Page 556 - Handbook Of Integral Equations

then
b

(ϕ 1 , ϕ 2 )=0, (ϕ, ψ) ≡ ϕ(x)ψ(x) dx.
a
The characteristic values of a symmetric kernel are real.
The eigenfunctions can be normalized; namely, we can divide each characteristic function by its
norm. If several linearly independent eigenfunctions correspond to the same characteristic value, say,
ϕ 1 (x), ... , ϕ n (x), then each linear combination of these functions is an eigenfunction as well, and
these linear combinations can be chosen so that the corresponding eigenfunctions are orthonormal.
Indeed, the function
ϕ 1 (x)
ψ 1 (x)= , ϕ 1 = (ϕ 1 , ϕ 1 ),
ϕ 1
has the norm equal to one, i.e., ψ 1 = 1. Let us form a linear combination αψ 1 + ϕ 2 and choose α
so that
(αψ 1 + ϕ 2 , ψ 1 )=0,
i.e.,
(ϕ 2 , ψ 1 )
α = – = –(ϕ 2 , ψ 1 ).
(ψ 1 , ψ 1 )
The function
αψ 1 + ϕ 2
ψ 2 (x)=
αψ 1 + ϕ 2
is orthogonal to ψ 1 (x) and has the unit norm. Next, we choose a linear combination αψ 1 +βψ 2 +ϕ 3 ,
where the constants α and β can be found from the orthogonality relations

(αψ 1 + βϕ 2 + ϕ 3 , ψ 1 )=0, (αψ 1 + βψ 2 + ϕ 3 , ψ 2 )=0.

For the coefficients α and β thus defined, the function

αψ 1 + βψ 2 + ϕ 2
ψ 3 =
αψ 1 + βϕ 2 + ϕ 3
is orthogonal to ψ 1 and ψ 2 and has the unit norm, and so on.
As was noted above, the eigenfunctions corresponding to distinct characteristic values are
orthogonal. Hence, the sequence of eigenfunctions of a symmetric kernel can be made orthonormal.
In what follows we assume that the sequence of eigenfunctions of a symmetric kernel is or-
thonormal.
We also assume that the characteristic values are always numbered in the increasing order of
their absolute values. Thus, if
λ 1 , λ 2 , ... , λ n , ... (1)

is the sequence of characteristic values of a symmetric kernel, and if a sequence of eigenfunctions

ϕ 1 , ϕ 2 , ... , ϕ n , ... (2)
corresponds to the sequence (1) so that

b

ϕ n (x) – λ n K(x, t)ϕ n (t) dt = 0, (3)
a
then
b 1 for i = j,
ϕ i (x)ϕ j (x) dx = (4)
a 0 for i ≠ j,

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 539