Exact observation of a
wave equation on
non-cylindrical domains
Bernhard
Haak
and
Trung Hoang
University of Bordeaux
Motivation
Consider an observed abstract non-autonomous
equation
 0
 x (t) + A(t)x(t) = 0
x(0)
= x0

y (t) = C (t)x(t)
on one side and, for all s ≥ 0,
 0
 x (t) + A(s)x(t) = 0
x(0)
= x0

y (t) = C (t)x(t)
on the other. Can one transfer properties?
(1)
(2.s)
Non-cylindrical domain
Let s > 0 of class C 2 , s(0) = 1 with ks 0 k∞ < 1. On
Ω = (x, t) ∈ R2 : t ≥ 0 and 0 ≤ x ≤ s(t) ,
consider a wave equation with
Dirichlet boundary condition.

utt − uxx = 0
(x, t) ∈ Ω



u(0, t) = u(s(t), t) = 0
t≥0
u(x, 0) = g (x)
x ∈ [0, 1]



ut (x, 0) = f (x)
x ∈ [0, 1]
x = s(t)
t
Ω
1
x
Series representation
Aim:
u(x, t) :=
X
n∈Z
An e 2πin ϕ(t+x)) − e 2πin ϕ(t−x))
Series representation
Aim:
u(x, t) :=
X
An e 2πin ϕ(t+x)) − e 2πin ϕ(t−x))
n∈Z
where ϕ satisfies
ϕ(t + s(t)) − ϕ(t − s(t)) = 1.
and is of class C 2 .
Series representation
Aim:
u(x, t) :=
X
An e 2πin ϕ(t+x)) − e 2πin ϕ(t−x))
n∈Z
where ϕ satisfies
ϕ(t + s(t)) − ϕ(t − s(t)) = 1.
and is of class C 2 .
Assume that additionally ϕ0 > 0. Then
kf kL2 (a,b) ∼ kf kL2 (a,b,ϕ0 dx) and properties of u can be
reduced to classical Fourier series by a change of var.
Abel’s equation 1
Let α(t) = t + s(t), β(t) = t − s(t) and γ = α ◦ β −1 . Then
γ : [−1, ∞) −→ [+1, ∞) is strictly increasing.
Reformulated problem:
(Abel)
ϕ◦γ =ϕ+1
on
[−1, ∞).
Abel’s equation 1
Let α(t) = t + s(t), β(t) = t − s(t) and γ = α ◦ β −1 . Then
γ : [−1, ∞) −→ [+1, ∞) is strictly increasing.
Reformulated problem:
ϕ◦γ =ϕ+1
(Abel)
(easy) case:
lim s 0 (t)
= s ∈ (0, 1). Then
[−1, ∞).
on
lim γ 0
=
1−s
1+s
=: ` > 1.
Abel’s equation 1
Let α(t) = t + s(t), β(t) = t − s(t) and γ = α ◦ β −1 . Then
γ : [−1, ∞) −→ [+1, ∞) is strictly increasing.
Reformulated problem:
ϕ◦γ =ϕ+1
(Abel)
on
(easy) case: lim s 0 (t) = s ∈ (0, 1). Then lim γ 0 =
We follow ideas of Lévy, Szekeres, ...
In the easy case ψ = `ϕ satisfies
(Schröder)
and ψ 0 ◦ ϕ =
`
ϕ0
ψ◦ϕ=`·ψ
· ψ 0 suggests to study
∞
Y
γ 0 (γ (n) (x))
P(x) =
`
n=0
on
[−1, ∞).
1−s
1+s
=: ` > 1.
[−1, ∞).
Abel’s equation 2
The product converges uniformly on [−1, ∞). Finally,
Z x
ψ(x) :=
P(t) dt + C
−1
solves (Schröder) for some (unique) C .
Abel’s equation 2
The product converges uniformly on [−1, ∞). Finally,
Z x
ψ(x) :=
P(t) dt + C
−1
solves (Schröder) for some (unique) C .
(less easy) case:
lim s 0 (t)
= 0, so
lim γ 0
=
1−s
1+s
= 1.
Abel’s equation 2
The product converges uniformly on [−1, ∞). Finally,
Z x
ψ(x) :=
P(t) dt + C
−1
solves (Schröder) for some (unique) C .
(less easy) case: lim s 0 (t) = 0, so lim γ 0 =
γ 0 (t) = 1 + o(t −δ ). Then whenever
ϕ(x) :=
1−s
1+s
= 1. Assume
γ (n) (x) − γ (n) (x0 )
n→∞ γ (n+1) (x0 ) − γ (n) (x0 )
lim
exists, is solves (Abel). This is hard to check / calculate.
Abel’s equation 3
Let ψ = g · ϕ0 . Then
(ψ ◦ γ)(x) =
g (γ(x))
ψ(x)
g (x)γ 0 (x)
Abel’s equation 3
Let ψ = g · ϕ0 . Then
g (γ(x))
ψ(x)
g (x)γ 0 (x)
(ψ ◦ γ)(x) =
Assume γ 0 (x) = 1 + ax −δ + o(x −δ ), a > 0 . Then
∞
Y
g (xn )γ 0 (xn )
P(x) =
g (xn+1 )
converges for g (x) = γ(x)1−δ ,
n=0
where xn = γ (n) (x). Then
Z
ϕ(x) := C
solves (Abel).
1
x
P(t)
dt.
γ(t)1−δ
Explicit examples
- linear moving boundary: s(t) = 1 + εt for ε ∈ (0, 1)
+ε −1
ϕ(t) = ln( 11−ε
)
ln(1+εt)
Explicit examples
- linear moving boundary: s(t) = 1 + εt for ε ∈ (0, 1)
+ε −1
ϕ(t) = ln( 11−ε
)
- parabolic boundary s(t) =
ϕ(t) =
1
2ε
√
ln(1+εt)
1 + εt
for ε ∈ (0, 2)
p
ε2 + 4εt + 4
Explicit examples
- linear moving boundary: s(t) = 1 + εt for ε ∈ (0, 1)
+ε −1
ϕ(t) = ln( 11−ε
)
- parabolic boundary s(t) =
ϕ(t) =
1
2ε
√
ln(1+εt)
1 + εt
for ε ∈ (0, 2)
p
ε2 + 4εt + 4
1
- shrinking domain s(t) = 1+εt
for ε ∈ (0, 1)
6
5
4
3
2
1
0
0
1
2
3
4
5
6
Series representation
Call s admissible if ϕ ∈ C 2 and ϕ0 > 0.
Series representation
Call s admissible if ϕ ∈ C 2 and ϕ0 > 0. Recall:
X (Series)
u(x, t) :=
An e 2πin ϕ(t+x)) −e 2πin ϕ(t−x))
n∈Z
Series representation
Call s admissible if ϕ ∈ C 2 and ϕ0 > 0. Recall:
X (Series)
u(x, t) :=
An e 2πin ϕ(t+x)) −e 2πin ϕ(t−x))
n∈Z
Let F (x) =
h(x) :=
1
2
·
Rx
f (s) ds and
g (x) + F (x)
−g (−x) + F (−x)
0
for
for
0≤x ≤1
−1 ≤ x < 0
Series representation
Call s admissible if ϕ ∈ C 2 and ϕ0 > 0. Recall:
X (Series)
u(x, t) :=
An e 2πin ϕ(t+x)) −e 2πin ϕ(t−x))
n∈Z
Let F (x) =
h(x) :=
1
2
·
Rx
f (s) ds and
g (x) + F (x)
−g (−x) + F (−x)
0
for
for
0≤x ≤1
−1 ≤ x < 0
P
Develop h(t) = An bn (t) with bn (t) = e 2πin ϕ(t) in
H = L2 (−1, 1; ϕ0 (t) dt).
Series representation
Call s admissible if ϕ ∈ C 2 and ϕ0 > 0. Recall:
X (Series)
u(x, t) :=
An e 2πin ϕ(t+x)) −e 2πin ϕ(t−x))
n∈Z
Let F (x) =
h(x) :=
1
2
·
Rx
f (s) ds and
g (x) + F (x)
−g (−x) + F (−x)
0
for
for
0≤x ≤1
−1 ≤ x < 0
P
Develop h(t) = An bn (t) with bn (t) = e 2πin ϕ(t) in
H = L2 (−1, 1; ϕ0 (t) dt).P
Important Lemma:
n2 |An |2 ∼ k(g , f )kH 1 ×L2
Point observations (1)
Let
m(t) =
M(t) =
0
min{ϕ
(x) : x ∈ [t − s(t), t + s(t)]}
max{ϕ
0
(x) : x ∈ [t − s(t), t + s(t)]}.
and
Point observations (1)
Let
m(t) =
M(t) =
0
min{ϕ
(x) : x ∈ [t − s(t), t + s(t)]}
max{ϕ
0
and
(x) : x ∈ [t − s(t), t + s(t)]}.
Theorem: If s is admissible the
(g , f )2 1
H ×L
0
Z
2
∼
0
γ(0) ux (0, t)2 dt
with explicit constants in terms of m(.) ad M(.).
In particular, with the observations C ψ = ψx (0) the
wave eq. is exactly observable in time τ if and only if
τ ≥ γ(0).
Point observations (2)
Theorem: For admissible s(.),
(g , f )2 1
∼
H ×L
0
2
Z
0
γ −1 (0) ux (s(t), t)2 dt
with explicit constants in the double
inequality.
In particular, with the observations
C (t)ψ = ψx (s(t)) the wave eq. is exactly observable in time τ if and only
if τ ≥ γ −1 (0).
Point observations (2)
t
Theorem: For admissible s(.),
(g , f )2 1
∼
H ×L
0
2
Z
0
γ −1 (0) ux (s(t), t)2 dt
γ(0)
γ −1 (0)
with explicit constants in the double
inequality.
In particular, with the observations
C (t)ψ = ψx (s(t)) the wave eq. is exactly observable in time τ if and only
if τ ≥ γ −1 (0).
1
x
Point observations (3)
Theorem: If s monotonic and admissible boundary
curve. Assume additionally that ϕ0 is strictly decreasing
if s(·) is increasing or that ϕ0 is strictly increasing if
s(·) is decreasing, respectively.
Then
Z a+γ(−a)
(g , f )2 1
ux (a, t)2 dt
∼
H0 ×L2
with explicit constants.
0
Point observations (3)
Theorem: If s monotonic and admissible boundary
curve. Assume additionally that ϕ0 is strictly decreasing
if s(·) is increasing or that ϕ0 is strictly increasing if
s(·) is decreasing, respectively.
Then
Z a+γ(−a)
(g , f )2 1
ux (a, t)2 dt
∼
H0 ×L2
0
with explicit constants.
Remark: in case s(t) = 1, and hence ϕ(x) = x we have
X
u(x, t) =
an e iπnt sin nπx .
n∈Z
So point observation at x=a is not possible when
a ∈ Q since then infinitely many terms in the sum
vanish, independently of the leading coefficient.
Simultaneous obs.
t
v
u

utt − uxx = 0
(x, t) ∈ Ω




u(
0
,
t)
=
u(s(t),
t)
=0
t≥0




u(x, 0) = g (x)
x ∈ [0, 1]


 u (x, 0) = f (x)
x ∈ [0 , 1 ]
t
vtt − vxx = 0
−1≤x ≤0




v
(−
1
,
t)
=
v
(
0
,
t)
=0
t≥0




e
v
(x,
0
)
=
g
(x)
x
∈
[−
1
, 0]



e
vt (x, 0) = f (x)
x ∈ [−1, 0]
x
−1
1
We observe t 7→ ux (0, t) + vx (0, t), the joint force at
x =0
Simultaneous obs
Theorem: Let s(·) satisfy either the situation (easy
case) or the (less easy case) in the study of Abel’s
equation.
Moreover assume that ϕ0 is bounded on R+ . Let (u, v )
be the solution to the simultaneous wave equation.
Then, for all λ > 0 there exists τ0 > 2 such that for all
τ ≥ τ0
Z τ
2
2
e
ux (0, t)+vx (0, t)2 dt
λ (g , f ) H 0 ×L + (e
g , f ) H 0 ×L
≤
1
1
1
2
0
Simultaneous obs
Theorem: Let s(·) satisfy either the situation (easy
case) or the (less easy case) in the study of Abel’s
equation.
Moreover assume that ϕ0 is bounded on R+ . Let (u, v )
be the solution to the simultaneous wave equation.
Then, for all λ > 0 there exists τ0 > 2 such that for all
τ ≥ τ0
Z τ
2
2
e
ux (0, t)+vx (0, t)2 dt
λ (g , f ) H 0 ×L + (e
g , f ) H 0 ×L
≤
1
1
1
2
Thank you.
0