An infinite dimensional KAM theorem with application to two dimensional completely resonant beam equation

0.21 MB·

by  Jiansheng Geng

An infinite dimensional KAM theorem with application to
two dimensional completely resonant beam equation
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Jiansheng Geng, Shidi Zhou

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a Departmentof Mathematics,Nanjing University,Nanjing 210093, P.R.China
J Email: [email protected]; [email protected]

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Abstract
D
. InthispaperweconsiderthecompletelyresonantbeamequationonT2 withcubic
h nonlinearity on a subspace of L2(T2) which will be explained later. We establish an
t
a abstract infinite dimensional KAM theorem and apply it to the completely resonant
m
beamequation. We provetheexistence ofaclassofWhitney smoothsmallamplitude
[ quasi-periodic solutions corresponding to finite dimensional tori.

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Mathematics Subject Classification: Primary 37K55; 35B10
v
5 Keywords: KAM theory; Hamiltonian systems; Beam equation; Birkhoff normal form
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5 1 Introduction
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.
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0
Inthispaperweconsiderthetwodimensionalcompletely resonantbeamequationwith
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1 cubic nonlinearity on a subspace of L2(T2):
: U
v
i
X
u +∆2u+u3 = 0 u= u(t,x),t R,x T2 (1.1)
r tt x ∈ ∈
a
Here t is time and x is the spatial variable. The subspace is defined by
U
= u= u φ , φ (x) = ei n,x (1.2)
n n n h i
U { }
n∈XZ2odd
where the integer set Z2 is defined as
odd
Z2 = n = (n ,n ) :n 2Z 1,n 2Z (1.3)
odd { 1 2 1 ∈ − 2 ∈ }
This idea comes from the work by M.Procesi [29] and we will explain it later in section 2.
The solution of ”real” completely resonant beam equation (not on Z2 , just on Z2) will
odd
be handled in our forthcoming paper.
∗This work is partially supported by NSFCgrant 11271180.
1
The infinite dimensional KAM theory with applications to Hamiltonian PDEs has at-
tracted great interests since 1980s. Starting from the remarkable work [6,19,31], a lot of
achievements have been made in 1-dimensional Hamiltonian PDEs about the existence
of quasi-periodic solutions by the methods of KAM theory. For these work, just refer
to [5,12,13,17,18,20-25,32]. But when people turn to the higher dimensional case, the
multiplicity of eigenvalues became a great obstacle because it leads to much more compli-
cated small divisor conditions and measure estimates. The first breakthrough comes from
Bourgain’s work [3] in 1998. In this work, the cumbersome second Melnikov condition is
avoided due to the application of the method of multiscale analysis, which essentially is a
Nash-Moser iterative procedureinstead of Newtonian iteration beingwidely used in KAM
theory. Following this idea, a lot of important work has been made in higher dimensional
case (refer to [1,2,4,30]).
However, despite the advantage of avoiding the difficulty of the second Melnikov con-
ditions, therearealsodrawbacks ofmultiscale analysis methods. For example, wecouldn’t
see the linear stability of the small-amplitude solutions and it couldn’t show us a descrip-
tion of the normal form, which is fundamental in knowing the dynamical structure of
an equation. For these reasons, KAM approach is also expected in dealing with higher
dimensional equations. The first work comes from Geng and You [14] in 2006, which
established the KAM theorem solving higher dimensional beam equations and nonlocal
smooth Schro¨dinger equations with Fourier multiplier. They used the “zero-momentum
condition” to avoid the multiplicity of eigenvalues and the regularity property to do the
measure estimate. Later in 2010, a remarkable work [8] by Eliasson and Kuksin dealt
with quite general case: higher dimensional Schro¨dinger equations with convolutional
type potential and without “zero-momentum condition”. To overcome the multiple eigen-
values they studied the distribution of integer points on a sphere and got a normal form
with block-diagonal structure, and conducted the measure estimates by developing the
technique named “Lipschitz domain”. Motivated by their method, the quasi-periodic so-
lutions of completely resonant Schro¨dinger equation on 2-dimension torus was developed
by Geng, Xu and You [11] in 2011, with a very elaborate construction of tangential sites.
Inthispaper,theydefinestheconception of“To¨plitz-Lipschitz” conditionandprovedthat
the perturbation satisfies “To¨plitz-Lipschitz” condition. Later, in [26,27] C.Procesi and
M.Procesi extended this result to higher dimensional case. For other work about higher
dimensional equation, just refer to [7,9,10,15,16,28,29].
Let us turn to beam equation now. In [15] Geng and You got the quasi-periodic
solutions of beam equation in high dimension with typical constant potential and the
nonlinearity is independent on the spatial variable x. Recently, in [7] Eliasson, Grebert
and Kuksin got the quasi-periodic solutions of beam equation having typical constant
potential in higher dimensional case, and with an elaborate but quite general choice of
tangential sites in the sense of probability. they allow that their normal form contain
hyperbolic terms which is cumbersome in solving homological equations. Motivated by
their work, we want to consider the completely resonant beam equation (1.1). In our case,
there are no outer parameters and only the amplitude provides parameters. Compared
with the case of typical constant potential, although we have “zero-momentum condition”
here, but when doing the normal form before KAM procedure, some terms still couldn’t
be eliminated because of the loss of outer parameters. We could only get a block-diagonal
normal form with finite dimensional block. As a consequence, our normal form is always
2
related to the angle variable θ, so here the linear stability is not available. Compared
with [11], our convenience is that we have regularity property here and needn’t verify the
complicated ”To¨eplitz-Lipschitz condition” at each step. But except for this, our normal
form structure and KAM iteration is similar to that in [11].
Now we state the choice of tangential sites. Let S = i Z2 : 1 j b here b 2.
{ j ∈ odd ≤ ≤ } ≥
We say S is admissible if it satisfies the following conditions.
Proposition 1 (Structure of S)
1 Any three of them are not vertices of a rectangle.
(cid:13)
2 For any n Z2 S, there exists at most one triple i,j,m with i,j S,m Z2 S
(cid:13) ∈ odd\ { } ∈ ∈ odd\
such that
n m+i j = 0
− −
n 2 m 2+ i2 j 2 = 0
(
| | −| | | | −| |
and if it exists, we say (n,m) are resonant in the first type and denote all such n by .
1
L
3 For any n Z2 S, there exists at most one triple i,j,m with i,j S,m Z2 S
(cid:13) ∈ odd\ { } ∈ ∈ odd\
such that
n+m i j = 0
− −
n 2+ m 2 i2 j 2 = 0
(
| | | | −| | −| |
and if it exists, we say (n,m) are resonant in the second type and denote all such n by .
2
L
4 Any n Z2 S shouldn’t be in and at the same time. It means that = .
(cid:13) ∈ odd\ L1 L2 L1∩L2 ∅
(Here means l2 norm)
|·|
The proof of the existence of admissible sets is postponed in the Appendix, which is a
modification of [11].
Now we could state the main theorem.
Theorem 1 Let S = (i ,i , ,i ) Z2 be an admissible set. There exists a Cantor
1 2 ··· b ⊆ odd
set of positive measure, s.t. ξ = (ξ ,ξ , ,ξ ) , equation (1.1) admits a small-
1 2 b
C ∀ ··· ∈ C
amplitude real-valued quasi-periodic solution
b
u(t,x) = ξj(eiωjtφij +e−iωjtφ¯ij)+O(|ξ|32)
jX=1q
Theoutlineof this paperis as follows: Insection 2westate somepreliminaries and the
abstract KAM theorem. In section 3 we deal with the normal form before KAM iteration.
In section 4 we conduct one step of KAM iteration: solving homological equation and
verifingthenewnormalformandperturbation. Insection 5weproveuniformconvergence
and get the invariant torus. In section 6 we complete the measure estimate. The choice
of tangential sites is put into the appendix.
3
2 Preliminaries and statement of the abstract KAM theo-
rem
InthissectionweintroducesomenotationsandstatetheabstractKAMtheoremwhich
allows the existence of some terms dependent on θ in the normal form part.
To simplify, we only consider the subspace Z2 (defined in (1.3)) instead of Z2. Given
odd
b points i ,i , ,i (b 2) in Z2 , denoted by S, which should be an admissible
{ 1 2 ··· b} ≥ odd
set(defined in Propositon 1), and let Z2 be the complementary set of S in Z2 . Denote
1 odd
z = (z ) with its conjugate z¯= (z¯ ) . We introduce the weighted norm as follows:
n n Z2 n n Z2
∈ 1 ∈ 1
z = z n aeρn a,ρ > 0 (2.1)
a,ρ n | |
k k | || |
nX∈Z21
Here n = n 2+ n 2,n = (n ,n ) Z2. DenoteaneighborhoodofTb I = 0 z =
| | | 1| | 2| 1 2 ∈ 1 ×{ }×{
0 z¯= 0 by
}×{ p}
D(r,s) = (θ,I,z,z¯): Imθ <r, I < s2, z < s, z¯ < s
a,ρ a,ρ
| | | | k k k k
(cid:8) (cid:9)
Here means the sup-norm of complex vectors.
|·|
Let α = α ,β = β , α ,β N with only finitely many non-vanishing
{ n}n∈Z21 { n}n∈Z21 n n ∈
components. Denote zαz¯β = n Z2znαnz¯nβn and let
∈ 1
Q
F(θ,I,z,z¯)= F (ξ)ei k,θ Ilzαz¯β (2.2)
klαβ h i
k,l,α,β
X
where ξ Rb is the parameter set. k = (k , ,k ) Zb and l = (l , ,l ) Nb,
1 b 1 b
Il = Il1 ∈ IOlb.⊆Denote the weighted norm of F by ··· ∈ ··· ∈
1 ··· b
F = sup F ekrs2l zα z¯β (2.3)
D(r,s), klαβ | | ||
k k O ξ∈O,kzka,ρ<s,kz¯ka,ρ<skXlαβ| |O | || |
F = sup ∂4F (2.4)
| klαβ|O ξ | ξ klαβ|
∈O0≤Xd≤4
where the derivatives with respect to ξ are in the sense of Whitney.
To a function F we define its Hamiltonian vector field by
X = (F , F ,i F , i F ) (2.5)
F I − θ { zn}n∈Z21 − { z¯n}n∈Z21
and the associated weighted norm is
1
X := F + F
k FkD(r,s),O k IkD(r,s),O s2k θkD(r,s),O
1
+ F n a¯enρ+ F n a¯enρ (2.6)
s  k znkD(r,s),O| | | | k z¯nkD(r,s),O| | | | 
nX∈Z21 nX∈Z21 
 
where a¯ > 0 is a constant and we need a¯ > a to measure the regularity property of the
perturbation at each iterative step.
4
The normal form has the following form:
H = N + + + ¯
0
A B B
N = ω(ξ),I + Ω z z¯
n n n
h i n Z2
P∈ 1
= a (ξ)ei(θi θj)z z¯
n − n m
A
n 1
P∈L
= a (ξ)e i(θi+θj)z z
n − n m
B
n 2
¯= P∈L a (ξ)ei(θi+θj)z¯ z¯
n n m
B
n 2
P∈L
where ξ is the parameter. For each n or n , the 3-triple (m,i,j) is uniquely
1 2
∈ O ∈ L ∈ L
determined.
Forthisunperturbedsystem,it’seasytoseethatitadmitsaspecialsolution(θ,0,0,0)
→
(θ+ωt,0,0,0) correspondingto an invariant torus in thephasespace. Ourgoal is to prove
that, after removing some parameters, the perturbed system H = H +P still admits in-
0
variant torus provided that X is sufficiently small. To achieve this goal, we
k PkDa,ρ(r,s),O
require that Hamiltonian H satisfies some conditions:
(A1) Nondegeneracy: The map ξ ω(ξ) is a C4 diffeomorphism between and its
→ W O
image (C4 means C4 in the sense of Whitney).
W
(A2) Asymptotics of normal frequencies:
Ω = ε p n 2+Ω˜ p > 0 (2.7)
n − n
| |
here Ω˜ is a C4 function of ξ, and Ω˜ = O(n ι) ι > 0
n W n | |−
(A3) Melnikov conditions: Let
A = Ω n Z2 ( )
n n ∈ 1\ L1∪L2
and
Ω +ω a
A = n i n n
n am Ωm+ωj ! ∈ L1
Ω ω a
An = na¯−m i Ωm n ωj ! n ∈ L2
−
Then we assume that there exists γ,τ > 0, such that
k,ω γ k = 0
|h i| ≥ kτ 6
| |
det( k,ω +A ) γ
| h i n | ≥ kτ
| |
|det(hk,ωi+An⊗I2±I2⊗An′)| ≥ kγτ k 6= 0
| |
(A4) Boundedness: + + ¯+P is real analytic in each variable θ,I,z,z¯and Whitney
A B B
smooth in ξ. And we have
X + X + X < 1, X < ε (2.8)
k AkDa,ρ(r,s),O k BkDa,ρ(r,s),O k B¯kDa,ρ(r,s),O k PkDa,ρ(r,s),O
5
(A5) Zero-momentum condition:
The normal form part + + ¯+P satisfy the following condition:
A B B
+ + ¯+P = ( + + ¯+P) (ξ)ei k,θ Ilzαz¯β
klαβ h i
A B B A B B
k Zb,Xl Nb,α,β
∈ ∈
we have
b
( + + ¯+P) =0 = k i + (α β )n = 0
klαβ j j n n
A B B 6 ⇒ −
jX=1 nX∈Z21
Now we state our abstract KAM theorem, and as a corollary, we get Theorem 1.
Theorem 2 Assume that the Hamiltonian H = N + + + ¯+P satisfies condition
A B B
(A1) (A5). Let γ > 0 be sufficiently small, then there exists ε > 0 and a,ρ > 0 such
−
that if X < ε, the following holds: There exists a Cantor subset with
meas(k PkD)a,=ρ(r,Os),(Oγς) (ς is a positive constant) and two maps which are aOnγal⊆ytOic in θ
γ
O \O
and C4 in ξ.
W
Φ :Tb D (r,s), ω˜ : Rb
γ a,ρ γ
×O → O →
where Φ is ε -close to the trivial embedding Φ : Tb Tb 0,0,0 and ω˜ is ε-close
γ16 0 ×O → ×{ }
to the unperturbed frequency ω, such that ξ and θ Tb, the curve t Ψ(θ+ω˜t,ξ) is
γ
∀ ∈ O ∈ →
a quasi-periodic solution of the Hamiltonian equation governed by H = N+ + + ¯+P.
A B B
3 Normal Form
Consider the equation (1.1). The linear operator ∆ has eigenvalues λ = n 2and
n
corresponding eigenfunctions φn = 21πeihn,xi. By scaling−u→ ε12u, (1.1)becomes | |
u +∆2u+εu3 = 0 (3.1)
tt
Now introduce v = u and (3.1) is turned into
t
u = v
t
v = ∆2u εu3 (3.2)
t
− −
Let q = √12((−∆)12u−i(−∆)−12v) and (3.2) becomes
1 1 1 q+q¯ 3
iqt = ∆q+ε ( ∆)−2 ( ∆)−2( ) (3.3)
− − √2 − − √2
(cid:18) (cid:19)
Write it in the form of Hamiltonian equation q = i∂H and we get the Hamiltonian
t ∂q¯
1 1 1 4
H = ∆q,q + ε ( ∆)−2(q+q¯) dx (3.4)
2h− i 4 T2 −
Z (cid:16) (cid:17)
6
where , is the inner product in L2(T2). Notice that in Z2 the origin is avoided so
h· ·i odd
( ∆)−21 is well defined. (That is why we use it instead of the whole Z2) Now expand q
−
into Fourier series
q = q φ (3.5)
n n
n∈XZ2odd
so the Hamiltonian becomes (justify ε if necessary)
1
H = λ q 2 + ε (q q q q +q¯q¯q¯q¯)
n n i j k l i j k l
| | λ λ λ λ
n∈XZ2odd i+j+Xk+l=0 i j k l
p 1
+ 4ε (q q q q¯ +q¯q¯q¯ q )
i j k l i j k l
λ λ λ λ
i+j+k l=0 i j k l
X−
p 1
+ 6ǫ (q q q¯q¯) (3.6)
i j k l
λ λ λ λ
i+j k l=0 i j k l
−X−
p
Now we state the normal form theorem of H.
Propsition 3.1 Let S be admissible. For Hamiltonian function (3.6), there exists a sym-
plectic transformation Φ satisfying
H Φ = ω,I + Ω z z¯ + + + ¯+P (3.7)
n n n
◦ h i A B B
nX∈Z21
where
2 1
ω = ε 4λ + ξ +4 ξ i S (3.8)
i − i λ2 i λ λ j ∈
i j S,j=i i j
∈X6
1
Ω = ε 4λ +4 ξ n Z2 (3.9)
n − n λ λ j ∈ 1
j n
j S
X∈
and
ξ ξ
= 4 i j ei(θi θj)z z¯ (3.10)
− n m
A λ λ λ λ
nX∈L1 pi j n m
p ξ ξ
= 4 i j ei( θi θj)z z (3.11)
− − n m
B λ λ λ λ
nX∈L2 pi j n m
p ξ ξ
¯ = 4 i j ei(θi+θj)z¯ z¯ (3.12)
n m
B λ λ λ λ
nX∈L2 pi j n m
p
P = O(ε2 I 2+ε2 I z 2 +εξ 12 z 3 +ε2 z 4 +ε2 ξ 3
| | | |k ka,ρ | | k ka,ρ k ka,ρ | |
+ ε3|ξ|25kzka,ρ +ε4|ξ|2kzk2a,ρ +ε5|ξ|23kzk3a,ρ) (3.13)
7
Proof : We construct a Hamiltonian function F to induce Φ = X1 which is the time-1
F
map of F. For convenience, we define three sets as below:
S = (i,j,n,m) : 1 : i j +n m =0
1
{ (cid:13) − −
2 : i2 j 2 + n 2 m 2 = 0
(cid:13) | | −| | | | −| | 6
3 : # i,j,n,m S 2 (3.14)
(cid:13) { }∩ ≥ }
and similarly
S = (i,j,n,m) : 1 : i+j +n+m =0
2
{ (cid:13)
2 : i2+ j 2 + n 2+ m 2 = 0
(cid:13) | | | | | | | | 6
3 : # i,j,n,m S 2 (3.15)
(cid:13) { }∩ ≥ }
S = (i,j,n,m) : 1 : i+j +n m =0
3
{ (cid:13) −
2 : i2+ j 2 + n 2 m 2 = 0
(cid:13) | | | | | | −| | 6
3 : # i,j,n,m S 2 (3.16)
(cid:13) { }∩ ≥ }
we define F as
iε
F = q q¯q q¯
i j n m
λ λ +λ λ
i j n m
XS1 − −
iε
+ (q q q q q¯q¯q¯ q¯ )
i j n m i j n m
6(λ +λ +λ +λ ) −
i j n m
XS2
2iε
+ (q q q q¯ q¯q¯q¯ q ) (3.17)
i j n m i j n m
3(λ +λ +λ λ ) −
i j n m
XS3 −
(3.6) is put into (set z = q ,z¯ = q¯ ,n / S)
n n n n
∈
1
H Φ = λ q 2+ λ z 2+ε q 4
◦ n| n| n| n| λ2| n|
nX∈S nX∈/S nX∈S n
1 1
+ 4ε q 2 q 2+4ε q 2 z 2
i j i n
λ λ | | | | λ λ | | | |
i j i n
i,j∈XS,i6=j i∈SX,n∈/S
1 1
+ 4ε q q¯z z¯ +4ε (q q z¯ z¯ +q¯q¯z z )
i j n m i j n m i j n m
λ λ λ λ λ λ λ λ
i j n m i j n m
nX∈L1 nX∈L2
+ O εq pz 3 +ε z 4 +ε2 q 6+ε2 q 5pz 3 +ε2 q 4 z 2 +ε2 q 3 z 3
| ||| ||a,ρ || ||a,ρ | | | | k ka,ρ | | || ||a,ρ | | || ||a,ρ
(cid:16) (cid:17)
Here we need to state a fact: For four points n,m,i,j Z2 , it could never satisfy
∈ odd
n 2+ m 2+ i2 j 2 = 0. If not, we assume n= (n ,n ),m = (m ,m ),i = (i ,i ),j =
1 2 1 2 1 2
| | | | | | −| |
(j ,j ) and in each one the first component is odd and the second component is even.
1 2
Then we have
n 2+ m 2+ i 2 j 2 = (n 2+ m 2+ i 2 j 2)
1 1 1 1 2 2 2 2
| | | | | | −| | − | | | | | | −| |
8
The right one can be divided by 4 but the left one couldn’t, which is a contradiction. By
this fact we know that the set
(i,j,n,m) (Z2 )4 : 1 :i+j +n m = 0
{ ∈ odd (cid:13) −
2 : i2 + j 2 + n 2 m 2 = 0 (3.18)
(cid:13) | | | | | | −| | }
is empty.
Introduce the action-angle variable in the tangential sites:
q = I +ξ eiθj, q¯ = I +ξ e iθj (3.19)
j j j j j j −
q q
so we have
1
H Φ = λ (I +ξ )+ λ z 2+ε (I +ξ )2
◦ i i i n| n| λ2 i i
Xi∈S nX∈/S Xi∈S i
1 1
+ 4ε (I +ξ )(I +ξ )+4ε (I +ξ )z 2
i i j j i i n
λ λ λ λ | |
i j i n
i,j∈XS,i6=j i∈SX,n∈/S
1
+ 4ε (I +ξ )(I +ξ )ei(θi θj)z z¯
i i j j − n m
λ λ λ λ
i j n m
nX∈L1 q
p 1
+ 4ε (I +ξ )(I +ξ )ei( θi θj)z z
i i j j − − n m
λ λ λ λ
i j n m
nX∈L2 q
p 1
+ 4ε (I +ξ )(I +ξ )ei(θi+θj)z¯ z¯
i i j j n m
λ λ λ λ
i j n m
nX∈L2 q
+ O(ε|ξ|12kpzk3a,ρ +εkzk4a,ρ +ε2|ξ|3+ε2|ξ|52kzka,ρ +ε2|ξ|2kzk2a,ρ+ε2|ξ|32kzk3a,ρ)
2ε 1
= (λ + ξ +4ε ξ )ω
i λ2 i λ λ j i
i S i j S,i=j i j
X∈ ∈X6
1
+ (λ +4ε ξ )z 2
n i n
λ λ | |
i n
nX∈/S Xi∈S
1
+ 4ε ξ ξ ei(θi θj)z z¯
i j − n m
λ λ λ λ
nX∈L1 i j n mq
p 1
+ 4ε ξ ξ ei( θi θj)z z
i j − − n m
λ λ λ λ
i j n m
nX∈L2 q
p 1
+ 4ε ξ ξ ei(θi+θj)z¯ z¯
i j n m
λ λ λ λ
nX∈L2 i j n mq
+ O(εI 2p+εI z 2 +εξ 12 z 3 +ε z 4 +ε2 ξ 3
| | | |k ka,ρ | | k ka,ρ k ka,ρ | |
+ ε2|ξ|25kzka,ρ +ε2|ξ|2kzk2a,ρ +ε2|ξ|32kzk3a,ρ)
By scaling in variables:
ξ ε3ξ, I ε5I, z ε52z, z¯ ε52z¯
→ → → →
9
and scale time t ε9t we get the Hamiltonian function as follows:
→
H = ω,I + Ωz,z + + + ¯+P (3.20)
h i h i A B B
where
2 1
ω = ε 4λ + ξ +4 ξ (3.21)
i − i λ2 i λ λ j
i j S,j=i i j
∈X6
1
Ω = ε 4λ +4 ξ (3.22)
n − n j
λ λ
j n
j S
X∈
ξ ξ
= 4 i j ei(θi θj)z z¯ (3.23)
− n m
A λ λ λ λ
nX∈L1 pi j n m
p ξ ξ
= 4 i j ei( θi θj)z z (3.24)
− − n m
B λ λ λ λ
nX∈L2 pi j n m
p ξ ξ
¯ = 4 i j ei(θi+θj)z¯ z¯ (3.25)
n m
B λ λ λ λ
nX∈L2 pi j n m
P = O(ε2|Ip|2+ε2|I|kzk2a,ρ +ε|ξ|12kzk3a,ρ +ε2kzk4a,ρ +ε2|ξ|3+ε3|ξ|52kzka,ρ
+ ε4 ξ 2 z 2 +ε5 ξ 32 z 3 ) (3.26)
| | k ka,ρ | | k ka,ρ
Now we verify that the normal form (3.7) (3.13) satisfy condition (A1) (A5).
− −
Verifying(A1): By (3.8) we get
∂ω
= (a ) (3.27)
ij 1 i,j b
∂ξ ≤ ≤
where a = 2 if i = j and a = 4 if i = j. It’s easy to see that this matrix is
ij λ2i ij λiλj 6
non–degenerate.
Verifying(A2): By (3.9), just take p = 4,ι = 2.
Verifying(A3): Recall the definition in condition (A3), we only verify the most com-
plicated case:
det(hk,ωi+An ⊗I2−I2⊗An′) (3.28)
′
where n,n . We verify two facts: (3.28) is a polynomial of parameter ξ with
1 2
∈ L ∪L
degree 4 and it couldn’t be equivalently zero. For the former one, notice that λI +A
⊗
I I B = (λI +A) I I B (here means determinant) and using the formula
2 2
− ⊗ ⊗ − ⊗ |·|
A I I B = (A B )2+ A(tr(B))2+ B (tr(A))2 (A + B )tr(A)tr(B)
| ⊗ ± ⊗ | | |−| | | | | | ± | | | |
then we get it. For the latter one, it’s the same as that in [11]. By this, we could get
|∂ξ4 det(hk,ωi+An⊗I2±I2⊗An′) |> c|k|
(cid:0) 1 (cid:1)
So by excluding parameters with measure O(γ4), we have
γ
| det(hk,ωi+An⊗I2±I2⊗An′) | > k τ k 6= 0
| |
(cid:0) (cid:1)
For the verification of (A4) and (A5), just refer to [14].
10