A note on the instability of spatially quasi-periodic states of the Ginzburg-Landau equation

31 March 1997 PHYSICS LETTERS A ELSEVIER

Physics Letters A 228 (1997) 53-58

A note on the instability of spatially quasi-periodic states of the Ginzburg-Landau equation Ognyan Christov 1 Faculty of Mathematics and lnformatics, Sofia University, 5 J. Bouchier, 1164 Sofia, Bulgaria Received 29 October 1996; revised manuscript received 18 December 1996; accepted for publication 14 January 1997 Communicated by A.E Fordy

Abstract

Bridges and Rowlands [Proc. R. Soc. A 444 (1994) 347] recently studied the stability of stationary quasi-periodic states of the Ginsburg-Landau equation. They obtained the surprising result that the sign of the KAM determinant is related to the stability type of the toms in the time- dependent problem. In this note we show that the Kolmogorov determinant is negative and, hence, only linear instability can occur for these tori. @ 1997 Elsevier Science B.V.

which leads to the cubic GL equation

1. Introduction

The classic Ginzburg-Landau (GL) equation is defined by (see Ref. [ 1 ] ) 6.~ ,~'

f = f [l~xl 2 - v ( l ~ } 2)]dx Xl

or

0q~

O2q~

at - Ox2 + V ' ( ] ~ 1 2 ) ~ "

(1)

Here the function q0(x, t) is a complex-valued function of the real variables x and t. It is seen that the functional ~" is gauge invariant, i.e., it is SO(2) invariant with action eq~ = exp(ie)q~ for all e C SO(2). The most studied form of the function V ( u ) is V(u) = u-

½u2,

1 E-mail: [email protected].

(2)

~xx + V'( Iq012)q0 = 0.

(4)

This equation has a Hamiltonian structure. Introduce the coordinates ( q , p ) , defined by cp = ql + i q e , p = qx. The symplectic form is the standard one: oJ = dpl A dql + dp2 A dq2 and the Hamiltonian reads H = ~ 1 (p2 q_ p22) q_ g1 V ( q 2I + q2).

(5)

Due to the SO(2) invariance, there exists a second integral, F = p2ql - plq2.

0375-9601/97/$17.00 (~) 1997 Elsevier Science B.V. All rights reserved.

PIIS0375-9601(97)00062-5

(3)

The GL equation is important in superconductor physics and the cubic GL equation particularly is studied in near critical hydrodynamic stability problems (see Ref. [1] and references therein). The stationary states of the GL equation satisfy the following real fourth-order integrable system,

X2

04~ or-

3~ 02,/, 0--'7 = ~ + ~ -l~12q~"

(6)

O. Christov/Physics Letters A 228 (1997) 53-58

54

Note that when V(u) = u - au2/2, with a = + l , the above Hamiltonian system is the truncated normal form for the Harniltonian-Hopf bifurcation (see Ref. [2] ). From now on we consider V in the form (2). We introduce more suitable coordinates via symplectic transformation (see for example Ref. [3] ), ql = r cos Opl = Pr cos 0 - Po sin 0, r

(7)

q2 = r sin 0p2 = pr sin 0 + Po cos 0. r Then w = do', o" = p~ dr + po dO and the Hamiltonian (5) and the second integral become H = ~(Pr I 2 + P2o/r2) + -lr2 - 1F4,

F=po =f.

(8)

Then the spatially quasi-periodic (QP) states of the stationary GL equation can be parametrized by (11,12) - the action variables for (8) with frequencies (Wl,O~2) and frequency map with determinant A K = 69(wl, w2)/69(11,12). In a recent paper Bridges and Rowlands [ 1 ] studied the linear stability and instability of these quasiperiodic states. They proved that when zl ~c < 0 the spatially QP state is unstable and A K > 0 is a necessary but not sufficient condition for linear stability. Therefore, ,5 X < 0 (or equivalently the sign of the Kolmogorov determinant is negative) is a sufficient condition for a given spatially QP state to be linearly unstable. Bridges and Rowlands prove that A X < 0 for f near zero. Doelman et al. [5] prove that all the QP states of the GL equation are unstable using a topological argument in which the Kolmogorov determinant does not appear. Here an alternative proof is given: we prove that A K < 0 for all spatially QP states of the GL equation and therefore by Theorem 1 of Ref. [ 1 ] these states are linearly unstable solutions of the time-dependent GL equation. The main result of the paper is the following.

Theorem 1. The Kolmogorov determinant for every spatially QP state of the stationary cubic GL equation is negative, 69112 69116912 a2H 02H

a116912 al~

In order to prove that the Kolmogorov determinant is negative for all spatially QP states, we use the method of Horozov [ 6]. The determinant is expressed via certain Abelian integrals which are studied as solutions of Picard-Fuchs equations. We refer to Refs. [3,4] for more details on the structure of integrable Hamiltonian systems and KAM theory. See also Refs. [6,7] and references therein for more details on methods for checking KAM conditions. A relation of the stability character to the matrix (02H/Oli69Ij) is known from Poincar6 [8]. Let an integrable system be subjected to a small perturbation. Then the periodic solutions of the integrable system can be continued analytically and this matrix appears in the expression for the characteristic exponents, hence influences stability (see also Ref. [3] ). The paper is organized as follows. In Section 2 the action variables are defined and some simple but important reductions are made. In Section 3 the PicardFuchs equations are derived for the functions introduced in Section 2 and their asymptotic behaviour is studied. The theorem is proved in Section 4.

2. Action variables Suitable action variables It, 12 for the stationary GL equation with Hamiltonian (8) are derived in Ref. [ 1 ]. To put them in a form suitable for the later analysis, consider the map J : R 4 ~ R 2 : (ql,q2,Pl,P2) --~ ( f , h) where H = h. The critical values of the map J are parametrized by

( f , h ) = ( 4-V/-~ - r6, r 2 0 ~< r ~<. 1.

3r4/4), (10)

This curve forms the boundary of a slice through a swallowtail (cf. Fig. 1 ). Denote by Ur the set of regular values of J (Fig. 1). In order to explain Fig. 1 we introduce the reduced system po = f , 1 2 H = ~Pr + Vf(r),

I 692H 692H ) det

Hence, the spatially quasi-periodic states of the cubic GL equation are linearly unstable.

< 0.

(9)

where Vf = r2/2 - r4/4 + f2 /2r2 is the effective potential. Then, the critical values (10) (and the corresponding branches in Fig. 1 ) refer to a saddle (s) (or

O. Christov/Physics Letters A 228 (1997) 53-58 deal/2 ---- f

¢(h,f)

55

ydz Z

(13)

3'

Denote by/)(11, I2) the Hamiltonian function of the system in action coordinates. The following lemma is proved in Refs. [6,1].

~LO3

f

Lemma 2.

Ur ( i n t e r i o r ) .

the maximum of Vf) and a center (c) (the minimum of Vf) of the reduced system (see also Ref. [ 1] ). For the points ( f , h) E U~ the level surface determined by the equations H = h, F = f is a torus Th,f. We choose a basis 3'1,3'2 of the homology group Hi (Th,f, Z) with the following representations. For 3"! take the curve on Th,f defined by fixing r, Pr and letting 0 run through [ 0, 2rr]. For 3'2 fix 0 and let r, Pr make one circle on the curve given by the equation

02¢

r2

Oh = f

__dz

r h , f = { ( y , z ) :y 2 = 2 h z - z 2 + z 3 / 2 - f

2}

(which exists for all ( h , f ) E Ur) by 3". Then,

4= 0

Next we show that the entries of D can be represented in terms of elliptic integrals. Differentiate (13) twice, formally, to get the following expressions,

(12)

For later use we make a change of variables in the integral defining 12. Put z = r 2. Denote the oval of the curve

_ _

D = ¢ h h ¢ f f -- ¢ 2 f < O.

O2¢

r2 - r4/2 + f2/r2 = 2h.

02¢

Y

/ z dz

rl

where r2 > rl are the two roots of the equation

a122

in/Jr • So, to prove the theorem it is sufficient to show that the Hessian D of ¢ is negative,

'2=SPrdr=2fq2h-r2+r4/2-f2/r2dr, Y2

allal2

Obviously we have

3'

(ll)

a2/:/

02¢ 82¢ OhOf af 2

p2 + Vf(r) = 2h.

II = 27rf,

a2/:/

Oh2 ahaf = det (

J

Now, following Ref. [4] we can define the action coordinates by the formulae lj = f:0 o-, j = 1,2, where ois the canonical one-form: tr = Pr dr +Po dO. Straightforward computations give

al~ o12

det

(2~)2 \ Oh / Fig. 1. The set

al 2

(e¢) 4

T

a2¢

_S.~ = -

O2¢

7'

/

=:

} dz

/ dz

3'

~'

~y _ f 2

ZY3"

Y

dz

7' (14)

The differential forms containing y-3 have poles along y. The standard way to avoid the poles on the integration path is to consider I'h, f as a complex curve. Topologically it is a torus with one point removed. Deforming the cycle 3' into the cycle y', homological to % on which the functions y and z have neither poles nor zeros, the function ¢ ( h , ) can be defined, using

56

O. Christov/Physics Letters A 228 (1997) 5 3 - 5 8

Cauchy's theorem [9], by the integral (13) on y' instead of y. After the above consideration, it is clear that the derivatives are well defined. We will henceforth denote y' by y. For the further analysis we define the following functions,

z j dz y3 , j = 0,1 . . . . .

wj( h, f) =

3. P i c a r d - F u c h s

equations

Lemma 4. Suppose f -- 0. Then the functions wo, w I satisfy the following system of Picard-Fuchs equations, dw0

•,,

4n)--d-~ = 5wl + 4 ( 7 h - 2)w0,

4h(1 (15)

dwl

( 1 - 4h)--;=--, = 5 w l - w0.

(17)

(In

It follows from the form of H and a result in Ref. [10] that w0 4= 0 in Ur. The next lemma gives a representation of D as a quadratic form in wow~ .

Lemma 3. The determinant D has the following representation,

Proof We follow Ref. [6]. Differentiating the expressions (13) with respect to h we obtain dwk _ - 3 f zk+l dh j --7- dz. 3' Put f = 0; then transform w0 in the following way,

fZw2.

D = - ~ (4hw0 - Wl) -

(16)

zz + 2hz dz wo = / Y-~5dz = f z3/2- y5

Proof. It is elementary to see that

020 wt,

OhOf = fwo.

Of 2

+ W 1 --

- -2hwo

1

2h dwo --+ 3 dh

1 dwl 3 dh'

3"

Off we make the following transformation,

020

3"

= ½/z3dzy5

a20

~Oh For

3"

f z 7 2d z .

/ z ~ - T d z = 2 / z dz3/2 y5 dz 3"

3"

e/'zd(Y 2-2hz+z

=~ j

3"

Now

y-~

2)

3'

z2

fTdz=~f

4( =~ -

d(z3/2)y3

/

zdy -3+

hdWO a w l ) dh

dh

3' 3'

3'

2 / " d(Y 2 + f 2 - 2 h z + z

=3 J

7

3'

= -

2)

4( = ~

hdWO wo +

dh

dw1) dh

'

This gives

hwo +

Hence, iemma. []

Off

=

(Wl

--

4hwo)/3 which proves the

It is clear from (16) that D does not depend on the sign of f ; therefore it is sufficient to prove Theorem 1 for f /> 0 only. The proof of Theorem 1 is given in Section 4, after introducing the Picard-Fuchs equations for wo and wl.

7w0 = dw---!- 4h dw° . dh dh

(18)

In the same manner we transform wl and again using integration by parts we obtain for wl the expression 15wl = -4h-~h° + 4 ( 1 -

,.

dWl

.~n)--~ - 4wo.

Now solving (18), (19) for the system (17).

(19)

dwo/dh, dwl/dh we get

O. Christov/Physics Letters A 228 (1997) 53-58

We also need the function Wl(h,0) 0-(h) = wo(h,O)"

(20)

Straightforward calculations show that 0- satisfiesthe Riccatiequation 4h(l-

d04h) ~--~ = -5o-2 + 8(1 - h ) 0 - - 4 h .

(21)

57

The asymptotic behaviour of the functions 0-, oq, can be found easily from the function e ( P ) . The following result is crucial for the proof of the theorem. Lemma 5. (From Ref. [6].) l i m p ~ _ 2 0 ( p ) = ~7 , l l• m p ~ z Q ( p ) = 1. The function O(P) is strictly decreasing in the interval [ - 2 , 2]. From the above lemma we get immediately

When f = 0 the expression for D factors, Lemma 6. W2

D=-~0"0"1,

o'l(h) = 0 - ( h ) - 4 h .

lim 0 - ( h ) -- 5, s

(26)

h~0

From (21), one can also get the Riccati equations for 0-1.

lira o - ( h ) = 1,

(27)

h---*l/4

We need also some additional functions both for the study of o-, 0-1 and for the case f > 0. Before introducing them we put the family of c u r v e s I'h, f into a normal form,

~Lt~p(h,O) = - 2 ,

Fp={(u,u)

Then using ( 2 3 ) - ( 2 5 ) we obtain

~C 2:u2=2(u3-3u+p)},

(22)

by the transformations z = t + 2/3, y = o~u, t =/3u where

Proof Note that

2 ~Lm0"(h) = ILmc/3lim O ( p ( h , 0 ) ) + -~ h

=2

13 = 2v/l - 3h/3,

oe2 =/33/4.

(23)

p(h,f)

=

h---*l/4

= ~1x

36h - 2 7 f 2 - 8

we get (22). In these variables the integrals wo(h, f ) , wj (h, f ) become

lira /3 lira O ( p ( h , O ) ) + ~ 2

h~l/4

h~l/4

k

I+2=I.[5]

W1 = ~

4. P r o o f o f t h e m a i n t h e o r e m

We start with the case f = 0.

__/3/2 UU 3du

y~p)

h ---*0

(24)

4V/( 1 - 3h) 3

wo = ~/3 fdu T,3'

0

_gx 7 + ~2= ~ , 8

lim 0 " ( h ) =

If we put

lim p ( h , 0 ) = 2.

h---*l/4

"

Lemma 7. The functions 0-(h) and o'1 (h) are positive when h E (0, 1/4).

y(p)

We introduce the new functions O 0 ( p ) ----

~,

01(p) =

y[p)

/

udu C3

Proof Consider first 0-(h). Suppose h0 E (0, 1/4) is a zero of o-(h). Then using the Riccati equation (21 ) we have

,

y(p)

and their ratio

o-I(h0)-

O(P ) = O~(p) /Oo(p). In this notation we have 2 o'(h) = f l p ( p ( h , O ) ) + ~.

(25)

- -

1

1 - 4h0

<0.

But limh_.l/4O'(h ) = 1, so there exists h i > h0,0-(hl) = 0 and d0-(hl)/dh > 0 - a contradiction. It is obvious that h0 can not be a double zero. Therefore, o'(h) > 0 when h E (0, 1/4).

o. Christov/PhysicsLetters A 228 (1997) 53-58

58

The inequality O-l(h) > 0 can be obtained in the following way, o-l(h) = o-(h) - 4 h

Therefore OG/Ofl < 0 for p < 2 and G is a decreasing function of/3. Then

=flO(p(h,O)) + ] -4h

> / 3 + ~ - 4h

aG aG O---f~(p,fl) < ~-/~(2, fl) = - 3 x 1 + 2 x 3 / 2 = 0 .

(28)

Using the expression for/3 we obtain the function for the right-hand side of (28),

G(p, fl) < G(p, 1/3) = - O z - p + p/2 + 1 b u t - 1 < p/2 < 1 and - 7 / 5 which we obtain

< -p

< - 1 , from

rl(h) = ]v/1 - 3 h + 2 _ 4h,

-O+p/2 and the derivative o f r/(h) is negative in (0, 1/4), so r/(h) > r / ( 1 / 4 ) = 0. Hence, t r l ( h ) > 0 when h E (0, 1/4). []


<0

Hence, G(p,/3) < 0. This finishes the proof of the lemma and together with it that of Theorem 1. []

Remark. The same result is also true for the potential

Corollary. For f = 0, D is negative.

V(u)

= u2 -

u).

Finally we turn to the case f > 0.

Lemma 8. For h E (0, 1 / 4 ) , f > 0 we have the Acknowledgement

representation 4

D = -~a6G(P, fl),

(29)

where

G(p, fl) = _ Q 2 ( p ) _ 3 t i p ( p ) + 3p/3/2 + 1.

(30)

The functions f l ( h ) , p ( h , f ) map the set Ur fq { f > 0} diffeomorphically onto the set

I am grateful to T. Bridges for drawing my attention to this problem and for his suggestions which led to improvement of the manuscript. Also I am grateful to E. Horozov for firsthand information. This work is partially supported by the contract MM 523/95 with the Ministry of Science and Technologies o f Republic Bulgaria.

Vr = ( ( p , fl) :/3 E ( 1 / 3 , 2 / 3 ) , p C ( - 2 , 2 ) } .

References

The proof is a straightforward computation using (16), ( 2 3 ) - ( 2 5 ) .

[ 1] T.J. Bridges and G. Rowlands, Proc. R. Soc. A 444 (1994) 347. [2] J.C. van der Meer, Lecture notes in mathematics, Vol. 1160. The Hamiltonian-Hopfbifurcation (Springer, Berlin, 1985). [3] V. Arnold, V. Kozlov and A. Neistadt, Dynamical systems 111,Encyclopedia of mathematical sciences ( Springer, Berlin, 1988). [4] V. Arnold, Mathematical methods of classical mechanics (Springer, Berlin, 1980). [5] A. Doelman, R.A. Gardner and C.K.R.T. Jones, Proc. R. Soc. Edinburgh 125A (1995) 501. [6] E. Horozov, J. Reine Angew. Math. 408 (1990) 114. [7] D. Dragnev, Phys. Lett. A 215 (1996) 260. [ 8 ] H, Poincar~,Les mdthodesnouvellesde la m6caniquec61este, Vol. 1 (Gauthier-Villars, Paris, 1892) [9] A. Hurvitz and R. Courant, Funktionenteorie (Springer, Berlin, 1964). [10l S. Chow and J. Sanders, J. Diff. Equ. 64 (1986) 51.

Lemma 9. For all (p,/3) E Vr the function G(p,/3) is negative.

Proof. We have aG

- - = - 3 0 + 3p/2,

0/3

a aG

@a/3

- -301 + 3 / 2 > 0,

since O' < 0. Hence, aG/?/3 is an increasing function ofp.