Dynamics of the whale equation
x
) − α1 xy = f1 (x, y)
k1
y
y 0 = r2 y(1 − ) − α2 xy = f2 (x, y)
k2
x0 = r1 x(1 −
Recall:
The intrinsic growth rates are r1 = 0.05, r2 =0.08
In absence of competition, k1 , k2 are maximum sustainable population.
α = competitive rate.
Note: When y= 400,000 =⇒
y 0 = −α1 x · 400, 000 = −(10−8 · 400, 000) x
|
{z
}
decrease rate proportional to x when y=
400,000
One rate we are interested in is the steady-state = no change (x 0 = y 0 = 0).
We solved for this already when α1 = α2 =α.
x
) − α1 y]
k1
y
f2 (x, y) = 0 = y[r2 (1 − ) − α2 x]
k2
f1 (x, y) = 0 = x[r1 (1 −
y
x'=y'=0
y'=0
x'=0
x
We do have a steady-state (equilibrium) where x 0 = y 0 = 0, x > 0, y > 0, also
(0,0)is an equilibrium.
Is there any value of α for which there is no such equilibrium? YES!
37
r1
r2
If we increase α then the points (0, ), ( , 0) on axes move.
α
α
0.08
0.05
< 400, 000 and
< 150, 000
α
α
Let’s graph the vector field:
y
x'>0, y'<0
x'<0, y'<0
y'=0
x'=0
x
x'>0, y'>0
x'<0, y'>0
y'=0
x'=0
Moves to y = 400,000, x= 0
If we start on equilibrium, stay there.
What if we start off on the equilibrium?
Consider the vector field and the equations will tell us how population changes.
Now compare without competition: continuous dynamic vs. discrete dynamics.
Discrete time
(FIGURE)
38
y
)
k
y 0 = ry(1 −
r∆t 2
y = αyn − βyn2
k n
yn+1 = yn + r∆tyn −
For simplicity we take β = 1
yn+1 = αyn − yn2
Predator-Prey Problem - (Foxes - Rabbits)
Let x = the number of foxes and y = the number of rabbits
Rabbits eat grass and foxes depend on rabbits.
ẋt = ax : without predators, the prey population grows without bound
ẏt = −cy : without prey, the predators die off
ẋt = ax − γxy : predator increases with prey encounters
ẏt = −cy + δxy : prey decreases with predator encounters
c
a
The equilibrium solution is: x = 0, y = 0 and x = , y =
δ
γ
Local stability (linear stability)
• At (x0 , y0 ) = (0, 0).
We solve the linear system of ODE’s:
ẋt = ax
ẏt = −cy
The general form of the solution is:
x
y
!
=e
σt
c1
c2
!
Substitute this solution into the system:
!
c
c
σ 1 eσt = eσt 1
c2
c2
a−σ
0
0
−c − σ
!
39
!
!
a 0
0 −c
!
c1 σt
e =0
c2
In order to have a nontrivial solution we must have:
a−σ
0
det
0
−c − σ
!
= 0 =⇒ (a − σ)(−c − σ) = 0
σ is an eigenvalue of the linear system: σ 1,2 = a, −c and next we find the corresponding eigenvectors.
For σ = a
a−a
0
0
−c − a
!
c1
c2
!
=0
c1
c2
!
Thus, an eigenvector corresponding to σ = a is
For σ = −c
a+c
0
0
−c + c
!
c1
c2
!
Thus, an eigenvector corresponding to σ = −c is
=
1
0
!
=0
c1
c2
!
=
0
1
!
So, the homogeneous solutions are:
k1
!
!
0 −ct
1 at
e
e + k2
1
0
where k1 , k2 are arbitrary constants determined from the initial conditions.
In fact we are interested in the behavior of this solution as t → ∞, i.e. the sign of
σ1 , σ 2 .
Note:
x = k1 eat
y = k2 e−ct
40
(0, 0) is ”attracting” in the y direction and ”repelling” in the x direction =⇒ (0, 0)
is a saddle point =⇒ (0, 0) is unstable.
Another view :
x = 0 + X(t)
y = 0 + Y (t)
where X(t), Y (t) are small perturbations. Then
Ẋ = aX − γ(0 + X)(0 + Y )
Ẏ = −cX + δ(0 + X)(0 + Y )
c a
• At (x0 , y0 ) = ( , ).
δ γ
c
+ X(t)
δ
a
y = + Y (t)
γ
x=
Then
c
a
c
Ẋ = a( + X) − γ( + X)( + Y )
δ
δ
γ
a
c
a
Ẏ = −c( + Y ) + δ( + X)( + Y )
γ
δ
γ
41
Next, neglect the nonlinear terms from the previous equations:
−cγ
Y
δ
aδ
Ẏ =
X
γ
Ẋ =
Again we have to look for a solution of the form:
X
Y
!
c1
c2
= eσt
!
Substitute this solution into the system:
σ
!
c1 σt
c
e = eσt 1
c2
c2

−σ

 aδ
γ
!

0

 aδ
γ
−cγ 
!
c1 σt
δ 
e =0

c2
−σ
−cγ 
δ 

0
For nontrivial solution we have to have

−cγ 
−σ
√

2
δ 
det  aδ
 = 0 =⇒ σ + ca = 0 =⇒ σ1,2 = ±i ca
−σ
γ
√
√
Then eσ1,2 t = cos(t ca) ± i sin(t ca). Next we find the corresponding eigenvectors
for σ1,2 .
√
For σ1 = i ca
√
−cγ 
!
−i ca
c


1
δ
 aδ
√  c =0
2
−i ca
γ

√
Thus, an eigenvector corresponding to σ 1 = −i ca is
√
For σ2 = −i ca
√
i ca

 aδ
γ
−cγ 
!
c1
δ 
√  c =0
2
i ca
42
c1
c2
!
=
√ !
iγ c/δ
√
a
!
√ !
c1
−iγ c/δ
√
Thus, an eigenvector corresponding to σ 2 = i ca is
=
a
c2
Therefore the solution is, for a certain choice of constants,
√
X
Y
We know that
!
 √
h √
√ i
γ c/δ ieit ca − ie−it ca
√ i 
= √ h √
a eit
ca
+ e−it
ca
√
√
ca
√
+ e−it ca
= cos t ca
2
√
√
it
ca
√
e
− e−it ca
= sin t ca
2i
eit
Then
√
√
2γ c
X=
cos t ca
δ
√
√
Y = 2 a sin t ca
Y
X
What if there is a finite amount of grass?
x
) − γxy
k
ẏt = −cy + δxy
ẋt = ax(1 −
43
The equilibrium solution is: (x0 , y0 ) = (0, 0), (k, 0),
Local stability (linear stability)
Write:
c a
c
, (1 − )
δ γ
δk
x = x0 + ξ
y = y0 + η
where ξ, η small
Solve the following system
d
dt
ξ
η
!


!
2x0
ξ
)
−
γy
−γx
a(1
−
0
0

=
k
η
δy0
−c + δx0
|
d
dt
{z
A
ξ
η
!


!
a
−
x
−γx
ξ
0
0

= k
η
δy0
0
}
In order to have a nontrivial solution we must have

Thus

a
−
x
−
λ
−γx
a
0
0
 = 0 =⇒ −λ(− x0 − λ) + γδx0 y0 = 0
det  k
k
δy0
−λ
λ=
a
− x0 ±
k
s
a
x0
k
2
2
− 4(γδ)x0 y0
Let’s consider the equilibrium solution (x 0 , y0 ) = (k, 0). Then
−a
−γk
A=
0 −c + δk
!
=0
In order to have a nontrivial solution we must have
A =
−a − λ
−γk
0
−c + δk − λ
!
= 0 =⇒ (−a − λ)(−c + δk − λ) = 0
=⇒ λ = −a, λ = −c + δk
If k >
c
=⇒ δk > c =⇒ solution unstable.
δ
44
(FIGURE)
General Price/ Quantity Produced model
Pt = f (P, Q) = a1 P + b1 Q + c1 P 2 + d1 Q2 + f1 P 3 + e1 Q3 + · · ·
Qt = g(P, Q) = a2 P + b2 Q + c2 P 2 + d2 Q2 + f2 P 3 + e2 Q3 + · · ·
We could find equilibrium solutions P 0 , Q0 such that f (P0 , Q0 ) = 0, g(P0 , Q0 ) = 0
and rewrite equations as:
(P − P0 )t = A1 (P − P0 ) + B1 (Q − Q0 ) + C1 (Q − Q0 )2 + D1 (P − P0 )2 +
F1 (P − P0 )(Q − Q0 ) + G1 (P − P0 )3 + H1 (Q − Q0 )3
(Q − Q0 )t = A2 (P − P0 ) + B2 (Q − Q0 ) + C2 (Q − Q0 )2 + D2 (P − P0 )2 +
F2 (P − P0 )(Q − Q0 ) + G2 (P − P0 )3 + H2 (Q − Q0 )3
Then we could write the deviation from the equilibrium: p = P −P 0 and q = Q−Q0 .
Therefore we can rewrite the above equations:
pt = A1 p + B1 q + C1 q 2 + D1 p2 + F1 pq + G1 p3 + H1 q 3
qt = A2 p + B2 q + C2 q 2 + D2 p2 + F2 pq + G2 p3 + H2 q 3
Example
ṗ = −p3 + αp − q
q̇ = p
We find the equilibrium solution p0 , q0 by solving the system ṗ = 0, q̇ = 0. From
here p0 = 0,q0 = 0.
Let’s graph the vector field −→ graph ṗ and q̇. When ṗ = 0 =⇒ q = −p 3 + αp
45
Local analysis
• If α = 0
Linearize:
ṗ = −q
q̇ = p
Equivalently in the matrix form:
ṗ
q̇
!
=
0 −1
1 0
!
p
q
!
0 −1
where A =
1 0
!
Then, we find the eigenvalues by solving
the following equation :
!
−σ −1
det(A − σI) = 0 =⇒ det
= 0 =⇒ σ = ±i. Note that we have pure
1 −σ
imaginary eigenvalues , so the eigenvectors will be
(FIGURE)
• If α < 0
46
p
q
!
cos t
=
sin t
!
Linearize:
ṗ = αp − q
q̇ = p
Equivalently in the matrix form:
ṗ
q̇
!
=
α −1
1 0
!
!
p
where A =
q
α −1
1 0
!
Then, we find the eigenvalues by solving
! the following equation :
α − σ −1
det(A−σI) = 0 =⇒ det
= 0 =⇒ (α−σ)(−σ)+1 = −σα+σ 2 +1 = 0.
1
−σ
So the solutions of this equation are:
√
α ± α2 − 4
σ=
2
Then
√
|α| ± Re( α2 − 4)
Re(σ) =
2
2
2
If α − 4 > 0 =⇒ α < −2, then α − 4 < α2 =⇒
Re(σ1 ) = −|α| + ( something less than α) < 0
Re(σ1 ) = −|α| − ( something less than α) < 0
=⇒ stable solution
If α2 − 4 < 0 =⇒ the eigenvalues are complex numbers:
√
−|α| ± i 4 − α2
σ=
2
The difference between these two cases
2
Let’s reconsider the case when
√ α > 4 and for simplicity take α = −3. So the
−3 ± 5
eigenvalues are: σ1,2 =
2
The eigenvectors for σ1 are given by:
α − σ1 −1
1
−σ1
47
!
c1
c2
!
=0
By solving this system we get the eigenvector for σ 1
c1
c2
!
=
1
α − σ1
!
Similarly we find the eigenvectors for σ 2
c1
c2
!
σ2
=
1
!
√
5
3
< 0. Therefore
So, for α = −3 we have α − σ1 = −3 + −
2
2
√ 

! 
!  3
1
5
√ 
σ2
1
= 3
= − 2 − 2  and
5
α
−
σ
1
1
− −
1
2
2
Note: the slope of ṗ = 0 near zero gives q ≈ αp
(FIGURE)
It moves ”quickly” along eigenvector into equilibrium, and away from eigenvector,
follows the vector field.
Now let’s reconsider the case √
when α 2 < 4 and for simplicity take α = −1. So, the
−1 ± 3
. The corresponding eigenvectors are:
eigenvalues are: σ1,2 =
2
√ 
!  1
3
c1
−
+
i

=
2
2  for σ1
c2
1
√ 
!  1
3
c1
= − 2 − i 2  for σ2
c2
1
48
q
spiral in towards
equilibrium point
p
• If α > 0
Same analysis as in the previous case, will give us the eigenvalues
√
α ± α2 − 4
σ=
2
Here Re(σ1,2 ) > 0
If α2 < 4
√
4 − α2
α
=⇒ the solution is unstable.
=⇒ σ = ± i
2
2
If α2 > 4
The eigenvalues are real positive. Therefore the eigenvectors are:
√


! 
! 
1
α
α2 − 4
√
1
σ2
 and
= 2 −
= α
α2 − 4 
2
1
α
−
σ
1
+
1
2
2
(FIGURE)
49
Note: for ṗ = 0 =⇒ q = αp − p3 . On this curve the local maximum, minimum is
α
p2 =
3
What if p and q grow to be large?
ṗ < 0 for q > αp − p3 , q could continue to grow until p < 0.
There is no other equilibrium than (0, 0) =⇒ limit cycle.
Examples
• Pendulum (oscillator)
θ̈ + k sin θ = 0
θ3
θ̈ + k θ −
3!
!
=0
• Van der Pol
ẍ + µ(x2 − 1)x + x = 0
• Bifurcation
Contrast cubic vs. quadratic
ẏ = αy − y 3
(FIGURE)
50
ẏ = αy − y 2
y
y'=0
Stable
S
S
Stable
y
y
y'=
- y2
Our Approach for continuous dynamical systems
ẋ = f (x, y)
ẏ = g(x, y)
−→ Find equilibrium points : (x0 , y0 ) by solving the system: f (x0 , y0 )=0, g(x0 , y0 )=0
−→ Linearize about x0 , y0 (local analysis)
x = x0 + X , y = y0 + Y where X, Y are small.
So we have the following linear system:
Ẋ = a11 X + a12 Y
Ẏ = a21 X + a22 Y
Or equivalently in the matrix form:
Ẋ
Ẏ
!
!
X
=B
where B =
Y
a11 a12
a21 a22
The solution of this system is:
X
Y
!
= k1
!
!
c 1 r1 t
d
e + k 2 1 e r2 t
c2
d2
51
!
!
c1
d
, 1
c2
d2
where r1 , r2 are the eigenvalues of B and
!
are the eigenvectors of B.
Equivalently
B x̄ = rx̄
c
B 1
c2
!
c
=r 1
c2
!
d
,B 1
d2
!
d
=r 1
d2
!
If r1 , r2 yield solutions that grow (decay) in time , then (x 0 , y0 ) is unstable (stable).
What about discrete systems?
xn+1 = f (xn , yn )
yn+1 = g(xn , yn )
Again determine equilibrium, (x0 , y0 ) and linearize about this point xn+1 = x0 +
Xn+1 , yn+1 = y0 + Yn+1 .
Or equivalently in the matrix form:
Xn+1
Yn+1
!
Xn
=B
Yn
!
If all eigenvalues of B are |r| > 1 =⇒ unstable (expansion)
If all eigenvalues of B are |r| < 1 =⇒ stable (contraction)
We have B x̄ = rx̄ for r eigenvalues and x̄ eigenvector.
First, look at what happens if we iterate, when
Xn
Yn
!
= k1
!
c1
c
+ k2 2
d1
d2
!
and |r1 | < 1 and |r2 | < 1.
Note that
Xn
Yn
!
is a linear combination of eigenvectors.
So
Xn+1
Yn+1
!
c
= k1 B 1
d1
!
c
+ k2 B 2
d2
52
!
= k 1 r1
!
c1
c
+ k 2 r2 2
d1
d2
!
In general
Xn+j
Yn+j
!
=
c1
d1
k1 r1j
!
+
c2
d2
k2 r2j
!
j
j
We know that
! |r1 | < 1 and |r2 | < 1 =⇒ r1 and r2 become smaller. Therefore as
X
j → ∞,
→0
Y
Compare also to the continuous case:


Xt+∆t − Xt
!
Ẋ
X
X


t
=B
=⇒  Y ∆t− Y  = B
t+∆t
t
Y
Yt
Ẏ
∆t
!
!
!
Xt
Xt
Xt+∆t
+ B∆t
=
=⇒
Yt
Yt
Yt+∆t =
!
!
!
Xt+∆t
=⇒
=
Yt+∆t =
!
!
1 0
+ B∆t
0 1
Xt
Yt
!
Thus
Ax̄ = (1 + r∆t)x̄ = r̂discrete x̄
where x̄ is an eigenvector, 1 + r∆t < 0 and r̂ discrete = 1 + rcontinuous ∆t.
Example
Recall the logistic model (the whale problem)
ẏ = r2 y(1 −
y
)
k2
Note that for y = k2 we have a stable equilibrium.
For the discrete model we have
u̇n = αun (1 − un )
The equilibrium solution is given by:
un+1 = un + (α − 1)un − αu2n
For the equilibrium we have to have
(α − 1)un − αu2n = 0 so
α−1
un = 0, un =
α
53
α−1
Linearize system about
:
α
α−1
un+1 =
+ Un+1 where Un+1 is a small deviation.
α
So we have
α−1
α−1
α−1
+ Un+1 = α
+ Un 1 −
− Un
α
α
α
α−1
1
=α
+ Un
− Un
α
α
α−1
=
+ Un − (α − 1)Un − αUn2
α
=⇒ Un+1 = 2Un − αUn = (2 − α)Un ( linearized )
Let A be 2 − α. So 2 − α is an eigenvalue and 1 is an eigenvector.
If |2 − α| > 1 =⇒ periodic, chaotic solution
If |2 − α| < 1 =⇒ stable
Next linearize system about 0: Un+1 = αUn
Discrete logistic model
yn+1 = αyn (1 − yn )
v.s continuous
y
ẏ = ry(1 − )
k
So
y(t)
y(t + ∆t) − y(t)
= ry(t) 1 −
∆t
k
y(t + ∆t) − y(t) = (∆t)ry(t) 1 −
y(t)
k
r∆t
y(t)2
k
r∆t
y(t + ∆t) = (1 + r∆t)y(t) 1 −
y(t)
k(1 + r∆t)
r
yn+1 = (1 + r)yn 1 −
yn for ∆t = 1
k(1 + r)
y(t + ∆t) = y(t)(1 + r∆t) −
54
Taking α = 1 + r we get
yn+1 = αyn 1 −
r
yn = αyn (1 − yn )
kα
Tracing the dynamics
(FIGURES)
The equilibrium solution is given by
yn+1 = yn =⇒ αyn (1 − yn ) − yn = 0 =⇒ y0 = 0 and y0 =
Is it stable ?
Let
r
α−1
=
.
α
r+1
yn+1 = y0 + δn+1
yn = y 0 + δ n
We have
yn+1 = f (yn ) = f (y0 ) + f 0 (y0 )(yn − y0 ) =⇒ y0 + δn+1 = f (y0 ) + f 0 (y0 )(yn − y0 )
y0 = f (y0 )
f (y0 ) = αy0 (1 − y0 )
Thus δn+1 = |f 0 (y0 )|δn .
Next we have
|f 0 (y0 )| = |α − 2αy0 | =

|α|
|α − 2α + 2| = | − α + 2|
for y0 = 0
α−1
for y0 =
.
α
If | − α + 2| < 1 =⇒ 1 < α < 3 =⇒ δn+1 is larger than δn =⇒ y0 = 0 is
α−1
unstable, y0 =
is stable.
α
55
If | − α + 2| > 1 =⇒ α > 3 =⇒ both equilibrium solutions are unstable
If |α| < 1, y0 = 0 is stable, y0 =
α−1
is unstable.
α
The periodic solution has two cycles:
xn+2 = xn
xn+2 = αxn+1 (1 − xn+1 ) = α[αxn (1 − xn )(1 − αxn (1 − xn ))]
and from here
xn = xn+2 = α2 xn (1 − xn ) − α3 x2n (1 − x2n )
xn [1 − α2 xn (1 − xn ) + α3 x2n (1 − x2n )] = 0
xn [1 − α + αxn ][1 + α − α(1 + α)xn + α2 x2n ] = 0
p
(α(1 + α))2 − 4(1 + α)α2
2α2
p
(1 + α) ± (1 + α2 ) − 4(1 + α)
=
2
p 2α
(1 + α) ± (1 + α)(−3 + α)
=
2α2
=⇒ xn =
α(1 + α) ±
Note that xn is real for α > 3. For stability we must have 3 < α < 1 +
√
xn+2 = f 2 (xn ) = xn α2 (1 − xn )[1 − αxn (1 − xn )]
From linearization:
df 2
|x∗ = α2 [(1 − xn )(1 − αxn (1 − xn )) − xn (1 − αxn (1 − xn ))
dx
+ xn (1 − xn )[−α(1 − 2xn )]]|xn =x∗
Consider p and q the roots. Then
xn+1 =
d
f (f (xn )) = f 0 (f (p))f 0 (p) = f 0 (q)f 0 (p) since f (p) = q
dx
We have f 0 (q) = r(1 − 2q) and f 0 (p) = r(1 − 2p). Then
xn+1 = r(1 − 2q)r(1 − 2p) = r 2 [1 − (p + q) + 4pq] = 4 + 2r − r 2
For stability we must have |4 + 2r − r 2 | < 1.
56
6.
Measures of chaos
Lyapunov exponent
x0 + δ0 −→ x(t) + δ(t)
In the discrete case we have x(t) + δn =⇒ |δn | = |δ0 |enλ
If λ > 0 −→ expansion
If λ < 0 −→ contraction ( attraction to stable behavior)
From the above equation
λ=
1
δn
1
f n (x0 + δ0 ) − f n (x0 )
1
· ln
= · ln
= · ln[(f n )0 (x0 )]
n
δ0
n
δ0
n
But
n 0
(f ) (x0 ) =
n−1
Y
f 0 (xi )
i=1
Therefore
λ=
X
1 n−1
·
ln f 0 (xi )
n i=1
If xn → x0 ( stable fixed point), then |f 0 (xi )| < 1 as xn → x0 =⇒ ln f 0 (xi ) < 0
X
1 n−1
ln f 0 (xi ) < 0
λ= ·
n i=1
Thus δn is decreasing with n.
Eigenvalues and eigenvectors for Lorenz
ẋ = −γ(x − y)
ẏ = −y + rx − xz
ż = −bz + xy
Find the equilibrium solutions by solving the system: ẋ = 0, ẏ = 0, ż = 0 Thus
the equilibrium solutions are:
(x0 , y0 , z0 ) = (0, 0, 0)
q
(x0 , y0 , z0 ) = ( b(r − 1),
q
q
b(r − 1), r − 1)
q
(x0 , y0 , z0 ) = (− b(r − 1), − b(r − 1), r − 1)
57
Let


Find the eigenvalues
−γ γ
0


A =  r −1 0 
0
0 −b
det(A − σI) = 0
−γ − σ
r
0
Equivalently
γ
0
−1 − σ
0
0
−b − σ
=0
(−b − σ) [(−γ − σ)(−1 − σ) − rγ] = 0
h
i
=⇒ (−b − σ) σ 2 + (1 + γ)σ + 1 − rγ = 0
The eigenvalues are
σ1 = −b and σ2,3 =
−(γ + 1) ±
p
(γ + 1)2 − 4(1 − rγ)
2
If 1 − rγ < 0, then (γ + 1)2 − 4(1 − rγ) > (γ + 1)2 =⇒ equilibrium solution unstable.
Problem - discrete example
Consider the spread of a disease through a population of 100,000. 30% of the
population is immune. Susceptibles become infected by intersection with those
infected with the disease. Those that are infected can spread the disease for 2 weeks
after they become infected. Once a person had the disease they cannot get it again.
No one dies from the disease.
Determine a model for susceptibles, infected, and immune.
How long does the disease last?
At what time is the number of people newly infected the largest?
I(1) = 18 (week one, 18 newly infected)
I(2) = 40 (week two, 18 newly infected)
Let S(t) = the number of susceptibles at time t
J(t) = the number of immune at time t
I(t) = the number of new infected at time t
58
J(1) = 30,000
The change in the number of susceptibles is
S(k + 1) = S(k) − αS(k) [I(k) + I(k − 1) + I(k − 2)]
The disease is contagious for two weeks.
The changes in the number of infected and immune are:
I(k + 1) = αS(k) [I(k) + I(k − 1) + I(k − 2)]
J(k + 1) = J(k) + αS(k)I(k)
The initial population is 100,000
How long does the disease last?
The time until S = 0. Now consider equivalent vector fields:
S(k + 1) − S(k) = ∆S < 0
I(k + 1) − I(k) = αS(k) [I(k) + I(k − 1) + I(k − 2)]
If we just had αS(k)I(k) ( not contagious for 2 weeks) then ∆I(k) = (αS(k)−1)I(k).
If αS(k) − 1 < 0 =⇒ ∆I(k) < 0
S (k)
S (k) = 1
I (k)
I (k)
S (k)
k
k
Cubic nonlinearities
Consider
ẏ = y − y 3
59
If y 1 =⇒ y grows (y y 3 )
If y large =⇒ y decays (y y 3 )
Stability of periodic behavior
Divergence of trajectories
Convergence:
Start at an initial point x0 , follow the dynamics =⇒ x(t) another initial point
x0 + δ0 =⇒ later x(t) + δ(t)
lim λ = lim
t→0
Obvious: ẋ = αx =⇒ x = x0
eαt
t→0
log δ(t)
δ0
t
=⇒ x(t) + δ(t) = (x0 + δ0 )eαt
(FIGURES)
Box dimension
dbox = lim
→0
ln N ()
ln 1
L
For a curve of length L the equation N () =
is the number of segments of length
to cover L.
A
The equation N () = 2 is the number of boxes with side to cover A.
• Expensive way
Cover a region with boxes of side , count the number of boxes that are occupied,
let → 0
• Better way
Run trajectories for long time:
60
(FIGURE)
Count the number Nx () of points in the ball. This is a point wise dimension: as increases Nx ???d
Average Nx over many x =⇒ C(x)???d
Logistic map
(FIGURE)
61