Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 8 Ordinary differential equation
Lecture 5 Introduction of ODE
1
1. Introduction (differential equation)
- A great many applied problems involve rates, that is, derivatives.
An equation containing derivatives is called a differential equation.
- If it contains partial derivatives, it is called a partial differential
equation; otherwise it is called an ordinary differential equation.
ex 1) Newton’s equation
dv
d 2r
F  ma  m
m 2
dt
dt
ex 2) Heat transfer
dQ
dT
 kA
dt
dx
ex 3) RLC circuit
VR  RI
dI
VL  L
dt
VC 
dI
q
d 2I
dI I dV
L  RI   V  L 2  R  
dt
C
dt
dt C dt
2
q
C
- order of a differential equation : order of the highest derivative in the eq
y  xy 2  1 :1st order
d 2r
m 2  kr : 2nd order
dt
- (non)Linear differential equation
a0 y  a1 y  a2 y  a3 y    b linear equation
Here,an and b are eitherconstantor a functionof x.
y  cot y, yy  1 , y2  xy,
nonlinearequation
3
Note 1
: A solution of a differential equation (in the variable x and y) is a
relation between x and y which, if substituted into the differential
equation, gives an identity.
 If you come up with a function to give an identity, that should
be a solution of the differential equation.
Example 1)y  sin x  C is a solutionof thedifferential equation(DE), y  cos x
x
x
x
Example 2) DE y  y  Solution y  e  y  Ae  Be
In order to verify if your solutions are correct, put the solutions into
the equations and check the identity.
4
Note 2
- First order DE  one arbitrary integration constant (IC)
- Second order DE  two ICs
- N-th order DE  # of ICs is n
- General solution with arbitrary IC
- Particular solution determined by the boundary condition or initial cond
Example 3)Find the distance which an object falls under gravity in t
seconds if it starts from rest.
d 2x
1 2

g

x

gt  v0t  x0
2
dt
2
(generalsolut ion)
Wit h t heinit ialcondit ion(v0  x0  0), x 
1 2
gt (particular solut ion)
2
5
Example 4) Find the solution which passes through the origin and (ln2, 3/4
y  y
y  y  y  Aex  Be x
T o satisfy the given condition,
0  A  B
1


A


B

.
2
 3  Aeln 2  Beln 2  2 A  1 B
2
4
 y  12 (e x  e  x )  sinh x
6
2. Separable equations
- Separable equation
ex)
dy  f ( x)dx
y terms in one side and x terms in the other side
 the equation is separable.
Example 1) Radioactive substance decay rate
dN
 N
dt
dN
 dt  ln N  t  const.
N
N  N 0 e t for N  N 0 at t  0
7
Example 2)
Solve xy  y  1.
y
1
dy
dx
 , or

y 1 x
y 1 x
ln y  1  ln x  const.  ln x  ln a  lnax
y  1  ax
8
- Orthogonal trajectories: ex) lines of force intersect the equipotential
curves at right angles.
y  1  ax  y  a
y 
y  1 orthogonal trajectory
x
    y  
(negative reciprocals)
x
y 1
( y  1)dy   xdx
1
2
y 2  y   12 x 2  C
x 2   y  1  2C  1
2
9
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 8 Ordinary differential equation
Lecture 6 First order ODE
10
3. Linear first-order equations
- Linear first-order equation
y  Py  Q, where P, Q are functionsof x.
First,let Q  0.
y  Py  0 or
dy
  Py
dx
dy
  Pdx, ln y    Pdx  c
y
ye


Pdx  c

 Ae

Pdx
 Ae I , where I   Pdx
11
yeI  A
differenti
ating


 
d
dI
yeI  ye I  yeI
 ye I  yeI P  e I ( y  Py)  QeI
dx
dx
yeI   QeI dx  c, or 
 where I   Pdx

y  e I  QeI dx  ce I 

12
Example 1)Solve x 2 y  2xy  1/ x.
For y  Py  Q,
x 2 y  2 xy  1 / x  y 
yeI   QeI dx  c, or 
 where I   Pdx
y  e  I  QeI dx  ce I 

2
1
2
1
y  3 . Here, P   , Q  3 .
x
x
x
x
1
 2
I     dx  2 ln x, e I  e  2 ln x  2 ,
x
 x
1
1 1
x4
5
ye  y 2   2 3 dx   x dx 
 c,
x
x x
4
I
y
1
 cx2 .
2
4x
13
Example 2)Radium decays to radon which decays to polonium. If at
t=0, a sample is pure radium, how much radon does it
contain at time t (created and simultaneously decay)?
N_0 = # of radium atoms at t=0
N_1= # of radium at time t
N_2 = # of radon atoms at time t,
lambda_1, lambda_2 = decay constants for Ra and Rn.
dN1
 N1 , N1  N 0e  1t .
dt
dN2
dN2
 1 N1  2 N 2 , or
 2 N 2  1 N1  1 N 0e  1t .
dt
dt
For y  Py  Q,
Here, P  2 ,
Q  1 N 0e  1t
yeI   QeI dx  c, or 
 where I   Pdx
y  e  I  QeI dx  ce I 

14
I   2 dt  2t ,
N 2e2 t   1 N 0e  1t e2 t dt  c  1 N 0  e2  1 t dt  c 
Since N 2 (t  0)  0, 0 
N2 
1 N 0 
e
2  1
2
 1 t
 c, if
1  2 .
1 N 0
N
 c or c   1 0 .
2  1
2  1
1 N 0   t   t

e  e .
2  1
1
2
15
4. Other methods for first-order
equations
1) Bernoulli equation
y  Py  Qyn
It is not linear,
but, is easily reduced to a linear equation by making the change of the v
z  y 1 n
z  1  n  y  n y
Multiplying theoriginalequation by 1  n  y  n ,
1  n y  n  y  Py  1  n  y  n Qyn 
1  n y  n y  1  n Py1 y  1  n Q
Using z  y1 n ,
z  1  n Pz  1  n Q : linear first - order equation
16
2) Exact equations; integrating factors
P x, y dx  Q x, y dy is an exact differential, if
P Q

.
y x
cf.   A  0, where A  ( Ax , Ay )  ( P, Q).
  A  0  F  A or F    A  dr    Ax dx  Ay dy.
For F ( x, y ),
P
F
F
,Q 
, Pdx  Qdy  dF.
x
y
Pdx  Qdy  0
or y  
P
P Q
is called exact, if

.
Q
y x
Pdx  Qdy  dF  0  F x, y   const.
17
ex. 1)
xdy  ydx  0 is not exact.
xdy  ydx 1
y
y
 y

dy

dx

d

0
is
exact.

 const.
 
2
2
x
x
x
x
x
1
: integrating factor
x2
ex. 2)
y  Py  Q  eI y  eI Py  e I Q, where e I : integrating factor
18
3) Homogeneous equations
Px, y dx  Qx, y dy  0 where P, Q : homogenousfunctions
It is a homogenousequation.
cf. homogeneous function: f (tx, ty, tz )  t n f ( x, y, z ),
ex) f ( x)  z 2 ln(x / y ), f x   x n f  y / x 
P x, y dx  Q x, y dy  0
y 
dy
P  x, y 
 y

 f 
dx
Q  x, y 
x
- The above equation can be reduced to a separable equation in variable v
19
Prob. 8)


ydy   x  x 2  y 2 dx (homogeneous eq.)
y  vx  dy  xdv  vd
P uttingthisinto theoriginaleq.,




vx( xdv  vdx)   x  x 2  v 2 x 2 dx  x  1  1  v 2 dx


vxdv  v 2 dx   1  1  v 2 dx


vxdv   1  v 2  1  v 2 dx
Separatingthe variables,
v
1
dv   dx
x
1  v2  1  v2
20
4) Change of variables
: If a differential equation contains some combination of the
variable x, y, we try replacing this combination by a new variable.
cf. Problem 11. y  cosx  y  Hint : u  x  y  u  1  y
u  1  y  1  cosu
du
 dx
1  cosu
21
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 8 Ordinary differential equation
Lecture 7 Second order ODE I
22
5. Second order linear equations with constant coefficients and
zero right-hand side
d2y
dy
a2 2  a1
 a0 y  0 : homogenous
dx
dx
d2y
dy
a2 2  a1
 a0 y  f  x  : inhomogenous
dx
dx
1) Auxiliary equation
Example 1.
y  5 y  4 y  0
dy
d  dy  d 2 y
2
Let Dy 
 y, D y     2  y. differential operator
dx
dx  dx  dx


T hen, y  5 y  4 y  0  D 2 y  5Dy  4 y  0 or D 2  5D  4 y  0.
D  4D  1 y  D  1D  4 y  0
23
T o solve this, we will first solve
D  1 y  0
DE
and D  4  y  0 solving
 these
separable

 y  c1e  4 x , y  c2 e  x
 T hegeneralequation of theDE is y  c1e  4 x  c2 e  x .
y  c1eax  c2ebx is thegeneralsolutionof ( D  a)(D  b) y  0, if a  b.
Comment. Here, we can see that solving the second-order linear
differential equation (y’’+my’+ny=0) is quite similar to solving the
second order normal equation (D2+mD+n=0). We know that there
are three cases for the solutions of the second order equation, two
real numbers, single real number, and two complex numbers. The
first case of DE corresponds to the equation with the two real
solutions. How about the others cases?
24
2) Equal roots of the auxiliary equation
D  a D  a y  0
One solutionis y  c1eax , then theotheris ?
u  D  a  y
D  a D  a y  0  D  a u  0  u  Aeax
D  a y  Aeax , or y  ay  Aeax (first order linear equation)
ye ax   e ax Aeax dx   Adx  Ax  B
y   Ax  Beax is thegeneralsolutionof ( D  a)(D  a) y  0.
25
3) Complex conjugate roots of the auxiliary equations
- The roots of auxiliary equations are complex.

y  Ae i x  Be i x  ex Aeix  Beix

 ex c1 sin x  c2 cos x   cex sin x   
Example 2) y  6 y  9 y  0.
D
2

 6D  9 y  D  3D  3y  0  y   Ax  Be3x
26
Example 3) motion of a mass oscillating at the end of a spring
d2y
d2y
k
m 2  ky  2   2 y for  2 
dy
dy
m


D 2 y   2 y  D 2   2 y  0  D   i
y  Aeit  Be it  c1 sin t  c2 cost  c sin t    : simple harmonicmotion
‘We can determine two unknown constants using initial conditions.’
Example 4. Initial condition: at t=0, y=-10, y’=0
y  c1 cost  c2 sin t
  10  c1  0  c2  1,
For t he initial condit ion, 
 c1  0, c2  10.
 0  c  1  c   0
1
2

y  10cost
27
Example 5. Considering the friction,
dy
d2y
(l  0)
m 2  ky  l
dt
dt
k
dy
d2y
 2  2b   2 y  0, for  2  ,
m
dt
dt
2b 
l
( 0)
m
D 2  2bD   2  0  D  b  b 2   2
28
- overdamtped if
y  Ae t
b2   2
  b  b 2   2

 Be t , where 
  b  b 2   2
- criticallydamped if
b2   2
y  ( A  Bt)e  bt
- underdamped or oscillatory if
y  ce bt sin  t   , where
b2   2
b 2   2  i
d 2I
dI I
cf. L 2  R   0
dt
dt C
29
- Underdamped oscillator
Undamped
Underdamped
Envelope
30
- Critically/over-damped
31
REVIEW
d2y
dy
a2 2  a1
 a0 y  0 :
dx
dx
homogenous
dy
d  dy  d 2 y
2
Dy 
 y, D y     2  y
dx
dx  dx  dx
 a2 D 2 y  a1Dy  a2 y  0  D 2 y  a1Dy  a2 y  0  ( D 2  a1D  a2 ) y  0
y  c1eax  c2ebx is thegeneralsolutionof ( D  a)(D  b) y  0, if a  b.
y   Ax  Beax is thegeneralsolutionof ( D  a)(D  a) y  0.

y  Ae i  x  Be i  x  ex Aeix  Be ix

 ex c1 sin  x  c2 cos x   cex sin x   
is t hesolut ionof ( D  a )(D  b)  0, where a    i , b    i .
32
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 8 Ordinary differential equation
Lecture 8 Second order ODE II
33
6. Second-order linear equations with constant coefficient and righthand side not zero
1) solution of an inhomogeneous DE
D
2

 5D  4 y  cos2 x
yc  Ae x  Be 4 x : complementary (generalsolutionof homogenousequation)
y p  101 sin 2 x : particularsolutionof inhomogeouns equation
D
D


 5D  4 y p  cos2 x 
  D 2  5D  4 y  y   cos2 x
p
c

2
 5D  4 yc  0

2


y  y p  yc  Ae x  Be 4 x  101 sin 2 x
“The general solution of an inhomogeneous DE is the combination of y_c
34
2) Inspection of particular solutions
: To find a simple particular solution, we may be able to guess and veri
y  2 y  3 y  5  y p 
5
3
y  6 y  9 y  8e x  y p  Aex  2e x
y  y  2 y  ex  y p  Aex??
not simple.
35
3) Successive integration of two first-order equations
y  y  2 y  e x  not simple.
D  1D  2y  e x
u  D  2 y  D  1u  e x
or u  u  e x
 first integration of thefirst order DE
I    dx   x, ue x   e  x e x dx  x  c1  u  xex  c1e x .
D  2y  xex  c1e x
or y  2 y  xex  c1e x .
 secondintegration of thefirst order DE
I   2dx  2 x


ye2 x   e 2 x xe x  c1e x dx  13 xe3 x  19 e3 x  13 c1e3 x  c2
 13 xe3 x  c1e3 x  c2  y  13 xe x  c1e x  c2 e  2 x
36
4) Exponential right-hand side
d2y
dy
cx
cx










a

a
y

F
x

ke

D

a
D

b
y

F
x

ke
1
0
dx2
dx
First, suppose that c is not equal to either a or b. Solving the DE by
the successive integration of two first-order equation gives the
particular solution, ecx.
ex) D  1D  5 y  7e 2 x (1,-5 2)
y p  Ce 2 x


yp  4 yp  5 y p  C 4e 2 x  8e 2 x  5e 2 x  7e 2 x  C  1
y  Aex  Be5 x  e 2 x
 Ce cx

cx
D  a D  by  ke   Cxecx
 2 cx
 Cx e

if c is not equal to eithera or b
if c equals a or b (a  b)
if c  a  b
Backing to the previous DE,
y  y  2 y  e x
c  a or b  y p  Cxex C  1/ 3
37
5) Use of complex exponentials
F x  sin x or cos x  eix
y  y  2 y  4 sin 2 x  Y   Y   2Y  4e 2ix
YR  YR  2YR  Re 4e 2ix  4 cos 2 x,
YI YI  2YI  Im 4e 2ix  4 sin 2 x
Yp  Ce 2ix
 4  2i  2Ce 2ix  4e2ix
4
4 2i  6  1
C

 i  3
2i  6
40
5
Yp   15 i  3e 2ix   15 i  3(cos2 x  i sin 2 x)
Here, theimaginarypart is thesolution we want.
 y p   15 cos2 x   53 sin 2 x (imaginarypart)
38
k sin x
T o solveD  a D  b  y  
,
k cosx
first solve
D  a D  b y  keix ,
then take thereal and imaginarypart as thesolution.
39
6) Method of undetermined coefficients
The method of assuming an exponential solution and determining the
constant factor C is an example of the method of undetermined
coefficients.
 Ce cxQn x 
if c is not equal to eithera or b

D  a D  by  ecx Pn x    CxecxQn x  if c equals a or b (a  b)
 2 cx
 Cx e Qn x  if c  a  b

Qn is thepolynomina
l of thesame degree as Pn with undetermined coefficient.
Example) Solve y  y  2 y  x 2  x
y p  Ax2  Bx  C
yp  yp  2 y p  2 A  2 Ax  B  2 Ax2  2 Bx  2C  x 2  x


y p   12 x 2  1
40
7) Several terms on the right-hand side; principle of superposition
 

y  y  2 y  D  1D  2 y  e x  4 sin 2 x   x 2  x

D  1D  2y  e x   y p1  13 xex
D  1D  2y  4 sin 2 x  y p 2   15 cos2 x  53 sin 2 x
D  1D  2y  x 2  x  y p3   12 x 2  1


y p  y p1  y p 2  y p 3  13 xe x  15 cos 2 x  53 sin 2 x  12 x 2  1
- Solve a separate equation and add the solutions.  principle of supe
(working only for linear equations)
41
8) Forced vibrations (steady state motion)
d2y
dy
2

2
b


y  F sin t ( F  const.)
2
dt
dt
d 2Y
dY
2
i t
i t

2
b


Y

Fe

Y

Ce
p
dt2
dt
 
2

 2bi   2 Cei t  Fei t





F
 2   2  2bi  F
i
C 2


re
2
   2  2bi 
 2   2  4b 2

C 
Yp 



F
2
 

2 2
 4b 2
F
2
 

2 2
 4b 2
, angle of C  
ei  t    y p 

F
2
 

2 2
 4b 2
sin  t   .
42
9) Resonance
yp 

sin  t   .
F
2
 

2 2
 4b 
2
1) Given  , themaximumof y p occurs at    
2) Given  , themaximumof y p occurs at  2   2  2b 2
43
10) Use of Fourier series in Finding particular solutions

d2y
dy
a2 2  a1  a0 y  f x    cn einx
dx
dx
n  
d2y
dy
a2 2  a1  a0 y  cn einx , and thenuse theprincipleof superposition.
dx
dx
2
Example) d y  2 dy  10y  f t , where f t   1, 0  t  

dt2
dt
0,   t  2
For auxiliaryequation,
D 2  2 D  10  0  D  1  3i
yc  e t  A cos3t  B sin 3t 
44
Fourier series expansion
f t  




1 1 it

e  eit  13 e3it  e3it   .
2 i
d2y
dy
1 ikt
1) First term:

2

10
y

e
dt2
dt
ik




1
1
1 10  k 2  2ik
y  Ce  C 

ik 10  k 2  2ik ik 10  k 2 2  4k 2
ikt
yp 


1
1 9  2i it 1 9  2i  it
1 1  6i 3it
1 1  6i  3it

e 
e 
e 
e 
20 i 85
i 85
3i 37
3i 37
1 2 9  eit  e  it 
4  eit  e  it  2 1  e3it  e  3it  12  e3it  e  3it 





 

  


20  85  2i   85 
2  3 37 
2i

37
2




1
2
9 sin t  2 cost   2 sin 3t  6 cos3t   .

20 85
3 37
45
PROBLEM
5-38. Solve the RLC circuit equation with V=0. Write the conditions and
solutions for overdamped, critically damped, and underdamped electrical
oscillations.


6-11 & 6-25 y  2 y  10y  100cos 4 x
( D  3)(D  1) y  16x 2e  x
6-42
 x,
y  9 y  
0,
0  x  1, 

 1  x  0.
46
7. Other second-order equations
Case (a) : dependentvariabley missing, a2 y  a1 y  f ( x)
y  p, y  p  a2 p  a1 p  f ( x) (first order DE)
Case b  : independent variablex missing,
dp dp dy
dp
y  p, y 

p
(first order DE with p as independent variabley)
dx dy dx
dy
47
Example 1. (plus or minus sign much be chosen correctly
at each stage of the motion so that the
retarding force opposes the motion.)
2
d 2 y  dy 
m 2  l    ky  0 l  0
dt
 dt 
2
d 2 y  dy 
 specialcase : 4 2  2   y  0 (t is missing.)
dt
 dt 
dy
d2y
dp
 p,

p
dt
dt 2
dy
dp
dp 1
1
4p
 2 p 2  y  0 or
 p   yp 1. (Bernoulliequation)
dy
dy 2
4
z  p2 ,
dz
dp
dz
 2p

 z   12 y (first order linear equation)
dy
dy
dy
ze  y   12  ye y dy   12 e  y  y  1  c
z   12  y  1  ce y
(describe the motion…) 48
Case (c) y  f  y   0
T o solve this, multiplyby y : yy  f  y  y  0
1
2
or
ydy  f  y dy  0
y2   f  y dy  const.
2
d
Example) m x  F x 
dt 2
dv
dx
m v  F x 
or m vdv F x dx
dt
dt
1 2
m v   F x dx  const. (conservation of energy)
2
49
d2y
dy
Case d  : a2 x

a
x
 a0 y  f ( x) (Euler or Cauchy equation)
1
2
dx
dx
x  ez
2
2
dy dy
d 2 y dy
2 d y
x

and x
 2 
2
dx dz
dx
dz
dz
 d 2 y dy 
dy
 a2  2    a1
 a0 y  f e z
dz 
dz
 dz
 
(secondorder ' linear'DE)
50
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 8 Ordinary differential equation
Lecture 9 Laplace transform
51
8. The Laplace transform

 pt
- Laplace transformL f   0 f t e dt  F  p .
cf. Laplace transform are useful in solving differential equati
Example 1. f(t)=1

F  p    1 e  pt dt  
0
1  pt
e
p


0
1
,
p
p  0.
Example 2. f(t)=e^(-at)

F  p   e
0
 at
e
 pt

dt   e a  p t dt 
0
1
,
pa
Re p  a   0.
52
- Some properties of Laplace transform
1) L f t   g t   

f t   g t e  pt dt
0



0
0
  f t e  pt dt   g t e  pt dt  L f   L g .
2) Lcf t    cf t e dt  c  f t e pt dt  cL f .

0
 pt

0
53
Example 3. Let us verify L3. L(cos at)
  
Start with L e
 at

0
e
 at
f t   eiat  cosat  sin at
e
 pt

dt   e a  p t dt 
0
1
,
pa
Re p  a   0.
a  ia
 
T akingLaplacetransform, L eiat  F  p  
1
p  ia
 2
,
2
p  ia p  a
Lcos at  i sin at  Lcos at  iLsin at 
p
a

i
.
2
2
2
2
p a
p a
Lcos at  i sin at 
Re p  ia   0.
p
a

i
,
p2  a2
p2  a2
Re p  ia   0.
Using theabove results,
Lcos at  i sin at  Lcos at  i sin at  2 Lcos at  2
Lcos at  i sin at  Lcos at  i sin at  2iLsin at  2i
p
p2  a2
a
p2  a2
: L3
: L4
54
Example 4. Let us verify L11. L(t sin at)

Lcosat   e  pt cosat dt 
0
p
.
2
2
p a
Differentiate the above relation with respect toa,


0
e  pt  t sin atdt 
p 2a 
p
2
a

2 2
or


0
e  pt t sin at dt 
p
2 pa
2
a

2 2
.
: L11
55
9. Solution of differential equations by Laplace transforms
- Laplace transforms can reduce an linear DE to an algebraic equation
and so simply solving it. Also since Laplace transforms automatically
use given values of initial conditions, we find immediately a desired
particular solution.



 pt
 pt


L y    y t e dt  e y t    p  y t e  pt dt
0
0
0
  y 0  pL y   pY  y0 .
Here L y   Y .
Using the above result,
L y  pL y  y0  p 2 L y   py0  y0  p 2Y  py0  y0 .
Note) The relations already include the initial conditions.
56
Example 1.
y  4 y  4 y  t 2e2t
for y0  0, y0  0.
1) T akingthe Laplacetransforms of each termin theequation,
 p Y  py
2



2  2t


0  y0  4 pY  y0   4Y  L t e
2
( L6).
3
 p  2
2) Using the initial condition,
p
2

 4p  4 Y 
2
2

Y

.
3
5
 p  2
 p  2
3) Using the inverse transform ( L6).
2t 4e 2t t 4e 2t
y

.
4!
12
57
y  4 y  sin 2t
Example 2.
for y0  10, y0  0.
T akingthe Laplacetransforms of each termin theequation,
 p Y  py


0  y0  4Y  L sin 2t  
2
2
.
2
p 4
Using the initial condition,
p
2

 4 Y  10 p 
2

2
p 4
Y
10 p
2

.
2
2
2
p 4 p 4


Using the inverse transform ( L 4, L17),
y  10cos2t 
1
sin 2t  2t cos2t   10cos2t  1 sin 2t  1 t cos2t.
8
8
4
58
y  2 y  z  0,
Example 4.
z  y  2z  0, for y0  1, z0  0.
T akingthe Laplacetransforms of each termin theequation, L( z )  Z 
 pY  y0   2Y  Z  0,  pZ  z0   Y  2Z  0.
Using the initial condition,
 p  2Y  Z  1,
1)
Y   p  2Z  0.
 p  2  1Y  p  2
2
or
Y
p2
 p  22  1
Using the inverse transform ( L14),
2) Sim iarly, Z 
1
 p  22  1
y  e 2t cost.
 z  e 2t sin t.
Alternatively we can use theoriginalequation with theresult of 1), y  2 y  z  0.
59
10. Convolution
- Method to write a formula for y
Example 1.
Ay  By  Cy  f (t ),
y0  y0  0.
Ap2Y  BpY  CY  L f   F  p   Y 
T  p 
1
1

Ap2  Bp  C A p  a  p  b 
1
F  p .
2
Ap  Bp  C
LT of some function,ex. L7, L8
In this case, y is the inverse Fourier Transform of a product of two
functions whose inverse transforms we know.
60
Example 2.
y  3 y  2 y  et ,
y0  y0  0.
 
p 2Y  3 pY  2Y  L et
  
 
1
t
t
 2t
t
L
e

L
e

e
L
e
 G p H  p .
2
p  3p  2
 Y
t

G p H  p   L  g t   h d   Y  L y 
 0

 y   g t   h d
t
0
In t his example,



y   g  ht   d   e   e  2 e  t   d
t
t
0
0
t



 e  t  1  e  d  e  t   e 

0


t
0
 e  t t  e  t  1  te  t  e  2t  e  t .
61
- Fourier Transform of a Convolution
g1  , g2   Fourier
Transform

 f1 x, f 2 x
1
g1    g 2   
2
 1 


2



2
 1 


2



2
 1 


 2 
2

 

0
0


0
0

 ix
 

0
e



f1 v e
 i v
1

2



f 2 u e  iu du
e  i v  u  f1 v  f 2 u dvdu
e  ix f1  x  u  f 2 u dxdu
x  v  u,
dx  dv
  f x  u  f u du dx.
2
 0 1

 Convolution : f1 * f 2  


f1 x  u  f 2 u du
62
g1  g 2 
1
2
g1  g 2 &
 1
 2



 1
f1 * f 2e  ix dx 
 Fourier transform of f1 * f 2 .
 2
1
f1 * f 2 : a pair of Fourier transforms.
2
g1 * g 2 & f1  f 2 : a pair of Fourier transforms.
63
11. The Dirac delta function
- Impulse: impulsive force f(t), t=t_0 to t=t_1

t1
t0
v1
dv
f (t )dt   m dt   mdv  mv1  v0 
t0
v0
dt
t1
- We are not interested in the shape of f(t). What we think
important is the value of the integration of f(t) during t_1 – t_0 = t.
64
- The above functions have the same integration value, 1.
In case of t  0, f t  is called ' Delta function'
65
- Laplace Transform of a  Function
 t0 , a  t  b
a  t  t  t0 dt  0, otherwise.

b
P rove:
  t  t  t dt   t   t  t dt   t  n  
b
b
a
0
0
a
0
0
in the Figure
66
- Example 2.
L t  a     t  a e pt dt  e pa ,

0
a  0.
- Example 3.
y   2 y   t  t0 ,
p
2

y0  y0  0.
  2 Y  L t  t0   e  pt 0
e  pt 0
1
Y 2

y

sin  t  t0 , t  t0 ( L3, L 28).
p  2

67
- Fourier Transform of a  Function
g   
1
2


0
1
 x  a  
2
 x   e  ix dx 


e  i  x  a d .

1  i a
e .
2
useful in quantum mechanics
68
- Another physical application of  functions
 Point mass (or charge) m x  a , q x  a 
Example 4. 2 at x=3, -5 at x=7, and 3 at x=-4
2 x  3  5x  7  3x  4
69
- Derivative of the  function




 x  x  a dx   x  x  a       x  x  a dx   a .


Sim ilarly,



 x  n  x  a dx   1n  n  a .
70
- Some formulas involving  functions

1, x  a
a. u x  a   

0, x  a
b. ux  a    x  a .
a.   x     x  and   x  a    a  x ;
b.   x     x  and   x  a    a  x ;
c.  ax 
1
 x , a  0;
a
d .   x  a  x  b  
   f  x   
i
1
 x  a    x  b , a  b;
ab
  x  xi 
f  xi 
if f  xi   0 and f  xi   0.
71
-  functions in 2 or 3 dimensions


   x, y, z  x  x   y  y dxdy   x , y 
0
 


0
o
o

    x, y, z  x  x   y  y  z  z dxdydz  x , y , z 
0
  
0
0
o
o
o
 r  r0    3 r  r0    x  x0   y  y0  z  z0 
   f r , ,  r  r d
0


Spherical coord.
 r  r0     0    0 
f r , , 
d  f r0 , 0 ,0 
r 2 sin 
   f r , , z  r  r d
0


f r , , z 
 r  r0     0   z  z 0 
r
Cylindrical coord.
d  f r0 , 0 , z 0 
72

er
 4 r  :
2
r
2
1
1
 e 
         r2   4 r 
r
r
 r 
2 
er
er
1 2


d



e
d


volume r 2
closed surface r 2 r
 0  0 r 2 r sin dd  4 .
cf. divergence theorem
cf .
  D  q r 
73
12. A brief introduction to Green Functions
- Example 1
y   2 y  f t , y0  y0  0

f t    f t  t   t dt .
0
d2
2



G
t
,
t


G t , t    t   t .
2
dt

y t    G t , t  f t dt 
0

 d2
 d2
2
2
y   y   2    y   2     G t , t  f t dt 
 dt

 dt
0
2


0

 d2
2
 2   G t , t  f t dt     t  t  f t dt   f t 
0
 dt

74