MATHEMATICAL APPENDIX
The aggregate dynamics of the economy can be conveniently described by a system of
deterministic differential equations involving the variables C 1 , C 2 , N 11 and N 12 .
Maximization of utility by the two representative households results in the familiar Euler
equations (11) and (21). As in the text, we denote the two constant growth rates of C 1 and C 2
as  C1 and  C 2 where  C 2   C1 because  2   1 . The striking aspect of equations (11)
and (21) is that consumption growths do not depend on the number of intermediates, N s .
To study the dynamic behavior of N 11 and N 12 , we must solve the system of
simultaneous differential equations (14) and (20). The equality between w 1 and the marginal
product of labor implies
w 1  ( 1  )  Y( 1 L
/1 ) .
(A1)
After some manipulation, the interest rate, given by equations (7) and (8) can be written as
r1  ( 1 / ) (1
(A2)

) Y
 1 ( N 1/
) .
Hence, aggregate income, w 1L 1  r1  1 N 1 , equals Y 1   2Y 1 . It follows that the Western
household’s budget constraint in equation (14) becomes
(A3)
 N 11  Y 1   2Y 1  C 1  r1  N 12
 (1   2 ) Y 1  C 1  r1  N 12
 (1   ) (1   ) Y 1  C 1  r1  N 12 .
If we substitute for Y 1 and r1 from equations (5) and (8) into (A3) and also use equation (7),
we get a formula for the Western household’s budget constraint
(A4)
 N 11  [ 1  (1   ) /  ]  N 11  ( 1/ )  N 12  C 1 .
By a similar process of substitution into equation (20), using equations (5), (6), (7), (8) and
(18), we can also get a formula for the Eastern household’s budget constraint
(A5)
(    ) N
 [  2

( 1  )  / 1N
1 2
]
1 2
 [  2  ( 1   ) /  ]  N 11    N 11  C 2 .
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We now must solve the system of simultaneous differential equations (A4) and (A5). We can
substitute ( 2 /  1 )  (A4) into equation (A5) to get
N 12  r1 N 12  a  C 1  b  C 2 ,
(A6)
where a and b are defined as
a    / [ (    )  ] ,
b  1 / (  
) .
(A5) is a first-order, linear differential equation in N 12 . The general solution of this equation is
(A7)
N 12 (t )  ( constant ) e r 1 t  [a / (r1  
C1
)]  C 1 (t )  [b / (r1  
C2
) ]  C 2 (t ) .
We assume that the production function is sufficiently productive to ensure growth in
consumption, but not so productive as to yield unbounded utility
r1   2  r1  (1   ) .
(A8)
The first part of this condition guarantees that  C 2  0 . The second part ensures that the
attainable utility is bounded and implies that r1　  C 2  0 . The transversality condition for the
dynamic optimization by the Eastern household implies
lim {N 12 (t )  e rｔ
(A9)
1
t 
}  0.
If we substitute for N 12 (t ) from equation (A7) into the transversality condition in equation
(A9), we get
(A10)
lim
t 
{ constant  [a / ( r1  
 [ b / ( r1  
C2
C1
)]  C 1 (T 0 )  e  ( r1  C 1 ) t
)]  C 2 (T 0 )  e  ( r1  C 2 ) t }  0 .

Since C 1 (T 0 ) and C 2 (T 0 ) are finite and r1　C2

 0 , r1　C1
 0 , the second and
the third terms in the braces converge toward zero. Hence, the transversality condition requires
the constant to be zero. The solution of N 12 becomes
2
(A11)
r/ 1( 
N 1 2(t )  a[
C
1
C
) ] t1  b
( ) r [ 
1 /C (
2C
t ).2 ]
If we substitute N 12 (t ) from equation (A11) into equation (A4), we get
(A12)
N 11  [ r1  (1   ) /  ]  N 11  f  C 1  g  C 2 ,
where f and g are defined as
f  a  r1 / [   (r1  
g  b 1r/ [ 
C1
(1r 
) ]  1/  ,
C
2
) ] .
(A12) is a first-order, linear differential equation in N 11 . By repeating the same procedure as
above, we can solve for N 11 as
(A13)
N 11 (t )  { f / [r1  (1   ) /   
C1
)]}  C 1 (t )
 { g / [ r1 ( 1   ) /    C 2 ) ]}  C 2 ( t )
Finally, if we add (A11) and (A13) together, we get
N 1 (t )  N 11 (t )  N 12 (t )
(A14)
 m  C1 (t )  n  C 2(t )
where m and n are defined as
m  b / [r1  (1   ) /   
C1
n  b / [r1  (1   ) /   
],
C2
].
  C1  0 and r1　The transversality conditions imply that r1　  C 2  0 , and hence
[r1  (1   ) /  ]　  C1  0 , [r1  (1   ) /  ]　  C 2  0. Since a , b , f , g , m , n >
0, the dynamic analysis of N 12 , N 11 and N 1 as described in the text can be derived from the
following three theorems of real numbers:
Let A, B, C be real numbers and functions of time, t . A(T 0 ), B(T 0 ), C (T 0 )  0 at t  T 0 .
Theorem 1: If A  B  C , and both B and C grow at positive constant rates with
/AB
 / B initially and A / A declines monotonically toward
B / B  C / C , then A
B / B as t   .
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( )
Theorem 2: If A  B  C , and both B and C grow at positive constant rates with
/ A B
 / B  C / C , then A
 / B initially and A / A declines monotonically. Let
B
 / A  0 initially, then A / A declines until at the some point, t  T  T , A
 / A  0 and A
A
1
0
begins to decline from then on for t  T 1 ; A will eventually decline until at t  T 2  T 1 , when
B  C , A  0 . Immediately after t  T , A / A  C / C and then declines monotonically
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toward C / C as t   .
Theorem 3: If A  B  C , and both B and C grow at positive constant rates with
 / A  C / C initially and A / A rise monotonically toward
/B A
B / B  C / C , then B
B / B as t   .
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