IX.7 ADDITIONAL INTEGRAL TRANSFORMSvps/ME505/IEM/09 07.pdf · 898 Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 6.7.1 Solution of 3-D

Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 897

IX.7 ADDITIONAL INTEGRAL TRANSFORMS

6.7.1 Solution of 3-D Heat Equation in Cylindrical Coordinates

6.7.2 Mellin Transform

6.7.3 Legendre Transform

6.7.4 Jacobi and Gegenbauer Transform Solution of generalized heat conduction problem

6.7.5 Laguerre Transform

6.7.6 Hermite Transform

6.7.7 Hilbert and Stiltjes Transform

6.7.8 Z Transform


6.7.1 Solution of 3-D Heat Equation in Cylindrical Coordinates

( )u r, ,z,tθ 2 2 2

2 2 2 2

u 1 u 1 u u g 1 ur r k tr r z αθ

∂ ∂ ∂ ∂ ∂+ + + + =

∂ ∂∂ ∂ ∂ (1)

1) z -variable a) Case of 0 z K≤ ≤ (Finite Cylinder) Use Finite Fourier Transform in z -variable: ( ){ }k ku z u=T (2) with the inverse

{ }-1kuT ( )k k

ku Z z= ∑

( )2

2k k k2

u zu

zω

∂ = − ∂

T (3)

b) Case of z 0≥ (Finite Cylinder) Use the corresponding standard, sine or cosine Fourier Transform in z -variable. c) Case of z−∞ < < ∞ (Finite Cylinder) Use Finite Integral Transform in z -variable. ( ){ } ( )F u z u ω= (4)

( ) ( )

22

2

u zF u

zω ω

∂ = − ∂

(5)

1r

r

0

z

1r

r

0

z K=

z


2) θ -variable

a) Case of 1 2θ θ θ≤ ≤ (Piece of Cake) Boundary conditions have to be set at 1θ θ= and 2θ θ= .

Reduce to interval 00 θ θ≤ ≤ and then use Finite Fourier Transform in θ variable:

( ){ }n nu uθ = T (6) { }-1

nu =T Operational property:

( )2

2n n n2

uu

θη

θ ∂ = −

∂

T (7)

b) Case of 0 2θ π≤ ≤ (full rotation) There no boundary conditions in this case for 0 2θ π= = , but solution has to be 2π -periodic:

( ) ( )u u 2θ θ π= +

Construct integral transform base on the Fourier series on the interval 0 2θ π≤ ≤ (see Table):

( ) ( )u u 2θ θ π= + ( ) ( )0 n nn 1

a a cos n b cos nθ θ∞

=

= + + ∑

where the Fourier coefficients are defined by:

( )2

00

1a u d2

π

θ θπ

= ∫

( ) ( )2

n0

1a u cos n dπ

θ θ θπ

= ∫

( ) ( )2

n0

1b u sin n dπ

θ θ θπ

= ∫

Substitute them into the Fourier series and rearrange:

( )u θ ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2

n 10 0 0

1 1u s ds u s cos ns ds cos n u s sin ns ds sin n2

π π π

θ θπ π

∞

=

= + +

∑∫ ∫ ∫

( )u θ ( ) ( ) ( ) ( ) ( ) ( )2 2

n 10 0

1 1u s ds u s cos ns cos n sin ns sin n ds2

π π

θ θπ π

∞

=

= + +

∑∫ ∫

( )u θ ( ) ( ) ( )2 2

n 10 0

1 1u s ds u s cos n s ds2

π π

θπ π

∞

=

= + −

∑∫ ∫

1θ θ=

θ

2θ θ=

r

( )r ,θ

0 θ


Then integral transform and its inverse can be defined as:

( ){ }0 u θT 0u ( )2

0

1 u s ds2

π

π= ∫ (8)

( ){ }n u θT nu ( ) ( )2

0

1 u s cos n s dsπ

θπ

= − ∫ n 1,2,...= (9)

{ }-1nuT ( )u θ n

n 0u

∞

=

= ∑ (10)

Operational properties of this integral transform:

2

0 2

uθ

∂ ∂

T ( )22

20

u s1 ds2 s

π

π∂

=∂∫

( ) 2

0

u s1 02 s

π

π∂

= = ∂ (11)

2

n 2

uθ

∂ ∂

T ( ) ( )22

20

u s1 cos n s dss

π

θπ

∂= − ∂∫

( ) ( )2

0

u s1 cos n s ds

π

θπ

∂ = − ∂

∫

( ) ( ) ( ) ( )2 2

00

u s u s1 1cos n s d cos n ss s

π π

θ θπ π

∂ ∂ = − − − ∂ ∂

∫

( ) ( )2

0

1n sin n s d u sπ

θπ

= − − ∫

( ) ( ) ( )22

00

1 1u sin n s n u s d sin n sππ

θ θπ π = − − + − ∫

( ) ( )2

2

0

1n u s cos n s dsπ

θπ

= − − ∫

2nn u= − (12)

Example:


Application of integral transforms in z and θ variables to the Heat Equation (1):

For the finite solid cylinder (combination of cases 1a and 2b) the consecutive application of the integral transforms (2) and (8,9) together with the operational properties (7) and (11,12) yields:

2 2

k ,n k ,n k ,n k ,n2k ,n k k ,n2 2

u u g u1 n 1u ur r k tr r

ωα

∂ ∂ ∂+ − − + =

∂ ∂∂

(14)

where n 0,1,2,...= ( )k 0 ,1,2,...= ( k 0= , I a case of N-N b.c.’s)

Equation (14) defines the order of the Hankel transform for elimination of the differential operator in r -variable (see p.388 and p.393). Therefore, the transformation of equation (14) should be consequently performed with the Hankel transforms of all non-negative integer orders: nH n 0,1,2,...=

Thus, the Hankel transform of equation (14) is: for solid cylinder 10 r r≤ ≤ :

k ,n,m k ,n,m2 2m k ,n,m k k ,n,m

g u1u uk t

λ ωα∂

− − + =∂

(15)

and for r 0≥ :

k ,n k ,n2 2k ,n k k ,n

g u1u uk t

λ ωα∂

− − + =∂

(16)

Now, equation (15) is an ordinary differential equation which subject to initial condition can be solved by variation of parameter or by the Laplace Transform to find the transformed solution:

( )k ,n,mu t

or ( )k ,nu t ,λ

(17)

Then the solution of the IBVP for the Heat Equation (1) can be found by the consecutive application of the inverse transforms:

( ) ( ) ( )( )

( )n mk ,n ,m k2

k n m n m

Fourier Series (10)

J ru r, ,z,t u t, Z z

J r

λθ θ

λ

=

∑ ∑ ∑

( ) ( ) ( )( )

( )nk ,n ,m k2

k n 0 n m

Fourier Series (10)

J ru r, ,z,t u t, d Z z

J r

λθ θ λ λ

λ

∞

=

∑ ∑ ∫

(18)

1r

r

0

z K=

z


( ) ( )1u r , Uθ θ=

θ

0 2π

( )u r,θ

Example: ( )u r,θ 2 2

2 2 2

u 1 u 1 u 0r rr r θ

∂ ∂ ∂+ + =

∂∂ ∂ (1’)

( ) ( )1u r , Uθ θ= Dirichlet Apply Integral Transform (8-9) with operational properties (11-12):

2 2n n

n2 2

u u1 n u 0r rr r

∂ ∂+ − =

∂∂

n 0,1,2,...= (14’)

( ){ } ( ) ( )n 1 n 1 nu r , u r , Uθ θ θ= =

T Apply FHT-I ( )I

nH with operational property on p.389 (Dirichlet): ( ) ( ) ( )2

1 m,n n 1 m,n 1 n m,n m,nr J r U u 0λ λ θ λ θ+ − =

where m,nλ are positive roots of ( )n 1J r 0λ = . Transformed Solution:

( ) ( ) ( )1 n 1 m,n 1n,m n

m,n

r J ru U

λθ θ

λ+=

Inverse Hankel Transform:

( ) ( ) ( )( )

n n,mn n,m 2

m 1n n,m

J ru r, u

J r

λθ θ

λ

∞

=

= ∑

Inverse Integral Transform (Fourier series):

( ) ( )nn 0

u r, u r,θ θ∞

=

= ∑ ( ) ( ) ( )( )

n 1 m,n 1 n m,n1 n 2

n 0 m 1 m,n n m,n

J r J rr U

J r

λ λθ

λ λ

∞ ∞+

= =

= ∑∑

( ) ( ) ( )( )

n 1 m,n 1 n m,n1 n 2

n 0 m 1 2m,n 1n 1 m,n 1

J r J rr U

r J r2

λ λθ

λλ

∞ ∞+

= =+

= ∑∑

( ) ( )

( )n m,nn

n 0 m 11 m,n n 1 m,n 1

J rU2r J r

λθλ λ

∞ ∞

= = +

= ∑∑

r

( )r ,θ

0 θ


Same Example: ( )u r,θ 2 2

2 2 2

u 1 u 1 u 0r rr r θ

∂ ∂ ∂+ + =

∂∂ ∂ (1’)

( ) ( )1u r , Uθ θ= Dirichlet Apply Integral Transform (8-9) with operational properties (11-12):

2 2n n

n2 2

u u1 n u 0r rr r

∂ ∂+ − =

∂∂

n 0,1,2,...= (14’)

( ){ } ( ) ( )n 1 n 1 nu r , u r , Uθ θ θ= =

T

22 2n n

n2

u ur r n u 0

rr∂ ∂

+ − =∂∂

n 0,1,2,...= (14’)

It is a Cauchy-Euler Equation for n 1,2,...= with the general solution: n n

n 1 2u c r c r−= +

and 2

2 n n2

u ur r 0

rr∂ ∂

+ =∂∂

n 0= (14’’)

with the general solution: 0 1 2u c c ln r= + Because solution has to be bounded: 0 1u c=

nn 1u c r=

Apply boundary condition: ( ) ( )0 1 0u r , Uθ θ=

( )0 0u U θ=

( ) ( )n 1 nu r , Uθ θ=

( )n

n n n1

ru Ur

θ=

( ) ( )nn 0

u r, u r,θ θ∞

=

= ∑ ( )n

nn 0 1

rUr

θ∞

=

=

∑

r

( )r ,θ

0 θ


6.7.2 Mellin Transform ( )u x : → , p∈

( ){ }u xM ( )u p= ( ) p 1

0

u x x dx∞

−= ∫ (19)

( ){ }1 u p−M ( )u x= ( )

c ip

c i

1 u p x dp2 iπ

+ ∞−

− ∞

= ∫ (20)

Examples: 1) { }nxe−M nx p 1

0

e x dx∞

− −= ∫

t p 1p

0

1 e t dtn

∞− −= ∫ t nx=

( )p

pn

Γ=

Operational properties: 1) ( ){ }u x′M ( ) ( )p 1 u p 1= − − − ( ){ }u x′′M ( )( ) ( )p 1 p 2 u p 2= − − −

( ) ( ){ }nu xM ( ) ( )

( ) ( )n p1 u p n

p nΓ

Γ= − −

−

2) ( )2dx u x

dx

M ( ) ( )2 21 p u p= −

3) ( )u t dt ∫x

0

M ( )1 u p 1p

= − +

4) Convolutions Applications: Mellin transform is used for solution of differential and integral

equations, evaluation of fractional integrals and derivatives, to summation of infinite series

Examples: 1) (Debnah, p.218)

2 2

22 2

u u ux x 0xx y

∂ ∂ ∂+ + =

∂∂ ∂ ( )x 0,∈ ∞ , ( )y 0,1∈

( )u x,0 0=

( )a 0 x 1

u x,10 x 1

≤ ≤= >

Solution:

( ) ( ) ( )n

n

n 1

1au x, y x sin n yn

π ππ

∞−

=

−= ∑

x

ya


6.7.3 Legendre Transform 6.7.4 Jacobi and Gegenbauer Transform Solution of generalized heat conduction problem 6.7.5 Laguerre Transform 6.7.6 Hermite Transform 6.7.7 Hilbert and Stiltjes Transform 6.7.8 Z Transform REFERENCES 1. J.V.Beck, K.D.Cole et al. Heat Conduction Using Green’s Functions. Hemisphere Publishing, 1992. 2. D.G. Duffy. Green’s Functions with Aplications, Chapman&Hall/CRC, 2001.



Les Modeles Globaux PPT 1st Radiation School Odeillo SLW-1 SPECTRAL MODEL

η

( )b bE Tη

1∆

C( )absorption - line blackbodydistribution function F C

( )F C0a1 0

η

( )η g

absorptioncross - sectionC = C T

SLW -1spectralmodel

C

0C

1C

1 0a =1- a

( )

0 0 g ba = F C ,T = T,T = T

gray gas

clear gas

0∆

κP

⋅P 1 1κ = a κ

R 1κ = κ

Planck -mean absorption coefficient

Rosseland-mean absorption coefficient

Physical meaning of SLW -1 parameters

∂+

∂1

R 1 P bI = -κ I κ Is

∂∂

0I = 0s

SSLLWW--11 RRTTEE::

1 1a , κ

with efficiently definedgray gas parameters


ηC , 2cm

molecule

0C

1C


CONSTRUCTION OF THE EFFICIENT SLW-1 SPECTRAL MODEL

0 1 1 1

Alternatively, two parameters must be defined in the SLW -1 spectral model : either cross - sections C and C , or gray gas coefficient κ and its weight a

1C

0C

CηC

η

0κ = 0

1 1κ = NYC

( )0 0 g ba = F C ,T ,T

−1 0a = 1 a

Total net radiative flux in the isothermal plane layer at temperature T bounded by black cold walls :

( ) ( ) ( ) ( ){ }−=

− ⋅ ⋅ − − ∑n

SLW n b j 3 j 3 jj 1

q x = 2πI T a E κ x E κ L xSLW-n

SLW-1 ( ) ( ) ( ) ( ){ }− − ⋅ ⋅ − − SLW 1 b 1 3 1 3 1q x = 2πI T a E κ x E κ L x

( ) ( )− −=

− → ∑M 2

SLW n m SLW 1 mm 0

q x q x minLEAST-SQUARES OPTIMIZATION


CONSTRUCTION OF THE SLW-1 MODEL FOR OTHER GEOMETRIES

L0 x

Plane-Parallel Layer

Objective

find the SLW-1 parameters a1, κ1 with the help of the equivalent plane layer

L

0

x

y

xL

M

0

2-D Cartesian

3-D Cartesian

z

y

xL

M

0

H


Rr

0

Infinite Cylinder

z

Rr

0

z H=

Rr

0

Finite Cylinder

Sphere


SLW P -1 Approximation (equation for gray gases) :

( ) ( ) ( )j j j j j∇⋅∇ 2 2bG r - 3κ G r = -12πa κ I T

( ) ( )ˆ ˆ∫j j4π

G r = I r,s s dΩ

( ) ( )∇j jj

1q r = - G r3κ

( ) ( )j jj

∇⋅∇1Q r = G r

3κ

irradiation (radiation intensity integrated over all directions)

radiative flux vector

( ) ( )⋅∇j jj

1q r = - n G r3κ

net radiative flux

divergence of radiative flux

( ) ( )j j j j b= κ G r - 4πa κ I T

( ) ( ) ( )j j j j j− ⋅∇ +w w b w2 - ε 2 n G r κ G r = 4πa κ I T

ε 3

boundarycondition

j =1,...,n

( )jG⋅∇n rw

n

( )jG∇ rw

( )jG r

r

rw

0

radiative transferis modelled as a diffusion process


P-1 APPROXIMATION FOR PLANE LAYER

L0 x

( ) ( ) ( )2

2 2j j j j j b2

d G x - 3κ G x = -12πa κ I Tdx

( )

∑n

j j bj

j=1j j

-4 3πa κ I LQ x = cosh 3κ x -23 32sinh κ L + 3cosh κ L

2 2

( )

∑n

j bj

j=1j j

4πa I Lq x = cosh 3κ x -23 32sinh κ L + 3cosh κ L

2 2

( ) ( )j j j2 d- G 0 + κ G 0 = 03 dx Cold

blackwalls

j =1,...,n

( ) ( )j j j2 d G L + κ G L = 03 dx

boundarycondition

Analytical Solution :


P-1 APPROXIMATION FOR INFINITE CYLINDER

Rr

0

SLW -10

SLW -1

( ) ( ) ( ) − 2 2

j j j j j b1 d dr G r 3κ G r = -12πa κ I Tr dr dr

( ) ( )( ) ( ) ( )∑

nj b

1 jj=1 1 j 0 j

4πa I Tq r = I 3κ r

2I 3κ R + 3I 3κ R

Coldblackwall

j =1,...,n

( ) ( )j j j2 d G R + κ G R = 03 dx

( ) ( )r r0 1where I and I are the modified Bessel functions

( ) ( )( ) ( ) ( )∑

nj j b

0 jj=1 1 j 0 j

-12πa κ I TQ r = I 3κ r

2 3I 3κ R +3I 3κ R

boundarycondition



P-1 APPROXIMATION FOR SPHERE

R r0

L 0.92 R=

SLW -1SLW -10

( ) ( ) ( ) − 2 2 2

j j j j j b21 d dr G r 3κ G r = -12πa κ I T

dr drr

Coldblackwall

j =1,...,n

( ) ( )j j j2 d G R + κ G R = 03 dr

( ) ( )

( ) ( )( )

∑2n jj j b

j=1j j j j

sinh 3κ r-6πa κ I T RQ r =

3 1 rκ - sinh 3κ R + 3κ cosh 3κ R2 R

( ) ( )

( ) ( )( ) ( )

∑n j j jj b

2 2j=1

j j j j

3κ cosh 3κ r sinh 3κ r2πa I T Rq r = -

3 1 r rκ - sinh 3κ R + 3κ cosh 3κ R2 R

boundarycondition



Documents

IX.7 ADDITIONAL INTEGRAL TRANSFORMSvps/ME505/IEM/09 07.pdf · 898 Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 6.7.1 Solution of 3-D