EE ngineering Mathematics II ngineering Mathematics II
Prof. Dr. Yong-Su Na g (32-206, Tel. 880-7204)
Text book: Erwin Kreyszig, Advanced Engineering Mathematics, 9th Edition, Wiley (2006)
Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations
(PDEs) (PDEs)
12 1 Basic Concepts 12.1 Basic Concepts
12.2 Modeling: Vibrating String, Wave Equations
12.3 Solution by Separating Variables. Use of Fourier Seriesy p g
12.4 D’Alembert’s Solution of the Wave Equation. Characteristics 12.5 Heat Equation: Solution by Fourier Series
12.6 Heat Equation: Solution by Fourier Integrals and Transforms 12.7 Modeling: Membrane, Two-Dimensional Wave Equation
12.8 Rectangular Membrane. Double Fourier Series
12.9 Laplacian in Polar Coordinates. Circular Membrane.
Fourier-Bessel Series
12.10 Laplace’s Equation in Cylindrical and Spherical Coordinates. Potential
l f b l f
12.11 Solution of PDEs by Laplace Transforms
2
Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations
(PDEs) (PDEs)
12 1 Basic Concepts 12.1 Basic Concepts
12.2 Modeling: Vibrating String, Wave Equations
12.3 Solution by Separating Variables. Use of Fourier Seriesy p g
12.4 D’Alembert’s Solution of the Wave Equation. Characteristics 12.5 Heat Equation: Solution by Fourier Series
12.6 Heat Equation: Solution by Fourier Integrals and Transforms 12.7 Modeling: Membrane, Two-Dimensional Wave Equation
12.8 Rectangular Membrane. Double Fourier Series
12.9 Laplacian in Polar Coordinates. Circular Membrane.
Fourier-Bessel Series
12.10 Laplace’s Equation in Cylindrical and Spherical Coordinates. Potential
l f b l f
12.11 Solution of PDEs by Laplace Transforms
3
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
K
∂u 2 2 2
z Heat Equation(열전도방정식) :
• σρ
c K u t c
u = ∇ =
∂
∂ 2 2 2
,
) y Diffusivit (Thermal
:
2 열확산계수
c
•
•
•
ty) Conductivi (Thermal
: 열전도도 K
비열 :
σ
밀도 : 물체의
ρ
•
z 문제
물체의 밀도 ρ :
• 1차원 열전도방정식 : 2
2 2
x c u t
u
∂
= ∂
∂
∂
• 경계조건(Boundary Condition) :
• 초기조건(Initial Condition) :
( ) ( )
0 ,
t u L,
t0 ( 모든
t에 대하여 )
u
= =
( )x f ( )x ( x L)
u
, 0 = , 0 ≤ ≤
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
z 1단계. 열전도방정식으로부터 두 개의 상미분방정식의 유도
( ), ( ) ( ) 22 ''( ) ( ), u F( ) ( )x G t t
G x u F
y G x F y
x
u( )= ( ) ( ) ⇒ ∂ = ( ) ( ) ∂ = ( ) ( )
( ) ( ) 2 ( ) ( ) ( ) ( )
(
2)
2
'' ''
,
,
x p F t
t G G x F c t
G x F
t G x t F
t G x x F
y G x F y
x u
=
=
⇒
=
∴
∂
⇒ ∂
( ) ( ) ( ) ( ) ( )
( ) ( )
( )
( )
2
p
x F t
G t c
G x F c t
G x
F = ⇒ = = −
∴
( ) ( ) 0, ( ) ( ) 0
''
+ 2 = + 2 2 =
∴ F x p F x G t c p G t
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
z 2단계. 경계조건의 만족
( ) ( ) ( ) ( ) ( ) ( )
( )0 ( ) 0
0 ,
, 0 0
, 0
=
=
⇒
=
=
=
= L F F
t G L F t
L u t
G F t
u 을 적용
경계조건
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) 0 ( ) cos sin
''
0 0
, 0 ''
2 2
+
=
⇒
= +
=
=
= +
∴
px B
px A
x F x
F p x F
L F F
x F p x F
( ) ( ) ( )
( )0 0, ( ) cos sin 0 sin 0 ( : )
sin cos
0
=
⇒
=
⇒
= +
=
=
=
+
=
⇒
= +
L n p n
pL B
pL B
pL A
L F A
F
px B
px A
x F x
F p x F
π 정수
( ) ( ) sin ( 1, 2, 3, ")
∴ = = x n =
L x n
F x
F n π
( ) n ( ) n Gn( )t Bne n t L
cp cn t
G t
G
+
λ 2= 0 ,
λ= =
π⇒ =
−λ 2( )
i
nπx λ 2t (1 2 3
) ( ), = sin
e−(n
= 1 , 2 , 3 , "
)L x B n
t x
un n π λn t
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
z 3단계. 전체 문제에 대한 해. 푸리에 급수
( ) ( ) sin 2 ⎟
⎠
⎜ ⎞
⎝
⎛ =
=
=∑∞ u x,t ∑∞ B nLxe− cnL
x,t
u n n π λn t λn π
1
1 = ⎝ ⎠
= n L L
n
( )x = ∑∞ Bn n x = f ( )x
u ,0 sin
: π
충족
초기조건의 ( ) ( )
∫ ( )
∑
=
⇒
=
L n
n n
L dx x x n
L f B
L f
1
2 sin
,
적용 π 급수
사인 푸리에
충
L L 0
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
Ex. 1 Sinuroidal Initial Temperaturep
( )
( ), . 50 C ?
C 0
, C 80 sin
100
cm 80
D
D D
얼마인가 시간은
걸리는 데
떨어지는 로
최고온도가 막대의
이때 구하라
를 온도
때 유지될 로
양끝이 이고
초기온도가 구리막대의
길이 절연된 측면이
t x u
πx
C sec cm cal 0.095
, C gm cal 0.092 , cm gm 8.92
:
밀도 3 비열 D 열전도도 D
데이터 물리적
대한 구리에
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
Ex. 1 Sinuroidal Initial Temperature
( )
( ), . 50 C ?
C 0
, C 80 sin
100
cm 80
D
D D
얼마인가 시간은
걸리는 데
떨어지는 로
최고온도가 막대의
이때 구하라
를 온도
때 유지될 로
양끝이 이고
초기온도가 구리막대의
길이 절연된 측면이
t x u
πx
p
C sec cm cal 0.095
, C gm cal 0.092 , cm gm 8.92
:
밀도 3 비열 D 열전도도 D
데이터 물리적
대한 구리에
의하여 초기조건에
( ) ( )
[ ]
n n
K c
B B x B
x x f
B n x
u
2 2 2
3 2 1
1
870 9 95
0
0
, 100
sin80
80 100 sin 0
,
∞
=
⎤
⎞ ⎡
⎛
=
=
=
=
⇒
=
=
=∑
π
π
π "
[ ]
( ) xe . t
t x u
K L
c
001785 0
1 2
2 1 2
2 2
sin80 100 ,
sec 001785 .
80 0 870 . 158 9 . 1
sec 158 cm
. 92 1 . 8 092 . 0
95 . 0
−
−
=
∴
=
⋅
=
⎥⎦ ⇒
⎢⎣ ⎤
= ⎡
= ⋅
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛=
=
π σρ λ λ π
( ) 80
[ ]sec 6.5[ ]min 001785 388
0 5 . 0 ln
50
100e−0.001785t = ⇒ t = = [ ]≈ [ ]
001785 .
−0
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
Ex. 4 Bar with Insulated Ends. Eigenvalue 0
경계조건을 양 끝이 단열되었을 때의 조건으로 바꾸고 열전도 방정식의 해를 구하라.
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
Ex. 4 Bar with Insulated Ends. Eigenvalue 0
경계조건을 양 끝이 단열되었을 때의 조건으로 바꾸고 열전도 방정식의 해를 구하라.
( ) ( ) (모든 에대하여)
경계조건
이다 기울기는
없다면 열흐름이
어 단열되 끝이
양
비례한다 에
기울기 온도의
전열속도는 의하면
실험에 물리적인
0 0
0 :
. 0
.
) (Gradient
t t
L u t
u ( ) ( ) (모든 에대하여)
경계조건 : ux 0,t = 0, ux L,t = 0 t
( )t F ( ) ( )G t u ( )L t F ( ) ( )L G t F ( ) F ( )L
u ( )0, = '( ) ( )0 = 0, ( ), = '( ) ( )= 0 ⇒ '( )0 = '( )= 0
( ) ( )
( ) ( ) n
px Bp
px Ap
x F' px
B px A
x F
L F F
t G L F t
L u t
G F
t
ux x
π cos
sin
sin
cos
0 0
0 ,
, 0 0
, 0
+
−
=
⇒ +
=
⇒
( ) ( )
( )x n x
F
L p n
p B
pL Bp
pL Ap
L F Bp
F n
π
π cos
, 0
0 cos
sin '
, 0 0
'
=
∴
=
=
=
⇒
= +
−
=
=
=
∴
( )x L Fn = cos
∴
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
( ) ( ) ( )
L e cn
L x A n
t G x F t
x
un n n n π λn t λn π
cos
, 2
⎞
⎛
⎟⎠
⎜ ⎞
⎝
⎛ =
=
=
∞
∞
−
( ) ( )
( ) A n x f ( )
L e cn
L x A n
t x u t
x
u n
n
t n
n
n n
π λ π π λ
0
cos
, ,
0 0
2
∑
∑
∑ = ⎜⎝⎛ = ⎟⎠⎞
=
∞
∞
=
∞ −
=
초기조건으로부터 ( ) ( )
( )x dx A f ( )x n x dx
f A
x L f
A x
u
L L
n
n
cos π 2
1
cos
0 ,
0
0
∫
∫
∑
=
=
⇒
=
=
=
초기조건으로부터
( ) ( ) dx
x L L f
A dx
x L f
A , n cos
0 0
0 ∫ ∫
⇒
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
z Steady Two-Dimensional Heat Problems. Laplacs’s Equation
• 2차원 열전도 방정식 :
⎟⎟ ⎠
⎜⎜ ⎞
⎝
⎛
∂ + ∂
∂
= ∂
∂
∂
2 2 2
2 2
y u x
c u t u
• ∂
) (
0 0
,
) (Steady
2 2
2 라플라스 방정식
무관 시간에
열전도가
∂ =
∂ +
=
∇
⇒
∂ =
⇒ ∂
u u u
t 정상적 u
z Boundary Value Problem, BVP(경계값 문제)
) (
0
2 2 = 라플라스 방정식
+ ∂
= ∂
∇
⇒ u x y
•
•
경우 주어지는
값이 의
위에서
문제 : Dirichlet C u
경우 주어지는
값이 의
법선도함수 문제 : 위에서
Neumann u
u
C ∂
•
•
경우 주어지는
값이 의
법선도함수 위에서
문제 :
Neumann
u n
C n
= ∂
주어지고 값이
의 한부분에서는 의
문제 문제
경계값
혼합 , Robin : C u ,
경우 주어지는
값이 의
부분에서 나머지
의
C un
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
z Dirichlet Problem in a Rectangular R y
u = f(x)
x u f(x)
u = 0
u = 0 u = 0
b
0
R
( ) ( ) ( )
d G F d d G
F y d
G x F y x
u , = ⇒ 2 2 = − 2 2 변수분리법 적용
u = 0 x
a 0
( ) ( ) ( )
dy k G d G dx
F d F
dy y dx
y
−
=
−
=
⇒ 2 2 2 2
2 2
1 1
( ) ( )
( ) F ( ) n G( ) G ( ) A B
n F k
a F F
y n y
π n
π π π
2 1 i
0 0
:
2
⎟⎞
⎜⎛
=
= 때 −
일
경계조건 변의
오른쪽 및
왼쪽
( ) ( ) ( ) ( )
( )
G
n e
B e
A y G y G a x
x n F x a F
k n
n
a n a
n n
n
π π
0 0 :
, 2 , 1 ,
sin
,
=
= +
=
=
=
⎟ =
⎠
⎜ ⎞
⎝
=⎛
∴
경계조건 아랫변에서의
때
일 "
( )
y n x
n
a y A n
e e
A y G B
A a n
y n a
y n n n
n n
π π
π π
π
sinh 2
0
⎟⎟⎠=
⎜⎜ ⎞
⎝
⎛ −
=
⇒
= +
∴ −
( ) ( ) ( )
a y n a
x A n
y G x F y x
un n n n π π
sinh sin
* ,
= =
⇒
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
y n x
n
∞
∞ π π
( ) ( )
( )x b f ( )x u
a y n a
x A n
y x u y
x u
n n
n n ∑
∑
=
=
= ∞
=
∞
=1 1
, :
sinh sin
* ,
, π π
경계조건 위변에서
( ) ( )
a b n a
x A n
x f b x u
n n
∑
=
=
⇒ ∞
=1
sinh
sin
* ,
π π
( ) ( ) dx
a x x n
f a
b a n
A a dx
x x n
a f a
b A n
bn = n = ∫a ⇒ n = a∫
⇒
0 0
sin sinh
* 2
2 sin sinh
*
π
π π
π
12
12.5 Heat Equation: Solution by Fourier.5 Heat Equation: Solution by Fourierqq yy Series
Series
((열전도방정식 열전도방정식))
PROBLEM SET 12.5
HW 13 23 25 34 HW: 13, 23, 25, 34
12
12.6 Heat Equation: Solution by Fourier.6 Heat Equation: Solution by Fourierqq yy Integrals and Transforms
Integrals and Transforms
((열전도방정식 열전도방정식))
z Bars of Infinite Length
• Heat Equation: 2
u
u ∂
∂
2 2
x c u
t u
∂
= ∂
∂
∂
• Initial Condition:
( )
x f( )
xu ,0 =
• u( )x,t = F( ) ( )x G t ⇒ F ''+ p2F = 0, G + c2 p2G = 0
( ) ( ) 2 2t
• ( ) ( )
( ) ( ) c p t
t p c
e px B
px A
FG p
t x u
e t
G px
B px
A x
F
2 2 2
2
sin cos
; ,
, sin cos
−
−
+
=
=
⇒
= +
=
12
12.6 Heat Equation: Solution by Fourier.6 Heat Equation: Solution by Fourierqq yy Integrals and Transforms
Integrals and Transforms
((열전도방정식 열전도방정식))
z Use of Fourier Integralsg
( )= ∫∞ ( ) = ∞∫( + ) −
0 0
2
sin 2
cos
; ,
,t u x t p dp A px B px e dp x
u c p t
12
12.6 Heat Equation: Solution by Fourier.6 Heat Equation: Solution by Fourierqq yy Integrals and Transforms
Integrals and Transforms
((열전도방정식 열전도방정식))
z Use of Fourier Integrals
( )= ∫∞ ( ) = ∞∫( + ) −
0 0
2
sin 2
cos
; ,
,t u x t p dp A px B px e dp x
u c p t
( ) (∞∫ ) ( )
기 건 터
g
( ) ( ) ( )
( ) ∫ ( ) ( ) ∫ ( ) ( ) ∫ ∫ ( ) ( )
∫
∞ ∞
∞
∞
⎥⎤
⎢⎡ −
=
⇒
=
=
⇒
= +
=
dp dv pv px v
f x
u pvdv
v f p
B pvdv v
f p
A
x f dp px B
px A
x u
0
1 cos 0
1 sin 1 cos
sin cos
0 , 초기조건으로부터
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ∫ ( ) ∫ ( )
∫ ∫
∫
∫
∞ ∞
−
∞
−
∞
−
∞
−
⎥⎦
⎢ ⎤
⎣
⎡ −
=
∴
⎥⎦
⎢⎣
⇒
⇒
dv dp pv px e
v f t
x u
dp dv pv px v
f x
u pvdv
v f p
B pvdv v
f p
A
t p c
0
1 cos ,
cos 0
, sin
, cos
2 2
π
π π
π
∞
− ⎣0 ⎦
π
12
12.6 Heat Equation: Solution by Fourier.6 Heat Equation: Solution by Fourierqq yy Integrals and Transforms
Integrals and Transforms
((열전도방정식 열전도방정식))
z Use of Fourier Integrals
( )= ∫∞ ( ) = ∞∫( + ) −
0 0
2
sin 2
cos
; ,
,t u x t p dp A px B px e dp x
u c p t
( ) (∞∫ ) ( )
기 건 터
g
( ) ( ) ( )
( ) ∫ ( ) ( ) ∫ ( ) ( ) ∫ ∫ ( ) ( )
∫
∞ ∞
∞
∞
⎥⎤
⎢⎡ −
=
⇒
=
=
⇒
= +
=
dp dv pv px v
f x
u pvdv
v f p
B pvdv v
f p
A
x f dp px B
px A
x u
0
1 cos 0
1 sin 1 cos
sin cos
0 , 초기조건으로부터
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ∫ ( ) ∫ ( )
∫ ∫
∫
∫
∞ ∞
−
∞
−
∞
−
∞
−
⎥⎦
⎢ ⎤
⎣
⎡ −
=
∴
⎥⎦
⎢⎣
⇒
⇒
dv dp pv px e
v f t
x u
dp dv pv px v
f x
u pvdv
v f p
B pvdv v
f p
A
t p c
0
1 cos ,
cos 0
, sin
, cos
2 2
π
π π
π
∞
− ⎣0 ⎦
π
∫ −
∞e−s2 bsds = e b2 2 2
cos
0
π 을 적용하면
공식
( ) ( ) ( )
( )
∫
∞
∞
∞
− ⎭⎬⎫
⎩⎨
⎧− −
= dv
t c
v v x
t f t c
x u
1
exp 4 2
, 1 2
2
π
( )
∫∞
−
+ −
= 1 f x 2cz t e z2dz
π
