Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations Ch. 12 Partial Differential Equations

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, 9^th 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 ,

,

0 ( 모든

에 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

'' ''

,

,

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 G_n( )t B_ne ^{n t} L

cp cn t

G t

G

 +

= 0 ,

= =

⇒ =

( )

ⁱ

^{1 2 3}

^{, = sin}

^{= 1 , 2 , 3 , "}

L x B n

t x

u_{n 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 n}_L^{xe− cn}_L

x,t

u _{n n} π _λ^{n t} λ_n 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 : u_x 0,t = 0, u_x 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

u_{x 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 F_n = 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

u_{n n n n} π _λ^{n t} λ_n 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 u_n

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

u_{n 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

b_n ⁼ _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:

( )

( )

u ,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