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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).
Journal of Mathematical Physics, Analysis, Geometry, 2017, Vol. 13, No. 2 137
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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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