Travelling waves for a nonlocal KdV-Burgers equation Sabine - - PowerPoint PPT Presentation

Travelling waves for a nonlocal KdV-Burgers equation

Sabine Hittmeir

University of Vienna joint work with:

Franz Achleitner, Carlota Cuesta, Christian Schmeiser

Anacapri, September 2015

Outline

Motivation Nonlinear conservation laws with nonlocal diffusion Travelling waves for the fractional KdV-Burgers equation

Motivation

The inviscid Burgers equation ∂tu + ∂xu2 = 0 (1) has shock solutions u(x, t) = φ(x − ct) = φ(ξ) of the form φ(ξ) =

• φ−

ξ < 0 φ+ ξ > 0

Motivation

The inviscid Burgers equation ∂tu + ∂xu2 = 0 (1) has shock solutions u(x, t) = φ(x − ct) = φ(ξ) of the form φ(ξ) =

• φ−

ξ < 0 φ+ ξ > 0 For {φ−, φ+, c} the Rankine-Hugoniot condition (RHC) has to hold −c(φ+ − φ−) + φ2

+ − φ2 − = 0 ,

i.e. c = φ+ + φ−

Motivation

The inviscid Burgers equation ∂tu + ∂xu2 = 0 (1) has shock solutions u(x, t) = φ(x − ct) = φ(ξ) of the form φ(ξ) =

• φ−

ξ < 0 φ+ ξ > 0 For {φ−, φ+, c} the Rankine-Hugoniot condition (RHC) has to hold −c(φ+ − φ−) + φ2

+ − φ2 − = 0 ,

i.e. c = φ+ + φ− Both cases φ− > φ+ and φ− < φ+ provide solutions to (1). How to obtain uniqueness?

Travelling waves for the viscous Burgers equation

∂tu + ∂xu2 = ∂2

xu ,

The travelling wave equation for u(t, x) = φ(ξ) with ξ = x − ct reads h(φ(ξ)) := −c(φ(ξ) − φ−) + φ2(ξ) − φ2

− = φ′(ξ)

The RHC is equivalent to h(φ+) = h(φ−) = 0.

Travelling waves for the viscous Burgers equation

∂tu + ∂xu2 = ∂2

xu ,

The travelling wave equation for u(t, x) = φ(ξ) with ξ = x − ct reads h(φ(ξ)) := −c(φ(ξ) − φ−) + φ2(ξ) − φ2

− = φ′(ξ)

The RHC is equivalent to h(φ+) = h(φ−) = 0.

Travelling waves for the viscous Burgers equation

∂tu + ∂xu2 = ∂2

xu ,

The travelling wave equation for u(t, x) = φ(ξ) with ξ = x − ct reads h(φ(ξ)) := −c(φ(ξ) − φ−) + φ2(ξ) − φ2

− = φ′(ξ)

The RHC is equivalent to h(φ+) = h(φ−) = 0.

Φ Φ Φ hΦ

Ξ ΦΞ

We obtain the entropy condition φ− > φ+ .

Travelling waves for the KdV-Burgers equation

∂tu + ∂xu2 = ∂2

xu + τ∂3 xu ,

where τ > 0 . The travelling wave equation reads h(φ) = φ′ + τφ′′ and as before we have the Rankine Hugoniot and entropy condition. For phase plane analysis the system is linearised around φ±: φ′ ψ′

• =
• 1

2φ±−c τ

− 1

τ

φ ψ

Travelling waves for the KdV-Burgers equation

∂tu + ∂xu2 = ∂2

xu + τ∂3 xu ,

where τ > 0 . The travelling wave equation reads h(φ) = φ′ + τφ′′ and as before we have the Rankine Hugoniot and entropy condition. For phase plane analysis the system is linearised around φ±: φ′ ψ′

• =
• 1

2φ±−c τ

− 1

τ

φ ψ

Travelling waves for the KdV-Burgers equation

∂tu + ∂xu2 = ∂2

xu + τ∂3 xu ,

where τ > 0 . The travelling wave equation reads h(φ) = φ′ + τφ′′ and as before we have the Rankine Hugoniot and entropy condition. For phase plane analysis the system is linearised around φ±: φ′ ψ′

• =
• 1

2φ±−c τ

− 1

τ

φ ψ

Eigenvalues for the linearised systems show: φ− : saddle point φ+: stable node for τ ≤ 1/(φ− − φ+) =: τ∗ stable spiral for τ > τ∗ Travelling wave solutions are monotone for τ ≤ τ∗

• scillatory as ξ → ∞ for τ > τ∗

for existence proof see Bona, Schonbeck 1985

Eigenvalues for the linearised systems show: φ− : saddle point φ+: stable node for τ ≤ 1/(φ− − φ+) =: τ∗ stable spiral for τ > τ∗ Travelling wave solutions are monotone for τ ≤ τ∗

• scillatory as ξ → ∞ for τ > τ∗

for existence proof see Bona, Schonbeck 1985

The fractional KdV-Burgers equation

Kluwick, Cox, Exner, Grinschgl (2010) 2d shallow water flow of an incompressible fluid with high Reynolds-number

Interaction equation for the pressure p = p(t, x)

∂tp + ∂x(p − p2) = A∂xD1/3p + W ∂3

xp

where D1/3p(t, x) = 1 Γ(2/3) x

−∞

∂yp(t, y) (x − y)1/3 dy

Nonlinear conservation laws with nonlocal diffusion

∂tu + ∂xu2 = ∂xDαu , (2) where Dαu = dα x

−∞

∂yu(t, y) (x − y)α dy , 0 < α < 1 , dα = 1 Γ(1 − α) An alternative representation of ∂xDα: F(∂xDαu)(k) = −Λ(k) u(t, k) where Λ(k) = (aα − ibαsgn(k))|k|α+1 with aα = sin(απ/2) > 0 , bα = cos(απ/2) > 0.

Nonlinear conservation laws with nonlocal diffusion

∂tu + ∂xu2 = ∂xDαu , (2) where Dαu = dα x

−∞

∂yu(t, y) (x − y)α dy , 0 < α < 1 , dα = 1 Γ(1 − α) An alternative representation of ∂xDα: F(∂xDαu)(k) = −Λ(k) u(t, k) where Λ(k) = (aα − ibαsgn(k))|k|α+1 with aα = sin(απ/2) > 0 , bα = cos(απ/2) > 0.

The Cauchy problem

∂tu + ∂xu2 = ∂xDαu, u(0, x) = u0(x) (3) The semigroup generated by ∂xDα is given by the convolution with K(t, x) = F−1e−Λ(k)t(x). Mild formulation of (3) u(t, x) = K(t, .) ∗ u0(x) − t K(t − τ, .) ∗ ∂xu2(τ, .)(x)dτ. Theorem (Feller 1971): For 0 < α < 1, the kernel K is nonnegative: K(t, x) ≥ 0 for all t > 0, x ∈ R.

The Cauchy problem

∂tu + ∂xu2 = ∂xDαu, u(0, x) = u0(x) (3) The semigroup generated by ∂xDα is given by the convolution with K(t, x) = F−1e−Λ(k)t(x). Mild formulation of (3) u(t, x) = K(t, .) ∗ u0(x) − t K(t − τ, .) ∗ ∂xu2(τ, .)(x)dτ. Theorem (Feller 1971): For 0 < α < 1, the kernel K is nonnegative: K(t, x) ≥ 0 for all t > 0, x ∈ R.

The Cauchy problem (II)

Theorem (Droniou, Gallouet, Vovelle 2003) If u0 ∈ L∞, then there exists a unique solution u ∈ L∞((0, ∞) × R) of (3) satisfying the mild formulation (4) almost everywhere. In particular u(t, .)∞ ≤ u0∞, for t > 0. Moreover, the solution satisfies u ∈ C ∞((0, ∞) × R).

Travelling wave solutions

Introducing ξ = x − ct we obtain the travelling wave problem −cφ′ + (φ2)′ = (Dαφ)′ , φ(±∞) = φ± , , Integrating the equation from −∞ gives h(φ) = Dαφ = dα ξ

−∞

φ′(y) (ξ − y)α dy (4) where as above h(φ) := −c(φ − φ−) + φ2 − φ2

−

and we have the Rankine-Hugoniot and entropy condition.

Travelling wave solutions

Introducing ξ = x − ct we obtain the travelling wave problem −cφ′ + (φ2)′ = (Dαφ)′ , φ(±∞) = φ± , , Integrating the equation from −∞ gives h(φ) = Dαφ = dα ξ

−∞

φ′(y) (ξ − y)α dy (4) where as above h(φ) := −c(φ − φ−) + φ2 − φ2

−

and we have the Rankine-Hugoniot and entropy condition.

Travelling wave solutions (II)

The equation is of Abel’s type. A well known transformation leads to φ − φ− = Iα(h(φ)) := d1−α ξ

−∞

h(φ(y)) (ξ − y)1−α dy . (5) Equivalence holds if φ ∈ C 1

b (R) is monotone.

The linearizations h′(φ−)v = Dαv , v = h′(φ−)Iαv , have solutions v(ξ) = beλξ, b ∈ R, where λ = h′(φ−)1/α. Indeed these are the only solutions: N (Dα − h′(u−)) = span

• eλξ

Travelling wave solutions (II)

The equation is of Abel’s type. A well known transformation leads to φ − φ− = Iα(h(φ)) := d1−α ξ

−∞

h(φ(y)) (ξ − y)1−α dy . (5) Equivalence holds if φ ∈ C 1

b (R) is monotone.

The linearizations h′(φ−)v = Dαv , v = h′(φ−)Iαv , have solutions v(ξ) = beλξ, b ∈ R, where λ = h′(φ−)1/α. Indeed these are the only solutions: N (Dα − h′(u−)) = span

• eλξ

Travelling wave solutions - Local existence

Lemma There exists a unique solution φ satisfying φ − φ− ∈ H2((−∞, ξε]) with φ(ξε) = φ− − ε and ξε = log ε/λ. Idea of the proof: Introduce the perturbation ¯ φ(ξ) = φ(ξ) − φ− + eλξ and use fixed point argument involving Fourier transform.

Travelling wave solutions - Local existence

Lemma There exists a unique solution φ satisfying φ − φ− ∈ H2((−∞, ξε]) with φ(ξε) = φ− − ε and ξε = log ε/λ. Idea of the proof: Introduce the perturbation ¯ φ(ξ) = φ(ξ) − φ− + eλξ and use fixed point argument involving Fourier transform.

Travelling wave solutions - Continuation principle

Lemma Let φ ∈ C 1

b ((−∞, ξ0]) be a (continuation of a) solution of the

travelling wave equation (TWE) as constructed above. Then there exists a δ > 0, such that it can be extended uniquely to C 1

b ((−∞, ξ0 + δ)).

• Proof. Writing the TWE as

φ(ξ) = f (ξ) + d1−α ξ

ξ0

h(φ(y)) (ξ − y)1−α dy , and considering f (ξ) = φ− + d1−α ξ0

−∞

h(φ(y)) (ξ − y)1−α dy as given inhomogenity, local existence of a smooth solution for ξ close to ξ0 is a standard result for Volterra integral equation (see e.g. Linz 1985).

Travelling wave solutions - Continuation principle

Lemma Let φ ∈ C 1

b ((−∞, ξ0]) be a (continuation of a) solution of the

travelling wave equation (TWE) as constructed above. Then there exists a δ > 0, such that it can be extended uniquely to C 1

b ((−∞, ξ0 + δ)).

• Proof. Writing the TWE as

φ(ξ) = f (ξ) + d1−α ξ

ξ0

h(φ(y)) (ξ − y)1−α dy , and considering f (ξ) = φ− + d1−α ξ0

−∞

h(φ(y)) (ξ − y)1−α dy as given inhomogenity, local existence of a smooth solution for ξ close to ξ0 is a standard result for Volterra integral equation (see e.g. Linz 1985).

Travelling wave solutions - Monotonicity

Lemma Let φ ∈ C 1

b (−∞, ξ0] be (a continuation of) the solution

constructed above. Then φ is nonincreasing.

Travelling wave solutions - Monotonicity

Lemma Let φ ∈ C 1

b (−∞, ξ0] be (a continuation of) the solution

constructed above. Then φ is nonincreasing.

• Proof. Let φm be the value, for which h′(φm) = 0 and

h′ < 0 in (φ+, φm) , h′ > 0 in (φm, φ−]

Φ Φm Φ hΦ

(i) φ′ < 0 as long as φ ≥ φm: Assume to the contrary that φ(ξ∗) ≥ φm , φ′(ξ∗) = 0 , φ′ < 0 in (−∞, ξ∗) . This leads to a contradiction, since 0 = φ′(ξ∗) = d1−α ξ∗

−∞

h′(φ(y))φ′(y) (ξ∗ − y)1−α dy < 0 . Here we used ξ

−∞ h(φ(y)) (ξ−y)1−α dy =

∞

h(φ(ξ−y)) y 1−α

dy

(ii) φ cannot become increasing for φ < φm. Assume the contrary φ(ξ∗) < φm , φ′ > 0 in (ξ∗, ξ∗ + δ) , φ′ ≤ 0 in (−∞, ξ∗] , where δ is small enough s.t. φ(ξ∗ + δ) < φm. Then Dαφ(ξ∗ + δ) = dα ξ∗+δ

−∞

φ′(y) (ξ∗ + δ − y)α dy = dα ξ∗

−∞

φ′(y) (ξ∗ + δ − y)α dy + dα ξ∗+δ

ξ∗

φ′(y) (ξ∗ + δ − y)α dy > dα ξ∗

−∞

φ′(y) (ξ∗ − y)α dy = Dαφ(ξ∗) . But on the other hand we know > h(φ(ξ∗ + δ)) − h(φ(ξ∗)) = Dαφ(ξ∗ + δ) − Dαφ(ξ∗) > 0 , leading again to a contradiction. Therefore φ′ cannot get positive.

(ii) φ cannot become increasing for φ < φm. Assume the contrary φ(ξ∗) < φm , φ′ > 0 in (ξ∗, ξ∗ + δ) , φ′ ≤ 0 in (−∞, ξ∗] , where δ is small enough s.t. φ(ξ∗ + δ) < φm. Then Dαφ(ξ∗ + δ) = dα ξ∗+δ

−∞

φ′(y) (ξ∗ + δ − y)α dy = dα ξ∗

−∞

φ′(y) (ξ∗ + δ − y)α dy + dα ξ∗+δ

ξ∗

φ′(y) (ξ∗ + δ − y)α dy > dα ξ∗

−∞

φ′(y) (ξ∗ − y)α dy = Dαφ(ξ∗) . But on the other hand we know > h(φ(ξ∗ + δ)) − h(φ(ξ∗)) = Dαφ(ξ∗ + δ) − Dαφ(ξ∗) > 0 , leading again to a contradiction. Therefore φ′ cannot get positive.

(ii) φ cannot become increasing for φ < φm. Assume the contrary φ(ξ∗) < φm , φ′ > 0 in (ξ∗, ξ∗ + δ) , φ′ ≤ 0 in (−∞, ξ∗] , where δ is small enough s.t. φ(ξ∗ + δ) < φm. Then Dαφ(ξ∗ + δ) = dα ξ∗+δ

−∞

φ′(y) (ξ∗ + δ − y)α dy = dα ξ∗

−∞

φ′(y) (ξ∗ + δ − y)α dy + dα ξ∗+δ

ξ∗

φ′(y) (ξ∗ + δ − y)α dy > dα ξ∗

−∞

φ′(y) (ξ∗ − y)α dy = Dαφ(ξ∗) . But on the other hand we know > h(φ(ξ∗ + δ)) − h(φ(ξ∗)) = Dαφ(ξ∗ + δ) − Dαφ(ξ∗) > 0 , leading again to a contradiction. Therefore φ′ cannot get positive.

Travelling wave solutions - Boundedness

Lemma Let φ ∈ C 1

b (−∞, ξ0] be (a continuation of) the solution from

• above. Then

φ+ < φ < φ−.

• Proof. Suppose φ(ξ∗) = φ+ for some finite ξ∗. Then due to the

monotonicity we obtain the contradiction 0 = h(φ+) = dα ξ∗

−∞

φ′(y) (ξ∗ − y)α dy < 0 .

Travelling wave solutions - Boundedness

Lemma Let φ ∈ C 1

b (−∞, ξ0] be (a continuation of) the solution from

• above. Then

φ+ < φ < φ−.

• Proof. Suppose φ(ξ∗) = φ+ for some finite ξ∗. Then due to the

monotonicity we obtain the contradiction 0 = h(φ+) = dα ξ∗

−∞

φ′(y) (ξ∗ − y)α dy < 0 .

Travelling waves - Existence result

Theorem Then there exists a decreasing solution φ ∈ C 1

b (R) of the

travelling wave problem (4). It is (up to a shift) unique among all φ ∈ φ− + H2((−∞, 0)) ∩ C 1

b (R).

Asymptotic stability of travelling waves

We change to the moving coordinates (t, ξ) ∂tu + ∂ξ(u2 − cu) = ∂ξDαu , u(0, ξ) = u0(ξ) (6) We fix the shift in the travelling wave φ such that

• R

(u(t, ξ) − φ(ξ))dξ = 0 The perturbation U = u − φ satisfies ∂tU + ∂ξ((2φ − c)U) + ∂ξU2 = ∂ξDαU (7) We test the equation with U and denote U ˙

Hs = |k|s

UL2 1 2 d dt U2

L2 +

• R

φ′U2dξ ≤ −aαU2

˙ H(1+α)/2

Asymptotic stability of travelling waves

We change to the moving coordinates (t, ξ) ∂tu + ∂ξ(u2 − cu) = ∂ξDαu , u(0, ξ) = u0(ξ) (6) We fix the shift in the travelling wave φ such that

• R

(u(t, ξ) − φ(ξ))dξ = 0 The perturbation U = u − φ satisfies ∂tU + ∂ξ((2φ − c)U) + ∂ξU2 = ∂ξDαU (7) We test the equation with U and denote U ˙

Hs = |k|s

UL2 1 2 d dt U2

L2 +

• R

φ′U2dξ ≤ −aαU2

˙ H(1+α)/2

Stability of travelling waves (II)

We introduce the primitive W (t, ξ) = ξ

−∞

U(t, η)dη Integration of (7) gives, ∂tW + (2φ − c)∂ξW + (∂ξW )2 = ∂ξDαW (8) We derive for J(t) = 1 2(W 2

L2 + γU2 L2)

the estimate d dt J + λ(W H1)

• W 2

˙ H

1+α 2

+ γU2

˙ H

1+α 2

• ≤ 0

where λ(W H1) = aα 2 − L(W H1) γ∗ W H1

Stability of travelling waves (II)

We introduce the primitive W (t, ξ) = ξ

−∞

U(t, η)dη Integration of (7) gives, ∂tW + (2φ − c)∂ξW + (∂ξW )2 = ∂ξDαW (8) We derive for J(t) = 1 2(W 2

L2 + γU2 L2)

the estimate d dt J + λ(W H1)

• W 2

˙ H

1+α 2

+ γU2

˙ H

1+α 2

• ≤ 0

where λ(W H1) = aα 2 − L(W H1) γ∗ W H1

Stability result

Theorem Let φ be a travelling wave solution as before. Let u0 be an initial datum for (6) such that W0(ξ) = ξ

−∞(u0(η) − φ(η))dη satisfies

W0 ∈ H1 and let α > 1/2. If W0H1 is small enough, then the Cauchy problem for equation (6) with initial datum u0 has a unique global solution converging to the travelling wave in the sense that lim

t→∞

∞

t

u(τ, ·) − φ2

L2dτ = 0 .

The fractional KdV-Burgers equation

∂tu + ∂xu2 = ∂xDαu + τ∂3

xu ,

x ∈ R , t ≥ 0 (9) with τ > 0. Travelling wave equation (TWE) h(φ) = Dαφ + τφ′′ , (10) where h(φ) := −c(φ − φ−) + φ2 − φ2

− .

Rankine-Hugoniot condition: c = φ+ + φ− Entropy condition φ− > φ+

The fractional KdV-Burgers equation

∂tu + ∂xu2 = ∂xDαu + τ∂3

xu ,

x ∈ R , t ≥ 0 (9) with τ > 0. Travelling wave equation (TWE) h(φ) = Dαφ + τφ′′ , (10) where h(φ) := −c(φ − φ−) + φ2 − φ2

− .

Rankine-Hugoniot condition: c = φ+ + φ− Entropy condition φ− > φ+

TWs for fKdV-Burgers - Local Existence

The linearisation about ξ = −∞ (or φ = φ−) h′(φ−)v = Dαv + τv ′′ , has solutions of the form v(ξ) = beλξ, b ∈ R, where λ > 0 is the positive real root of P(z) = τz2 + zα − h′(φ−) . Assumption: N

• τ∂2

ξ + Dα − h′(φ−)Id

• = span{eλξ}

in H4(R) Lemma There exists a unique solution φ satisfying φ − φ− ∈ H4((−∞, ξε]) with φ(ξε) = φ− − ε and ξε = log ε/λ.

TWs for fKdV-Burgers - Local Existence

The linearisation about ξ = −∞ (or φ = φ−) h′(φ−)v = Dαv + τv ′′ , has solutions of the form v(ξ) = beλξ, b ∈ R, where λ > 0 is the positive real root of P(z) = τz2 + zα − h′(φ−) . Assumption: N

• τ∂2

ξ + Dα − h′(φ−)Id

• = span{eλξ}

in H4(R) Lemma There exists a unique solution φ satisfying φ − φ− ∈ H4((−∞, ξε]) with φ(ξε) = φ− − ε and ξε = log ε/λ.

TWs for fKdV-Burgers - Local Existence

The linearisation about ξ = −∞ (or φ = φ−) h′(φ−)v = Dαv + τv ′′ , has solutions of the form v(ξ) = beλξ, b ∈ R, where λ > 0 is the positive real root of P(z) = τz2 + zα − h′(φ−) . Assumption: N

• τ∂2

ξ + Dα − h′(φ−)Id

• = span{eλξ}

in H4(R) Lemma There exists a unique solution φ satisfying φ − φ− ∈ H4((−∞, ξε]) with φ(ξε) = φ− − ε and ξε = log ε/λ.

TWs for fKdV-Burgers - Continuation principle

Lemma Let φ ∈ C 3

b ((−∞, ξ0]) be a solution of (10) as above. Then

∃δ > 0, s.t. φ can be extended uniquely to C 3

b ((−∞, ξ0 + δ)).

Idea of Proof. Write the equation as a system of fractional differential equations of orders α, 1 − α and use local Lipschitz continuity as Jafari and Daftardar-Gejji 2006.

TWs for fKdV-Burgers - Continuation principle

Lemma Let φ ∈ C 3

b ((−∞, ξ0]) be a solution of (10) as above. Then

∃δ > 0, s.t. φ can be extended uniquely to C 3

b ((−∞, ξ0 + δ)).

Idea of Proof. Write the equation as a system of fractional differential equations of orders α, 1 − α and use local Lipschitz continuity as Jafari and Daftardar-Gejji 2006.

TWs for fKdV-Burgers - Boundedness

The key quantity for boundedness is the energy functional H(φ) = φ h(y)dy = −c φ2 2 + φ3 3 + Aφ , with A = cφ− − φ2

− (11)

Lemma Let φ ∈ C 3

b ((−∞, ξ0]) be a solution of the TWE. Then the

solution is bounded for ξ ∈ (−∞, ξ0) by ¯ φ < φ(ξ) < φ− , where ¯ φ = 3φ+ − φ− 2 < φ+ (12) is the second root of H(φ) − H(φ−) φ − φ− = 0 .

TWs for fKdV-Burgers - Boundedness

The key quantity for boundedness is the energy functional H(φ) = φ h(y)dy = −c φ2 2 + φ3 3 + Aφ , with A = cφ− − φ2

− (11)

Lemma Let φ ∈ C 3

b ((−∞, ξ0]) be a solution of the TWE. Then the

solution is bounded for ξ ∈ (−∞, ξ0) by ¯ φ < φ(ξ) < φ− , where ¯ φ = 3φ+ − φ− 2 < φ+ (12) is the second root of H(φ) − H(φ−) φ − φ− = 0 .

Proof of boundedness. We first derive an energy type of estimate by multiplying the TWE by φ′ and integrating w.r.t. ξ: H (φ(ξ)) − H (φ−) = τ 2 (φ′(ξ))2 + ξ

−∞

φ′(y) Dαφ(y)dy . (13) The first term on the RHS is clearly nonnegative. Also the second term is nonnegative, since ξ

−∞

φ′(y) Dαφ(y)dy

!

= dα 2 ξ

−∞

φ′(y) ξ

−∞

φ′(x) |x − y|α dx dy = dα 2 ξ

−∞

φ′(y) y

−∞

φ′(x) (y − x)α dx dy + dα 2 ξ

−∞

φ′(y) ξ

y

φ′(x) |x − y|α dx dy

Proof of boundedness. We first derive an energy type of estimate by multiplying the TWE by φ′ and integrating w.r.t. ξ: H (φ(ξ)) − H (φ−) = τ 2 (φ′(ξ))2 + ξ

−∞

φ′(y) Dαφ(y)dy . (13) The first term on the RHS is clearly nonnegative. Also the second term is nonnegative, since ξ

−∞

φ′(y) Dαφ(y)dy

!

= dα 2 ξ

−∞

φ′(y) ξ

−∞

φ′(x) |x − y|α dx dy = dα 2 ξ

−∞

φ′(y) y

−∞

φ′(x) (y − x)α dx dy + dα 2 ξ

−∞

φ′(y) ξ

y

φ′(x) |x − y|α dx dy

Proof of boundedness. We first derive an energy type of estimate by multiplying the TWE by φ′ and integrating w.r.t. ξ: H (φ(ξ)) − H (φ−) = τ 2 (φ′(ξ))2 + ξ

−∞

φ′(y) Dαφ(y)dy . (13) The first term on the RHS is clearly nonnegative. Also the second term is nonnegative, since ξ

−∞

φ′(y) Dαφ(y)dy

!

= dα 2 ξ

−∞

φ′(y) ξ

−∞

φ′(x) |x − y|α dx dy = dα 2 ξ

−∞

φ′(y) y

−∞

φ′(x) (y − x)α dx dy + dα 2 ξ

−∞

φ′(y) ξ

y

φ′(x) |x − y|α dx dy

To see this, we observe that by changing the order of integration ξ

−∞

φ′(y) ξ

y

φ′(x) (x − y)α dx dy = ξ

−∞

φ′(x) x

−∞

φ′(y) (x − y)α dy dx .

Ξ x Ξ y Ξ x y

Employing an extension φ′

E ∈ L2(R) of φ′ to R so that φ′ E(y) = 0 for

y > ξ we can deduce ξ

−∞

φ′(y) Dαφ(y)dy = dα 2

• R

φ′

E(x)

• R

φ′

E(y)

|x − y|α dy dx ≥ 0 , (14) where the last inequality was shown by Lieb and Loss 1997.

To see this, we observe that by changing the order of integration ξ

−∞

φ′(y) ξ

y

φ′(x) (x − y)α dx dy = ξ

−∞

φ′(x) x

−∞

φ′(y) (x − y)α dy dx .

Ξ x Ξ y Ξ x y

Employing an extension φ′

E ∈ L2(R) of φ′ to R so that φ′ E(y) = 0 for

y > ξ we can deduce ξ

−∞

φ′(y) Dαφ(y)dy = dα 2

• R

φ′

E(x)

• R

φ′

E(y)

|x − y|α dy dx ≥ 0 , (14) where the last inequality was shown by Lieb and Loss 1997.

We have H (φ(ξ)) − H (φ−) = τ 2 (φ′(ξ))2 + ξ

−∞

φ′(y) Dαφ(y)dy ≥ 0 Upper bound φ ≤ φ−: Suppose φ(ξ∗) = φ− for some ξ∗ < ∞, then ξ∗

−∞

φ′(y) Dαφ(y)dy = 0, implying φ′(ξ) = 0 for all ξ ∈ (−∞, ξ∗] (see Lieb, Loss). Therefore non constant solution is always below φ−.

Lower bound: We use the nonnegativity of H (φ) − H (φ−) = −c 2(φ2 − (φ−)2) + 1 3(φ3 − (φ−)3) + A(φ − φ−) ≥ 0 . Since φ − φ− < 0 in (−∞, ξ0], we obtain the condition H(φ) − H(φ−) φ − φ− = −c 2(φ + φ−) + 1 3(φ2 + φφ− + (φ−)2) + A ≤ 0 and this implies exactly the lower bound.

TWs for fKdV-Burgers - Far-field behaviour

Lemma Let φ be the TW solution from above. Suppose that lim

ξ→∞ φ = φ0 ∈ R .

Then φ0 = φ+.

• Proof. We argue by contradiction and assume that φ0 = φ+, then

h(φ(ξ)) → h(φ0) = 0 and Iαh(φ(ξ)) → ±∞ . Use the integrated TWE to show contradiction...

TWs for fKdV-Burgers - Far-field behaviour

Lemma Let φ be the TW solution from above. Suppose that lim

ξ→∞ φ = φ0 ∈ R .

Then φ0 = φ+.

• Proof. We argue by contradiction and assume that φ0 = φ+, then

h(φ(ξ)) → h(φ0) = 0 and Iαh(φ(ξ)) → ±∞ . Use the integrated TWE to show contradiction...

TWs for fKdV-Burgers - Far-field behaviour ctd.

Lemma Let φ be a solution as above. Then there exists a constant φ0 ∈ R such that lim

ξ→∞ φ(ξ) = φ0.

Idea of the proof. We rewrite τφ′′ + Dα

ξ0φ + φ = q(φ, ξ)

(15) for ξ ≥ ξ0, where q(φ, ξ) = −dα ξ0

−∞

φ′(y) (ξ − y)α dy + h(φ(ξ)) + φ(ξ) .

TWs for fKdV-Burgers - Far-field behaviour ctd.

Lemma Let φ be a solution as above. Then there exists a constant φ0 ∈ R such that lim

ξ→∞ φ(ξ) = φ0.

Idea of the proof. We rewrite τφ′′ + Dα

ξ0φ + φ = q(φ, ξ)

(15) for ξ ≥ ξ0, where q(φ, ξ) = −dα ξ0

−∞

φ′(y) (ξ − y)α dy + h(φ(ξ)) + φ(ξ) .

We can now write down the solution implicitly by applying Laplace transform methods (see e.g. Gorenflo, Mainardi) to obtain a ’variations

• f constants’ representation:

φ(ξ) = φ(ξ0) v(ξ) − φ′(ξ0) v ′(ξ) − ξ

ξ0

q(φ(ξ − s), ξ − s)v ′(s)ds where the function v and its derivatives are uniformly bounded and have polynomial decay implying the integrability of the term with the inhomogeneity q as well as the decay of φ towards a constant.

Travelling waves - Existence result

Theorem Assume that the exponential functions are the only solutions to the linearised TWE. There exists a travelling wave solution φ ∈ C 3

b (R) of

the travelling wave problem (4), which is (up to a shift) unique among all φ ∈ φ− + H4((−∞, 0)) ∩ C 3

b (R).

• F. Achleitner, S. Hittmeir, C. Schmeiser: On nonlinear conservation laws

with a nonlocal diffusion term, J. Diff. Equ. 250, pp. 2177-2196 (2011)

• F. Achleitner, C. Cuesta, S. Hittmeir: Travelling waves for a non-local

Korteweg-de Vries-Burgers equation, J. Diff. Equ. 257, No. 3, pp. 720-758 (2014)

Thank you for your attention!

• F. Achleitner, S. Hittmeir, C. Schmeiser: On nonlinear conservation laws

with a nonlocal diffusion term, J. Diff. Equ. 250, pp. 2177-2196 (2011)

• F. Achleitner, C. Cuesta, S. Hittmeir: Travelling waves for a non-local

Korteweg-de Vries-Burgers equation, J. Diff. Equ. 257, No. 3, pp. 720-758 (2014)

Thank you for your attention!