Page 530 - Advanced engineering mathematics

510 CHAPTER 15 Special Functions and Eigenfunction Expansions

To prove equation (15.2), begin with the fact that the eigenvalues and corresponding
eigenfunctions satisfy the Sturm-Liouville differential equation:

rϕ + (q + λ n p)ϕ n = 0
n
and

rϕ + (q + λ m p)ϕ m = 0.
m
Multiply the first equation by ϕ m and the second by ϕ n and subtract to obtain

rϕ ϕ m − rϕ ϕ n + (λ n − λ m )pϕ n ϕ m = 0.
n m
Write this equation as
d
r ϕ m ϕ − ϕ n ϕ = (λ m − λ n )pϕ n ϕ m .

dx n m
This means that
b
(λ m − λ n ) p(x)ϕ n (x)ϕ m (x)dx
a
b

= r ϕ m yϕ − ϕ n ϕ m a
n

=r(b) ϕ m (b)ϕ (b) − ϕ n (b)ϕ (b) −r(a) ϕ m (a)ϕ (a) − ϕ n (a)ϕ (a) .

n m n m
This gives us
b
(λ m − λ n ) p(x)ϕ n (x)ϕ m (x)dx
a

=r(b) ϕ m (b)ϕ (b) − ϕ n (b)ϕ (b) −r(a) ϕ m (a)ϕ (a) − ϕ n (a)ϕ (a) . (15.3)

n m n m
Equation (15.2) is therefore verified if we can show that the right side of equation (15.3)
is zero. This is done by examining the boundary conditions accompanying each type of Sturm-
Liouville problem.
Suppose first that the problem is regular. Each eigenfunction must satisfy the boundary
condition at a:

A 1 ϕ n (a) + A 2 ϕ (a) = 0,
n

A 1 ϕ m (a) + A 2 ϕ (a) = 0,
m
with not both A 1 and A 2 zero. Think of these boundary conditions at a as a homogeneous system
of two algebraic equations in two unknowns A 1 and A 2 . The fact that there is a nontrivial solution
for these numbers means that the determinant of the system is zero:

ϕ n (a) ϕ (a)
n
= 0.
ϕ m (a)ϕ (a)

m
The same argument applies to the regular boundary condition at b:

ϕ n (b) ϕ (b)
n
= 0.
ϕ m (b) ϕ (b)

m
This shows that both terms in square brackets on the right side of equation (15.3) are zero. Then
b
(λ m − λ n ) p(x)ϕ n (x)ϕ m (x)dx = 0.
a
Since λ n and λ n were assumed to be distinct eigenvalues, this means that
b
p(x)ϕ n (x)ϕ m (x)dx = 0,
a
proving conclusion (3) for the regular Sturm-Liouville problem.
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October 14, 2010 15:20 THM/NEIL Page-510 27410_15_ch15_p505-562