II. The Reproduction Number
 I 
Case I. Age class i is isolated: i  ai i  i  .
 Ni 
In this case, we may obtain the reproduction number for group i via either the endemic or
disease-free equilibrium. The endemic equilibrium Vi *  (ui* , xi* , yi* , zi* ) can be solved as:
ui* 

1
1 
 
  *
 * *


, xi*  1 
xi , zi 
xi* ,
 1 
1 
 , yi 
0i
 
  
 0i          
where 0i 
 ai i
.
(   )(   )
Alternatively, consider the disease-free equilibrium. In this case, the next generation
matrix is:
 a11

 (   )(   )

0



0

K1  
0




0


0

a11
 
0
0
0
0
0
0
0
0
 a2  2
(   )(   )
0
a2  2
 
0
0
0
0
 an  n
(   )(   )
0
0
0
0
0
0
0

0 

0 

0 

.
0 


an  n 
  

0 
Because groups are isolated, this is a diagonal matrix. The non-zero elements,
 ai i
0i 
, are the sub-population reproduction numbers. The meta(   )(   )
population reproduction number is the dominant eigenvalue of K1, which is
0  max{01 , 02 , , 0 n }.
 Ij
Case II. Age classes are not isolated: i  ai i  j cij 
N
 j

 .

In this case, the endemic equilibrium Vi *  (ui* , xi* , yi* , zi* ) generally cannot be solved
explicitly. The component yi* satisfies the following system of quadratic equations:
 

(   )(   )  * 
*
*
0i 1  1 

 yi   j cij y j  yi , i  1, 2,

 
  
n.
Now we consider the disease-free equilibrium. In this case, the next generation matrix is:
  a11c11
 (   )(   )

0


  a2  2 c21
 (   )(   )
K2  
0



  an  n cn1
 (   )(   )

0

a11c11
 
0
 a11c12
(   )(   )
a2  2 c21
 
0
 a2  2c22
(   )(   )
an  n cn1
 
0
 an  n cn 2
(   )(   )
0
0
0
a11c12
 
0
 a11c1n
(   )(   )
a2  2c22
 
0
 a2  2 c2 n
(   )(   )
an  n cn 2
 
0
 an  n cnn
(   )(   )
0
0
0
a11c1n
 
0
a2  2 c2 n
 
0











an  n cnn 
  

0 
 0 is the dominant eigenvalue of the matrix K2. Note that this matrix has n zero
eigenvalues. The others are given by an n by n matrix,
 01c11

K3  
 c
 0 n n1
01c1n 

,
0 n cnn 
which highlights the importance of mixing. Because K3 is a positive matrix, its dominant
eigenvalue must be positive. The probabilities of transmission upon contact with
infectious people in our model reflect differences in susceptibility. If these probabilities
also or only reflected differences in infectiousness, the off-diagonal elements would be of
the form
 ai ij cij
, which cannot be written as 0i  cij. While one cannot
(   )(   )
generally obtain explicit formulae for dominant eigenvalues when n>3, one can always
calculate them numerically. When n=2, for example, the dominant eigenvalue of matrix
K3 is
1
A  D  ( A  D)2  4 BC  , where


2
A  R 01c11 , B  R 01c12 , C  R 02c21 , and D  R 0c22 .
R0 