94
2. Partial Differential Equations
Section 2.1 Separation of Variables
Section 2.2 Classical PDEs and Boundary Value Problems (只教不考)
Section 2.3 Heat Equation
Section 2.4 Laplace’s Equation
Section 2.5 Nonhomogeneous PDEs (只考前三個解法)
Section 2.6 Higher Dimensional PDEs
Section 2.7 PDEs in Polar Coordinates
Section 2.8 PDEs in Cylindrical Coordinates (只教不考)
Section 2.9 PDEs in Spherical Coordinates (只教不考)
Section 2.10 Solving PDEs by Laplace Transforms (只教不考)
Section 2.11 Solving PDEs by Fourier Transforms (只教不考)
[1] D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value Problem
(metric version), 9th edition, Cengage Learning, 2017.
[2] http://djj.ee.ntu.edu.tw/DE.htm
95
Solving PDEs
Introduction of
the Method
Separation of
Variables
Basic Process (Sec. 2.1)
Superposition (Sec. 2.1, Sec. 2.4)
Nonhomogeneous Cases (Sec. 2.5)
(Two Methods)
Rectangular
Coordinate
Examples
Problem Classification (Sec. 2.2)
Heat Equation (Sec. 2.3)
Wave Equation
Laplace Equation (Sec. 2.4)
Higher-Dimensional Case (Sec. 2.6)
Polar and Spherical Coordinates (Sections 2.7, 2.8, 2.9)
Integral Transform Methods (Sec. 2.10 and Sec. 2.11)
96
2.1 Boundary-Value Problem in
Rectangular Coordinates
Use the methods of
(1) separation of variables
(2) the Laplace / Fourier transforms
Sections 2.1 ~ 2.9
Sections 2.10 and 2.11
to solve the PDE problem.
D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage Learning, 2017, Section 12.1.
97
linear second order partial differential equation for two independent
variables
2
2
2



u
u
 C u2  D u  E u  Fu  G
A 2 B
xy
x
y
x
y
7 terms
B2 − 4AC > 0 : hyperbolic, B2 − 4AC = 0 : parabolic
B2 − 4AC < 0 : elliptic
homogeneous : G(x, y) = 0, nonhomogeneous : G(x, y)  0
Linear: A, B, C, D, E, F, and G are independent of u
2.1.1 Superposition Principle
【Theorem 2.1.1】 Superposition Principle
If u1, u2, …., uk are solutions of a homogeneous linear partial
differential equation, then
u  c1u1  c2u2    ck uk
is also a solution of the homogeneous linear partial differential
equation.
(Proof): If
 2u1
 2u1
 2u1
u
u
A 2 B
 C 2  D 1  E 1  Fu1  0
 x y
x
y
x
y
 2u 2
 2u 2
 2u2
u
u
A 2 B
 C 2  D 2  E 2  Fu2  0
 x y
x
y
x
y
98
99
then
 2 (c1u1  c2u2 )
 2 (c1u1  c2u2 )
 2 (c1u1  c2u2 )
A
B
C

2
2
x
y


x
y
 (c1u1  c2u2 )
 (c1u1  c2u2 )
D
E
 F (c1u1  c2u2 )
x
y
  2u1

 2u1
 2u1
 u1
 u1
 c1  A 2  B
C 2  D
E
 Fu1 
x
y
x
y




y
 x

  2u 2

 2u 2
 2u 2
 u2
 u2
 c2  A 2  B
C 2  D
E
 Fu2 




x
y
x
y
y
 x

 c1 0  c2 0  0
2.1.2 Method of Separation of Variables
解 PDE with BVP (or IVP) 的方法
(1) method of separation of variables
若 PDE 當中有對 x 及對 y 的偏微分，
假設解為 u(x, y) = X(x)Y(y)
(2) using the Fourier transform (or Fourier cosine transform,
Fourier sine transform) (see Sections 2.10 and 2.11)
共通的精神： PDE
ODE
100
101
Method of Separation of Variables 的流程
(Step 1) 假設解為 u(x, y) = X(x)Y(y)
解法關鍵
(Step 2) 將 u(x, y) = X(x)Y(y) 代入 PDE，把 PDE 變成
“ function of X ” = “ function of Y ” = 
的型態
 被稱為 real separation constant
Steps 3, 4, 5 要分成不同的 Cases 來解
除了trivial 的情形外，所有可能的 cases 都要考慮
(Step 3) 將 function of X =  的解算出，即為 X(x)
註：(a) 如果有等於零的 boundary (initial) conditions，
也要在這一步考慮
(See the Examples in Sections 2.3, 2.4, and 2.5)
(b) 有時，先解 Y(y) 會比較容易
(視 boundary (initial) conditions 而定)
(c) 在這一步中，有的時候，會得出  的限制
(Step 4) 將 function of Y =  的解算出，即為 Y(y)
需注意的地方和 Step 3 相同
(Step 5) u(x, y) = X(x)Y(y)
102
(Step 6) 將所有可能的解全部加起來
103
(Step 7) 用非零的 boundary (initial) conditions 將 coefficients 求出
註： 這一步經常會用到 Fourier series, Fourier cosine series
或 Fourier sine series
※ 若沒有 boundary (initial) conditions，Steps 6, 7 可以省略
Rules:
x 的 BVP (IVP) 簡單
先算 X(x)
y 的 BVP (IVP) 簡單
先算 Y(y)
沒有 BVP (IVP)
先算 X(x) 或 Y(y) 皆可
 u ( x, y )  u ( x, y )

0
2
2
x
y
u  L, y   0
u  0, y   0
2
104
2
u  x,0   f  x 
u
 g  x
y y 0
先算 X(x)
 2u ( x, y )  2 u ( x, y )

2
x
y 2
u  0, y   f ( y )
 u  x, y   0
y
y 0
u  L, y   0
 u  x, y 
0
y
yH
先算 Y(y)
105
Note: Separation of variables 的方法其實未必可以得出 PDE 所有的解
有些解無法用 X(x)Y(y) 來表示
Separation of variables 的主要好處是比其他方法簡單
106
[Example 1]
u 2  4 u
y
x 2
Step 1 假設解為 u(x, y) = X(x)Y(y)
Step 2 將 u(x, y) = X(x)Y(y)
(解法關鍵)
2

u
u

4
代入
y
x 2
X   x  Y  y   4 X  x  Y   y 
X   x  Y   y 

4X  x Y  y


令 X  x   Y  y   
4X  x Y  y
X   x   4 X  x   0
real separation constant
(解法關鍵)
Y   y   Y  y   0
(The detail can be reviewed from the PowerPoint in DE1)
X   x   4 X  x   0
Case 1 for Steps 3, 4, 5
Step 3-1
Y   y   Y  y   0
107
 0
X   x   0
m2  0
auxiliary function
roots : 0, 0
X  x   c1  c2 x
Step 4-1 Y   y   0
Y  y   c3
Step 5-1 u  x, y   X  x  Y  y    c1  c2 x  c3  A1  B1 x
A1  c1c3
B1  c2c3
108
Case 2 for Steps 3, 4, 5   0
為了方便起見，令
   2
Step 3-2 X   x   4 X  x   0
roots of the auxiliary function: 2, 2
2
X  x   d1e 2 x  d 2 e 2 x
通常將解改寫成 X  x   c4 cosh(2 x )  c5 sinh(2 x )
Step 4-2
Y  y 
2
Y  y
Y   y    2Y  y   0
Y  y    Y  y   0
2
Y  y   c6 e
2y
Step 5-2 u  x, y   X  x  Y  y   A2 e
2y
2y
cosh(2 x )  B2 e
A2  c4c6
sinh(2 x )
B2  c5c6
109
Case 3 for Step 3   0
為了方便起見，令    2
Step 3-3 X   x   4 2 X  x   0 roots of the auxiliary function: j2, j2
X  x   c7 cos(2 x )  c8 sin(2 x )

Step 4-3 Y  y    2
Y  y
Step 5-3 u  x, y   A3e
Y   y    2Y  y   0
 2 y
cos(2 x )  B3e
 2 y
Y  y   c9 e
 2 y
sin(2 x )
若要處理 boundary conditions，或著想得到 general solution，
要將所有可能的解都加起來
Step 6
2y
2y
u  x, y   A1  B1 x   [ A2, e cosh(2 x )  B2, e sinh(2 x )]
 0
  [ A3, e
 0
 2 y
cos(2 x )  B3, e
 2 y
sin(2 x )]  是任意實數
(註：nonseparable 的解在這一步得到)
110
附錄三： Hyperbolic Function
x
x
e

e
sinh  x  
2
jx
 jx
e

e
比較： sin  x  
2j
x
x
e

e
cosh  x  
2
jx
 jx
e

e
cos  x  
2
sinh  x  e x  e  x

tanh  x  
cosh  x  e x  e  x
cosh  x  e x  e  x

coth  x  
sinh  x  e x  e  x
sech  x  
1
 x 2 x
cosh  x  e  e
csch  x  
1
 x 2 x
sinh  x  e  e
111
5
5
2
sinh(x)
1
0
0
tanh(x)
0
cosh(x)
-1
-5
-2
0
2
3
2
-2
0
2
2
coth(x)
1
1
0
sech(x)
2
-2
0
2
-2
-2
0
2
csch(x)
1
0
-1
-1
-2
-2
3
0
-1
-3
-5
-2
-2
0
2
-3
-2
0
2
d sinh  ax   a cosh  ax 
dx
sinh  0   0
d cosh  ax   a sinh  ax 
dx
cosh  0   1
d tanh  ax   a sech 2  ax 
dx
sinh  0   1
d coth  ax   a csch 2  ax 
dx
cosh  0   0
d sech  ax   a sech  ax  tanh  ax 
dx
sin  ix   i sinh  x 
d csch  ax   a csch  ax  coth  ax 
dx
cos  ix   cosh  x 
112
Section 2.2 Classical PDEs and Boundary-
113
Value Problems
2.2.1 本節綱要
(1) one-dimensional heat equation (或簡稱為 heat equation)
2
u
2
Generally,
k
u


t
k  u2  u
k
>
0
t
x
(2) one-dimensional wave equation (或簡稱為 wave equation)
2
2
2
2 2

u
2  u

u
a
u


Generally,
a
 2
2
2
t

x
t
(3) two-dimensional form of Laplace’s equation (或簡稱為 Laplace’s
equation ) 2
Generally,  2u  0
 u   2u  0
x 2 y 2
D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage Learning, 2017, Section 12.2.
114
名詞：
heat equation,
wave equation,
(page 117)
Laplace’s equation, (page 120)
Laplacian,
(page 121)
Dirichlet condition, (page 123)
Neumann condition, (page 123)
Robin condition
(page 115)
(page 123)
本節的重點：七大名詞，和它們所對應的公式
115
2.2.2 One-Dimensional Heat Equation
2

k u2  u
t
x
由來：熱傳導的理論
u(x, t): temperature, t: time, x: location
Fig. 2.2.1
heat equation 別名：diffusion equation
From D. G. Zill and Michael R. Cullen,
Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage
Learning, 2017, Section 12.2.
116
2

k u2  u
t
x
Example:
u(x, t): temperature,
t: time, x: location
u(x, t0)
Fig. 2.2.2
x0
x-axis
 u ( x, t )
0
x0 的溫度將上升
 t x x0
2.2.3 One-Dimensional Wave Equation
2
2

u

u
a

x 2 t 2
「拉像皮筋」的模型
2
u(x, t): height, t: time, x: location
(1) 剛開始時
Fig. 2.2.3
f(x)
u(x, t) = 0
when x = 0
From D. G. Zill and Michael R. Cullen,
Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage
Learning, 2017, Section 12.2.
u(x, t) = 0
when x = L
wave equation 別名： telegraph equation
117
118
(2) 手放開之後產生振動
振動
Fig. 2.2.4
From D. G. Zill and Michael R. Cullen,
Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage
Learning, 2017, Section 12.2.
 Wave equation 其他的應用：
Theory of high-frequency transmission line
Fluid mechanics (流體力學)
Acoustics (聲學)
Elasticity (彈力學)
Microwave engineering (電波工程)
119
2.2.4 Two-Dimensional Form of Laplace’s Equation
 2u   2u  0
x 2 y 2
溫度隨著位置而變化的模型
u(x, y): temperature,
x, y: location
120
121
Laplace’s Equation 亦可用 Laplacian 表示, 2u(x, y) = 0
Laplacian: 2
2
2

u

 u  x, y   2  u2
x y
2
2
2
2

u

u

 u  x, y, z   2  2  u2
x y z
2
122
Modification
加上外力，或與外界的交互作用
例：heat equation 的 modification
2

k u2  h  u  um   u
t
x
例：wave equation 的 modification
2
2

u

u
a

F
x
t
u
u

,
,
,


t
x 2
t 2
2
 Laplace’s Equation 的其他應用
Static displacement of membranes
Edge detection (邊緣偵測)
Microwave engineering (電波工程)
2.2.6 Boundary Conditions 或 Initial Conditions
Dirichlet condition
u = ……..
(沒微分)
Neumann condition
u  
n
(有微分)
Robin condition
u  hu  
n
(混合)
h is a constant
123
2.3 Heat Equation
124
This section can also be viewed as an example of Section 2-1
 2u u
(1)
k 2  , 0  x  L, t  0
x
t
(2)
u (0, t )  0, u ( L, t )  0, t  0
(3)
u ( x, 0)  f ( x), 0  x  L.
Solution:
(Step 1)
u  x, t   X  x  T  t 
kX   x  T  t   X  x  T   t 
D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage Learning, 2017, Section 12.3.
125
kX   x  T  t   X  x  T   t 
(Step 2)
X '' T '

 
X
kT
X ''  X  0
T ' k T  0.
(4)
(5)
(6)
(From Zero Boundary Conditions)
u (0, t )  X (0)T (t )  0 and u ( L, t )  X ( L)T (t )  0.
Since for a nontrivial solution, T(t) cannot be zero,
X(0) = 0 and X(L) = 0.
We have
X ''  X  0, X (0)  0, X ( L)  0.
T ' k T  0.
(7)
(i) X ''  X  0, X (0)  0, X ( L)  0.
(ii) T ' k T  0.
Case 1 for Steps 3, 4, 5  = 0
X ''  0
X ( x)  c1  c2 x
From X (0)  0, X ( L)  0
u  x, t   X  x  T  t   0
(trivial solution)
c1  c2  0
X ( x)  0
126
(i) X ''  X  0, X (0)  0, X ( L)  0.
(ii) T ' k T  0.
Case 2 for Steps 3, 4, 5  < 0
Set    2
X ''  2 X  0
X ( x)  c1e x  c2 e  x
X ( x)  c3 cosh  x   c4 sinh  x 
Note:
x
 x
x
 x
e

e
e

e
c3 cosh  x   c4 sinh  x   c3
 c4
2
2
c c
c c
 3 4 e x  3 4 e  x
2
2
From X(0) = 0, c3 = 0
From X(L) = 0, c4 sinh  L   0 , c4 = 0
X ( x)  0
(trivial solution)
u  x, t   X  x  T  t   0
127
(i) X ''  X  0, X (0)  0, X ( L)  0.
(ii) T ' k T  0.
Case 3 for Steps 3, 4, 5  > 0
Set    2
X ( x)  c1 cos  x  c2 sin  x.
X ''  2 X  0
X (0)  0,
X ( L)  0.
c1  0,
c2 sin  L  0
  n / L,   n 2 2 / L2 .
X ( x)  c2 sin( nx / L),   n 2 2 / L2 .
2 2
n
T ' k 2 T  0.
L
un  x, t   X ( x)T (t )  An e
T (t )  c3e
 k ( n 2 2 / L2 ) t
n = 1, 2, 3, ….
 k ( n 2 2 / L2 ) t
n
si n
x,
L
n = 1, 2, 3, ….
128
n
(Step 6) u ( x, t )   n 1 un  x, t    An e  k ( n  / L )t sin
x,
L
n 1


2 2
2
(Step 7) From the boundary condition, u ( x, 0)  f ( x)
n
x  f  x
L
n 1
From Fourier sine series (page 131 in 附錄四)

u ( x, 0)   An sin

g  x    bn sin n x
p
n 1
p
2
bn   g  x  sin n xdx
p 0
p
2 L
n
An   f ( x) sin
xdx.
0
L
L
Therefore,
2 2 2
2   L
n
n

u ( x, t )     f ( x) sin
xdx e  k ( n  / L )t sin
x.
0
L n 1 
L
L

129
130
附錄四 Review for Fourier Series and Fourier Cosine / Sine
Series
Fourier Series
a0  
f  x      an cos n x  bn sin n x 
2 n1 
p
p 
p
1
a0   f  x  dx
p p
p
1
an   f  x  cos n xdx
p p
p
p
1
bn   f  x  sin n xdx
p p
p
131
f(x) is even
Fourier Series
Fourier cosine series (或 cosine series)
a0 
f  x     an cos n x
2 n1
p
p
2
an   f  x  cos n xdx
p 0
p
p
a0  2  f  x  dx
p 0
f(x) is odd
Fourier sine series (或 sine series)

f  x    bn sin n x
p
n 1
p
2
bn   f  x  sin n xdx
p 0
p
Section 2.4 Laplace’s Equation
2.4.1 Section 2.4 綱要
 2u   2u  0
x 2 y 2
0 < x < a,
0 < y < b,
(使用method of separation of variables 來解)
0
「問題 1」 u
x x0
u  x,0   0
「問題 2」 u  0, y   0
u  x,0   0
u
0
x xa
for 0 < y < b,
u  x, b   f  x 
for 0 < x < a
u  a, y   0
for 0 < y < b,
u  x, b   f  x 
for 0 < x < a
D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage Learning, 2017, Section 12.5.
132
 u u 0
x 2 y 2
2
133
2
0 < x < a,
0 < y < b,
「問題 3」 u  0, y   F  y  u  a, y   G  y  for 0 < x < a
u  x,0   f  x 
u  x, b   g  x 
※ 特別注意 “superposition principle”
for 0 < y < b,
134
2.4.2 Solutions for Laplace’s Equations (挑戰解解看)
 2u   2u  0
x 2 y 2
0 < x < a,
0 < y < b,
u
0
x x0
u
0
x xa
u  x,0   0
u  x, b   f  x  for 0 < x < a
for 0 < y < b,
Step 1 假設解為 u(x, y) = X(x)Y(y)
 2u   2u  0
得出
Step 2 代入
x 2 y 2
X   x  Y  y   X  x  Y   y   0
令
X   x 
Y   y 

 
X  x
Y  y
得出2 個 ODEs
X   x    X  x   0
X   x 
Y   y 

X  x
Y  y
Y   y   Y  y   0
135
Steps 3, 4, 5 的前處理
(1) 因為 x 的 boundary condition 較簡單，所以先解 X(x)
(2) 分成  = 0,  < 0 ,  > 0 三個 cases
 0 for all 0 < y < b,
(3) 由於 u
x x0
X  x  Y  y 
 X   0Y  y   0
x
x 0
Y(y) 不可為 0 (否則 u(x, y) = X(x)Y(y) = 0)
所以 X   0   0
同理，由 u
0
x xa
X  a   0
同理，由 u  x,0   0
Y 0  0
X   x    X  x   0
X 0  0
Y   y   Y  y   0
Y 0  0
136
X  a   0
Case 1 of Steps 3, 4, 5:  = 0
Step 3-1
X   x   0
solution: X  x   c1  c2 x
由 boundary conditions
X  0  0
X  a  0
X  x   c1
Step 4-1 Y   y   0
Y(0) = 0
solution: Y  y   c3  c4 y
根據 boundary condition Y(0) = 0, c3 = 0
Y  y   c4 y
Step 5-1
u  x, y   X  x  Y  y   c1c4 y  A0 y
A0  c1c4
c2  0
137
Case 2 of Steps 3, 4, 5:  < 0
令  = −2
Step 3-2
X  0  0
X   x    2 X  x   0
solution:
X  a   0
X  x   d 2e x  d3e x
可改寫成 X  x   d 4 cosh  x   d5 sinh  x 
由 boundary conditions X   0   0
以及 d cosh  x    sinh  x  ,
dx
X  a   0
d sinh  x    cosh  x 
dx
d5  0
d5  0
d 4 sinh  a   d5 cosh  a   0
d4  0
因此， case 2 得出 trivial solution u(x, y) = X(x)Y(y) = 0
u  x, b   f  x  將無法滿足
 < 0 時無解
(不需再算 Steps 4-2, 5-2)
X  x  0
138
Case 3 of Steps 3, 4, 5:  > 0
令  = 2
Step 3-3
X   x    2 X  x   0
X 0  0
X  a  0
solution: X  x   c1 cos( x)  c2 sin( x)
由 boundary conditions X   0   0
c2  0
c1 sin( a)  c2 cos( a)  0
X  a   0
c1  any nonzero constant
n 是任意整數
  n
a
c2  0
再次注意：不可直接判斷成 c1 = 0 and c2 = 0
應該看看是否有適當的 , 讓第二個式子等於零
X n  x   c1 cos n x
a
2 2
n
n 是任意正整數     2
a
2
Step 4-3 Y   y   n 2 Y  y   0
a
Y(0) = 0
2
solution:
2
Yn  y   d3e
n y
a
 d 4e
  n 2
2
since
139
2
a
 n y
a
 
 
經常改寫為 Yn  y   c3 cosh n y  c4 sinh n y
a
a
c3  0
根據 boundary condition Y(0) = 0
 
Yn  y   c4 sinh n y
a
Step 5-3
u  x, y   X  x  Y  y 
 


  
 c1 cos n x c4 sinh n y  An cos n x sinh n y
a
a
a
a
n 是任意正整數
An  c1c4

140
Step 6 把所有可能的解，全部加起來

  
u  x, y   A0 y   An cos n x sinh n y
a
a
n 1

Q: 為什麼 n 是從 1 加到 ，而非由  加到 ？
討論：既然 n 是任意整數，那為什麼 n 是從1 加到 ，
而非由   加到  ？
因為
 


cos n x  cos n x ,
a
a
 


sinh n y  sinh n y ,
a
a
sinh  0   0
可證明
 B cos  na x  sinh  na y 

n 
n
 
 
  A cos  n x  sinh  n y 
a
a



  cos n x  Bn sinh n y  B n sinh n y 
a
a
a


n 1

n 1
n
An  Bn  B n
141
142
  

Step 7 u  x, y   A0 y   An cos n x sinh n y
a
a
n 1

nonzero boundary condition: u  x, b   f  x 
   
也就是說，2A b 和 A sinh  n b  (n = 1, 2, …., )
a

f  x   A0b   An cos n x sinh n b
a
a
n 1
0
n
是 f(x) 的 Fourier cosine series 的 coefficients
a
2
2 A0b   f ( x )dx
a 0
a
A0  1  f ( x ) dx
ab 0
 
a
n

2
An sinh
b   f ( x ) cos n xdx
a
a 0
a
a
n xdx
2
(
)
cos
An 
f
x

a
n

a sinh
b 0
a
 
143
2.4.3 Laplace’s Equations with Dirichlet Problem
 2u   2u  0
x 2 y 2
0 < x < a,
u  0, y   0
u  a, y   0
0 < y < b,
u  x,0   0
u  x, b   f  x 
0 < x < a,
0 < y < b,
用 method of separation of variables，經過計算得出

u  x, y    An sinh n y sin n x
a
a
n 1
a
n xdx
2
sin
An 
f
x



a
a sinh n b 0
a
可自行練習解解看
144
2.4.4 Superposition Principle
Dirichlet Problem 可分解成兩個子問題
 2u   2u  0
x 2 y 2
0 < x < a,
u  0, y   F  y 
u  a, y   G  y 
for 0 < y < b,
u  x,0   f  x 
u  x, b   g  x 
for 0 < x < a,
0 < y < b,
當四個邊界都不為零時，很難直接用 separation of variable 的方法
解出來
2
2

u

u1
1
子問題 1

0
2
2
x
y
0 < x < a,
0 < y < b,
u1  0, y   0
u1  a, y   0
for 0 < y < b,
u1  x,0   f  x 
u1  x, b   g  x 
for 0 < x < a,
2
2

u

u2
子問題 2
2
 2 0
2
x
y
0 < x < a,
0 < y < b,
u2  0, y   F  y 
u2  a, y   G  y 
for 0 < y < b,
u2  x,0   0
u 2  x, b   0
for 0 < x < a,
假設 u1(x, y), u2(x, y) 分別是子問題 1, 子問題 2 的解
則 u (x, y) = u1(x, y) + u2(x, y) 是原來問題的解
145
146
當 u (x, y) = u1(x, y) + u2(x, y)
u (0, y) = u1(0, y) + u2(0, y) = 0 + F(y) = F(y)
u (a, y) = u1(a, y) + u2(a, y) = 0 + G(y) = G(y)
u (x, 0) = u1(x, 0) + u2(x, 0) = f(x) + 0 = f(x)
u (x, b) = u1(x, b) + u2(x, b) = g(x) + 0 = g(x)
=
Fig. 2.5.1
+
From D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage Learning, 2017, Section 12.5.



子問題 1 的解 u1  x, y    An cosh n y  Bn sinh n y sin n x
a
a
a
n 1
147
 
a
2
An   f  x  sin n x dx
a 0
a
 
 
a
n x dx  A cosh n b 
1
2

sin
Bn 
g
x


n
 a 0
a
a 

n
sinh
b
a
 
子問題 2 的解 u  x, y     A cosh n x  B sinh n x sin n y
b
b
b
A  2  F  y  sin  n y  dy
b
b
1
 2 G  y  sin n y dy  A cosh n a 
B 




 b 

b
b

n
sinh  a 
b

2
n 1
n
n
b
n
0
b
n
原來問題的解 u1  x, y   u2  x, y 
0
n
148
2.4.5 Sections 2.1~2.4需要注意的地方
(1) Method of separation of variables 解 PDE 的過程雖然長，但是把
握住講義 pages 101-103 的 7 個 steps，就大致上沒問題。
(2) 注意，
若 boundary conditions 出現 u(0, y) = 0, u(L, y) = 0,
最後的解總是和 sine 有關
X  x   c2 sin n x
L
若 boundary conditions 出現
u
0
x x0
週期為 2L/n
u
0
x x L
最後的解總是和 cosine 或 constant 有關
X  x   c1
or
X n  x   c1 cos n x
L
週期也為 2L/n
(3) 經驗足夠後，看到 u(x, y) 的 boundary conditions
出現 u(a, y) = 0
就知道 X(a) = 0 ，
看到 u(x, b) = 0
0
看到 u
x xa
看到 u
0
y y b
就知道 Y(b) = 0 。
就知道 X'(a) = 0，
就知道 Y'(b) = 0
(4) 要熟悉 cosh(x), sinh(x) 的性質
149
(5) Method of separation of variables 在計算上容易出錯的地方
(以 講義 pages 134-142 Laplace equations 為例)
(a) X   x 
Y   y 

 
X  x
Y  y
(b) Steps 3, 4, 5 要考慮所有 cases
(c) 不可直接由 c1 = 0 及 c1 cos  x  c2 sin  x  0 判斷 c1 = c2 = 0
因為  可以是 n/L, 如講義 page 138 所述
(d) 在 Step 6，要將所有可能的解加起來，才是 u(x, t) 的一般解
如講義 page 140 所述
150
151
2.5 Nonhomogeneous Boundary-Value Problems
2
2
2
u
u



A 2 B
 C u2  D u  E u  Fu  G
xy
x
y
x
y
Nonhomogeneous:
G0
Key ideas: Separate the original problem into two or more problems
D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage Learning, 2017, Section 12.6.
2
2
2
u
u



A 2 B
 C u2  D u  E u  Fu  G
xy
x
y
x
y
Method 1 u  x, y   v  x, y    x 
Method 2 u  x, y   v  x, y     y 
Method 3 u  x, y   v  x, y   1  x    2  y 
Method 4 u  x, y   v  x, y     x, y 
(Method 4 只教不考)
Extra Methods: Expansion by Fourier series, Fourier cosine series,
or Fourier sine series
2.6.1 Method 1
153
2
2
2

u

u

A 2 B
 C u2  D u  E u  Fu  G
xy
x
y
x
y
Method 1
u  x, y   v  x, y     x 
[Constraint]: G is independent of y
 2 v ( x, y )
 2 v ( x, y )
 2 v ( x, y )
 v ( x, y )
 v ( x, y )




 Fv( x, y )
A
B
C
D
E
2
2




x
y
x
y
x
y
 A   x   D   x   F  x   G  x 
Problem A: (ODE for (x))
A   x   D   x   F  x   G  x 
Problem B: (homogeneous PDE for v(x, y))
 2 v ( x, y )
 2 v ( x, y )
 2 v ( x, y )
v( x, y )
v( x, y )




 Fv( x, y )  0
A
B
C
D
E
2
2
xy
x
y
x
y
154
[Example 1]
 2u
u
Solve k 2  r  , 0  x  1, t  0
x
t
u (0, t )  0, u (1, t )  u1 , t  0
r and u1 are nonzero constants
u ( x, 0)  f ( x), 0  x  1
(Solution): Since G = r,
u (0, t )  0, u (1, t )  u1
independent of t
constants
Method 1 can be applied.
u ( x, t )  v( x, t )   ( x),
 2 v ( x, t )
d 2 ( x)
v( x, t ) d ( x)
k

k

r


2
2
dx
dt
x
t
 2 v ( x, t )
d 2 ( x)
v( x, t )
k
k
r 
2
2
dx
x
t
Problem A
Problem B
k ( x)  r  0,
 (0)  0,  (1)  u1
 2 v v
k 2  , 0  x  1, t  0
x
t
v(0, t )  0, v(1, t )  0, t  0
v( x, 0)  f ( x)  ( x), 0  x  1
(i) For Problem A
k ( x)  r  0,
 (0)  0,  (1)  u1
 ( x)   r x 2  c1 x  c0
2k
 ( x)   r / k
c0  0,  r  c1  u1
2k
 ( x)   r x 2  r  u1 x
2k
2k
 (0)  0,  (1)  u1


155
156
(ii) For Problem B, from Section 2-3

v( x, t )   An e
 kn 2 2t
sin  n x 
n 1


2
where An  20  f  x   r x  r  u1 x  sin  n x  dx
2k
2k


1
Therefore,



2
 kn 2 2t
r
r
u ( x, t )  
x 
 u1 x   An e
sin  n x 
2k
2k
n 1
2.6.2 Methods 2 and 3
2
2
2

u

u

A 2 B
 C u2  D u  E u  Fu  G
xy
x
y
x
y
Method 2
u  x, y   v  x , y     y 
[Constraint]: G is independent of x
Method 3 u  x, y   v  x, y    1  x    2  y 
[Constraint]: G = G1(x) + G2(y)
G1(x) is independent of y
G2(y) is independent of x
157
158
A  u2  B  u  C  u2  D u  E u  Fu  G
xy
x
y
x
y
u  x, y   v  x, y    1  x    2  y 
2
2
2
 2 v ( x, y )
 2 v ( x, y )
 2 v ( x, y )
 v ( x, y )
 v ( x, y )
A

B

C

D

E
 Fv( x, y )
2
2

x

y

x

y
x
y
 A 1 x   D 1  x   F 1  x   C 2  y   E 2  y   F 2  y   G1  x   G2  y 
Problem A: (ODE for 1(x))
A 1 x   D 1  x   F 1  x   G1  x 
Problem B: (ODE for 2(x))
C 2  y   E 2  y   F 2  y   G2  y 
Problem C: (homogenous PDE for v(x, y))
 2 v ( x, y )
 2 v ( x, y )
 2 v ( x, y )
v( x, y )
v( x, y )




 Fv( x, y )  0
A
B
C
D
E
2
2
xy
x
y
x
y
2.6.3 Method 4 (只教不考)
Method 4
u  x, t   v  x , t     x, t 
Constraint of Method 4: Not applicable for Laplace’s equation.
Method 1 can be applied to the wave equation and Laplace’s
equation, but Method 4 cannot.
Example:
 2u
u
k 2  F ( x , t )  , 0  x  L, t  0
x
t
u (0, t )  u0  t  , u ( L, t )  u1  t  , t  0
u ( x, 0)  f ( x), 0  x  L,
159
u
u
k 2  F ( x , t )  , 0  x  L, t  0
x
t
u (0, t )  u0  t  , u ( L, t )  u1  t  , t  0
2
u ( x, 0)  f ( x), 0  x  L,
Set
u  x, t   v  x , t     x, t 
 2u  2 v  2
Since
 2 2
2
x
x
x
 2u
u
k 2  F ( x, t ) 
x
t
u (0, t )  u0  t 
u v 
 
t t t
 2v
 2
v 
k 2  k 2  F ( x, t )  
t t
x
x
v  0, t    0,t   u0  t 
u ( L, t )  u1  t 
v  L, t    L, t   u1  t 
u ( x, 0)  f ( x)
v( x, 0)  ( x, 0)  f ( x)
160
Therefore, after setting u  x, t   v  x, t     x, t  , we separate
 2u
u
k 2  F ( x , t )  , 0  x  L, t  0
x
t
u (0, t )  u0  t  , u ( L, t )  u1  t  , t  0
u ( x, 0)  f ( x), 0  x  L,
into two sub-problems:
 2
Problem A: k 2  0,   0, t   u0  t  ,   L, t   u1  t 
x
 2v
v
Problem B: k 2  G ( x, t )  , 0  x  L, t  0
x
t
v(0, t )  0, v( L, t )  0, t  0
v( x, 0)  f ( x)  ( x, 0), 0  x  L

G
(
x
,
t
)

F
(
x
,
t
)

where
t 
Guess: The solution of Problem B v  x, t    vn  t  sin n x
L
n 1
161
 2
Problem A: k 2  0,   0, t   u0  t  ,
x
  x, t   c1  t  x  c0  t 
  L, t   u1  t 
c0  t   u0  t 
  0,t   u0  t 
  L, t   u1  t 
c1  t  L  u0  t   u1  t 
  x, t   u0  t   x  u1  t   u0  t  
L
162
To Solve Problem B:
 2v
v
k 2  G ( x, t )  , 0  x  L, t  0
x
t
v(0, t )  0, v( L, t )  0, t  0
v( x, 0)  f ( x)  ( x, 0), 0  x  L

where G ( x, t )  F ( x, t ) 
t
An assumption can be applied
(from the associated homogeneous PDE).

v  x, t    vn  t  sin n x
L
n 1

G  x, t    Gn  t  sin n x
L
n 1
Try to solve vn(t) and Gn(t).


 k n 2 2 v t  G t  sin n x  v t sin n x



n 
n  
n 
2
L
L
L

n 1 
n 1
163
Summary for the Process of Method 4
164
(Step 1) Use u  x, t   v  x, t     x, t  to separate the original problem
into two sub-problems.
(Step 2) Solve Problem A
(Step 3) Use the associated homogeneous PDE to express the solution
of Problem B by Fourier sine series
(Step 4) Expand Gn(t) to solve vn(t)
(Step 5) Use u(x, 0) to solve the unknowns of vn(t)
(Step 6) Add the solutions of Problems A and B and obtain u(x, t).
[Example 2]
 2u u
, 0  x  1, t  0

2
x
t
u (0, t )  cos t , u (1, t )  0, t  0
u ( x, 0)  0, 0  x  1.
(Solution):
(Step 1)
u  x, t   v  x, t     x, t 
Problem A:
 2
 0,
2
x
Problem B:
  0, t   cos t ,  1, t   0
 2 v  v

 , 0  x  1, t  0
2
x
t
t
v(0, t )  0, v(1, t )  0, t  0
v( x, 0)    x, 0  , 0  x  1
165
166
(Step 2)
Problem A:
 2
 0,
2
x
  0, t   cos t ,  1, t   0
  x, t   c1  t  x  c0  t 
Solution:
Problem B:
  x, t    0  cos t  x  cos t  1  x  cos t
 2v
v

1

x
sin
t

, 0  x  1, t  0


2
x
t
v(0, t )  0, v(1, t )  0, t  0
v( x, 0)  x  1, 0  x  1
We can guess that the solution of Problem B is

v  x, t    vn  t  sin n x
n 1
Problem B:
 2v
v

1

x
sin
t

, 0  x  1, t  0


2
x
t
v(0, t )  0, v(1, t )  0, t  0
v( x, 0)  x  1, 0  x  1
(Step 3) From the associated homogeneous PDE
 2 v v
 , 0  x  1, t  0
2
x
t
v(0, t )  0, v(1, t )  0, t  0
v( x, 0)  x  1, 0  x  1
v  x, t   X  x  T  t 
X   x  T  t   X  x  T   t 
X   x    X  x   0
T   t   T  t   0
X   x  T   t 

 
X  x  T t 
X 0  0
X 1  0
167
X   x    X  x   0
X 0  0
X 1  0
168
After checking the three cases, the non-trivial solution exists only when
  n 2 2  0
In this case,
X   x   n 2 2 X  x   0
X  x   c sin n x
Therefore, the solution of Problem B should has the following form:

v  x, t    vn  t  sin n x
n 1
to be solved
169
 2v
v

1

x
sin
t

, 0  x  1, t  0
 
2
x
t
v(0, t )  0, v(1, t )  0, t  0
v( x, 0)  x  1, 0  x  1

v  x, t    vn  t  sin n x
n 1
to be solved
(Step 4) First, express the non-homogeneous term (1x)sint as

1  x  sin t   Gn  t  sin n x
n 1
From the Fourier sine series (附錄四)
1
2
Gn  t    1  x  sin t sin n xdx  2 sin t
n
1 0
1

1  x  sin t   n2 sin t sin n x
n 1

f  x    bn sin n x
p
n 1
p
2
bn   f  x  sin n xdx
p 0
p
 2v
v



1
x
sin
t
 
x 2
t

v  x, t    vn  t  sin n x
n 1

1  x  sin t   n2 sin t sin n x
Since
n 1


 2 v  x, t   v  t   n 2 2 sin n x



n
x 2
n 1
 v  x, t   v  t  sin n x

n
t
n 1
we have


 v  t   n    n2 sin t  sin n x   v  t  sin n x
n 1
2
2
n
n 1
n
vn  t   n 2 2 vn  t   2sin t
n
2 2
 n 2 2t
n

sin
t

cos
t
vn  t   2
 Cn e
4 4
n  n   1
 2 2

 n 2 2t

n

sin
t
cos
t
 sin n x
 Cn e
v  x, t     2
4
4

n 1 
 n  n   1


170
 2v
v

1

x
sin
t

, 0  x  1, t  0
 
2
x
t
v(0, t )  0, v(1, t )  0, t  0
v( x, 0)  x  1, 0  x  1
 2 2

 n 2 2t

n

sin
t
cos
t
 sin n x
 Cn e
v  x, t     2
4
4

n 1 
 n  n   1

(Step 5) To determine Cn, we can apply v(x, 0) = x–1
 


2
x 1   
 Cn  sin n x
4
4

n 1  n  n   1


From the Fourier sine series

1
2
2
C
x
n

xdx


2

1
sin



n
0
n
n  n 4 4  1
Cn 
2
2

n  n 4 4  1 n
171
 2 2

 n 2 2t

n

sin
t
cos
t
 sin n x
 Cn e
v  x, t     2
4
4

n 1 
 n  n   1

2
2
Cn 

n  n 4 4  1 n

 2 2
 n 2 2t
 n 2 2t 
te
e
 sin n x

v  x, t   2   n  sin t 4 cos
4
 n 1 
n 
n  n   1



(Step 6)
u  x, t   v  x , t     x, t 
 2 2
 n 2 2t
 n 2 2t 
te
e
 sin n x

u  x, t   1  x  cos t  2   n  sin t 4 cos
4
n 
 n 1 
n  n   1



172
2.6 Higher-Dimensional Problems
173
Modifying the method in Section 2-1 just a little.
Two-dimensional heat equation
  2u  2u   u
k 2  2 
.
 x y  t
Two-dimensional wave equation
  2u  2u   2u
a  2  2  2 .
y  t
 x
2
u ( x, y, t )  X ( x)Y ( y )T (t )
 2u
 2u
u
X
''
YT
,
XY
''
T
,
and


 XYT '.
2
2
x
y
t
D. G. Zill and Michael R. Cullen, Differential Equations-with Boundary-Value
Problem (metric version), 9th edition, Cengage Learning, 2017, Section 12.8.
174
[Example 1] Temperatures in a Plate
  2u  2u  u
k  2  2   , 0  x  b, 0  y  c, t  0
y  t
 x
u (0, y, t )  0, u (b, y, t )  0, 0  y  c, t  0
u ( x, 0, t )  0, u ( x, c, t )  0, 0  x  b, t  0
u ( x, y, 0)  f ( x, y ), 0  x  b, 0  y  c.
From D. G. Zill and Michael R. Cullen, Differential
Equations-with Boundary-Value Problem (metric version),
9th edition, Cengage Learning, 2017, Section 12.8.
175
  2u  2u  u
k 2  2 
y  t
 x
u ( x, y, t )  X ( x)Y ( y )T (t ),
k ( X '' YT  XY '' T )  XYT '
Divided by XYT
k ( X ''  Y '' )  T '
X
Y
T
Set
X ''   Y ''  T '
X
Y kT
X ''   Y ''  T '  
X
Y kT
X ''  X  0
Y '' T '

 .
Y kT
Y ''
 
Y
Y   Y  0
T'
   
kT
T ' k (   )T  0.
u (0, y, t )  0, u (b, y, t )  0,
X (0)  0, X (b)  0,
u ( x, 0, t )  0, u ( x, c, t )  0
Y (0)  0, Y (c)  0
X ''  X  0, X (0)  0, X (b)  0
Y '' Y  0, Y (0)  0, Y (c)  0.
T ' k (   )T  0.
There are 3 cases for X:  = 0,  < 0, and  > 0.
m 2 2
There is non-trivial solution for X only when m 
0
2
b
m
x
In this case, X ( x)  c2 sin
b
There are 3 cases for Y:  = 0,  < 0, and  > 0.
n 2 2
There is non-trivial solution for Y only when n  2  0
c
n
y
In this case, Y ( y )  c4 sin
c
176
m 
m
b2
2
2
n 
n
.
2
c
2
2
2 2
2 2
m
n

T ' k ( 2  2 )T  0
b
c
 k [( m / b )2  ( n / c )2 ]t
T (t )  c5e
.
T ' k (   )T  0
u ( x, y, t )  X ( x)Y ( y )T (t ),
umn ( x, y, t )  Amn e


 k [( m / b )2  ( n / c )2 ]t
u ( x, y, t )   Amn e
m 1 n 1
m
n
x sin
y,
sin
b
c
 k [( m / b )2  ( n / c )2 ]t
sin
m
n
x sin
y.
b
c
177


m
n
x sin
y.
b
c
0  x  b, 0  y  c.
u ( x, y, t )   Amn e  k [( m / b )  ( n / c ) ]t sin
m 1 n 1
2
2
u ( x, y, 0)  f ( x, y )
 
m
n
Amn sin
x sin
y  f  x, y 

b
c
m 1 n 1

 
n y  sin m x  f  x, y 
A
sin

  mn

c
b
m 1  n 1

From the Fourier sine series along the x-axis

g  x    bn sin n x
p
n 1
p
2
bn   g  x  sin n xdx
p 0
p
 
 2 b
n

m xdx

A
y
f
x
y
sin
,
sin


  mn
 b 0
c
b
 n 1

From the Fourier sine series along the y-axis


 
c
b
2
2
Amn    f  x, y  sin m x dx sin n y dy
c 0b 0
b
c
178


u ( x, y, t )   Amn e  k [( m / b )  ( n / c ) ]t sin
2
2
m 1 n 1
m
n
x sin
y.
b
c
where


 
c b
4
Amn    f  x, y  sin m x dx sin n y dy
bc 0 0
b
c
[Example 2]
  2u  2u  2u   u
k 2  2  2 
.
 x y z  t
179
Double Sine Series (Two Dimensional Sine Series)


  

f  x, y    Bm ,n sin m x sin n y
b
b
m 1 n 1
where

  
c b
4
Bm,n    f  x, y  sin m x sin n y dxdy
bc 0 0
b
c
180