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ENTROPY FORMULATION FOR FORWARD-BACKWARD PARABOLIC EQUATION - - PowerPoint PPT Presentation

ENTROPY FORMULATION FOR FORWARD- BACKWARD PARABOLIC EQUATION A. Terracina ENTROPY FORMULATION FOR FORWARD-BACKWARD PARABOLIC EQUATION A.Terracina University La Sapienza, Roma, Italy 25/06/2012 ENTROPY FORMULATION FOR Collaboration


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ENTROPY FORMULATION FOR FORWARD- BACKWARD PARABOLIC EQUATION

  • A. Terracina

ENTROPY FORMULATION FOR FORWARD-BACKWARD PARABOLIC EQUATION

A.Terracina University La Sapienza, Roma, Italy 25/06/2012

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ENTROPY FORMULATION FOR FORWARD- BACKWARD PARABOLIC EQUATION

  • A. Terracina

Collaboration

Corrado Mascia Flavia Smarrazzo Alberto Tesei

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ENTROPY FORMULATION FOR FORWARD- BACKWARD PARABOLIC EQUATION

  • A. Terracina

Introduction

Forward-backward parabolic equation ut = ∆φ(u) (1) where the function φ ∈ Liploc(R) is decreasing in some interval.

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Introduction

Forward-backward parabolic equation ut = ∆φ(u) (1) where the function φ ∈ Liploc(R) is decreasing in some interval. Example 1 Model of phase separation φ′(u) > 0 if u ∈ (−∞, b) ∪ (a, ∞), φ′(u) < 0 if u ∈ (b, a);

u v φ(u) B A c b a d

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  • A. Terracina

Example 2 Model of image processing (1D), Perona–Malik equation φ(u) =

u 1+u2 .

In this case the instability region is unbounded.

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Example 2 Model of image processing (1D), Perona–Malik equation φ(u) =

u 1+u2 .

In this case the instability region is unbounded. Example 3 Model of population dynamic, Padron (Comm. Partial Differential Equations 1998) φ(u) = ue−u u ≥ 0.

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Example 2 Model of image processing (1D), Perona–Malik equation φ(u) =

u 1+u2 .

In this case the instability region is unbounded. Example 3 Model of population dynamic, Padron (Comm. Partial Differential Equations 1998) φ(u) = ue−u u ≥ 0. Problems are ill–posed.

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Example 2 Model of image processing (1D), Perona–Malik equation φ(u) =

u 1+u2 .

In this case the instability region is unbounded. Example 3 Model of population dynamic, Padron (Comm. Partial Differential Equations 1998) φ(u) = ue−u u ≥ 0. Problems are ill–posed. Hollig (Trans. Amer. Math. Soc. 83) φ piecewise linear, there is an infinite number of solutions of the Neumann boundary problem.

a ! u v B A b d !

" +

! c

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  • A. Terracina

IDEA : Introducing a viscous regularization that gives a good formulation in analogy with first order conservation laws

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IDEA : Introducing a viscous regularization that gives a good formulation in analogy with first order conservation laws Problem is ill-posed since some relevant physical terms are neglected

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Phase transition, Cahn–Hilliard equation ut = ∆(φ(u) − δ∆u).

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Phase transition, Cahn–Hilliard equation ut = ∆(φ(u) − δ∆u). An analogous approximation for the Perona–Malik equation (1D)

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Phase transition, Cahn–Hilliard equation ut = ∆(φ(u) − δ∆u). An analogous approximation for the Perona–Malik equation (1D) Model of population dynamic, Padron ut = (φ(u) + ǫut)xx.

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Phase transition

Viscous Cahn–Hilliard Novick Cohen (1988) Gurtin (1996) based on microforce balance ut = ∆(φ(u) − δ∆u + ǫut)

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Phase transition

Viscous Cahn–Hilliard Novick Cohen (1988) Gurtin (1996) based on microforce balance ut = ∆(φ(u) − δ∆u + ǫut) In the following δ = 0, φ is of cubic type.

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Novick Cohen-Pego (Trans. Amer. Math. Soc. 1991) study the viscosity problem            ut = ∆v in Ω × (0, T] =: QT

∂v ∂ν = 0

in ∂Ω × (0, T] u = u0 in Ω × {0}, (2) where v := φ(u) + ǫut (ǫ > 0) , (3) is the chemical potential, Ω ⊆ I Rn is bounded , ∂Ω regular, T > 0.

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Partial differential equation in (2) can be rewritten ut = −1 ǫ (I − (I − ǫ∆)−1)φ(u) that corresponds to the Yosida approximation of the operator ∆.

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Partial differential equation in (2) can be rewritten ut = −1 ǫ (I − (I − ǫ∆)−1)φ(u) that corresponds to the Yosida approximation of the operator ∆. Moreover v = (I − ǫ∆)−1φ(u).

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Partial differential equation in (2) can be rewritten ut = −1 ǫ (I − (I − ǫ∆)−1)φ(u) that corresponds to the Yosida approximation of the operator ∆. Moreover v = (I − ǫ∆)−1φ(u). Using the standard theory of ODE in the Banach spaces we have

Theorem

(Novick Cohen-Pego) Given u0 ∈ L∞(Ω), ǫ > 0 there exists a unique solution (uǫ, vǫ) defined in (0, Tǫ), uǫ ∈ C 1([0, Tǫ), L∞(Ω)).

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A priori estimates

For every g ∈ C 1(R) such that g ′ ≥ 0 G(u) = u g(φ(s)) ds + c.

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A priori estimates

For every g ∈ C 1(R) such that g ′ ≥ 0 G(u) = u g(φ(s)) ds + c. Then [G(uǫ)]t = div

  • g(vǫ)∇vǫ
  • − g ′(vǫ)|∇vǫ|2+

−1 ǫ

  • g(φ(uǫ)) − g(vǫ)
  • (φ(uǫ) − vǫ) .
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A priori estimates

For every g ∈ C 1(R) such that g ′ ≥ 0 G(u) = u g(φ(s)) ds + c. Then [G(uǫ)]t = div

  • g(vǫ)∇vǫ
  • − g ′(vǫ)|∇vǫ|2+

−1 ǫ

  • g(φ(uǫ)) − g(vǫ)
  • (φ(uǫ) − vǫ) .

Integrating in Ω and using boundary condition we have d dt

G(uǫ(x, t)) dx ≤ 0 that gives a priori estimates in L∞. Moreover choosing g(u) ≡ u we have

  • QT
  • |∇v ǫ|2 + ǫ|∂tuǫ|2

dxdt ≤ C2 .

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Entropy formulation

In analogy with conservation laws we characterize an entropy solution

  • f problem

           ut = ∆φ(u) in Ω × (0, T] = QT

∂φ(u) ∂ν

= 0 in ∂Ω × (0, T] u = u0 in Ω × {0}, (4) as that obtained as limit of the solutions of problem (2) when ǫ → 0+.

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Entropy formulation

In analogy with conservation laws we characterize an entropy solution

  • f problem

           ut = ∆φ(u) in Ω × (0, T] = QT

∂φ(u) ∂ν

= 0 in ∂Ω × (0, T] u = u0 in Ω × {0}, (4) as that obtained as limit of the solutions of problem (2) when ǫ → 0+. For every ǫ > 0 and g ∈ C 1(R), g ′ ≥ 0 we have

  • QT
  • G(uǫ)ψt − g(v ǫ)∇v ǫ · ∇ψ − g ′(v ǫ)|∇v ǫ|2ψ
  • ≥ 0

(5) for every ψ ∈ C ∞

0 (QT), ψ ≥ 0.

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Entropy formulation

In analogy with conservation laws we characterize an entropy solution

  • f problem

           ut = ∆φ(u) in Ω × (0, T] = QT

∂φ(u) ∂ν

= 0 in ∂Ω × (0, T] u = u0 in Ω × {0}, (4) as that obtained as limit of the solutions of problem (2) when ǫ → 0+. For every ǫ > 0 and g ∈ C 1(R), g ′ ≥ 0 we have

  • QT
  • G(uǫ)ψt − g(v ǫ)∇v ǫ · ∇ψ − g ′(v ǫ)|∇v ǫ|2ψ
  • ≥ 0

(5) for every ψ ∈ C ∞

0 (QT), ψ ≥ 0.

The idea is to pass in the limit in (5) to characterize an entropy solution of (4).

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Plotnikov’s results

Study of the singular limit, Plotnikov (J. Math. Sci. 1993).

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Plotnikov’s results

Study of the singular limit, Plotnikov (J. Math. Sci. 1993). Using previous a priori estimate we deduce that there exist two subsequence {uǫn}, {v ǫn} and a couple (u, v) u ∈ L∞(QT), v ∈ L∞(QT) ∩ L2((0, T); H1(Ω)) such that for every T > 0: uǫn ∗ ⇀ u in L∞(QT) , v ǫn ∗ ⇀ v in L∞(QT) , v ǫn ⇀ v in L2((0, T), H1(Ω)) .

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Plotnikov’s results

Study of the singular limit, Plotnikov (J. Math. Sci. 1993). Using previous a priori estimate we deduce that there exist two subsequence {uǫn}, {v ǫn} and a couple (u, v) u ∈ L∞(QT), v ∈ L∞(QT) ∩ L2((0, T); H1(Ω)) such that for every T > 0: uǫn ∗ ⇀ u in L∞(QT) , v ǫn ∗ ⇀ v in L∞(QT) , v ǫn ⇀ v in L2((0, T), H1(Ω)) . Unfortunately this is not enough to pass to the limit in (5).

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Plotnikov’s results

Study of the singular limit, Plotnikov (J. Math. Sci. 1993). Using previous a priori estimate we deduce that there exist two subsequence {uǫn}, {v ǫn} and a couple (u, v) u ∈ L∞(QT), v ∈ L∞(QT) ∩ L2((0, T); H1(Ω)) such that for every T > 0: uǫn ∗ ⇀ u in L∞(QT) , v ǫn ∗ ⇀ v in L∞(QT) , v ǫn ⇀ v in L2((0, T), H1(Ω)) . Unfortunately this is not enough to pass to the limit in (5). Plotnikov gives a characterization of the Young measure ν(x, t) associate to the sequence {uǫn} proving that this is given by superposition of Dirac measures, ν(x,t)(τ) =

2

  • i=0

λi(x, t)δ(τ − βi(v(x, t))) a.e in QT

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where βi(v), i = 0, 1, 2 are the three branches of the graph v = φ(u)

u v φ(u) B A c b a d β₁(v) β₂(v) β₀(v)

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where βi(v), i = 0, 1, 2 are the three branches of the graph v = φ(u)

u v φ(u) B A c b a d β₁(v) β₂(v) β₀(v)

Moreover, 0 ≤ λi ≤ 1 e 2

i=0 λi(x, t) = 1.

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where βi(v), i = 0, 1, 2 are the three branches of the graph v = φ(u)

u v φ(u) B A c b a d β₁(v) β₂(v) β₀(v)

Moreover, 0 ≤ λi ≤ 1 e 2

i=0 λi(x, t) = 1.

It could be proved that a subsequence of {v ǫn} converges strongly to the function v and (u, v) satisfies equation ut = ∆v in the weak sense,

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where βi(v), i = 0, 1, 2 are the three branches of the graph v = φ(u)

u v φ(u) B A c b a d β₁(v) β₂(v) β₀(v)

Moreover, 0 ≤ λi ≤ 1 e 2

i=0 λi(x, t) = 1.

It could be proved that a subsequence of {v ǫn} converges strongly to the function v and (u, v) satisfies equation ut = ∆v in the weak sense, meanwhile u satisfies ut = ∆(φ(u)) in the sense of measured valued solutions.

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Letting ǫn → 0+ in the viscous entropy inequality

  • QT
  • G(uǫn)ψt − g(v ǫn)∇v ǫn · ∇ψ − g ′(v ǫn)|∇v ǫn|2ψ
  • dxdt+

G(u0(x))ψ(x, 0)dx ≥ 0

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Letting ǫn → 0+ in the viscous entropy inequality

  • QT
  • G(uǫn)ψt − g(v ǫn)∇v ǫn · ∇ψ − g ′(v ǫn)|∇v ǫn|2ψ
  • dxdt+

G(u0(x))ψ(x, 0)dx ≥ 0 we have

  • QT
  • G(u)ψt − g(v)∇v · ∇ψ − g ′(v)|∇v 2|ψ
  • dxdt+

G(u0(x))ψ(x, 0) dx ≥ 0 (6) where G(u) = 2

i=0 λiG(βi(v)).

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Entropy solution

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Entropy solution

Definition

Given u0 ∈ L∞(Ω) an entropy solution of problem forward–backward (4) is given by the functions λi ∈ L∞(QT), i = 0, 1, 2, u ∈ L∞(QT), v ∈ L∞(QT) ∩ L2((0, T), H1(Ω)). Such that (i) 2

i=0 λi = 1, λi ≥ 0, u = 2 i=0 λiβi(v);

(ii) ut = ∆v in the weak sense; (iii) u and v satisfy (6) for every g ∈ C 1(R), g ′ ≥ 0 (entropy condition).

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Entropy solution

Definition

Given u0 ∈ L∞(Ω) an entropy solution of problem forward–backward (4) is given by the functions λi ∈ L∞(QT), i = 0, 1, 2, u ∈ L∞(QT), v ∈ L∞(QT) ∩ L2((0, T), H1(Ω)). Such that (i) 2

i=0 λi = 1, λi ≥ 0, u = 2 i=0 λiβi(v);

(ii) ut = ∆v in the weak sense; (iii) u and v satisfy (6) for every g ∈ C 1(R), g ′ ≥ 0 (entropy condition). Problems: Existence in a stronger sense? Uniqueness? Study of the evolution of the different phases.

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Two phase entropy solution

Case n = 1. Let Ω = (−L, L), u0 ≤ b in (−L, 0), u0 ≥ a in (0, L), initial data in the two stable phases.

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Two phase entropy solution

Case n = 1. Let Ω = (−L, L), u0 ≤ b in (−L, 0), u0 ≥ a in (0, L), initial data in the two stable phases. φ(u0) ∈ BC([−L, L]), φ(u0) ∈ C 1([−L, 0]), φ(u0) ∈ C 1([0, L])

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Two phase entropy solution

Case n = 1. Let Ω = (−L, L), u0 ≤ b in (−L, 0), u0 ≥ a in (0, L), initial data in the two stable phases. φ(u0) ∈ BC([−L, L]), φ(u0) ∈ C 1([−L, 0]), φ(u0) ∈ C 1([0, L]) We search a solution in which the two stable phases are separated by an interface ξ, ξ(0) = 0,

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Two phase entropy solution

Case n = 1. Let Ω = (−L, L), u0 ≤ b in (−L, 0), u0 ≥ a in (0, L), initial data in the two stable phases. φ(u0) ∈ BC([−L, L]), φ(u0) ∈ C 1([−L, 0]), φ(u0) ∈ C 1([0, L]) We search a solution in which the two stable phases are separated by an interface ξ, ξ(0) = 0, V1 := {(x, t) ∈ QT | − L ≤ x < ξ(t), t ∈ (0, T)}, V2 := QT \ V 1

x t

  • L

L ξ(t) V2 V1

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An entropy solution is a triple of functions (ξ, u, v) such that : (a) ξ ∈ C 1((0, T)), ξ(0) = 0, u ∈ L∞(QT), v ∈ L∞(QT) ∩ L2((0, T) ; H1((−L, L)); (b) u, v satisfy u = βi(v) in Vi (i = 1, 2) (v = φ(u)); (c) ut = vxx in the weak sense, entropy condition, boundary and initial condition.

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Determinate conditions for the interface.

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Determinate conditions for the interface.

Theorem

(Evans–Portilheiro Math. Models Methods Appl. Sci. (2004)) Let u, v, ξ a two phase entropy “piecewise regular” solution then (i) Rankine-Hugoniot condition: ξ′ = −[vx] [u] a.e on γ := {(ξ(t), t) : t ∈ (0, T))}. (ii) entropy condition: ξ′ [G(u)] ≥ −g(v)[vx] a.e. su γ .

where [h] := h(ξ(t)+, t) − h(ξ(t)−, t).

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Admissibility condition for the interface

Choosing properly the function g we can select admissibility conditions for the interface γ = (ξ(t), t)

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Admissibility condition for the interface

Choosing properly the function g we can select admissibility conditions for the interface γ = (ξ(t), t)

Theorem

(a) ξ′(t) > 0 = ⇒ φ(u(ξ(t), t)) = A ; (b) ξ′(t) < 0 = ⇒ φ(u(ξ(t), t)) = B.

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Admissibility condition for the interface

Choosing properly the function g we can select admissibility conditions for the interface γ = (ξ(t), t)

Theorem

(a) ξ′(t) > 0 = ⇒ φ(u(ξ(t), t)) = A ; (b) ξ′(t) < 0 = ⇒ φ(u(ξ(t), t)) = B. Condition for the phase change.

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Admissibility condition for the interface

Choosing properly the function g we can select admissibility conditions for the interface γ = (ξ(t), t)

Theorem

(a) ξ′(t) > 0 = ⇒ φ(u(ξ(t), t)) = A ; (b) ξ′(t) < 0 = ⇒ φ(u(ξ(t), t)) = B. Condition for the phase change. We can pass from phase 1 to phase 2 only if v = B

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If v ∈ (A, B) phase does not change

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Piecewise linear case

φ(u) =      φ−(u) if u ≤ b φ0(u) if b < u < a φ+(u) if u ≥ a , where φ±(u) := α± u + β± , φ0(u) := A(u − b) − B(u − a) a − b .

a ! u v B A b d !

" +

! c

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Theorem

(Mascia, T., Tesei, Arch. Rat. Mech 2009) There exists at most a unique two phase entropy solution

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Theorem

(Mascia, T., Tesei, Arch. Rat. Mech 2009) There exists at most a unique two phase entropy solution

Theorem

(Mascia, T., Tesei) Suppose that one of the following conditions is satisfies i) α−u0(0−) + β− = α+u0(0+) + β+ ∈ (A, B); ii) α−u′

0(0−) = α+u′ 0(0+).

Then there exists τ > 0 such that the two phase problem has solution in R × (0, τ).

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Theorem

(Mascia, T., Tesei, Arch. Rat. Mech 2009) There exists at most a unique two phase entropy solution

Theorem

(Mascia, T., Tesei) Suppose that one of the following conditions is satisfies i) α−u0(0−) + β− = α+u0(0+) + β+ ∈ (A, B); ii) α−u′

0(0−) = α+u′ 0(0+).

Then there exists τ > 0 such that the two phase problem has solution in R × (0, τ). PROOF (idea) Local existence using two auxiliary problems: moving boundary problem and steady boundary problem.

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Extension in time of the solution

We can extend in time the solution until a first time τ such that ξ′(τ) = 0 and φ(u(ξ(τ), τ)) = A or B (critical case).

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Extension in time of the solution

We can extend in time the solution until a first time τ such that ξ′(τ) = 0 and φ(u(ξ(τ), τ)) = A or B (critical case). Is it possible to obtain the solution with a sequence of solutions of moving boundary problems that alternate in time with solutions of steady boundary problems?

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Extension in time of the solution

We can extend in time the solution until a first time τ such that ξ′(τ) = 0 and φ(u(ξ(τ), τ)) = A or B (critical case). Is it possible to obtain the solution with a sequence of solutions of moving boundary problems that alternate in time with solutions of steady boundary problems? It is necessary to study the critical case

Theorem

(T. Siam J. Mat. Anal. 2011) Let (ξ, u) be a solution of the two phase problem in Qτ. Let t1 < τ such that in (t1, τ) the solution is given by the solution either of the moving boundary problem or of the steady boundary problem. Then there exists t2 > τ such that the solution of the two phase problem can be extended in (0, t2).

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Nolinear φ

Smarrazzo, T. (Discrete and Continuous dynamical systems A.) Rankine–Hugoniot has to be understood in a weak sense, the same for the entropy condition (theory of vector filelds with divergence–measure). We obtain the same compatibility condition for the two phase problem.

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ENTROPY FORMULATION FOR FORWARD- BACKWARD PARABOLIC EQUATION

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Nolinear φ

Smarrazzo, T. (Discrete and Continuous dynamical systems A.) Rankine–Hugoniot has to be understood in a weak sense, the same for the entropy condition (theory of vector filelds with divergence–measure). We obtain the same compatibility condition for the two phase problem. Uniqueness

Theorem

There exists at most one two–phase solution of problem . Let (u1, v 1, ξ1) and (u2, v 2, ξ2) be two two–phase solutions then we have

  • QT
  • (u1 − u2)ψt − (v 1

x − v 2 x )ψx

  • dxdt = 0

(7) for every test function ψ ∈ H1(QT) ∩ C(QT) such that ψ(·, T) ≡ 0.

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Choosing ψ(x, t) := t

T

{v 1(x, s) − v 2(x, s)} ds, (8) equation reads

  • QT

(u1−u2)(v 1−v 2)dxdt=

  • QT

(v 1

x −v 2 x )

  • t

T

(v 1

x (x, s) − v 2 x (x, s)ds

  • dxdt.

(9) Concerning the right-hand side of (9) we have

  • QT

(v 1

x − v 2 x )

t

T

(v 1

x (x, s) − v 2 x (x, s) ds

  • dxdt

(10) = 1 2 ω2

ω1

T d dt t

T

v 1

x (x, s) − v 2 x (x, s) ds

2 dtdx ≤ 0.

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ENTROPY FORMULATION FOR FORWARD- BACKWARD PARABOLIC EQUATION

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Existence

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Existence Solutions of the approximation problems ut = (φ(u) + ǫut)xx converge to the solution of the two phase problem ?

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Existence Solutions of the approximation problems ut = (φ(u) + ǫut)xx converge to the solution of the two phase problem ? Suppose that one of the following conditions is satisfied

  • there exists δ > 0 such that φ(u0) < B in (0, δ);
  • there exists δ > 0 such that φ(u0) > A in (−δ, 0));

then there exists τ > 0 such that uǫn converges strongly to the solution of the two phase problem with fixed boundary (ξ ≡ 0) in (−L, L) × (0, τ).

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1) Maximum principle for the viscous problem

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1) Maximum principle for the viscous problem If φ(u0(·)) ≤ B in (−L, x0) and v ǫ(x0, ·) ≤ B in (0, t0) then φ(uǫ), v ǫ ≤ B in (−L, x0) × (0, t0). Moreover if u0(·) ≤ b (−L, x0) then uǫ ≤ b in (−L, x0) × (0, t0).

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ENTROPY FORMULATION FOR FORWARD- BACKWARD PARABOLIC EQUATION

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1) Maximum principle for the viscous problem If φ(u0(·)) ≤ B in (−L, x0) and v ǫ(x0, ·) ≤ B in (0, t0) then φ(uǫ), v ǫ ≤ B in (−L, x0) × (0, t0). Moreover if u0(·) ≤ b (−L, x0) then uǫ ≤ b in (−L, x0) × (0, t0). 2) A priori estimates

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1) Maximum principle for the viscous problem If φ(u0(·)) ≤ B in (−L, x0) and v ǫ(x0, ·) ≤ B in (0, t0) then φ(uǫ), v ǫ ≤ B in (−L, x0) × (0, t0). Moreover if u0(·) ≤ b (−L, x0) then uǫ ≤ b in (−L, x0) × (0, t0). 2) A priori estimates v ǫk

t L2(Qτǫk ) + v ǫk x L∞(0,τǫk ;L2((−L,L)) ≤ C

(11) where τǫk is the maximum time in which the two phases are separated by the stationary interface.

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1) Maximum principle for the viscous problem If φ(u0(·)) ≤ B in (−L, x0) and v ǫ(x0, ·) ≤ B in (0, t0) then φ(uǫ), v ǫ ≤ B in (−L, x0) × (0, t0). Moreover if u0(·) ≤ b (−L, x0) then uǫ ≤ b in (−L, x0) × (0, t0). 2) A priori estimates v ǫk

t L2(Qτǫk ) + v ǫk x L∞(0,τǫk ;L2((−L,L)) ≤ C

(11) where τǫk is the maximum time in which the two phases are separated by the stationary interface. Using (11) we can pass to the limit strongly and prove that there exists τ > 0 such that τǫk ≥ τ for k large enough.