Multiscale analysis
Example 1
Heat conduction + slow radiation
ut + u = uxx −∞ < x < ∞ t > 0
u(x, 0) = f (x)
Exact
u(x, t) = e−t v(x, t)
⇒
vt = vxx
v=
Z
∞
G(x, ξ, t) f (ξ)dξ
−∞ |
{z
(x−ξ)2
e √ 4t
4πt
−
}
Compare to a regular perturbation expansion, for 1
u0 (x, 0) = f (x)
u0t = u0xx
u1t − u1xx = −u0 u1 (x, 0) = 0
⇒ u1 = −tu0
u2t − u2xx = −u1 u2 (x, 0) = 0
⇒ u2 = −t
2 u1 =
t2
2
u0
Note: to verify the above solutions:
u1xx = −tu0xx
u1t = −tu0t − u0
t
1
t
u2t = − u1t − u1
u2xx = − u1xx
2
2
2
t
1
u2t − u2xx = − (u1t − u1xx ) − u1
2
2
1
t
= − (−u0 ) + tu0 = tu0 = −u1
2
2
etc., so
t2
u0 + . . .
2
What happens as t → ∞, if we only keep a few terms? These terms blow-up as
t → ∞.
If we keep all the terms,
u ∼ u0 + −tu0 + 2
uexact = e−t v(x, t)
2 t2
= (1 − t +
. . .)u0 .
2
But in general we want only a few terms! These terms which grow in time are
secular terms. To avoid such terms, try multiple scales - the presence of t in the
exact solution suggests a new time scale t = τ .
u(t, τ, x), with τ = t is introduced, then the time derivative is u t ⇒ ut + uτ ,
treating τ as independent of t.
⇒ new equation: ut + uτ + u = uxx ,
37
Regular perturbation method:
u ∼ u0 (t, τ, x) + u1 (x, t, τ ) . . .
u0t = u0xx
u(x, 0, 0) = f (x)
u0 = C(τ )v(x, t), C(0) = 1
v(x, t) is the solution of the 1-D heat equation with i.c. f (x)
⇒
u1t − u1xx = −u0τ − u0
u1t − u1xx = −C 0 (τ )v(x, t) − C(τ )v(x, t)
⇒ u1 = t[−u0τ − u0 ] = −tv(x, t)[Ct + C]
Again we have a form of equation which gives a solution which would grow in time
unless C 0 + C = 0 ⇒ C = e−τ
⇒ u0 = −e−τ v(x, t)
⇒ u1t − u1xx = 0
u1 (x, 0) = 0 ⇒ u1 = 0
⇒ un+1 t − un+1 xx = 0, un+1 (x, 0) = 0 ⇒ un+1 = 0
so u = u0 = exact solution!
38
n = 1, 2 . . .
Multiscale Analysis: nonlinear ode’s
Van der Pol (oscillator) equation:
ẋ3
dx
)=0
(ẋ =
)
3c
dt
For 1 this is a weakly nonlinear equation:
If we were to use a regular perturbation expansion,
ẍ + x − (ẋ −
0(1) :
x ∼ x0 + x1 + . . .
ẍ0 + x0 = 0 ⇒ x0 = A cos t + B sin t
ẋ3
(−A sin t + B cos t)3
0() :
ẍ1 + x1 = ẋ0 − 0 = −A sin t + B cos t −
3c
3c
Since the right hand side includes terms with sin t, cos t, the solution for x 1 is
x1 =
C
sin t +
D cos }t
|
{z
homogeneous solution
+ F
t sin t +
Gt cos }t
|
{z
particular solution
Note that this solution grows in time, which is incorrect.
x
15
10
5
5
10
15
20
t
-5
-10
-15
The terms t sin t, t cos t are called “secular” terms. This suggests that the form
we found for x0 is incorrect.
What we want is a solution which behaves as x 0 does in the 0(1) equation, but
which has a different behavior in the 0() equation. In particular, A, B should vary
in the 0() equation. To accomplish this we introduce a “slow” time τ = t, and
look for a solution x(t, τ ). Then we allow x to have a behavior that may change on
2 time scales. We treat these 2 time scales, t and τ = t, as independent variables.
Then ẋ = xt + xτ dτ
dt = xt + xτ
ẍ = xtt + 2xtτ + 2 xτ τ
Then the leading order equation is
0(1) :
0() :
x0tt + x0 = 0
x0 = A(τ ) sin t + B(τ ) cos t
|
{z
x0 =x0 (t,τ )
}
x0 3t
3c
= 2A0 (τ ) cos t − 2B 0 (τ ) sin t − A(τ ) cos t + B(τ ) sin t
1
− [−A3 cos3 t + 3A2 B cos2 t sin t − 3AB 2 sin2 t cos t + B 3 sin3 t]
3c
x1tt + x1 = −2x0τ t + x0t −
39
Now use the trigonometric identities:
cos3 t = 3/4 cos t + 1/4 cos 3t
sin3 t = 3/4 sin t − 1/4 sin 3t
cos t sin2 t = 1/4 cos t − 1/4 cos 3t
sin t cos2 t = 1/4 sin t − 1/4 sin 3t
and we find that
0() :
1 3
1
A (τ ) cos t + AB 2 cos t
4c
4c
1
1
−2B 0 (τ ) sin t + B(τ ) sin t − A2 (τ )B(τ ) sin t − B 3 sin t + ( ) cos 3t + (
4c
4c
x1tt + x1 = 2A0 (τ ) cos t − A(τ ) cos t +
To avoid secular terms of the form t sin t, t cos t in the particular solution for x 1 ,
we must set the coefficients of cos t and sin t = 0. That is;
B
1
A(A2 + B 2 )
8c
1
B 0 (τ ) = 1/2B − (A2 + B 2 )B
8c
A0 (τ ) = 1/2A −
√
2 c
θ
A
Note that one steady-state solution is A = B = 0;
another is A = R cos θ, B = R sin θ
√
R = 2 c, (c > 0), θ = const
Later, we will conduct a linear stability analysis to determine when we see these
solutions.
The procedure used to avoid secular terms is known as the Fredholm alternative,
or use of a solvability condition. This condition states that in order to find bounded
solutions, the right hand side of an equation must be orthogonal to homogeneous
solutions of the left hand side. Specifically, we have the equation for x 1 on above:
1
1
ẍ1 + x1 = (2A0 (τ ) − A(τ ) + A3 (τ ) + A(τ )B 2 (τ )) cos t
4
4
1 2
1
0
+(−2B (τ ) + B(τ ) − A (τ )B(τ ) − B 3 (τ )) sin t
4
4
+(
) cos 3 +
+
(
) sin 3t
≡ f (τ ) cos t + g(τ ) sin t + h(τ ) cos 3t + j(τ ) sin 3t
The homogeneous solutions are solutions of ẍ 1 + x1 = 0:
i.e. cos t, sin t
In order to find bounded solutions of x 1 (no secular terms) we must have that
the right hand side (f (τ ) cos t + g(τ ) sin t + h(τ ) cos 3t + j(τ ) sin 3t) is orthogonal to
cos t and sin t, i.e.
⇒
Z
2π
(f (τ ) cos t + g(τ ) sin t + h(τ ) cos 3t + j(τ ) sin 3t) cos tdt
0
=
Z
2π
0
cos2 tdtf (τ ) + g(τ ) · 0 + h(τ ) · 0 + j(τ ) · 0 = 0
40
) sin 3t
and
Z
⇒ f (τ ) = 0
2π
(f (τ ) cos t + g(τ ) sin t + h(τ ) cos 3t + j(τ ) sin 3t) sin t dt
0
= f (τ ) · 0 + g(τ )
⇒ g(τ ) = 0
Z
2π
0
sin2 tdt + h(τ ) · 0 + j(τ ) · 0 = 0
The conditions f (τ ) = 0 and g(τ ) = 0 give the equations for A(τ ), B(τ ) above.
Next, we reconsider this problem from the point of view of the
Lindstedt - Poincare method. The idea is to look for a solution with an undetermined frequency, while allowing multiple time scales. Note that this method is
used in expectation of an oscillating solution.
Method: let x = x0 (ωt, τ ), ω ∼ ω0 + ω1 + . . .
τ = t
Then, ẋ0 ⇒ ωx0t , ẍ0 ⇒ ω 2 x0tt , x0t = x0t + x0τ
Substituting in the VanderPol equation:
(ω02 + 2ω0 ω1 + 2 ω1 )x0tt + x0 = [(ω0 + ω1 )x0t − (ω0 + ω1 )3
x0 3t
] − 2ω0 x0tτ + 0(2 )
3c
Then, to leading order, 0(1) : ω02 x0tt + x0 = 0
If we look for 2π-periodic solutions, to leading order, then ω 0 = 1 and x0 =
A(τ ) sin t + B(τ ) cos t. At the next order,
1
1 3
A + AB 2 ) cos t
4c
4c
1
1
+(−2B 0 (τ ) + B − A2 B − B 3 ) sin t − 2ω1 x0tt + (
4c
4c
0() : x1tt + x1 = (2A0 (τ ) − A +
) cos 3t + (
) sin 3t
Again to avoid secular terms of the form t sin t, t cos t in the particular solution
for x, we set the coefficients of cos t, sin t = 0. Then
A0 =
B0 =
1
1
A − A(A2 + B 2 ) − Bω1
2
8c
1
1
B − B(A2 + B 2 ) + ω1 A
2
8c
√
Again, the steady state solutions are A = B = 0, A = 2 c cos θ for ω1 =
√
√
0, |A| = |B| = 2 c, B = 2 c sin θ, θ = const and the oscillatory solution is
A = R cos(ω1 τ + θ), B = R(sin(ω1 τ + θ)),
|
{z
in general there is
a relationship between amplitude
and the frequency correction
}
41
Note that the Lindstedt-Poincare
includes a frequency correction
√
√
⇒ x0 = 2 c cos(ω1 τ + θ) sin t + 2 c sin(ω1 τ + θ) cos t
√
= 2 c sin(t + ω1 t + θ)
x00
Phase plane:
2π-periodic
solution
x0
θ in general is a constant which would
be determined by the initial condismall perturbation tions.
We will be interested in long term
to 2π-periodic
behavior, so set θ = 0 without loss
solution
of generality.
√
Note also if we expand x0 about = 0, i.e. x0 = 2 c sin(t + θ) + 0(), we get
the result from multiscale analysis, which did not use an expansion for the frequency.
Linear stability
Checking the linear stability of these solutions:
For A = B = 0, let
A=0+ρ
B =0+r
where ρ and r are small deviations from zero. Substituting in the equations for A
and B yields
1
ρ0 (τ ) = ρ(τ ) − ω1 r(τ ) + 0(ρ3 ) + 0(ρr 2 )
2
1
0
r (τ ) = r(τ ) + ω1 ρ(τ ) + 0(r 3 ) + 0(rρ2 )
2
Neglecting the terms 0(ρ3 ), 0(ρr 2 ),!0(r 3 )," 0(rρ2 ) as small
compared
to ρ(τ ), r(τ ) (ρ(τ ) #
!
0
1
ρ(τ )
ρ(τ )
−ω1
2
=
1, r(τ ) 1) we find that
1
r(τ )
r(τ )
ω1
2
So
, and
ρ(τ )
r(τ )
c1
c2
!
!
= eστ
c1
c2
!
, where σ is an eigenvalue of the matrix
"
1
2
ω1
−ω1
1
2
#
is an eigenvector.
Since the eigenvalues are the roots of det
"
1
2
−σ
ω1
−ω1
1
2 −σ
#
=0
1
1
⇒ ( − σ) + ω12 = 0, ⇒ σ = ± iω,
2
2
!
!
!
1
c1
ρ(τ )
c1
e( 2 ±iω1 )τ which grows in time, so that
eστ =
then
=
c2
r(τ )
c2
A = B = 0 is unstable (even if ω1 = 0). So we conclude that the zero solution
is unstable.
42
Now let’s look at the linear stability of non-zero solution:
A = R cos ω1 t B = R sin ω1 t
We could try
A = R cos ω1 t + ρ B = R sin ω1 t + r
Equations: (linearize about ρ, r = 0)
1
(R cos ω1 t + ρ)
2
1
[R cos ω1 t + ρ) [(
)2 + (
)2
|
{z
}
8c
these terms are
a bit messy
(R cos ω1 t + ρ)0 =
−
−ω1 (R sin ω1 t + r)
A little “cleaner”
A = (R + ρ) cos(ω1 t + φ)
⇒
B = (R + ρ) sin(ω1 t + φ)
0
ρ cos(ω1 t + φ) + (R + ρ)[sin(ω1 t + φ)(ω1 + φ0 ] =
1
1
[(R + ρ) cos(ω1 t + φ)] − [(R + ρ) cos(ω1 t + φ)][(R + ρ)2 ]
2
8c
−ω1 (R + ρ) sin(ω1 t + φ)
(Keeping linear terms in ρ and φ)
Rτ < 0
Coefficient of cos(ω1 t + φ) :
R
Rτ < 0
Rτ > 0
c
Rτ > 0
,
,
1
1
1
1
ρ =
R−
R3 + ρ − [ρR2 + 2ρR2 ]
2
8c
2
8c
1
1
= ρ − [3ρ]
2
2
√
( using R = 2 c (c > 0)), so ρ decays
0
Rτ < 0
Coefficient of sin(ω1 t + φ) :
−Rφ0 sin ω1 t = 0 ⇒ neutral stability of angle φ
Rτ > 0
Here we choose ω1 = 0, otherwise we
get an additional contribution when solving for x1 , y1 .
This is a typical pitchfork bifurcation, when (ω 1 = 0)
Rewriting the equation for R:
Rt = λR − R3
( rescale time as = t)
we see there is a bifurcation point: (λ
= 0}, R
= 0})
| {z
| {z
λc
bifurcation parameter
λ<0
Rc
equilibrium points
R=0
√
R = 0, ± λ
λ>0
43
Rewrite equation using a perturbation expansion about the critical point (λ c , Rc ),
Rt = f (R, λ) = f (Rc , λc ) + fR (Rc , λc )(R − Rc )
fλλ (Rc , λc )
(λ − λc
2
)2
+ fλ (Rc , λc )(λ − λc )
+ fRλ (Rc , λc )(λ − λc )(R − Rc ) . . .
= fλλ (Rc , λc )
(R − Rc )3
(λ − λc )2
+ fRRR (Rc , λc )
...
2
6
Note: if fR , fλ 6= 0 at the critical pt., which by definition gives f (R c , λc ) = 0, then
R ∼ Rc −
fλ (R1 , λc )
(λ − λc )
fR (Rc , λc )
is the equilibrium solution (locally)
There is 1 solution, unless fR = 0 at Rc , λc In this case we get a different equation:
fλ = 0
e.g. for pitchfork fR = 0
which yields the equations:
fRR = 0
fλλ = 0 fλR = 1
fRRR = −6
|R|
6fRλ
(λ − λc )(R − Rc )
fRRR
p
1
R − Rc = λ − λc
(R − Rc )2 = − (λ − λc )
−1
⇒ (R − Rc )3 = −
λ
Contrast with a saddle node bifurcation
Ṙ = λ − R2 ⇒ fλ = 1
fRR = −2
1
(ẏ = −y) ⇒ (R − Rc )2 = − λ − λc
−1
−|R|
Now there are only 2 solutions (not R = 0) with
different stability. The name “saddle node” comes
from the “augmented system”:
R
Ṙ = λ − R2 ,
√
linearizing about y = 0, R = λ,
λ
The stability analysis shows a saddle node:
√
ρ̇1 = +2 λ ρ1 (increases in 1 direction)
ẏ = −y
√
y = 0 + ρ 2 , R = − λ + ρ1 ,
ρ̇2 = −ρ2 , (decrease in another)
The bifurcation above is a branch of saddle
points!
44
Hopf bifurcation
eigenvalues σ
Imσ
λ<0
xt = λx − wy − x(x2 + y 2 )
yt = λx + wx − y(x2 + y 2 )
σ1
Linearize about (x, y) = (0, 0):
x = 0 + ρ1
y = 0 + ρ2
Reσ
σ2
λ>0
λ=0
ρ1 t
=
ρ2 t
λ −w
w λ
!
ρ1
ρ2
!
⇒
ρ1
ρ2
!
=e
σt
c1
c2
⇒ (λ − σ)2 + w2 = 0 ⇒ σ = λ ± iw
For λ > 0
Re(σ) > 0, solution is oscillatory
λ < 0,
eλ±iwt
↑
45
w 6= 0
ρ 1 , ρ2 → 0
!