Green’s Function for Heat Conduction in a Slab X55
with Type 5 Boundary Conditions
by
Donald E. Amos
Abstract The Green’s function is the principal tool in construction of the general solution to the
classical heat conduction problem. The solution is presented in terms of the internal heat generation,
initial temperature and integrals which reflect the physical influence of the boundary. In the current
literature ( http://Exact.unl.edu ) the common boundary conditions are presented as Types 1,2,3,4, and 5
ranging from a specified temperature (Type 1) to the most general form (Type 5) where input energy
(flux), heat loss to the surroundings, heat storage on a boundary layer and conduction into the material
are considered. Since the driving energy for the Green’s function is internal, the homogeneous form of
the boundary condition is used to define the Green’s function. The thrust of this work is to derive the
Green’s function, labeled X55, for a slab with Type 5 boundary conditions on both faces.
In the general solution with a Type 5 boundary, the usual integrals emerge, but an extra term which
accounts for the release (or absorption) of energy stored in the boundary layer also appears. The results
are used to construct the solution to an X55T0 slab problem where fluxes are the energy sources at the
Type 5 boundaries x=0 and x=L. Another problem, describing the cooling of a boundary layer, is
constructed to utilize only the extra term where the energy source is the heat stored in a boundary layer
and the slab acts as a heat sink. Both solutions agree with the direct Laplace transform solutions. Finally,
a closed system where there is no heat loss or gain is considered. The initial temperature differences of
the slab and boundary layers provide the driving force for a redistribution of energy. The transient
temperature distribution and equilibrium temperatures are calculated and agree with known results.
1. Introduction
The Green’s function is the principal tool in construction of the general solution to the classical heat
conduction problem. In [2], the solution is presented in terms of the internal heat generation, initial
temperature and integrals which reflect the physical influence of the boundary. In [1, p 49] and
( http://Exact.unl.edu ) the common boundary conditions are presented as Types 1,2,3,4, and 5 ranging
from constant temperature (Type 1) to the most general form (Type 5) where input energy (flux), heat
loss to the surroundings, heat storage on the boundary and conduction into the material are considered.
In mathematical terms,
T
T
k
 f B  hT  
h  0,   0
(GX.1.1)
ni
t
where f B  0 is an input flux, h is the heat transfer coefficient for losses to the exterior,  is the product
of density, layer thickness and specific heat and the flux derivative is an interior normal derivative. The
sub-cases can be summarized by
Type 5 Boundary condition:   0, h  0
Type 4 Boundary condition:   0, h  0
(GX.1.2)
Type 3 Boundary condition:   0, h  0
Type 2 Boundary condition:   0, h  0
Type 1 Boundary condition: f B  0,   0, h  
In the Type 1 condition, a zero boundary temperature is produced by taking the limit.
The range of problems solved by classical Green’s functions with boundary Types 1-5 is outlined in [2].
For the slab on [0,L] with Type 5 boundaries, the problem to be solved is given by
1
 2T
qv ( x, t ) 1 T

, 0  x  L , t>0
k
 t
x
T
T
k
 h0T  f 0 (t )   0
, at x=0 , h0  0, ,  0  0
x
t
2
(GX.1.3)
(GX.1.4)

.
T
T
k
 hLT  f L (t )   L
, at x=L , hL  0, ,  L  0
x
t
T ( x,0)  f ( x ) , 0<x<L,
T(0,0)=T0 , T(L,0)=TL .
The solution in terms of the corresponding Green’s function is [2]
T ( x ', t ) 
(GX.1.5)
+


k

k
t
L
 
L
G( x, x ', t   )qv ( x, )dxd   G ( x, x ', t  0) f ( x )dx
0 0
t

G(0, x ', t   ) f 0 ( )d 
0
 L
 0
k
G(0, x ', t  0)T0  
k

k
t

0
G( L, x ', t   ) f L ( )d
0
G( L, x ', t  0)TL 
Notice that the usual integrals emerge, but extra terms in  0 and  L which account for the release (or
absorption) of energy stored in the boundary layers also appear. It was noted above that one can have
mixed boundary types in one problem. In the general solution [2], integrands over subsets of the
boundary may look quite different depending on the type which applies there. This point is illustrated in
Section 5, equation (GX.5.3).
The Green’s function corresponding to the Type 5 boundary condition is derived in Section 2. Section 3
is devoted to sub-cases X51 and X15 because the Type 1 condition cannot be obtained by setting h or 
to zero. In Section 4, the main results of Section 2 are used to construct the solution to an X55T0 slab
problem where fluxes are the energy source in the Type 5 boundaries x=0 and x=L. This solution agrees
with the direct solution in [3]. In Section 5 a problem is constructed to utilize only one extra term of
(GX.1.5) where the energy source is the heat stored in a boundary layer and the slab acts as a heat sink.
This solution is compared with the direct (Laplace transform) solution. Finally, a closed system where
there is no heat loss or gain is considered in Section 6. The initial temperature differences in the slab and
boundary layers provide the driving force for a redistribution of energy. The transient temperature
distribution and equilibrium temperatures are calculated and agree with known results.
2. Green’s Function for X55 Slab
The differential equation for the slab is
 2G 1 G
(GX.2.1)

, 0  x  L , t>0
x 2  t
with 55 boundary conditions expressing how the internal heat generation is distributed to the external
region or stored on a surface film with only heat capacity [2]. It is common practice to use the
homogenous form of the boundary condition for the Green’s function since the energy source is interior
to the material,
G
G
k
 h0G   0
, at x=0 , h0  0, ,  0  0
x
t
(GX.2.2)
.
G
G
k
 hLG   L
, at x=L , hL  0, ,  L  0
x
t
2
These are the Type 5 boundary conditions in the nomenclature of [1, p. 49] and http://Exact.unl.edu .
The initial temperature distribution is taken to be zero
G( x,0)  0 ,
(GX.2.3)
0  x  L,
except in the neighborhood of x  x ' where the heat source has the form
2
1
(GX.2.4)
Gs ( x, x ', t ) 
e ( x  x ') /(4 t )
4 t
The Laplace transform is generally the method of choice for these linear problems with constant
parameters where the boundary conditions vary with time. Then
 2G
p
(GX.2.5)
,
 q2G , 0  x  L , q 
2

x
and we form the solution from the source term and the solutions of this equation,
G( x, x ', p )  Gs ( x, x ', p )  C1 sinh qx  C2 cosh qx
(GX.2.6)
or
1 q x x '
G ( x, x ', p ) 
e
 C1 sinh qx  C2 cosh qx .
(GX.2.7)
2 q
We apply this equation to the boundary conditions at x=0 ( x  x '  x ' x) and x=L ( x  x '  x  x ')
G
 h0G   0 pG , at x=0 ,
x
 1  qx '

kq  qx '
e
 kqC1  P0 
e
 C2  , P0  h0   0 p
2 q
 2 q

k
(GX.2.8)
G
 hLG   L pG , at x=L ,
x
 1  q( L  x ')

kq  q( L  x ')
e
 kqC1 cosh qL  kqC2 sinh qL  PL 
e
 C1 sinh qL  C2 cosh qL 
2 q
 2 q

k
Then
AC1  BC2  R0 , P0  h0   0 p
CC1  DC2  RL , PL  hL   L p
A  kq
(GX.2.9)
,
B=  P0
C   kq cosh qL  PL sinh qL  , D=  kq sinh qL  PL cosh qL 
R0 
e qx '
e q( L  x ')
 PL  kq 
 P0  kq  , RL  
2 q
2 q
The solution is
(GX.2.10)
C1  ( DR0  BRL ) / Dˆ
,
C2  ( ARL  CR0 ) / Dˆ
,
Dˆ  AD - BC
q x x '
Dˆ e
( DR0  BRL )
( ARL  CR0 )
G( x, x ', p ) 

sinh qx 
cosh qx
2 q Dˆ
Dˆ
Dˆ
.
The algebra which follows expands this transform, collects like terms, and rewrites G in a more
classical form. The final result is presented in (GX.2.31). We start with the coefficients:
3
DR0  BRL 
(GX.2.11)

e qx '
e q( L  x ')
 PL  kq
 P0  kq kq sinh qL  PL cosh qL  P0
2 q
2 q
 P0  kq  PL  kq eq( L x ')   P0  kq  PL  kq e q( L x ')  P  PL  kq e q( L x ')
2 q
ARL  CR0  
(GX.2.12)  

2 q
2
2
0
2 q
kqe q( L  x ')
e qx '
 PL  kq 
 P0  kq kq cosh qL  PL sinh qL
2 q
2 q
kqe q( L  x ')
 P  kq  ( PL  kq) eq( L x ')  ( PL  kq) e q( L x ') 
 PL  kq  0

2 q
2 q 
2
2
 P0  kq ( PL  kq) eq( L x ')   P0  kq ( PL  kq) e q( L x ') 
2 q
2 q
2
2
kq
 PL  kq e q( L x ')
2 q
Dˆ ( p )  AD - BC  kq  kq sinh qL  PL cosh qL   P0  kq cosh qL  PL sinh qL  
  P0 PL  (kq)2  sinh qL  kq( P0  PL )cosh qL)


(GX.2.13)

1
1
P0 PL  ( kq)2  kq( P0  PL )  eqL   P0 PL  (kq)2  kq( P0  PL )  e  qL


2
2

1
1
( P0  kq)( PL  kq) eqL  ( P0  kq)( PL  kq) e qL
2
2
For x  x ' ,
(GX.2.14)
(GX.2.15)
Dˆ ( p )  q( x ' x ) ( P0  kq) ( PL  kq) q( L  x ' x ) ( P0  kq) ( PL  kq)  q( L  x ' x )
e

e

e
2 q
2 q
2
2 q
2
U
V
W
( DR0  BRL )sinh qx  eq( L  x ' x )  e q( L  x ' x )  e q( L  x ' x )
2
2
2

U q( L  x ' x ) V  q( L  x ' x ) W  q( L  x ' x )
e
 e
 e
2
2
2
( ARL  CR0 )cosh qx  
U q( L  x ' x ) V  q( L  x ' x ) W1  q( L  x ' x )
e
 e

e
2
2
2

U q( L  x ' x ) V  q( L  x ' x ) W1  q( L  x ' x )
e
 e

e
2
2
2
(GX.2.16)
U
(GX.2.17)
( P0  kq) ( PL  kq)
( P  kq) ( PL  kq)
( P  kq)
( P  kq )
, V 0
, W   P0 L
, W1 =  kq L
2 q
2
2 q
2
2 q
2 q
( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
Ve q( L  x ' x )  Ueq( L x ' x ) 
(W  W1 )  q( L  x ' x ) (W1  W )  q( L  x ' x )
e

e
2
2
4
Dˆ ( p )  q( x ' x )
e
 ( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
2 q

( P0  kq) ( PL  kq) q( L  x ' x ) ( P0  kq) ( PL  kq)  q( L  x ' x )
e

e
2 q
2
2 q
2
 Ueq( L  x ' x )  Ve q( L  x '  x ) 
(GX.2.18)

(W  W1 )  q( L  x '  x ) (W1  W )  q( L  x '  x )
e

e
2
2
( P0  kq) ( PL  kq) q( L  x ' x ) (W1  W )  q( L  x ' x )
e

e
 Ueq( L  x ' x )
2 q
2
2
 ( P  kq) ( PL  kq)

(W  W1 )  q( L  x '  x )
+  0
 V  e q( L  x '  x ) 
e
2 q
2
2



(GX.2.19)
1  ( P0  kq)( PL  kq) 
W1  W ( P0  kq) ( PL  kq)
 V  0,

,


2 q 
2
2
2 q
2

W1  W
( P  kq ) ( PL  kq)
 0
2
2 q
2
1
2 q
(GX.2.20)
e q( x ' x ) Dˆ ( p )  ( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 

( P0  kq) ( PL  kq) q( L  x ' x ) ( P0  kq) ( PL  kq)  q( L  x ' x )
e

e
2 q
2
2 q
2

( P0  kq) ( PL  kq) q( L  x ' x ) ( P0  kq) ( PL  kq)  q( L  x ' x )
e

e
2 q
2
2 q
2
We use e z  cosh z  sinh z on the numerator
1  q( x ' x ) ˆ
e
D( p )  ( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
2 q
 ( P  kq) ( PL  kq) ( P0  kq) ( PL  kq) 
 0

 cosh q( L  x ' x )
2
2 q
2
 2 q

(GX.2.21)
 ( P  kq) ( PL  kq) ( P0  kq) ( PL  kq) 
+  0

 sinh q( L  x ' x )
2
2 q
2
 2 q

 ( P  kq) ( PL  kq) ( P0  kq) ( PL  kq) 
 0

 cosh q( L  x ' x )
2
2 q
2
 2 q

 ( P  kq) ( PL  kq) ( P0  kq) ( PL  kq) 
 0

 sinh q( L  x ' x )
2
2 q
2
 2 q

(GX.2.22)

sinh qL 
Dˆ ( p )  qD( p )  q k (kq sinh qL  PL cosh qL)  P0 (k cosh qL  PL
)
q


5
Dˆ ( p )  q( x ' x )
e
 ( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
2 q
(GX.2.23)

1
P0 PL  ( kq)2  cosh q( L  x ' x ) +
kq( P0  PL )sinh q( L  x ' x )

2 q
2 q

1
P0 PL  ( kq)2  cosh q( L  x ' x ) 
kq( P0  PL )sinh q( L  x ' x )

2 q
2 q
1
1
Now we can write the transform explicitly by dividing by Dˆ ( p ) in the form
(GX.2.24)

sinh qL 
Dˆ ( p )  qD( p )  q k (kq sinh qL  PL cosh qL)  P0 (k cosh qL  PL
)
q


(GX.2.25)
  P P  ( kq)2  cosh q( L  x ' x ) +  kq( P  P )  sinh q( L  x ' x ) 
0
L

1
  0 L

G( x, x ', p )  

2
   P0 PL  ( kq)2  cosh q( L  x ' x )   kq( P0  PL )  sinh q( L  x ' x )  2 q D( p )

 

Now, D( p ) is an analytic function in the whole p plane because each of the terms
sinh qL
q sinh qL, cosh qL,
(GX.2.26)
q
can be represented as a power series in p which has an infinite radius of convergence( analytic
functions) while P0 and PL are only linear functions of p. However the term   /(2 q2 ) looks like it
  has the estimate
may have a pole at p=0 ( p   q2 ) , but the following analysis shows that the term
O(q2 ) which makes the numerator an analytic function of p in the whole p plane. The analysis is
  P0 PL  ( kq)2  1  q 2 ( L  x ' x )2 / 2  O ( q 4 )  +  kq( P0  PL )   q( L  x ' x )  O ( q 3 )  



 1
 



2
   P0 PL  ( kq)2  1  q 2 ( L  x ' x ) 2 / 2  O ( q 4 )    kq( P0  PL )   q( L  x ' x )  O ( q 3 )   2 q




 
(GX.2.27)


 2( kq)2  q 2  P0 PL ( L  x ' x ) 2 / 2  P0 PL ( L  x ' x )2 / 2  O ( q 4 )


  q 2  k ( P0  PL )  ( L  x ' x )    k ( P0  PL )   ( L  x ' x )   O ( q 2 )




 21
 2k 2  2h0 hL ( L  x ') x  2khL ( L  x ')  2kh0 x
This analysis is further confirmed by the following manipulations
(GX.2.28)
6

 1

2
 2 q

  P0 PL  ( kq )2  cosh q( L  x ') cosh qx   P0 PL  ( kq ) 2  cosh q( L  x ') cosh qx 



 



2
2
 P P  ( kq )  sinh q( L - x ')sinh qx   P0 PL  ( kq )  sinh q( L - x ')sinh qx 



1
  0 L

G ( x, x ', p )  

2
 +  kq( P0  PL ) sinh q( L  x ') cosh qx   kq( P0  PL )  sinh q( L  x ') cosh qx
 2 q D ( p )


 +  kq( P0  PL ) cosh q( L  x ')sinh qx   kq( P0  PL )  cosh q( L  x ')sinh qx



 2(kq)2 cosh q( L  x ')cosh qx  2 P0 PL sinh q( L - x ')sinh qx 
1



(GX.2.29) G( x, x ', p )  
+2kqPL sinh q( L  x ')cosh qx  2kqP0 cosh q( L  x ')sinh qx  2 q2 D( p )



(GX.2.30)
 kq  kq cosh q( L  x ')  PL sinh q( L  x ') cosh qx
G( x, x ', p )  2 
  P0  kq cosh q( L  x ')  PL sinh q( L  x ') sinh qx

1

2
 2 q D( p )
 2  kq cosh q( L  x ')  PL sinh q( L  x ')   kq cosh qx  P0 sinh qx 
1
2 q D( p )
2
Finally,

sinh q( L  x ')  
sinh qx  1
G( x, x ', p )   k cosh q( L  x ')  PL
k cosh qx  P0


q
q   D( p )


We have demonstrated (2 ways) that we have the quotient of two analytic functions. Any singularities
can only come from the zeros of the denominator, D( p ) . Therefore we can apply the residue theorem in
(GX.2.31)
the classical way and sum the residues of poles from G( x, x ', p)e pt . We assume from previous history
for finite thicknesses that the poles are on the negative real axis at p   n2 / L2 . Therefore,
  i
1
G( x, x ', p )e pt   Re sidue of G( x, x ', p )e pt at p   n2 / L2
(GX.2.32) G( x, x ', t ) 

2 i   i
n
The residue computation for the ratio of two analytic functions f ( z ) / g ( z ) at a simple pole z  z0
where g ( z0 )  0 is f ( z0 ) / g '( z0 ) . Then, using p   2 / L2 , q  i  / L, sinh iz  i sin z, and
cosh iz  cos z the result for x  x ' is (β is written for βn )
G( x, x ', t )  G0 ( x, x ', t )  
n 1
(GX.2.33)
 (  , P0 , x ) (  , PL , L  x ')
W ( )

 (  , P, x )   k cos  x / L  LP(  )

e
2
t / L2
, x  x'
sin  x / L 



where  n , n  1, is a non-zero solution of D( p   2 / L2 )  D(  )  0 , (GX.2.24), with
(GX.2.34)


D(  )  (k 2 / L)   LP0 (  ) PL (  ) /  sin   P0 (  )  PL (  ) k cos 
or
(GX.2.35)
tan  
 P0 (  )  PL (  ) k 
(k 2 / L) 2  LP0 (  )PL (  )
and
7
W (  )   D( p )
(GX.2.36)
p  2 / L2

 cos  

1
sin 
tan 
 V (  )cos   
 V (  )
U (  ) L
U (  ) L
2

2 



U (  )  k 2  2   0 PL (  )   L P0 (  )   kL  P0 (  )  PL (  )  +L2 P0 (  ) PL (  ) /  2
V (  )  k 2 L  L3P0 (  ) PL (  ) /  2  2k ( L   0 )
P0 (  )  h0   0  2 / L2
PL (  )  hL   L  2 / L2
Here P (  ) in the definition of  is either P0 (  ) or PL (  ) .
The derivative computation can be found in [3]. It looks like   0 is a solution of the eigen-equation
and there is a pole at p=0. The analysis presented in [3] shows that
D(0)  k ( h0  hL )  Lh0hL  0 .
(GX.2.37)
Therefore, p=0 is a pole if and only if hL  0 and h0  0 . Then,
D( p)
(GX.2.38)
p 0
 [k 2 L  k ( 0   L )]/ 
and the n=0 term is
(GX.2.39)
k


G0 ( x, x ', t )   kL   ( 0   L )

0


h0  hL  0 
.
h0  0 or hL  0
The complete solution has to include the case where x  x ' . The analysis can be repeated, but the
problem for x  x ' is not essentially different from that for x  x ' . That is, we map the region
x '  x  L onto 0  x  x ' so that L  0 and x'  L  x ' . The mapping y=L-x does the job and the
contents of the APPENDIX confirms this approach. Thus we exchange subscripts and replace x and x '
by L-x and L- x ' respectively. Notice that the eigen-equation and the denominator are symmetric in
P0 (  ) and PL (  ) and the exchange does not change any value. Then
G( x, x ', t )  G0 ( x, x ', t )  
(GX.2.40)
n 1
 (  , P0 , x ') (  , PL , L  x )
W ( )
e
2
t / L2
, x  x'
sin  x / L 




where P (  ) is either P0 (  ) or PL (  ) . The complete solution is then a combination of (GX.2.33),
(GX.2.39) and (GX.2.40).

 (  , P, x )   k cos  x / L  LP(  )
3. Slabs X51, X15 and Sub-cases
The Type 5 boundary condition
T
T
 f B  hT  
ni
t
and the corresponding Green’s function boundary condition
G
G
k
 hG  
(GX.3.2)
ni
t
(GX.3.1)
k
8
has two parameters h and  [1, p 50] which lead to sub-cases of Types 4, 3, and 2 by setting h or  or
both to zero, making a total of 10 different slab problems. The Green’s function for Type 1 for a zero
boundary temperature is derived by setting  to zero and taking h   . We can delineate the cases by
considering XI5, I=5,4,3,2,1; then XI4, I=4,3,2,1; etc., making a total of 15 cases. The cases for X5I;
I=4,3,2,1; X4I, I=3,2,1; etc., lead to 10 more cases for a total of 25 cases with the subscripts defined
according to (GX.1.2)
We consider the Green’s function for X51 because it requires taking a limit and all sub cases X41, X31,
X21 can be obtained by setting h0 or  0 or both to zero. X11 is obtained from  0  0 and  L  0 with
h0   and hL   .
X51 is define as a Type 5 boundary at x=0 and a constant zero temperature at x=L. This means that we
set  L  0 and take hL   in (GX.2.33) and (GX.2.40). The limit is obtained first by dividing the
numerator and denominator by hL . Then for x  x ' in (GX.2.33)
PL  hL ,
(GX.3.4)
 (  , PL , L  x ')
hL
L
sin  (1  x '/ L)

P0 (  )  h0   0  2 / L2
,

W ( )
1  
L2 P0 (  )  sin   L3P0 (  ) 
 W0 (  )    2 0  kL+
L

cos





2
hL
2  
 2  
 


tan  
 P0 (  )  PL (  ) k 
 ( k
2

/ L)   LP0 (  ) PL (  )
2
and we have
G( x, x ', t )   L
n 1
 tan  
sin  (1  x '/ L)  (  , P0 , x )  2 t / L2
e
,

W0 (  )

 (  , P0 , x )   k cos  x / L  LP0 (  )

(GX.3.5)
tan  
k 
LP0 (  )
x  x'
sin  x / L 



k 
, P0 (  )  h0   0  2 / L2
LP0 (  )

1  
L2 P0 (  )  sin   L3P0 (  ) 
W0 (  )    2 0  kL+
L

cos





2
2  
 2  
 


And similarly for x  x ' ,
(GX.3.6)
G( x, x ', t )   L
n 1
sin  (1  x / L)  (  , P0 , x ')  2t / L2
e
,

W0 (  )
x  x'
If in addition we take  0  0 and h0   , we get the Green’s function for the X11 case
2
2
2
(GX.3.7)
G( x, x ', t )   sin  x / L sin  x '/ L e t / L ,   n , n  1,2,3,...
L n 1
9
We do the case for X15 so that the sub-cases X14, X13, X12 can be obtained by setting hL or  L or
both to zero. X15 is obtained from X55 by setting  0  0 and h0  
G ( x, x ', t )   L
n 1
sin  x / L  (  , PL , L  x ')  2 t / L2
e
,

WL (  )

 (  , PL , x )   k cos  x / L  LPL (  )

(GX.3.8)
WL (  ) 
sin  x / L) 




1  
L2 PL (  )  sin   L3PL (  ) 
2


kL
+
L

cos






L
2
2  
 2  
 


tan  
k 
, P (  )  hL   L  2 / L2 ,
LPL (  ) L
and
(GX.3.9)
x  x'
G( x, x ', t )   L
n 1
sin  x '/ L  (  , PL , L  x )  2t / L2
e
,

WL (  )
x  x'
4 Application to the X55T0 Problem
The differential equation for the X55T0 slab on (0,L) is
 2T 1 T
(GX.4.1)

, 0  x  L , t>0
x 2  t
with 55 boundary conditions expressing how input fluxes f0 and f L are conducted into the region,
conducted or radiated into an exterior medium, and stored on a surface layer with only heat capacity
T
T
k
 h0T  f 0 (t )   0
, at x=0 , h0  0, ,  0  0
x
t
(GX.4.2)
.
T
T
k
 hLT  f L (t )   L
, at x=L , hL  0, ,  L  0
x
t
The initial temperature distribution (T0) is taken to be zero
T ( x,0)  0 ,
(GX.4.3)
0  x  L.
This problem was solved in [3]. The application here gives an independent check on both the Green’s
function and the solution in [3]. The theory for the solution of this problem using the Green’s function is
presented in [2],
(GX.4.4) T ( P ', t ) 

t
k 
G( P, P ', t   ) f B ( P, )dS ( P )d 
0 S

k
 G( P, P ', t  0)T ( P,0) dS ( P) .
S
Since T ( P,0)  0 on S from (GX.4.3) only the first term is applicable. Then, in one dimension, the
surface integral is just sum of the evaluations at the endpoints. From (GX.2.33) and (GX.2.40) we get
10
T ( P ', t ) 
T ( x ', t ) 

k

k
t

G (0, x ', t   ) f 0 ( )d 
0
t
t


k

G ( L, x ', t   ) f L ( )d
0
t
 G0 (0, x ', t   ) f0 ( )d  k  G0 ( L, x ', t   ) f L ( )d 
0
0

(GX.4.5)
+

k

k

n 1

n 1
2
2
k (  , PL , L  x ')
f 0 ( ) e ( t  ) / L d

W ( )
0
t
2
2
k (  , P0 , x ')
f L ( ) e ( t  ) / L d

W ( )
0
t

 (  , P, x )   k cos  x / L  LP(  )

P0 (  )  h0   0  2 / L2
,
sin  x / L 



PL (  )  hL   L  2 / L2
k


h0  hL  0 

G0 ( x, x ', t )   kL   ( 0   L )


0
h0  0 or hL  0

where we have used the expression for x  x ' for G (0, x ', t   ) and the expression for x  x ' for
G( L, x ', t   ) . Equation (GX.4.5) agrees with the solution from [3]. If h0  hL  0 , then this becomes
an X44T0 case and the leading terms for n=0 from the two formulations match also.
5. Cooling of a Boundary Layer
In [2], the general theory for the solution of a heat conduction problem was derived using Type 5
boundary conditions. The components of the solution follow the classical forms for volumetric heat
generation, initial temperature and input flux on the boundary, but an extra term appears which comes
from the release of heat from a boundary layer. This boundary layer has heat storage properties, acting
like a source or sink, with infinite conductive properties and assumes the temperature of the slab on the
boundary. It is the purpose of this example to construct a problem which uses only this extra term as the
solution and then verify the result by direct solution of the problem. The problem is expressed by
 2T
(GX.5.1)
x 2
with 14 boundary conditions
(GX.5.2)
k

1 T
 t
, 0  x  L , t>0
T (0, t )  0, t  0
T
T
  L
, at x=L ,  L  0
x
t
(Type 1)
(Type 4) .
T ( x,0)  0 0  x<L , T(L,0)=TL ,
This problem is of Type 14 since we have a zero temperature at x=0 and a Type 5 condition with hL  0
at x=L with no external driving energy f B  0 . These equations represent the cooling of the boundary
layer initially at temperature TL using the slab as a heat sink. The heat energy contained in the boundary
11
layer is  (TL  0) . The zero in this formula is the ambient temperature. We write the formula from [2]
for the solution, taking into account the different types at x=0 and x=L.
t
(GX.5.3) T ( P ', t )   

T ( P, )
0 S1
G( P, P ', t   )

dS1d 
ni
k
 G( P, P ', t  0)T ( P,0) dS4 ( P) .
S4
When we apply this to an interval, surface integrals are just the integrands at end points. With
T (0, )  0 we get
t
T ( x ', t )   T (0, )
(GX.5.4)
0

 L
G(0, x ', t   )

d  L  G( L, x ', t  0)T ( L,0) 
x
k
.
TLG( L, x ', t  0)
k
In the development we have G for X15 from (GX.3.9)
G( x, x ', t )   L
n 1
sin  x '/ L  (  , PL , L  x )  2 t / L2
e
,

WL (  )

 (  , PL , x )   k cos  x / L  LPL (  )

(GX.5.5)
tan  
x  x'
sin  x / L 



k 
, PL (  )  hL   L  2 / L2
LPL (  )

1  
L2 PL (  )  sin   L3PL (  ) 
WL (  )    2 L  kL+
L

cos





2
2  
 2  
 


For the X14 geometry, we set hL  0 and take the case for x  x ' since G is evaluated at x=L,
sin  x '/ L  (  , PL , L  x )  2 t / L2
G ( x, x ', t )   L
e
,

WL (  )
n 1

sin  (1  x / L) 
(GX.5.6)  (  , PL , L  x )   k cos  (1  x / L)  LPL (  )




tan  
kL
 L  
, WL (  ) 

1
sin 
( L  kL) L
  L L cos   , PL (  )   L  2 / L2

2


Then, for x=L,
T ( x ', t ) 
 L
k
TLG ( L, x ', t  0) 
(GX.5.7)
 2 LTL 
n 1
 L
k
TL  L
n 1
sin  x '/ L
k

WL (  )
e
2
2
sin  x '/ L
e t / L
( L  kL)sin    L  cos 
Solution of (GX.5.1) and (GX.5.2) by Laplace Transform
To verify the solution, we solve the initial problem by the Laplace transform.
12
2
t / L2
 2T
(GX.5.8)
x
2
 q2T , 0  x  L , q=
p

T (0, p )  0,
k
(GX.5.9)
T
  L  pT ( L, p )  TL  ,  L  0
x
T ( x,0)  0 0  x<L , T(L,0)=TL ,
We take a solution which satisfies the boundary condition at x=0
(GX.5.10)
T ( x, p )  C sinh(qx )
and apply the boundary condition at x=L,
kCq cosh ql   L pC sinh qL   LTL
(GX.5.11)
or
C
(GX.5.12)
 LTL
kq cosh ql   L p sinh qL
and
sinh qx
q
(GX.5.13)
.
T ( x, p ) 
k cosh ql   Lq sinh qL
Now we have the ratio of two analytic functions in the p plane. The zeros of the denominator give
contributions to the residues at p   2 / L2
 LTL
D( p )  k cosh qL   L q sinh qL
D '( p )  [kL sinh qL   L ( qL cosh qL  sinh qL)]/(2 q)

(GX.5.14)
L
[( kL   L )sinh qL   L qL cosh qL]/( qL)
2
D(  2 / L2 )  k cos   ( L / L)  sin   0
tan  
kL
 L 
The inversion becomes
(GX.5.15) T ( x, t )  
 LTL L
, p   2 / L2 , qL  i 
sin  x / L

e
2
t / L2
L
[( kL   L )sin    L  cos  ]
2
which agrees with the Green’s function approach in (GX.5.7).
n 1
2 LTL sin  x / L e t / L
( kL   L )sin    L  cos 
2

n 1
2
6. Temperatures in a Slab with No Gain or Loss of Heat
In this section we are interested in constructing a problem which will test the validity of G0 and verify
transient terms for consistency. This means that we take h0  hL  0 which implies no loss of energy
from the system. Further, if we take the input fluxes at the boundaries to be zero, f0  f L  0 , there is
13
no gain of energy to the system either. The Green’s function solution from [2], with the only non-zero
terms coming from the initial temperatures T0 or TS or TL , is
L
T ( x ', t )  TS  G( x, x ', t )dx 
(GX.6.1)
0
 0T0
k
G(0, x ', t ) 
 LTL
k
G( L, x ', t ) .
Note that with h0  hL  0 , G0 in the Green’s function from (GR2.33) and (GR2.40) is not zero, but
has the value
k
G0 
(GX.6.2)
.
kL   ( a   b )
Steady State
Since there is no gain or loss of heat, a system with unequal initial temperatures in the slab and boundary
layers must come to an equilibrium temperature, TE for t   . This energy balance expresses the fact
that the initial energy must be equal to the final energy with all components at the equilibrium
temperature. Then, if the slab is initially at temperature TS on (0,L) and the boundary layers initially at
temperatures T0 and TL , the energy balance is
(GX.6.3)
initial energy= 0T0   cLTS   LTL  final energy=TE ( 0   cL   L )
TE  ( 0T0  (k /  ) LTS   LTL ) /( kL /    0   L )
Since the transient part vanishes for t   , G  G0 in each of the terms of T and we get the steady state
or equilibrium temperature form (GX.2.33) and (GX.2.40)
kL

k

k
T ( x ', ) 
TS  0
T0  L
TL
kL   ( 0   L )
k kL   ( 0   L )
k kL   ( 0   L )
(GX.6.4)
 0T0
kLTS
 LTL
=

+
 TE
kL   ( 0   L ) kL   ( 0   L ) kL   ( 0   L )
Thus, we have verified that the Green’s function approach using G0 gives the correct result. Further,
since TE is a convex linear combination of the initial temperatures, TE lies between the minimum and
maximum of the three initial temperatures.
Transient Part
The mathematical problem can be stated as
 2T 1 T
(GX.6.5)

, 0  x  L , t>0
x 2  t
with boundary and initial conditions
T
T
k
  0
, at x=0 ,  0  0
(Type 4)
x
t
(GX.6.6)
k
T
T
  L
, at x=L ,  L  0
x
t
(Type 4)
T (0,0)  T0 , T ( x,0)  TS , 0<x<L , T(L,0)=TL
14
The Green’s function solution for this problem is given by (GX.6.1). We continue by investigating the
transient part of the solution
.
To simplify the algebra, we drop the summation in (GX.2.33) and (GX.2.40) and pick up the
coefficients which depend only on x and x ' . Then, we have
P0 (  )   0  2 /L2
 0
k
(GX.6.7)
 L
k
L

G(0, x ', t ) :
 0
G( L, x ', t ) :
 L
G( x, x ', t )dx :
k
PL (  )   L  2 / L2
 (  , P0 ,0) (  , PL , L  x ')   0 (  , PL , L  x ')
 (  , PL ,0) (  , P0 , x ')   L (  , P0 , x ')
k
x'
L'
0
x'
 (  , PL , L  x ')   (  , P0 , x )dx   (  , P0 , x ')   (  , PL , L  x )dx
0
where we have used the expressions for x  x ' and x  x' for the integration. We want to compute the
terms of (GX.6.7)
x'
L'


 (  , PL , L  x ')   (  , P0 , x )dx   (  , P0 , x ')   (  , PL , L  x )dx  TS
0
x'


(GX.6.8)
+ 0T0 (  , PL , L  x ')   LTL (  , P0 , x ')
The computation is:
x'

x'
 (  , P0 , x )dx  k  cos  x / Ldx   0
0
0

(GX.6.9)
L'

kL


L
x'

sin  x / Ldx
0
sin  x '/ L   0 (cos  x '/ L  1)
L
 (  , PL , L  x )dx  k  cos  (1  x / L)dx   L
x'
x'

kL


L
L

sin(1  x / L)dx
x'
sin(1  x '/ L)   L 1  cos  (1  x '/ L) 
x'
 (  , PL , L  x ')   (  , P0 , x )dx 
0
 [k cos  (1  x '/ L)   L
(GX.6.10)
L

L
sin(1  x '/ L)][
kL

sin  x '/ L   0 cos  x '/ L]   0 (  , PL , L  x ')
 (  , P0 , x ')   (  , PL , L  x )dx 
x'
 [k cos  x '/ L   0

L
sin  x '/ L][
kL

sin(1  x '/ L)   L cos  (1  x '/ L)]   L (  , P0 , x ')
Now we pick out the coefficients of the product of the sines and cosines in each expression to get
15
x'
L'
0
x'
 (  , PL , L  x ')   (  , P0 , x )dx   (  , P0 , x ')   (  , PL , L  x )dx 
 k 2L


  2 0 L   cos  (1  x '/ L)sin  x '/ L  cos  x '/ L sin  (1  x '/ L)  
L 
 
(GX.6.11)
 k [ 0   L ] cos  (1  x '/ L)cos  x '/ L  sin  x '/ L sin  (1  x '/ L)  
  0 (  , PL , L  x ')   L (  , P0 , x ')
 k 2L


  2 0 L  sin   k [ 0   L ]cos    0 (  , PL , L  x ')   L (  , P0 , x ')
L 
 
Now the terms involving sin  and cos  are a form of the eigen equation for this case where
h0  hL  0 and the combination is zero.
 k 2L

L2
2




sin


k
[



]cos



D(  )  0


0 L
0
L
L 
2
 
Then, the final sum is
(GX.6.12)
x'
L'
0
x'
 (  , PL , L  x ')   (  , P0 , x )dx   (  , P0 , x ')   (  , PL , L  x )dx 
(GX.6.13)
  0 (  , PL , L  x ')   L (  , P0 , x ')
and the transient part is
  0 (T0  TS ) (  , PL , L  x ')   L (TL  TS ) (  , P0 , x ') e t / W (  )
2
(GX.6.14)
n 1
.
sin  x / L 




The final solution with both the steady state and transient terms in (GX2.33) and (GX2.40) is

 (  , P, x)   k cos  x / L  LP(  )
T ( x ', t )  TE  
 0 (T0  TS ) (  , PL , L  x ')   L (TL  TS ) (  , P0 , x ') e
W ( )
n 1

 (  , P, x )   k cos  x / L  LP(  )

tan  
(GX.6.15)
W ( ) 
sin  x / L 



 0   L  kL
k 2 L2   2 0 L  2 
 cos 
1
sin 
 V (  )cos   
U (  ) L
2

2



tan 
 V (  )
U (  ) L



U (  )  k 2  2   0 PL (  )   L P0 (  )   kL  P0 (  )  PL (  )  +L2 P0 (  ) PL (  ) /  2
V (  )  k 2 L  L3P0 (  ) PL (  ) /  2  2k ( L   0 )
P0 (  )   0  2 / L2
PL (  )   L  2 / L2
16
2
t
It is interesting to note that if all three initial temperatures are the same, T0  TS  TL  V , we have
equilibrium and there is no gradient to change the distribution. Consequently, we get the known result
T ( x ', t )  TE  V
(GX.6.16)
7. References
[1] Cole, KD, Beck, JV, et. al. (2010), Heat Conduction Using Green’s Functions, 2nd Ed., CRC Press
Boca Raton, 643pp
[2] Amos, DE (2014) Theory of Heat Conduction with a Type 5 Boundary Condition,
http://nanohub.org/resources/20365
[3] Amos, DE, (2014) Heat Conduction in a Slab X55T0 and Sub-cases,
http://nanohub.org/resources/20381
APPENDIX
Algebra for X55 Green’s Function for x > x'
The object here is to do the algebra for the x  x ' case and compare it to the results obtained in
(GX2.33) and (GX2.40) with the mapping of x '  x  L onto the region 0  x  x ' with y=L-x . The
source term for x  x ' requires x  x '  ( x  x ') . Then, for x  x ' , (GX2.14) - (GX2.17) become
(GX2.14A)
(GX2.15A)
Dˆ  q( x  x ') ( P0  kq) ( PL  kq) q( L  x  x ') ( P0  kq) ( PL  kq)  q( L  x  x ')
e

e

e
2 q
2 q
2
2 q
2
U
V
W
( DR0  BRL )sinh qx  eq( L  x ' x )  e q( L  x ' x )  e q( L  x ' x )
2
2
2

U q( L  x ' x ) V  q( L  x ' x ) W  q( L  x ' x )
e
 e
 e
2
2
2
( ARL  CR0 )cosh qx  
U q( L  x ' x ) V  q( L  x ' x ) W1  q( L  x ' x )
e
 e

e
2
2
2

U q( L  x ' x ) V  q( L  x ' x ) W1  q( L  x ' x )
e
 e

e
2
2
2
(GX2.16A)
U
( P0  kq) ( PL  kq)
( P  kq) ( PL  kq)
( P  kq)
( P  kq )
, V 0
, W   P0 L
, W1 =  kq L
2 q
2
2 q
2
2 q
2 q
(GX2.17A)
( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
(W  W1 )  q( L  x ' x ) (W1  W )  q( L  x ' x )
e

e
2
2
Now we add the source term (GX2.14A) to this sum
Ve q( L  x ' x )  Ueq( L x ' x ) 
17
Dˆ  q( x  x ')
e
 ( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
2 q

( P0  kq) ( PL  kq) q( L  x  x ') ( P0  kq) ( PL  kq)  q( L  x  x ')
e

e
2 q
2
2 q
2
 Ueq( L  x ' x )  Ve q( L  x '  x ) 
(GX2.18A)

( P0  kq) ( PL  kq) q( L  x  x ') ( P0  kq) ( PL  kq) q( L  x '  x )
e

e
2 q
2
2 q
2
+
where
( P0  kq) ( PL  kq)  q( L  x '  x ) ( P0  kq) ( PL  kq)  q( L  x '  x )
e

e
2 q
2
2 q
2
W1  W ( P0  kq) ( PL  kq)

,
2
2 q
2
(GX2.19A)
(W  W1 )  q( L  x '  x ) (W1  W )  q( L  x '  x )
e

e
2
2
W1  W
( P  kq) ( PL  kq)
 0
2
2 q
2
 W1  W ( P0  kq) ( PL  kq)   q( L  x  x ')

0
 2
e
2 q
2


Finally, the analog to (GX2.20) is
Dˆ  q( x  x ')
e
 ( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
2 q

(GX2.20A)
( P0  kq) ( PL  kq) q( L  x  x ') ( P0  kq) ( PL  kq) q( L  x ' x )
e

e
2 q
2
2 q
2
( P0  kq) ( PL  kq)  q( L  x ' x ) ( P0  kq) ( PL  kq)  q( L  x ' x )
e

e
2 q
2
2 q
2
Now, if we take the expression for x<x' (GX2.20),
1  q( x '  x ) ˆ
e
D  ( DR0  BRL )sinh qx  ( ARL  CR0 )cosh qx 
2 q
+
(GX2.20)

( P0  kq) ( PL  kq) q( L  x ' x ) ( P0  kq) ( PL  kq)  q( L  x ' x )
e

e
2 q
2
2 q
2
( P0  kq) ( PL  kq) q( L  x ' x ) ( P0  kq) ( PL  kq)  q( L  x ' x )
e

e
2 q
2
2 q
2
exchange the subscripts, replace x' with L-x' and x with L-x , we get (GX2.20A). That is, exchanging
the subscripts with the mapping y  L  x gets the expression for x>x' . This change carries over to the
inversion shown in (GX2.33) and (GX2.40).

18