A note on the Cauchy problem for −D2

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A note on the Cauchy problem for

− D ² ₀ + 2x ₁ D ₀ D ₂ + D ₁ ² + x ³ ₁ D ₂ ² + P 2

j=0 b _j D _j

Tatsuo Nishitani

^∗

Abstract

In this note, we improve a previously proved non-solvability result of the Cauchy problem in the Gevrey class for a homogeneous second order diﬀerential operator with polynomial coeﬃcients. We prove that the Cauchy problem for the same operator is not locally solvable for any lower order term in the Gevrey class of order greater than 5, lowering the previous Gevrey order 6.

1 Introduction

In [9] considering a second order operator inR¹⁺²

(1.1) Pmod=−D₀²+ 2x1D0D2+D₁²+x³₁D₂², x= (x0, x^′) = (x0, x1, x2) we have proved

Theorem 1.1. ([9, Theorem 1.1])The Cauchy problem forP_mod+P2

j=0b_jD_j is not locally solvable at the origin in the Gevrey class of ordersfor anyb0, b1, b2∈ Cifs >6. In particular the Cauchy problem for Pmod isC^∞ ill-posed near the origin for anyb0, b1, b2∈C.

Recall that the Gevrey class of order s, denoted by γ^(s)(Rⁿ), is the set of allf(x)∈C^∞(Rⁿ) such that for any compact set K⊂Rⁿ, there existC >0, h >0 such that

(1.2) |∂_x^αf(x)| ≤Ch^|^α^||α|!^s, x∈K, α∈Nⁿ. We say that the Cauchy problem forP =Pmod+P2

j=0bjDj is locally solvable in γ^(s) at the origin if for any Φ = (u0, u1)∈(γ^(s)(R²))² there exists a neigh- borhoodUΦ, may depend on Φ, of the origin such that the Cauchy problem (1.3)

(

P u= 0 in UΦ,

D₀^ju(0, x^′) =u_j(x^′), x^′ ∈U_Φ∩ {x₀= 0}, j= 0,1

∗Department of Mathematics, Osaka University: [email protected]

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has a solution u(x)∈ C²(UΦ). In this note we remark that one can improve Theorem 1.1 so that

Theorem 1.2. The Cauchy problem forP_mod+P2

j=0b_jD_jis not locally solvable inγ^(s) at the origin for anyb0, b1, b2∈Cifs >5.

Before explaining a special role played by P_mod in studying of the Cauchy problem for differential operators with non-effectively hyperbolic characteristics, we give a short introduction to the general context. Let P be a differential operator of order m with principal symbol p(x, ξ). At a singular point ρ of p = 0, the Hamilton map F_p(ρ) is defined as the linearization at ρ of the Hamilton vector fieldHp. The first fundamental result is

Theorem 1.3. ([5], [4]) Let ρ be a singular point of p= 0 and assume that Fp(ρ)has no non-zero real eigenvalues. If the Cauchy problem forP isC^∞well posed it is necessary that

(1.4) ImP_sub(ρ) = 0, P_sub(ρ)≤Tr⁺F_p(ρ)/2 where Tr⁺Fp(ρ) = P

|µj| and µj are the eigenvalues of Fp(ρ), counted with multiplicity, andPsub is the subprincipal symbol of P.

We call (1.4) the Ivrii-Petkov-H¨ormander condition (IPH condition, for short).

If the strict inequality holds in (1.4) we call it the strict Ivrii-Petkov-Hörmander condition (strict IPH condition, for short). For the sufficiency of the IPH condition, we assume that the set of singular points Σ ofp= 0 is a C^∞ manifold and the following conditions are satisfied:

(1.5)







Fp has no non-zero real eigenvalues at each point of Σ, Near each point of Σ,pvanishes exactly of order 2 and the rank ofPn

j=0dξj∧dxj is constant.

According to the spectral type ofFp(ρ), two diﬀerent possible cases may arise KerF_p²(ρ)∩ImF_p²(ρ) ={0},

(1.6)

KerF_p²(ρ)∩ImF_p²(ρ)̸={0}. (1.7)

We say thatpis of spectral type 1 (resp. type 2) near ρ∈Σ if there is a conic neighborhoodV ofρsuch that (1.6) (resp. (1.7)) holds inV ∩Σ. We say that there is no transition (of spectral type) if for any ρ∈ Σ one can ﬁnd a conic neighborhoodV ofρsuch that either (1.6) or (1.7) holds inV ∩Σ.

Theorem 1.4. ([10, Theorem 5.1])Assume (1.5)and that there is no transition and no bicharacteristic tangent to Σ. Then the Cauchy problem for P is C^∞ well posed under the strict IPH condition.

The principal symbol p= −ξ₀²+ 2x1ξ0ξ2+ξ²₁+x³₁ξ²₂ of Pmod is a typical example of spectral type 2 with tangent bicharacteristics (note that there is no

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tangent bicharacteristic ifp is of spectral type 1, see [6]). The set of singular point ofp= 0 is Σ ={x1= 0, ξ0=ξ1= 0} which is aC^∞ manifold on whichp vanishes exactly of order 2 (note that we are working near (0, e3),e3= (0,0,1)) and a tangent bicharacteristic is given explicitly by

(1.8) (x₁, x₂) = (−x²₀/4, x⁵₀/80), (ξ₀, ξ₁, ξ₂) = (0, x³₀/8,1)

which is parametrized byx0. The operator Pmod shows how the situation be- comes to be complicated when a tangent bicharacteristic exists. We now give some such results: The Cauchy problem forPmodis not locally solvable inγ^(s) fors >5, in particular ill-posed in C^∞ while the Cauchy problem for general second order operatorP of spectral type 2 satisfying Psub= 0 on Σ (note that the subprincipal symbol ofPmodis identically zero) is well posed in the Gevrey class of order 1< s≤5 ([1] see also [10]). The Cauchy problem forPmod+SD2

with 0̸=S∈Cis not locally solvable inγ^(s) fors >3 (Proposition 2.1 below) while the Cauchy problem for general second order operatorP of spectral type 2 with codim Σ = 3 is well posed in the Gevrey class of order 1< s <3 for any lower order term even if a tangent bicharacteristic exists ([2]).

2 A family of exact solutions

First recall

Lemma 2.1. ([10, Proposition 8.1])The Cauchy problem forP_mod+P1 j=0b_jD_j is not locally solvable inγ^(s)at the origin for anyb0, b1∈Cifs >5.

Thus in order to prove Theorem 1.2 it suﬃces to prove the following result which also improves [2, Theorem 1.3].

Proposition 2.1. The Cauchy problem for Pmod+P2

j=0bjDj is not locally solvable inγ^(s) at the origin for anyb0, b1∈Cand0̸=b2∈Cifs >3.

To prove Proposition 2.1 we repeat the proof of [2, Theorem 1.3] with obvious minor chnges. Look for a family of solutions to (Pmod+P2

j=0bjDj)Uλ = 0 in the form

(2.1) Uλ(x) =e^iξ⁰^λx⁰Vλ(x^′), Vλ(x^′) =e^±^λ⁵^x²⁻^i(b¹^/2)x¹u(λ²x1), ξ0=ξ0(λ) that is, we look foru(x) satisfying

(2.2) u^′′(x) = x³+ 2ξ0x+b2λ−ξ₀²λ⁻²+b0ξ0λ⁻³−b²₁λ⁻⁴/4 u(x).

To study solutions to (2.2) we consider

(2.3) u^′′(x) = (x³+a₂x+a₃)u(x), x∈C, a_j∈C, j= 2,3.

LetY0(x;a),a= (a2, a3) be the solution given in [11, Chapter 2] to (2.3) which has asymptotic representation

(2.4) Y0(x;a)≃x⁻^3/4 1 +

X∞ N=1

BNx⁻^N/2

e⁻^E(x,a)

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asxtends to inﬁnity in any closed subsector of the open sector |argx|<3π/5 where

E(x, a) =2

5x^5/2+a2x^1/2

andAN,BN are polynomials in (a2, a3). Let ω=e^2πi/5 and set (2.5) Yk(x;a) =Y0(ω⁻^kx;ω⁻^2ka₂, ω⁻^3ka₃), k= 0,1,2,3,4 which are also solutions to (2.3). Recall that [11, Chpater 17]

Yk(x;a) =Ck(a)Yk+1(x;a)−ωYk+2(x;a)

whereCk(a) are entire analytic ina= (a2, a3) andCk(a2, a3) =C0(ω⁻^2ka2, ω⁻^3ka3).

Choose

u(x) =Y0(x;a) =C0(a)Y1(x;a)−ωY2(x;a), a₂= 2ξ₀, a₃=b₂λ−ξ₀²λ⁻²+b₀ξ₀λ⁻³−b²₁λ⁻⁴/4 (2.6)

which solves (2.2). We require thatV_λ(x^′) is bounded as λ→ +∞ when |x^′| remains in a bounded set. Note (2.5) and

(ω⁻¹x)^5/2=−i|x|^5/2, 2ω⁻²ξ₀(ω⁻¹x)^1/2=−2iξ₀|x|^1/2, (ω⁻²x)^5/2=i|x|^5/2, 2ωξ0(ω⁻²x)^1/2= 2iξ0|x|^1/2 (2.7)

C0(2ξ0, a3) = 0, (2.8)

−Re(iξ0) =Imξ0<0.

(2.9)

Instead of solving directly the “Stokes equation” (2.8) we go rather indirectly.

Let us consider

H(β) =p²+x²+iβx³

as an operator inL²(R) with the domainD(H(β)) =D(p²)∩D(x³). Here p² denotes the self-adjoint realization of−d²/dx² deﬁned inH²(R) and byD(x³) we mean the domain of the maximal multiplication operator by the functionx³. Proposition 2.2. [3, Corollary 2.16, Lemma 3.1]) Let k ∈ N0 and ϵ > 0 be given. Then there is aB >0such that for |β|< B,Reβ >0,H(β)has exactly one eigenvalueEk(β)near2k+ 1. Such eigenvalues are analytic functions of β for|β|< B,Reβ >0, and admit an analytic continuation across the imaginary axis to the whole sector|β|< B,|argβ|< ^5π₈ −ϵ and uniformly asymptotic to the following formal Taylor expansion in powers of β²

(2.10)

X∞ j=0

a2jβ^2j, a0= 2k+ 1 nearβ = 0in any closed subsector in |argβ|< ^5π₈ −ϵ.

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For the proof we refer to [3], [12]. Now we have

Lemma 2.2. ([2, Lemma 6.1])Assume thatReβ >0andE(β)is an eigenvalue of the problem

(2.11) −u^′′(x) + (x²+iβx³)u(x) =E(β)u(x) that is(2.11)has a solution0̸=u∈D(H(β)). Then we have

(2.12) C0

−ω²

3 β⁻⁸⁵, ω³ 2

27β⁻¹²⁵ +β⁻²⁵E(β) = 0

whereβ⁻^j/5= (β⁻^1/5)^j and the branchβ^±^1/5 is chosen such that|argβ^±^1/5|<

π/10.

(2.13)







2ξ₀=−ω² 3 β⁻⁸⁵, a₃=ω³2

27β⁻¹²⁵ +β⁻²⁵E(β) then (2.8) is satisﬁed hence we have

u(λ²x) =Y0(λ²x; 2ξ0, a3) =−ωY2(λ²x; 2ξ0, a3).

Therefore we look forξ₀ satisfying (2.13) and (2.9). Pluggingξ₀=−ω²β⁻^8/5/6 into the second equation, (2.13) is reduced to

2

27ω³β⁻¹²⁵ +E(β)ω³β⁻²⁵ + 1

36ω⁴β⁻¹⁶⁵λ⁻² +b0

6 ω²β⁻^8/5λ⁻³=b₂λ−b²₁ 4λ⁻⁴. (2.14)

2.1 Solving the equation (2.14)

Solve (2.14) with respect toβ where E(β) is one ofEk(β) which is analytic in

|argβ|<5π/8−ϵand admits a uniform asymptotic expansion (2.10) there by Proposition 2.2. Putζ=β⁻^2/5= (β⁻^1/5)² so that the equation leads to (2.15) ω³ζ⁶+27

2 E(ζ⁻^5/2)ω³ζ+3

8ω⁴ζ⁸λ⁻²±9

4b0ω²ζ⁴λ⁻³=±27

2 b2λ−27 8 b²₁λ⁻⁴ and we look for a solutionζ(λ) to (2.15) verifying

(2.16) |argζ(λ)|< π/4−ϵ, Imω²ζ(λ)⁴>0

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where the second requirement comes from (2.9). Assume that (2.17) 0<argb2< π or −π≤argb2<−π/2

and denote 27b2/2 by Afor notational simplicity and look forζ(λ) in the form ( ζ(λ) =e⁻^πi/5A^1/6 1 +λ⁻^5/6z

λ^1/6 (0<argA < π), ζ(λ) =e^2πi/15A^1/6 1 +λ⁻^5/6z

λ^1/6 (−π≤argA <−π/2).

It is clear that ζ(λ) veriﬁes (2.16) provided that z is bounded and λ is large.

Note thatE(ζ⁻^5/2) is analytic in|argζ|< π/4−ϵfor large|ζ|and veriﬁes E(ζ⁻^5/2)−a0≤C|ζ|⁻⁵

with somea₀= 2k+ 1,k∈Nuniformly in|argζ|< π/4−ϵwhen|ζ| → ∞. We insertζ into (2.15) to get

Aλ(1 +λ⁻^5/6z)⁶+λ^1/6H(z) +d1λ⁻^2/3(1 +λ⁻^5/6z)⁸ +d2λ⁻^7/3(1 +λ⁻^5/6z)⁴=Aλ+d3λ⁻⁴, di∈C (2.18)

whereH(z) is analytic in|z|< B forλ≥R and

|H(z)−a₀| ≤Cλ⁻^5/6, λ≥R.

Note thatB can be chosen to be arbitrarily large takingRlarge. Thus one can write (2.18) as

(2.19) 6Az+a₀+λ⁻^5/6F(z, λ) = 0

where F(z, λ) is analytic in|z| < B and bounded uniformly in |λ| ≥ R. We may assume that |a0/6A|< B/2 taking R large as noted above. By Rouch´e’s thoerem we conclude that the equation (2.19) has a solutionz(λ) with|z|< B for any |λ| ≥R₁. Returning to β (ζ = β⁻^2/5) we conclude that (2.14) has a solution of the form

( β(λ) =iA⁻^5/12λ⁻^5/12 1 +λ⁻^5/6z(λ)

(0<argA < π), β(λ) =e⁻^πi/3A⁻^5/12λ⁻^5/12 1 +λ⁻^5/6z(λ)

(−π≤argA <−π/2) where|z(λ)|< B forλ≥R1. Plugging thisβ(λ) into (2.13) we get Proposition 2.3. Assume (2.17). Then there existsξ0=ξ0(λ) such that

( 2ξ₀=c λ^2/3(1 +λ⁻^5/6z(λ)), Imc <0, C₀(2ξ₀, b₂λ−ξ₀²λ⁻²+b₀ξ₀λ⁻³−b²₁λ⁻⁴/4) = 0 where|z(λ)|< B forλ > R.

If we write

(2.20) 2ξ0=λ^2/3a(λ), b2λ−ξ₀²λ⁻²+b0ξ0λ⁻³−b²₁λ⁻⁴/4 =λb(λ) it is clear that|a(λ)|,|b(λ)|< M with someM >0 forλ≥R1.

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2.2 Asymptotics of Y

0

(x; aλ

^2/3

, bλ) for large | x | and λ

Recalling (2.20) we study howY0(x;aλ^2/3, bλ) behaves for large|x|and largeλ where|a|,|b| ≤M is assumed. In what followsf =oa(1) means that there are positive constantsCa>0 andδa>0 such that

|f| ≤Caλ⁻^δ^a, λ→+∞.

We make the asymptotic representation (2.4) slightly precise.

Proposition 2.4. Let ρ >1/3 be given. Then one can write

Y0(x;aλ^2/3, bλ)≃(1 +pρ(x, λ))e^R^ρ^(x,λ)x⁻^3/4exp{−Eρ(x;a, b, λ)}, Y0^′(x;aλ^2/3, bλ)≃(−1 +pρ(x, λ))e^R^ρ^(x,λ)x^3/4exp{−Eρ(x;a, b, λ)} asxtends to inﬁnity in any closed subsector of

Sλ={x;|argx|<3π/5,|x|> λ^ρ} where

Eρ(x;a, b, λ) =2

5x^5/2+aλ^2/3x^1/2−bλx⁻^1/2+rρ(x, λ) andr_ρ is a polynomial inx⁻^1/2 such that

|rρ(x, λ)|=Cλ^4/3⁻^3ρ/2(1 +oρ(1)), x∈Sλ

andpρ(x, λ),Rρ(x, λ)are holomorphic in Sλ and in any closed subsector ofSλ

|pρ(x, λ)| ≤Cρλ⁻^2(ρ⁻^1/3), |Rρ(x, λ)| ≤Cρλ⁻¹, x∈Sλ

andR_ρ(x, λ)→0,p_ρ(x, λ)→0 as|x| → ∞,x∈S_λ.

Proof. We follow Sibuya [11, Chapter 2] and [2, Proposition 2.3] with needed modiﬁcations.

Lemma 2.3. Assume thatY0(x;aλ^2/3, bλ)veriﬁes (2.21) Y0(x;aλ^2/3, bλ) =−ωY2(x;aλ^2/3, bλ) and

(2.22) Ima=−δ(1 +oa(1))

with some δ > 0. Let X > 0. There exist ℓ, c >0, C >0 such that for any 0≤µ <5/6 we have

(d/dx)^kY0(λ²x, aλ^2/3, bλ)≤C(1 +o_µ(1))λ^ℓ

×exp

−δλ^5/3|x|^1/2(1 +oa(1)) +Cµ(λ^µ+λ⁻^5/3+3µ) fork= 0,1 and|x| ≥λ⁻^2µX.

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Proof. In Proposition 2.4 chooseρ= 2(1−µ) and estimateY0in x≤ −λ⁻^2µX ﬁrst. Recall that forx <0 we have

Y0(x;aλ^2/3, bλ) =−ωY2(x;aλ^2/3, bλ) =−ωY0(e^πi/5|x|;ωaλ^2/3, ω⁻¹bλ).

Denote

ϕ⁻(x, λ) =E_ρ(e^πi/5λ²|x|;ωa, ω⁻¹b, λ), ϕ⁺(x, λ) =E_ρ(λ²x;ωa, ω⁻¹b, λ) then we have

ϕ⁻(x, λ) = 2

5iλ⁵|x|^5/2+iaλ^5/3|x|^1/2+ib|x|⁻^1/2+r_ρ(e^πi/5λ²|x|, λ).

(2.23)

Sincer_ρ(e^πi/5λ²|x|, λ)≤C_ρλ⁻^5/3+3µ for|x| ≥λ⁻^2µX and then

(2.24) −Reϕ⁻(x, λ)≤ −δλ^5/3|x|^1/2(1 +oa(1)) +Cρ(λ^µ+λ⁻^5/3+3µ).

Forx≥λ⁻^2µX >0 note that

−Reϕ⁺(x, λ)≤ −cλ⁵x^5/2+Cλ^5/3x^1/2+C_ρ(λ^µ+λ⁻^5/3+3µ)

=−cλ⁵⁻^4µx^1/2(1 +oa(1)) +Cρ(λ^µ+λ⁻^5/3+3µ) (2.25)

forµ <5/6. Then the assertion follows from (2.24) and (2.25).

Lemma 2.4. Assume that (2.21) and (2.22) hold with some δ > 0 and that µ <5/6. Let0< X1< X2<1. Then there exist C >0,ℓandc >0such that

Y0(λ²x;aλ^2/3, bλ)≥Cλ^ℓe⁻^cλ^5/3−µ, λ⁻^2µX1≤ −x≤λ⁻^2µX2. Proof. It is clear from (2.23) that there exists C1>0 such that

−λ^5/3⁻^µ/C1≤ −Reϕ⁻(x, λ)≤ −C1λ^5/3⁻^µ whenλ⁻^2µX1≤ −x≤λ⁻^2µX2.

Lemma 2.5. Under the same assumptions as in Lemma 2.3 there exist c >0, A >0 such that

(d/dx)^kY0(λ²x, aλ^2/3, bλ)≤C_µA^k+1(1 +k³+λ^5/2)^ke^cλ^5/6, k∈N for|x| ≥λ⁻^2µX.

Proof. We ﬁrst estimate |(d/dx)^kY0(λ²x;aλ^2/3, b)| in x ≤ −λ⁻^2µX. From Proposition 2.4 withρ= 2(1−µ) we have

Y0(λ²x;aλ^2/3, bλ) =C(1 +pµ(x))λ⁻^3/2x⁻^3/4e⁻^ϕ⁻^(x,λ)+R^µ^(x)

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wherepµ(x) andRµ(x) are holomorphic and bounded in|x|> λ⁻^2µX,|argx|<

3π/5. Since|x|⁻¹≤√

Xλ^2µ ≤√

Xλ^5/3 we have

|d^k(−ϕ⁻(x, λ) +Rµ(x))/dx^k|

≤CµA^kk!(λ⁵|x|^3/2+λ^5/3|x|⁻^1/2+Cµ|x|⁻^3/2)|x|¹⁻^k

≤C_µA^kk!(1 +λ⁵|x|^3/2+λ^5/2)λ^5(k⁻^1)/3

≤CµA^kk!λ^5k/3(λ^10/3|x|^3/2+λ^5/6), |x| ≥λ⁻^2µX, k≥1.

Therefore it follows that forx≤ −λ⁻^2µX

|x|^3k/2e⁻^cλ^5/3^|^x^|^1/2≤C^k+1λ⁻^5kk^3k we conclude that

|d^ke⁻^ϕ⁻^(x,λ)+R^µ^(x)/dx^k| ≤CA^k(λ^5/2+k³)^ke^cλ^5/6

which proves the assertion forx ≤ −λ⁻^2µX. For x≥λ⁻^2µX it is enough to repeat the same arguments noting that (2.25) and 5−4µ >5/3.

Corollary 2.1. Assume that Y0(x;aλ^2/3, bλ)veriﬁes (2.21) and (2.22). Then there existc >0,A >0,C >0 such that for anyϵ >0there is λϵ such that

(d/dx)^kY0(λ²x, aλ^2/3, bλ)≤CϵA^k+1(1 +k³+λ^5/2)^ke^cλ^5/6, k∈N forλ⁻^5/3+ϵ≤ |x|,λ≥λϵ.

Proof. Chooseµ= 5/6−ϵ/2 in Lemma 2.5.

Next we estimateY0(λ²x;aλ^2/3, bλ) for|x| ≤λ⁻^5/3+ϵ.

Lemma 2.6. ([2, Lemma 6.5, Lemma 6.7]) Assume thaty(x, λ)satisﬁes (2.26) y^′′(x, λ) = (x³+aλ^2/3x+bλ)y(x, λ), |a|, |b| ≤M.

Then there arec >0,C >0 andℓ_i>0such that for any T >0 we have (d/dx)^ky(x, λ)≤C^k+1(k+λ^1/3+|x|)^3k/2λ^ℓ¹(1 +T)^ℓ²e^cλ^5/6^(1+λ^−1/3^T⁾^5/2

×

|y(T, λ)|+|y^′(T, λ)| , |x| ≤T, k∈N, λ≥1.

Proposition 2.5.Assume thatY0(x;aλ^2/3, bλ)veriﬁes(2.21)and(2.22). Then there areℓ,c >0,A >0,C >0 such that for anyϵ >0 there isλ_ϵ such that

(d/dx)^kY0(λ²x;aλ^2/3, bλ)≤CA^k+1λ^ℓ(k³+λ⁴)^ke^cλ^5/6+ϵ, k∈N forλ≥λϵ.

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Proof. Applying Lemma 2.3 with µ= 5/6−ϵ/2. we have

(d/dx)^kY0(±λ^1/3+ϵ)≤Cλ^ℓe^c¹^λ^5/6, λ≥λ_ϵ, k= 0,1.

SinceY0(x) =Y0(x, aλ^2/3, bλ) satisﬁes (2.26), choosingT =λ^1/3+ϵ in Lemma 2.6 we get

(d/dx)^kY0(x)≤C^k+1λ^ℓ(k+λ^1/3+ϵ)^3k/2e^cλ^5/6+3ϵ, |x| ≤λ^1/3+ϵ, k∈N forλ≥λϵ. This proves that

(d/dx)^kY0(λ²x)≤C^k+1λ^ℓ(λ²k^3/2+λ^5/2+2ϵ)^ke^cλ^5/6+3ϵ

for |x| ≤ λ⁻^5/3+ϵ. Since λ^k(5/2+2ϵ)e⁻^λ^5/6+3ϵ ≤ k^3k and λ²k^3/2 ≤C(λ⁴+k³) combining Corollary 2.1 with the above obtained estimates we conclude the assertion.

Recalling

V_λ(x^′) =e^iλ⁵^x²⁻^i(b¹^/2)x¹Y0(λ²x₁;λ^2/3a(λ), λb(λ))

withλ^2/3a(λ) = 2ξ₀ andb(λ) =b₂+b₀ξ₀λ⁻⁴−ξ₀²λ⁻³−b²₁λ⁻⁵/4 one has Lemma 2.7. There exist c >0 and A > 0 such that for any ϵ > 0 there are C >0 andλϵ such that

(2.27) ∂_x^k

1Vλ(x^′)≤CA^k(k!)³e^cλ^4/3, k∈N, λ≥λϵ.

Proof. Noting λ^4k ≤ C^k(k!)³e^3λ^4/3 the assertion follows from Proposition 2.5.

3 Proof of Theorem 1.2

First assume thatb2̸= 0 satisﬁes (2.17). Following Section 2.1 we have a family of exact solutions {Uλ} satisfying (Pmod+P2

j=0bjDj)Uλ = 0. We show that {U_λ}does not satisfy apriori estimates derived fromγ^(s)local solvability of the Cauchy problem ifs >3. Leth >0 and a compact setK be ﬁxed and denote by γ^(s),h₀ (K) the set of all f(x^′) ∈ γ^(s)(R²) such that suppf ⊂ K and (1.2) holds with someC >0 for all α∈N². Note that γ₀^(s),h(K) is a Banach space with the norm

sup

α,x

|∂^α_xf(x^′)| h^|^α^||α|!^s.

Proposition 3.1(Holmgren). DenoteDϵ={x∈R³| |x^′|²+|x0|< ϵ}. There existsϵ0>0 such that forϵ satisfying0< ϵ < ϵ0 ifu(x)∈C²(Dϵ)satisﬁes

(

P_mod+P2

j=0b_jD_j

u= 0 in D_ϵ,

D₀^ju(0, x^′) = 0 (j= 0,1), x∈Dϵ∩ {x0= 0} thenu(x)≡0in Dϵ.

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Lemma 3.1.(e.g.[8, Proposition 4.1, Theorem 4.2], [7])Assume that the Cauchy problem forP_mod+P2

j=0b_jD_jis locally solvable inγ^(s)at the origin. Then there existsδ >0 such that for any 0< ϵ1< δand any (uj(x^′))∈γ₀^(s),h({|x^′| ≤ϵ1}) there is a unique solution u(x) ∈ C²(Dδ) to the Cauchy problem (1.3) with U_(u_j_(x′))=Dδ and for any compact setL⊂Dδ there existsC >0such that (3.1) |u(x)|C²(L)≤C

X1 j=0

sup

α,x^′

|∂_x^α′uj(x^′)| h^|^α^||α|!^s holds.

Sinceλ^5k ≤k!^se^sλ^5/s and U_λ(0, x^′) =V_λ(x^′),D₀U_λ(0, x^′) =ξ₀(λ)λV_λ(x^′) it is clear from Lemma 2.7 that one can ﬁndc₁>0,C >0 such that

(3.2)

X1 j=0

sup

α,x^′

|∂_x^α_′D₀^jU_λ(0, x^′)|

h^|^α^||α|^s^|^α^| ≤Ce^c¹^λ^max^{5/s,4/3}, s≥3.

On the other hand thanks to Lemma 2.4 and Proposition 2.3 there isc₀ > 0 such that

(3.3) Uλ(x0,−λ⁻^2µ,0)≥Cλ^ℓe^c⁰^λ^5/3^x⁰⁻^cλ^5/3−2µ, x0>0

where µ is chosen such that 0 < µ <5/6. Let χ(x^′) ∈ γ^(s)(R²) be such that χ(x^′) = 0 for |x^′| ≥ √ϵ1 and χ(x^′) = 1 for |x^′| ≤ √ϵ2 < √

ϵ1. Since Φλ = χ(x^′)(U_λ(0, x^′), D₀U_λ(0, x^′)) ∈ γ^(s),h₀ ({|x^′| ≤ ϵ₁}) there is a unique solution u_λ(x)∈C²(D_δ) to the Cauchy problem (1.3) with Cauchy data Φ_λ(x^′) which satisﬁes (3.2). Thanks to Proposition 3.1 we see thatu_λ =U_λ in D_ϵ₂. Take a compact set L ⊂ D_ϵ₂ such that the interior of L contains (x₀,−λ⁻^2µ,0) with small x0 > 0 and large λ. If s > 3 hence max{5/s,4/3} < 5/3 the inequalities (3.2) and (3.3) are not compatible contradicting Lemma 3.1, which proves Theorem 1.2.

Whenb2 ̸= 0 does not satisfy (2.17) we make a change of local coordinates (x₀, x₁, x₂)→(−x₀, x₁,−x₂) such thatP_mod+P2

j=0b_jD_j will be (3.4) Pmod−b0D0+b1D1−b2D2.

in the new local coordinates. Since the local solvability in γ^(s)at the origin is invariant under (analytic) change of local coordinates and−b2obviously satisﬁes (2.17), we conclude the same assertion also in this case.

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A note on the Cauchy problem for −D2