PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 1 of 30
Problem Sheet 1 – Non Linearity, Chaos and Complexity Solutions
Sheet 1 Question 1
(i) Particle motion in B field
m
dv
 qv  B
dt
Normalise v * 
dr
v
dt
v
, t  t *T
v0
r  r*L
B  B* B0
sub in
m
dv* v0
*
dt T
 q v0 v*  B* B0
1
1
 qB 
which is normalised if T   0  

 m 
dv*
qB
 T . 0 v *  B*
*
dt
m
also
dr* L
 v*v0 ie:
dt * T
v0 
so
L
T
L  v0T 
v0

solving the equations yields circular motion about B with frequency  , radius L.
Frequency is independent of velocity (particle energy), whereas gryroradius (L) depends on
velocity.
(ii)
Wave equation (ID here)
1  2  2
 2
c 2 t 2
x
Normalise:
1  2 *  0  2 * 0

c 2 t *2 T 2
x*2 L2
which is normalised (dimensionless) if
 2 *  2 *

t *2
x 2
L
 c.
T
Therefore, c is characteristic velocity of all structures regardless of length scale and is
independent of amplitude  . Solutions are of the form   f  x  ct   g  x  ct  .
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
(iii)
Page 2 of 30
Conservation of quantity Q with number density n
  nQ 
 .  n Q v  ,
t
where Q is carried by "particles" of density n.
Normalise
  n*Q *  1 1
1 1 L
Q0  .  n*Q *v* 
Q0
*
3
t
T L
L L3 T
then:
ie
v* 
v
v
 L
v0  T 
 * *
n Q   * .  n*Q* v*  .
* 
t
There is no characteristic scale if v0 
and timescales are conserved.
L
equation just specifies that structures on all length
T
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 3 of 30
Sheet 1 Question 2
F  F0  F1M  F2 M 2  F3 M 3  F4 M 4
can always be written as
2
3
4
F  F0 ' F2 '  M  M 0   F3 '  M  M 0   F4 '  M  M 0 
since both are general polynormals up to degree 4 then M  M  M 0 is the required
transformation.
(i)
For symmetry F3  0 .
We then have (dropping 's)
F  M   F0   T  Tc  M 2   M 4
extrema
2
F
 2 T  Tc  M  4  M 3  2 M  T  Tc   2  M 
M
ie:
at M  0 or M 2  
Tc  T  .
2
But M is real so:
 Tc  T 
is an extreme for T  Tc
2
M 
look for minima
2 F
 2 T  Tc   12 M 2 .
2
M
M  0:
M 
min for T  Tc
max for T  Tc .
  Tc  T 
2
6   Tc  T 
2F
 2  T  Tc   12 
 4  T  Tc 
2
M
2
min for T  Tc
pitchfork bifurcation at T  Tc
max for T  Tc
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
M
M 
Page 4 of 30
 Tc  T 
2
M 0
T
Tc
F(M)
F
T  Tc
T  Tc


M
M
As we go from T  Tc to T  Tc system "falls" into one of the potential walls – which one is
determined by fluctuations at T  Tc .
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
(ii)
Page 5 of 30
Asymmetric, now F3    0
F
 2 T  Tc  M  3 M 2  4  M 3
M
extrema now
M  0, M 
F
 0  M 2 T  Tc   3 M  4  M 2 
M
 9
3 
2
 4.2 T  Tc  .4  
2.4 
Two real values of M when
9 2  32 T  Tc 
M
write M as
3  3  2   c2
8
.
Consider
2 F
 2 T  Tc   6 M  12  M 2
M 2
M 0
For
is min for
M 0
extrema
T  Tc .
given
by
2 T  Tc   3 M  4  M 2  0
 F
 3 M  8 M 2 ,
2
M
2
or
2F
 M 3  2   c2
2
M
Then in addition to M  0 solution
 2   c2
2 real M  0 roots,
 2   c2
M
3
8
one max, one min
32
T  Tc 
9
 T  Tc .
 c2 
 2   c2 - M imaginary
no max/min.
Also at   0
3  3
M
8
2
c
T  Tc
ie:
M 0
6
M
8
(a)
(b)
which
gives
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 6 of 30
(b)
2F
is    ve root hence
 0 is a min
M 2
(a)
is inflexion. Finally, for  c2  0 2 real roots, both min and M  0 is max
graphically
 c2
2
0
 
2
c
 ve root
M
2
T  Tc
M 0
minima
 ve root
T  Tc
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 7 of 30
F
T  Tc
 c2   2

o
M
T  Tc
F
 c2   2

o
M
F
T  Tc
c  0

o
M
F
T  Tc

o
M
 – going from T  Tc
o – going from T  Tc
}
}
Now fluctuations are unimportant.
hysteresis
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
iii) Van der Vaal
Expand for bm  1
using
2
3


 bM   bM 
ln 1  bm    bM  

.....

 

 2   3 


Substitute into F
2
3
3
4
T
aM 2
2
 bM   bM 
 bM   bM  
F    bM  
 
   bM   
 
   MT 
b 
2
 2   3 
 2   3  
b2
TM 4
 bT a 
M2
   M 3 T  b3
6
12
 2 2
then
Tc 
 T  Tc  
a
.
b
bT  a b 
a
 T  
2
2
b
Page 8 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 9 of 30
Sheet 1 Question 3
dq
 sin q
dt
(i)
q  n
sin q  0
fixed points
n integer
linearize about fixed points
q t   q   q
d q
 sin  q   q   sin q cos  q  cos q sin  q  0
dt
sin  q   q,cos  q  0 as  q is small
then
s  ve
s  ve
d q
N
  1  q
dt
solution is of form  q   q0 e st
for n even
for n odd
–
–
unstable
stable
Phase plane analysis
stable
 unstable
sin q

flow arrows
dq
 ve
dt
dq
for
 ve
dt

 ve q for
q increases with time
ve
q decreases with time
q
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
ii)
Page 10 of 30
dq
  q   q2
dt
fixed points
q  q2  0
q    q   0
q  0 or
ie:
q

.

Stability
q t   q   q t 
Sub in
d
2
 q     q   q     q   q 
dt
  q   q 2   q   2  q   0   q 2 
but
q  q  0
neglect as small
2
d  q 
  q   2  q  ,
dt
then, assuming that
 q   q0 e st
So,
we will have
s  ve for   2  q  0
s  ve for   2  q  0 .
Take
,   0
then
q  0 is s  ve , ie: unstable (repellor)
q

is s  ve , ie: stable (attractor)

Phase plane – sketch
dq
vz q
dt
 q   q2
0


q
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Problem Sheet 2 – Non Linearity, Chaos and Complexity Solutions
Sheet 2 Question 1.
i)
Undamped oscillator
d 2x
  2 sin x .
2
dt

Can integrate this once
dx
dt
d 2 x dx
dx

  2 sin x
2
dt dt
dt
2

1  dx 
2
    cos x  E  constant.
2  dt 
To obtain the dynamics – obtain fixed points, phase plane, etc.
first write as two coupled first order DE
dx
y
dt
dy
  2 sin x
dt
fixed points y  0, sin x  0 or x  n .
Stability
Linearize
then
use
y  y  y
x  x x
 y
d x
y
dt
d y
  2 sin  x   x 
dt
  2 sin  n   x 
sin  A  B   sin A cos B  cos A sin B
sin  n   x   sin n cos  x  cos n sin  x
0
cos  n    1
so
n
d x
y
dt
and sin  x   x
since  x small
d y
n
  2  1  x .
dt
Sufficiently simple to go direct to second order DE
Page 11 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 12 of 30
d 2 x
N
  2  1  x for which we know solutions of form  x  Aet  Be  t .
2
dt
ie:
Then n even
d 2 x
  2 x
dt 2
 x  Aeit  Be  it ,
n odd
d 2 x
  2 x
2
dt
 x  Aet  Be t .
So, n even are centre fixed points
x
d x
 i Aeit  i Be t
dt
is oscillatory and  y 
i
recall i  e 2 and i  e
 y   Ae
So,
- out of phase

i wt  2

 i
2
(complex numbers x  iy  rei )
  Be
A  0, B  0

y

with  x
2
t
n odd

 i wt  2
 x  Ae  Be
x
 t
 y   Aet   Be t
A  0,
B0
Saddle point
A  0, B  0
Separatrix has lines given by
 y  Aet
t


x
Aet
 y  Be t
t  

  .
x
Be t
Topology: constant of the motion defines the phase plane orbits: and
E
A, B  0
y2
  2 cos x has symmetry in y and x
2
Phase plane: see lecture notes and handouts for sketch.
Separatrix has x    cos x  1 when y  0 , Ec   2 on the separatrix.
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
ii)
Page 13 of 30
Damped oscillator
d 2x
dx

  2 sin x  0
2
dt
dt
Now we will have first order DE:
dx
y
dt
dy
  2 sin x   y .
dt
Fixed point
y  0,  2 sin x  0 ,
ie: as undamped case
y  0,
x  n .
Stability analysis
y  y
So
x  x  x
d x
y
dt
d y
n
    y   2  1  x (as before – same identities).
dt
Now more complicated – solve using general formula as in lectures (given in detail here).
We write
 x 
x   
 y 
then pair of equations are just
d x
 J  x
dt
a b
J

c d
where we use notation
d x
 axb y
dt
d y
cxd y
dt
We then have solutions of the form
 x  C1 e St u   C2 e St u 
1
 0
J 2
   1 n   


PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
where the eigenvalues s are solutions of
as
b
0
c
d s
ie:
0   a  s  d  s   bc  s 2  s  a  d   ad  bc
thus
s
1
a  d  
2
here, this is s 
a  d 
2
 4  ad  bc 
1
n
   2  4  2  1
2
.
Two cases:
n odd
n even
1
   2  4 2
2
1
s     2  4 2
2
s 
n odd:
s are real, distinct.
1 
4 2
s       1  2
2 




for   ve or  ve
s are real and of opposite sign – saddle points (as before).
n even:
s may be complex
s 
1 
4 2
    1  2
2 




complex if 4 2   2 otherwise real.
For
  0  decay to stable fixed point
  0  growth – unstable fixed point
If 4 2   2 these are spiral.
Note that if   0 we have
s  
s  i
n odd – saddle and
n even- circle fixed points
So, essentially here, circle points  spiral fixed points for 4 2   2 .
Page 14 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 15 of 30
Topology
Look for symmetries in original DE.
d 2x
dx

  2 sin x  0
2
dt
dt
2
d x
dx
x   x  2   1 
  2 sin x  1  0
dt
dt
Same equation 
t  t
x   x is this symmetry by reflection? Check what happens to y (below).
d 2x
dx
 1 2   1    2 sin x  0
dt
dt
2
t  t is    ,
ie: damping and increasing t  growth and decreasing t
Sufficient to sketch one of these and note that
dx
y
so x   x gives y   y rotational symmetry.
dt
See course handout for sketch
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Sheet 2 Question 2
Lotka-Volterra
In our original notation
dx
    y  x
dt
dy
     x  y
dt
   y  x  0
Fixed points
    x  y  0
ie: x  0, y  0
x



y  0, or x 

x  0 or y 


, y .


Stability – linearise
x  x  x
y  y  y
d x
   x   x     y   y  x   x 
dt
  x   y x      y   x   x y    x y
0
d x
     y   x   x y
dt
d y
   y   y     x   x  y   y 
dt
  y   x y   y     x    x   y     x y
0
d y
     x   y   y x
dt
again – can use formula but shown in full here: write in the form
then in notation of notes
 a b      y 
J

c d    y
 x
d
 x  J  x
dt

  x    
Page 16 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
with eigenvalues
1
a  d  
2
s 
a  d 
2
 4  ad  bc 
Consider two fixed points

J
0
x  0, y  0
s 
1
   
2
   
2
0
 
 4   
 2  2   2  4      
s 
1
          
2
s  
ie:
s  
saddle point.
Consider fixed point
x


y


 
 


0 


 0
J
 
 
S 
1 
      
 0  4 
 

2 
      
  
ie: wholly imaginary – centre fixed point.
Topology: no t symmetry since
t  t


dx
    y  x
dt
dy
     x  y
dt
2
Page 17 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 18 of 30
Similarly, no symmetries in x  y except change of sign in  ,  ,  ,  – unrealistic.
Phase plane:
y

x
C   ln R   R    F   ln F 
dC  dR
dR
dF
1 dF


F

dt R dt
dt
dt
F dt
     F    R       F    R 
 0.
Hence C is a constant and different values of C specify trajectories (closed) about the centre fixed
point.
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Sheet 2 Question 3
Proof of existence of a limit cycle:
dx
dy
 x  y  x  x2  2 y 2  ,  x  y  y  x2  y2 
dt
dt
given
convert to plane polar coordinates r ,  use
x  r cos 
y  r sin 
dx
dy
dr
dy
dx
d
y
r
x y
 r2
dt
dt
dt
dt
dt
dt
and
x
then
r2
d
 x  x  y  y ( x 2  y 2 )  y  x  y  x( x 2  2 y 2 )




dt
 x 2  y 2  xy 3  r 2  r 4 cos  sin 3 
r
dr
 x  x  y  x ( x 2  2 y 2 )   y  x  y  y ( x 2  y 2 ) 
dt
 x2  y 2  x4  3 y2 x2  y4
 x 2  y 2  ( x 2  y 2 )2  x 2 y 2
 r 2  r 4  r 4 cos2  sin 2  .
Identity:
sin  A  B   sin A cos B  cos A sin B
sin 2 A  2sin A cos A
d
1
 r 2  r 4 sin 2  sin 2
dt
2
dr
 1

r  r 2  r 4  1  sin 2 2 
dt
 4

r2
Giving
now
r
dr
 1

 r 2  r 4  1  sin 2 2   r 2 1  r 2 B 
dt
 4

B
Bracket B is bounded 1,
5
4

Page 19 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 20 of 30
B
5
4
hence
dr
0
dt
dr
r 0
0
dt
for any 
r 
1

Minimum value of B  1 has
5
dr
4
has
 0 for r 
4
dt
5
Maximum
B
If
r  1,
If
r
dr
 0 for r  1
dt
dr
0
dt
r 1
4 dr
, 0
5 dt
r
4
5
orbits are attracted into the annulus for any 
and
d
 0 in annulus
dt
therefore, limit cycle.
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 21 of 30
Problem Sheet 3 – Non Linearity, Chaos and Complexity Solutions
Sheet 3 Question 1
Lyapunov exponent.
For a general map xn 1  f  xn 
This has iterates
x1... xn
initial condition x0 so x1  f  x0  ,
x2  f  x1  , etc.
For initially neighbouring points x0  x0   0 , x0 with  0  1 .
After one iterate x1  f  x0   f  x0   0   f  x0    0
df
 x0  ... by Taylor expansion.
dx
Now, two points separated by 1 after one iterate, i.e.
df
x1  x1  1  f  x0   0   f  x0    0  x0   ...
so
dx
1   0 f   x0  to first order in  0 .
Generally, for j th iterate we have x j  x j   j thus  j   j 1 f   x j 1  provided  j  1
Then,
xn  xn   n  xn   n 1 f   xn 1 
 xn   n 2 f   xn 2  f   xn 1 
 xn   0 f   x0  f   x1  .... f   xn 1 
or
xn 1  xn 1 0 f   x0  ... f   xn 
n 1
xn  xn   0  f   x j 
j 0
Now write
f  x j   e
 
ln  f  x j 
and neglecting signs of f  we can write
 n 1

xn  xn   0 exp   ln f   x j  
 j 0

and hence Lyapurov exponent defined as:
  lim
n 
which is a measure of exponential divergence
If   0
1 n 1
 ln f   n j 
n j 0
xn  xn   0e n
then xn  xn for large n, converging – this is attractor (attractive fixed point).
If   0 – exponential divergence for large n. repellor (repulsive fixed point).
0 j n.
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 22 of 30
Sheet 3 Question 2
The map
xn 1 
xn
a
0 xa
xn 1 
1  xn 
1  a 
a  x 1
where 0  a  1 .
x  0 and
Consider fixed points
x in the range  a,1
xn 1
ie: x 
1
1 x
1  a 
x  ax  1  x
or  2  a  x  1
x
xn
a
thus fixed points x  0
x
hence unstable for all 0  a  1 :  xn 1 
1
1
.
2  a
Stability
Linearize
xn  x   xn
sub into
xn 1 
x   xn 1 
 xn 1 
1  xn 
1  a 
1  x   xn 
x   xn 1 
ie:
xn 1  x   xn 1 .
1 a
1  x    xn
1  a  1  a 
 xn
 xn

1  a  a  1
 a  1
n 1
 x0
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 23 of 30
M 2  x
Find "folding points" such that M 2  x   0 or M 2  x   1 .
M 2  x  0
Clearly, M 2  x   0 for M  x   0 or 1 ie: M  x   0
xR  0
xR  a
or
M 2  x  1
Since M  a   1 we seek xR such that M  xR   a .
Two possibilities
0 xa
a  x 1
x
x
a R
xR  a 2
a
a
1 x
1  xR
M  x 
a
x R  1  a 1  a 
1 a
1 a
M  x 
Sketch:
here a 
1
2
thus
a
(try it!).
2
1 a
1  a 1  a  
2
a2 
1
Same topology as symmetric
case (stretching and folding)
just asymmetric.
xR
a
xR
1
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 24 of 30
Lyapunov exponent for M  x 
Fixed point is in the range  a,1
so
so
M  x 
1  x 
1  a 
dM
1

dx 1  a
and
dM
1
dx
 1 
hence   ln 
1  a 
0  a 1
 0
Special cases
exponential divergence
a  0 and a  1
a0
M  x
Now M  x   1  x
1
fixed point x  1  x
1
2
dM
gradient
 1 everywhere.
dx
x
1
2

1
2

1
n
Lyapunov exponent   ln 1  0
  0 is marginally stable –
1
now M  x   x 
2
1
for any
0  x  1, x 
write x0  x  
2
M  x0   1  x    x1
M 2  x0   M  x1   1  1  x     x    x0
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
M 2  x0   x0
hence
M  x
these are period two orbits
1
graphically
x1
x0
1
or by simply calculating M 2  x   1  1  x   x
M 2  x

1
x0
This
M
x0
2
is
 x  x
a
return
map
1
a=1
M ( x )  x again, a return m
Note that
dM d ( x )

 1 so Lyapunov exponent   ln 1  0 marginally stable
dx
dx
true for both orbits of M  x, a  1 and of M 2  x, a  0  [period 2 orbits of M]
Page 25 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 26 of 30
Sheet 3 Question 3
We have
dg
 g g  eR
dt
dR
 b g   FR
dt
and from Lotka-Volterra equations
dF
    R  F
dt
fast growing grass g  B
then we assume the grass is enslaved to the rabbits –
dg
0
dt
giving
g g  eR  0
g
eR
g
dR eb

R   FR      F  R
dt
g
where  
eb
g
which are the original Lotka-Volterra equations so dynamics of foxes and rabbits are the
same and the grass is enslaved to rabbits.
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Problem Sheet 4 – Non Linearity, Chaos and Complexity Solutions
Sheet 4 Question 1
(a) B  0 case
F  M    T  Tc  M 2   M 4
M  0, M  
minima
Thus, if we normalise M to some M
M*  
M* 
 Tc  T 
2
M
M
 Tc  T 
1  
2  M 2  Tc 
1 
Two dimensionless groups
 Tc
2  M 2
2 
B  B0 case
T
.
Tc
F  M    T  Tc  M 2   M 3   M 4 extrema at M  0 and
M  3  9 2  32 T  Tc 
8
Normalise M to M
M
M
M* 

3
9 2
M 

8 M  8 M

*
2

32 Tc 
T 

1  .
2 
8 M  Tc  
3 dimensionless groups
1 
3
8 M
2 
32 Tc
2
8 M
3 
T
.
Tc
.
Page 27 of 30
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 28 of 30
(b) Microscopic model
Quantity
dimension
what it is
M 
 
1/2
 L  T 
1/2
 M
m
c
Magnetization/spin

 L
Spin separation
L0
 L
box size
t
T 
time step

 M c  T 
B0
 M 
1
average charge in magnetization
due to random fluctuations per
spin
externally applied field
c
1/2

M 

 since Tesla 

1/2


L
T






In absence of B0
N 5
R3
2 groups
With applied B0
N 6
R3
3 groups
These are:
1 

t
m
2 
L0

3 
B0
,
m
so in absence of applied B0 we have  1 and  2 only. With applied B0 we have  3 as well.
Then we can identify

T
t 
m
Tc
 Tc
L
 0
2

2 M

B
3
 0.

8 M m
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 29 of 30
Sheet 4 Question 2
Fireflies
Fly around at random, and each has a "clock" to tell it when to flash
cycle length  c
firefly flashes as t=12 say……
all start at random time  s
flash duration  d
Quantity
diversion
c
T 
cycle length
s
T 
average start time
d
T 
duration
R
 L
interaction radius
Nf

L0
 L
Size of box
t
T 
timestep
1

N
N 9
No of flashes to reset
 L  T 
v
R2
what it is
speed
number of fireflies
7 parameters
(most are trivial)
PX3910 Nonlinearity Chaos Complexity solutions to problems [S. C. Chapman]
Page 30 of 30
There are some 'trivial' and 'non-trivial' parameters here.
Trivial
1 
R
L0
2 
1  1
fireflies all see each other
vt
L0
2 1
fireflies cross box in one timestep
3 
R
vt
3  1
fireflies rush past each other
4 
d
c
4 1
fireflies 'always switched on'
5 
s
c
if
– only relevant if no synchronization –
otherwise system 'forgets' initial phase
c
t

7  d
t
6 
need
 6 , 7  1
to resolve the dynamics
Thus, to realise the 'interesting' dynamics on computer we need
 1  1,
 2  1,
 3  1,
 4  1,
 6,7  1 .
In this case these are 'trivial'.
Non-trivial parameters
For synchronization a firefly must see N f flashes within R – at least 'some of the time'.
Let number of flashes seen with R be 

R2 N  d
L20  c
number within R
fraction of these ‘on’
want   N f for synchronization.
Thus, non-trivial parameters are  1   ,  2  N f and for synchronization   N f .