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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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 898
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 899
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 900
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:
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 901
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 902
( ) ( )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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 903
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 904
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 905
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.
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 906
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 907
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 908
ηC , 2cm
molecule
0C
1C
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 909
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 910
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 911
Rr
0
Infinite Cylinder
z
Rr
0
z H=
Rr
0
Finite Cylinder
Sphere
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 912
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
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 913
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 :
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 914
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
Analytical Solution :
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 915
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
Analytical Solution :
Chapter IX The Integral Transform Methods IX.7 Additional Integral Transforms August 5 2017 916