Diagonal symmetrizers for hyperbolic operators with triple characteristics

Abstract

Symmetrizers for hyperbolic operators are obtained by diagonalizing the Bézoutian matrix of the principal symbols and its derivatives. Such diagonal symmetrizers are applied to the Cauchy problem for hyperbolic operators with triple characteristics. In particular, the Ivrii’s conjecture concerned with strongly hyperbolic operators with triple effectively hyperbolic characteristics is proved for differential operators with time dependent coefficients, also for third order differential operators with two independent variables with analytic coefficients.

1 Introduction

This paper is devoted to the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{ll} D_t^mu+\sum _{j=0}^{m-1}\sum _{|\alpha |+j\le m}a_{j,\alpha }(t,x)D_x^{\alpha }D_t^ju=0,\\ D_t^ju(0,x)=u_j(x),\quad j=0,\ldots ,m-1 \end{array}\right. \end{aligned}$$

(1.1)

where $t\ge 0$, $x\in {\mathbb {R}}^n$ and the coefficients $a_{j,\alpha }(t,x)$ are real valued $C^{\infty }$ functions in a neighborhood of the origin of ${\mathbb {R}}^{1+n}$ and $D_x=(D_{x_1},\ldots ,D_{x_n})$, $D_{x_j}=(1/i)(\partial /\partial x_j)$ and $D_t=(1/i)(\partial /\partial t)$. The problem is $C^{\infty }$ well-posed near the origin for $t\ge 0$ if one can find a $\delta >0$ and a neighborhood U of the origin of ${\mathbb {R}}^n$ such that (1.1) has a unique solution $u\in C^{\infty }([0,\delta )\times U)$ for any $u_j(x)\in C^{\infty }({\mathbb {R}}^n)$. We assume that the principal symbol p is hyperbolic for $t\ge 0$, that is

$$\begin{aligned} p(t,x,\tau ,\xi )=\tau ^m+\sum _{j=0}^{m-1}\sum _{|\alpha |+j=m}a_{j,\alpha }(t,x)\xi ^{\alpha }\tau ^j \end{aligned}$$

has only real roots in $\tau$ for $(t,x)\in [0,\delta ')\times U'$ and $\xi \in {\mathbb {R}}^n$ with some $\delta '>0$ and a neighborhood $U'$ of the origin which is necessary in order that the Cauchy problem (1.1) is $C^{\infty }$ well-posed near the origin for $t\ge 0$ ([17, 20]).

In this paper we are mainly concerned with the case that the multiplicity of the characteristic roots is at most 3. This implies that it is essential to study operators P of the form

$$\begin{aligned} P=D_t^3+\sum _{j=1}^3a_j(t,x,D)\langle {D}\rangle ^jD_t^{3-j} \end{aligned}$$

(1.2)

which is differential operator in t with coefficients $a_{j}\in S^0$, classical pseudodifferential operator of order 0, where $\langle {D}\rangle = \mathrm{Op}((1 + |\xi |^2)^{1/2})$. One can assume that $a_1(t,x,D)=0$ without loss of generality and hence the principal symbol has the form

$$\begin{aligned} p(t, x, \tau , \xi ) = \tau ^3 -a(t, x, \xi )|\xi |^2\tau - b(t, x, \xi )|\xi |^3. \end{aligned}$$

With $U={^t}(D_t^2u,\langle {D}\rangle D_tu,\langle {D}\rangle ^2u)$ the equation $P u=f$ is reduced to

$$\begin{aligned} D_tU=A(t,x,D)\langle {D}\rangle U+B(t,x,D)U+F \end{aligned}$$

(1.3)

where $A, B\in S^0$, $F={^t}(f,0,0)$ and

$$\begin{aligned} A(t,x,\xi )= \begin{bmatrix}0&{}a&{}b\\ 1&{}0 &{}0\\ 0&{}1&{}0 \end{bmatrix}. \end{aligned}$$

Let S be the Bézoutiant of p and $\partial p/\partial \tau$, that is

$$\begin{aligned} S(t,x,\xi )= \begin{bmatrix}3&{}0&{}-a\\ 0&{}2a&{}3b\\ -a&{}3b&{}a^2 \end{bmatrix} \end{aligned}$$

(1.4)

then S is nonnegative definite and symmetrizes A, that is SA is symmetric which is easily examined directly, though this is a special case of a general fact (see [15, 28]). Then one of the most important works would be to obtain lower bound of $(\mathrm{Op}(S)U,U)$. The sharp Gårding inequality ([8, 18]) gives a lower bound

$$\begin{aligned} {\mathsf {Re\,}}\left( \mathrm{Op}(S)U,U\right) \ge -C\Vert \langle {D}\rangle ^{-1/2}U\Vert ^2 \end{aligned}$$

which is, in general, too weak to study the Cauchy problem for general weakly hyperbolic operator P, in particular the well posed Cauchy problem with loss of derivatives, although applying this symmetrizer many interesting results are obtained by several authors, see for example [1, 5, 6, 16, 19, 29]. In these works one of the main points is how one can derive a suitable lower bound of $\mathrm{Op}(S)$ from the hyperbolicity condition assumed on p, that is

$$\begin{aligned} \Delta =4\,a(t,x,\xi )^3-27\,b(t,x,\xi )^2\ge 0, \quad (t,x,\xi )\in [0,T)\times U\times {\mathbb {R}}^n. \end{aligned}$$

(1.5)

In this paper we employ a new idea which is to diagonalize S by an orthogonal matrix T so that $T^{-1}ST=\Lambda =\mathrm{diag}\,(\lambda _1,\lambda _2,\lambda _3)$ where $0\le \lambda _1\le \lambda _2\le \lambda _3$ are the eigenvalues of S and reduce the equation to that of $V=T^{-1}U$; roughly

$$\begin{aligned} D_tV=A^T\langle {D}\rangle V+B^TV,\quad A^T=T^{-1}AT \end{aligned}$$

(1.6)

where $\Lambda$ symmetrizes $A^T$. For general nonnegative definite symmetric S it seems that we have nothing new but our S is a special one which is the Bézoutiant of hyperbolic polynomial p and $\partial p/\partial \tau$. Indeed, as we will see in Sect. 2, one has

$$\begin{aligned} \frac{\Delta }{a}\preceq \lambda _1\preceq a^2,\quad \lambda _2\simeq a,\quad \lambda _3\simeq 1. \end{aligned}$$

Since (1.6) is a symmetrizable system with a diagonal symmetrizer $\Lambda$, a natural energy will be

$$\begin{aligned} {\mathsf {Re}}\,\left( \mathrm{Op}(\Lambda )U,U\right) =\sum _{j=1}^3\left( \mathrm{Op}\left( \lambda _j\right) U_j,U_j\right) \end{aligned}$$

and it could be expected that scalar operators $\mathrm{Op}(\lambda _j)$ reflect the hyperbolicity condition (1.5) quite directly.

If $p=0$ has a triple characteristic root $\tau =0$ at $(0,x,\xi )$ such that $(0,x,\tau ,\xi )$ is effectively hyperbolic (see Sect. 4, also [7, 9]) then one sees that $\partial _ta(0,x,\xi )>0$ and hence $\partial _t^3\Delta (0,x,\xi )>0$, which follows from (1.5), so that essentially a and $\Delta$ are polynomials in t of degree 1 and 3 respectively. Then we see that $\Delta /a$ behaves like a second order polynomial in t which is nonnegative for $t\ge 0$. Finding a finite number of functions $\phi _j$ satisfying $\partial _t\phi _j>0$ and $\phi _j^2\preceq \Delta /a$, $\phi _j\big |\partial _t\Delta \big |\preceq \Delta$, $\phi _j\big |\partial _ta\big |\preceq a$ we estimate the weighted energy ${\mathsf {Re}}\,\big (\mathrm{Op}(\phi _j^{N_j}\Lambda )U,U\big )$ with a suitable $N_j\in {\mathbb {R}}$. In Sect. 4 this procedure is carried out for operators of order m with effectively hyperbolic critical points with time dependent coefficients, and it is proved that the Cauchy problem is $C^{\infty }$ well-posed for any lower order term. In Sect. 5 the same assertion is proved for third order operators with two independent variables with analytic coefficients, so that Ivrii’s conjecture is proved for these operators (Theorems 4.1 and 5.1).

In Sect. 3, admitting the existence of such a weight function $\phi _j$, we explain how to derive energy estimates. To do so we need to estimate the derivatives of $\Lambda$ and $A^T$, essentially those of $\lambda _j$, which is done in Sect. 2.

In the last section we show that the same idea is applicable to hyperbolic operators with general triple characteristics, utilizing a homogeneous third order operator with two independent variables (Theorem 6.1).

2 Daiagonal symmetrizers

Consider

$$\begin{aligned} p(\tau ,t,X)=\tau ^3-a(t,X)\tau -b(t,X) \end{aligned}$$

where a(t, X) and b(t, X) are real valued and $C^{\infty }$ in $(t,X)\in (-c,T)\times W$ with bounded derivatives of all order where W is an open set in ${\mathbb {R}}^l$ such that ${\bar{X}}\in W$ and $c>0$ is some positive constant. Assume

$$\begin{aligned} \Delta (t,X)=4\,a(t,X)^3-27\,b(t,X)^2\ge 0,\; (t,X)\in [0,T)\times W, \;\;a(0,\bar{X})=0 \end{aligned}$$

(2.1)

that is, $p(\tau ,t,X)=0$ has only real roots for $(t,X)\in [0,T)\times W$ and has a triple root $\tau =0$ at $(0,{\bar{X}})$. Moreover assume that there is no triple root in $t>0$;

$$\begin{aligned} a(t,X)>0, \quad (t,X)\in (0,T)\times W. \end{aligned}$$

(2.2)

Denote

$$\begin{aligned} S(t,X)= \begin{bmatrix}3&{}0&{}-a\\ 0&{}2\,a&{}3\,b\\ -a&{}3\,b&{}\,a^2 \end{bmatrix},\qquad A(t,X)= \begin{bmatrix}0&{}a&{}b\\ 1&{}0&{}0\\ 0&{}1&{}0 \end{bmatrix} \end{aligned}$$

(2.3)

then S is nonnegative definite and S(t, X)A(t, X) is symmetric. Let

$$\begin{aligned} 0\le \lambda _1(t,X)\le \lambda _2(t,X)\le \lambda _3(t,X) \end{aligned}$$

be the eigenvalues of S(t, X).

2.1 Behavior of eigenvalues

We show

Proposition 2.1

There exist a neighborhood ${\mathcal U}$ of $(0,\bar{X})$ and $K>0$ such that

$$\begin{aligned}&\Delta /\left( 6a+2a^2+2a^3\right) \le \lambda _1\le \left( 2/3+Ka\right) \,a^2, \end{aligned}$$

(2.4)

$$\begin{aligned}&(2-Ka)\,a\le \lambda _2\le (2+Ka)\,a, \end{aligned}$$

(2.5)

$$\begin{aligned}&3\le \lambda _3\le 3+Ka^2 \end{aligned}$$

(2.6)

for $(t,X)\in {\mathcal U}\cap \{t>0\}$.

Corollary 2.1

There exists a neighborhood ${\mathcal U}$ of $(0,\bar{X})$ such that

$$\begin{aligned} \lambda _i(t,X)\in C^{\infty }\left( {\mathcal U}\cap \{t>0\}\right) ,\quad i=1,2,3. \end{aligned}$$

Proof

Recalling $a(0,\bar{X})=0$ from Proposition 2.1 one can choose ${\mathcal U}$ such that

$$\begin{aligned} \lambda _1<\lambda _2<\lambda _3\quad \text {in}\;\;{\mathcal U}\cap \{t>0\} \end{aligned}$$

then the assertion follows immediately from the Implicit function theorem. $\square$

Remark 2.1

It may happen $\Delta (t,X)=0$ for $t>0$ so that $p(\tau ,t,X)=0$ has a double root $\tau$ at (t, X) while $\lambda _i(t,X)$ are smooth there.

Proof of Proposition 2.1:

Denote $q(\lambda )=\mathrm{det}\,(\lambda I-S)$;

$$\begin{aligned} q(\lambda )=\lambda ^3-\left( 3+2a+a^2\right) \lambda ^2+\left( 6a+2a^2+2a^3-9b^2\right) \lambda -\Delta . \end{aligned}$$

(2.7)

Let $\mu _1\le \mu _2$ be the roots of $q_{\lambda }=\partial q/\partial \lambda =0$ and hence

$$\begin{aligned} \lambda _1\le \mu _1\le \lambda _2\le \mu _2\le \lambda _3. \end{aligned}$$

It is easy to see $\mu _1=a(1+O(a))$ and $\mu _2=2+O(a)$ which gives

$$\begin{aligned} \lambda _1\le a(1+O(a)),\quad \lambda _3\ge 2+O(a). \end{aligned}$$

(2.8)

In the $(\lambda ,\eta )$ plane, the tangent line of the curve $\eta =q(\lambda )$ at (0, q(0)) intersects with $\lambda$ axis at $(\Delta /q_{\lambda }(0),0)$ and hence

$$\begin{aligned} \lambda _1(t,X)\ge \Delta /q_{\lambda }(0). \end{aligned}$$

Since $q_{\lambda }(0)\le 6a+2a^2+2a^3$ the left inequality of (2.4) is obvious. Compute $q(\delta a^2)$ with $\delta >0$. Since $2a^3-9b^2=\Delta /2+9b^2/2\ge 0$ one has

$$\begin{aligned} q\left( \delta a^2\right)&\ge \delta ^3a^6-\delta ^2a^4\left( 3+2a+a^2\right) +\delta a^2\left( 6a+2a^2\right) -4a^3+27b^2\\&\ge a^3\left\{ (6\delta -4)+\delta (2-3\delta )a-2\delta ^2a^2+\delta ^2(\delta -1)a^3\right\} . \end{aligned}$$

Here we take $\delta =2/3+Ka$ then noting $a(0,\bar{X})=0$ one can choose a neighborhood ${\mathcal U}$ of $(0,\bar{X})$ such that

$$\begin{aligned} q(\delta a^2)\ge a^4\left\{ K-3K\delta a-2\delta ^2 a+\delta ^2(\delta -1)a^2\right\} >0 \end{aligned}$$

for $(t,X)\in {\mathcal U}\cap \{t>0\}$. This proves that $\lambda _1\le \delta a^2$ and hence the right inequality of (2.4). Turn to $\lambda _2$. Consider $q(\delta a)$ with $\delta >0$ again. Note

$$\begin{aligned} q(\delta a)&\ge a^2\left\{ \delta ^3a-\delta ^2\left( 3+2a+a^2\right) +\delta (6+2a)-4a\right\} +27b^2\\&\ge a^2\left\{ 3\delta (2-\delta )+\left( \delta ^3-2\delta ^2+2\delta -4\right) a-\delta ^2 a^2\right\} \end{aligned}$$

and choose $\delta =2-Ka$ which gives

$$\begin{aligned} q(\delta a)\ge a^3\left\{ 6K-\left( 3K^2+2K+K\delta ^2-\delta ^2\right) a\right\} . \end{aligned}$$

Therefore for any $K>0$ one can find ${\mathcal U}$ such that $q(\delta a)>0$ in ${\mathcal U}\cap \{t>0\}$. Since one can assume $\delta a<\lambda _3$ by (2.8) then $\delta a\in (\lambda _1,\lambda _2)$ which proves the left inequality of (2.5). Repeating similar arguments one gets

$$\begin{aligned} q(\delta a)\le a^2\left\{ 3\delta (2-\delta ) +\left( 4+2\delta +\delta ^3-2\delta ^2\right) a+\delta (2-\delta )a^2\right\} \end{aligned}$$

because $27b^2\le 4 a^3$ in $t\ge 0$. Taking $\delta =2+Ka$ one has

$$\begin{aligned} q(\delta a)\le a^3\left\{ ( 8-6K)+\left( 2K+\delta ^2 K-\delta Ka-3K^2\right) a\right\} . \end{aligned}$$

Fixing any $K>4/3$ one can find ${\mathcal U}$ such that $q(\delta a)<0$ in ${\mathcal U}\cap \{t>0\}$. Since $\lambda _1<\delta a$ thanks to (2.4) one concludes $(2+Ka)a\in (\lambda _2,\lambda _3)$ which shows the right inequality of (2.5). Finally we check (2.6). It is easy to see that

$$\begin{aligned} q(3)=a^2(-3+2a)<0 \end{aligned}$$

in ${\mathcal U}\cap \{t>0\}$ if ${\mathcal U}$ is small so that $3\le \lambda _3$ in ${\mathcal U}\cap \{t>0\}$. Note that

$$\begin{aligned} q(\delta )\ge \delta \left\{ \delta (\delta -3)+(6-2\delta )a+\left( 2-\delta ^2\right) a^2\right\} -4a^3 \end{aligned}$$

where we take $\delta =3+Ka^2$ so that

$$\begin{aligned} q\left( 3+Ka^2\right)&=a^2\left\{ 3(3K-1)-(6K+4)a+3\left( K^2-k\right) a^2\right\} \\&\quad \,+Ka^4\left\{ (3K-1)-2Ka+\left( K^2-K\right) a^2\right\} . \end{aligned}$$

Thus fixing any $K>1/3$ one can find ${\mathcal U}$ such that $q(3+Ka^2)>0$ in ${\mathcal U}\cap \{t>0\}$. Since $3+Ka^2>\lambda _2$ which proves the right inequality of (2.6). $\square$

2.2 Behavior of eigenvectors

If we write $n_{ij}$ for the (i, j)-cofactor of $\lambda _kI-S$ then $^t(n_{j1},n_{j2},n_{j3})$ is, if non-trivial, an eigenvector corresponding to $\lambda _k$. We take $k=1$, $j=3$ and hence

$$\begin{aligned} \begin{bmatrix}a(2\,a-\lambda _1)\\ 3\,b(\lambda _1-3)\\ \left( \lambda _1-3\right) \left( \lambda _1-2\,a\right) \end{bmatrix}=\begin{bmatrix}\ell _{11}\\ \ell _{21}\\ \ell _{31} \end{bmatrix} \end{aligned}$$

is an eigenvector corresponding to $\lambda _1$ and therefore

$$\begin{aligned} \mathbf{t}_1=\begin{bmatrix}t_{11}\\ t_{21}\\ t_{31} \end{bmatrix}=\frac{1}{d_1}\begin{bmatrix}\ell _{11}\\ \ell _{21}\\ \ell _{31} \end{bmatrix},\quad d_1=\sqrt{\ell _{11}^2+\ell _{21}^2+\ell _{31}^2} \end{aligned}$$

is a normalized eigenvector corresponding to $\lambda _1$. Thanks to Proposition 2.1 and $b=O(a^{3/2})$ it is clear that there is $C>0$ such that

$$\begin{aligned} a/C\le d_1\le C\,a,\quad \text {in}\quad {\mathcal U}\cap \left\{ t>0\right\} . \end{aligned}$$

(2.9)

Similarly choosing $k=2, j=2$ and $k=3, j=1$

$$\begin{aligned} \begin{bmatrix}-3\,ab\\ \left( \lambda _2-3\right) \left( \lambda _2-a^2\right) -a^2\\ 3\,b\left( \lambda _2-3\right) \end{bmatrix}=\begin{bmatrix}\ell _{12}\\ \ell _{22}\\ \ell _{32} \end{bmatrix},\quad \begin{bmatrix}\left( \lambda _3-2a\right) \left( \lambda _3-a^2\right) -9b^2\\ -3ab\\ -a(\lambda _3-2a) \end{bmatrix}=\begin{bmatrix}\ell _{13}\\ \ell _{23}\\ \ell _{33} \end{bmatrix} \end{aligned}$$

are eigenvectors corresponding to $\lambda _2$ and $\lambda _3$ respectively and

$$\begin{aligned} \mathbf{t}_j=\begin{bmatrix}t_{1j}\\ t_{2j}\\ t_{3j} \end{bmatrix}=\frac{1}{d_j}\begin{bmatrix}\ell _{1j}\\ \ell _{2j}\\ \ell _{3j} \end{bmatrix},\quad d_j=\sqrt{\ell _{1j}^2+\ell _{2j}^2+\ell _{3j}^2} \end{aligned}$$

are normalized eigenvectors corresponding to $\lambda _j$, $j=2,3$. Thanks to Proposition 2.1 there is $C>0$ such that

$$\begin{aligned} a/C\le d_2\le C\,a,\quad 1/C\le d_3\le C. \end{aligned}$$

(2.10)

Denote $T=(\mathbf{t}_1,\mathbf{t}_2, \mathbf{t}_3)=(t_{ij})$ then T is an orthogonal matrix, ${^t}TT=I$, smooth in $(t,X)\in {\mathcal U}\cap \{t>0\}$ which diagonalizes S;

$$\begin{aligned} \Lambda =T^{-1}ST={^t}TST=\begin{bmatrix}\lambda _1&{}0&{}0\\ 0&{}\lambda _2&{}0\\ 0&{}0&{}\lambda _3 \end{bmatrix}. \end{aligned}$$

Note that $\Lambda$ symmetrizes $A^T=T^{-1}AT$;

$$\begin{aligned} ^{t}\left( \Lambda A^T\right) =^{t}\!\!\left( \,^{t}TSAT\right) =^{t}\!\!T\,^{t}\!(SA)T={^t}TSAT=\Lambda A^T. \end{aligned}$$

Denote $A^T=({\tilde{a}}_{ij})$. Since $\Lambda A^T$ is symmetric ${\tilde{a}}_{ij}$ satisfies

$$\begin{aligned} {\tilde{a}}_{21}=\frac{\lambda _1{\tilde{a}}_{12}}{\lambda _2},\quad {\tilde{a}}_{31}=\frac{\lambda _1{\tilde{a}}_{13}}{\lambda _3},\quad {\tilde{a}}_{32}=\frac{\lambda _2{\tilde{a}}_{23}}{\lambda _3} \end{aligned}$$

(2.11)

which shows, in particular, that ${\tilde{a}}_{21}=O(a){\tilde{a}}_{21}$, ${\tilde{a}}_{31}=O(a^2){\tilde{a}}_{13}$ and ${\tilde{a}}_{32}=O(a){\tilde{a}}_{23}$. Finally in view of (2.9), (2.10) and Proposition 2.1 it is easy to check that

$$\begin{aligned} T=\left( \mathbf{t}_1,\mathbf{t}_2,\mathbf{t}_3\right) =\begin{bmatrix}O(a)&{}O\left( a^{3/2}\right) &{}O(1)\\ O\left( \sqrt{a}\right) &{}O(1)&{}O(a^{5/2})\\ O(1)&{}O\left( \sqrt{a}\right) &{}O(a) \end{bmatrix} \end{aligned}$$

(2.12)

near $(t,X)=(0,{\bar{X}})$.

2.3 Smoothness of eigenvalues

First recall [29, Lemma 3.2]

Lemma 2.1

Assume (2.1). Then

$$\begin{aligned} |\partial _X^{\alpha }a|\preceq \sqrt{a},\quad |\partial _X^{\alpha }b|\preceq a,\quad |\partial _tb|\preceq \sqrt{a} \end{aligned}$$

for $|\alpha |=1$ and $(t,X)\in (0,T)\times W$.

We show

Lemma 2.2

For $|\alpha |=1$ one has

$$\begin{aligned} |\partial _X^{\alpha }\lambda _1|\preceq a^{3/2},\quad |\partial _X^{\alpha }\lambda _2|\preceq \sqrt{a},\quad |\partial _X^{\alpha }\lambda _1|\preceq \sqrt{a}. \end{aligned}$$

(2.13)

Proof

Since

$$\begin{aligned} \partial _X^{\alpha }q(\lambda )=-\partial _X^{\alpha }\left( 2a+a^2\right) \lambda ^2+\partial _X^{\alpha } \left( 6a+2a^2+2a^3-9b^2\right) \lambda -\partial _X^{\alpha }\left( 4a^3-27b^2\right) \end{aligned}$$

it follows from Lemma 2.1 that

$$\begin{aligned} \big |\partial _X^{\alpha }q\left( \lambda _j\right) \big |&\preceq |\partial _X^{\alpha }a|\lambda _j^2+\left( |\partial _X^{\alpha }a|+|b||\partial _X^{\alpha }b|\right) \lambda _j+\left( a^2|\partial _X^{\alpha }a|+|b||\partial _X^{\alpha }b|\right) \\&\preceq |\partial _X^{\alpha }a|\lambda _j+a^{5/2}\preceq \sqrt{a}\,\lambda _j+a^{5/2}. \end{aligned}$$

From $q_{\lambda }(\lambda _j)\partial _X^{\alpha }\lambda _j+\partial _X^{\alpha }q(\lambda _j)=0$ one has

$$\begin{aligned} |\partial _X^{\alpha }\lambda _j|\preceq \frac{\sqrt{a}\,\lambda _j+a^{5/2}}{|q_{\lambda }\left( \lambda _j\right) |}. \end{aligned}$$

Noting $q_{\lambda }(\lambda _j)=\prod _{k\ne j}(\lambda _j-\lambda _k)$ one sees

$$\begin{aligned} q_{\lambda }\left( \lambda _j\right) \simeq a \;\;\text {for}\;\;j=1,2,\quad q_{\lambda }\left( \lambda _3\right) \simeq 1 \end{aligned}$$

(2.14)

thanks to Proposition 2.1 and hence the assertion. $\square$

Next estimate $\partial _t\lambda _j$.

Lemma 2.3

Assuming (2.1) one has

$$\begin{aligned} |\partial _t\lambda _1|\preceq a,\quad |\partial _t\lambda _2|\preceq 1,\quad |\partial _t\lambda _3|\preceq 1. \end{aligned}$$

(2.15)

Proof

Repeating the same arguments in the proof of Lemma 2.2 one has

$$\begin{aligned} |\partial _tq\left( \lambda _j\right) |\preceq |\partial _ta|\lambda _j+a^2|\partial _ta|+|b||\partial _tb|\preceq \lambda _j+a^2 \end{aligned}$$

which proves the assertion. $\square$

3 How to apply diagonal symmetrizers

Taking

$$\begin{aligned} Pu=\partial _t^3u-a(t,x)\partial _x^2\partial _tu-b(t,x)\partial _x^3u \end{aligned}$$

with one space variable $x\in {\mathbb {R}}$, we explain how to apply diagonal symmetrizers constructed in preceding sections. Assume that

$$\begin{aligned} \Delta (t,x)=4\,a(t,x)^3-27\,b(t,x)^2\ge 0\quad (t,x)\in [0,T)\times W \end{aligned}$$

(3.1)

and $a(0,0)=0$ such that $p(\tau ,0,0,1)=0$ has the triple root $\tau =0$ where W is an open interval containing the origin. In what follows we work in a region where $a(t,x)>0$. With $U=(\partial _t^2u,\partial _x\partial _tu,\partial _x^2u)$ the equation $Pu=f$ is reduced to

$$\begin{aligned} \partial _tU=A(t,x)\partial _xU+F,\quad A= \begin{bmatrix} 0&{}a&{}b\\ 1&{}0&{}0\\ 0&{}1&{}0\\ \end{bmatrix},\quad F=\begin{bmatrix}f\\ 0\\ 0\\ \end{bmatrix}. \end{aligned}$$

(3.2)

Then S given by (2.3) symmetrizes A, and T given by (2.12) diagonalizes S. So we set $V=T^{-1}U$ and rewrite the equation (3.2) to

$$\begin{aligned} \partial _tV=A^T\partial _xV+\left( \left( \partial _tT^{-1}\right) T-A^T\left( \partial _xT^{-1}\right) T\right) V+T^{-1}F \end{aligned}$$

(3.3)

where $A^T=T^{-1}AT$. To simplify notation let us write (3.3) with $f=0$ as

$$\begin{aligned} \partial _tV={\mathcal A}\partial _xV+{\mathcal B}V \end{aligned}$$

with ${\mathcal A}=A^T$ and ${\mathcal B}=(\partial _tT^{-1})T-{\mathcal A}(\partial _xT^{-1})T$.

3.1 Energy with scalar weight

Consider an energy with a scalar weight $\phi (t,x)>0$ with $\partial _t\phi =1$ and $|\partial _x\phi |\preceq 1$;

$$\begin{aligned} \left( \phi ^{-N} \Lambda V,V\right) =\int \phi ^{-N}\langle {\Lambda V,V}\rangle dx=\sum _{j=1}^3\int \phi ^{-N}\lambda _j |V_j|^2\,dx \end{aligned}$$

where $\langle {V,W}\rangle$ stands for the inner product in ${\mathbb {C}}^3$ and $N>0$ is a positive parameter. In what follows we assume that V(t, x) has small support in x. Note that

$$\begin{aligned} \frac{d}{dt}\,\left( \phi ^{-N} \Lambda V,V\right) =&-N\left( \phi ^{-N-1}\Lambda V,V\right) +\left( \phi ^{-N}\left( \partial _t\Lambda \right) V,V\right) \\&+2\,{\mathsf {Re}}\,\left( \phi ^{-N}\Lambda \left( {\mathcal A}\partial _xV+{\mathcal B}V\right) ,V\right) . \end{aligned}$$

Since $\Lambda {\mathcal A}$ is symmetric and hence

$$\begin{aligned} 2\,{\mathsf {Re}}\,\left( \phi ^{-N}\Lambda {\mathcal A}\partial _x V,V\right) =N\left( \phi ^{-N-1}\left( \partial _x\phi \right) \Lambda {\mathcal A}V,V\right) -\left( \phi ^{-N}\partial _x\left( \Lambda {\mathcal A}\right) V,V\right) . \end{aligned}$$

(3.4)

As for a scalar weight $\phi$ we assume

$$\begin{aligned} \phi ^2\,a\preceq \Delta ,\qquad \phi \big |\partial _t\Delta \big |\preceq \Delta ,\qquad \phi \big |\partial _ta\big |\preceq a. \end{aligned}$$

(3.5)

Lemma 3.1

The assumption (3.5) implies

$$\begin{aligned} \phi ^2\preceq \lambda _1,\qquad \phi \big |\partial _t\lambda _1\big |\preceq \lambda _1,\qquad \phi \big |\partial _t\lambda _2\big |\preceq \lambda _2. \end{aligned}$$

(3.6)

Proof

In view of Proposition 2.1 the assertion $\phi ^2\preceq \lambda _1$ is clear. Note that from $|\partial _tq(\lambda _i)|\preceq (|\partial _ta|+|b||\partial _tb|)\lambda _i+|\partial _t\Delta |$ and Lemma 2.1 it follows that

$$\begin{aligned} \big |\partial _t\lambda _i\big |\preceq \frac{\left( |\partial _ta|+a^2\right) \lambda _i+|\partial _t\Delta |}{|q_{\lambda }\left( \lambda _i\right) |}. \end{aligned}$$

Taking (2.14) into account one has

$$\begin{aligned} |\partial _t\lambda _i|\preceq \frac{|\partial _ta|}{a}\lambda _i+a\,\lambda _i+\frac{|\partial _t\Delta |}{a},\quad i=1,2 \end{aligned}$$

which implies $\phi |\partial _t\lambda _1|\preceq \lambda _1$ thanks to (3.5) and (2.4). As for $\lambda _2$ noting that $|\partial _t\Delta |\preceq a^2$ by Lemma 2.1 the assertion follows immediately from (3.5) and (2.5). $\square$

3.2 Estimate of energy, terms $\langle {(\partial _t\Lambda )V,V}\rangle$, $\langle {(\partial _x\phi )\Lambda {\mathcal A}V,V}\rangle$

Thanks to (3.6), $\big |\phi ^{-N}\langle {(\partial _t\Lambda )V,V}\rangle \big |$ is bounded by $N\phi ^{-N-1}\langle {\Lambda V,V}\rangle$ taking N large. On the other hand, from Lemma 3.2 below it follows that

$$\begin{aligned} \Lambda {\mathcal A}=\begin{bmatrix} O\left( \lambda _1\sqrt{a}\right) &{}O\left( \lambda _1\right) &{}O\left( \lambda _1\sqrt{a}\right) \\ O\left( a^2\right) &{}O\left( a^{3/2}\right) &{}O(a)\\ O\left( a^{3/2}\right) &{}O(a)&{}O\left( a^{5/2}\right) \\ \end{bmatrix}. \end{aligned}$$

Recalling that $\Lambda {\mathcal A}$ is symmetric it is clear that $\big |\langle {\Lambda {\mathcal A}V,V}\rangle \big |$ is bounded by

$$\begin{aligned} \sqrt{a}\left( \lambda _1|V_1|^2+a|V_2|^2+a^2|V_3|^2\right) +\lambda _1\,|V_1|\,|V_2| +\lambda _1\sqrt{a}\,|V_1|\,|V_3| +a|V_2||V_3|. \end{aligned}$$

Since $\lambda _1\preceq \sqrt{\lambda _1}\,a$ and hence $\lambda _1|V_1|\,|V_2|\preceq \sqrt{a}\,(\lambda _1|V_1|^2+a\,|V_2|^2)$ it follows that

$$\begin{aligned} \big |N\phi ^{-N-1}\left( \partial _x\phi \right) \langle {\Lambda {\mathcal A}V,V}\rangle \big |\le CN\sqrt{a}\,|\partial _x\phi |\,\sum _{j=1}^3\phi ^{-N-1}\lambda _j|V_j|^2 \end{aligned}$$

(3.7)

with some $C>0$. In a small neighborhood of (0, 0) where a is enough small one can bound the right-hand side by $N\phi ^{-N-1}\langle {\Lambda V,V}\rangle$.

3.3 Estimate of energy, term $\langle {\Lambda {\mathcal B}V,V}\rangle$

Recall

$$\begin{aligned} \langle {\Lambda {\mathcal B}V,V}\rangle =\langle {\Lambda \left( \partial _tT^{-1}\right) TV,V}\rangle -\langle {\Lambda {\mathcal A}\left( \partial _xT^{-1}\right) TV, V}\rangle . \end{aligned}$$

Applying Lemmas 2.2 and 2.3 we estimate $(\partial _tT^{-1})T$ and $(\partial _xT^{-1})T$. First note that

$$\begin{aligned} \left( \partial _tT^{-1}\right) T=\left( \partial _t\left( {}^{t}T\right) \right) T=\left( \langle {\partial _t \mathbf{t}_i,\mathbf{t}_j}\rangle \right) \end{aligned}$$

and $\langle {\partial _t \mathbf{t}_i,\mathbf{t}_j}\rangle =-\langle {\mathbf{t}_i, \partial _t\mathbf{t}_j}\rangle =-\langle { \partial _t \mathbf{t}_j,\mathbf{t}_i}\rangle$ so that $(\partial _tT^{-1})T$ is antisymmetric. Note that

$$\begin{aligned} \langle {\partial _t \mathbf{t}_i,\mathbf{t}_j}\rangle =\frac{1}{d_id_j}\sum _{k=1}^3\partial _t\ell _{ki}\,\cdot {\bar{\ell }}_{kj}=\frac{1}{d_i}\sum _{k=1}^3\partial _t\ell _{ki}\,\cdot {\bar{t}}_{kj} \end{aligned}$$

(3.8)

because $\sum _{k=1}^3\ell _{ki}\,{\bar{\ell }}_{kj}=0$ if $i\ne j$. Thanks to Proposition 2.1 and Lemmas 2.3, 2.1 it follows that

$$\begin{aligned} \begin{aligned} |\ell _{11}|&\preceq a^2,\quad |\ell _{21}|\preceq a^{3/2},\quad |\ell _{31}|\preceq a,\\ |\ell _{12}|&\preceq a^{5/2},\quad |\ell _{22}|\preceq a,\quad |\ell _{32}|\preceq a^{3/2},\\ |\ell _{13}|&\preceq 1,\quad |\ell _{23}|\preceq a^{5/2},\quad |\ell _{33}|\preceq a \end{aligned} \end{aligned}$$

(3.9)

and that

$$\begin{aligned} \begin{aligned} |\partial _t\ell _{11}|&\preceq a,\quad |\partial _t\ell _{21}|\preceq \sqrt{a},\quad |\partial _t\ell _{31}|\preceq 1,\\ |\partial _t\ell _{12}|&\preceq a^{3/2},\quad |\partial _t\ell _{22}|\preceq 1,\quad |\partial _t\ell _{32}|\preceq \sqrt{a},\\ |\partial _t\ell _{13}|&\preceq 1,\quad |\partial _t\ell _{23}|\preceq a^{3/2},\quad |\partial _t\ell _{33}|\preceq 1. \end{aligned} \end{aligned}$$

(3.10)

Therefore taking (2.12), (3.8) and (3.10) into account one obtains

$$\begin{aligned} \left( \partial _tT^{-1}\right) T=\begin{bmatrix}0&{}O\left( 1/\sqrt{a}\right) &{}O(1)\\ O\left( 1/\sqrt{a}\right) &{}0&{}O\left( \sqrt{a}\right) \\ O(1)&{}O(\sqrt{a})&{}0\\ \end{bmatrix}. \end{aligned}$$

(3.11)

In order to estimate $\big |\phi ^{-N}\langle {\Lambda (\partial _tT^{-1})TV,V}\rangle \big |$, noting $\Lambda \simeq \mathrm{diag}(\lambda _1,a,1)$ and $\lambda _1\preceq a^2$, it suffices to estimate

$$\begin{aligned} \sqrt{a}\,|V_1|| V_2|+|V_1||V_3|+\sqrt{a}\,|V_2||V_3|. \end{aligned}$$

Note that

$$\begin{aligned} \sqrt{a}\,|V_1||V_2|\preceq \phi ^{-1}\lambda _1 |V_1|^2+a\frac{\phi }{\lambda _1} |V_2|^2 \preceq \phi ^{-1}\left( \lambda _1 |V_1|^2+a |V_2|^2\right) \end{aligned}$$

because $\phi /\lambda _1\preceq 1/\phi$ by (3.6). As for $|V_1||V_3|$ one has

$$\begin{aligned} |V_1||V_3|\preceq \phi ^{-1}\lambda _1|V_1|^2+\left( \phi /\lambda _1\right) |V_3|^2 \preceq \phi ^{-1}\left( \lambda _1|V_1|^2+|V_3|^2\right) . \end{aligned}$$

Finally since $\sqrt{a}|V_2||V_3|\preceq a |V_2|^2+|V_3|^2$ one concludes that $\big |\phi ^{-N}\langle {\Lambda (\partial _tT^{-1})TV,V}\rangle \big |$ is bounded by $N\phi ^{-N-1}\langle {\Lambda V,V}\rangle$ taking N large.

Turn to $\big |\phi ^{-N}\langle {\Lambda {\mathcal A}(\partial _xT^{-1})TV, V}\rangle \big |$. From Proposition 2.1 and Lemmas 2.2, 2.1 one has

$$\begin{aligned} \begin{aligned} |\partial _x\ell _{11}|&\preceq a^{3/2},\quad |\partial _x\ell _{21}|\preceq a,\quad |\partial _x\ell _{31}|\preceq \sqrt{a},\\ |\partial _x\ell _{12}|&\preceq a^{2},\quad |\partial _x\ell _{22}|\preceq \sqrt{a},\quad |\partial _x\ell _{32}|\preceq a,\\ |\partial _x\ell _{13}|&\preceq 1,\quad |\partial _x\ell _{23}|\preceq a^{2},\quad |\partial _x\ell _{33}|\preceq \sqrt{a} \end{aligned} \end{aligned}$$

(3.12)

from which one concludes

$$\begin{aligned} \partial _xT^{-1}={^t}\left( \partial _xT\right) =\begin{bmatrix} O\left( \sqrt{a}\right) &{}O(1)&{}O\left( 1/\sqrt{a}\right) \\ O(a)&{}O\left( 1/\sqrt{a}\right) &{}O(1)\\ O(1)&{}O(a^2)&{}O\left( \sqrt{a}\right) \\ \end{bmatrix} \end{aligned}$$

(3.13)

and hence

$$\begin{aligned} \left( \partial _xT^{-1}\right) T=\begin{bmatrix}0&{}O(1)&{}O(\sqrt{a})\\ O(1)&{}0&{}O(a)\\ O(\sqrt{a})&{}O(a)&{}0\\ \end{bmatrix}. \end{aligned}$$

(3.14)

Here note that

Lemma 3.2

One has

$$\begin{aligned} {\mathcal A}=\begin{bmatrix} O\left( \sqrt{a}\right) &{}O(1)&{}O(\sqrt{a})\\ O(a)&{}O(\sqrt{a})&{}O(1)\\ O\left( a^{5/2}\right) &{}O(a)&{}O\left( a^{5/2}\right) \\ \end{bmatrix},\;\; \partial _x{\mathcal A}=\begin{bmatrix} O(1)&{}O(1/\sqrt{a})&{}O(1)\\ O(\sqrt{a})&{}O(1)&{}O(1/\sqrt{a})\\ O(a)&{}O(\sqrt{a})&{}O(\sqrt{a})\\ \end{bmatrix}. \end{aligned}$$

Proof

With ${\mathcal A}=({\tilde{a}}_{ij})$ it is clear that

$$\begin{aligned} {\tilde{a}}_{ij}=t_{1i}\,a\,t_{2j}+t_{1i}\,b\,t_{3j}+t_{2i}t_{1j}+t_{3i}t_{2j}. \end{aligned}$$

(3.15)

Since $t_{2i}t_{1j}=O(\sqrt{a})$ unless $(i,j)=(2,3)$ and $t_{3i}t_{2j}=O(\sqrt{a})$ unless $(i,j)=(1,2)$ the first assertion follows from (2.12), (3.15) and (2.11). Note that $\partial _x\big (1/d_j\big )=O(1/a^{3/2})$ for $j=1,2$ and $\partial _x\big (1/d_3\big )=O(1)$ then the second assertion follows from (3.13), (3.15) and (2.12). $\square$

From Lemma 3.2 and (3.14) one has

$$\begin{aligned} {\mathcal A}\left( \partial _xT^{-1}\right) T=\begin{bmatrix}O(1)&{}O(\sqrt{a})&{}O(a)\\ O\left( \sqrt{a}\right) &{}O(a)&{}O\left( a^{3/2}\right) \\ O(a)&{}O\left( a^{5/2}\right) &{}O(a^2)\\ \end{bmatrix}. \end{aligned}$$

Then it is clear that $\big |\langle {\Lambda {\mathcal A}(\partial _xT^{-1})TV, V}\rangle \big |$ is bounded by

$$\begin{aligned} a^{3/2}|V_1||V_2|+a|V_1||V_3| +a^{3/2}\, |V_2||V_3|+\lambda _1|V_1|^2+a^2\left( |V_2|^2+|V_3|^2\right) \end{aligned}$$

where

$$\begin{aligned} a^{3/2}|V_1||V_2|\preceq & {} a\,\phi ^{-1}\lambda _1|V_1|^2+\frac{a^2\phi }{\lambda _1}|V_2|^2\preceq a\,\phi ^{-1}\left( \lambda _1|V_1|^2+a|V_2|^2\right) ,\\ a\,|V_1||V_3|\preceq & {} a\,\phi ^{-1}\lambda _1|V_1|^2+a\,\frac{\phi }{\lambda _1}|V_3|^2\preceq a\,\phi ^{-1}\left( \lambda _1|V_1|^2+|V_3|^2\right) ,\\ a^{3/2}\,|V_2||V_3|\preceq & {} a\,\left( a|V_2|^2+|V_3|^2\right) \end{aligned}$$

hence $\big |\phi ^{-N}\langle {\Lambda {\mathcal A}(\partial _xT^{-1})TV, V}\rangle \big |$ is bounded by $N\phi ^{-N-1}\langle {\Lambda V,V}\rangle$ taking N large.

3.4 Estimate of energy, term $\langle {\partial _x(\Lambda {\mathcal A})V,V}\rangle$

Write

$$\begin{aligned} \langle {\partial _x\left( \Lambda {\mathcal A}\right) V,V}\rangle =\langle {\left( \partial _x\Lambda \right) {\mathcal A}V,V}\rangle +\langle {\Lambda \left( \partial _x{\mathcal A}\right) V,V}\rangle \end{aligned}$$

and estimate each term on the right-hand side. To estimate the first term note

Lemma 3.3

One has

$$\begin{aligned} \big |\partial _x\lambda _1\big |/\lambda _1\preceq 1/\sqrt{a}+1/\sqrt{\Delta }\preceq 1/\left( \phi \sqrt{a}\right) . \end{aligned}$$

(3.16)

Proof

Recall

$$\begin{aligned} \partial _x\lambda _1=\left\{ \partial _x\left( 2a+a^2\right) \lambda _1^2-\partial _x\left( 6a+2a^2+2a^3-9b^2\right) \lambda _1+\partial _x\Delta \right\} /q_{\lambda } \left( \lambda _1\right) \end{aligned}$$

where Lemma 2.1 shows that

$$\begin{aligned} \frac{|\partial _x\lambda _1|}{\lambda _1}\preceq \frac{\sqrt{a}\,\lambda _1^2+\sqrt{a}\,\lambda _1}{\lambda _1\,a}+\frac{|\partial _x\Delta |}{\lambda _1\,a}\preceq \frac{1}{\sqrt{a}}+\frac{|\partial _x\Delta |}{\lambda _1\,a}\preceq \frac{1}{\sqrt{a}}+\frac{|\partial _x\Delta |}{\Delta } \end{aligned}$$

because $1/\lambda _1\preceq a/\Delta$ by Proposition 2.1. Noting that $\Delta \ge 0$ in a neighborhood of $x=0$ we see that $|\partial _x\Delta |\preceq \sqrt{\Delta }$, hence the first inequality. The second inequality follows from the first one thanks to the assumption (3.5). $\square$

Since $|\partial _x\lambda _2|/\lambda _2=O(1/\sqrt{a})$ and $|\partial _x\lambda _3|/\lambda _3=O(1)$ by Lemma 2.2 it follows from 3.3 that $\phi (\partial _x\Lambda )=\phi \Lambda (\Lambda ^{-1}\partial _x\Lambda )=\mathrm{diag}\big (O(\lambda _1/\sqrt{a}),O(\phi \sqrt{a}),O(\phi )\big )$ for $\lambda _2\simeq a$ then by Lemma 3.2 and (3.13) one sees

$$\begin{aligned} \phi \left( \partial _x\Lambda \right) {\mathcal A}=\begin{bmatrix} O\left( \lambda _1\right) &{}O\left( \lambda _1/\sqrt{a}\right) &{}O\left( \lambda _1\right) \\ O\left( \phi \,a^{3/2}\right) &{}O\left( \phi \,a\right) &{}O\left( \phi \,\sqrt{a}\right) \\ O\left( \phi \, a^{5/2}\right) &{}O\left( \phi \, a\right) &{}O\left( \phi \, a^{5/2}\right) \end{bmatrix}. \end{aligned}$$

Therefore to estimate $\phi \langle {(\partial _x\Lambda ){\mathcal A}V,V}\rangle$ it suffices to bound

$$\begin{aligned} \left( \frac{\lambda _1}{\sqrt{a}}+a^{3/2}\phi \right) |V_1||V_2|+\left( \lambda _1+a^{5/2}\phi \right) |V_1||V_3|+\phi \sqrt{a}|V_2||V_3|. \end{aligned}$$

Since $\lambda _1\preceq a\sqrt{\lambda _1}$ and $\phi \preceq \sqrt{\lambda _1}$ this is bounded by $\langle {\Lambda V,V}\rangle$ and hence

$$\begin{aligned} \big |\phi ^{-N}\langle {\left( \Lambda ^{-1}\partial _x\Lambda \right) {\mathcal A}V,\Lambda V}\rangle \big |\preceq \phi ^{-N-1}\langle {\Lambda V,V}\rangle . \end{aligned}$$

(3.17)

As for $\big |\phi ^{-N}\langle {(\partial _x{\mathcal A})V,\Lambda V}\rangle \big |$ it follows from Lemma 3.2 that

$$\begin{aligned} \Lambda \left( \partial _x{\mathcal A}\right) =\begin{bmatrix} O\left( \lambda _1\right) &{}O\left( \lambda _1/\sqrt{a}\right) &{}O\left( \lambda _1\right) \\ O\left( a^{3/2}\right) &{}O(a)&{}O(\sqrt{a})\\ O(a)&{}O(\sqrt{a})&{}O(\sqrt{a}) \end{bmatrix} \end{aligned}$$

hence repeating similar arguments it is easy to see that

$$\begin{aligned} \big |\phi ^{-N}\langle {\Lambda \left( \partial _x{\mathcal A}\right) V,V}\rangle \big |\preceq \phi ^{-N-1}\langle {\Lambda V,V}\rangle \end{aligned}$$

(3.18)

which is bounded by $N\phi ^{-N-1}\langle {\Lambda V,V}\rangle$ taking N large.

Remark 3.1

It shoud be remarked that the condition (3.5), assumed in this section, are stated in terms of $\Delta$ and a, where a is constant times the discriminant of $\partial p/\partial \tau$, without any reference to characteristic roots $\tau _i$.

We conclude this section with an important remark. To obtain energy estimates it suffices to find a finite number of pairs $(\phi _j,\omega _j)$, where $\phi _j$ is a scalar weight satisfying (3.5) in subregion $\omega _j$ of which union covers a neighborhood of (0, 0), such that one can collect such estimates obtained in $\omega _j$. In the following sections this observation is carried out for hyperbolic operators with effectively hyperbolic critical points with coefficients depending on t, also for third order hyperbolic operators with effectively hyperbolic critical points with two independent variables.

4 Hyperbolic operators with effectively hyperbolic critical points with time dependent coefficients

In [9], Ivrii and Petkov proved that if the Cauchy problem (1.1) is $C^{\infty }$ well posed for any lower order term then the Hamilton map $F_p$ has a pair of non-zero real eigenvalues at every critical point ([9, Theorem 3]). Here the Hamilton map $F_p$ is defined by

$$\begin{aligned} F_p(X,\Xi )=\begin{bmatrix} \displaystyle {\frac{\partial ^2 p}{\partial X\partial \Xi }}&{}\displaystyle {\frac{\partial ^2 p}{\partial \Xi \partial \Xi }}\\ \displaystyle {-\frac{\partial ^2 p}{\partial X\partial X}}&{}\displaystyle {-\frac{\partial ^2 p}{\partial \Xi \partial X}} \end{bmatrix},\quad X=(t,x),\;\;\Xi =(\tau ,\xi ) \end{aligned}$$

and a critical point $(X,\Xi )$ is the point where $\partial p/\partial X=\partial p/\partial \Xi =0$. Note that $p(X,\Xi )=0$ at critical points by the homogeneity in $\Xi$ so that $(X,\Xi )$ is a multiple characteristic and $\tau$ is a multiple characteristic root. A critical point where the Hamilton map $F_p$ has a pair of non-zero real eigenvalues is called effectively hyperbolic ([7]). In [10], Ivrii has proved that if every critical point is effectively hyperbolic, and p admits a decomposition $p=q_1q_2$ with real smooth symbols $q_i$ near the critical point then the Cauchy problem is $C^{\infty }$ well-posed for every lower order term. In this case the critical point is effectively hyperbolic if and only if the Poisson bracket $\{q_1,q_2\}$ does not vanish. He has conjectured that the assertion would hold without any additional condition.

If a critical point $(X,\Xi )$ is effectively hyperbolic then $\tau$ is a characteristic root of multiplicity at most 3 ([9, Lemma 8.1]). If every multiple characteristic root is at most double, the conjecture has been proved in [10,11,12,13, 21, 25, 28]. When there exists an effectively hyperbolic critical point $(X,\Xi )$ such that $\tau$ is a triple characteristic root, the conjecture is not yet proved in general. For more details about the subsequent progress on proving the conjecture, see [2, 3, 26, 29].

If $(t, x,\tau ,\xi )$ with $t>0$ is effectively hyperbolic critical point then $\tau$ must be a double characteristic root ([9, Lemma 8.1]). Therefore a main concern is double or triple characteristic roots at $(0, x, \xi )$. If $\tau$ is a triple characteristic root at $(0, x, \xi )$ then $(0, x,\tau , \xi )$ is always a critical point. On the other hand, for double characteristic root $\tau$, the point $(0, x, \tau , \xi )$ is not necessarily a critical point. Here is a simple example

$$\begin{aligned} P=\left( D_t^2-t^{\ell }D_x^2\right) \left( D_t+c\,D_x\right) ,\qquad \ell \in {\mathbb {N}},\quad x\in {\mathbb {R}},\;\;t\ge 0 \end{aligned}$$

where $c\in {\mathbb {R}}$. Let $c\ne 0$ then it is clear that $\tau =0$ is a double characteristic root at (0, 0, 1). If $\ell =1$ then $\partial _tp(0,0,0,1)=-c\ne 0$ and hence (0, 0, 0, 1) is not a critical point. If $\ell \ge 2$ then (0, 0, 0, 1) is a critical point and $F_p$ has non-zero real eigenvalues there if and only if $\ell =2$. Let $c=0$ then $\tau =0$ is a triple characteristic root at (0, 0, 1) hence (0, 0, 0, 1) is a critical point and $F_p$ has non-zero real eigenvalues there if and only if $\ell = 1$.

Now we restrict ourselves to the case that the coefficients depend only on t and consider the Cauchy problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle {Pu=D_t^mu+\sum \nolimits _{j=0}^{m-1}\sum \nolimits _{j+|\alpha |\le m}a_{j,\alpha }(t)D_x^{\alpha }D_t^ju=0},\;(t,x)\in [0,T)\times {\mathbb {R}}^n\\ D_t^ku(0,x)=u_k(x),\;\;x\in {\mathbb {R}}^n,\quad k=0,\ldots ,m-1 \end{array}\right. } \end{aligned}$$

(4.1)

where $a_{j,\alpha }(t)$ ($j+|\alpha |=m$) are real valued and $C^{\infty }$ in $(-c,T)$ with some $c>0$ and the principal symbol p is hyperbolic for $t\ge 0$, that is

$$\begin{aligned} p(t,\tau ,\xi )=\tau ^m+\sum _{j=0}^{m-1}\sum _{j+|\alpha |=m}a_{j,\alpha }(t)\xi ^{\alpha }\tau ^j \end{aligned}$$

(4.2)

has only real roots in $\tau$ for any $(t,\xi )\in [0,T)\times {\mathbb {R}}^n$. Denoting by ${\mathcal S}({\mathbb {R}}^n)$ the set of all rapidly decreasing functions on ${\mathbb {R}}^n$ we have

Theorem 4.1

If every critical point $(0,\tau ,\xi )$, $\xi \ne 0$ is effectively hyperbolic then there exists $\delta >0$ such that for any $a_{j,\alpha }(t)$ with $j+|\alpha |\le m-1$, which are $C^{\infty }$ in a neighborhood of $[0,\delta ]$, one can find $N\in {\mathbb {N}}$, $C>0$ such that for any $u_j(x)\in {\mathcal S}({\mathbb {R}}^n)$ $(j=0,\ldots ,m-1)$ there exists a unique solution $u(t,x)\in C^{\infty }([0,\delta ];{\mathcal S}({\mathbb {R}}^n))$ to the Cauchy problem (4.1) with $T=\delta$ satisfying

$$\begin{aligned} \sum _{k=0}^{m-1}\Vert \langle {D}\rangle ^{m-1-k}D_t^ku(t,\cdot )\Vert \le C\sum _{k=0}^{m-1}\Vert \langle {D}\rangle ^{N-k}u_k\Vert \end{aligned}$$

for $0\le t\le \delta$.

Remark 4.1

Under the assumption of Theorem 4.1 it follows that p has necessarily non-real characteristic roots in the $t<0$ side near $(0,\xi )$, that is P would be a Tricomi type operator. Indeed from [9, Lemma 8.1] it follows that $F_p(0,\tau ,\xi )=O$ if all characteristic roots are real in a full neighborhood of $(0,\xi )$ and $\tau$ is a triple characteristic root.

4.1 Triple effectively hyperbolic characteristic

Assume that $p(t,\tau ,\xi )$ has a triple characteristic root ${\bar{\tau }}$ at $(0,{\bar{\xi }})$, $|{\bar{\xi }}|=1$ and $(0,{\bar{\tau }},{\bar{\xi }})$ is effectively hyperbolic. As we see later, without restrictions one may assume that $m=3$ and p has the form

$$\begin{aligned} p\left( t,\tau ,\xi \right) =\tau ^3-a(t,\xi )|\xi |^2\tau -b(t,\xi )|\xi |^3 \end{aligned}$$

(4.3)

where $a(t,\xi )$ and $b(t,\xi )$ are homogeneous of degree 0 in $\xi$ and satisfy

$$\begin{aligned} \Delta (t,\xi )=4\,a(t,\xi )^3-27\,b(t,\xi )^2\ge 0, \quad (t,\xi ) \in [0,T) \times {\mathbb {R}}^n. \end{aligned}$$

(4.4)

The triple characteristic root of $p(0,\tau ,{\bar{\xi }})=0$ is $\tau =0$ and

$$\begin{aligned} \mathrm{det}\,\left( \lambda -F_p(0,0,{\bar{\xi }})\right) =\lambda ^{2n}\left( \lambda ^2-\{\partial _ta(0,{\bar{\xi }})\}^2\right) \end{aligned}$$

hence $(0,0,{\bar{\xi }})$ is effectively hyperbolic if and only if

$$\begin{aligned} \partial _t a(0,{\bar{\xi }})\ne 0. \end{aligned}$$

(4.5)

Since $a(0,{\bar{\xi }})=0$ and $\partial _ta(0,{\bar{\xi }})\ne 0$ there is a neighborhood ${\mathcal U}$ of $(0,{\bar{\xi }})$ in which one can write

$$\begin{aligned} a(t,\xi )=e_1(t,\xi )\left( t+\alpha (\xi )\right) \end{aligned}$$

(4.6)

where $e_1>0$ in ${\mathcal U}$. Note that $\alpha (\xi )\ge 0$ near ${\bar{\xi }}$ because $a(t,\xi )\ge 0$ in $[0,T)\times {\mathbb {R}}^n$.

Lemma 4.1

There exists a neighborhood ${\mathcal U}$ of $(0,{\bar{\xi }})$ in which one can write

$$\begin{aligned} \Delta (t,\xi )=e_2(t,\xi )\left\{ t^3+a_1(\xi )t^2+a_2(\xi )t+a_3(\xi )\right\} \end{aligned}$$

where $e_2>0$ and $a_j({\bar{\xi }})=0$.

Proof

Thanks to the Malgrange preparation theorem it suffices to show

$$\begin{aligned} \partial _t^k\Delta \left( 0,{\bar{\xi }}\right) =0,\;\;k=0,1,2,\quad \partial _t^3\Delta \left( 0,{\bar{\xi }}\right) \ne 0. \end{aligned}$$

It is clear that $\partial _t^ka^3=0$ at $(0,{\bar{\xi }})$ for $k=0,1,2$ and $\partial _t^3a(0,{\bar{\xi }})\ne 0$. Since $\Delta =4a^3-27b^2$ and $b(0,{\bar{\xi }})=0$ it is enough to show $\partial _tb(0,{\bar{\xi }})=0$. Suppose $\partial _tb(0,{\bar{\xi }})\ne 0$ and hence

$$\begin{aligned} b\left( t,{\bar{\xi }}\right) =t\left( b_1+tb_2(t)\right) \end{aligned}$$

where $b_1\ne 0$. Since $a(t,{\bar{\xi }})=c\,t$ with $c>0$ then $\Delta (t,{\bar{\xi }})=4\,c^3\,t^3-27b(t,{\bar{\xi }})^2\ge 0$ is impossible. This proves the assertion. $\square$

Lemma 4.2

There exist a neighborhood U of ${\bar{\xi }}$ and a positive constant $\varepsilon >0$ such that for any $\xi \in U$ one can find $j\in \{1,2,3\}$ such that

$$\begin{aligned} \varepsilon ^{-1}\,|\nu _j(\xi )|\ge \alpha (\xi ) \end{aligned}$$

where $\nu _j(\xi )$ are roots of $t^3+a_1(\xi )t^2+a_2(\xi )t+a_3(\xi )=0$.

Proof

In view of Lemma 4.1 one can write

$$\begin{aligned} \Delta&=4a^3-27b^2=4e_1^3(t+\alpha )^3-27b^3=4e_1^3\left\{ (t+\alpha )^3-{\hat{b}}^2\right\} \\&=e_2\left\{ t^3+a_1(\xi )t^2+a_2(\xi )t+a_3(\xi )\right\} \end{aligned}$$

where ${\hat{b}}=3\sqrt{3}\,b/(2e_1^{3/2})$ and hence

$$\begin{aligned} (t+\alpha )^3-{\hat{b}}^2=E\left\{ t^3+a_1(\xi )t^2+a_2(\xi )t+a_3(\xi )\right\} \end{aligned}$$

with $E=e_2/(4e_1^3)$. Choose $I=[-\delta _1,\delta _1]$, $\delta _1>0$ and a neighborhood $U_1$ of ${\bar{\xi }}$ such that $I\times U_1\subset {\mathcal U}$ and denote

$$\begin{aligned} \sup _{(t,\xi )\in I\times U_1, 0\le k\le 2}\big |\partial _t^kE(t,\xi )\big |=C. \end{aligned}$$

Write ${\hat{b}}(t,\xi )={\hat{b}}_0(\xi )+{\hat{b}}_1(\xi )t+{\hat{b}}_2(\xi )t^2+{\hat{b}}_3(t,\xi )t^3$ and set

$$\begin{aligned} \sup _{\xi \in U_1}\alpha (\xi )+\sup _{\xi \in U_1,0\le k\le 2}|{\hat{b}}_k(\xi )|+\sup _{(t,\xi )\in I\times U_1}|{\hat{b}}_3(t,\xi )|=B. \end{aligned}$$

(4.7)

Choose a neighborhood $U\subset U_1$ of ${\bar{\xi }}$ such that

$$\begin{aligned} 4\,B\sqrt{\alpha (\xi )}<1\quad \text {for}\;\;\xi \in U \end{aligned}$$

which is possible because $\alpha ({\bar{\xi }})=0$. Choose $\varepsilon =\varepsilon (B,C)>0$ such that

$$\begin{aligned} 6\left( 1-\varepsilon C\left( 1+\varepsilon B+\varepsilon ^2B^2/6\right) \right) -\frac{9}{2}\left( 1+\varepsilon ^2C(1+\varepsilon B)\right) ^2/\left( 1-\varepsilon ^3 C\right) >1 \end{aligned}$$

(4.8)

and prove that if there exists $\xi \in U$ such that

$$\begin{aligned} \big |\nu _j(\xi )\big |<\epsilon \,\alpha (\xi ),\quad j=1,2,3 \end{aligned}$$

(4.9)

we would have a contradiction. We omit to write $\xi$ for simplicity. Recall

$$\begin{aligned} (t+\alpha )^3-{\hat{b}}^2=E\prod (t-\nu _j)\ge 0\quad (t\ge 0) \end{aligned}$$

(4.10)

and hence, taking $t=0$ one has $\alpha ^3-{\hat{b}}_0^2=E(0)|\nu _1\nu _2\nu _3|<C\varepsilon ^3\alpha ^3$. This shows

$$\begin{aligned} \sqrt{1-C\varepsilon ^3}\,\alpha ^{3/2}\le |{\hat{b}}_0|\le \alpha ^{3/2}. \end{aligned}$$

(4.11)

Differentiating (4.10) by t and putting $t=0$ one has

$$\begin{aligned} |3\alpha ^2-2{\hat{b}}_0{\hat{b}}_1|\le C\varepsilon ^3\alpha ^3+3C\varepsilon ^2\alpha ^2. \end{aligned}$$

This gives

$$\begin{aligned} 3\alpha ^2\left( 1-C\varepsilon ^2(1+\varepsilon B)\right) \le 2|{\hat{b}}_0{\hat{b}}_1|\le 3\alpha ^2\left( 1+C\varepsilon ^2(1+\varepsilon B)\right) . \end{aligned}$$

In view of (4.11) one has

$$\begin{aligned} \frac{3}{2}\alpha ^{1/2}\left( 1-C\varepsilon ^2(1+\varepsilon B)\right) \le |{\hat{b}}_1|\le \frac{3}{2}\frac{\alpha ^{1/2}}{\sqrt{1-C\varepsilon ^3}}\left( 1+C\varepsilon ^2(1+\varepsilon B)\right) . \end{aligned}$$

(4.12)

Differentiating (4.10) twice by t and putting $t=0$ on has

$$\begin{aligned} \big |6\alpha -\left( 2{\hat{b}}_1^2+4{\hat{b}}_0{\hat{b}}_2\right) \big |\le C\,\varepsilon \,\alpha \left( 6+6\varepsilon \,B+\varepsilon ^2\,B^2\right) \end{aligned}$$

which proves $\big |4{\hat{b}}_0{\hat{b}}_2+2{\hat{b}}_1^2\big |\ge 6\,\alpha \big (1-\varepsilon C-\varepsilon ^2CB-\varepsilon ^3CB^2/6\big )$. Using (4.12) one obtains

$$\begin{aligned} |4{\hat{b}}_0{\hat{b}}_2|&\ge 6\alpha \left( 1-\varepsilon C\left( 1+\varepsilon B+\varepsilon ^2B^2/6\right) \right) \\&\quad \,-\frac{9}{2}\alpha \left( 1+\varepsilon ^2C\left( 1+\varepsilon B\right) \right) ^2/\left( 1-\varepsilon ^3 C\right) \end{aligned}$$

where the right-hand side is greater than $\alpha$ by (4.8). On the other hand from (4.7) and (4.11) we have

$$\begin{aligned} 4B\,\alpha ^{3/2}\ge 4\,\alpha ^{3/2}\,|{\hat{b}}_2|\ge 4\,|{\hat{b}}_0{\hat{b}}_1|> \alpha \end{aligned}$$

and hence $4\,B\sqrt{\alpha }> 1$ which contradicts with (4.9). $\square$

Denote $\Delta /e_2$ by ${\bar{\Delta }}$;

$$\begin{aligned} {\bar{\Delta }}(t,\xi )=\Delta /e_2=t^3+a_1(\xi )t^2+a_2(\xi )t+a_3(\xi ). \end{aligned}$$

Lemma 4.3

There is a neighborhood $V\subset U$ of ${\bar{\xi }}$ where one can write either

$$\begin{aligned} {\bar{\Delta }}=\big |t-\nu _2(\xi )\big |^2\left( t-\nu _1(\xi )\right) ,\quad \nu _1(\xi )\;\;\text {is real and}\;\; \nu _1(\xi )\le 0 \end{aligned}$$

(4.13)

or

$$\begin{aligned} {\bar{\Delta }}=\prod _{k=1}^3\left( t-\nu _k(\xi )\right) ,\quad \nu _k(\xi )\;\;\text {are real and}\;\; \nu _k(\xi )\le 0. \end{aligned}$$

(4.14)

Proof

Let $\nu _j(\xi )$, $j=1,2,3$ be the roots of ${\bar{\Delta }}(t,\xi )=0$. Since $\nu _j({\bar{\xi }})=0$ one can assume $|\nu _j(\xi )|<\delta _1$ in V. Since $a_j(\xi )$ are real we have two cases; one is real and other two are complex conjugate or all three are real. For the former case denoting the real root by $\nu _1(\xi )$ we have (4.13) where $\nu _1(\xi )\le 0$ because ${\bar{\Delta }}\ge 0$ for $0\le t\le \delta _1$. In the latter case, if two of them coincide, denoting the remaining one by $\nu _1(\xi )$ one has (4.13). If $\nu _j(\xi )$ are different each other then we have (4.14) since ${\bar{\Delta }}\ge 0$ for $0\le t\le \delta _1$. $\square$

4.2 Key proposition

Thanks to Lemma 4.3 we have either (4.13) or (4.14). As was observed in [22] (see also [27]) in order to obtain energy estimates it is important to consider not only real zeros of $\Delta$ but also ${\mathsf {Re}}\,\nu _j(\xi )$, the real part of $\nu _j(\xi )$ for non-real zeros. Define

$$\begin{aligned} \psi (\xi )=&\max {\left\{ 0,{\mathsf {Re}}\,\nu _j(\xi )\right\} } =\max {\left\{ 0,{\mathsf {Re}}\,\nu _2(\xi )\right\} },\\ \phi _1=&t, \;\;\omega _1(\xi )=[0,\psi (\xi )/2],\quad \phi _2=t-\psi (\xi ),\;\;\omega _2(\xi )=\left[ \psi (\xi )/2,\delta \right] \end{aligned}$$

with small $\delta >0$. If $\psi (\xi )\le 0$, we have $\phi _1=\phi _2=t$ and $\omega _2=[0,\delta ]$. The next proposition is the key to applying the arguments in Sect. 3 to operators with triple effectively hyperbolic characteristics.

Proposition 4.1

There exist a neighborhood U of ${\bar{\xi }}$, positive constants $\delta >0$ and $C>0$ such that

$$\begin{aligned} \phi _j^2\,a\le C\Delta ,\quad |\phi _j|\,|\partial _t\Delta |\le C\Delta ,\quad |\phi _j|\le C\,a \end{aligned}$$

(4.15)

for any $\xi \in U$ and $t\in \omega _j(\xi )$, $j=1,2$.

Proof

Thanks to (4.6) and Lemma 4.1 it suffices to prove (4.15) for ${\bar{\Delta }}$ and $t+\alpha$ instead of $\Delta$ and a. Note that

$$\begin{aligned} \Big |\frac{\partial _t{\bar{\Delta }}}{\bar{\Delta }}\Big |&\le \frac{1}{t+|\nu _1(\xi )|}+2\,\frac{|t-{\mathsf {Re}}\,\nu _2(\xi )|}{|t-\nu _2(\xi )|^2}\le \frac{1}{\phi _1}+\frac{2}{|\phi _2|},\\ \Big |\frac{\partial _t{\bar{\Delta }}}{\bar{\Delta }}\Big |&\le \sum _{k=1}^3\frac{1}{t+|\nu _k(\xi )|}\le \frac{3}{\phi _1} \end{aligned}$$

in the case (4.13) and (4.14) respectively. Since $|\phi _2|\ge t=\phi _1$ in $\omega _1(\xi )$ and $t=\phi _1\ge |\phi _2|$ in $\omega _2(\xi )$ it is easy to see

$$\begin{aligned} |\phi _j|\,|\partial _t{\bar{\Delta }}|\le 3\,{\bar{\Delta }}\quad \text {in}\;\;\omega _j(\xi ) \end{aligned}$$

for both cases. Similarly noting $t+\alpha (\xi )\ge t$ it is clear that

$$\begin{aligned} |\phi _j|\le t+\alpha \quad \text {in}\;\;\omega _j(\xi ). \end{aligned}$$

Therefore it rests to prove $\phi _j^2\,(t+\alpha )\le C{\bar{\Delta }}$ in $\omega _j(\xi )$. First we study the case (4.13). From Lemma 4.2 either $\varepsilon ^{-1}|\nu _1|\ge \alpha$ or $\varepsilon ^{-1}|\nu _2|\ge \alpha$ holds. First assume that $\varepsilon ^{-1}|\nu _1|\ge \alpha$ and hence $t+|\nu _1|\ge \varepsilon (t+\alpha )$ then

$$\begin{aligned} \frac{{\bar{\Delta }}}{|\phi _j|(t+\alpha )}\ge \varepsilon \,\frac{|t-\nu _2|^2}{|\phi _j|}\ge \varepsilon \,\frac{|t-{\mathsf {Re}}\,\nu _2|^2}{|\phi _j|}=\varepsilon \,\frac{\phi _2^2}{|\phi _j|}\ge \varepsilon |\phi _j|,\quad t\in \omega _j. \end{aligned}$$

Next assume $\varepsilon ^{-1}|\nu _2|\ge \alpha$. If $0<{\mathsf {Re}}\,\nu _2\le |{\mathsf {Im}}\,\nu _2|$ one has for $t\ge 0$

$$\begin{aligned} | t-{\mathsf {Re}}\,\nu _2|+|{\mathsf {Im}}\,\nu _2|&\ge t-{\mathsf {Re}}\,\nu _2+|{\mathsf {Im}}\,\nu _2|\ge t,\\ | t-{\mathsf {Re}}\,\nu _2|+|{\mathsf {Im}}\,\nu _2|&\ge |{\mathsf {Im}}\,\nu _2|\ge |\nu _2|/2\ge \frac{\varepsilon }{2}\alpha \end{aligned}$$

which also holds for ${\mathsf {Re}}\,\nu _2\le 0$ clearly. Then we see that

$$\begin{aligned} |t-\nu _2|\ge \frac{1}{2}\left( | t-{\mathsf {Re}}\,\nu _2|+|{\mathsf {Im}}\,\nu _2|\right) \ge \frac{\varepsilon }{2\left( 2+\varepsilon \right) }\left( t+\alpha \right) . \end{aligned}$$

Therefore it follows that

$$\begin{aligned} \frac{{\bar{\Delta }}}{|\phi _j|(t+\alpha )}\ge \frac{\varepsilon }{2(2+\varepsilon )}\,\frac{t\,|t-\nu _2|}{|\phi _j|}\ge \frac{\varepsilon }{2(2+\varepsilon )}\,\frac{|\phi _1\phi _2|}{|\phi _j|}\ge \frac{\varepsilon }{2(2+\varepsilon )}|\phi _j|,\quad t\in \omega _j. \end{aligned}$$

If ${\mathsf {Re}}\,\nu _2>|{\mathsf {Im}}\,\nu _2|$ noting that, for $t\in \omega _1$

$$\begin{aligned} |t-{\mathsf {Re}}\,\nu _2|\ge {\mathsf {Re}}\,\nu _2/2\ge |\nu _2|/4\ge \varepsilon \alpha /4,\quad |t-{\mathsf {Re}}\,\nu _2|\ge t \end{aligned}$$

one has $(4+\varepsilon )|t-{\mathsf {Re}}\,\nu _2|\ge \varepsilon (t+\alpha )$ in $\omega _1$. Hence

$$\begin{aligned} \frac{{\bar{\Delta }}}{t(t+\alpha )}\ge \frac{|t-{\mathsf {Re}}\,\nu _2|^2}{t+\alpha }\ge \frac{\varepsilon }{4+\varepsilon }\,|t-{\mathsf {Re}}\,\nu _2|\ge \frac{\varepsilon }{4+\varepsilon }\,t,\quad t\in \omega _1. \end{aligned}$$

For $t\in \omega _2$ note that

$$\begin{aligned} t\ge {\mathsf {Re}}\,\nu _2/2\ge |\nu _2|/4\ge \varepsilon \alpha /4 \end{aligned}$$

and hence $(4+\varepsilon )\,t\ge \varepsilon (t+\alpha )$. Thus one has

$$\begin{aligned} \frac{{\bar{\Delta }}}{|t-{\mathsf {Re}}\,\nu _2|(t+\alpha )}\ge \frac{t\,|t-{\mathsf {Re}}\,\nu _2|}{(t+\alpha )}\ge \frac{\varepsilon }{4+\varepsilon }|t-{\mathsf {Re}}\,\nu _2|,\quad t\in \omega _2 \end{aligned}$$

which proves the assertion for the case (4.13). In the case (4.14) note that

$$\begin{aligned} {\bar{\Delta }}=\prod _{k=1}^3\left( t+|\nu _k|\right) . \end{aligned}$$

If $\varepsilon ^{-1}|\nu _j|\ge \alpha$ then $t+|\nu _j|\ge \varepsilon (t+\alpha )$ and hence it is clear that

$$\begin{aligned} {\bar{\Delta }}\ge \varepsilon \,t^2\,(t+\alpha ) \end{aligned}$$

which shows the assertion. Thus the proof of Proposition 4.1 is completed. $\square$

4.3 Energy estimates

Let P be a differential operator of order 3 with coefficients depending on t. After Fourier transform in x the equation $Pu=f$ reduces to

$$\begin{aligned} D_t^3{\hat{u}}+\sum _{j+|\alpha |\le 3, j\le 2}a_{j,\alpha }(t)\xi ^{\alpha }D_t^j{\hat{u}}={\hat{f}} \end{aligned}$$

(4.16)

where ${\hat{u}}(t,\xi )$ stands for the Fourier transform of u(t, x) with respect to x. With

$$\begin{aligned} E(t,\xi )=\exp {\left( \frac{i}{3}\int _0^t \sum _{|\alpha |=1}a_{2,\alpha }(s)\xi ^{\alpha }\,ds\right) } \end{aligned}$$

it is clear that ${\hat{v}}=E(t,\xi ){\hat{u}}$ satisfies

$$\begin{aligned} D_t^3{\hat{v}}-a(t,\xi )|\xi |^2D_t{\hat{v}}-b(t,\xi )|\xi |^3{\hat{v}} +\sum _{j=1}^3b_j(t,\xi )|\xi |^{j-1}D_t^{3-j}{\hat{v}}=E{\hat{f}} \end{aligned}$$

(4.17)

where $b_j(t,\xi )=0$ for $|\xi |\le 1$ can be assumed since energy estimates for $|\xi |\le 1$ is easily obtained. Since

$$\begin{aligned} \sum _{k=0}^{\ell }\big |\left( |\xi |+1\right) ^{\ell -k}\partial _t^k{\hat{u}}(t)\big |^2\le C_{\ell }\sum _{k=0}^{\ell }\big |\left( |\xi |+1\right) ^{\ell -k}\partial _t^k{\hat{v}}(t)\big |^2 \end{aligned}$$

in order to obtain energy estimates for ${\hat{u}}$ one can assume that ${\hat{u}}$ satisfies (4.17) from the beginning. With $U={^t}\big (D_t^2{\hat{u}},|\xi |D_t{\hat{u}}, |\xi |^2{\hat{u}}\big )$ the equation (4.17) can be written

$$\begin{aligned} \begin{aligned} \frac{\partial }{\partial t}U&=i\begin{bmatrix}0&{}a(t,\xi )&{}b(t,\xi )\\ 1&{}0&{}0\\ 0&{}1&{}0\\ \end{bmatrix}\!|\xi |\,U\\&\quad +i\begin{bmatrix} b_1(t,\xi )&{}b_2(t,\xi )&{}b_3(t,\xi )\\ 0&{}0&{}0\\ 0&{}0&{}0\\ \end{bmatrix}U+\begin{bmatrix}iE{\hat{f}}\;\\ 0\\ 0\\ \end{bmatrix}\\&=iA|\xi |U+BU+F. \end{aligned} \end{aligned}$$

(4.18)

Let $S(t,\xi )$ and $T(t,\xi )$ be defined in Sect. 2 with $X=\xi$ such that $T^{-1}ST=\Lambda =\mathrm{diag}\,(\lambda _1,\lambda _2,\lambda _3)$. With $V=T^{-1}U$ one has

$$\begin{aligned} \partial _tV&=iA^T|\xi |V+\left( B^T+(\partial _tT^{-1})T\right) V+T^{-1}F\\&=i{\mathcal A}|\xi |V+{\mathcal B}V+{\tilde{F}} \end{aligned}$$

where ${\mathcal A}=T^{-1}AT$ and ${\mathcal B}=T^{-1}BT+ (\partial _tT^{-1})T$. Thanks to Proposition 4.1 we have candidates for scalar weights in each $\omega _j$. To simplify notation denote

$$\begin{aligned} t_0(\xi )=0,\quad t_1(\xi )=\psi (\xi )/2,\quad t_2(\xi )=\psi (\xi ),\quad t_3(\xi )=\delta \end{aligned}$$

and following [22] (also [27]) introduce three subintervals $\Omega _j=[t_{j-1}(\xi ),t_j(\xi )]$ and scalar weights $\varphi _j$, $j=1,2,3$

$$\begin{aligned} \varphi _1(t,\xi )=t,\quad \varphi _2(t,\xi )=\psi (\xi )-t,\quad \varphi _3(\xi )=t-\psi (\xi ). \end{aligned}$$

Note that $\omega _1=\Omega _1$, $\omega _2=\Omega _2\cup \Omega _3$ and $\varphi _j=|\phi _2|$ in $\Omega _j$, $j=2,3$. Thanks to Proposition 4.1 and Lemma 3.1 one has

$$\begin{aligned} \varphi _j^2\le C\lambda _1,\;\;\varphi _j|\partial _t\lambda _1|\le C\lambda _1,\;\;\varphi _j\le C\lambda _2,\quad t\in \Omega _j(\xi ),\quad j=1,2,3 \end{aligned}$$

(4.19)

where C is independent of $\xi \in U$. Consider the following energy in $\Omega _j(\xi )$;

$$\begin{aligned} {\mathcal E}_j=g_j\langle {\Lambda V,V}\rangle =g_j\sum _{k=1}^3\lambda _k(t,\xi )|V_k(t,\xi )|^2,\quad g_j=\varphi _j(t,\xi )^{2(-1)^jN-1}. \end{aligned}$$

Since ${\mathsf {Re}}\,\langle {i\Lambda {\mathcal A}|\xi |V,V}\rangle =0$ and $\partial _t\varphi _j=(-1)^{j-1}$ one has

$$\begin{aligned} \frac{d}{dt}{\mathcal E}_j=&-\left( 2N-(-1)^j\right) \varphi _j^{-1}{\mathcal E}_j+g_j\langle {\left( \partial _t\Lambda \right) V,V}\rangle \\&+2g_j{\mathsf {Re}}\,\langle {\Lambda {\mathcal B}V,V}\rangle +2g_j{\mathsf {Re}}\,\langle {\Lambda {\tilde{F}},V}\rangle . \end{aligned}$$

Repeating the same arguments as in Sect. 3 one can estimate

$$\begin{aligned} \big |g_j\langle {\Lambda (\partial _tT^{-1})TV,V}\rangle \big |+ \big |g_j\langle {\left( \partial _t\Lambda \right) V,V}\rangle \big |\le C_1\varphi _j^{-1}{\mathcal E}_j \end{aligned}$$

in $\Omega _j$. It rests to estimate $|g_j\langle {\Lambda B^TV,V}\rangle |$. Let $C'$ be a bound of all entries of $B^T$. Since $0\le \lambda _1\le \lambda _2\le \lambda _3$ then $|g_j\langle {\Lambda B^TV,V}\rangle |$ is bounded by

$$\begin{aligned} C'g_j\sum _{k=1}^3\sum _{l=1}^3\lambda _k|V_l|\,|V_k|&\le C'g_j\sum _{k=1}^3\lambda _k|V_k|^2+2\,C'g_j\lambda _2|V_1||V_2|\\&\quad \, +2\,C'g_j\lambda _3\left( |V_1||V_3|+|V_2||V_3|\right) . \end{aligned}$$

Thanks to (4.19) one has

$$\begin{aligned} 2\lambda _2|V_1||V_2|\le \sqrt{\lambda _2}\varphi _j|V_1|^2+\sqrt{\lambda _2}\frac{\lambda _2}{\varphi _j}|V_2|^2 \le C\sqrt{\lambda _2}\,\varphi _j^{-1}\left( \lambda _1|V_1|^2+\lambda _2|V_2|^2\right) \end{aligned}$$

and

$$\begin{aligned} 2\lambda _3|V_1||V_3|&\le \varphi _j|V_1|^2+\varphi _j^{-1}\lambda _3^2|V_3|^2\le C\,\varphi _j^{-1}\left( \lambda _1|V_1|^2+\lambda _3|V_3|^2\right) ,\\ 2\lambda _3|V_2||V_3|&\le \varphi _j^{1/2}|V_2|^2+\varphi _j^{-1/2}|V_3|^2\le C\sqrt{\varphi _j}\varphi _j^{-1}\left( \lambda _2|V_2|^2+\lambda _3|V_3|^2\right) \end{aligned}$$

and therefore there exists $C_2$ such that

$$\begin{aligned} |g_j\langle {\Lambda B^TV,V}\rangle |\le C_2\,\varphi _j^{-1}{\mathcal E}_j\quad \text {in}\;\;\Omega _j. \end{aligned}$$

(4.20)

Noting that

$$\begin{aligned} 2\big |{\mathsf {Re}}\,\langle {\Lambda {\tilde{F}},V}\rangle \big |\le \varphi _j\langle {\Lambda {\tilde{F}},{\tilde{F}}}\rangle +\varphi _j^{-1}\langle {\Lambda V,V}\rangle \end{aligned}$$

and $|\langle {\Lambda {\tilde{F}},{\tilde{F}}}\rangle |\le C''\Vert {\tilde{F}}\Vert ^2=C''\Vert F\Vert ^2$ we obtain

Lemma 4.4

Let $j=1$ or 3. There exist $N_0$ and $C>0$ such that for any $N\ge N_0$ and any $U(t,\xi )$ verifying $\partial _t^kU(t_{j-1}(\xi ),\xi )=0$, $k=0,1,\ldots , N$ one has

$$\begin{aligned} \Vert U(t)\Vert ^2+N\int _{t_{j-1}}^t\varphi _j^{-2N}(s)\Vert U(s)\Vert ^2ds\le C\int _{t_{j-1}}^t\varphi _j^{-2N}(s)\Vert F(s)\Vert ^2ds \end{aligned}$$

for $t\in \Omega _j(\xi )$.

With $\langle {\xi }\rangle =|\xi |+1$ it follows from Lemma 4.4 that

Corollary 4.1

Let $j=1$ or 3. There is $N_0$ such that for any $L\in {\mathbb {N}}$ there exists $C_L>0$ such that for any U with $\partial _t^kU(t_{j-1}(\xi ),\xi )=0$, $k=0,1,\ldots ,N+L$ one has

$$\begin{aligned}&\sum _{k=0}^L\left\| \langle {\xi }\rangle ^{L-k}\partial _t^kU(t)\right\| ^2+N\sum _{k=0}^L\int _{t_{j-1}}^t\varphi _j^{-2N}(s)\left\| \langle {\xi }\rangle ^{L-k}\partial _t^kU(s)\right\| ^2ds\nonumber \\&\quad \le C_L\sum _{k=0}^L\int _{t_{j-1}}^t\varphi _j^{-2N}(s)\left\| \langle {\xi }\rangle ^{L-k}\partial _t^kF(s)\right\| ^2ds \end{aligned}$$

(4.21)

for $t\in \Omega _j(\xi )$ and $N\ge N_0$.

For the subinterval $\Omega _2(\xi )$ the argument in Sect. 3 shows again

Lemma 4.5

There exist $N_0\in {\mathbb {N}}$ and $C>0$ such that one has

$$\begin{aligned} \varphi _2^{2N-1}(t)\Vert U(t)\Vert ^2+N\int _{t_1}^t\varphi _2^{2N}(s)\Vert U(s)\Vert ^2ds\nonumber \\ \le C\Vert U(t)\Vert ^2+C\int _{t_1}^t\varphi _2^{2N}(s)\Vert F(s)\Vert ^2ds. \end{aligned}$$

(4.22)

for $t\in \Omega _2(\xi )$ and $N\ge N_0$.

Corollary 4.2

There exists $N_0$ such that for any $L\in {\mathbb {N}}$ there is $C_L$ such that

$$\begin{aligned}&\varphi _2^{2N-1}(t)\sum _{k=0}^L\left\| \langle {\xi }\rangle ^{L-k}\partial _t^kU(t)\right\| ^2+N\sum _{k=0}^L\int _{t_1}^t\varphi _2^{2N}(s)\left\| \langle {\xi }\rangle ^{L-k}\partial _t^kU(s)\right\| ^2ds\nonumber \\&\quad \le C_L\sum _{k=0}^L\left\| \langle {\xi }\rangle ^{L-k}\partial _t^kU(t_1)\right\| ^2+C_L\sum _{k=0}^L\int _{t_1}^t\varphi _2^{2N}(s)\left\| \langle {\xi }\rangle ^{L-k}\partial _t^kF(s)\right\| ^2ds \end{aligned}$$

(4.23)

for $t\in \Omega _2$ and $N\ge N_0$.

Since energy estimates in each subinterval $\Omega _j$ is obtained, repeating the same arguments as in [22, 27, Sect. 6] one can collect the energy estimates in $\Omega _j$ yielding energy estimates of $U(t,\xi )$ in the whole interval $[0,\delta ]$.

Proposition 4.2

Assume that p has a triple characteristic root ${\bar{\tau }}$ at $(0,{\bar{\xi }})$, $|\bar{\xi }|=1$ and $(0,{\bar{\tau }},{\bar{\xi }})$ is effectively hyperbolic. Then there exist $\delta >0$ and a conic neighborhood U of ${\bar{\xi }}$ such that for any $a_{j,\alpha }(t)$ with $j+|\alpha |\le 2$ one can find $N_0\in {\mathbb {N}}$ such that for any $q\in {\mathbb {N}}$ with $q\ge N_0$ there is $C>0$ such that

$$\begin{aligned} \sum _{k=0}^{q+3}\big |\langle {\xi }\rangle ^{q+2-k}\partial _t^k{\hat{u}}(t,\xi )\big |^2\le C\sum _{k=0}^q\int _0^t\big |\langle {\xi }\rangle ^{N_0+q-k}\partial _t^k{\hat{f}}(s,\xi )\big |^2ds \end{aligned}$$

(4.24)

for $(t,\xi )\in [0,\delta ]\times U$ and for any ${\hat{u}}(t,\xi )$ with $\partial _t^k{\hat{u}}(0,\xi )=0$, $k=0,1,2$ and ${\hat{f}}(t,\xi )$ with $\partial _t^k{\hat{f}}(0,\xi )=0$, $k=0,\ldots ,q+N_0$ satisfying (4.16).

4.4 Remarks on double characteristics

Assume that P is a differential operator of order 2 and the principal symbol p has a double characteristic root ${\bar{\tau }}$ at $(0,{\bar{\xi }})$, $|\xi |=1$. After Fourier transform in x the equation $Pu=f$ reduces to

$$\begin{aligned} D_t^2{\hat{u}}+\sum _{j+|\alpha |\le 2, j\le 1}a_{j,\alpha }(t)\xi ^{\alpha }D_t^j{\hat{u}}={\hat{f}} \end{aligned}$$

(4.25)

Making similar procedure in Sect. 4.3 one can assume that the principal symbol p has the form

$$\begin{aligned} p(t,\tau ,\xi )=\tau ^2-a(t,\xi )|\xi |^2,\quad a(0,{\bar{\xi }})=0 \end{aligned}$$

(4.26)

so that ${\bar{\tau }}=0$ is a double characteristic root.

If $\partial _ta(0,{\bar{\xi }})\ne 0$ one can write

$$\begin{aligned} a(t,\xi )=e_1(t,\xi )\left( t+\alpha (\xi )\right) ,\quad \alpha ({\bar{\xi }})=0 \end{aligned}$$

in some neighborhood ${\mathcal U}$ of $(0,{\bar{\xi }})$ where $e_1>0$ and $\alpha (\xi )\ge 0$ near ${\bar{\xi }}$. In this case we choose $\varphi _1=t$, $\Omega _1=[0,\delta ]$ so that $\varphi _2$ is not needed.

If $(0,0,{\bar{\xi }})$ is a critical point (hence $\partial _ta(0,{\bar{\xi }})= 0$) and effectively hyperbolic then

$$\begin{aligned} \partial _t^2a\left( 0,{\bar{\xi }}\right) \ne 0. \end{aligned}$$

(4.27)

Indeed, assuming $a(0,{\bar{\xi }})=\partial _ta(0,{\bar{\xi }})=0$ it is easy to see

$$\begin{aligned} \mathrm{det}\,\left( \lambda -F_p(0,0,{\bar{\xi }})\right) =\lambda ^{2n}\left( \lambda ^2-2\,\partial _t^2a(0,{\bar{\xi }})\right) \end{aligned}$$

which shows that $\partial _t^2a(0,{\bar{\xi }})\ne 0$ if $(0,0,{\bar{\xi }})$ is effectively hyperbolic. From the Malgrange preparation theorem one can write, in some neighborhood ${\mathcal U}$ of $(0,{\bar{\xi }})$

$$\begin{aligned} a(t,\xi )=e_2(t,\xi )\left( t^2+a_1(\xi )t+a_2(\xi )\right) =e_2\prod _{k=1}^2\left( t-{\nu }_k(\xi )\right) \end{aligned}$$

where $e_2>0$ and $a_i({\bar{\xi }})=0$. Note that if ${\mathsf {Re}}\,{\nu }_1(\xi )\ne {\mathsf {Re}}\,{\nu }_2(\xi )$ then ${\nu }_i(\xi )$ is necessarily real and ${\nu }_i(\xi )\le 0$. In the case that either ${\mathsf {Re}}\,{\nu }_1(\xi )\ne {\mathsf {Re}}\,{\nu }_2(\xi )$ or ${\mathsf {Re}}\,{\nu }_1(\xi )={\mathsf {Re}}\,{\nu }_2(\xi )\le 0$ we take $\varphi _1=t$, $\Omega _1=[0,\delta ]$ and $\varphi _2$ is absent. In the case ${\mathsf {Re}}\,{\nu }_1(\xi )={\mathsf {Re}}\,{\nu }_2(\xi )=\psi (\xi )> 0$ so that

$$\begin{aligned} a(t,\xi )=e_2\left\{ (t-\psi (\xi ))^2+\left( {\mathsf {Im}}\,\nu _1(\xi )\right) ^2\right\} \end{aligned}$$

we take $\varphi _1=\psi (\xi )-t$, $\Omega _1(\xi )=[0,\psi (\xi )]$, $\varphi _2=t-\psi (\xi )$, $\Omega _2(\xi )=[\psi (\xi ),\delta ]$. Repeating similar arguments as in [22, 23] one obtains

Proposition 4.3

Assume that p has a double characteristic root ${\bar{\tau }}$ at $(0,{\bar{\xi }})$, $|\bar{\xi }|=1$ and $(0,{\bar{\tau }}, {\bar{\xi }})$ is effectively hyperbolic if it is a critical point. Then one can find $\delta >0$ and a conic neighborhood U of ${\bar{\xi }}$ such that for any $a_{j,\alpha }(t)$ with $j+|\alpha |\le 1$ one can find $N_0\in {\mathbb {N}}$ such that for any $q\in {\mathbb {N}}$ with $q\ge N_0$ there is $C>0$ such that

$$\begin{aligned} \sum _{k=0}^{q+2}\big |\partial _t^k{\hat{u}}(t,\xi )\big |^2\le C\sum _{k=0}^q\int _0^t\big |\langle {\xi }\rangle ^{N_0+q-k}\partial _t^k{\hat{f}}(s,\xi )\big |^2ds \end{aligned}$$

(4.28)

for $(t,\xi )\in [0,\delta ]\times U$ and for any ${\hat{u}}(t,\xi )$ with $\partial _t^k{\hat{u}}(0,\xi )=0$, $k=0,1$ and ${\hat{f}}(t,\xi )$ with $\partial _t^k{\hat{f}}(0,\xi )=0$, $k=0,\ldots ,q+N_0$ satisfying (4.25).

4.5 Proof of Theorem 4.1

We turn to the Cauchy problem (4.1). First note that, after Fourier transform in x, the equation is reduced to

$$\begin{aligned} \left\{ \begin{array}{ll} P\left( t,D_t,\xi \right) {\hat{u}}=D_t^m{\hat{u}}+\sum _{j=0}^{m-1}\sum _{|\alpha |\le m-j}a_{j,\alpha }(t)\,\xi ^{\alpha }D_t^j{\hat{u}}=0,\\ D_t^k{\hat{u}}(0,\xi )={\hat{u}}_k(\xi ),\;\;\xi \in {\mathbb {R}}^n,\quad k=0,\ldots ,m-1. \end{array}\right. \end{aligned}$$

(4.29)

Proposition 4.4

Assume that every critical point $(0,\tau ,\xi )$, $\xi \ne 0$ is effectively hyperbolic. Then there exists $\delta >0$ such that for any $a_{j,\alpha }(t)$ with $j+|\alpha |\le m-1$ one can find $N_0, N_1\in {\mathbb {N}}$ and $C>0$ such that

$$\begin{aligned} \sum _{k=0}^{m-1}\big |\langle {\xi }\rangle ^{m-1-k}\partial _t^k{\hat{u}}(t,\xi )\big |^2\le C\sum _{k=0}^{N_0}\int _0^t\big |\langle {\xi }\rangle ^{N_0-k}\partial _t^k{\hat{f}}(s,\xi )\big |^2ds \end{aligned}$$

for $(t,\xi )\in [0,\delta ]\times {\mathbb {R}}^n$ and for any ${\hat{u}}(t,\xi )$ with $\partial _t^k{\hat{u}}(0,\xi )=0$, $k=0,\ldots ,m-1$ and ${\hat{f}}(t,\xi )$ with $\partial _t^k{\hat{f}}(0,\xi )=0$, $k=0,\ldots ,N_1$ satisfying $P(t,D_t,\xi ){\hat{u}}={\hat{f}}$.

Proof

Let ${\bar{\xi }}\ne 0$ be arbitrarily fixed. Write $p(0,\tau ,{\bar{\xi }})=\prod _{j=1}^r\big (\tau -\tau _j)^{m_j}$ where $\sum m_j=m$ and $\tau _j$ are real and different each other, where $m_j\le 3$ which follows from the assumption. There exist $\delta >0$ and a conic neighborhood U of ${\bar{\xi }}$ such that one can write

$$\begin{aligned} p(t,\tau ,\xi )&=\prod _{j=1}^rp^{(j)}(t,\tau ,\xi ),\\ p^{(j)}(t,\tau ,\xi )&= \tau ^{m_j}+a_{j,1}(t,\xi )\tau ^{m_j-1}+\cdots + a_{j,m_j}(t,\xi ) \end{aligned}$$

for $(t,\xi )\in (-\delta ,\delta )\times U$ where $a_{j,k}(t,\xi )$ are real valued, homogeneous of degree k in $\xi$ and $p^{(j)}(0,\tau ,{\bar{\xi }})=(\tau -\tau _j)^{m_j}$. If $(0,\tau _j,{\bar{\xi }})$ is a critical point of p, and necessarily $m_j\ge 2$, then $(0,\tau _j,{\bar{\xi }})$ is a critical point of $p^{(j)}$ and it is easy to see

$$\begin{aligned} F_p\left( 0,\tau _j,{\bar{\xi }}\right) =c_j\,F_{p^{(j)}}\left( 0,\tau _j,{\bar{\xi }}\right) \end{aligned}$$

with some $c_j\ne 0$ and hence $F_{p^{(j)}}(0,\tau _j,{\bar{\xi }})$ has non-zero real eigenvalues if $F_{p}(0,\tau _j,{\bar{\xi }})$ does and vice versa. It is well known that one can write, in some conic neighborhood U of ${\bar{\xi }}$ that

$$\begin{aligned} P=P^{(1)}P^{(2)}\cdots P^{(r)}+R \end{aligned}$$

where $P^{(j)}$ are differential operators in t of order $m_j$ with coefficients which are poly-homogeneous symbol in $\xi$ and R is a differential operators in t of order at most $m-1$ with $S^{-\infty }$ (in $\xi$) 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

$$\begin{aligned} \sum _{k=0}^{q+m_j}\big |\langle {\xi }\rangle ^{q+m_j-k}\partial _t^k{\hat{u}}(t)\big |^2&\le C\sum _{k=0}^q\left\{ \big |\langle {\xi }\rangle ^{q-k}\partial _t^q\left( P^{(j)}{\hat{u}}\right) (t)\big |^2\right. \\&\quad \left. +\,\int _0^t\big |\langle {\xi }\rangle ^{N+q-k}\partial _t^k\left( P^{(j)}{\hat{u}}\right) (s)\big |^2ds\right\} \end{aligned}$$

in some conic neighborhood of ${\bar{\xi }}$ and for $j=1,\ldots ,r$. Then by induction on $j=1,\ldots , r$ one obtains

$$\begin{aligned} \sum _{k=0}^{q+m}\big |\langle {\xi }\rangle ^{q+m-k}\partial _t^k{\hat{u}}(t)\big |^2\le C\sum _{k=0}^q\left\{ \big |\langle {\xi }\rangle ^{q-k}\partial _t^kh(t)\big |^2 +\int _0^t\big |\langle {\xi }\rangle ^{rN+q-k}\partial _t^kh(s)\big |^2ds\right\} \end{aligned}$$

where $h(t)={\hat{f}}(t)-R\,{\hat{u}}(t)$. Note that for any $k, l\in {\mathbb {N}}$ there is $C_{k,l}$ such that

$$\begin{aligned} \big |\partial _t^k(R\,{\hat{u}})(t)\big |\le C_{k,l}\langle {\xi }\rangle ^{-l}\sum _{j=0}^{k+m-1}\big |\langle {\xi }\rangle ^{k+m-1-j}\partial _t^j{\hat{u}}(t)\big |^2. \end{aligned}$$

Therefore one concludes that

$$\begin{aligned} \sum _{k=0}^{q+m}\big |\langle {\xi }\rangle ^{q+m-k}\partial _t^k{\hat{u}}(t)\big |&\le C_1\sum _{k=0}^{q+m}\int _0^t\big |\langle {\xi }\rangle ^{q+m-k}\partial _t^k{\hat{u}}(s)\big |^2ds\\&\quad +C_2\sum _{k=0}^{q+1}\int _0^t\big |\langle {\xi }\rangle ^{rN+q+1-k}\partial _t^k{\hat{f}}(s)\big |^2ds. \end{aligned}$$

Then the assertion follows from the Gronwall’s lemma. Finally applying a compactness arguments one can complete the proof. $\square$

Proof of Theorem 4.1:

Let $u_j(x)\in {\mathcal S}({\mathbb {R}}^n)$ and hence ${\hat{u}}_j(\xi )\in {\mathcal S}({\mathbb {R}}^n)$. From $P{\hat{u}}=0$ one can determine $\partial _t^k{\hat{u}}(0,\xi )$ successively from ${\hat{u}}_j(\xi )$. Take $N\ge N_1+m$ and define

$$\begin{aligned} {\hat{u}}_N(t,\xi )=\sum _{k=0}^N\frac{t^k}{k!}\partial _t^k{\hat{u}}(0,\xi ) \end{aligned}$$

which is in $C^{\infty }({\mathbb {R}};{\mathcal S}({\mathbb {R}}^n))$. With ${\hat{f}}=-P{\hat{u}}_N$ it is clear that $\partial _t^k{\hat{f}}(0,\xi )=0$ for $k=0,\ldots ,N_1$. Apply Proposition 4.4 to the following Cauchy problem

$$\begin{aligned} P{\hat{w}}=-P{\hat{u}}_N={\hat{f}}(t,\xi ),\quad \partial _t^k{\hat{w}}(0,\xi )=0,\;k=0,\ldots ,m-1 \end{aligned}$$

to obtain

$$\begin{aligned} \sum _{k=0}^{m-1}\big |\langle {\xi }\rangle ^{m-1-k}\partial _t^k{\hat{w}}(t,\xi )\big |^2\le C\sum _{k=0}^{N_0}\int _0^t\big |\langle {\xi }\rangle ^{N_0-k}\partial _t^k{\hat{f}}(s,\xi )\big |^2ds. \end{aligned}$$

Since it is clear that

$$\begin{aligned} \sum _{k=0}^{N_0}\big |\langle {\xi }\rangle ^{N_0-k}\partial _t^k{\hat{f}}(s,\xi )\big |^2\le C_{N_0,N_1} \sum _{j=0}^{m-1}\big |\langle {\xi }\rangle ^{N+N_0-j}{\hat{u}}_j(\xi )\big |^2 \end{aligned}$$

for $0\le s\le \delta$ then noting that ${\hat{u}}={\hat{w}}+{\hat{u}}_N$ is a solution to the Cauchy problem (4.29) one obtains

$$\begin{aligned} \sum _{k=0}^{m-1}\big |\langle {\xi }\rangle ^{m-1-k}\partial _t^k{\hat{u}}(t,\xi )\big |^2\le C'\sum _{j=0}^{m-1}\big |\langle {\xi }\rangle ^{N+N_0-j}{\hat{u}}_j(\xi )\big |^2. \end{aligned}$$

Therefore, by a Paley-Wiener Theorem we prove the $C^{\infty }$ well-posedness of the Cauchy problem (4.1). $\square$

5 Third order hyperbolic operators with effectively hyperbolic critical points with two independent variables

In this section we consider the Cauchy problem for third order operators with two independent variables in a neighborhood of the origin for $t\ge 0$;

$$\begin{aligned} \left\{ \begin{array}{ll} D_t^3u+\sum _{j=0}^{2}\sum _{j+k\le 3}a_{j,k}(t,x)D_x^kD_t^ju=0,\\ D_t^ju(0,x)=u_j(x),\quad j=0,1,2 \end{array}\right. \end{aligned}$$

(5.1)

where the coefficients $a_{j,k}(t,x)$ ($j+k=3$) are assumed to be real valued real analytic in (t, x) in a neighborhood of the origin and the principal symbol p

$$\begin{aligned} p\left( t,x,\tau ,\xi \right) =\tau ^3+\sum _{j=0}^{2}\sum _{j+k=3}a_{j,k}(t,x)\xi ^k\tau ^j \end{aligned}$$

has only real roots in $\tau$ for $(t,x)\in [0,T')\times U$ with some $T'>0$ and a neighborhood U of the origin.

Theorem 5.1

If every critical point $(0,0,\tau ,1)$ is effectively hyperbolic then there exist $T>0$, $\delta >0$ such that for any $a_{j,k}(t,x)$ with $j+k\le 2$, which are $C^{\infty }$ in a neighborhood of $\Omega =\{(t,x)\mid |x|\le \delta (T-t), 0\le t\le T\}$, and for any $n\in {\mathbb {N}}$ one can find $Q\in {\mathbb {N}}$, $C>0$ such that for any $u_j(x)\in C^{\infty }({\mathbb {R}})$ $(j=0,1)$ there exists a unique solution $u(t,x)\in C^{\infty }(\Omega )$ to the Cauchy problem (5.1) satisfying

$$\begin{aligned} \sum _{k+l\le n}\int _{\Omega }\big |D_t^kD_x^lu(t,x)\big |^2dxdt\le C\sum _{j=0}^2\sum _{l\le Q}\int _{|x|\le \delta T}\big |D_x^lu_j(x)\big |^2dx. \end{aligned}$$

5.1 Key proposition, x dependent case

Assume that p has a triple characteristic root ${\bar{\tau }}$ at (0, 0, 1) hence $(0,0,{\bar{\tau }},1)$ is a critical point. Making a suitable change of local coordinates $t=t'$, $x=x(t',x')$ such that $x(0,x')=x'$ one can assume that $a_{2,1}(t,x)=0$ so that

$$\begin{aligned} p\left( t,x,\tau ,\xi \right) =\tau ^3-a(t,x)\xi ^2\tau -b(t,x)\xi ^3. \end{aligned}$$

(5.2)

Since the triple characteristic root is now ${\bar{\tau }}=0$ hence $b(0,0)=a(0,0)=0$ and the hyperbolicity condition implies that

$$\begin{aligned} \Delta (t,x)=4\,a(t,x)^3-27\,b(t,x)^2\ge 0,\quad (t,x)\in [0,T')\times U. \end{aligned}$$

(5.3)

Note that $\partial _xa(0,0)=\partial _tb(0,0)=\partial _xb(0,0)=0$ which follows from Lemma 2.1 then it is clear that

$$\begin{aligned} \mathrm{det}\left( \lambda I- F_p\left( 0,0,0,1\right) \right) =\lambda ^2\left( \lambda ^2-\left( \partial _ta(0,0)\right) ^2\right) . \end{aligned}$$

This implies $\partial _ta(0,0)\ne 0$ since (0, 0, 0, 1) is assumed to be effectively hyperbolic.

A key proposition corresponding to Proposition 4.1 is obtained by applying similar arguments as in Sect. 4.1 together with some observations on non-negative real analytic functions with two independent variables given in [22, Lemma 2.1] (see also [27]). We just give a sketch of the arguments. From the Weierstrass’ preparation theorem there is a neighborhood of (0, 0) where one can write

$$\begin{aligned} \Delta (t,x)=e_2(t,x)\left\{ t^3+a_1(x)t^2+a_2(x)t+a_3(x)\right\} \end{aligned}$$

where $e_2>0$ and $a_j(x)$ are real valued, real analytic with $a_j(0)=0$. Denote

$$\begin{aligned} {\bar{\Delta }}(t,x)=t^3+a_1(x)t^2+a_2(x)t+a_3(x) \end{aligned}$$

then the next lemma corresponds to Lemma 4.3

Lemma 5.1

There exists $\delta >0$ such that, in each interval $0<\pm \,x<\delta$, one can write

$$\begin{aligned} {\bar{\Delta }}(t,x)=\big |t-\nu _2(x)\big |^2\left( t-\nu _1(x)\right) \end{aligned}$$

(5.4)

where $\nu _1(x)$ is real valued with $\nu _1(x)\le 0$ or

$$\begin{aligned} {\bar{\Delta }}(t,x)=\prod _{k=1}^3\left( t-\nu _k(x)\right) \end{aligned}$$

(5.5)

where $\nu _k(x)$ are real valued with $\nu _k(x)\le 0$, in both cases $\nu _j(x)$ are expressed as convergent Puiseux series;

$$\begin{aligned} \nu _k(x)=\sum _{j\ge 0}C_{k,j}^{\pm }(\pm \,x)^{j/p_j},\quad \left( \text {for some}\;\;p_j\in {\mathbb {N}}\right) \end{aligned}$$

on $0<\pm x<\delta$. In all cases there is $C>0$ such that

$$\begin{aligned} \big |d{\mathsf {Re}}\,\nu _j(x)/dx\big |\le C,\quad 0<|x|<\delta . \end{aligned}$$

(5.6)

Next we show a counterpart of Lemma 4.2. Note that one can write

$$\begin{aligned} a(t,x)=e_1(t,x)\left( t+\alpha (x)\right) \end{aligned}$$

where $e_1>0$ and $\alpha (x)$ is real analytic with $\alpha (0)=0$ and $\alpha (x)\ge 0$ in $|x|<\delta$.

Lemma 5.2

There exist $\varepsilon >0$ and $\delta >0$ such that one can find $j^{\pm }\in \{1,2,3\}$ such that

$$\begin{aligned} \varepsilon ^{-1}|\nu _{j^{\pm }}(x)|\ge \alpha (x),\quad 0<\pm x<\delta . \end{aligned}$$

Recall that, choosing a smaller $\delta >0$ if necessary, one can assume that any two of ${\mathsf {Re}}\,\nu _j(x)$, ${\mathsf {Im}}\,\nu _j(x)$, $\nu (x)\equiv 0$ are either different or coincide in each interval $0<\pm x<\delta$. Denote

$$\begin{aligned} \psi (x)=\max {\left\{ 0,{\mathsf {Re}}\,\nu _2(x)\right\} },\quad |x|<\delta \end{aligned}$$

and define

$$\begin{aligned} {\left\{ \begin{array}{ll} \phi _1=t,&{}\Omega _1=\left\{ (t,x)\mid |x|<\delta , \; 0\le t\le \psi (x)/2\right\} \\ \phi _2=t-\psi (x),&{}\Omega _2=\{(t,x)\mid |x|<\delta , \;\psi (x)/2\le t\le T\} \end{array}\right. } \end{aligned}$$

with a small $T>0$. If $\psi =0$ then $\phi _1=\phi _2=t$ and $\Omega _2=\{(t,x)\mid |x|<\delta , \; 0\le t\le T\}$. Now we have a key proposition corresponding to Proposition 4.1;

Proposition 5.1

There exist $\delta >0$, $T>0$ and $C>0$ such that

$$\begin{aligned} \phi _j^2\,a\le C\Delta ,\quad |\phi _j|\,|\partial _t\Delta |\le C\Delta ,\quad |\phi _j|\le C a \end{aligned}$$

(5.7)

for $(t,x)\in \Omega _j$ and $j=1,2$.

5.2 Energy estimates, x dependent case

With $U={^t}(D_t^2u,D_xD_tu, D_x^2u)$ the equation (5.1) can be written

$$\begin{aligned} \frac{\partial }{\partial t}U&=\begin{bmatrix}0&{}a(t,x)&{}b(t,x)\\ 1&{}0&{}0\\ 0&{}1&{}0\\ \end{bmatrix}\!\frac{\partial }{\partial x}U\nonumber \\&\quad +i\begin{bmatrix} b_1(t,x)&{}b_2(t,x)&{}b_3(t,x)\\ 0&{}0&{}0\\ 0&{}0&{}0\\ \end{bmatrix}U+\begin{bmatrix}i{ f}\;\\ 0\\ 0\\ \end{bmatrix}\nonumber \\&=A(t,x)\partial _xU+BU+F. \end{aligned}$$

(5.8)

Let T(t, x) be the orthonormal matrix introduced in Sect. 2.2 such that with $V=T^{-1}U$ the equation (5.8) becomes

$$\begin{aligned} \partial _tV={\mathcal A}\partial _xV+{\mathcal B}V+{\tilde{F}} \end{aligned}$$

(5.9)

where

$$\begin{aligned} {\mathcal A}=T^{-1}AT,\quad {\mathcal B}=\left( \partial _tT^{-1}\right) T-{\mathcal A}\left( \partial _xT^{-1}\right) T+T^{-1}BT,\quad {\tilde{F}}=T^{-1}F. \end{aligned}$$

Let $\omega$ be an open domain in ${\mathbb {R}}^2$ and let $g\in C^1(\omega )$ be a positive scalar function. Denote by $\partial \omega$ the boundary of $\omega$ equipped with the usual orientation. Let

$$\begin{aligned} G(V)=g\langle {\Lambda V,V}\rangle dx+g\langle {\Lambda {\mathcal A}V,V}\rangle dt \end{aligned}$$

then one has

$$\begin{aligned}&2\,{\mathsf {Re}}\,\int _{\omega }g\langle {\Lambda V,{\mathcal B}V+{\tilde{F}}}\rangle \,dxdt=-\int _{\partial \omega }G(V)\nonumber \\&\quad -\int _{\omega }\left\{ (\partial _tg)\langle {\Lambda V,V}\rangle +g\langle {\left( \partial _t\Lambda \right) V,V}\rangle \right\} \,dxdt\nonumber \\&\quad +\int _{\omega }\left\{ \left( \partial _xg\right) \langle {\Lambda {\mathcal A}V,V}\rangle +g\langle {\partial _x\left( \Lambda {\mathcal A}\right) V,V}\rangle \right\} \,dxdt. \end{aligned}$$

(5.10)

Here make a remark on the boundary term. Denote

$$\begin{aligned} \tau _{max}=\max _{(t,x)\in [0,T]\times U}\big |\tau _j(t,x)\big | \end{aligned}$$

where $\tau _j(t,x)$, $j=1,2,3$ are characteristic roots of $p(t,x,\tau ,1)$.

Lemma 5.3

Let $\Gamma : [a,b]\ni x\mapsto (f(x),x)$ be a space-like curve, that is

$$\begin{aligned} 1>\tau _{max}\,\big |f'(x)\big |,\quad x\in [a,b]. \end{aligned}$$

Then one has

$$\begin{aligned} \int _{\Gamma }G(V)\ge 0. \end{aligned}$$

Proof

Since $g(t,x)>0$ is scalar function it suffices to prove

$$\begin{aligned} \langle {\Lambda \left( f(x),x\right) V,V}\rangle +\langle {f'(x)\Lambda (f(x),x){\mathcal A}(f(x),x)V,V}\rangle \ge 0,\;\;\forall V\in {\mathbb {C}}^3. \end{aligned}$$

To simplify notation we denote $\Lambda (f(x),x)$ and ${\mathcal A}(f(x),x)$ by just $\Lambda$ and ${\mathcal A}$. Noting that $\Lambda {\mathcal A}={^t}\!{\mathcal A}\Lambda$ one has

$$\begin{aligned} \big |\langle {f'\Lambda {\mathcal A}V,V}\rangle \big |=\big |f'\langle {\Lambda V,{\mathcal A}V}\rangle \big |\le \langle {\Lambda V,V}\rangle ^{1/2}\langle {\Lambda {\mathcal A}V,{\mathcal A}V}\rangle ^{1/2}|f'|. \end{aligned}$$

Therefore to prove the assertion it suffices to show

$$\begin{aligned} \langle {\Lambda {\mathcal A}V,{\mathcal A}V}\rangle |f'|^2\le \langle {\Lambda V,V}\rangle \end{aligned}$$

that is, the maximal eigenvalue of ${|f'|^2}({^t\!}{\mathcal A}\Lambda {\mathcal A})$ with respect to $\Lambda$ is at most 1. From $^t\!{\mathcal A}\Lambda =\Lambda {\mathcal A}$ it follows that

$$\begin{aligned} \mathrm{det}\left( \lambda \Lambda -|f'|^2\left( ^t\!{\mathcal A}\Lambda {\mathcal A}\right) \right)&=\mathrm{det}\left( \lambda \Lambda -|f'|^2\Lambda {\mathcal A}^2\right) \big )\\&=\left( \mathrm{det}\,\Lambda \right) \,\mathrm{det}\left( \lambda I-|f'|^2{\mathcal A}^2\right) . \end{aligned}$$

Since $\tau _j$ are the eigenvalues of ${\mathcal A}$ we see that the eigenvalues of $|f'|^2{\mathcal A}^2$ is at most $|f'(x)|^2\,\tau _{max}^2<1$ hence the assertion. $\square$

Define $\Omega =\{(t,x)\mid |x|\le \delta (T-t), 0\le t\le T\}$ and

$$\begin{aligned} \begin{aligned} \varphi _1&= t,\quad \omega _1=\left\{ (t,x)\mid |x|\le \delta (T-t), 0\le t\le \psi (x)/2\right\} ,\\ \varphi _2&= \psi (x)-t,\quad \omega _2=\left\{ (t,x)\mid |x|\le \delta (T-t), \psi (x)/2\le t\le \psi (x)\right\} ,\\ \varphi _3&= t-\psi (x),\quad \omega _3=\left\{ (t,x)\mid |x|\le \delta (T-t), \psi (x)\le t\le T\right\} \end{aligned} \end{aligned}$$

(5.11)

where $\delta>0, T>0$ are small such that the lines $|x|=\delta (T-t)$, $0\le t\le T$ are space-like. Thanks to Proposition 5.1 and Lemma 5.1 it follows that

$$\begin{aligned} \varphi _j^2\le C\lambda _1,\;\;\varphi _j|\partial _t\lambda _1|\le C\lambda _1,\;\;\varphi _j\le C\lambda _2,\;\;\big |\partial _x\varphi _j\big |\le C,\quad (t,x)\in \omega _j. \end{aligned}$$

(5.12)

Apply (5.10) with

$$\begin{aligned} g=g_j=\varphi _j^{2(-1)^jN-1},\;\; G(V)= G_j(V)=g_j\langle {\Lambda V,V}\rangle dx+g_j\langle {\Lambda {\mathcal A}V,V}\rangle dt \end{aligned}$$

and $\omega =\omega _j$ then from the arguments in Sect. 3 one obtains

Lemma 5.4

There exist $N_0$, $C>0$ such that

$$\begin{aligned} C\int _{\omega _j}\varphi _jg_j\Vert F\Vert ^2dxdt\ge -\int _{\partial \omega _j}G_j(V)+N\int _{\omega _j}\varphi _j^{-1}g_j\langle {\Lambda V,V}\rangle dxdt \end{aligned}$$

for $N\ge N_0$.

Repeating the same arguments as in [22, Lemma 3.1] (also [27]) one has

Proposition 5.2

There exists $C>0$ such that for every $n\in {\mathbb {N}}$ one can find $N_1$ such that

$$\begin{aligned}&\sum _{k+\ell \le n}\int _{\omega _j}g_j\varphi _j\Vert \partial _t^k\partial _x^{\ell }U\Vert ^2dxdt-\sum _{\ell \le n}\int _{\partial \omega _j}G_j(T^{-1}\partial _x^{\ell }U)\\&\quad \le C\sum _{k+\ell \le n}\int _{\omega _j}g_j\varphi _j\Vert \partial _t^k\partial _x^{\ell }LU\Vert ^2dxdt \end{aligned}$$

for any $N\ge N_1$.

Following the same arguments as in [22, 27] one can collect energy estimates in each $\omega _j$ to obtain

Proposition 5.3

Assume that p has a triple characteristic root ${\bar{\tau }}$ at (0, 0, 1) and $(0,0,{\bar{\tau }},1)$ is effectively hyperbolic. Then there exist $T>0, \delta >0$ such that for any $a_{j,k}(t,x)\in C^{\infty }(\Omega )$, $j+k\le 2$ one can find $C>0$ and $Q\in {\mathbb {N}}$ such that

$$\begin{aligned} \sum _{k+\ell \le n}\int _{\Omega }\Vert \partial _t^k\partial _x^{\ell }U\Vert ^2dxdt \le C\sum _{k+\ell \le Q}\int _{\Omega }\Vert \partial _t^k\partial _x^{\ell }LU\Vert ^2dxdt. \end{aligned}$$

(5.13)

for any $U(t,x)\in C^{\infty }(\Omega )$ with $\partial _t^kU(0,x)=0$, $k=0,\ldots , Q$.

5.3 Proof of Theorem 5.1

To complete the proof of Theorem 5.1, study the remaining case that p has a double characteristic root at (0, 0, 1).

Proposition 5.4

Assume that p has a double characteristic root ${\bar{\tau }}$ at (0, 0, 1) such that $(0,0,{\bar{\tau }},1)$ is effectively hyperbolic if it is a critical point. Then the same assertion as in Proposition 5.3 holds.

We give a sketch of the proof. Assume that p has a double characteristic root ${\bar{\tau }}$ at (0, 0, 1) and hence, after a suitable change of local coordinates, one can write

$$\begin{aligned} p\left( t,x,\tau ,\xi \right) =\left( \tau -b(t,x)\xi \right) \left( \tau ^2-a(t,x)\xi ^2\right) =p_1p_2 \end{aligned}$$

(5.14)

where $p_1=\tau -b(t,x)\xi$, $p_2=\tau ^2-a(t,x)\xi ^2$ and $a(0,0)=0$, $b(0,0)\ne 0$.

If (0, 0, 0, 1) is a critical point of p and hence $\partial _ta(0,0)=0$ then $F_p=c\,F_{p_2}$ at (0, 0, 0, 1) with some $c\ne 0$ and $\mathrm{det}\big (\lambda I-F_{p_2}(0,0,0,1)\big )=\lambda ^2\big (\lambda ^2-2\,\partial _t^2a(0,0)\big )$ which shows $\partial _t^2a(0,0)\ne 0$. From the Weierstrass’ preparation theorem one can write

$$\begin{aligned} a(t,x)=e_2(t,x)\left( t^2+2{\tilde{a}}_1(x)t+{\tilde{a}}_2(x)\right) =e_2(t,x)\Delta _2(t,x) \end{aligned}$$

where $e_2>0$, ${\tilde{a}}_j(0)=0$ and $\Delta _2$ takes the form, in each $0<\pm x<\delta$, either (5.4) or (5.5) where $\nu _1$ is absent. If $\Delta _2$ has the form (5.5) or (5.4) with ${\mathsf {Re}}\,\nu _2(x)\le 0$ it suffices to take $\varphi _1=t$, $\omega _1=\Omega$ ($\varphi _2$ is not needed). If $\Delta _2$ takes the form (5.4) with ${\mathsf {Re}}\,\nu _2(x)=\psi (x)> 0$ we choose $\varphi _1=\psi (x)-t$, $\omega _1=\{(t,x)\mid |x|\le \delta (T-t), 0\le t\le \psi (x)\}$ and $\varphi _2=t-\psi (x)$, $\omega _2=\{(t,x)\mid |x|\le \delta (T-t), \psi (x)\le t\le T\}$.

If $\partial _ta(0,0)\ne 0$ one can write

$$\begin{aligned} a(t,x)=e_1(t,x)\left( t+\alpha (x)\right) ,\quad \alpha (0)=0 \end{aligned}$$

where $e_1>0$ and $\alpha (x)\ge 0$ in a neighborhood of $x=0$. In this case we take $\varphi _1=t$, $\omega _1=\Omega$ ($\varphi _2$ is not needed).

Denote

$$\begin{aligned} P_2=D_t^2-a(t,x)D_x^2,\quad P_1=D_t-b(t,x)D_x. \end{aligned}$$

Then repeating the same arguments as in [22, 27] one has

$$\begin{aligned} \int _{\omega _j}\varphi _jg_j|P_2u|^2dxdt\ge -\int _{\partial \omega _j}G^{(2)}_j(u) +N\int _{\omega _j}\varphi _jg_j\left( |D_tu|^2+|D_xu|^2\right) dxdt \end{aligned}$$

(5.15)

for $N\ge N_0$ with $G^{(2)}_j(u)=g_j\big (|\partial _tu|^2+a|\partial _xu|^2\big )dx+ag_j\big (\partial _xu\cdot \overline{\partial _tu}+\overline{\partial _xu}\cdot \partial _tu\big )dt$. On the other hand, since $P_1$ is a first order differential operator with a real valued b(t, x) it is easy to see that

$$\begin{aligned} \int _{\omega _j}\varphi _jg_j|P_1u|^2dxdt&\ge -\int _{\partial \omega _j}G^{(1)}_j(u) +N\int _{\omega _j}\varphi _j^{-1}g_j|u|^2dxdt \nonumber \\&\ge -\int _{\partial \omega _j}G^{(1)}_j(u) +N\int _{\omega _j}\varphi _jg_j|u|^2dxdt \end{aligned}$$

(5.16)

for $N\ge N_0$ with $G^{(1)}_j(u)=g_j|u|^2dx+bg_j|u|^2dt$. Inserting $u=P_1u$ in (5.15) and $u=P_2u$ in (5.16) respectively and adding them one obtains

$$\begin{aligned}&\int _{\omega _j}\varphi _jg_j\left( |P_2P_1u|^2+|P_1P_2u|^2\right) dxdt\ge -\int _{\partial \omega _j}{\tilde{G}}_j(u)\nonumber \\&\quad +N\int _{\omega _j}\varphi _jg_j\left( |D_tP_1u|^2+|D_xP_1u|^2+|P_2u|^2\right) dxdt. \end{aligned}$$

(5.17)

In view of $b(0,0)\ne 0$ and $a(0,0)=0$ it is easy to see that $D_t^2$, $D_xD_t$ and $D_x^2$ are linear combinations of $D_tP_1$, $D_xP_1$ and $P_2$ with smooth coefficients modulo first order operators. Since one can write

$$\begin{aligned} P=P_2P_1+\sum _{i+j\le 2}b_{i,j}(t,x)D_x^iD_t^j=P_1P_2+\sum _{i+j\le 2}{\tilde{b}}_{i,j}(t,x)D_x^iD_t^j \end{aligned}$$

one concludes from (5.17) that

$$\begin{aligned} \int _{\omega _j}\varphi _jg_j|Pu|^2dxdt\ge -\int _{\partial \omega _j}{\tilde{G}}_j(u) +N\sum _{i+j\le 2}\int _{\omega _j}\varphi _jg_j|D_t^iD_x^ju|^2dxdt. \end{aligned}$$

The rest of the proof is parallel to that of Proposition 5.3.

Proof of Theorem 5.1:

Note that (5.13) implies

$$\begin{aligned} \sum _{k+\ell \le n+2}\int _{\Omega }|\partial _t^k\partial _x^{\ell }u|^2dxdt \le C\sum _{k+\ell \le Q}\int _{\Omega }|\partial _t^k\partial _x^{\ell }Pu|^2dxdt. \end{aligned}$$

Then applying an approximation argument with the Cauchy-Kowalevsky theorem one can conclude the proof of Theorem 5.1.

We restrict ourselves to third order operators in Theorem 5.1 because it seems to be hard to apply the same arguments as in Sect. 4.5 to this case. $\square$

6 Example of third order homogeneous equation with general triple characteristics

To show that the same arguments in the previous sections can be applicable to hyperbolic operators with more general triple characteristics, we study the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{ll} D_t^3u-a(t,x)D_tD_x^2u-b(t,x)D_x^3u=0,\;\;U\cap \{t> s\}\\ D_t^ju(s,x)=u_j(x),\quad j=0,1,2 \end{array}\right. \end{aligned}$$

(6.1)

in a full neighborhood of (0, 0) in ${\mathbb {R}}^2$. The Cauchy problem (6.1) is uniformly $C^{\infty }$ well posed near the origin if one can find a neighborhood U of (0, 0) and a small $\epsilon >0$ such that for any $|s|<\epsilon$ and any $u_j(x)\in C^{\infty }(\Omega \cap \{t=s\})$ there is a unique solution to (6.1). Assume that there is a neighborhood $U'$ of (0, 0) such that $p(t,x,\tau ,1)=0$ has only real roots in $\tau$ for $(t,x)\in U'$, that is

$$\begin{aligned} \Delta (t,x)=4\,a(t,x)^3-27\,b(t,x)^2\ge 0, \quad (t,x)\in U' \end{aligned}$$

(6.2)

which is necessary for the Cauchy problem (6.1) to be well posed near the origin ([17, 20]).

Theorem 6.1

Assume (6.2) and that there exists $C>0$ such that

$$\begin{aligned} a^3\le C\Delta , \qquad |\partial _tb|\le C\sqrt{a}\,|\partial _ta| \end{aligned}$$

(6.3)

holds in a neighborhood of (0, 0). Then the Cauchy problem (6.1) is uniformly $C^{\infty }$ well posed near the origin.

Note that if both a(t, x) and b(t, x) are independent of t, Theorem 6.1 is a very special case of [30, Theorem 1.1]. If a(t, x) and b(t, x) are independent of x this is also a special case of [4, Theorem 2] and [19, Theorem 1]. Inparticular, in [19], it is proved that, under the first condition of (6.3) (the second condition is a consequence of the first, in this case), the Bézout matrix itself behaves as if it were diagonal.

We give a rough sketch of the proof. If $\Delta (0,0)>0$ then p is strictly hyperbolic near the origin and the assertion is clear so $\Delta (0,0)=0$ is assumed from now on and hence $a(0,0)=0$ by assumption (6.3). Then p has a triple characteristic root at (0, 0, 1). Thanks to Proposition 2.1 the eigenvalues $\lambda _j$ of the Bézoutian matrix satisfy

$$\begin{aligned} \lambda _1\simeq a^2,\quad \lambda _2\simeq a,\quad \lambda _3\simeq 1 \end{aligned}$$

for $\Delta /a\ge a^2/C$. Let $\phi$ be a scalar function satisfying $\phi >0$ and $\partial _t\phi >0$ in $\omega$ and consider the following energy;

$$\begin{aligned} \left( \phi ^{-N}e^{-\gamma t} \Lambda V,V\right) =\int _{\omega } \phi ^{-N}e^{-\gamma t}\langle {\Lambda V,V}\rangle dx \end{aligned}$$

(6.4)

where $N>0, \gamma >0$ are positive parameters. Note that

$$\begin{aligned}&\frac{d}{dt}\,\left( e^{-\gamma t}\phi ^{-N}\, \Lambda V,V\right) \\&\quad = -\left( \left( N\phi ^{-1}\partial _t\phi +\gamma \right) e^{-\gamma t}\phi ^{-N}\,\Lambda V,V\right) \\&\qquad +\left( e^{-\gamma t}\phi ^{-N}(\partial _t\Lambda )V,V\right) +2\,{\mathsf {Re}}\,\left( e^{-\gamma t}\phi ^{-N}\Lambda \left( {\mathcal A}\partial _xV+{\mathcal B}V\right) ,V\right) \end{aligned}$$

and

$$\begin{aligned} \left( N\phi ^{-1}\partial _t\phi +\gamma \right) \langle {\Lambda V,V}\rangle =N\sum _{j=1}^3\phi ^{-1}\partial _t\phi \lambda _j |V_j|^2+\gamma \sum _{j=1}^3\lambda _j |V_j|^2. \end{aligned}$$

(6.5)

Since the hyperbolicity condition $4a^3\ge 27b^2$ is assumed in a full neighborhood of (0, 0) then Lemma 2.1 now states

$$\begin{aligned} |\partial _ta|\preceq \sqrt{a},\quad |\partial _xa|\preceq \sqrt{a},\quad |\partial _xb|\preceq a,\quad |\partial _tb|\preceq a. \end{aligned}$$

(6.6)

Suppose that a scalar weight $\phi$ satisfying the following property is obtained;

$$\begin{aligned} \phi \,|\partial _ta|\preceq a\,\partial _t\phi ,\qquad \phi \preceq \partial _t\phi \quad \text {in}\quad \omega . \end{aligned}$$

(6.7)

Since $|\partial _tq(\lambda _j)|\preceq (|\partial _ta|+|b||\partial _tb|)\lambda _j+|\partial _t\Delta |$ and $|\partial _t\Delta |\preceq a^2\,|\partial _ta|$ by (6.3) then

$$\begin{aligned} |\partial _t\lambda _j|\preceq \frac{\left( |\partial _ta|+a^{5/2}\right) \lambda _j+a^2\,|\partial _ta|}{|q_{\lambda }\left( \lambda _j\right) |} \end{aligned}$$

(6.8)

and hence $|\partial _t\lambda _j|\preceq (|\partial _ta|/a)\lambda _j+a^{3/2}\,\lambda _j+a\,|\partial _ta|$ for $j=1,2$ which shows that

$$\begin{aligned} \phi \,|\partial _t\lambda _j|\preceq \partial _t\phi \,\left( \lambda _j+a^2\right) . \end{aligned}$$

Thus $\phi \,|\partial _t\lambda _j|\preceq \partial _t\phi \,\lambda _j$ for $j=1,2,3$ since $\lambda _2\simeq a$. The case $j=3$ is clear from (6.7). This proves that $\big |\phi ^{-N}\langle {(\partial _t\Lambda )V,V}\rangle \big |$ is bounded by (6.5) taking N large. Next assume that a scalar weight $\phi$ satisfies

$$\begin{aligned} \sup _{(t,x)\in \omega _{\epsilon }}\sqrt{a}\,|\partial _x\phi |\left( \partial _t\phi \right) ^{-1}=0\quad \left( \epsilon \rightarrow 0\right) \end{aligned}$$

(6.9)

where $\omega _{\epsilon }$ is a family of regions converging to (0, 0) as $\epsilon \rightarrow 0$. Recall (3.4);

$$\begin{aligned} 2\,{\mathsf {Re}}\,\left( \phi ^{-N}\Lambda {\mathcal A}\partial _x V,V\right) =N\left( \phi ^{-N-1}(\partial _x\phi )\Lambda {\mathcal A}V,V\right) -\left( \phi ^{-N}\partial _x\left( \Lambda {\mathcal A}\right) V,V\right) . \end{aligned}$$

Thanks to (3.7) the term $N\big |\phi ^{-N-1}(\partial _x\phi )\langle {\Lambda {\mathcal A}V,V}\rangle \big |$ is bounded by (6.5) in $\omega _{\epsilon }$ for enough small $\epsilon$ in virtue of the assumption (6.9). Consider the term $\langle {\partial _x(\Lambda {\mathcal A})V,V}\rangle =\langle {(\partial _x\Lambda ){\mathcal A}V,V}\rangle +\langle {(\partial _x{\mathcal A})V,\Lambda v}\rangle$. From (3.16) one has $|\partial _x\lambda _1|\preceq a^{3/2}$. Since $|\partial _x\lambda _2|=O(\sqrt{a})$ and $|\partial _x\lambda _3|=O(1)$ it follows from Lemmas 3.2

$$\begin{aligned} \left( \partial _x\Lambda \right) {\mathcal A}=\begin{bmatrix} O\left( a^2\right) &{}O\left( a^{3/2}\right) &{}O\left( a^2\right) \\ O\left( a^{3/2}\right) &{}O(a)&{}O\left( \sqrt{a}\right) \\ O\left( a^{3/2}\right) &{}O( a)&{}O\left( a^{5/2}\right) \end{bmatrix} \end{aligned}$$

and then $|\langle {(\partial _x\Lambda ){\mathcal A}V,V}\rangle |$ is bounded by

$$\begin{aligned}&a^2|V_1|^2+a|V_2|^2+a^{5/2}|V_3|^2+a^{3/2}|V_1||V_2|+a^{3/2}|V_1||V_3|+\sqrt{a}|V_2||V_3|\\&\quad \preceq a^2|V_1|^2+a|V_2|^2+|V_3|^3. \end{aligned}$$

As for $\big |\langle {(\partial _x{\mathcal A})V,\Lambda V}\rangle \big |$, thanks to Lemma 3.2 one has

$$\begin{aligned} \Lambda \left( \partial _x{\mathcal A}\right) =\begin{bmatrix} O\left( a^2\right) &{}O\left( a^{3/2}\right) &{}O\left( a^2\right) \\ O\left( a^{3/2}\right) &{}O(a)&{}O\left( \sqrt{a}\right) \\ O( a)&{}O( a)&{}O\left( \sqrt{a}\right) \end{bmatrix} \end{aligned}$$

hence $|\langle {\Lambda (\partial _x{\mathcal A})V,V}\rangle |$ is bounded by

$$\begin{aligned}&a^2|V_1|^2+a|V_2|^2+\sqrt{a}|V_3|^2+a^{3/2}|V_1||V_2|+a|V_1||V_3|+\sqrt{a}|V_2||V_3|\\&\quad \preceq a^2|V_1|^2+a|V_2|^2+|V_3|^3. \end{aligned}$$

Then $|(e^{-\gamma t}\phi ^{-N}\Lambda {\mathcal A}\partial _x V,V)|$ is bounded by (6.5) taking $\epsilon >0$ small and $\gamma$ large.

Turn to ${\mathsf {Re}}\,\big (e^{-\gamma t}\phi ^{-N}\Lambda {\mathcal B}V,V\big )$ where ${\mathcal B}=(\partial _tT^{-1})T-{\mathcal A}(\partial _xT^{-1})T$. Using (6.8) it is easy to see

$$\begin{aligned} \begin{aligned} |\partial _t\ell _{11}|\preceq&a|\partial _ta|+a^2,\;\; |\partial _t\ell _{21}|\preceq |\partial _tb|+a^{5/2},\;\; |\partial _t\ell _{31}|\preceq |\partial _ta|+a,\\ |\partial _t\ell _{12}|\preceq&a^{2},\;\; |\partial _t\ell _{22}|\preceq |\partial _ta|+a^{3/2},\;\; |\partial _t\ell _{32}|\preceq |\partial _tb|+a^2,\\ |\partial _t\ell _{13}|\preceq&|\partial _ta|+a^{5/2},\;\; |\partial _t\ell _{23}|\preceq a^{2},\;\; |\partial _t\ell _{33}|\preceq |\partial _ta|. \end{aligned} \end{aligned}$$

(6.10)

Noting $|\partial _tb|\preceq \sqrt{a}\,|\partial _ta|$ these estimates improve (3.11) to

Lemma 6.1

Let $(\partial _tT^{-1})T=({\tilde{t}}_{ij})$ then

$$\begin{aligned} \left( {\tilde{t}}_{ij}\right) =\begin{bmatrix}0&{}O\left( a+|\partial _ta|/\sqrt{a}\right) &{}O\left( |\partial _ta|+a^{5/2}\right) )\\ O\left( a+|\partial _ta|/\sqrt{a}\right) &{}0&{}O\left( \sqrt{a}|\partial _ta|+a^2\right) \\ O\left( |\partial _ta|+a^{5/2}\right) &{}O\left( \sqrt{a}|\partial _ta|+a^2\right) &{}0\\ \end{bmatrix}. \end{aligned}$$

In order to estimate $\big |\phi ^{-N}\langle {(\partial _tT^{-1})TV,\Lambda V}\rangle \big |$ recalling $\Lambda \simeq \mathrm{diag}(a^2,a,1)$ it suffices to estimate

$$\begin{aligned} |\phi ^{-N}|\left( a\,{\tilde{t}}_{21}|V_1|| V_2|+{\tilde{t}}_{31}|V_1||V_3|+{\tilde{t}}_{32}|V_2||V_3|\right) . \end{aligned}$$

Note that $a\,{\tilde{t}}_{21}\preceq a^2+\sqrt{a}\,|\partial _ta|$ and

$$\begin{aligned} a^2|V_1||V_2|\preceq & {} \sqrt{a}\,\left( a^2|V_1|^2+a|V_2|^2\right) ,\\ \sqrt{a}\,|\partial _ta||V_1||V_2|\preceq & {} \phi ^{-1}\partial _t\phi \,a^{3/2} |V_1||V_2| \preceq \phi ^{-1}\partial _t\phi \left( a^2 |V_1|^2+a |V_2|^2\right) . \end{aligned}$$

As for ${\tilde{t}}_{31}|V_1||V_3|\preceq (|\partial _ta|+a^{5/2})|V_1||V_3|$ one has

$$\begin{aligned} a^{5/2}\,|V_1||V_3|\preceq & {} a^{3/2}\left( a^2|V_1|^2+|V_3|^2\right) ,\\ |\partial _ta|\,|V_1||V_3|\preceq & {} \phi ^{-1}\partial _t\phi \,a|V_1||V_3| \preceq \phi ^{-1}\partial _t\phi \left( a^2|V_1|^2+|V_3|^2\right) . \end{aligned}$$

Noting ${\tilde{t}}_{32}|V_2||V_3|\preceq (\sqrt{a}|\partial _ta|+a^2)|V_2||V_3|\preceq a\,|V_2||V_3|\preceq \sqrt{a}(a|V_2|^2+|V_3|^2)$ one concludes that $\big |\phi ^{-N}\langle {(\partial _tT^{-1})TV,\Lambda V}\rangle \big |$ is bounded by (6.5) taking N large and $\gamma \ge 1$. Consider $|\phi ^{-N}\langle {(\partial _xT^{-1})TV,\Lambda {\mathcal A}V}\rangle$. From Lemma 3.2 it follows that

$$\begin{aligned} \Lambda {\mathcal A}=\begin{bmatrix} O\left( a^{5/2}\right) &{}O\left( a^2\right) &{}O\left( a^{5/2}\right) \\ O\left( a^2\right) &{}O\left( a^{3/2}\right) &{}O(a)\\ O\left( a^{5/2}\right) &{}O(a)&{}O\left( a^{5/2}\right) \\ \end{bmatrix} \end{aligned}$$

because $\Lambda {\mathcal A}$ is symmetric and $\Lambda \simeq \mathrm{diag}(a^2,a,1)$. Taking (3.14) into account this shows that

$$\begin{aligned} \Lambda {\mathcal A}\left( \partial _xT^{-1}\right) T=\begin{bmatrix} O\left( a^2\right) &{}O\left( a^{5/2}\right) &{}O\left( a^3\right) \\ O\left( a^{3/2}\right) &{}O\left( a^2\right) &{}O\left( a^{5/2}\right) \\ O(a)&{}O(a^{3/2})&{}O\left( a^2\right) \\ \end{bmatrix}. \end{aligned}$$

Thus $|\langle {(\partial _xT^{-1})TV,\Lambda {\mathcal A}V}\rangle |$ is bounded by

$$\begin{aligned}&a^2\sum _{j=1}^3|V_j|^2+a^{3/2}|V_1||V_2|+a|V_1||V_3|+a^{3/2}|V_2||V_3|\\&\quad \preceq a^2|V_1|^2+a|V_2|^2+|V_3|^2 \end{aligned}$$

so that $|{\mathsf {Re}}\,\big (e^{-\gamma t}\phi ^{-N}\Lambda {\mathcal B}V,V\big )|$ is bounded by (6.5) taking N and $\gamma$ large.

Therefore to obtain energy estimates it suffices to find pairs $(\phi _j,\omega _{j,\epsilon })$ of scalar weight $\phi _j$ and subregion $\omega _{j,\epsilon }$ such that (6.7) and (6.9) are verified and $\cup \, \omega _{j,\epsilon }$ covers a neighborhood of (0, 0) for any fixed small $\epsilon >0$. Since the choice of $(\phi _j,\omega _{j,\epsilon })$ is exactly same as in [22] (also [27]) we only mention how to choose $\phi _j$ and $\omega _{j,\epsilon }$ ($\phi$ is denoted by $\rho$ in [22, 27]).

Following [22] one can define a real valued function $\alpha (t,x)$

$$\begin{aligned} \alpha (t,x)=x^{n}\prod _{i\in I_1}\left( t-t_i(x)\right) \prod _{i\in I_2}|t-t_i(x)|e(t,x) \end{aligned}$$

(6.11)

so that $a(t,x)=\alpha (t,x)^2$ where $t_i(x)$ has a convergent Puiseux expansion in $0<\pm x<\delta$ with small $\delta$ and ${\mathsf {Im}}\,t_i(x)\ne 0$ if $i\in I_2$. We choose all distinct $t_k(x)$ in (6.11) and rename them as $t_1(x),\ldots ,t_m(x)$. Taking $\delta$ small one can assume that

$$\begin{aligned} {\mathsf {Re}}\,t_{\mu _1}(x)\le&{\mathsf {Re}}\,t_{\mu _2}(x)\le \cdots \le {\mathsf {Re}}\,t_{\mu _m}(x),\quad 0<x<\delta ,\\ {\mathsf {Re}}\,t_{\nu _1}(x)\le&{\mathsf {Re}}\,t_{\nu _2}(x)\le \cdots \le {\mathsf {Re}}\,t_{\nu _m}(x),\quad -\delta<x<0. \end{aligned}$$

Define $\sigma _j(x)$ by

$$\begin{aligned} \sigma _j(x)={\mathsf {Re}}\,t_{\mu _j}(x)\;\; \text {for}\;\; x>0, \quad \sigma _j(x)={\mathsf {Re}}\,t_{\nu _j}(x)\;\; \text {for}\;\; x<0 \end{aligned}$$

so that $\sigma _1(x)\le \cdots \le \sigma _m(x)$ in $|x|<\delta$. Define

$$\begin{aligned} s_j(x)=\frac{\sigma _j(x)+\sigma _{j+1}(x)}{2},\;1\le j\le m-1,\;s_0(x)=-3t^*(x),\;s_m(x)=3t^*(x) \end{aligned}$$

with

$$\begin{aligned} t^*(x)=\left( \sum _{j=1}^m|t_j(x)|^2\right) ^{1/2} \end{aligned}$$

where the sum is taken over all distinct $t_i(x)$ in (6.11). Denote by $\omega _j^{\pm }$ and $\omega (T)$ the subregions defined by

$$\begin{aligned} \omega _j=&\left\{ (t,x)\mid |x|\le {\bar{\delta }}(T-t), s_{j-1}(x)\le t\le s_j(x)\right\} \;\; (j=1,\ldots ,m),\\ \omega ^{\pm }_j=&\omega _j\cap \{t > rless \sigma _j\},\;\;\omega (T)=\left\{ (t,x)\mid |x|\le {\bar{\delta }}(T-t), s_m(x)\le t\le T\right\} \end{aligned}$$

for small ${\bar{\delta }}>0$, $T>0$. Here ${\bar{\delta }}>0$ and $T>0$ play the role of $\epsilon >0$ in (6.9).

For $\omega _j$, $j=1,\ldots ,m$ we take $\phi =\phi _j^{\pm }(t,x)=\pm (t-\sigma _j(x))$. For $\omega (T)$ we take $\phi =\phi _{m+1}=t-s_m(x)$ if $n\ge 1$. Turn to $\omega (T)$ with $n=0$. Without restrictions one can assume $\alpha >0$ and $\partial _t\alpha >0$ in $\omega (T)$ (see [22, Lemma 2.2]) and we take $\phi =\alpha (t,x)$.