null controllability of one-dimensional parabolic...

Null controllability of one-dimensional parabolic equations by theflatness approach

Lionel RosierCAS, Ecole des Mines, Paris

9th Workshop on Control of Distributed Parameter SystemsBeijing, China

June 29 - July 3, 2015

1 / 30

This talk is based on joint work with:

Philippe Martin (CAS, Ecole des Mines, Paris)

Pierre Rouchon (CAS, Ecole des Mines, Paris)

2 / 30

Outline

1 Flatness method for 1D heat equation

2 Reachable states

3 Parabolic equation with discontinuous coefficients

4 Heat equation on cylinders

5 Numerics

3 / 30

Boundary control of the heat equation

Let Ω ⊂ RN be a bounded smooth open set, Γ0 ⊂ ∂Ω be a (nonempty) openset, and T > 0.

We are concerned with the null controllability problem:given θ0, find a function u = u(t , x) s.t. the solution of

θt −∆θ = 0 (t , x) ∈ (0,T )× Ω,

∂θ

∂ν= 1Γ0 u(t , x) (t , x) ∈ (0,T )× Ω,

θ(0, x) = θ0(x), x ∈ Ω

satisfiesθ(T , x) = 0 x ∈ Ω

Huge literature...

4 / 30

Null controllability of the heat equation

Duality methods (observability estimate for the adjoint eq.)Fattorini-Russell ’71, Luxembourg-Korevarr ’71, Dolecki ’73 (1D, usingbiorthogonal families and complex analysis)Lebeau-Robbiano ’95, Imanuvilov-Fursikov 96’ (ND, ∀(Ω, Γ0,T ), usingCarleman estimates)

Direct methodsJones ’77, Littman ’78 (construction of a fundamental solution withcompact support in time, Γ0 = ∂Ω)Littman-Taylor 2007 (solution of ill-posed problems)Laroche-Martin-Rouchon 2000 (approximate controllability using a flatnessapproach)

Here, we shall revisit the flatness approach, obtain the null controllability,and show its relevance to numerics.

5 / 30

What is Flatness?

Roughly: parameterisation of the trajectory by some (flat) output;introduced in 1995 by M. Fliess, J. Levine, Ph. Martin, P. Rouchon for(linear or nonlinear) ODE; very useful for motion planning of mechanicalsystemsMethod applied then by Laroche-Martin-Rouchon in 2000 to derive theapproximate controllability of (i) the 1D heat eq; (ii) the beam equation;(iii) the linearized KdV equation.The first control problem considered reads:

θt − θxx = 0, x ∈ (0, 1)

θx (t , 0) = 0, θx (t , 1) = u(t),

θ(0, x) = θ0.

They proved in 2000 that for “nice” initial data decomposed as

θ0(x) =∑i≥0

yix2i

(2i)!

with

|yi | ≤ Ci!s

R i , i ≥ 0

with s ∈ (1, 2), C, R > 0, the system can be driven to 0 with a controlu(t) that is Gevrey of order s.

6 / 30

Flat output, trajectory, control,...

Take y = θ(t , 0) as output. It is flat, in the sense that the map θ → y is abijection between appropriate spaces of functions.

Seek a formal solution (analytic in x) in the form

θ(t , x) =∑i≥0

ai (t)x i

i!

Plugging this sum in the heat eq. gives∑i≥0

[ai+2 − ai′ ]

x i

i!= 0 (′ = d/dt), and

henceai+2 = a′i , i ≥ 0.

Since a0(t) = θ(t , 0) = y(t) and a1(t) = 0, we arrive to

a2i+1 = 0, a2i = y (i), i ≥ 0,

θ(t , x) =∑i≥0

y (i)(t)x2i

(2i)!, u(t) = θx (t , 1) =

∑i≥1

y (i)(t)(2i − 1)!

7 / 30

Gevrey functions

Since θ(t , x) =∑

i≥0 y (i)(t) x2i

(2i)!, it remains to find y ∈ C∞([0,T ]) s.t. the

series converges and

y (i)(0) = yi , y (i)(T ) = 0, i ≥ 0.

Impossible to do with an analytic function, but possible with a functionGevrey of order s > 1!

Definition: a function y ∈ C∞([0,T ]) is Gevrey of order s ≥ 0 if thereexist R,C > 0 such that

|y (p)(t)| ≤ Cp!s

Rp , ∀p ∈ N, ∀t ∈ [0,T ]

The larger s, the less regular y is (s = 1 ⇐⇒ y ∈ Cω)

8 / 30

First main result (Automatica, 2014)

Theorem

Let θ0 ∈ L2(0, 1) and T > 0. Pick τ ∈ (0,T ) and s ∈ (1, 2). There existsy ∈ C∞([τ,T ]) Gevrey of order s on [τ,T ] such that, setting

u(t) =

0 if 0 ≤ t ≤ τ∑

i≥0y(i)(t)

(2i−1)!if τ < t ≤ T ,

the solution θ of

θt − θxx = 0, x ∈ (0, 1)

θx (t , 0) = 0, θx (t , 1) = u(t),

θ(0, x) = θ0(x)

satisfies θ(T , .) = 0.

9 / 30

Proof: Step 1 (free evolution)

We apply a null control to smooth out the state and reach the class ofstates for which Laroche-Martin-Rouchon’ result is valid.

Decomposing the initial state θ0 as a Fourier series of cosinesθ0(x) =

∑n≥0 cn

√2 cos(nπx), we obtain

θ(τ, x) =∑n≥0

cne−n2π2τ√

2 cos(nπx) =∑i≥0

yix2i

(2i)!

Lemma

|yi | ≤ Ci!τ i ∀i ≥ 0

for some constant C > 0, so that x → θ(τ, x) is Gevrey of order 1/2.

10 / 30

Step 2: Design of the control

Proposition

(Flatness property) Let s ∈ (1, 2) and y ∈ C∞([t1, t2]) (−∞ < t1 < t2 <∞)be Gevrey of order s on [t1, t2]. Let

θ(t , x) :=∑i≥0

x (2i)

(2i)!y (i)(t).

Then θ is Gevrey of order s in t and s/2 in x on [t1, t2]× [0, 1], and it solvesthe ill-posed problem

θt − θxx = 0, (t , x) ∈ [t1, t2]× [0, 1],

θ(t , 0) = y(t),

θx (t , 0) = 0.

It remains to design a function y ∈ C∞([τ,T ]) Gevrey of order s ∈ (1, 2) suchthat

y (i)(τ) = yi , y (i)(T ) = 0, ∀i ≥ 0.

This is Laroche-Martin-Rouchon’ result.

11 / 30

Step 2: Design of the control (2)

Let

y(t) =∑i≥0

yi(t − τ)i

i!

Since |yi | ≤ Ci!/τ i , y is analytic on [τ, τ + R] if R < τ .(Actually, y can be extended to (τ,+∞) as an analytic function)

For any s ∈ (1, 2), we introduce the “Gevrey step function”

φs(t) =

1 if t ≤ 0

e−(1−t)−κ

e−(1−t)−κ+e−t−κ if 0 < t < 10 if t ≥ 1.

where κ = (s − 1)−1. Then φs is Gevrey of order s.

For y , we pick s ∈ (1, 2), 0 < R ≤ T − τ (where 0 < τ < T ), and set

y(t) := φs(t − τ

R)y(t), t ∈ [τ,T ].

12 / 30

Reachable states

We are interested in describing the states θ1 that can be reached at timeT from 0 (as in Fattorini-Russell ’71):

θt − θxx = 0, x ∈ (0, 1), t ∈ (0,T )

θ(t , 0) = h0(t), θ(t , 1) = h1(t), t ∈ (0,T ) [h0, h1 ∈ L2(0,T )]

θ(0, x) = 0

θ(T , x) = θ1(x)

θ1(x) =∑

n≥1 cn sin(nπx) is a reachable state if

∃ε > 0 s.t.∞∑

n=1

|cn|n−1e(1+ε)nπ <∞ (Fattorini-Russell, 1971)

∞∑n=1

|cn|2ne2nπ <∞ (Ervedoza-Zuazua, 2011)

Both (FR) and (EZ) imply that θ1 has to satisfy the condition

θ(2p)1 (0) = θ

(2p)1 (1) = 0 ∀p ∈ N.

Very conservative!! (no nontrivial polynomial function concerned!!)

13 / 30

Notation: H(Ω) denotes the set of (complex) analytic functions in the domainΩ ⊂ C.

Theorem

1 If θ1 ∈ H(z; |z − 12 | <

R2 with R > R0 := e(2e)−1

∼ 1.2, then θ1 isreachable from 0 in any time T > 0.

2 Conversely, any reachable state belongs to

H(z = x + iy ; |x − 12|+ |y | < 1

2)

Examples (1) Any polynomial function is reachable!!(2) Let

θ1(x) :=1

(x − 12 )2 + a2

·

Then θ1 is

reachable if |a| > R0/2 ∼ 0.6

not reachable if |a| < 0.5

Method of proof: flatness approach + (new) Borel-Ritt theoremWe have similar results with only one control, of Dirichlet or Neumann type.

14 / 30

Variable coefficients

Consider now the equation

(a(x)θx )x + b(x)θx + c(x)θ − ρ(x)θt = 0

where a, b, c, ρ ∈ L1(0, 1).

Alessandrini-Escauriaza (2006) proved the null controllability of thisequation (with internal or Dirichlet boundary control) whena, b, c, ρ ∈ L∞(0, 1) with

a(x) > K > 0 and ρ(x) > K > 0 a.e. in (0, 1)

Method of proof: Lebeau-Robbiano ’95 + complex variable analysis.

We shall see that this result can be extended to parabolic equationswith singular or degenerate coefficients by using the flatness approach.(for degenerate eq., we refer to Cannarsa-Martinez-Vancostenoble2004,...)

15 / 30

Main result (SICON, to appear)

Let a, b, c, ρ with

a(x) > 0 and ρ(x) > 0 for a.e. x ∈ (0, 1)

(1a,

ba, c, ρ) ∈ [L1(0, 1)]4

∃K ≥ 0,c(x)

ρ(x)≤ K for a.e. x ∈ (0, 1)

∃p ∈ (1,∞], a1− 1p ρ ∈ Lp(0, 1).

Theorem

Let (a, b, c, ρ) be as above, and (α0, β0) 6= (0, 0), (α1, β1) 6= (0, 0).Let θ0 ∈ L1

ρ(x)dx (0, 1) and T > 0. Pick τ ∈ (0,T ) and s ∈ (1, 2− p−1). Thenthere exists a control h = h(t) Gevrey of order s on [0,T ] such that thesolution θ of

(a(x)θx )x + b(x)θx + c(x)θ − ρ(x)θt = 0, x ∈ (0, 1)

α0θ(t , 0) + β0(aθx )(t , 0) = 0,

α1θ(t , 1) + β1(aθx )(t , 1) = h(t),

θ(0, x) = θ0(x)

satisfies θ(T , .) = 0.16 / 30

Examples

(a(x)θx )x − θt = 0, with a(x) > 0 a.e. and

a, 1/a ∈ L1(0, 1)

Possible: a(x) ∼ (x − x0)r with−1 < r < 0 (singular) or0 < r < 1 (weakly degenerate)

at a single point x0 ∈ [0, 1], or at a sequence of points. Ex:

a(x) = | sin(x−1)|r , −1 < r < 1

Transmission pb. for the heat eq. (piecewise constant coef.)

ρ0θt = a0θxx , 0 < x < X

ρ1θt = a1θxx , X < x < 1

θ(t ,X−) = θ(t ,X +)

a0θx (t ,X−) = a1θx (t ,X +)

17 / 30

Sketch of the proof

Step 1. Using changes of variables, we can put the system in thecanonical form

θxx − ρ(x)θt = 0, x ∈ (0, 1)

α0θ(t , 0) + β0θx (t , 0) = 0,

α1θ(t , 1) + β1θx (t , 1) = h(t),

θ(0, x) = θ0(x)

where ρ ∈ Lp(0, 1), 1 < p ≤ ∞.

Step 2. In the time interval (0, τ), we apply a null control to smoothout the state, while in the interval (τ,T ) we apply a non-trivial controlto reach 0 at time t = T .

The trajectory reads then

θ(x , t) =∑n≥0

e−λn ten(x), x ∈ (0, 1), t ∈ [0, τ ],

θ(x , t) =∑i≥0

y (i)(t)gi (x), x ∈ (0, 1), t ∈ [τ,T ].

18 / 30

Sketch of the proof (2)

For t ∈ (0, τ), θ(x , t) =∑

n≥0 e−λn ten(x).(en, λn) is the nth pair of eigenfunction/eigenvalue for

−e′′n = λn ρ en, x ∈ (0, 1)

α0en(0) + β0e′n(0) = 0,

α1en(1) + β1e′n(1) = 0,

For t ∈ (τ,T ), θ(x , t) =∑

i≥0 y (i)(t)gi (x), where the generatingfunction gi is defined inductively as follows:

g′′0 = 0, x ∈ (0, 1)

α0g0(0) + β0g′0(0) = 0,

β0g0(0)− α0g′0(0) = 1

for i = 0, and gi , for i ≥ 1, is the solution to the Cauchy problem

g′′i = ρ gi−1, x ∈ (0, 1)

gi (0) = 0,

g′i (0) = 0

We can prove ||gi ||W 2,p(0,1) ≤ C/(R i (i!)2−1/p)

19 / 30

Sketch of the proof (3)

To ensure that the two expressions of θ agree at t = τ , we have to relatethe eigenfunctions en to the generating functions gi .

We haveen(x) = ζn

∑i≥0

(−λn)igi (x) (∗)

with ζn ∈ R.

For ρ ≡ 1 and (α0, β0, α1, β1) = (0, 1, 0, 1) [Neumann control at x = 1],(∗) is nothing but the classical Taylor expansion of cos(nπx) aroundx = 0:

cos(nπx) =∑i≥0

(−1)i (nπx)2i

(2i)!

20 / 30

Null controllability of the heat equation on cylinders (Automatica 2014)

Consider the control system

(S)

θt −∆θ = 0, (t , x) ∈ (0,T )× Ω∂θ∂ν

(t , x ′, 1) = u(t , x ′), (t , x ′) ∈ (0,T )× ω∂θ∂ν

(t , x) = 0 (t , x) ∈ (0,T )× ∂Ω \ ω × 0θ(0, x) = θ(x), x ∈ Ω := ω × (0, 1)

Model for the temperature along a rod which is controlled by taking as inputthe heat flux at one extremity.

Theorem

Let Ω = ω × (0, 1) ⊂ RN−1 × R, θ0 ∈ L2(Ω) and T > 0 be given. Pick anyτ ∈ (0,T ) and any s ∈ (1, 2). Then there exists a sequence (yj )j≥0 offunctions in C∞([τ,T ]) which are Gevrey of order s on [τ,T ] and such that,setting

u(t , x ′) =

0 if 0 ≤ t ≤ τ,∑

i,j≥0 e−λj ty(i)

j (t)

(2i−1)!ej (x ′) if τ ≤ t ≤ T ,

we haveθ(T , .) = 0.

Here, (ej , λj ) denotes the jth pair of eigenfunction/eigenvalue for theNeumann Laplacian on ω ⊂ RN−1.

21 / 30

Numerical approximation

Assume given T > 0, τ ∈ (0,T ), s ∈ (1, 2), and θ0 ∈ L2(Ω) decomposed as

θ0(x ′, xN) =∑j,n≥0

cj,nej (x ′)√

2 cos(nπxN).

The exact solution θ of the previous control problem such that θ(T , .) = 0reads

θ(t , x ′, xN) =∑j,n≥0

cj,ne−(λj +n2π2)tej (x ′)√

2 cos(nπxN), 0 ≤ t ≤ τ,

θ(t , x ′, xN) =∑j≥0

e−λj tej (x ′)∑i≥0

y (i)j (t)

x2iN

(2i)!, τ ≤ t ≤ T ,

where

yj (t) = φ(t)∑n≥0

cj,ne−n2π2t , τ ≤ t ≤ T ,

φ(t) = φs(t − τT − τ ), τ ≤ t ≤ T .

In practice, only partial sums can be computed. They prove to giveaccurate approximations of both the trajectory and the control.Exponentially small errors!

22 / 30

Numerical simulations (N=1) Trajectory

Initial state: θ0 := 1(1/2,1)(x)− 1(0,1/2)(x)

Parameters: τ = 0.3, R = 0.2, T = τ + R = 0.5, s = 1.6

0 0.1 0.2 0.3 0.4 0.5 0

0.5

1

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

space x

time t

Fig.1. θ(t , x)

Computations by Philippe Martin

23 / 30

Numerical simulations (N=1) Control

Initial state: θ0 := 1(1/2,1)(x)− 1(0,1/2)(x)

Parameters: τ = 0.3, R = 0.2, T = τ + R = 0.5, s = 1.6

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−25

−20

−15

−10

−5

0

5

10

15

Fig. 2. u(t) (blue) and ||u(t)||L2(0,t) (green)

Computations by Philippe Martin

24 / 30

Numerical simulations (N=2) Trajectory

Initial state: θ0 :=(1(1/2,1)(x1)− 1(0,1/2)(x1)

)(1(0,1/2)(x2)− 1(1/2,1)(x2)

)Parameters: τ = 0.05, R = 0.25, T = τ + R = 0.3, s = 1.65

Computations by Philippe Martin 25 / 30

Numerical simulations (N=2) Control

Initial state: θ0 :=(1(1/2,1)(x1)− 1(0,1/2)(x1)

)(1(0,1/2)(x2)− 1(1/2,1)(x2)

)Parameters: τ = 0.05, R = 0.25, T = 0.35, s = 1.65

00.05

0.10.15

0.20.25

0.3 0

0.2

0.4

0.6

0.8

1

−20

−10

0

10

20

x1

Control effort u

t

Fig. 4. u(t , x1)

Computations by Philippe Martin

26 / 30

Transmission pb. for the heat eq. (piecewise constant coef.)

ρ0θt = a0θxx , 0 < x < X

ρ1θt = a1θxx , X < x < 1

θ(t ,X−) = θ(t ,X +)

a0θx (t ,X−) = a1θx (t ,X +)

Parameters: X = 1/2, (a0, ρ0, a1, ρ1) = (10/19, 15/8, 10, 1/8)

27 / 30

Numerical simulations (N=1, discontinuous coefficients) Trajectory

Initial state: θ0 := 12 1(1/2,1)(x)− 1

2 1(0,1/2)(x)

Parameters: τ = 0.3, T = 0.35, s = 1.6,(a0, ρ0, a1, ρ1) = (10/19, 15/8, 10, 1/8)

Fig.1. θ(t , x)

Computations by Philippe Martin

28 / 30

Concluding remarks

The flatness approach allows to prove in a very simple way the nullcontrollability of the heat equation (in cylinders), with explicit controlsand trajectories easy to approximate.

Extension to general parabolic equations with discontinuouscoefficients done.Future direction of research

Extension to any pair (Ω, Γ0) in 2DExact controllability results for linear/nonlinear equationsNumerical investigation of the cost of the control in terms of:T (time control), τ = T − R (free evolution), s (Gevrey regularity)

29 / 30

Thank you!

30 / 30