#  $C^{\infty}$ well-posedness of higher order hyperbolic pseudo-differential
equations with multiplicities

Claudia Garetto  Claudia Garetto: School of Mathematical Sciences Queen Mary
University of London Mile End Road, E1 4UJ, London United Kingdom E-mail
address<EMAIL_ADDRESS>and  Bolys Sabitbek  Bolys Sabitbek: School of
Mathematical Sciences Queen Mary University of London Mile End Road, London,
E1 4NS United Kingdom E-mail address<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study higher order hyperbolic pseudo-differential equations
with variable multiplicities. We work in arbitrary space dimension and we
assume that the principal part is time-dependent only. We identify sufficient
conditions on the roots and the lower order terms (Levi conditions) under
which the corresponding Cauchy problem is $C^{\infty}$ well-posed. This is
achieved via transformation into a first order system, reduction into upper-
triangular form and application of suitable Fourier integral operator methods
previously developed for hyperbolic non-diagonalisable systems. We also
discuss how our result compares with the literature on second and third order
hyperbolic equations.

###### Key words and phrases:

Hyperbolic equations, multiplicities, lower order terms

###### 2020 Mathematics Subject Classification:

Primary 35L25; 35L30; Secondary 46E35;

The authors were supported by the EPSRC grant EP/V005529/2. The authors
declare to have no conflicts of interest.

##  1\. Introduction

In this paper, we study the Cauchy problem for hyperbolic equations of order
$m$:

$\displaystyle\begin{cases}D_{t}^{m}u-\sum_{j=0}^{m-1}A_{m-j}(t,x,D_{x})D_{t}^{j}u=f(t,x),&t\in[0,T],\,x\in\mathbb{R}^{n},\\\
D_{t}^{k-1}u(0,x)=g_{k}(x),&k=1,\ldots,m,\end{cases}$ (1)

where $A_{m-j}(t,x,D_{x})$ is a pseudo-differential operator of order $m-j$,
$D_{t}=i^{-1}\partial_{t}$ and $D_{x}=i^{-1}\partial_{x}$. We assume that the
principal part of these operators, denoted by $A_{(m-j)}$ is independent of
the special variable $x$, i.e.

$A_{m-j}(t,x,D_{x})=A_{(m-j)}(t,D_{x})+(A_{m-j}(t,x,D_{x})-A_{(m-j)}(t,D_{x})),$

for all $j=0,\cdots,m-1$. We work under the hypothesis that the roots
$\lambda_{i}(t,\xi)$ of the characteristic polynomial

$\tau^{m}-\sum_{j=0}^{m-1}A_{(m-j)}(t,\xi)\tau^{j}=\prod_{i=1}^{m}(\tau-\lambda_{i}(t,\xi)).$

are real-valued functions in $C^{1}([0,T],S^{1}(\mathbb{R}^{2n}))$, hence the
equation is hyperbolic with multiplicities or also called weakly hyperbolic.
The right-hand side $f(t,x)$ is assumed to be continuous in $t$ and smooth in
$x$.

The main aim of this paper is to establish under which conditions on the roots
and the lower order terms (Levi conditions) the Cauchy problem (1) is
$C^{\infty}$ well-posed. We want our conditions to allow variable
multiplicities, i.e., up to order $m$.

In the sequel we give a brief and non exhausting overview of the existing
literature on weakly hyperbolic equations.

The well-posedness of the Cauchy problem for weakly hyperbolic equations has
been a challenging problem in mathematics since a long time. Unlike strictly
hyperbolic equations, weakly hyperbolic equations are characterised by their
principal part having real, but not necessarily distinct, roots. This
distinction results in more complex and unpredictable behaviour, as
illustrated by the following second-order example in one space dimension:

$\begin{cases}\partial_{t}^{2}u-a(t)\partial_{x}^{2}u+b(t)\partial_{x}u=0,\\\
u(0,x)=g_{1}(x),\quad\partial_{t}u(0,x)=g_{2}(x).\end{cases}$ (2)

In [23] Oleinik investigated the $C^{\infty}$ well-posedness of the Cauchy
problem (2) and obtained the sufficient condition

$t|b(t)|^{2}\leq Ca(t)+\partial_{t}a(t),$

which holds uniformly in $t\in[0,T]$ for some positive constant $C$. The
understanding of this Cauchy problem was further refined by Colombini and
Spagnolo in the celebrated paper [10], where they constructed a $C^{\infty}$
function $a(t)\geq 0$, such that the homogeneous problem (2) (i.e., with
$b=0$) is not $C^{\infty}$ well-posed.

Additional insights into the $C^{\infty}$ well-posedness of second-order
hyperbolic equations have been provided by many authors. Nishitani in [21]
explored necessary and sufficient conditions on lower order terms for
equations with a single space variable and analytic coefficients. Colombini,
De Giorgi, and Spagnolo in [6], and Colombini, Jannelli, and Spagnolo in [8],
focused on equations with coefficients dependent solely on time, examining
both $C^{\infty}$ and Gevrey well-posedness. In [7] Colombini, Gramchev, Orrù,
and Tagilalatela have obtained sufficient conditions for the $C^{\infty}$
well-posedness of the following third order Cauchy problem:

$\begin{cases}\partial_{t}^{3}u+a_{2,1}(t)\partial_{t}^{2}\partial_{x}u+a_{1,2}(t)\partial_{t}\partial_{x}u+a_{0,3}(t)\partial_{x}^{3}u+\text{lower
order terms}=f(t,x),\\\
u(0,x)=g_{1}(x),\quad\partial_{t}u(0,x)=g_{2}(x),\quad\partial_{t}^{2}u(0,x)=g_{3}(x).\end{cases}$
(3)

Their research particularly focuses on cases where the characteristics roots
have constant multiplicity and the coefficients in the principal part are of
class $C^{2}$. In contrast, Wakabayashi in [30] explored similar problems but
with double characteristics and analytic coefficients. For further results on
third order hyperbolic equations we refer the reader to [9] and [11] and the
recent extension by Nishitani in [22]. Passing now to arbitrary higher order
equations, the majority of the known results of $C^{\infty}$ well-posedness
hold for specific classes of equations:

  * -

equations with constant coefficients or with principal part with constant
coefficients (see [16, 20, 28, 29])

  * -

equations with constant multiplicities or multiplicities up to order two (see
[4, 5, 13, 15, 24, 25])

  * -

homogenous equations with coefficients dependent on one variable (either space
or time) (see [9, 11, 26])

To our knowledge general results for non-homogenous higher order hyperbolic
equations with $(t,x)$-dependent coefficients and variable multiplicities in
any space dimension are still missing. This paper is the first attempt to
provide results of $C^{\infty}$ well-posedness for a wider class of higher
order hyperbolic equations, without restrictive assumptions on the
multiplicities or the space dimension. The main idea is to transform the
higher order equation into a first order system of pseudo-differential
equations and then apply our knowledge of non-diagonalisable hyperbolic
systems developed in [17, 18] to formulate suitable Levi conditions on the
lower order terms that will guarantee $C^{\infty}$ well-posedness. Under this
point of view this paper is the natural application of [17, 18] to higher
order hyperbolic equations. Note that to better control the lower order terms
we work here on equations that are only time-dependent in the principal part.
This choice leads to a specific formulation of the Levi conditions on the
lower order terms (in line with recent results on $x$-dependnet hyperbolic
equations in [27]) however it still allows us to have variable multiplicities
of any order, which is a big achievement in comparison to the previous results
in this field.

###  1.1. Main result

Before stating our main result let us recall some basic definitions and fix
some notations. Let us fix the notations and definitions of symbol classes.

We say that a function $a(x,\xi)\in
C^{\infty}(\mathbb{R}^{n}\times\mathbb{R}^{n})$ belongs to
$S^{m}_{1,0}(\mathbb{R}^{n}\times\mathbb{R}^{n})$ if there exist constants
$C_{\alpha,\beta}$ such that

$\forall\alpha,\beta\in\mathbb{N}_{0}^{n}\quad:\quad|\partial_{x}^{\alpha}\partial_{\xi}a(x,\xi)|\leq
C_{\alpha,\beta}\langle\xi\rangle^{m-\beta}\quad\forall(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}.$

We denote by $C([0,T],S^{m}_{0,1}(\mathbb{R}^{n}\times\mathbb{R}^{n}))$ the
space of all symbols $a(t,x,\xi)\in S^{m}(\mathbb{R}^{n}\times\mathbb{R}^{n})$
which are continuous with respect to $t$. If there is no question about the
domain under consideration, we will abbreviate the symbol classes by
$S_{1,0}^{m}$ and $C([0,T],S^{m}_{0,1}(\mathbb{R}^{n}\times\mathbb{R}^{n}))$,
respectively, or simply by $S^{m}$ and $CS^{m}$.

Going back to the Cauchy problem (1) we will make use of the following
notations:

$\begin{split}b_{j}&=A_{m-j+1}(t,x,\xi)\langle\xi\rangle^{j-m},\\\
b_{(j)}&=A_{(m-j+1)}(t,\xi)\langle\xi\rangle^{j-m},\\\
d_{m,k}&=\sum_{j=k}^{m}(b_{j}-b_{(j)})\omega_{j,k},\end{split}$

for $j,k=1,\cdots,m$, where we set $\omega_{k,k}=1$ and

$\displaystyle\omega_{j,k}=\sum_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{k}=j-k}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\cdots\lambda_{k}^{\alpha_{k}}\langle\xi\rangle^{k-j}.$

for $j>k$. We then define the transformation matrix

$\displaystyle T(t,\xi)=\begin{pmatrix}1&0&0&\ldots&0&0&0\\\
\omega_{2,1}&1&0&\ldots&0&0&0\\\ \omega_{3,1}&\omega_{3,2}&1&\ldots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
\omega_{m-2,1}&\omega_{m-2,2}&\omega_{m-2,3}&\ldots&1&0&0\\\
\omega_{m-1,1}&\omega_{m-1,2}&\omega_{m-1,3}&\ldots&\omega_{m-1,m-2}&1&0\\\
\omega_{m,1}&\omega_{m,2}&\omega_{m,3}&\ldots&\omega_{m,m-2}&\omega_{m,m-1}&1\end{pmatrix}.$

which we will prove to be invertible in Proposition 2.1. The matrix $T^{-1}$
is employed to define the following symbols for $i>j$:

$e_{i,j}=\sum_{j<k\leq i}(T^{-1})_{i,k},D_{t}\omega_{k,j}.$

We are now ready to state our main result and its immediate corollary.

#### Main Theorem

_Let $n\geq 1$ and $m\geq 2$ and consider the Cauchy problem_

$\displaystyle\begin{cases}D_{t}^{m}u-\sum_{j=0}^{m-1}A_{m-j}(t,x,D_{x})D_{t}^{j}u=f(t,x),&t\in[0,T],\,x\in\mathbb{R}^{n},\\\
D_{t}^{k-1}u(0,x)=g_{k}(x),&k=1,\ldots,m,\end{cases}$

_where $A_{m-j}$ is a pseudo-differential operator of order $m-j$ with
principal part $A_{(m-j)}$ independent of $x$ and $f(t,x)$ is in
$C([0,T],C^{\infty}(\mathbb{R}^{n}))$. Assume that the characteristic
polynomial associated to this equation has $m$ real roots
$\lambda_{i}(t,\xi)\in C^{1}([0,T],S^{1}(\mathbb{R}^{2n}))$. If the Levi
conditions_

$\begin{split}e_{i,j}&\in C([0,T],S^{j-i}(\mathbb{R}^{2n})),\\\
d_{m,k}-e_{m,k}&\in C([0,T],S^{k-m}(\mathbb{R}^{2n})),\end{split}$

_for $i>j$, $2\leq i\leq m-1$, and for all $k=1,\ldots,m-1$, then the Cauchy
problem is $C^{\infty}$ well-posed._

#### Corollary

_If the the roots $\lambda_{j}$ are constant for $j=1,\cdots,m-2$ and_

$\begin{split}d_{m,k}&\in C([0,T],S^{k-m}(\mathbb{R}^{2n})),\\\
d_{m,m-1}-D_{t}\lambda_{m-1}\langle\xi\rangle&\in
C([0,T],S^{-1}(\mathbb{R}^{2n})),\end{split}$

_for $k=1,\cdots,m-2$, then the Cauchy problem is $C^{\infty}$ well-posed._

Note that the formulation of the main theorem is clear in view of the results
on non-diagonalisable systems proven in [17, 18], however the corollary allow
us to formulate easy examples of higher order hyperbolic equations whose
Cauchy problem is $C^{\infty}$ well-posed. Indeed, the Levi conditions above
given in terms of symbol order can be translated into conditions on the lower
order terms. As an explanatory example, consider the third order equation in
space dimension one with characteristic roots

$\begin{split}\lambda_{1}&=\xi,\\\ \lambda_{2}&=a(t)\xi,\\\
\lambda_{3}&=b(t)\xi.\end{split}$

and lower order terms

$a_{0,2}(t,x)D_{x}^{2}+a_{1,1}(t,x)D_{x}D_{t}+a_{2,0}(t,x)D_{t}^{2}+a_{0,1}(t,x)D_{x}+a_{1,0}(t,x)D_{t}+a_{0,0}(t,x).$

The Levi conditions are fulfilled imposing

$\begin{split}a_{1,1}(t,x)+(1+a(t))a_{2,0}(t,x)-D_{t}a(t)&=0,\\\
a_{0,2}(t,x)+a_{1,1}(t,x)+a_{2,0}(t,x)&=0,\\\
a_{0,1}(t,x)+a_{1,0}(t,x)&=0.\end{split}$

Note that the specific choice of lower order terms is balanced by variable
multiplicities and non-restrictive assumptions on $a(t)$ and $b(t)$, which
indeed are simply required to be of class $C^{1}$. Analogously, one can
consider the fourth order example with characteristic roots

$\lambda_{1}=\xi,\,\,\lambda_{2}=-\xi,\,\,\,\lambda_{3}=\sqrt{a(t)}\xi,\,\,\,\lambda_{4}=-\sqrt{a(t)}\xi,$

where $a(t)\geq 0$ and lower order terms fulfilling the Levi conditions

$\displaystyle a_{2,1}(t,x)+\sqrt{a(t)}a_{3,0}(t,x)=D_{t}\sqrt{a(t)},$
$\displaystyle a_{1,2}(t,x)+a_{3,0}(t,x)=0,$ $\displaystyle
a_{2,1}(t,x)+a_{0,3}(t,x)=0,$ $\displaystyle a_{0,2}(t,x)+a_{2,0}(t,x)=0,$
$\displaystyle a_{0,1}(t,x)+a_{1,0}(t,x)=0,$ $\displaystyle
a_{1,3}=a_{3,1}=a_{1,1}=0.$

We refer the reader to Example 3.5 and Example 4.5 for more details.

The paper is organised as follows. In Section 2 we collect some preliminaries
needed to state and prove our main theorem. In particular, we transform the
$m$-order equation into a system of first order pseudo-differential equations
and we reduce it into upper-triangular form while a Schur decomposition that
will define the matrix $T(x,\xi)$. We also recall the $C^{\infty}$ result for
non-diagonalisable hyperbolic systems proven in [17]. Section 3 is dedicated
to the proof of our main theorem and corollary and to some explanatory
examples. The paper ends with Section 4 where we discuss how our result
compares with well-known $C^{\infty}$ well-posdenss results for second and
third order hyperbolic equations and we provide further examples.

##  2\. Preliminaries

It is standard procedure when dealing with higher order equations to transform
the equation into a system of first order pseudo-differential equations. In
detail, making use of

${u_{k}=D_{t}^{k-1}\langle D_{x}\rangle^{m-k}u},$

for $k=1,\ldots,m$, where $\langle D_{x}\rangle$ is the pseudo-differential
operator with symbol $\langle\xi\rangle=(1+|\xi|^{2})^{\frac{1}{2}}$, the
Cauchy problem (1) can be re-written as

$\begin{split}D_{t}U&=A(t,D_{x})U+B(t,x,D_{x})U+F,\\\ U(0,x)&=(\langle
D_{x}\rangle^{m-1}g_{1},\langle D_{x}\rangle^{m-2}g_{2},\cdots
g_{m})^{T}\end{split}$ (4)

where $U=(u_{1},u_{2},\cdots,u_{m})^{T}$,

$\displaystyle A(t,\xi)=$
$\displaystyle\begin{pmatrix}0&\langle\xi\rangle&0&\dots&0&0\\\
0&0&\langle\xi\rangle&\dots&0&0\\\ 0&0&0&\ddots&0&0\\\
\vdots&\vdots&\vdots&\dots&\vdots&\vdots\\\ 0&0&0&\dots&0&\langle\xi\rangle\\\
b_{(1)}&b_{(2)}&b_{(3)}&\dots&b_{(m-1)}&b_{(m)}\end{pmatrix},$ (5)
$\displaystyle B(t,x,\xi)=$ $\displaystyle\begin{pmatrix}0&0&0&\dots&0&0\\\
0&0&0&\dots&0&0\\\ 0&0&0&\ddots&0&0\\\
\vdots&\vdots&\vdots&\dots&\vdots&\vdots\\\ 0&0&0&\dots&0&0\\\
b_{1}-b_{(1)}&b_{2}-b_{(2)}&b_{3}-b_{(3)}&\dots&b_{m-1}-b_{(m-1)}&b_{m}-b_{(m)}\end{pmatrix},$
(6)

with

$\begin{split}b_{j}&=A_{m-j+1}(t,x,\xi)\langle\xi\rangle^{j-m},\\\
b_{(j)}&=A_{(m-j+1)}(t,\xi)\langle\xi\rangle^{j-m},\end{split}$

for $j=1,\cdots,m$, and $F=(0,0,\cdots,0,f)^{T}$. The matrix $A$ is in
Sylvester form and has the roots $\lambda_{j}(t,\xi)$’s as eigenvalues. In the
rest of the paper we will therefore focus on the Cauchy problem (4). As a
first step to prove $C^{\infty}$ well-posedness we will reduce the matrix $A$
into upper-triangular form. This is a known as Schur decomposition. In our
particular case the transformation matrix $T$ is defined by symbols of order
$0$ as proven in the following proposition.

###  2.1. Schur decomposition

######  Proposition 2.1.

Let $A(t,\xi)$ be a $m\times m$ matrix valued symbol $A(t,\xi)$ as defined in
(5) with real eigenvalues $\lambda_{1},\ldots,\lambda_{m}$. There exists a
unitary $m\times m$ matrix valued symbol $T(t,\xi)$ such that $J=T^{-1}AT$ is
an upper triangular matrix with diagonal elements
$\lambda_{1},\ldots,\lambda_{m}$. More precisely,

$\displaystyle
T^{-1}AT=\begin{pmatrix}\lambda_{1}&\langle\xi\rangle&0&\cdots&0&0\\\
0&\lambda_{2}&\langle\xi\rangle&\cdots&0&0\\\ 0&0&\lambda_{3}&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&\lambda_{m-1}&\langle\xi\rangle\\\
0&0&0&\cdots&0&\lambda_{m}\end{pmatrix},$

with

$\displaystyle T(t,\xi)=\begin{pmatrix}1&0&0&\ldots&0&0&0\\\
\omega_{2,1}&1&0&\ldots&0&0&0\\\ \omega_{3,1}&\omega_{3,2}&1&\ldots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
\omega_{m-2,1}&\omega_{m-2,2}&\omega_{m-2,3}&\ldots&1&0&0\\\
\omega_{m-1,1}&\omega_{m-1,2}&\omega_{m-1,3}&\ldots&\omega_{m-1,m-2}&1&0\\\
\omega_{m,1}&\omega_{m,2}&\omega_{m,3}&\ldots&\omega_{m,m-2}&\omega_{m,m-1}&1\end{pmatrix}.$

$T(x,\xi)$ has determinant $1$ and the inverse matrix $T^{-1}$ is lower
triangular with entries $1$ on the diagonal and

$\displaystyle(T^{-1})_{i,j}=-\omega_{i,j}-\sum_{k=j+1}^{i-1}\omega_{i,k}(T^{-1})_{k,j},$

for $i>j$, where

$\displaystyle\omega_{j,k}=\sum_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{k}=j-k}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\cdots\lambda_{k}^{\alpha_{k}}\langle\xi\rangle^{k-j},$

for $j>k$, with $\alpha_{1},\cdots\alpha_{k}\in\mathbb{N}_{0}$.

###### Proof of Proposition 2.1.

This is a constructive proof. We aim to identify a matrix $T$ satisfying the
equation $AT=TJ$, where $A(t,\xi)$ is defined in (5) and

$\displaystyle
J(t,\xi)=\begin{pmatrix}\lambda_{1}(t,\xi)&\langle\xi\rangle&0&\dots&0&0\\\
0&\lambda_{2}(t,\xi)&\langle\xi\rangle&\dots&0&0\\\
0&0&\lambda_{3}(t,\xi)&\dots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\dots&\lambda_{m-1}(t,\xi)&\langle\xi\rangle\\\
0&0&0&\dots&0&\lambda_{m}(t,\xi)\end{pmatrix}.$

The relation $AT=TJ$ can be expressed as follows:

$\displaystyle
A\begin{pmatrix}T_{1}&T_{2}&T_{3}&\dots&T_{m-1}&T_{m}\end{pmatrix}=\begin{pmatrix}T_{1}&T_{2}&T_{3}&\dots&T_{m-1}&T_{m}\end{pmatrix}J$
$\displaystyle=\begin{pmatrix}\lambda_{1}T_{1}&T_{1}\langle\xi\rangle+\lambda_{2}T_{2}&T_{2}\langle\xi\rangle+\lambda_{3}T_{3}&\dots&T_{m-2}\langle\xi\rangle+\lambda_{m-1}T_{m-1}&T_{m-1}\langle\xi\rangle+\lambda_{m}T_{m}\end{pmatrix},$

where $T_{j}$, for $j=1,\ldots,m$, denotes the $j$-th column vector of the
matrix $T$. We denote the entries of $T_{j}$ with $\omega_{h,j}$, with
$h=1,\cdots,m$. To determine the matrix $T$, we systematically solve the
following sequence of equations:

$\displaystyle(A-\lambda_{1}I)T_{1}$ $\displaystyle=0,$
$\displaystyle(A-\lambda_{2}I)T_{2}$ $\displaystyle=T_{1}\langle\xi\rangle,$
$\displaystyle(A-\lambda_{3}I)T_{3}$ $\displaystyle=T_{2}\langle\xi\rangle,$
$\displaystyle\vdots$ $\displaystyle(A-\lambda_{m-1}I)T_{m-1}$
$\displaystyle=T_{m-2}\langle\xi\rangle,$ $\displaystyle(A-\lambda_{m}I)T_{m}$
$\displaystyle=T_{m-1}\langle\xi\rangle.$

This will lead us to formulate $\omega_{h,j}$ in terms of the eigenvalues of
$A$ and powers of $\langle\xi\rangle$.

Step 1. Solving the equation $(A-\lambda_{1}I)T_{1}=0$ yields:

$\displaystyle-\lambda_{1}\omega_{1,1}+\langle\xi\rangle\omega_{2,1}=0,\quad$
$\displaystyle\Longrightarrow\quad\omega_{1,1}=1\quad\text{ and
}\quad\omega_{2,1}=\lambda_{1}\langle\xi\rangle^{-1},$
$\displaystyle-\lambda_{1}\omega_{2,1}+\langle\xi\rangle\omega_{3,1}=0,\quad$
$\displaystyle\Longrightarrow\quad\omega_{3,1}=\lambda_{1}^{2}\langle\xi\rangle^{-2},$
$\displaystyle-\lambda_{1}\omega_{3,1}+\langle\xi\rangle\omega_{4,1}=0,\quad$
$\displaystyle\Longrightarrow\quad\omega_{4,1}=\lambda_{1}^{3}\langle\xi\rangle^{-3},$
$\displaystyle\vdots$
$\displaystyle-\lambda_{1}\omega_{m-1,1}+\langle\xi\rangle\omega_{m,1}=0,\quad$
$\displaystyle\Longrightarrow\quad\omega_{m,1}=\lambda_{1}^{m-1}\langle\xi\rangle^{-m+1},$

and the validity of the last term

$b_{(1)}\omega_{1,1}+b_{(2)}\omega_{2,1}+b_{(3)}\omega_{3,1}+\cdots+b_{(m-1)}\omega_{m-1,1}+\omega_{m,1}(b_{(m)}-\lambda_{1})=0$

is ensured by $\det(A-\lambda_{1}I)=0$. Hence, we can write the first column
vector $T_{1}$ as

$\displaystyle
T_{1}=\begin{pmatrix}1&\omega_{2,1}&\omega_{3,1}&\ldots&\omega_{m-1,1}&\omega_{m,1}\end{pmatrix}^{T},$

where the coefficients $\omega_{j,1}$ are given by

$\displaystyle\omega_{j,1}=\lambda_{1}^{j-1}\langle\xi\rangle^{1-j},$

for $j=2,\ldots,m$.

Step 2. In a similar manner, solving the equation
$(A-\lambda_{2}I)T_{2}=T_{1}\langle\xi\rangle$ yields:

$\displaystyle-\lambda_{2}\omega_{1,2}+\langle\xi\rangle\omega_{2,2}=\langle\xi\rangle,\quad$
$\displaystyle\Longrightarrow\quad\omega_{1,2}=0\quad\text{ and
}\quad\omega_{2,2}=1$
$\displaystyle-\lambda_{2}\omega_{2,2}+\langle\xi\rangle\omega_{3,2}=\lambda_{1}\quad$
$\displaystyle\Longrightarrow\quad\omega_{3,2}=(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1},$
$\displaystyle-\lambda_{2}\omega_{3,2}+\langle\xi\rangle\omega_{4,2}=\lambda_{1}^{2}\langle\xi\rangle^{-1}\quad$
$\displaystyle\Longrightarrow\quad\omega_{4,2}=(\lambda_{1}^{2}+\lambda_{1}\lambda_{2}+\lambda_{2}^{2})\langle\xi\rangle^{-2},$
$\displaystyle\vdots$
$\displaystyle-\lambda_{2}\omega_{m-1,2}+\langle\xi\rangle\omega_{m,2}=\lambda_{1}^{m-1}\langle\xi\rangle^{2-m}\quad$
$\displaystyle\Longrightarrow\quad\omega_{m,2}=\sum_{\alpha_{1}+\alpha_{2}=m-2}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\langle\xi\rangle^{2-m},$

and the validity of the last term

$b_{(1)}\omega_{1,2}+b_{(2)}\omega_{2,2}+b_{(3)}\omega_{3,2}+\cdots+b_{(m-1)}\omega_{m-1,2}+\omega_{m,2}(b_{(m)}-\lambda_{2})=\lambda_{1}^{m-1}\langle\xi\rangle^{2-m},$

is confirmed by employing the characteristic equation

$\tau^{m}-\sum_{j=1}^{m}b_{(j)}\langle\xi\rangle^{m-j}\tau^{j-1}=\prod_{j=1}^{m}(\tau-\lambda_{j}),$

and writing each $b_{(j)}$ for $j=1,\ldots,m$ in terms of the characteristic
roots $\lambda_{1},\ldots,\lambda_{m}$. Hence, the second column vector
$T_{2}$ is can be written as

$\displaystyle
T_{2}=\begin{pmatrix}0&1&\omega_{3,2}&\omega_{4,2}&\ldots&\omega_{m-1,2}&\omega_{m,2}\end{pmatrix}^{T},$

where

$\displaystyle\omega_{j,2}=\sum_{\alpha_{1}+\alpha_{2}=j-2}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\langle\xi\rangle^{2-j},$

for $j=3,\ldots,m$.

Step 3. Continuing with this process, by solving the equation
$(A-\lambda_{3}I)T_{3}=T_{2}\langle\xi\rangle$, we determine the third column
vector $T_{3}$ as

$\displaystyle
T_{3}=\begin{pmatrix}0&0&1&\omega_{4,3}&\omega_{5,3}&\ldots&\omega_{m-1,3}&\omega_{m,3}\end{pmatrix}^{T},$

where for $j=4,\ldots,m$, the coefficients $\omega_{j,3}$ are given by

$\displaystyle\omega_{j,3}=\sum_{\alpha_{1}+\alpha_{2}+\alpha_{3}=j-3}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\lambda_{3}^{\alpha_{3}}\langle\xi\rangle^{3-j}.$

Analogously, from the equation
$(A-\lambda_{j}I)T_{j}=T_{j-1}\langle\xi\rangle$ we get the entries of $T_{j}$
with $\omega_{j,j}=1$ and

$\displaystyle\omega_{j,k}=\sum_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{k}=j-k}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\cdots\lambda_{k}^{\alpha_{k}}\langle\xi\rangle^{k-j},$

for $j>k$, with $\alpha_{1},\cdots\alpha_{k}\in\mathbb{N}_{0}$.

Step m-1: In the penultimate step, solving the equation
$(A-\lambda_{m-1}I)T_{m-1}=T_{m-2}\langle\xi\rangle$ yields the column vector
$T_{m-1}$ as

$\displaystyle
T_{m-1}=\begin{pmatrix}0&0&0&\ldots&1&\omega_{m,m-1}\end{pmatrix}^{T},$

where the coefficient $\omega_{m,m-1}$ is determined by

$\displaystyle\omega_{m,m-1}$
$\displaystyle=\sum_{\alpha_{1}+\alpha_{2}+\dots+\alpha_{m-1}=1}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\lambda_{3}^{\alpha_{3}}\cdots\lambda_{m-1}^{\alpha_{m-1}}\langle\xi\rangle^{-1}$
$\displaystyle=(\lambda_{1}+\lambda_{2}+\lambda_{3}+\cdots+\lambda_{m-1})\langle\xi\rangle^{-1}.$

Step m. In the final step, solving
$(A-\lambda_{m}I)T_{m}=T_{m-1}\langle\xi\rangle$ allows us to determine the
column vector $T_{m}$ as

$\displaystyle T_{m}=\begin{pmatrix}0&0&0&\ldots&0&1\end{pmatrix}^{T}.$

Given that $T$ is a unitary matrix with determinant $1$, it is invertible. The
matrix $T^{-1}$ is lower-triangular, identically $1$ on the diagonal and with
entries

$\displaystyle(T^{-1})_{i,j}=-\omega_{i,j}-\sum_{k=j+1}^{i-1}\omega_{i,k}(T^{-1})_{k,j},$

for $i>j$. This concludes the proof. ∎

######  Remark 2.2.

Note that the entries of the matrices $T$ and $T^{-1}$ are polynomials in the
roots $\lambda_{i}$, $i=1,\cdots,n$, they are also symbols of order $0$,
independent of $x$ and of class $C^{1}$ with respect to $t$.

######  Remark 2.3.

Proposition 2.1 holds also when the matrix $A$ depends on $x$. However, in
this case the equality $J=T^{-1}AT$, valid at the level of the symbol
matrices, can be transferred to the level of the operators by adding a matrix
of remainders terms of order $0$. This is a direct consequence of the symbolic
calculus.

######  Example 2.4.

As explanatory examples, we write down explicitly $T$ and $T^{-1}$ when
$m=2,3,4$.

  * •

For the case $m=2$, the matrix $T$ and its inverse $T^{-1}$ are given by

$\displaystyle T=\begin{pmatrix}1&0\\\
\lambda_{1}\langle\xi\rangle^{-1}&1\end{pmatrix}\quad\text{ and }\quad
T^{-1}=\begin{pmatrix}1&0\\\
-\lambda_{1}\langle\xi\rangle^{-1}&1\end{pmatrix}.$

  * •

For the case $m=3$, we have

$\displaystyle T=\begin{pmatrix}1&0&0\\\
\lambda_{1}\langle\xi\rangle^{-1}&1&0\\\
\lambda_{1}^{2}\langle\xi\rangle^{-2}&(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}&1\end{pmatrix}$

and

$\displaystyle T^{-1}=\begin{pmatrix}1&0&0\\\
-\lambda_{1}\langle\xi\rangle^{-1}&1&0\\\
\lambda_{1}\lambda_{2}\langle\xi\rangle^{-2}&-(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}&1\end{pmatrix}.$

  * •

For the case $m=4$,

$\displaystyle T=\begin{pmatrix}1&0&0&0\\\
\lambda_{1}\langle\xi\rangle^{-1}&1&0&0\\\
\lambda_{1}^{2}\langle\xi\rangle^{-2}&(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}&1&0\\\
\lambda_{1}^{3}\langle\xi\rangle^{-3}&(\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{1}\lambda_{2})\langle\xi\rangle^{-2}&(\lambda_{1}+\lambda_{2}+\lambda_{3})\langle\xi\rangle^{-1}&1\end{pmatrix},$

and

$\displaystyle T^{-1}=\begin{pmatrix}1&0&0&0\\\
-\lambda_{1}\langle\xi\rangle^{-1}&1&0&0\\\
\lambda_{1}\lambda_{2}\langle\xi\rangle^{-2}&-(\lambda_{2}+\lambda_{1})\langle\xi\rangle^{-1}&1&0\\\
-\lambda_{1}\lambda_{2}\lambda_{3}\langle\xi\rangle^{-3}&(\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}+\lambda_{2}\lambda_{3})\langle\xi\rangle^{-2}&-(\lambda_{1}+\lambda_{2}+\lambda_{3})\langle\xi\rangle^{-1}&1\end{pmatrix}.$

We conclude this section by recalling the $C^{\infty}$ well-posedness result
proven for hyperbolic systems which in general are not diagonalisable but have
principal part in upper triangular form in [17]. This result will be applied
to our system after transformation via the matrix $T$.

###  2.2. $C^{\infty}$ well-posdenss of non-diagonalisable hyperbolic systems

######  Theorem 2.5 (Garetto-Jäh-Ruzhansky in [17]).

Let $n\geq 1$, $m\geq 2$, and let

$\displaystyle\begin{cases}D_{t}V=P(t,x,D_{x})V+L(t,x,D_{x})V+F(t,x),&(t,x)\in[0,T]\times\mathbb{R}^{n},\\\
V(0,x)=V_{0}(x),&x\in\mathbb{R}^{n},\end{cases}$ (7)

where $P(t,x,D_{x})\in(CS^{1})^{m\times m}$ is an upper-triangular matrix of
pseudo-differential operators of order $1$ and
$L(t,x,D_{x})\in(CS^{0})^{m\times m}$ is a matrix of pseudo-differential
operators of order $0$, continuous with respect to $t$. Hence, if

  * the lower order terms $\ell_{i,j}$ belong to $C([0,T],\Psi^{j-i})$ for $i>j$,

$V_{k}^{0}\in H^{s+k-1}(\mathbb{R}^{n})$ and $F_{k}\in C([0,T],H^{s+k-1})$ for
$k=1,\ldots,m$, then (7) has a unique anisotropic Sobolev solution $V$, i.e.
$V_{k}\in C([0,T],H^{s+k-1})$ for $k=1,\ldots,m$.

From Theorem 2.5 it follows immediately that the Cauchy problem is
$C^{\infty}$ well-posed when we choose right-hand side and initial data that
are smooth with respect to $x$ and compactly supported.

##  3\. $C^{\infty}$ well-posedness of the Cauchy problem for higher order
hyperbolic equations with time-dependent principal part

This section is devoted to the proof of our main result: the $C^{\infty}$
well-posedness of the hyperbolic Cauchy problem

$\begin{cases}D_{t}^{m}u-\sum_{j=0}^{m-1}A_{m-j}(t,x,D_{x})D_{t}^{j}u=f(t,x),&t\in[0,T],x\in\mathbb{R}^{n},\\\
D_{t}^{k-1}u(0,x)=g_{k}(x),&k=1,2,\ldots,m.\end{cases}$ (8)

where the principal part $A_{(m-j)}$ of the operators $A_{m-j}$ is
$x$-independent. We assume that the characteristic polynomial

$\tau^{m}-\sum_{j=0}^{m-1}A_{(m-j)}(t,\xi)\tau^{j}=\prod_{j=1}^{m}(\tau-\lambda_{j}(t,\xi)),$
(9)

has real-valued roots $\lambda_{1},\ldots,\lambda_{m}\in
C^{1}([0,T],S^{1}(\mathbb{R}^{2n}))$. By Section 2, we will equivalently study
the first order system

$\begin{cases}D_{t}U=A(t,D_{x})U+B(t,x,D_{x})U+F(t,x),&(t,x)\in[0,T]\times\mathbb{R}^{n},\\\
U|_{t=0}=U_{0},&x\in\mathbb{R}^{n},\end{cases}$ (10)

where $U_{0}=(u_{1}(0),u_{2}(0),\ldots,u_{m}(0))^{T}$,
$F(t,x)=(0,0,\ldots,f(t,x))^{T}$, with the operators $A(t,D_{x})$ and
$B(t,x,D_{x})$ given by

$\displaystyle A(t,D_{x})=$ $\displaystyle\begin{pmatrix}0&\langle
D_{x}\rangle&0&\dots&0&0\\\ 0&0&\langle D_{x}\rangle&\dots&0&0\\\
0&0&0&\ddots&0&0\\\ \vdots&\vdots&\vdots&\dots&\vdots&\vdots\\\
0&0&0&\dots&0&\langle D_{x}\rangle\\\
b_{(1)}&b_{(2)}&b_{(3)}&\dots&b_{(m-1)}&b_{(m)}\end{pmatrix}$

and

$\displaystyle B(t,x,D_{x})=\begin{pmatrix}0&0&\dots&0&0\\\ 0&0&\dots&0&0\\\
0&0&\ddots&0&0\\\ \vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&\dots&0&0\\\
b_{1}-b_{(1)}&b_{2}-b_{(2)}&\dots&b_{m-1}-b_{(m-1)}&b_{m}-b_{(m)}\end{pmatrix}.$

The entries of $A(t,D_{x})$ and $B(t,x,D_{x})$ are given by

$\displaystyle b_{j}=A_{m-j+1}(t,D_{x})\langle D_{x}\rangle^{j-m}\quad$
$\displaystyle\text{ and }\quad b_{(j)}=A_{(m-j+1)}(t,D_{x})\langle
D_{x}\rangle^{j-m}$

for $j=1,2,\dots,m$

###  3.1. Transformation to upper-triangular form

We now employ Proposition 2.1 to transform the system (8) into upper-
triangular form, i.e., we set $U=TV$. This is a significant step as it allows
us to apply Theorem 2.5 and to find the conditions which guarantee anisotropic
Sobolev well-posedness. In detail, we can write

$\displaystyle D_{t}V=T^{-1}ATV+T^{-1}BTV-T^{-1}D_{t}TV+T^{-1}F,$

and $V|_{t=0}=T^{-1}|_{t=0}U_{0}=V_{0}$. Passing now at the symbol level, we
have that the principal operator $(T^{-1}AT)(t,D_{x})$ has symbol

$\displaystyle
P=T^{-1}AT=\begin{pmatrix}\lambda_{1}&\langle\xi\rangle&0&\cdots&0&0\\\
0&\lambda_{2}&\langle\xi\rangle&\cdots&0&0\\\ 0&0&\lambda_{3}&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&\lambda_{m-1}&\langle\xi\rangle\\\
0&0&0&\cdots&0&\lambda_{m}\end{pmatrix},$

in upper-triangular form. The zero-order operator $T^{-1}BT-T^{-1}D_{t}T$ can
be written as $BT-T^{-1}D_{t}T$ since $T^{-1}B=B$. It follows that we do not
create any remainder terms in this composition of these operators and that
$T^{-1}BT-T^{-1}D_{t}T$ has symbol matrix

$D(t,x,D_{x})-E(t,D_{x})$

where

$\displaystyle D=T^{-1}BT=\begin{pmatrix}0&0&0&\cdots&0&0\\\
0&0&0&\cdots&0&0\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&0&0\\\ d_{m,1}&d_{m,2}&d_{m,3}&\cdots&d_{m,m-1}&d_{m,m}\\\
\end{pmatrix},$ $\displaystyle
E=T^{-1}D_{t}T=\begin{pmatrix}0&0&0&\ldots&0&0\\\ e_{2,1}&0&0&\ldots&0&0\\\
e_{3,1}&e_{3,2}&0&\ldots&0&0\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
e_{m-1,1}&e_{m-1,2}&e_{m-1,3}&\ldots&0&0\\\
e_{m,1}&e_{m,2}&e_{m,3}&\ldots&e_{m,m-1}&0\end{pmatrix}.$ $\displaystyle
D_{t}V=P(t,D_{x})V+L(t,x,D_{x})V+T^{-1}F,$ (11)

where $P(t,D_{x})=T^{-1}AT$ and $L(t,x,D_{x})=(D-E)(t,x,D_{x})$. By direct
computations we obtain the following auxiliary result.

######  Proposition 3.1.

The entries of the matrices $D$ and $E$ are given by

$\displaystyle d_{m,k}=\sum_{j=k}^{m}(b_{j}-b_{(j)})\omega_{j,k}$

for $k=1,\ldots,m$ and

$\displaystyle e_{i,j}=\sum_{j<k\leq i}(T^{-1})_{i,k},D_{t}\omega_{k,j},$

for $i>j$, where

$\displaystyle\omega_{k,j}=\sum_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{j}=k-j}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\cdots\lambda_{j}^{\alpha_{j}}\langle\xi\rangle^{j-k}.$

###### Proof.

By definition of the matrices $T$, $T^{-1}$ and $B$, we immediately see that
$D=T^{-1}BT=BT$ and the only non-zero row is the $m$-th row. Hence,

$\displaystyle
d_{m,k}=\sum_{j=k}^{m}(b_{j}-b_{(j)})\omega_{j,k}=b_{k}-b_{(k)}+\sum_{j=k+1}^{m}(b_{j}-b_{(j)})\omega_{j,k},$

for $k=1,\ldots,m$. By definition of $T^{-1}$ and $T$ we have that $E$ is a
strictly lower triangular matrix. From Proposition 2.1, we have that

$e_{i,j}=\sum_{j<k\leq i}(T^{-1})_{i,k},D_{t}\omega_{k,j},$

for $i>j$, where

$\displaystyle\omega_{k,j}=\sum_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{j}=k-j}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\cdots\lambda_{j}^{\alpha_{j}}\langle\xi\rangle^{j-k}.$

∎

We now apply Theorem 2.5, to the system (11) with initial condition $V_{0}$.
We immediately obtain anisotropic Sobolev well-posedness of any order and
therefore $C^{\infty}$ well-posedness if the $(i,j)$-entry of the matrix $L$
is given by pseudodifferential operators of order $j-i$ below the diagonal,
i.e., when $i>j$. Since $L(t,x,D_{x})=(D-E)(t,x,D_{x})$ this means to request

$\begin{split}e_{i,j}&\in C([0,T],S^{j-i}(\mathbb{R}^{2n})),\\\
d_{m,k}-e_{m,k}&\in C([0,T],S^{k-m}(\mathbb{R}^{2n})),\end{split}$ (12)

for $i>j$, $2\leq i\leq m-1$, and for all $k=1,\ldots,m-1$.

We have therefore proven the following theorem.

######  Theorem 3.2.

Let $n\geq 1$ and $m\geq 2$ and consider the Cauchy problem

$\displaystyle\begin{cases}D_{t}^{m}u-\sum_{j=0}^{m-1}A_{m-j}(t,x,D_{x})D_{t}^{j}u=f(t,x),&t\in[0,T],\,x\in\mathbb{R}^{n},\\\
D_{t}^{k-1}u(0,x)=g_{k}(x),&k=1,\ldots,m,\end{cases}$

where $A_{m-j}$ is a pseudo-differential operator of order $m-j$ with
principal part $A_{(m-j)}$ independent of $x$ and $f(t,x)$ is in
$C([0,T],C^{\infty}(\mathbb{R}^{n}))$. Assume that the characteristic
polynomial associated to this equation has $m$ real roots
$\lambda_{i}(t,\xi)\in C^{1}([0,T],S^{1}(\mathbb{R}^{2n}))$. If the Levi
conditions

$\begin{split}e_{i,j}&\in C([0,T],S^{j-i}(\mathbb{R}^{2n})),\\\
d_{m,k}-e_{m,k}&\in C([0,T],S^{k-m}(\mathbb{R}^{2n})),\end{split}$

for $i>j$, $2\leq i\leq m-1$, and for all $k=1,\ldots,m-1$, then the Cauchy
problem is $C^{\infty}$ well-posed.

A careful analysis of the Levi condition on $e_{i,j}$ leads us to other two
immediate corollaries of Theorem 3.2. It suffices to notice that the terms
$D_{j}\lambda_{h}$, $h=1,\cdots,j$, appear in the definition of $e_{i,j}$
since

$e_{i,j}=\sum_{j<k\leq i}(T^{-1})_{i,k},D_{t}\omega_{k,j}=\sum_{j<k\leq
i}(T^{-1})_{i,k}D_{t}\biggl{(}\sum_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{j}=k-j}\lambda_{1}^{\alpha_{1}}\lambda_{2}^{\alpha_{2}}\cdots\lambda_{j}^{\alpha_{j}}\langle\xi\rangle^{j-k}\biggr{)},$

for $i>j$. Hence, if the roots $\lambda_{1},\lambda_{2},\cdots,\lambda_{j}$
are constant (in time) than the entries $e_{i,j}$ vanish and trivially fulfil
our Levi conditions. In addition,

$e_{m,m-1}=(T^{-1})_{m,m}D_{t}\omega_{m,m-1}=D_{t}\omega_{m,m-1}=\sum_{h=1}^{m-1}D_{t}\lambda_{h}\langle\xi\rangle=D_{t}\lambda_{m-1}\langle\xi\rangle.$

######  Corollary 3.3.

If the the roots $\lambda_{j}$ are constant for $j=1,\cdots,m-2$ and

$\begin{split}d_{m,k}&\in C([0,T],S^{k-m}(\mathbb{R}^{2n})),\\\
d_{m,m-1}-D_{t}\lambda_{m-1}\langle\xi\rangle&\in
C([0,T],S^{-1}(\mathbb{R}^{2n})),\end{split}$

for $k=1,\cdots,m-2$, then the Cauchy problem is $C^{\infty}$ well-posed.

To illustrate our result we discuss in detail the $m=3$ toy model.

###  3.2. Third-Order Hyperbolic Equations with $(t,x)$-dependent lower order
terms

Let us consider the Cauchy problem

$\begin{cases}D_{t}^{3}u-\sum_{j=0}^{2}A_{3-j}(t,x,D_{x})D_{t}^{j}u=f(t,x),&t\in[0,T],x\in\mathbb{R}^{n},\\\
D_{t}^{k-1}u(0,x)=g_{k}(x),&k=1,2,3.\end{cases}$ (13)

where the principal part $A_{(3-j)}$ of the operators $A_{3-j}$ is
$x$-independent. Moreover, the characteristic polynomial

$\tau^{3}-\sum_{j=0}^{2}A_{(3-j)}(t,\xi)\tau^{j}=\prod_{j=1}^{3}(\tau-\lambda_{j}(t,\xi)),$
(14)

has the real-valued roots $\lambda_{1},\lambda_{2},\lambda_{3}\in
C^{1}([0,T],S^{1}(\mathbb{R}^{2n}))$.

By setting $U=(u_{1},u_{2},u_{3})^{T}$ with $u_{k}=D_{t}^{k-1}\langle
D_{x}\rangle^{3-k}u$ for $k=1,\cdots,3$ the equation above is reduced to

$\begin{cases}D_{t}U=A(t,D_{x})U+B(t,x,D_{x})U+F(t,x),&(t,x)\in[0,T]\times\mathbb{R}^{n},\\\
U|_{t=0}=U_{0},&x\in\mathbb{R}^{n},\end{cases}$ (15)

where $U_{0}=(u_{1}(0),u_{2}(0),u_{3}(0))^{T}$, $F(t,x)=(0,0,f(t,x))^{T}$,
with the operators $A(t,D_{x})$ and $B(t,x,D_{x})$ given by

$A(t,D_{x})=\begin{pmatrix}0&\langle D_{x}\rangle&0\\\ 0&0&\langle
D_{x}\rangle\\\ b_{(1)}&b_{(2)}&b_{(3)}\end{pmatrix}$

and

$B(t,x,D_{x})=\begin{pmatrix}0&0&0\\\ 0&0&0\\\
b_{1}-b_{(1)}&b_{2}-b_{(2)}&b_{3}-b_{(3)}\end{pmatrix}.$

The entries of $A(t,D_{x})$ and $B(t,x,D_{x})$ are given by

$\displaystyle b_{1}=A_{3}(t,x,D_{x})\langle D_{x}\rangle^{-2}\quad$
$\displaystyle\text{ and }\quad b_{(1)}=A_{(3)}(t,D_{x})\langle
D_{x}\rangle^{-2},$ $\displaystyle b_{2}=A_{2}(t,x,D_{x})\langle
D_{x}\rangle^{-1}\quad$ $\displaystyle\text{ and }\quad
b_{(2)}=A_{(2)}(t,D_{x})\langle D_{x}\rangle^{-1},$ $\displaystyle
b_{3}=A_{1}(t,x,D_{x})\langle D_{x}\rangle^{0}\quad$ $\displaystyle\text{ and
}\quad b_{(3)}=A_{(1)}(t,D_{x})\langle D_{x}\rangle^{0}.$

By implementing the transformation $U=TV$ on our system with

$T=\begin{pmatrix}1&0&0\\\ \lambda_{1}\langle\xi\rangle^{-1}&1&0\\\
\lambda_{1}^{2}\langle\xi\rangle^{-2}&(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}&1\end{pmatrix}\,\,\,\text{
and }\,\,T^{-1}=\begin{pmatrix}1&0&0\\\
-\lambda_{1}\langle\xi\rangle^{-1}&1&0\\\
\lambda_{1}\lambda_{2}\langle\xi\rangle^{-2}&-(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}&1\end{pmatrix},$

we are led to

$D_{t}V=T^{-1}ATV+T^{-1}BTV-T^{-1}(D_{t}T)V+F.$

Note that $(T^{-1}AT)(t,D_{x})=P(t,D_{x})$ where

$\displaystyle
P=T^{-1}AT=\begin{pmatrix}\lambda_{1}(t,\xi)&\langle\xi\rangle&0\\\
0&\lambda_{2}(t,\xi)&\langle\xi\rangle\\\
0&0&\lambda_{3}(t,\xi)\end{pmatrix},$

and

$(T^{-1}BT)(t,x,D_{x})-T^{-1}(D_{t}T)(t,D_{x})=L(t,x,D_{x})=D(t,x,D_{x})-E(t,D_{x}),$

where

$\displaystyle D(t,x,\xi)=T^{-1}BT(t,x,\xi)=\begin{pmatrix}0&0&0\\\ 0&0&0\\\
d_{3,1}&d_{3,2}&d_{3,3},\end{pmatrix},$

with

$\begin{split}d_{3,1}&=b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}+(b_{3}-b_{(3)})\lambda_{1}^{2}\langle\xi\rangle^{-2},\\\
d_{3,2}&=b_{2}-b_{(2)}+(b_{3}-b_{(3)})(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1},\\\
d_{3,3}&=b_{3}-b_{(3)},\end{split}$

and

$\begin{split}E(t,\xi)&=(T^{-1}D_{t}T)(t,\xi)\\\ &=\begin{pmatrix}0&0&0\\\
D_{t}\lambda_{1}\langle\xi\rangle^{-1}&0&0\\\
-D_{t}\lambda_{1}\langle\xi\rangle^{-2}(\lambda_{1}+\lambda_{2})+2\lambda_{1}D_{t}\lambda_{1}\langle\xi\rangle^{-2}&D_{t}(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}&0\end{pmatrix}.\end{split}$

The Levi conditions (12) formulated in Theorem 3.2 are given by

$\begin{split}e_{2,1}&\in C([0,T],S^{-1}(\mathbb{R}^{2n})),\\\
d_{3,2}-e_{3,2}&\in C([0,T],S^{-1}(\mathbb{R}^{2n})),\\\ d_{3,1}-e_{3,1}&\in
C([0,T],S^{-2}(\mathbb{R}^{2n})).\end{split}$

This means

$\begin{split}e_{2,1}&=D_{t}\lambda_{1}\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{2n})),\\\
d_{3,2}-e_{3,2}&=b_{2}-b_{(2)}+(b_{3}-b_{(3)})(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}-D_{t}(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{2n})),\\\
d_{3,1}-e_{3,1}&=b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}+(b_{3}-b_{(3)})\lambda_{1}^{2}\langle\xi\rangle^{-2}+D_{t}\lambda_{1}\langle\xi\rangle^{-2}(\lambda_{1}+\lambda_{2})\\\
&-2\lambda_{1}D_{t}\lambda_{1}\langle\xi\rangle^{-2}\in
C([0,T],S^{-2}(\mathbb{R}^{2n})).\end{split}$

######  Remark 3.4.

Since the roots $\lambda_{i}$ are symbols of order $1$, the Levi conditions
above imply that $\lambda_{1}$ is constant. The equation can have multiplicity
up to order $3$ as illustrated by the following example.

######  Example 3.5.

Let us work in $\mathbb{R}$ and set

$\begin{split}\lambda_{1}&=\xi,\\\ \lambda_{2}&=a(t)\xi,\\\
\lambda_{3}&=b(t)\xi.\end{split}$

The corresponding characteristic polynomial is

$(\tau-\xi)(\tau-b(t)\xi)(\tau-a(t)\xi)=\tau^{3}-(a(t)+b(t)+1)\xi\tau^{2}+(a(t)b(t)+a(t)+b(t))\xi^{2}\tau-a(t)b(t)\xi^{3}.$

Hence, we are considering the operator

$D_{t}^{3}-(a(t)+b(t)+1)D_{x}D_{t}^{2}+(a(t)b(t)+a(t)+b(t))D_{x}^{2}D_{t}+a(t)b(t)D_{x}^{3}$

with lower order terms

$a_{0,2}(t,x)D_{x}^{2}+a_{1,1}(t,x)D_{x}D_{t}+a_{2,0}(t,x)D_{t}^{2}+a_{0,1}(t,x)D_{x}+a_{1,0}(t,x)D_{t}+a_{0,0}(t,x).$

Hence,

$\begin{split}b_{1}-b_{(1)}&=(a_{0,2}(t,x)\xi^{2}+a_{0,1}(t,x)\xi+a_{0,0}(t,x))\langle\xi\rangle^{-2},\\\
b_{2}-b_{(2)}&=(a_{1,1}(t,x)\xi+a_{1,0}(t),x)\langle\xi\rangle^{-1},\\\
b_{3}-b_{(3)}&=a_{2,0}(t,x).\end{split}$

The first Levi condition

$d_{32}-D_{t}\lambda_{2}\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{2n})),$

where
$d_{32}=b_{2}-b_{(2)}+(b_{3}-b_{(3)})(\lambda_{1}+\lambda_{2})\langle\xi\rangle^{-1}$.
Then we have

$\displaystyle d_{32}-D_{t}\lambda_{2}\langle\xi\rangle^{-1}$
$\displaystyle=b_{2}-b_{(2)}+(b_{3}-b_{(3)})(\lambda_{1}(t,\xi)+\lambda_{2}(t,\xi))\langle\xi\rangle^{-1}-D_{t}\lambda_{2}\langle\xi\rangle^{-1}$
$\displaystyle=(a_{1,1}(t,x)\xi+a_{1,0}(t))\langle\xi\rangle^{-1}+a_{2,0}(t,x)(\xi+a(t)\xi)\langle\xi\rangle^{-1}-D_{t}a(t)\xi\langle\xi\rangle^{-1}$
$\displaystyle=((a_{1,1}(t,x)+(1+a(t)a_{2,0}(t,x)-D_{t}a(t))\xi+a_{1,0}(t,x))\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{2n})).$

This can be obtained by setting

$a_{1,1}(t,x)+(1+a(t))a_{2,0}(t,x)-D_{t}a(t)=0.$

The second Levi condition is written as

$\begin{split}d_{31}&=b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}+(b_{3}-b_{(3)})\lambda_{1}^{2}\langle\xi\rangle^{-2}\\\
&=b_{1}-b_{(1)}+(a_{1,1}(t)\xi+a_{1,0}(t))\xi\langle\xi\rangle^{-2}+a_{2,0}(t)\xi^{2}\langle\xi\rangle^{-2}\\\
&\in C([0,T],S^{-2}(\mathbb{R}^{2n})),\end{split}$

i.e.,

$\begin{split}&b_{1}-b_{(1)}+(b_{2}-b_{(2)})\xi\langle\xi\rangle^{-1}+(b_{3}-b_{(3)})\xi^{2}\langle\xi\rangle^{-2}\\\
&=(a_{0,2}(t,x)\xi^{2}+a_{0,1}(t,x)\xi+a_{0,0}(t,x))\langle\xi\rangle^{-2}+(a_{1,1}(t,x)\xi+a_{1,0}(t,x))\xi\langle\xi\rangle^{-2}+a_{2,0}(t,x)\xi^{2}\langle\xi\rangle^{-2}\\\
&=(a_{0,2}(t,x)+a_{1,1}(t,x)+a_{2,0}(t,x))\xi^{2}\langle\xi\rangle^{-2}+(a_{0,1}(t,x)+a_{1,0}(t,x))\xi\langle\xi\rangle^{-2}+a_{0,0}(t,x)\langle\xi\rangle^{-2}\\\
&\in C([0,T],S^{-2}(\mathbb{R}^{2n})).\end{split}$

This can be easily obtained by setting

$\begin{split}a_{0,2}(t,x)+a_{1,1}(t,x)+a_{2,0}(t),x&=0,\\\
a_{0,1}(t,x)+a_{1,0}(t,x)&=0.\end{split}$

Summarising, we choose lower order terms fulfilling

$\begin{split}a_{1,1}(t,x)+(1+a(t))a_{2,0}(t,x)-D_{t}a(t)&=0,\\\
a_{0,2}(t,x)+a_{1,1}(t,x)+a_{2,0}(t,x)&=0,\\\
a_{0,1}(t,x)+a_{1,0}(t,x)&=0.\end{split}$

Note that in this example we have multiplicity $1$ when $1\neq a(t)\neq b(t)$,
multiplicity $2$ when $a(t)=1\neq b(t)$, $a(t)\neq 1=b(t)$, $a(t)=b(t)\neq 1$
and multiplicity $3$ when $a(t)=b(t)=1$.

##  4\. Comparison with previous $C^{\infty}$ well-posedness results

In this section we compare our result with the few results of $C^{\infty}$
well-posedness known for higher order hyperbolic equations with multiplicities
and we provide some explanatory examples. We begin by saying that a general
treatment of $m$-order hyperbolic equations with time and space dependent
coefficients is still missing. Note that the recent result obtained in [27] is
valid only for equations with space dependent coefficients in space dimension
1. The focus of this paper is to allow variable multiplicities of any order
$m$, to work in any space dimension and to allow $(t,x)$-dependent lower order
terms. Let us start by analysing second order hyperbolic equations.

###  4.1. $C^{\infty}$ well-posdeness of second order hyperbolic equations

In her seminal paper [23] Oleinik consider second order hyperbolic operators
of the type

$Lu=u_{tt}-\sum_{i,j=1}^{n}(a_{ij}(t,x)u_{x_{j}})_{x_{i}}+\sum_{i=1}^{n}[(b_{i}(t,x)u_{x_{i}})_{t}+(b_{i}(t,x)u_{t})_{x_{i}}]\\\
+c(t,x)u_{t}+\sum_{i=1}^{n}d_{i}(t,x)u_{x_{i}}+e(t,x)u$

with smooth and bounded coefficients, i.e., coefficients in
$B^{\infty}([0,T]\times\mathbb{R}^{n})$, the space of smooth functions with
bounded derivatives of any order $k\geq 0$. She proves that the corresponding
Cauchy problem is $C^{\infty}$ well-posed provided that the lower order terms
fulfil a specific Levi condition known nowadays as Oleinik’s condition: there
exist $A,C>0$ such that

$t\biggl{[}\sum_{i=1}^{n}d_{i}(t,x)\xi_{i}\biggr{]}^{2}\leq
C\biggl{\\{}A\sum_{i,j=1}^{n}a_{ij}(t,x)\xi_{i}\xi_{j}-\sum_{i,j=1}^{n}\partial_{t}a_{ij}(t,x)\xi_{i}\xi_{j}\biggr{\\}},$

for all $t\in[0,T]$ and $x,\xi\in\mathbb{R}^{n}$. Note that Oleinik’s
condition is automatically fulfilled when the coefficients of the principal
part are independent of $t$ and the $d_{i}$’s vanish identically. Let us
consider the equation

$D_{t}^{2}u-a^{2}(t)D_{x}^{2}+ia_{0,1}(t,x)D_{x}u+ia_{1,0}(t,x)D_{t}u+a_{0,0}(t,x)u=f(t,x)$

where $a\geq 0$ is of class $C^{1}$ with respect to $t$. This equation has
roots $\lambda_{1}(t,\xi)=-a(t)\xi$ and $\lambda_{2}(t,\xi)=a(t)\xi$ which
coincide when $a$ vanishes. We assume for simplicity, that all the equation
coefficients are real valued. Making use of Oleinik’s formulation we get
equivalently the equation $Lu=-f$ where

$Lu=u_{tt}-a^{2}(t)u_{xx}-a_{0,1}u_{x}-a_{1,0}u_{t}-a_{0,0}(t,x)u.$

Hence, according to Oleinik the Cauchy problem for the equation above is
$C^{\infty}$ well-posed when

$ta_{0,1}^{2}(t,x)\leq C(Aa^{2}(t)-2a(t)\partial_{t}a(t)),$ (16)

for all $x\in\mathbb{R}$ and $t\in[0,T]$.

Let us now express our Levi conditions for the equation

$D_{t}^{2}u-a^{2}(t)D_{x}^{2}+ia_{0,1}(t,x)D_{x}u+ia_{1,0}(t,x)D_{t}u+a_{0,0}(t,x)u=f(t,x).$

We have to request

$d_{2,1}-e_{2,1}\in C([0,T],S^{-1}(\mathbb{R}^{2})).$

Since

$\displaystyle T^{-1}BT$ $\displaystyle=\begin{pmatrix}0&0\\\
b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}&b_{2}-b_{(2)}\end{pmatrix},$
$\displaystyle T^{-1}D_{t}T$ $\displaystyle=\begin{pmatrix}1&0\\\
-\lambda_{1}\langle\xi\rangle^{-1}&1\end{pmatrix}\begin{pmatrix}0&0\\\
D_{t}\lambda_{1}\langle\xi\rangle^{-1}&0\end{pmatrix}=\begin{pmatrix}0&0\\\
D_{t}\lambda_{1}\langle\xi\rangle^{-1}&0\end{pmatrix}.$

it follows that

$\displaystyle
b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}-D_{t}\lambda_{1}\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{2})).$

where

$\displaystyle b_{1}-b_{(1)}$
$\displaystyle:=(ia_{0,1}(t,x)\xi+a_{0,0}(t,x))\langle\xi\rangle^{-1},$
$\displaystyle b_{2}-b_{(2)}$ $\displaystyle:=ia_{1,0}(t,x).$

This can be equivalently formulated as

$(ia_{0,1}(t,x)+ia_{1,0}(t,x)a(t)-D_{t}a(t))\xi\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{2})).$

Our Levi condition is fulfilled by choosing $a_{0,1}$ and $a_{1,0}$ such that

$a_{0,1}(t,x)+a_{1,0}(t,x)a(t)-\partial_{t}a(t)=0.$ (17)

Note that (17) does not imply in general (16). Indeed, assuming that
$a_{1,0}(t,x)=0$ then we end up with

$a_{0,1}(t,x)=\partial_{t}a(t).$

In order for (16) to hold we need

$\begin{split}t(\partial_{t}a(t))^{2}&\leq
C(Aa^{2}(t)-2a(t)\partial_{t}a(t)),\\\ t(\partial_{t}a(t))^{2}&\leq
Ca(t)(Aa(t)-2\partial_{t}a(t)).\end{split}$

This is not true for $a(t)=t$. Indeed, it would lead to

$t\leq Ct(At-2)=CAt^{2}-2Ct\quad\Leftrightarrow\quad(1+2C)t\leq
CAt^{2}\quad\Leftrightarrow\quad\frac{1}{t}\leq\frac{CA}{1+2C},$

which does not hold uniformly on $(0,T]$. This example shows that our Levi
conditions allow us to handle second order hyperbolic equations that do not
fulfil Oleinik’s condition.

######  Example 4.1.

As an explanatory example let us consider the following characteristic roots:

$\displaystyle\lambda_{1}=t^{2\alpha}\xi,\quad\text{ and
}\quad\lambda_{2}=-t^{2\beta}\xi.$

Hence, we consider the following operator

$D_{t}^{2}+(t^{2\alpha}-t^{2\beta})D_{x}D_{t}-t^{2\alpha+2\beta}D_{x}^{2},$

with, for the sake of simplicity, $t$-dependent lower order terms

$a_{0,1}(t)D_{x}+a_{0,0}(t)+a_{1,0}(t)D_{t}.$

The Levi condition $d_{2,1}-e_{2,1}\in C([0,T],S^{-1}(\mathbb{R}^{2}))$ can be
written as

$b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}-D_{t}\lambda_{1}\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{2})).$

and holds if

$\displaystyle a_{0,1}(t)+t^{\alpha}a_{1,0}(t)$
$\displaystyle=D_{t}t^{\alpha}.$

In the sequel we give few examples of coefficients fulfilling the Levi
condition above and the corresponding second order operator. Note that no
assumptions are required on $a_{0,0}(t)$ a part from the standard continuity.

  * •

Let $a_{0,1}=0$ and $a_{1,0}=i^{-1}\alpha t^{-1}$. This leads to

$D_{t}^{2}u+(t^{2\alpha}-t^{2\beta})D_{t}D_{x}u-t^{2\alpha}D_{x}^{2}u-i^{-1}\alpha
t^{-1}D_{t}u+a_{0,0}(t)u=0.$

  * •

Let $a_{1,0}=0$ and $a_{0,1}=i^{-1}\alpha t^{\alpha-1}$. In this case, we get

$D_{t}^{2}u+(t^{2\alpha}-t^{2\beta})D_{t}D_{x}u-t^{2\alpha}D_{x}^{2}u-i^{-1}\alpha
t^{\alpha-1}D_{x}u+a_{0,0}(t)u=0.$

  * •

Let $a_{0,1}=i^{-1}\frac{\alpha}{2}t^{\alpha-1}$ and
$a_{1,0}=i^{-1}\frac{\alpha}{2}t^{-1}$. This case leads to

$D_{t}^{2}u+(t^{2\alpha}-t^{2\beta})D_{t}D_{x}u-t^{2\alpha}D_{x}^{2}u-i^{-1}\frac{\alpha}{2}t^{\alpha-1}D_{x}u-i^{-1}\frac{\alpha}{2}t^{-1}D_{t}u+a_{0,0}(t)u=0.$

######  Example 4.2.

As another example let us consider the characteristic roots:

$\displaystyle\lambda_{1}=\cos^{2}(t)\xi,\quad\text{ and
}\quad\lambda_{2}=1+\sin^{2}(t)\xi,$

where the double multiplicity is represented by

Hence, we consider the following operator

$D_{t}^{2}-2D_{x}D_{t}+\cos^{2}(t)(1+\sin^{2}(t))D_{x}^{2},$

with lower order terms

$a_{0,1}(t,x)D_{x}+a_{0,0}(t,x)+a_{1,0}(t,x)D_{t}$

will fulfil the Levi condition

$\displaystyle a_{0,1}(t)+\cos(t)a_{1,0}(t)$ $\displaystyle=D_{t}\cos(t)$

if $a_{1,0}(t)=0$, then it leads to $a_{0,1}(t)=i\sin(t)$.

We conclude this subsection with a higher-dimensional example. Note that many
results available in the literature hold only in space dimension 1, e.g., [26,
27], so our result has he technical advantage of being formulated in any space
dimension.

######  Example 4.3.

The second order operator with principal part

$D_{t}^{2}-a_{1}(t)D_{x_{1}}D_{t}-a_{2}(t)D_{x_{2}}D_{t}+a_{1}(t)a_{2}(t)D_{x_{1}}D_{x_{2}}$

has characteristic roots $\lambda_{1}(t,\xi)=a_{1}(t)\xi_{1}$ and
$\lambda_{1}(t,\xi)=a_{2}(t)\xi_{2}$ which coincides when
$a_{1}(t)=a_{2}(t)=0$. Writing the lower order terms in operator form as

$\sum_{j=1}^{2}a^{(j)}_{0,1}(t)D_{x_{j}}+a_{1,0}(t)D_{t}+a_{0,0}(t,x),$

we can now formulate the Levi condition

$b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}-D_{t}\lambda_{1}\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{4})),$

as

$\biggl{(}\sum_{j=1}^{2}a^{(j)}_{0,1}(t)\xi_{j}+a_{1,0}(t)\lambda_{1}(t,\xi)-D_{t}\lambda_{1}(t,\xi)\biggr{)}\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{4})).$

This leads to

$\biggl{(}\sum_{j=1}^{2}a^{(j)}_{0,1}(t)\xi_{j}+a_{1,0}(t)a_{1}(t)\xi_{1}-D_{t}a_{1}(t)\xi_{1}\biggr{)}\langle\xi\rangle^{-1}\in
C([0,T],S^{-1}(\mathbb{R}^{4}))$

which is fulfilled by choosing $a^{(2)}_{0,1}=0$ and

$a^{(1)}_{0,1}(t)+a_{1,0}(t)a_{1}(t)=D_{t}a_{1}(t)$

###  4.2. $C^{\infty}$ well-posdeness of third order hyperbolic equations

Our result requires a limited level of regularity on the coefficients. Indeed,
we need continuity with respect to $t$ and that the roots $\lambda_{i}(t,\xi)$
are of class $C^{1}$ with respect to $t$. Note that in [7] the coefficients of
the principal part are assumed to be of class $C^{2}$ and all the
coefficients, also the lower order terms, need to be time-dependent only. We
can therefore provide bigger generality under this point of view. As an
explanatory example, let us consider a third order operator with
characteristic roots:

$\displaystyle\lambda_{1}=0,\quad\lambda_{2}=a(t)\xi,\quad\lambda_{3}=b(t)\xi,$

where $a,b\in C^{1}([0,T])$. This is an example of variable multiplicity,
since we have multiplicity $3$ when $a(t)=b(t)=0$, multiplicity $2$ when
$a(t)=b(t)\neq 0$, $a(t)=0\neq b(t)$ and $a(t)\neq 0=b(t)$, and multiplicity
$1$ otherwise. We can therefore consider the operator

$D_{t}^{3}-(a(t)+b(t))D_{x}D_{t}^{2}+a(t)b(t)D_{x}^{3},$

with lower order terms

$a_{0,2}(t,x)D_{x}^{2}+a_{1,1}(t,x)D_{x}D_{t}+a_{2,0}(t,x)D_{t}^{2}+a_{0,1}(t),xD_{x}+a_{1,0}(t,x)D_{t}+a_{0,0}(t,x).$

Note that for the sake of simplicity we work in space dimension $1$. Our Levi
conditions on the lower order terms are formulated as follows:

$\begin{split}a_{1,1}(t,x)+a(t)a_{2,0}(t,x)-D_{t}a(t)&=0,\\\
a_{0,2}(t,x)+a_{1,1}(t,x)+a_{2,0}(t,x)&=0,\\\
a_{0,1}(t,x)+a_{1,0}(t,x)&=0.\end{split}$

Note that if we choose $a_{2,0}=f(t)+g(x)$ then the first Levi condition holds
when

$\begin{split}a_{1,1}(t,x)+a(t)g(x)&=0,\\\ a(t)f(t)-D_{t}a(t)&=0.\end{split}$

######  Example 4.4.

Let us select the characteristic roots:

$\lambda_{1}=\xi,\quad\lambda_{2}=\cos(t)\xi,\quad\lambda_{3}=(1+\sin^{2}(t))\xi,$

where the variable multiplicities are represented by

Hence, we are considering the operator

$\displaystyle
D_{t}^{3}+(\cos(t)\sin^{2}(t)+2\cos(t)+1+\sin^{2}(t))D_{x}^{2}D_{t}$
$\displaystyle-(\cos(t)+\sin^{2}(t)+2)D_{x}D_{t}^{2}+\cos(t)(1+\sin^{2}(t))D_{x}^{3},$

with lower order terms

$a_{0,2}(t,x)D_{x}^{2}+a_{1,1}(t,x)D_{x}D_{t}+a_{2,0}(t,x)D_{t}^{2}+a_{0,1}(t,x)D_{x}+a_{1,0}(t,x)D_{t}+a_{0,0}(t,x).$

Then Levi conditions on the lower order terms:

$\displaystyle a_{1,1}(t,x)+(1+\cos(t))a_{2,0}(t,x)$ $\displaystyle=i\sin(t),$
$\displaystyle a_{0,2}(t,x)+a_{1,1}(t,x)+a_{2,0}(t,x)$ $\displaystyle=0,$
$\displaystyle a_{0,1}(t,x)+a_{1,0}(t,x)$ $\displaystyle=0.$

Note that by choosing $a_{1,1}=0$ and an interval $[0,T]$ where
$(1+\cos(t))\neq 0$ then both $a_{2,0}$ and $a_{0,2}$ are time-dependent and
can be expressed as

$a_{2,0}=i\frac{\sin(t)}{1+\cos(t)}$

with $a_{0,2}=-a_{2,0}$.

###  4.3. $C^{\infty}$ well-posedness of fourth order hyperbolic equations

We conclude this section with an explanatory fourth order example.

######  Example 4.5.

Let us work in $\mathbb{R}$ and set

$\lambda_{1}=\xi,\,\,\lambda_{2}=-\xi,\,\,\,\lambda_{3}=\sqrt{a(t)}\xi,\,\,\,\lambda_{4}=-\sqrt{a(t)}\xi,$

where $a(t)\geq 0$. The corresponding characteristic polynomial is

$(\tau^{2}-\xi^{2})(\tau^{2}-a(t)\xi^{2})=\tau^{4}-(a(t)+1)\xi^{2}\tau^{2}+a(t)\xi^{4}.$

Hence, we consider the operator

$D_{t}^{4}-(a(t)+1)D_{x}^{2}D_{t}^{2}+a(t)D_{x}^{4}$

with lower order terms

$\displaystyle b_{1}-b_{(1)}$
$\displaystyle=(a_{0,3}(t,x)\xi^{3}+a_{0,2}(t,x)\xi^{2}+a_{0,1}(t,x)\xi+a_{0,0}(t,x))\langle\xi\rangle^{-3},$
$\displaystyle b_{2}-b_{(2)}$
$\displaystyle=(a_{1,3}(t,x)\xi^{3}+a_{1,2}(t,x)\xi^{2}+a_{1,1}(t,x)\xi+a_{1,0}(t,x))\langle\xi\rangle^{-2},$
$\displaystyle b_{3}-b_{(3)}$
$\displaystyle=(a_{2,1}(t,x)\xi+a_{2,0}(t,x))\langle\xi\rangle^{-1},$
$\displaystyle b_{4}-b_{(4)}$ $\displaystyle=a_{3,1}(t,x)\xi+a_{3,0}(t,x).$

The first Levi condition

$\displaystyle d_{43}-e_{43}$
$\displaystyle=b_{3}-b_{(3)}+(b_{4}-b_{(4)})\lambda_{3}\langle\xi\rangle^{-1}-D_{t}\lambda_{3}\langle\xi\rangle^{-1}$
$\displaystyle=(a_{2,1}\xi+a_{2,0}+(a_{3,1}\xi+a_{3,0})\sqrt{a(t)}\xi-
D_{t}\sqrt{a(t)}\xi)\langle\xi\rangle^{-1},$

is fulfilled if

$\displaystyle a_{2,1}(t,x)+\sqrt{a(t)}a_{3,0}(t,x)=D_{t}\sqrt{a(t)},$
$\displaystyle a_{3,1}(t,x)=0.$

The second Levi condition

$\displaystyle d_{42}-e_{42}$
$\displaystyle=b_{2}-b_{(2)}+(b_{4}-b_{(4)})(\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{1}\lambda_{2})\langle\xi\rangle^{-2}$
$\displaystyle=(a_{1,3}\xi^{3}+a_{1,2}\xi^{2}+a_{1,1}\xi+a_{1,0}+(a_{3,1}\xi+a_{3,0})\xi^{2})\langle\xi\rangle^{-2}$

is fulfilled if

$\displaystyle a_{1,3}(t,x)+a_{3,1}(t,x)=0,$ $\displaystyle
a_{1,2}(t,x)+a_{3,0}(t,x)=0,$ $\displaystyle a_{1,1}(t,x)=0.$

The third Levi condition

$\displaystyle d_{41}-e_{41}$
$\displaystyle=b_{1}-b_{(1)}+(b_{2}-b_{(2)})\lambda_{1}\langle\xi\rangle^{-1}+(b_{3}-b_{(3)})\lambda_{1}^{2}\langle\xi\rangle^{-2}+(b_{4}-b_{(4)})\lambda_{1}^{3}\langle\xi\rangle^{-3}$
$\displaystyle=(a_{0,3}\xi^{3}+a_{0,2}\xi^{2}+a_{0,1}\xi+a_{0,0}+a_{1,3}\xi^{4}+a_{1,2}\xi^{3}+a_{1,1}(t,x)\xi^{2}+a_{1,0}\xi)\langle\xi\rangle^{-3}$
$\displaystyle+(a_{2,1}\xi^{3}+a_{2,0}\xi^{2}+a_{3,1}\xi^{4}+a_{3,0}\xi^{3})\langle\xi\rangle^{-3}$

should be fulfilled if

$\displaystyle a_{0,3}(t,x)+a_{1,2}(t,x)+a_{2,1}(t,x)+a_{3,0}(t,x)=0,$
$\displaystyle a_{0,2}(t,x)+a_{1,1}(t,x)+a_{2,0}(t,x)=0,$ $\displaystyle
a_{0,1}(t,x)+a_{1,0}(t,x)=0,$ $\displaystyle a_{3,1}(t,x)+a_{1,3}(t,x)=0.$

In conclusion, we have identified the following Levi conditions on the lower
order terms:

$\displaystyle a_{2,1}(t,x)+\sqrt{a(t)}a_{3,0}(t,x)=D_{t}\sqrt{a(t)},$
$\displaystyle a_{1,2}(t,x)+a_{3,0}(t,x)=0,$ $\displaystyle
a_{2,1}(t,x)+a_{0,3}(t,x)=0,$ $\displaystyle a_{0,2}(t,x)+a_{2,0}(t,x)=0,$
$\displaystyle a_{0,1}(t,x)+a_{1,0}(t,x)=0,$ $\displaystyle
a_{1,3}=a_{3,1}=a_{1,1}=0.$

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