boundary integral operator and its...

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!