Real Ginzburg-Landau equation
Consider the scalar reaction-diffusion equation:
ut = uxx + λu + u3 + g(u) = 0
with conditions: u is spatially periodic, u(x, 0) = h(x)
Note that u = 0 is the basic solution. Here we have no boundary conditions to select particular
solutions otherwise, but we can consider possible solutions which may bifurcate from u = 0.
We consider the linearized problem (i.e. linearized about u = 0) which is
ut = uxx + λu
and if we look for solutions of the form e σt φ(x), with φ(x) spatially periodic, we get
φxx + (λ − σ)φ = 0
and spatially periodic solutions are given by e ikx with σ − k 2 + λ. To determine whether these
solutions or u = 0 might be stable (or whether they might grow) we look at the neutral stability
curve (NSB) σ = 0 which is just λ = k 2 . For λ > 0 we find that there will be some values of k for
which σ > 0.
λ
f (λ, k) = 0
k
(λ0 , k0 )
That is, there is a band of wave numbers for which we find that σ = 0 from the linear stability
analysis about u = 0. Therefore, if we consider values of λ near the minimum of the NSB, λ = 2 λ2 ,
we will be considering wave numbers with the form k = κ. Furthermore, this suggests that
σ = O(2 ). Then we would expect that our solution depends on slow time and space scales ξ = x
and τ = 2 t.
Then we look for an expansion of the form:
u ∼ u1 (x, t, ξ, τ ) + 2 u2 (x, t, ξ, τ ) + . . .
Then we get the form of the equations:
u1t − u1xx = 0
u2t − u2xx = 2u1xξ = 0
u3t − u3xx = −u1τ + u1ξξ = 0 + 2u2xξ + u1 − u31
If we look for periodic solutions, we can attempt to solve with a Fourier transform. After careful
inspection, and using the initial condition, we conclude that to leading order u 1 is constant with
respect to x and t. Then to leading order we have that u 1 = a(ξ, τ ).
If we continue to solve the sequence of equations, we get u 2 = 0, and satisfying the solvability
condition for u3 , we get
1
aτ = aξξ + a − a3
This is the Ginzburg-Landau equation. Note that in some sense this is a less interesting case in
that to leading order the solution does not depend on the fast scales t and x. In fact, for k 1 we
are essentially considering a long wave solution (small wave number). We will consider solutions of
this types in other contexts.
Part of the reason for this case being less interesting is the fact that we have a scalar reaction
diffusion equation. What is more typical, and realistic, is that reaction-diffusion models are systems, related to the fact that there are two components which are interacting. In the next section
we consider a scalar equation which captures many of the features found in systems of reactiondiffusion systems, as well as other applications including convection, sedimentation, and elasticity.
Swift-Hohenberg equation
Consider the Swift-Hohenberg equation:
ut = −(∂x2 + q 2 )2 u + λu − u3 = 0
together with conditions of solutions that are bounded at ±∞.
Note: u ≡ 0 is a basic solution:
To test the linear stability, consider u = e σt+ikx . Plug in and linearize about u = 0 to get
σ + (q 2 − k 2 )2 − λ = 0
which is the dispersion relation for cellular behavior (standing wave-type oscillation).
λ
f (λ, k) = 0
k
(λ0 , k0 )
Looking for the neutral stability curve (NSB) σ = 0 gives
λ = (q 2 − k 2 )2 .
Then if we consider the stability of u as a function of λ, at λ = 0 there is a change from stability
to instability of the basic solution. For λ > 0 we suspect the existence of some other solution. In
this case there are no boundaries, so for any λ > 0 there is a continuum of unstable states with
wave number k. As we will see, this leads to a Ginzburg-Landau equation for the amplitude of the
disturbance.
In contrast to the previous example, notice that the band of unstable wave numbers is now
centered at k = q, that is, we expect to see spatial oscillations with O(1) wave numbers, giving us
variation on the fast spatial scale. The minimum of the NSB indicates the critical value of λ c and
wave number kc , that is, the minimum value of λ for which u = 0 is unstable, and the wave number
that we expect to see once λ > λc .
Again, if we pursue a solution near the critical values, λ = 2 λ2 , we expect σ = O(2 ) and
perturbation to kc will look like kc + O(). Thus we expect to use the slow spatial and temporal
scales ξ = x and τ = t.
2
Then we introduce
u ∼ u1 (x, t, ξ, τ ) + 2 u2 (x, t, ξ, τ ) + . . .
and get the form of the equations:
u1t + Lu1 ≡ u1t + u1xxxx + 2q 2 u1xx + q 2 u1 = 0
Lu2 = −4q 2 u1xξ − 4u1xxxξ
Lu3 = −4q 2 u2xξ − 4u2xxxξ − u1τ + λ2 u1 − u31 − 2q 2 u1ξξ − −6u1xxξξ
Then we solve these equations to get:
u1 = A(ξ, τ )eikc x + c.c.
keeping only the periodic part (bounded at ±∞).
At the next order, we find that the right hand side of the equation for u 2 vanishes, since kc = q.
Then we can set u2 = 0 since it has the same form as u1 .
Then at O(3 ) the right hand side is:
−(Aeikc x + c.c.)τ + λ2 (Aeikc x + c.c.) − (Aeikc x + c.c.)3 − 2q 2 (Aeikc x + c.c.)ξξ − 6(Aeikc x + c.c.)xxξξ =
(−Aτ + λ2 A − 3|A|2 A − 2q 2 Aξξ + 6q 2 Aξξ )eikc x + c.c. − A3 e3ikc x + c.c.
Setting the coefficient of eikc x to zero to eliminate secular terms (i.e. get only periodic solutions),
we get the Ginzburg-Landau equation for A:
Aτ eikc x = +λ2 A − 3|A|2 A + 4q 2 Aξξ
“Brusselator” system:
ut = A − (B + 1)u + u2 v + D1 vxx
(1)
2
vt = Bu − u v + D2 vxx

The basic solution: u = u0 = A

0≤x≤1
 periodic boundary condition 
or infinite domain
v = v0 = B/A (simple equilibrium)
Linear stability of basic state:
u = A+U
,
v = B/A + V
(U, V small)
U
D1 0
U
U̇
B − 1 A2
,
+
⇒
=
V x
0 D2
V
−B −A2
V̇
Consider stability to oscillatory states:
U
= eσt+ikx
Re σ > 0?
V
U
B − 1 − σ − k 2 D1
A2
=0
⇒
V
−B
−A2 − k 2 D2 − σ
U
U
0 = M
− σI
V
V
3
(2)
(3)
σ is eigenvalue of M , det (M − σI) = 0
Look for σ = 0 ⇒ stability boundary for basic state / standing waves
(Im(σ) = 0)
(See Figure 1)
σ = 0,
det (M ) = 0
⇒
2
(B − 1 − k D1 )(−A2 − k 2 D2 ) + A2 B = 0
D1 D2 k 4 + k 2 D1 A2 − D2 (B − 1) + A2 = 0
q
2 − D (B − 1) ±
−
D
A
(D1 A2 − D2 (B − 1))2 − 4A2 D1 D2
1
2
⇒ k2 =
2D1 D2
For k real U = eikx , not decaying in space
2
D1 A2 − D2 (B − 1) > 4A2 D1 D2
(4a)
(4b)
(4c)
2
D1 A − D2 (B − 1) < 0
Together with (4a), we get critical bifurcation parameter B c , and critical wave # kc
D1 D2 kc4 + kc2 D1 A2 − D2 (Bc − 1) + A2 = 0
2
D1 A2 − D2 (Bc − 1) − 4A2 D1 D2 = 0
(D1 , D2 , A are constants, fixed)
kc – critical wave #, Bc – bifurcation parameter
For A = 1, D1 = D2 = 1
B
σ>0
σ<0
4
(kc , Bc )
0
1
5
K
Now do a bifurcation analysis near bifurcation – critical values k c and Bc ( near σ = 0!)
B ∼ Bc + B1 + 2 B2
u ∼ A + U1 + 2 U2 + 3 U3 . . . = A + U
v ∼ B/A + V1 + 2 V2 + 3 U3 . . . = B/A + V
Let
Uj = Uj (x, X, τ )
Vj = Vj (x, X, τ )
4
X = x
τ = 2 t
(5)
(no t dependence – we looked for standing waves; Im(σ) = 0 above critical curve (neutral stability
curve))
(−A − k 2 D2 )(B − 1 − k 2 D) + A2 B > 0
(For k, B above the curve)
Equation for U, V
B 2
U + 2AU V + U 2 V + D1 Uxx
A
Vt = −BU − A2 V − 2AU V − U 2 V + D2 Vxx − B/A U 2
Ut = (B − 1)U + A2 V +
When we introduce slow and fast time scales:
Uxx −→ Uxx + 2UxX + UXX
Ut = 2 Uτ
For A = 1, D1 = D2 = 1
ux = Ux + UX
ut = 2 Uτ
vx = Vx + VX
vt = 2 Vτ
(6)
Now consider the resulting sequence of equations:
(substituting
U1
U1
U1
D1 0
Bc − 1 A 2
=
0
=
L
+
0() :
0 D2
V1 xx
V1
−Bc −A2
V1
Let
U1
V1
=
f1
f2
R(X, τ )eikc x + c.c.
this is the same equation
is eigenvector of
we saw in the linear
f1
f1
stability analysis with
Mc for σ = 0
=0 ⇒
⇒ Mc
f2
f2
k = kc
(0 eigenvalue)
B = Bc
Mc = M (B = Bc , k = kc )
f1
f2
=
1
−
(Bc −1−kc2 D)
A2
!
(7)
0
1
U2 A
V2
}|
z
{
U2
D1 0
U2
Bc − 1 A 2
=
+
V2 xx
0 D2
V2
−Bc −A2
L@
0(2 )
−B1 U1
−
+B1 V1
D1 0
0 D2
2ik2 RX
5
f1
f2
eikc x + U V, U 2 terms
(8)
Solution to Homogeneous equation: M c
f1
f2
eikc x = 0
Solvability condition:
Right hand side must be orthogonal to solution to homogeneous ADJOINT problem.
Mc∗
g1
g2
g1
g2
Mc∗
=0
1
=
Bc −1−D1 kc2
Bc
=
!
Bc − 1 − D1 kc2
−Bc
A2
−A2 − D2 k12
eigenvector for Mc∗ ,
with eigenvalue 0
(9)
Explicitly
2π/k
Z c
g1
−B1 U1
D1 f1
ikc x
e±ikc x
−e
2ikc RX
·
0=
g2
B1 V1
D2 f2
0
D1 f1
g1
∗
+e−ikc x + 2ikc RX
·
e±ikc x dx
D2 f2
g2
g1
D1 f1
= 0! =⇒ B1 = 0
·
g2
D2 f2
(10)
(also, no t time scale necessary)
This is not a “lucky” combination.
In general for diffusion equations and nonlinear reaction the solvability condition is automatically satisfied!
Note:
2π/k
R c
(U V ) ·
0
2π/k
R c
(U 2 )
·
0
Then, with B1 = 0
U2
V2
g1
g2
e±ikc x
e±ikc x
g1
g2
=



dx = 







dx = 

2π/k
R c
(
) e±3ikc x dx
0
2π/k
R c
(
) e±ikc x dx
0
∗ −ikc x
RX eikc x +
RX
e
d1
R2 e2ikc x + Rc∗2 e−2ikc x
+
d2
h1
+
|R|2
h2
Then
3
0( ) :
U2
D1 0
=−
L
V2 Xx
0 D2
B
2U2 V1
−B2 U1
2U1 U2
+
+ 2A
2U2 V1
+B2 U1
A −2U1 U2
U12 V1
U1
+
+
−U12 V1
V1 τ
U3
V3
6
Solvability condition:
2π/k
Z c
0
2A
+
=
=
2U1 U2
−2U1 U2
=
2U2 V1
2U2 V1
=
U12 V1
−U12 V1
=
U1
V1
↓


RXX 













R








+
other terms
2 


|R| R 




 coefficients depend on specific application
. ↓ &
2 
2
R
=
aR

|R| R 
τ
xx + bR − c|R |R








Ginzburg-Landau
2 



|R| R 











Rτ





+c.c.
Xx
−B2 U
B2 U
B
A
U2
V2
|
all terms
g1
e−ikc x dx
·
with eikc x
g2
{z
}
=
τ
Consider a general operator:
ut = Lu + nonlinear terms
(Think of Lu = D∂x2 u+ other linear terms like λu . . .)
Then, if we look for solutions u = Aeσt eikx we get an equation for σ:
σu = Lu = f (λ, k)u ⇒ σ = f (k, λ)
Then, for σ = 0, we have a curve in the λ − k plane given by f (k, λ) = 0, and in general it will
have a minimum at some point (λ0 , k0 ). We will take λ0 = 0 without loss of generality.
Then, locally (near λ0 , k0 ), we have (λ − λ0 ) = a(k − k0 )2 a quadratic curve.
λ
f (λ, k) = 0
k
(λ0 , k0 )
In terms of σ, this curve is σ = λ − a(k − k 0 )2 = 0 (locally)
∂σ =0
∂k k=k0
and
7
(1)
Now, let’s consider the problem near the critical point λ ∼ 0 + λ 1 + 2 λ2 and look at the sequence
of equations we get for Uj , where u ∼ U1 + U2 .
First, we have U1 = A(X, T )eik0 x + c.c. where X, T are “slow” scales.
Since the leading order equation is just
LU1 = 0
(this is the linear problem which we already solved above).
Then the equation for U2 is
LU2 = f2 = −2D (∂x ∂X ) U1 + |other{zterms} +λ1 U1
(e.g. U12 )
Solvability requires that hf2 , wi = 0 where w is the solution of L∗ w = 0 the adjoint problem.
In general w = Beik0 x or w = Ce−ik0 x so hw, U12 i = 0 (think of ha, bi as
left with
R 2π/k0
0
a · b dx), and we are
h−2D∂x ∂X U1 , wi + hλ1 U1 , wi = 0
If
h−2D∂x ∂X U1 , wi = 0,
then
λ1 = 0
otherwise λ1 6= 0
Claim: h−2D∂x ∂X U1 , wi is proportional to
∂σ =0
∂k k=k0
( from (1))
so λ1 = 0, and the correct scaling for T is T = 2 t.
⇒ λ ∼ 2 λ2
To show this, note the following:
Take the inner product of w with the equation for σ
hw, σui = σhw, ui = hw, Lui.
Specifically, Lu = −Dk 2 u+ (other terms not involving k)−u so
∂σ hw,
vi
=
hw,
−2k
D ui = hw, −2k0 Dui
∂k k=k0
k=k0
But,
h−2D∂x ∂X U1 , wi = AX h−2Dik0 U1 , wi
so the claim (2) is shown.
= AX e−σt ih−2Dk0 u, wi
∂σ hw, ui = 0
= AX e−σt i ∂k k=ko
8
(2)