MODELING OF LOW-CONTRAST PHOTONIC CRYSTALS WITH COUPLED-MODE EQUATIONS By Dmitri Agueev SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE AT MCMASTER UNIVERSITY HAMILTON, ONTARIO SEPTEMBER 2004 c Copyright by Dmitri Agueev, 2004
89
Embed
MODELING OF LOW-CONTRAST PHOTONIC CRYSTALS WITH … · 2004. 9. 15. · • We classify wave resonances and coupled-mode equations for low-contrast cubic-lattice photonic crystals
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
As the scattering of light is considered to be elastic, we introduce the following energy
and momentum constraint on the resonating waves
Definition 2.1 The wave vector k(n,m,l)out with a non-empty triple (n,m, l) is said to be
resonant with the wave vector kin, if |k(n,m,l)out | = |kin|, such that |ω(k
(n,m,l)out )| = |ω(kin)|.
The resonance conditions are illustrated in Figure 2.2.
From our result ∆k = G, the resonance condition is (k + G)2 = k2, or
2k ·G +G2 = 0(2.7)
11
Figure 2.2: The points on the right of the sphere are reciprocal lattice points of thecrystal. The vector k is drawn in the direction of the incident beam and it terminatesat any reciprocal lattice point. We draw a sphere of radius k about the origin ofk. A diffracted beam will be formed if this sphere intersects any other point in thereciprocal lattice. The sphere as drawn intersects a point connected with the end ofk by reciprocal lattice vector G. The diffracted beam is in the direction k′ = k + G.This construction is due to P.P. Ewald. The figure is taken from [Ki]
This is the central result in the theory of elastic scattering in a periodic lattice. The
identical result occur in the theory of the electron energy band structure of crystals
[Ki]. Notice that if G is a reciprocal lattice vector, −G is also one; thus we can
equally well write (2.7) as 2k ·G = G2.
The Brillouin zone gives a vivid geometrical interpretation of the resonance con-
dition 2k ·G = G2 or
k · (12G) = (
1
2G)2(2.8)
See the Figure 2.3.
12
We construct a plane normal to the vector G at the midpoint; any vector k from
the origin to the plane will satisfy the resonance condition. The plane thus described
forms a part of the zone boundary. An incident wave on the crystal will be diffracted if
it’s wavevector has the magnitude and direction required by (2.8), and the diffracted
wave will be in the direction of the vector k−G.
The set of planes that are the perpendicular bisectors of the reciprocal lattice vectors
are of particular importance in the theory of the wave propagation in crystals, because
a wave whose wavevector drawn from the origin terminates on any of these planes will
satisfy the condition for diffraction. These planes divide the Fourier space of the crys-
tal into bits and pieces, as shown on the Figure 2.4. The central square is a primitive
cell of the reciprocal lattice. It is called the first Brillouin zone. The first Brillouin
zone of the cubic lattice in two dimensions is shown on the Figure 2.4. The Brillouin
construction exhibits all the incident wavevectors which can be Bragg-reflected by
the crystal. As can be seen from the picture there can be several resonances for one
wave.
We prove a simple lemma that we will need later.
Lemma 2.2 The number of waves resonant to the incident kin is less then the numberof nodes of the reciprocal lattice G lying inside the sphere of the radius 2|kin| centeredat the origin.
Proof. The proof is obvious from the geometrical picture, as the longest vector G
that gives resonance is in the direction of kin.
We consider here a simple cubic crystal, where the fundamental lattice vectors
and reciprocal lattice vectors are all orthogonal [Ki]:
x1,2,3 = ae1,2,3, k1,2,3 = k0e1,2,3, k0 =2π
a,
13
where e1,2,3 are unit vectors in R3. The coordinate axes (x, y, z) are oriented along
the axes of the simple cubic crystal. The set of resonant Bloch waves is given by the
The resonant set S can be non-empty for (p, q, r) /∈ Z3, which corresponds to oblique
Bloch waves. For instance, two oblique waves are resonant on the (x, y)-plane if
kin =π
a(p, q, 0), k
(n,m,0)out =
π
a(p+ 2n, q + 2m, 0).(2.17)
where (n,m) ∈ Z2 are arbitrary and (p, q) ∈ R2 are taken on the straight line:
np+mq = −(n2 +m2).(2.18)
Similarly, three oblique waves can be resonant on the (x, y)-plane if
kin =π
a(p, q, 0), k
(n1,m1,0)out =
π
a(p+2n1, q+2m1, 0), k
(n2,m2,0)out =
π
a(p+2n2, q+2m2, 0),
(2.19)
where (n1,m1) ∈ Z2 and (n2,m2) ∈ Z2 are arbitrary subject to the constraint: m1n2 6=
m2n1, while (p, q) take rational values:
p =m1(n
22 +m2
2)−m2(n21 +m2
1)
m2n1 −m1n2
, q =n1(n
22 +m2
2)− n2(n21 +m2
1)
n2m1 − n1m2
.(2.20)
In general case, two oblique waves (2.17) or three oblique waves (2.19) may have
resonances with other Bloch waves in three-dimensional photonic crystals.
2.3.4 Three-dimensional resonances of counter-propagatingwaves
When (p, q, r) ∈ Z3+, the resonant sets S include eight coupled waves for fully three-
dimensional Bragg resonance:
kin =π
a(p, q, r), k
(−p,−q,−r)out =
π
a(−p,−q,−r),
17
k(−p,0,0)out =
π
a(−p, q, r), k
(0,−q,0)out =
π
a(p,−q, r),
k(0,0,−r)out =
π
a(p, q,−r), k
(−p,−q,0)out =
π
a(−p,−q, r),
k(−p,0,−r)out =
π
a(−p, q,−r), k
(0,−q,−r)out =
π
a(p,−q,−r).(2.21)
The resonance condition for the three-dimensional Bragg resonance takes the form:
ϕ = arctan
(q
p
), θ = arctan
(√p2 + q2
r
),√p2 + q2 + r2k0 = 2k,(2.22)
The incident wave kin is directed along the diagonal of the (px, qy, rz)-cell of the
cubic lattice crystal and the wavelength is λ = 2a/√p2 + q2 + r2. The eight waves
(2.21) can be coupled with some other resonant waves, such that dim(S) ≥ 8 for
(p, q, r) ∈ Z3+. For instance, dim(S) = 8 for (p, q, r) = (1, 1, 1) and (p, q, r) = (2, 1, 1),
but dim(S) = 10 for (p, q, r) = (2, 2, 1) and dim(S) = 16 for (p, q, r) = (3, 2, 1).
18
Figure 2.3: Reciprocal lattice points near the point O at the origin of the reciprocallattice. The reciprocal lattice vector GC connects points OC; and GD connects OD.Two planes 1 and 2 are drawn which are perpendicular bisectors of GC and GD,respectively. Any vector from the origin to the plane 1, such as k1, satisfies thediffraction condition k1 · (1
2GC) = (1
2GC)2. Any vector from the origin to the plane 2
, such as k2, satisfies the diffraction condition k2 · (12GD) = (1
2GD)2. Figure is taken
from [Ki].
19
Figure 2.4: Square reciprocal lattice with reciprocal lattice vectors shown as blacklines. The dashed lines are perpendicular bisectors of the reciprocal lattice vectors.The central square is the smallest volume about the origin which is bounded entirelyby dashed lines. The square is the Wigner-Seitz primitive cell of the reciprocal lattice.It is called the first Brillouin zone. Figure is taken from [Ki].
Chapter 3
Derivation of coupled-modeequations
The dispersion surface ω = ω(k) for the Bloch waves (2.4) in the periodic photonic
crystal is defined by the profile of the refractive index n(x). We shall consider the
asymptotic approximation of the dispersion surface ω = ω(k) in the limit when the
photonic crystal is low-contrast, such that the refractive index n(x) is given by
n(x) = n0 + εn1(x),(3.1)
where n0 is a constant and ε is small parameter. It is proved in [K1] that the
Bloch waves (2.4) are smooth functions of ε, such that the asymptotic solution of
the Maxwell equations (2.1) as ε → 0 takes the form of the perturbation series ex-
pansions:
E(x, t) = E0(x, t) + εE1(x, t) + O(ε2).(3.2)
The leading-order term E0(x, t) consists of free transverse waves (2.2) with wave
vectors k(n,m,l)out , given by (2.6), such that the asymptotic form (3.2) represents the
Bloch wave (2.4) as ε 6= 0.
20
21
Coupled-mode equations are derived by separating resonant free waves from non-
resonant free waves in the Bloch wave (2.4), where the resonant set S with N =
dim(S) <∞ is defined by (2.9). Let E0(x, t) be a linear superposition of N resonant
waves with wave vectors kj at the same frequency ω:
E0(x, t) =N∑
j=1
Aj(X, T )ekjei(kjx−ωt), X =
εx
k, T =
εt
ω,(3.3)
where ω and kj are related by the same dispersion equation (2.3), Aj(X, T ) is the
envelope amplitude of the jth resonant wave (2.2) and (X, T ) are slow variables. The
slow variables represent a deformation of the dispersion surface ω = ω(kj) for free
waves (2.3) due to the low-contrast periodic photonic crystal. The degeneracy in the
polarization vector is neglected by the assumption that the incident wave is linearly
polarized with the polarization vector ein = ekin. The triple Fourier series (2.5) for
the cubic-lattice crystal is simplified as follows:
n1(x) = n0
∑(n,m,l)∈Z3
αn,m,leik0(nx+my+lz),(3.4)
where α0,0,0 = 0. The Fourier coefficients αn,m,l satisfy the constraints:
αn,m,l = α−n,−m,−l,
due to the reality of n1(x),
αn,m,l = αm,n,l = αn,l,m = αl,m,n,
due to the crystal isotropy in the directions of x,y,z-axes, and
due to the crystal symmetry with respect to the origin (0, 0, 0) (the latter property
can be achieved by a simple shift of (x, y, z)). It follows that all coefficients αn,m,l for
(n,m, l) ∈ Z3 are real-valued.
22
It follows from (2.1), (3.1), and (3.2) that the first-order correction term E1(x, t)
solves the non-homogeneous linear problem:
∇2E1 −n2
0
c2∂2E1
∂t2= 2
n20ω
c2∂2E0
∂T∂t− 2k (∇ · ∇X)E0 +
2n0n1(x)
c2∂2E0
∂t2(3.5)
+2
n0
∇ (∇n1 · E0) ,
where ∇X = (∂X , ∂Y , ∂Z) and the second equation (2.1) has been used. The right-
hand-side of the non-homogeneous equation (3.5) has resonant terms, which are par-
allel to the free-wave resonant solutions of the homogeneous problem. The resonant
terms lead to the secular growth of E1(x, t) in t, unless they are identically zero.
The latter conditions define the coupled-mode equations for amplitudes Aj(X, T ),
j = 1, ..., N in the general form:
2ik2
(∂Aj
∂T+
(kj
k· ∇X
)Aj
)+∑k 6=j
aj,kAk = 0, j = 1, ..., N,(3.6)
where the elements aj,k1≤j,k≤N are related to the Fourier coefficients αn,m,l(n,m,l)∈Z3
at the resonant terms (n,m, l) ∈ S.
The explicit forms of the coupled-mode equations (3.6) are derived below for a
number of examples of Bloch waves resonances.
3.1 Coupled-mode equations for one-dimensional
resonance
The lowest-order Bragg resonance for two counter-propagating waves (2.14) occurs
for r = 1, when
k1 =π
a(0, 0, 1), k2 =
π
a(0, 0,−1).(3.7)
Let A1 = A+(Z, T ) and A2 = A−(Z, T ) be the amplitudes of the right (forward) and
left (backward) propagating waves, respectively. The envelope amplitudes are not
23
modulated across the (X, Y )-plane, since the coupled-mode equations for A± are es-
sentially one-dimensional. The polarization vectors are chosen in the x-direction, such
that ek1 = ek2 = (1, 0, 0) and E0 = (E0,x(z, Z, T )e−iωt, 0, 0). The non-homogeneous
equation (3.5) at the x-component of the solution E1 at e−iωt takes the form:
∇2E1,x + k2E1,x = −2ik2 ∂
∂TE0,x − 2k
∂2
∂Z∂zE0,x −
2k2n1(x)
n0
E0,x +2
n0
∂2n1(x)
∂x2E0,x.
By removing the resonant terms at e±ikz, the coupled-mode equations for amplitudes
A±(Z, T ) take the form:
i
(∂A+
∂T+∂A+
∂Z
)+ αA− = 0,(3.8)
i
(∂A−∂T
− ∂A−∂Z
)+ αA+ = 0,(3.9)
where α = α0,0,1 = α0,0,−1. The coupled-mode equations (3.8)–(3.9) can be defined
on the interval 0 ≤ Z ≤ Lz for T ≥ 0, where the end points at Z = 0 and Z = Lz
are the left and right (x, y)-planes, which cut a slice of the photonic crystal. The
linear system (3.8)–(3.9) is reviewed in [SS]. The nonlinear coupled-mode equations
are derived in [BS, SSS] and analyzed recently in [GWH, PSBS, PS].
3.2 Coupled-mode equations for two-dimensional
resonance
3.2.1 Coupled-mode equations for four counter-propagatingwaves
The lowest-order resonance for four counter-propagating waves (2.16) occurs for p =
q = 1, when
k1 =π
a(1, 1, 0), k2 =
π
a(1,−1, 0),(3.10)
k3 =π
a(−1, 1, 0), k4 =
π
a(−1,−1, 0).
24
Let A1 = A+(X, Y, T ) and A4 = A−(X, Y, T ) be the amplitudes of the counter-
propagating waves along the main diagonal of the (x, y) plane, whileA2 = B+(X, Y, T )
and A3 = B−(X, Y, T ) be the amplitudes of the counter-propagating waves along
the anti-diagonal of the (x, y)-plane. The envelope amplitudes are not modulated
in the Z-direction, since the coupled-mode equations for A± and B± are essentially
two-dimensional. The polarization vectors are chosen in the z-direction, such that
ekj= (0, 0, 1), 1 ≤ j ≤ 4, which corresponds to the TE mode of the photonic crystal,
such that E0 = (0, 0, E0,z(x, y,X, Y, T )e−iωt). The non-homogeneous equation (3.5)
at the z-component of the solution E1 at e−iωt takes the form:
∇2E1,z + k2E1,z = −2ik2 ∂
∂TE0,z − 2k
∂2
∂X∂xE0,z − 2k
∂2
∂Y ∂yE0,z
− 2k2n1(x)
n0
E0,z +2
n0
∂2n1(x)
∂z2E0,z.(3.11)
By removing the resonant terms at ei√2(±kx±ky)
, the coupled-mode equations for am-
plitudes A±(X, Y, T ) and B±(X, Y, T ) take the form:
i
(∂A+
∂T+∂A+
∂X+∂A+
∂Y
)+ αA− + β (B+ +B−) = 0,(3.12)
i
(∂A−∂T
− ∂A−∂X
− ∂A−∂Y
)+ αA+ + β (B+ +B−) = 0,(3.13)
i
(∂B+
∂T+∂B+
∂X− ∂B+
∂Y
)+ β (A+ + A−) + αB− = 0,(3.14)
i
(∂B−
∂T− ∂B−
∂X+∂B−
∂Y
)+ β (A+ + A−) + αB+ = 0,(3.15)
where α = α1,1,0 = α−1,−1,0 = α1,−1,0 = α−1,1,0 and β = α0,1,0 = α1,0,0 = α0,−1,0 =
α−1,0,0. The coupled-mode equations (3.12)–(3.15) can be defined in the domain
(X, Y ) ∈ D and T ≥ 0, where D is a domain on the (x, y)-plane of the photonic
crystal. The system has not been previously studied in literature, to the best of our
knowledge.
25
3.2.2 Second and higher order resonance for four counter-propagating waves in two-dimensional cubic crystal
The coupled-mode equations (3.12)–(3.15) for four counter-propagating waves de-
rived in the first order Bragg resonance. We show here that the higher order Bragg
resonances it 2D crystal can give the same coupled-mode system of four counter-
propagating waves.
Consider second order Bragg resonance on the (x, y) plane with kin = 2πa
(1, 0), as
can be seen in Figure 2.4 there are three lattice vectors G that are in resonance with
kin = 2πa
(1, 0)
G1 =2π
a(2, 0)
G2 =2π
a(1,−1)
G3 =2π
a(1, 1)
Hence there are four resonant counter-propagating waves, that are given by wave-
vectors
k =2π
a(1, 0)
k−G1 =2π
a(−1, 0)
k−G2 =2π
a(0, 1)
k−G3 =2π
a(0,−1)
It can be seen from the Figure 3.1 that another basis cell can be use for the same
crystal. We introduce new dual lattice k′1,k′2 , which is rotated about the basis k1,k2
.
k′1 = k1 + k2
26
Figure 3.1: The same two-dimensional cubic crystal can be obtained by translationof two basis cells. The basis cell we the smallest volume is called primitive.
k′2 = k2 − k1
The triple Fourier series for n(x) is now
n(x) =∑G′
α′G′eiG′x
Where G′ lies in the new dual lattice G′ = nk′1 +mk′2 .
Now the derivation of coupled-mode equations for four counter-propagating waves
in the first Bragg resonance can be repeated to obtain the coupled-mode equations
for four counter-propagating waves in the second Bragg resonance in 2D crystals with
obvious change α→ α′.
Let A1 = A+(X, Y, T ) and A2 = A−(X, Y, T ) be the amplitudes of the counter-
propagating waves in x direction, while A3 = B+(X, Y, T ) and A4 = B−(X, Y, T ) be
the amplitudes of the counter-propagating waves in y direction. The envelope am-
plitudes are not modulated in the Z-direction, since the coupled-mode equations for
A± and B± are essentially two-dimensional. We obtain the coupled-mode equations
27
for amplitudes A±(X, Y, T ) and B±(X, Y, T ):
i
(∂A+
∂T+∂A+
∂X
)+ α′A− + β′ (B+ +B−) = 0,(3.16)
i
(∂A−∂T
− ∂A−∂X
)+ α′A+ + β′ (B+ +B−) = 0,(3.17)
i
(∂B+
∂T+∂B+
∂Y
)+ β′ (A+ + A−) + α′B− = 0,(3.18)
i
(∂B−
∂T− ∂B−
∂Y
)+ β′ (A+ + A−) + α′B+ = 0,(3.19)
where α′ = α′1,1,0 = α′−1,−1,0 = α′1,−1,0 = α′−1,1,0 and β′ = α′0,1,0 = α′1,0,0 = α′0,−1,0 =
α′−1,0,0. The explicit formula for α′ in terms of α is
α′n,m = αn−m,n+m
We conclude the treatment of two-dimensional photonic crystals with a simple
algebraic theorem which is almost obvious from geometric construction on Figure 2.4.
Theorem 3.1 All orthogonal counter propagating resonant waves in 2D cubic crystaloccur for kin = 2π
a(p+s
2, p−s
2), where p, s are integers.
Proof. It is clear that the only parallel resonant waves allowed in photonic crystal
are counter propagating waves. In order to find two orthogonal pairs of counter-
propagating waves we have to solve the system
|k| = |k +G|
k ⊥ (k +G)
Denote k = (k1, k2), G = (n,m), the system transforms to
k21 + k2
2 = (n+ k1)2 + (m+ k2)
2
k1(k1 + n) + k2(k2 +m) = 0
28
Expressing m in terms of n from the second equation we get
(n+ k1)2 = k2
2
m = −k1(k1 + n) + k22
k2
This gives us two sets of solutions
n1 = k2 − k1, m1 = −k2 − k1
n2 = −k2 − k1, m2 = −k2 + k1
It is easy to see that this two sets describe two orthogonal vectors G1 and G2. For
this vectors to lie on the reciprocal lattice we require
k1 =p+ s
2
k2 =p− s
2,
where p and s are integers. The theorem is now proven.
The theorem gives us necessary and sufficient condition for existence of four orthog-
onal counter-propagating resonant waves.
It is not clear for what kin, in general, there are only four non-orthogonal counter
propagating waves in cubic crystal. The obvious necessary condition for that is
kin = 2πa
(p, s) where p, s ∈ Z2.
3.3 Coupled-mode equations for two oblique waves
Two oblique resonant waves on the (x, y)-plane are defined by the resonant wave
vectors (2.17) under the constraint (2.18). Assuming that e1 = e2 = (0, 0, 1), the
Maxwell equations can be reduced to the same form (3.11), where the resonant terms
29
are eliminated at the wave vectors k1 = kin and k2 = k(n,m,0)out . The coupled-mode
equations for amplitudes A1,2(X, Y, T ) take the form:
i
(∂A1
∂T+
p√p2 + q2
∂A1
∂X+
q√p2 + q2
∂A1
∂Y
)+ αA2 = 0,(3.20)
i
(∂A2
∂T+
p+ 2n√p2 + q2
∂A2
∂X+
q + 2m√p2 + q2
∂A2
∂Y
)+ αA1 = 0.(3.21)
where α = αn,m,0 = α−n,−m,0. Coupled-mode equations (3.20)–(3.21) for two oblique
waves can not be reduced to the one-dimensional system (3.8)–(3.9), since the char-
acteristics in the system (3.20)–(3.21) are no longer parallel.
The coupled-mode equations for three oblique resonant waves (2.19) can be de-
rived similarly, subject to the resonance condition (2.20). Three characteristics along
the wave vectors k1 = kin, k2 = k(n1,m1,0)out , and k3 = k
(n2,m2,0)out belong to the same
(X, Y )-plane. The stationary transmission problem for the three oblique waves is
hence a boundary-value problem on the (X, Y )-plane with three (linearly dependent)
characteristic coordinates. Oblique interaction of three oblique resonant Bloch waves
in a hexagonal crystal was considered numerically in [SGS].
3.4 Coupled-mode equations for six counter-propagating
waves in 3D photonic crystal
The lowest-order resonance for six counter-propagating waves occurs for p = 2, q =
r = 0, when
k1 =2π
a(1, 0, 0)
k2 =2π
a(−1, 0, 0)
k3 =2π
a(0, 1, 0)
30
k4 =2π
a(0,−1, 0)
k5 =2π
a(0, 0, 1)
k6 =2π
a(0, 0,−1)
Let A1 = A+(X, Y, Z, T ) and A2 = A−(X, Y, Z, T ) be the amplitudes of the counter-
propagating waves in x direction, while A3 = B+(X, Y, Z, T ) and A4 = B−(X, Y, Z, T )
be the amplitudes of the counter-propagating waves in y direction, A5 = C+(X, Y, Z, T )
and A6 = C−(X,Y, Z, T ) be the amplitudes of the counter-propagating waves in z
direction. This resonance is fully three-dimensional.
Substituting into the non-homogeneous equation (3.5) and removing resonant
terms for solution E1 at ei±kx, ei±ky, ei±kz the coupled-mode equations for the am-
plitudes A±(X, Y, ZT ), B±(X, Y, Z, T ) and C±(X, Y, Z, T ) take the form:
where a+(y), a−(y), b+(x), b−(x) are given amplitudes of the incident waves at the
left, right, bottom and top boundaries of the crystal. The geometry of the problem
is shown on the Figure 4.1.
To prove existence and uniqueness of the solution of the system (4.2)–(4.5) with
boundary condition (4.6) we use the following facts [KF]
• Let A be a continuous map of complete metric space R into itself such that
some power of it B = An is a contraction; then the equation Au = u has a
unique solution.
• The space R of continuous vector functions v(x, y) on the compact domain with
the norm ρ(v1,v2) = maxx,y,i |vi1(x, y)− vi
2(x, y)| is complete.
Theorem 4.1 There exist a unique solution of the system (4.2)–(4.5) in the domainD that satisfy boundary condition (4.6).
37
ξ1
ξ2
ξ3
ξ4
P
Q
a+
a−
b+
b−
Figure 4.1: Four counter-propagating waves on the plane. The direction of the char-acteristics is shown with the arrowed lines. The characteristics that are going throughpoints P and Q are shown with dashed lines. The wave a+ propagates along ξ1, a−along ξ2, b+ along ξ3 and b− along ξ4. The boundary data for this particular geometryof domain D is shown. For this case we will find the explicit solution of linear problemby separation of variables in the next chapter.
Proof. The system (4.2)–(4.5) with boundary condition (4.6) is equivalent to
a+(x, y) = a+(0, y) +
∫ x
0
(Ωa+ + αa− + β(b+ + b−)) dx
a−(x, y) = a−(L, y) +
∫ x
L
(αa+ + Ωa− + β(b+ + b−)) dx
b+(x, y) = a+(x, 0) +
∫ y
0
(β(a+ + a−) + Ωb+ + αb−) dy
b−(x, y) = b−(x,H) +
∫ y
H
(β(a+ + a−) + αb+ + Ωb−) dy
38
Hence, each component of the vector fieldv1
v2
v3
v4
=
a+(x, y)
a−(x, y)
b+(x, y)
b−(x, y)
is expressed by a corresponding integral over the characteristic. To construct the
solution we view the the integral system as a fixed point of the iteration problem.
v1n+1(x, y) = a+(0, y) +
∫ x
0
(Ωv1n + αv2
n + β(v3n + v4
n)) dx
v2n+1(x, y) = a−(L, y) +
∫ x
L
(αv1n + Ωv2
n + β(v3n + v4
n)) dx
v3n+1(x, y) = a+(x, 0) +
∫ y
0
(β(v1n + v2
n) + Ωv3n + αv4
n) dy
v4n+1(x, y) = b−(x,H) +
∫ y
H
(β(v1n + v2
n) + αv3n + Ωv4
n) dy
or symbolically
vn+1 = Avn where A is the integral operator
To apply the contraction mapping principle we want to show that some power of
the map A is a contraction map. Let v1 and v2 be two continuous vector fields on
D = (x, y) : 0 ≤ x ≤ L, 0 ≤ y ≤ H , define metrics
ρ(v1,v2) = maxx,y,i
|vi1(x, y)− vi
2(x, y)|
Hence, we use the maximum norm ‖v‖ of a continuous vector field v, that is, the
largest value of v attained in the closed domain D for any component of v in D. We
have
Av1 − Av2 =
∫ x
0(Ω(v1
1 − v12) + α(v2
1 − v22) + β(v3
1 − v32) + β(v4
1 − v42)) dx∫ x
L(α(v1
1 − v12) + Ω(v2
1 − v22) + β(v3
1 − v32) + β(v4
1 − v42)) dx∫ y
0(β(v1
1 − v12) + β(v2
1 − v22) + Ω(v3
1 − v32) + α(v4
1 − v42)) dy∫ y
H(β(v1
1 − v12) + β(v2
1 − v22) + α(v3
1 − v32) + Ω(v4
1 − v42)) dy
39
Taking the absolute value of each component we obtain
|Av1 − Av2|(x, y) ≤
x
L− x
y
H − y
M‖v1 − v2‖
Where M = |Ω|+ |α|+ |2β|. Hence
|A2v1 − A2v2|(x, y) ≤
x2
2
(L−x)2
2
y2
2
(H−y)2
2
M2‖v1 − v2‖
And finally
|Anv1 − Anv2|(x, y) ≤
xn
n!
(L−x)n
n!
yn
n!
(H−y)n
n!
Mn‖v1 − v2‖
For any value of M , there exists a number N such that
‖ANv1 − ANv2‖ ≤ θ‖v1 − v2‖, θ < 1
Hence, the map AN is contraction and as A is a continuous map the solution of
Av = v exists and unique.
The previous result can be easily generalized to the nonlinear problem
i∂u1
∂x+ F 1(x, y,u) = 0,(4.7)
−i∂u2
∂x+ F 2(x, y,u) = 0,(4.8)
i∂u3
∂y+ F 3(x, y,u) = 0,(4.9)
−i∂u4
∂y+ F 4(x, y,u) = 0.(4.10)
40
defined on the rectangle,
D = (x, y) : 0 ≤ x ≤ L, 0 ≤ y ≤ H,(4.11)
with the boundary conditions:
u1(0, y) = u1(y), u2(L, y) = u2(y),(4.12)
u3(x, 0) = u3(x), u4(x,H) = u4(x),
Theorem 4.2 Let u1(y), u2(y), u3(x), u4(x) be continuous and F(x, y,u) is continu-ous and satisfy Lipschitz condition in it’s “functional“ argument
‖F(x, y;u1)− F(x, y;u2)‖ ≤M‖u1 − u2‖
Then, there exists a unique solution of the the problem (4.7)–(4.12).
Proof. The proof essentially repeats the one for the linear case. The nonlinear
system is equivalent to
u1(x, y) = u1(0, y) +
∫ x
0
F 1(x, y;u) dx
u2(x, y) = u2(L, y) +
∫ x
L
F 2(x, y;u) dx
u3(x, y) = u3(x, 0) +
∫ y
0
F 3(x, y;u) dy
u4(x, y) = u4(x,H) +
∫ y
H
F 4(x, y;u) dy
Setting w = (u1(0, y), u2(L, y), u3(x, 0), u4(x,H))T and defining the integral map A
as
Au =
w1 +
∫ x
0F 1(x, y;u) dx
w2 +∫ x
LF 2(x, y;u) dx
w3 +∫ y
0F 3(x, y;u) dy
w4 +∫ y
HF 4(x, y;u) dy
41
we find that
|Au1−Au2|(x, y) ≤
|∫ x
0(F 1(x, y;u1)− F 1(x, y;u2)) dx|
|∫ x
L(F 2(x, y;u1)− F 2(x, y;u2) dx|
|∫ y
0(F 3(x, y;u1)− F 3(x, y;u2) dy|
|∫ y
HF 4(x, y;u1)− F 4(x, y;u2)) dy|
≤
x
L− x
y
H − y
M‖u1−u2‖
Hence,
|Anu1 − Anu2|(x, y) ≤
xn
n!
(L−x)n
n!
yn
n!
(H−y)n
n!
Mn‖u1 − u2‖
and An is a contraction map.
It is clear that under inclusion of the cubic Kerr nonlinearities into the Maxwell
equations (2.1), i.e by decomposition of the refractive index n = n(x, |E|2) into the
linear and nonlinear parts [SS]:
n(x, |E|2) = n0 + εn1(x) + εn2(x)|E|2,
the vector function F(x, y;u) in the coupled-mode system is the sum of linear and
cubic terms of uk and uk. The set of admissible vector functions in the statements and
proofs of both theorems is the set of continuous vector functions on D, and thus there
is no upper bound for ‖u‖ and Kerr nonlinearity is not Lipschitz in its functional
argument. We can obtain the existence and uniqueness result by restricting the set
of admissible vector functions to bounded (in the maximum norm) by the doubled
norm of the boundary data 2‖w‖ and continuous on D and considering small enough
D so that the map Au gives us again an admissible vector function.
Theorem 4.3 Let u1(y), u2(y), u3(x), u4(x) be continuous and F(u) be the sum oflinear and cubic terms of uk and uk representing cubic Kerr nonlinearity and
max(L,H)(2a+ 8b‖w‖2) ≤ 1
42
where constants a, b are the sums of absolute values of coefficients of F in linear andcubic parts correspondingly. Then, there exists a unique solution of the the problem(4.7)–(4.12).
Proof. We denote w = (u1(y), u2(y), u3(x), u4(x))T and restrict the set of admissible
vector functions u, such that u is continuous on D and ‖u‖ ≤ 2‖w‖. Defining the
integral map A as
Au =
w1 +
∫ x
0F 1(u) dx
w2 +∫ x
LF 2(u) dx
w3 +∫ y
0F 3(u) dy
w4 +∫ y
HF 4(u) dy
we find that
‖Au‖ ≤ ‖w‖+ amax(L,H)‖u‖+ bmax(L,H)‖u‖3,
where a, b are the sums of absolute values of coefficients of F in linear and cubic parts
correspondingly. Under condition that
max(L,H)(2a+ 8b‖w‖2) ≤ 1
the set of admissible functions is mapped by A into itself. As vector function F (u)
is Lipschitz on the set of admissible functions and the equation u = Au has only one
solution.
Thus, for the finite norm boundary data there is always a unique solution of the Kerr
nonlinear boundary-value problem given that domain is small enough.
43
4.2 Existence and uniqueness of solution for N waves
Let us consider now the case of N waves in resonance. The characteristic coordinates
are introduced from the set of resonant wave vectors
∂
∂ξj=
kj,x
k
∂
∂X+
kj,y
k
∂
∂Y+
kj,z
k
∂
∂Z, j = 1, . . . , N,
such that characteristic lines are taken in the direction of wave vectors kj’s.
We will consider now two dimensional spatial space X = (X, Y ) for simplicity. The
only parallel resonant waves allowed in crystal are counter propagating waves: vectors
(ki,x, ki,y) and (kj,x, kj,y) could be in the opposite directions but any other vector
(kl,x, kl,y) should be linearly independent of them. Hence, generally characteristics fill
the whole plane.
We consider now the general problem of N-wave resonant propagation in the
convex crystal. Due to convexity the characteristics for any point inside the domain
lie entirely in the domain. Clearly the same family of characteristics can intersect
more than one face of the crystal as shown on the Figures 4.2,4.3. As can be seen
from the Figure 4.1, we only needed one wave profile at each face.
We consider the following boundary conditions: on avery face of the crystal we
give a profile A′i(x, y) of the wave Ai whose characteristics incidents on the face, we
require the continuity of profile on the verges (edges for 3D).
The boundary-value problem for the stationary transmission can be written in the
form
∂ai
∂ξi+ F i(x, y; a) = 0, i = 1, . . . , N(4.13)
in the given convex domain D, with the boundary conditions described above, such
that profiles a′i(x, y) are continuous (on the verges as well as on the faces) and
44
a+
a−
a−a+
b+
b+
b−
b−
ξ1
ξ2
ξ3
ξ4
P
Q
Figure 4.2: Modified boundary value problem for four counter-propagating waves onthe plane. The direction of the characteristics is shown with the arrowed lines. Thecharacteristics that are going through points P and Q are shown with dashed lines.The wave a+ propagates along ξ1, a− along ξ2, b+ along ξ3 and b− along ξ4. Theboundary data for this particular geometry of domain D is shown.
F(x, y, a) is continuous and satisfy Lipschitz condition in it’s “functional“ argument
‖F(x, y; a1)− F(x, y; a2)‖ ≤M‖a1 − a2‖
Theorem 4.4 The general transmission problem (4.13) has a unique solution.
Proof. The proof essentially repeats the one for the four counter-propagating waves.
For any point (x, y) in the domain D we find the N characteristics going through it,
we parametrize all characteristics so that ξi = 0 on the boundary of the domain
from which the characteristic goes inside the domain. For the convex domain D the
45
ξ1
ξ2
ξ3
P
a
a
a
a
a
1
1
2
2
3
Figure 4.3: The case of three non-parallel families of characteristics. The boundarydata for this particular geometry of domain is shown.
nonlinear system is equivalent to
ai(x, y) = a′i(x(0), y(0)) +
∫ ξi
0
F i(x(ξi), y(ξi); a) dξi, i = 1, . . . .N,
where integrals are taken along characteristics. Setting bi = a′i(x, y) we define the
integral map A as
Aa =
b1(x(0), y(0)) +
∫ ξi
0F 1(x(ξ1), y(ξ1); a) dξ1
...
bN(x(0), y(0)) +∫ ξN
0FN(x(ξN), y(ξN); a) dξN
46
we find that
|Aa1 − Aa2| ≤
|∫ ξ1
0(F 1(x, y; a1)− F 1(x, y; a2)) dξ
1|...
|∫ ξN
0(FN(x, y; a1)− FN(x, y; a2)) dξ
N |
≤
|∫ ξ1
0dξ1|
...
|∫ ξN
0dξN |
M‖a1 − a2‖
≤
ξ1
...
ξN
M‖a1 − a2‖
Hence,
|Ana1 − Ana2| ≤
(ξ1)n
n!...
(ξN )n
n!
Mn‖a1 − a2‖
and An is a contraction map. Hence, the equation Aa = a has a unique solution.
The theorems proved can be used for numerical solutions of the coupled-mode
systems through iteration of the integral map un+1 = Aun. The analytical solutions
for linear case is obtained in the next chapter.
4.3 Multi-symplectic structure of the coupled-mode
equations
Following [Br], a multi-symplectic system on the plane in canonical form in real
coordinates is
Mux +Kuy = ∇h(u), u ∈ R2n,
where M and K are any 2n× 2n skew-symmetric matrices.
47
An important part of the theory of conservative systems is the Noether theory
that relates symmetries and conservation laws. When a conservative system has
a Hamiltonian structure the symplectic operator gives a natural correspondence be-
tween symmetries and invariants or conserved densities ([O]). However in the classical
setting there is only one symplectic operator and therefore there is no relation be-
tween symplecticity and the fluxes of a conservation law. In the multi-symplectic
framework such a connection is possible and leads to a new and useful decomposition
of the Noether theory.
In its simplest setting, finite-dimensional Hamiltonian systems, the connection
between symmetry and conservation laws can be stated as follows. Suppose u ∈ R2n
and
Jut = ∇h(u)
with J the usual unit symplectic operator on R2n(0 −InIn 0
)
Suppose there exists a one-parameter Lie group G(ε) acting symplectically on R2n
which leaves the Hamiltonian functional invariant. Let
v =d
dε[G(ε)u]
∣∣∣∣ε=0
and suppose
Jv = ∇p(u)(4.14)
for some functional p(u). Then ∂tp = 0. A proof of this result is given in [O]. The
above result connects the action of the symmetry group with the conserved densities
but in the case of classical Hamiltonian system it does not establish a connection
48
between the action of the symmetry and the fluxes. In the expression (4.14) it is
the action of the symplectic operator, on the infinitesimal action of the Lie group,
that generates the gradient of the conserved quantity. Therefore its generalization to
include fluxes is clear: act on V with each element in the family of skew-symmetric
operators in the multi-symplectic structure to obtain all the components of the con-
servation law. This leads to a new decomposition of the Noether theory.
Let u ∈ R2n and consider the multi-symplectic Hamiltonian system
M(u)ux +K(u)uy + L(u)uz = ∇h(u)(4.15)
We say that the system (4.15) is invariant with respect to the action of one parameter
Lie group G(ε) if
h(G(ε) · u) = h(u)(4.16)
DG(ε)∗M(G(ε) · u)DG(ε) = M(u)
DG(ε)∗K(G(ε) · u)DG(ε) = K(u)
DG(ε)∗L(G(ε) · u)DG(ε) = L(u)
where DG(ε) is Jacobian with respect to ε and ∗ indicates formal adjoint.
Theorem 4.5 (Bridges, 1997) Let the Hamiltonian system (4.15) be invariant withrespect to the action of a one-parameter Lie group G(ε) with generator v. Supposethere exist a solution u(x, y, z) and functionals p(u), q(u), r(u) such that
M(u)v = ∇p(u), K(u)v = ∇q(u), L(u)v = ∇r(u).
Then p, q and r, evaluated at the solution u of (4.15), satisfy the conservation law
∂p
∂x+∂q
∂y+∂r
∂z= 0
49
Proof. The proof of this result is straightforward. Differentiating the first equation
of (4.16) with respect to ε and setting to zero results in
0 =d
dεh(G(ε) · u)
∣∣∣∣ε=0
= (∇h(u),v)
= (M(u)ux +K(u)uy + L(u)uz,v)
= −(ux,M(u)v)− (uy, K(u)v)− (uz, L(u)v)
= −(ux,∇p)− (uy,∇q)− (uz,∇r)
= −∂p∂x
− ∂q
∂y− ∂r
∂z
proving the claim.
We illustrate the multi-symplectic Noether theorem on the example of four counter
propagating waves in crystal (4.18)–(4.21).
The system for stationary transmission of four counter-propagating waves can be
parameters (λ, µ) are arbitrary, and vectors (u+, u−)T and (v+, v−)T solve the two
uncoupled ODE systems:(i∂ξ α
α −i∂ξ
)(u+
u−
)= βΓ−1
(1 1
1 1
)(u+
u−
)(5.21)
and (i∂η α
α −i∂η
)(v+
v−
)= βΓ
(1 1
1 1
)(v+
v−
),(5.22)
where Γ = λ/µ. The boundary conditions for (5.21)–(5.22) follow from (5.13) as
follows:
u+(0) = 1, u−(Lξ) = 0,(5.23)
and
v+(0) = v−(Lη) = 0.(5.24)
The homogeneous problem (5.22) and (5.24) defines the spectrum of Γ, while the
inhomogeneous problem (5.21) and (5.23) defines a unique particular solution (5.19)–
(5.20). The general solution of the problem (5.8)–(5.11) with the boundary values
(5.13) is thought to be a linear superposition of infinitely many particular solutions, if
the convergence and completeness of the decomposition formulas can be proved [St].
We first give solutions of the two problems above and then consider the orthogonality
and completeness of the generalized Fourier series.
Lemma 5.3 All eigenvalues Γ of the homogeneous problem (5.22) and (5.24) aregiven by non-zero roots of the characteristic equation:
R =
k ∈ C :
(k − α
k + α
)2
e−2ikLη = 1, Re(k) ≥ 0, k 6= 0
,(5.25)
such that
Γ =α2 + k2
2αβ.(5.26)
59
Let α > 0. Then, the roots k ∈ R are all located in the first open quadrant of k ∈ C.Moreover, all roots are simple and there exists C > 0 and N ∈ N such that only oneroot k ∈ R is located in each rectangle:
D+n =
k ∈ C :
π(4n− 1)
2Lη
< k <π(4n+ 1)
2Lη
, 0 < Im(k) < C
, n ≥ N
and
D−n =
k ∈ C :
π(4n+ 1)
2Lη
< k <π(4n+ 3)
2Lη
, 0 < Im(k) < C
, n ≥ N.
Proof. The general solution of the ODE system (5.22) with the use of (5.26) is
found explicitly as follows:(v+
v−
)= ck
(α− k
α+ k
)eikη + c−k
(α+ k
α− k
)e−ikη.(5.27)
The coefficients ck and c−k satisfy the relations due to the boundary conditions (5.24):
ckc−k
=k + α
k − α=k − α
k + αe−2ikLη ,(5.28)
from which the characteristic equation (5.25) for roots k ∈ C follows. The symmetric
roots k and (−k) correspond to the same Γ and v±(η). The root k = 0 corresponds
to the zero solution for v±(η). Therefore, the roots k = 0 and Re(k) ≤ 0 are excluded
from the definition of R. The characteristic equation (5.25) is equivalent to the
modulus equation:
|k − α||k + α|
=∣∣eikLη
∣∣ .When α > 0, the left-hand-side equals to one at Re(k) = 0 and is smaller than one
for Re(k) > 0. The right-hand-side equals to one at Im(k) = 0 and is larger than one
for Im(k) < 0. Therefore, roots k ∈ R may only occur in the first open quadrant of
k ∈ C.
All roots k ∈ R are simple, since
f ′(k) = −2i(k + α)
(k − α)
[(k2 − α2)Lη + 2iα
]6= 0
60
for f(k) = (k − α)2e−2ikLη − (k + α)2 = 0, as the values of k2 − α2 for k ∈ R are
located in upper half-plane of the complex plane.
The characteristic equation (5.25) split into two sets of roots R+ and R−, such that
R+ ∪R− = R, where
R± =
k ∈ C :
k − α
k + αe−ikLη = ±1, Re(k) > 0
.(5.29)
We consider the set k ∈ R+ and rewrite it in the form f(k) = g(k), where
f(k) = eikLη − 1, g(k) = − 2α
k + α.
The function f(k) has a zero at
k = kn =2πn
Lη
, n ≥ 1.
Let us consider the domain
D+n =
k ∈ C :
π(4n− 1)
2Lη
< k <π(4n+ 1)
2Lη
, −C < Im(k) < C
, n ≥ N
for some large C > 0 and N such that π(4n−1)2Lη
> α, that surrounds a simple zero of
f(k) at k = kn. It is clear that for sufficiently large C |f(k)| > |g(k)| on the boundary
of D+n . By the Rouche’s Theorem, the function f(k) + g(k) has the same number of
zeros inside D+n as f(k) does, i.e. only one zero. We can change D+
n to D+n as all
k ∈ R lie in the first open quadrant. The same analysis applies to the second set
k ∈ R− in the domain D−n .
Roots k ∈ R and (−k) ∈ R are shown on Figure 5.1 from the numerical solution
of the characteristic equation (5.25) for α = 1 and Lη = 20. In agreement with
Lemma 5.3, all roots k ∈ R are isolated points in the first open quadrant, which
61
accumulate to the real axis of k at infinity. The standard analysis of analytic functions
at infinity leads to the asymptotic formula for distribution of large roots k in the
domain |k| > k0 1:
k+n =
2πn
Lη
+iα
πn+ O
(1
n2
), k−n =
π(1 + 2n)
Lη
+2iα
π(1 + 2n)+ O
(1
n2
),(5.30)
where n is large positive integer. The leading order of the asymptotic approximation
(5.30) is also shown on Figure 5.1 by dotted curves. The two sets in (5.30) correspond
to the splitting k ∈ R± in (5.29).
−5 −2.5 0 2.5 5
−0.1
0
0.1
Re(k)
Im(k)
Figure 1: Roots k ∈ R and (−k) ∈ R of the characteristic equation (4.30) for α = 1 and
Lη = 20. Dark dots show roots of R+ and bright dots show roots of R−. The dotted curves
show the leading-order asymptotic approximation (4.37).
27
Figure 5.1: Roots k ∈ R and (−k) ∈ R of the characteristic equation (5.25) for α = 1and Lη = 20. Dark dots show roots of R+ and bright dots show roots of R−. Thedotted curves show the leading-order asymptotic approximation (5.30).
The eigenfunction v(η) = v+(η) + v−(η) is symmetric (anti-symmetric) with re-
spect to η = Lη/2 for k ∈ R+ (k ∈ R−). Moreover, explicit formulas for v(η) follow
62
from (5.27) and (5.28):
k ∈ R+ : v(η) = c+ cos k
(Lη
2− η
),
k ∈ R− : v(η) = c− sin k
(Lη
2− η
),
where (c+, c−) are normalization constants. Asymptotic solutions (5.30) correspond
to two sets of eigenfunctionscos(πnη), sin
(π(2n+ 1)η
2
), η =
2η
Lη
− 1,(5.31)
which solve the homogeneous Neumann’s problem on the normalized interval −1 ≤
η ≤ 1.
Lemma 5.4 Let Γ be an eigenvalue of the problem (5.22) and (5.24). There existsa unique solution of the non-homogeneous problem (5.21) and (5.23) for this Γ.
Proof. A general solution of the ODE system (5.21) is found explicitly as follows:(u+
u−
)= dk
(α2 + k2 − 2β2
λk(α2 + k2) + 2β2
)eiαλkη + d−k
(α2 + k2 − 2β2
−λk(α2 + k2) + 2β2
)e−iαλkη,
(5.32)
where
λk =
√4β2
α2 + k2− 1.(5.33)
The relation (5.33) satisfies the determinant equation (5.15), such that D(0, αλk, k) =
0. Using the boundary conditions (5.23), coefficients dk and d−k are found uniquely,
We show that u0 6= 0. The equation u0 = 0 can be rewritten in the form:
(λk − 1)2
(λk + 1)2= e2iαλkLξ .(5.35)
63
By analysis of Lemma 5.3, it is clear that non-zero roots of the characteristic equation
(5.35) may exist only in the first and in the third open quadrant of λk ∈ C for α > 0.
On the other hand the map k → λk given by (5.33) maps first quadrant onto second
and third quadrant onto forth one. Thus there could not be a solution of (5.34)
except when k is mapped into λ = 0:√
4β2
α2+k2 − 1 = 0. In this case k ∈ R must be
purely real or imaginary which is not the case. Therefore, the relation (5.33) leads to
contradiction, which proves that u0 6= 0.
Solutions of the non-homogeneous problem (5.21) and (5.23) with the normaliza-
tion u+(0) = u0 6= 0 can be written explicitly by eliminating dk and d−k from the
implicit form (5.32):(u+
u−
)= λk(α
2+k2)
(1
0
)cosαλk(Lξ−ξ)+i
(2β2
α2 + k2 − 2β2
)sinαλk(Lξ−ξ).(5.36)
When the representation (5.19) is used for α+(η) = a+(0, η).
The function α+(η) is expanded as series of scalar eigenfunctions v(η) = v+(η)+v−(η),
defined for roots k ∈ R. This decomposition is only possible if the set of eigenfunctions
v(η) is orthogonal and complete.
Lemma 5.5 There exists a set of normalized and orthogonal eigenfunctions vj(η) fordistinct roots k = kj ∈ R, according to the inner product:∫ Lη
0
vi(η)vj(η)dη = δi,j.(5.37)
Proof. The set of adjoint eigenvectors to the problem (5.22) and (5.24) with respect
to the standard inner product in L2([0, Lη]) is given by the vectors (v−, v+)T . As a
result, the scalar eigenfunctions vj(η) for distinct roots k = kj satisfy the orthogo-
nality relations (5.37) with i 6= j. The scalar eigenfunction v(η) is found from (5.27)
64
and (5.28) in the explicit form:
v(η) = c0 (k cos kη + iα sin kη) ,(5.38)
where c0 is a normalization constant. Integrating v2(η) on η ∈ [0, Lη], we confirm that
the eigenfunctions vj(η) can be normalized by the inner product (5.37) with i = j,
under the constraint:
(k2 − α2)Lη + 2iα 6= 0.(5.39)
Since the values of k2 − α2 for k ∈ R are located in upper half-plane of the complex
plane, the constraint (5.39) is met for α > 0 ( constraint (5.39) is the condition that
all roots k ∈ R are simple).
Proposition 5.6 Any continuously differentiable complex-valued function f(η) on0 ≤ η ≤ Lη is uniquely represented by the series of eigenfunctions:
f(η) =∑
all kj∈R
cjvj(η), cj =
∫ Lη
0
f(η)vj(η)dη,(5.40)
and the series converges to f(η) uniformly on 0 ≤ η ≤ Lη.
Proof. It follows from (5.22) and (5.24) that the scalar eigenfunction v(η) solves
the second-order boundary-value problem:
v′′ + k2v = 0,(5.41)
such that
iv′(0) + αv(0) = 0, −iv′(Lη) + αv(Lη) = 0.(5.42)
The Sommerfeld radiation boundary conditions (5.42) explain why the spectrum of
the formally self-adjoint operator (5.41) is complex-valued. The statement of Propo-
sition follows from Expansion Theorem [CL, p.303], since the theorem’s condition is
65
satisfied: A2,4 = 1, where A2,4 is the determinant of the second and fourth columns
of the matrix A, associated with the boundary conditions:
A =
(α i 0 0
0 0 α −i
).
The key idea here is that asymptotically the eigenvalue problem is the homogeneous
Neumann’s problem. As a result, the Fourier series of asymptotic eigenfunctions
(5.31) approximates the series expansion (5.40) for large roots k = k±n uniformly on
η ∈ [0, Lη]. The uniform convergence of (5.40) follows from that of the Fourier series
[St].
Using separation of variables and convergence of series of eigenfunctions, we sum-
marize the existence and uniqueness results on the generalized Fourier series solutions
of the linear boundary-value problem (5.8)–(5.11) and (5.13) with Ω = 0.
Proposition 5.7 Let the set cj be uniquely defined by the series (5.40) for f(η) =α+(η). There exists a unique continuously differentiable solution of the boundary-value problem (5.8)–(5.11) and (5.13) with Ω = 0 in the domain (5.12):
a+(ξ, η) =∑
all kj∈R
cju+j(ξ)
u+j(0)(v+j(η) + v−j(η)) ,(5.43)
a−(ξ, η) =∑
all kj∈R
cju−j(ξ)
u+j(0)(v+j(η) + v−j(η)) ,(5.44)
b+(ξ, η) = −∑
all kj∈R
cju+j(ξ) + u−j(ξ)
Γju+j(0)v+j(η),(5.45)
b−(ξ, η) = −∑
all kj∈R
cju+j(ξ) + u−j(ξ)
Γju+j(0)v−j(η).(5.46)
We illustrate the generalized Fourier series solutions (5.43)–(5.46) with two ex-
amples: (i) a single term of the generalized Fourier series and (ii) a constant input
function α+(η) = α+. For both examples, we compute the integral invariants for
66
the incident (Iin), transmitted (Iout), reflected (Iref) and diffracted (Idif) waves from
their definitions:
Iin =
∫ Lη
0
|a+(0, η)|2dη, Iout =
∫ Lη
0
|a+(Lξ, η)|2dη,
Iref =
∫ Lη
0
|a−(0, η)|2dη, Idif =
∫ Lξ
0
(|b+(ξ, Lη)|2 + |b−(ξ, 0)|2
)dξ.
Let the transmittance T , reflectance R and diffractance D be defined from the rela-
tions:
T =Iout
Iin
, R =Iref
Iin
, D =Idif
Iin
.
The integral invariants satisfy the balance identity:
R + T +D = 1,
which follows from integration of the balance equation:
∂
∂ξ
(|a+|2 − |a−|2
)+
∂
∂η
(|b+|2 − |b−|2
)= 0.
First, we consider a single term of the Fourier series solutions (5.43)–(5.46). The
transmittance and reflectance for k ∈ R are found from (5.36) in the explicit form:
Tk =
∣∣∣∣ λk(α2 + k2)
λk(α2 + k2) cosαλkLξ + 2iβ2 sinαλkLξ
∣∣∣∣2 ,Rk =
∣∣∣∣ (α2 + k2 − 2β2) sinαλkLξ
λk(α2 + k2) cosαλkLξ + 2iβ2 sinαλkLξ
∣∣∣∣2 ,while the diffractance is found from the balance identity as Dk = 1− Tk −Rk. These
integral invariants of the stationary transmission for α = 1 and Lξ = Lη = 20 are
shown on Figure 5.2 for β = 0.25 and on Figure 5.3 for β = 0.75. In the first case,
when α2 > 4β2, there is a stop band at Ω = 0, such that all modes are fully reflected,
except for small losses due to diffraction. In the second case, when α2 < 4β2, there is
no stop band at Ω = 0, such that transmittance and diffractance are large for smaller
values of |k| and become negligible for larger values of |k|.
67
0 1 2 3 4 50
2
4x 10−15
Re(k)
T
0 1 2 3 4 5
0.98
0.99
1
Re(k)
R
0 1 2 3 4 50
0.01
0.02
Re(k)
D
Figure 2: Transmittance (T ), reflectance (R), and diffractance (D) versus Re(k) for the roots
k ∈ R, when α = 1, β = 0.25, and Lξ = Lη = 20.
28
Figure 5.2: Transmittance (Tk), reflectance (Rk), and diffractance (Dk) versus Re(k)for the roots k ∈ R, when α = 1, β = 0.25, and Lξ = Lη = 20.
68
0 1 2 3 4 50
0.2
0.4
0.6
0.8
Re(k)
T
0 1 2 3 4 50
0.2
0.4
0.6
0.8
1
Re(k)
R
0 1 2 3 4 50
0.2
0.4
0.6
0.8
1
Re(k)
D
Figure 3: Transmittance (T ), reflectance (R), and diffractance (D) versus Re(k) for the roots
k ∈ R, when α = 1, β = 0.75, and Lξ = Lη = 20.
29
Figure 5.3: Transmittance (Tk), reflectance (Rk), and diffractance (Dk) versus Re(k)for the roots k ∈ R, when α = 1, β = 0.75, and Lξ = Lη = 20.
69
Next, we consider a constant input function:
α+(η) = α+, η ∈ [0, Lη],(5.47)
when cj can be found from (5.40):
cj =4iαα+
kj[L(k2j − α2) + 2iα]
, kj ∈ R+,
and cj = 0 for kj ∈ R−. The solution surfaces |a±(ξ, η)|2 and |b±(ξ, η)|2 in the domain
(5.12) are shown for α = 1, Lξ = Lη = 20, and α+ = 1 on Figure 5.4 for β = 0.25
and on Figure 5.5 for β = 0.75. We can see from the figures that the boundary
conditions (5.13) and (5.47) are satisfied by the truncated generalized Fourier series
(5.43)–(5.46) with only 30 first terms.
Parseval’s identity can not be applied to eigenfunctions vj(η), because the inner
product (5.37) is not the standard inner product in L2([0, Lη]). As a result, the
energy spectrum of Iout, Iref , and Idif can not be decomposed into a superposition
of the squared amplitudes |cj|2. Nevertheless, the numerical values for T , R, and D
can be found from numerical integration of the solution surfaces (5.47)–(5.47). The
numerical values are
β = 0.25 : T ≈ 3× 10−15, R ≈ 0.9853, D ≈ 0.0147,
β = 0.75 : T ≈ 0.7394, R ≈ 0.0133, D ≈ 0.2473,
such that T +R+D ≈ 1. When α2 > 4β2, there exists a stop band at Ω = 0 and the
incident wave is reflected from the photonic crystal with energy loss of 1.5% due to
diffraction. When α2 < 4β2, there is no stop band at Ω = 0 and the incident wave is
transmitted along the photonic crystal with energy loss of 26% due to reflection and
diffraction.
70
01
23
45 0
5
10
15
200
0.2
0.4
0.6
0.8
1
1.2
1.4
η
ξ
|a+|2
01
23
45 0
5
10
15
200
0.2
0.4
0.6
0.8
1
1.2
1.4
η
ξ
|a−|2
01
23
45 0
5
10
15
20
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
η
ξ
|b+|2
01
23
45 0
510
15200
0.05
0.1
0.15
0.2
0.25
0.3
0.35
η
ξ
|b−|2
Figure 4: Solution surfaces |a±|2 and |b±|2 on the domain D for α = 1, β = 0.25, Lξ = Lη = 20,
and α+ = 1.
30
Figure 5.4: Solution surfaces |a±|2(ξ, η) and |b±|2(ξ, η) on the domain (5.12) forα = 1, β = 0.25, Lξ = Lη = 20, and α+ = 1. Symmetries: |a+|2(ξ, Lη/2 − η) =|a+|2(ξ, Lη/2 + η) and |a−|2(ξ, Lη/2− η) = |a−|2(ξ, Lη/2 + η) (i.e |a+|2 |a−|2 are sym-metric with respect to η = Lη/2) and |b+|2(ξ, Lη/2− η) = |b−|2(ξ, Lη/2 + η) hold asexpected.
71
0
5
10
15
20
05
1015
200
0.5
1
1.5
2
2.5
ξ
η
|a+|2
05
1015
20
05
1015
200
0.05
0.1
0.15
0.2
0.25
ξη
|a−|2
05
1015
20
05
1015
200
0.2
0.4
0.6
0.8
1
1.2
1.4
ξη
|b+|2
05
1015
20
05
1015
200
0.2
0.4
0.6
0.8
1
1.2
1.4
ξη
|b−|2
Figure 5: Solution surfaces |a±|2 and |b±|2 on the domain D for α = 1, β = 0.75, Lξ = Lη = 20,
and α+ = 1.
31
Figure 5.5: Solution surfaces |a±|2(ξ, η) and |b±|2(ξ, η) on the domain (5.12) forα = 1, β = 0.75, Lξ = Lη = 20, and α+ = 1. Symmetries: |a+|2(ξ, Lη/2 − η) =|a+|2(ξ, Lη/2 + η) and |a−|2(ξ, Lη/2− η) = |a−|2(ξ, Lη/2 + η) (i.e |a+|2 |a−|2 are sym-metric with respect to η = Lη/2) and |b+|2(ξ, Lη/2− η) = |b−|2(ξ, Lη/2 + η) hold asexpected.
72
5.3 Transmission of two oblique waves
The stationary transmission of two oblique waves in the coupled-mode equations
(3.20)–(3.21) becomes diagonal in the characteristic coordinates (ξ, η):
ξ =
√p2 + q2
2(pm− nq)((q + 2m)X − (p+ 2n)Y ) ,
η =
√p2 + q2
2(pm− nq)(−qX + pY ) .
After the separation of variables (4.1), the linear coupled-mode equations (3.20)–
(3.21) reduce to the PDE system:
i∂a1
∂ξ+ Ωa1 + αa2 = 0,(5.48)
i∂a2
∂η+ αa1 + Ωa2 = 0.(5.49)
The coordinate lines ξ = ξ0 are parallel to the wave vector k2 = k(n,m,0)out , while the
coordinate lines η = η0 are parallel to the wave vector k1 = kin. The problem (5.48)–
(5.49) is defined in a bounded domain on the plane (ξ, η). We consider the same
rectangle D, defined by (5.12). When the incident wave is illuminated in the wave k1
but not in the wave k2, the linear system (5.48)–(5.49) is completed by the boundary
conditions:
a1(0, η) = α1(η), a2(ξ, 0) = 0.(5.50)
The boundary value problem of type (5.48)–(5.50) is referred to as Goursat problem
or characteristic Cauchy problem in the terminology of [CH]. The linear dispersion
relation Ω = Ω(Kξ, Kη), where (Kξ, Kη) are Fourier wave numbers, is given explicitly
as (Ω− Kξ +Kη
2
)2
= α2 +
(Kξ −Kη
2
)2
.(5.51)
73
Two surfaces of the dispersion relation (5.51) correspond to the two oblique resonant
waves. Although the stop band is not the band gap in the characteristic coordinates
(ξ, η) , there exist the reference frame on the plane (ξ, η) that moves with constant
velocity in time T , such that Kξ + Kξ = const , where exists a stop band in the
dispersion relation (5.51). We set Ω = 0, and write the formal solution of the system
(5.48)–(5.49) as follows:
a1(ξ, η) = iα
∫ ξ
0
a2(ξ, η) dξ + a1(0, η)
a2(ξ, η) = iα
∫ η
0
a1(ξ, η) dη
Interchanging integrals, we reduce the system to the Volterra type integral equation
a1(ξ, η) = −α2
∫ ξ
0
∫ η
0
a1(ξ, η) dη dξ + a1(0, η)
Hence the solution a1(ξ, η), a2(ξ, η) exist and unique. We consider solutions of the
system (5.48)–(5.49) at Ω = 0 by using the Fourier transform:
a1(ξ, η) =
∫ ∞
−∞kc(k)eiα(k−1ξ+kη)dk,(5.52)
a2(ξ, η) =
∫ ∞
−∞c(k)eiα(k−1ξ+kη)dk.(5.53)
It follows from the boundary conditions (5.50) that
kc(k) =α
2π
∫ Lη
0
α1(η)e−iαkηdη, k ∈ R(5.54)
and
0 =
∫ ∞
−∞c(k)eiαk−1ξdk, 0 ≤ ξ ≤ Lξ.(5.55)
Interchanging integrals, we reduce the constraint (5.55) to the form:
0 =α
2πi
∫ Lη
0
α1(η)
(∫ ∞
−∞
sinα(kη − k−1ξ)
kdk
)dη, 0 ≤ ξ ≤ Lξ.
74
The inner integral is zero for ξ > 0 and η > 0, due to the table integral 3.871 on p.
474 of [GR]. Therefore, the constraint (5.55) is satisfied and a unique solution of the
problem (5.48)–(5.50) exists in the form (5.52)–(5.54).
We illustrate the Fourier transform solution (5.52)–(5.53) with the constant input
function:
α1(η) = α1, η ∈ [0, Lη],
when c(k) can be found from (5.54):
c(k) =α1
2πi
1− e−iαkLη
k2, k ∈ R.
Evaluating Fourier integrals (5.52)–(5.53) with the help of the table integral 3.871 on
p. 474 of [GR], we find the explicit solution of the stationary problem:
a1(ξ, η) = α1J0(2α√ξη), a2(ξ, η) =
iα1√η
√ξJ1(2α
√ξη),(5.56)
where J0,1(z) are Bessel functions [GR]. Figure 6 shows the solution surfaces |a1(ξ, η)|2
and |a2(ξ, η)|2 in the domain (5.12) for α = 1, Lξ = Lη = 10, and α1 = 1. The inte-
gral invariants for the stationary transmission follow from integration of the balance
equation:
∂
∂ξ|a1|2 +
∂
∂η|a2|2 = 0.(5.57)
We define the incident (Iin), transmitted (Iout), and diffracted (Idif) intensities by
Iin =
∫ Lη
0
|a1(0, η)|2dη, Iout =
∫ Lη
0
|a1(Lξ, η)|2dη, Idif =
∫ Lξ
0
|a2(ξ, Lη)|2dξ.
The transmittance (T ) and diffractance (D) are defined by the same relations (5.2)
and the balance identity T +D = 1 follows from integration of the balance equation
(5.57). The numerical values for T and D are found from numerical integration of
75
the solution surfaces (5.58) as follows:
T ≈ 0.032, D ≈ 0.968,
such that T + D ≈ 1. These values show that the incident wave is diffracted to
the oblique resonance wave, such that only 3.2% of the wave energy remains in the
transmitted wave.
76
0
5
10
0
5
10
0
0.5
1
ξη
|a1|2
0
5
10
0
5
10
0
50
100
ξη
|a2|2
Figure 6: Solution surfaces |a1|2 and |a2|2 on the domain D for α = 1, Lξ = Lη = 10, and
α1 = 1.
32
Figure 5.6: Solution surfaces |a1|2(ξ, η) and |a2|2(ξ, η) on the domain (5.12) for α = 1,Lξ = Lη = 10, and α1 = 1.
Chapter 6
Summary and open problems
We have shown that the coupled-mode equations can be used for analysis and model-
ing of resonant interaction of Bloch waves in low-contrast cubic-lattice three-dimensional
photonic crystals.
We have proved existence and uniqueness of solution for the N -wave coupled-mode
system boundary value problem in any convex domain for a linear and non-linear
cases.
The analytical solutions for the linear stationary transmission problem are found
by using separation of variables and generalized Fourier series.
Non-stationary transmission problems are also of interest, and very few analytical
results are available on local and global well-posedness of the non-stationary nonlinear
coupled-mode equations.
Finally, numerical integration of the coupled-mode equations using multi-symplecticity
seems to be very promising.
77
Bibliography
[AP] D. Agueev and D. Pelinovsky [2004] Wave resonances in low-contrast photonic
crystals. SIAM J. Appl. Math., to be published.
[AS] A. Arraf and C.M. de Sterke [1998] Coupled-mode equations for quadratically
nonlinear deep gratings, Phys. Rev. E 58, 7951–7958.
[AJ1] N. Akozbek and S. John [1998] Optical solitary waves in two- and three-
dimensional nonlinear photonic band-gap structures, Phys. Rev. E 57, 2287–
2319.
[AJ2] N. Akozbek and S. John [1998] Self-induced transparency solitary waves in a
doped nonlinear photonic band gap materials, Phys. Rev. E 58, 3876–3895.
[BF1] A. Babin and A. Figotin [2001] Nonlinear photonic crystals: I. Quadratic non-
linearity, Waves in Random Media 11, R31–R102.
[BF2] A. Babin and A. Figotin [2002] Nonlinear photonic crystals: II. Interaction
classification for quadratic nonlinearities, Waves in Random Media 12, R25–
R52.
[BS] N. Bhat and J.E. Sipe [2001] Optical pulse propagation in nonlinear photonic
crystals, Phys. Rev. E 64, 056604
[Br] T. J. Bridges [1997] Multi-symplectic structures and wave propagation. Math.
Proc. Camb. Phil. Soc. 121, 147–190.
78
79
[BR] T. J. Bridges and S. Reich [2001] Multi-symplectic integrators: numerical
schemes for Hamiltonian PDEs that conserve symplecticity. Phys. Lett. A 284,
184–193.
[CL] E.A. Coddington and N. Levinson, Theory of Ordinary Differential Equations,
(McGraw-Hill, New York, 1955).
[CH] R. Courant and D. Hilbert [1962] Methods of Mathematical Physics. Interscience
Publisher.
[E] M. S. P. Eastham [1973] The Spectral Theory of Periodic Differential Equations.
Scottish Acad. Press, Edinburgh, London.
[FK1] A. Figotin and P. Kuchment [1994] Band-gap structure of the spectrum of
periodic and acoustic media. I. Scalar model. SIAM J. Appl. Math. 56, 68–88.
[FK2] A. Figotin and P. Kuchment [1994] Band-gap structure of the spectrum of
periodic and acoustic media. II. Two-dimensional photonic crystals. SIAM J.