chapter 13 Partial differential equations
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Transcript chapter 13 Partial differential equations
Mathematical methods in the physical sciences 3nd edition Mary L. Boas
Chapter 13 Partial differential equations
Lecture 13 Laplace, diffusion, and wave equations
1
1. Introduction (partial differential equation)
ex 1) Laplace equation
2
2
2
ux, y, z 0 2 u 2 u 2 u 0
x
y
z
2
: gravitational potential, electrostatic potential, steady-state
temperature with no source
2u f x, y, z
ex 2) Poisson’s equation:
: with sources (=f(x,y,z))
ex 3) Diffusion or heat flow equation
2u x, y , z , t
1
u x, y, z, t ( : diffusivit y)
2 t
2
1 2u
ex 4) Wave equation u 2 2
v t
2
ex 5) Helmholtz equation 2 F k 2 F 0
: space part of the solution of either the diffusion or the wave equation
3
2. Laplace’s equation: steady-state temperature in a rectangular plate
(2D)
In case of no heat source
2T 2T
T 0 or
2 0
2
x
y
2
T o solve this equation, we try a solution of the form,
T x, y X x Y y ,
X ( x) : a function only of x,
Y ( y ) : a function only of y.
"Separation of variables"
4
2T 2T
2 XY 2 XY
2 0
0
2
2
2
x
y
x
y
d2X
d 2Y
1 d 2 X 1 d 2Y
XY
Y
X 2 0
0,
2
2
2
dx
dy
X dx
Y dy
1 d2X
1 d 2Y
2
X dx
Y dy2
t above equation with the variable- separated sides,
T o satisfy he
1 d 2Y
1 d2X
2
(separationconstant k 0)
k
.
const
Y dy2
X dx2
X k 2 X and Y k 2Y .
sin kx,
X
coskx,
e ky ,
Y
ky
e ,
e ky sin kx
e ky sin kx
T XY
ky
e coskx
ky
e coskx
5
e ky sin kx
e ky sin kx
General solut ion: T XY
ky
e coskx
ky
e coskx
1) In the current problem, boundary conditions are
i ) T 0 as y . discard e ky
ii) T 0 when x 0. discard coskx
iii) T 0 when x 10. sin 10k 0 k n / 10
iv) T 10 when y 0.
nx
nx
T exp[ny / 10]sin
T bn sin
100.
10
10
n 1
6
nx
nx
T exp[ny / 10] sin
T bn sin
100.
10
10
n 1
Using theFourier series,
10
2 l
nx
2 10
nx
10
nx
bn f ( x) sin
dx 100sin
dx 20
cos
l 0
l
10 0
10
n
10 0
400
, odd n,
200
n
1 1 n
n
0, even n.
T
400
x 1 3y / 10 3x
sin
.
exp[ny / 10] sin e
10 3
10
ex.
At x 5 and y 5, T 26.1
7
2) How about changing the boundary condition? Let us consider a
finite plate of height 30 cm with the top edge at T=0.
T=0 at 30 cm
In this case, e^ky can not be discarded.
e ky ae ky beky
T 0 at y 30 12 e k 30 y 12 e k 30 y ( thatis, a 12 e30 k , b 12 e 30 k ) sinh k 30 y
T Bn sinh
n 1
n
30 y sin nx
10
10
nx
nx
100 Bn sinh 3n sin
bn sin
, where bn Bn sinh 3n .
10
10
n 1
n 1
Ty 0
T
odd n
1
n
400
30 y sin nx
sinh
10
10
n sinh 3n
8
- To be considered I
e kx sin ky
kx
e sin ky
We could have another general solution,T XY
, for k 2 .
kx
e cosky
kx
e cosky
This is correct, but makes the problem more complicated.
(Please check the boundary condition.)
- To be considered II
In case that the two adjacent sides are held at 100 (ex. C=D=100),
the solution can be the combination of C=100 solutions (A, B, D: 0)
and D=100 (A, B, C: 0) solutions.
9
-. Summary of separation of variables.
1)
2)
3)
4)
5)
A solution is a product of functions of the independent variables.
Separate partial equation into several independent ordinary equation.
Solve the ordinary differential eq.
Linear combination of these basic solutions
Boundary condition (boundary value problem)
10
3. Diffusion or heat flow equation; heat flow in a bar or slab
- Heat flow : 2u
u F x, y, z T t
T 2 F
1
2
F
1 u
.
2 t
"Separation of variables"
dT
1
1 1 dT
2 F 2
k 2 .
dt
F
T dt
1 2
1 1 dT
2
F k 2 ,
k
F
2 T dt
F k F ,
2
2
dT
2 2
k 2 2 t
k T T e
dt
cf. “Why do we need to choose –k^2, not +k^2?”
11
Let’s take a look at one example.
At t=0, T=0 for x=0 and T=100 x=l.
From t=0 on, T=0 for x=l.
In case of this 1 - D problem,
sin kx,
1 2
d 2F
2
2
F k 2 k F 0 F
F
dx
coskx
e k t sin kx
u 2 2
k t
coskx
e
2
2
For T(x=0)=0 and T(x=l)=100 at t=0, the initial steady-state
temperature distribution:
d 2u0
u0 0 (no heat source)
0
2
dx
100
u0 ax b u0
x
l
2
12
1) u 0 when x 0. discard cos kx
- Using Boundary condition
2)u 0 when x l. sin kl 0 kl n
ue
n / l 2 t
2
nx
nx
sin
u bn e n / l t sin
l
l
n 1
At t 0, u u0
u bn sin
n 1
nx
100
u0
x.
l
l
100 2l 1
1n 1 200 1
bn
l n
n
n 1
u
200 / l 2 t
x 1 2 / l 2 t
2x 1 3 / l 2 t
3x
e
sin
e
sin
e
sin
.
l 2
l
3
l
13
For some variation, when T0,
we need to consider uf.as the final state, maybe a linear
In this case, we can write down the solution simply like this.
u bn e n / l t sin
n 1
2
nx
uf .
l
nx
nx
u0 bn sin
u f u0 u f bn sin
.
l
l
n 1
n 1
14
4. Wave equation; vibrating string
node
Under the assumption that the string is not stretched,
2 y 1 2 y
x 2 v 2 t 2
x=0
x=l
y X x T t
1 d2X
1 1 d 2T
2
2
2 2
const
.
k
X
k
X
0
and
T
k
v T 0.
2
2
2
X dx
v T dt
frequency(sec-1 ),
wavelength
2 angular frequency(radians)
k
2
2
wave number.
v
v
15
sin kx,
X
coskx,
sin kx sin t
sin kvt,
sin kx cost
T
y XT
, where kv.
coskvt,
coskx sin t
coskx cost
Boundarycondition: y 0 at x 0, x l.
nvt
nx
sin
sin
l
l
y
sin nx cos nvt .
l
l
16
nvt
nx
sin
sin
l
l
y
sin nx cos nvt
l
l
nvt
nx
sin
cos
l
l
y
sin nx sin nvt
l
l
1) case 1
For y (t 0) f ( x) and y ' 0, sin nvt / l should be discarded.
y bb sin
n 1
y0 bn sin
n 1
nx
nvt
cos
.
l
l
nx
f x .
l
After finding the Fourier series coefficients,
y
8h x
vt 1 3x
3vt
sin
cos
sin
cos
9
2
l
l
l
l
17
nvt
nx
sin
sin
l
l
y
sin nx cos nvt .
l
l
2) case 2
For y (t 0) 0 and y0 ' 0, cosnvt / l should be discarded.
y Bn sin
n 1
nx
nvt
sin
.
l
l
nv
nx
nx
y
B
sin
b
sin
V x (Use Fourier series expansion.)
n
n
l
l
l
t t 0 n 1
n 1
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3) Eigenfunctions
nx
nvt
sin
: a characteristic function or eigenfunction
l
l
n nv / l 2 n : characteristic frequency
y sin
A vibration with a pure freuquencyis called thenormalmode of vibration.
first harmonic, fundamental second harmonic
third
fourth
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Mathematical methods in the physical sciences 2nd edition Mary L. Boas
Chapter 13 Partial differential equations
Lecture 14 Using Bessel equation
20
5. Steady-state temperature in a
cylinder
For this problem, cylindrical coordinate (r, , z) is more us
1 u 1 2u 2u
u
2 0
r 2
2
r r r r
z
2
u Rr Z z
P utting this into the equation, and dividing with u RZ ,
1 1 d dR 1 1 d 2 1 d 2 Z
0
r
2
2
2
R r dr dr r d
Z dz
21
1 1 d dR 1 1 d 2 1 d 2 Z
0
r
2
2
2
R r dr dr r d
Z dz
In order to say that a term is constant,
1) function of only one variable
2) variable does not elsewhere in the equation.
- 1st step
1 1 d dR 1 1 d 2
1 d 2Z
k 2
r
2
2
2
R r dr dr r d
Z dz
e kz
1 d 2Z
2
k ,Z
.
2
kz
Z dz
e
Because u ~ 0 at z ,
u ~ e kz .
22
- 2nd step
1 1 d dR 1 1 d 2
r d dR 1 d 2
r 2
2
k 0
r 2k 2 0
r
r
2
2
2
R r dr dr r d
R dr dr d
Here,
sin n ,
1 d 2
2
n
,
2
cosn .
d
- 3rd step
r d dR 2
d dR
2 2
2 2
2
2
2 2
2
r
n k r 0 r r
k r n R 0 r R rR k r n R 0
R dr dr
dr dr
" Bessel equation": x xy x 2 p 2 y x 2 y xy x 2 p 2 y 0
Solut ion of the Bessel equations: J n kr (Bessel),
N n kr (Neumannor Weber).
Here, we take only R ~ J n (kr), because N n kr at r 0.
- Boundarycondition: Rr 1 J n k 0.
J n kr sin ne kr
u RZ
, where k is a zero of J n .
J kr cosne kr
n
23
In azimuthalsymmetry,n 0 J 0 km 0 ( J n : oscillating function)
u cm J 0 km r e k m z .
n 1
Boundarycondition: u z 0 cm J 0 km r 100.
n 1
Here, using theorthogonal
ity of the Bessel function,
1
0
rJ 0 k r cm J 0 km r dr 100 rJ 0 k r dr
1
0
n 1
Only if m, we have the non - zero termin the left side.
cm r J 0 km r dr 100rJ 0 km r dr.
1
2
0
1
0
left term: rJ 0 km r dr 12 J12 km .
1
2
0
J p (a ) 0 J p (b)
0, if a b
0 xJ p axJ p bxdx 1 J p2 1 a 1 J p2 1 a 1 J p21 a , if a b
2
2
2
1
24
cm rJ 0 km r dr 100rJ 0 km r dr.
1
2
0
1
0
For the right term:
rJ 0 km r
d
xJ1 x xJ0 x (recusion relation)
dx
1 d
km rJ1 km r km rJ 0 km r ,
For x km r ,
km dr
1 d
rJ1 km r
km dr
1
1
1
rJ
k
r
dr
rJ
k
r
J1 km .
1 m
0 0 m
km
k
m
0
1
Combiningtwo results,
cm r J 0 km r dr 100rJ 0 km r dr cm 12 J12 km
1
0
cm
2
1
0
100J1 km
km
100J1 km
2
200
2
km
J1 km km J1 km
25
If thegiven temperatureof thebase of thecylinderis f r , , 1, but an cosn bn sin n .
u J n kmn r amn cosn bmn sin n e k mn z .
m 1 n 0
u z 0 J n kmn r amn cosn bmn sin n f r , .
m 1 n 0
Because of theorthogonality,
1 2
0 0
f r , J k r cosrdrd a
1 2
a 12 J21 k .
Similarly, b
2
J
0 0
2
1
1 2
k
0 0
J k r cos2 rdrd
2
f r , J k r sinrdrd .
26
Bessel’s equation
- Bessel’s equation
1) Equation and solution
- named equation which have been studied extensively.
- “Bessel function”: solution of a special differential equation.
xxy x 2 p 2 y x 2 y xy x 2 p 2 y 0
- being something like damped sines and cosines.
- many applications.
ex) problems involving cylindrical symmetry (cf. cylinder function); motion
of pendulum whose length increases steadily; small oscillations of a flexible
chain;
railway transition curves; stability of a vertical wire or beam;
Fresnel integral in optics; current distribution in a conductor; Fourier series
for the arc of a circle.
27
Bessel’s equation
xxy x 2 p 2 y x 2 y xy x 2 p 2 y 0
- Solution :
2n p
n
1
x
p 1 x
J p x
,
where
p
x
e dx p 1!,
0
n 0 n 1n p 1 2
cosp J p x J p x
N p x
(Neumann or Weber)
p 0.
sin p
generalsolution y AJ p x BN p x
- Graph
28
Bessel’s equation
2) Recursion relations
4)
d p
x J p x x p J p 1 x
dx
d p
x J p x x p J p 1 x
dx
2p
J p 1 x J p 1 x
J p x
x
J p 1 x J p 1 x 2 J p x
5)
J p x
1)
2)
3)
p
p
J p x J p 1 x J p x J p 1 x
x
x
29
Bessel’s equation
3) Orthogonality
cf. Comparison
J p x , N p x
consider just J p x for one value of p
x a, b,, J p x 0
x 1, J p ax 0, J p bx 0,
2
For y sin nx, y n y 0 For J p ax, xxy a 2 x 2 p 2 y 0
sin x, cos x
consider just sin x
x nx, sin x 0
x 1, sin nx 0
1
sin nx sin mxdx 0,
0
for n m,
xJ axJ bxdx 0,
1
0
p
p
for a b
30
Bessel’s equation
xxy x 2 p 2 y x 2 y xy x 2 p 2 y 0
xxy a 2 x 2 p 2 y 0
J p (a) 0 J p (b)
0, if a b
0 xJ p axJ p bxdx 12 J p21 a 12 J p21 a 12 J p21 a , if a b
1
31
6. Vibration of a circular membrane (just like drum)
1 2 z
z 2 2
v t
2
z F x, y T t 2 F k 2 F 0 and T k 2v 2T 0
1 F 1 2 F
In polarcoordinate,
k 2 F 0.
r
2
2
r r r r
F Rr . P utting this form into the above equation, and then dividing with F ,
1 1 d dR 1 1 d 2
k2 0
r
2
2
r R dr dr r d
1 d 2
d 2
2
2
n
n
0 (simple harmonic equation)
2
2
d
d
r
d dR
2
2 2
r
n k r R 0 (Bessel's equation)
dr dr
sin n sin kvt
sin n sin kvt
sin n coskvt
J n kr
z Rr T t J n kr
cosn coskvt
cosn sin kvt
cosn coskvt
32
sin n sin kvt
sin n coskvt
z Rr T t J n kr
cosn sin kvt
cosn coskvt
Boundarycondit ion: z (r 1) 0 J n k 0.
kmn : zerosof J n (double series)
Characteristic freuqency of thenormal modes: mn kmn v / 2
cf. They are not integral multiples of the fundamental as is true for the
string (characteristics of the bessel function). This is why a drum is less
musical than a violin.
33
ex) z J n kmn r cosn coskmn vt
34
(m,n)=(1,0)
(2,0)
(1,1)
American Journal of Physics, 35, 1029 (1967)
35
Mathematical methods in the physical sciences 2nd edition Mary L. Boas
Chapter 13 Partial differential equations
Lecture 15 Using Legendre equation
36
7. Steady-state temperature in a sphere
- Sphere of radius 1 where the surface of upper half is 100, the other is 0 degree
1 2 u
1
u
1
2u
u 2 r
0
2
sin
2 2
2
r r r r sin
r sin
2
1 d 2 dR 1 1 d
d 1 1 d 2
For u Rr ,
0
r
sin
R dr dr sin d
d sin 2 d 2
sin m
1 d 2
2
m ,
d 2
cosm
1 d 2 dR 1 1 d
d
m2
0.
r
sin
R dr dr sin d
d sin 2
r l
1 d 2 dR
R
r
k
R dr dr
1 / r l 1
1 d
d
m2
sin
2 k 0 (Associat ed Legendre equat ion k l l 1)
sin d
d sin
Pl m cos (Legendre polynomial
).
37
1) For the interior of the sphere, no divergence. T herefore,discard 1/ r l 1.
sin m
u r Pl cos
cosm
l
m
2) Due to the azimuthal symmetry,m 0
u cl r l Pl cos
l 0
100, 0 / 2
3) ur 1 cl Pl cos
/2
l 0
0,
0, 1 x 0,
or ur 1 cl Pl cos 100 f x , where f x
l 0
1, 0 x 1.
In thiscase, we can expandthisin a series of Legendrepolynomina
ls
f x 12 P0 x 34 P1 x 167 P3 x 11
P x (Here, x cos )
32 5
c P cos 100 P cos
l 0
1
2
l l
0
3
4
P1 cos 167 P3 cos 11
P cos
32 5
1
3
7
c0 100 , c1 100 , c2 100 ,.
2
4
16
5
u x 100 12 P0 cos 34 rP1 cos 167 r 3 P3 cos 11
r
P5 cos
32
38
Legendre’s equation
- Legendre’s equation
1) Equation and solution
i. .m 0 (Legendre equation)
l
l
1 d 2
1 x 2 y 2 xy l l 1 y 0 Solution : Pl x
x 1 .
2!l! dx
ii. m 0 (Associated Legendre equation)
m2
1 x y 2 xy l l 1
y0
2
1 x
2
Solution : Pl x 1 x
m
For
m l , Pl m 0
2 m/2
m
d
Pl x l m l .
dx
1 d
d
m2
cf .
l l 1 0 (Associated Legendre equation).
sin
sin d
d sin 2
39
Legendre’s equation
1 d
d
m2
cf .
l l 1 0 (Associated Legendre equation)
sin
sin d
d sin 2
1
d d cos
d d cos
m2
l l 1 0
sin
sin d cos x d
d cos d sin 2
d 2
d
m2
sin
2 l l 1 0
d cos x
d cos sin
d
d
m2
2
l l 1 0
1 cos
d cos x
d cos 1 cos2
d 2
d
m2
1 cos
2 cos
l l 1 0
d cos x 2
d cos 1 cos2
2
d 2
d
m2
1 x
2x
l l 1 0
2
2
dx
dx 1 x
2
40
Legendre’s equation
- Legendre polynomials
- Associated Legendre polynomials
P0 x 1
Pl m x 1
m
P1 x x
P2 x
P10 x x
P11 x 1 x 2
1
cf . x 3P1 x 2 P3 x
5
1/ 2
P2 2 x
3
1
P11 x P11 x
2
1 2
3x 1
2
1
P3 x 5 x 3 3x
2
1
P4 x 35x 4 30x 2 3
8
1
P5 x 63x 5 70x 3 15x
8
1
P6 x
231x 6 315x 4 105x 2 5
16
l m! P m x
l m! l
1 2
P2 x
24
1
P21 x P21 x
6
1
P20 x 3 x 2 1
2
P21 x 3 x 1 x 2
P22 x 3 1 x 2
1/ 2
41
Legendre’s equation
2) Orthogonality
2
2
P
x
dx
P
cos
sin d
1 l
0 l
1
P
1
1
l
m
2
.
2l 1
2
2 l m!
Pl m cos sin d
.
2l 1 l m!
x 2 dx 0
cf .x cos dx sin d
42
8. Poisson’s equation
F 0 (F : conservative) F V
1) For gravit ational force, F 4G 2V 4G : Poisson equation
cf. 0 2V 0 : Laplace equation
2) For electrostatic force, E 4 2V 4 (in Gaussian unit ).
cf. 0 2V 0
In general, 2u x, y, z f x, y, z
u
1
4
f x, y, z
x x y y z z
2
2
2
dxdydz.
2 w 0 (Laplaceequation)
2 u w 2u 2 w f u w : solutionof Poissonequation
43
Example 1
2V 4
V x, y , z
1
4
4 x, y, z
x x y y z z
2
2
2
dxdydz
q
x y z a
2
2
2
.
In sphericalcoordinat e,
Vq
q
r 2ar cos a
2
2
.
r l m
sin m
Basic solution of Laplaceequation
Pl cos
r l 1Pl cos
r l 1
cosm
1) zero at infinite discard r l
2) symmet ricabout z - axis solutionis indep.of . m 0, cosm 1
V Vq cl r l 1Pl cos
l
grounded sphere
44
Boundarycondition: V 0 for r R.
Vr R
Using
cl R l 1Pl cos 0.
q
R 2 2aR cos a 2
l
R l Pl cos
q
,
l 1
2
2
a
R 2aR cos a
l
q
R l Pl cos
q
cl R l 1Pl cos
l 1
a
l
l
From this relation, cl R
V
qRl
qR2l 1
l 1 or cl l 1 .
a
a
R 2l 1r l 1Pl cos
q
2
2
a l 1
r 2ar cos a
l
q
q
r 2ar cos a
2
2
q
r 2ar cos a
2
l 1
2
q ( R / a )
R / a P cos
2
r
l
l
l
l 1
R / a q
r R / a 2r R / a cos
2
2
2
2
‘Method of the images’
.
In thesecond term,charge: R / a q, position: 0,0,R 2 / a
45
cf. Electric multipoles
monopole
dipole
quadrupole
octopole
2) Expansion for the potential of an arbitrary localized charge distribution
V r
1
4 0
1
rdr
r r
2
r
r
2
2
r r r r 2rr cos r 1 2 cos
r
r
2
2
r r
r r r 1 , where 2 cos
r r
1
1
1 1
3
5
1 / 2
1 1 2 3
r r r
r 2
8
16
46
2
2
3
3
1
1 1 r r
5 r r
3 r r
1 2 cos 2 cos 2 cos
r r
r 2 r r
8 r r
16 r r
2
3
1 r
r
r
2
3
1 cos 3 cos 1 / 2 5 cos 3 cos / 2
r r
r
r
1 r
Pn cos
r n 0 r
n
V r
-
1
4 0
n
n
n
n
=
=
=
=
0
1
2
3
Legendrepolynomina
l
r r P cos rd multipoleexpansion
n 0
:
:
:
:
1
n 1
n
n
monopole contribution
dipole
quadrupole
octopole
47