2. Some analytical models of nonlinear physical
systems
(a) Discrete Dynamical Systems:
1. Inverted Double Pendulum: Consider the double
pendulum shown:
P
k1 , k2  linear torsional spring
2
x
A
1
O
P  external (conservative) load
k2
l
k1
l  length of each rod
l
g
m mass of each rod
y
1
Inverted Double Pendulum –
Equation of motion
The equations of motion can be determined by
using Lagrange’s equations:
d T
T V
nc
( )

 Qi ,
dt qi
qi qi
i  1,2,3,......
Here,
• 𝑞𝑖 - are the generalized coordinates,
• T and V- are the respective kinetic and potential
Energies for the system,
𝑛𝑐
• 𝑄𝑖 -- the generalized forces due to nonconservative effects.
2
Inverted Double Pendulum –
Equations of motion
For the double pendulum- kinetic energy:
2
2
T  T1  T2  [(mv G1
 IG112 )  (mv G2
 IG222 )] / 2
T1  (ml2 / 3)12 / 2;
v G2  v A  v G2/ A ;
v A  1k  r OA
r OA  l(cos 1 i  sin 1 j ); v A  1l(  sin 1 i  cos 1 j );
v G2/ A  2 k  r G2/ A ;
r G2/ A  l(cos 2 i  sin 2 j ) / 2;
v G2/ A  2l(  sin 2 i  cos 2 j ) / 2;
 T2  (ml2 / 12)22 / 2  ml2 [12  22 / 4  12 cos( 2  1 ) / 2] / 2
3
Inverted Double Pendulum –
Equations of motion
Potential energy:
V  V1  V2  mglcos 1 / 2  mg[lcos 1  lcos 2 / 2]
[k112  k 2 (2  1 )2 ] / 2
The work done by the external force P in a virtual
displacement from straight vertical position is:
W  P i   r B ;
r B  l(cos 1 i  sin 1 j )  l(cos 2 i  sin 2 j )
 r B  1l(  sin 1 i  cos 1 j )  2l(  sin 2 i  cos 2 j )
 W  P[1lsin 1  2lsin 2 ]
The generalized forces are :
 Q1  Plsin 1;
Q2  Plsin 2 ;
4
Inverted Double Pendulum –
Equations of motion:
Equation for 1:
d T
(
)  ml2 [1 / 3  1  2 cos( 2  1 ) / 4
dt 1
 2 (1  2 )sin(2  1 ) / 4]
T
 ml221 sin(2  1 ) / 4
1
V
 3mglsin 1 / 2  (k1  k 2 )1  k 22
1
 ml2 [41 / 3  2 cos( 2  1 ) / 4  22 sin( 2  1 ) / 4]
 (k1  k 2 )1  k 22  3mglsin 1 / 2  Plsin 1
5
Inverted Double Pendulum –
Equations of motion:
Equation for 2:
d T
(
)  ml2 [2 / 3  1 cos( 2  1 ) / 4
dt 2
 1(1  2 )sin(2  1 ) / 4]
T
 ml221 sin(2  1 ) / 4
2
V
 mglsin 2 / 2  k 21  k 22
1
 ml2 [2 / 3  1 cos( 2  1 ) / 4  12 sin( 2  1 ) / 4]
 k 2 1  k 2 2  mglsin 2 / 2  Plsin 2
6
Inverted Double Pendulum –
Equations of motion:
We now consider a simplified version with k1= k2=k
Let k  k / ml2 , P  Pl / ml2 , M  mgl / 2ml2
Then, the equations are:
Equation for 1:
[41 / 3  2 cos(2  1 ) / 4  22 sin(2  1 ) / 4]
 2k1  k2  (P  3M)sin 1
Equation for 2:
[2 / 3  1 cos(2  1 ) / 4  12 sin(2  1 ) / 4]
 k1  k2  (P  M)sin 2
7
Discrete dynamical systems…….
2. Inverted Double Pendulum with Follower Force:
Consider the same system as in last example,
except that the force P changes direction
depending on the orientation of the body on which
it acts.
P

The force P now acts at
an angle  to the rod AB and
2
B
always maintains this
l
x
direction relative to the rod
A
1 k2
regardless of the position in
space of the system during
l
g
its oscillations.
k1
O
y
8
Inverted Double Pendulum with Follower….–
Equations of motion
Note that the only change is in the effect of the external force P.
The potential and kinetic energy expressions remain the same.
So,
potential energy:
V  V1  V2  mglcos 1 / 2  mg[lcos 1  lcos 2 / 2]
[k112  k 2 (2  1 )2 ] / 2
The work done by the external force P in a virtual
displacement from straight vertical position is:
W  P[cos(2  ) i  sin(2  ) j ]   r B;
r B  l(cos 1 i  sin 1 j )  l(cos 2 i  sin 2 j )
9
Inverted Double Pendulum with Follower–
Equations of motion
The virtual displacement is:
 r B  1l( sin 1 i  cos 1 j )  2l(  sin 2 i  cos 2 j )
So,the virtual work done is 
W  Pl[1 sin(1  2  )  2 sin ]
Thus, the generalized forces are:
Q1  Plsin(1  2  ); Q2  Plsin .
The resulting equations of motion are:
ml2 [41 / 3  2 cos(2  1 ) / 4  22 sin(2  1) / 4]
(k1  k 2 )1  k 22  3mglsin 1 / 2  Plsin(1  2  )
10
Inverted Double Pendulum with Follower–
Equations of motion
and
ml2 [2 / 3  1 cos(2  1) / 4  12 sin(2  1) / 4]
k 21  k 22  mglsin 2 / 2  Plsin 
In the reduced case, with equal springs etc., the
equations are:
[41 / 3  2 cos(2  1 ) / 4  22 sin(2  1 ) / 4]
2k1  k2  P sin(1  2   )  3Msin 1
[2 / 3  1 cos(2  1 ) / 4   sin(2  1 ) / 4]
2
1
k1  k2  PSin  Msin 2
11
Discrete dynamical systems…….
3. Dynamics of a Bouncing Ball: Consider a
ball bouncing above a horizontal table. The table
oscillates vertically in a specified manner (here
we assume harmonic oscillations).
ball
The motion of the ball,
mg
g
during free flight, is
Y(t)
governed by
table
X (t )  A sin(t )
ground
my  mg or y   g
Integrating once gives:
y=y0  g (t  t0 )
12
Dynamics of a Bouncing Ball …….
y=y0 -g(t-t 0 )
Integrating once gives:
1
Integrating again, we get: y  y0  y0 (t  t0 )  g (t  t0 ) 2
2
The position of the ball has to remain above the
table  y(t)  X(t), t  t 0
Finally, we have the law of
ball
interaction between the
mg
g
Y(t)
table and the ball:
table
X (t )  A sin(t )
If we assume simple law of
impact, the relative velocities
before and after impact are related
ground by coefficient of restitution.
13
Dynamics of a Bouncing Ball …….
Thus, we have the relation:
V(t i )-X(t i )=e[X(t i )  U(t i )]
where V(t i ) - velocity of the ball immediately after impact
and U(t i ) - velocity of the ball just before impact
ball
Y(t)
mg
g
table
ground
These relations provide a
complete description of
motion of the ball. Note
that, given an initial
condition, we have to
piece together the motion
in forward time
14
Dynamics of a Bouncing Ball …….
Let us now proceed in a systematic manner. For
nonlinear analysis, it is always advisable to nondimensionalize equations: So we define
Time : t  t / T , whereT  2 
ball
Acceleration units : g 2
mg
g
Y(t)
Veloctiy units : ( g 2)T   g 
table
ground
Pos. units : ( g  )T  2 2 g  2
So, X (t )  (2 2 g  2 ) X ( )
 A sin(t )  A sin(2 )

2 A
 X ( )  sin(2 );  
2
g
15
Dynamics of a Bouncing Ball …….
Ball motion: z(t)= (22g/2)Z(),
dz
22g dZ d
22g dZ 
g dZ
( 2 )
( 2 )
( )
dt

d dt

d 2
 d
d2 z
g d2 Z d
g d2 Z 
g d2 Z
( ) 2
( ) 2
( ) 2
2
dt
 d dt
 d 2
2 d
ball
Y(t)
mg
g
table
Now,
z   g  Z   2 .
Integrating,
Z ( )  Z0  2(   0 ),
ground
Z ( )  Z 0  Z 0 (   0 )  (   0 )2
16
Dynamics of a Bouncing Ball …….
Thus, we have
Z ( )  Z0  2(   0 ),
Ball motion: Z   2;
Z ( )  Z0  Z0 (   0 )  (   0 )2
Table motion:
ball
Y(t)
mg
g
table
X( ) 

sin(2)
2
(1)
(2)
(3)
Initially: ball starts at time 0
when it is in contact with the
table, and just about to
leave 
ground X(  )  Z(  )  Z   sin(2 ) (4)
0
0
0
0
2
17
Dynamics of a Bouncing Ball …….
Ball velocity at =0: dZ
d
0
 Z0  W0  X( 0 )
 Z0  W0   cos(20 )
(5)
(Here W0  0 is relative velocity of the ball, an unknown)
ball
Y(t)
mg
g
table
Next collision at time 1 > 0
when Z()  X()  W()  0
Now, W()  Z0  Z0 (  0 )

(  0 ) 
sin(2)
2
ground Using (3)-(5) 
2
(6)
18
Dynamics of a Bouncing Ball …….

W( ) 
[sin(20 )  sin(2)]
2
 [W0   cos(20 )](   0 )  (  0 )2
Then, the time instant 1 is defined by
ball
Y(t)
mg
g
table

W( 1) 
[sin(20 )  sin(21)]
2
 [W0   cos(20 )](1  0 )
( 1  0 )2  0
(7)
(this is a relation in W’0, 1
and 0, and it depends on )
ground
19
Dynamics of a Bouncing Ball …….
When just about to contact at this time instant 1 the
relative velocity is : W1  W (1 )  Z(1)  X( 1)
dW()

  cos(21)  [W0   cos(20 )]  2(1  0 )
d 
1
On impact, the ball relative velocity changes:
ball
Y(t)
mg
(8)




W1  eW1
g
(e  coefficient of restitution)
or:
table
W1  e[{cos(20 )
 cos(21)}  W0  2(1  0 )]
(9)
ground
20
Dynamics of a Bouncing Ball …….
One can write the equations now in a more compact
form:
and

[sin(2i )  sin(2i1)]
2
 [Wi   cos(2i )](i1  i )  (i1  i )2  0
W ( i1) 
Wi1  e[ {cos(2i )  cos(2i1)}  Wi
ball
Y(t)
(A)
mg
 2(i1  i )]
(B)
g
table
Knowing (I,W’i), equations
(A) and (B) can be used to
compute (I+1,W’i+1), thus
generating the trajectory.
ground
21
(b) Continuous Dynamical Systems
1. Rotating Thermosyphon: Consider a closed
circular tube in a vertical plane. The tube is filled with
a liquid of constant properties, except for variation of
its density with

temperature in
r
cooling
buoyancy and
centrifugal terms, i.e. in
R

α
body forces. One part
of the loop is heated,
heating and the other cooled.
The tube is spun about
the vertical axis.
22
Rotating Thermosyphon…
There are many ways to develop a model for the
system. If the tube radius ‘r’ is much smaller than the
torus radius ‘R’, one can assume that there is
negligible flow in the

radial direction. Another
r
cooling
approach is to average
the velocity and
R
temperature over the

α
tube radius. Then, the
equations for fluid
heating
motion are:
23
Rotating thermosyphon: equations…
 1  (ρV)

0
t R 
(1)
Continuity:
Here, V – average flow velocity at any section ;
 - density of the fluid and (V) is independent of .
Momentum:

2
r
cooling
R

 (ρV) 1 (ρV)
1 p


  g cos 
t
R 
R 
2 w
2



R cos  sin 
(2)
α
r
heating
Here, p – fluid
pressure at a section,
w- shear stress at the
wall
24
Rotating thermosyphon:
Equations…
Energy:
2
 T V (T )  k  T 2
 r C p  
 2 2  r  q( )

 t R   R 
2
(3)
where Cp- specific heat of the fluid, T – mean fluid
temperature at a section,

r
k – thermal conductivity,
cooling
q’ – applied heat source
per unit length.
R

α
Remark: Here viscous
dissipation term is
heating
neglected
25
Rotating thermosyphon:
Equations…
Simplification and nondimensionalization:
Integrating the momentum eqn. (2) along the
2
2
2
2
2

(ρV)
1

(ρV)
1

p
loop
d 
d  
d   g cos  d

0
R
t
0

cooling
R


2

r

0
α
R
 
0

0
2
2 w
d    2 R  cos  sin  d
r
0
Shear stress:
 w  f V 2 / 2
(4)
(5)
Friction factor:
heating f  16 / Re, Re  2 Vr / 
26
Rotating thermosyphon:
Equations…
Simplification:
Introduce the variation of density with temperature
in buoyancy and centrifugal terms, and use
periodicity of variables (eqn. (4))
V
32V  g


2
t
(2r )
2
2
 (T  T ) cos d
r
0
2
2R

  (T  Tr ) cos  sin  d
2
0
(6)
where -kinematic viscosity, Tr-reference temperature
 - coefficient of thermal expansion
27
Rotating thermosyphon:
Equations…
Non-dimensional variables:
32 t

,
2
(2 r )
(2 r ) 2
U
V,
32 R
2048 R 2
T 
,
4
 g (2r )
T  Tr

,
T
q '( )
Ra * r
1
Q ( ) 

( )
2
8 C pT
Pr R 8192
Here Pr  C p / k  Pr andtl number
Ra* 
2  ( 2r )3 q '  C p
k 
2
 Modified Rayleigh number
28
Rotating thermosyphon:
Equations…
Non-dimensional equations:
The resulting momentum and energy equations are:
U
 U 

2
2
  cos d     cos sin  d
0
(7)
0

 2


U
 Q ( )
2



(8)
r 2
where    R / g ,   ( ) / 8 Pr
R
2
2
(
r
/
R
)
Note:  - combination of geometric parameter
and the Prandtl number Pr. Pr > 1 for ordinary
fluids (air, water etc.) and <1 for liquid metals.
29
Rotating thermosyphon: equations…
Solution approach: The solutions to (7) and (8) can
be expressed as:

( , )   [ Bn ( )sin(n ) Cn ( ) cos(n )]
n1
where as, the externally imposed heat flux can be
(9)
represented in a Fourier series as:

Q( )   [ An sin(n ) Dn cos(n )]
n1
(10)
Substituting (9) and (10) in equations (7) and (8), and
collecting the appropriate Fourier coefficients gives an infinite
set of ordinary differential equations In the unknowns U(),
Bn(), and Cn().
30
Rotating thermosyphon: equations…
Solution approach…: It turns out that only five of
these equations are independent-master equations.
The remaining equations are linear equations for the
remaining variable (called “slave variables” and “slave
equations”). Equation (8) gives:

  [ Bn ( ) sin(n ) Cn ( ) cos(n )]
n 0




2
 [ B ( ) sin(n ) C ( ) cos(n )]
n 0
n
n
 2

  [ Bn ( ) sin(n ) Cn ( ) cos(n )]
 U n 0


  [ An sin(n ) Dn cos(n )]
n1
31
Rotating thermosyphon: equations…
Solution approach…: Collecting terms of different n’s give:
n  0 : C0   C0
n  1: B1   B1  UC1  A1
(sin  )
C1   C1  UB1  D1
(cos  )
n  2 : B2  4 B2  2UC2  A2
(sin 2 )
C2  4 C2  2UB2  D2
(cos 2 )
n  3:
B3  9 B3  3UC3  A3 (sin 3 )
C3  9 C3  3UB3  D3
n  p:
(cos 3 )
B p   p 2  B p  pUC p  Ap
C p   p 2  C p  pUB p  D p
p  3, 4,5, 6,...........
(sin p )
(cos p )
(11)
32
Rotating thermosyphon:
Equations…
Solution approach…: Now, considering equation (7)
we get:
2 
U
 U    [ Bn ( ) sin(n ) Cn ( ) cos( n )]cos  d

0 n 1
2

0

 [ B ( ) sin(n ) C ( ) cos(n )]cos sin  d
n 1
n
n
Evaluating the two integral terms on the right-hand
side, it is clear that only coefficients of
cos( ) and sin(2 ) will survive. Thus, we get
U  U  C1  B2
(12)
Equations (11) and (12) govern the dynamics.
33
(b) Continuous Dynamical Systems
2. Buckling of Elastic Columns: Consider a thin
beam that is initially straight. O xyz is coordinate
system with x-y plane coinciding with undisturbed
neutral axis of the beam. Let EI is the bending stiffness
V
X
34
Buckling of elastic…..
V(s) – vertical displacement of the centroidal axis,
X – distance measured along the centroidal axis
from left end.
Define: x  X / L; u  V / L;   L2P / 2EI;   L3K / 3EI;
Then, the strain energy of the system is:

1
u2
V
(u, ,  ) 
(
)dx
2
2 0 1  u

X

  (1 1  u2 )dx
0
1 2
1 2
 u (0)  u ( )
2
2
35
(b) Continuous Dynamical Systems
3. Thin rectangular plates: Consider a thin plate
that is initially flat. Oxyz is coordinate system with x-y
plane coinciding with undisturbed middle surface of
the plate. Let h – plate thickness. The equations of
motion for the plate, for
y, v
moderately large
displacements – von Karman
z, w
equations.
In here, we give a short
review of the derivation of
a
x, u these equations.
36
Thin rectangular plates…..
Consider a differential plate element: The equations
2

N

N

u
xy
x
in the three directions are:

 h
(1)
Ny
F
z
Nxy
Nx
Nxy
Ny
y
x
N xy
x
y
N y
t 2
2v

 h 2
x
y
t
(2)
2
 2 M xy
Nx  2 M x  M y

2
2
2
x y
x
y

w

w

(N x
)  (N y
)
x
x
y
y

w

w
2 w
 ( N xy
)  ( N xy
)   h 2  F (3)
x
y
y
x
t
37
Thin rectangular plates…..
The constitutive equations for a linearly elastic and
isotropic material are:  N x  Eh(1   2 ) 1 [ux  w x2 / 2
F
z
Nxy
Nx
Ny
x
(4)
 N y  Eh(1   2 ) 1 [v y  w 2y / 2
Nx
Nxy
y
  v y   w 2y / 2]  N xi
Ny
  ux   w x2 / 2]  N yi
(5)
i
 N xy  Gh[u y  v x  w x w y ]  N xy
(6)
 M x   D( w xx   w yy )
(7)
In these expressions, N, Ni – forces, M - moments
38
Thin rectangular plates…..
The constitutive equations for a linearly elastic and
isotropic material are:  M y   D( w yy   w xx ) (8)
 M xy   D(1   ) w xy
(9)
Ny
F
z
D  Eh3 /12(1   2 )
Nxy
Nx
Nx
Nxy
Ny
y
G  2 E /(1   );
x
Also u,v,w – displacements
Substituting the force –
displacement relations in
the dynamic equations give:
39
Thin rectangular plates…..
The dynamic equations for a plate made of linearly
elastic and isotropic material are then simplified by
introducing a stress function  such that:
2 
2 
2 
Nx  2 , Ny 
, N xy  
2
xy
y
x
(10)
Then, (1) and (2) are automatically satisfied if inplane inertia terms are neglected. Furthermore, the
expressions (7)-(9) can be substituted in (3) to get
equation for transverse displacement. Also, a
compatibility condition is (gives an equation for ):
(uxyy  v yxx )  (u y  v x ) xy  0
(11)
40
(b) Continuous Dynamical Systems
4. Flow between concentric rotating cylinders:
Consider two concentric cylinders with radii a, b;
Let 1, 2 – angular
2 b
velocities of inner and outer
1
a
cylinders; let (ur,uθ,uz) –
velocity components in a
cylindrical coordinate system;
p – pressure at a point;
We now define the equations
of motion for the system.
41
Flow between concentric rotating….
Equations of motion for the system: In cylindrical
coordinate system the NS equations are
2 b
a
1
Dur u2
ur 2 u
1 p


 ( ur  2  2
),
Dt
r
 r
r
r 
Du ur u
u 2 ur
1 p


 ( u  2  2
)
Dt
r
r 
r
r 
Duz
1 p

 ( uz ),
(1)
Dt
 z
where
D 



  ur
 u
 uz
Dt t
r
r
z
42
Flow between concentric rotating….
Equations of motion for the system: and
2

2
2
 2 
 2 2  2
rr r 
r
z
2 b
a
1
There is also the equation for mass
conservation:
ur ur 1 u uz
 

 0.
(2)
r
r r 
z
The basic flow is defined by:
ur  0, uz  0, u  V(r)  r(r) and p  P(r)
1 dP V 2
or

 dr
r
and
DD V  0
(3)
43
Flow between concentric rotating….
The basic flow…: Equations (3) have solution of
the form
2
(r)  A  B / r
2 b
2



1
2 1 
where A  1
, B  1R1
a
2
2
1 
1 
and
  2 / 1,   R1 / R2
(4)
Let us consider perturbations to
the basic flow: u  (ur,V  u ,uz ), p  P  p
We can then obtain linearized
equations about the basic flow.
44
Flow between concentric rotating….
Linearized equations about the basic flow:
ur
ur
ur 2 u
1 p

 2u  
 ( ur  2  2
)
t

 r
r
r 
u
u
u 2 ur
1 p

 (D V)ur  
 ( u  2  2
)
t

r 
r
r 
uz
uz
1 p


 ( uz )
t

 z
ur ur 1 u uz
and
 

0
r
r r 
z
45