Bifurcations
Nonlinear Boundary Value Problem
v 00 + λf (v) = 0, for v(x; λ)
v(0) = v(π) = 0
0<x<π
(f (v) = v + av k )
Assume f (0) = 0, e.g.
v = 0 is a solution ∀λ
v = 0 is the basic solution
Linearize about basic solution
v 00 + λf 0 (0)v = 0
(expand f in a Taylor series about v = 0)
v(0) = v(π) = 0
If λ 6= λn =
n2
f 0 (0)
If λ = λn =
n2
f 0 (0) ,
for n = 1, 2, . . .
v = 0 is solution
∃ nonzero solutions, eigenfunction’s
vn ∼ sin nx
As a general rule, bifurcation points of nonlinear problem are found among e.v.’s
of the linear problem ( linearized about the basic solution)
Nonlinear analysis in the neighborhood of n th eigenvalue
λ = λn0 + λn1 + 2 λn2
(λn0 = λn )
2
v = 0 + v1 + v2
Substituting

(v100 + 2 v200 . . .) + (λn0 + λn1


 0
 f (0)(v1 + v2 )
2
+ λn2 . . .) 

f 00 (0)
 +
(v1 + . . .)2 + . . .

2
|
{z
}
=0
assuming f 00 (0) = 0
O() :
O(2 ) :
Lv1 = v100 + λn0 f 0 (0)v1 = 0 = v100 + n2 v1
Lv2 = −λn1 f 0 (0)v1 − λn0
f 00 (0) 2
v
2 } 1
| {z
by assumption
=0
O(3 )
Lv3 = −λn1 f 0 (0)v2 − λn2 f 0 (0)v1 −
vj (0) = vj (π) = 0
46
∀j
λn0 f 000 (0)v13
3!









0() :
2
O( ) :
v1 = A sin nx
v200 + n2 v2 = −λn1 f 0 (0)A sin nx
Using the Fredholm alternative, we must have the right hand side of the equation
for v2 orthogonal to the homogeneous solution, that is,
Z
π
{λn1 f 0 (0)A sin nx}A sin nxdx = 0 for v2 to exist
0
⇒ λ n1 = 0
(operator is self-adjoint)
f 000 (0) 3
v1 (λn1 = 0)
6
again the Fredholm alternative must hold
Z π
λn f 000 (0) 3
−λn2 f 0 (0)v1 − 0
v1 A sin nxdx = 0
6
0
for v3 to exist
Z
Z π
f 000 (0) 3 π 4
sin nxdx = 0
A
sin2 nxdx + λn0
λn2 f 0 (0)A
6
0
0
Then Lv3 = −λn2 f 0 (0)v1 − λn0
⇒
000
⇒ λ n2
Since
λ n0
Then
λ n2
R
−λn0 f 6(0) A2 0π sin4 nx dx
R
=
f 0 (0) 0π sin2 nxdx
R
n2
n2 f 000 (0)A2 0π sin4 nx
R
= 0
λ n2 = −
f (0)
6(f 0 (0))2 0π sin2 nx
> 0 if f 000 (0) < 0
λn2 < 0 if f 000 (0) > 0
v ∼ A sin nx, λ − λn ∼ 2 λn2
A
A
λn
Depending on
f 000 (0),
λ
we have either
supercritical
λn
or
subcritical
bifurcation
General principle:
If λn has multiplicty 1, it is a bifurcation point with 1 branch
If multiplicity 6= 1, we don’t know # of branches,
If odd multiplicty, bifurcation does occur. If even, we don’t know if bifurcation
occurs.
Definition of We have used a small parameter 1 in this expansion. But this parameter
does not appear in the original problem. How can we define this parameter?
47
λ
One way is to define is in terms of the amplitude of the bifurcating solution
v ∼ v1 + 2 v2 . . .
= (v, v1 ) = (v1 , v1 ) + 2 (v2 , v1 ) . . .
with the conditions
(v1 , v1 ) = 1
(vj , v1 ) = 0, ∀j ≥ 2
2
⇒A =
π
2
⇒
λ n2
v1 =
−n2 f 000 (0) π2
=
6(f 0 (0)2

r
2
sin nx
π


R
 π sin4 nxdx 

 0

R π
2


sin
nxdx
| 0
{z
}
3π π
/2
8
n2 f 000 (0) π2 3
6(f 0 (0))2 4
n2 f 000 (0)
=−
4π(f 0 (0))2
=−
Alternatively, one can define in terms of the distance from the critical value
λ − λn0 , so that
=
q
λ − λ n0
setting λn2 = 1. Then the amplitude A is given in terms of the equation,
R
and v ∼
p
n2 f 000 (0)A2 0π sin4 nx
R
1=−
6(f 0 (0))2 0π sin2 nx
λ − λn0 A sin nx.
Either of these definitions of gives the same result. Note that in either case, is
an artificial parameter introduced for purposes of bookkeeping, and can be written
in terms of the actual parameters of the problem.
Aside
In the above analysis, we have assumed that f 00 (0) = 0, which led to λ1 = 0. If
6= 0, then we have to return to the O( 2 ) equation; in particular, let’s look at
the solvability condition:
f 00 (0)
0
Aλn1 f (0)
Z
π
0
f 00 (0)
sin nxdx + A λn0
2
2
2
Z
π
sin3 nxdx = 0
0
Let’s consider the second integral
Z
π
3
sin nxdx
0
(
= 0 n even
6= 0 n odd
Then for n odd, we must choose A and λn1 in order to satisfy the solvability condition.
⇒ A=
λn1 f 0 (0)
λ n0 f
00 (0)
2
Rπ
0
Rπ
0
48
sin2 nxdx
sin3 nxdx
Note that in this case A is in terms of λ n1 , so that it is appropriate to define
= (λ − λn0 ). Then v ∼ (λ − λn0 )A sin nx, with A given above with λn1 = 1.
Stability
Related to the previous problem:
−ut + uxx + λf (u) = 0
u(0, t) = u(π, t) = 0
f (0) = f 00 (0) = 0
u(x, 0) = H(x)
f 0 (0) > 0
The steady state solution is given previously.
A
several bifurcating sol’s
...
0 λ1
λ2
λ3
λ
We test stablility of the steady state solution(s), using linear stability analysis.
First test the stability of the basic solution u = 0.
Linearize about u = 0
u = 0 + δw, (small perturbation)
⇒
−δwt + δwxx + λf (δw) = 0
−δwt + δwxx + λ[f (0) +δf 0 (0)w +
| {z }
δ 2 00
f (0)w2 . . .] = 0
2
=0
where again f (u) is expanded about basic solution u = 0.
Linearizing:
−wt + wxx + λf 0 (0)w = 0
w = 0 at x = 0, π
Also, we write δH(x) = h(x) so w| t=0 = H(x)
since the set of eigenfunctions
sin nx’s is complete, h(x) can be written
h(x) =
X
w(x, t) =
From the i.c., wn = hn
Substituting into linearized equation
⇒
⇒
hn sin nx
X
wn eσn t sin nx
−σn − n2 + λf 0 (0) = 0
σn = −n2 + λf 0 (0)
"
n2
+λ
= f 0 (0) − 0
f (0)
= f 0 (0) [−λn + λ]
49
#
For f 0 (0) > 0, σn < 0 for λ < λn
v = 0 is stable ⇔ all σn ’s < 0
so v = 0 is stable for λ < λ1
A
...
S
U
λ1
v=0
stable
λ2
U
λ
v=0
unstable
Stability of bifurcating solution
First, linearize about vn
u = vn + 0(2 ) +δw
|
Substituting,
{z
v
}
−δwt + (vn + 0(2 ))xx + δwxx + λ[f (v) + δwf 0 (v)] + nonlinear terms = 0
Since vxx + λf (v) = 0 (steady-state problem)
This leaves
−wt + wxx + λf 0 (v)w = 0
expanding w and σn about ψ
z
}|
{
w = Σhn eσn t (ψ0 + ψ1 + . . .)
σn = σn0 + σn1 + 2 σn2 + . . .
λ = λn0 + λn1 + 2 λn2 + . . .
Substituting
−σn ψ + ψxx + λf 0 (vn + 0(2 ))ψ = 0
f 000 (0)
f 0 (vn + 0(2 )) ∼ f 0 (0) +
(vn + . . .)2
2
To leading order
Lψ0 = −σn0 ψ0 + λn f 0 (0)ψ0 + ψ0 xx = 0
ψ0 ∼ sin jx where j 2 = −σn0 + λn0 f 0 (0)
σn0 = λn0 f 0 (0) − j 2 = n2 − j 2
If n > 1, then for j = 1, σn0 > 0
⇒ for n > j, vn is unstable
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If n = 1 then n2 − j 2 = 0
for j = 1, and < 0 for j > 1. Then we must look
at the next terms, to see if this is stable at n = 1.
A
unstable branches
?
S
λ1
U
?
U
U
λ2 U
...
U
U
λ3 U
U
λ
stability
yet unknown
0() :
0(2 ) :
Lψ1 = σn1 ψ0
⇒ σn1 (ψ0 , ψ1 ) = 0, so σn1 = 0
f 000 (0) 2
Lψ2 = [σn2 − λn0
vn − λn2 f 0 (0)]ψ0
2
Using the Fredholm alternative
(ψ0 , ψ0 ) · [σn2 − λn2 f 0 (0)] −
⇒ σn2 − λn2 f 0 (0) −
λn f 000 (0) 2
(vn ψ0 , ψ0 ) = 0
2
λn f 000 (0) (vn2 ψ0 , ψ0 )
2
(ψ0 , ψ0 )
Since we are only worried about the case n = 1 we need v n2 = A2 sin2 x
(for normalization)
A=
π
2
2
(sin x(sin jx), sin jx)
3/4
=
= 3/2π for j = 1
(sin jx, sin jx)
π/2
3λn0 f 000 (0)
⇒
σn2 = λn2 f 0 (0) +
4π
ψ0 = sin jx
2 000
f (0)
Recall λn2 = − n4πf
0 (0) and λn0 =
(ψ0 , ψ0 ) =
n2
f 0 (0)
2 000
f (0)
(n = 1 in this case)
Substituting ⇒ σn2 = n2πf
0 (0)
000
If f (0) < 0, n = 1, j = 1, the solution is stable in the supercritical case
If f 000 (0) > 0, the solution is unstable in subcritical case
A
S
A
S
λ1
U
U
λ2
U
S
λ
000
U
λ1
U
λ2
U
λ
000
f (0) < 0
f (0) > 0
This means that for f 000 (0) < 0, as t → ∞
51
v → A sin x for λ = λ1 + 2 λ12
q
2
π
Slow time evolution
Now consider λ > λ1
(close to λ1 ). We trace the evolution of arbitrary
initial condition to its eventual steady-state.
For the linear stability analysis, we see that if the initial condition is u| t=0 =
P
h(x) = Σhn sin nx, then we write u = An un , so that
un ∼ eσn t sin nx where σn is the growth or decay rate
σn = f 0 (0)(λ − λn )
Let λ = λ10 + 2 (definition of ), with (λ10 = 1/f 0 (0), λ12 = 1)
σ1 = f 0 (0)2 1
0
σn = f (0)(
( and > 0)
| {z }
0(1) & <0
so for h(x) =
+2 )
λ 1 − λn
for
n≥2
+ h2 sin 2x + h3 sin 3x + . . .
h1 sin x
| {z }
|
this term grows
{z
}
these terms decay
If we try a regular perturbation expansion for the solution, (taking into account
nonlinear terms)
u ∼ u1 + 2 u2 . . .
Lu1 = −u1t + u1xx + u1 = 0
Lu2 = 0
Lu3 =
for f 00 (0) = 0
|f 000 (0)|
(u1 )3
6
|
{z
}
−u1 f 0 (0)
causes secular terms
X
u1 = A1 sin x +
An eσn t sin nx
n=2
From the leading order equation, we see that there is no growth rate except
possibly the 1st term σ1 = 0
σn < 0 for n ≥ 2
2
Previously we saw that h1 sin x grows like e t
This suggests using a 2-time method, with slow time τ = 2 t
Let u ∼ Σj uj
|f 000 (0)|
(u1 )3 − f 0 (0)u1 + u1τ
6X
u1 = A1 (τ ) +
An (τ )eσn t sin nx
so Lu3 =
n=2
A1 (0) = h1
h
so Lu3 = sin x − |f
000 (0)|
6
3
4
|
{z
these decay
An (0) = hn
i
}
A31 − f 0 (0)A1 + A1τ + decay
52
To avoid secular terms, we set the coefficients of sin x = 0
A1τ = −
|f 000 (0)| 3
A1 + f 0 (0)A1
8
dA1
A1 (1 −
|f 000 (0)| 2
8f 0 (0) A1 )
= f 0 (0)dτ
Using partial fractions
√
√
1
1
c
c
1
√
√
+
−
=
|f 000 (0)| 2
A1 2 (1 − cA1 ) 2 (1 + cA1 )
A1 (1 − 8f 0 (0) A1 )
1
c=
|f 000 (0)|
8f 0 (0)
and integrating
Z
dA1
+
A1
Z
ln A1 −
⇒
q
√
√
1
1
ln(1 − cA1 ) − ln(1 + cA1 ) = f 0 (0)τ + K
2
2
A1
1−
√
√
Z
1
c dA1
c dA1
1
√
√
−
= f 0 (0)τ
2 (1 − cA1 )
2 (1 + cA1 )
cA21
= K 1 ef
0 (0)τ
⇒
A21
0
= K12 e2f (0)τ
2
1 − cA1
⇒
A21
K12 e2f
=
1+
(K1 = eK )
0 (0)τ
K12 |f 000 (0)| 2f 0 (0)τ
8f 0 (0) e
We can solve for K1 using initial conditions.
Review:
Recall the choice of the bifurcation parameter: λ = λ 1 + 2 :
1) For λ ∼ λ1 + 2 λ12 with v ∼ A sin x, and A is in terms of λ12 .
2) We also took the slow time to be 2 τ , to get the term Aτ to enter at 0(3 ),
and thus kill secular terms
So we’ve related 2 parameters: a measure of the distance from the critical
bifurcation point and the slow time scale
53