z) of the AKNS equations (7) and (8) by the formula

Φ(t, x, z) = M(t, x, z)e−iθ(t,x,z)σ3.

Formula (52) follows from (7) and (RH5). Indeed, by substituting the last for-

mula into equation (7), we ﬁnd

Mt+ iz[σ3, M ] + HM = 0.(53)

138 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

Using (RH5), we put

M(t, x, z) = I+m(t, x)

z+ o(z−1),

where

m(t, x) = lim

z→∞ z(M(t, x, z)−I).

This asymptotics and equation (53) give

H(t, x) = −i[σ3, m(t, x)],

and hence

E(t, x) = −4im12 =−lim

z→∞ 4izM12(t, x, z).

Taking into account the well-known fact that a(z) can have multiple zeroes

or inﬁnitely many zeroes with limit points on the contour Σ or zeroes on Σ

(the so-called spectral singularities), to avoid the complexities, we propose below

another formulation of the matrix RH problem. We use the ideas proposed in

[6, 7, 21] and introduce another set of solutions of the AKNS equations with

suitable asymptotic behavior of the solutions in the vicinity of the origin (z= 0).

5. Basic Solutions with Well-Controlled Asymptotic Behavior

at z= 0

We will use here the gauge AKNS equations deﬁned by (11), (12) and (13),

(14). Let ˆ

X(t, x, λ) be a product of the matrices

ˆ

X(t, x, λ) = ˆ

W(t, x, λ)ˆ

Φ(t, λ),(54)

where ˆ

W(t, x, λ) satisﬁes the x-equation (12) for all tand ˆ

W(t, 0, λ) = I,

and ˆ

Φ(t, λ) satisﬁes the t-equation (11) for x= 0 under the initial condition

ˆ

Φ(0, λ) = I. By the same way as in Lemmas 3.2 and 3.3, we prove the existence

of representations

ˆ

Φ(t, λ)=e−itλσ3+ iλ

t

Z

−t

ˆ

K0(t, τ )e−iτλσ3dτ (55)

and

ˆ

W(t, x, λ)=eixµ(λ)σ3+

x

Z

−x

ˆ

L0(x, y, t)eiyµ(λ|σ3dy + iλ

x

Z

−x

ˆ

M0(x, y, t)eiyµ(λ|σ3dy (56)

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 139

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

with some matrix functions ˆ

K0(t, τ ), ˆ

L0(x, y, t), ˆ

M0(x, y, t) which have the same

symmetry properties as in Lemmas 2–5: ˆ

L0(x, y, t)=Λˆ

L0(x, y, t)Λ−1, etc. These

representations guarantee the analyticity of ˆ

Xin z∈C\ {0}and the following

asymptotics as z→0:

ˆ

X[1](t, x, z)eiθ(t,x,z)=1

0+ O(z),Im z≥0, z →0;

ˆ

X[2](t, x, z)e−iθ(t,x,z)=0

1+ O(z),Im z≤0, z →0.

Due to Lemma 1, the matrix (54) ˆ

X(t, x, λ) is a compatible solution of the gauge

AKNS Eqs. (11), (12). Then, due to Sec. 2, the matrix

X(t, x, λ) = g(t, x)ˆ

X(t, x, λ) (57)

with gdeﬁned by (15)–(17) is a solution of the AKNS system of Eqs. (7), (8).

The matrix Xis analytic in z∈C\ {0}and its columns have the asymptotics:

X[1](t, x, z)eiθ(t,x,z)=g(t, x)1

0+ O(z),Im z≥0, z →0; (58)

X[2](t, x, z)e−iθ(t,x,z)=g(t, x)0

1+ O(z),Im z≤0, z →0.(59)

Finally, Xis bounded in zfor any compact set of the complex plane except the

vicinity of the origin and for any ﬁxed tand x.

Let ˆ

Z(t, x, λ) be a product of the matrices

ˆ

Z(t, x, λ) = ˆ

Ψ(t, x, λ) ˆw(x, λ),(60)

where ˆ

Ψ(t, x, λ) satisﬁes the t-equation (11) for all xand ˆ

Ψ(0, x, λ) = I, and

ˆw(x, λ) satisﬁes the x-equation (12) with t= 0 under the initial condition

ˆw(l, λ) = eilµ(λ)σ3, where µ(λ) = λ+1

4λ. If l=∞, then the initial condition

takes the form

lim

x→∞ ˆw(x, λ)e−ixµ(λ)σ3=I.

It is easy to see that due to Lemma 1, the matrix ˆ

Z(t, x, λ) is a compatible

solution of the gauge AKNS Eqs. (11), (12).

Again we prove the existence of representations

ˆ

Ψ(t, x, λ)=e−itλσ3+ iλ

t

Z

−t

ˆ

KΨ(t, τ, x)e−iτ λσ3dτ

140 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

and

ˆw(x, λ)=eixµ(λ)σ3+

2l−x

Z

x

ˆ

Lw(x, y)eiyµ(λ|σ3dy + iλ

2l−x

Z

x

ˆ

Mw(x, y)eiyµ(λ|σ3dy

with some matrix functions ˆ

KΨ(t, τ, x), ˆ

Lw(x, y), ˆ

Mw(x, y) which have the men-

tioned above symmetry properties. These representations guarantee the analytic-

ity of the columns ˆ

Z[1](t, x, z) and ˆ

Z[2](t, x, z) of the matrix (60) in the domains

Im µ(z)≥0 and Im µ(z)≤0, respectively, and give the following asymptotics as

z→0:

ˆ

Z[1](t, x, z)eiθ(t,x,z)=1

0+ O(z),Im µ(z)≥0, z →0; (61)

ˆ

Z[2](t, x, z)e−iθ(t,x,z)=0

1+ O(z),Im µ(z)≤0, z →0.(62)

Due to Section 2, the matrix

e

Z(t, x, λ) = g(t, x)ˆ

Z(t, x, λ) (63)

with gdeﬁned by (15), (16), (17) is a solution of the AKNS equations (7), (8).

The columns e

Z[1](t, x, z) and e

Z[2](t, x, z) of the matrix e

Zare analytic in the

domains Im µ(z)≥0 and Im µ(z)≤0, respectively, and these columns have the

asymptotics:

e

Z[1](t, x, z)eiθ(t,x,z)=g(t, x)1

0+ O(z),Im z≥0, z →0; (64)

e

Z[2](t, x, z)e−iθ(t,x,z)=g(t, x)0

1+ O(z),Im z≤0, z →0.(65)

Thus the construction of the compatible solutions (57) and (63) with well-

controlled behavior (58), (59) and (64), (65) as z→0 is ﬁnished.

Introduce the corresponding spectral functions. They are generated by the

linear dependence of the above solutions. Thus, for Im µ(λ) = 0, the matrices

e

Z(t, x, λ) and X(t, x, λ) are linear dependent:

e

Z(t, x, λ) = X(t, x, λ)Te

Z(λ), T e

Z(λ) = e

Z(0,0, λ) = ˆw(0.λ) = α?

0(λ)β0(λ)

−β?

0(λ)α0(λ).

(66)

The spectral functions α0(λ), α?

0(λ) = α0(λ) and β0(λ), β?

0(λ) = β0(λ) are deﬁned

by the gauge x-equation (12) with t= 0 which, in turn, is determined by the

function g(0, x) and the initial functions E0(x), ρ0(x) and N0(x). These spectral

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 141

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

functions are analytic in z∈D+and, due to (61), (62), have the asymptotic

behavior:

α0(z) = 1 + O(z), β0(z) = O(z), z →0, z ∈D+\ {0}.(67)

The functions α?

0(z) and β?

0(z) are analytic in z∈D−and have the asymptotics:

α?

0(z) = 1 + O(z), β?

0(z) = O(z), z →0, z ∈D−\ {0}.(68)

Formulas (67), (68) yield that there exists ε > 0 and a set Oε={z∈C:|z|< ε}

such that

α0(z)6= 0, z ∈D+∩Oε\ {0};α?

0(z)6= 0, z ∈D−∩Oε\ {0}.

The relation

Z(t, x, λ) = X(t, x, λ)TZ(λ),Im µ(λ)=0,

where TZ(λ) = X−1(0,0, λ)Z(0,0, λ) = g−1(0,0)w(0, λ), generates the spectral

functions

TZ(λ) = g−1(0,0) α(λ)−β?(λ)

β(λ)α?(λ)=: αg(λ)−β?

g(λ)

βg(λ)α?

g(λ),(69)

where g−1(0,0) is the unitary matrix, and the entries of the matrix w(0, λ) were

deﬁned in (37).

Finally, the relation

Y(t, x, λ) = X(t, x, λ)TY(λ), λ ∈R\ {0},

deﬁnes the transition matrix TY(λ) = X−1(0,0, λ)Y(0,0, λ) = g−1(0,0)Φ(0, λ),

which has the form

TY(λ) = g−1(0,0) A(λ)B(λ)

−B(λ)A(λ)=: Ag(λ)Bg(λ)

−Bg(λ)Ag(λ),Im λ= 0,(70)

where the entries of the matrix Φ(0, λ) were deﬁned in (38).

The spectral functions deﬁned by ˆw(0, λ) and w(0, λ) are related each other

by the formula

ˆw(0, λ) = g−1(0,0)w(0, λ),Im µ(λ)=0.(71)

Indeed, the matrices w(x, λ) and g(0, x) ˆw(x, λ) are the solutions of (8). Therefore

they are linear dependent, i.e., w(x, λ) = g(0, x) ˆw(x, λ)C(λ), where

C(λ)=e−ilµ(λ)σ3g−1(0, l)eilµ(λ)σ3. Without loss of generality, we put N0(l) =

142 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

1. Then ρ0(l) = 0, g−1(0, l) = e−χ(0,l)σ3where χ(0, l) =

l

R0

h(0, x)dx +χ(0,0).

Since χ(0,0) is a free parameter, we can put χ(0, l) = 0. Then ˆw(x, λ) =

g−1(0, x)w(x, λ),and hence (71) is valid. Finally, from (66) and (69), we ﬁnd

α?

0(λ)β0(λ)

−β?

0(λ)α0(λ)=αg(λ)−β?

g(λ)

βg(λ)α?

g(λ).(72)

All the introduced compatible solutions X,Y,Z,e

Zof the AKNS equations

(7), (8) and the spectral functions deﬁned by T(λ) (34), Te

Z(λ) (66), TZ(λ) (69),

TY(λ) (70) allow us to formulate a new matrix RH problem which is regular,

i.e, without poles. Moreover, this formulation does not require any additional

assumptions on the absence of spectral singularities.

6. Regular Matrix Riemann–Hilbert Problem and a Solution of

the Mixed Problem for the Maxwell–Bloch Equations

Due to (39), a(z) = 1 + O(z−1) and a?(z) = 1 + O(z−1) as z→ ∞.

Fig. 2. The domains OR,Oεand the oriented contour Γ. OR={z∈C:|z|>

R},Oε={z∈C:|z|< ε},C+

R={z∈C|z|=R, arg z∈(0, π)},C−

R={z∈

C|z|=R, arg z∈(π, 2π)},C+

ε={z∈C|z|=ε, arg z∈(0, π)},C−

ε={z∈

C|z|=ε, arg z∈(π, 2π)}, Γ := R∪Cε∪CR=R∪C+

ε∪C−

ε∪C+

R∪C−

R.

Hence there exists R > 0 and a set OR={z∈C:|z|> R}such that

a(z)6= 0, z ∈Ω+∩OR;a?(z)6= 0, z ∈Ω−∩OR.

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 143

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

Deﬁne the new contour Γ which contains the real axis Rorientated from left to

right and two circles of the radii εand Roriented clockwise, i.e., Γ := R∪Cε∪

CR(Fig. 2).

Now, using the vectors Y[1], Y[2], Z[1], Z[2] from Sec. 3 and the vectors (58),

(59), (64), (65), we can deﬁne a new matrix

Mreg (t, x, z) = Z[1](t, x, z)eiθ(t,x,z)Y[2](t, x, z)

a(z)e−iθ(t,x,z), z ∈Ω+∩OR,

(73)

= X[1](t, x, z)eiθ(t,x,z)e

Z[2](t, x, z)

α?

g(z)e−iθ(t,x,z)!, z ∈D+∩Oε,

(74)

=X[1](t, x, z)eiθ(t,x,z)X[2](t, x, z)e−iθ(t,x,z), ε < |z|< R,

(75)

=e

Z[1](t, x, z)

αg(z)eiθ(t,x,z)X[2](t, x, z)e−iθ(t,x,z), z ∈D−∩Oε,

(76)

=Y[1](t, x, z)

a?(z)eiθ(t,x,z)Z[2](t, x, z)e−iθ(t,x,z), z ∈Ω−∩OR,

(77)

which satisﬁes the following properties:

•Mreg (t, x, z) is analytic in z∈C\Γ and continuous up to the set

R\ {−R, −ε, ε, R};RRH1

•Mreg

−(t, x, λ) = Mreg

+(t, x, λ)Jreg (t, x, λ), λ ∈Γ\ {−R, −ε, 0, ε, R}RRH 2

•Mreg (t, x, z) is bounded in the vicinity of the points {−R, −ε, 0, ε, R};

RRH3

•Mreg (t, x, z) = M0(t, x) + O(z),|z| → 0. RRH4

•Mreg (t, x, z) = I+O(z−1),|z| → ∞. RRH5

Formulas (73)–(77) give the jump matrices on the circles:

Jreg (t, x, λ) =

Ag(λ)

a(λ)−Bg(λ)

a(λ)e−2iθ(t,x,λ)

−βg(λ)e2iθ(t,x,λ)αg(λ)

, λ ∈C+

R; (78)

=1−r?

g(λ)e−2iθ(t,x,λ)

0 1 , λ ∈C+

ε; (79)

144 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

Jreg (t, x, λ) = 1 0

rg(λ)e2iθ(t,x,λ)1, λ ∈C−

ε; (80)

=

α?

g(λ)β?

g(λ)e−2iθ(t,x,λ)

B?

g(λ)

a?(λ)e2iθ(t,x,λ)A?

g(λ)

a?(λ)

, λ ∈C−

R(81)

and on the real line:

Jreg (t, x, λ) =

1

|α?

g(λ)|2r?

g(λ)e−2iθ(t,x,λ)

rg(λ)e2iθ(t,x,λ)1

, λ ∈(−ε, ε)\ {0};

(82)

=1 0

0 1, ε < |λ|< R Im λ= 0;

(83)

=

1

|a(λ)|2−r(λ)e−2iθ(t,x,λ)

−r(λ)e2iθ(t,x,z)1

, λ ∈(−∞,−R)∪(R, ∞),

(84)

where

rg(λ) = βg(λ)

αg(λ), r?

g(λ) = β?

g(λ)

α?

g(λ), r(λ) = b(λ)

a(λ), r(λ) = b(λ)

a(λ).

The matrix Mreg (t, x, z) is bounded in the vicinity of the points

{−R, −ε, 0, ε, R}by the construction. Further, at the point z= 0, there ex-

ist nontangential limits

lim

z→0±i0 Mreg (t, x, z) = M0(t, x),

where M0(t, x) = g(t, x). Hence, Mreg (t, x, z ) has a continuation up to a contin-

uous function at the point z= 0 . Indeed, taking into account (58), (59), (61),

(62), (67), (68) and (72), we have that

lim

z→0+i0 Mreg (t, x, z) = lim

z→0+i0 X[1](t, x, z)eiθ(t,x,z)e

Z[2](t, x, z)

α0(z)e−iθ(t,x,z)

=g(t, x)1 0

0 1;

and

lim

z→0−i0 Mreg (t, x, z) = lim

z→0−i0 e

Z[1](t, x, z)

α0(z)eiθ(t,x,z)X[2](t, x, z)e−iθ(t,x,z)

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 145

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

=g(t, x)1 0

0 1.

Asymptotics at the inﬁnity is evident in view of (29), (30), (31), (32), (39), (40).

Theorem 3. Let the functions E(t, x),N(t, x)and ρ(t, x)be a solution

of the mixed problem (5)–(6) to the Maxwell–Bloch equations. Then there exists

the matrix M(t, x, z)which solves the regular Riemann–Hilbert problem RRH 1–

RRH5. The matrix M(t, x, z)gives the ﬁeld E(t, x)by the relation

E(t, x) = −lim

z→∞ 4izM12(t, x, z).(85)

The entries N(t, x)and ρ(t, x)of the density matrix F(t, x)are deﬁned as follows:

F(t, x) = M0(t, x)σ3M−1

0(t, x), M0(t, x) = lim

z→0±i0 M(t, x, z).(86)

P r o o f. The existence of the matrix M(t, x, z) follows from the above con-

siderations. We only need to prove equations (85) and (86). The matrix M(t, x, z)

deﬁnes the solution Φ(t, x, z) of the AKNS equations (7) and (8) by the formula

Φ(t, x, z) = M(t, x, z)e−iθ(t,x,z)σ3.

Formulas (85) follow from (7) and (RRH5). Indeed, substituting the last formula

into Eq. (7), we can ﬁnd

Mt+ iz[σ3, M ] + HM = 0.(87)

Using (RRH5), we put

M(t, x, z) = I+m(t, x)

z+ o(z−1),

where

m(t, x) = lim

z→∞ z(M(t, x, z)−I).

This asymptotics and Eq. (87) give

H(t, x) = −i[σ3, m(t, x)],

and hence

E(t, x) = −4im12 =−lim

z→∞ 4izM12(t, x, z).

Further, since lim

z→0±i0 Mreg (t, x, z) = M0(t, x), then the x-equation for M(t, x, z ),

Mx+i

4zMσ3= iz[σ3, M ] + HM +iF

4zM,

gives F(t, x) = M0(t, x)σ3M−1

0(t, x).

Thus the mixed initial boundary value problem in the quarter xt-plane to the

Maxwell–Bloch equations without spectral broadening is completely linearizable.

146 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

7. The Study of the Matrix Riemann–Hilbert Problem

It remains to consider the matrix Riemann–Hilbert problem independently of

its origin. Namely, let the conjugation contour Γ = R∪Cε∪CRand all spectral

functions be given. Let us consider the following problem:

Find a 2×2matrix M(t, x, z)such that the conditions RRH1–RRH5are satisﬁed

by the jump matrices given by (78)–(84).

Let tand xbe ﬁxed. We look for the solution M(t, x, z) of the RH problem

in the form

M(t, x, z) = I+1

2πiZ

Γ

P(t, x, s)[I−J(t, x, s)]

s−zds, z /∈Γ.(88)

The Cauchy integral (88) provides all properties of the RH problem (cf. [8]) if

and only if the matrix Q(t, x, λ) := P(t, x, λ)−Isatisﬁes the singular integral

equation

Q(t, x, z)− K[Q](t, x, z ) = R(t, x, z), z ∈Γ.(89)

The singular integral operator Kand the right-hand side R(t, x, z) are as follows:

K[Q](t, x, z) := 1

2πiZ

Γ

Q(t, x, s)[I−J(t, x, s)]

s−z+

ds,

R(t, x, z) := 1

2πiZ

Γ

I−J(t, x, s)

s−z+

ds.

We consider this integral equation in the space L2(Γ) of the 2 ×2 matrix complex

valued functions Q(z) := Q(t, x, z), z∈Γ. The norm of Q∈L2(Γ) is given by

||Q||L2(Γ) =

Z

Γ

tr(Q∗(z)Q(z))|dz|

1/2

=

2

X

j,l=1 Z

Γ

|Qjl (z))|2|dz|

1/2

.

The operator Kis deﬁned by the jump matrix J(t, x, z) and the generalized

function 1

s−z+

= lim

z0→z,z0∈side+

1

s−z0.

For the unique solvability of this integral equation the conjugation contour and

the jump matrices on its non real part should be Schwartz symmetric. The

contour Γ is symmetric with respect to the real axis, and the jump matrices

satisfy the relations:

J(t, x, λ)|λ∈C+

ε=J∗−1(t, x, λ)|λ∈C−

ε, J(t, x, λ)|λ∈C+

R=J∗−1(t, x, λ)|λ∈C−

R,

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 147

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

which can be easily veriﬁed under the clockwise orientation of the contours C±

ε

and C±

R. We recall that the asterisk is the Hermitian conjugation. It means that

the Schwartz symmetry principle is performed. Further, the matrix

J(t, x, λ) + J∗(t, x, λ)

is positive deﬁnite on the real axis (λ∈R). Then Theorem 9.3 from [31] (p. 984)

guarantees the L2invertibility of the operator I d−K (Id is the identical operator).

The function R(t, x, z) belongs to L2(Γ) because I−J(t, x, z)∈L2(Γ) when z∈

Γ, and the Cauchy operator

C+[f](z) := 1

2πiZ

Γ

f(s)

s−z+

ds =f(z)

2+ p.v. 1

2πiZ

Γ

f(s)

s−zds

is bounded in the space L2(Γ) [26]. Therefore, the singular integral equation (89)

has a unique solution Q(t, x, z)∈L2(Γ) for any ﬁxed x, t ∈R, and formula (88)

gives the solution of the above RH problem.

The uniqueness can be proved in the same way as in [8] (p. 194–198). Thus

the next theorem is valid.

Theorem 4. Let a contour Γand a jump matrix J(t, x, z ) (78)–(84) sat-

isfy the Schwartz symmetry principle. Let I−J(t, x, .)∈L2(Γ) ∩L∞(Γ). Then

for any ﬁxed and real t, x the regular RH problem RRH 1–RRH 5has the unique

solution M(t, x, z)given by (88).

Theorem 5. Let M(t, x, z)be the solution of the RH problem RRH 1–

RRH5given by Theorem 4. If M(t, x, z)is absolutely continuous (smooth) in t

and x, then Φ(t, x, z)(z∈C\Γ) satisﬁes the AKNS equations

Φt=−(izσ3+H(t, x))Φ,(90)

Φx=izσ3+H(t, x)+iF(t, x)

4zΦ (91)

almost everywhere (point-wise) with respect to tand x. The matrix H(t, x)is

given by

H(t, x) = −i[σ3, m(t, x)], m(t, x) = 1

πZ

Γ

(I+Q(t, x, λ))(J(t, x, λ)−I)dλ,

(92)

where Q(t, x, λ)is the unique solution of the singular integral equation (89).

The matrix F(t, x) = M(t, x, 0)σ3M−1(t, x, 0) is Hermitian and it has the struc-

ture

F(t, x) = N(t, x)ρ(t, x)

ρ∗(t, x)−N (t, x).

148 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

P r o o f. The matrix Φ(t, x, z ) := M(t, x, z)e(−izt+iη(z)x)σ3is analytic in z∈

C\Γ and has a jump across the contour Γ,

Φ−(t, x, λ)=Φ+(t, x, λ)J0(λ), λ ∈Γ,

where J0(λ) := e(iλt−iη(λ)x)σ3Jreg (t, x.λ)e(−iλt+iη(λ)x)σ3is independent on tand x.

This relation implies:

dΦ−(t, x, λ)

dt Φ−1

−(t, x, λ) = dΦ+(t, x, λ)

dt Φ−1

+(t, x, λ),

dΦ−(t, x, λ)

dx Φ−1

−(t, x, λ) = dΦ+(t, x, λ)

dx Φ−1

+(t, x, λ)

for λ∈Γ. The relations obtained mean that the matrix logarithmic derivatives

Φt(t, x, z)Φ−1(t, x, z) and Φx(t, x, z)Φ−1(t, x, z ) are analytic in z∈C\ {0}with

exception of self-intersection points of the contour Γ. The matrix M(t, x, z ) and

its derivative Mt(t, x, z ) are analytic in z∈C\Γ, and the Cauchy integral (88)

gives the asymptotic formula

M(t, x, z) = I+m±(t, x)

z+ O(z−2), z → ∞, z ∈C±.

Hence,

Φt(t, x, z)Φ−1(t, x, z) = −izσ3+ i[σ3, m+(t, x)] + O(z−1)

=−izσ3+ i[σ3, m−(t, x)] + O(z−1), z → ∞,

where

m−(t, x) = m+(t, x) = m(t, x) = i

2πZ

Γ

P(t, x, z)[I−J(t, x, z)]dz.

Since M(t, x, z) is bounded up to the boundary, then z= 0, the end points

and self-intersection points are removable singularities for Φt(t, x, z)Φ−1(t, x, z).

Therefore, by Liouville’s theorem, this derivative is a polynomial

U(z) := Φt(t, x, z)Φ−1(t, x, z) = −izσ3−H(t, x),

where H(t, x) := −i[σ3, m(t, x)] = 0q(t, x)

p(t, x) 0 . Using the Schwartz sym-

metries of the jump matrix J(t, x, z), we show that U(z) = σ2U(z)σ2, where σ2=

0−i

i 0 . These reductions imply H(t, x) = −H∗(x, t), i.e., q(t, x) = −p(t, x),

and we put q(t, x) := E(t, x)/2. Thus, Φ(t, x, z) satisﬁes equation (90), and a

scalar function E(t, x) is deﬁned by (92). The function E(t, x) is smooth in tand

Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 149

M.S. Filipkovska, V.P. Kotlyarov, and E.A. Melamedova (Moskovchenko)

x, because the matrix M(x, t, z ) and hence m(t, x) are smooth in tand xby

supposition. In the same way as above, we ﬁnd that Φx(x, t, λ)Φ−1(x, t, λ) is a

rational matrix function

V(z) := Φx(x, t, λ)Φ−1(x, t, λ)=izσ3+H(t, x) + iF(t, x)

4z

because the two asymptotics are true:

Φx(t, x, z)Φ−1(t, x, z)=izσ3+H(t, x) + O(z−1), z → ∞

and

Φx(t, x, z)Φ−1(t, x, z) = iF(t, x)

4z+F0(t, x) + O(z), z →0,

where F(t, x) = M(t, x, 0)σ3M−1(t, x, 0) and F0(t, x) is some matrix. Moreover,

the previous relations give that F0(t, x)≡H(t, x). Thus we can see that the

matrix Φ(x, t, z) satisﬁes two diﬀerential equations (90) and (91). Their compat-

ibility (Φxt(x, t, z)=Φtx (x, t, z)) gives the identity in z,

Ux(z)−Vt(z)+[U(z), V (z)] = 0, U =−izσ3−H, V = izσ3+H+iF

4z,

i.e.,

Ht(t, x) + Hx(t, x) + izσ3+H(t, x),izσ3+H(t, x) + iF(t, x)

4z= 0.

This identity is equivalent to the system of matrix equations:

Ht(t, x) + Hx(t, x) =1

4[σ3, F (t, x)],(93)

Ft(t, x) =[F(t, x), H(t, x)].(94)

Using the Schwartz symmetries of the jump matrix J(t, x, z ), we can ﬁnd that

F(t, x) is a Hermitian and traceless matrix, and we put

F(t, x) = N(t, x)ρ(t, x)

ρ∗(t, x)−N (t, x).

Matrix equations (93) and (94) are equivalent to the scalar equations (5). Thus,

we have proved that the matrix Φ(t, x, z) satisﬁes Eqs. (90) and (91), which

coincide with the AKNS Eqs. (7) and (8), and the scalar functions E(t, x), N(t, x),

ρ(t, x) satisfy the MB equations (5) due to the compatibility of Eqs. (90) and

(91).

150 Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2

Maxwell–Bloch Equations and Matrix Riemann–Hilbert Problems

The diﬀerentiability of the matrix M(t, x, z) (88) with respect to tand xcan

be proved if we assume that the boundary and initial conditions exponentially

vanish at inﬁnity or have a ﬁnite support. Then the introduced spectral functions

will be analytic in some strip along the real axis of the complex plane. Further,

using a factorization of the jump matrices into upper / lower triangular matrices,

one can deform the contour into a complex plane in the neighborhoods of zero and

inﬁnity in such a way that on the deformed contour the new jump matrices will

be exponentially close to the identity matrices when z→0 and z→ ∞. Outside

these points, the new contour coincides with Γ. This enables us to diﬀerentiate

(many times) the singular integral Eq. (89) with respect to the parameters tand

xunder the integral sign, and thereby to prove the smoothness of the matrices

Q(t, x, λ) = P(t, x, λ)−Iand M(t, x, z) with respect to tand x. As a consequence,

we obtain the smoothness of the matrix F(t, x) = M(t, x, 0)σ3M−1(t, x, 0) and,

due to (92), the smoothness of the matrix H(t, x), i.e., the smoothness of the

solution E(t, x), N(t, x), ρ(t, x) of the Maxwell–Bloch equations.

It remains only to show that these functions give exactly the solution of the

mixed problem. In proving, the deformations of the matrix RH problem RRH1–

RRH5 into Rieman–Hilbert problems can be used. One of the problems should

be in the one-to-one correspondence with the initial functions while another RH

problem should be in the one-to-one correspondence with the boundary condition.

These deformations can be done in the same way as in [21]. The so-called global

relation (sf. [10,12,13]) does not appear in our consideration because the Maxwell–

Bloch equations are the PDEs of the ﬁrst order.

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