Fred Cooper- New class of Compactons in Generalized Korteweg-DeVries Equations and Global Relations

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New class of Compactons in Gener- alized Korteweg-DeVries Equations and Global Relations Talk by Fred Cooper (NSF, SFI) at UIUC Complex Systems conference, May 16 -19, 2005 Collaborators: Avinash Khare (Inst. of Physics, Bhubaneswar) and Avadh Saxena (LANL) 1

Transcript of Fred Cooper- New class of Compactons in Generalized Korteweg-DeVries Equations and Global Relations

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New class of Compactons in Gener-alized Korteweg-DeVries Equations andGlobal Relations

Talk by Fred Cooper (NSF, SFI) atUIUC Complex Systems conference,

May 16 -19, 2005Collaborators: Avinash Khare (Inst. of 

Physics, Bhubaneswar) and AvadhSaxena (LANL)

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for self-focusing... seemed indepen-dent of choice of trial wave function,using time dependent variational prin-ciple.

• – Today will show that previous re-

sults can be DERIVED WITHOUTrecourse to trial functions, using sim-ilar variational methods (Minimiza-tion of ACTION).

• Find analytically new two parameter

class of compactons in GeneralizedK-dV equtaions

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OUTLINE OF TALK

• We discuss two generalizations of theK-dV equation by Rosenau and Hy-man (RH) and Cooper Shepard andSodano (CSS)

• We find a new two parameter familyof compact solitary wave solutions toboth equations of the form

AZ γ (β (x + ct)) (1)

, γ  continuous, where

(Z )2 = 1 − Z 2q (2)

and q is continuous. γ  and q are re-lated to the powers of nonlinearity inthe equation of motion.

• We derive for the CSS equation anexact relation for all solitary wave so-lutions that the Height, Width and

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velocity are related in a simple fash-ion.

• We explicity determine the Energyand Momentum for all these solu-tions and verify the relationship

E/P  = c/r (3)

for all the compacton solutions.

• We determine the domain of stabilityfor the new solutions.

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HISTORY OF COMPACTONS

• Discovered originally in extension of the KdV equation by Rosenau andHyman [1].

K (m, n) : ut+(um)x+(un)xxx = 0,

(4)

• m = n ≤ 3 the solutions are

[cos(ξ)]2/(m−1), (5)

where ξ = a(x

−ct).

−π/2

≤ξ

≤π/2, zero elsewhere.

• We will show here that for m = 2n−1 with n continuous variable 1 <n ≤ 3 t

u(x, t) = Acnγ 

(β (x−vt); k2

= 1/2)(6)

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•We then will find solutions for ALLm, n with 1 < n ≤ 3 and m, n con-tinuous.

• The RH equations were not derivablefrom a Hamiltonian.

• Cooper, Shepard and Sodano (CSS)[3] considered instead

K ∗(l, p) : ut = uxul−2 + α (2uxxxu p+

4 pu p−1uxuxx + p( p − 1)u p−2(u

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Lagrangian Hamiltonian Dynamics→VARIATIONAL FORMULATION

L(l, p) = 

1

2ϕxϕt − (ϕx)l

l(l − 1)

+α(ϕx) p(ϕxx)2

dx, (8)• SAME class of solitary wave solu-

tions when l = m + 1 and p = n−1.

• Using TRIAL WAVE FUNCTIONS

uv(x, t) = A(t)exp[−β (t)|x + q(t)|γ 

](9)

q = r( p,l)E 

P (10)

r( p,l) = ( p + l + 2)/( p + 6−

l).

• When l = p + 2 the width did notdepend on the amplitude or velocity

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•We will show using Hamilton’s equa-tions this result is exact for

u(x, t) = AZ [β (x + q(t))] (11)

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Solitary Waves in Rosenau Hyman equa-tion

• Let u = f (y) where y = x − vt

vf  =d

dy(f m) +

d3

dy3(f n). (12)

• Integrating twice

n

n + 1vf n+1 =

n

n + mf n+m +

1

2[d(f n)

dy]2

+C 1f n + C 2 (13)

• compactons are solutions with C 1 =C 2 = 0.

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Circular Function solutions.

• RH equation

(f )2 =2v

n(n + 1)f 3−n− 2

n(n + m)f m−n+2.

(14)

• Choose Ansatz for m = n

f  = A cos2/(m−1)(β (x−vt)); (15)

−π/2 ≤ βy ≤ π/2., and f  = 0elsewhere. (m, n) = (2, 2)

β  = 1/4; A = 43

v (16)

• For m = n, β  = constant, in-dependent of the Amplitude whichdepends on v

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Elliptic function solutions of RH equa-tion.

• Solutions of the form cnm.

• (m, n) = (3, 2)

1

3vf −1

5f 

3

= (f )2

(17)f  = Acn2(β (x − vt); k2).

A = 10β 2; , k2 = 1/2; β 4 =v

60.

(18)

• (m, n) = (5, 3).v

2− 1

4f 4 = 3(f )2. (19)

f  = Acn(βy; k2),

A2 = 6β 2; k2 = 1/2 ; β  = (v

18)1/4.(20)

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•Case (m, n) = (4, 2)

(f )2 =vf 

3− f 4

6(25)

• Put in standard Form:

f  = AZ 2(βy) (26)

(ξ = βy)

(dZ 

dξ)2 =

2v − A3Z 6

24Aβ 2(27)

• ChooseA = (2v)1/3; β  =

(2v)1/3

2√

6(28)

±  Z 0

dz√1

−z6

= ξ; 0 ≤ Z ≤ 1.

(29)

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•Simplifying:

Z (ξ) =

1 − cn(2(3)1/4ξ)

(1 +√

3) + (√

3 − 1)cn(2(3)1/4y)

1

(30)

k

2

=

1

2 −

√3

4 . (31)

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•NEW Solutions ANSATZ: FIND val-ues of a and m, n so that

f  = AZ a(βy) (32)

leads to the differential equation

dZ (ξ)

2= 1 − Z 2q(ξ) (33)

Here we allow for m,n,q to be con-tinous variables.

• Conditions are

m = q(n − 1) + 1; a = 2/(n − 1)

Am−1 = vn + m

n + 1;

a2β 2A2 =2

n(n + m)Am

−n+2. (34)

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β 2 =(n − 1)2

2(n + m)

(n + m)v

n + 1

(m−n)/(m−1)

(35)1 < n ≤ 3.

• Special solutions: q = 1 (m = n), circular compactons with width in-dependent of  velocity, q = 2 solu-tions of the form cnγ (k2 = 1/2), so-lutions with q = 3, leading to Weirstrass

functions.• we will show that stability requires

(n − 1)(q − 1) < 4 (36)

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CSS equations

• CSS equation of motion

K ∗(l, p) : ut = uxul−2

+α2uxxxu p + 4 pu p−1uxuxx + p( p − 1)u

• can be derived from  L(x, t), dxdt(Least Action)

L(l, p) = 

1

2ϕxϕt − (ϕx)l

l(l

−1)

+ α(ϕx) p(ϕxx

(38)u(x) = ϕx(x). (39)

RH set (m, n) corresponds to the CSSset (l − 1, p + 1).)

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SOLITARY WAVES

• Solitary Wave Ansatz:

u(x, t) = f (y) = f (x + ct), (40)

cf  = f f l−2 + α2f f  p + 4 pf  p−1f f 

+ p( p − 1)f  p−2f 3 . (41)

• Integrate Twice

c

2f 2

−f l

l(l − 1)−αf 2

f  p

= C 1f +C 2.(42)

C 1 and C 2 are zero for compactonsWell behaved : : l > 1 and f f  p →0, f 2f  p−1

→0 at edges where f 

→0.

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•CSS equation

αf 2 =c

2f 2− p − f l− p

l(l − 1). (43)

For finite f  at the edges, we musthave p

≤2, l

≥p.

• Compare RH equation

(f )2 =2v

n(n + 1)f 3−n− 2

n(n + m)f m−n+2

(44)

l = m + 1 and p = n − 1 equa-tions identical in form, differing co-efficients.

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Special Cases and Relations:

• l = p + 2 (m = n)

u(x, t) = A cos2/p[β (x + ct)] (45)

for noninteger p.

E  = 2P  p + 2

q. (46)

• l = 2 p + 2

f  = Acn

γ 

(βy; k

2

= 1/2) (47)

α(f )2 =c

2f 2− p− f  p+2

(2 p + 2)(2 p + 1).

(48)

γ  = 4l − 2

;

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c2 = A

l

−2

l(l − 1)

β 4α2 =2c

l(l − 1)(

l − 2

4)2 (49)

• Nonsingular behavior condition:

2 < l ≤ 6. (50)

• The rest of the story:

f  = AZ a(ξ = βy) (51)

(Z )2 = 1 − Z 2q (52)

• This leads to continuous ql = pq + 2; a = 2/p

Al−2 = l(l − 1)c

2; αa2β 2A2 =

c

2A2− p.

(53)

β 2 =c

2αa2A p (54)

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•For well behaved solutions we need0 < p ≤ 2.

• We will show

E/P  = c/r (55)

and stable for

 p(q − 1) < 4 (56)

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Conservation laws and canonical struc-ture

• Canonical form for KdV

ut = ∂ xδH 

δu= {u, H } (57)

H  = 

[(πϕ) − L] dx

(ϕx)l

l(l − 1)− α(ϕx) p(ϕxx)2

dx,

=  u

l

l(l − 1) − αu p(ux)2 dx. (58)

(59)

• Poisson bracket structure

{u(x), u(y)} = ∂ xδ(x − y). (60)• Conservation of Mass

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ut = ∂ xδH δu

(61)

M  = 

u(x, t)dx (62)

• Momentum Conservation Multiply eq.61 by u(x, t)

∂ t(u2

2) = ∂ x

ul

l+ α{( p − 1)u pu2

x + 2u p+1u

(63)

(1/2) 

u2(x, t)dx = P  (64)

• P  is the generator of the space trans-lations:

{u(x, t), P } =∂u

∂x. (65)

• (i) φ(x, t) → φ(x, t) + c1; (ii) x →x + c2 nd (iii) t → t + c.

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Energy-Momentum relationship

• ACTION

Γ = 

Ldt, (66)

L(l, p) =  1

2ϕxϕt − (ϕx)l

l(l − 1)+ α(ϕx) p(ϕxx

(67)

• Generic Solitary Wave

φx = AZ (β (x + q(t))) = u, (68)• Using

φt = φxq. (69)  1

2ϕxϕtdx = P q (70)

P  =1

2

 u2(x, t)dx (71)

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•Now have point ”Particle” Lagrangian

L = P q − H  (72)

H  = 

dx

ul

l(l − 1)− αu p(βx)(ux)2

(73)

• Usingux = βAZ [β (x + q(t))] (74)

H  = C 1(l)Al

βl(l−

1)−αβA p+2C 2( p)

(75)where

C 1(l) = 

Z l(z)dz; C 2( p) = 

[Z (z)]2Z  p(z)(76)

• Since H  is independent of  q,P  = −∂H 

∂q= 0, (77)

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P  is conserved.

• Rewrite A in terms of P  :

P  =1

2

 dxu2 =

A2

2β C  (78)

C  = 

dzZ 2(z) (79)

A2 =2βP 

C (80)

H  = C 3(l)P l/2β (l−2)/2

−C 4( p)P ( p+2)/2β ( p+

(81)where

C 3(l) =C 1(l)

l(l − 1)[

2

C ]l/2; C 4 = αC 2( p)[

2

C ]( p+

(82)

• Key Point: exact solutions minimizethe Hamiltonian with respect to β  .

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H  =A2c

β 

 10

dZ √1 − Z 2q

[2Z a( pq+2)−Z 2a]

(92)

• Evalueat in terms of the Beta func-

tion B(µ, ν ) by substituting t = Z 

2q

.

H  =A2c

2βq

(6 + p − l)

(l + p + 2)B

 p + 4

2 pq,

1

2

.

(93)

P  =A2

2βqB

 p + 4

2 pq,

1

2

. (94)

Using a = 2/p, and a(l − 2) = 2q

H/P  = c/r (95)

Stability of Solutions

• The stability problem at q = 1 wasstudied by Dey and Karpman.

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•The result of detailed analysis is thatthe criteria for Linear Stability is equiv-alent to the condition,

∂P 

∂c> 0. (96)

P  = A2

2βqB

 p + 42 pq

, 12

. (97)

Al−2 = l(l−1)c

2; αa2β 2A2 =

c

2A2− p.

(98)

• Deduce

 p(q − 1) < 4 (99)

The requirement for non-singular so-lutions is that 0 < p

≤2.

0 < p < 4/(q − 1). (100)

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Analysis of Lyapunov stability fol-lowing [10] [11] [12] leads to the samerestrictions on p.

References

[1] P. Rosenau and J.M. Hyman, Phys. Rev. Lett. 70, 564 (1993).

[2] P. Rosenau Phys. Lett. A 275, 193 (2000).

[3] F. Cooper, H. Shepard, and P. Sodano Phys. Rev. E 48, 4027 (1993).

[4] A. Khare and F. Cooper, Phys Rev. E 48, 4843 (1993).

[5] F. Cooper, J. Hyman, and A. Khare, Phys Rev. E 64, 026608. (2001).

[6] A. Das, Integrable Models (World Scientific Lecture Notes in Physics,Singapore, 1989) Vol.30.

[7] F.Cooper, H. Shepard, C. Lucheroni, and P. Sodano, Physica D68

(1993), 344. hep-ph/9210234

[8] F. Cooper, C. Lucheroni, H. Shepard, and P. Sodano, Phys. Lett. A173, 33 (1993).

[9] P.F. Byrd and M.D. Friedman ”Handbook of Elliptic Integrals for En-gineers and Scientists” 2nd Edition (Springer, Berlin, 1971).

[10] B. Dey and A. Khare, Phys. Rev. E 58, R2741 (1998).

[11] V. I. Karpman, Phys. Lett. A. 210, 77, 1996.

[12] V. I. Karpman, Phys. Lett. A. 215, 254, 1996.

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