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化工應用數學
授課教師: 郭修伯
Lecture 8 Solution of Partial Differentiation Equations
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Solution of P.D.E.s
– To determine a particular relation between u, x, and y, expressed as u = f (x, y), that satisfies
• the basic differential equation• some particular conditions specified
– If each of the functions v1, v2, …, vn, … is a solution of a linear, homogeneous P.D.E., then the function
is also a solution, provided that the infinite series converges and the dependent variable u occurs once and once only in each term of the P.D.E.
1
nvv
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Method of solution of P.D.E.s
• No general formalized analytical procedure for the solution of an arbitrary partial differential equation is known.
• The solution of a P.D.E. is essentially a guessing game.• The object of this game is to guess a form of the
specialized solution which will reduce the P.D.E. to one or more total differential equations.
• Linear, homogeneous P.D.E.s with constant coefficients are generally easier to deal with.
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Example, Heat transfer in a flowing fluid
An infinitely wide flat plate is maintained at a constant temperature T0. The plate is immersed in an infinately wide the thick stream of constant-density fluid originally at temperature T1. If the origin of coordinates is taken at the leading edge of the plate, a rough approximation to the true velocity distribution is:
Turbulent heat transfer is assumed negeligible, and molecular transport of heat is assumed important only in the y direction. The thermal conductivity of the fluid, k is assumed to be constant. It is desired to determine the temperature distribution within the fluid and the heat transfer coefficient between the fluid and the plate.
00 zyx VVyV
B.C.T = T1 at x = 0, y > 0T = T1 at x > 0, y = T = T0 at x > 0, y = 0T0
x
y
T1 T1dx
dy
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Input - output = accumulation
y
Tk
yx
CTVx const. properties
2
2
y
T
CV
k
x
T
x
yVx
2
2
y
T
y
A
x
T
C
kA
T = T1 at x = 0, y > 0T = T1 at x > 0, y = T = T0 at x > 0, y = 0
Heat balance on a volume element of length dx and height dy situated in the fluid :
Input energy rate: y
TkdxCTdyVx
Output energy rate:
dyy
Tkdx
yy
TkdxdxCTdyV
xCTdyV xx
10
1
TT
TT
2
2
yy
A
x
= 0 at x = 0, y > 0 = 0 at x > 0, y = = 1 at x > 0, y = 0 http://ebooks.edhole.com
2
2
yy
A
x
B.C. = 0 at x = 0, y > 0 = 0 at x > 0, y = = 1 at x > 0, y = 0
Assume: fx
yf
n
Compounding the independent
variables into one variable
= 0 at = = 1 at = 0
Replace y and x in the P.D.E by
dyy
dxxd
dd
d
dd
d
d
x
n
d
d
x
ny
xd
d
x n
1
nx
y
d
d
xyd
d
y n
1
dy
d
d
d
xd
d
yxd
d
xyy nnn
2
2
2
2 111
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2
2
yy
A
x
2
2
3
1
d
dA
xd
d
x
nn
In order to eliminate x and y, we choose n = 1/3
03
2
2
2
d
d
Ad
d
AB
d
d
9exp
3
10
13
0 9exp
TT
TTd
ABd
= 0 at = = 1 at = 0
dA
dA
dB
0
3
0
3
1
0
9exp
1
9exp
dA
dA
TT
TT
0
3
3
10
1
9exp
9exp
= 1 at = 0
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0
10 )(
y
TkTThLocal heat transfer coefficient
y
TTy
T
10
d
d
xyd
d
y n
1
10
1
TT
TT
yd
dT
y
T
d
d
x
TT
y
T
31
10
d
A
B
0
3
9exp
1
AB
d
d
9exp
3
dA
A
x
TT
y
T
0
3
3
31
10
9exp
9exp
0
10 )(
y
TkTTh
d
Ax
kh
0
33
1
9exp
= 0 at y = 0
31
43.0
kx
Ckh
C
kA
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Separation of variables: often used to determine the solution of a linear P.D.E.
Suppose that a slab (depending indefinately in the y and z directions) at an initial temperature T1 has its two faces suddenly cooled to T0. What is the relation between temeprature, time after quenching, and position within the slab?
2R
x
dx
Since the solid extends indefinately in the y and z direction, heat flows only in the x direction. The heat-conduction equation:
2
2
x
T
t
T
Boundary condition:
0,2
0,0
0,
0,0
0
0
0
1
tRxatTT
txatTT
xtatTT
xtatTT
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Dimensionless:2
2
x
T
t
T
0,2
0,0
0,
0,0
0
0
0
1
tRxatTT
txatTT
xtatTT
xtatTT 01
0
TT
TT
2
2
xt
0,20
0,00
0,0
0,01
tRxat
txat
xtat
xtat
Assume: )(tgxf Separation of variables
)(
)(
2
2
tgxfx
tgxft
2
2
xt
)()()()( tgxftgxf
)(
)(
)(
)(
tg
tg
xf
xf
independent of t independent of xhttp://ebooks.edhole.com
0)()(
0)()(
tgtg
xfxf
)(
)(
)(
)(
tg
tg
xf
xf
when 0
tCetg
xBxAxf
)(
cossin)(
when = 0
0
00
)(
)(
Ctg
BxAxf
)(tgxf texBxA cossin
)(tgxf00 BxA
A0, B0, A, B, and have to be chosen to satisfy the boundary conditions.
0,20
0,00
0,0
0,01
tRxat
txat
xtat
xtat
texBxABxA cossin00Superposition:
000 BA
0B
R
n
2
n is an integer
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texBxABxA cossin00
tR
n
eR
xnA
2
2sin
The constant has to be determined.But no single value can satisfy the B.C.
0,01 xtatB.C.
More general format of the solution (by superposition):
1
2
2sin
n
tR
n
n eR
xnA
0,01 xtat
1 2sin1
nn R
xnA
dxR
xnA
R
xmdx
R
xm R
nn
R
2
01
2
0 2sin
2sin
2sin
dxR
xmAdx
R
xn
R
xmAdx
R
xm R
nn
R
n
R
2
0
2
1
2
0
2
0 2sin
2sin
2sin
2sin
Orthogonality property
ndx
R
xn
RA nR
n
2])1(1[
2sin
1 2
0
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nA n
n
2])1(1[
1
2
2sin
n
tR
n
n eR
xnA
01
0
TT
TT
1
4
10
1 2
22
2sin
)1(12
n
tR
nn
eR
xn
nTT
TT
The representation of a function by means of an infinite series of sine functions is known as a “Fourier sine series”.
More about the “Orthogonal Functions”Two functions m(x) and n(x) are said to be “orthogonal” with respect to the weighting function r(x) over interval a, b if:
b
a nm dxxxxr 0)()()(
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dxR
xmAdx
R
xn
R
xmAdx
R
xm R
nn
R
n
R
2
0
2
1
2
0
2
0 2sin
2sin
2sin
2sin
R
xm
2sin
R
xn
2sin
and are orthogonal with respect to the weight function
(i.e., unity) over the interval 0, 2R when m n.
Each term is zero except when m = n.
Back to our question, we had two O.D.E.s and the solutions are :
0)()(
0)()(
tgtg
xfxf
tCetg
xBxAxf
)(
cossin)( where shows!
R
xn
2sin
0)()( xfxf 00 xat
xCxf sin)( Rxat 20
R
nn 2
These values of are called the “eigenvalues” of the equation, and the
correponsing solutions, are called the “eigenfunctions”.
n
xsin http://ebooks.edhole.com
Sturm-Liouville Theory
• A typical Sturm-Liouville problem involves a differential equation defined on an interval together with conditions the solution and/or its derivative is to satisfy at the endpoints of the interval.
• The Strum-Liouville differential equation:
• In Strum-Liouville form:
0)]()([)( yxPxQyxRy
0)]()([])([ xpxqyxr eigenvalue
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• The regular problem on [a,b]
• The periodic problem on [a,b]
• The singular problem on [a,b]
0)()(
0)()(
21
21
byBbyB
ayAayA
)()(
)()(
byay
byay
0)()(0)( 21 byBbyBar
0)()(0)( 21 ayAayAbr0)()( brar
A Strum-Liouville differential equation with boundary conditions at each end point x = a and x = b which satisfy one of the following forms:
has solutions, the eigenfunctions m(x) and n(x) which are orthogonal provided that the eigenvalues, m and n are different.
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If the eigenfunctions, (x) result from a Strum-Liouville differential equation and nemce be orthogonal. The formal expansion of a general solution f(x) can be written in the form:
0
)()(n
nn xAxf
The value of An can be obtained by making use of the orthogonal properties of the functions (x)
b
a
b
an
nnmm dxxAxxrdxxfxxr0
)()()()()()(
b
a
b
a nmn
nm dxxxxrAdxxfxxr )()()()()()(0
Each term is zero except when m = n.
b
a
n
b
a
n
n
dxxxr
dxxfxxr
A2)]()[(
)()()(
0, 1, 2…… are eigenfunctionshttp://ebooks.edhole.com
Steady-state heat transfer with axial symmetry
0sinsin
12
T
r
Tr
r
0cot22
2
2
22
TT
r
Tr
r
Tr
Assume: )(grfT
)(
)(
grfT
grfr
T
0)()(cot)()()()(2)()(2 grfgrfgrfrgrfr
)1(cot22
llg
gg
f
frfr Dividing by fg and separate variables
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)1(cot22
llg
gg
f
frfr 0)1(cot
0)1(22
gllgg
fllfrfr
0)1(22 fllfrfr
nArf set
0)1(2)1( nnn ArllnArArnn 0)1(2)1( llnnn
1 ll BrArf
0)1(cot gllgg cosmset
ddm sin
0)1(2)1( 2 gllgmgmLegendre’s equation of order l
Solved by the method of Frobenius
0
)(n
cnnmamgset
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0)1(2)1( 2 gllgmgm
0
)(n
cnnmamg
0)1()(2)1)(()1(00
1
0
22
n
cnn
n
cnn
n
cnn mallmacnmmacncnm
比較係數2cm 0)1( 0 acc 10 orc
1cm 0)1( 1 cac 00 1 aorc
and
ss alcslcsacscs )1)(()1)(2( 2
)()()( mBQmAPmg ll where Pl(m) is the “Legendre polynomial”
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)(grfT 1 ll BrArf
)()()( mBQmAPmg ll
cosm
)(cos)( 1 ll
ll
l PrBrAT
superposition
0
1 )(cos)(l
ll
ll
l PrBrAT
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Unsteady-state heat transfer to a sphere
t
TT
12
t
T
r
T
rr
T
12
2
2
A sphere, initially at a uniform temperature T0 is suddenly placed in a fluid medium whose temperature is maintained constant at a value T1. The heat-transfer coefficient between the medium and the sphere is constant at a value h. The sphere is isotropic, and the temperature variation of the physical properties of the material forming the sphere may be neglected. Derive the equation relating the temperature of the sphere to the radius r and time t.
independent of and
Boundary condition:
0,00
0,
0,0
1
0
tratr
T
rtatTT
rtatTT
022
1 44)(0
rratrr
TkrTThq rrs
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Assume: )(tgrfT
t
T
r
T
rr
T
12
2
2
gfgfr
gf 12
g
g
f
f
rf
f 12
0)()(
0)()(2)( 22
tgtg
rfrrfrrfr
g
g
f
f
rf
f 12
0)()(2)( 22 rfrrfrrfr Bessel’s equation see next slide...
rJrcrJrcrf 2
12
1
22
12
1
1)(
if 0
rccrf 1)( 43 if = 0
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• Bessel’s equation of order
– occurs in studies of radiation of energy and in other contexts, particularly those in cylindrical coordinates
– Solutions of Bessel’s equation• when 2 is not an integer
• when 2 is an integer
– when = n + 0.5
– when = n + 0.5
0)( 222 yxyxyx
0
22 )1(!2
)1()(
n
nn
n
xnn
xJ )()()( 21 xJcxJcxy
)()()( 5.025.01 xJcxJcxy nn
0)()( tgtg
tectg 5)( if 0
if = 0
6)( ctg http://ebooks.edhole.com
rcrcr
rJrcrJrcrf
cossin21
)( 212
12
1
22
12
1
1
tectg 5)(
)(tgrfT
Dr
CerBrAr
T t
1
cossin21
rccrf 1)( 43
6)( ctg
0,00
0,
0,0
1
0
tratr
T
rtatTT
rtatTTB.C.
B = C = 0
D = T1
0)( 1 rrr
TkTTh
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1sin21
TerAr
T t
0,00 rtatTTB.C.
ter
rA
r
rA
r
T
2
sincos2
0)( 1 rrr
TkTTh
0)( 1 rrr
TkTTh
tt er
rA
r
rAkerA
rh
2
0
0
0
00
0
sincos2sin
21
hrk
krr
0
00tan
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1sin21
TerAr
T t
hrk
krr
0
00tan
0,00 rtatTT
B.C.
0,00 rtatTT
rAr
TT n
n
sin21
10
rr
TTA
n
n
sin21
10
11
10 sin
sin21
21
n
tn
n
n
n
Ter
rr
TT
rT n
More general format of the solution (by superposition):
or
11sin
21
n
tnn
n
TerAr
T n http://ebooks.edhole.com
11 sin
21
n
tnn
n
nerAr
TT
If the constants An can be determined by making use of the properties of orthogonal function?
0)()(2)( 22 rfrrfrrfr solution of the form rr
r n
n
n
sin21
)(
orthogonal
100
00
0
0
2
2
0 102
2
cossin2
sin21
sin21
)]()[(
)()()(
)(0
0
TTr
rr
r
drrr
r
drTTrr
r
dxxxr
dxxfxxr
rA
n
n
n
nn
r
n
n
r
n
n
b
a
n
b
a
n
n
11
sin21
TerAr
Tn
tnn
n
n
where
100
00
0
cossin2)( TT
rr
r
rrA
n
n
n
nnn
and
hrk
krr n
n0
00tan
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Free Ebooks Download
Mba Ebooks
By Edhole
Mba ebooks
Free ebooks downloadhttp://ebooks.edhole.com