on the determination of the principal coefficient from ... · the kdv equation is used to describe...

On the determination of the principal coefficient from boundarymeasurements in a KdV equation

Eduardo Cerpa

Universidad Tecnica Federico Santa Marıa, Valparaıso Chile.

Joint work with Lucie Baudouin, Emmanuelle Crepeau and Alberto Mercado.

Groupe de Travail Controle - Paris VISeptember 30, 2011

E. Cerpa (USM) Inverse problem for KdV September 2011 1 / 20

Presentation of the problem

The KdV equation with non-constant coefficient a = a(x) is given asyt + a(x)yxxx + yx + yyx = g, ∀(x, t) ∈ (0, L)× (0, T ),y(0, t) = g0(t), y(L, t) = g1(t), ∀t ∈ (0, T ),

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

where the initial data y0, the source term g, and the functions g0, g1, g2 are assumed tobe known.

The KdV equation is used to describe approximately long waves in water of relativelyshallow depth.

In this context, the principal coefficient a = a(x) is related with the shape of the bottomof the channel where the water wave propagates. See Samer Israwi’s PhD thesisunder supervision of David Lannes (Bordeaux, 2010).

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

yt + a(x)yxxx + yx + yyx = g, ∀(x, t) ∈ (0, L)× (0, T ),y(0, t) = g0(t), y(L, t) = g1(t), ∀t ∈ (0, T ),

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

In general, in an inverse problem, we want to get some information of a system(conductivity, diffusion, shape of the bottom of a channel, etc) from partial knowledge(observations/measurements) of the solution.

Inverse ProblemCan we recover a = a(x) from some partial knowldege of y = y(x, t)?

Inverse Problem (Uniqueness)Given some boundary measurements meas(y), is there a unique a = a(x) associated?

i.e. meas(y) = meas(y) =⇒ a = a?

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

Inverse Problem (Stability)

‖a− a‖X ≤ C‖meas(y)−meas(y)‖Y ?

Inverse Problem (Reconstruction)Given some measurement meas(y), is it possible to reconstruct the coefficienta = a(x)?

In this talk, we are concerned with the stability problem (and consequently with theuniqueness problem!).

Remark: This kind of inverse problem is called a single-measurement IP

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

Calderon’s Problem, multi-measurement IP

It is the problem of recovering the conductivity γ in a body Ω from current-voltagemeasurements

div (γ∇u) = 0, Ωu = f, ∂Ω

The Dirichlet-to-Neumann map is defined by:

Λγf = γ∂u

Is the operatorγ → Λγ

invertible?

div (γ∇u) = 0, Ωu = f, ∂Ω

Λγf = γ∂u

invertible?

div (γ∇u) = 0, Ωu = f, ∂Ω

Λγf = γ∂u

invertible?

Questions about the Inverse Problem:1 Uniqueness. If Λγ1 = Λγ2 then γ1 = γ2.2 Reconstruction.3 Stability. If Λγ1 is close to Λγ2 , then γ1 and γ2 are close (in a suitable sense).4 Partial data. If Γ ⊂ ∂Ω and if Λγ1 |Γ = Λγ2 |Γ for all boundary voltages f , thenγ1 = γ2.

Starting from the work of Calderon in 1980, the inverse conductivity problem has beenstudied intensively. In the case where n ≥ 3 and all conductivities are in C2(Ω), thefollowing positive results are some examples of what can be proven.

Theorem. (Sylvester-Uhlmann 1987). Uniqueness.

Theorem. (Nachman 1988) There is an algorithm for reconstructing γ from Λγ .

Theorem. (Alessandrini 1988). Logarithmic stability.

Theorem. (Kenig-Sjostrand-Uhlmann 2007) Partial data. Assume that Ω is convexand Γ is any open subset of ∂Ω. If Λγ1 |Γ = Λγ2 |Γ for all f ∈ H

12 (∂Ω), and if

γ1|∂Ω = γ2|∂Ω, then γ1 = γ2.

12 (∂Ω), and if

γ1|∂Ω = γ2|∂Ω, then γ1 = γ2.

12 (∂Ω), and if

γ1|∂Ω = γ2|∂Ω, then γ1 = γ2.

Back to recovering the main coefficient in KdVyt + a(x)yxxx + yx + yyx = g, ∀(x, t) ∈ (0, L)× (0, T ),y(0, t) = g0(t), y(L, t) = g1(t), ∀t ∈ (0, T ),

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

We hope to get only boundary measurements:

‖yx(0, t)− yx(0, t)‖, ‖yxx(0, t)− yxx(0, t)‖

or‖yxx(L, t)− yxx(L, t)‖

For parabolic equations, there appears measurements like

‖y(x, T0)− y(x, T0)‖

for a given T0 ∈ (0, T ).

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

‖yx(0, t)− yx(0, t)‖, ‖yxx(0, t)− yxx(0, t)‖

‖y(x, T0)− y(x, T0)‖

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L),

‖yx(0, t)− yx(0, t)‖, ‖yxx(0, t)− yxx(0, t)‖

‖y(x, T0)− y(x, T0)‖

The approach

1 The Bukhgeim-Klibanov-Malinsky method.2 Carleman estimate for the linearized equation.

BUKHGEIM, KLIBANOV, 1981; KLIBANOV, MALINSKY 1991: Inverse problems withCarleman estimates.

PUEL, YAMAMOTO 1996; YAMAMOTO, 1999; IMANUVILOV, YAMAMOTO 2001: Waveequation.

IMANUVILOV, YAMAMOTO 1998, BENABDALLAH, GAITAN,LE ROUSSEAU:2007Parabolic equations.

BAUDOUIN, PUEL 2002; CARDOULIS,CRISTOFOL, GAITAN 2008;MERCADO,OSSES, ROSIER 2008: Schrodinger equation.

EGGER, ENGL, KLIBANOV, 2005; BOULAKIA, GRANDMONT, OSSES, 2009:Nonlinear equations.

BELLASSOUED, YAMAMOTO 2006; BELLASSOUED, CHOULLI, 2009 : Logarithmicstability for the wave equation and the Schrodinger equation

The approach

BMK methodWe follow ideas of Bukhgeim, Klibanov (1981), and Klibanov, Malinsky (1991).

If we set:

u = y − y and

σ = a− athen u solves the following KdV equation:

ut + a(x)uxxx + (1 + y)ux + yxu+ uux = σyxxx, ∀(x, t) ∈ (0, L)× (0, T ),u(0, t) = 0, u(L, t) = 0, ux(L, t) = 0 ∀t ∈ (0, T ),u(x, 0) = 0, ∀x ∈ (0, L).

Then z = ut satisfies the following equation:zt + a(x)zxxx + (1 + y)zx + yxz = fσ, ∀(x, t) ∈ (0, L)× (0, T ),z(0, t) = 0, z(L, t) = 0, zx(L, t) = 0 ∀t ∈ (0, T ),z(x, 0) = σ(x)y0,xxx(x), ∀x ∈ (0, L),

wherefσ = σ(x)yxxxt − yxtu− ytux.

If we set:

u = y − y and

If we set:

u = y − y and

BMK methodThen z = ut satisfies the following equation:

zt + a(x)zxxx + (1 + y)zx + yxz = fσ, ∀(x, t) ∈ (0, L)× (0, T ),z(0, t) = 0, z(L, t) = 0, zx(L, t) = 0 ∀t ∈ (0, T ),z(x, 0) = σ(x)y0,xxx(x), ∀x ∈ (0, L),

We would like to have an estimate like

‖z(x, 0)‖X ≤ C‖fσ‖Y + (boundary terms)

where C can be chosen small !!

We shall need y0,xxx(x) bounded by below by a positive constant!

Here Carleman estimates will be used. Time-singular weights are required!

Remark: This kind of inequality is called observability in control theory

BMK method - Heat equations

zt − a(x)zxx = fσ, ∀(x, t) ∈ (0, L)× (0, T ),z(0, t) = 0, z(L, t) = 0 ∀t ∈ (0, T ),z(x, 0) = σ(x)y0,xx(x), ∀x ∈ (0, L),

What happens for the heat equation?

Observability‖z(x, 0)‖X ≤ C‖fσ‖Y + (boundary terms),

can not be proved for parabolic equation.

Instead, one gets ‖z(x, T0)‖X ≤ C‖fσ‖Y + (boundary terms).

We use the equation

‖z(x, T0)‖ = ‖ut(x, T0)‖ = ‖σR(x, T0) + a(x)uxx(x, T0)‖≥ ‖σR(x, T0)‖ − ‖a(x)uxx(x, T0)‖

and we have to add an observation like ‖yxx(x, T0)− yxx(x, T0)‖!

We use the equation

BMK method - Wave equation

ztt − a(x)zxx = fσ, ∀(x, t) ∈ (0, L)× (0, T ),z(0, t) = 0, z(L, t) = 0 ∀t ∈ (0, T ),z(x, 0) = σ(x)y0,xx(x), ∀x ∈ (0, L),

What happens for wave equation?

Extend the solution to (−T, T ) by using the symmetry under the change ofvariable t→ (T − t).Use Carleman inequalities on (−T, T ).

The time t = 0 is not singular and you get

‖z(x, 0)‖X ≤ C‖fσ‖Y + (boundary terms),

with C small.

BMK method - Schrodinger equation

izt + a(x)zxx = fσ, ∀(x, t) ∈ (0, L)× (0, T ),z(0, t) = 0, z(L, t) = 0 ∀t ∈ (0, T ),z(x, 0) = σ(x)y0,xx(x), ∀x ∈ (0, L),

What happens for the Schrodinger equation?

The Carleman weight is singular at t = 0 and t = T .

Extend the solution to (−T, T ) by defining the solution for negative time as

z(x, t) := −z(x,−t), fσ(x, t) = −fσ(x,−t)

Use Carleman inequalities on (−T, T ).

The time t = 0 is not singular anymore and you get

with C small.

Remark: The method needs Re(y0(x)) to be zero.

z(x, t) := −z(x,−t), fσ(x, t) = −fσ(x,−t)

with C small.

z(x, t) := −z(x,−t), fσ(x, t) = −fσ(x,−t)

with C small.

z(x, t) := −z(x,−t), fσ(x, t) = −fσ(x,−t)

with C small.

z(x, t) := −z(x,−t), fσ(x, t) = −fσ(x,−t)

with C small.

z(x, t) := −z(x,−t), fσ(x, t) = −fσ(x,−t)

with C small.

BMK method - KdV equation

What happens for KdV equation?

Not parabolic neither hyperbolic.

From a control point of view, in some cases it is parabolic and in others hyperbolic.

KdV has only one time-derivative and so the change t→ T − t is not adequate.

But it has the symmetry t→ T − t and x→ L− x, which allows to define thesolution for negative times.

Carleman estimate on (−T, T )× (0, L).

Time t = 0 is not singular any more for Carleman and therefore

is obtained with C small.

Remark: Some symmetry conditions have to be imposed.

BMK method - Extension for negative time

Symmetric extension to (0, L)× (−T, T ) of g defined on (0, L)× (0, T ):

gs(x, t) =

g(x, t) if x ∈ [0, L], t ∈ [0, T ],

g(L− x,−t) if x ∈ [0, L], t ∈ [−T, 0).

Anti-symmetric extension to (0, L)× (−T, T ) of g defined on (0, L)× (0, T ):

ga(x, t) =

g(x, t) if x ∈ [0, L], t ∈ [0, T ],

−g(L− x,−t) if x ∈ [0, L], t ∈ [−T, 0).

Defining v = zs, we obtain:

vt + a(x)vxxx + (1 + ys)vx + (yx)av = faσ , ∀x ∈ (0, L), t ∈ (−T, T ),v(0, t) = 0, v(L, t) = 0, ∀t ∈ (−T, T ),

vx(L, t) =

0, ∀t ∈ (0, T ),

−zx(0,−t), ∀t ∈ (−T, 0).

v(x, 0) = σ(x)y0,xxx(x), ∀x ∈ (0, L).

gs(x, t) =

g(x, t) if x ∈ [0, L], t ∈ [0, T ],

g(L− x,−t) if x ∈ [0, L], t ∈ [−T, 0).

ga(x, t) =

g(x, t) if x ∈ [0, L], t ∈ [0, T ],

−g(L− x,−t) if x ∈ [0, L], t ∈ [−T, 0).

vx(L, t) =

0, ∀t ∈ (0, T ),

−zx(0,−t), ∀t ∈ (−T, 0).

v(x, 0) = σ(x)y0,xxx(x), ∀x ∈ (0, L).

gs(x, t) =

g(x, t) if x ∈ [0, L], t ∈ [0, T ],

g(L− x,−t) if x ∈ [0, L], t ∈ [−T, 0).

ga(x, t) =

g(x, t) if x ∈ [0, L], t ∈ [0, T ],

−g(L− x,−t) if x ∈ [0, L], t ∈ [−T, 0).

vx(L, t) =

0, ∀t ∈ (0, T ),

−zx(0,−t), ∀t ∈ (−T, 0).

v(x, 0) = σ(x)y0,xxx(x), ∀x ∈ (0, L).

The solution of

vx(L, t) =

0, ∀t ∈ (0, T ),

−zx(0,−t), ∀t ∈ (−T, 0).

v(x, 0) = σ(x)y0,xxx(x), ∀x ∈ (0, L).

satisfies a Carleman estimate which allows to prove

‖v(x, 0)‖X ≤ C‖fσ‖Y + (boundary terms)

with C small.

Carleman estimates - Lv = vt + avxxx = f

Any v ∈ L2(−T, T ;H3 ∩H10 (0, L)) and a weight function φ(x, t) = β(x)

(T+t)(T−t) .

w = e−λφv, and Lφw = e−λφL(eλφw)

where λ is a large parameter to be chosen later.

The obtained Carleman estimate is an inequality like

λ5‖w‖2L2φ

+ λ3‖wx‖2L2φ

+ λ‖wxx‖2L2φ

λ‖wt‖2L2

φ≤ C ‖Lφw‖2L2

φ+B.D.(w)

Note that w(−T, 0) = 0, and therefore

‖w(0, x)‖2L2φ

∫wwt ≤

∫ ∫|w|2

)1/2(1

∫ ∫|wt|2

∫ ∫|w|2

)1/2 (‖Lφw‖2L2

φ+B.D.(w)

(‖Lφw‖2L2

φ+B.D.(w)

)E. Cerpa (USM) Inverse problem for KdV September 2011 17 / 20

(T+t)(T−t) .

λ5‖w‖2L2φ

+ λ3‖wx‖2L2φ

+ λ‖wxx‖2L2φ

λ‖wt‖2L2

φ+B.D.(w)

‖w(0, x)‖2L2φ

∫wwt ≤

∫ ∫|w|2

)1/2(1

∫ ∫|wt|2

∫ ∫|w|2

)1/2 (‖Lφw‖2L2

φ+B.D.(w)

(‖Lφw‖2L2

φ+B.D.(w)

(T+t)(T−t) .

λ5‖w‖2L2φ

+ λ3‖wx‖2L2φ

+ λ‖wxx‖2L2φ

λ‖wt‖2L2

φ+B.D.(w)

‖w(0, x)‖2L2φ

∫wwt ≤

∫ ∫|w|2

)1/2(1

∫ ∫|wt|2

∫ ∫|w|2

)1/2 (‖Lφw‖2L2

φ+B.D.(w)

(‖Lφw‖2L2

φ+B.D.(w)

(T+t)(T−t) .

λ5‖w‖2L2φ

+ λ3‖wx‖2L2φ

+ λ‖wxx‖2L2φ

λ‖wt‖2L2

φ+B.D.(w)

‖w(0, x)‖2L2φ

∫wwt ≤

∫ ∫|w|2

)1/2(1

∫ ∫|wt|2

∫ ∫|w|2

)1/2 (‖Lφw‖2L2

φ+B.D.(w)

(‖Lφw‖2L2

φ+B.D.(w)

(T+t)(T−t) .

λ5‖w‖2L2φ

+ λ3‖wx‖2L2φ

+ λ‖wxx‖2L2φ

λ‖wt‖2L2

φ+B.D.(w)

‖w(0, x)‖2L2φ

∫wwt ≤

∫ ∫|w|2

)1/2(1

∫ ∫|wt|2

∫ ∫|w|2

)1/2 (‖Lφw‖2L2

φ+B.D.(w)

(‖Lφw‖2L2

φ+B.D.(w)

Carleman estimates for KdV.

Rosier [2004]. Null control of the surface of a water wave by means of awavemaker at the left end-point.

Glass-Guerrero [2008]. Cost of the null control of KdV by means of a control at theleft end-point.

Both papers prove Carleman estimates with one parameter λ > 0.

For us, it is important a second parameter. Look at one dominating term:

∫∫φ4x(−axφx − 5aφxx + 4a2φxx)|w|2

This impose bad conditions of kind ‖ax/a‖L∞ ≤M .

Solution is to choose φ such that φxx ≈ s2ϕ with a second parameter s > 0.

Main Result.

yx(L, t) = g2(t), ∀t ∈ (0, T ),y(0, x) = y0(x), ∀x ∈ (0, L).

Data (g, gk, y0) fixed and regular enough!

TheoremLet |y0,xxx(x)| ≥ δ > 0, also symmetric wrt L/2. Let

Σ =a symmetric wrt L/2

/a ≥ a0 > 0, ‖a‖W3,∞ ≤M1, and ‖y(a)‖W1,∞(Q) ≤M2

There exists a constant C = C(L, T, a0,M1,M2, δ) > 0 such that for any a, a ∈ Σ:

C‖a− a‖L2(0,L) ≤ ‖yx(0, t)− yx(0, t)‖H1(0,T ) + ‖yxx(0, t)− yxx(0, t)‖H1(0,T )

+ ‖yxx(L, t)− yxx(L, t)‖H1(0,T )

Thank you for your attention!

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