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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



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.



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



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



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



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



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



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



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



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)()()(



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



• 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.



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



)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



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”


)(grfT 1 ll BrArf

)()()( mBQmAPmg ll

cosm

)(cos)( 1 ll

ll

l PrBrAT

superposition

0

1 )(cos)(l

ll

ll

l PrBrAT


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


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


• 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


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


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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