Boundary Integral Operator and Its...
Transcript of 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!