BV solutions of the Jin-Xin model Stefano Bianchini, IAC(CNR) Roma - - PowerPoint PPT Presentation

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BV solutions of the Jin-Xin model

Stefano Bianchini, IAC(CNR) Roma http://www.iac.cnr.it/ September 17, 2004

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We consider the (special) Jin-Xin relaxation model [Jin-Xin ’95]

• ut + vx

= vt + Λ2ux =

1 (F(u) − v)

u, v ∈ Rn, Λ ∈ R, (1) F : Rn → Rn smooth.

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We consider the (special) Jin-Xin relaxation model [Jin-Xin ’95]

• ut + vx

= vt + Λ2ux =

1 (F(u) − v)

u, v ∈ Rn, Λ ∈ R, (1) F : Rn → Rn smooth. Scaling: x → x/Λ, v → Λv = ⇒ Λ = 1,

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We consider the (special) Jin-Xin relaxation model [Jin-Xin ’95]

• ut + vx

= vt + Λ2ux =

1 (F(u) − v)

u, v ∈ Rn, Λ ∈ R, (1) F : Rn → Rn smooth. Scaling: x → x/Λ, v → Λv = ⇒ Λ = 1, Diagonalizing 2F − = u − v, 2F + = u + v, we obtain the BGK model

• F −

t − F − x

=

1 (M−(u) − F −)

F +

t

+ F +

x

=

1 (M+(u) − F +)

F −, F + ∈ Rn, (2) where u = F − + F +, M−(u) = u−F(u)

2

, M+(u) = u+F(u)

2

.

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General settings Equation (1) can be written as ut + A(u)ux = (uxx − utt), u ∈ Rn, (3) with A(u) = DF(u).

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General settings Equation (1) can be written as ut + A(u)ux = (uxx − utt), u ∈ Rn, (3) with A(u) = DF(u). The above equation is meaningful even if A(u) is not a Jacobian matrix.

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General settings Equation (1) can be written as ut + A(u)ux = (uxx − utt), u ∈ Rn, (3) with A(u) = DF(u). The above equation is meaningful even if A(u) is not a Jacobian matrix. Assumptions:

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General settings Equation (1) can be written as ut + A(u)ux = (uxx − utt), u ∈ Rn, (3) with A(u) = DF(u). The above equation is meaningful even if A(u) is not a Jacobian matrix. Assumptions: 1) A(u) strictly hyperbolic and −1 + c ≤ λi(u) ≤ 1 − c, c > 0;

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General settings Equation (1) can be written as ut + A(u)ux = (uxx − utt), u ∈ Rn, (3) with A(u) = DF(u). The above equation is meaningful even if A(u) is not a Jacobian matrix. Assumptions: 1) A(u) strictly hyperbolic and −1 + c ≤ λi(u) ≤ 1 − c, c > 0; 2) the initial data (u0, u0,t) are sufficiently smooth and with total variation less than δ0 1: u0L∞, u0,tL∞ ≤ δ0, u0,xL1, u0,txL1 ≤ δ0.

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Existence and stability theorem. Under the above assump- tions, there exists a global solution (u, ut) of (3), defined for all t ≥ 0, such that u(t)L∞, u(t)L∞ ≤ Cδ0, ux(t)L1, utx(t)L1 ≤ Cδ0. (4)

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Existence and stability theorem. Under the above assump- tions, there exists a global solution (u, ut) of (3), defined for all t ≥ 0, such that u(t)L∞, u(t)L∞ ≤ Cδ0, ux(t)L1, utx(t)L1 ≤ Cδ0. (4) Moreover, u(t) − ˆ u(s)L1 + ut(t) − ˆ ut(s)L1 ≤ L

• |t − s| +
• (u0 + u0,t) − (ˆ

u0 + ˆ u0,t)

• L1
• + Le−t/u0,t − ˆ

u0,tL1 + L

• 2u0,tx − ˆ

u0,txL1 + 3u0,txx − ˆ u0,txxL1

• .

(5)

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Convergence theorem. As → 0, the solution u(t) with initial data (u0, u0,t) converges to a unique limit u(t) in L1

loc.

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Convergence theorem. As → 0, the solution u(t) with initial data (u0, u0,t) converges to a unique limit u(t) in L1

loc.

The function u(t) has uniformly bounded total variation and gen- erates a Lipschitz continuous semigroup u(t) = St−su(s), u(t) − ˆ u(s)L1 ≤ L

• |t − s| + u(τ) − ˆ

u(τ)L1

• , t, s ≥ τ > 0. (6)

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Convergence theorem. As → 0, the solution u(t) with initial data (u0, u0,t) converges to a unique limit u(t) in L1

loc.

The function u(t) has uniformly bounded total variation and gen- erates a Lipschitz continuous semigroup u(t) = St−su(s), u(t) − ˆ u(s)L1 ≤ L

• |t − s| + u(τ) − ˆ

u(τ)L1

• , t, s ≥ τ > 0. (6)

Moreover, u(t) − (u0 + u0,t)L1 ≤ Lt. (7)

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Convergence theorem. As → 0, the solution u(t) with initial data (u0, u0,t) converges to a unique limit u(t) in L1

loc.

The function u(t) has uniformly bounded total variation and gen- erates a Lipschitz continuous semigroup u(t) = St−su(s), u(t) − ˆ u(s)L1 ≤ L

• |t − s| + u(τ) − ˆ

u(τ)L1

• , t, s ≥ τ > 0. (6)

Moreover, u(t) − (u0 + u0,t)L1 ≤ Lt. (7) This semigroup is defined on a domain D containing all the func- tion with sufficiently small total variation, and can be uniquely identified by a relaxation limiting Riemann Solver, i.e. the unique Riemann solver compatible with (3).

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Remarks.

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Remarks.

• u0,t /

∈ L1, so that the variable vx = −ut is not well defined (even if A(u) = DF(u)).

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Remarks.

• u0,t /

∈ L1, so that the variable vx = −ut is not well defined (even if A(u) = DF(u)). However, set ˜ u(t, x) = u(t, x) + e−t/u0,t(x),

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Remarks.

• u0,t /

∈ L1, so that the variable vx = −ut is not well defined (even if A(u) = DF(u)). However, set ˜ u(t, x) = u(t, x) + e−t/u0,t(x), which satisfies .

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Remarks.

• u0,t /

∈ L1, so that the variable vx = −ut is not well defined (even if A(u) = DF(u)). However, set ˜ u(t, x) = u(t, x) + e−t/u0,t(x), which satisfies ˜ ut + A(u)˜ ux = (˜ uxx − ˜ utt)

• ˜

ux,˜ ut∈L1

.

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Remarks.

• u0,t /

∈ L1, so that the variable vx = −ut is not well defined (even if A(u) = DF(u)). However, set ˜ u(t, x) = u(t, x) + e−t/u0,t(x), which satisfies ˜ ut + A(u)˜ ux = (˜ uxx − ˜ utt)

• ˜

ux,˜ ut∈L1

+ e−t/(A(u)u0,tx − u0,txx)

• exponential decay of initial data

.

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Remarks.

• u0,t /

∈ L1, so that the variable vx = −ut is not well defined (even if A(u) = DF(u)). However, set ˜ u(t, x) = u(t, x) + e−t/u0,t(x), which satisfies ˜ ut + A(u)˜ ux = (˜ uxx − ˜ utt)

• ˜

ux,˜ ut∈L1

+ e−t/(A(u)u0,tx − u0,txx)

• exponential decay of initial data

. The initial data for ˜ u are (u0 + u0,t, 0):

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Remarks.

• u0,t /

∈ L1, so that the variable vx = −ut is not well defined (even if A(u) = DF(u)). However, set ˜ u(t, x) = u(t, x) + e−t/u0,t(x), which satisfies ˜ ut + A(u)˜ ux = (˜ uxx − ˜ utt)

• ˜

ux,˜ ut∈L1

+ e−t/(A(u)u0,tx − u0,txx)

• exponential decay of initial data

. The initial data for ˜ u are (u0 + u0,t, 0): in BV estimates it is important not ut ∈ L1 but utx ∈ L1.

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• Green kernel for relaxation. [Zeng ’99, Hanouzet-Natalini-

SB ’04]

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• Green kernel for relaxation. [Zeng ’99, Hanouzet-Natalini-

SB ’04] Consider the 2 × 2 system

• ut + vx

= vt + ux = λu − v

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• Green kernel for relaxation. [Zeng ’99, Hanouzet-Natalini-

SB ’04] Consider the 2 × 2 system

• ut + vx

= vt + ux = λu − v Then for the Green kernel Γ(t, x) =

• Γ11

Γ12 Γ21 Γ22

• ,

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• Green kernel for relaxation. [Zeng ’99, Hanouzet-Natalini-

SB ’04] Consider the 2 × 2 system

• ut + vx

= vt + ux = λu − v Then for the Green kernel Γ(t, x) =

• Γ11

Γ12 Γ21 Γ22

• ,

Γ11(t, x) = e−(x2−λt)2/(4(1−λ2)(1+t)) 2

• (1 − λ2)(1 + t)

+ exp.dec., h.o. terms,

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• Green kernel for relaxation. [Zeng ’99, Hanouzet-Natalini-

SB ’04] Consider the 2 × 2 system

• ut + vx

= vt + ux = λu − v Then for the Green kernel Γ(t, x) =

• Γ11

Γ12 Γ21 Γ22

• ,

Γ11(t, x) = e−(x2−λt)2/(4(1−λ2)(1+t)) 2

• (1 − λ2)(1 + t)

+ exp.dec., h.o. terms, Γ21(t, x) = ∂ ∂x e−(x2−λt)2/(4(1−λ2)(1+t)) 2

• (1 − λ2)(1 + t)

+ exp.dec., h.o. terms,

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The kernels Γ12, Γ22 show that only vx influences u:

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The kernels Γ12, Γ22 show that only vx influences u: Γ12(t, x) = ∂ ∂x

  e−(x2−λt)2/(4(1−λ2)(1+t))

2

• (1 − λ2)(1 + t)

+ exp.dec, h.o. terms

   ,

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The kernels Γ12, Γ22 show that only vx influences u: Γ12(t, x) = ∂ ∂x

  e−(x2−λt)2/(4(1−λ2)(1+t))

2

• (1 − λ2)(1 + t)

+ exp.dec, h.o. terms

   ,

Γ22(t, x) = ∂ ∂x

   ∂

∂x e−(x2−λt)2/(4(1−λ2)(1+t)) 2

• (1 − λ2)(1 + t)

+ ”,”

   + exp.dec.

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The kernels Γ12, Γ22 show that only vx influences u: Γ12(t, x) = ∂ ∂x

  e−(x2−λt)2/(4(1−λ2)(1+t))

2

• (1 − λ2)(1 + t)

+ exp.dec, h.o. terms

   ,

Γ22(t, x) = ∂ ∂x

   ∂

∂x e−(x2−λt)2/(4(1−λ2)(1+t)) 2

• (1 − λ2)(1 + t)

+ ”,”

   + exp.dec.

This result is important when studying decay to an equilibrium state (¯ u, ¯ v) = (0, F(0) = 0), because by Duhamel formula

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The kernels Γ12, Γ22 show that only vx influences u: Γ12(t, x) = ∂ ∂x

  e−(x2−λt)2/(4(1−λ2)(1+t))

2

• (1 − λ2)(1 + t)

+ exp.dec, h.o. terms

   ,

Γ22(t, x) = ∂ ∂x

   ∂

∂x e−(x2−λt)2/(4(1−λ2)(1+t)) 2

• (1 − λ2)(1 + t)

+ ”,”

   + exp.dec.

This result is important when studying decay to an equilibrium state (¯ u, ¯ v) = (0, F(0) = 0), because by Duhamel formula

• u(t)

v(t)

• = Γ(t)∗
• u(0)

v(0)

• +

t

0 Γ(t − s) ∗

• F(u(s)) − A(0)u(s)
• ≈Gx(t−s)∗u(s)2

ds

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• The dependence w.r.t. u0 + u0,t can be easily seen with the

example ut = (uxx − utt). with initial data u(0) = 0, ut(0) = −1.

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• The dependence w.r.t. u0 + u0,t can be easily seen with the

example ut = (uxx − utt). with initial data u(0) = 0, ut(0) = −1. The solution is 1 − e−t/, which converges to u(t) ≡ 1, t > 0.

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• The dependence w.r.t. u0 + u0,t can be easily seen with the

example ut = (uxx − utt). with initial data u(0) = 0, ut(0) = −1. The solution is 1 − e−t/, which converges to u(t) ≡ 1, t > 0. The hyperbolic limit → 0 has the ”initial data” lim

t→0+ u(t) = 1 = lim →0 u0 + ut,0.

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BV estimates in the conservative case

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BV estimates in the conservative case We assume A(u) = DF(u), = 1 and u0,t ∈ L1.

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BV estimates in the conservative case We assume A(u) = DF(u), = 1 and u0,t ∈ L1. Differentiating w.r.t. x the BGK scheme (2)

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BV estimates in the conservative case We assume A(u) = DF(u), = 1 and u0,t ∈ L1. Differentiating w.r.t. x the BGK scheme (2)

  

f−

t − f− x

= −I+A(u)

2

f− + I−A(u)

2

f+ f+

t

+ f+

x

=

I+A(u) 2

f− − I−A(u)

2

f+ f± = F ±

x .

(8)

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BV estimates in the conservative case We assume A(u) = DF(u), = 1 and u0,t ∈ L1. Differentiating w.r.t. x the BGK scheme (2)

  

f−

t − f− x

= −I+A(u)

2

f− + I−A(u)

2

f+ f+

t

+ f+

x

=

I+A(u) 2

f− − I−A(u)

2

f+ f± = F ±

x .

(8) Differentiating (2) w.r.t. t we obtain

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BV estimates in the conservative case We assume A(u) = DF(u), = 1 and u0,t ∈ L1. Differentiating w.r.t. x the BGK scheme (2)

  

f−

t − f− x

= −I+A(u)

2

f− + I−A(u)

2

f+ f+

t

+ f+

x

=

I+A(u) 2

f− − I−A(u)

2

f+ f± = F ±

x .

(8) Differentiating (2) w.r.t. t we obtain

  

g−

t − g− x

= −I+A(u)

2

g− + I−A(u)

2

g+ g+

t + g+ x

=

I+A(u) 2

g− − I−A(u)

2

g+ g± = F ±

t .

(9)

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BV estimates in the conservative case We assume A(u) = DF(u), = 1 and u0,t ∈ L1. Differentiating w.r.t. x the BGK scheme (2)

  

f−

t − f− x

= −I+A(u)

2

f− + I−A(u)

2

f+ f+

t

+ f+

x

=

I+A(u) 2

f− − I−A(u)

2

f+ f± = F ±

x .

(8) Differentiating (2) w.r.t. t we obtain

  

g−

t − g− x

= −I+A(u)

2

g− + I−A(u)

2

g+ g+

t + g+ x

=

I+A(u) 2

g− − I−A(u)

2

g+ g± = F ±

t .

(9) Our aim: f±(0)L1, g±(0)L1 ≤ δ0 = ⇒ f±(t), g±(t) ∈ L1(R).

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Center manifold of travelling profiles

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Center manifold of travelling profiles We study the ODE −σux + A(u)ux = (1 − σ2)uxx,

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Center manifold of travelling profiles We study the ODE −σux + A(u)ux = (1 − σ2)uxx, which can be written as the first order system

    

ux = p (1 − σ2)px = (A(u) − σI)p σx = (10)

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Center manifold of travelling profiles We study the ODE −σux + A(u)ux = (1 − σ2)uxx, which can be written as the first order system

    

ux = p (1 − σ2)px = (A(u) − σI)p σx = (10) Close to any equilibrium (0, 0, λi(0)), one can find a center man- ifold of travelling profiles:

11

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Center manifold of travelling profiles We study the ODE −σux + A(u)ux = (1 − σ2)uxx, which can be written as the first order system

    

ux = p (1 − σ2)px = (A(u) − σI)p σx = (10) Close to any equilibrium (0, 0, λi(0)), one can find a center man- ifold of travelling profiles: p = vi˜ ri(u, vi, σ), ˜ λi = ˜ ri, A(u)˜ ri, |˜ ri(u)| = 1. (11)

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We can parameterize by the the kinetic component fi:

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+.

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+. The center manifold for the kinetic components f−, f+ is (12) (13)

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+. The center manifold for the kinetic components f−, f+ is f− = (12) (13)

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+. The center manifold for the kinetic components f−, f+ is f− = (1 − σ)vi˜ ri(u, vi, σ) (12) (13)

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+. The center manifold for the kinetic components f−, f+ is f− = (1 − σ)vi˜ ri(u, vi, σ)= f−

i ˜

ri

• u,

f−

i

1 − σ, σ

• (12)

(13)

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+. The center manifold for the kinetic components f−, f+ is f− = (1 − σ)vi˜ ri(u, vi, σ)= f−

i ˜

ri

• u,

f−

i

1 − σ, σ

• = f−

i ˜

r−

i (u, f− i , σ)

(12) (13)

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+. The center manifold for the kinetic components f−, f+ is f− = (1 − σ)vi˜ ri(u, vi, σ)= f−

i ˜

ri

• u,

f−

i

1 − σ, σ

• = f−

i ˜

r−

i (u, f− i , σ)

(12) f+ = (1 + σ)˜ ri

• u, (1 + σ)vi

1 + σ , σ

• (13)

12

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We can parameterize by the the kinetic component fi: ut = f− − f+ = −σux = f− + f+ ux = 1 1 − σf− = 1 1 + σf+. The center manifold for the kinetic components f−, f+ is f− = (1 − σ)vi˜ ri(u, vi, σ)= f−

i ˜

ri

• u,

f−

i

1 − σ, σ

• = f−

i ˜

r−

i (u, f− i , σ)

(12) f+ = (1 + σ)˜ ri

• u, (1 + σ)vi

1 + σ , σ

• = f+

i ˜

r+

i (u, f+ i , σ)

(13)

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Identification of a travelling profile: u(¯ x), σ and vi(¯ x),

13

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Identification of a travelling profile: u(¯ x), σ and vi(¯ x),

x x x u(x− t) σ u

−

f ,f ux

+

x −

13

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Identification of a travelling profile: u(¯ x), σ and f−

i (¯

x),

x x x u(x− t) σ u

−

f ,f ux (1− )u

x

σ

+

x −

14

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Identification of a travelling profile: u(¯ x), σ and f+

i (¯

x),

x x x u(x− t) σ u

−

f ,f ux (1+ )ux (1− )u

x

σ σ

+

x −

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Decomposition in travelling profiles

16

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Decomposition in travelling profiles We decompose (f−, g−) and (f+, g+) separately:

16

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Decomposition in travelling profiles We decompose (f−, g−) and (f+, g+) separately: (14)

16

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Decomposition in travelling profiles We decompose (f−, g−) and (f+, g+) separately:

• f−

=

• i f−

i ˜

r−

i (u, f− i , σ− i )

g− =

• i g−

i ˜

r−

i (u, f− i , σ− i )

(14)

16

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Decomposition in travelling profiles We decompose (f−, g−) and (f+, g+) separately:

• f−

=

• i f−

i ˜

r−

i (u, f− i , σ− i )

g− =

• i g−

i ˜

r−

i (u, f− i , σ− i )

σ−

i = θi

• −g−

i

f−

i

• ,

(14)

16

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Decomposition in travelling profiles We decompose (f−, g−) and (f+, g+) separately:

• f−

=

• i f−

i ˜

r−

i (u, f− i , σ− i )

g− =

• i g−

i ˜

r−

i (u, f− i , σ− i )

σ−

i = θi

• −g−

i

f−

i

• ,

(14) where θi is the cutoff function

λi θi

− i −

−g /f

i

.

16

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Decomposition in travelling profiles We decompose (f−, g−) and (f+, g+) separately:

• f−

=

• i f−

i ˜

r−

i (u, f− i , σ− i )

g− =

• i g−

i ˜

r−

i (u, f− i , σ− i )

σ−

i = θi

• −g−

i

f−

i

• ,

(14) where θi is the cutoff function

λi θi

− i −

−g /f

i

. Similarly for (f+, g+):

16

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Decomposition in travelling profiles We decompose (f−, g−) and (f+, g+) separately:

• f−

=

• i f−

i ˜

r−

i (u, f− i , σ− i )

g− =

• i g−

i ˜

r−

i (u, f− i , σ− i )

σ−

i = θi

• −g−

i

f−

i

• ,

(14) where θi is the cutoff function

λi θi

− i −

−g /f

i

. Similarly for (f+, g+):

• f+

=

• i f+

i ˜

r+

i (u, f+ i , σ+ i )

g+ =

• i g+

i ˜

r+

i (u, f+ i , σ+ i )

σ+

i

= θi

 −g+

i

f+

i

  ,

(15)

16

SLIDE 71

To find travelling profiles, we look separately to the t, x deriva- tives of F −, F +, and try to fit n travelling profiles into F − and n into F +.

17

SLIDE 72

To find travelling profiles, we look separately to the t, x deriva- tives of F −, F +, and try to fit n travelling profiles into F − and n into F +.

F− x t

17

SLIDE 73

To find travelling profiles, we look separately to the t, x deriva- tives of F −, F +, and try to fit n travelling profiles into F − and n into F +.

F− x t

18

SLIDE 74

To find travelling profiles, we look separately to the t, x deriva- tives of F −, F +, and try to fit n travelling profiles into F − and n into F +.

F− x t (x− t) φ σ

19

SLIDE 75

To find travelling profiles, we look separately to the t, x deriva- tives of F −, F +, and try to fit n travelling profiles into F − and n into F +.

F− x t (x− t) φ σ

We obtain thus 2n travelling waves: n for F − and n for F +.

19

SLIDE 76

Equation for the components f±

i , g± i

are of the form:

20

SLIDE 77

Equation for the components f±

i , g± i

are of the form:

      

f−

j,t − f− j,x

= −

1+˜ λ−

j

2

f−

j + 1−˜ λ+

j

2

f+

j

+ ς−

f,j(t, x)

f+

j,t + f+ j,x

=

1+˜ λ−

j

2

f−

j − 1−˜ λ+

j

2

f+

j

+ ς+

f,j(t, x)

(16)

20

SLIDE 78

Equation for the components f±

i , g± i

are of the form:

      

f−

j,t − f− j,x

= −

1+˜ λ−

j

2

f−

j + 1−˜ λ+

j

2

f+

j

+ ς−

f,j(t, x)

f+

j,t + f+ j,x

=

1+˜ λ−

j

2

f−

j − 1−˜ λ+

j

2

f+

j

+ ς+

f,j(t, x)

(16)

      

g−

i,t − g− i,x

= −1+˜

λ−

i

2

g−

i + 1−˜ λ+

i

2

g+

i

+ ς−

g,i(t, x)

g+

i,t + g+ i,x

=

1+˜ λ−

i

2

g−

i − 1−˜ λ+

i

2

g+

i

+ ς+

g,i(t, x)

(17)

20

SLIDE 79

Equation for the components f±

i , g± i

are of the form:

      

f−

j,t − f− j,x

= −

1+˜ λ−

j

2

f−

j + 1−˜ λ+

j

2

f+

j

+ ς−

f,j(t, x)

f+

j,t + f+ j,x

=

1+˜ λ−

j

2

f−

j − 1−˜ λ+

j

2

f+

j

+ ς+

f,j(t, x)

(16)

      

g−

i,t − g− i,x

= −1+˜

λ−

i

2

g−

i + 1−˜ λ+

i

2

g+

i

+ ς−

g,i(t, x)

g+

i,t + g+ i,x

=

1+˜ λ−

i

2

g−

i − 1−˜ λ+

i

2

g+

i

+ ς+

g,i(t, x)

(17) with ς±

f , ς± g sources of total variation for F ± x , F ± g

and

20

SLIDE 80

Equation for the components f±

i , g± i

are of the form:

      

f−

j,t − f− j,x

= −

1+˜ λ−

j

2

f−

j + 1−˜ λ+

j

2

f+

j

+ ς−

f,j(t, x)

f+

j,t + f+ j,x

=

1+˜ λ−

j

2

f−

j − 1−˜ λ+

j

2

f+

j

+ ς+

f,j(t, x)

(16)

      

g−

i,t − g− i,x

= −1+˜

λ−

i

2

g−

i + 1−˜ λ+

i

2

g+

i

+ ς−

g,i(t, x)

g+

i,t + g+ i,x

=

1+˜ λ−

i

2

g−

i − 1−˜ λ+

i

2

g+

i

+ ς+

g,i(t, x)

(17) with ς±

f , ς± g sources of total variation for F ± x , F ± g

and ˜ λ−

i = ˜

λi

• u,

f−

i

1 − σ−

i

, σ−

i

• ,

˜ λ+

i

= ˜ λi

 u,

f+

i

1 + σ−

i

, σ+

i

  .

20

SLIDE 81

After some computations, one obtains the source terms of the form

21

SLIDE 82

After some computations, one obtains the source terms of the form |ς±

f,i|, |ς± g,i| ≤ C

• j=k

(|f−

j | + |g− j |)(|f+ k | + |g+ k |) + C

• j

|g−

j f+ j

− f−

j g+ j |

+ C

• j
• |f−

j + f+ j |2 + |g− j + g+ j |2

• χ

  

f+

j

f−

j

≇ 1

  

+ C

• j

(f−

j 2 L1 + f+ j 2 L1)|f− j − f+ j |χ{f− j · f+ j

< 0} + C

• j

(f−

j 2 L1 + f+ j 2 L1)|g− j − g+ j |χ{g− j · g+ j

< 0}. (18)

21

SLIDE 83

After some computations, one obtains the source terms of the form |ς±

f,i|, |ς± g,i| ≤ C

• j=k

(|f−

j | + |g− j |)(|f+ k | + |g+ k |) + C

• j

|g−

j f+ j

− f−

j g+ j |

+ C

• j
• |f−

j + f+ j |2 + |g− j + g+ j |2

• χ

  

f+

j

f−

j

≇ 1

  

+ C

• j

(f−

j 2 L1 + f+ j 2 L1)|f− j − f+ j |χ{f− j · f+ j

< 0} + C

• j

(f−

j 2 L1 + f+ j 2 L1)|g− j − g+ j |χ{g− j · g+ j

< 0}. (18) Prove that the source terms are quadratic w.r.t. f±L1, g±L1.

21

SLIDE 84

Different types of source terms:

22

SLIDE 85

Different types of source terms:

• interaction of different families:
• j=k

(|f−

j | + |g− j |)(|f+ k | + |g+ k |);

22

SLIDE 86

Different types of source terms:

• interaction of different families:
• j=k

(|f−

j | + |g− j |)(|f+ k | + |g+ k |);

• interaction of the same family:

C

• j

|g−

j f+ j

− f−

j g+ j |;

22

SLIDE 87

Different types of source terms:

• interaction of different families:
• j=k

(|f−

j | + |g− j |)(|f+ k | + |g+ k |);

• interaction of the same family:

C

• j

|g−

j f+ j

− f−

j g+ j |;

• energy type terms:
• j

(|f−

j + f+ j |2 + |g− j + g+ j |2)χ{f+ j /f− j ≇ 1};

22

SLIDE 88

Different types of source terms:

• interaction of different families:
• j=k

(|f−

j | + |g− j |)(|f+ k | + |g+ k |);

• interaction of the same family:

C

• j

|g−

j f+ j

− f−

j g+ j |;

• energy type terms:
• j

(|f−

j + f+ j |2 + |g− j + g+ j |2)χ{f+ j /f− j ≇ 1};

• L1 decay terms:
• j

|f−

j − f+ j |χ{f− j · f+ j

< 0} +

• j

|g−

j − g+ j |χ{g− j · g+ j

< 0}.

22

SLIDE 89

Interaction of the same family

23

SLIDE 90

Interaction of the same family Consider the 2 × 2 system

  

F −

t − F − x

=

1−F(u) 2

− F − F +

t

− F +

x

=

1+F(u) 2

− F + u = F − + F +, |F′(u)| ≤ 1 − c. (19)

23

SLIDE 91

Interaction of the same family Consider the 2 × 2 system

  

F −

t − F − x

=

1−F(u) 2

− F − F +

t

− F +

x

=

1+F(u) 2

− F + u = F − + F +, |F′(u)| ≤ 1 − c. (19) Let F ±

x = f±, F ± t

= g±, so that (same for g±)

• f−

t − f− x

= −1+λ

2 f− + 1−λ 2 f+

f+

t

+ f+

x

=

1+λ 2 f− − 1−λ 2 f+

λ(u) = F′(u). (20)

23

SLIDE 92

Interaction of the same family Consider the 2 × 2 system

  

F −

t − F − x

=

1−F(u) 2

− F − F +

t

− F +

x

=

1+F(u) 2

− F + u = F − + F +, |F′(u)| ≤ 1 − c. (19) Let F ±

x = f±, F ± t

= g±, so that (same for g±)

• f−

t − f− x

= −1+λ

2 f− + 1−λ 2 f+

f+

t

+ f+

x

=

1+λ 2 f− − 1−λ 2 f+

λ(u) = F′(u). (20) Construct a functional which bounds

+∞

• R |f−(t, x)g+(t, x) − g−(t, x)f+(t, x)|dxdt.

23

SLIDE 93

We can rewrite the integrand as ,

24

SLIDE 94

We can rewrite the integrand as f−g+ − g−f+ = ,

24

SLIDE 95

We can rewrite the integrand as f−g+ − g−f+ =f−f+

• −g−

f− + g+ f+

• ,

24

SLIDE 96

We can rewrite the integrand as f−g+ − g−f+ =f−f+

• −g−

f− + g+ f+

• = F −

x F + x

 −F −

t

F −

x

+ F +

t

F +

x

 ,

24

SLIDE 97

We can rewrite the integrand as f−g+ − g−f+ =f−f+

• −g−

f− + g+ f+

• = F −

x F + x

 −F −

t

F −

x

+ F +

t

F +

x

 ,

= strengths of waves × difference in speed.

24

SLIDE 98

We can rewrite the integrand as f−g+ − g−f+ =f−f+

• −g−

f− + g+ f+

• = F −

x F + x

 −F −

t

F −

x

+ F +

t

F +

x

 ,

= strengths of waves × difference in speed. Remark. This is not a Glimm functional, it is the interaction term.

24

SLIDE 99

We can rewrite the integrand as f−g+ − g−f+ =f−f+

• −g−

f− + g+ f+

• = F −

x F + x

 −F −

t

F −

x

+ F +

t

F +

x

 ,

= strengths of waves × difference in speed. Remark. This is not a Glimm functional, it is the interaction term. Since it holds g− + g+ = f− − f+, the condition g−/f− = g+/f+ implies that the solution is a travelling profile, replacing σx = 0.

24

SLIDE 100

We can rewrite the integrand as f−g+ − g−f+ =f−f+

• −g−

f− + g+ f+

• = F −

x F + x

 −F −

t

F −

x

+ F +

t

F +

x

 ,

= strengths of waves × difference in speed. Remark. This is not a Glimm functional, it is the interaction term. Since it holds g− + g+ = f− − f+, the condition g−/f− = g+/f+ implies that the solution is a travelling profile, replacing σx = 0. For simplicity we assume in the following λ = 0.

24

SLIDE 101

Consider the system (20), and construct the scalar variables P −−(t, x, y) = f−(t, x)g−(t, y) − f−(t, y)g−(t, x) P −+(t, x, y) = f+(t, x)g−(t, y) − f−(t, y)g+(t, x) P +−(t, x, y) = f−(t, x)g+(t, y) − f+(t, y)g−(t, x) P ++(t, x, y) = f+(t, x)g+(t, y) − f+(t, y)g+(t, x)

25

SLIDE 102

Consider the system (20), and construct the scalar variables P −−(t, x, y) = f−(t, x)g−(t, y) − f−(t, y)g−(t, x) P −+(t, x, y) = f+(t, x)g−(t, y) − f−(t, y)g+(t, x) P +−(t, x, y) = f−(t, x)g+(t, y) − f+(t, y)g−(t, x) P ++(t, x, y) = f+(t, x)g+(t, y) − f+(t, y)g+(t, x) which satisfy the system

              

P −−

t

+ div((−1, −1)P −−) =

P +−+P −+ 2

− P −− P −+

t

+ div((−1, 1)P −+) =

P −−+P ++ 2

− P −+ P +−

t

+ div((1, −1)P +−) =

P −−+P ++ 2

− P +− P ++

t

+ div((1, 1)P ++) =

P +−+P −+ 2

− P ++ (21) for x ≥ y and the boundary conditions P −+(t, x, x) + P +−(t, x, x) = 0, P ++(t, x, x) = P −−(t, x, x) = 0.

25

SLIDE 103

x y x=y P++ P+− P−+ P−−

26

SLIDE 104

x y x=y P++ P+− P−+ P−−

We may read the boundary conditions as follows: a particle P −+ hits the boundary and bounce back as P +− but with opposite sign.

26

SLIDE 105

x y x=y P++ P+− P−+ P−−

We may read the boundary conditions as follows: a particle P −+ hits the boundary and bounce back as P +− but with opposite sign. We are interested in an estimate of the flux of P −+ through the boundary {x = y}, which is given by

+∞

• R |P −+(t, x, x)|dxdt =

+∞

• R |f−g+ − g−f+|dxdt.

26

SLIDE 106

Flux through the boundary

27

SLIDE 107

Flux through the boundary A very simple situation is the 2 × 2 system

  

f−

t − f− x

=

f+−f− 2

f+

t

+ f+

x

=

f−−f+ 2

x ≥ 0, with boundary condition f+(x = 0) + f−(x = 0) = 0.

27

SLIDE 108

Flux through the boundary A very simple situation is the 2 × 2 system

  

f−

t − f− x

=

f+−f− 2

f+

t

+ f+

x

=

f−−f+ 2

x ≥ 0, with boundary condition f+(x = 0) + f−(x = 0) = 0. We want to estimate

∞

|f−(t, 0)|dt, (22) i.e. the total amount of particles which hit the boundary and bounce back with the opposite sign.

27

SLIDE 109

t x x

28

SLIDE 110

t x x

29

SLIDE 111

t x x

30

SLIDE 112

x t t x

31

SLIDE 113

x t t x

32

SLIDE 114

x t t x

33

SLIDE 115

x t t x

We can rewrite the integral (22) as particles with speed -1 − particles with speed 1.

33

SLIDE 116

x t t x

We can rewrite the integral (22) as particles with speed -1 − particles with speed 1. After some time we expect that the solution has almost forgotten the initial data so that particles with speed -1 ≃ particles with speed 1.

33

SLIDE 117

We consider the solution (f−, f+) with initial data (0, δ(x)) as

• f−(t, x)

f+(t, x)

• =
• f−,0(t, x)

f+,0(t, x)

• +
• f−,1(t, x)

f+,1(t, x)

• +
• f−,2(t, x)

f+,2(t, x)

• ,

34

SLIDE 118

We consider the solution (f−, f+) with initial data (0, δ(x)) as

• f−(t, x)

f+(t, x)

• =
• f−,0(t, x)

f+,0(t, x)

• +
• f−,1(t, x)

f+,1(t, x)

• +
• f−,2(t, x)

f+,2(t, x)

• ,

where

• f−,0

t

− f−,0

x

= −f−,0 f+,0

t

+ f+,0

x

= −f+,0 (0, δ(x)),

34

SLIDE 119

We consider the solution (f−, f+) with initial data (0, δ(x)) as

• f−(t, x)

f+(t, x)

• =
• f−,0(t, x)

f+,0(t, x)

• +
• f−,1(t, x)

f+,1(t, x)

• +
• f−,2(t, x)

f+,2(t, x)

• ,

where

• f−,0

t

− f−,0

x

= −f−,0 f+,0

t

+ f+,0

x

= −f+,0 (0, δ(x)),

  

f−,1

t

− f−,1

x

=

f−,0+f+,0 2

− f−,1 f+,1

t

+ f+,1

x

=

f−,0+f+,0 2

− f+,1 (0, 0),

34

SLIDE 120

We consider the solution (f−, f+) with initial data (0, δ(x)) as

• f−(t, x)

f+(t, x)

• =
• f−,0(t, x)

f+,0(t, x)

• +
• f−,1(t, x)

f+,1(t, x)

• +
• f−,2(t, x)

f+,2(t, x)

• ,

where

• f−,0

t

− f−,0

x

= −f−,0 f+,0

t

+ f+,0

x

= −f+,0 (0, δ(x)),

  

f−,1

t

− f−,1

x

=

f−,0+f+,0 2

− f−,1 f+,1

t

+ f+,1

x

=

f−,0+f+,0 2

− f+,1 (0, 0),

  

f−,2

t

− f−,2

x

=

f−,1+f+,1 2

+ f+,2−f−,2

2

f+,2

t

+ f+,2

x

=

f−,1+f+,1 2

− f−,2−f+,2

2

(0, 0).

34

SLIDE 121

x x t t

35

SLIDE 122

x x t t

36

SLIDE 123

x x t t

Explicitly f−,0(t, x) = 0, f+,0(t, x) = e−tδ(x − t),

36

SLIDE 124

x x t t

Explicitly f−,0(t, x) = 0, f+,0(t, x) = e−tδ(x − t), f−,1(t, x) = e−t 2 χ{0 ≤ x ≤ t}, f+,1(t, x) = − e−t 2 χ{0 ≤ x ≤ t} + t 2e−tδ(x − t).

36

SLIDE 125

The flux of f±,0, f±,1 at x = 0 is 1 + 1/2, the source term for f±,2 has total mass of 1/2.

37

SLIDE 126

The flux of f±,0, f±,1 at x = 0 is 1 + 1/2, the source term for f±,2 has total mass of 1/2. We thus proved that after 1 + 1/2 boundary flux, the L1 norm has become 1/2 of the initial L1 norm.

37

SLIDE 127

The flux of f±,0, f±,1 at x = 0 is 1 + 1/2, the source term for f±,2 has total mass of 1/2. We thus proved that after 1 + 1/2 boundary flux, the L1 norm has become 1/2 of the initial L1 norm. We thus can estimate the flux as flux of f±,0 + f±,1 loss of L1 norm = 1 + 1/2 1 − 1/2 = 3.

37

SLIDE 128

The flux of f±,0, f±,1 at x = 0 is 1 + 1/2, the source term for f±,2 has total mass of 1/2. We thus proved that after 1 + 1/2 boundary flux, the L1 norm has become 1/2 of the initial L1 norm. We thus can estimate the flux as flux of f±,0 + f±,1 loss of L1 norm = 1 + 1/2 1 − 1/2 = 3. We conclude that

∞

|f−(t, 0)|dt ≤ 3(f−(t = 0)L1 + f+(t = 0)L1). (23)

37

SLIDE 129

The flux of f±,0, f±,1 at x = 0 is 1 + 1/2, the source term for f±,2 has total mass of 1/2. We thus proved that after 1 + 1/2 boundary flux, the L1 norm has become 1/2 of the initial L1 norm. We thus can estimate the flux as flux of f±,0 + f±,1 loss of L1 norm = 1 + 1/2 1 − 1/2 = 3. We conclude that

∞

|f−(t, 0)|dt ≤ 3(f−(t = 0)L1 + f+(t = 0)L1). (23) Similarly we can estimate

+∞

• R |f−g+ − g−f+|dxdt ≤ 3
• α,β=+−

P αβ(t = 0)L1({x>y}).

37