Green’s Function For Radial Heat Conduction in Two-Region Composite Cylinders
With Perfect Boundary Contact
by
Donald E. Amos
Albuquerque, NM
Abstract This paper presents the derivation of the Green’s function for composite cylinders 0<r<a and
r>a in perfect contact on the surface r=a. Because the source function can be in either region, there are
two pairs of functions which define the Green’s function. Each pair is the solution to a two-region
conduction problem with zero initial temperatures and continuity of temperature and flux on the cylinder
r=a. These pairs are used in conjunction with a general formula to get the solution to other problems
where the cylinders are in perfect contact, but may have non-zero initial conditions and/or possibly a
distribution of internal heat sources. The Green’s function approach and a direct approach agree when
applied to three problems with known solutions. A fourth problem illustrates a complication where the
Laplace transform of the general solution is more useful.
Keywords Two Regions Heat Conduction Unsteady State Laplace Transform
1.0 Introduction
The main thrust of this paper is to derive the radial Green’s function G(r,r',t) for adjacent cylinders in
perfect contact on the common surface r=a. The inner cylinder, 0  r  a , is denoted as Region 1 and the
outer cylinder, a  r   , is denoted as Region 2
In the notation of [7, Chapter 2], we are solving the heat conduction problem labeled (R0C0). In order to
apply the concept of a Green’s function to a problem, one has to use these functions in conjunction with
the general formula [6].
t

j  t
j
Tj (P', t)     G j (P, P', t   )q j (P, )dVd    G 3j  j (P, P', t   )q 3 j(P, )dVd 
k j  0 Vj

0 V3 j

  G jj (P, P', t  0)f j (P)dV 
k 3- j j
G
(P, P', t  0)f 3 j (P)dV
j  1,2
k j 3 j V3 j
t

G jj (P, P', t   )
T (P, ) 
  j   Tj (P, )
 G jj (P, P', t   ) j
dS jd
n i
n i 
0 S j  SC 

k  t
G j (P, P', t   )
T (P, ) 
 3- j j   T3 j (P, ) 3 j
 G 3j  j (P, P', t   ) 3 j
dS3 jd
k j 0 S3 j  SC 
n i
n i

The indices j and 3-j with j=1,2 can only have the values of 1 or 2. Here V1 and V2 are the volumes of
Regions 1 and 2 with surfaces S1 and S 2 ; Sc is the contact boundary and S j  Sc , j  1,2 are external
(1.1)
j
3 j
Vj
boundaries. q1, q2 and f1, f 2 are heat source and initial temperature distributions respectively. The
operators  / ni are internal normal derivatives. Since the Green’s function source can be in either
region, the construction requires the solution of two problems each of which satisfy the conditions on
the common boundary and each yielding two functions. The complete Green’s function therefore
consists of two pairs (G11( r, r ', t ), G21 ( r, r ', t )) and (G12 ( r, r ', t ), G22 ( r, r ', t )) where the superscript
designates the region where the source occurs and the subscript designates the region where the function
applies.
1
For this application, Region 1 has no external boundary and the external boundary for Region 2 is at
infinity where we select zero values for all temperatures. Consequently, (1.1) reduces to only the terms
involving the source distributions q1, q2 and the initial temperatures f1, f 2 .
Laplace transforms of the Green’s functions are used in most of the manipulations because the
transforms contain well documented handbook functions.
2. Conduction Equations for the Green’s Function
We solve the two-region problem of heat conduction in concentric cylinders with a heat source and
perfect boundary contact. The inner cylinder we designate as Region 1, 0<r<a, and the outer cylinder as
Region 2, a  r   . The conduction equations for the Green’s functions are
1 G1
1 G2
 2G1 
, 0<r<a ,  2G2 
, a<r< , G2 ( r, t )  0 for r  
(2.1)
1 t
 2 t
with boundary conditions, expressing continuity of temperature and flux at r=a, as
G
G
G1( a, t )  G2 ( a, t ) , k1 1 ( a, t )  k 2 2 ( a, t ) .
(2.2)
r
r
The initial conditions are
(2.3)
G j ( r,0)  0 , j  1,2
except possibly at the source point where r  r ' . The energy comes from the point source solution
which can appear in either Region 1 or Region 2.
3. Green’s Function for the Source in Region 1 (G11( r, r ', t ), G21 ( r, r ', t ))
For the sake of simplicity, we suppress the superscripts for the analysis of this section, and set
G1( r, r ', t )  G11( r, r ', t ) and G2 ( r, r ', t )  G21 ( r, r ', t ) . With the source in Region 1 along the cylindrical
surface r  r ' , the source solution of the heat equation is [ 8, p.533], [9, Chapter XIII]
2
2
B
rr '
1
S1( r, r ', t )  1 e ( r  r ' ) /(41t ) I 0 (
) , B1 
(3.1)
.
2t
21t
21
We use the Laplace transform with parameter p and write the solution in Region 1 as the sum of the
source solution and regular solutions of the conduction equation (2.1) which are finite for r=0. In Region
2 we use only regular solutions of the conduction equation which vanish for r   . Then, the (Laplace)
transformed equations from Section 2 lead to modified Bessel functions and we select the appropriate
solutions to form the equations
G1( r, r ', p )  S1( r, r ', p )  AI 0 (q1r ) , 0  r  a
q j  p / j
(3.2)
G2 ( r, r ', p )  BK 0 (q2 r ) ,
ar
where the transform of (3.1) is [9, Chapter XIII]
 I (q r ')K0 (q1r ) r  r ' 
(3.3)
S1( r, r ', p )  B1  0 1

 I 0 (q1r )K0 (q1r ') r  r ' 
and I 0 and K0 are modified Bessel functions of zero order. The next step is to apply the boundary
conditions for continuity of temperature and flux,
G
G
G1(a, p )  G2 ( a, p ) , k1 1 ( a, p)  k2 2 ( a, p)
(3.4)
r
r
2
and solve for A and B. Since a> r and a  r ' in Region 1, we use the first form of S1 for S1( a, r ', p )
with (3.2) in (3.4)
B1I 0 (q1r ')K 0 (q1a )  AI 0 (q1a )  BK 0 (q2a )
(3.5)
 q1k1B1I 0 (q1r ')K1(q1a )  q1k1 AI1(q1a )  q2k2 BK1(q2a )
or
AI 0 ( q1a )  BK 0 (q2a )   B1I 0 (q1r ')K 0 (q1a )
(3.6)
.
q1k1 AI1(q1a )  q2k2 BK1(q2a )  q1k1B1I 0 (q1r ')K1(q1a )
Then
 K 0 ( q1a )
 K 0 ( q2a )
I 0 ( q1a )
 K 0 ( q1a )
(3.7)
A  B1I 0 ( q1r ')
q1k1K1( q1a ) q2k2 K1(q2a )
,
I 0 ( q1a )
 K 0 ( q2a )
q1k1I1( q1a ) q2k2 K1(q2a )
B  B1I 0 ( q1r ')
q1k1I1(q1a ) q1k1K1(q1a )
I 0 ( q1a )
 K 0 ( q2a )
q1k1I1(q1a ) q2k 2 K1(q2a )
or
A  B1I 0 (q1r ')   q2k2 K1(q2a )K 0 (q1a )  q1k1K1(q1a )K 0 (q2a ) / D
(3.8)
B  B1I 0 (q1r ')  q1k1K1(q1a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q1a ) / D 
k1
B1I 0 (q1r ') / D
a
D  q2k2 K1( q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a )
where we have used the Wronskian of the I and K functions
I1( z )K0 ( z )  I 0 ( z )K1( z )  1/ z
(3.9)
in the computation of B. The solution (3.2) becomes
r'  r  a
r  r'  a
 I ( q r ') K 0 ( q1r ) 
G1( r, r ', p )  B1  0 1
  B1I 0 ( q1r ) I 0 ( q1r ')C ,
 I 0 ( q1r ) K 0 ( q1r ') 
(3.10)
k
G2 ( r, r ', p )  1 B1K 0 ( q2 r ) I 0 ( q1r ') / D ,
a
ar
C    q2k 2 K1( q2a ) K 0 ( q1a )  q1k1K1( q1a ) K 0 ( q2a )  / D
D  q2k 2 K1( q2a ) I 0 ( q1a )  q1k1I1( q1a ) K 0 ( q2a )
or
(3.11)
 F ( r, r ', p )
G1( r, r ', p )   1
 F1( r ', r, p )
G2 ( r, r ', p ) 
r'  r  a
r  r'  a
k1
B1I 0 ( q1r ')K 0 (q2r ) / D ,
a
where
3
ar
qj 
p / j
p /  j , j  1,2
D  q2k2 K1(q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a ) , q j 
(3.12)
.
 q k K (q a )  I 0 (q1a )K 0 (q1r )  I 0 (q1r )K 0 (q1a )  
F1( r, r ', p )  B1I 0 ( q1r ')  2 2 1 2
/ D
  q1k1K 0 ( q2a )  I1( q1a )K 0 (q1r )  I 0 (q1r )K1(q1a ) 
Now we follow the procedure of [9, Chapter XII, Section 128] and invert the Laplace transform by the
complex inversion integral. Then with G as G1 or G2 ,
  i
1
G( r, r ' t ) 
G( r, r ', p )e pt dp ,
2 i  i
(3.13)
  0.
The indication from [9] is that the integrand has no poles, but it is clear from (3.11) that G1 and G2 have
branch points at p=0. Therefore we select the branch cut along the negative real axis and use Cauchy’s
theorem to close the contour about the cut with a classical keyhole contour. Then with G as G1 or G2
we have
0
G ( r, r ' t ) 
(3.14)

2
G ( r, r ', p )e pt

2 i 
pw e
2 i
1
  0 2 i
wei dw  lim


2

G( r, r ', p )e pt
2 i 0

 G( r, r ', p)e
pt

pw e
2  i
we
 i
 p  e
 iei d
i
.
dw  0
Now, the integrand is real on the positive real axis. Therefore, by the reflection principle, the values on
the top and bottom of the cut where p  w2ei and p  w2ei , 0  w  , are complex conjugates of
one another and the difference is 2i times the imaginary part on p  w2ei . Then, (3.14) becomes
(3.15) G( r, r ' t )  

Im G( r, r ', p )
pw e

2
2 i
we
 w2 t
0
1
dw  lim
  0 2

 G( r, r ', p)e

pt
 p  e
i
 ei d .
The task now is to explicitly compute the real and imaginary parts of G1 and G2 on p  w2ei and
evaluate the limit   0 . The details of the algebra are relegated to APPENDIX A and we jump to the
final result.
The Solution for G11( r, r ', t ) and G21 ( r, r ', t ) with the Source in Region 1
With the essentials presented in APPENDIX A, the final solution, with the proper superscript attached,
is
r'
r
 J 0 ( 1w ) J 0 ( 1w )
4
k
k
a
a e w2 t dw , r  a
G11( r, r ' t )  21 22 B1 
(3.16)
r'  a
 a
( D12  D22 )w
0

G21 ( r, r ', t ) 
(3.17)
2 k1
B1
 a 0
r' 
r
r 
J 0 ( 1w )  D1J 0 ( 2 w )  D2Y0 ( 2 w ) 
ra
a 
a
a   w2 t
e
dw ,
2
2
r'  a
D1  D2
D1( w)   2 J 0 ( 1w)Y1( 2 w)  1J1( 1w)Y0 ( 2 w),
D2 ( w)   2 J 0 ( 1w) J1( 2 w)  1J 0 ( 2 w) J1( 1w)
B1 
1
21
The asymptotic expressions
,
 j  k j /  j  j  a /  j , j  1,2
4
2
for z  0

z
show that 1/ ( D12  D22 )w  O( w), D1 / ( D12  D22 )   O( w) and D2 / ( D12  D22 )  =O(w3 ) for w  0






making each integrand integrable at w=0.
J 0 ( z ) 1, J1( z )
z / 2, Y0 ( z )
2
ln( z / 2), Y1( z )

4. Green’s Function for the Source in Region 2 (G12 ( r, r ', t ), G22 ( r, r ', t ))
We take the source in Region 2, a  r '   , on the cylindrical surface r  r ' ,
2
2
B
rr '
1
S2 ( r, r ', t )  2 e ( r  r ' ) /(4 2 t ) I 0 (
) , B2 
(4.1)
2t
2 2t
2 2
and solve the heat conduction problem in the manner of Section 3. We use the Laplace transform and
write the solution in Region 2 as the sum of the source solution and regular solutions of the conduction
equation which vanish for r   . In Region 1 we use only regular solutions of the conduction equation
which are finite at r=0. Then, proceeding as before with the superscript suppressed, we have
G1( r, r ', p )  AI 0 ( q1r ) ,
0ra ,
(4.2)
G2 ( r, r ', p )  S2 ( r, r ', p )  BK 0 ( q2r ) ,
a  r ', r  
qj 
p /  j , j  1,2
where [9, Chapter XIII ]
 I (q r ')K0 (q2r ) r  r ' 
S2 ( r, r ', p )  B2  0 2

 I 0 (q2r )K0 (q2r ') r  r ' 
and I 0 and K0 are modified Bessel functions of zero order. The next step is to apply the boundary
conditions
G
G
G1(a, p )  G2 ( a, p ) , k1 1 ( a, p)  k2 2 ( a, p)
(4.4)
r
r
and solve for A and B. Since a  r and a  r ' in Region 2, we use the second form of S 2 for
(4.3)
S2 ( a, r ', p ) with (4.2) in (4.4),
(4.5)
B2 I 0 (q2a )K 0 (q2 r ')  BK 0 (q2a )  AI 0 (q1a )
q2k2 B2 I1(q2a )K 0 (q2 r ')  q2k2 BK1(q2a )  q1k1AI1(q1a )
Solving for A and B with the help of the Wronskian
(4.6)
gives
I 0 ( z )K1( z )  I1( z )K0 ( z )  1/ z
G1( r, r ', p ) 
(4.7)
k2
B2 K 0 ( q2 r ') I 0 ( q1r ) / D ,
a
 F ( r, r ', p )
G2 ( r, r ', p )   2
 F2 ( r ', r, p )
r  a  r'
r  r'  a
r'  r  a
where
5
.
p /  j , j  1,2
D  q2k2 K1(q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a ) , q j 
(4.8)
 q k I ( q a )  I 0 ( q2 r ')K1(q2a )  I1(q2a )K 0 (q2r ')  
F2 ( r, r ', p )  B2 K 0 ( q2 r )  2 2 0 1
/ D
  q1k1I1( q1a )  I 0 ( q2 r ')K 0 (q2a )  I 0 (q2a )K 0 (q2r ') 
.
Again, we follow Section 3 and invert the Laplace transform by the complex inversion integral using the
keyhole contour. The solution with the proper designation (G12 ( r, r ', t ), G22 ( r, r ', t )) is
(4.9) G 2j ( r, r ' t )  
2



 pw e
Im G j ( r, r ', p )
0
2 i
1
  0 2
we  w t dw  lim
2

 G j (r, r ', p )e
pt
 p  e
i

 ei d
2 i
for j=1 or 2. The details for F2 ( r, r ', p ) on p  w e , 0  w   and p   ei , -      are
contained in APPENDIX B.
The Solution for G12 ( r, r ', t ) and G22 ( r, r ', t ) with the Source in Region 2

J 0 ( 1wr / a )  D1J 0 ( 2 wr '/ a )  D2Y0 ( 2 wr '/ a )  e  w t
2k 2

B2
dw,
 a 0
D12  D22
2
G12 ( r, r ' t )
(4.10)
ra
D1( w)   2 J 0 ( 1w)Y1( 2 w)  1J1( 1w)Y0 ( 2 w)
D2 ( w)   2 J 0 ( 1w) J1( 2 w)  1J 0 ( 2 w) J1( 1w)
 
 B2 
 0
G22 ( r, r ', t )  


 B2 
 0
(4.11)
 2 J 0 ( 1w) ( r ', w,  2 )  1J1( 2 w) ( r ', w,  2 )  we w t dw

 2
2
•  D1J 0 ( 2 wr / a )  D2Y0 ( 2 wr / a ) 

 D1  D2
r  r'
 2 J 0 ( 1w) ( r, w,  2 )  1J1( 2 w) ( r, w,  2 )  we w t dw

 2
2
•  D1J 0 ( 2 wr '/ a )  D2Y0 ( 2 wr '/ a ) 

 D1  D2
r  r'
2
, ra
2

r 
r
 ( r, w,  2 )   J 0 ( 2 w )Y0 ( 2 w)  J 0 ( 2 w)Y0 ( 2 w ) 
a
a 



r
a
r
a


 ( r, w,  2 )   J1( 2 w)Y0 ( 2 w )  J 0 ( 2 w )Y1( 2 w) 
1
,  j  k j /  j  j  a /  j , j  1,2
2 2
Notice that  (a, w,  2 )  0 , and because  (a, w,  2 ) is a Wronskian,  (a, w,  2 )  2 /( 2w) .
B2 
5. Applications of the Two Region Radial Green’s Functions
In Applications I and II, we have a common geometry with perfect boundary contact. In application I,
we consider the problem of the cooling of Region 1 by Region 2, which we take initially to be at zero
temperature. There are no heat generation terms, q1  q2  0 . In Application II, we take the initial
temperatures to be zero, f1  f 2  0 , and heat the media with a uniform source over Region 1,
q1  Q1 >0 , q2  0 . In Application III, we take the material properties of both regions to be the same,
6
which gives a one region infinite medium. It is verified that the solution for the Green’s function is just
the source term. In application IV, we take an initial temperature distribution which is uniform over both
Region 1 and Region 2 with no heat generation. Since there is no gradient, the temperature does not
change and the solution should reflect this. The analytics, however, become cumbersome and this is a
case where the Laplace transform is better suited to solving the problem.
Application I
We apply the radial Green’s function for cylinders 0  r  a and a  r   in perfect contact on the
surface r=a to the cooling of the inner cylinder 0  r  a
 2T1 
1 T1
, 0ra
1 t
T1 (a, t )  T2 (a, t )
(5.1)
T1 ( r ,0)  V
 2T2 
k1
1 T2
, ar
 2 t
T1
T
(a, t )  k1 1 (a, t )
r
r
T2 ( r ,0)  0
lim T2 ( r , t )  0
r 
This problem can be solved by a direct application of the Laplace transform and the solution is given in
Carslaw and Jaeger [9, Chapter XII, Section 128].
The general solution of a two-region problem is expressed in terms of the Green’s function by (1.1). We
designate Region 1 as 0  r  a and Region 2 as a  r   . In the context of this problem,
G11( P, P ', t  0)  G11( r, r ', t ) and G21 ( P, P ', t  0)  G21 ( r, r ', t ) from Section 3. Region 1 has no S1
boundary, the S 2 boundary values at infinity are zero and there is no heat generation. We are left with
only the contributions from the initial temperature distributions. Then with
f1( P )  V and f 2 ( P )  0
(5.2)
and j=1 in (1.1) we have
(5.3) T1 (P', t)  V  G11 (P,P', t  0)dV+
V1
k 21
G 2 (P,P', t  0)f 2 (P)dV  V  G11(r,r ', t  0)dV .

k1 2 V2
V1
To get T1 (P', t)  T1(r ', t) , we need to integrate the Green’s function over the interval 0  r  a . The
integration, applied to (3.16), is ( dV  2 rdr )
a
(5.4)
r
2 a 2
2 a 2
2  J 0 ( 1w )rdr 

wJ
(

w
)

J1( 1w)
2 1 1 1
a

w

w
1


0
1
and from (3.15) we have the solution for Region 1:
T1( r ', t )  B1V
(5.5)
4k1k2 2 a
 2a 2  1
2

0
r'
J 0 ( 1w ) J1( 1w)
 w2 t
a
e
dw
( D12  D22 )w2
D1( w)   2 J 0 ( 1w)Y1( 2 w)  1J1( 1w)Y0 ( 2 w)
D2 ( w)   2 J 0 ( 1w) J1( 2 w)  1J 0 ( 2 w) J1( 1w)
 j  k j /  j  j  a /  j , j  1,2
7
Our goal now is to bring this formula into the form shown in Carslaw and Jaeger [9, Chapter XII,
Section 128]. We multiply the numerator and denominator by 1 2 and change the variable with
w  v 1 . Then with D1 and D2 redefined, we have
T1( r ', t ) 
(5.6)
V
4k1k2 2 a 2 1 2
21  2a 2
1
1

J 0 ( vr ') J1( va )
 ( D 2  D 2 )v 2
1
0
e1v t dv,
2
r'  a
2
D1( v )   k2 1 J 0 ( va )Y1( va )  k1  2 J1( va )Y0 ( va )
 = 1/ 2
D2 ( v )   k2 1 J 0 ( va ) J1( va )  k1  2 J 0 ( va ) J1( va )
The final form is
(5.7)
T1( r ', t ) 
4Vk1k2 2
a
2

J (vr ') J (va )
 ( D0 2  D12 )v 2 e
0
1
1v 2 t
dv,
r'  a
2
which agrees with the solution presented in Carslaw and Jaeger [9, Chapter XII].
In order to solve for the temperatures in Region 2, we need the Green’s function with the source in
Region 2. To this end we take j=2 in (1.1). Again, since there is no heat source and the solution vanishes
at infinity, we get only the contribution from the initial temperature in Region 1. Then, with f1( P )  V ,
and f 2 ( P )  0 (1.1) becomes
k1 2
k1 2 V
2
2
T
(P',
t)

G
(P,P',
t

0)f
(P)dV+
G
(P,P',
t

0)f
(P)dV

G12 (P,P', t)dV .
2
2
2
1
1



(5.8)
k 21 V1
k 21 V1
V2
The Green’s function for Region 1 with the source in Region 2 is taken from Section 4, equation (4.10).
Again, to get T2 ( r ', t ) we need to integrate over 0<r<a. The integral is given in (5.4). The result is
2k 2
k1 2V 2 a 2
T2 ( r ', t ) 
B2
a
k 21  1


J1( 1w)  D1J 0 ( 2 wr '/ a )  D2Y0 ( 2 wr '/ a ) e  w
2
t
dw,
( D12  D22 )w
0
r'  a
D1   2 J 0 ( 1w)Y1( 2 w)  1J1( 1w)Y0 ( 2 w )
(5.9)
.
D2   2 J 0 ( 1w) J1( 2 w)  1J 0 ( 2 w) J1 ( 1w)
 j  k j /  j  j  a /  j , j  1,2
In order to get this to look like the formula in [9, Chapter XII], we need to change the variable
w  v 1 and multiply the denominator by 1 2 and the numerator by 1 2 . This manipulation
redefines D1 and D2 and puts a compensatory factor
1 2 in the coefficient to give

J ( va )  D1J 0 ( vr ')  D2Y0 ( vr ') e 1v t
2k
k  V 2 a 2
T2 ( r ', t )  2 B2 1 2
1 2  1
dv,
2
2
a
k 21  1
(
D

D
)
v
1
2
0
2
(5.10)
D1( v )   k2 1 J 0 ( va )Y1( va )  k1  2 J1( va )Y0 ( va )
r'  a
D2 ( v )   k2 1 J 0 ( va ) J1( va )  k1  2 J 0 ( va ) J1( va )
 = 1/ 2
The coefficient reduces to
(5.11)
2k V  2
2k2
k12V 2 a 2
B2
12  1
a
k 21  1

8
and this formula agrees with the [9, Chapter XII].
Application II
In this application, also taken from Carslaw & Jaeger [9, Chapter XII, Section 128], we apply the
general formula (1.1) to a problem with the same geometry and Green’s function. In this problem, we
have perfect contact between cylinders with zero initial conditions, but there is constant heat production
q1  Q1 of heat units per unit area per unit length of cylinder in Region 1 and no production in Region 2,
q2  0 . Then (1.1) becomes
T1 (P', t) 
(5.12)

1  t

1
   G (P, P', t   )q1 (P, )dVd +   G 2 (P, P', t   )q 2 (P, )dVd 
k1  0 V1
0 V2

1Q1
k1
t
1
1
t
.
  G (r,r ', t   )dVd
1
1
0 V1
The integration on r and the change of variables w  v 1 are the same as that in Part I. Therefore
(replace V with 1Q1 / k1 in (5.7))
t 
4k k   Q
T1( r ', t )  1 22 2 1 1 
k1 0
a
(5.13)
J 0 ( vr ') J1( va )
 ( D 2  D 2 )v 2
1
0
e1v
2
( t  )
dvd ,
r'  a
2
D1( v )   k 2 1 J 0 ( va )Y1( va )  k1  2 J1( va )Y0 ( va )
 = 1/ 2
D2 ( v )   k2 1 J 0 ( va ) J1( va )  k1  2 J 0 ( va ) J1( va )
The integration on  produces

T1( r ', t ) 
(5.14)
4Q1k2 2 (1  e1v t ) J 0 (vr ') J1(va )
a 2
2

( D12  D22 )v 4
0
dv,
r'  a
and this agrees with [9, Chapter XII].
To get T2 ( r ', t ) for Region 2, we take j=2 in (1.1),
T2 (P', t) 
(5.15)

2  t
t

2
2
G
(P,
P',
t


)q
(P,

)dVd

+
G
(P,
P',
t


)q
(P,

)dVd

  2

2
1
1
0 V
k 2  0 V2

1
 2Q1
k2
t
  G (P,P', t   )dVd
2
1
0 V1
Since the integration on r is the same as Part I, we use the form (5.10) from Part I (
by
 2Q1
):
k2
2k 2
2 a 2  2Q1
T2 (r ', t) 
B2
1 2 
a
1 k2
0
t
(5.16)


0
k1 2V
is replaced
k 21
J1 (va)  D1J 0 ( vr ')  D2Y0 ( vr ')  e 1v
(D12  D22 )v
D1 (v)   k 2 1 J 0 (va)Y1 ( va)  k1  2 J1 (va)Y0 ( va)
D2 (v)   k 2 1 J 0 (va)J1 ( va)  k1  2 J 0 ( va)J1 (va)
9
2
( t  )
r'  a
 = 1/ 2
dvd ,
or

(1  e1v t )J1 (va)  D1J 0 ( vr ')  D2Y0 ( vr ') 
2k 2
2 a 2  2Q1 1 2
(5.17) T2 (r ', t) 
B2
dv
a
 1 k2
1 0
(D12  D22 )v3
and this agrees with [9, Chapter XII] since the coefficients are the same:
2k 2
2 a 2  2Q1 1 2 2Q1  2
(5.18)
.
B2

a
 1 k2
1

By using two different applications of the Green’s functions, we have provided a partial check on the
j
k 3- j j
coefficients
in the general formula (1.1) and a partial check on the formulas for T1( P ', t )
and
kj
k j3 j
2
and T2 ( P ', t ) .
Application III
In this application, we take the physical properties of both regions to be the same, 1  2   ,
k1  k2  k to confirm that the reduction yields the Green’s function for the whole plane, 0  r   .
The solution is given by the source solution,
 r 2  r '2   rr ' 
1
S ( r, r ', t ) 
exp  
(5.19)
 I 0 
.

4 t
4

t
2

t




In this case, the parameters become 1   2    k /  and  1   2    a /  . The Wronskian
J1( z )Y0 ( z )  J 0 ( z )Y1( z )  2 /( z )
helps in the reduction process. In all of the formulas in (3.16), (3.17), (4.10) and (4. 11) the reduction
comes down to
2
D1( w) 
, D2 ( w)  0
(5.20)
 w
and the integral
(5.21)
1
2


0
2
r
r'
J 0 ( w ) J 0 ( w )we w t dw ,
a
a
which can be evaluated from a table of Hankel transforms to yield (5.19).
Application IV
This example is instructive because, in using both terms of the initial distribution, we give credence to
the formula and at the same time show, as we have observed in previous papers [3],[4],[5],[6], that
working with the Laplace transform may be a better approach than a direct evaluation. We consider the
problem with
(5.22)
f1( P )  V , 0  r  a ,
f 2 ( P )  V , a  r< ,
q1  q2  0
and try to apply (1.1). Since there is no gradient in the initial distribution, the temperature does not
change and the solution to the conduction problem is
10
T1( r ', t )  V , 0  r'  a
(5.23)
T2 ( r ', t )  V , a  r'< .
,
To see what the difficulties are, we apply (1.1) using (3.16) and (3.17) with the temperature for r>a
limited to a finite interval a  r  R in order to keep the integration finite,
a
T1( r ', t )  2V 
(5.24)
R
G11( r, r ', t )rdr
0
Using the relations
(5.25)
 J 0 ( z) zdz  zJ1( z )
k
 2V 2 1  G21 ( r, r ', t )rdr .
k1 2 a
 Y0 ( z ) zdz  zY1( z )
,
the integrals are
a
(5.26)

0
R
r
a2
J 0 ( 1w )rdr 
J1( 1w)
a
 1w
r
r 

 D1J 0 ( 2 w a )  D2Y0 ( 2 w a )  rdr
(5.27)
a
a  
R
a  
R



D1  RJ 0 ( 2 w )  aJ 0 ( 2 w)   
D2  RY0 ( 2 w )  aY0 ( 2 w)  


 2w  
a
a
  2w  

and with (3.17) we have an indeterminate form
r' 
R
R 
 J 0 ( 1w )  D1J 0 ( 2 w )  D2Y0 ( 2 w ) 
2 ak1
a 
a
a   w2 t
B1 lim R 
e
dw
(5.28)
 a 2 R 0
( D12  D22 )w

to evaluate. Basically, f 2 ( r ) does not have the proper behavior at infinity to give a convergent integral.
However, there is a way around this complication. If we work with the Laplace transform,
a
(5.29)
T1( r ', p )  2V  G11( r, r ', p )rdr  2V
0
(5.30)

k21
G21 ( r, r ', t )rdr

k1 2 a
 F ( r, r ', p )
G11( r, r ', p )   1
 F1( r ', r, p )
r'  r  a
r  r'  a
k1
B1I 0 ( q1r ')K 0 (q2r ) / D ,
ar
a
D  q2k2 K1(q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a ) , q j  p /  j , j  1,2
G21 ( r, r ', p ) 
(5.31)
,
 q k K (q a )  I 0 (q1a )K 0 (q1r )  I 0 (q1r )K 0 (q1a )  
F1( r, r ', p )  B1I 0 ( q1r ')  2 2 1 2
/ D
  q1k1K 0 ( q2a )  I1( q1a )K 0 (q1r )  I 0 (q1r )K1(q1a ) 
there is no problem with the integration because of the exponential decay of the K functions, but one has
to go through the inversion process again.
The integration on G11 takes the form
a
(5.32)

G11( r, r ', p )rdr
while the integration on
a
0
r'
  F1( r ', r, p )rdr   F1( r, r ', p )rdr
0
G21
r'
is simple
11


(5.33)

G21 ( r, r ', p )rdr
a
k B I (q r ')
 1 1 0 1  K 0 ( q2r )rdr .
a
D
a
In each of these cases, we use the integrals
(5.34)
 I0 ( z) zdz  zI1( z)
 K0 ( z) zdz   zK1( z ) .
The exponential decay of the K functions guarantees the convergence. We need the integrals
a
(5.35)

r'
r'
a
K 0 ( q1r )rdr  K1( q1r ')  K1(q1a ) ,
q1
q1
r'
(5.36)

I 0 ( q1r )rdr 
0
r'
I1(q1r ') ,
q1
a

I 0 (q1r )rdr 
r'


K 0 ( q2 r )rdr 
a
a
r'
I1(q1a )  I1(q1r ')
q1
q1
a
K1(q2a ) .
q2
Then, for r  r ' substitution and reduction using the Wronskian (3.9 ) gives
1 


 q2k2 K1( q2a )  I 0 ( q1a ) K1( q1r ')  q r '  
a

  I (q r ')K (q a ) 1  
r'
0 1
 1 1
 / D .
(5.37)
 F1(r, r ', p)rdr  B1 q1 I 0 (q1r ') 

 I1( q1a ) K1( q1r ')  
r'
  q1k1K 0 ( q2a ) 
 
  I1( q1r ') K1( q1a )  

For the integral where r '  r we have
r'
r'
 q2k 2 K1( q2a )  I 0 ( q1a ) K 0 ( q1r ')  I 0 ( q1r ') K 0 ( q1a )  
 F1( r ', r, p )rdr  B1  I 0 (q1r )rdr  q1k1K0 (q2a )  I1(q1a )K 0 (q1r ')  I 0 (q1r ')K1(q1a ) / D
0
0
(5.38)
 q k K ( q a )  I 0 ( q1a ) K 0 ( q1r ')  I 0 ( q1r ') K 0 ( q1a ) 
r'
 B1 I1( q1r ')  2 2 1 2
/D
q1
  q1k1K 0 ( q2a )  I1( q1a ) K 0 ( q1r ')  I 0 ( q1r ') K1( q1a )  
and, with the Wronskian again, (5.37) and (5.38) combine to give the sum in (5.32)

 I ( q a ) I ( q r ')  
q2k2 K1( q2a )  0 1  0 1  

a
q1r '  
 q1r '
r' 
1
G
(
r
,
r
',
p
)
rdr

B
(5.39)
1

/D
 1
q
1
 I (q a ) 
0

 q k K (q a ) 1 1
 1 1 0 2  q r' 

 1 


After noting
q2k2 / q12  k21 /(  2 p )  k21 /( 2 q2 ) ,
(5.40)
we combine (5.39) with

(5.41)
k21
k  k B I (q r ') a
k
G21 ( r, r ', p )rdr  2 1 1 1 0 1
K1(q2a )  2 1 B1I 0 (q1r ')K1(q2a ) / D

k1 2 a
k1 2 a
D
q2
 2q2
to get the main result for (5.29),
a

T1( r ', p )
k
  G11( r ', r, p )rdr  2 1  G21 ( r, r ', p )rdr
2V
k1 2 a
(5.42)
0
2
 B1 ( q2k2 / q1 ) K1( q2a ) I 0 ( q1a )  ( q1k1 / q12 ) K 0 ( q2a ) I1( q1a )  / D  B1 / q12


where we have used the definition of D in (5.31) in the reduction.
Then the transform becomes
T1( r ', p)  2VB1 / q12  V / p
(5.43)
12
since B1  1/(21) and q1 
p / 1 . The inversion gives what one expects
T1( r ', t )  V
(5.44)
, 0  r'  a .
The result for T2 ( r ', t ) should be similar.
References
[1] Abramowitz S, Stegun IA (1965) Handbook of Mathematical Functions, AMS 55, Dover
Publications Inc., New York, 1046pp
[2] Amos DE (2006) Handbook of Integrals Related to Heat Conduction and Diffusion,
http://nanohub.org/resources/13874
[3] Amos DE, Beck JV, de Monte F (2011) Transient Heat Conduction in Adjacent Quadrants
Separated by a Thermal Resistance, http://nanohub.org/resources/12465
[4] Amos DE (2011) Transient Heat Conduction in Adjacent Materials Heated on Part of the
Common Boundary, http://nanohub.org/resources/12390
[5] Amos, DE (2012), Green’s Functions For Heat Conduction in Adjacent Materials,
http://nanohub.org/resources/12856
[6] Amos, DE (2012), Theory of Heat Conduction for Two-region Problems Using Green’s Functions,
http://nanohub.org/resources/13671
[7] Amos, DE (2012), 1-D Green’s Functions For Heat Conduction Between Semi-infinite Slabs
With Perfect and Imperfect Boundary Contact, http://nanohub.org/resources/15237.
[8] Cole DC, Beck JV, Haji-SheikhA, Litkouhi B (2010) Heat Conduction Using Green’s Functions,
2nd Ed., CRC Press, 643p.
[9] Carslaw HS, Jaeger JC (1948) Conduction of Heat in Solids, Oxford Univ Press, London, 386pp
APPENDIX A
Calculation of the integrands for (3.15)
Evaluation of F1( r, r ', p ) and G2 ( r, r ', p ) on the contour p  w2ei , 0  w  
Because of the symmetry in the cases for r  r ' and r  r ' we consider only the case for r  r ' . We take
F1( r, r ', p ) ,
 q k K ( q a )  I 0 ( q1a ) K 0 ( q1r )  I 0 ( q1r ) K 0 ( q1a )  
F1( r, r ', p )  B1I 0 ( q1r ')  2 2 1 2
/ D ,
  q1k1K 0 ( q2a )  I1( q1a ) K 0 ( q1r )  I 0 ( q1r ) K1( q1a )  
and start by applying the analytic continuation relations
(A.1)
(A.2)
K0 ( zei / 2 ) 
K1( zei / 2 ) 

2

2
i   J 0 ( z )  iY0 ( z ) =
  J1( z )  iY1( z )
,

 iJ 0 ( z )  Y0 ( z )
2
I 0 ( zei / 2 )  J 0 ( z ) ,
13
I1( zei / 2 )  iJ1( z )
and the Wronskian
(A.3)
J1( z )Y0 ( z )  J 0 ( z )Y1( z )  2 /( z )
to F1( r, r ', p ) . Then with the definitions,
 j  kj / j ,
(A.4)
 j  a/  j
we have
D  q2k2 K1(q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a ) 

w
2
 D1  iD2 
(A.5)
,
, j  1,2
 iw 2 J 0 ( 1w)   J1( 2 w)  iY1( 2 w) 


2  iw1J1( 1w)   J 0 ( 2 w)  iY0 ( 2 w) 


1
2 D1  iD2

D  w D12  D22
.
D1   2 J 0 ( 1w)Y1( 2 w)  1J1( 1w)Y0 ( 2 w)
D2   2 J 0 ( 1w) J1( 2 w)  1J 0 ( 2 w) J1( 1w)
The first term of F1( r, r ', p ) in braces is

q2k2 K1(q2a ) p  w e    2iwK1(i 2w)   2iw   J1( 2w)  iY1( 2w)
2 i
2
 I 0 (q1a)K0 (q1r )  I 0 (q1r )K0 (q1a ) p  w e
2 i
(A.6)

r
r 
 J 0 ( 1w )Y0 ( 1w)  J 0 ( 1w)Y0 ( 1w ) 
2
a
a 



2
 ( r, w,  1 ) (note  is real and  ( a, w,  1)  0)
q2k2 K1(q2a )  I 0 (q1a )K0 (q1r )  I 0 (q1r )K0 (q1a ) p  w e
2 i
and the second term is
q1k1K0 (q2a ) p  w e 
2 i
 i1wK 0 (i 2 w)  1w
 I1(q1a )K0 (q1r )  I 0 (q1r )K1(q1a ) p  w e
2 i
(A.7)
 i

2
  2 w
2
4
Y1( 2 w )  i 2w
2
4
J1( 2w )
  J 0 ( 2w)  iY0 ( 2w)

r
r

J1( 1w)Y0 ( 1w )  J 0 ( 1w )Y1( 1w) 

2
a
a


=  i  ( r, w,  1 )
2
q1k1K0 (q2a )  I1(q1a )K0 (q1r )  I 0 (q1r )K1(q1a )
p  w2 ei
.
(note  is real and  ( a, w,  1 )  2 /( 1w)
 1w
2
4
Y0 ( 2 w )  i1w
2
4
J 0 ( 2 w )
Then, with (A.6) and (A.7), F1( r, r ', p ) becomes
F1(r, r ', p) p  w e
2 i
 B1J 0 ( 1wr '/ a )
(A.8)
 B1J 0 ( 1wr '/ a )
 2   2 wY1( 2 w) ( r, w,  1 )  i 2 wJ1( 2 w) ( r, w,  1 ) 

 / D( w)
4  1wY0 ( 2 w) ( r, w,  1 )  i1wJ 0 ( 2 w) ( r, w,  1 ) 
 2 w   2Y1( 2 w) ( r, w,  1 )  1Y0 ( 2 w) ( r, w,  1 )
 2 D  iD
1
2


4  i   2 J1( 2 w) ( r, w,  1 )  1J 0 ( 2 w) ( r, w,  1 )   w D12  D22


or
14
 D1   2Y1( 2 w) ( r, w,  1 )  1Y0 ( 2 w) ( r, w,  1 ) 



F1(r, r ', w2ei )  B1 J 0D(21wrD'/2a) , 2  D   J ( w) (r, w, )   J ( w) (r, w, ) ,
1
(A.9)
+iB1
2
2
2 1
2
1
1 0
2
1
J 0 ( 1wr '/ a )   D1   2 J1( 2 w) ( r, w,  1 )  1J 0 ( 2 w) ( r, w,  1 )  

,
D12  D22 2   D2   2Y1( 2 w) ( r, w,  1 )  1Y0 ( 2 w) ( r, w,  1 )  
and finally with
r
a
r
a
 ( r, w,  1 )  J 0 ( 1w )Y0 ( 1w)  J 0 ( 1w)Y0 ( 1w )
(A.10)
r
a
r
a
 ( r, w,  1 )  J1( 1w)Y0 ( 1w )  J 0 ( 1w )Y1( 1w)
we get
r'
Im F1( r, r ', p )
p  w 2 ei
  B1
 J 0 ( 1w a )  D1  2 J1( 2 w) ( r, w,  1 )  1J 0 ( 2 w) ( r, w,) 1  


2 D12  D22   D2  2Y1( 2 w) ( r, w,  1 )  1Y0 ( 2 w) ( r, w,  1 )  


r'
(A.11)
  B1
  B1
where
(A.12)
 J 0 ( 1w a )  2 D1J1( 2 w)   2 D2Y1( 2 w) ( r, w,  1 ) 


2 D12  D22   1D1J 0 ( 2 w)  1D2Y0 ( 2 w) ( r, w,  1 ) 


r '  A( w) ( r, w,  1 )  B( w) ( r, w,  1 ) 
J 0 ( 1w ) 

2
a 
D12  D22


A( w)   2  D1J1( 2 w)  D2Y1( 2 w) , D1( w)   2 J 0 ( 1w)Y1( 2w)  1J1( 1w)Y0 ( 2w)
B( w)  1  D1J 0 ( 2 w)  D2Y0 ( 2 w) , D2 ( w)   2 J 0 ( 1w) J1( 2 w)  1J 0 ( 2w) J1( 1w)
Further reduction with (A.10) and the Wronskian (A.3) gives
2  J ( w)
A( w)   2  D1J1( 2 w)  D2Y1( 2 w) = 1 2 1 1
 2 w
(A.13)
B( w)  1  D1J 0 ( 2 w)  D2Y0 ( 2 w) =
A( w) ( r, w,  1 )  B( w) ( r, w,  1 ) 
21 2 J 0 ( 1w)
 2 w
41 2
r
J 0 ( 1w )
a
  1 2 w
2
2
and with
 j  k j /  j  j  a /  j , j  1,2
we have
(A.14)
Im F1( r, r ', p )
p  w2 ei
r'
r
J 0 ( 1w ) J 0 ( 1w )
2 k1k2
a
a
  B1
 a 2 w2
D12  D22
For G2 ( r, r ', p ) on p  w2ei we have
(A.15)
G2 ( r, r ', p ) 
1
B1I 0 ( q1r ') K 0 ( q2 r ) / D ,
1
Then,
15
a  r  .
.
r'  
r
r 
 J 0 ( 1w ) i   J 0 ( 2 w )  iY0 ( 2 w ) 
a 2 
a
a 
and the denominator from (A.5) gives the product
I 0 (q1r ')K 0 (q2r ) p  w e 
(A.16)
(A.17)
2 i
1
r' 
r
r
r
r 

J 0 ( 1w )   D1Y0 ( 2 w )  D2 J 0 ( 2w )  i   D1J 0 ( 2w )  D2Y0 ( 2w )   /( D12  D22 ) .
w
a 
a
a
a
a 

Then
(A.18)
Im G2 ( r, r ', p )
2 i
pw e

k1 1
r' 
r
r 
B1 J 0 ( 1w )   D1J 0 ( 2 w )  D2Y0 ( 2 w )  /( D12  D22 )
a w
a 
a
a 
Evaluation of F1( r, r ', p ) and G2 ( r, r ', p ) on the contour p   ei ,       for   0
We have completed the combination of the contours along the top and bottom of the branch, but we still
need to evaluate the line integral around the origin with the radius  and take  to zero,
1
lim
  0 2
(A.19)

 F1(r, r ', p)e
pt
 p  e
i

 ei d .
This requires the evaluation of
 q k K (q a )  I 0 (q1a )K 0 (q1r )  I 0 (q1r )K 0 (q1a )  
F1( r, r ', p )  B1I 0 (q1r ')  2 2 1 2
/ D

q
k
K
(
q
a
)
I
(
q
a
)
K
(
q
r
)

I
(
q
r
)
K
(
q
a
)


1 1
0 1
0 1
1 1
 11 0 2

(A.20)
D  q2k2 K1( q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a )
with p   ei for small  . The power series representations for the I and K functions are needed to get
the proper limits:
K0 ( z )   ln( z / 2)    O( z 2 ln z ) , I 0 ( z )  1  O( z 2 ) , I1( z )  z / 2  O( z 3 )
(A.21)
.
1
K1( z )   ( z / 2)ln( z / 2)  ( z / 2)(  1/ 2)  O( z 3 ln z)
z
The limit lim  ln b  0 in the form lim  ln b   0 helps in the evaluation. Then, the only terms
 0
 0
which give a non-zero contribution for   0 are
D  q2k2 K1(q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a )
(A.22)
F1( r, r ', p )
k2 / a
B1I 0 ( q1r ')q2k2 K1(q2a ) I 0 (q1a )K 0 (q1r ) / D

 O  B1( k2 / a) K 0 ( q1r ) /( k2 / a)   O ln(  ei / 2 1r / a) / 2

One can see clearly now that the limit in (A.15) is
(A.23)
1
lim
  0 2

 F1( r, r ', p)e
pt

 p  e
i
 ei d  0 .
Similarly for G2 ( r, r ', p ) , it is clear that  in the integrand dominates the logarithmic term in K0 (q2r ) ,
(A.24)
G2 ( r, r ', p ) 
1
B1I 0 ( q1r ') K 0 ( q2 r ) / D ,
1
and
16
ar
1
lim
  0 2
(A.25)

 G2 ( r, r ', p)e
pt

 p  e
i
 ei d  0 .
APPENDIX B
Calculation of the integrands for (4.9)
Evaluation of F2 ( r, r ', p ) on the contour p  w2ei , 0  w  
Because of the symmetry of F2 ( r, r ', p ) in the cases for r  r ' and r  r ' we consider only the case
for r  r ' and note that the computation of D is the same as in APPENDIX A:
w
D =
 D1  iD2 
2
1
2 D1  iD2
D1   2 J 0 ( 1w)Y1( 2 w)  1J1( 1w)Y0 ( 2 w)
(B.1)

D  w D12  D22
D2   2 J 0 ( 1w) J1( 2 w)  1J 0 ( 2 w) J1( 1w)
We need the Bessel function relations for the values on the top of the cut:
(B.2)
K0 ( zei / 2 ) 
K1( zei / 2 ) 

2

2
i   J 0 ( z )  iY0 ( z ) =
  J1( z )  iY1( z )

 iJ 0 ( z )  Y0 ( z )
2
.
, I 0 ( zei / 2 )  J 0 ( z ) , I1( zei / 2 )  iJ1( z )
and the Wronskian
J1( z )Y0 ( z )  J 0 ( z )Y1( z )  2 /( z )
is helpful in some reductions. Then,
 q k I ( q a )  I 0 ( q2 r ')K1(q2a )  I1(q2a )K 0 (q2 r ')  
F2 ( r, r ', p )  B2 K 0 ( q2 r )  2 2 0 1
/ D

q
k
I
(
q
a
)
I
(
q
r
')
K
(
q
a
)

I
(
q
a
)
K
(
q
r
')


1
1
1
1
0
2
0
2
0
2
0
2


The first term in braces is
(B.3)
q1k1I1(q1a ) p  w e   1iwI1(i 1w)  1wJ1( 1w)
2 i
 I 0 (q2r ')K0 (q2a)  I 0 (q2a)K0 (q2r ') p  w e
2 i

(B.4)

r'
r' 
 J 0 ( 2 w )Y0 ( 2 w)  J 0 ( 2 w)Y0 ( 2 w ) 
2
a
a 

   ( r ', w,  2 ) (note  is real and  ( a, w,  2 )  0)
2

q1k1I1(q1a )  I 0 (q2r ')K0 (q2a )  I 0 (q2a )K0 (q2r ') p  w2ei  2 1wJ1( 1w) (r ', w,  2 )
and the second term is
q2k2 I 0 (q1a ) p  w2 ei  i 2wI 0 (i 1w)  i 2wJ 0 ( 1w)
 I 0 (q2 r ')K1(q2a )  I1(q2a )K 0 (q2r ') p  w e
2 i
(B.5)
 i

r'
r'

J
(

w
)
Y
(

w
)

J
(

w
)Y1( 2 w) 
1
2
0
2
0
2

2
a
a


=  i  ( r ', w,  2 ) (note  is real and  ( a, w,  2 )  1/( 2 w))
2
q2k2 I 0 (q1a ) I 0 (q2 r ')K1(q2a )  I1(q2a )K 0 (q2r ') p  w e 
2 i
17


2
 2 wJ 0 ( 1w) ( r ', w,  2 )
Then, F2 ( r, r ', p ) becomes
F2 (r, r ', p ) p  w e
2 i
(B.6)
=
=
or
  2 wJ 0 ( 1w) ( r ', w,  2 ) 
i   J 0 ( 2 wr / a )  iY0 ( 2 wr / a )  
 / D( w)
2
2  1wJ1( 2 w) ( r ', w,  2 ) 

  iJ 0 ( 2 wr / a )  2 D1  iD2
B2  2 J 0 ( 1w) ( r ', w,  2 )  1J1( 2 w) ( r ', w,  2 )  

2
2  Y0 ( 2 wr / a )   w D12  D22
w

2
B2
 2 J 0 ( 1w) ( r ', w,  2 )  1J1( 2w) ( r ', w,  2 )
D12  D22
Im F2 ( r, r ', p )
=
(B.7)
 B2

2
B2
p  w2 ei
  D1Y0 ( 2 wr / a )  D2 J 0 ( 2 wr / a )



 i   D1J 0 ( 2 wr / a )  D2Y0 ( 2 wr / a ) 

 2 J 0 ( 1w) ( r ', w,  2 )  1J1( 2w) ( r ', w,  2 )
  D1J 0 ( 2wr / a )  D2Y0 ( 2wr / a )
D12  D22
r'
r' 

 ( r ', w,  2 )   J 0 ( 2 w )Y0 ( 2 w)  J 0 ( 2 w)Y0 ( 2 w ) 
a
a 



r'
a
r'
a


 ( r ', w,  2 )   J1( 2 w)Y0 ( 2 w )  J 0 ( 2 w )Y1( 2 w) 
For G1 we have
G1( r, r ', p ) 
k2
B2 K 0 ( q2 r ') I 0 ( q1r ) / D
a
G1(r, r ', p) p  w e
2 i
(B.8)

k2

B2 J 0 ( 1wr / a ) i   J 0 ( 2 wr '/ a )  iY0 ( 2 wr '/ a )  / D
a
2
=
 k2
B2 J 0 ( 1wr / a) Y0 ( 2 wr '/ a )  iJ 0 ( 2 wr '/ a )
2 D1  iD2
 w D12  D22
2a
k2

B2 J 0 ( 1wr / a )  D1Y0 ( 2 wr '/ a )  D2 J 0 ( 2 wr '/ a )  /( D12  D22 ) 
aw
i
k2
B2 J 0 ( 1wr / a )  D1J 0 ( 2 wr '/ a )  D2Y0 ( 2 wr '/ a )  /( D12  D22 )
aw
and
(B.9)
ImG1( r, r ', p )
2 i
pw e

k2
J ( wr / a )
B2 0 2 1 2 D1J 0 ( 2wr '/ a )  D2Y0 ( 2wr '/ a)
aw
D1  D2
Evaluation of F2 ( r, r ', p ) and G1( r, r ', p ) on the contour p   ei , -     
An inspection of (B.3) with the relations in (A.21) shows that the only terms which will produce a nonzero contribution for   0 are
18
k 2 / a for   0
D  q2k2 K1( q2a ) I 0 (q1a )  q1k1I1(q1a )K 0 (q2a )
(B.10)
F2 ( r, r ', p )
B2 K 0 ( q2 r )q2k 2 I 0 (q1a ) I 0 (q2r ')K1(q2a ) / D  O  K 0 (q2r )   O  ln(q2r ) / 2 

 O ln(  ei / 2 2 r / a ) / 2

Therefore, we have
(B.11)
1
lim
  0 2

 F2 (r, r ', p)e
pt
 p  e
i

 ei d  0
Similarly for G1( r, r ', p ) , it is clear that  in the integrand dominates the logarithmic term in K0 (q2r ') ,
k
G1( r, r ', p )  2 B2 K 0 ( q2 r ') I 0 ( q1r ) / D ,
(B.12)
a
and
(B.13)
1
lim
  0 2

 G1(r, r ', p)e

pt
 p  e
19
i
 ei d  0 .