chapter 8 Ordinary differential equation
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Transcript chapter 8 Ordinary differential equation
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 , y2 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 lnax
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
ye
Pdx c
Ae
Pdx
Ae I , where I Pdx
11
yeI A
differenti
ating
d
dI
yeI ye I yeI
ye 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
x4
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 2e2 t 1 N 0e 1t e2 t dt c 1 N 0 e2 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
Px, y dx Qx, 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 cosx 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 4D 1 y D 1D 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 Beax 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 ex Aeix Beix
ex c1 sin x c2 cos x cex sin x
Example 2) y 6 y 9 y 0.
D
2
6D 9 y D 3D 3y 0 y Ax Be3x
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 Aeit Be it c1 sin t c2 cost 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 cost 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 10cost
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 a1Dy a2 y 0 ( D 2 a1D a2 ) y 0
y c1eax c2ebx is thegeneralsolutionof ( D a)(D b) y 0, if a b.
y Ax Beax is thegeneralsolutionof ( D a)(D a) y 0.
y Ae i x Be i x ex Aeix Be ix
ex c1 sin x c2 cos x cex 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 1D 2y e x
u D 2 y D 1u 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 2y 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 c1e3 x c2 y 13 xe x c1e 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 1D 5 y 7e 2 x (1,-5 2)
y p Ce 2 x
yp 4 yp 5 y p C 4e 2 x 8e 2 x 5e 2 x 7e 2 x C 1
y Aex Be5 x e 2 x
Ce cx
cx
D a D by 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 2Ce 2ix 4e2ix
4
4 2i 6 1
C
i 3
2i 6
40
5
Yp 15 i 3e 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 solveD a D b y
,
k cosx
first solve
D a D b y keix ,
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 by 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
yp yp 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 1D 2 y e x 4 sin 2 x x 2 x
D 1D 2y e x y p1 13 xex
D 1D 2y 4 sin 2 x y p 2 15 cos2 x 53 sin 2 x
D 1D 2y 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 eit 13 e3it e3it .
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 : yy f y y 0
1
2
or
ydy f y dy 0
y2 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
,
pa
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) Lcf 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
,
pa
Re p a 0.
a ia
T akingLaplacetransform, L eiat F p
1
p ia
2
,
2
p ia p a
Lcos at i sin at Lcos at iLsin at
p
a
i
.
2
2
2
2
p a
p a
Lcos at i sin at
Re p ia 0.
p
a
i
,
p2 a2
p2 a2
Re p ia 0.
Using theabove results,
Lcos at i sin at Lcos at i sin at 2 Lcos at 2
Lcos at i sin at Lcos at i sin at 2iLsin at 2i
p
p2 a2
a
p2 a2
: L3
: L4
54
Example 4. Let us verify L11. L(t sin at)
Lcosat e pt cosat dt
0
p
.
2
2
p a
Differentiate the above relation with respect toa,
0
e pt t sin atdt
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 y0 p 2 L y py0 y0 p 2Y py0 y0 .
Note) The relations already include the initial conditions.
56
Example 1.
y 4 y 4 y t 2e2t
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 2Y Z 1,
1)
Y p 2Z 0.
p 2 1Y p 2
2
or
Y
p2
p 22 1
Using the inverse transform ( L14),
2) Sim iarly, Z
1
p 22 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 et ,
y0 y0 0.
p 2Y 3 pY 2Y L et
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 ht 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
ix
0
e
f1 v e
i v
1
2
f 2 u e iu du
e i v u f1 v f 2 u dvdu
e ix 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 ix 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 mv1 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 ix 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 5x 7 3x 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 1n n a .
70
- Some formulas involving functions
1, x a
a. u x a
0, x a
b. ux 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;
ab
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 dd 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