Date post: | 29-Dec-2015 |
Category: |
Documents |
Upload: | evelyn-willis-francis |
View: | 222 times |
Download: | 2 times |
Boundary-Value Problems in Other Coordinates
CHAPTER 14
Ch14_2
Contents
14.1 Problems in Polar Coordinates14.2 Problems in Cylindrical Coordinates14.3 Problems in Spherical Coordinates
Ch14_3
14.1 Problems in Polar Coordinates
Laplacian in the Polar CoordinatesWe already know that
urr
uy
uyr
ru
yu
urr
ux
uxr
ru
xu
xy
yxrryrx
cossin
sincos
Besides,
tan , ,sin ,cos 222
Ch14_4
Thus
(1)
(2)
urr
ur
u
r
xu
rr
u
x
u
cossin2sinsin
cossin2cos
2
2
2
2
2
2
22
2
2
urr
ur
u
rxu
rr
u
x
u
cossin2sin
sincossin2cos
2
2
2
2
2
2
22
2
2
Ch14_5
Adding (1) and (2) we have
(3) 011
concern only wesection, In this
11
2
2
22
2
2
2
22
22
u
rru
rr
u
u
rru
rr
uu
Ch14_6
Example 1
Solve Laplace’s Equation (3) subject to u(c,) = f(), 0 < < 2.
Solution Since (r, + 2) is equivalent to (r, ), we must have u(r, ) = u(r, + 2). If we seek a product function u = R(r)(), then (r, + 2) = (r, ).
Ch14_7
Example 1 (2)
Introducing the separation constant , we have
We are seeking a solution of the form
(6)
(5) 0"
(4) 0'"
"'"
2
2
RrRRr
RrRRr
)2()0( ,0
Ch14_8
Example 1 (3)
Of the three possible general solutions of (5):(7)
(8)
(9)
we can dismiss (8) as an inherently non-periodic unless c1 = c2 = 0. Similarly (7) is non-periodic unless c2 = 0. The solution = c1 0 can be assigned any period and so = 0 is an eigenvalue.
0 hsinhcos 221 cc
0 )( 21 cc
0 sincos 221 cc
Ch14_9
Example 1 (4)
When we take = n, n = 1, 2, …, (9) is 2-periodic. The eigenvalues of (6) are then 0 = 0 and n = n2, n = 1, 2, …. If we correspond 0 = 0 with n = 0, the eigenfunctions are
When n = n2, n = 0, 1, 2, … the solutions of (4) are
,...2,1,sincos)(
;0,)(
21
1
nncnc
nc
(11) ,...2,1,)(
(10) 0,ln)(
43
43
nrcrcrR
nrccrRnn
Ch14_10
Example 1 (5)
Note we should define c4 = 0 to guarantee that the solution is bounded at he center of the plate (r = 0). Finally we have
(12) )sincos(),(
gives principleion superposit The
,...2,1,)sincos(
0,
10
00
nnn
n
nnn
n
BnArAru
nnBnAru
nAu
Ch14_11
Example 1 (6)
Applying the boundary condition at r = c, we get
(13) )( is,that
,,2
series,Fourier full ain ofexpansion an as
)sincos()(
2
00
00
10
dfA
bBcaAca
A
f
BnAcAf
nnn
nnn
nnn
n
(15) sin)(1
(14) cos)(1
2
0
2
0
dnfc
B
dnfc
A
nn
nn
Ch14_12
Example 2
Find the steady-state temperature u(r, ) shown in Fig 14.3.
Ch14_13
Example 2 (2)
Solution The boundary-value problem is
crruru
ucu
cru
rru
rr
u
0,0) ,( ,0)0 ,(
0 ,) ,(
0 ,0 ,011
0
2
2
22
2
Ch14_14
Example 2 (3)
and (16)
(17)
The boundary conditions translate into (0) = 0 and () = 0.
"'"
constant, separation and )()( Defining2
RrRRr
rRu
022 RRrRr
02
Ch14_15
Example 2 (4)
Together with (17) we have
(18)
The familiar problem possesses n = n2 and eigenfunctions () = c2 sin n, n = 1, 2, … Similarly, R(r) = c3rn and
un = R(r)() = An rn sin n
0)( 0,)( 0,
Ch14_16
Example 2 (5)
Thus we have
1
0
0
0 0
10
1
sin)1(12
),(
)1(12 ,sin
2
sin ,sin),(
n
nn
n
nnn
n
n
nn
n
nn
ncr
nu
ru
nc
uAdnucA
cAunrAru
Ch14_17
14.2 Problems in Polar Coordinates and Cylindrical Coordinates: Bessel Functions
Radial Symmetry The two-dimensional heat and wave equations expressed in polar coordinated are, in turn
(1)
where u = u(r, , t). The product solution is defined as u = R(r)()T(t). Here we consider a simpler problems that possesses radial symmetry, that is, u is independent of .
and 11
2
2
22
2
tuu
rru
rr
uk
2
2
2
2
22
22 11
t
uu
rru
rr
ua
Ch14_18
In this case, (1) take the forms, in turn,
(2)
where u = u(r, t).
and 1
2
2
tu
ru
rr
uk
2
2
2
22 1
t
uru
rr
ua
Ch14_19
Example 1
Find the displacement u(r, t) of a circular membrane of radius c clamped along its circumference if its initial displacement is f(r) and its initial velocity is g(r). See Fig 14.7.
Ch14_20
Example 1 (2)
Solution The boundary-value problem is
crrgtu
rfru
ttcu
tcrt
uru
rr
ua
t
0 ),( ),()0 ,(
0 ,0) ,(
0 ,0 ,1
0
2
2
2
22
Ch14_21
Example 1 (3)
Substituting u = R(r)T(t) into the PDE, then
(3)
The two equations obtained from (3) are(4)
(5)
This problem suggests that we use only = 2 > 0, > 0.
22
1
Ta
TR
Rr
R
02 rRRRr
022 TaT
Ch14_22
Example 1 (4)
Now (4) is the parametric Bessel differential equation of order v = 0, that is, rR” + R’ + 2rR = 0. The general solution is
(6)
The general solution of (5) is T = c3 cos at + c4 sin at
Recall that Y0(r) − as r 0+ and so the implicit assumption that the displacement u(r, t) should be bounded at r = 0 forces c2 = 0 in (6).
)()( 0201 rYcrJcR
Ch14_23
Example 1 (5)
Thus R = J0(r). Since the boundary condition u(c, t) = 0 implies R(c) = 0,we must have c1J0(c) = 0. We rule out c1 = 0, so
J0(c) = 0 (7)
If xn = nc are the positive roots of (7) then n = xn/c and so the eigenvalues are n = n
2 = xn2/c2 and the eigenfunc
tions are c1J0(nr). The product solutions are
(8)),()sincos( 0 rJtaBtaARTu nnnnnn
Ch14_24
Example 1 (6)
where we have done the useful relabeling of the constants. The superposition principle gives
(9)Setting t = 0 in (9) and using u(r, 0) = f(r) give
(10)This is recognized as the Fourier-Bessel expansion of f on the interval (0, c).
10 )()sincos(),(
nnnnnn rJtaBtaAtru
10 )()(
nnn rJArf
Ch14_25
Example 1 (7)
Now we have
(11)Next differentiating (9) with respect to t, set t = 0, and use ut(r, 0) = g(r):
c
nn
n drrfrrJcJc
A0 02
12 )()(
)(
2
(12) )()()(
2
then,)()(
0 021
2
10
c
nnn
n
nnnn
drrgrrJcJca
B
rJBarg
Ch14_26
Standing Waves
The solution (8) are called standing waves. For n = 1, 2, 3, …, they are basically the graph of J0(nr) with the time-varying amplitude
An cos nt + Bn sin nt The zeros of each standing wave in the interval (0, c) are the roots of J0(nr) = 0 and correspond to the set of points of a standing wave where there is no motion. This set is called a nodal line.
Ch14_27
As in Example 1, the zeros of standing waves are determined from
J0(nr) = J0(xnr/c) = 0Now from Table 5.2 and for n = 1, the first positive root of
J0(x1r/c) = 0 is 2.4r/c = 2.4 or r = cSince the desired interval is (0, c), the last result has
no nodal line. For n = 2, the roots of J0(x2r/c) = 0 are 5.5r/c = 2.4 and 5.5r/c = 5.5We have r = 2.4c/5.5 that has one nodal line. See Fig 14.8.
Ch14_28
Fig 14.8
Ch14_29
Laplacian in Cylindrical Coordinates
See Fig 14.10. We havex = r cos , y = r sin , z = z
and
2
2
2
2
22
22 11
z
uu
rru
rr
uu
Ch14_30
Fig 14.10
Ch14_31
Example 2
Find the steady-state temperature shown in Fig 14.11.
Ch14_32
Example 2 (2)
SolutionThe boundary conditions suggest that the temperature u has radial symmetry. Thus
20 ,)4 ,( ,0)0 ,(
40 ,0) ,2(
40 ,20 ,01
0
2
2
2
2
rururu
zzu
zrz
uru
rr
u
Ch14_33
Example 2 (3)
Using u = R(r)Z(z) and separation constant,
(13)
(14)(15)
For the choice = 2 > 0, > 0, the solution of (14) is R(r) = c1 J0(r) + c2 Y0(r)
Since the solution of (15) is defined on [0, 2], we haveZ(z) = c3 cosh z + c4 sinh z
2
1
ZZ
R
Rr
R
02 rRRRr 02 ZZ
Ch14_34
Example 2 (4)
As in Example 1, the assumption that u is bounded at r = 0 demands c2 = 0. The condition u(2, z) = 0 implies R(2) = 0. Then
J0(2) = 0 (16)defines the eigenvalues n = n
2. Last, Z(0) = 0 implies c3 = 0. Hence we have
R(r) = c1 J0(r), Z(z) = c4 sinh z,
10
0
)(sinh) ,(
and ),(sinh)()(
nnnn
nnnn
rzJAzru
rzJAzZrRu
Ch14_35
Example 2 (5)
2
0 021
20
100
0
)()2(2
24sinh
then,)(4sinh
thus,,4When
drrrJJ
uA
rJAu
uuz
nn
nn
nnnn
Ch14_36
Example 2 (6)
For the last integral, using t = nr and d[tJ1(t)]/dt = tJ0(t), then
10
10
1
0
1
02
0 121
20
)()2(4sinh
sinh),(
)2(4sinh
obtain we
,)2(
)()2(2
4sinh
nn
nnn
n
nnnn
nnnnnn
rJJz
uzru
Ju
A
Ju
dtttJdtd
J
uA n
Ch14_37
14.3 Problems in Spherical Coordinates: Legendre Polynomials
Laplacian in Spherical Coordinates See Fig 14.15. We knew that
(1)and
(2)We shall consider only a few of the simpler problems that are independent of the azimuthal angle .
cos ,sinsin ,cossin rzryrz
u
r
u
r
u
rru
rx
uu 22
2
22
2
222
22 cot1
sin
12
Ch14_38
Fig 14.15
Ch14_39
Example 1
Find the steady-state temperature u(r, ) shown in Fig 14.16.
Ch14_40
Example 1 (2)
Solution The problem is defined as
' cot"'2"
then),()(Let
0 ),() ,(
0 ,0 ,0cot1
2
222
2
22
2
RrRRr
rRu
fcu
cru
rr
uru
rr
u
Ch14_41
Example 1 (3)
and so (2)
(3)After letting x = cos , 0 , (3) becomes
(4)
This is a form of Legendre’s equation. Now the only solutions of (4) that are continuous and have continuous derivatives on [-1, 1] are the Legendre polynomials Pn(x) corresponding to 2 = n(n+1), n = 0, 1, 2, ….
02 22 RRrR
0sincossin 2
11,02)1( 22
22 x
dxd
xdx
dx
Ch14_42
Example 1 (4)
Thus we take the solutions of (3) to be = Pn(cos )
When = n(n + 1), the solution of (2) is R = c1 rn + c2 r –(n+1)
Since we again expect u to be bounded at r = 0, we define c2 = 0. Hence,
0
0
)(cos)( ,At
)(cos),( and ,)(cos
nn
nn
nn
nnn
nnn
PcAfcr
PrAruPrAu
Ch14_43
Example 1 (5)
Therefore Ancn are the coefficients of the Fourier-Legendre series (23) of Sec 12.5:
00
0
)(cossin)(cos)(2
12
),(
thus
,sin)(cos)(2
12
nn
n
n
nnn
Pcr
dPfn
ru
dPfc
nA