Non null controllability of the Grushin equation in 2D

0.65 MB·

by  Armand Koenig

Non null controllability of the Grushin operator
in 2D
Armand KOENIG∗
Université Côte d’Azur
7 January 24, 2017
1
0
2

n
a
Abstract
J

3
We are interested in the exact null controllability of the equation ∂ f −
2 t
∂2f −x2∂2f = 1 u, with control u supported on ω. We show that, when
x y ω
] ω does not intersect a horizontal band, the considered equation is never
P
null-controllable. The main idea is to interpret the associated observability
A
inequality as an L2 estimate on entire functions, which Runge’s theorem
.
h
disproves. To that end, we study in particular the first eigenvalue of the
t
a operator −∂2 + (nx)2 with Dirichlet conditions on (−1,1) and we show a
m x
quite precise estimation it satisfies, even when n is in some complex domain.
[

1
v
7
6
Contents
4
6
0
1 Introduction 2
.
1
0
2 Outline of the proof 3
7
1
: 3 Estimates for the holomorphy default operators 7
v
i
X
4 Spectral analysis of the Fourier components 14
r
a
5 Proof of the non null-controllability of the Grushin equation 20
∗ThisworkwaspartiallysupportedbytheERCadvancedgrantSCAPDE,seventhframework
program, agreement no. 320845
AMS 2010 subject classification: 93B05, 93B07, 93C20, 35K65, 30A10, 34L15
Keywords: null controllability, observability, degenerate parabolic equations
1
6 Conclusion and open problems 24
Appendix A The stationary phase theorem 25
Appendix B Newton’s method 27
Appendix C Two variations on the non null controllability of the
Grushin equation 28
1 Introduction
1.1 The problem of controllability of the Grushin equation
We are interested in the following equation, where T = R\2πZ, Ω = (−1,1)×T
and ω is an open subset of Ω:
∂ f −∂2f −x2∂2f = 1 u on Ω
t x y ω
f| = 0
∂Ω
It is a control problem with state f and control u supported on ω. More precisely,
we are interested in the exact null controllability of this equation:
Definition 1. We say that the Grushin equation is null-controllable on ω in time
T > 0 if for all f in L2(Ω), there exists u in L2([0,T]×ω) such that the solution
0
f of:
∂ f −∂2f −x2∂2f = 1 u
t x y ω
f| = 0 (1)
∂Ω
f(0) = f
0
satisfies f(T) = 0.
We show in this paper that if ω does not intersect a horizontal band, then the
answer is negative whatever T is:
Theorem 2. Let [a,b] be a non trivial segment of T and ω = (−1,1)×(T\[a,b]).
Let T > 0. The Grushin equation in not null-controllable in ω in time T.
1.2 Bibliographical comments
This equation has previously been studied on (−1,1) × (0,1), and some results
already exist for different controllability sets. Controllability holds for large time
but not in small time if ω = (a,b)×(0,1) with 0 < a < b, as shown by Beauchard,
2
Cannarsa and Guglielmi [3], and holds in any time if ω = (0,a) × (−1,1) with
0 < a, as shown by Beauchard, Miller and Morancey [4].
The controllability of the Grushin equation is part of the larger field of the
controllability of degenerate parabolic partial differential equations of hypoelliptic
type. Forthenon-degeneratecase, controllabilityisknownsince1995toholdwhen
Ω is any bounded C2 domain, in any open control domain and in arbitrarily small
time [13, 11]. For parabolic equations degenerating on the boundary, the situation
is well understood in dimension one [8] and in dimension two [9]. For parabolic
equations degenerating inside the domain, in addition to the already mentioned
two articles on the Grushin equation, we mention articles on Kolmogorov-type
equations [2, 7], and quadratic hypoelliptic equations on the whole space [5, 6].
2 Outline of the proof
2.1 Observability
As usual in controllability problems, we focus on the following observability in-
equality on the adjoint system, which is equivalent by a duality argument to the
controllability (definition 1): there exists C > 0 such that for all f in L2(Ω), the
0
solution f of:
∂ f −∂2f −x2∂2f = 0
t x y
f| = 0 (2)
∂Ω
f(0) = f
0
satisfies:
(cid:90) (cid:90)
|f(T,·)|2 ≤ C |f|2 (3)
Ω [0,T]×ω
The rest of this article is devoted to finding a counterexample to the previous
observability inequality when ω = (−1,1)×(T\[a,b]) = (−1,1)×ω .
y
2.2 Fourier components: eigenvalues and eigenfunctions
To find such a counterexample, we look for solutions of the Grushin equation (2)
that concentrate near x = 0. To that end, we remark that, denoting v an
n,k
eigenfunction of the operator −∂2 +(nx)2 with Dirichlet boundary conditions on
x
(−1,1) associated with eigenvalue λ , Φ (x,y) = v (x)einy is an eigenfunction
n,k n,k n,k
of the Grushin operator −∂2 −x2∂2 with eigenvalue λ . In addition, we expect
x y n,k
that the first eigenfunction v = v of −∂2 + (nx)2 on (−1,1) is close to the
n n,0 x
first eigenfunction of the same operator on R, that is v ∼ ( n )1/4e−nx2/2, and
n 4π
that the associated eigenvalue λ is close to n. So it is natural to look for a
n
3
counterexample of the observability inequality (3) among the linear combinations
of Φ (x,y) = v (x)einy for n ≥ 0.
n n
The previous observations on v and λ were already mentioned by Beauchard,
n n
Cannarsa and Guglielmi [3], and we will strengthen their results on that subject
in section 4: we will prove that λ −n = γ(n)e−n with γ holomorphic on domains
n
{|arg(z)| < θ,|z| > r(θ)} and satisfying estimates of the form |γ(z)| ≤ C (1 +
θ
|z|)3/2 on those domains, and that v (x) satisfies properties similar to γ .
n n
2.3 The toy model
Let’s write the observability inequality on functions of the form (cid:80)a v (x)einy
n n
(noting ω = T\[a,b] so that ω = (−1,1)×ω ):
y y
(cid:90) 1 (cid:90) T (cid:90)
(cid:88) (cid:88)
|a |2e−2λnT ≤ C a a v v e−(λn+λm)tdt ei(n−m)ydy (4)
n n m n m
n n,m −1 0 y∈ωy
Now let’s proceed by heuristic arguments to see what we can expect from the
previous estimates on the eigenvalues λ and the eigenvectors v . We imagine that
n n
in the previous inequality, λ = n and (cid:82)1 v v = √1 (nm)1/4(cid:82) e−(n+m)x2/2dx =
√ n −1 n m 4π R
2(√nm)1/4, which does not decay very fast off-diagonal, so we further imagine
n+m
(cid:82)1
that v v = 1. Then, with these approximations, the previous observability
−1 n m
inequality reads:
(cid:90)
(cid:88) (cid:88)
|a |2e−2nT ≤ C a a e−(n+m)t+i(n−m)ydtdy
n n m
n n,m [0,T]×ωy (5)
(cid:90) (cid:12)(cid:88) (cid:12)2
= C (cid:12) a e−nt+iny(cid:12) dtdy
(cid:12) n (cid:12)
[0,T]×ωy
This suggests that the controllability problem of the Grushin equation (1) is
similar to the following model control problem: let’s consider the Hilbert space
{(cid:80) a einy,(cid:80)|a |2 < +∞}, |∂ | the unbounded operator on this space with do-
n≥0 n n y
main {(cid:80)a einy,(cid:80)n2|a |2 < +∞} defined by |∂ |((cid:80)a einy) = (cid:80)na einy. Then
n n y n n
the controllability of the operator |∂ | on an open set ω = T\[a,b] in time T is
y
equivalent to the previous “simplified” observability inequality (5). Since this is
rather easy while showing the main idea of the proof of the non controllability of
Grushin’s equation, let’s prove now that this model is not null controllable on ω :
y
Theorem 3. Let [a,b] be a nontrivial segment of T, ω = T \ [a,b] and T > 0.
y
The operator |∂ | is not null controllable on ω in time T.
y y
Incidentally, this is an answer to a specific case of an open problem mentioned
by Miller [15, section 3.3] and again by Duyckaerts and Miller [10, remark 6.4].
4
Proof. The right hand side of the observability inequality (5) suggests to make
the change of variables z = e−t+iy, whose Jacobian is equal to |z|2, so if we note
D = {e−T < |z| < 1,arg(z) ∈ ω }, the right hand side of the observability
y
inequality (5) is equal to1 C(cid:82) |(cid:80)a zn−1|2d|z|.
D n
About the left hand side, we first note that by writing the integral on a disk
D = D(0,r) of znz¯m in polar coordinates, we find that the functions z (cid:55)→ zn are
orthogonal on any disk. So, again by writing the integral in polar coordinates,
for all entire functions f(z) = (cid:80)a zn with f(0) = 0: (cid:82) |f(z)|2|z|−2d|z| =
n D(0,e−T)
2π(cid:80)|a |2(cid:82)e−T |r|2n−1dr = 2π(cid:80)|a |2e−2nT 1 ≤ 2π(cid:80)|a |2e−2nT.
n 0 n 2n n
So the observability inequality (5) implies the following “holomorphic observ-
ability inequality”: for some C(cid:48) > 0 and for all functions f holomorphic in the unit
disk D(0,1) which are in L2(D(0,1)):
(cid:90) (cid:90)
|f|2 ≤ C(cid:48) |f|2 (6)
D(0,e−T) D
We will disprove it thanks to Runge’s theorem, whose proof can be found in
Rudin’s famous textbook [16, theorem 13.9]. More specifically, we will need the
following special case:
Property 4 (Runge’s theorem). Let U be a connected and simply connected open
set of C, and let f be an holomorphic function on U. There exists a sequence (f )
n
of polynomials that converges uniformly on every compact subsets of U to f.
Let θ ∈ T non-adherent to ω (for instance θ = (a + b)/2)). We choose in
y
the previous theorem U = C \ eiθR (see figure 1) and f(z) = 1. Since z (cid:55)→ 1
+ z z
is bounded on D, f is uniformly bounded on D and the right hand side of the
n
(cid:82)
holomorphic observability inequality (6) |f |2d|z| stays bounded. But since
D n
z (cid:55)→ 1 has infinite L2 norm on any nontrivial sector {|z| < r,arg(z) ∈ ω }, the left
z (cid:82) y
hand side |f |2d|z| tends to infinity.
D(0,e−T) n
2.4 From the toy model to the Grushin equation: holomor-
phy default operators
We aim to show that the heuristics described previously are actually good. To that
end, denoting λ = n+ρ and z(t,y) = e−t+iy, we rewrite the real observability
n n
inequality (3) applied to f(t,x,y) = (cid:80)a e−λntv (x)einy as:
n n
1We denote d|z| the Lebesgue measure on C ; that is if f is integrable on R2, (cid:82) f(z)d|z|=
(cid:82) C
f(x+iy)dxdy.
R2
5
1
ω
y
θ
D
D(0,e−T)
Figure 1: In yellow, the domain D, in red, the disk D(0,e−T). The thick outer
circular arc is the subset ω of T (cid:39) U. The controllability of the model operator
y
|∂ | on ω in time T would imply the the control of the L2(D(0,e−T))-norm of
y y
entire functions by their L2 norm on D.
(cid:90) (cid:18)(cid:90) (cid:19)
(cid:88)
| v (x)a z(T,y)n|z(T,y)|ρn|2dy dx
n n
x∈(−1,1) y∈T
(cid:32) (cid:33)
(cid:90) (cid:90)
(cid:88)
≤ C | v (x)a z(t,y)n|z(t,y)|ρn|2dtdy dx
n n
x∈(−1,1) (t,y)∈[0,T]×ωy
We then write |z(t,y)|ρn = (cid:80)(−t)j 1ρj, and for all entire functions f =
j! n
(cid:80)a zn, H (f)(z) = (cid:80)v (x)ρja zn, so that the previous inequality implies
n v(x)ρj n n n
that for all entire functions f = (cid:80)a zn:
n
 
(cid:12) (cid:12)2
(cid:90) (cid:90) (cid:12)(cid:88) (−T)j (cid:12)
 (cid:12)(cid:12) j! Hv(x)ρj(f)(z(T,y))(cid:12)(cid:12) dy dx
x∈(−1,1) y∈T(cid:12) (cid:12)
j
 
(cid:12) (cid:12)2
(cid:90) (cid:90) (cid:12)(cid:88) (−t)j (cid:12)
≤ C  (cid:12)(cid:12) j! Hv(x)ρj(f)(z(t,y))(cid:12)(cid:12) dtdy dx (7)
x∈(−1,1) (t,y)∈[0,T]×ωy (cid:12) j (cid:12)
Since the operators H describe how far this inequality is from the observ-
v(x)ρj
ability inequality (5) of the model, we will call them holomorphy default operators,
andwillproveinsection3estimatesonthoseoperators. Thoseestimateswillallow
6
us to prove that if we modify slightly the counterexample to the observability of
the model, only the term j = 0 in the previous inequality matters, so that the
observability inequalities of the model (5) and of the Grushin operator (3) can be
disproved essentially the same way.
3 Estimates for the holomorphy default operators
In this section, we prove estimates on holomorphy default operators, i.e. opera-
tors on entire functions of the form H((cid:80)a zn) = (cid:80)γ a zn. We start with the
n n n
following easy estimate:
Property 5. Let r > 0 and (γ ) be a bounded sequence of complex numbers. We
n
write H ((cid:80)a zn) = (cid:80)γ a zn. Then for all entire functions f, |H (f)| ≤
γ n n n γ L2(D(0,r))
sup|γ ||f| .
n L2(D(0,r))
Proof. This is a consequence of the already mentioned fact that the functions
z (cid:55)→ zn are orthogonal on D(0,r).
But we will need estimates on other domains than disks, so we use another
approach. Let’s look at the easy example of the operator H((cid:80)a zn) = (cid:80)na zn.
n n
Since we have H(f)(z) = zf(cid:48)(z), thanks to the Cauchy integral formula, we have
for all domains U of C, δ > 0 and Uδ = {z,distance(z,U) < δ}, |H(f)| ≤
L∞(U)
δ−1sup |z||f| . Under suitable hypotheses on (γ ) and U, we will prove
z∈U L∞(Uδ) n
similar estimates for H((cid:80)a zn) = (cid:80)γ a zn.
n n n
TheideaistousetheCauchyintegralformulatowriteH(f)(z)asaconvolution
operator on the unit circle: H(f)(z) = (cid:72) 1K(z/ζ)f(ζ)dζ with K(z(cid:48)) = (cid:80)γ z(cid:48)n.
∂U ζ n
ThenweextendthesumdefiningK thankstoPoisson’ssummationformula, which
allows us to change the integration path in the expression of H(f) as a convolution
on the unit circle.
3.1 Symbols
The class of sequences (γ ) which will allow us to do all we described just before
n
are the restriction to N of the following functions:
Definition 6. Let r : (0,π/2) → R be an non-decreasing function, and for θ
+
in (0,π/2), let U = {|z| > r(θ),|arg(z)| < θ} and. We note S(r) the set of
θ,r(θ)
holomorphic functions on each U that have sub-exponential growth on those
θ,r(θ)
domains.
We endow S(r) with the topology defined by the seminorms p for all θ ∈
θ,(cid:15)
(0,π/2) and (cid:15) > 0 defined by p (γ) = sup |γ(z)e−(cid:15)|z||.
θ,(cid:15) z∈U
θ,r(θ)
7
In the rest of this article, when we write S(r) it is implicitly assumed that r is
a non-decreasing function from (0,π/2) to R .
+
Example 7. • Every bounded holomorphic function on the half plane {(cid:60)(z) ≥
0} is in S(0). For instance, z (cid:55)→ e−z is in S(0).
• Every polynomial is in S(0).
• For all s > 0, z (cid:55)→ zs is in S(0).
• More generally, if γ is holomorphic on every domain U and has at most
θ,r(θ)
polynomial growth on those domains, γ is in S(r).
Let’s remind that if Ω is an open subset of C, O(Ω) is the space of holomorphic
functions in Ω with the topology of uniform convergence in every compact subset
of Ω. With this notation, here are a few simple facts about S(r), which will be
useful later:
Property 8. The space S(r) enjoys the following properties:
• The topology of S(r) is stronger than topology of uniform convergence on
every compact.
• For all compact K of (cid:83)U , and all j ∈ N, the seminorm γ (cid:55)→ |γ(j)|
θ,r(θ) L∞(K)
is continuous on S(r).
• For all z in the domain of definition of γ, the punctual evaluation at z , i.e.
0 0
γ (cid:55)→ γ(z ), is continuous on S(r).
0
• If γ is in S(r), then γ (cid:55)→ γ γ is continuous from S(r) to S(r).
0 0
(cid:83)
Proof. • Let K be a compact of U . By the Borel-Lebesgue
0<θ<π/2 θ,r(θ)
property, there is a finite number of θ in (0,π/2), say θ ,...,θ such that
1 k
K ⊂ (cid:83)k U . By noting R = sup |z|, we then have |u| ≤
j=1 θk,r(θk) z∈K L∞(K)
sup p (γ)eR. This proves the first fact.
1≤j≤k θj,1
• We remind that if Ω is an open subset of C, j is a natural number and
K a compact subset of Ω then Cauchy’s integral formula implies that the
seminorm on O(Ω): f (cid:55)→ |f(j)| is continuous. Thus, the second point
L∞(K)
is a consequence of the first.
• Since {z } is compact, the third point is a direct consequence of the second
0
point (or the first).
• In order to prove the fourth point, we write for z ∈ U : |γ (z)γ(z)| ≤
θ,r(θ) 0
p (γ )p (γ)e(cid:15)|z|, so p (γ γ) ≤ p (γ )p (γ).
θ,(cid:15)/2 0 θ,(cid:15)/2 θ,(cid:15) 0 θ,(cid:15)/2 0 θ,(cid:15)/2
8
Property 9. We have the following continuous injections between spaces S(r):
• If r(cid:48) ≥ r, then denoting U(cid:48) = (cid:83)U , the restriction map γ ∈ S(r) (cid:55)→
θ,r(cid:48)(θ)
γ ∈ S(r(cid:48)) is continuous.
|U(cid:48)
• Let θ in (0,π/2) and a > r(θ ). Define r(cid:48)(θ) by r(cid:48)(θ) = 0 if |θ| < θ and
0 0 0
r(cid:48)(θ) = r(θ) otherwise. Then γ (cid:55)→ γ(·+a) is continuous.
Proof. • For readability, let’s write U = U and U(cid:48) = U . To prove
θ θ,r(θ) θ θ,r(cid:48)(θ)
the first point, we simply remark that r(cid:48) ≥ r implies U(cid:48) ⊂ U , so, we have:
θ θ
|γ(z)e−(cid:15)|z|| ≤ |γ(z)e−(cid:15)|z|| .
z∈Uθ(cid:48) z∈Uθ
• We let the reader check that it makes sense when looking at the domain of
definition (drawing a picture can help). The continuity is a consequence of:
|γ(z +a)e−(cid:15)|z|| ≤ e(cid:15)a|γ(z +a)e−(cid:15)|z+a||.
3.2 Holomorphic Fourier transform of a symbol and convo-
lution kernel
We mentioned we wanted to apply Poisson’s summation formula to the sum
(cid:80)γ(n)zn. In order to do that, we need some information on the Fourier transform
of γ, the first of which being the existence of it.
We suppose in this subsection that for some θ in (0,π/2), r(θ ) = 0 (so that
0 0
r(θ) = 0 for 0 < θ ≤ θ ). Then we define the Fourier transform γˆ of γ for ξ
0
with negative imaginary part by γˆ(ξ) = (cid:82)+∞γ(x)e−ixξdx. We first prove that this
0
holomorphic Fourier transform can be extended on a bigger domain than the lower
half-plane, then we prove under more hypotheses some estimates on it.
Property 10. Let γ in S(r). The holomorphic Fourier transform γˆ of γ, can be
holomorphically extended on C\iR .
+
Proof. Let φ in (0,π/2), let θ in (φ,π/2) and r > r(θ). We make a change of
1
contour in the integral defining γˆ(ξ): let c the path [0,r ] ∪ {r eiϕ,−φ ≤ ϕ ≤
1 1
0}∪e−iφ[r ,+∞) (see figure 2). We have for ξ in {(cid:61)(ξ) < 0}∩eiφ{(cid:61)(ξ) < 0}:
1
(cid:90) +∞
γˆ(ξ) = γ(x)e−ixξdx
0
(cid:90)
= γ(z)e−izξdz
c
(cid:90) r1 (cid:90) −φ
= γ(x)e−ixξdx+ γ(r eit)e−ieitξir eitdt
1 1
0 0
(cid:90) +∞
+ γ(e−iφx)e−ie−iφξxe−iφdx
r1
9
θ eiφ{(cid:61)(ξ) < 0}
U
θ,r(θ) φ
r
1
φ
{(cid:61)(ξ) < 0}
Figure 2: In the left figure: in red, a part of the domain of definition of γ, and
in blue, a integration path that allows us to extend γˆ. In the right figure: in red,
the a priori domain of definition of γˆ, in yellow, the domain we extend γˆ to, when
choosing the blue integration path of the left figure.
The first two terms can be extended holomorphically on C, while the third can
be extended holomorphically on eiφ{(cid:61)(ξ) < 0}. So, taking φ → π/2, γˆ can be
extended holomorphically on {(cid:61)(ξ) < 0}∪i{(cid:61)(ξ) < 0}. By taking the path c(cid:48) the
symmetricofcwithrespecttotherealline, γˆ canalsobeextendedholomorphically
on {(cid:61)(ξ) < 0}∪−i{(cid:61)(ξ) < 0}.
Property 11. Let (cid:15) > 0. There exists C > 0 and η > 0 such that for all γ in
S(r) satisfying p(γ) := sup |γ(z)| < +∞ and for all ξ in {−ireiθ,r >
|z|<1,|arg(z)|<θ0 |z|
(cid:15),|θ| < 2θ } (see figure 3):
0
|γˆ(ξ)| ≤ C(p(γ)+p (γ))|ξ|−2
θ0,η
Proof. The proof is mostly redoing the calculation of the proof of the previous
property, but this time using the increased regularity (at 0) to get the stated
estimate.
Let ξ = −irei2θ with |θ| < θ and r > (cid:15). Thanks to the proof of the previous
0
property, we have γˆ(ξ) = (cid:82) γ(z)e−izξdz = e−iθ(cid:82)+∞γ(e−iθx)e−eiθxrdx. We
e−iθR+ 0
then write |γˆ(ξ)| ≤ (cid:82)1p(γ)xe−cos(θ)rxdx + (cid:82)+∞p (γ)eηxe−cos(θ)rxdx, which is
0 1 θ0,η
true for all η > 0.
For the first term of the right hand side, we make the change of variables
x(cid:48) = cos(θ)rx, sothat(cid:82)1xe−cos(θ)rxdx ≤ 1 (cid:82)rcos(θ)x(cid:48)e−x(cid:48)dx ≤ 1 Γ(2).
0 (rcos(θ))2 0 (rcos(θ))2
10