TRANSIENT HEAT CONDUCTION IN ADJACENT MATERIALS
HEATED ON PART OF THE COMMON BOUNDARY
Donald E. Amos
Albuquerque, NM 87110
Abstract This paper considers a classical linear, transient heat conduction problem set in Regions 1 and
2 defined by the half planes x>0 and x<0. These regions are in perfect contact along the boundary x=0,
-   y   , are initially at zero temperature and have distinct, constant physical properties. The energy
source is an infinitely thin, semi-infinite uniform source of heat along the lower half of the boundary
x  0, -   y  0 between the two materials. In a three dimensional setting we have a uniform halfplane heat source sandwiched between two semi-infinite slabs.
The solution for temperatures in each half plane is obtained by a classical application of the Laplace
transform. The results are line integrals in the complex plane in terms of iterated co-error functions of a
complex variable. Quadratures are required for numerical evaluation and computation of error functions
of a complex variable are discussed. Temperatures along the boundary are of special interest, but
because of slow convergence near the origin, special formulae which accelerate the convergence and
display the asymptotic behavior for y to zero are developed.
A solution for a uniform source on a finite segment of the common boundary can be constructed from
the semi-infinite source solution. By translating the endpoints of these source segments and using
superposition, some interesting solutions for step source patterns distributed along the boundary x=0 are
possible.
The special case where corresponding physical properties are the same in both half planes is known.
Numerical experiments compare the general solution for the semi-infinite source with this special case.
Keywords Two regions
Heat Conduction
Unsteady State
Laplace Transform
1. Introduction
The main problem is to find the transient temperature distribution in Region 1and Region 2 of the (x,y)
plane defined by x>0 and x<0 respectively. These half-plane regions have distinct, constant physical
properties, zero initial temperatures and perfect contact along their common boundary, x=0. An
infinitely thin, uniform heat source is sandwiched between the half planes along the lower half of the
common boundary x  0, -   y  0 . In a three dimensional setting, we are looking at adjacent, semiinfinite slabs sharing a common, uniform heat source on half of their boundary with no variation in the y
or z directions. Figure 1 shows a typical cross section of the slab geometry. In the past, there was high
motivation to only explore solutions which lead to computation in real arithmetic because only real
algorithms were available to compute the special functions and paths in the complex plane were used
only for theoretical purposes. The emphasis here is on a much neglected mathematical approach to a
class of problems where we do not avoid paths in the complex plane, but take advantage of their
desirable properties. More recently numerical evaluation of a number of special functions of a complex
variable has shown up in the literature and computational libraries, which will mitigate the need for only
real arithmetic. As always, these classical problems serve as benchmarks in code verification work to
assess accuracies of more complex numeric codes and may have direct physical applications. For code
verification work it is extremely helpful to have accurate tabulated results.
1
The mathematical problem can be stated as follows:
(1.1)
 2T1  2T1 1 T1


,
x 2 y 2 1 t
(1.2)
 2T2  2T2
1 T2
 2 
,    x  0, -  y  , t  0
2
 2 t
x
y
(1.3)
(1.4)
(1.5)
(1.6)
 k1
0  x  , -  y  , t  0
q
T1 
T
(0 , y, t )  k2 2 (0 , y, t )   0
x
x
0
  y  0
0 y
T1 (0 , y, t )  T2 (0 , y, t ),
 y 
T j (, y, t ) is finite, T j (, y, t ) is finite for j  1, 2
T j ( x, , t ) is finite, T j (, y, t ) is finite for j  1, 2
T j ( x, y,0)  0, j  1, 2
q0 is the constant source strength per unit area on the lower half of the plane x  0 , -   y  0 ,
  z   with normal flux vectors residing in the (x,y) plane. Equations (1.1) and (1.2) are the
diffusion equations for the half-planes and equations (1.3) and (1.4) are the boundary conditions along
the boundary between regions. Equation (1.3) is an energy balance stating that the heat generated along
the lower half of the boundary   y  0 flows out to the adjacent regions and that the fluxes across
the upper half of the boundary 0  y   are continuous. Equation (1.4) requires the temperatures all
along the common boundary to be equal. Equations (1.5) reflect physical realities, and equations (1.6)
define the initial temperatures in both regions to be zero.
In Section 2 the problem stated in equations (1.1) through (1.6) is solved in terms of contour integrals in
the complex plane. The results of Section 3 further refine the solution in terms of real quadratures and
compute special formulae for the boundary temperatures along x=0. Section 4 displays the known
solution for the special case where the corresponding half plane properties are equal and there is
symmetry about x=0. In Section 5, the solution of Section 2 is manipulated to produce the solution for a
source on a segment of finite length (on x=0). This solution is further manipulated to derive solutions for
patterns of step sources along the boundary x=0. Section 6 describes the numerical computation and
compares the results with the formulas for the equal parameter case of Section 4. APPENDIX A
provides the details for verifying the solution presented in Section 2. Details associated with the
computations in Section 4 are discussed in APPENDIX B and the symmetry associated with the finite
segment source described in Section 5 is displayed in APPENDIX C.
2
2. Solution for a Semi-infinite Source Along x=0, -<y<0
We start the solution by applying the Laplace transform to all equations. In particular,
LT j ( x, y , t )  T j ( x, y , p )
(2.1)
2
2T j  Tj
p

 T j , j  1,2
2
2
j
x
y
and follow the procedure outlined in [2]. The separation of variables in (2.2) leads to the solutions
T j  X j ( x )Y j ( y )
(2.2)
Xj e
(2.3)
 x  2 q j 2
, Y j  e  i y
p /  j , j  1,2 . Now we form the solution by summing
where  is the separation parameter and q j 
on the parameter 

T j ( x, y , p ) 
(2.4)

Aj ( )e
 x q2j   2
eiy d  , q j 
p / j
j  1,2 .

The symbolism



is interpreted as a complex Fourier transform [5, Chapter 4] where the path of
integration is taken in a strip of analyticity which may include the real axis. The exact nature is made
clear in (2.6).
The Laplace transform of
q
T
T
 k1 1 (0 , y, t )  k2 2 (0 , y, t )   0
x
x
0
 y 0
0 y
is
 k1
q
T1 
T
q( y )
(0 , y, p )  k2 2 (0 , y, p ) 
, q( y )   0
x
x
p
0
 y 0
0 y
and

(2.5)
 A( ) k
1

q12   2  k2 q22   2  eiy d  

q( y )
p
.
Then, with the inverse Fourier transform, we have

0
1
q0
 i y
i y
A( )  k1 q12   2  k 2 q22   2  
q
(
y
)
e
dy


 e dy

 2 p 
2

p
(2.6)
.
q0

,
Im( )  0
2 p( i )
Notice that the integral evaluation on y requires that  have a positive imaginary part in order for the
integral to exist. This means that the path of integration in (2.4) and (2.5) is not on the real  axis, but on
a line above the axis. Consequently this inversion and the original sums on  should be interpreted as
complex, two-sided Fourier transforms [5, Chapter 4] where the path of integration  is defined to be a
straight line parallel to the real axis in a strip of analyticity with Im(  )  0 ,
 (  )
T j ( x, y , p ) 
q0
1
e j
d ,
2 p  k1 q12   2  k2 q22   2 ( i )
 j ( )  x q    iy , q j  p /  j ,
2
j
2
3
j  1,2
.
We have the Laplace transform of the solution. At this point, the classical approach would be to apply
the complex inversion integral
  i
1
1
L T j ( x, y , p ) 
e pt T j ( x, y, p )dp,
 >0
2 i  i


to obtain a solution. This means identifying poles and branch points, selecting branch cuts and applying
Cauchy’s theorem to close the contour around the singularities. These cuts are generally taken in the left
half p-plane, Re(p)<0, so that pt gives exponential convergence for t>0. If the cuts are chosen properly,
the arcs connecting the inversion path   ip, -<p< to the extended contour vanish and only the
residues at the poles and the contributions from the two sides of the each branch cut remain. In the
classical cases, the contours on each side of a cut are conjugates whose difference yields a purely
imaginary result. Division by 2 i makes the result real. Examples of this procedure are contained
throughout [6], especially in Chapter XI and Appendix I. In the case at hand, the inversion would
involve the integration parameter  , a complication one would rather not have. In the developments to
follow, we get a direct inversion from tables, which limits the solution to one integral, but requires
integration in the complex plane with complex arguments to obtain the solution. We do not avoid
complex contours because the functions involved can be computed in the complex plane from accurate,
numerically stable algorithms. Further, the integrals for T j ( x, y, t ) converge exponentially over a large
portion of the (x,y) plane.
With p chosen to be real and positive, we rescale  with  
get our candidate for the Laplace transform
T j ( x, y , p ) 
(2.7)
q0
2
k
C
1
p 
p (u  i ) , -  u   ,   0 to
 (  ) p
1
12   2  k2
1 e j
d,
3/ 2
 22   2 ( i  ) p
 j (  )  x  2j   2  iy  ,  j  1/  j ,
j  1,2
where C is the path C :   u  i , -   u  ,   0 . This integrand has a pole at   0 and
branch points at   i 1 , i 2 with cuts on the imaginary axis extending from each branch point to
i . Since we do not know which parameter 1 or  2 is the smaller or larger, we work with
m  min  1, 2  and  M  max  1, 2  . The presence of the pole at   0 divides the strip between
the branch points into two strips of analyticity with imaginary parts 0  Im(  )   m and
 m  Im(  )  0 . Consequently, in order to keep the path of integration in the strip of analyticity
dictated by (2.6), C is further restricted to the path between the pole at   0 and the nearest branch
point,   i  m (See Figure2). With this change of variable from  to  , we have separated out the
terms containing p making the inversion of the Laplace transform possible from tables
 (  ) p 

 j ( )
e j

L1 
(2.8)
  2 t ierfc(v j ) , v j 
3/ 2
2 t
 p



and the final candidate for the solution becomes
q t
T j ( x, y , t )  0
f j (  )d  ,
(2.9)
 
C
f j ( ) 
ierfc( v j )
k1     k2
2
1
2
1
, vj  ( x
   (i  )
2
2
2
4
 2j   2  iy  ) /(2 t ),  j  1/  j, j  1,2 .
The path C and its relation to m and  M is shown in Figure 2.
In the remainder of this section, we analyze the integral along C and show that for all x  0 and
y  0 the solution can be expressed in terms of contours labeled C ( / 4) and C ( / 4) in Figures 3 and
5. However if we break the problem down into the cases
y>0 with x  y >0 and y > x >0 (Figures 2 and 3)
y<0 with x  y >0 and y > x >0 (Figures 4 and 5)
,
we can get better rates of convergence by modifying the contours from C ( / 4) and C ( / 4) to
C ( ) and C ( ) as shown in Figures 2-5. If the derivation or logic is not of interest, one can skip to
(2.20) for the final results.
To assess the convergence of the integral for u   , we use the asymptotic form [1, 7.2.14]
2

 2e  z
n
i erfc( z )  O 
for z  , n  0 , arg z  3 / 4
(2.10)
  (2 z )n 1 


with z  v j , n=1 ,    u  i to get
 e Re( v2j ) 
 for v j  , arg v j  3 / 4
ierfc( v j )  O 
(2.11)
2  v 2
j


2
 ierfc(v j ) 
 e Re( v j ) 
 =O 
f j ( )  O 
 ,  
2
 4 







2
and compute Re(v j ) ,
 j 2   x 2 (  2j   2 )  y 2  2  2i x y   2j   2  (4t )
(2.12)

Re j 2 



 2j x  ( x 2  y 2 )(u 2   2 )  4 x y u  (4t ),
2
u  
.
Thus, for x  y , we have Re(v 2j )  0 and arg( v j )   / 4 . Thus we get exponential convergence for
u   , regardless of the sign of y. For x  y and y positive, we still get exponential convergence.
We follow the case x  y with y positive and defer the analysis for the cases x  y until later.
5
The integrand here is not very much different from that in [2] in that it is the iterated co-error function
which will govern the convergence of these integrals for x and y not zero. One can see from this analysis
that this classical procedure produces only the solution in the wedges, x  y  0 , closest to the x-axis as
it did in [2]. However, following [2] the idea is to deform the path of integration (using Cauchy’s
theorem) into a region of the  plane where the integrals not only converge faster, but give the analytic
continuation of the parameters into the remainder of the (x,y) plane. In [2] the path of integration C is Vshaped with the vertex at the origin formed from the lines   rei (  ) and   rei . In this treatment,
we need a modified path C, not only to avoid the pole at the origin, but to stay in a region of analyticity
above the line C . To be specific, we choose C to be a tangentially smooth parabola connecting the
straight lines   rei (  ) and   rei . The result is a parabolic spline C whose vertex i  is (defined to
be) half way between the pole at   0 and the first branch point   i  m . Then C, also called C ( ) , is
defined by C :   B(u) where
2
2
   m / 2  min( 1 ,  2 ) / 2
u tan  /(4  )    0  u  2  / tan 
B (u )  u  i 
,
(2.13)

u  2  / tan 
  u  
u tan 


We also further restrict  by 0     to make C pass below the vertex of C. C ( ) is continuous with
a continuous first derivative, symmetric about u=0 with a value of B (0)  i  , and has the form of
  rei (  ) and   rei for u  u  2 / tan  . Figure 2 shows the geometry and the paths of
integration.
Now the analysis of [2] for the convergence along the path C applies completely for r   or
u  r cos    . The goal is to apply Cauchy’s theorem to obtain more rapidly convergent integrals. In
order to apply Cauchy’s theorem, we need to close the open region between C and C ( ) with first and
second quadrant arcs CR1 :   Rei , sin -1 ( / R)     and CR2 :   Rei (  ) , sin -1 ( / R)    
giving the closed contour C  CR1  C  CR2 and take the limit R   (See Figure 2). To assess the
magnitude of these line integrals on CR1 and CR2 as a function of R, we examine

i
  f j (  )   Rei i Re d and
(2.14)


  f (  ) 
j
 Rei (   )
( i ) Rei (  )d
 e Re( v j ) 
f j (  ) =O 
 ,   .
 4 


2
  sin ( / R) ,
1
Now we turn our attention to  j   j (  ) /(2 t ) and establish the behavior of e
 Re( 2j )
for R   in
(2.14). We first look at the arc C in the first quadrant. On C with   Rei we have
1
R
1
R
 j 2   x 2 (  2j   2 )  y 2  2  2i x y   2j   2  (4t )
(2.15)

Re j 2 


2
j

2
j

x  ( x 2  y 2 )  2i x y   2
2
x  ( x 2  y 2 )cos 2  2 x y sin 2  R2 (4t ),
2
 (4t )
R
For the second quadrant arc, CR2 , the values of  are   Rei (  ) corresponding to   Rei in the first
quadrant. In Section 3, it is shown that the values of v j (  ) for  in the left half plane are conjugates of
the values of v j (  ) for the corresponding values of  in the right half plane. Therefore, the squares are
conjugates of one another and both have the same real parts. Consequently, (2.15) applies on CR2 also.
6
Now, if x  y  0 , ( x 2  y 2 )  0 in (2.15) and we want cos2  0 to keep the term from becoming
negative. This implies 0       / 4 where cos2  0 and sin2  0 . This makes Re j 2   0 .
Taking    / 4 in cos2 and    in sin 2  2sin  cos  makes the exponential larger and
f j (  )  O(e
4 x y R cos 
/ R 4 ) at worst for y>0 and this establishes that the integrals in (2.14) with
4 x y R cos 
/ R3 ) along CR1 and CR2 vanish for R   . Therefore, the
integrands bounded by O( e
difference of integrals on C  C vanish and the C ( ) contour is equivalent to the C contour with the
restriction that x  y  0 and 0     / 4 . In complex variables, a change in the path of integration
often leads to an analytic continuation of the parameters into nearby regions. In this case we get an
extension to the case where x  y  0 .
If y>|x|, the term ( x 2  y 2 ) in (2.15) is negative and selecting  / 4       / 2 puts 2 in the second
quadrant where cos2  0 and sin2  0 and again Re{ 2j }  0 . But in order to apply Cauchy’s theorem
directly to the case where    / 4 , we would have to make the path   Rei connecting C and C ( )
go through the region sin 1      / 4 where ( x 2  y 2 )cos2 is negative for y>|x| and Re  j 2  could
be negative. This gives exponential divergence for R   .
There is a way out of this dilemma by noticing that if    / 4 , cos 2  0 , and the term ( x 2  y 2 )cos2
is zero and there are no restrictions on either |x| or y, other than they both be positive. We could stop
here and say that we have the complete solution for y>0, but we want to take advantage of the higher
rate of convergence offered by the path where    / 4 and the term ( x2  y 2 )cos2  0 contributes to
the exponential convergence. The way is to apply Cauchy’s theorem to the sector composed of C ( / 4) ,
C ( ) and the arc CR1 :   Rei ,  / 4       / 2 in the first quadrant and the arc
C
2
R
:   Rei (  ) ,  / 4       / 2 in the second quadrant with R   (See Figure 3), taking
advantage of the unrestricted case    / 4 . But the observation made in the first sentence of the
previous paragraph establishes what we need, Re{ 2j }  0 . This establishes that C ( / 4) and C ( ) are
equivalent with the restriction y  x  0 since each of the integrals on CR1 and CR2 in Figure 3 goes to
zero as R   .
As in [2], we choose  away from its extremes, so that  is well into a region of analyticity of f j (  ) ,
(2.16)
 / 6 x  y  0

 / 3 0  x  y
to completely define the paths of integration. Then, for y>0
1
T j ( x, y, t ) /( q0 t )   f j (  )d  , y  0 .
(2.17)

C
Equation (2.17) completes the analysis for y>0.
For y<0 and |x|>|y|, we close the C contour into the lower half plane (See Figure 4) to get convergence
of the integrals. We opt for symmetry and choose the conjugate contours in the lower half plane which
7
includes C ( ) and the arcs CR3 :   Rei , -    0 and CR4 :   Rei , -       which are
in the fourth and third quadrants respectively. C ( ) is indeed the conjugate of C ( ) which is defined in
(2.13). But we still need to close the ends where the contour crosses the strip 0  Im(  )   .
We choose the vertical straight lines    R  iv , 0  v   and denote these by
V
1
 VR1 :   R  iv , 0  v    and V2  VR2 :    R  iv , 0  v    . The values of  for the
integrals in the quadrants 3 and 4 are conjugates like they were in the upper half plane, but for y<0 and
x  y the estimate (2.15), Re{ 2j }  0 , remains the same because y is replace by  y and  is
replaced by  . The conclusion is that the integrals along arcs CR4 and CR3 vanish as R   . That
leaves us with the integrals over the vertical contours VR1 and VR2 . We analyze the integral in the first
quadrant on VR1
0
J ( R)   f j ( R  iv, y  0)idv .
(2.18)

Since the dominant behavior of f j (  ) is expressed by e
 v 2j
,
) 
e
f j (R  iv )  O 
,
 R4 


2
and we need to re-examine Re  j  in this context. Then with x  y ,   R  iv and y replaced by
 Re( v 2j
 y , (2.15) becomes
Re j 2 

2
j

x  ( x 2  y 2 )( R 2  v 2 )  4 x y Rv  (4t ),
2
R  , 0  v   .
It is clear that for x  y and R sufficiently large, the negative term involving y will not be dominant
and Re j 2   0 for R   . Then, J(R) goes to zero as R   . A similar analysis also holds for the
second quadrant contour VR2 where  2j (  ) is the conjugate of the corresponding function in the first
quadrant (See Section 3) and consequently has the same real part.
The results above establish that the integrals vanish along the arcs VR1 , CR3 , CR4 and VR2 for R   . The
complete contour, composed of C  VR1  CR3  C  CR4  VR2 , includes the pole at   0 and for
x  y the residue theorem with R   gives
8
T j ( x, y , t ) /( q0 t )  G j ( x, t ) 
(2.19)
G j ( x, t ) 
1

 f (  )d  ,
j
y0
C
2ierfc[ x /(2  j t )]
k1 / 1  k 2 /  2
where C is the conjugate of C .
For y<0 and y  x (Figure 5) the argument is the same as it was for y>0 (Figure 3), namely that
   / 4 only requires that |x| and |y| be positive in order to get exponential convergence along the path
C ( / 4) . Cauchy’s theorem is then applied to the contour C( ) - CR5 - C( / 4) - CR6 . As a consequence,
the integrals on the contours CR5 and CR6 , which form the closure of the sector, vanish making C ( )
equivalent to C( /4) .
The final result is
(2.20a)
T j ( x, y, t ) /( q0
G j ( x, t ) 
1
   f j (  )d 
 C
t)  
G j ( x, t )  1 f j (  )d 

 C
2ierfc[ x /(2  j t )]
k1 / 1  k2 /  2
vj  ( x
f j ( ) 
y0
y0
ierfc(v j )
k1     k2
2
1
2
 2j   2  iy  ) /(2 t ),  j  1/  j
1
   ( i  )
2
2
2
j  1,2
or in terms of the parameter u,
(2.20b)
T j ( x, y, t ) /(q0
 1 
   f j (  )  B ( u ) B '(u )du

t )   

G ( x , t )  1
  f j (  )  B (u ) B '(u)du
 j
 

y0
y0
where
2
2
  min( 1 ,  2 ) / 2
u tan  /(4  )    0  u  2  / tan 
B (u )  u  i 
,

u  2  / tan 
  u  
u tan 


u tan 2  /(2  )  0  u  2  / tan 
B '(u )  1  i 
,

u  2  / tan 
 sgn(u ) tan  
 / 6 x  y  0
.

 / 3 0  x  y
Notice that the residue at the pole   0 produced the one-dimensional solution G j ( x, t ) which is what
one expects for y   where the solution is one dimensional. If the sources q0 had been placed on the
positive y-axis, this analytical process would have produced (2.9) with   0 and C would be in the
9
strip of analyticity defined by  m  Im(  )  0 . The results, as expected, are the same formulas with y
replaced by –y.
3. Reduction of Contour Integrals to Real Integrals
The solution above is written in terms of complex integrations. Since the solution is real, we suspect that
the integrations can be reduced to real arithmetic. This is indeed the case and it is easy to show with the
following manipulations. We carry out the computation on the case for y>0. First write

0

1
1
1
  f j (  ) B '(u)du    f j (  ) B '(u)du    f j (  ) B '(u)du .


  B(u )


  B(u )

  B(u )
0
Then, replacing u with –u for the integral on the negative u-axis, we get



1
1
1
B '( u )du    f j (  )
  f j (  ) B '(u)du    f j (  )


  B(u )


  B( u )
0
  B(u )
B '(u )du .
0
Now, we show that the first integrand on the right is the complex conjugate of the second and hence
their sum is twice the real part of the second integrand. First note that B ( u ) is a number in the second
quadrant corresponding to the number B (u ) in the first quadrant, and B '( u ) is the conjugate of B '(u ) .
Let  r  u  iv, u  0, v  0 be a number in the right half plane. Then, the corresponding  in the left
half plane is  l  u  iv  (u  iv ), u  0, v  0 . Now the relations follow easily,
 r2  u 2  v 2  2uvi,  l2  u 2  v 2  2uvi and
 l2   r2
i  r  v  ui,
 i  l  v  ui
and  i  l  i  r
i  r   v  ui,
i  l   v  ui
and
il  i r
From these, we can see (using the reflection principle) that (for y<0, y is replaced by  y and for y>0, y
is replaced by |y|)
2
v j (l )  ( x
 2j   r  (i  r ) y ) /(2 t )  ( x  2j   r 2  (i  r ) y ) /(2 t )  v j (  r ), y  0
v j (l )  ( x
 2j   r  (i  r ) y ) /(2 t )  ( x  2j   r 2  (i  r ) y ) /(2 t )  v j (  r ),
(3.1)
2
1
1

i  l  i  r
y0
and f j (  l )  f j (  r ) .
 
After noting that B '(-u )  B '(u ) , we can write the integrals for y>0 and y<0 as
2


 Re f ( B(u), y  0)B '(u)du
,
j
0
2


 Re f ( B(u), y  0)B '(u)du
j
.
0
Before we take the final step to write the real solution, we note one more simplification. We show that
f j (  , y  0)   f j (  , y  0) and, as a consequence,
(3.2a)
2




 Re f j ( B(u), y  0)B(u) du  
0
2


 Re f ( B(u), y  0)B(u)du
j
0
or
(3.2b)
1
1
f (  , y  0)d     f (  , y  0)d  .


j
C
j
C
10
To show this, we need to examine ierfc( v j ) and, in particular, v j :

vj  x

 2j   2  iy  /(2 t ),
 j  1/  j j  1,2 .
Now, using the reflection principle and elementary manipulations of conjugates, we have for  on C
with y<0 (y=-|y|),
2


v j (  , y  0)   x  2j    i y   /(2 t )   x  2j   2  (i y  )  /(2 t )




.

= x

 2j   2  i y  /(2 t )  v j (  , y  0)
With the reflection principle again,
ierfc(v j )  ierfc(v j )
and with ( i  )  (i  ) in the denominator of f j (  ) we get
f j (  , y  0)   f j (  , y  0) .
(3.3)
Now we have one complex number as the negative conjugate of another, w   z . Then
Re( w)   Re( z )   Re( z ) , and the integral result (3.2) follows.
Since we expect the solutions to be continuous on y=0, the formulas for y<0 and y>0 in each half plane
must be the same and we get
T j ( x,0, t ) /(q0 t )  G j ( x, t ) / 2 , j  1,2
(3.4)
which is half of the one dimensional solution for y   ,
T j ( x, , t ) /(q0 t )  G j ( x, t ) , j  1,2 .
(3.5)
The complete solution from (2.20) is now


2
y 0

0 Re f j ( x, y, B(u ))B(u )du



T j ( x, y, t ) /( q0 t )  
G j ( x, t ) / 2
y0
(3.6a)


G j ( x, t )  2 Re f j ( x, y , B(u )) B(u )du
y 0

 0
where
2ierfc[ x /(2  j t )]
ierfc(v j )
1
f j ( x, y ,  ) 
G j ( x, t ) 
k1 / 1  k2 /  2
k1 12   2  k2  22   2 ( i  )
vj  ( x
 2j   2  iy  ) /(2 t ),  j  1/  j
j  1,2
2
2
  min( 1 ,  2 ) / 2
u tan  /(4  )    0  u  2  / tan 
B (u )  u  i 
,

u  2  / tan 
  u  
u tan 


2u tan 2  /(4  )  0  u  2  / tan 
.
B '(u )  1  i 

u  2  / tan 
 sgn(u ) tan  
With a change of variables, u  w /  M , M  max(1,2 ) , we get the dimensionless form
11


2

0 Re f j ( x, y, B(w))B(w) dw



T j ( x, y , t ) /( q0  M t / k M )  
G j ( x) / 2
(3.6b)


G j ( x )  2 Re f j ( x, y , B ( w)) B ( w) dw

 0

where, with   M  1/ 2 ,



G j ( x) 
2ierfc  xC j 
f j ( x, y ,  ) 
C1K1  C2 K 2
v j  ( x C 2j   2  i y ) ,
Cj 

ierfc(v j )
K1 C    K 2
2
1
2
y 0
y0
y0
1
C   ( i )
2
2
2
x  x /(2  M t ) , y  y /(2  M t )
j  1,2
k
M
, K j  j ,  M  max(1 , 2 ) , k M  max(k1 , k2 ) , j  1,2
j
kM
( w 2 / 2) tan 2   1/ 2  0  w  1/ tan 
B( w)  w  i 
,

w  1/ tan 
  w  
w tan 


 w tan 2   0  w  1/ tan 
B '( w)  1  i 

w  1/ tan 
sgn( w) tan  
Actually, k M can be either k1 or k2 , but we chose to normalize on  M rather than  m  min(1, 2 ) so
that the smallest branch point would always be at   i and the minimum of the spline would be
halfway between singularities at B(0)  i / 2 .
Computation on the Boundary, x=0
In an application, the temperatures along the boundary x=0 would be of special interest. The formulae in
(3.6) are computationally good for x=0 and relatively large y. However, when x=0 and y is zero or
small, we lose the exponential decay from ierfc( v j ) which dominates when y is larger. In this case, the
integral is O (1/  ) with a truncation error at   R of only O (1/ R) . In this section we start with
(3.6b) for y>0 and accelerate the convergence of the integral by subtracting off the asymptotic form of
f j (0, y,  ) for    and adding it back on. The integrand formed from the subtraction accelerates the
2
convergence from a truncation error at   R of O (1/ R) to O(1/ R3 ) and in the process creates an addon integral which can be evaluated in closed form. This closed form gives the asymptotic behavior for
x=0 and y  0 and can be used in place of (3.6) in the neighborhood of the origin where the
convergence of the integrals is poor. For the case where y<0, the formula for y  0 in (3.6b) throws the
computation back to the case where y>0 because the y position in f j ( x, y,  ) is positive in both integrals.
We rewrite (3.6b) for y>0 ( y  0 ) in the form
(3.7)
T (0, y, t ) /( q0  M t / k M ) 
2

where
12
I1 ( y ) 
2



Re I 2 ( y )
u



I1 ( y )   Re f j (0, y, B(u )) B(u ) du
I 2 ( y )   f j (0, y, rei )ei dr
0
r0
u  1/ tan  , r0  1/ sin  ,    / 3 .
   / 3 comes from (2.16 ) because x=0<y. Now I1( y ) is the same as that in (3.6b) up to the spline knot
at u  u (or r  r0 ) and no new computation is needed (See Section 6). I 2 ( y ) is (3.6b) for
u  u expressed in terms of r instead of u (See Figure 3). We accelerate the convergence of I 2 ( y ) by
noting the asymptotic form of
ierfc( i y )
1
ierfc( i y )
(3.8)
f j (0, y,  ) 

,
  .
2
2
2
2 ( i )
( K1  K 2 )( i 2 )
K1 C1    K 2 C2  
Now we subtract this asymptotic form from f j (0, y,  ) . With a little algebra, we can manipulate this
difference into a suitable computational form which minimizes losses of significance in the subtraction:
K1C12
K 2C22



  C12   2   C22   2
1
1
H ( )  


.
 K1 C12   2  K 2 C22   2 ( K1  K 2 ) 
( K1  K 2 )  K1 C12   2  K 2 C22   2 


Thus we have
I 2 ( y )  I 21 ( y )  I 22 ( y )
where

 ierfc( i y )

I 21 ( y )   
H ( ) 
ei dr
(  i )
  rei
r0 
(3.9)

 ierfc( i y ) 
1
I 22 ( y ) 
ei dr , r0  1/ sin 


2

K1  K 2 r0 
i
  rei


For the integral Re I 21 ( y ) , we compute the quadrature as we do in (3.6)(See Section 6). For    ,

with y small, the integrand of I 21 ( y ) is O 1/ 
4
 with a truncation error at   R of O 1/ R  , a
3
considerable improvement over O 1/ R  without the acceleration.
For I 22 ( y ) we manipulate this into a form which is recognizable in terms of known functions from [3].
Then, (3.9) with the explicit parameterization becomes

1
ierfc( r yc )
I 22 ( y ) 
dr , r0  1/ sin  , c  -iei  e i / 6
(3.10)
2

( K1  K 2 )c r0
r
and with the recurrence relation
ierfc( z )   z erfc( z )  e z / 
2
we get


c
erfc( r yc )
1


(
yc
)
dr 
(3.11) I 22 ( y ) 


( K1  K 2 ) 
r


r0
2
e ( ycr ) 

i
 i / 6
r r 2 dr  , r0  1/ sin  , c  -ie  e .

0

A change of variables, w  ( r yc ) , in the first integral and r  r0 w in the second gives
13



c
1
 w0 erfc( w)
2 
 i / 6
(3.12) I 22 ( y ) 
dw

E
(
w
)
, w0  r0c y .

 , r0  1/ sin  , c  e
3
/
2
0

( K1  K 2 ) 
2r0 
 r0 w0 w


Now the first integral has been studied in [3], is denoted by G( w0 ) , and has a functional form
G( z )  F ( z )   / 2  ln(2 z ) ,   Euler constant=0.5772156649015328606...
(3.13)
where

G( z )  
z
z
erfc( w)
erf ( w)
dw , F ( z )  
dw .
w
w
0
In the range of interest we have here, (say z  4 ) we can compute F(z) for (3.13) by its power series,
(1)k
z 2k

 k  0 k ! (2k  1)2
and for z  4 we can compute G ( z ) from its asymptotic expansion
F ( z) 
(3.14)
G( z ) 
(3.15)
1
2z 
N
 (1)
k 0
k
2z

(1/ 2)k
(1/ 2)
Ek  3 / 2 ( z 2 )  RN , RN  2( N N1)1 E N  5 / 2 ( z 2 ) , z   .
2k
z
z
Finally,
I 22 ( y ) 
(3.16)
c
 w0G( w0 )  ierfc( w0 ) , r0  1/ sin  , c  e i / 6 , w0  r0c y
( K1  K 2 )r0
 1 N

c
(1/ 2)k
( 1)k
Ek  3 / 2 ( w02 )  , w0  


2k
( K1  K 2 )r0  2  k 1
w0

The final computational form is
2
2
2
(3.17) T (0, y, t ) /( q0  M t / k M )  I1 ( y )  Re I 21 ( y )  Re I 22 ( y ) , y  y /(2  M t ), y  0







where Re I 21 ( y ) is computed by quadrature on the real part of the integrand in (3.9) and Re I 22 ( y ) is




computed from (3.16) for w0  4 and w0  4 .
While (3.17) gives a way to compute the temperatures accurately for small y, a further analysis of (3.17)
for small y can be done. We know that for y=0, (3.6b) gives
1
T (0,0, t ) /(q0  M t / k M )  G j (0) / 2 
 (C1K1  C2 K 2 )
and any perturbation from this must come from the y terms of (3.17). The dominant term for small y is
( y /(2  M t )ln y /(2  M t /( K1  K 2 ) from the logarithmic part of cw0G(w0 ) /  r0 ( K1  K2 ) in (3.16)


and G( w0 ) from (3.13). Therefore, incorporating the coefficient (2 /  ) from Re I 22 ( y ) in (3.17) and
taking only the dominant term, we get
(3.18)
T (0, y, t ) / q0
t
(1/  )

 y ln y  , y  0 .
 (k1 / 1  k2 /  2 ) (k1  k2 )
The logarithmic behavior, y ln y , shows that the slope is infinite for y  0 . For equal parameters,
(3.18) is, to first order terms, the same as (4.5). Computationally, a combination of (3.17) or (3.18) for
small y and (3.6b) for larger y covers the region of interest on x=0 (See TABLE 2).
14
There is a technical detail which should be addressed. In [3], the formula (3.13) was derived for real z.
In writing down (3.13) we have invoked a theorem in complex variables called the permanence of
functional forms which, applied to the real line, states that:
A formula, which applies on a segment of the real axis and is computed from functions that are analytic
in a region that includes this segment, also applies to all analytic continuations (i.e. In simple terms, the
same formula holds for complex z).
This principle has also been invoked when we changed variables in the integrals defining G( w0 ) and
E3 / 2 ( w02 ) by taking c to be real as well as the change of variables in going from (2.6) to (2.7) with p real
and positive.
4. Special Case for Equal Parameters in the Left and Right Half Planes
In this case,
1   2   ,
k1  k2  k ,
and the problem reduces to a single region with a line of sources along the negative y-axis. This problem
is also symmetric about the y-axis yielding a half plane problem for x>0 with the flux boundary
conditions
y0
 0
T
(4.1)
.
Q  k

x x  0 q0 / 2 y  0
The solution of this half plane problem is given in [2, Eq. (4.7)] for the upper half plane. In this context
(rotation of the half-plane by 900 ), this solution can be written as
 1  c x
y
I1 (
,
,T ) ,
T  1/ t ,
y 0

2  2 
 2k 
  t
T ( x, y, t ) / q0  
ierfc( x /[2  t ])
y0
(4.2a)
 2k
 t
y
1  c x
ierfc( x /[2  t ]) 
I1 (
,
,T ) , y  0

2k 
2  2 
 k
or, in terms of dimensionless parameters,
y
 1 c x
I1 (
,
,1)

2

2

t
2

t

1
T ( x, y, t ) /( q0  t / k )   ierfc( x /[2  t ])
(4.2b)
2

y
1 c x
I1 (
,
,1)
ierfc( x /[2  t ]) 
2 
2 t 2 t

y0
y0 .
y0
where the I1c function is defined by

e a w
(4.3)
I (a, b,T )   2 erfc  bw dw , a  0 , b  0 , T  0 .
w
T
and its properties are given in [3, Chapters 2 & 3, Folder 10b]. Notice that for y=0, T(x,0,t) is one half
that for y   because in (4.2a),
2
2
c
1
15

e x w /[4 ]
lim I (
,
, T )  lim 
erfc y w / 4 dw  0 .
y 
y 
w2
2  2 
T
c
1
(4.4)
x
y
2
2


The cases for the temperature along the boundary x=0 are of special interest. For x=0 and y>0

y
1  c
1  1
T (0, y, t ) / q0 
I1 (0,
,T ) 
erfc y w / 4 dw
2k 
2k  T w2
2 
and integration by parts gives
y

1 t 
T (0, y, t ) / q0 
E1 ( y 2 /[4 t ])  , y  0 .
 erfc( y /[2  t ]) 
2k  
2 t 



For x=0 and y<0,
y
t
1  c

I1 (0,
,T ) , y  0
k  2k 
2 
T (0, y, t ) / q0 
y

t
1 t 

E1 ( y 2 /[4 t ]) 
 erfc( y /[2  t ]) 
k  2k  
2 t 

=
y

t 
E1 ( y 2 /[4 t ]) 
1  erf y /[2  t ] 
2k  
2 t 

This last form is best for computation for y<0 because all terms are positive and cannot suffer losses of
significance due to subtraction of nearly equal numbers. The final form is

=
(4.5)
 1

2 
 1
T (0, y, t ) / q0  t / k  
2 
 1

 2 



y


E1 ( y 2 /[4 t ])  ,
 erfc( y /[2  t ]) 
2 t 


y 0
,
y0
y


E1 ( y 2 /[4 t ])  , y  0
1  erf y /[2  t ] 
2 t 




5. Solution for a Segment Source Along x=0, a<y<b
We start with the solution for the problem of a uniform source on ( x  0,    y  0) sandwiched
between two half planes by restating (3.6) in the form
(5.1)
 T j 1 ( x, y , t )
T j ( x, y , t )  
T j 2 ( x, y, t )
y 0
y0
where
T j1 ( x, y , t ) 
2


 Re f ( B(u))B '(u)du ,
j
y 0
0
T j 2 ( x, y , t )  G j ( x , t )  T j1 ( x , y , t )
16
y0
.
Now we translate the origin to the points (0,b) and (0,a), a<b, for two new solutions where the heat
sources are x  0,   y  b and x  0,   y  a :
T j1 ( x, y  b, t ) applies for y  b
T j 2 ( x, y  b, t ) applies for y  b
(5.2)
T j1 ( x, y  a, t ) applies for y  a
T j 2 ( x, y  a, t ) applies for y  a .
Notice that the flux boundary condition is satisfied by both solutions for y<a where the arguments y-a
and y-b are negative. Therefore the difference T j 2 ( x, y  b, t )  T j 2 ( x, y  a, t ) , when applied to the flux
equation, is equal to q0  q0 =0, y<a. This observation also applies to the difference
T j1 ( x, y  b, t )  T j1 ( x, y  a, t ) for y>b where the arguments y-b and y-a are positive and the flux
condition gives 0-0=0. For the interval a<y<b, we have the difference T j 2 ( x, y  b, t )  T j1 ( x, y  a, t )
where the first term with a negative value for y-b satisfies the flux condition equal to q0 and the second
term with a positive value for y-a satisfies the flux condition equal to zero giving q0  0  q0 . The
temperature boundary condition T1 (0, y, t )  T2 (0, y, t ) =0 along x=0 is satisfied by both solutions and
hence the differences.
Then, the construction
(5.3)
 T j1 ( x, y  b, t )  T j1 ( x, y  a, t )

T j ( x, y, t , a, b)   T j 2 ( x, y  b, t )  T j1 ( x, y  a, t )
 T ( x, y  b, t )  T ( x, y  a, t )
j2
 j2
y b
a  y b
ya
,
j  1,2
gives the complete solution for the temperatures in the plane with a heat source of strength q0 on x=0,
a<y<b, sandwiched between materials of different properties. The symmetry about y=(a+b)/2 is
displayed in APPENDIX C.
As a final note, some interesting source patterns with different source strengths along x=0 can be
constructed by superposition of solutions with disjoint or overlapping sources. The components of these
superpositions are obtained from (5.3) or (5.1). For example, we can construct a temperature distribution
for a segment of sources on x=0 and c<y<d with a source strength q1 by
 T j1 ( x , y  d , t )  T j1 ( x , y  c, t )
yd

T j ( x, y, t , c, d , q1 )   T j 2 ( x, y  d , t )  T j1 ( x, y  c, t )
c yd
(5.9)
 T ( x , y  d , t )  T ( x , y  c, t )
yc
j2
 j2
where the dependence on the source strength q1 is also shown. The expression for the source q0 on the
interval a<y<b would be written as T j ( x, y, t , a, b, q0 ) and the superposition would yield several results
depending on the relation of the source segments. For example, if c<d<a<b or c<a<d<b or a<c<d<b, the
source intervals are disjoint in the first case, overlap in the second case and the source interval for (c,d)
is a subset of the interval (a,b) in the third case. Superposition gives the solutions for a source on x=0
with strengths
17
0
yb
0
yb
0
yb
q0
a yb
q0
d  yb
q0
d  yb
0
q1
d  ya
c yd
q0  q1
q1
a yd
c ya
q0  q1
q0
c yd.
a yc
0
yc
0
yc
0
ya
6. Numerical Experiments
In order to compute the general solution (3.6), we have to compute the integrand inside of a quadrature
using complex arithmetic and return only the real part to the integrator. The computational procedure for
the quadrature and iterated co-error function is the same as that in [2]. Briefly, the whole computation
2
relies on the subroutine ZERF [3] which computes erf ( z ) , erfc( z ) or e z erfc( z ) for complex z in double
precision arithmetic. On most computers unit round-off in double precision is O(1015 ) , but in
computation, some precision is always lost and the accuracy returned in ZERF is O(1013 ) . In order to
minimize losses of significance, the function ierfc(z) is computed from the relations

2
e z 
1  z  e z erfc( z ) 
 ierfc( z ) 





2
(3 / 2) m
1 e z 

(1) m
2 

z 2m
2  z m 0

ierfc( z )  2( z )  ierfc(  z )
2
(6.1)
Re( z )  0
Re ( z )  0
z 7
z 7
where (-z) is in the right half plane and ierfc(-z) is computed from the first expression. The quadrature is
carried out in subroutine DQUAD8 [3, Folder 21] which, at its most basic level, implements the Gauss
8-point formula in an adaptive fashion. In the mode used here, one specifies a stepping interval and
quadratures over this interval length are summed until the last quadrature, which approximates the
truncation error, satisfies a relative error test which was set at O(1012 ) .
The main numerical experiments consisted of comparing the general solution (3.6) with the special case
(4.2) where the parameters are the same for both half planes and the solution is known. Some results are
shown in TABLE 1. Most all of the relative errors shown are O(1012 ) or better. The entries for
x  0.00D  00 and y  .10 D  04 reflect the loss of exponential convergence when both x and y are
extremely small with an integrand closer to O ( 
2
) and a truncation error at   R of O( R1 ) . The
maximum value set in the integration routine was R=100, giving a truncation error O(102 ) which is
consistent with the results of TABLE 1. The results for x=0 and y=0 are highly accurate because of the
special formulae used in (3.6) and (4.2) for y=0. In general, the numerical experiments on (3.6) indicated
that losses of significance occur as the values of T(x,y,t) approach unit round-off (large x or large y), not
values of great interest.
Temperatures along the boundary x =0, especially near the origin where the heat source starts, are of
particular interest. It was noted in the previous paragraph that the convergence of (3.6) was slow near
the origin, but the analysis leading to (3.17) and the simple approximation in (3.18) were derived to over
come this difficulty, at least for x=0. The accuracy of (3.17) near y=0 however is limited by a truncation
error of O(1/ R3 ) and with R=100 we get O(106 ) since the exponential decay is practically absent.
TABLE 2 shows a comparison of (3.6), (3.17), and (3.18) along the boundary for x=0 and y>0 (the
18
values for y<0 are obtained by subtracting these values from G j (0, t ) according to (3.6)). The relative
errors (REL ERR) in columns 4 and 6 are computed by means of
(3.17)-(3.6)
(3.17)-(3.18)
.
REL ERR=
or REL ERR=
(3.17)
(3.17)
This table shows that, overall, (3.17) gives the best results for all values with an overall accuracy of
O(106 ) . From the discussion of TABLE 1 above, we expect the agreement between (3.6) and (3.17) to
be only O(102 ) near y=0 with increasing accuracy of (3.6) and (3.17) as y increases because the
exponential decay terminates the quadrature before the limit of R=100 is reached. Changing from (3.17)
to (3.6) near y=1 is probably a good strategy. On the other hand, we expect the agreement between
(3.17) and (3.18) to be good for small y and diverge for larger values of y. Relative errors at y=0 were
less than full precision at O(109 ) because the values for (3.17) were computed using the quadrature,
while the values for (3.6) and (3.18) are accurate to unit round-off at y=0. The entries marked with (*) in
TABLE 2 were computed from the asymptotic expansion in (3.16) with a value of w0 just over 4, the
break point between the series and the asymptotic form. This break point seems to be fairly optimal
since increasing it diminishes the accuracy of the power series; on the other hand, decreasing it
diminishes the accuracy of the asymptotic expansion. Details on the computation with the asymptotic
expansion are contained in APPENDIX B.
Numerical evaluations of (4.5) for values along the line x=0 were also checked against the general
solution (3.6b) with corresponding parameters. A sample of the computations is shown in TABLE 3.
As a final check, the integral in (A.2), which must vanish to satisfy the boundary condition (1.3), was
evaluated for a variety of parameters to verify that, while not zero, it was on the order of unit round-off,
O(1015 ) . Sample values are shown in TABLE 4.
TABLE 1
COMPARISON OF (3.6) WITH (4.2)
α1 = 0.50 α2= 0.50 k1 = 0.75 k2= 0.75 q 0 = 1.00
t =
1.00
y
0.10D+01
0.50D+00
0.10D-04
0.00D+00
-0.10D-04
-0.50D+00
-0.10D+01
-0.10D+02
x = 0.00
0.25D-01 0.31D-13
0.78D-01 0.52D-14
0.26D+00 0.48D-02
0.27D+00 0.21D-15
0.27D+00 0.48D-02
0.45D+00 0.61D-15
0.51D+00 0.13D-14
0.53D+00 0.21D-15
x = 0.25
0.24D-01 0.42D-13
0.72D-01 0.56D-14
0.19D+00 0.12D-14
0.19D+00 0.44D-15
0.19D+00 0.17D-14
0.31D+00 0.11D-14
0.36D+00 0.23D-14
0.38D+00 0.29D-15
x = 0.50
0.20D-01 0.28D-13
0.58D-01 0.10D-13
0.13D+00 0.42D-15
0.13D+00 0.21D-15
0.13D+00 0.84D-15
0.21D+00 0.26D-14
0.24D+00 0.21D-14
0.26D+00 0.21D-15
x = 0.75
0.16D-01 0.50D-13
0.42D-01 0.16D-13
0.87D-01 0.21D-14
0.87D-01 0.32D-15
0.87D-01 0.17D-14
0.13D+00 0.48D-14
0.16D+00 0.47D-14
0.17D+00 0.16D-15
x = 1.00
0.11D-01 0.42D-13
0.28D-01 0.21D-13
0.56D-01 0.32D-14
0.56D-01 0.37D-15
0.56D-01 0.27D-14
0.83D-01 0.70D-14
0.10D+00 0.43D-14
0.11D+00 0.25D-15
y
0.10D+01
0.50D+00
0.10D-04
0.00D+00
-0.10D-04
-0.50D+00
-0.10D+01
-0.10D+02
x = 0.00
0.70D-01 0.12D-13
0.15D+00 0.25D-14
0.37D+00 0.48D-02
0.38D+00 0.30D-15
0.38D+00 0.48D-02
0.60D+00 0.74D-15
0.68D+00 0.13D-14
0.75D+00 0.30D-15
t = 2.00
x = 0.25
x = 0.50
0.67D-01 0.13D-13 0.61D-01 0.12D-13
0.14D+00 0.25D-14 0.13D+00 0.42D-14
0.30D+00 0.78D-14 0.23D+00 0.30D-14
0.30D+00 0.37D-15 0.23D+00 0.36D-15
0.30D+00 0.80D-14 0.23D+00 0.30D-14
0.45D+00 0.74D-15 0.34D+00 0.16D-14
0.53D+00 0.15D-14 0.40D+00 0.18D-14
0.60D+00 0.19D-15 0.47D+00 0.00D+00
x = 0.75
0.52D-01 0.17D-14
0.10D+00 0.62D-14
0.18D+00 0.62D-15
0.18D+00 0.31D-15
0.18D+00 0.62D-15
0.25D+00 0.26D-14
0.30D+00 0.18D-15
0.36D+00 0.00D+00
x = 1.00
0.43D-01 0.30D-13
0.81D-01 0.89D-14
0.13D+00 0.13D-14
0.13D+00 0.63D-15
0.13D+00 0.17D-14
0.19D+00 0.40D-14
0.22D+00 0.60D-14
0.27D+00 0.21D-15
y
0.10D+01
0.50D+00
0.10D-04
0.00D+00
-0.10D-04
-0.50D+00
-0.10D+01
-0.10D+02
x = 0.00
0.11D+00 0.65D-14
0.22D+00 0.13D-14
0.46D+00 0.48D-02
0.46D+00 0.24D-15
0.46D+00 0.48D-02
0.71D+00 0.47D-15
0.81D+00 0.82D-15
0.92D+00 0.12D-15
t = 3.00
x = 0.25
x = 0.50
0.11D+00 0.61D-14 0.10D+00 0.70D-14
0.21D+00 0.29D-14 0.19D+00 0.25D-14
0.38D+00 0.26D-13 0.31D+00 0.14D-14
0.38D+00 0.15D-15 0.31D+00 0.35D-15
0.38D+00 0.26D-13 0.31D+00 0.14D-14
0.56D+00 0.10D-14 0.44D+00 0.13D-14
0.65D+00 0.10D-14 0.52D+00 0.13D-14
0.76D+00 0.15D-15 0.63D+00 0.00D+00
x = 0.75
0.92D-01 0.10D-13
0.16D+00 0.35D-14
0.25D+00 0.22D-15
0.25D+00 0.22D-15
0.25D+00 0.66D-15
0.35D+00 0.96D-15
0.41D+00 0.17D-14
0.51D+00 0.44D-15
x = 1.00
0.79D-01 0.14D-13
0.13D+00 0.48D-14
0.20D+00 0.82D-15
0.20D+00 0.41D-15
0.20D+00 0.14D-14
0.27D+00 0.22D-14
0.33D+00 0.31D-14
0.40D+00 0.27D-15
19
TABLE 2
REFINEMENT OF COMPUTATION ALONG X=0 FOR THE PARAMETERS OF TABLE 1
USING (3.6), (3.17) AND (3.18)
t =
*
*
Y
0.00D+00
0.10D-05
0.10D-04
0.10D-03
0.10D-02
0.10D-01
0.10D+00
0.10D+01
0.20D+01
0.30D+01
0.40D+01
0.50D+01
(3.6)
0.188063D+00
0.187329D+00
0.187318D+00
0.187210D+00
0.186147D+00
0.176788D+00
0.124069D+00
0.630483D-02
0.777063D-04
0.192294D-06
0.811085D-10
0.537165D-14
(3.17)
0.188063D+00
0.188060D+00
0.188037D+00
0.187853D+00
0.186446D+00
0.176781D+00
0.124069D+00
0.630483D-02
0.777063D-04
0.192294D-06
0.811084D-10
0.537166D-14
t =
*
Y
0.00D+00
0.10D-05
0.10D-04
0.10D-03
0.10D-02
0.10D-01
0.10D+00
0.10D+01
0.20D+01
0.30D+01
0.40D+01
0.50D+01
(3.6)
0.265962D+00
0.264923D+00
0.264913D+00
0.264805D+00
0.263737D+00
0.253977D+00
0.194631D+00
0.249986D-01
0.172431D-02
0.580579D-04
0.860876D-06
0.531425D-08
(3.6)
0.376126D+00
0.374659D+00
0.374648D+00
0.374540D+00
0.373470D+00
0.363413D+00
0.297450D+00
0.695508D-01
0.126097D-01
0.168364D-02
0.155413D-03
0.957348D-05
Y
0.00D+00
0.10D-05
0.10D-04
0.10D-03
0.10D-02
0.10D-01
0.10D+00
0.10D+01
0.20D+01
0.30D+01
0.40D+01
0.50D+01
(3.6)
0.460659D+00
0.458862D+00
0.458851D+00
0.458743D+00
0.457671D+00
0.447482D+00
0.377683D+00
0.113834D+00
0.297965D-01
0.651932D-02
0.113625D-02
0.153639D-03
(3.18)
0.188063D+00
0.188060D+00
0.188039D+00
0.187868D+00
0.186597D+00
0.178291D+00
0.139201D+00
0.188063D+00
0.482244D+00
0.887461D+00
0.136479D+01
0.189573D+01
REL ERR
0.12D-09
0.81D-06
0.81D-05
0.81D-04
0.81D-03
0.85D-02
0.12D+00
0.29D+02
0.62D+04
0.46D+07
0.17D+11
0.35D+15
REL ERR
0.12D-09
0.39D-02
0.39D-02
0.35D-02
0.20D-02
0.13D-03
0.37D-10
0.89D-12
0.37D-11
0.14D-09
0.46D-08
0.95D-06
(3.18)
0.265962D+00
0.265959D+00
0.265937D+00
0.265766D+00
0.264496D+00
0.256189D+00
0.217099D+00
0.265962D+00
0.560142D+00
0.965360D+00
0.144268D+01
0.197363D+01
REL ERR
0.12D-09
0.85D-06
0.85D-05
0.85D-04
0.85D-03
0.88D-02
0.12D+00
0.96D+01
0.32D+03
0.17D+05
0.17D+07
0.37D+09
(3.18)
0.376126D+00
0.376123D+00
0.376102D+00
0.375931D+00
0.374661D+00
0.366354D+00
0.327264D+00
0.376126D+00
0.670307D+00
0.107552D+01
0.155285D+01
0.208379D+01
REL ERR
0.12D-09
0.80D-06
0.80D-05
0.80D-04
0.80D-03
0.82D-02
0.10D+00
0.44D+01
0.52D+02
0.64D+03
0.10D+05
0.22D+06
(3.18)
0.460659D+00
0.460656D+00
0.460634D+00
0.460463D+00
0.459193D+00
0.450886D+00
0.411796D+00
0.460659D+00
0.754840D+00
0.116006D+01
0.163738D+01
0.216833D+01
REL ERR
0.12D-09
0.74D-06
0.74D-05
0.74D-04
0.75D-03
0.76D-02
0.90D-01
0.30D+01
0.24D+02
0.18D+03
0.14D+04
0.14D+05
0.20000D+01
(3.17)
0.376126D+00
0.376123D+00
0.376099D+00
0.375901D+00
0.374362D+00
0.363373D+00
0.297450D+00
0.695508D-01
0.126097D-01
0.168364D-02
0.155413D-03
0.957348D-05
t =
REL ERR
0.12D-09
0.39D-02
0.38D-02
0.34D-02
0.16D-02
0.39D-04
0.96D-11
0.19D-11
0.76D-10
0.18D-07
0.57D-07
0.20D-05
0.10000D+01
(3.17)
0.265962D+00
0.265958D+00
0.265935D+00
0.265744D+00
0.264271D+00
0.253944D+00
0.194631D+00
0.249986D-01
0.172431D-02
0.580579D-04
0.860876D-06
0.531424D-08
t =
Y
0.00D+00
0.10D-05
0.10D-04
0.10D-03
0.10D-02
0.10D-01
0.10D+00
0.10D+01
0.20D+01
0.30D+01
0.40D+01
0.50D+01
0.50000D+00
REL ERR
0.12D-09
0.39D-02
0.39D-02
0.36D-02
0.24D-02
0.11D-03
0.14D-10
0.62D-12
0.19D-11
0.13D-11
0.76D-10
0.20D-08
0.30000D+01
(3.17)
0.460659D+00
0.460656D+00
0.460631D+00
0.460429D+00
0.458852D+00
0.447476D+00
0.377683D+00
0.113834D+00
0.297965D-01
0.651932D-02
0.113625D-02
0.153639D-03
20
REL ERR
0.12D-09
0.39D-02
0.39D-02
0.37D-02
0.26D-02
0.15D-04
0.11D-09
0.42D-12
0.13D-11
0.16D-11
0.27D-11
0.12D-10
TABLE 4
TESTING BOUNDARY INTEGRAL
(3.8) FOR A ZERO VALUE
TABLE 3
COMPARISON OF (3.6) WITH
DIMENSIONLESS FORM (4.7)
y= y /(2 αt)
y
0.1000D+01
0.7500D+00
0.5000D+00
0.2500D+00
0.0000D+00
-0.2500D+00
-0.5000D+00
-0.7500D+00
-0.1000D+01
-0.1000D+02

η=T(0,y,t)/ q 0 αt/k
η
0.9457D-02
0.2293D-01
0.5216D-01
0.1143D+00
0.2821D+00
0.4498D+00
0.5120D+00
0.5413D+00
0.5547D+00
0.5642D+00

ρ = 0.25  = 3π/8 y= y /(2 α Mt)
y
0.5000D+00
0.1000D+01
0.1500D+01
0.2000D+01
0.2500D+01
0.3000D+01
0.3500D+01
0.4000D+01
0.4500D+01
0.5000D+01
REL ERROR
0.3265D-13
0.2844D-13
0.1104D-13
0.2306D-14
0.0000D+00
0.6170D-15
0.1301D-14
0.1231D-14
0.6004D-15
0.0000D+00
INTEGRAL
0.2838D-14
0.2834D-14
0.2878D-14
0.2921D-14
0.2904D-14
0.2679D-14
0.2408D-14
0.2732D-14
0.2704D-14
0.2511D-14
Acknowledgement The author is indebted to Dr. James V. Beck who posed this problem and first
obtained (3.4) by simply observing that the sum of formulations with sources on the positive and
negative y-axes is one-dimensional and the x-axis is common to both formulations.
References
[1] Abramowitz S, Stegun IA (1965) Handbook of Mathematical Functions, AMS 55, Dover
Publications Inc., New York, 1046 pp
[2] Amos DE, Beck JV, de Monte F (2011) Transient Heat Conduction in Adjacent Quadrants
Separated by a Thermal Resistance, https://thermalhub.org/resources/470.
[3] Amos DE (2006) Handbook of Integrals Related to Heat Conduction and Diffusion,
https://thermalhub.org/resources/459.
[4] Beck JV, Cole KD, Haji-Sheikh A, Litkouhi B (1992) Heat Conduction Using Green’s Functions,
Hemisphere Press, Washington D.C., 523 pp
[5] Morse PM, Feshbach H (1953) Methods of Theoretical Physics, Part I, McGraw-Hill, New York,
997 pp
[6] Carslaw HS, Jaeger JC, (1948) Conduction of Heat in Solids, Oxford Univ Press, London, 386pp
[7] Amos DE (1990), Computation of Exponential Integrals of a Complex Argument, ACM Trans Math
Software, Vol 16, No 2, pp 169-177
[8] Amos DE (1990), ALGORITHM 683, A Portable Subroutine for Exponential Integrals of a Complex
Argument, ACM Trans Math Software, Vol 16, No 2, pp 178-182
21
APPENDIX A
VERIFICATION OF T1 (x,y,t) and T2 (x,y,t) AS THE PROBLEM SOLUTION
The manipulations that led to the solution (3.6) were, in some places, assumed to be valid; namely early
in the derivation where exchanges of integrals and derivatives were made. To establish that we have the
solution to the problem, we apply standard theorems on differentiation and taking limits under integral
signs and note that t ierfc( v j ) satisfies the heat equations. It remains to verify that the boundary
conditions are satisfied. It is apparent that for x=0, the temperatures for j=1 and j=2 are the same
since v j at x=0 does not depend on j and (1.3) is satisfied. We apply the flux boundary conditions to the
solution in the form of (2.20) with (3.2b) to obtain
 1
y0
   f j (  , y )d 
 C
(A.1)
T j ( x, y, t ) /( q0 t )  
G ( x, t )  1 f (  , y )d 
y0
j
 j
 C
and
 1 erfc( iy  /(2 t )
y0
d
 
i
T1 
T2 
 C
(A.2)
 k1
(0 , y, t )  k2
(0 , y, t )  q0 
x
x
 1 erfc( i y  /(2 t )
d y  0
1   C
i
We get the result we seek if the integrals above are zero. Because |y|>|x|=0,    / 3 and we close the
contour C ( ) with an arc of a circle CR :   Rei ,  / 3     -  / 3  2 / 3 . By Cauchy’s
theorem the integral over the closed contour C ( )  CR is zero because this contour is in a region of
analyticity and does not enclose any singularities. If we show that the integral over CR goes to zero as
the radius goes to infinity then the value of the contour integral over C ( ) is also equal to zero. We need
to verify that we have the exponential decay in the erfc function on the circular arc CR ,
erfc(v j )  erfc( i y Rei /(2 t ))
for R   . Then, from (2.10) with n=0, we can estimate erfc from
 e Re( v2j ) 
 for v j  
(A.3)
erfc(v j )  O 
  vj 


provided that arg( v j )  3 / 4 . Now v j  i y Rei /(2 t ) giving arg( v j )     / 2 and on the circular
arc CR , we have  / 3  arg(v 2j )  2     / 3 . This puts v 2j in the right half plane where Re(v 2j )  0
and arg( v j )   / 4 (more precisely, arg( v j )   / 6 ). This gives the integrand an exponential decay in
R. Therefore the integral over the circular arc CR goes to zero as R   , making the integral on C ( )
equal to zero. Then (A.2) becomes (1.3) and the boundary condition is satisfied.
22
Initial Conditions for t  0
The final part of the verification procedure examines the way in which the initial conditions are
satisfied. While direct evaluation at t=0 is not possible, the limiting form T j ( x, y, 0)  0 , j  1, 2 is
considered an acceptable replacement for T j ( x, y, 0)  0 , especially for semi-infinite regions (e.g.
T ( x, t )  T0 erfc[ x /(2  t )] , 0  x   for the semi-infinite slab initially at zero temperature). Thus, with
this understanding, (3.6) satisfies the initial conditions.
In summary, we have verified that the differential equations, boundary conditions, and initial conditions
as stated in (1.1)- (1.6) are satisfied by (3.6).
Uniqueness of the solution can be surmised from a standard energy argument. Assume that two distinct
solutions exist with the same boundary conditions and consider the difference D  0 , t>0. Because of
the linearity of the problem, D satisfies the diffusion equations with internal boundary conditions which
express continuity of temperature and flux on x=0. Since there is no energy source, D cannot change
from its initial value of zero. Consequently the two solutions must be equal, contrary to the assumption
that they were distinct. A mathematical version of uniqueness can be found in [6] with citations to the
literature that discuss the assumptions necessary to make solutions unique.
APPENDIX B
COMPUTATION OF EXPONENTIAL INTEGRALS AND EXPANSION (3.16)
The asymptotic expansion in (3.16) requires the evaluation of a sequence of exponential integrals,
Ek  3 / 2 ( z ) , k  1,..., N . It is well known that generation of sequences of exponential integrals must
proceed up and down from the index closest to z in order to have a stable computation [1, Section 5.3,
Examples 4 & 5]. Specifically   [ z ]  v, 0  v  1, where the brackets denote the integer function.
To apply this to (3.16), we take v  1/ 2 with the starting value for the sequence to be   [ z ]  1/ 2 .
Since the magnitude of z in these exponential integrals must be relatively large in order to use the
asymptotic expansion,  is fairly large, and we compute E ( z ) and recur downward from index
 toward index 1/2 with the recurrence
(B.1a)
E 1 ( z )   e z  (  1) E ( z )  / z ,
   ,  1,...,3/ 2,1/ 2 .
and, if necessary, upward from index  with
(B.1b)
E 1 ( z )   e  z  zE ( z )  /  ,
   ,  1,...
In order to start the sequence, we need to compute E ( z ) . The continued fraction [1, Eq.5.1.22], while
written for integer indices, actually applies for non-integer values and converges best for relatively large
z and   0 . We compute the continued fraction in the form
(B.2)
E ( z )  e z /( z  R1 )
,
Rn 
23
(  n  1)
1  n /( z  Rn 1 )
,
n  N , N  1,...,1 .
N=20 and R21  0 seem to suffice with a minimum value of z  w02  16 in (3.16). A strategy for
smaller values of z would include the power series for z  2 and (B.2) for larger z. Note that the
continued fraction can be applied for any   0 ; it is just a question of the amount of computation that
one is willing to do to get convergence. The procedure developed in Reference [7] and implemented in
reference [8] could easily be modified to compute exponential integrals of half odd orders.
It is the nature of many asymptotic expansions that term magnitudes decrease as the index increases and
then, at some point, the terms start to increase in magnitude. That is, there is a minimum term at index
KMIN. In many cases, as it is in (3.16), the error created by truncation of the series is close to the
magnitude of the next term. As z increases, the minimum term at index KMIN will drop below some
pre-assigned tolerance, say TOL. In this case, one can stop the sum at an index K  KMIN and have an
error no worse than TOL. If z is not large enough, then KMIN will be reached, the magnitude of the
minimum term will be larger than TOL, and one has to be satisfied with the partial sum up to KMIN-1
since the next term at index KMIN is an estimate of the error (which is smaller than the values of terms
KMIN-1 or KMIN+1). If the last index K    1/ 2 (corresponding to E ( z ) ) is reached before KMIN
is reached, then one can continue generating terms by applying (B.1b) for the sequence of exponential
integrals.
In order to get as many significant digits as possible, TOL, which measures the absolute error, can be
replace by REL with the relative error estimate
Term  Accumulated Sum * REL .
Thus, there are two ways to exit a loop when computing the asymptotic expansion. They are applied in
the order:
(a) when a computed term exceeds the previous term in magnitude. The best result possible is the
sum of KMIN-1 terms which has an error bound equal to the magnitude of the next term
(minimum term),
(b) when a term satisfies the relative error test, producing the requested error REL( i.e. when z is
large enough).
APPENDIX C
SYMMETRY OF (5.3) ABOUT y=(a+b)/2
Notice that if we replace y by ( a  b) / 2  v , v  (b - a ) / 2  0 ( ie. y<a) in the expression for y<a in
(5.3), we get
(C.1)
T j 2 ( x,( a  b) / 2  v, t )  T j 2 ( x,(b  a) / 2  v, t ) 
 G j ( x, t )  T j1 ( x, ( a  b) / 2  v , t )  G j ( x, t )  T j1( x, (b  a) / 2  v , t ) 
  T j 1 ( x , ( a  b) / 2  v , t )  T j 1 ( x , ( b  a ) / 2  v , t )
which is the expression for y  ( a  b) / 2  v  b in (5.3). Similarly, for the interval a<y<b,
24
Tj 2 ( x, y  b, t )  T j1( x, y  a, t )  G j ( x, t )  T j1( x, y  b , t )  T j1( x, y  a, t ) .
(C.2)
Now if we replace y by ( a  b) / 2  v , 0  v  (b - a ) / 2 , (a<y <(a+b)/2) we get
T j 2 ( x,( a  b) / 2  v, t )  T j1 ( x,(b  a ) / 2  v, t ) 
= G j ( x, t )  T j1 ( x, (a  b) / 2  v , t )  T j1 ( x,(b  a ) / 2  v, t )
=G j ( x, t )  T j1 ( x,(b  a ) / 2  v, t )  T j1 ( x , ( a  b) / 2  v , t )
= T j 2 ( x, v  (b  a ) / 2, t )  T j1 ( x,(b  a ) / 2  v, t )
(C.3)
which is the expression in (5.3) with y replaced by y  ( a  b) / 2  v ((a+b)/2<y<b). Since the
expressions for T j ( x, y , t , a, b) are the same for y  ( a  b) / 2  v with y inside or outside of the interval
a<y<b, we have symmetry. This implies that the derivative T j ( x,( a  b) / 2, t , a, b) / y is zero.
We can also show symmetry by computing T j ( x,( a  b) / 2, t , a, b) / y directly. Then from the middle
expression for T j ( x, y , t , a, b) on a<y<b
T j ( x, y, t , a, b)  T j 2 ( x, y  b, t )  T j1 ( x, y  a, t )  G j ( x, t )  T j1( x, b  y , t )  T j1( x, y  a, t )
we have
(C.4)
and for y=(a+b)/2 we get
T j1 ( x, v, t )
T j1 ( x, v, t )

T j ( x, y , t , a, b) 

y
v
v
v b y
v ya

T j ( x,( a  b) / 2, t , a, b)  0 .
y
25