11.5. Heat Equation: Solution by Fourier Series

11.5. Heat Equation: Solution by Fourier Series

u: temperature, c²: thermal diffusivity ρ: density, σ: specific heat

u(t,x,y,z)

 Simple case: temp. distribution in a long thin bar or wire: u(t,x)

B.C.s: u(0,t) = 0, u(L,t) = 0 for all t I.C.: u(x,0) = f(x) (f(0) = f(L) = 0) - Same procedure as Section 11.3

First Step: Two ODEs let u(x,t) = F(x)G(t)

Second Step: Satisfying the B.C.’s



 





= ρσ

∇

∂ =

∂ K

c u

t c

u _{2 2 2}

x = 0 x = L

2 2 2

x c u t u

∂

= ∂

∂

∂ Parabolic Eqn. (B²-4AC=0)

0 G p c G

0 F p '' F

2 2 2

= +

= +



) , 2 , 1 n ( 1 B for L x

sin n )

x ( F

0 ) L ( F ) 0 ( F : s ' C . B apply px

sin B px cos A )

x ( F 0

F p '' F

n 2

=  π =

 =

=

=

← +

 =

= +

0 ) t ( G ) L ( F ) t , L ( u

0 ) t ( G ) 0 ( F ) t , 0 ( u

=

=

=

=

(For n = 0 ?  F = 0 no interest)

Third Step: Solution of the Entire Problem - General solution:

I.C.:

Ex.1~3)

Ex. 4) Bar with insulated ends.: B.C.’s: u_x(0,t) = F’(0)G(t) = 0, u_x(L,t) = F’(L)G(t) = 0 )

, 2 , 1 n ( e

B ) t ( L G

0 cn G

G ²_{n n}   _n = _n ^2nt = 



 



λ = π

= λ

+ ^−λ

) , 2 , 1 n ( e

L x sin n B

) t , x (

u

_{n n}



^2nt

= 



 



 π

=

^−λ

Eigenfunctions with eigenvalues λn

 

 



 λ = π

 

 



 π

=

=  

^∞

=

λ

∞ −

=

L

e cn L x

sin n B

) t , x ( u )

t , x (

u

_n

1 n

t n

1 n

n

2 n

) , 2 , 1 n ( L dx

x sin n

) x ( L f

B 2 );

x ( f L x

sin n B

) 0 , x (

u

^L

n 0 1

n

n

 = = π = 



 



 π

= 

^∞



=

(Fourier sine series)

(all the terms  0 as time increases)

) , 2 , 0 n ( 1 A for L x

cos n )

x ( F

0 ) L ( ' F ) 0 ( ' F : s ' C . B apply px

sin B px cos A )

x ( F 0

F p '' F

n 2

=  π =

 =

=

=

← +

 =

= +

(n includes 0 !!!)



 



λ = π



 



 π

=

=

 

^∞

=

λ

∞ −

= L

e cn L x

cos n A

) t , x ( u )

t , x (

u _n

0 n

t n

0 n

n

2 n

) , 2 , 1 , 0 n L , ( cn

L e x cos n

A )

t , x (

u

_{n n}



^2nt

λ

_n

= π = 



 



 π

=

^−λ

) , 2 , 1 n ( L dx

x cos n ) x ( L f

A 2 , dx ) x ( L f

A 1

);

x ( f L x

cos n A

A L x

cos n A

) 0 , x ( u

L n 0

L 0 0

1 n

n 0

0 n

n

= 

= π

=

=



 



 π

+

=



 



 π

=









^∞

=

∞

=

(Fourier cosine series)

Steady-State 2D Heat Flow



 



 =

∂

= ∂

∂ + ∂

∂

 ∂



 





∂ + ∂

∂

= ∂

∇

∂ =

∂ 0

t 0 u

y u x

: u state steady

y at u x

c u u t c

u

2 2 2

2 2

2 2

2 2 2

2 

Elliptic Eqn. (B²-4AC<0) I.C.:

From G_n(0) = 0 at y=0:

0 ) a ( F , 0 ) 0 ( F 0

dx kF F d

dy k G d G

1 dx

F d F ) 1

y ( G ) x ( F ) y , x ( u

2 2

2 2 2

2

=

=

←

= +

−

=

−

 =

=

) , 2 , 1 n ( a x

sin n )

x ( F

F = _n = π = 



a / y n n a

/ y n n n

2 2

2

e B e

A ) y ( G ) y ( G 0

a G n dy

G

d _{π − π}

+

=

 =

=



 



 π

−

a y sinh n

a A y sinh n A

2 ) y (

G_n = _n π = ^*_n π

a x sin n

a y sinh n

A ) y , x (

u_n = ^*_n π π





^∞

=

∞

=

π

= π

=

1 n

* n 1

n

n a

x sin n

a y sinh n

A )

y , x ( u )

y , x ( u



^_^{ π }_^{ π π =}



^π

= ^∞

=

a 0

* n 1

n

*

n dx

a x sin n ) x ( a f

2 a

b sinh n

A a ;

x sin n

a b sinh n

A )

b , x ( u

y

x u=f(x)

u=0 u=0 0 u=0

R

a b

Dirichlet Problem in a Rectangle R

B.C.:

11.6. Heat Equation: Solution by Fourier Integrals and Transforms

- In the case of infinite bars: Fourier series  Fourier integrals

(laterally insulated, infinite ends), u(x,0) = f(x) (-∞ < x < ∞)

Use of Fourier Integrals

- f(x): nonperiodic function (p multiples of a fixed number)  use of Fourier integrals A & B: functions of p, then A(p) & B(p)

Determination of A(p) and B(p) from the Initial Condition

2 2 2

t c u t u

∂

= ∂

∂

∂

0 G p c G

0 F p '' F

2 2 2

= +

= +

 ( )

^{c p t}

t p c

2 2

2 2

e px sin B px cos A )

p

; t , x ( u

e ) t ( G

; px sin B px

cos A )

x ( F

−

−

+

=

= +

=

( ^{A ( p ) cos px B ( p ) sin px} ) ^{e dp}

dp ) p

; t , x ( u )

t , x (

u

^{c p t}

0 0

2

− 2

∞

∞



 ^{= +}

=

( )



 





= π

= π



 −

 π

= +

=





 



∞

∞

−

∞

∞

−

∞ ∞

∞

−

∞

pvdv sin

) v ( 1 f

) p ( B ,

pvdv cos

) v ( 1 f

) p ( A

dp dv ) pv px

cos(

) v ( 1 f

) x ( f dp px sin ) p ( B px cos ) p ( A )

0 , x (

u 0 0

dv dp ) pv px

cos(

e )

v ( 1 f

) t , x (

u

0

t p c^{2 2}



₋^∞_∞

_ ^ 

^{∞ −}

⁻ _ ^

= π















0^{∞ − 2} = π ^−b2

s e

ds 2 bs 2 cos

e





 



 = = −

t c 2

v b x

, t p c s

let

^{2 2 2}







− −

= π



0^{∞ −e cos(px} −^pv)dp _{2c t} ^{exp (x}_4c^2v_t⁾²

t p c^{2 2}

t dv c 4

) v x exp (

) v ( t f

c 2 ) 1 t , x (

u

₂



₋^∞_∞

_ ^{  − −}

2

_ ^{ }

= π

( ^{let z = ( v − x ) /( 2 c t )} ) ^{u ( x , t ) = 1} _π 

₋^∞_∞

^{f ( x + 2 cz t ) e}

^−z2

^dz

Ex. 1)

Use of Fourier Transforms

Ex. 2) Temperature in the infinite bar

- Fourier transform w.r.t. x and resulting ODE in t

( ^{let uˆ = ℑ ( u )} )

t dx uˆ

t ue 2

dx 1 e

2 u ) 1

u (

uˆ w c )

u ( c ) u (

iwx iwx

t t

2 2 xx

2 t

∂

= ∂

∂

∂

= π

= π ℑ

−

= ℑ

= ℑ





₋^∞_∞ ⁻ ₋^∞_∞ ⁻

Ex. 3) Convolution method

t w c 2

2

2

e 2

) w ( C ) t , w ( uˆ

uˆ w t c

uˆ

= −

−

∂ =

∂

= 

= C(w) fˆ(w) )

0 , w ( uˆ : . C .

I uˆ(w,t) = fˆ(w)e^−c2w2t

dv dw e

e )

v ( 2 f

1

dw e

e ) w ( 2 fˆ

) 1 t , x ( u : Inversion

) wv wx ( i t w c

iwx t w c

2 2

2 2

 



∞

∞

−

∞

∞

−

−

−

∞

∞

−

−





= π

= π





 





= π 

−^∞∞

−

dv

e ) v ( 2 f

) 1 w (

fˆ

^iwv



dv dw ) wv wx

cos(

e )

v ( 1 f

) t , x (

u ⁼ _π 

₋^∞_∞

^ _ 

₋^∞_∞ ^−c2w2t

⁻ _ ^



₋^∞_∞ ⁻

= π fˆ(w)e e dw 2

) 1 t , x ( u : from

Starting ^{c2w2t iwx}



 



^{use u(x,t)} = ^{f *g(x)} =



₋^∞_∞ ^f(p)g(x −^p)dp =



₋^∞_∞ ^{fˆ(w)gˆ(w)eiwvdw}

t w c^{2 2}

2 e ) 1

w (

gˆ ⁻

= π

(Imaginary part: odd func)

 

 



 ℑ

⁻

= e

⁻

 ℑ ( e

⁻

) = 2 c t e

⁻

= 2 c t 2 π gˆ ( w ) a

2 ) 1

e

(

^{ax2 w2/4a x2/4c2t 2 c2w2 2}







 = tc a

4

1 ₂

t c 4 / x 2

2

e

2

2 t c 2 ) 1

w ( gˆ of

Inverse

⁻

 π



₋^∞_∞

_ ^{  − −} _ ^{ }

= π

= dp

t c 4

) p x exp (

) p ( t f

c 2 ) 1 x )(

g

* f ( ) t , x (

u

₂

2

Ex. 4) Fourier sine transform applied to the heat equation

A laterally insulated bar from x=0 to infinity, (u(x,0) = f(x), u(0,t) = 0)

 



∞ ∞ −

− ∞

= π













= π

 =

−

∂ =

∂

0 0

t w c

s 0 t

w c s

s

s 2 s 2

dw dp wx sin e

wp sin ) p ( 2 f

) t , x ( u

dp wp sin ) p ( 2 f

) w ( fˆ e

) w ( fˆ ) t , w ( uˆ

uˆ w t c

uˆ

2 2 2

2

) 0 ( 2wf ))

x ( f ( w

)) x ( '' f

( ² _S

S = − ℑ + π

ℑ

←

(See Ex. 2 of Sec. 10.10 for the Fourier transform)

Example) Heating a semi-infinite slab

I.C.: u(x,0) = 0, B.C.’s: u(0,t) = 1, u(∞,t) = 0

2 2

x u t

u

∂ α ∂

∂ =

∂

t 4

x

= α η

∞

= η

=

= η

= η =

η η +



η α

∂ =

∂ η

− η

∂ =

∂

at 0 u , 0 at

1 u : s . C . B with d 0

2 du d

u d

t 4

1 d

u d x

; u d du t 2 1 t

u

2 2

2 2 2

2

) ( erf 1

d ) 2 exp(

1 d

) exp(

d ) exp(

1 ) ( u

C d

) exp(

C u ), exp(

C then

d , let du

0

2

0

2 0

2

0 2

2 1

2 1

η

−

= η η

π −

− η =

η

−

η η

− −

= η

+ η η

−

= η

−

= η ψ

= ψ

 





η

∞ η

η

Define a new variable:

error function (See Ex. 2 of Sec. 10.10 for the integration of special function)