On the Control of some Galpern-Sobolev Equations
Diego Souza (Uni-Graz)
Joint work with F. W. Chaves-Silva (Nice)
A conference in the honor of Jean-Michel Coron June 20th-24th, 2016
On the Control of some Galpern-Sobolev Equations Diego Souza - - PowerPoint PPT Presentation
On the Control of some Galpern-Sobolev Equations Diego Souza (Uni-Graz) Joint work with F. W. Chaves-Silva (Nice) A conference in the honor of Jean-Michel Coron June 20th-24th, 2016 Outline Introduction Barenblatt-Zheltov-Kochina
Diego Souza (Uni-Graz)
Joint work with F. W. Chaves-Silva (Nice)
A conference in the honor of Jean-Michel Coron June 20th-24th, 2016
A partial differential equation is of Sobolev-Galpern type if the highest order terms contain derivatives in both space and time coordinates. A typical example is M∂tu + Lu = f with L and M are linear partial differential operators in the spatial variable of order 2m and l ≤ 2m, respectively (independent of t).
A partial differential equation is of Sobolev-Galpern type if the highest order terms contain derivatives in both space and time coordinates. A typical example is M∂tu + Lu = f with L and M are linear partial differential operators in the spatial variable of order 2m and l ≤ 2m, respectively (independent of t).
We will focus on the following equations:
ut − ∆ut − ∆u = f
ut − uxxt + ux = f
We are interested in the null controllability of the BZK equation:
in Q := (0, T) × Ω, u = 0
Σ := (0, T) × ∂Ω, u(0) = u0 in Ω, i.e, given u0, find f such that the solution of BZK satisfies: u(T) = 0. Here, Ω ⊂ RN and ω ⊂ Ω is a nonempty subset.
in Q := (0, T) × Ω, u = 0
Σ := (0, T) × ∂Ω, u(0) = u0 in Ω,
The BZK equation can be decomposed as
in Q, vt + v − u = f1ω in Q, u = 0
Σ, v(0) = u0 − ∆u0 in Ω. We have: v(T) = 0 iff u(T) = 0.
The null controllability for BZK system is equivalent to the observability inequality for the adjoint solutions, i.e ψ(·, 0)2
L2(Ω) ≤ C
|ψ|2, where
in Q, −ψt + ψ = ϕ in Q, ϕ = 0
Σ, ψ(T) = ψT in Ω.
Remark For a given (fixed) ω ⊂ Ω, null controllability fails for system BZK. This is due to:
µj 1+µj : µj ∈ σ(−∆)};
What can we do?
Remark For a given (fixed) ω ⊂ Ω, null controllability fails for system BZK. This is due to:
µj 1+µj : µj ∈ σ(−∆)};
What can we do?
Make the control in the second equation of BZK move on time, i.e. ω = ω(t). The set ω(t) covers the whole domain Ω in its motion. This strategy has been previously used for other models:
Benjamin-Bona-Mahony equation on a periodic domain, J. Differential Equations 254 (2013), 141–178.
Wave Equation with Moving Control, SIAM J. Control Optim., 51 (1) (2013), 660–684.
Viscoelasticity with Moving Control, J. Math. Pures Appl., 101 (9) (2014), 198–222.
Make the control in the second equation of BZK move on time, i.e. ω = ω(t). The set ω(t) covers the whole domain Ω in its motion. This strategy has been previously used for other models:
Benjamin-Bona-Mahony equation on a periodic domain, J. Differential Equations 254 (2013), 141–178.
Wave Equation with Moving Control, SIAM J. Control Optim., 51 (1) (2013), 660–684.
Viscoelasticity with Moving Control, J. Math. Pures Appl., 101 (9) (2014), 198–222.
With moving controls, BZK equation and BZK system read
in Q, u = 0
Σ, u(·, 0) = u0 in Ω and
in Q, vt + v − u = f1ω(t) in Q, u = 0
Σ, v(·, 0) = v0 in Ω, respectively. And, still, we want to find a control f such that u(T) = v(T) = 0.
For the 1D case, using moment method, Q. Tao et al.1, showed the null controllability
control, J. Math. Analysis and Appl., 418 (2)(2014)
The adjoint system of BZK reads
in Q, −ψt + ψ = ϕ in Q, ϕ = 0
Σ, ψ(T) = ψT in Ω. Null controllability of BZK system is equivalent to ψ(·, 0)2
L2(Ω) ≤ C
T
|ψ|2dxdt.
Theorem (Chaves-Silva & S.) Given ψT ∈ L2(Ω), the solution (ϕ, ψ) of the adj. system of BZK satisfies:
ρ1(x, t)(|∇ϕ|2 + |ϕ|2)dxdt +
ρ2(x, t)|ψ|2dxdt +
ρ3(t)(|∇ϕt|2 + |ϕt|2)dxdt ≤ C T
ρ4(x, t)|ψ|2dxdt, where ρi, i = 1, . . . , 4 are appropriate weights.
Three main difficulties appear:
1 Carleman inequalities for the Laplace operator and ODE equations with a
moving control region2;
2 We must have the same weight functions in the Carleman for both equations. 3 Eliminate a local integral of ϕ.
Fortunately, we can handle all these difficulties.
viscoelasticity with a moving control, J. Math. Pures Appl., 101 (9) (2014), 198–222.
Theorem (Chaves-Silva & S.) Given v0 ∈ L2(Ω), there exists f ∈ L2(Q) such that the associated solution (u, v) of BZK system satisfies: v(T) = u(T) = 0.
Corollary Given u0 ∈ H1
0(Ω) ∩ H2(Ω), there exists f ∈ L2(Q) such that the associated solution
u of BZK equation satisfies u(T) = 0.
Similar ideas (but not the same!) can be used to study the controllability of
in Q, u = 0
Σ, u(0) = u0 in Ω, where A is a regular enough vector function.
The BBM equation can be decomposed as
in Q, vt + ∇ · (A(x, t)u) = f1ω in Q, u = 0
Σ, v(·, 0) = v0 in Ω. We have: v(T) = 0 iff u(T) = 0.
The null controllability for BBM system is equivalent to the observability inequality for the adjoint solutions, i.e ψ(·, 0)2
L2(Ω) ≤ C
|ψ|2, where
in Q, −ψt = ϕ in Q, ϕ = 0
Σ, ψ(T) = ψT in Ω.
Control Optim., 39 (2001), 1677–1696.
equation with space-dependent potential, Mathematishe Annalen 325 (2003), 543–582.
With moving controls, BBM system reads
in Q, vt + ∇ · (A(x, t)u) = f1ω(t) in Q, u = 0
Σ, v(·, 0) = v0 in Ω. And, still, we want to find a control f such that u(T) = v(T) = 0.
The adjoint system of BBM reads
in Q, −ψt = ϕ in Q, ϕ = 0
Σ, ψ(T) = ψT in Ω. Null controllability of BZK system is equivalent to ψ(·, 0)2
L2(Ω) ≤ C
T
|ψ|2dxdt.
Theorem (Chaves-Silva & S.) Given ψT ∈ L2(Ω), the solution (ϕ, ψ) of the adj. system of BZK satisfies:
ρ∗
1(x, t)(|∇ϕ|2 + |ϕ|2)dxdt +
ρ∗
2(x, t)|ψ|2dxdt
+
ρ∗
3(t)(|∇ϕt|2 + |ϕt|2)dxdt
≤ C T
ρ∗
4(x, t)|ψ|2dxdt,
where ρ∗
i , i = 1, . . . , 4 are appropriate weights.
Key ingredients:
M∂tu + Lu = f1ω?