Solve Integral Equation by Basis Function Expansions
The eigenfunction is an integral function and difficult to solve in closed form. A general
strategy for solving the eigenfunction problem in (4) is to convert the continuous eigen-analysis
problem to an appropriate discrete eigen-analysis task [24]. In this paper, we use basis function
expansion methods to achieve this conversion.
Let { j (t )} be the series of Fourier functions. For each j, define 2j-1  2 j  2j. We
expand each genetic variant profile X i (t ) as a linear combination of the basis function  j :
T
X i (t )   Cij  j (t ).
(7)
j 1
Define the vector-valued function X (t )  [ X 1 (t ), , X N (t )]T and the vector-valued
function  (t )  [1 (t ),,T (t )]T . The joint expansion of all N genetic variant profiles can be
expressed as
X (t )  C (t ) .
(8)
where the matrix C is given by
C11,  C1T 
C  


CN 1  CNT .
.
In matrix form we can express the variance-covariance function of the genetic variant profiles as
1 T
X ( s ) X (t )
N
1
  T ( s )C T C (t ).
N
R ( s, t ) 
(9)
Similarly, the eigenfunction  (t ) can be expanded as
T
T
j 1
j1
(t )   b j  j (t ) and D4(t )   4j b j  j (t ) or
(t )  (t )T b and D 4(t )  (t )T S 0 b
(10)

1

1
where b  [b1 ,..., bT ]T and S0  diag ( 14 ,..., T4 ) . Let S  diag ((1  14 ) 2 ,..., (1  T4 ) 2 ). Then,
we have
(t )  D 4(t )  (t )T S 2 b.
(11)
Substituting expansions (9) and (11) of variance-covariance R(s,t) and eigenfunction  (t ) into
the functional eigenequation (6), we obtain
(t ) T
1 T
C Cb  T (t ) S  2 b ,
N
(12)
Since equation (12) must hold for all t, we obtain the following eigenequation:
1 T
C Cb  S  2 b ,
N
(13)
which can be rewritten as
[S (
S(
1 T
C C ) S ][ S 1b]  [ S 1b] , or
N
1 T
C C ) Su  u ,
N
(14)
where u  S 1b . Thus, b  Su and (t )  (t )T b is a solution to eigenequation (6).
Note that  u j , u j  1 and  u j , uk  0, for k  j. Therefore, we obtain a set of orthonormal
eigenfunctions with an inner product of two functions defined in equation (4), as shown in
equation (15):
||  j ||2  bTj S 2b j  uTj SS 2 Su j  1 and   j ,k   bTj S 2bk  uTj uk  0 .
(15)