Weakly dissipative symmetrizer 191
Because of |p|≤c, we deduce that necessarily q =
b. Since then |p| = c, the energy
E
p
is convex but not strictly. Presumably, the corresponding homogeneous
IBVP is not strongly well-posed in L
2
(Ω).
Finally, we consider the (UKL) case (7.1.17). The same choice as above
yields the same formula (7.1.20), where the quadratic form is now negative-
definite. Hence the IBVP is strictly dissipative in the Friedrichs sense. Notice
that in the simple case a<0,
b = 0, the naive choice p = 0 yields weak but not
strong dissipativeness; the corrector p
d
∂
t
u∂
d
u in the energy thus plays a rather
important role.
7.2 Weakly dissipative symmetrizer
We turn to the construction of a weakly dissipative symbolic symmetrizer. We
recall the properties to fulfil:
i) The map (τ,η) → K(τ,η)isC
∞
on Re τ ≥ 0, η ∈ R
d−1
, |τ | + |η| =0,and
is homogeneous of degree zero,
ii) the matrix Σ(τ,η):=K(τ, η)A
d
is Hermitian,
iii) the Hermitian form x
∗
Σ(τ,η)x is non-positive on kerB,
iv) there exists a number c
0
> 0 such that, for every (τ,η), the inequality
Re M ≥ c
0
(Re τ)I
n
holds in the sense of Hermitian matrices, where
M = M (τ,η):=−(ΣA)(τ, η)=K(τ,η)(τI
n
+ iA(η)).
We assume throughout this section that the operator L is constantly hyper-
bolic, and that the Lopatinski˘ı condition holds true everywhere in the closed
hemisphere Re τ ≥ 0, |τ |
2
+ |η|
2
= 1, but at some elliptic points of the boundary
Re τ = 0. In particular, n =2p. As explained in the previous section, we ask that
the Lopatinski˘ı determinant ∆ vanish in a non-degenerate way: its differential
does not vanish simultaneously. Note that we do not allow
4
∆ to vanish at some
glancing point.
We may anticipate that a dissipative symmetrizer will degenerate somehow
at boundary points P
0
where the Lopatinski˘ı condition fails. For we already now
that Σ(τ, η), restricted to E
−
(τ,η), is positive when Re τ>0. By continuity, the
restriction of Σ(P
0
)toE
−
(P
0
) must be non-negative. Since its restriction to kerB
is non-positive, we see that Σ(P
0
) vanishes on E
−
(P
0
) ∩ kerB. This immediately
implies that E
−
(P
0
) ∩ kerB is contained in the kernel of the restrictions of Σ(P
0
)
to both E
−
(P
0
)andkerB. In other words, it holds that
w
∗
Σ(P
0
)v =0, ∀v ∈ E
−
(P
0
) ∩ kerB, ∀w ∈ E
−
(P
0
)+kerB. (7.2.21)
Actually, noticing that x
d
→ v
∗
Σv is non-increasing along the flow of A (because
of Re (ΣA) ≤ 0
n
), we deduce the more remarkable constraint that v
∗
Σ(P
0
)v
vanishes identically over the Krylov space of E
−
(P
0
) ∩ kerB for A(P
0
). Recall
4
Thecase(n, p)=(2, 1) with a constantly hyperbolic operator is isomorphic to the system (1.2.17),
for which the analysis is explicit and similar to that of Section 7.1.1.