A Sturm Liouville problem involves the differential equation,
(p(x)y′)′ + (λq(x)+ r (x))y = 0,x ∈ [a,b], p (x ) ≥ 0 (17.8.38)
(17.8.38)
where we assume that q
(x)
≥ 0 for x ∈
[a,b]
and is positive except for finitely many
points. Also, assume it is continuous. Probably, you could generalize this to assume less
about q if this is of interest. There will also be boundary conditions at a,b. These are
typically of the form
Also we assume here that a and b are finite numbers. In the example, the constants C_{i}
are given and λ is called the eigenvalue while a solution of the differential equation and
given boundary conditions corresponding to λ is called an eigenfunction.
There is a simple but important identity related to solutions of the above differential
equation. Suppose λ_{i} and y_{i} for i = 1,2 are two solutions of 17.8.38. Thus from the
equation, we obtain the following two equations.
∫
′ ′ b b
((p(x)y1)y2 − (p(x)y2)y1)|a + (λ1 − λ2) a q(x)y1(x)y2(x)dx = 0 (17.8.42)
(17.8.42)
We have been purposely vague about the nature of the boundary conditions because of a
desire to not lose generality. However, we will always assume the boundary
conditions are such that whenever y_{1} and y_{2} are two eigenfunctions, it follows
that
((p(x)y′1)y2 − (p (x )y′2)y1)|ba = 0 (17.8.43)
(17.8.43)
In the case where the boundary conditions are given by 17.8.39, and 17.8.40, we obtain
17.8.43. To see why this is so, consider the top limit. This yields
p (b)[y′(b)y2 (b)− y′(b)y1(b)]
1 2
However we know from the boundary conditions that
′
C3y1 (b)+ C4y1 (b) = 0
C3y2 (b)+ C4y′2 (b) = 0
and that from 17.8.40 that not both C_{3} and C_{4} equal zero. Therefore the determinant of
the matrix of coefficients must equal zero. But this implies
[y′1(b)y2(b)− y′2(b)y1(b)] = 0
which yields the top limit is equal to zero. A similar argument holds for the lower limit.
Note that y_{1},y_{2}satisfy different differential equations because of differenteigenvalues.
From now on the boundary condition will be conditions L,
ˆL
,L
(y(a),y′(a))
= 0,
ˆL
(y (b),y′(b))
= 0 which imply that if y_{i} correspond to two different
eigenvalues,
((p(x)y′1)y2 − (p (x )y′2)y1)|ba = 0 (*)
(*)
and if α is a constant, if L
(y (a) ,y′(a))
= 0,
ˆL
(y(b),y′(b))
= 0, then also
L(αy (a),αy′(a)) = 0, ˆL(αy (b),αy′(b)) = 0
For example, maybe one wants to say that y is bounded at a,b.
With the identity 17.8.42 here is a result on orthogonality of the eigenfunctions.
Proposition 17.8.1Suppose y_{i}solves the boundary conditions and the differentialequation for λ = λ_{i}where λ_{1}≠λ_{2}. Then we have the orthogonality relation
∫
b
a q(x)y1(x)y2(x)dx = 0. (17.8.44)
(17.8.44)
In addition to this, if u,v are two solutions to the differential equation corresponding to asingle λ, 17.8.38, not necessarily the boundary conditions, (same differential equation)then there exists a constant, C such that
W (u,v)(x)p (x) = C (17.8.45)
(17.8.45)
for all x ∈
[a,b]
. In this formula, W
(u,v)
denotes the Wronskiangiven by
( )
det u (x) v(x) . (17.8.46)
u′(x) v′(x)
(17.8.46)
Proof:The orthogonality relation, 17.8.44 follows from the fundamental assumption,
17.8.43 and 17.8.42.
This is obtained from the one of interest by letting λ = 0.
Criterion 17.8.2Suppose we are able to find functions, u and v such that they solvethe differential equation, 17.8.47and u solves the boundary condition at x = a while vsolves the boundary condition at x = b. Assume both are in L^{2}
(a,b)
and W
(u,v)
≠0. Itfollows that both are in L^{2}
(a,b,q)
, the L^{2}functions with respect to the measure q
(x)
dx.Thus
∫
b
(f,g)L2(a,b,q) ≡ a f (x)g (x)q (x )dx
If p
(x)
> 0 on
[a,b]
it is typically clear from the fundamental existence and
uniqueness theorems for ordinary differential equations that such functions u
and v exist. (See any good differential equations book or Problem 10 on Page
2349.)
However, such functions might exist even if p vanishes at the end points.
Lemma 17.8.3Assume Criterion 17.8.2. A function y is a solution to the boundaryconditions along with the equation,
(p(x)y′)′ + r(x)y = g (17.8.48)
(17.8.48)
if
∫ b
y(x) = a G (t,x)g (t)dt (17.8.49)
(17.8.49)
where
{
c−1(v(x)u(t)) if t < x
G (t,x) = c−1(v(t)u(x)) if t > x . (17.8.50)
(17.8.50)
where c is the constant of Proposition 17.8.1which satisfies p
(x)
W
(u,v)
(x)
= c.
Proof: Why does y solve the equation 17.8.48 along with the boundary
conditions?
∫ ∫
1 x 1 b
y(x) = c a g(t)u(t)v(x)dt+ c x g(t) v(t)u (x )dt
Differentiate
∫ x
y′(x) = 1g (x)u (x)v(x)+ 1 g(t)u(t) v′(x)dt
c c a
1 1∫ b ′
− cg(x)v(x)u (x) + c g(t)v(t)u (x )dt
x
∫ ∫
= 1 xg (t)u (t)v′(x)dt+ 1 b g(t)v (t)u′(x)dt
c a c x
Then
∫ x ∫ b
p(x)y′(x) = 1 g(t)u(t)p (x )v′(x)dt+ 1 g(t)v(t)p(x)u′(x)dt
c a c x
Then
(p (x)y′(x))
^{′} =
1g(x)p(x)u (x)v′(x)− 1 g(x)p(x)v(x)u′(x)
c c
∫ ∫
1 x ′ ′ 1 b ′ ′
+c a g(t)u(t)(p (x) v(x)) dt+ c x g (t)v(t)(p(x)u (x )) dt
From the definition of c, this equals
1 ∫ x ′ 1∫ b ′
= g(x)+ c g(t)u(t)(p (x) v′(x)) dt+ c g (t)v(t)(p(x)u′(x )) dt
a x
1∫ x 1∫ b
= g (x)+ - g(t)u (t)(− r (x )v(x)) dt+- g (t)v(t)(− r(x)u (x))dt
c a c x
( )
1 ∫ x 1 ∫ b
= g(x)− r (x) c a g(t)u(t)v(x)dt+ c x g(t)v (t)u (x)dt
= g(x)− r(x)y (x)
Thus
′
(p(x)y′(x))+ r (x)y (x ) = g(x)
so y satisfies the equation. As to the boundary conditions, by assumption,
( )
′ 1 ∫ b ′ 1 ∫ b
ˆL (y(b),y (b)) = ˆL v(b)c a g (t)u (t)dt,v (b)c a g(t)u(t) dt = 0
because v satisfies the boundary condition at b. The other boundary condition is exactly
similar. ■
Now in the case of Criterion 17.8.2, y is a solution to the Sturm Liouville
eigenvalue problem, 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 xq(t)y(t)u(t)v (x)dt
c a
∫
− λ b
+ c x q(t)y(t)v(t)u (x)dt, (17.8.51)
(17.8.51)
Letting μ =
1λ
, this is of the form
∫ b
μy (x) = G (t,x)q(t)y(t) dt (17.8.52)
a
(17.8.52)
where
{ − 1
G (t,x) = − c− 1(v(x)u(t)) if t < x . (17.8.53)
− c (v(t)u (x )) if t > x
(17.8.53)
Could μ = 0? If this happened, then from Lemma 17.8.3, we would have that y = 0 is
a solution of 17.8.48 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 on q
(t)
. Thus we are not interested in this case. It follows from 17.8.53
that G :
[a,b]
×
[a,b]
→ ℝ is continuous and symmetric, G
(t,x)
= G
(x,t)
.
{
− c−1(v(t)u(x)) if x < t
G (x,t) ≡ − c−1(v(x)u (t)) if x > t
= G (t,x )
Also we see that for f ∈ C
([a,b])
, and
∫
b
w (x) ≡ a G (t,x)q(t)f (t)dt,
Lemma 17.8.3 implies w is a solution to the boundary conditions and the equation
(p(x)y′)′ +r (x) y = − q(x)f (x) (17.8.54)
(17.8.54)
Theorem 17.8.4Suppose u,v are given in Criterion 17.8.2. Then there exists asequence of functions,