nonlinear waves in layered media: solutions of the kdv ... · the transient wave in a dissipative...

Nonlinear waves in layered media: solutions of theKdV-Burgers equation

ALEXEY SAMOKHIN

Institute of Control Sciences, RAS

Local and Nonlocal Geometry of PDEs and Integrability

dedicated to the 70th birthday of Joseph Krasil’shchik.

Triest

October 9, 2018

ALEXEY SAMOKHIN Nonlinear waves in layered media: solutions of the KdV-Burgers equation

Equation

The generalized KdV-Burgers (KdV-B) equation

ut(x , t) = γ(x)uxx(x , t) + 2u(x , t)ux(x , t) + uxxx(x , t). (1)

It is related to the viscous and dispersive medium. The layeredmedia consist of layers with both dispersion an dissipation(modelled by KdV-B) and layers without dissipation (KdV).

γ(x) = 0⇒ KdV ⇒ travelling waves solutions (TWS) aresolitons;γ(x) = const > 0⇒ KdV-Burgers ⇒ TWS are shock waves.

Initial value-boundary problem for the KdV-Burgers :

u(x , 0) = u0(x , a, s) = 6a2 sech2(a(x + s)), ux(±∞, t) = 0 : (2)

This is the initial placement of a solitonu0(x , a, s) = 6a

2 sech2(4a3t + a(x + s))∣∣t=0

moving to the left.For numerical computations x ∈ [a, b] for appropriately large a, b.


We use the KdV-B equation to model a behavior of the solitonwhich, while moving in non-dissipative medium encounters abarrier with dissipation. The modelling included the case of a finitewidth dissipative layer as well as a wave passing from anon-dissipative layer into a dissipative one.

Thus we consider two possibilities for γ(x).

1 The two-layer case: γ(x) = α(1− θ(x)), the Heaviside stepfunction; or γ(x) = α(1− tanh(x)), its smooth analog;

2 The three-layer case: γ(x) = α(θ(x − β)− θ(x + β)), aΠ-form density of viscosity, or γ(x) = α sech2(βx), its smoothanalog with (numerically) compact support.


Soliton in 2-layer medium

This case models a passage from non-dissipative half-space to adissipative one. We take γ(x) = α(1− θ(x)) to present a singleboundary separating these half-spaces.

It is natural to expect each solution to behave as the one of theKdV at the right half-space and as solution of KdV-B at the leftone.

The process of transition from the soliton to the correspondentsolution of the KdV-B is predictable. The transient wave in adissipative media becomes a solitary shock which loses speed anddecays to become nonexistent at t → +∞; and a reflected wave isseen in the non-dissipative half-space, see figures for result ofnumeric modelling.


Step obstacles

Figure: Step obstacles γ(x) = α(1− θ(x)) (left)and γ(x) = 2α(1− tanh(x)) (right); α = 6


Figure: Left: Soliton (a = 0.5) passing a step-like obstacle, t = 0.Right: t = 12


Figure: Left: Soliton (a = 0.5) passing a step-like obstacle (α = 6),t = 18.Right: t = 30


Soliton (a = 0.5) at stair obstacle(α = 6)

Push


Step-tri.aviMedia File (video/avi)

Soliton (a = 0.5) at tanh-type obstacle (α = 6)

Push


tanh-tri.aviMedia File (video/avi)

Three layers

This case models a dissipative barrier between two non-dissipativehalf-spaces. A soliton solution of the KdV equation, meeting alayer with dissipation, transforms somewhat similarly to a ray oflight in the air crossing a semi-transparent plate. We model abarrier by γ(x) = α(θ(x − β)− θ(x + β)) (a Π-form density with acompact support or by γ(x) = a sech2(ax), which is negligiblysmall far from origin.

It is natural to expect each solution to behave as one of the KdVoutside the vicinity of the origin and as a solution of KdV-B nearthe origin.

The numerical experiments reveal that, after the initial shock, weget a soliton of a smaller amplitude to the left of the dissipativearea and a reflected wave moving to the right.

Π - type obstacle

Figure: Left: Π - type viscose layer; α = 1, β = 2. Right: Soliton(a = 1) passing a thin viscose Π-type layer (α = 2.5, β = 2); t = 1

Figure: Left: Soliton (a = 1) passing a thin viscose Π-type layer(α = 2.5, β = 2); t = 3. Right: Soliton (a = 1) passing a thin viscoseΠ-type layer (α = 2.5, β = 2); t = 5

SolitonPi-type obstacle 12-high, location: −2 < x < 2

Push


13.aviMedia File (video/avi)

Refraction

Figure: Left: Soliton (initially a = 1) becomes that of a = 0.4 afterpassing a thin viscose Π-type layer (α = 6, β = 2); t = 50. Right:

Refraction coefficient k = sin(s)sin(r) =V1V2

=4a214a22

equals the ratio of

amplitudes6a216a22

, so k ≈ 60.95 ≈ 6.3.

The absolute filter is impossible,but...

Push

Figure: Π-type obstacle, location: −20 < x < 0


widePi.aviMedia File (video/avi)

Soliton-like obstacle

Since the dissipation distribution a sech2(ax) has a numericallycompact support the transition process differs, in its form, onlyslightly from that for the Π-type obstacle.

Figure shows the soliton (initially a = 1⇒ amplitude= 6) passinga thin viscose layer (3 sech2(3x)); t = 20 The refraction coefficienthere is k ≈ 64.3 ≈ 1.4

An a priori estimate for refraction

Both height and velocity of a solitonu = 6a2 sech2(4a3t + a(x + s)) are proportional to a2.For a soliton, a2 is a conserved quantity:

a2 =1

48

(∫ +∞−∞

u2 dx

) 23

.

So

∂

∂ta2 =

1

32

(∫ +∞−∞

u2 dx

) 13(

2

∫ +∞−∞

u · ut dx)

=1

16

(∫ +∞−∞

u2 dx

) 13(∫ +∞−∞

u · (γuxx + 2uux + uxxx) dx).


Integrating by part and using u(±∞, t) = 0

a = −√

6

36

∫ +∞−∞

(γ(x)u)xux dx

In particular, for a Π-type barrier, γ(x) = γ = const and

a = a0 −√

6γ

36

∫ t0dt

∫ β−β

u2x dx ≈ a0 −√

6γβ

18

∫ t0u2x (β, t) dt

at β → 0; here β is the half-width of the barrier; the decline of thesoliton’s amplitude is monotone. See also the graph of u2x on figure8 (left).


KdV-Burgers asymptotic

The KdV-Burgers equation considered below is of the form

ut(x , t) = k · γ(x)uxx(x , t) + 2u(x , t)ux(x , t) + uxxx(x , t), (3)

k — a small parameter.

Put u(x , t) = w(x , t) + k · v(x , t). For w we obtain the KdVequation

wt(x , t) = 2w(x , t)wx(x , t) + wxxx(x , t), (4)

so a soliton, w(x , t) = 6a2 sech2(a(4a2t + x) + s) is its solution


For v(x , t) we get the following linear 3-d order equation

vt(x , t) =

2w(x , t)vx(x , t) + 2v(x , t)wx(x , t) + vxxx(x , t) + γ(x)wxx(x , t)

or

vt = 2·6a2 sech2(a(4a2t+x)+s)vx+2v(6a2 sech2(a(4a2t+x)+s))x+ vxxx + γ(x)(6a

2 sech2(a(4a2t + x) + s))xx ,


Example

Take KdV-B:

∂

∂tu (x , t) =

3 sech2(3 x))∂2

∂x2u (x , t) + 2 u (x , t)

∂

∂xu (x , t) +

∂3

∂x3u (x , t)

with initial-boundary conditions for the first addendum v of theform:

v(x , 0) = 0, v(0, t) = 6 sech(4t − 6)2,vx(20, t) = 0, vx(0, t) = 6(sech(4t − 6))2)x

Its numeric solution is presented below by the graphs of the firstapproximation sum u1(x , t) = w(x , t) + k · v(x , t). The picturesare similar to ones obtained above numerically for k = 1.


Figure: Left: Graph of u2x (x , 5) in case γ(x) = 12 for −2 ≤ x ≤ 2.Right: First approximation to breather, u1 = w + k · v , t = 1.5


Figure: Left: First approximation to breather, u1 = w + k · v , t = 1.7Right: t = 1.9


Figure: Left: First approximation to breather, u1 = w + k · v , t = 2.Right: t � 1

As t → +∞, the value of soliton and its derivatives becomenegligible at any fixed point x . So we have an approximateequation vt ≈ vxxx , whose linear basis consists ofe−λt−(λ/2)x · e±i(

√3λ/2)x as it may be seen on the enlarged

graph, figure 10 (right).


Inverse motion

TWS solutions for KdV-B with γ(x) = g are shock waves. Theyhave a form

u =3

50g2 sech2(

g

10(−Vt+s+x))+( 3

25g tanh(

g

10(−Vt+s+x))−V

2

If u = 0 at t → −∞ is required, the sole shock wave is

3

50g2 sech2(

3

125g3t+

1

10gx+s)+

3

25g2 tanh((

3

125g3t+

1

10gx+s)+

3

25g2

As it moves from the right it passes the the stair boundary andenters the area without dissipation.Next slide show that the quasi-harmonic oscillations develop, of akind Known for KdV

A shock wave enter nondissipative area

Push

Figure: Π-type obstacle, location: 0 < x


inverse.aviMedia File (video/avi)

Inference

Thank youfor your attention

nonlinear waves in layered media: solutions of the kdv ... · the transient wave in a dissipative...

Documents