Chapter 7 Partial Differential Equations
I.
Introduction
(i) What are the boundary value problem (BVP) and
the initial value problem (IVP)?
IVP  A problem with the conditions given only at one point.
BVP  A problem with the conditions given at all the boundaries.
(ii) What are the transient and steady-state problems?
Transient

Steady-state 

 0 in the D.E.
t

 0 in the D.E.
t
(iii) Are the types of the partial differential equations?
Quadratic form
 Matrices
 Parabolic  b 2  4ac  0

2
ax 2  bxy  cy 2  dx  ey  f   Elliptic  b  4ac  0
Hyperbolic  b 2  4ac  0

Standard form of the 2nd-order PDEs:
au xx
 Parabolic  b 2  4ac  0

2
 bu xy  cu yy  du x  eu y  fu  g   Elliptic  b  4ac  0
Hyperbolic  b 2  4ac  0

Parabolic

1-D transient heat conduction problem s or diffusion prob.
Elliptic  2-D steady-state heat conduction problems (Laplace Eq.)
Hyperbolic  Wave equations (It is hard to be solved.)
(a)
g=0
g ≠ 0
(b)
 Homogeneous
 Nonhomogeneous
Linear problems  a, b, c, d, e and f are the functions of
the indep. variables.
Nonlinear problems 
a, b, c, d, e and f are the functions of
the dep. variable.
II. Schemes of solving PDE
(a) Separation of variables  Linear, homogeneous and SLS.
(b) Method of eigenfunction expansion  Fourier series
 Sturm-Liouville system (SLS)
(一)
1-D diffusion prob.
 2 u u

x 2 t
0 < x < L, t > 0
with the boundary conditions
a1u  b1
u
 p1 ( t )
x
at x = 0
a 2u  b2
u
 p 2 (t)
x
at x = L
and the initial condition
u(x, 0) = f1 (x)
(1)
where a12  b12  0 and a 22  b 22  0 .
(I) Dirichlet B.C. (Constant temperature B.C.):
b1  0
(II) Neumann B.C. (Constant heat flux B.C.): a1  0
(III) Robin B.C. (Convective B.C.): a12  b12  0
(A) Separation of variables
Assume that the b.C.’s and I.C. are
u(0, t) = 0
(2)
u(L, t) = 0
(3)
u(x, 0) = 1
(4)
and
Assume that the solution u(x, t) is
u(x, t) = X(x)T(t)
(5)
Substitution of Eq.(5) into Eq.(1) yields
1 d 2 X 1 dT

 2 (How do you know it?)
2
X dx
T dt
where λ is the separation constant.
Thus we can obtain
X = Acos(λx) + Bsin(λx) 
 A=0
A 2  B2  0
∵
B ≠ 0
∴   n 
n
, n = 1, 2,…
L

 n are eigenvalues
 Infinite
Its corresponding eigenfunctions X n are
X n  sin(  n x )  sin(
nx
)
L
Tn is
Tn  e n t
2
u(x, t) = c1X1T1  c2 X2T2    
nx ( nL )2 t
c n sin(
)e

L
n 1

=

u(x, 0) =
 c n sin(
n 1
nx
)
L
=1

cn 
2 L
nx
sin(
)dx

L 0
L

cn 
2
[1  (1) n ]
n
 Fourier-sine series
(B) Method of eigenfunction expansion
Its corresponding S-L S. is
d 2 2
 0
dx 2
with
(0)  (L)  0
Its general solution is
( x )  A cos(x )  B sin( x )

(0)  0
∴   n 


A 2  B2  0
A=0
n
, n = 1, 2,…
L
B ≠ 0)
(∵
Its corresponding eigenfunctions n (x) are
n ( x )  sin(  n x )  sin(
nx
)
L
Thus u(x, t) obtained from the Fourier-sine series expansion can be written as

u(x, t) =
c
n 1
n
( t ) sin(
nx
)
L
(6)
Substitution of Eq.(6) into Eq.(1) yields
 2n c n ( t ) 
dc n ( t )
dt


u(x, 0) =

c n ( 0) 
c
n 1
n
(0) sin(
cn (t )  Ee n t
2
nx
) = 1
L
2
[1  (1) n ] = A
n
(二)
Non-homogeneous problem
 2 u u

x 2 t
0 < x < L, t > 0
(1)
with the boundary conditions
u(0, t) = 0
(2)
u(L, t) = f1
(7)
and the initial condition
u(x, 0) = f(x)
(8)
Sol:
u(x, t) is assumed as
u(x, t) = G(x) + v(x, t)
↗
↖
steady-state part
transient part
(9)
Substitution of Eq. (9) into Eqs. (1), (2) and (7) yields
d 2G  2 v v


dx 2 x 2 t
Let
d 2G
H
dx 2
 2 v v
H 2 
x
t
u(0, t) = G(0) + v(0, t) = 0
 G(0) = 0
and v(0, t) = 0
(10)
u(L, t) = G(L) + v(L, t) = f1
 G(L) = f1
and v(L, t) = 0
Thus we can obtain
G(x) = f1 x/L
and

v(x, t) =
c
n 1
n
( t ) sin(
nx
)
L
(11)
In addition, the constant H is also assumed as

d
H=

n 1
dn 
n
sin(
nx
)
L
(12)
2H
[1  (1) n ]
n
Substitution of Eqs.(11) and (12) into Eq.(10) yields
dc n ( t )
dt
d
 2n t
c
(
t
)

Ee
 2n
n

n
d n  2n c n ( t ) 


c n ( 0)  E 
u(x, 0) = G(x) +  c n (0) sin(
n 1

(三)
E
dn
2n
nx
) = f(x)
L
2 L
nx
dn
[
f
(
x
)
G(x)]
sin(
)
dx

L 0
L
2n
Multi-dimensional parabolic diffusion problem
2u 
u
t
(13)
2u 
1
 h 2 h 3 u
 h1h 3 u
[
(
)
(
)
h1h 2 h 3 x1 h1 x1 x 2 h 2 x 2

 h1h 2 u
(
)]
x 3 h 3 x 3
(14)
where  2 u is the Laplacian of u. h1 , h 2 and h 3 are the scalar coefficients and
are determined as follows.
(i) Rectangular coordinate system
2-D:
 2 u  2 u u


x 2 y 2 t
for 0 < x < L, 0 < y < L, t > 0
 Parabolic type or elliptic type?
with the boundary conditions
u(0, y) = u(L, y) = 0
u(x, 0) = 0
at y = 0
u(x, L) = 1
at y = L
at x = 0, L
(15)
and the initial condition
u(x, y, 0) = 0
Assume that the solution u(x, y, t) is
u(x, y, t) = X(x)Y(y)T(t)
Substitution of Eq.(5) into Eq.(1) yields
1 d 2 X 1 d 2 Y 1 dT


X dx 2 Y dy 2 T dt
 Can you use the separation of variables to solve this prob.?
(16)
However, can you use the separation of variables to solve this prob. with u(x,
L) = 0 at y = L?


1 d 2X
 2n
2
X dx
1 d 2Y
 2m
2
Y dy
and
∴   n 
n
,
L
  m 
n = 1, 2,…
m
,
L
m = 1, 2,…
Its corresponding eigenfunctions X n and Ym are
X n  sin(  n x )  sin(
nx
)
L
Ym  sin(  m y)  sin(

u(x, t) =

 c
n 1 m 1
mn
my
)
L
my
nx [( mL ) 2 ( nL ) 2 ]t
sin(
) sin(
)e
L
L
my
nx
c
sin(
)
sin(
)
u(x, y, 0) =  mn
L
L
n 1 m 1


=1
4 L L
my
nx
sin(
)
sin(
)dxdy
L 0 0
L
L

c mn 

c mn 
4
[1  (1) m ][1  (1) n ]
2
mn 
(ii) Cylindrical coordinate system: (x, y, z) → (r,  , z)
x = rcos(  )

y = rsin(  )
z=z
x 2
y
z
)  ( )2  ( )2  1
r
r
r
h1  h r 
(
h2  h 
(
x 2
y
z
)  ( )2  ( )2  r



h3  h z 
(
x 2
y
z
)  ( )2  ( )2  1
z
z
z
Thus Eq. (14) can be written as
1  u 1  2 u  2 u
u
[ (r ) 

]

r r r
r 2 z 2
t
(17)
1-D diffusion problem 
1  u
u
(r ) 
r r r
t ,
with u(a, t) = 0 and u(r, 0) = f
(18) 
0 < r < a, t > 0

(18)
Is one b.c. missed?
 2 u 1 u u


r 2 r r t
(19)
Obviously, the point at r = 0 is a regular-singular point.
Assume
u(r, t) = R(r)T(t)
Then
d 2 R dR
r 2 
 2 rR  0
dr
dr
(20)
and
dT
 2 T  0
dt
Thus R(r) and T(t) can be obtained as
 AJ (r )  BY0 (r )
R (r )   0
C  D ln r
0
0
(21)
and
Ee   t
T( t )  
 F
2
0
0
(22)
Substitution of Eqs. (21) and (22) into Eq. (20) yields
u (r, t )  (C  D ln r )F  [AJ 0 (r )  BY0 (r )]Ee   t
2
At r = 0
 D=B=0

u (r, t )  C*  A*J 0 (r )e  t
2
 u(a, t) = 0 =

C*  0

n a  zn
C*  A*J 0 (a )e   t
and
2
J 0 ( a )  0
where z n is the n-th root of J 0 (z)  0

u (r, t )   A*n J 0 (
n 1
2
rz n
)Ee  t
a
rz n
)dr
0
*
a
 An  a
rz 2
0 r[J 0 ( an )] dr
a
f  rJ 0 (
(iii) Spherical coordinate system: (x, y, z) → (r,  ,  )
x = rcos(  )sin(  )

(
h2  h 
(
(
z = r cos(  )
x 2
y
z
)  ( )2  ( )2  1
r
r
r
h1  h r 
h3  h 
y = rsin(  )sin(  )
x 2
y
z
)  ( ) 2  ( ) 2  r sin( )



x 2
y
z
)  ( )2  ( )2  r



Thus Eq. (14) can be written as
1  2 u
1  2u
1 
u
u
[
(
r
)


(sin(

)
)]

r 2 r
r sin 2 () 2 sin( ) 

t
(23)
1-D diffusion problem 
1  2 u
u
(
r
)

r 2 r
r
t ,
with u(a, t) = 0 and u(r, 0) = f
(18) 
0 < r < a, t > 0

(24)
Is one b.c. missed?
 2 u 2 u u


r 2 r r t
Obviously, the point at r = 0 is a regular-singular point.
(25)
Assume
u(r, t) = R(r)T(t)
(26)
d 2R
dR
2 2
r

2
r


r R 0
dr 2
dr
(27)
Then
2
and
dT
 2 T  0
dt
Eq. (27) is difficult to be solved.
(四)
One-dimensional hyperbolic diffusion problem
 2u  2u
c

x 2 t 2
2
for -∞ < x < ∞, t > 0
with
u(x, 0) = f(x)
and
du
( x ,0 )  g ( x )
dt
The Fourier transform of this problem with respective to x is
d 2~
u
 2 c 2 ~
u 0
2
dt
with
and
~
~
u ( x,0)  f ()
d~
u ( 0) ~
 g ()
dt
where

~
u (, t )   u(x, t )eix dx


~
f ()   f (x)e ix dx

and

~
g ()   g(x)e ix dx

~
It is easy to obtain the solution of u (, t ) as
1 ~ i ~ ict 1 ~ i ~ ict
~
u (, t )  ( f 
g )e
 (f 
g )e
2
c
2
c
~
The inverse transform of u (, t ) is
  ~ i ~ ict 1 ~ i ~ ict ix
u ( x , t )   [( f 
g )e
 (f 
g )e ]e d
4 
c
2
c

  ~ i ~ i( x ct ) 1 ~ i ~ i( x  ct )
[( f 
g )e
 (f 
g )e
]d



4
c
2
c
Thus we can obtain
1
1
u ( x, t )  [f ( x  ct )  f ( x  ct )]  [g( x  ct )  g( x  ct )]
2
2
or
u ( x, t )  ( x  ct )   ( x  ct )
Assume g(x) = 0, then we can obtain
1
u ( x, t )  [f ( x  ct )  f ( x  ct )]
2