Outline

Boundary Integral Operator and ItsApplications

Bingyu ZhangUniversity of Cincinnati

at

The 9th Workshop on Control of Distributed ParameterSystems

Beijing, China

July 3, 2015

Outline

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarter plane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

Outline

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarter plane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

Outline

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarter plane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

Outline

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarter plane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-homogeneous boundary value problems

ut = P(x ,D)u + f (x , t) u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1.

Ω ⊂ Rn, Γ = ∂Ω

P(x ,D) : 2m−th order operator,

Bj (x ,D) is sj−th order operator,

0 ≤ sj ≤ 2m − 1, j = 1,2, · · ·m − 1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-homogeneous boundary value problems

utt = P(x ,D)u + f (x , t), x ∈ Ω, t ∈ (0,T )

u(x ,0) = φ(x), ut (x ,0) = ψ(x),

Bj(x ,D) = gj(x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogeneous boundary value problems

ut = P(x ,D)u + f (x , t) u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj(x ,D) = 0, (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1.

utt = P(x ,D)u + f (x , t), x ∈ Ω, t ∈ (0,T )

u(x ,0) = φ(x), ut (x ,0) = ψ(x),

Bj(x ,D) = 0, (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Question

If the homogeneous boundary value problem admits a solution

u ∈ C([0,T ); Hs(Ω)),

what are

the optimal regularity conditions on boundary data gj

for the non-homogenous boundary value problem to have asolution u ∈ C([0,T ); Hs(Ω))?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-Homogeenous Boundary Value Problems and Applications

Vol. I, Vol. II and Vol. III

( 1968 French edition, 1972 English edition)

J.-L. Lions and E. Magnets

Optimal Control of Systems Governed by Partial DifferentialEquations

( 1968 French edition, 1971 English edition)

J.-L. Lions

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-Homogeenous Boundary Value Problems and Applications

Vol. I, Vol. II and Vol. III

( 1968 French edition, 1972 English edition)

J.-L. Lions and E. Magnets

Optimal Control of Systems Governed by Partial DifferentialEquations

( 1968 French edition, 1971 English edition)

J.-L. Lions

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

ut = P(x ,D)u + f (x , t) u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj , (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1.

Strategy to study non-homogeneous boundary value problems

(1) Study the homogeneous boundary value problem

(2) Homogenization

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

ut = P(x ,D)u + f (x , t) u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj , (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1.

Strategy to study non-homogeneous boundary value problems

(1) Study the homogeneous boundary value problem

(2) Homogenization

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

ut = P(x ,D)u + f (x , t) u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj , (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1.

Strategy to study non-homogeneous boundary value problems

(1) Study the homogeneous boundary value problem

(2) Homogenization

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J. L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J. L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J. L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J. L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: homogeneous Dirichlet boundary

ut = ∆u, u|t=0 = φ(x), u|Γ = 0

φ ∈ H1(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: homogeneous Dirichlet boundary

ut = ∆u, u|t=0 = φ(x), u|Γ = 0

φ ∈ H1(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Dirichlet boundary

ut = ∆u, u|t=0 = φ(x), u|Γ = g

φ ∈ H1(Ω)

andg ∈ L2(0,T ; H

32 (Γ)) ∩ H

34 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Dirichlet boundary

ut = ∆u, u|t=0 = φ(x), u|Γ = g

φ ∈ H1(Ω)

andg ∈ L2(0,T ; H

32 (Γ)) ∩ H

34 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Dirichlet boundary

ut = ∆u, u|t=0 = φ(x), u|Γ = g

φ ∈ H1(Ω)

andg ∈ L2(0,T ; H

32 (Γ)) ∩ H

34 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Dirichlet boundary

ut = ∆u, u|t=0 = φ(x), u|Γ = g

φ ∈ H1(Ω)

andg ∈ L2(0,T ; H

32 (Γ)) ∩ H

34 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: homogeneous Neumann Boundary

ut = ∆u, u|t=0 = φ(x),∂u∂ν

∣∣∣∣Γ

= 0

φ ∈ H1(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: homogeneous Neumann Boundary

ut = ∆u, u|t=0 = φ(x),∂u∂ν

∣∣∣∣Γ

= 0

φ ∈ H1(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Neumann Boundary

ut = ∆u, u|t=0 = φ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω),

g ∈ L2(0,T ; H12 (Γ)) ∩ H

14 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Neumann Boundary

ut = ∆u, u|t=0 = φ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω),

g ∈ L2(0,T ; H12 (Γ)) ∩ H

14 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Neumann Boundary

ut = ∆u, u|t=0 = φ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω),

g ∈ L2(0,T ; H12 (Γ)) ∩ H

14 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Heat equation: non-homogeneous Neumann Boundary

ut = ∆u, u|t=0 = φ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω),

g ∈ L2(0,T ; H12 (Γ)) ∩ H

14 (0,T ; L2(Γ)).

=⇒ u ∈ C([0,T ]; H1(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ L2(0,T ; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H32 (Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H

32 (Γ)) ∩ H

32 (0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ L2(0,T ; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H32 (Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H

32 (Γ)) ∩ H

32 (0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ L2(0,T ; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H32 (Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H

32 (Γ)) ∩ H

32 (0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ L2(0,T ; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H32 (Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H

32 (Γ)) ∩ H

32 (0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

Hidden regularity (Sharp trace regularity):

∂u∂ν

∣∣∣∣Γ

∈ L2(0,T ; L2(Γ))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Dirichlet boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

Hidden regularity (Sharp trace regularity):

∂u∂ν

∣∣∣∣Γ

∈ L2(0,T ; L2(Γ))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

Observerbility (L. F. Ho, 1986):

∣∣∣∣∂u∂ν

∣∣∣∣L2(Γ×(0,T ))

≥ C(‖φ‖H1(Ω) + ‖ψ‖L2(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x), u|Γ = 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

Observerbility (L. F. Ho, 1986):

∣∣∣∣∂u∂ν

∣∣∣∣L2(Γ×(0,T ))

≥ C(‖φ‖H1(Ω) + ‖ψ‖L2(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ L2(0,T ; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ L2(0,T ; H1(Ω))

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2((0,T )× Γ)

u ∈ H−12 (0,T ; H1(Ω) ∩ H

12 (0,T ; L2(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2((0,T )× Γ)

u ∈ H−12 (0,T ; H1(Ω) ∩ H

12 (0,T ; L2(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2((0,T )× Γ)

u ∈ H−12 (0,T ; H1(Ω) ∩ H

12 (0,T ; L2(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2((0,T )× Γ)

u ∈ C([0,T ]; H35−ε(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2((0,T )× Γ)

u ∈ C([0,T ]; H35−ε(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2((0,T )× Γ)

u ∈ C([0,T ]; H35−ε(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

Lasiecka & Triggiani (1980’s):

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= g

φ ∈ H1(Ω), ψ ∈ L2(Ω), g ∈ L2((0,T )× Γ)

u ∈ C([0,T ]; H35−ε(Ω)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

Conjecture (Trace regularity)

u|Γ ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

Conjecture (Trace regularity)

u|Γ ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Wave equation: Neumann boundary

utt = ∆u, u|t=0 = φ(x), ut |t=0 = ψ(x),∂u∂ν

∣∣∣∣Γ

= 0

φ ∈ H1(Ω), ψ ∈ L2(Ω) =⇒ u ∈ C([0,T ]; H1(Ω))

Conjecture (Trace regularity)

u|Γ ∈ L2(0,T ; H1(Γ)) ∩ H1(0,T ; L2(Γ)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Lasiecka & Triggiani (1989):

True if Ω = (0,∞);

True if Ω = Rn+ (n > 1), φ and ψ have compact support in

Ω.;

u|Γ ∈ H34 (ΣT ), but u|Γ /∈ H

34 +ε(ΣT ), ∀ ε > 0 if Ω = Rn

+.

u ∈ H34−ε(Γ× (0,T )) if Ω is parallelepiped;

u ∈ H23 (Ω× (0,T )) if Ω is a sphere.

u ∈ H35 (Γ× (0,T )) in general.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Lasiecka & Triggiani (1989):

True if Ω = (0,∞);

True if Ω = Rn+ (n > 1), φ and ψ have compact support in

Ω.;

u|Γ ∈ H34 (ΣT ), but u|Γ /∈ H

34 +ε(ΣT ), ∀ ε > 0 if Ω = Rn

+.

u ∈ H34−ε(Γ× (0,T )) if Ω is parallelepiped;

u ∈ H23 (Ω× (0,T )) if Ω is a sphere.

u ∈ H35 (Γ× (0,T )) in general.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Lasiecka & Triggiani (1989):

True if Ω = (0,∞);

True if Ω = Rn+ (n > 1), φ and ψ have compact support in

Ω.;

u|Γ ∈ H34 (ΣT ), but u|Γ /∈ H

34 +ε(ΣT ), ∀ ε > 0 if Ω = Rn

+.

u ∈ H34−ε(Γ× (0,T )) if Ω is parallelepiped;

u ∈ H23 (Ω× (0,T )) if Ω is a sphere.

u ∈ H35 (Γ× (0,T )) in general.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Lasiecka & Triggiani (1989):

True if Ω = (0,∞);

True if Ω = Rn+ (n > 1), φ and ψ have compact support in

Ω.;

u|Γ ∈ H34 (ΣT ), but u|Γ /∈ H

34 +ε(ΣT ), ∀ ε > 0 if Ω = Rn

+.

u ∈ H34−ε(Γ× (0,T )) if Ω is parallelepiped;

u ∈ H23 (Ω× (0,T )) if Ω is a sphere.

u ∈ H35 (Γ× (0,T )) in general.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Lasiecka & Triggiani (1989):

True if Ω = (0,∞);

True if Ω = Rn+ (n > 1), φ and ψ have compact support in

Ω.;

u|Γ ∈ H34 (ΣT ), but u|Γ /∈ H

34 +ε(ΣT ), ∀ ε > 0 if Ω = Rn

+.

u ∈ H34−ε(Γ× (0,T )) if Ω is parallelepiped;

u ∈ H23 (Ω× (0,T )) if Ω is a sphere.

u ∈ H35 (Γ× (0,T )) in general.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Lasiecka & Triggiani (1989):

True if Ω = (0,∞);

True if Ω = Rn+ (n > 1), φ and ψ have compact support in

Ω.;

u|Γ ∈ H34 (ΣT ), but u|Γ /∈ H

34 +ε(ΣT ), ∀ ε > 0 if Ω = Rn

+.

u ∈ H34−ε(Γ× (0,T )) if Ω is parallelepiped;

u ∈ H23 (Ω× (0,T )) if Ω is a sphere.

u ∈ H35 (Γ× (0,T )) in general.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Lasiecka & Triggiani (1989):

True if Ω = (0,∞);

True if Ω = Rn+ (n > 1), φ and ψ have compact support in

Ω.;

u|Γ ∈ H34 (ΣT ), but u|Γ /∈ H

34 +ε(ΣT ), ∀ ε > 0 if Ω = Rn

+.

u ∈ H34−ε(Γ× (0,T )) if Ω is parallelepiped;

u ∈ H23 (Ω× (0,T )) if Ω is a sphere.

u ∈ H35 (Γ× (0,T )) in general.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Nonlinear wave equations

Nonlinear Schrodinger equations

Nonlinear equations of the KdV type

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Nonlinear wave equations

Nonlinear Schrodinger equations

Nonlinear equations of the KdV type

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Nonlinear wave equations

Nonlinear Schrodinger equations

Nonlinear equations of the KdV type

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Strichartz smoothing

Kato smoothing and sharp Kato smoothing

Dispersive smoothing

Bourgain smoothing

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Strichartz smoothing

Kato smoothing and sharp Kato smoothing

Dispersive smoothing

Bourgain smoothing

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Strichartz smoothing

Kato smoothing and sharp Kato smoothing

Dispersive smoothing

Bourgain smoothing

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Strichartz smoothing

Kato smoothing and sharp Kato smoothing

Dispersive smoothing

Bourgain smoothing

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Strichartz smoothing

Kato smoothing and sharp Kato smoothing

Dispersive smoothing

Bourgain smoothing

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The Korteweg-de Vries equation

ut + uux + uxxx = 0, −∞ < x , t <∞.

G. de Vries (1866-1934) and D. J. Korteweg (1848-1941)

On the Change of Form of Long Waves advancing in aRectangular Canal and on a New Type of Long Stationary Waves

Philosophical Magazine, 5th series, 36, 1895, pp. 422–443

One of the most intensively studied nonlinear PDEs in the lastfive decades.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The Korteweg-de Vries equation

ut + uux + uxxx = 0, −∞ < x , t <∞.

G. de Vries (1866-1934) and D. J. Korteweg (1848-1941)

On the Change of Form of Long Waves advancing in aRectangular Canal and on a New Type of Long Stationary Waves

Philosophical Magazine, 5th series, 36, 1895, pp. 422–443

One of the most intensively studied nonlinear PDEs in the lastfive decades.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The Korteweg-de Vries equation

ut + uux + uxxx = 0, −∞ < x , t <∞.

G. de Vries (1866-1934) and D. J. Korteweg (1848-1941)

On the Change of Form of Long Waves advancing in aRectangular Canal and on a New Type of Long Stationary Waves

Philosophical Magazine, 5th series, 36, 1895, pp. 422–443

One of the most intensively studied nonlinear PDEs in the lastfive decades.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The Korteweg-de Vries equation

ut + uux + uxxx = 0, −∞ < x , t <∞.

G. de Vries (1866-1934) and D. J. Korteweg (1848-1941)

On the Change of Form of Long Waves advancing in aRectangular Canal and on a New Type of Long Stationary Waves

Philosophical Magazine, 5th series, 36, 1895, pp. 422–443

One of the most intensively studied nonlinear PDEs in the lastfive decades.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Discovery of Solitons

Inverse scattering transform

Advances of its mathematical theories due toapplications of harmonic analysis

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Discovery of Solitons

Inverse scattering transform

Advances of its mathematical theories due toapplications of harmonic analysis

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Discovery of Solitons

Inverse scattering transform

Advances of its mathematical theories due toapplications of harmonic analysis

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarterplane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The Cauchy problem

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ R

Question

For what values of s, the Cauchy problem is well-posed in thespace Hs(R)?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The Cauchy problem

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ R

Question

For what values of s, the Cauchy problem is well-posed in thespace Hs(R)?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The Cauchy problem

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ R

Question

For what values of s, the Cauchy problem is well-posed in thespace Hs(R)?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x , t ∈ Ris well-posed in the space Hs(R) for

Bona and Smith (1975) : s ≥ 2

Kato (1979 ): s > 32

Kenig, Ponce and Vega (1989) : s > 98

Kenig, Ponce and Vega (1991): s > 34

Bourgain (1993): s ≥ 0

Kenig, Ponce and Vega (1993): s > −58

Kenig, Ponce and Vega (1996): s > −34

Christ, Colliander and Tao (2003): s = −34

Conjecture: s ≥ −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarterplane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The KdV equation posed on a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞

u(x ,0) = φ(x), u(0, t) = h(t)

Well-posedness

Existence + Uniqueness + Continuous Dependence

(φ,h) ∈ Hs(R+)× Hs′loc(R+)→ u ∈ C(0,T ; Hs(R+)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The KdV equation posed on a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞

u(x ,0) = φ(x), u(0, t) = h(t)

Well-posedness

Existence + Uniqueness + Continuous Dependence

(φ,h) ∈ Hs(R+)× Hs′loc(R+)→ u ∈ C(0,T ; Hs(R+)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona and Wither (1983, SIAM J. Math. Anal.)Let k ≥ 1 be given integer. For any

φ ∈ H3k+1(R+) and h ∈ Hk+1loc (R+)

satisfying certain standard compatibility conditions, theIBVP admits a unique solution u ∈ L∞loc(R+; H3k+1(R+)).

Bona and Wither (1989, Diff. & Integral Eqns.)The solution map is continuous from the space

H3k+1(R+)× Hk+1loc (R+)

to the space

L∞loc(R+; H3k+1(R+)) ∩ L2loc(R+; H3k+2(R+)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona and Wither (1983, SIAM J. Math. Anal.)Let k ≥ 1 be given integer. For any

φ ∈ H3k+1(R+) and h ∈ Hk+1loc (R+)

satisfying certain standard compatibility conditions, theIBVP admits a unique solution u ∈ L∞loc(R+; H3k+1(R+)).

Bona and Wither (1989, Diff. & Integral Eqns.)The solution map is continuous from the space

H3k+1(R+)× Hk+1loc (R+)

to the space

L∞loc(R+; H3k+1(R+)) ∩ L2loc(R+; H3k+2(R+)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Question

For given s ∈ R with φ ∈ Hs(R+), what is the optimal valueof s′ such that when h ∈ Hs′

loc(R+) one has that the solutionu ∈ C([0,T ]; Hs(R+))?

Answer: s′ = (s + 1)/3.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Question

For given s ∈ R with φ ∈ Hs(R+), what is the optimal valueof s′ such that when h ∈ Hs′

loc(R+) one has that the solutionu ∈ C([0,T ]; Hs(R+))?

Answer: s′ = (s + 1)/3.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Definition

The IVBP is said to be well-posed in the space Hs(R+) if for agiven compatible pair

(φ,h) ∈ Hs(R+)× Hs+1

3loc (R+)

the IVBP admits a unique solution u ∈ C([0,T ]; Hs(R+)) whichdepends on (φ,h) continuously.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

For what values of s, the IBVP is well-posed in the spaceHs(R+)?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

How to solve the IBVP using the harmonic analysis basedapproach?Convert the IBVP to an equivalent integral equation

u(t) = W0(t)φ+ Wbdr (t)h −∫ t

0W0(t − τ)(uux )(τ)dτ

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

How to solve the IBVP using the harmonic analysis basedapproach?Convert the IBVP to an equivalent integral equation

u(t) = W0(t)φ+ Wbdr (t)h −∫ t

0W0(t − τ)(uux )(τ)dτ

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

u(t) = W0(t)φ+ Wbdr (t)−∫ t

0 W0(t − τ)(uux )(τ)dτ

w(x , t) = Wbdr (t)h solveswt + wx + wxxx = 0, 0 ≤ x , t <∞

w(x ,0) = 0, w(0, t) = h(t).

v(x , t) = W0(t)φ solvesvt + vx + vxxx = 0, 0 ≤ x , t <∞

v(x ,0) = φ(x), v(0, t) = 0.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

u(t) = W0(t)φ+ Wbdr (t)−∫ t

0 W0(t − τ)(uux )(τ)dτ

w(x , t) = Wbdr (t)h solveswt + wx + wxxx = 0, 0 ≤ x , t <∞

w(x ,0) = 0, w(0, t) = h(t).

v(x , t) = W0(t)φ solvesvt + vx + vxxx = 0, 0 ≤ x , t <∞

v(x ,0) = φ(x), v(0, t) = 0.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

u(t) = W0(t)φ+ Wbdr (t)−∫ t

0 W0(t − τ)(uux )(τ)dτ

Extend the integral equation posed on the quarter planeR+ × R+ to the whole plane R × R.

How?

Find explicit integral representations of W0(t)φ andWbdr (t)h.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

u(t) = W0(t)φ+ Wbdr (t)−∫ t

0 W0(t − τ)(uux )(τ)dτ

Extend the integral equation posed on the quarter planeR+ × R+ to the whole plane R × R.

How?

Find explicit integral representations of W0(t)φ andWbdr (t)h.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

u(t) = W0(t)φ+ Wbdr (t)−∫ t

0 W0(t − τ)(uux )(τ)dτ

Extend the integral equation posed on the quarter planeR+ × R+ to the whole plane R × R.

How?

Find explicit integral representations of W0(t)φ andWbdr (t)h.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uxxx = 0, 0 < x , t <∞

u(x ,0) = 0, u(0, t) = h(t)u(x , t) = Wbdr(t)h

Wbdr (t)h = [Ub(t)h] (x) + [Ub(t)h] (x)

where, for x , t ≥ 0,

[Ub(t)h] (x) =1

2π

∫ ∞1

eit(µ3−µ)e−(√

3µ2−4+iµ2

)xh(µ)dµ

withh(µ) = (3µ2 − 1)

∫ ∞0

e−iξ(µ3−µ)h(ξ)dξ.

Boundary integral operator

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uxxx = 0, 0 < x , t <∞

u(x ,0) = 0, u(0, t) = h(t)u(x , t) = Wbdr(t)h

Wbdr (t)h = [Ub(t)h] (x) + [Ub(t)h] (x)

where, for x , t ≥ 0,

[Ub(t)h] (x) =1

2π

∫ ∞1

eit(µ3−µ)e−(√

3µ2−4+iµ2

)xh(µ)dµ

withh(µ) = (3µ2 − 1)

∫ ∞0

e−iξ(µ3−µ)h(ξ)dξ.

Boundary integral operator

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uxxx = 0, 0 < x , t <∞

u(x ,0) = 0, u(0, t) = h(t)u(x , t) = Wbdr(t)h

Wbdr (t)h = [Ub(t)h] (x) + [Ub(t)h] (x)

where, for x , t ≥ 0,

[Ub(t)h] (x) =1

2π

∫ ∞1

eit(µ3−µ)e−(√

3µ2−4+iµ2

)xh(µ)dµ

withh(µ) = (3µ2 − 1)

∫ ∞0

e−iξ(µ3−µ)h(ξ)dξ.

Boundary integral operator

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The integral representation of W0(t)φ is too complicated!

For the Cauchy problem of the KdV equation posed on thewhole line R:

ut + ux + uxxx = 0, u(x ,0) = ψ(x), −∞ < x , t <∞

u(x , t) = WR(t)ψ = 12πi

∫∞−∞ ei(ξ3−ξ)teixξ

∫∞−∞ e−iyξψ(y)dydξ.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The integral representation of W0(t)φ is too complicated!

For the Cauchy problem of the KdV equation posed on thewhole line R:

ut + ux + uxxx = 0, u(x ,0) = ψ(x), −∞ < x , t <∞

u(x , t) = WR(t)ψ = 12πi

∫∞−∞ ei(ξ3−ξ)teixξ

∫∞−∞ e−iyξψ(y)dydξ.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

The integral representation of W0(t)φ is too complicated!

For the Cauchy problem of the KdV equation posed on thewhole line R:

ut + ux + uxxx = 0, u(x ,0) = ψ(x), −∞ < x , t <∞

u(x , t) = WR(t)ψ = 12πi

∫∞−∞ ei(ξ3−ξ)teixξ

∫∞−∞ e−iyξψ(y)dydξ.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = 0

, u(x , t) = W0(t)φ(x)

Rewrite W0(t)φ in terms of WR(t)ψ and Wbdr (t)h.

Benefits:

Estimates on WR(t)ψ are known; one only needs to workon the boundary integral operator Wbdr (t)h.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = 0

, u(x , t) = W0(t)φ(x)

Rewrite W0(t)φ in terms of WR(t)ψ and Wbdr (t)h.

Benefits:

Estimates on WR(t)ψ are known; one only needs to workon the boundary integral operator Wbdr (t)h.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = 0

, u(x , t) = W0(t)φ(x)

Rewrite W0(t)φ in terms of WR(t)ψ and Wbdr (t)h.

Benefits:

Estimates on WR(t)ψ are known; one only needs to workon the boundary integral operator Wbdr (t)h.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Non-homogenization

ut + ux + uxxx = 0, u(x ,0) = φ(x), u(0, t) = 0, x , t ∈ R+

(i) Solve

vt + vx + vxxx = 0, v(x ,0) = φ(x), x , t ∈ R

(ii) Let q(t) := v(0, t) and solve

zt +zx +zxxx = 0, z(0, t) = 0, z(0, t) = q(t), x , t ∈ R+

(iii) u(x , t) = v(x , t)− z(x , t)

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Non-homogenization

ut + ux + uxxx = 0, u(x ,0) = φ(x), u(0, t) = 0, x , t ∈ R+

(i) Solve

vt + vx + vxxx = 0, v(x ,0) = φ(x), x , t ∈ R

(ii) Let q(t) := v(0, t) and solve

zt +zx +zxxx = 0, z(0, t) = 0, z(0, t) = q(t), x , t ∈ R+

(iii) u(x , t) = v(x , t)− z(x , t)

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Non-homogenization

ut + ux + uxxx = 0, u(x ,0) = φ(x), u(0, t) = 0, x , t ∈ R+

(i) Solve

vt + vx + vxxx = 0, v(x ,0) = φ(x), x , t ∈ R

(ii) Let q(t) := v(0, t) and solve

zt +zx +zxxx = 0, z(0, t) = 0, z(0, t) = q(t), x , t ∈ R+

(iii) u(x , t) = v(x , t)− z(x , t)

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Non-homogenization

ut + ux + uxxx = 0, u(x ,0) = φ(x), u(0, t) = 0, x , t ∈ R+

(i) Solve

vt + vx + vxxx = 0, v(x ,0) = φ(x), x , t ∈ R

(ii) Let q(t) := v(0, t) and solve

zt +zx +zxxx = 0, z(0, t) = 0, z(0, t) = q(t), x , t ∈ R+

(iii) u(x , t) = v(x , t)− z(x , t)

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

W0(t)φ = WR(t)φ−Wbdr (t)q

withq(t) = WR(t)φ

∣∣∣x=0

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

∫ t

0W0(t − τ)f (τ)dτ =

∫ t

0WR(t − τ)f (τ)dτ −Wbdr (t) p

with

p(t) =

∫ t

0W0(t − τ)f (τ)dτ

∣∣∣∣x=0

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

u(t) = W0(t)φ+ Wbdr (t)h −∫ t

0W0(t − τ)(uux )(τ)dτ

u(t) = WR(t)φ+

∫ t

0WR(t − τ)(uux )(τ)dτ + Wbdr (t)(h − q − p)

withq(t) = WR(t)φ

∣∣∣x=0

,

p(t) =

∫ t

0W0(t − τ)(uux )(τ)dτ

∣∣∣∣x=0

.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Key to work

ut + ux + uxxx = 0, u(x ,0) = φ(x),−∞ < x , t <∞

supx∈R‖u(x , ·)‖Hs′

loc(R)≤ C‖φ‖Hs(R).

vt + vx + vxxx = 0, 0 < x , t <∞v(x ,0) = 0, v(0, t) = h(t)

supt∈(0,T )

‖v(·, t)‖Hs(R+) ≤ C‖h‖Hs∗loc (R+).

Key: s′ = s∗ (= s+13 ).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Key to work

ut + ux + uxxx = 0, u(x ,0) = φ(x),−∞ < x , t <∞

supx∈R‖u(x , ·)‖Hs′

loc(R)≤ C‖φ‖Hs(R).

vt + vx + vxxx = 0, 0 < x , t <∞v(x ,0) = 0, v(0, t) = h(t)

supt∈(0,T )

‖v(·, t)‖Hs(R+) ≤ C‖h‖Hs∗loc (R+).

Key: s′ = s∗ (= s+13 ).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

Key to work

ut + ux + uxxx = 0, u(x ,0) = φ(x),−∞ < x , t <∞

supx∈R‖u(x , ·)‖Hs′

loc(R)≤ C‖φ‖Hs(R).

vt + vx + vxxx = 0, 0 < x , t <∞v(x ,0) = 0, v(0, t) = h(t)

supt∈(0,T )

‖v(·, t)‖Hs(R+) ≤ C‖h‖Hs∗loc (R+).

Key: s′ = s∗ (= s+13 ).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona, Sun and Zhang [2001, Trans. Amer. Math. Soc.]

The IVBP is well-posed in the space Hs(R+) for s > 34 .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Colliander and Kenig [2002, Comm. on PDE]

For any φ ∈ Hs(R+) and h ∈ H(s+1)/3(R+) with 0 ≤ s ≤ 1,the IVBP admits a solution u ∈ C([0,T ]; Hs(R+)).

The result has been extended to the case of s > −34 by

Justin Holmer [2006, Comm. PDEs].

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Colliander and Kenig [2002, Comm. on PDE]

For any φ ∈ Hs(R+) and h ∈ H(s+1)/3(R+) with 0 ≤ s ≤ 1,the IVBP admits a solution u ∈ C([0,T ]; Hs(R+)).

The result has been extended to the case of s > −34 by

Justin Holmer [2006, Comm. PDEs].

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona, Sun and Zhang [2006, Dynamics of PDEs]

The IBVP is well-posed in Hs(R+) for any s > −34 .

Can s be smaller than −34?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona, Sun and Zhang [2006, Dynamics of PDEs]

The IBVP is well-posed in Hs(R+) for any s > −34 .

Can s be smaller than −34?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona, Sun and Zhang [2008, Ann. I.H. Poincare - AN]

The IBVP is well-posed for φ ∈ Hsw (R+) and

h ∈ H(s+1)/3(R+) for any s > −1.

Here Hsw (R+) is the weighted Sobolev space:

Hsw (R+) := v ∈ Hs(R+); eνxv ∈ Hs(R+).

Conjecture

The IBVP is well-posed in Hs(R+) for any s > −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona, Sun and Zhang [2008, Ann. I.H. Poincare - AN]

The IBVP is well-posed for φ ∈ Hsw (R+) and

h ∈ H(s+1)/3(R+) for any s > −1.

Here Hsw (R+) is the weighted Sobolev space:

Hsw (R+) := v ∈ Hs(R+); eνxv ∈ Hs(R+).

Conjecture

The IBVP is well-posed in Hs(R+) for any s > −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Cauchy problemNon-homogeneous boundary value problems in a quarter plane

ut + ux + uux + uxxx = 0, 0 < x , t <∞u(x ,0) = φ(x), u(0, t) = h(t)

Bona, Sun and Zhang [2008, Ann. I.H. Poincare - AN]

The IBVP is well-posed for φ ∈ Hsw (R+) and

h ∈ H(s+1)/3(R+) for any s > −1.

Here Hsw (R+) is the weighted Sobolev space:

Hsw (R+) := v ∈ Hs(R+); eνxv ∈ Hs(R+).

Conjecture

The IBVP is well-posed in Hs(R+) for any s > −1.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Bubnov, 1980

ut + uux + uxxx = f , u(x ,0) = 0, x ∈ (0,L).

a1uxx (0, t) + a2ux (, t) + a3u(0, t) = 0b11uxx (L, t) + b12ux (L, t) + b13u(L, t) = 0,b22ux (L, t) + b23u(L, t) = 0.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Bubnov, 1980

ut + uux + uxxx = f , u(x ,0) = 0, x ∈ (0,L).

a1uxx (0, t) + a2ux (, t) + a3u(0, t) = 0b11uxx (L, t) + b12ux (L, t) + b13u(L, t) = 0,b22ux (L, t) + b23u(L, t) = 0.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Set

F1 =a3

a1− a2

22a1

, F3 = b12b23 − b13b22,

F2 =b12b23

b11b22− b13

b11−

b223

2b222.

Assume

if a1b11b22 6= 0, then F1 > 0, F2 > 0;

if b11 6= 0, b22 6= 0, a1 = 0, then F2 > 0, a2 = 0, a3 6= 0;

if b11 = 0, a1 6= 0, b22 6= 0, then F1 > 0, F3 6= 0;

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Set

F1 =a3

a1− a2

22a1

, F3 = b12b23 − b13b22,

F2 =b12b23

b11b22− b13

b11−

b223

2b222.

Assume

if a1b11b22 6= 0, then F1 > 0, F2 > 0;

if b11 6= 0, b22 6= 0, a1 = 0, then F2 > 0, a2 = 0, a3 6= 0;

if b11 = 0, a1 6= 0, b22 6= 0, then F1 > 0, F3 6= 0;

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Set

F1 =a3

a1− a2

22a1

, F3 = b12b23 − b13b22,

F2 =b12b23

b11b22− b13

b11−

b223

2b222.

Assume

if a1b11b22 6= 0, then F1 > 0, F2 > 0;

if b11 6= 0, b22 6= 0, a1 = 0, then F2 > 0, a2 = 0, a3 6= 0;

if b11 = 0, a1 6= 0, b22 6= 0, then F1 > 0, F3 6= 0;

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Set

F1 =a3

a1− a2

22a1

, F3 = b12b23 − b13b22,

F2 =b12b23

b11b22− b13

b11−

b223

2b222.

Assume

if a1b11b22 6= 0, then F1 > 0, F2 > 0;

if b11 6= 0, b22 6= 0, a1 = 0, then F2 > 0, a2 = 0, a3 6= 0;

if b11 = 0, a1 6= 0, b22 6= 0, then F1 > 0, F3 6= 0;

· · · · · ·

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uux + uxxx = f , u(x ,0) = 0, x ∈ (0,L).a1uxx (0, t) + a2ux (, t) + a3u(0, t) = 0b11uxx (L, t) + b12ux (L, t) + b13u(L, t) = 0,b22ux (L, t) + b23u(L, t) = 0.

Bubnov (1980):

if f ∈ H1loc(R+; L2(0,L)), there exists T > 0 such that

u ∈ L2(0,T ; H3(0,L)), ut ∈ L∞(0,T ; L2(0,L))∩L2(0,T ; H1(0,L)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uux + uxxx = f , u(x ,0) = 0, x ∈ (0,L).a1uxx (0, t) + a2ux (, t) + a3u(0, t) = 0b11uxx (L, t) + b12ux (L, t) + b13u(L, t) = 0,b22ux (L, t) + b23u(L, t) = 0.

Bubnov (1980):

if f ∈ H1loc(R+; L2(0,L)), there exists T > 0 such that

u ∈ L2(0,T ; H3(0,L)), ut ∈ L∞(0,T ; L2(0,L))∩L2(0,T ; H1(0,L)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uux + uxxx = f , u(x ,0) = 0, x ∈ (0,L).a1uxx (0, t) + a2ux (, t) + a3u(0, t) = 0b11uxx (L, t) + b12ux (L, t) + b13u(L, t) = 0,b22ux (L, t) + b23u(L, t) = 0.

Bubnov (1980):

if f ∈ H1loc(R+; L2(0,L)), there exists T > 0 such that

u ∈ L2(0,T ; H3(0,L)), ut ∈ L∞(0,T ; L2(0,L))∩L2(0,T ; H1(0,L)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L).a1uxx (0, t) + a2ux (, t) + a3u(0, t) = 0b11uxx (L, t) + b12ux (L, t) + b13u(L, t) = 0,b22ux (L, t) + b23u(L, t) = 0.

Kato smoothing:

f ∈ L2(0,T ; L2(0,L)) =⇒ u ∈ C([0,T ]; L2(0,L)) ∩ L2(0,T ; H1(0,L)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L).a1uxx (0, t) + a2ux (, t) + a3u(0, t) = 0b11uxx (L, t) + b12ux (L, t) + b13u(L, t) = 0,b22ux (L, t) + b23u(L, t) = 0.

Kato smoothing:

f ∈ L2(0,T ; L2(0,L)) =⇒ u ∈ C([0,T ]; L2(0,L)) ∩ L2(0,T ; H1(0,L)).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(i) u(0, t) = 0,u(1, t) = 0,ux (1, t) = 0

(ii) uxx (0, t) + aux (0, t) + bu(0, t) = 0,ux (1, t) = 0,u(1, t) = 0

a > b2/2;

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(i) u(0, t) = 0,u(1, t) = 0,ux (1, t) = 0

(ii) uxx (0, t) + aux (0, t) + bu(0, t) = 0,ux (1, t) = 0,u(1, t) = 0

a > b2/2;

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(i) u(0, t) = 0,u(1, t) = 0,ux (1, t) = 0

(ii) uxx (0, t) + aux (0, t) + bu(0, t) = 0,ux (1, t) = 0,u(1, t) = 0

a > b2/2;

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(iii) u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0

b < c2/2;

(iv) uxx (0, t) + a1ux (0, t) + a2u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0,

a2 > a21/2, b < c2/2

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(iii) u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0

b < c2/2;

(iv) uxx (0, t) + a1ux (0, t) + a2u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0,

a2 > a21/2, b < c2/2

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(iii) u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0

b < c2/2;

(iv) uxx (0, t) + a1ux (0, t) + a2u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0,

a2 > a21/2, b < c2/2

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(iii) u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0

b < c2/2;

(iv) uxx (0, t) + a1ux (0, t) + a2u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0,

a2 > a21/2, b < c2/2

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(iii) u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0

b < c2/2;

(iv) uxx (0, t) + a1ux (0, t) + a2u(0, t) = 0,uxx (1, t) + bu(1, t) = 0,ux (1, t) + cu(1, t) = 0,

a2 > a21/2, b < c2/2

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

KdV on a bounded domain

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L)

B1u = h1(t), B2u = h2(t), B3u = h3(t)

Bk u =2∑

j=0

akj∂jxu(0, t) + bkj∂

jxu(L, t), k = 1,2,3,

Under what conditions on akj , bkj , is the IBVP well-posed?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(I)

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

B1u = u(0, t),B2u = u(L, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(I)

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

B1u = u(0, t),B2u = u(L, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(II)

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

B1u = u(0, t),B2u = uxx (L, t) + b21ux (L, t) + b20u(L, t) + a21ux (0, t) + a20u(0, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(III)

ut + ux + uux + uxxx = f , u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

B1u = uxx (0, t) + a11ux (0, t) + b10u(0, t) + b11ux (L, t) + b10u(L, t),B2u = u(L, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(IV )

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

B1u = uxx (0, t) + a11ux (0, t) + b10u(0, t) + b11ux (L, t) + b10u(L, t),B2u = uxx (L, t) + b21ux (L, t) + b20u(L, t) + a21ux (0, t) + a20u(0, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(I)

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).B1u = u(0, t),B2u = u(L, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

Theorem 1 ( Capistrano-Filho, Sun, Zhang): The IBVP is locallywell-posed in Hs(0,L) for any s > −1 with φ ∈ Hs(0,L),

h1, h2 ∈ Hs+1

3loc (R+), h3 ∈ H

s3

loc(R+).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(II)

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).B1u = u(0, t),B2u = uxx (L, t) + b21ux (L, t) + b20u(L, t) + a21ux (0, t) + a20u(0, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

Theorem 2 (CSZ): The IBVP is locally well-posed in Hs(0,L) for anys > −1 with φ ∈ Hs(0,L),

h1 ∈ Hs+1

3loc (R+), h2 ∈ H

s−13 (R+), h3 ∈ H

s3

loc(R+).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(II)

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).B1u = u(0, t),B2u = uxx (L, t) + b21ux (L, t) + b20u(L, t) + a21ux (0, t) + a20u(0, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

Theorem 2 (CSZ): The IBVP is locally well-posed in Hs(0,L) for anys > −1 with φ ∈ Hs(0,L),

h1 ∈ Hs+1

3loc (R+), h2 ∈ H

s−13 (R+), h3 ∈ H

s3

loc(R+).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(III)

ut + ux + uux + uxxx = f , u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

B1u = uxx (0, t) + a11ux (0, t) + b10u(0, t) + b11ux (L, t) + b10u(L, t),B2u = u(L, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

Theorem 3 (CSZ): The IBVP is locally well-posed in Hs(0,L) for anys > −1 with φ ∈ Hs(0,L),

h1 ∈ Hs−1

3loc (R+), h2 ∈ H

s+13 (R+), h3 ∈ H

s3

loc(R+).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

(IV )

ut + ux + uux + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

B1u = uxx (0, t) + a11ux (0, t) + b10u(0, t) + b11ux (L, t) + b10u(L, t),B2u = uxx (L, t) + b21ux (L, t) + b20u(L, t) + a21ux (0, t) + a20u(0, t),B3u = ux (L, t) + b30u(L, t) + a30u(0, t).

Theorem 4 (CSZ): The IBVP is locally well-posed in Hs(0,L) for anys > −1 with φ ∈ Hs(0,L),

h1 ∈ Hs−1

3loc (R+), h2 ∈ H

s−13 (R+), h3 ∈ H

s3

loc(R+).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarterplane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

The key ingredients of the proofs

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

Kato smoothing

φ ∈ L2(0,L) =⇒ u ∈ C([0,T ]; L2(0,L)) ∩ L2(0,T ; H1(0,L)).

Sharp Kato smoothing (Hidden regularities):

φ ∈ L2(0,L) =⇒ ∂ jxu ∈ L∞x (0,L; H

1−j3 (0,T )), j = 0,1,2.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

The key ingredients of the proofs

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

Kato smoothing

φ ∈ L2(0,L) =⇒ u ∈ C([0,T ]; L2(0,L)) ∩ L2(0,T ; H1(0,L)).

Sharp Kato smoothing (Hidden regularities):

φ ∈ L2(0,L) =⇒ ∂ jxu ∈ L∞x (0,L; H

1−j3 (0,T )), j = 0,1,2.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

Kato smoothing

f ∈ L1(0,T ; L2(0,L)) =⇒ u ∈ C([0,T ]; L2(0,L))∩L2(0,T ; H1(0,L)).

Sharp Kato smoothing (Hidden regularities):

f ∈ L1(0,T ; L2(0,L)) =⇒ ∂ jxu ∈ L∞x (0,L; H

1−j3 (0,T )), j = 0,1,2.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

Kato smoothing

f ∈ L1(0,T ; L2(0,L)) =⇒ u ∈ C([0,T ]; L2(0,L))∩L2(0,T ; H1(0,L)).

Sharp Kato smoothing (Hidden regularities):

f ∈ L1(0,T ; L2(0,L)) =⇒ ∂ jxu ∈ L∞x (0,L; H

1−j3 (0,T )), j = 0,1,2.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

f ∈ L1x (0,L; L2(0,T )) =⇒ u ∈ C([0,T ]; H1(0,L)).

Hidden regularities:

f ∈ L1x (0,L; L2(0,T ) =⇒ ∂2

x u ∈ L∞x (0,L; L2(0,T )).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

f ∈ L1x (0,L; L2(0,T )) =⇒ u ∈ C([0,T ]; H1(0,L)).

Hidden regularities:

f ∈ L1x (0,L; L2(0,T ) =⇒ ∂2

x u ∈ L∞x (0,L; L2(0,T )).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Outline

1 Introduction

2 KdV equationCauchy problemNon-homogeneous boundary value problems in a quarterplane

3 KdV on a bounded domainSmoothing propertiesBoundary integral operator

4 Conclusion remarks

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

u = Wbdr (t)~h

The bridge to access Fourier analysis tools!

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

u = Wbdr (t)~h

The bridge to access Fourier analysis tools!

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).

u = Wbdr (t)~h

The bridge to access Fourier analysis tools!

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

vt + vxxx = 0, v(x ,0) = φ(x), x ∈ R, t ∈ R

v = W (t)φ, φ is the extension of φ from (0,L) to R.

u(x , t) = W (t)φ−Wbdr (t)~q

q1(t) = B1v , q2(t) = B2v , q3(t) = B3v .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

vt + vxxx = 0, v(x ,0) = φ(x), x ∈ R, t ∈ R

v = W (t)φ, φ is the extension of φ from (0,L) to R.

u(x , t) = W (t)φ−Wbdr (t)~q

q1(t) = B1v , q2(t) = B2v , q3(t) = B3v .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

vt + vxxx = 0, v(x ,0) = φ(x), x ∈ R, t ∈ R

v = W (t)φ, φ is the extension of φ from (0,L) to R.

u(x , t) = W (t)φ−Wbdr (t)~q

q1(t) = B1v , q2(t) = B2v , q3(t) = B3v .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

vt + vxxx = 0, v(x ,0) = φ(x), x ∈ R, t ∈ R

v = W (t)φ, φ is the extension of φ from (0,L) to R.

u(x , t) = W (t)φ−Wbdr (t)~q

q1(t) = B1v , q2(t) = B2v , q3(t) = B3v .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

vt + vxxx = 0, v(x ,0) = φ(x), x ∈ R, t ∈ R

v = W (t)φ, φ is the extension of φ from (0,L) to R.

u(x , t) = W (t)φ−Wbdr (t)~q

q1(t) = B1v , q2(t) = B2v , q3(t) = B3v .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

wt + wxxx = f , v(x ,0) = 0, x ∈ R, t ∈ R

w =

∫ t

0W (t − τ)f (τ)dτ, f is the extension of f from (0,L) to R.

u(x , t) =

∫ t

0W (t − τ)f (τ)dτ −Wbdr (t)~p

p1(t) = B1w , p2(t) = B2w , p3(t) = B3w .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

wt + wxxx = f , v(x ,0) = 0, x ∈ R, t ∈ R

w =

∫ t

0W (t − τ)f (τ)dτ, f is the extension of f from (0,L) to R.

u(x , t) =

∫ t

0W (t − τ)f (τ)dτ −Wbdr (t)~p

p1(t) = B1w , p2(t) = B2w , p3(t) = B3w .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

wt + wxxx = f , v(x ,0) = 0, x ∈ R, t ∈ R

w =

∫ t

0W (t − τ)f (τ)dτ, f is the extension of f from (0,L) to R.

u(x , t) =

∫ t

0W (t − τ)f (τ)dτ −Wbdr (t)~p

p1(t) = B1w , p2(t) = B2w , p3(t) = B3w .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

wt + wxxx = f , v(x ,0) = 0, x ∈ R, t ∈ R

w =

∫ t

0W (t − τ)f (τ)dτ, f is the extension of f from (0,L) to R.

u(x , t) =

∫ t

0W (t − τ)f (τ)dτ −Wbdr (t)~p

p1(t) = B1w , p2(t) = B2w , p3(t) = B3w .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Homogeneous boundary value problem

ut + uxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0.

wt + wxxx = f , v(x ,0) = 0, x ∈ R, t ∈ R

w =

∫ t

0W (t − τ)f (τ)dτ, f is the extension of f from (0,L) to R.

u(x , t) =

∫ t

0W (t − τ)f (τ)dτ −Wbdr (t)~p

p1(t) = B1w , p2(t) = B2w , p3(t) = B3w .

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Non-homogenization

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0,

u(x , t) = W (t)φ−Wbdr (t)~qvt + vxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1v = 0, B2v = 0, B3v = 0,

v(x , t) =∫ t

0 W (t − τ)f (τ)dτ −Wbdr (t)~p

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Non-homogenization

ut + uxxx = 0, u(x ,0) = φ(x), x ∈ (0,L),

B1u = 0, B2u = 0, B3u = 0,

u(x , t) = W (t)φ−Wbdr (t)~qvt + vxxx = f , u(x ,0) = 0, x ∈ (0,L),

B1v = 0, B2v = 0, B3v = 0,

v(x , t) =∫ t

0 W (t − τ)f (τ)dτ −Wbdr (t)~p

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Smootng properties of the boundary integral operator

u(x , t) := [Wbdr~h](x , t), ~h = (h1,h2,h3.)

Kato smoothing,Sharp Kato smoothing,Bourgain smoothing

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Smootng properties of the boundary integral operator

u(x , t) := [Wbdr~h](x , t), ~h = (h1,h2,h3.)

Kato smoothing,Sharp Kato smoothing,Bourgain smoothing

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Smootng properties of the boundary integral operator

u(x , t) := [Wbdr~h](x , t), ~h = (h1,h2,h3.)

Kato smoothing,Sharp Kato smoothing,Bourgain smoothing

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Smootng properties of the boundary integral operator

u(x , t) := [Wbdr~h](x , t), ~h = (h1,h2,h3.)

Kato smoothing,Sharp Kato smoothing,Bourgain smoothing

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Explicit integral representation boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).su + uxxx = 0, x ∈ (0,L)

B1u(s) = h1(s), B2u(s) = h2(s), B3u(s) = h3(s).

u(x , s) = [R(s,A)~h](x , s)

u(x , t) =∫ γ+i∞γ−i∞ eist [R(s,A)

~h](x , s)ds

Appropriate contour transformation

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Explicit integral representation boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).su + uxxx = 0, x ∈ (0,L)

B1u(s) = h1(s), B2u(s) = h2(s), B3u(s) = h3(s).

u(x , s) = [R(s,A)~h](x , s)

u(x , t) =∫ γ+i∞γ−i∞ eist [R(s,A)

~h](x , s)ds

Appropriate contour transformation

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Explicit integral representation boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).su + uxxx = 0, x ∈ (0,L)

B1u(s) = h1(s), B2u(s) = h2(s), B3u(s) = h3(s).

u(x , s) = [R(s,A)~h](x , s)

u(x , t) =∫ γ+i∞γ−i∞ eist [R(s,A)

~h](x , s)ds

Appropriate contour transformation

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Explicit integral representation boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).su + uxxx = 0, x ∈ (0,L)

B1u(s) = h1(s), B2u(s) = h2(s), B3u(s) = h3(s).

u(x , s) = [R(s,A)~h](x , s)

u(x , t) =∫ γ+i∞γ−i∞ eist [R(s,A)

~h](x , s)ds

Appropriate contour transformation

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Smoothing propertiesBoundary integral operator

Explicit integral representation boundary integral operator

ut + uxxx = 0, u(x ,0) = 0, x ∈ (0,L),

B1u = h1(t), B2u = h2(t), B3u = h3(t).su + uxxx = 0, x ∈ (0,L)

B1u(s) = h1(s), B2u(s) = h2(s), B3u(s) = h3(s).

u(x , s) = [R(s,A)~h](x , s)

u(x , t) =∫ γ+i∞γ−i∞ eist [R(s,A)

~h](x , s)ds

Appropriate contour transformation

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-homogeneous boundary value problems

∂tu = P(x ,D)u + f (x , t) u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

P(x ,D) : m−th order operator, Bj (x ,D) is sj−th order operator,

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J.L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J.L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J.L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

J.L. Lions and E. Magenes (1972):

Curious enough, the non-homogeneous boundary valueproblems do not seem to have undergone any systematic studyfor the cases considered in this Chapter, even for second-orderhyperbolic operators (except, of course, where only “regular”data on the boundary are considered and where a “loss ofregularity ” in the results is accepted ...).

lt was therefore necessary to take up this question from thebeginning. We start with the proof of regularity results, which(although far from optimal; see the Remarks of section 6) seemto be new.

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-homogeneous boundary value problems

∂tu = P(x ,D)u, u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

where P(x ,D) is m−th oder operator and Bj (x ,D) is sj−th oderoperator,

Question

What are the optimal regularity conditions on boundary data gj for theIBVP to have a solution u ∈ C([0,T ); Hs(Ω))?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-homogeneous boundary value problems

∂tu = P(x ,D)u, u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

where P(x ,D) is m−th oder operator and Bj (x ,D) is sj−th oderoperator,

Question

What are the optimal regularity conditions on boundary data gj for theIBVP to have a solution u ∈ C([0,T ); Hs(Ω))?

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-homogeneous boundary value problems

∂tu = P(x ,D)u, u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j =,1,2, · · · ,m − 1,

Answer

gj ∈ H rj ,s∗j (Γ× (0,T )) := L2(0,T ; Hs∗

j (Γ)) ∩ H rj (0,T ; L2(Γ))

with

rj ≥12−

s − sj − 1/2m

, s∗j ≥m − 1

2+ s − sj , j = 1,2, · · · ,m

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

To solve

∂tu = P(x ,D)u + f (x , t), u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

Step One: solve∂tv = P(x ,D)v , v(x ,0) = 0, x ∈ Ω, t ∈ (0,T )

Bj (x ,D)v = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

v(x , t) := Wbdr (t)~g

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

To solve

∂tu = P(x ,D)u + f (x , t), u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

Step One: solve∂tv = P(x ,D)v , v(x ,0) = 0, x ∈ Ω, t ∈ (0,T )

Bj (x ,D)v = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

v(x , t) := Wbdr (t)~g

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Step Two: solve

∂tw = P(x ,D)w + f (x , t), w(x ,0) = φ(x), x ∈ Rn t ∈ (0,T )

w(x , t) = W (t)φ+

∫ t

0(W (t − τ)f (τ)dτ)

Step Three: set

qj (x , t) = Bj (x ,D)w(x , t), j = 1,2, . . . ,m − 1

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Step Two: solve

∂tw = P(x ,D)w + f (x , t), w(x ,0) = φ(x), x ∈ Rn t ∈ (0,T )

w(x , t) = W (t)φ+

∫ t

0(W (t − τ)f (τ)dτ)

Step Three: set

qj (x , t) = Bj (x ,D)w(x , t), j = 1,2, . . . ,m − 1

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Step Two: solve

∂tw = P(x ,D)w + f (x , t), w(x ,0) = φ(x), x ∈ Rn t ∈ (0,T )

w(x , t) = W (t)φ+

∫ t

0(W (t − τ)f (τ)dτ)

Step Three: set

qj (x , t) = Bj (x ,D)w(x , t), j = 1,2, . . . ,m − 1

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Step Three: set

qj (x , t) = Bj (x ,D)w(x , t), j = 1,2, . . . ,m − 1

and computeWbdr (t)~q.

z(x , t) = w(x , t)−Wbdr (t)~q solves∂tz = P(x ,D)z + f (x , t), u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D)z = 0, (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Step Three: set

qj (x , t) = Bj (x ,D)w(x , t), j = 1,2, . . . ,m − 1

and computeWbdr (t)~q.

z(x , t) = w(x , t)−Wbdr (t)~q solves∂tz = P(x ,D)z + f (x , t), u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D)z = 0, (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Conclusion:

u(x , t) = W (t)φ+

∫ t

0W (t − τ)f )τ)dτ −Wbdr (t)~q + Wbdr (t)~g

solves∂tu = P(x ,D)u + f (x , t), u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Conclusion:

u(x , t) = W (t)φ+

∫ t

0W (t − τ)f )τ)dτ −Wbdr (t)~q + Wbdr (t)~g

solves∂tu = P(x ,D)u + f (x , t), u(x ,0) = φ(x), x ∈ Ω, t ∈ (0,T )

Bj (x ,D) = gj (x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Non-homogenization

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

The key to work:

∂tv = P(x ,D)v , v(x ,0) = 0, x ∈ Ω, t ∈ (0,T )

Bj(x ,D)v = gj(x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

is solvable in Hs(Ω) for gj ∈ H rj ,s∗j (Γ× (0,T )) with the optimalregularities

rj =12

+s − sj − 1/2

m, s∗j =

m2

+s−sj−12, j = 1,2, · · · ,m−1

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

The key to work:

∂tv = P(x ,D)v , v(x ,0) = 0, x ∈ Ω, t ∈ (0,T )

Bj(x ,D)v = gj(x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

is solvable in Hs(Ω) for gj ∈ H rj ,s∗j (Γ× (0,T )) with the optimalregularities

rj =12

+s − sj − 1/2

m, s∗j =

m2

+s−sj−12, j = 1,2, · · · ,m−1

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

The key to work:

∂tv = P(x ,D)v , v(x ,0) = 0, x ∈ Ω, t ∈ (0,T )

Bj(x ,D)v = gj(x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

is solvable in Hs(Ω) for gj ∈ H rj ,s∗j (Γ× (0,T )) with the optimalregularities

rj =12

+s − sj − 1/2

m, s∗j =

m2

+s−sj−12, j = 1,2, · · · ,m−1

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

The key to work:

∂tv = P(x ,D)v , v(x ,0) = 0, x ∈ Ω, t ∈ (0,T )

Bj(x ,D)v = gj(x , t), (x , t) ∈ Γ× (0,T ), j = 1,2, · · · ,m − 1,

is solvable in Hs(Ω) for gj ∈ H rj ,s∗j (Γ× (0,T )) with the optimalregularities

rj =12

+s − sj − 1/2

m, s∗j =

m2

+s−sj−12, j = 1,2, · · · ,m−1

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

Homogenization vs Non-homogenization

Homogenization

Access to Functional analysis toolsBoundary regularity may not be optimal

Non-homogenization

Access to Harmonic analysis toolsBoundary regularities are optimalMay not work

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

From the Above

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

From the Above

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

From the Above

IntroductionKdV equation

KdV on a bounded domainConclusion remarks

THANK YOU VERYMUCH!