Transcript Green`s function methods

Impulsive Methods
• The big picture
– Principle of Superposition
– Overview of two methods.
• Impulse superposition
– Green’s function for underdamped oscillator
• Exponential driving force
– Green’s function for an undamped oscillator
• Solution for constant force
• Step function method
– Why it works?
– Undamped example
• Purely time dependent force as Green’s integral
•
– Equivalence to double integral solution
Non-Quiescent initial conditions
Impulsive Methods
1
What will we do in this chapter?
We develop the impulse (Green’s
function) method for getting solutions for
the harmonic oscillator with an arbitrary
time dependent driving force. We do this
using two techniques. In the first method
we write the solution as a superposition
of solutions with zero initial
displacement but velocities given by the
impulses acting on the oscillator due to
the external force. An arbitrary driving
force is written as a sum of impulses, The
single impulse x(t) responses are added
together in the form of a continuous
integral.
Another way to get this integral is based
on adding the responses of the oscillator
to a rectangular driving force by
considering this as the superposition of a
positive and negative step function. We
next take the limit where the time base of
the rectangle becomes infinitesimal and
compute the response of the driving force
to this “impulse” .
Our solutions will be for quiescent initial
conditions (zero initial displacement and
velocity) We conclude by extending the
treatment to different initial conditions.
This basic Green’s function method is
used in nearly all branches of physics.
Impulsive Methods
2
The big picture
F (t )
We are going to work out a general
expression for the response of a damped
mass-spring system to an arbitrary force as
a function of time making some very
clever uses of Superposition. We will view
the force as a sum of rectangular
infinitesimal impulses and add the x(t)
solutions for each impulsive force. For an
initially quiescent oscillator each impulse
produces a solution equivalent to a free
oscillator with initial velocity of v0 =
FDt/m. The solution becomes a sum
(integral) over such impulse responses.
An alternative method of solution is to
solve for the x(t) response of each impulse
by viewing an impulse as the sum of a step
function and an inverted step function. The
difference of these step function responses
is related to the derivative of the step
function response.
t
f (t )
impulse
t0  
t0
t0
t0  
(t  t0 )
Impulsive Methods
Step function
t0
3
Impulse Superposition Method
F (t ')
D
F (t )
F (ti )
t
t
Consider an underdamped harmonic oscillator at x  0
and at rest at t  0 subjected to an sharp impulse
delivered at t  t '. Immediately after the impulse the
oscillator aquires a velocity of
t ' D
t ' D F (t )
F (t ') D
v  lim D  a (t ) dt  
dt 
t'
t'
m
m
But as D   we still have x(t'+D)  0 . At a time
t  t ' we just have the familiar solution for an
F (t ')D
underdamped oscillator with x o  0 and v 0 
:
m
1 F (t ') D
x (t ) 
sin  (t  t ')  exp    t  t ') 
 m
If we had a series of impulses of duration D at t i  at a
time t than max t i  , superposition gives us:
N 1 F (ti ) D
x(t )   i 1
sin  (t  ti )  exp    t  ti ) 
 m
ti
tN
In the D   limit the solution becomes the integral:
sin  (t  t ')  exp    t  t ') 
1 t
dt
'
F
(
t
')
m 0

This result is called a Green's Integral for the undamped
oscillator. This is a powerful result since it represents the
solution to the differential equation for an arbitrary force
mx  2m  x  kx  F (t ).
The only catch is that solution is for an initially
quiescent oscillator x(0)  x(0)  0. We sometimes
write this as an integral over a "Green's" function
x(t ) 
1 
dt ' F (t ') G (t , t ') where
m 0
1    t t '
G (t , t ') 
e
sin 1  t  t '  (t  t ')
m1
x(t)=
Impulsive Methods
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Example: exponential driving force
One then combines factors such as exp(-i1t )
F (t )  F0e  t (t )
x(t)=
Fo t
   t  t '   t '
dt
'

(
t
')
e
 sin 1  t  t ' 


m1
Fo e   t t
   t ' i1  t t '
   t '  i1  t t '

dt
'
e

e
m1 2i 0
  i1 t '

Fo e   t  e  i1t e


 
m1 2i        i1  0
   i1 t '  t
F e e e
 o


m1 2i        i1  0
 i1t

 

e
1
Fo e  e

m1 2i        i1

   i1 t '
 i1t
t  e
e
1
Fo e 

m1 2i       t ' i1t '

t
 i1t

   i1 t '
Reduction is tedious but straightforward

Fo
m       12
2

x(t)=
 e  t 



e   t  cos  t     sin  t  

1
1 

1



Sometimes difficult to work problems analytically but
often easy to integrate Green’s functions on the computer!
t
t
and exp(+i1t ) into sines and cosine.







We show the response
to an exponential with
moderate and very
high damping. At high
damping the x(t)
response is nearly the
same as the driving
force except for the
clear x0 = 0 initial
condition.
1.4
1.2
  0.21
exp( t )
x(t )
1
0.8
0.6
0.4
0.2
0
1.2
0
10
20
30
0
10
20
30
1
0.8
40
50
540
50
  1
0.6
0.4
0.2
0
Impulsive Methods
A
simpler
example
Consider solving the solving the undamped
oscillator subject to a constant force
x -  2 x  Fo / m and a quiescent initial state
using Green's function techniques. We can get
build the Green's integral from the homogenous
v
solution: x(t )  x0 cos  t  0 sin  t for x 0 =0

and v0  F D / m . For x(0)  x(0)  0, the
response to an impulse delivered at t' is:
v
F (t ')D
x(t )  0 sin  (t  t ') ; v0 
. Hence by

m
1 t
superposition x(t ) 
dt ' F (t ') sin  (t  t ')

0
m
We can check this by turning the damping off
of our damped general solution:
1 t
   t t '
x(t ) 
dt
'
F
(
t
')
e
sin 1  t  t '

0
m1
If  =0, then 1   and
1 t
x(t ) 
dt ' F (t ') sin   t  t '  as before
m 0
For the case of a constant force F  F0 (t ')
Fo t
sin (t-t')
x(t ) 
dt
'
m 0

Fo t
F0
sin (t-t')
t 't

d

t
'

cos

(
t

t
')


t ' 0
m 0
2
m 2
1  cos  t
Thus x(t )  F0
.
m 2
Check by solution of the inhomogeneous DE
with a constant particular sol.
First solve: x   2 x  F0 / m
xH  A sin  t  B cos  t
F /m
F
xP  0 2  x  A sin  t  B cos  t  0 2

m
Solve for A and B using the initial conditions
F
x(0)  x(0)  0  x(0)  B  0 2  0
m
F
x(0)   A  0 Thus x  0 2 1  cos  t 
m
This checks our Green's solution!
Impulsive Methods
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A simple non-oscillator example
The Green's function approach works for other
inhomogenous linear differential equations.
An example is v   v  f (t ) . This describes
a mass subjected to a viscous drag force of
FDrag   m  v and a time dependent force of
F (t )  m f (t ). We wish to find a Green's
integral describing v(t ) for an abtrary f (t ).
As before our technique extends only to
the quiescent initial condition v(0) = 0 or
release from rest. To obtain the Green's integral
using the impulse method we need to find the
the homogeneous solution for v(t) for an
initial velocity v o that we will ultimately set
to the impulse f D where D is the time
duration of the impulse.
For v   v  0, the auxillary equation for
v  exp(r ) is just r    0  r  - 
Hence v  Ae-  t ; v(0)  A, thus v  v0 e   t
Hence the response of a quiescent mass to a
single impulse of duration D delivered at t'
would be v(t )  f (t ')De-  (t t ') or for an
t
arbitrary f (t ') : v(t )   dt ' f (t ')e -  (t t ')
0
As an example, we apply this solution to a mass
released form rest in gravity. Here f (t ')  g .
t
t
0
0
v(t )   dt ' f (t ')e-  (t t ')  ge   t  dt ' e  (t ')
t
 exp(  t ') 
g
v(t )  ge   t 

1  exp    t  


 
0 
This is indeed the solution we obtained for this
problem in Physics 225. Recall the terminal
velocity vterm  g /  
Impulsive Methods
7
An alternative (step function) method
It is also possible to obtain a Green’s integral through the response of the
system to a unit step function. We will state the technique, give an
example, and argue why it works.
We start with a linear differential equation like A x  Bx  Cx  f (t ) 
F (t )
m
(1) We start by solving the differential equation A X  BX  CX  1 subjected
to the initial condition X(0)  0
X (0)  0 . (This solution will be a sum of
a homogeneous (transient) and particular solution.
(2) Let X(t) be the solution to part (1) and let X(t) =
dX
dt
(3) The Green's function is just G(t,t')=X(t-t') (t-t')
t
 And the Green's integral is simply x(t )   dt ' f (t ') X (t - t ')
-
Impulsive Methods
8
An example of the step function method
Consider the special case of an undamped
F (t )
m
We first solve: X   2 X  1
X H  A sin  t  B cos  t
1
1
X P  2  X  A sin  t  B cos  t  2
oscillator x   2 x 

Why does this work? We again write the force
as a sum of square impulses.
F (t )
t

Each square impulse can be written as a
Solve for A and B using the initial conditions superposition of an upright and inverted step
1
function: (t-t0)
X (0)  X (0)  0  X (0)  B 
X (0)   A  0 Thus X=
and X 
1

2

2
0
f (t )
1  cos  t 
d 1
 sin  t
1

cos

t
Thus:
 
 2
dt  


t0
t
x(t )   dt ' f (t ') X (t - t ')
t0  
t0
(t  t0 )
-
1 t
sin (t-t')
 x(t )   dt ' F (t ')
as before!
m -

Impulsive Methods
t0  
t0
unit step
function
9
Why did our simple prescription work?
f (ti )
D
We write the force as a sum of upright
and inverted unit step functions
multiplied by the force in the center
t
t1 t2 t3
ti
f (t )  LimD0  f (ti )  (t  ti )  (t  ti  D ) 
i
Let X(t) be the solution to A X  BX  CX  (t )
(or the solution to A X  BX  CX  1 for t  0)
By superposition:
x(t)  LimD0  f (ti )  X (t  ti )  X (t  ti  D ) 
ti t
 dX 
LimD0  X (t  ti )  X (t  ti  D )   
D

 dt  t t '
 X (t  ti )D
Thus x(t )   f (ti ) X (t  ti ) D   dt ' f (t ') X (t - t ')
t
ti t
-
We write the response as a sum of
responses for each rectangular impulse.
These response differences are just the
time derivatives in the infinitesimal limit.
The sum in the limit of small D becomes
an integral but only forces ahead of time
considered are included.
Impulsive Methods
10
f (t ) 
F (t )
m
Another Non-Oscillator Example
These two forms do not look alike but we can
integrate the Green solution by parts:
t0
t
t0
t0
Consider a free particle subjected to an external x(t )  0 dt ' (t - t ') f (t ')  0 udv
force satisfying the Diff Eq: x  F (t ) / m  f (t ). Let dv  f (t ') dt ' and u  -(t '- t )
t'
Assume x(0)  x(0)  0 and f (t  t0 )  0. We
v(t )   f (t '') dt '' ; du  dt '
0
already know the solution to this problem is a
t '  t0
t0
t'
t0


double integral of the form:
0 udv  (t - t ')0 f (t '')dt ''  0 vdu
t
t'
0
0
x(t )   dt ' dt '' f (t '') . Since
f (t  t0 )  0, we can write this solution as
t0
t ' 0
t0
t'
0
0
0
x(t )  (t - t0 )  f (t '')dt ''   dt '  f (t '')dt ''
Hence we essentially can write the single
t0
t0
t'
Green's integral as the familiar double integral
x(t )  (t  t0 )  dt '' f (t ')   dt ' dt '' f (t '')
0
0
0
solution for a purely time dependent force.
Let us try to solve this using Green's functions. We could have also obtained the Green's integral
from impulse method. In absence of external force
We solve X  1 for X (0)  X (0)  0. The
F (t ')D
x(t )  v  (t - t ') 
(t - t ')  f (t ') D  (t - t ')
solution is X (t )  t 2 / 2. The Green function
m
t
is then G (t - t ')  X (t - t ')  (t - t ')(t  t ')
By superposition : x(t )   dt '(t - t ') f (t ')
0
t0
and x(t )   dt ' (t - t ') f (t ')
0
Impulsive Methods
11
Non-quiescent initial conditions
This is clearly the correct solution when f (t ')  0.
Lets consider solving x   2 x  f (t ) with
We can also check the solution for the below case:
x(0)  0 ; x(0)  v0 and the driving acceleration
f (t )
f0
f(t) first becomes non-zero at t > 0. We cannot
D 0
t
write solution in the form x(t)=  dt' G(t-t') f(t')
0
D
since if f(t)=0 this form gives x(t)=0 when in
v
reality x(t) = 0 sin  t. The solution is to
t
t0

consider

t
0
dt' G(t-t') f(t') as the "particular"
solution and the complete solution as
t
v0
x(t )  sin  t   dt ' G (t - t ') f (t ')

0
where G(t-t') is the Green's function for
the initial condition x(0)=x(0)=0. Hence
x(t ) 
v0

sin  t 
1
t
dt '


0
f (t ') sin  (t  t ')
x(t ) 
v0

sin  t 
x(t  t0 ) 
v0

1

t0 D
t0
sin  t 
dt ' f 0 sin  (t  t ')
f0D

sin  (t  t0 )
We can now try to check this answer by starting
our clock at t  t - t0 . At t  0 the mass is at
v
x(t  0)  0 sin  t0 and x(t  0)  v0 cos  t0

After the impulse is applied x(t  D) 
and x(t  D)  f 0 D  v0 cos  t0 since
F  t
f0D 
v
m
v0

12
sin  t0
Checking our solution in impulsive limit
We can think of x(t  D) and x(t  D) as a new set of "initial conditions"
that starts the oscillator at t  t0 or t  0 The oscillator moves according to:
v0
f 0 D  v0 cos  t0
x(t )  sin  t0 cos  t 
sin  t


where you can easily verify the "initial" conditions.
We can re-arrange the formula as:
v0
f0D
x   sin  t0 cos  t  cos  t0 sin  t  
sin  t

x
x
v0

v0

sin   t  t0  
sin  t 
f0D

f0D


sin  t . Using t  t - t0 we have
sin   t - t0  which equals our Green's solution.
Impulsive Methods
13