2 A proof of Proposition 1.2

Notes on tangent bicharacteristics and transition of the spectral type of the Hamilton map

Tatsuo Nishitani

****

1 Introduction

In [4, 5, 6] we have studied the Cauchy problem for differential operatorsP with double characteristics when the spectral type of the Hamilton mapF_p changes, more precisely, assuming that the principal symbol p(x, ξ) vanishes exactly of order 2 on aC^∞ manifold Σ with codim Σ = 3 and

rank Xn j=0

dξ_j∧dx_j

Σ

= constant, (1.1)

the spectral type of Fpchanges across a submanifoldS of Σ with codimension 1.

(1.2)

Under these assumptions, after a conjugation with a Fourier integral operator if needed, one can write near any pointρ∈Σ

p(x, ξ) =−ξ₀²+ϕ1(x, ξ^′)²+ϕ2(x, ξ^′)², x= (x0, x^′) = (x0, x1, . . . , xn) wheredϕ1 anddϕ2are linearly independent at ρand Σ ={ξ0= 0, ϕ1= 0, ϕ2= 0}. Since (1.1) is equivalent to that the rank of ({ϕi, ϕj})i,j=0,1,2is constant on Σ withϕ0=ξ0the rank is either 0 or 2 for the matrix is anti-symmetric. If the rank is 0 then Σ is an involutive manifold and the spectral type is unchanged (see [1], [2] for the Cauchy problem in this case). Under the assumptions rank(dξ∧dx) = 2 on Σ and (1.2), following [4], one can assume without restrictions that (1.3) {ξ₀, ϕ₂}>0, {ξ₀, ϕ₁}=O(|ϕ|)

near ρ. Here and in what follows f =O(|ϕ|), ϕ = (ϕ₁, ϕ₂) means thatf is a linear combination ofϕ₁ andϕ₂ near the reference point. We first recall

∗Department of Mathematics, Osaka University, Machikaneyama 1-1, Toyonaka, 560-0043, Osaka, Japan

Lemma 1.1. ([5, Lemma 1.2])If the spectral structure ofFp changes acrossS then we have

{ξ0, ϕ2}²− {ϕ1, ϕ2}²= 0 on S.

Therefore we have one of the following cases;

(i) {ξ₀, ϕ₂}²− {ϕ₁, ϕ₂}² <0 in Σ\S, that is pis noneffectively hyperbolic in Σ\S withKerF_p²∩ImF_p²={0} and noneffectively hyperbolic on S with KerF_p²∩ImF_p²̸={0},

(ii){ξ₀, ϕ₂}²− {ϕ₁, ϕ₂}²>0inΣ\S, that ispis effectively hyperbolic inΣ\S and noneffectively hyperbolic on S with KerF_p²∩ImF_p²̸={0},

(iii) {ξ0, ϕ2}²− {ϕ1, ϕ2}² changes the sign across S, that is p is effectively hyperbolic in the one side ofΣ\S, noneffectively hyperbolic in the other side with KerF_p²∩ImF_p² = {0} and noneffectively hyperbolic on S with KerF_p²∩ImF_p²̸={0}.

We have also assumed that the spectral type ofFp changessimplyacrossS, that is

(1.4) {ξ₀, ϕ₂}²− {ϕ₁, ϕ₂}²= −θ²or θ²or θ

+c₁ϕ₁+c₂ϕ₂

near ρ ∈ S according to the case (i), (ii) and (iii) respectively where S is given by {ξ₀ = 0, ϕ₁ = 0, ϕ₂ = 0, θ = 0} and dξ₀, dϕ₁, dϕ₂, dθ are linearly independent. Since{ξ₀, ϕ₂}²− {ϕ₁, ϕ₂}²={ξ₀+ϕ₁, ϕ₂}{ξ₀−ϕ₁, ϕ₂} we have either {ξ₀ +ϕ₁, ϕ₂} = 0 or {ξ₀−ϕ₁, ϕ₂} = 0 on S. Since the arguments are completely parallel we may assume that {ξ0 −ϕ1, ϕ2} = 0 on S hence {ξ0, ϕ2}={ϕ1, ϕ2}>0 by (1.3). Now one can write

(1.5) {ξ0−ϕ1, ϕ2}= −θ²or θ²or θ

+c1ϕ1+c2ϕ2

near ρ for the case (i), (ii) and (iii) respectively, if we change the defining functionθ, if necessary.

Proposition 1.1. ([4]) Assume the case(i).

(1) If there is a bicharacteristic tangent toS atρthen{ξ0−ϕ1, θ}(ρ) = 0.

(2) Assume {ξ₀−ϕ₁, θ} = 0near ρ on S. If{{ξ₀−ϕ₁, ϕ₂}, ϕ₂}(ρ)̸= 0there is a bicharacteristic tangent to S atρ.

Proposition 1.2. ([5, Proposition 2.1])Assume the case (ii).

(1) If{ξ0−ϕ1, θ}(ρ)̸= 0there is a bicharacteristic tangent to Σ atρ.

(2) Assume {ξ0−ϕ1, θ} = 0near ρ on S. If{{ξ0−ϕ1, ϕ2}, ϕ2}(ρ)̸= 0there is a bicharacteristic tangent to S atρ.

In the proof of Proposition 1.2 in [5] there was a missing case to be examined, so in the next section we simplify slightly the whole proof including the missing case. To exclude the case with bicharacteristics tangent to Σ (orS) in our study of the Cauchy problem we introduce the conditions

(1.6) {ξ0−ϕ1, θ}(ρ)̸= 0 and

(1.7) {ξ0−ϕ1, θ}= 0, {{ξ0−ϕ1, ϕ2}, ϕ2}= 0 on S.

For a coordinates free expression of the condition (1.6) see [4, Lemma 12.3]. In [4] the Cauchy problem for the case (i) is discussed under the condition either (1.6) or (1.7) and we studied the Cauchy problem for the case (ii) in [5] assuming the condition (1.7).

Proposition 1.3. Assume the case (iii)and {ξ₀−ϕ₁, θ} = 0near ρon S. If {ξ₀−ϕ₁, ϕ₂}, ϕ₂}(ρ)̸= 0 there is a bicharacteristic tangent to S atρ.

Assuming (1.7) and some additional condition the Cauchy problem for the case (iii) is discussed in [6].

Non existence of tangent bicharacteristics is reflected in

Lemma 1.2. ([5, Lemma 2.2])Assume{{ξ0−ϕ1, ϕ2}, ϕ2}= 0onS. Then one can write

{ξ₀−ϕ₁, ϕ₂}= −θ²or θ²or θ

+c₀θϕ₁+c₁ϕ²₁+c₂ϕ₂ according to the case(i),(ii)and(iii) respectively.

Proof. Since (1.5) is independent of the choice of defining function θ, replacing θby

θ+ {ϕ2, θ}

{ϕ1, ϕ2}ϕ1− {ϕ1, θ} {ϕ1, ϕ2}ϕ2

one can assume that{θ, ϕ_j}=O(|ϕ|),j = 1,2. Thus

{{ξ0−ϕ1, ϕ2}, ϕ2}=c1{ϕ1, ϕ2}+O(|ϕ|) =O(|(θ, ϕ)|)

which implies thatc1=O(|(θ, ϕ)|) and hence the result. □ Lemma 1.3. ([5, Lemma 2.3])Assume (1.7). Then we have

{ξ0−ϕ1, θ}=c0θ+c1ϕ²₁+c2ϕ2.

Proof. Note that{ξ0−ϕ1, θ}=αθ+βϕ1+γϕ2where one can assume{θ, ϕj}= O(|ϕ|) as above. Therefore it follows from Lemma 1.2 that

{θ,{ξ₀−ϕ₁, ϕ₂}}=O(|ϕ|), {ξ₀−ϕ₁,{θ, ϕ₂}}=O(|(θ, ϕ)|).

Then from the Jacobi identity it follows thatβ=O(|(θ, ϕ)|) and hence {ξ0−ϕ1, θ}=αθ+c0θϕ1+c1ϕ²₁+c2ϕ2

which proves the assertion. □

Here we give simple examples. Consider

(1.8) p_±(x, ξ) =−ξ₀²+ξ²₁+ (x0+x1±x1x²₂)²ξ²_n=−ξ₀²+ϕ²₁+ϕ²₂ (n≥3) near (0, ξ) = (0, en) where Σ = {ξ0 = ξ1 = 0, x0+x1±x1x²₂ = 0}. Since {ξ₀, ϕ₁}= 0,{ξ₀, ϕ₂}= 1 and{ξ₀, ϕ₂}²− {ϕ₁, ϕ₂}²=∓2x²₂(1±x²₂/2)ξ_n² we see that the spectral type ofH_p±changes acrossS= Σ∩{x₂= 0}which corresponds to the case (i) and (ii) in Lemma 1.1. It is clear that {ξ₀−ϕ₁, x₂} = 0 and {{ξ₀ −ϕ₁, ϕ₂}, ϕ₂} = {∓x²₂ξ_n, ϕ₂} = 0 hence the condition (1.7) is clearly satisfied withθ=x₂. Next consider

(1.9) p(x, ξ) =−ξ₀²+ξ₁²+ (x0+x1−x1x2)²ξ²_n=−ξ²₀+ϕ²₁+ϕ²₂ (n≥3).

Since{ξ0, ϕ2}²− {ϕ1, ϕ2}²= 2x2(1−x2/2)ξ_n² the spectral type changes across S = Σ∩ {x2 = 0} which corresponds to the case (iii). It is clear that (1.7) is satisfied withθ=x2.

2 A proof of Proposition 1.2

Proposition 2.1. Assume the case (ii). If {ξ0−ϕ1, θ}(ρ)̸= 0 then there is a bicharacteristic tangent toΣatρ(which is not tangent to S).

We follow the arguments given in [3]. Since the assumption is independent of the choice of defining functionθone can assume that{θ, ϕj}=O(|ϕ|),j= 1,2.

To simplify notations let us set Ξ0 =ξ0−ϕ1, X0 =x0 and extend them to a full symplectic coordinates (X,Ξ). Switching the notation from (X,Ξ) to (x, ξ) one can write

p=−ξ₀(ξ₀+ 2ϕ₁) +ϕ²₂ where we have from (1.3) that

(2.1) {ξ₀, ϕ₁}=O(|ϕ|), {ξ₀, ϕ₂}=θ²+O(|ϕ|), {θ, ϕ_j}=O(|ϕ|), j= 1,2.

The assumption is now{ξ0, θ}(ρ)̸= 0 hence one can writeθ=e(x0−ν(x^′, ξ^′)) withe̸= 0. Thus one can assumeθ=x0+ν(x^′, ξ^′). Noting thatdξ0,dx0,dϕ1, dϕ2 are linearly independent at ρin view of (2.1) one can take

ξ0, x0, ϕ1, ϕ2, ψ1, . . . , ψr, r+ 4 = 2(n+ 1) to be a system of local coordinates aroundρ. Thus

(2.2) ν =O(|(ϕ1, ϕ2, ψ1, . . . , ψr)|).

Note that we can assume that ψ_j are independent of x₀ taking ψ_j(0, x^′, ξ^′) as newψ_j such that{ξ₀, ψ_j}= 0. Moreover we can assume that

(2.3) {ψ_j, ϕ_k}=O(|ϕ|), k= 1,2, j = 1, . . . , r

takingψj− {ψj, ϕ1}ϕ2/{ϕ2, ϕ1} − {ψj, ϕ2}ϕ1/{ϕ1, ϕ2} as newψj. Consider the Hamilton equations

(2.4) dx/ds=∂p/∂ξ, dξ/ds=−∂p∂x.

Letγ(s) = (x(s), ξ(s)) be a solution to the Hamilton equations and we consider ξ0(s),x0(s),ϕj(γ(s)),θ(γ(s)),ψj(γ(s)) and recall that

d

dsf(γ(s)) ={p, f}(γ(s)).

Let us change the parameter fromstot= 1/sso that we have d/ds=−tD, D=t(d/dt)

and hencetD(t^pF) =t^p+1(DF+pF) forp∈N. Let us introduce new unknowns;

(2.5)

( ξ0(s) =t⁴Ξ0(t), x0(s) =tX0(t), ϕ1(γ(s)) =t²Φ1(t), ϕ₂(γ(s)) =t³Φ₂(t), ψ_j(γ(s)) =t²Ψ_j(t).

Recallδ={ϕ1, ϕ2}(ρ)>0 and

{ξ₀, ϕ₂}=θ²+κϕ₁+Cϕ₂=x²₀+ 2νx₀+ν²+κϕ₁+Cϕ₂. Let us set

V = (X0,Φ1,Ξ0,Φ2,Ψ), Ψ = (Ψ1, . . . ,Ψr) then, taking (2.2) and (2.3) into account, it is not difficult to see

(2.6)













DX0=−X0+ 2Φ1+tG(t, V), DΦ1=−2Φ1+ 2δΦ2+tG(t, V),

DΞ0=−4Ξ0+ 2κΦ1Φ2+ 2Φ2X₀²+tG(t, V), DΦ2=−3Φ2+ 2κΦ²₁+ 2δΞ0+ 2Φ1X₀²+tG(t, V), DΨj =−2Ψj+tG(t, V)

whereG(t, V) denotes a smooth function in (t, V) such thatG(t,0) = 0.

Let us define the class of formal series int and log 1/t E={ X

0≤j≤i

tⁱ(log 1/t)^jVij|Vij ∈C^N}

in which we look for our formal solutions to the reduced Hamilton equations (2.6).

Lemma 2.1. There exists a formal solutionV ∈ Esatisfying (2.6)withΦ1(0)̸= 0,X₀(0)̸= 0.

Proof. Let us set

(2.7)











X0=P

0≤j≤itⁱ(log 1/t)^jβ⁽⁰⁾_ij , Ξ0=P

0≤j≤itⁱ(log 1/t)^jα⁽⁰⁾_ij Φ1=P

0≤j≤itⁱ(log 1/t)^jβ_ij⁽¹⁾, Φ2=P

0≤j≤itⁱ(log 1/t)^jα⁽¹⁾_ij

Ψ =P

tⁱ(log 1/t)^jγ^(k).

Equating the constant terms of both sides of (2.6) one has

−β⁽⁰⁾₀₀ + 2β⁽¹⁾₀₀ = 0,

−2β₀₀⁽¹⁾+ 2δα⁽¹⁾₀₀ = 0,

−4α⁽⁰⁾₀₀ + 2κβ₀₀⁽¹⁾α⁽¹⁾₀₀ + 2α⁽¹⁾₀₀(β₀₀⁽⁰⁾)²= 0,

−3α⁽¹⁾₀₀ + 2κ(β₀₀⁽¹⁾)²+ 2δα⁽⁰⁾₀₀ + 2β₀₀⁽¹⁾(β₀₀⁽⁰⁾)²= 0,

−2γ₀₀^(k)= 0.

Setting b =β⁽¹⁾₀₀ we see β₀₀⁽⁰⁾ = 2b, α⁽¹⁾₀₀ =δ⁻¹b from the first and the second equation. It follows from the third equation that

2α⁽⁰⁾₀₀ =κδ⁻¹b²+ 4δ⁻¹b³=δ⁻¹b(κb+ 4b²).

Inserting these into the fourth equation we have

−3δ⁻¹b+ 3κb²+ 12b³= 3b −1

δ +κb+ 4b²

= 0.

Let us study

(2.8) −1

δ +κb+ 4b²= 0.

Sinceδ >0 it is clear that this equation has nonzero real roots b=b(κ, δ), one is positive and the other one is negative. Let us choose one of suchb. Then

V = β₀₀⁽⁰⁾, β₀₀⁽¹⁾, α⁽⁰⁾₀₀, α⁽¹⁾₀₀, γ₀₀^(k)

= (2b, b, bδ⁻²/2, δ⁻¹b,0)

is uniquely determined. We look for a formal solution to (2.6) in the formV+V, V ∈ E^# where

E^#={ X

1≤i,0≤j≤i

tⁱ(log 1/t)^jV_ij |V_ij ∈C^N}. Let us denote

V^I =^t(X₀,Φ₁,Ξ₀,Φ₂), V^II = Ψ.

Then (2.6) becomes (2.9)

(

DV^I =AIV^I+FIt+GI(t, V), DV^II =−2V^II+FIIt+GII(t, V) where

GJ(t, V) = X

2≤i,0≤j≤i

GJ,ijtⁱ(log 1/t)^j, GJ,ij=GJ,ij(Vpq|p≤i−1)

andFJ are constant vectors. Making a more precise look onAI we see

AI =







−1 2 0 0

0 −2 0 2δ

8δ⁻¹b² 2κδ⁻¹b −4 2δ⁻¹ 8b² 2(κb+δ⁻¹) 2δ −3







where we have used (2.8). Then we have

Lemma 2.2. A_I has real eigenvalues 1, −6. Other real eigenvalues ofA_I are non positive.

Proof. We have

|λ−A_I|=

λ+ 1 −2 0 0

0 λ+ 2 0 −2δ

−8δ⁻¹b² −2κδ⁻¹b λ+ 4 −2δ⁻¹

−8b² −2(κb+δ⁻¹) −2δ λ+ 3

= (λ−1)(λ+ 6) λ²+ 5λ+ 8−4κδb where we have used−4b²δ=κbδ−1. Noting that

1−κbδ= 4b²δ≥0

it is clear that real roots ofλ²+ 5λ+ 8−4κδb = 0, if exist, are less than or

equal to−1. □

Proof of Lemma 2.1: Note that (2.9) implies that

(iV_ij−(j+ 1)V_ij+1) =AV_ij+δ_i1δ_j0F+G_ij whereG_ij= 0 fori= 0,1. Then we have

(

(I−A)V11= 0, (I−A)V10=V11+F.

ChooseV11∈Ker (I−A) so that

F+V₁₁∈Im (I−A).

Then one can takeV₁₀̸= 0 so that

(I−A)V₁₀=F+V₁₁

since Ker (I−A)̸={0}by Lemma 2.2. We turn to the case i≥2;

(2.10) (iI−A)Vij = (j+ 1)Vij+1+Gij.

Withj=i, (2.10) becomes to

(iI−A)Vii=Gii(Vpq|p≤i−1).

SinceiI−Ais non singular for i≥2 by Lemma 2.2 one has V_ii= (iI−A)⁻¹G_ii(V_pq|p≤i−1).

Recurrently one can solveV_ij by V_ij = (iI−A)⁻¹

(j+ 1)V_ij+1+G_ij(V_pq|p≤i−1)

forj=i−1,i−2, . . . ,0. This proves the assertion. □ Proof of Proposition 2.1: Repeating a similar but much simpler arguments as in [3] we can conclude that there is a solution to (2.6) which is asymptotic to the formal solution given in Lemma 2.1. Thus we get a solution (x(s), ξ(s)) to the Hamilton equations (2.4). From (2.5) we have, withξ₀=ϕ₀, that

dϕj

dx0

x₀=0= dϕj

dx .dx0

ds

x₀=0= 0, dθ dx0

= 1

hence it is clear that thus obtained bicharacteristic (x(s), ξ(x)) is tangent to Σ

but not toS. □

Proposition 2.2.Assume{ξ₀−ϕ₁, θ}= 0nearρonS. If{{ξ₀−ϕ₁, ϕ₂}, ϕ₂}(ρ)̸= 0there is a bicharacteristic tangent to S atρ.

Proof. As in the proof of Proposition 2.1 setting Ξ0=ξ0−ϕ1,X0=x0we extend them to a full symplectic coordinates (X,Ξ) and then switch the notation from (X,Ξ) to (x, ξ). Now the assumption reads

{ξ0, ϕ1}=O(|ϕ|), {ξ0, ϕ2}={−θ²or θ²or θ}+O(|ϕ|), {θ, ϕj}=O(|ϕ|), {ξ0, θ}=O(|(θ, ϕ)|).

(2.11)

Sincedξ0,dx0,dϕ1,dϕ2,dθare linearly independent atρin view of (2.11) one can take

ξ₀, x₀, ϕ₁, ϕ₂, θ, ψ₁, . . . , ψ_r, r+ 5 = 2(n+ 1)

to be a system of local coordinates aroundρ. After that the proof is a repetition of [5, Proposition 4.3] (similar to the proof of Proposition 2.1). □ Combining Propositions 2.1 and 2.2 we conclude the proof of Proposition 1.2.

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