Download Lecture 3

Ordinary differential
equations (ODE’s)
definitions of ODE, initial (boundary)
conditions, general and particular solutions
of an ODE
integration of some 1st order diff. equations
separable 1st order ODE's
integration of some 2nd order ODE’s using the
method of integrating multipliers;
1st integral and its relation to the energy
conservation law in mechanics
diff. eq. for a harmonic oscillator
(particle in a parabolic potential well)
Literature:
All you wanted to know about Mathematics,
but were afraid to ask, L.Lyons - Vol. 1, Chapt. 5
Other reading:
Mathematical Methods for Science Students, G.Stephenson – Chapt.21
Mathematical Methods in Physical Sciences, M.L.Boas – Chapt. 8
Methods of solving some 2nd order ODE’s
General form is
it also demands
2 initial condition
involving dx and x
 d 2 x dx


 2 , , x, t   0
 dt dt

dt
or
d 2x
 dx


F
,
x
,
t


2
dt
 dt

_______________________________________________________________
Equation of motion of a conservative dynamical system
Differential
equation:
supplemented with
boundary conditions
d 2 x(t )
m
 F [ x(t )]
2
dt

depends only on an unknown function x,
but neither on its derivative nor t directly.
dx(0)
x(0)  x0 ;
 v0
dt
That is, force in a mechanical system
described by this differential form of
Newton's equation depends only on the
coordinate of a particle, but not on its
velocity or time.
Potential forces:
Force
dU potential energy
F ( x)  
dx
x
 dx F ( x)  {U ( x)  U (0)}
0
Method of an integrating multiplier
Differential equation
of 2nd order
d 2 x(t )
m
 F [ x(t )]
2
dt
dx
Let us multiply both sides by
and, then, integrates both sides over t.
dt
t
t
dx d 2 x
dx
m dt
 2   dt
 F[ x(t )]
dt dt
dt
0
0
.
Note that:
d
dt
 dx  2 
dx d  dx 
dx d 2 x
 
 2
 2
2

dt dt  dt 
dt dt
 dt  
2
dx d 2 x
d  1  dx  
 2 
 
 
dt dt
dt  2  dt  
2
t
dx d 2 x
d  1  dx  
0 dt  dt  dt 2  0 dt  dt  2  dt   


t
for any x(t)
2

t
1  dx(t ) 
1  dx(0) 





2  dt 
2  dt 
2
x (t )
dx
0 dt  dt  F[ x(t )]  x (0)dx F ( x)  U [ x(t )]  U [ x(0)]
nothing but
dx
change of dx 
dt
dt
variables
Force is related to potential
energy as
dU
F 
dx
After substituting the results of these integrations into both sides:
2
2
m  dx(t ) 
m  dx(0) 

  
  {U [ x(t )]  U [ x(0)]}
2  dt 
2  dt 
2
2
m  dx(t ) 
m  dx(0) 

  U [ x(t )]  
  U [ x(0)]
2  dt 
2  dt 
for any value of variable t
for t=0
___________________________________________________
d 2x
m 2  F ( x)
dt
Differential equation
of the 2nd order
has the 1st integral (also called 'invariant')
2
m  dx 
E     U ( x)  const
2  dt 

x
U ( x)   dx F ( x)
0
The 1st integral of an ODE describing motion of a particle
subjected to a potential force is nothing but the total energy,
which conserves throughout the motion of a particle.
total
energy

kinetic
energy

potential
energy
How does the existence of 1st integral help to solve the
original ODE?
2
m  dx(t ) 

  U [ x(t )]  E  const
2  dt 
Now, 2nd order equation
d 2x
m 2  F ( x)
dt
can be reduced to a simpler 1st order ODE
dx
2
E0  U ( x)

dt
m
Moreover, this 1st order ODE can be solved using the method of
separation of variables
f (t )  
dx
2
f (t )
E  U ( x )  

dt
m
g ( x)
General solution
xE,x(0)(t)
can be found from
algebraic equation
x (t )

x (0)
g ( x) 
2
m
1
E  U ( x)
dx
2
 t
m
E  U ( x)
x(0)
and
E
are free parameters
Example: Particle in a parabolic potential well
kx2
U ( x) 
2

dU ( x)
F ( x)  
 kx
dx
 x0
x
Differential equation:
d 2x
k
 x
2
dt
m
depends only
on x, does not
contain t or
dx
derivatives
dt

We can apply the method
of an integrating multiplier
Initial conditions
x(0)   x0
dx (0)
0
dt
dx d x
dx  k 
0 dt  dt  dt 2  0 dt  dt    m x 
t
2
d x
k
 x
2
dt
m
x(0)   x0
dx (0)
0
dt
t
2
x 2 (t )  x02
dx(t )

0 dt  dt  [ x(t )]   x(0)xdx
2
x0
x (t )
t
t
t
dx d 2 x
d
dt



dt

0 dt dt 2 0 dt
 1  dx  2  1  dx(t )  2
 0
    
 2  dt   2  dt 

1  dx(t ) 
k x02  x 2 (t )

 
2  dt 
2m
2
2
m  dx  kx2
kx02
E
  
2  dt 
2
2
dx
k

x02  x 2
dt
m
1st integral
Separable 1st order ODE
x (t ) x
dx   x0 d (cos  )
 x0 sin  d
sin   cos   1
2
2
cos 0  1
x x
2
0
x ( 0 )   x0
x   x0 cos 
arccos
 x (t )
x0

0
t
dx

x
  arccos
x0

x0 d cos  
x0 1  cos 
2
  dt
2
0
arccos

 x (t )
x0

0
 x(t )
arccos

x0
k
m
d sin

sin

k
t
m
 x(t )
arccos

x0
 k 
x(t )   x0 cos
t 
 m 
k
t 
m
x0 x(t )
Periodic (oscillatory) solution:
T
0
t
 x0
_______________________________________________________________
Quick check: substitute solution
x(t )   x0 cost 
oscillation
frequency
k

m
into diff. equation
d 2x
k
2


x



x
2
dt
m
d cos t
  sin t
dt
d sin t
  cos t
dt
U (x )

period of oscillations
T   2
 x0
2
x0
m
T
 2

k
Example: Ball rolling of the top of a hill
kx2
U ( x)  
2
Home reading
d 2x k
 x
2
dt
m
v0

x
Initial conditions:
x(0)  0
F ( x)  
dU ( x)
 kx
dx
t
Apply
dx(0)
 v0
dt
 dt 
0
dx
 to both sides and use the initial conditions:
dt
x (t )
t
dx(t )
x 2 (t ) x 2 (0) x 2 (t )
0 dt  dt  x(t )  x(0)xdx  2  2  2
2
2
t
t
2


dx d x
d 1  dx 
1  dx(t ) 
v02
0 dt  dt  dt 2  0 dt  dt  2  dt    2  dt   2


2
1  dx(t ) 
k 2
v02
x (t ) 

 
2  dt 
2m
2
dx


dt
kx2
 v02
m
Separable 1st order ODE
k
t
m
t

0
x (t ) x
k
dt  
m
x ( 0 ) 0
dx
x2 
2
0

mv
k
2 

mv
0
 ln  x  x 2 
  ln
k 

mv02
k
Home reading
mv02
x x 
k 
ln
mv02
k
ln A  ln B  ln( A  B)
A
ln A  ln B  ln
B
exp(ln A)  A
2
2
0
mv
x x 

k
2
k
t
m
notations:
 k 
mv
 exp 
t 
k
 m 
2
0
x  x2   2    eDt

mv02
k
D
k
m
x      e  x    2e2 Dt  2 xe Dt  x 2
2
2
Dt
2
 2e2 Dt   2  Dt   Dt
x (t ) 
 e  e
Dt
2e
2
2
____________________________________________________
Note that this is a particular solution of the 2nd order ODE
2
d x k
2
 xD x
2
dt
m
2
Equivalent
to
d x k
 x0
2
dt
m