Observability for parabolic equations
from a measurable set in time
Kim Dang PHUNG , joint work Gengsheng Wang
Université d'Orléans.
27 September 2011, Orléans, GDRE CONEDP
EXACT CONTROL FOR THE HEAT EQUATION
Let Ω be a bounded smooth domain in Rn , ω ⊂ Ω, T > 0. Given
y0 ∈ L2 (Ω), there is (y, f ) satisfying

∂t y − ∆y = 1|ω f




 y=0
y (·, 0) = y0


y (·, T ) = 0



kf kL2 (Ω×(0,T )) ≤ c ky0 kL2 (Ω)
G. Lebeau, L. Robbiano,
in Ω × (0, T ) ,
on ∂Ω × (0, T ) ,
in Ω ,
in Ω ,
.
CPDE (1995).
A. Fursikov, O. Imanuvilov,
parabolic equations
Lecture Note Series (1996).
Also OK for
EQUIVALENTLY

in Ω × (0, T ) ,
 ∂t u − ∆u = 0
u=0
on ∂Ω × (0, T ) ,

2
u (·, 0) ∈ L (Ω) ,
ku (·, T )k2L2 (Ω)
Z Z
≤c
ω
0
T
|u (x, t)|2 dxdt .
ORIGINAL LEBEAU-ROBBIANO STRATEGY
Hölder type Observation for elliptic
1−α α
kwkH 1 (Ω×(γ,T −γ)) ≤ C kwkL2 (ω×(0,T ))
kwkH 1 (Ω×(0,T ))
f or any w ∈ H 2 (Ω × (0, T )) ,
=⇒
∂t2 + ∆ w = 0 and w|∂Ω = 0 .
Controllability in nite and innite dim
=⇒
Observability
SHORTCUT OF LEBEAU-ROBBIANO STRATEGY
Hölder type Observation for elliptic
=⇒
Sum of eigenfunctions estimate
=⇒
G. Lebeau, E. Zuazua,
L. Miller,
DCDS (2010).
Observability
ARMA (1998).
PLAN & AIM
I. A similar way for parabolic equations
II. Replace (0, T ) by a measurable set E in time, |E| > 0
III. Time optimal bang-bang control
NEW OBSERVABILITY INEQUALITY
a ∈ L∞ (0, T ; Lq (Ω)), q ≥ 2 and b ∈ L∞ (Ω × (0, T ))n

 ∂t u − ∆u + au + b · ∇u = 0 in Ω × (0, T ) ,
u=0
on ∂Ω × (0, T ) ,

u (·, 0) ∈ L2 (Ω) ,
Ω convex, ω ⊂ Ω, E ⊂ (0, T ) and |E| > 0
Z Z
ku (·, T )kL2 (Ω) ≤ c
ω
E
|u (x, t)| dxdt .
STRATEGY FOR PARABOLIC
Hölder type Observation from one point in time
C
ku (·, `)kL2 (Ω) ≤ Ce `
ku (·, `)kL2 (ω)
1−α ku (·, 0)kL2 (Ω)
=⇒
Observability for a measurable set in time
Z Z
ku (·, T )kL2 (Ω) ≤ c
|u (x, t)| dxdt
ω
E
α
Proof
Step I. Observation from one point in time
→here Ω convex or star-shaped domain
→interpolation inequality, parameter ε > 0
Step II. Usage of set of positive measure, E ⊂ (0, T ) and |E| > 0
→Construction of a sequence of points {`j }j≥1
→parameter z > 1
Step III. Choice of the parameters: ε and z
Observation from a point in time
ku (·, `)kL2 (Ω) ≤ Ce
C
`
ku (·, `)kL2 (ω)
1−α ku (·, 0)kL2 (Ω)
α
=⇒
ku (·, `)kL2 (Ω) ≤
C
1
Ce ` ku (·, `)kL1 (ω) + ε ku (·, 0)kL2 (Ω)
εγ
∀ε > 0
Set of positive measure
Let T > 0 and E ⊂ (0, T ) be a set of positive measure. Let L be
a point of density for E . Then for any z > 1, there exists
`1 ∈ (L, T ) such that the decreasing sequence {`j }j≥1 given by
`j+1 = L +
1
zj
(`1 − L)
converges to L and satises
`j − `j+1 ≤ 3 |E ∩ (`j+1 , `j )| .
End of proof 1/2
Take 0 < `j+2 < `j+1 ≤ t ≤ `j < T ,
ku (·, t)kL2 (Ω) ≤ ε ku (·, `j+2 )kL2 (Ω)
1
+ γ CeC/(t−`j+2 ) ku (·, t)kL1 (ω)
ε
We know that ku (·, `j )kL2 (Ω) ≤ c ku (·, t)kL2 (Ω) .
We integrate over E ∩ (`j+1 , `j )
|E ∩ (`j+1 , `j )| c ku (·, `j )kL2 (Ω)
≤ |E ∩ (`j+1 , `j )| ε ku (·, `j+2 )kL2 (Ω)
Z
C C/(`j+1 −`j+2 )
+ γ e
ku (·, t)kL1 (ω) dt
ε
E∩(`j+1 ,`j )
End of proof 2/2
We use `j
h
−C
εγ e
1
z j+2
`1 −L z(z−1)
i
h
−C
ku (·, `j )kL2 (Ω) − εγ+1 e
z j+2
1
`1 −L z(z−1)
i
ku (·, `j+2 )kL2 (Ω
Z
≤c
E∩(`j+1 ,`j )
Set d = C
h
ku (·, t)kL1 (ω) dt
1
1
`1 −L z(z−1)
We choose ε = e−dz
j
i
j+2
.
and (γ + 1) z 2 = (γ + 2)
e−d(γ+2)z ku (·, `j )kL2 (Ω) − e−d(γ+2)z
Z
≤c
ku (·, t)kL1 (ω) dt
j+2
ku (·, `j+2 )kL2 (Ω)
E∩(`j+1 ,`j )
We take j = 2j 0 and make the sum for j 0 = 1 until innity.
Proof of Observation from a point in time 1/3
−
Gλ (x, t) =
e
|x − x0 |2
4 (` − t + λ)
(` − t − λ)n/2
R
|∇u (x, t)|2 Gλ (x, t) dx
Nλ,u (t) = RΩ
2
Ω |u (x, t)| Gλ (x, t) dx
Proof of Observation from a point in time 2/3
Ω convex
1
2
d
dt
Z
2
Z
|u| Gλ dx +
Ω
Z
2
|∇u| Gλ dx =
Ω
d
1
Nλ,u (t) ≤
Nλ,u (t) +
dt
`−t+λ
u (∂t − ∆) uGλ dx
Ω
R
− ∆) u|2 Gλ dx
2
Ω |u| Gλ dx
Ω |(∂
Rt
Proof of Observation from a point in time 3/3
Step I. Make appear Br ⊂ ω
Z
Z
2
16λ n
0|
2 − |x−x
4λ
dx ≤
|u (`)|2 dx
− 2 λNλ,u (`) +
+1
|u (`)| e
r
4
Br
Ω
Step II. Estimate λNλ,u (`) +
n
λNλ,u (`) + ≤ C
4
n
4
"
#
R
2
|u
(0)|
dx
1
λ
+ 1 eC` ln eC (1+ ` ) RΩ
2
`
Ω |u (`)| dx
Step III. Choice of λ
1
16λ
= 2 C
2
r
"
#
R
2
λ
C (1+ 1` ) Ω |u (0)| dx
C`
+ 1 e ln e
R
2
`
Ω |u (`)| dx
Exact control and cost imply Bang-bang control
For any y0 ∈ L2 (Ω), there exists a control f ∈ L∞ (Ω × E) such
that the unique solution y of

 ∂t y − ∆y + a (x, t) y = 1|ω×E f
y=0

y (·, 0) = y0
in Ω × (0, T ) ,
on ∂Ω × (0, T ) ,
in Ω ,
satises
y (·, T ) = 0 and
kf kL∞ (Ω×E) ≤ c ky0 kL2 (Ω) .
Known result
Let Ω be a bounded smooth domain in Rn, ω ⊂ Ω. Given
y0 ∈ L2 (Ω) and M > 0, if y ∗ , 1|(0,T ∗ ) f ∗ is a couple (y, f )
satisfying

∂t y − ∆y = 1|ω f




 y=0
y (·, 0) = y0


y (·, T ∗ ) = 0



kf (·, t)kL2 (Ω) ≤ M
in Ω × (0, +∞) ,
on ∂Ω × (0, +∞) ,
in Ω ,
in Ω ,
a.e. t ∈ (0, +∞) ,
is the smallest of the admissible time among
and T ∗ > 0
all the couples (y, f ), then
kf ∗ (·, t)kL2 (Ω) = M
Further, y ∗ , 1|(0,T ∗ ) f ∗ is unique.
a.e. t ∈ (0, T ∗ ) .
NEW RESULT
Let Ω be a convex or star-shaped domain in Rn , ω ⊂ Ω, T > 0,
2
a ∈ L∞ (Ω × (0,
T )). Given y0 ∈ L (Ω) and M > 0, if
∗
∗
y , 1|(τ ∗ ,T ) f is a couple (y, f ) satisfying

∂t y − ∆y + a (x, t) y = 1|ω f




 y=0
y (·, 0) = y0


y (·, T ) = 0



kf (·, t)kL2 (Ω) ≤ M
in Ω × (0, T ) ,
on ∂Ω × (0, T ) ,
in Ω ,
in Ω ,
a.e. t ∈ (0, T ) ,
is the largest admissible time among all the
and τ ∗ ∈ [0, T )
couples (y, f ), then
kf ∗ (·, t)kL2 (Ω) = M
Further, y ∗ , 1|(τ ∗ ,T ) f ∗ is unique.
a.e. t ∈ (τ ∗ , T ) .
Proof by contradiction
If kf ∗ (·, t)kL2 (Ω) = M a.e. t ∈ (τ ∗ , T ) is false,
then ∃ ε > 0 and E ⊂ (τ ∗ , T ) , |E| > 0 such that
kf ∗ (·, t)kL2 (Ω) ≤ M − ε a.e. t ∈ E .
Aim : Contradiction with the optimality of τ∗
i.e., construction of a couple z, 1|(τ ∗ +δ,T ) v solution of

∂t z − ∆z + az = 1|ω×(τ ∗ +δ,T ) v




 z=0
z (·, 0) = y0


z (·, T ) = 0



kv (·, t)kL2 (Ω) ≤ M
in Ω × (0, T ) ,
on ∂Ω × (0, T ) ,
in Ω ,
in Ω ,
a.e. t ∈ (0, T ) ,
for a small δ > 0 with E ⊂ (τ ∗ + δo , T ) for a suciently small
δ o > 0.
Thank You!