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
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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!