Transcript 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.

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

v   vn
1
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.
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:
Vx  y
Vy  0 Vz  0
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.
y
dx T
1
T1
dy
x
T0
B.C.
T = T1 at x = 0, y > 0
T = 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:
Vx CTdy  kdx
T
y

T  
T  

 
  kdx dy
Output energy rate: Vx CTdy  Vx CTdydx  kdx
x
y y 
y  

 
Input - output = accumulation
Vx CT   T  const. properties

  k
x
y  y 
T
k   2T
 2

x Vx C  y
Vx  y
 A   2
  2
x y  y
 = 0 at x = 0, y > 0
 = 0 at x > 0, y = 
 = 1 at x > 0, y = 0




T  T1
T0  T1
A



k
 C
T A   2T
  2
x y  y



T = T1 at x = 0, y > 0
T = T1 at x > 0, y = 
T = T0 at x > 0, y = 0
 A   
  2
x y  y
2
Assume:
B.C.
 = 0 at x = 0, y > 0
 = 0 at x > 0, y = 
 = 1 at x > 0, y = 0



 y
 f  
n 
x 
Compounding the independent
variables into one variable
  f
 = 0 at  = 
 = 1 at  = 0
y
Replace y and x in the P.D.E by    n
x
d
d  
 

d 
d 
dx 
dy 
d
d  x
y 
 d 
ny d
n d

  n1

x d x
x d
x d
 d  1 d

 n
y d y x d
 2   1 d  1   d  1  d 2  d
  n

  n  2 
  n
2
y
y  x d  x y  d  x  d  dy
 A   2
  2
x y  y



n d
1 A d 2

 3n
x d x  d 2
In order to eliminate x and y, we choose n = 1/3
 = 0 at  = 
 = 1 at  = 0
d 2  2 d

0
2
d
3 A d
 3 
d

 B exp 
d
 9A 
 = 1 at  = 0

 
0
1
 d
B
0


0
 3 
d
exp 
9
A



 3 
T  T1
d 
d  B  exp 

T0  T1
 9A 

1

 3 
0 exp  9 A d
 3 
d
exp 


T  T1
 9A 


T0  T1
 3 
0 exp  9 A d

 T 

Local heat transfer coefficient h(T0  T1 )  k 
 y 0
T dT 

y d y

T  T1
T0  T1
 C 
h  0.43k 

 kx 
A
h
T

 T0  T1 
y
y
1
 d  1 d


y d y x n d
T T0  T1  d

1
y
x 3 d
 3 
d

 B exp 
d
 9A 
3
B
k
 C
k
3
1




3

d
x  exp 
0
 9A 
 3 

 exp 
T  T 
 9A 
 0 1 1
3

3
y



x
0 exp  9 A d
 T 
h(T0  T1 )  k  
 y 0 T
 = 0 at y = 0
1

 3 
0 exp  9 A d
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
Since the solid extends indefinately in the y and z direction, heat
flows only in the x direction. The heat-conduction equation:
T
 2T
 2
t
x
x
dx
Boundary condition: T  T1
at t  0, x  0
T  T0
at t  , x  0
T  T0
at x  0, t  0
T  T0
at x  2 R, t  0
T
 2T
 2
t
x
T  T1 at t  0, x  0
T  T0
at t  , x  0
T  T0
at x  0, t  0
T  T0
at x  2 R, t  0
Assume:   f xg (t )

 f x g (t )
t
 2
 f x g (t )
2
x

 2
 2
t
x
  1 at t  0, x  0
  0 at t  , x  0
  0 at x  0, t  0
  0 at x  2 R, t  0
Dimensionless:

T  T0
T1  T0
Separation of variables

 2
 2
t
x
f ( x) g (t )  f ( x) g (t )
f ( x) g (t )

 
f ( x) g (t )
independent of t
independent of x
f ( x)  f ( x)  0
g (t )   g (t )  0
f ( x) g (t )

 
f ( x) g (t )
when   0
f ( x)  A sin  x  B cos  x   f xg (t )
g (t )  Ce
 t


  Asin  x  B cos  x et
when  = 0
f ( x)  A0 x  B0   f xg (t )
  A0 x  B0
g (t )  C0
Superposition:


  A0 x  B0  Asin  x  B cos  x et
A0, B0, A, B, and  have to be chosen to satisfy the boundary conditions.




 1 at t  0, x  0
 0 at t  , x  0
 0 at x  0, t  0
 0 at x  2 R, t  0
A0  B0  0
B0

n
2R
n is an integer


n
  A0 x  B0  Asin  x  B cos  x et
nx   2 Rt

  A sin
e
2R 

  1 at t  0, x  0
B.C.
The constant has to be determined.
But no single value can satisfy the B.C.
More general format of the solution (by superposition):
n

n

x

 2 Rt   1
   An  sin
e
2R 

n 1


2R
0

2R 
mx    nx  
 mx 
 sin
dx  0  sin
  An  sin
 dx
2R 
2R  n1 
2R  


2R
0
at t  0, x  0
nx 

1   An  sin

2R 

n 1

Orthogonality property

2R
2R
 mx 
 mx  nx 
 mx 
sin
dx

A
sin
sin
dx

A




 dx

n
n  sin


0
0
2R 
2R 
2R 
2R 



n 1
1 2R 
nx 
2
An    sin
dx  [1  (1) n ]

R 0 
2R 
n
2
n
 nx   2 Rt
   An  sin
e
2R 

n 1

2
An  [1  (1) ]
n
n

n 2 2
T  T1 2 1  (1) 
nx   4 R 2 t
 
 sin
e
T0  T1  n 1
n
2R 


n
T  T0
T1  T0
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
 r ( x)
a
m
( x)n ( x)dx  0

2R
0

2R
2R
 mx 
 mx  nx 
 mx 
sin
dx

A
sin
sin
dx

A




 dx

n
n  sin


0
0
2R 
2R 
2R 
2R 



n 1
2
mx 
nx 


 sin
 and  sin

2R 
2R 


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 :
nx 

sin
where

 shows!
f ( x)  f ( x)  0 f ( x)  A sin  x  B cos  x
2R 

  t
g (t )   g (t )  0 g (t )  Ce
f ( x)  f ( x)  0
These values of
  0 at x  0
f ( x)  C sin  x
  0 at x  2R
n
n 
2R
n are called the “eigenvalues” of the equation, and the
correponsing solutions,
sin  x are called the “eigenfunctions”.
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:
y  R( x) y  [Q( x)  P( x)]y  0
• In Strum-Liouville form:
eigenvalue
[r ( x) y]  [q( x)  p( x)]  0
A Strum-Liouville differential equation with boundary conditions at
each end point x = a and x = b which satisfy one of the following
forms:
• The regular problem on [a,b]
A1 y(a)  A2 y(a)  0
B1 y(b)  B2 y(b)  0
• The periodic problem on [a,b]
y(a)  y(b)
y(a)  y(b)
• The singular problem on [a,b]
r (a)  0 B1 y(b)  B2 y(b)  0
r (b)  0 A1 y(a)  A2 y(a)  0
r (a)  r (b)  0
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:

f ( x)   Ann ( x)
n 0
The value of An can be obtained by making use of the orthogonal
properties of the functions (x)


r
(
x
)

(
x
)
f
(
x
)
dx

r
(
x
)

(
x
)
A

(
x
)
m
m
 n n  dx
a
a
 n 0


b
b
b

a
r ( x)m ( x) f ( x)dx   An  r ( x)m ( x)n ( x)dx
n 0
b
An 
 r ( x)n ( x) f ( x)dx
b
a
Each term is zero except when m = n.
a
b
2
r
(
x
)[

(
x
)]
dx
n

a
0,  1,  2…… are eigenfunctions
Steady-state heat transfer with axial symmetry
  2 T 
1  
T 
r

sin




  0


r  r  sin   
 
 2T
T  2T
T
r
 2r
 2  cot
0
2
r
r 

2
Assume: T  f r g ( )
T
 f r g ( )
r
T
 f r g ( )

r 2 f (r ) g ( )  2rf (r ) g ( )  f (r ) g( )  cotf (r ) g( )  0
Dividing by fg and separate variables
r 2 f   2rf 
g   cotg 

 l (l  1)
f
g
r 2 f   2rf   l (l  1) f  0
g   cotg   l (l  1) g  0
r 2 f   2rf 
g   cotg 

 l (l  1)
f
g
r 2 f   2rf   l (l  1) f  0
set f  Ar n
n(n 1) Arn  2nArn  l (l  1) Arn  0
n(n  1)  2n  l (l  1)  0
f  Arl  Br l 1
g   cotg   l (l  1) g  0
set m  cos 
dm   sin d
(1  m ) g   2mg   l (l  1) g  0
2

set g (m)   an m n c
n 0
Legendre’s equation of order l
Solved by the method of Frobenius
(1  m2 ) g   2mg   l (l  1) g  0

g ( m)   a n m n  c
n 0

(1  m ) (n  c)(n  c  1)an m
2
n 0
nc 2

 2m (n  c)an m
n 0
n  c 1

 l (l  1) an m nc  0
n 0
比較係數
mc 2
c(c 1)a0  0
c  0 or 1
mc 1
(c  1)ca1  0
c  0 or a1  0
and
(s  c  2)(s  c  1)as2  (s  c  l )(s  c  l  1)as
g (m)  APl (m)  BQl (m)
where Pl(m) is the “Legendre polynomial”
T  f r g ( )
f  Arl  Br l 1
g (m)  APl (m)  BQl (m)
T  ( Al r l  Bl r l 1 )Pl (cos )
m  cos 
superposition

T   ( Al r l  Bl r l 1 ) Pl (cos )
l 0
Unsteady-state heat transfer to a sphere
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.
1 T independent of  and 
T
 t
2
 2T 2 T 1 T


2
r
r r  t
Boundary condition:
T  T0
at t  0, r  0
T  T1
at t  , r  0
T
 0 at r  0, t  0
r
 T 
2
q  h(Ts  T1 )4r  k 
4

r
at r  r0
 r r0
 r 
2
 2T 2 T 1 T


2
r
r r  t
Assume: T  f r g (t )
f g 
f  2 f  1 g 


 
f r f  g
2
1
f g  fg 
r

f  2 f  1 g 


 
f r f  g
r 2 f (r )  2rf (r )  r 2 f (r )  0
g (t )  g (t )  0
r 2 f (r )  2rf (r )  r 2 f (r )  0
f (r )  c1r
1
2
J1
2
f (r )  c3  c4 1
r
 r  c2r
1
2
J1
Bessel’s equation
r
if
 0
if
 =0
2
see next slide...
• Bessel’s equation of order 
x 2 y  xy  ( x 2  2 ) y  0
– 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
y( x)  c1 J ( x)  c2 J  ( x)
(1) n
J ( x)   2 n 
x 2 n
n!(n   1)
n 0 2

• when 2 is an integer
– when  = n + 0.5 y( x)  c1 J n0.5 ( x)  c2 J n0.5 ( x)
– when  = n + 0.5
g (t )  g (t )  0
g (t )  c5et
if
 0
g (t )  c6
if
 =0
T  f r g (t )
f (r )  c1r
1
2
J1
2
 r  c2 r
1
2
J1
r 
2
1
r

c sin

2
1
 r  c2 cos  r
g (t )  c5et
f (r )  c3  c4 1
r
g (t )  c6
1
T  
r


2 
1
 A sin  r  B cos  r e t  C  D
r
 
B.C.
T  T0
at t  0, r  0
T  T1
at t  , r  0
T
 0 at r  0, t  0
r
 T 
h(T  T1 )  k 
 r r0

r


D = T1
B=C=0



B.C. T  T0
1
T  
r
2 
 A sin  r e t  T1
 
T 
 
r 
2  A  cos  r A sin  r  t

e

2



r
r
 

at t  0, r  0
 T 
h(T  T1 )  k 
 r r0
 r 
 T 
h(T  T1 )  k 
 r r0
 r 
 1
h 
 r0


2 
 t
 A sin  r0 e   k 

 




tan  r0 
2  A  cos  r0 A sin  r0  t
e


2



r0
r0
 

 r0 k
k  r0 h
1
T  
r


B.C.


r
k
  t
0
 A sin  r e
 T1 tan  r0 

 
k  r0 h T  T0
2
T  T0
1
T0  T1  
r

at t  0, r  0

 A sin  r
n

n  
2
at t  0, r  0


A
1

r

T0  T1
2 
sin n r
n  

More general format of the solution (by superposition):
1
T  

n 1 r





2  
n     1
 
   r

1
or T   

n 1 r








T0  T1
 sin n r e nt  T1


2 

sin n r 


n  





 A sin  r e nt  T
n
n
1
n  
2



1
T  T1   

n 1 r



 A sin  r e nt
n
n

n  
2


If the constants An can be determined by making use of the properties of orthogonal
function?
r 2 f (r )  2rf (r )  r 2 f (r )  0
1
r 
 r
 r ( x) ( x) f ( x)dx   
b
r0
n
An (r ) 
solution of the form

a
b
2
r
(
x
)[

(
x
)]
dx
n

a
 2 
n


r0

0
2
1
r 
 r
  
r0
0
2
1
 n (r ) 
r

sin

2
n

sin n r  T0  T1 dr
n 

2

2
sin n r  dr
n 

2

 sin n r0
cos n r0 

 T0  T1 
 r0

n
n 





n r
orthogonal

1
T  

n 1 r



 A sin  r e nt  T
n
n
1

n  
2


where
 2 
n
An (r )  

r0

 sin n r0
cos n r0 

 T0  T1 
 r0

n
n 


and
tan n r0 
n r0 k
k  r0 h
Equations involving three independent variables
The steady-state flow of heat in a cylinder is governed by Laplace’s equation in
cylindrical polar coordinates:
 2T 1 T 1  2T  2T

 2
 2 0
2
2
r
r r r 
z
There are three independent variables r, , z.
Assume: T  f r , g ( z )
2 f
1 f
1 2 f
g
g 2
g  fg   0
2
2
r
r r
r 
Separation of variables
1
f
  2 f 1 f 1  2 f   g 
2





v
 2
2
2

r
r

r
r


g


2
2

f

f

f
r 2 2  r  2  r 2v 2 f  0
r
r 
g ( z )  v 2 g ( z )  0
Two independent variable P.D.E.
OK,
g ( z)  Avevz  Bvevz
2 f
f  2 f
2 2
r

r


r
v f 0
2
2
r
r 
2
Assume: f r ,   F (r )G( )
r 2 F G  rF G  FG  r 2v 2 FG  0
Separation of variables
r 2 F   rF  2 2  G
r v 
 k2
F
G
r 2 F   rF   (r 2v 2  k 2 ) F  0
Bessel’s equation
G( )  k 2G( )  0
OK,
The solution of the Bessel’s equation:
G( )  Ak cosk  Bk sin k
F (r )  AJk (rv)  BYk (rv)
T  f r , g ( z )  F (r )G( ) g ( z )

 AJk (rv)  BYk (rv) Ak cosk  Bk sin k  Av evz  Bv evz

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.
V  D
a
U
r
horizontal component of fluid velocity
U0
b
z
r
Material balance:
U
Input:
2rUr  D
U
2rz
r

U
  U

2rz   D
2rz r
Output: 2rUr  2rUr z  D
z
r
r  r


  U
2rUr z   D 2rz r  0
z
r  r

B.C.

  U

2rUr z   D 2rz r  0 at z = 0, if r < a, U = U0
z
r  r

  2U 1 U
U
 D 2 
z
r r
 r



at z = 0, if r > a, U = 0
at r = 0, U = finite
at r = b,  D
U
2b   2b k U
r
z
Assume: U  f r g (z )
1



fg  D( f g  fg )
r
d2 f
df
2 2
r

r


r f 0
2
dr
dr
g ( z )   2 Dg ( z )  0
2
g

Dg
f  
f
1
f
r   2
Bessel’s equation
g ( z)  Ae
2
Dz
The solution of the Bessel’s equation: f (r )  A J 0 (r )  BY0 (r ) if   0
f (r )  A0  B0 ln r
if  = 0
U  f r g (z)
g ( z)  Ae
2
Dz
f (r )  A0
f (r )  A J 0 (r )
U  f (r ) g ( z )  A0  An J 0 ( n r )e
General form:

U  A0   An J 0 ( n r )e
n 1
 n2 Dz
 n2 Dz
The Laplace Transform
• It is defined of an improper integral and can be
used to transform certain initial value problems
into algebra problems.


 f ( s)   e  st f (t )dt
0
  f (s)  sF (s)  f (0)

 f (s)  s 2 F (s)  sf (0)  f (0)
• Laplace Transform table!
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:
T
 2T
 2
t
x
x
Boundary condition:
dx
T  T0
at t  0
The initial condition can use the Laplace transform method
T  T1 at x  0
 T 
    sT ( s)  T0
 t 
T
 2T Laplace transform
 2
t
x
T   sT  T0  0
  2T  d 2T ( s)
and   x 2   dx2


x and t are independent variables
d 2T ( s)
sT ( s)  T0  
dx2
Second order linear O.D.E.

s regards as constant
T  Ae
s

x
s
 Be

x
T0

s
The boundary condition: T  T1
at x  0
Laplace transform
T (s) 
s

T  Ae

x
s
 Be

T1
s
x
at x  0
T0 and

s
when x  , T remains finite 
s

T  Ae
T  T0
T  1
e
s


x
s

T
 0
s
x
T
 0
s
T (s) 
T (s) 
T1
s
at x  0
B.C.
T remains finite  B = 0
T1
s
at x  0
inverse transform
T1  T0
A
s
 x 
T  T1  T0 erfc
  T0
 2 t 
For unsteady-state one-dimensional heat conduction:
T
 2T
 2
t
x
x
Boundary condition:
dx
T  0 at t  0, x  0
H  constant
 T 
    sT (s)
 t 
and
the heat is concentrated at the surface initially
  2T  d 2T ( s)
  2
dx2
 x 
x and t are independent variables
d 2T ( s)
T
 2T Laplace transform
 2
sT ( s)  
t
x
dx2
Second order linear O.D.E.
T   sT  0

s regards as constant
T  Ae
s

x
s
 Be

x
The boundary condition: H  constant 

H   C pTdx
0
Laplace transform
x and t are independent variables

H
 C p  T dx
0
s
s


H
 C p  T dx
0
s
T  Ae

x


H
  Ae
0
sC p
s

x
dx
H
A

sC p
s


s
H
e 
T 
C p 
s
x
inverse transform
T 
H
C p 
e
x2

4t
t
For unsteady-state one-dimensional heat conduction:
T
 2T
 2
t
x
x
dx
Boundary condition:
T  0 at t  0, x  0
Q  k
T
x
at x  0
 T 
    sT (s)
 t 
and
the heat is supplied at a fixed rate
  2T  d 2T ( s)
  2
dx2
 x 
x and t are independent variables
d 2T ( s)
T
 2T Laplace transform
 2
sT ( s)  
t
x
dx2
Second order linear O.D.E.
T   sT  0

s regards as constant
T  Ae
s

x
s
 Be

x
The boundary condition: Q   k T
x
at x  0
Laplace transform
x and t are independent variables
Q
T
 k
s
x

T  Ae
s

at x  0

Q
s  
e
 A 

 ks



x
A
Q 
T 
s e
k
3
2

s

x
inverse transform

x
x 0
Q
ks

s
s

Q 
T
k
t
1
0


e

x2
4
d
Heat conduction between parallel planes
Consider the flow of heat between parallel planes maintained at different temperatures:
T0
T
 2T
 2
t
x
Boundary condition:
x
T1
T  T0
at t  0, x  0 The initial condition can use the Laplace
T  T0
at x  0
transform method
T  T1 at x  L
 T 
    sT ( s)  T0
 t 
T
 2T Laplace transform
 2
t
x
T   sT  T0  0
  2T  d 2T ( s)
and   x 2   dx2


x and t are independent variables
d 2T ( s)
sT ( s)  T0  
dx2
Second order linear O.D.E.

s regards as constant
T  Ae
s

x
s
 Be

x
T0

s
The boundary condition: T  T0
at x  0
T  T1 at x  L
Laplace transform
T ( s) 
T0
s
at x  0
T ( s) 

T  Ae
s

x
s
 Be

x
s
T1  T0 e
T 
s
e

s


x
L
T0
s
e
e
T0
s
T
T ( s)  1
s


s

s

Laplace transform
T ( s) 
at x  0
T1
s
at x  L
A B  0

T1
 Ae
at x  L
s
s

L
s
 Be

L
T0

s
x
L
T0

s
inverse transform
T  T0
x
2
nx 
n
 
( 1) sin
e
T1  T0
L n  0 n
L

n 2 2t
L2
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:
x
T
 2T
 2
t
x
T  0 at t  0, x  0 The initial condition can use the Laplace
transform method
T  T0 at x  0
T
 0 at x  L
x
  2T  d 2T ( s)
 T 
    sT (s) and   2  
dx2
 t 
 x 
x and t are independent variables
T
 2T Laplace transform
 2
t
x
T   sT  0
d 2T ( s)
sT ( s)  
Second order linear O.D.E.
dx2

s regards as constant
T  Ae
s

x
s
 Be

x
The boundary condition: T  T0
T
 0 at x  L
x
at x  0
Laplace transform
T ( s) 
T0
s
T ( s) 

T  Ae
s

x
s
 Be
 2
e
T0 
T 
s
s

L

Laplace transform
T ( s )
0
x
at x  0
T0
s
at x  0
x
T ( s )
0
x
 
 1e


s

s
2

e
x
L
at x  L

2

 1 e


e
2
s

L
s

L

e


at x  L
T0
A B 
s
A
s

s


e
s

L
B
s

s
e

L
0
x
inverse transform

 ( x  2nL) 
 (2nL  2 L  x)  
T  T0  (1) erfc
  erfc

2 t
 2 t 


n 0


n
Example
An 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:
 2 p 1 p 1 p


2
r
r r  t
p  p0
at t  0, r  0
where  is the hydraulic diffusivity
2rkh p
If the oil is removed at a constant rate q: q  lim
r 0
 r
where k is the permeability; h is the thickness of the formation; and  is the coefficient of
viscosity
 1 p  s
  2 p  d 2 p( s)
1
 p  dp ( s)




 

p
(
s
)

p
and
 2
0
2



dr
dr

 r 
 r 
 t  
r and t are independent variables
 2 p 1 p 1 p


2
r
r r  t
Laplace Transform
p0
d 2 p 1 dp s


p

dr2 r dr 

Second O.D.E
s
p0 
r p  rp  r  p    0
 

p0
d 2 p 1 dp s


p

dr2 r dr 

2
2
xr
s

p0 
2



x p  xp  x  p    0
s 

Modified Bessel’s equation
2
p  BK 0 ( x) 
p0
s
p  AI 0 ( x)  BK 0 ( x) 
p0
s
x  ; I 0 ( x)  
2rkh p
r 0
 r
The boundary condition: q  lim
x  ; K0 ( x)   ln x
Laplace transform
q
2rkh dp
 lim
s r 0  dr
B
 q
2rkhs
p  BK 0 ( x) 
p0
s
B
 q


2rkhs
s

p0
p   K 0 (r
)
s

s
s
inverse transform
q  t
 r2 
  exp(
p
)dt   p0
4rkh  0
4t

  r2 
q
  p0
p
Ei
4rkh  4t 
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).

 st
e
 f (t )dt  
 f (t )  F ( s)
0
The parameter s must be large enough to make the integral convergent at the upper
limit and t must be positive.


K
 st


e
(
K
)
dt


K

0
s

e
 st
(t )dt  
n
t   
n

0
0
e
0


 st

[ f (t )]dt  f (t )e


z
s n 1

 st 
0

(n  1)
s n 1
1
sa
The shifting theorem

 ( s)  e  st [ f (t )]dt
0
 f (t )   f (0)  s  f (t )  sF (s) 

 st
e
 [ f (t )]dt  f (t )e
0
n
z e dz
0

 
 st
at
at
e
(
e
)
dt


e



 st 
0
f (0)

 (  s )  e  st [ f (t )]dt
0
 f (t )   f (0)  s  f (t )   f (0)  ssF ( s) 
 f (t )  s 2 F ( s)  sf (0)  f (0)
f (0) 
Usually given as boundary conditions
  f n (t )  s n F (s)  s n1 f (0)  s n2 f (0)  ... sf ( n2) (0)  f ( n1) (0)
Only valid for continuous functions
 sin(t ) 

s2  2
Advantages about Laplace Transform:
(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)
The inverse transformation
• It must be convenient to convert the transform
back to a function of the independent variable:

1
F (s) 
f (t )
• Using partial fraction
1
A
B
C



2
( s  a)(s  b)
s  a s  b ( s  b) 2
• Laplace transform table
Example:
d2y
Solve
 4 y  3 , where y(0) = y’(0) = 1
2
dt
d2y
 4y  3
2
dt
1
3
y (t )  cos 2t  sin st  (1  cos 2t )
2
4
Laplace Transform
d 2 y 
  2    4 y    3 
 dt 
K
 K  
s

1
Y (s)   1 
s 

2

 s  4
1
 1 
 s 2  4   
  f (s)  s 2 F (s)  sf (0)  f (0)

s Y ( s)  sy (0)  y(0)  4Y ( s ) 
2
3
s
Inverse Laplace Transform
B.C.


3
s Y ( s )  s  1  4Y ( s ) 
s
2
1
s
1
3
Y ( s)  2
 2

s  4 s  4 s( s 2  4)


3
 s(s 2  4) 


Properties of the Laplace transform
• Differentiation of the transform

   st
d   st
F ( s )    e f (t )dt    e (t ) f (t )dt
ds  0
 0
• Integral of a function

e
 st
0

t


f (t )dt    f (t )     f (t )dt 
 0


t
 t
e 
1 
 st
    f (t )dt e  st dt     f (t )dt 

f
(
t
)
e
dt


0
0
0


 s 0
s

0
1
 F ( s)
s
 st