化工應用數學授課教師：郭修伯 Lecture 8 Solution of Partial Differentiation Equations

化工應用數學

授課教師：郭修伯

Lecture 8 Solution of Partial Differentiation Equations

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

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 x

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

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 eigenfunctions

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

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

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

Equations involving three independent variables The steady-state flow of heat in a cylinder is governed by Laplace’s equation in cylindrical polar coordinates:

011

2

2

2

2

22

2

z

TT

rr

T

rr

T

There are three independent variables r, , z.

Assume: )(, zgrfT

011

2

2

22

2

gfgf

rg

r

f

rg

r

f

22

2

22

2 111v

g

gf

rr

f

rr

f

f

Separation of variables

0)()(

0

2

222

2

2

22

zgvzg

fvrf

r

fr

r

fr

OK,

vzv

vzv eBeAzg )(

Two independent variable P.D.E.

0222

2

2

22

fvrf

r

fr

r

fr

Assume: )()(, GrFrf

0222 FGvrGFGFrGFr

2222

kG

Gvr

F

FrFr

Separation of variables

0)()(

0)(2

2222

GkG

FkvrFrFr

OK, kBkAG kk sincos)(

Bessel’s equation

The solution of the Bessel’s equation: )()()( rvBYrvAJrF kk

vz

vvz

vkkkk eBeAkBkArvBYrvAJ

zgGrFzgrfT

sincos)()(

)()()()(,

In the study of flow distribution in a packed column, the liquid tends to aggregate at the walls. If the column is a cylinder of radius b m and the feed to the column is distributed within a central core of radius a m with velocity U0 m/s, determine the fractional amount of liquid on the walls as a function of distance from the inlet in terms of the parameters of the system.

z

U0

a

b

r

U

r

UDV

horizontal component of fluid velocity

Material balance:

Input: zrr

UDrrU 22

Output: rzrr

UD

rzr

r

UDzrrU

zrrU

2222

022

rzrr

UD

rzrrU

z

022

rzrr

UD

rzrrU

z

r

U

rr

UD

z

U 12

2

B.C.at z = 0, if r < a, U = U0

at z = 0, if r > a, U = 0at r = 0, U = finite

at r = b, z

Ukbb

r

UD

22

Assume: )(zgrfU

)1

( gfr

gfDgf 2

1

f

fr

f

Dg

g

0)()(

0

2

222

22

zDgzg

frdr

dfr

dr

fdr

DzeAzg2

)(

Bessel’s equation

The solution of the Bessel’s equation: )()()( 00 rYBrJArf if 0

rBArf ln)( 00 if = 0

)(zgrfU

DzeAzg2

)(

Dznn

nerJAAzgrfU2

)()()( 00

0)( Arf

)()( 0 rJArf

General form:

1

00

2

)(n

Dznn

nerJAAU

The Laplace Transform

• It is defined of an improper integral and can be used to transform certain initial value problems into algebra problems.

• Laplace Transform table!

0

)()( dttfesf st

)0()()( fssFsf

)0()0()()( 2 fsfsFssf

The Laplace transform method for P.D.E.

• The Laplace transform can remove the derivatives from an O.D.E.

• The same technique can be used to remove all derivatives w.r.t. one independent variable from a P.D.E. provided that it has an open range.

• A P.D.E has two independent variables can use “the Laplace transform method” to remove one of them and yields an O.D.E..

• The boundary conditions which are not used to transform the equation must themselves be transformed.

For unsteady-state one-dimensional heat conduction:

0)( TsTst

T

2

2

2

2 )(

dx

sTd

x

T

and

2

2

x

T

t

T

Laplace transform

2

2

0

)()(

dx

sTdTsTs

Second order linear O.D.E.

x and t are independent variables

x

dx

2

2

x

T

t

T

Boundary condition:

0

0

1

0

xatTT

tatTT The initial condition can use the Laplace transform method

00 TTsTs

TBeAeT

xs

xs

0

s regards as constant

The boundary condition: 01 xatTT

Laplace transform

0)( 1 xats

TsT

s

TBeAeT

xs

xs

0

0)( 1 xats

TsTand

when x , T remains finite remains finite B = 0T

B.C.

s

TAeT

xs

0

0)( 1 xat

s

TsT

s

TTA 01

s

Te

s

TTT

xs

001

inverse transform 001

2T

t

xerfcTTT


)(sTst

T

2

2

2

2 )(

dx

sTd

x

T

and

2

2

x

T

t

T

Laplace transform

2

2 )()(

dx

sTdsTs


x

dx

2

2

x

T

t

T

Boundary condition:

onstantcH

xtatT

0,00

0 TsT xs

xs

BeAeT


the heat is concentrated at the surface initially


xs

AeT

The boundary condition:

0

TdxCHtantconsH p

Laplace transform

0

dxTCs

Hp


0

dxTCs

Hp

0dxAe

Cs

H xs

p

s

A

Cs

H

p

s

e

C

HT

xs

p

inverse transform

t

e

C

HT

t

x

p

4

2


)(sTst

T

2

2

2

2 )(

dx

sTd

x

T

and

2

2

x

T

t

T

Laplace transform

2

2 )()(

dx

sTdsTs


x

dx

2

2

x

T

t

T

Boundary condition:

0

0,00

xatx

TkQ

xtatT

0 TsT xs

xs

BeAeT


the heat is supplied at a fixed rate


xs

AeT

The boundary condition: 0

xatx

TkQ

Laplace transform

s

ks

QA

xs

esk

QT 2

3 inverse transform

0

xatx

Tk

s

Q


0

x

xs

es

Aks

Q

t

x

dek

QT

0

4

2

1

Heat conduction between parallel planes

Consider the flow of heat between parallel planes maintained at different temperatures:

T0

T1

x Boundary condition:

LxatTT

xatTT

xtatTT

1

0

0

0

0,0 The initial condition can use the Laplace transform method

2

2

x

T

t

T

0)( TsTst

T

2

2

2

2 )(

dx

sTd

x

T

and

2

2

x

T

t

T

Laplace transform

2

2

0

)()(

dx

sTdTsTs



00 TTsTs

TBeAeT

xs

xs

0



Laplace transform

0)( 0 xats

TsT

LxatTT 1

Laplace transform

Lxats

TsT 1)(

s

TBeAeT

xs

xs

0

0)( 0 xats

TsT

Lxats

TsT 1)( s

TBeAe

s

T

BA

Ls

Ls

01

0

s

T

ee

ee

s

TTT

Ls

Ls

xs

xs

001

inverse transform

001

0 2

22

sin)1(2

n

L

tnn e

L

xn

nL

x

TT

TT

Symmetrical heat conduction between parallel planes

Consider a wall of thickness 2L with a uniform initial temperature throughout, and let both faces be suddenly raised to the same higher temperature.

Boundary condition:

Lxatx

T

xatTT

xtatT

0

0

0,00

0

The initial condition can use the Laplace transform method

2

2

x

T

t

T

)(sTst

T

2

2

2

2 )(

dx

sTd

x

T

and

2

2

x

T

t

T

Laplace transform

2

2 )()(

dx

sTdsTs



0 TsT xs

xs

BeAeT


x


Laplace transform

0)( 0 xats

TsT

Lxatx

T

0

Laplace transform

Lxatx

sT

0)(

xs

xs

BeAeT

0)( 0 xats

TsT

0

0

Ls

Ls

es

Bes

A

s

TBA

Ls

Ls

xs

Ls

xs

Ls

ee

eeee

s

TT

22

22

0

11

inverse transform

00

2

)22(

2

)2()1(

n

n

t

xLnLerfc

t

nLxerfcTT

Lxatx

sT

0)(

ExampleAn extensive shallow oilfield is to be exploited by removing product at a constant rate from one well. How will the pressure distribution in the formation vary with time?Taking a radial coordinate r measured from the base of the well system, it is known that the pressure (p) follows the normal diffusion equation in the r direction:

t

p

r

p

rr

p

11

2

2

where is the hydraulic diffusivity

If the oil is removed at a constant rate q:r

prkhq

r

2lim

0

where k is the permeability; h is the thickness of the formation; and is the coefficient of viscosity

0,00 rtatpp

0

1)(

1psp

s

t

p

2

2

2

2 )(

dr

spd

r

p

and

r and t are independent variables

dr

spd

r

p )(

t

p

r

p

rr

p

11

2

2 Laplace Transform

0

2

2 1 pp

s

dr

pd

rdr

pd

Second O.D.E

0

2

2 1 pp

s

dr

pd

rdr

pd 0022

p

ps

rprpr

Modified Bessel’s equation

s

pxBKxAIp 0

00 )()(

s

rx

0022

s

ppxpxpx

)(; 0 xIx

s

pxBKp 0

0 )(

The boundary condition: r

prkhq

r

2lim

0

Laplace transform

dr

pdrkh

s

qr

2lim

0

rkhs

qB

2

xxKx ln)(; 0

s

pxBKp 0

0 )( srkhs

qB

2

s

psrK

sp 0

0 )(

inverse transform

00

2

)4

exp(4

pdtt

r

rkh

qp

t

0

2

44p

t

rEi

rkh

qp

The restriction on the use of Laplace transform to solve P.D.E. problems

• The problem must be of initial value type.• Th dependent variable and its derivative remain

finite as the transformed variable tends to infinity• The Laplace transform should be tried whenever a

variable has an open range and the method of separation should be used in all other cases.

• There are many P.D.E.s which cannot be solved by either method, and the numerical methods are recommended.

The Laplace Transform• A particular “operational method” of solving differential

equations.

• An O.D.E. is converted into an equivalent algebraic form which can be solved by the laws of elementary algebra.

• If f(t) is a continuous function of an independent variable t for all values of t greater than zero, then the integral with respect to t of the product of f(t) with e-st between the limits 0 and is defined as the Laplace transform of f(t).

)()()(0

sFtfdttfe st

The parameter s must be large enough to make the integral convergent at the upper limit and t must be positive.

s

KKdtKe st

0

)(

11

0

0

)1()(

nn

zn

nnst

s

n

s

dzeztdtte

as

edtee atatst

1

)(0

)0()()()0()(

)]([)()()]([0

0

0

fssFtfsftf

dttfesetfdttfe ststst

)0()0()()(

)0()()0()()0()(

)]([)()()]([

2

0

0

0

fsfsFstf

fssFsftfsftf

dttfesetfdttfe ststst

)0()0(...)0()0()()( )1()2(21 nnnnnn fsffsfssFstf

Only valid for continuous functions

Usually given as boundary conditions

The shifting theorem

22

)sin(

s

t

(1) Boundary conditions are introduced into the problem before solution of the equation.(2) The differential equation is reduced to an algebraic equation in terms of the operator s.Note:The operation described is only applicable to “initial value” problems. (i.e., the value of the function and its derivatives must be known when the independent variable is zero)

Advantages about Laplace Transform:

The inverse transformation

• It must be convenient to convert the transform back to a function of the independent variable:

• Using partial fraction

• Laplace transform table

)()(1 tfsF

22 )())((

1

bs

C

bs

B

as

A

bsas

Example:

Solve , where y(0) = y’(0) = 1 342

2

ydt

yd

342

2

ydt

yd

Laplace Transform

342

2

y

dt

yd

s

sYysysYs3

)(4)0()0()(2

)0()0()()( 2 fsfsFssf s

KK

B.C.

s

sYssYs3

)(41)(2 )4(

3

4

1

4)(

222

ssss

ssY

)4(

3

4

1

4)(

21

21

211

ssss

ssY

Inverse Laplace Transform

)2cos1(4

3sin

2

12cos)( tsttty

Properties of the Laplace transform

• Differentiation of the transform

• Integral of a function

00

)()()()( dttftedttfeds

dsF stst

)(1

)(1

)()(

)()()(

00

00

0

00

sFs

dtetfss

edttfdtedttf

dttftfdttfe

ststtstt

tst

Documents

化工應用數學 授課教師： 郭修伯 Lecture 8 Solution of Partial Differentiation Equations

化工應用數學授課教師：郭修伯 Lecture 8 Solution of Partial Differentiation Equations