76 Linear Cauchy problem with variable coefficients
we see that (ξ,λ)and(ν, n
0
) are not parallel, unless
ξ = 0. By assumption, either
n(x
0
,t
0
)or−n(x
0
,t
0
) belongs to Γ(x
0
,t
0
). Then we know from Theorem 1.5 of
Chapter 1 that the operator is constantly hyperbolic in the direction n,which
means that for all (ξ, λ) not parallel to n =(ν, n
0
), the roots σ of the polynomial
det(A(x
0
,t
0
,ξ+ σν)+(λ + σn
0
)I
n
)
are real with constant multiplicities. Because of (2.2.20) this shows that the roots
τ of
det( τ
A
0
(0, 0) +
A(0, 0,
ξ, τ ))
are real with constant multiplicities for
ξ = 0. This means that the transformed
operator
L is constantly hyperbolic at point (0, 0), and thus also in the neigh-
bourhood of (0, 0). Up to replacing
A
α
(x, t)by
A
α
( θ((x, t))(x, t) ) for all
α ∈{0, ···,d},whereθ is a smooth cut-off function, we can assume that
L
is globally defined, constantly hyperbolic, and has constant coefficients outside
some bounded ball. So we are led to show the result for
L instead of L,and
H := {(x,
t);
t =0} instead of H.
From now on we drop the tildas. The (hyperbolic system associated with)
operator L
∗
meets the assumptions of Theorem 2.3. Thus Theorem 2.6 applies
to L
∗
. This will enable us to apply the Holmgren principle to L.
We must show the existence of a neighbourhood N of (0, 0) such that
if u is a C
1
solution of Lu =0 in N and u(x, 0) = 0 for (x, 0) ∈ N then
u(x, t) = 0 for all (x, t) ∈ N . Without loss of generality, we can consider a conical
neighbourhood N , foliated by the hypersurfaces with boundary
H
θ
:= {(x, t); θ
3
x
2
− V
2
( t − θT )
2
+ θ
2
V
2
T
2
(1 − θ)=0,
0 ≤ t ≤ θT, x≤VT},θ∈ [0, 1) .
(The reader may refer to Fig. 2.3.) Choosing
V =max{|λ
j
(x, t, ξ)|; ξ =1},
where λ
j
(x, t, ξ) denote, as usual, the eigenvalues of A(x, t, ξ) (recall that
A(x, t, ξ) has been modified to be independent of (x, t) outside a bounded ball), it
is not difficult to show that all the hypersurfaces H
θ
are space-like, as H
0
⊂H.
To be precise, one can compute that a unit normal vector to H
θ
at point
(x, t)readsn =(ν, n
0
)=
N/
N where
N := ( θ
3
x, V
2
(θT − t) ). And thus the
matrix (A(x, t, ν)+n
0
I
n
) has eigenvalues
µ
j
=
1
N
V
2
(θT − t)+λ
j
(x, t, θ
3
x)
.
By definition of V ,wehave
|λ
j
(x, t, θ
3
x)|≤Vθ
3
x,