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Chapter 9
Solving Second order
differential equations
numerically, 2
Online lecture materials
•The online lecture notes by Dr. Tai-Ran
Hsu of San José State University,
http://www.engr.sjsu.edu/trhsu/Chapt
er%204%20Second%20order%20DEs.p
df
provides a very clear explanation of the
solutions and applications of some
typical second order differential
equations.
DSolve
•DSolve of Mathematica can
provide analytical solution to a
generic second order differential
equation. See
Math_built_in_2ODE.nb.
Typical second order, non-homogeneous
ordinary differential equations
n(x)
Typical second order, non-homogeneous
ordinary differential equations
n(x)
Guess:
After some algebra
Simple Harmonic pendulum as a special
case of second order DE
Force on the pendulum
for small oscillation,
Equation of motion (EoM)
𝐹𝜃 = 𝑚𝑎𝜃
𝑑𝑣𝜃
−𝑚𝑔𝑠𝑖𝑛𝜃=𝑚
𝑑𝑡
=
𝑑 𝑑𝑟
𝑚
𝑑𝑡 𝑑𝑡
l
≈
𝑑2
𝑚 2
𝑑𝑡
𝑑2 𝜃
𝑔𝜃
≈−
2
𝑑𝑡
𝑙
n(x)
𝑙𝜃
r
The period of the SHO is
given by
l
T  2
g
Simple Harmonic pendulum as a special
case of second order DE (cont.)
n(x)
𝑥≡𝑡
𝑢 𝑥 ≡ 𝜃(t)
𝑎≡0
𝑔
𝑏≡
𝑙
𝑛(𝑥) ≡ 0
𝑑 2 𝜃(𝑡)
𝑔𝜃
=−
2
𝑑𝑡
𝑙
Simple Harmonic pendulum as a special
case of second order DE (cont.)
𝑑 2 𝜃(𝑡)
𝑔𝜃
=−
2
𝑑𝑡
𝑙
Analytical solution:
Ω = 𝑔/𝑙 natural frequency of the pendulum;
𝜃0 and 𝜙 are constant determined by boundary conditions
Simple Harmonic pendulum with drag
force as a special case of second order DE
Drag force on a moving object, fd = - kv
For a pendulum, instantaneous velocity v = wl = l (dq/dt)
Hence, fd = - kl (dq/dt).
l
The net force on the forced pendulum along the
tangential direction
- kl (dq/dt).
dq
dq
 mgq  kl
;
dt
dt
d 2r
d2
d 2q
m 2  m 2  lq   ml 2 ;
d t
dt
dt
d 2r
d 2q
g
k dq
g
dq
k
Fq  m 2  2   q 
  q q
;q 
l
m dt
l
dt
m
d t
dt
Fq  mg sin q  kl
fd
r
Simple Harmonic pendulum with drag force
as a special case of second order DE (cont.)
n(x)
𝑥≡𝑡
𝑢 𝑥 ≡ 𝜃(t)
𝑎≡𝑞
𝑔
𝑏≡
𝑙
𝑛(𝑥) ≡ 0
l
fd
r
d 2q
dt 2
g
dq
k
  q q
;q 
l
dt
m
Analytical solution
Underdamped regime (small damping). Still oscillate,
but amplitude decay slowly over many period before
dying totally. q  t   q0 eqt /2 sin    t 2  q 2 



4 

g
the natural frequency of the system
l
Overdamped regime (very large damping), decay slowly
over several period before dying totally. q is dominated
2
 qt

q
by exponential term.
2
  t
 
q  t   q0 e
 2

4


Critically damped regime, intermediate between
under- and overdamping case.
q  t   q 0  Ct  e

qt
2
Critically damped
Overdamped
q  t   q0
q  t   q 0  Ct  e
 qt

q2
2 


t

 2

4


e

qt
2
Underdamped
0.2
q  t   q0 e
 qt /2

q2
2
sin    t  

4

0.1
2
0.1
0.2
4
6
8
10




See 2ODE_Pendulum.nb where
DSolve solves the three cases of a
damped pendulum analytically.
Adding driving force to the damped
oscillator: forced oscillator
- kl (dq/dt) + FD sin(Dt)
D frequency of
the applied force
dq
dq
Fq  mg sin q  kl
 FD sin   D t   mgq  kl
 FD sin   D t  ;
dt
dt
d 2r
d2
d 2q
Fq  m 2  m 2  lq   ml 2 ;
d t
dt
dt
d 2r
d 2q
dq
Fq  m 2  ml 2  mgq  kl
 FD sin   D t 
dt
d t
dt
d 2q
g
dq FD sin   D t 
k
  q q

;q 
2
l
dt
ml
m
dt
Analytical solution
q  t   q 0 sin   D t   
q0 
FD / (ml )

2
  2D
   q 
2
2
D
Resonance happens when  D    g / l
Forced oscillator: An example of
non homogeneous 2nd order DE
n(x)
𝑥≡𝑡
𝑢 𝑥 ≡ 𝜃(t)
𝑎≡𝑞
𝑔
𝑏≡
𝑙
𝐹𝐷 sin 𝛺𝐷 𝑡
𝑛 𝑥 ≡
𝑚𝑙
d 2q
dt 2
g
dq FD sin   D t 
  q q

l
dt
ml
Exercise: Forced oscillator
d 2q
dt 2
g
dq FD sin   D t 
  q q

l
dt
ml
Use DSolve to solve the forced oscillator. Plot on the same
graph the analytical solutions of q(t) for t from 0 to 10 T ,
where T= 2/𝛺, 𝛺 = 𝑔/𝑙, for 𝛺𝐷 = 0.01𝛺, 0.5𝛺, 0.99𝛺,
1.5𝛺, 4𝛺. Assume the boundary conditions q(t=0)=0;
dq/dt(t=0)=0; m=l=FD=1; q=0.
See forced_Pendulum.nb.
Second order Runge-Kutta (RK2) method
Consider a generic second order differential
equation.
d u  x
2
dx
2
 G u 
It can be numerically solved using second order
Runge-Kutta method. First, split the second order DE
into two first order parts:
v  x 
du  x 
dv  x 
dx
dx
 G u 
Algorithm
Set boundary conditions: u(x=x0)=u0, u’(x=x0)=v(x=x0)=v0.
calculate
calculate
1
u  ui  vi x
2
1
v  vi  G (u )x
2
calculate
ui 1  ui  v x
calculate
vi 1  vi  G  u  x
Translating the SK2 algorithm into the case of
simple
pendulum
2
2
g
d u  x
dx
 G u 
2
v  x 
dv  x 
G  u    q (t )
l
du  x 
dx
 G u 
dx
Set boundary conditions:
u(x=x0)=u0, u’(x=x0)=v(x=x0)=v0
1
u  ui  vi x
2
1
v  vi  G (u )x
2
ui 1  ui  v x
vi 1  vi  G  u  x
𝑑 𝜃(𝑡)
𝑔𝜃
=−
2
𝑑𝑡
𝑙
w (t ) 
dw t 
dq  t 
dt

gq  t 
dt
l
Set boundary conditions:
q(t=t0)= q0, q’ (t=t0)=w(t=t0)=w0
1
2
1  gq
w  wi   
2 l
q  qi  wi t

 t

qi 1  qi  wt
wi 1
 gq
 wi   
 l

 t

Exercise: Develop a code to implement SK2
for the case of the simple pendulum.
g
w
0

;q  0   0


Boundary conditions:
l
See pendulum_RK2.nb
Translating the SK2 algorithm into the case of
damped
pendulum
2
𝑑 2 𝜃(𝑡)
𝑔𝜃
d u  x
dx
2
v  x 
dv  x 
 G u 
𝑑𝑡 2
du  x 
w (t ) 
𝑑𝜃
=−
−𝑞
𝑙
𝑑𝑡
dq  t 
dt
dw  t 
gq

 qw  t 
dt
l
dx
g
 G  u  G  u    q (t )  qw (t )
l
dx
Set boundary conditions:
Set boundary conditions:
u(x=x0)=u0, u’(x=x0)=v(x=x0)=v0
q(t=t0)= q0, q’ (t=t0)=w(t=t0)=w0
1
1
q  qi  wi t
u  ui  vi x
2
2
g


w

q
(
t
)

t
 i 2l

1
1 g



v  vi  G (u )x
w  wi    q (t )  qw  t  w 
2
2 l
 1


1

q

t
 2



ui 1  ui  v x
qi 1  qi  wt
vi 1  vi  G  u  x
wi 1
 gq

 wi   
 qw  t
l


Exercise:
Develop a code to implement SK2 for the case of a
pendulum experiencing a drag force, with damping
coefficient q= 0.1* (4),  𝑔/𝑙, 𝑙 = 1.0 m.
Boundary conditions: θ 0 = 0.2; 𝜔 𝑡 = 0 = 0;
d 2q
dt 2
g
dq
  q q
l
dt
See pendulum_RK2.nb
Exercise:
Develop a code to implement SK2 for the case of a
forced pendulum experiencing no drag force but a
driving force𝐹𝐷 sin 𝛺𝐷 𝑡 ,  𝑔/𝑙, 𝑙 = 1.0 m,
m=1kg; 𝐹D = 1N; ΩD =0.99 ;
Boundary conditions: θ 0 = 0.0; 𝜔 𝑡 = 0 = 0;
d q
2
dt
2
g
dq FD sin   D t 
  q q

l
dt
ml
Exercise: Stability of the total
energy a SHO in RK2.
𝜔=
𝑑𝜃
,
𝑑𝑡
angular velocity. M=1kg; l=1m.
User your RK2 code to track the total energy for t running
g
w
0

;q  0   0
from t=0 till t=25T; T= 𝑔/𝑙. Boundary conditions:  
l
𝐸𝑖 should remain constant throughout all 𝑡𝑖 .
Exercise: Develop a RK2 code for
planetary motion in polar
y
coordinates
E
S
q
O
𝐿
𝑚𝑟 − 𝑚𝑟
𝑚𝑟 2
where
2
𝐺𝑀𝑚 𝑚𝑟 2 𝜃 = 𝐿
=− 2
𝑟
𝑑2𝑟
𝑑𝜃
𝑟 = 2 ;𝜃 =
𝑑𝑡
𝑑𝑡
x
Velocity verlet algorithm
for solving the Newton
second law
Newton’s second law
•Given a position-dependent force acting on
a particle, F(r), Newton second law
determines the acceleration of a particle, a.
d2r/dt2= F(r)/m
•This is a second order differential equation.
•Given F(r), we wish to know what are the
subsequent evolution of r(t), v(t) beginning
from the boundary values of r(0), v(0).
•Previously we solve the second order DE
using RK2 to obtain r(t), v(t).
Verlet algorithms
•The equation d2r/dt2=F(r)/m can be integrated
to obtain r(t), v(t), via a numerical scheme:
Verlet algoritm
•Three types: ordinary Verlet, velocity Verlet,
leap frog verlet.
•In the following, we shall denote the RHS as
F(r)/m → a
So that d2r/dt2=F(r)/m
↓
d2r/dt2= a
•See http://en.wikipedia.org/wiki/Verlet_integration
Velocity Verlet algorithm
Global error in velocity Verlet algorithm
•The global (cumulative) error in x over a
constant interval of time is given by
Δ𝑥 ~ O(Δt2)
•Because the velocity is determined in a noncumulative way from the positions in the
Verlet integrator, the global error in velocity
is also
Δv ~ O(Δt2)
Exercise: SHO
•Solve for x(t), v(t), for a simple harmonic
oscillator with constant k, mass m, initial
displacement x0, and velocity v0 =0, using
velocity Verlet algorithm.
•For SHO, F = -kx ⇒ a = -(k/m)x
d2x/dt2 = -(k/m)x
See verlet_algorithm_samples.nb
Exercise: 2D free-fall projectile
•Solve for x(t), y(t), v(t) of a 2D free-fall projectile
with initial speed v0=0 and launching angle q0
using velocity Verlet algorithm.
•For 2D projectile motion, 𝑭 = −𝑚𝑔𝑦 + 0𝑥.
a = 𝑭/m
ax𝑥 + ay𝑦 = 0𝑥 − 𝑔𝑦
d2x/dt2𝑥+ d2y/dt2𝑥= −𝑔𝑦
⇒ d2x/dt2 = 0, d2y/dt2 = −𝑔
See verlet_algorithm_samples.nb
Exercise: Scattering of a projectile charge
via Coulomb force
F=F𝑟 =
𝑘𝑞𝑄
𝑟
2
2
(𝑥−𝑥𝑄 ) +(𝑦−𝑦𝑄 )
=
𝑘𝑞𝑄
𝑟
2
𝑅
q
𝒓𝑞 = (𝑥, 𝑦)
𝑦
(projectile, with mass m)
R=(rq – rQ)=R𝑟, 𝑟 =
𝒓q−𝒓𝑄
𝒓q−𝒓𝑄
Q(Fixed)
𝒓𝑄 = 𝑥𝑄, 𝑦𝑄
rQ
o
𝑥
Notation for scattering of a projectile
charge via Coulomb force
• 𝒓𝑄 = 𝑥𝑄, 𝑦𝑄 ; 𝒓q = (𝑥, 𝑦)
•𝑭 =
𝑞𝑄
𝑘
𝑟;
2
2
(𝑥−𝑥𝑄 ) +(𝑦−𝑦𝑄 )
•⇒
• 𝑎𝑦 =
• 𝑎𝑥 =
1
𝑭∙𝑦
𝑚
1
𝑭∙𝑥
𝑚
=
=
𝒓q − 𝒓𝑄
𝑟=
𝒓q − 𝒓𝑄
𝑘
𝑞𝑄
𝑟
𝑚 (𝑥−𝑥𝑄 )2 +(𝑦−𝑦𝑄 )2
𝑘
𝑞𝑄
𝑟
2
2
𝑚 (𝑥−𝑥𝑄 ) +(𝑦−𝑦𝑄 )
∙𝑦 =
∙𝑥=
𝑑2𝑦
,
𝑑𝑡 2
𝑑2𝑥
.
2
𝑑𝑡
The equations required by Verlet
algorithm
• 𝒓𝑄 = 𝑥𝑄, 𝑦𝑄 ; 𝒓q = (𝑥, 𝑦)
𝑑2 𝑦
𝑑𝑡 2
=
𝒓q − 𝒓𝑄
𝑟=
𝒓q − 𝒓𝑄
𝑘𝑞𝑄
𝑟∙𝑦
,
2
2
𝑚 (𝑥−𝑥𝑄 ) +(𝑦−𝑦𝑄 )
𝑑2 𝑥 𝑘𝑞𝑄
𝑟∙𝑥
=
2
𝑑𝑡
𝑚 (𝑥 − 𝑥𝑄 )2 +(𝑦 − 𝑦𝑄 )2
See verlet_algorithm_2D_coulomb_scatterings.nb
Störmer-Verlet integration
algorithm
𝑥𝑛+1 = 2𝑥𝑛 − 𝑥𝑛−1 + 𝑎𝑛 ∆𝑡
2
𝑣𝑛+1 = (𝑥𝑛+1 −𝑥𝑛 )/∆𝑡
• Another variant of Verlet algoritm
• Use this for integrating dynamical system with
a velocity-dependent acceleration, such as
Lorentz force on a moving charge particle.
• The cumulative error in the velocity is larger
than that in velocity Verlet algorithm
Exercise: Charge moving in a
magnetic field
• A charge (mass m and charge q) moving with velocity v
=(vx, vy , vz) in a magnetic field B=(Bx, By , Bz) experiences a
velocity-dependent Lorentz force F=(Fx, Fy , Fz) = q v × B.
Develop a code based on the Störmer-Verlet integration
algorithm to simulate the dynamical path of the charge
particle moving through the magnetic field. Assume: q=+1
unit, mass m = 1 unit, initially located at (0,0,0), initial
velocity (v0x,v0y, v0z), v0x=v0y=0.1 unit, v0z =0.05 unit,
B=(0, 0, Bz), Bz = 0.1 unit. You should see a helical
trajectory circulating about the z-direction.
• verlet_algorithm_3D_coulomb_helix.nb