19.8. STURM LIOUVILLE PROBLEMS 547

so y satisfies the equation. As to the boundary conditions, by assumption,

L̂(y(b) ,y′ (b)

)= L̂

(v(b)

1c

∫ b

ag(t)u(t)dt,v′ (b)

1c

∫ b

ag(t)u(t)dt

)= 0

because v satisfies the boundary condition at b. The other boundary condition is exactlysimilar.

Now in the case of Criterion 19.8.2, y is a solution to the Sturm Liouville eigenvalueproblem, if and only if y solves the boundary conditions and the equation,(

p(x)y′)′+ r (x)y(x) =−λq(x)y(x) .

This happens if

y(x) =−λ

c

∫ x

aq(t)y(t)u(t)v(x)dt

+−λ

c

∫ b

xq(t)y(t)v(t)u(x)dt, (19.8.51)

Letting µ = 1λ

, this is of the form

µy(x) =∫ b

aG(t,x)q(t)y(t)dt (19.8.52)

where

G(t,x) ={−c−1 (v(x)u(t)) if t < x−c−1 (v(t)u(x)) if t > x

. (19.8.53)

Could µ = 0? If this happened, then from Lemma 19.8.3, we would have that y = 0 is asolution of 19.8.48 where the right side is −q(t)y(t) which would imply that q(t)y(t) = 0(since the left side is 0) for all t which implies y(t) = 0 for all t thanks to assumptions onq(t). Thus we are not interested in this case. It follows from 19.8.53 that G : [a,b]× [a,b]→R is continuous and symmetric, G(t,x) = G(x, t).

G(x, t) ≡{−c−1 (v(t)u(x)) if x < t−c−1 (v(x)u(t)) if x > t

= G(t,x)

Also we see that for f ∈C ([a,b]), and

w(x)≡∫ b

aG(t,x)q(t) f (t)dt,

Lemma 19.8.3 implies w is a solution to the boundary conditions and the equation(p(x)y′

)′+ r (x)y =−q(x) f (x) (19.8.54)

Theorem 19.8.4 Suppose u,v are given in Criterion 19.8.2. Then there exists a sequenceof functions, {yn}∞

n=1 and real numbers, λ n such that(p(x)y′n

)′+(λ nq(x)+ r (x))yn = 0, x ∈ [a,b] , (19.8.55)