Diagonal symmetrizers for hyperbolic operators with triple characteristics
aa r X i v : . [ m a t h . A P ] J un Diagonal symmetrizers for hyperbolic operatorswith triple characteristics
Tatsuo Nishitani ∗ Abstract
Symmetrizers for hyperbolic equations are obtained by diagonalizingthe B´ezoutian matrix of hyperbolic symbols. Such diagonal symmetrizersare applied to the Cauchy problem for hyperbolic operators with triplecharacteristics. In particular, the V.Ivrii’s conjecture concerned withtriple effectively hyperbolic characteristics is proved for differential op-erators with coefficients depending on the time variable, and for thirdorder differential operators with two independent variables with analyticcoefficients.
Keywords: Symmetrizer, B´ezoutiant, triple characteristic, Tricomi type, Cauchy problem.Mathematics Subject Classification 2010: Primary 35L30
This paper is devoted to the Cauchy problem(1.1) D mt u + P m − j =0 P | α | + j ≤ m a j,α ( t, x ) D αx D jt u = 0 ,D jt u (0 , x ) = u j ( x ) , j = 0 , . . . , m − t ≥ x ∈ R n and the coefficients a j,α ( t, x ) are real valued C ∞ functionsin a neighborhood of the origin of R n and D x = ( D x , . . . , D x n ), D x j =(1 /i )( ∂/∂x j ) and D t = (1 /i )( ∂/∂t ). The problem is C ∞ well-posed near theorigin for t ≥ δ > U of the origin of R n such that (1.1) has a unique solution u ∈ C ∞ ([0 , δ ) × U ). We assume thatthe principal symbol p is hyperbolic for t ≥
0, that is p ( t, x, τ, ξ ) = τ m + m − X j =0 X | α | + j = m a j,α ( t, x ) ξ α τ j has only real roots in τ for ( t, x ) ∈ [0 , δ ′ ) × U ′ and ξ ∈ R n with some δ ′ > U ′ of the origin which is necessary in order that the Cauchyproblem (1.1) is C ∞ well-posed near the origin for t ≥ ∗ Department of Mathematics, Osaka University: [email protected]
1n this paper we are mainly concerned with the case that the multiplicity ofthe characteristic roots is at most 3. This implies that it is essential to studyoperators P of the form(1.2) P = D t + X j =1 a j ( t, x, D ) D − jt h D i j which is differential operator in t with coefficients a j ∈ S , classical pseudodif-ferential operator of order 0, where h D i = Op((1 + | ξ | ) / ). One can assumethat a ( t, x, D ) = 0 without loss of generality and hence the principal symbolhas the form p ( t, x, τ, ξ ) = τ − a ( t, x, ξ ) | ξ | τ − b ( t, x, ξ ) | ξ | . With U = t ( D t u, D t h D i u, h D i u ) the equation P u = f is reduced to(1.3) D t U = A ( t, x, D ) h D i U + B ( t, x, D ) U + F where A, B ∈ S , F = t ( f, ,
0) and A ( t, x, ξ ) = a b . Let S be the B´ezoutiant of p and ∂p/∂τ , that is(1.4) S ( t, x, ξ ) = − a a b − a b a then S is nonnegative definite and symmetrizes A , that is SA is symmetricwhich is easily examined, though this is a special case of a general fact (see [10],[20]). Then one of the most important works would be to obtain lower boundof (Op( S ) U, U ). The sharp G˚arding inequality ([13], [6]) gives a lower bound Re (Op( S ) U, U ) ≥ − C kh D i − / U k which is, in general, too weak to study the Cauchy problem for general weaklyhyperbolic operator P , although applying this symmetrizer many interestingresults are obtained by several authors, see for example [11], [1], [14], [21]. Inthese works one of the main points is how one can derive a suitable lower boundof Op( S ) from the hyperbolicity condition assumed on p , that is(1.5) ∆ = 4 a ( t, x, ξ ) − b ( t, x, ξ ) ≥ , ( t, x, ξ ) ∈ [0 , T ) × U × R n . In this paper we employ a new idea which is to diagonalize S by an orthogonalmatrix T so that T − ST = D = diag ( λ , λ , λ ) where 0 ≤ λ ≤ λ ≤ λ arethe eigenvalues of S and reduce the equation to that of V = T − U ; roughly(1.6) D t V = A T h D i V + B T V, A T = T − AT D symmetrizes A T . For general nonnegative definite symmetric S it seemsthat we have nothing new but our S is a special one which is the B´ezoutiant ofhyperbolic polynomial p and ∂p/∂τ . Indeed, as we will see in Section 2, one has∆ a (cid:22) λ (cid:22) a , λ ≃ a, λ ≃ . Since (1.6) is a symmetrizable system with a symmetrizer D , a natural energywill be Re (cid:0) Op( D ) U, U (cid:1) = X j =1 (cid:0) Op( λ j ) U j , U j (cid:1) and it could be expected that scalar operators Op( λ j ) reflect the hyperbolicitycondition (1.5) quite directly. To apply this energy we need to estimate thederivatives of D and A T , essentially those of λ j , which is done also in Section2. We discuss how to derive energy estimates following this idea in Section 3.In Section 4 this procedure is carried out for hyperbolic operators with tripleeffectively hyperbolic characteristics with time dependent coefficients, and thatthe Cauchy problem for such operators is C ∞ well-posed for any lower orderterm is proved. In Section 5 the same question is discussed for third orderhyperbolic operators with two independent variables with analytic coefficients.In the last section, taking a homogeneous third order hyperbolic operator withtwo independent variables, we show that the same argument is applicable tothose with more general triple characteristics. Consider p ( τ, t, X ) = τ − a ( t, X ) τ − b ( t, X )where a ( t, X ) and b ( t, X ) are real valued and C ∞ in ( t, X ) ∈ ( − c, T ) × W withbounded derivatives of all order where W is an open set in R l such that ¯ X ∈ W and c > t, X ) = 4 a ( t, X ) − b ( t, X ) ≥ , ( t, X ) ∈ [0 , T ) × W, a (0 , ¯ X ) = 0that is, p ( τ, t, X ) = 0 has only real roots for ( t, X ) ∈ [0 , T ) × W and has a tripleroot τ = 0 at (0 , ¯ X ). Moreover assume that there is no triple root in t > a ( t, X ) > , ( t, X ) ∈ (0 , T ) × W. Denote(2.3) S ( t, X ) = − a a b − a b a , A ( t, X ) = a b then S is nonnegative definite and S ( t, X ) A ( t, X ) is symmetric. Let0 ≤ λ ( t, X ) ≤ λ ( t, X ) ≤ λ ( t, X )be the eigenvalues of S ( t, X ). 3 .1 Behavior of eigenvalues We show
Proposition 2.1.
There exist a neighborhood U of (0 , ¯ X ) and K > such that ∆ / (6 a + 2 a + 2 a ) ≤ λ ≤ (cid:0) / Ka (cid:1) a , (2.4) (2 − Ka ) a ≤ λ ≤ (2 + Ka ) a, (2.5) 3 ≤ λ ≤ Ka (2.6) for ( t, X ) ∈ U ∩ { t > } . Corollary 2.1.
There exists a neighborhood U of (0 , ¯ X ) such that λ i ( t, X ) ∈ C ∞ ( U ∩ { t > } ) , i = 1 , , . Proof.
Recalling a (0 , ¯ X ) = 0 from Proposition 2.1 one can choose U such that λ < λ < λ in U ∩ { t > } then the assertion follows immediately from the Implicit function theorem. Remark 2.1.
It may happen ∆( t, X ) = 0 for t > p ( τ, t, X ) = 0 havea double root τ at ( t, X ) while λ i ( t, X ) are smooth there.Proof of Proposition 2.1: Denote q ( λ ) = det ( λI − S );(2.7) q ( λ ) = λ − (3 + 2 a + a ) λ + (6 a + 2 a + 2 a − b ) λ − ∆ . Let µ ≤ µ be the roots of q λ = ∂q/∂λ = 0 and hence λ ≤ µ ≤ λ ≤ µ ≤ λ . It is easy to see µ = a (1 + O ( a )) and µ = 2 + O ( a ) which gives(2.8) λ ≤ a (1 + O ( a )) , λ ≥ O ( a ) . In the ( λ, η ) plane, the tangent line of the curve η = q ( λ ) at (0 , q (0)) intersectswith λ axis at (∆ /q λ (0) ,
0) and hence λ ( t, X ) ≥ ∆ /q λ (0) . Since q λ (0) ≤ a + 2 a + 2 a the left inequality of (2.4) is obvious. Compute q ( δa ) with δ >
0. Since 2 a − b = ∆ / b / ≥ q ( δa ) ≥ δ a − δ a (3 + 2 a + a ) + δa (6 a + 2 a ) − a + 27 b ≥ a n (6 δ −
4) + δ (2 − δ ) a − δ a + δ ( δ − a o . δ = 2 / Ka then noting a (0 , ¯ X ) = 0 one can choose a neighbor-hood U of (0 , ¯ X ) such that q ( δa ) ≥ a (cid:8) K − Kδa − δ a + δ ( δ − a (cid:9) > t, X ) ∈ U ∩ { t > } . This proves that λ ≤ δa and hence the rightinequality of (2.4). Turn to λ . Consider q ( δa ) with δ > q ( δa ) ≥ a n δ a − δ (3 + 2 a + a ) + δ (6 + 2 a ) − a o + 27 b ≥ a n δ (2 − δ ) + ( δ − δ + 2 δ − a − δ a o and choose δ = 2 − Ka which gives q ( δa ) ≥ a n K − (3 K + 2 K + Kδ − δ ) a o . Therefore for any
K > U such that q ( δa ) > U ∩ { t > } .Since one can assume δa < λ by (2.8) then δa ∈ ( λ , λ ) which proves the leftinequality of (2.5). Repeating similar arguments one gets q ( δa ) ≤ a n δ (2 − δ ) + (4 + 2 δ + δ − δ ) a + δ (2 − δ ) a o because 27 b ≤ a in t ≥
0. Taking δ = 2 + Ka one has q ( δa ) ≤ a n (8 − K ) + (2 K + δ K − δKa − K ) a o . Fixing any
K > / U such that q ( δa ) < U ∩ { t > } . Since λ < δa thanks to (2.4) one concludes (2 + Ka ) a ∈ ( λ , λ ) which shows theright inequality of (2.5). Finally we check (2.6). It is easy to see that q (3) = a ( − a ) < U ∩ { t > } if U is small so that 3 ≤ λ in U ∩ { t > } . Note that q ( δ ) ≥ δ n δ ( δ −
3) + (6 − δ ) a + (2 − δ ) a o − a where we take δ = 3 + Ka so that q (3 + Ka ) = a n K − − (6 K + 4) a + 3( K − k ) a o + Ka n (3 K − − Ka + ( K − K ) a o . Thus fixing any
K > / U such that q (3+ Ka ) > U ∩{ t > } .Since 3 + Ka > λ which proves the right inequality of (2.6).5 .2 Behavior of eigenvectors If we write n ij for the ( i, j )-cofactor of λ k I − S then t ( n j , n j , n j ) is, if non-trivial, an eigenvector corresponding to λ k . We take k = 1, j = 3 and hence a (2 a − λ )3 b ( λ − λ − λ − a ) = ℓ ℓ ℓ is an eigenvector corresponding to λ and therefore t = t t t = 1 d ℓ ℓ ℓ , d = q ℓ + ℓ + ℓ is a normalized eigenvector corresponding to λ . Thanks to Proposition 2.1 andrecalling b = O ( a / ) it is clear that there is C > a/C ≤ d ≤ C a, in U ∩ { t > } . Similarly choosing k = 2 , j = 2 and k = 3 , j = 1 − ab ( λ − λ − a ) − a b ( λ − = ℓ ℓ ℓ , ( λ − a )( λ − a ) − b − ab − a ( λ − a ) = ℓ ℓ ℓ are eigenvectors corresponding to λ and λ respectively and t j = t j t j t j = 1 d j ℓ j ℓ j ℓ j , d j = q ℓ j + ℓ j + ℓ j are normalized eigenvectors corresponding to λ j , j = 2 ,
3. Thanks to Proposi-tion 2.1 there is
C > a/C ≤ d ≤ C a, /C ≤ d ≤ C. Denote T = ( t , t , t ) = ( t ij ) then T is an orthogonal matrix, t T T = I , smoothin ( t, X ) ∈ U ∩ { t > } which diagonalizes S ; D = T − ST = t T ST = λ λ
00 0 λ . Note that D symmetrizes A T = T − AT ; t ( DA T ) = t ( t T SAT ) = t T t ( SA ) T = t T SAT = DA T . Remark 2.2.
Since DA T is symmetric it follows that if λ = 0 and 0 < λ ≤ λ then ˜ a = ˜ a = 0 so that A T is block triangular.6inally in view of (2.9), (2.10) and Proposition 2.1 it is easy to check that(2.11) T = (cid:0) t , t , t (cid:1) = O ( a ) O ( a / ) O (1) O ( √ a ) O (1) O ( a / ) O (1) O ( √ a ) O ( a ) near ( t, X ) = (0 , ¯ X ). First recall [21, Lemma 3.2]
Lemma 2.1.
Assume (2.1) . Then | ∂ αX a | (cid:22) √ a, | ∂ αX b | (cid:22) a, | ∂ t b | (cid:22) √ a for | α | = 1 and ( t, X ) ∈ (0 , T ) × W . We show
Lemma 2.2.
For | α | = 1 one has (2.12) | ∂ αX λ | (cid:22) a / , | ∂ αX λ | (cid:22) √ a, | ∂ αX λ | (cid:22) √ a. Proof.
Since ∂ αX q ( λ ) = − ∂ αX (2 a + a ) λ + ∂ αX (6 a + 2 a + 2 a − b ) λ − ∂ αX (4 a − b )it follows from Lemma 2.1 that (cid:12)(cid:12) ∂ αX q ( λ j ) (cid:12)(cid:12) (cid:22) | ∂ αX a | λ j + (cid:0) | ∂ αX a | + | b || ∂ αX b | (cid:1) λ j + (cid:0) a | ∂ αX a | + | b || ∂ αX b | (cid:1) (cid:22) | ∂ αX a | λ j + a / (cid:22) √ a λ j + a / . From the Implicit function theorem one has q λ ( λ j ) ∂ αX λ j + ∂ αX q ( λ j ) = 0 andhence | ∂ αX λ j | (cid:22) √ a λ j + a / | q λ ( λ j ) | . Noting q λ ( λ j ) = Q k = j ( λ j − λ k ) one has(2.13) q λ ( λ j ) ≃ a for j = 1 , , q λ ( λ ) ≃ ∂ t λ j . Lemma 2.3.
Assuming (2.1) one has (2.14) | ∂ t λ | (cid:22) a, | ∂ t λ | (cid:22) , | ∂ t λ | (cid:22) . Proof.
Repeating the same arguments in the proof of Lemma 2.2 one has | ∂ t q ( λ j ) | (cid:22) | ∂ t a | λ j + a | ∂ t a | + | b || ∂ t b | (cid:22) λ j + a which proves the assertion. 7 How to apply diagonal symmetrizers
To explain how to apply diagonal symmetrizers constructed in preceding sec-tions, study hyperbolic operators with one space variable, x ∈ R ; P u = ∂ t u − a ( t, x ) ∂ x ∂ t u − b ( t, x ) ∂ x u where we assume(3.1) ∆( t, x ) = 4 a ( t, x ) − b ( t, x ) ≥ t, x ) ∈ [0 , T ) × W and a (0 ,
0) = 0 such that p ( τ, , ,
1) = 0 has the triple root τ = 0 where W is an open interval containing the origin. In what follows we work in a regionwhere a ( t, x ) >
0. With U = ( ∂ t u, ∂ x ∂ t u, ∂ x u ) the equation P u = f is reducedto(3.2) ∂ t U = A ( t, x ) ∂ x U + F, A = a b , F = f . Then S given by (2.3) symmetrizes A , and T given by (2.11) diagonalizes S . Sowe set V = T − U and rewrite the equation (3.2) to(3.3) ∂ t V = A T ∂ x V + (cid:0) ( ∂ t T − ) T − A T ( ∂ x T − ) T (cid:1) V + T − F where A T = T − AT . To simplify notation let us write (3.3) with f = 0 as ∂ t V = A ∂ x V + B V with A = A T and B = ( ∂ t T − ) T − A ( ∂ x T − ) T . Consider an energy with a scalar weight φ ( t, x ) > ∂ t φ = 1 and | ∂ x φ | (cid:22) φ − N DV, V ) = Z φ − N h DV, V i dx where h V, W i stands for the inner product in C and N > V ( t, x ) has small support in x . Notethat ddt ( φ − N DV, V ) = − N (cid:0) φ − N − DV, V (cid:1) + (cid:0) φ − N ( ∂ t D ) V, V (cid:1) +2 Re (cid:0) φ − N D ( A ∂ x V + B V ) , V (cid:1) . Since D A is symmetric and hence2 Re ( φ − N D A ∂ x V, V ) = N (cid:0) φ − N − ( ∂ x φ ) D A V, V (cid:1) − (cid:0) φ − N ∂ x ( D A ) V, V ) . N φ − N − h DV, V i = N X j =1 φ − N − λ j | V j | . As for a scalar weight φ we assume(3.5) φ a (cid:22) ∆ , φ (cid:12)(cid:12) ∂ t ∆ (cid:12)(cid:12) (cid:22) ∆ , φ (cid:12)(cid:12) ∂ t a (cid:12)(cid:12) (cid:22) a. Lemma 3.1.
The assumption (3.5) implies (3.6) φ (cid:22) λ , φ (cid:12)(cid:12) ∂ t λ (cid:12)(cid:12) (cid:22) λ , φ (cid:12)(cid:12) ∂ t λ (cid:12)(cid:12) (cid:22) λ . Proof.
In view of Proposition 2.1 the assertion φ (cid:22) λ is clear. Note that from | ∂ t q ( λ j ) | (cid:22) ( | ∂ t a | + | b || ∂ t b | ) λ j + | ∂ t ∆ | and Lemma 2.1 it follows that (cid:12)(cid:12) ∂ t λ i (cid:12)(cid:12) (cid:22) (cid:0) | ∂ t a | + a (cid:1) λ i + | ∂ t ∆ || q λ ( λ i ) | . Taking (2.13) into account one has | ∂ t λ i | (cid:22) | ∂ t a | a λ i + a λ i + | ∂ t ∆ | a , i = 1 , φ | ∂ t λ | (cid:22) λ thanks to (3.5) and (2.4). As for λ noting that | ∂ t ∆ | (cid:22) a by Lemma 2.1 the assertion follows immediately from (3.5) and(2.5). h ( ∂ t D ) V, V i , h ( ∂ x φ ) D A V, V i Thanks to (3.6), (cid:12)(cid:12) φ − N h ( ∂ t D ) V, V i (cid:12)(cid:12) is bounded (3.4) taking N large. On theother hand, from Lemma 3.2 below it follows that D A = O ( λ √ a ) O ( λ ) O ( λ √ a ) O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) O ( a / ) . Recalling that D A is symmetric it is clear that (cid:12)(cid:12) h D A V, V i (cid:12)(cid:12) is bounded by √ a ( λ | V | + a | V | + a | V | ) + λ | V | | V | + λ √ a | V | | V | + a | V || V | . Since λ (cid:22) √ λ a and hence λ | V | | V | (cid:22) √ a ( λ | V | + a | V | ) it follows that(3.7) (cid:12)(cid:12) N φ − N − ( ∂ x φ ) h D A V, V i (cid:12)(cid:12) ≤ CN √ a | ∂ x φ | X k =1 φ − N − λ j | V j | with some C >
0. In a small neighborhood of (0 ,
0) where a is enough smallone can bound the right-hand side by (3.4).9 .3 Estimate of energy, term h D B V, V i Recall h D B V, V i = h ( ∂ t T − ) T V, DV i − h ( ∂ x T − ) T V, D A V i = h D ( ∂ t T − ) T V, V i − h D A ( ∂ x T − ) T V, V i because D A is symmetric. Applying Lemmas 2.2 and 2.3 we estimate ( ∂ t T − ) T and ( ∂ x T − ) T . First note that( ∂ t T − ) T = ( ∂ t ( t T )) T = ( h ∂ t t i , t j i )and h ∂ t t i , t j i = −h t i , ∂ t t j i = −h ∂ t t j , t i i so that ( ∂ t T − ) T is antisymmetric.Note that(3.8) h ∂ t t i , t j i = 1 d i d j X k =1 ∂ t ℓ ki · ¯ ℓ kj = 1 d i X k =1 ∂ t ℓ ki · ¯ t kj because P k =1 ℓ ki ¯ ℓ kj = 0 if i = j . Thanks to Proposition 2.1 and Lemmas 2.3,2.1 it follows that | ℓ | (cid:22) a , | ℓ | (cid:22) a / , | ℓ | (cid:22) a, | ℓ | (cid:22) a / , | ℓ | (cid:22) a, | ℓ | (cid:22) a / , | ℓ | (cid:22) , | ℓ | (cid:22) a / , | ℓ | (cid:22) a (3.9)and that | ∂ t ℓ | (cid:22) a, | ∂ t ℓ | (cid:22) √ a, | ∂ t ℓ | (cid:22) , | ∂ t ℓ | (cid:22) a / , | ∂ t ℓ | (cid:22) , | ∂ t ℓ | (cid:22) √ a, | ∂ t ℓ | (cid:22) , | ∂ t ℓ | (cid:22) a / , | ∂ t ℓ | (cid:22) . (3.10)Therefore taking (2.11), (3.8) and (3.10) into account one obtains(3.11) ( ∂ t T − ) T = O (1 / √ a ) O (1) O (1 / √ a ) 0 O ( √ a ) O (1) O ( √ a ) 0 . In order to estimate (cid:12)(cid:12) φ − N h D ( ∂ t T − ) T V, V i (cid:12)(cid:12) , noting D ≃ diag( λ , a,
1) and λ (cid:22) a , it suffices to estimate √ a | V || V | + | V || V | + √ a | V || V | . Note that √ a | V || V | (cid:22) φ − λ | V | + a φλ | V | (cid:22) φ − ( λ | V | + a | V | )10ecause φ/λ (cid:22) /φ by (3.6). As for | V || V | one has | V || V | (cid:22) φ − λ | V | + ( φ/λ ) | V | (cid:22) φ − ( λ | V | + | V | ) . Finally since √ a | V || V | (cid:22) a | V | + | V | one concludes that (cid:12)(cid:12) φ − N h D ( ∂ t T − ) T V, V i (cid:12)(cid:12) is bounded by (3.4) taking N large.Turn to (cid:12)(cid:12) φ − N h D A ( ∂ x T − ) T V, V i (cid:12)(cid:12) . From Proposition 2.1 and Lemmas 2.2,2.1 one has | ∂ x ℓ | (cid:22) a / , | ∂ x ℓ | (cid:22) a, | ∂ x ℓ | (cid:22) √ a, | ∂ x ℓ | (cid:22) a , | ∂ x ℓ | (cid:22) √ a, | ∂ x ℓ | (cid:22) a, | ∂ x ℓ | (cid:22) , | ∂ x ℓ | (cid:22) a , | ∂ x ℓ | (cid:22) √ a (3.12)from which one concludes(3.13) ∂ x T = O ( √ a ) O ( a ) O (1) O (1) O (1 / √ a ) O ( a ) O (1 / √ a ) O (1) O ( √ a ) and hence(3.14) ( ∂ x T − ) T = O (1) O ( √ a ) O (1) 0 O ( a ) O ( √ a ) O ( a ) 0 . Here note that
Lemma 3.2.
One has A = O ( √ a ) O (1) O ( √ a ) O ( a ) O ( √ a ) O (1) O ( a / ) O ( a ) O ( a / ) , ∂ x A = O (1) O (1 / √ a ) O (1) O ( √ a ) O (1) O (1 / √ a ) O ( a ) O ( √ a ) O ( √ a ) . Proof.
It is clear that(3.15) ˜ a ij = t i a t j + t i b t j + t i t j + t i t j . Since t i t j = O ( √ a ) unless ( i, j ) = (2 ,
3) and t i t j = O ( √ a ) unless ( i, j ) =(1 ,
2) the first assertion follows from (2.11) and (3.15). Note that ∂ x (cid:0) /d j (cid:1) = O (1 /a / ) for j = 1 , ∂ x (cid:0) /d (cid:1) = O (1) then the second assertion followsfrom (3.13), (3.15) and (2.11).From Lemma 3.2 and (3.14) one has A ( ∂ x T − ) T = O (1) O ( √ a ) O ( a ) O ( √ a ) O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) . Then it is clear that (cid:12)(cid:12) h D A ( ∂ x T − ) T V, V i (cid:12)(cid:12) is bounded by a / | V || V | + a | V || V | + a / | V || V | + λ | V | + a ( | V | + | V | )11here a / | V || V | (cid:22) a φ − λ | V | + a φλ | V | (cid:22) a φ − ( λ | V | + a | V | ) ,a | V || V | (cid:22) a φ − λ | V | + a φλ | V | (cid:22) a φ − ( λ | V | + | V | ) ,a / | V || V | (cid:22) a ( a | V | + | V | )since φ/λ (cid:22) /φ . Therefore (cid:12)(cid:12) φ − N h D A ( ∂ x T − ) T V, V i (cid:12)(cid:12) is bounded by (3.4)taking N large. h ∂ x ( D A ) V, V i Write h ∂ x ( D A ) V, V i = h ( ∂ x D ) A V, V i + h D ( ∂ x A ) V, V i and estimate each term on the right-hand side. To estimate the first term note Lemma 3.3.
One has (3.16) (cid:12)(cid:12) ∂ x λ (cid:12)(cid:12) /λ (cid:22) / √ a + 1 / √ ∆ (cid:22) / ( φ √ a ) . Proof.
Recall ∂ x λ = (cid:8) ∂ x (2 a + a ) λ − ∂ x (6 a + 2 a + 2 a − b ) λ + ∂ x ∆ (cid:9) /q λ ( λ )where Lemma 2.1 shows that | ∂ x λ | λ (cid:22) √ a λ + √ a λ λ a + | ∂ x ∆ | λ a (cid:22) √ a + | ∂ x ∆ | λ a (cid:22) √ a + | ∂ x ∆ | ∆because 1 /λ (cid:22) a/ ∆ by Proposition 2.1. Noting that ∆ ≥ x = 0 we see that | ∂ x ∆ | (cid:22) √ ∆, hence the first inequality. The secondinequality follows from the first one thanks to the assumption (3.5).Since | ∂ x λ | /λ = O (1 / √ a ) and | ∂ x λ | /λ = O (1) by Lemma 2.2 it followsfrom 3.3 that φ ( ∂ x D ) = φD ( D − ∂ x D ) = diag (cid:0) O ( λ / √ a ) , O ( φ √ a ) , O ( φ ) (cid:1) for λ ≃ a then by Lemma 3.2 and (3.13) one sees φ ( ∂ x D ) A = O ( λ ) O ( λ / √ a ) O ( λ ) O ( φ a / ) O ( φ a ) O ( φ √ a ) O ( φ a / ) O ( φ a ) O ( φ a / ) . Therefore to estimate φ h ( ∂ x D ) A V, V i it suffices to bound (cid:16) λ √ a + a / φ (cid:17) | V || V | + ( λ + a / φ ) | V || V | + φ √ a | V || V | . Since λ (cid:22) a √ λ and φ (cid:22) √ λ this is bounded by h DV, V i and hence(3.17) (cid:12)(cid:12) φ − N h ( D − ∂ x D ) A V, DV i (cid:12)(cid:12) (cid:22) φ − N − h DV, V i .
12s for (cid:12)(cid:12) φ − N h ( ∂ x A ) V, DV i (cid:12)(cid:12) it follows from Lemma 3.2 that D ( ∂ x A ) = O ( λ ) O ( λ / √ a ) O ( λ ) O ( a / ) O ( a ) O ( √ a ) O ( a ) O ( √ a ) O ( √ a ) hence repeating similar arguments it is easy to see that(3.18) (cid:12)(cid:12) φ − N h D ( ∂ x A ) V, V i (cid:12)(cid:12) (cid:22) φ − N − h DV, V i . Thus one can bound (cid:12)(cid:12) φ − N h ∂ x ( D A ) V, V i (cid:12)(cid:12) by (3.4) taking N large. Remark 3.1.
It shoud be remarked that the condition (3.5) assumed in thissection are stated in terms of ∆ and a , where a is constant times the discriminantof ∂p/∂τ , without any reference to characteristic roots τ i .We conclude this section with an important remark. To obtain energy esti-mates it suffices to find a finite number of pairs ( φ j , ω j ), where φ j is a scalarweight satisfying (3.5) in subregion ω j of which union covers a neighborhood of(0 , collect such obtained estimates in ω j . In the followingsections we carry out this observation for hyperbolic operators with triple effec-tively hyperbolic characteristics with coefficients depending on t , also for thirdorder hyperbolic operators with effectively hyperbolic characteristics with twoindependent variables, and prove the Ivrii’s conjecture for such operators. If the Cauchy problem (1.1) is C ∞ well posed for any lower order term then thecharacteristic roots are at most triple and the Hamilton map H p has non-zeroreal eigenvalues at every critical point ([7, Theorem 3]). Here the Hamilton map F p is defined by F p ( X, Ξ) = ∂ p∂X∂ Ξ ∂ p∂ Ξ ∂ Ξ − ∂ p∂X∂X − ∂ p∂ Ξ ∂X , X = ( t, x ) , Ξ = ( τ, ξ ) . In this paper we call ( X, Ξ) a critical point if p = 0 and ∂p/∂X = ∂p/∂ Ξ =0 at ( X, Ξ). A critical point where the Hamilton map H p has non-zero realeigenvalues is called effectively hyperbolic characteristic ([5]). In [8], V.Ivrii hasproved that if the characteristic roots are at most triple and every critical pointis effectively hyperbolic then the Cauchy problem is C ∞ well-posed for everylower order term under the additional condition such that in a neighborhood ofevery critical point, p admits a decomposition p = q q with real smooth symbol13 i vanishing at the point. In this case the Hamilton map has non-zero realeigenvalues if and only if the Poisson bracket { q , q } does not vanish. He hasconjectured that the assertion would hold without any additional condition. Formore details about the conjecture and subsequent progress on several questionsincluding the above conjecture, see [2], [3], [18], [21].Note that for any triple characteristic root τ at ( t, x, ξ ) with t ≥
0, ( t, x, τ, ξ )is a critical point and for any double characteristic root τ at ( t, x, ξ ) with t > t, x, τ, ξ ) is also a critical point (see [7, Lemma 8.1]) while for double character-istic roots τ at (0 , x, ξ ), the points (0 , x, τ, ξ ) are not necessarily critical points.Here is a simple example in R , x ∈ R and t ≥ P = ( D t − t ℓ D x )( D t + c D x ) , ℓ ∈ N where c ∈ R . Let c = 0 then it is clear that τ = 0 is a double characteristic rootat (0 , , ℓ = 1 then ∂ t p (0 , , ,
1) = − c = 0 and hence (0 , , ,
1) is not acritical point. If ℓ ≥ , , ,
1) is a critical point and F p has non-zeroreal eigenvalues there if ℓ = 2 while F p = O if ℓ ≥
3. Let c = 0 then τ = 0 isa triple characteristic root at (0 , ,
1) and (0 , , ,
1) is a critical point. If ℓ = 1then F p has non-zero real eigenvalues there while F p (0 , , ,
1) = O if ℓ ≥ t and consider the Cauchy problem(4.1) ( D mt u + P m − j =0 P j + | α |≤ m a j,α ( t ) D αx D jt u = 0 , ( t, x ) ∈ [0 , T ) × R n D kt u (0 , x ) = u k ( x ) , x ∈ R n , k = 0 , . . . , m − a j,α ( t ) are real valued and C ∞ in ( − c, T ) with some c > p is hyperbolic for t ≥
0, that is(4.2) p ( t, τ, ξ ) = τ m + m − X j =0 X j + | α | = m a j,α ( t ) ξ α τ j has only real roots in τ for any ( t, ξ ) ∈ [0 , T ) × R n . Theorem 4.1.
Assume that the characteristic roots p (0 , τ, ξ ) = 0 are at mosttriple for any ξ = 0 and every critical point (0 , τ, ξ ) is effectively hyperbolic.Then there exists δ > such that for any a j,α ( t ) with j + | α | ≤ m − theCauchy problem (4.1) with T = δ is C ∞ well-posed. Remark 4.1.
Assume that p has a triple characteristic root ¯ τ at (0 , ¯ ξ ), | ¯ ξ | = 1and F p (0 , ¯ τ , ¯ ξ ) = O then p has necessarily non-real characteristic roots in the t < , ¯ ξ ), that is P would be a Tricomi type operator. Indeed fromLemma [7, Lemma 8.1] it follows that F p (0 , ¯ τ , ¯ ξ ) = O if all characteristic rootsare real in a full neighborhood of (0 , ¯ ξ ). Assume that p ( t, τ, ξ ) has a triple characteristic root ¯ τ at (0 , ¯ ξ ), | ¯ ξ | = 1 and(0 , ¯ τ , ¯ ξ ) is effectively hyperbolic. As we see later, without restrictions one may14ssume that m = 3 and p has the form(4.3) p ( t, τ, ξ ) = τ − a ( t, ξ ) | ξ | τ − b ( t, ξ ) | ξ | where a ( t, ξ ) and b ( t, ξ ) are homogeneous of degree 0 in ξ and satisfy(4.4) ∆( t, ξ ) = 4 a ( t, ξ ) − b ( t, ξ ) ≥ , ( t, ξ ) ∈ [0 , T ) × R n . The triple characteristic root of p (0 , τ, ¯ ξ ) = 0 is τ = 0 anddet (cid:0) λ − F p (0 , , ¯ ξ ) (cid:1) = λ n (cid:0) λ − { ∂ t a (0 , ¯ ξ ) } (cid:1) hence (0 , , ¯ ξ ) is effectively hyperbolic if and only if(4.5) ∂ t a (0 , ¯ ξ ) = 0 . Since a (0 , ¯ ξ ) = 0 and ∂ t a (0 , ¯ ξ ) = 0 there is a neighborhood U of (0 , ¯ ξ ) in whichone can write(4.6) a ( t, ξ ) = e ( t, ξ )( t + α ( ξ ))where e > U . Note that α ( ξ ) ≥ ξ because a ( t, ξ ) ≥ , T ) × R n . Lemma 4.1.
There exists a neighborhood U of (0 , ¯ ξ ) in which one can write ∆( t, ξ ) = e ( t, ξ ) (cid:8) t + a ( ξ ) t + a ( ξ ) t + a ( ξ ) (cid:9) where e > and a j ( ¯ ξ ) = 0 .Proof. Thanks to the Malgrange preparation theorem it suffices to show ∂ kt ∆(0 , ¯ ξ ) = 0 , k = 0 , , , ∂ t ∆(0 , ¯ ξ ) = 0 . It is clear that ∂ kt a = 0 at (0 , ¯ ξ ) for k = 0 , , ∂ t a (0 , ¯ ξ ) = 0. Since∆ = 4 a − b and b (0 , ¯ ξ ) = 0 it is enough to show ∂ t b (0 , ¯ ξ ) = 0. Suppose ∂ t b (0 , ¯ ξ ) = 0 and hence b ( t, ¯ ξ ) = t (cid:0) b + tb ( t ) (cid:1) where b = 0. Since a ( t, ¯ ξ ) = c t with c > t, ¯ ξ ) = 4 c t − b ( t, ¯ ξ ) ≥ Lemma 4.2.
There exist a neighborhood U of ¯ ξ and a positive constant ε > such that for any ξ ∈ U one can find j ∈ { , , } such that ε − | ν j ( ξ ) | ≥ α ( ξ ) where ν j ( ξ ) are roots of t + a ( ξ ) t + a ( ξ ) t + a ( ξ ) = 0 . roof. In view of Lemma 4.1 one can write∆ = 4 a − b = 4 e ( t + α ) − b = 4 e (cid:8) ( t + α ) − ˆ b (cid:9) = e (cid:8) t + a ( ξ ) t + a ( ξ ) t + a ( ξ ) (cid:9) where ˆ b = 3 √ b/ (2 e / ) and hence( t + α ) − ˆ b = E (cid:8) t + a ( ξ ) t + a ( ξ ) t + a ( ξ ) (cid:9) with E = e / (4 e ). Choose I = [ − δ , δ ], δ > U of ¯ ξ such that I × U ⊂ U and denotesup ( t,ξ ) ∈ I × U , ≤ k ≤ (cid:12)(cid:12) ∂ kt E ( t, ξ ) (cid:12)(cid:12) = C. Write ˆ b ( t, ξ ) = ˆ b ( ξ ) + ˆ b ( ξ ) t + ˆ b ( ξ ) t + ˆ b ( t, ξ ) t and denote(4.7) sup ξ ∈ U α ( ξ ) + sup ξ ∈ U , ≤ k ≤ | ˆ b k ( ξ ) | + sup ( t,ξ ) ∈ I × U | ˆ b ( t, ξ ) | = B. Choose a neighborhood U ⊂ U of ¯ ξ such that4 B p α ( ξ ) < ξ ∈ U which is possible because α ( ¯ ξ ) = 0. Choose ε = ε ( B, C ) > (cid:0) − εC (1 + εB + ε B / (cid:1) − (cid:0) ε C (1 + εB ) (cid:1) / (1 − ε C ) > ξ ∈ U such that(4.9) (cid:12)(cid:12) ν j ( ξ ) (cid:12)(cid:12) < ǫ α ( ξ ) , j = 1 , , ξ for simplicity. Recall(4.10) ( t + α ) − ˆ b = E Y ( t − ν j ) ≥ t ≥ t = 0 one has α − ˆ b = E (0) | ν ν ν | < Cε α . This shows(4.11) p − Cε α / ≤ | ˆ b | ≤ α / . Differentiating (4.10) by t and putting t = 0 one has | α − b ˆ b | ≤ Cε α + 3 Cε α . This gives 3 α (cid:0) − Cε (1 + εB ) (cid:1) ≤ | ˆ b ˆ b | ≤ α (cid:0) Cε (1 + εB ) (cid:1) .
16n view of (4.11) one has(4.12) 32 α / (cid:0) − Cε (1 + εB ) (cid:1) ≤ | ˆ b | ≤ α / √ − Cε (cid:0) Cε (1 + εB ) (cid:1) . Differentiating (4.10) twice by t and putting t = 0 on has (cid:12)(cid:12) α − (2ˆ b + 4ˆ b ˆ b ) (cid:12)(cid:12) ≤ C ε α (6 + 6 ε B + ε B )which proves (cid:12)(cid:12) b ˆ b + 2ˆ b (cid:12)(cid:12) ≥ α (cid:0) − εC − ε CB − ε CB / (cid:1) . Using (4.12) oneobtains | b ˆ b | ≥ α (cid:0) − εC (1 + εB + ε B / (cid:1) − α (cid:0) ε C (1 + εB ) (cid:1) / (1 − ε C )where the right-hand side is greater than α by (4.8). On the other hand from(4.7) and (4.11) we have4 B α / ≥ α / | ˆ b | ≥ | ˆ b ˆ b | > α and hence 4 B √ α > /e by ¯∆;¯∆( t, ξ ) = ∆ /e = t + a ( ξ ) t + a ( ξ ) t + a ( ξ ) . Lemma 4.3.
There is a neighborhood V ⊂ U of ¯ ξ where one can write either (4.13) ¯∆ = (cid:12)(cid:12) t − ν ( ξ ) (cid:12)(cid:12) (cid:0) t − ν ( ξ ) (cid:1) , ν ( ξ ) is real and ν ( ξ ) ≤ or (4.14) ¯∆ = Y k =1 ( t − ν k ( ξ ) (cid:1) , ν k ( ξ ) are real and ν k ( ξ ) ≤ . Proof.
Let ν j ( ξ ), j = 1 , , t, ξ ) = 0. Since ν j ( ¯ ξ ) = 0 onecan assume | ν j ( ξ ) | < δ in V . Since a j ( ξ ) are real we have two cases; one isreal and other two are complex conjugate or all three are real. For the formercase denoting the real root by ν ( ξ ) we have (4.13) where ν ( ξ ) ≤ ≥ ≤ t ≤ δ . In the latter case, if two of them coincide, denoting theremaining one by ν ( ξ ) one has (4.13). If ν j ( ξ ) are different each other then wehave (4.14) since ¯∆ ≥ ≤ t ≤ δ .17 .2 Key proposition Thanks to Lemma 4.3 we have either (4.13) or (4.14). As was observed in [16](see also [19]) in order to obtain energy estimates it is important to considerthe curves t = Re ν j ( ξ ), the real part of ν j ( ξ ). Define ψ ( ξ ) = max (cid:8) , Re ν j ( ξ ) (cid:9) = max (cid:8) , Re ν ( ξ ) (cid:9) ,φ = t, ω ( ξ ) = [0 , ψ ( ξ ) / , φ = t − ψ ( ξ ) , ω ( ξ ) = [ ψ ( ξ ) / , δ ]with small δ >
0. If ψ ( ξ ) ≤
0, we have φ = φ = t and ω = [0 , δ ]. The nextproposition is the key to applying the arguments in Section 3 to operators withtriple effectively hyperbolic characteristics. Proposition 4.1.
There exist a neighborhood U of ¯ ξ , positive constants δ > and C > such that (4.15) φ j a ≤ C ∆ , | φ j | | ∂ t ∆ | ≤ C ∆ , | φ j | ≤ C a for any ξ ∈ U and t ∈ ω j ( ξ ) , j = 1 , .Proof. Thanks to (4.6) and Lemma 4.1 It suffices to prove (4.15) for ¯∆ and t + α instead of ∆ and a . Note that (cid:12)(cid:12)(cid:12) ∂ t ¯∆¯∆ (cid:12)(cid:12)(cid:12) ≤ t + | ν ( ξ ) | + 2 | t − Re ν ( ξ ) || t − ν ( ξ ) | ≤ φ + 2 | φ | , (cid:12)(cid:12)(cid:12) ∂ t ¯∆¯∆ (cid:12)(cid:12)(cid:12) ≤ X k =1 t + | ν k ( ξ ) | ≤ φ in the case (4.13) and (4.14) respectively. Since | φ | ≥ t = φ in ω ( ξ ) and t = φ ≥ | φ | in ω ( ξ ) it is easy to see | φ j | | ∂ t ¯∆ | ≤ ω j ( ξ )for both cases. Similarly noting t + α ( ξ ) ≥ t it is clear that | φ j | ≤ t + α in ω j ( ξ ) . Therefore it rests to prove φ j ( t + α ) ≤ C ¯∆ in ω j ( ξ ). First we study the case(4.13). From Lemma 4.2 either ε − | ν | ≥ α or ε − | ν | ≥ α holds. First assumethat ε − | ν | ≥ α and hence t + | ν | ≥ ε ( t + α ) then¯∆ | φ j | ( t + α ) ≥ ε | t − ν | | φ j | ≥ ε | t − Re ν | | φ j | = ε φ | φ j | ≥ ε | φ j | , t ∈ ω j . Next assume ε − | ν | ≥ α . If 0 < Re ν ≤ | Im ν | one has for t ≥ | t − Re ν | + | Im ν | ≥ t − Re ν + | Im ν | ≥ t, | t − Re ν | + | Im ν | ≥ | Im ν | ≥ | ν | / ≥ ε α Re ν ≤ | t − ν | ≥ (cid:0) | t − Re ν | + | Im ν | (cid:1) ≥ ε ε ) ( t + α ) . Therefore it follows that¯∆ | φ j | ( t + α ) ≥ ε ε ) t | t − ν || φ j | ≥ ε ε ) | φ φ || φ j | ≥ ε ε ) | φ j | , t ∈ ω j . If Re ν > | Im ν | noting that, for t ∈ ω | t − Re ν | ≥ Re ν / ≥ | ν | / ≥ εα/ , | t − Re ν | ≥ t one has (4 + ε ) | t − Re ν | ≥ ε ( t + α ) in ω . Hence¯∆ t ( t + α ) ≥ | t − Re ν | t + α ≥ ε ε | t − Re ν | ≥ ε ε t, t ∈ ω . For t ∈ ω note that t ≥ Re ν / ≥ | ν | / ≥ εα/ ε ) t ≥ ε ( t + α ). Thus one has¯∆ | t − Re ν | ( t + α ) ≥ t | t − Re ν | ( t + α ) ≥ ε ε | t − Re ν | , t ∈ ω which proves the assertion for the case (4.13).In the case (4.14) note that¯∆ = Y k =1 (cid:0) t + | ν k | (cid:1) . If ε − | ν j | ≥ α then t + | ν j | ≥ ε (1 + α ) and hence it is clear that¯∆ ≥ ε t ( t + α )which shows the assertion. Thus the proof of Proposition 4.1 is completed. Let P be a differential operator of order 3 with coefficients depending on t .After Fourier transform in x the equation P u = f reduces to(4.16) D t ˆ u + X j + | α |≤ ,j ≤ a j,α ( t ) ξ α D jt ˆ u = ˆ f where ˆ u ( t, ξ ) stands for the Fourier transform of u ( t, x ) with respect to x . With E ( t, ξ ) = exp (cid:16) i Z t X | α | =1 a ,α ( s ) ξ α ds (cid:17)
19t is clear that ˆ v = E ( t, ξ )ˆ u satisfies(4.17) D t ˆ v − a ( t, ξ ) | ξ | D t ˆ v − b ( t, ξ ) | ξ | ˆ v + X j =1 b j ( t, ξ ) | ξ | j − D − jt ˆ v = E ˆ f where b j ( t, ξ ) = 0 for | ξ | ≤ | ξ | ≤ ℓ X k =0 (cid:12)(cid:12) ( | ξ | + 1) ℓ − k ∂ kt ˆ u ( t ) (cid:12)(cid:12) ≤ C ℓ ℓ X k =0 (cid:12)(cid:12) ( | ξ | + 1) ℓ − k ∂ kt ˆ v ( t ) (cid:12)(cid:12) in order to obtain energy estimates for ˆ u one can assume that ˆ u satisfies (4.17)from the beginning. With U = t (cid:0) D t ˆ u, | ξ | D t ˆ u, | ξ | ˆ u (cid:1) the equation (4.17) can bewritten ∂∂t U = i a ( t, ξ ) b ( t, ξ )1 0 00 1 0 | ξ | U + i b ( t, ξ ) b ( t, ξ ) b ( t, ξ )0 0 00 0 0 U + iE ˆ f = iA | ξ | U + BU + F. (4.18)Let S ( t, ξ ) and T ( t, ξ ) be defined in Section 2 with X = ξ such that T − ST = D = diag ( λ , λ , λ ). With V = T − U one has ∂ t V = iA T | ξ | V + (cid:0) B T + ( ∂ t T − ) T (cid:1) V + T − F = i A| ξ | V + B V + ˜ F where A = T − AT and B = T − BT + ( ∂ t T − ) T . Thanks to Proposition 4.1we have candidates for scalar weights in each ω j . To simplify notation denote t ( ξ ) = 0 , t ( ξ ) = ψ ( ξ ) / , t ( ξ ) = ψ ( ξ ) , t ( ξ ) = δ and following [16] (also [19]) introduce three subintervals Ω j = [ t j − ( ξ ) , t j ( ξ )]and scalar weights ϕ j , j = 1 , , ϕ ( t, ξ ) = t, ϕ ( t, ξ ) = ψ ( ξ ) − t, ϕ ( ξ ) = t − ψ ( ξ ) . Note that ω = Ω , ω = Ω ∪ Ω and ϕ j = | φ | in Ω j , j = 2 ,
3. Thanks toProposition 4.1 and Lemma 3.1 one has(4.19) ϕ j ≤ Cλ , ϕ j | ∂ t λ | ≤ Cλ , ϕ j ≤ Cλ , t ∈ Ω j ( ξ ) , j = 1 , , C is independent of ξ ∈ U . Consider the following energy in Ω j ( ξ ); E j = g j h DV, V i = g j X k =1 λ k ( t, ξ ) | V k ( t, ξ ) | , g j ( t, ξ ) = ϕ − j N − j . Re h iD A| ξ | V, V i = 0 and ∂ t ϕ j = ( − j − one has ddt E j = − (2 N − ( − j ) ϕ − j E j + g j h ( ∂ t D ) V, V i + 2 g j Re h D B V, V i + 2 g j Re h D ˜ F , V i . Repeating the same arguments as in Section 3 one can obtain
Lemma 4.4.
Let j = 1 or . There exist N and C > such that for any N ≥ N and any U ( t, ξ ) verifying ∂ kt U ( t j − ( ξ ) , ξ ) = 0 , k = 0 , , . . . , N one has k U ( t ) k + N Z tt j − ϕ − Nj ( s ) k U ( s ) k ds ≤ C Z tt j − ϕ − Nj ( s ) k F ( s ) k ds for t ∈ Ω j ( ξ ) . With h ξ i = | ξ | + 1 it follows from Lemma 4.4 that Corollary 4.1.
Let j = 1 or . There is N such that for any L ∈ N thereexists C L > such that for any U with ∂ kt U ( t j − ( ξ ) , ξ ) = 0 , k = 0 , , . . . , N + L one has L X k =0 (cid:13)(cid:13) h ξ i L − k ∂ kt U ( t ) (cid:13)(cid:13) + N L X k =0 Z tt j − ϕ − Nj ( s ) (cid:13)(cid:13) h ξ i L − k ∂ kt U ( s ) (cid:13)(cid:13) ds ≤ C L L X k =0 Z tt j − ϕ − Nj ( s ) (cid:13)(cid:13) h ξ i L − k ∂ kt F ( s ) (cid:13)(cid:13) ds (4.20) for t ∈ Ω j ( ξ ) and N ≥ N . For the subinterval Ω ( ξ ) the argument in Section 3 shows again Lemma 4.5.
There exist N ∈ N and C > such that one has ϕ N − ( t ) k U ( t ) k + N Z tt ϕ N ( s ) k U ( s ) k ds ≤ C k U ( t ) k + C Z tt ϕ N ( s ) k F ( s ) k ds. (4.21) for t ∈ Ω ( ξ ) and N ≥ N . Corollary 4.2.
There exists N such that for any L ∈ N there is C L such that ϕ N − ( t ) L X k =0 (cid:13)(cid:13) h ξ i L − k ∂ kt U ( t ) (cid:13)(cid:13) + N L X k =0 Z tt ϕ N ( s ) (cid:13)(cid:13) h ξ i L − k ∂ kt U ( s ) (cid:13)(cid:13) ds ≤ C L L X k =0 (cid:13)(cid:13) h ξ i L − k ∂ kt U ( t ) (cid:13)(cid:13) + C L L X k =0 Z tt ϕ N ( s ) (cid:13)(cid:13) h ξ i L − k ∂ kt F ( s ) (cid:13)(cid:13) ds (4.22) for t ∈ Ω and N ≥ N . j is obtained, repeating the samearguments as in [16], [19, Section 6] one can collect the energy estimates in Ω j yielding energy estimates of U ( t, ξ ) in the whole interval [0 , δ ]. Proposition 4.2.
Assume that p has a triple characteristic root ¯ τ at (0 , ¯ ξ ) , | ¯ ξ | = 1 and (0 , ¯ τ , ¯ ξ ) is effectively hyperbolic. Then there exist δ > and a conicneighborhood U of ¯ ξ such that for any a j,α ( t ) with j + | α | ≤ one can find N ∈ N such that for any q ∈ N with q ≥ N there is C > such that (4.23) q +3 X k =0 (cid:12)(cid:12) h ξ i q +2 − k ∂ kt ˆ u ( t, ξ ) (cid:12)(cid:12) ≤ C q X k =0 Z t (cid:12)(cid:12) h ξ i N + q − k ∂ kt ˆ f ( s, ξ ) (cid:12)(cid:12) ds for ( t, ξ ) ∈ [0 , δ ] × U and for any ˆ u ( t, ξ ) with ∂ kt ˆ u (0 , ξ ) = 0 , k = 0 , , and ˆ f ( t, ξ ) with ∂ kt ˆ f (0 , ξ ) = 0 , k = 0 , . . . , q + N satisfying (4.16) . Assume that P is a differential operator of order 2 and the principal symbol p has a double characteristic root ¯ τ at (0 , ¯ ξ ), | ξ | = 1. After Fourier transform in x the equation P u = f reduces to(4.24) D t ˆ u + X j + | α |≤ ,j ≤ a j,α ( t ) ξ α D jt ˆ u = ˆ f Making similar procedure in Section 4.3 one can assume that the principalsymbol p has the form(4.25) p ( t, τ, ξ ) = τ − a ( t, ξ ) | ξ | , a (0 , ¯ ξ ) = 0so that ¯ τ = 0 is a double root. If ∂ t a (0 , ¯ ξ ) = 0 one can write a ( t, ξ ) = e ( t, ξ ) (cid:0) t + α ( ξ ) (cid:1) , α ( ¯ ξ ) = 0in some neighborhood U of (0 , ¯ ξ ) where e > α ( ξ ) ≥ ξ . In this casewe choose ϕ = t , Ω = [0 , δ ]. If (0 , , ¯ ξ ) is a critical point (hence ∂ t a (0 , ¯ ξ ) = 0)and effectively hyperbolic then(4.26) ∂ t a (0 , ¯ ξ ) = 0 . Indeed, assuming a (0 , ¯ ξ ) = ∂ t a (0 , ¯ ξ ) = 0 it is easy to seedet (cid:0) λ − F p (0 , , ¯ ξ ) (cid:1) = λ n (cid:0) λ − ∂ t a (0 , ¯ ξ ) (cid:1) which shows that ∂ t a (0 , ¯ ξ ) = 0. From the Malgrange preparation theorem onecan write, in some neighborhood U of (0 , ¯ ξ ) a ( t, ξ ) = e ( t, ξ ) (cid:0) t + a ( ξ ) t + a ( ξ ) (cid:1) = e Y k =1 (cid:0) t − ν k ( ξ ) (cid:1) e > a i ( ¯ ξ ) = 0. Note that if Re ν ( ξ ) = Re ν ( ξ ) then ν i ( ξ ) isnecessarily real and ν i ( ξ ) ≤
0. In the case that either Re ν ( ξ ) = Re ν ( ξ ) or Re ν ( ξ ) = Re ν ( ξ ) ≤ ϕ = t, Ω = [0 , δ ]. In the case Re ν ( ξ ) = Re ν ( ξ ) = ψ ( ξ ) > a ( t, ξ ) = e (cid:8) ( t − ψ ( ξ )) + ( Im ν ( ξ )) (cid:9) we take ϕ = ψ ( ξ ) − t , Ω ( ξ ) = [0 , ψ ( ξ )], ϕ = t − ψ ( ξ ), Ω ( ξ ) = [ ψ ( ξ ) , δ ].Repeating similar arguments as in [16], [17] one obtains Proposition 4.3.
Assume that p has a double characteristic root ¯ τ at (0 , ¯ ξ ) , | ¯ ξ | = 1 and (0 , ¯ τ , ¯ ξ ) is effectively hyperbolic if it is a critical point. Then onecan find δ > and a conic neighborhood U of ¯ ξ such that for any a j,α ( t ) with j + | α | ≤ one can find N ∈ N such that for any q ∈ N with q ≥ N there is C > such that (4.27) q +2 X k =0 (cid:12)(cid:12) ∂ kt ˆ u ( t, ξ ) (cid:12)(cid:12) ≤ C q X k =0 Z t (cid:12)(cid:12) h ξ i N + q − k ∂ kt ˆ f ( s, ξ ) (cid:12)(cid:12) ds for ( t, ξ ) ∈ [0 , δ ] × U and for any ˆ u ( t, ξ ) with ∂ kt ˆ u (0 , ξ ) = 0 , k = 0 , and ˆ f ( t, ξ ) with ∂ kt ˆ f (0 , ξ ) = 0 , k = 0 , . . . , q + N satisfying (4.24) . We turn to the Cauchy problem (4.1). First note that, after Fourier transformin x , the equation is reduced to(4.28) ( P ( t, D t , ξ )ˆ u = D mt ˆ u + P m − j =0 P | α |≤ m − j a j,α ( t ) ξ α D jt ˆ u = 0 ,D kt ˆ u (0 , ξ ) = ˆ u k ( ξ ) , ξ ∈ R n , k = 0 , . . . , m − . Proposition 4.4.
Assume that the characteristic roots of p (0 , τ, ξ ) are at mosttriple for any ξ = 0 and every critical point (0 , τ, ξ ) , ξ = 0 is effectively hyper-bolic. Then there exists δ > such that for any a j,α ( t ) with j + | α | ≤ m − onecan find N , N ∈ N and C > such that m − X k =0 (cid:12)(cid:12) h ξ i m − − k ∂ kt ˆ u ( t, ξ ) (cid:12)(cid:12) ≤ C N X k =0 Z t (cid:12)(cid:12) h ξ i N − k ∂ kt ˆ f ( s, ξ ) (cid:12)(cid:12) ds for ( t, ξ ) ∈ [0 , δ ] × R n and for any ˆ u ( t, ξ ) with ∂ kt ˆ u (0 , ξ ) = 0 , k = 0 , . . . , m − and ˆ f ( t, ξ ) with ∂ kt ˆ f (0 , ξ ) = 0 , k = 0 , . . . , N satisfying P ( t, D t , ξ )ˆ u = ˆ f .Proof. Let ¯ ξ = 0 be arbitrarily fixed. Write p (0 , τ, ¯ ξ ) = Q rj =1 (cid:0) τ − τ j ) m j where P m j = m and τ j are real and different each other, where m j ≤ δ > U of ¯ ξ such that onecan write p ( t, τ, ξ ) = r Y j =1 p ( j ) ( t, τ, ξ ) ,p ( j ) ( t, τ, ξ ) = τ m j + a j, ( t, ξ ) τ m j − + · · · + a j,m j ( t, ξ )23or ( t, ξ ) ∈ ( − δ, δ ) × U where a j,k ( t, ξ ) are real valued, homogeneous of degree k in ξ and p ( j ) (0 , τ, ¯ ξ ) = ( τ − τ j ) m j . If (0 , τ j , ¯ ξ ) is a critical point of p , andnecessarily m j ≥
2, then (0 , τ j , ¯ ξ ) is a critical point of p ( j ) and it is easy to see H p (0 , τ j , ¯ ξ ) = c j H p ( j ) (0 , τ j , ¯ ξ )with some c j = 0 and hence H p ( j ) (0 , τ j , ¯ ξ ) has non-zero real eigenvalues if H p (0 , τ j , ¯ ξ ) does and vice versa. It is well known that one can write, in someconic neighborhood U of ¯ ξ that P = P (1) P (2) · · · P ( r ) + R where P ( j ) are differential operators in t of order m j with coefficients which arepoly-homogeneous symbol in ξ and R is a differential operators in t of orderat most m − S −∞ (in ξ ) coefficients. Note that the principal symbol of P ( j ) is p ( j ) and hence the assumptions in Propositions 4.2 and 4.3 are satisfied.Therefore thanks to Propositions 4.2 and 4.3 we have q + m j X k =0 (cid:12)(cid:12) h ξ i q + m j − k ∂ kt ˆ u ( t ) (cid:12)(cid:12) ≤ C q X k =0 n(cid:12)(cid:12) h ξ i q − k ∂ qt ( P ( j ) ˆ u )( t ) (cid:12)(cid:12) + Z t (cid:12)(cid:12) h ξ i N + q − k ∂ kt ( P ( j ) ˆ u )( s ) (cid:12)(cid:12) ds o in some conic neighborhood of ¯ ξ and for j = 1 , . . . , r . Then by induction on j = 1 , . . . , r one obtains q + m X k =0 (cid:12)(cid:12) h ξ i q + m − k ∂ kt ˆ u ( t ) (cid:12)(cid:12) ≤ C q X k =0 n(cid:12)(cid:12) h ξ i q − k ∂ kt h ( t ) (cid:12)(cid:12) + Z t (cid:12)(cid:12) h ξ i rN + q − k ∂ kt h ( s ) (cid:12)(cid:12) ds o where h ( t ) = ˆ f ( t ) − R ˆ u ( t ). Note that for any k, l ∈ N there is C k,l such that (cid:12)(cid:12) ∂ kt ( R ˆ u )( t ) (cid:12)(cid:12) ≤ C k,l h ξ i − l k + m − X j =0 (cid:12)(cid:12) h ξ i k + m − − j ∂ jt ˆ u ( t ) (cid:12)(cid:12) . Therefore one concludes that q + m X k =0 (cid:12)(cid:12) h ξ i q + m − k ∂ kt ˆ u ( t ) (cid:12)(cid:12) ≤ C q + m X k =0 Z t (cid:12)(cid:12) h ξ i q + m − k ∂ kt ˆ u ( s ) (cid:12)(cid:12) ds + C q +1 X k =0 Z t (cid:12)(cid:12) h ξ i rN + q +1 − k ∂ kt ˆ f ( s ) (cid:12)(cid:12) ds. Then the assertion follows from the Gronwall’s lemma. Finally applying a com-pactness arguments one can complete the proof.24roof of Theorem 4.1: Let u j ( x ) ∈ C ∞ ( R n ) and hence ˆ u j ( ξ ) ∈ S ( R n ). From P ˆ u = 0 one can determine ∂ kt ˆ u (0 , ξ ) successively from ˆ u j ( ξ ). Take N ≥ N + m and define ˆ u N ( t, ξ ) = N X k =0 t k k ! ∂ kt ˆ u (0 , ξ )which is in C ∞ ( R ; S ( R n )). With ˆ f = − P ˆ u N it is clear that ∂ kt ˆ f (0 , ξ ) = 0 for k = 0 , . . . , N . Apply Proposition 4.4 to the following Cauchy problem P ˆ w = − P ˆ u N = ˆ f ( t, ξ ) , ∂ kt ˆ w (0 , ξ ) = 0 , k = 0 , . . . , m − m − X k =0 (cid:12)(cid:12) h ξ i m − − k ∂ kt ˆ w ( t, ξ ) (cid:12)(cid:12) ≤ C N X k =0 Z t (cid:12)(cid:12) h ξ i N − k ∂ kt (cid:0) P ˆ u N (cid:1) ( s, ξ ) (cid:12)(cid:12) ds. Since it is clear that N X k =0 (cid:12)(cid:12) h ξ i N − k ∂ kt (cid:0) P ˆ u N (cid:1) ( s, ξ ) (cid:12)(cid:12) ≤ C N ,N m − X j =0 (cid:12)(cid:12) h ξ i N + N − j ˆ u j ( ξ ) (cid:12)(cid:12) for 0 ≤ s ≤ δ then noting that ˆ u = ˆ w + ˆ u N is the solution to the Cauchy problem(4.28) one obtains m − X k =0 (cid:12)(cid:12) h ξ i m − − k ∂ kt ˆ u ( t, ξ ) (cid:12)(cid:12) ≤ C ′ m − X j =0 (cid:12)(cid:12) h ξ i N + N − j ˆ u j ( ξ ) (cid:12)(cid:12) . Therefore, by a Paley-Wiener Theorem we prove the C ∞ well-posedness of theCauchy problem (4.1). In this section we consider the Cauchy problem for third order hyperbolic op-erators in a neighborhood of the origin of R ;(5.1) D t u + P j =0 P j + k ≤ a j,k ( t, x ) D kx D jt u = 0 ,D jt u (0 , x ) = u j ( x ) , j = 0 , , t ≥ x ∈ R and the coefficients a j,k ( t, x ) ( j + k = 3) are real valued realanalytic in ( t, x ) in a neighborhood of the origin and the principal symbol pp ( t, x, τ, ξ ) = τ + X j =0 X j + k =3 a j,k ( t, x ) ξ k τ j τ for ( t, x ) ∈ [0 , T ) × U with some T > U of the origin. Theorem 5.1.
Assume that every critical point (0 , , τ, is effectively hyper-bolic. Then for any a j,k ( t, x ) with j + k ≤ which are C ∞ near (0 , the Cauchyproblem (5.1) is C ∞ well-posed near the origin for t ≥ . Assume that p has a triple characteristic root ¯ τ at (0 , ,
1) hence (0 , , τ,
1) is acritical point. Making a suitable change of local coordinates t = t ′ , x = x ( t ′ , x ′ )such that x (0 , x ′ ) = x ′ one can assume that a , ( t, x ) = 0 so that(5.2) p ( t, x, τ, ξ ) = τ − a ( t, x ) ξ τ − b ( t, x ) ξ . Since the triple characteristic root is ¯ τ = 0 hence b (0 ,
0) = a (0 ,
0) = 0 and thehyperbolicity condition implies that(5.3) ∆( t, x ) = 4 a ( t, x ) − b ( t, x ) ≥ , ( t, x ) ∈ [0 , T ) × U. Note that ∂ x a (0 ,
0) = ∂ t b (0 ,
0) = ∂ x b (0 ,
0) = 0 which follows from Lemma 2.1then it is clear thatdet (cid:0) λI − F p (0 , , , (cid:1) = λ (cid:0) λ − ( ∂ t a (0 , (cid:1) . This implies ∂ t a (0 , = 0 since (0 , , ,
1) is assumed to be effectively hyperbolic.A counterpart of key Proposition 4.1 is obtained by applying similar argu-ments as in Section 4.1 together with some observations on non-negative realanalytic functions with two independent variables given in [16, Lemma 2.1] (seealso [19]). We just give a sketch of the arguments. From the Weierstrass’ prepa-ration theorem there is a neighborhood of (0 ,
0) where one can write∆( t, x ) = e ( t, x ) (cid:8) t + a ( x ) t + a ( x ) t + a ( x ) (cid:9) where e > a j ( x ) are real valued, real analytic with a j (0) = 0. Denote¯∆( t, x ) = t + a ( x ) t + a ( x ) t + a ( x )then the next lemma is a counterpart of Lemma 4.3 Lemma 5.1.
There exists δ > such that, in each interval < ± x < δ , onecan write (5.4) ¯∆( t, x ) = (cid:12)(cid:12) t − ν ( x ) (cid:12)(cid:12) ( t − ν ( x )) where ν ( x ) is real valued with ν ( x ) ≤ or (5.5) ¯∆( t, x ) = Y k =1 (cid:0) t − ν k ( x ) (cid:1) here ν k ( x ) are real valued with ν k ( x ) ≤ where, in both cases, ν j ( x ) areexpressed as convergent Puiseux series; ν k ( x ) = X j ≥ C ± k,j ( ± x ) j/p j , ( p j ∈ N ) . In all cases there is
C > such that (5.6) (cid:12)(cid:12) d Re ν j ( x ) /dx (cid:12)(cid:12) ≤ C, < | x | < δ. Next we show a counterpart of Lemma 4.2. Note that one can write a ( t, x ) = e ( t, x ) (cid:0) t + α ( x ) (cid:1) where e > α ( x ) is real analytic with α (0) = 0 and α ( x ) ≥ | x | < δ . Lemma 5.2.
There exist ε > and δ > such that one can find j ± ∈ { , , } such that ε − | ν j ± ( x ) | ≥ α ( x ) , < ± x < δ. Recall that, choosing a smaller δ > Re ν j ( x ), Im ν j ( x ), ν ( x ) ≡ < ± x < δ . Denote ψ ( x ) = max (cid:8) , Re ν ( x ) (cid:9) , | x | < δ and define ( φ = t, Ω = { ( t, x ) | | x | < δ, ≤ t ≤ ψ ( x ) / } φ = t − ψ ( x ) , Ω = { ( t, x ) | | x | < δ, ψ ( x ) / ≤ t ≤ T } with a small T >
0. If ψ = 0 then φ = φ = t and Ω = { ( t, x ) | | x | < δ, ≤ t ≤ T } . Now we have a counterpart of Proposition 4.1; Proposition 5.1.
There exist δ > , T > and C > such that (5.7) φ j a ≤ C ∆ , | φ j | | ∂ t ∆ | ≤ C ∆ , | φ j | ≤ Ca for ( t, x ) ∈ Ω j and j = 1 , . x dependent case With U = t ( D t u, D x D t u, D x u ) the equation (5.1) can be written ∂∂t U = a ( t, x ) b ( t, x )1 0 00 1 0 ∂∂x U + i b ( t, x ) b ( t, x ) b ( t, x )0 0 00 0 0 U + if = A ( t, x ) ∂ x U + BU + F. (5.8) 27et T ( t, x ) be the orthonormal matrix introduced in Section 2.2 such that with V = T − U the equation (5.8) becomes(5.9) ∂ t V = A ∂ x V + B V + ˜ F where A = T − AT , B = ( ∂ t T − ) T − A ( ∂ x T − ) T + T − BT , ˜ F = T − F . Let ω be an open domain in R and let g ∈ C ( ω ) be a positive scalar function.Denote by ∂ω the boundary of ω equipped with the usual orientation. Let G ( V ) = g h DV, V i dx + g h D A V, V i dt then one has 2 Re Z ω g h DV, B V + ˜ F i dxdt = − Z ∂ω G ( V ) − Z ω n ( ∂ t g ) h DV, V i + g h ( ∂ t D ) V, V i o dxdt + Z ω n ( ∂ x g ) h D A V, V i + g h ∂ x ( D A ) V, V i o dxdt. (5.10)Here make a remark on the boundary term. Denote τ max = max ( t,x ) ∈ [0 ,T ] × U (cid:12)(cid:12) τ j ( t, x ) (cid:12)(cid:12) where τ j ( t, x ), j = 1 , , p ( t, x, τ, Lemma 5.3.
Let
Γ : [ a, b ] ( f ( x ) , x ) be a space-like curve, that is > τ max (cid:12)(cid:12) f ′ ( x ) (cid:12)(cid:12) , x ∈ [ a, b ] . Then one has Z Γ G ( V ) ≥ . Proof.
Since g ( t, x ) > h D ( f ( x ) , x ) V, V i + h f ′ ( x ) D ( f ( x ) , x ) A ( f ( x ) , x ) V, V i ≥ , ∀ V ∈ C . To simplify notation we denote D ( f ( x ) , x ) and A ( f ( x ) , x ) by just D and A .Noting that D A = t A D one has (cid:12)(cid:12) h f ′ D A V, V i (cid:12)(cid:12) = (cid:12)(cid:12) f ′ h DV, A V i (cid:12)(cid:12) ≤ h DV, V i / h D A V, A V i / | f ′ | . Therefore to prove the lemma it suffices to show h D A V, A V i| f ′ | ≤ h DV, V i that is, the maximal eigenvalue of | f ′ | ( t A D A ) with respect to D is at most 1.From t A D = D A it follows thatdet (cid:0) λD − | f ′ | ( t A D A ) (cid:1) = det (cid:0) λD − | f ′ | D A ) (cid:1) = (det D ) det (cid:0) λI − | f ′ | A (cid:1) . Since τ j are the eigenvalues of A we see that the eigenvalues of | f ′ | A is atmost | f ′ ( x ) | τ max < { ( t, x ) | | x | ≤ δ ( T − t ) , ≤ t ≤ T } and ϕ = t, ω = { ( t, x ) | | x | ≤ δ ( T − t ) , ≤ t ≤ ψ ( x ) / } ,ϕ = ψ ( x ) − t, ω = { ( t, x ) | | x | ≤ δ ( T − t ) , ψ ( x ) / ≤ t ≤ ψ ( x ) } ,ϕ = t − ψ ( x ) , ω = { ( t, x ) | | x | ≤ δ ( T − t ) , ψ ( x ) ≤ t ≤ T } (5.11)where δ > , T > | x | = δ ( T − t ), 0 ≤ t ≤ T arespace-like. Thanks to Proposition 5.1 and Lemma 5.1 it follows that(5.12) ϕ j ≤ Cλ , ϕ j | ∂ t λ | ≤ Cλ , ϕ j ≤ Cλ , (cid:12)(cid:12) ∂ x ϕ j (cid:12)(cid:12) ≤ C, ( t, x ) ∈ ω j . Apply (5.10) with g = g j = ϕ − j N − j , G ( V ) = G j ( V ) = g j h DV, V i dx + g j h D A V, V i dt and ω = ω j then from the arguments in Section 3 one obtains Lemma 5.4.
There exist N , C > such that C Z ω j ϕ j g j k F k dxdt ≥ − Z ∂ω j G j ( V ) + N Z ω j ϕ − j g j h DV, V i dxdt for N ≥ N . Repeating the same arguments as in [16, Lemma 3.1] (also [19]) one has
Proposition 5.2.
There exists
C > such that for every n ∈ N one can find N such that X k + ℓ ≤ n Z ω j g j ϕ j k ∂ kt ∂ ℓx U k dxdt − X ℓ ≤ n Z ∂ω j G j ( T − ∂ ℓx U ) ≤ C X k + ℓ ≤ n Z ω j g j ϕ j k ∂ kt ∂ ℓx LU k dxdt for any N ≥ N . Following the same arguments as in [16], [19] one can collect energy estimatesin each ω j to obtain Proposition 5.3.
Assume that p has a triple characteristic root ¯ τ at (0 , , and (0 , , ¯ τ , is effectively hyperbolic. Then there exist T > , δ > such thatfor any a j,k ( t, x ) ∈ C ∞ (Ω) , j + k ≤ one can find C > and Q ∈ N such that (5.13) X k + ℓ ≤ n Z Ω k ∂ kt ∂ ℓx U k dxdt ≤ C X k + ℓ ≤ Q Z Ω k ∂ kt ∂ ℓx LU k dxdt. for any U ( t, x ) ∈ C ∞ (Ω) with ∂ kt U (0 , x ) = 0 , k = 0 , . . . , Q . .3 Proof of Theorem 5.1 To complete the proof of Theorem 5.1, study the remaining case that p has adouble characteristic root at (0 , , Proposition 5.4.
Assume that p has a double characteristic root ¯ τ at (0 , , and (0 , , ¯ τ , is effectively hyperbolic if it is a critical point. Then the sameassertion as in Proposition 5.3 holds. We give a sketch of the proof. Assume that p has a double characteristicroot ¯ τ at (0 , ,
1) and hence, after a suitable change of local coordinates, onecan write(5.14) p ( t, x, τ, ξ ) = (cid:0) τ − b ( t, x ) ξ (cid:1)(cid:0) τ − a ( t, x ) ξ (cid:1) = p p where p = τ − b ( t, x ) ξ , p = τ − a ( t, x ) ξ and a (0 ,
0) = 0, b (0 , = 0. If(0 , , ,
1) is a critical point of p and hence ∂ t a (0 ,
0) = 0 then H p = c H p at(0 , , ,
1) with some c = 0 and det (cid:0) λI − F p (0 , , , (cid:1) = λ (cid:0) λ − ∂ t a (0 , (cid:1) which shows ∂ t a (0 , = 0. From the Weierstrass’ preparation theorem one canwrite a ( t, x ) = e ( t, x ) (cid:0) t + 2˜ a ( x ) t + ˜ a ( x ) (cid:1) = e ( t, x )∆ ( t, x )where e >
0, ˜ a j (0) = 0 and ∆ takes the form, in each 0 < ± x < δ , either(5.4) or (5.5) where ν does not occur. If ∆ has the form (5.5) or (5.4) with Re ν ( x ) ≤ ϕ = t , ω = Ω. If ∆ takes the form (5.4) with Re ν ( x ) = ψ ( x ) > ϕ = ψ ( x ) − t , ω = { ( t, x ) | | x | ≤ δ ( T − t ) , ≤ t ≤ ψ ( x ) } and ϕ = t − ψ ( x ), ω = { ( t, x ) | | x | ≤ δ ( T − t ) , ψ ( x ) ≤ t ≤ T } . If ∂ t a (0 , = 0 one can write a ( t, x ) = e ( t, x ) (cid:0) t + α ( x ) (cid:1) , α (0) = 0where e > α ( x ) ≥ x = 0. In this case we onlyneed ϕ = t , ω = Ω.Denote P = D t − a ( t, x ) D x , P = D t − b ( t, x ) D x . Then repeating the same arguments as in [16], [19] one has(5.15) Z ω j ϕ j g j | P u | dxdt ≥ − Z ∂ω j G (2) j ( u )+ N Z ω j ϕ j g j (cid:0) | D t u | + | D x u | (cid:1) dxdt for N ≥ N with G (2) j ( u ) = g j (cid:0) | ∂ t u | + a | ∂ x u | (cid:1) dx + ag j (cid:0) ∂ x u · ∂ t u + ∂ x u · ∂ t u (cid:1) dt .On the other hand, since P is a first order differential operator with a realvalued b ( t, x ) it is easy to see that Z ω j ϕ j g j | P u | dxdt ≥ − Z ∂ω j G (1) j ( u ) + N Z ω j ϕ − j g j | u | dxdt ≥ − Z ∂ω j G (1) j ( u ) + N Z ω j ϕ j g j | u | dxdt (5.16) 30or N ≥ N with G (1) j ( u ) = g j | u | dx + bg j | u | dt . Inserting u = P u in (5.15)and u = P u in (5.16) respectively and adding them one obtains Z ω j ϕ j g j (cid:0) | P P u | + | P P u | (cid:1) dxdt ≥ − Z ∂ω j ˜ G j ( u )+ N Z ω j ϕ j g j (cid:0) | D t P u | + | D x P u | + | P u | (cid:1) dxdt. (5.17)In view of b (0 , = 0 and a (0 ,
0) = 0 it is easy to see that D t , D t D x and D x are linear combinations of D t P , D x P and P with smooth coefficients modulofirst order operators. Since one can write P = P P + X i + j ≤ b i,j ( t, x ) D it D jx = P P + X i + j ≤ ˜ b i,j ( t, x ) D it D jx one concludes from (5.17) that Z ω j ϕ j g j | P u | dxdt ≥ − Z ∂ω j ˜ G j ( u ) + N X i + j ≤ Z ω j ϕ j g j | D it D jx u | dxdt. The rest of the proof is parallel to that of Proposition 5.3.Proof of Theorem 5.1: Note that (5.13) implies X k + ℓ ≤ n +2 Z Ω k ∂ kt ∂ ℓx u k dxdt ≤ C X k + ℓ ≤ Q Z Ω k ∂ kt ∂ ℓx P u k dxdt. Then applying an approximation argument with the Cauchy-Kowalevsky theo-rem one can conclude the proof of Theorem 5.1.We restrict ourselves to third order operators in Theorem 5.1 because itseems to be hard to apply the same arguments as in Section 4.5 to this case.
To show that the same arguments in the previous sections can be applicableto hyperbolic operators with more general triple characteristics, we study theCauchy problem(6.1) ( D t u − a ( t, x ) D t D x u − b ( t, x ) D x u = 0 , U ∩ { t > s } D jt u ( s, x ) = u j ( x ) , j = 0 , , full neighborhood of (0 ,
0) in R . The Cauchy problem (6.1) is (uniformly)well posed near the origin if one can find a neighborhood U of (0 ,
0) and a small ǫ > | s | < ǫ and u j ( x ) ∈ C ∞ (Ω ∩ { t = s } ) there is a unique31olution to (6.1). Assume that there is a neighborhood U ′ of (0 ,
0) such that p ( t, x, τ,
1) = 0 has only real roots in τ for ( t, x ) ∈ U ′ , that is(6.2) ∆( t, x ) = 4 a ( t, x ) − b ( t, x ) ≥ , ( t, x ) ∈ U ′ which is necessary for the Cauchy problem (6.1) to be well posed near the origin.Then one has Theorem 6.1.
Assume (6.2) and that there exists
C > such that (6.3) a ≤ C ∆ , | ∂ t b | ≤ C √ a | ∂ t a | holds in a neighborhood of (0 , . Then the Cauchy problem (6.1) is C ∞ wellposed near the origin. Note that if both a ( t, x ) and b ( t, x ) are independent of t , Theorem 6.1 is avery special case of [22, Theorem 1.1] and if a ( t, x ) and b ( t, x ) are independentof x then this is also a special case of [4, Theorem 2].We give a rough sketch of the proof. If ∆(0 , > p is strictly hyper-bolic near the origin and the assertion is clear so ∆(0 ,
0) = 0 is assumed fromnow on and hence a (0 ,
0) = 0 by assumption. Then p has triple characteristicroot at (0 , , λ j of the B´ezoutianmatrix satisfy λ ≃ a , λ ≃ a, λ ≃ /a ≥ a /C . Let φ be a scalar function satisfying φ > ∂ t φ > ω and consider the following energy;(6.4) ( φ − N e − γt DV, V ) = Z ω φ − N e − γt h DV, V i dx where N > , γ > ddt ( e − γt φ − N DV, V ) = − (cid:0) ( N φ − ∂ t φ + γ ) e − γt φ − N DV, V (cid:1) + (cid:0) e − γt φ − N ( ∂ t D ) V, V (cid:1) + 2 Re (cid:0) e − γt φ − N D ( A ∂ x V + B V ) , V (cid:1) where(6.5) ( N φ − ∂ t φ + γ ) h DV, V i = N X j =1 φ − ∂ t φλ j | V j | + γ X j =1 λ j | V j | . Since the hyperbolicity condition 4 a ≥ b is assumed in a full neighborhoodof (0 ,
0) then Lemma 2.1 now states(6.6) | ∂ t a | (cid:22) √ a, | ∂ x a | (cid:22) √ a, | ∂ x b | (cid:22) a, | ∂ t b | (cid:22) a. As for a scalar weight φ we assume(6.7) φ | ∂ t a | (cid:22) a ∂ t φ, φ (cid:22) ∂ t φ in ω. | ∂ t q ( λ j ) | (cid:22) ( | ∂ t a | + | b || ∂ t b | ) λ j + | ∂ t ∆ | and | ∂ t ∆ | (cid:22) a | ∂ t a | by (6.3) then(6.8) | ∂ t λ j | (cid:22) ( | ∂ t a | + a / ) λ j + a | ∂ t a || q λ ( λ j ) | and hence | ∂ t λ j | (cid:22) ( | ∂ t a | /a ) λ j + a / λ j + a | ∂ t a | for j = 1 , φ | ∂ t λ j | (cid:22) ∂ t φ (cid:0) λ j + a (cid:1) . Thus φ | ∂ t λ j | (cid:22) ∂ t φ λ j for j = 1 , , λ ≃ a and the case j = 3 is clearfrom (6.7). This proves that (cid:12)(cid:12) φ − N h ( ∂ t D ) V, V i (cid:12)(cid:12) is bounded by (6.5) taking N large. Next assume that(6.9) sup ( t,x ) ∈ ω ǫ √ a | ∂ x φ | ( ∂ t φ ) − = 0 ( ǫ → ω ǫ is a family of regions converging to (0 ,
0) as ǫ →
0. Thanks to (3.7)the term N (cid:12)(cid:12) φ − N − ( ∂ x φ ) h D A V, V i (cid:12)(cid:12) is bounded by (6.5) in ω ǫ for enough small ǫ in virtue of the assumption (6.9).Using (6.8) it is easy to see | ∂ t ℓ | (cid:22) a | ∂ t a | + a , | ∂ t ℓ | (cid:22) | ∂ t b | + a / , | ∂ t ℓ | (cid:22) | ∂ t a | + a, | ∂ t ℓ | (cid:22) a , | ∂ t ℓ | (cid:22) | ∂ t a | + a / , | ∂ t ℓ | (cid:22) | ∂ t b | + a , | ∂ t ℓ | (cid:22) | ∂ t a | + a / , | ∂ t ℓ | (cid:22) a , | ∂ t ℓ | (cid:22) | ∂ t a | . (6.10)Noting | ∂ t b | (cid:22) √ a | ∂ t a | these estimates improve (3.11) to Lemma 6.1.
Let ( ∂ t T − ) T = (˜ t ij ) then (˜ t ij ) = O ( a + | ∂ t a | / √ a ) O ( | ∂ t a | + a / )) O ( a + | ∂ t a | / √ a ) 0 O ( √ a | ∂ t a | + a ) O ( | ∂ t a | + a / ) O ( √ a | ∂ t a | + a ) 0 . In order to estimate (cid:12)(cid:12) φ − N h ( ∂ t T − ) T V, DV i (cid:12)(cid:12) recalling D ≃ diag( a , a,
1) itsuffices to estimate | φ − N | (cid:16) a ˜ t | V || V | + ˜ t | V || V | + ˜ t | V || V | (cid:17) . Note that a ˜ t (cid:22) a + √ a | ∂ t a | and a | V || V | (cid:22) √ a ( a | V | + a | V | ) , √ a | ∂ t a || V || V | (cid:22) φ − ∂ t φ a / | V || V | (cid:22) φ − ∂ t φ (cid:0) a | V | + a | V | (cid:1) . As for ˜ t | V || V | (cid:22) ( | ∂ t a | + a / ) | V || V | one has a / | V || V | (cid:22) a / (cid:0) a | V | + | V | (cid:1) , | ∂ t a | | V || V | (cid:22) φ − ∂ t φ a | V || V | (cid:22) φ − ∂ t φ (cid:0) a | V | + | V | (cid:1) . t | V || V | (cid:22) ( √ a | ∂ t a | + a ) | V || V | (cid:22) a | V || V | (cid:22) √ a ( a | V | + | V | ) oneconcludes that (cid:12)(cid:12) φ − N h ( ∂ t T − ) T V, DV i (cid:12)(cid:12) is bounded by (6.5) taking N large and γ ≥ | φ − N h ( ∂ x T − ) T V, D A V i . From Lemma 3.2 it follows that D A = O ( a / ) O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) O ( a / ) because D A is symmetric and D ≃ diag( a , a, D A ( ∂ x T − ) T = O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) . Therefore |h ( ∂ x T − ) T V, D A V i| is bounded by a X j =1 | V j | + a / | V || V | + a | V || V | + a / | V || V |(cid:22) a | V | + a | V | + | V | . Consider the term h ∂ x ( D A ) V, V i . From (3.16) one has | ∂ x λ | (cid:22) a / . Since | ∂ x λ | = O ( √ a ) and | ∂ x λ | = O (1) it follows from Lemmas 3.2( ∂ x D ) A = O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) O ( √ a ) O ( a / ) O ( a ) O ( a / ) and then |h ( ∂ x D ) A V, V i| is bounded by a | V | + a | V | + a / | V | + a / | V || V | + a / | V || V | + √ a | V || V |(cid:22) a | V | + a | V | + | V | . Consider (cid:12)(cid:12) h ( ∂ x A ) V, DV i (cid:12)(cid:12) . Thanks to Lemma 3.2 one has D ( ∂ x A ) = O ( a ) O ( a / ) O ( a ) O ( a / ) O ( a ) O ( √ a ) O ( a ) O ( a ) O ( √ a ) . Thus |h ( ∂ x D ) A V, V i| is bounded by a | V | + a | V | + √ a | V | + a | V || V | + a / | V || V | + √ a | V || V |(cid:22) a | V | + a | V | + | V | . Therefore to obtain energy estimates it suffices to find pairs ( φ j , ω j,ǫ ) ofscalar weight φ j and subregion ω j,ǫ such that (6.7) and (6.9) are verified and34 ω j,ǫ covers a neighborhood of (0 ,
0) for any fixed small ǫ >
0. Since the choiceof ( φ j , ω j,ǫ ) is exactly same as in [16] (also [19]) we only mention how to choose φ j and ω j,ǫ ( φ is denoted by ρ in [16] or [19]).Following [16] one can define a real valued function α ( t, x )(6.11) α ( t, x ) = x n Y i ∈ I ( t − t i ( x )) Y i ∈ I | t − t i ( x ) | e ( t, x )so that a ( t, x ) = α ( t, x ) where t i ( x ) has a convergent Puiseux expansion in0 < ± x < δ with small δ and Im t i ( x ) = 0 if i ∈ I . We choose all distinct t k ( x )in (6.11) and rename them as t ( x ) , . . . , t m ( x ). Taking δ small one can assumethat Re t µ ( x ) ≤ Re t µ ( x ) ≤ · · · ≤ Re t µ m ( x ) , < x < δ, Re t ν ( x ) ≤ Re t ν ( x ) ≤ · · · ≤ Re t ν m ( x ) , − δ < x < . Define σ j ( x ) by σ j ( x ) = Re t µ j ( x ) for x > , σ j ( x ) = Re t ν j ( x ) for x < σ ( x ) ≤ · · · ≤ σ m ( x ) in | x | < δ . Denote s j ( x ) = σ j ( x ) + σ j +1 ( x )2 , ≤ j ≤ m − , s ( x ) = − t ∗ ( x ) , s m ( x ) = 3 t ∗ ( x )with t ∗ ( x ) = (cid:0) m X j =1 | t j ( x ) | (cid:1) / where the sum is taken over all distinct t i ( x ) in (6.11). Denote by ω j and ω ( T )the subregions defined by ω j = { ( t, x ) | | x | ≤ ¯ δ ( T − t ) , s j − ( x ) ≤ t ≤ s j ( x ) } ( j = 1 , . . . , m ) ,ω ( T ) = { ( t, x ) | | x | ≤ ¯ δ ( T − t ) , s m ( x ) ≤ t ≤ T } . for small ¯ δ > T >
0. Here ¯ δ >
T > ǫ > ω j , j = 1 , . . . , m we take φ = φ j ( t, x ) = t − σ j ( x ). For ω ( T ) we take φ = φ m +1 = t − s m ( x ) if n ≥
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