6

Sturm-Liouville Eigenvalue Problems

6.1 Introduction
In the last chapters we have explored the solution of boundary value problems
that led to trigonometric eigenfunctions. Such functions can be used to repre-
sent functions in Fourier series expansions. We would like to generalize some
of those techniques in order to solve other boundary value problems. A class of
problems to which our previous examples belong and which have eigenfunc-
tions with similar properties are the Sturm-Liouville Eigenvalue Problems.
These problems involve self-adjoint (dierential) operators which play an im-
portant role in the spectral theory of linear operators and the existence of the
eigenfunctions we described in Section 4.3.2. These ideas will be introduced
in this chapter.
In physics many problems arise in the form of boundary value problems
involving second order ordinary dierential equations. For example, we might
want to solve the equation
a
2
(x)y

+a
1
(x)y

+a
0
(x)y = f(x) (6.1)
subject to boundary conditions. We can write such an equation in operator
form by dening the dierential operator
L = a
2
(x)
d
2
dx
2
+a
1
(x)
d
dx
+a
0
(x).
Then, Equation (6.1) takes the form
Ly = f.
As we saw in the general boundary value problem (4.20) in Section 4.3.2,
we can solve some equations using eigenvalue expansions. Namely, we seek
solutions to the eigenvalue problem
L =
186 6 Sturm-Liouville Eigenvalue Problems
with homogeneous boundary conditions and then seek a solution as an expan-
sion of the eigenfunctions. Formally, we let
y =

n=1
c
n

n
.
However, we are not guaranteed a nice set of eigenfunctions. We need an
appropriate set to form a basis in the function space. Also, it would be nice
to have orthogonality so that we can easily solve for the expansion coecients
as was done in Section 4.3.2. [Otherwise, we would have to solve a innite
coupled system of algebraic equations instead of an uncoupled and diagonal
system.]
It turns out that any linear second order operator can be turned into an
operator that possesses just the right properties (self-adjointedness to carry
out this procedure. The resulting operator is referred to as a Sturm-Liouville
operator. We will highlight some of the properties of such operators and prove
a few key theorems, though this will not be an extensive review of Sturm-
Liouville theory. The interested reader can review the literature and more
advanced texts for a more in depth analysis.
We dene the Sturm-Liouville operator as
L =
d
dx
p(x)
d
dx
+q(x). (6.2)
The Sturm-Liouville eigenvalue problem is given by the dierential equation
Lu = (x)u,
or
d
dx
_
p(x)
du
dx
_
+q(x)u +(x)u = 0, (6.3)
for x (a, b). The functions p(x), p

(x), q(x) and (x) are assumed to be

continuous on (a, b) and p(x) > 0, (x) > 0 on [a, b]. If the interval is nite
and these assumptions on the coecients are true on [a, b], then the problem
is said to be regular. Otherwise, it is called singular.
We also need to impose the set of homogeneous boundary conditions

1
u(a) +
1
u

(a) = 0,

2
u(b) +
2
u

(b) = 0. (6.4)
The s and s are constants. For dierent values, one has special types of
boundary conditions. For
i
= 0, we have what are called Dirichlet boundary
conditions. Namely, u(a) = 0 and u(b) = 0. For
i
= 0, we have Neumann
boundary conditions. In this case, u

(a) = 0 and u

(b) = 0. In terms of the

heat equation example, Dirichlet conditions correspond to maintaining a xed
temperature at the ends of the rod. The Neumann boundary conditions would
6.1 Introduction 187
correspond to no heat ow across the ends, or insulating conditions, as there
would be no temperature gradient at those points. The more general boundary
conditions allow for partially insulated boundaries.
Another type of boundary condition that is often encountered is the pe-
riodic boundary condition. Consider the heated rod that has been bent to
form a circle. Then the two end points are physically the same. So, we would
expect that the temperature and the temperature gradient should agree at
those points. For this case we write u(a) = u(b) and u

(a) = u

(b). Boundary
value problems using these conditions have to be handled dierently than the
above homogeneous conditions. These conditions leads to dierent types of
eigenfunctions and eigenvalues.
As previously mentioned, equations of the form (6.1) occur often. We now
show that Equation (6.1) can be turned into a dierential equation of Sturm-
Liouville form:
d
dx
_
p(x)
dy
dx
_
+q(x)y = F(x). (6.5)
Another way to phrase this is provided in the theorem:
Theorem 6.1. Any second order linear operator can be put into the form of
the Sturm-Liouville operator (6.2).
The proof of this is straight forward, as we shall soon show. Consider the
equation (6.1). If a
1
(x) = a

2
(x), then we can write the equation in the form
f(x) = a
2
(x)y

+a
1
(x)y

+a
0
(x)y
= (a
2
(x)y

+a
0
(x)y. (6.6)
This is in the correct form. We just identify p(x) = a
2
(x) and q(x) = a
0
(x).
However, consider the dierential equation
x
2
y

+xy

+ 2y = 0.
In this case a
2
(x) = x
2
and a

2
(x) = 2x = a
1
(x). The linear dierential
operator in this equation is not of Sturm-Liouville type. But, we can change
it to a Sturm Liouville operator.
In the Sturm Liouville operator the derivative terms are gathered together
into one perfect derivative. This is similar to what we saw in the rst chap-
ter when we solved linear rst order equations. In that case we sought an
integrating factor. We can do the same thing here. We seek a multiplicative
function (x) that we can multiply through (6.1) so that it can be written in
Sturm-Liouville form. We rst divide out the a
2
(x), giving
y

+
a
1
(x)
a
2
(x)
y

+
a
0
(x)
a
2
(x)
y =
f(x)
a
2
(x)
.
Now, we multiply the dierential equation by :
188 6 Sturm-Liouville Eigenvalue Problems
(x)y

+(x)
a
1
(x)
a
2
(x)
y

+(x)
a
0
(x)
a
2
(x)
y = (x)
f(x)
a
2
(x)
.
The rst two terms can now be combined into an exact derivative (y

if
(x) satises
d
dx
= (x)
a
1
(x)
a
2
(x)
.
This is formally solved to give
(x) = e
_
a
1
(x)
a
2
(x)
dx
.
Thus, the original equation can be multiplied by factor
(x)
a
2
(x)
=
1
a
2
(x)
e
_
a
1
(x)
a
2
(x)
dx
to turn it into Sturm-Liouville form.
In summary,
Equation (6.1),
a
2
(x)y

+a
1
(x)y

+a
0
(x)y = f(x), (6.7)
can be put into the Sturm-Liouville form
d
dx
_
p(x)
dy
dx
_
+q(x)y = F(x), (6.8)
where
p(x) = e
_
a
1
(x)
a
2
(x)
dx
,
q(x) = p(x)
a
0
(x)
a
2
(x)
,
F(x) = p(x)
f(x)
a
2
(x)
. (6.9)
Example 6.2. For the example above,
x
2
y

+xy

+ 2y = 0.
We need only multiply this equation by
1
x
2
e
_
dx
x
=
1
x
,
to put the equation in Sturm-Liouville form:
0 = xy

+y

+
2
x
y
= (xy

+
2
x
y. (6.10)
6.2 Properties of Sturm-Liouville Eigenvalue Problems 189
6.2 Properties of Sturm-Liouville Eigenvalue Problems
There are several properties that can be proven for the (regular) Sturm-
Liouville eigenvalue problem. However, we will not prove them all here. We
will merely list some of the important facts and focus on a few of the proper-
ties.
1. The eigenvalues are real, countable, ordered and there is a smallest eigen-
value. Thus, we can write them as
1
<
2
< . . . . However, there is no
largest eigenvalue and n ,
n
.
2. For each eigenvalue
n
there exists an eigenfunction
n
with n 1 zeros
on (a, b).
3. Eigenfunctions corresponding to dierent eigenvalues are orthogonal with
respect to the weight function, (x). Dening the inner product of f(x)
and g(x) as
< f, g >=
_
b
a
f(x)g(x)(x) dx, (6.11)
then the orthogonality of the eigenfunctios can be written in the form
<
n
,
m
>=<
n
,
n
>
nm
, n, m = 1, 2, . . . . (6.12)
4. The set of eigenfunctions is complete; i.e., any piecewise smooth func-
tion can be represented by a generalized Fourier series expansion of the
eigenfunctions,
f(x)

n=1
c
n

n
(x),
where
c
n
=
< f,
n
>
<
n
,
n
>
.
Actually, one needs f(x) L
2

[a, b], the set of square integrable functions

over [a, b] with weight function (x). By square integrable, we mean that
< f, f >< . One can show that such a space is isomorphic to a Hilbert
space, a complete inner product space.
5. Multiply the eigenvalue problem
L
n
=
n
(x)
n
by
n
and integrate. Solve this result for
n
, to nd the Rayleigh Quotient

n
=
p
n
dn
dx
|
b
a

_
b
a
_
p
_
dn
dx
_
2
q
2
n
_
dx
<
n
,
n
>
The Rayleigh quotient is useful for getting estimates of eigenvalues and
proving some of the other properties.
190 6 Sturm-Liouville Eigenvalue Problems
Example 6.3. We seek the eigenfunctions of the operator found in Example
6.2. Namely, we want to solve the eigenvalue problem
Ly = (xy

+
2
x
y = y (6.13)
subject to a set of boundary conditions. Lets use the boundary conditions
y

(1) = 0, y

(2) = 0.
[Note that we do not know (x) yet, but will choose an appropriate function
to obtain solutions.]
Expanding the derivative, we have
xy

+y

+
2
x
y = y.
Multiply through by x to obtain
x
2
y

+xy

+ (2 +x) y = 0.
Notice that if we choose (x) = x
1
, then this equation can be made a
Cauchy-Euler type equation. Thus, we have
x
2
y

+xy

+ ( + 2) y = 0.
The characteristic equation is
r
2
+ + 2 = 0.
For oscillatory solutions, we need + 2 > 0. Thus, the general solution is
y(x) = c
1
cos(

+ 2 ln |x|) +c
2
sin(

+ 2 ln |x|). (6.14)
Next we apply the boundary conditions. y

(1) = 0 forces c
2
= 0. This
leaves
y(x) = c
1
cos(

+ 2 ln x).
The second condition, y

(2) = 0, yields
sin(

+ 2 ln 2) = 0.
This will give nontrivial solutions when

+ 2 ln 2 = n, n = 0, 1, 2, 3 . . . .
In summary, the eigenfunctions for this eigenvalue problem are
y
n
(x) = cos
_
n
ln 2
ln x
_
, 1 x 2
6.2 Properties of Sturm-Liouville Eigenvalue Problems 191
and the eigenvalues are
n
= 2 +
_
n
ln 2
_
2
for n = 0, 1, 2, . . . .
Note: We include the n = 0 case because y(x) = constant is a solution
of the = 2 case. More specically, in this case the characteristic equation
reduces to r
2
= 0. Thus, the general solution of this Cauchy-Euler equation is
y(x) = c
1
+c
2
ln |x|.
Setting y

(1) = 0, forces c
2
= 0. y

(2) automatically vanishes, leaving the

solution in this case as y(x) = c
1
.
We note that some of the properties listed in the beginning of the section
hold for this example. The eigenvalues are seen to be real, countable and
ordered. There is a least one, = 2. Next, one can nd the zeros of each
eigenfunction on [1,2]. Then the argument of the cosine,
n
ln2
ln x, takes values
0 to n for x [1, 2]. The cosine function has n 1 roots on this interval.
Orthogonality can be checked as well. We set up the integral and use the
substitution y = ln x/ ln 2. This gives
< y
n
, y
m
> =
_
2
1
cos
_
n
ln 2
ln x
_
cos
_
m
ln 2
ln x
_
dx
x
=
ln 2

_

0
cos ny cos my dy
=
ln 2
2

n,m
. (6.15)
6.2.1 Adjoint Operators
In the study of the spectral theory of matrices, one learns about the adjoint of
the matrix, A

, and the role that self-adjoint, or Hermitian, matrices play in

diagonalization. also, one needs the concept of adjoint to discuss the existence
of solutions to the matrix problemy = Ax. In the same spirit, one is interested
in the existence of solutions of the operator equation Lu = f and solutions of
the corresponding eigenvalue problem. The study of linear operator on Hilbert
spaces is a generalization of what the reader had seen in a linear algebra course.
Just as one can nd a basis of eigenvectors and diagonalize Hermitian, or
self-adjoint, matrices (or, real symmetric in the case of real matrices), we will
see that the Sturm-Liouville operator is self-adjoint. In this section we will
dene the domain of an operator and introduce the notion of adjoint operators.
In the last section we discuss the role the adjpoint plays in the existence of
solutions to the operator equation Lu = f.
We rst introduce some denitions.
Denition 6.4. The domain of a dierential operator L is the set of all u
L
2

[a, b] satisfying a given set of homogeneous boundary conditions.

Denition 6.5. The adjoint, L

, of operator L satises
192 6 Sturm-Liouville Eigenvalue Problems
< u, Lv >=< L

u, v >
for all v in the domain of L and u in the domain of L

.
Example 6.6. As an example, we nd the adjoint of second order linear dier-
ential operator L = a
2
(x)
d
2
dx
2
+a
1
(x)
d
dx
+a
0
(x).
In order to nd the adjoint, we place the operator under an integral. So,
we consider the inner product
< u, Lv >=
_
b
a
u(a
2
v

+a
1
v

+a
0
v) dx.
We have to move the operator L from v and determine what operator is acting
on u in order to formally preserve the inner product. For a simple operator like
L =
d
dx
, this is easily done using integration by parts. For the given operator,
we will need to apply several integrations by parts to the individual terms.
We will consider the individual terms.
First we consider the a
1
v

term. Integration by parts yields

_
b
a
u(x)a
1
(x)v

(x) dx = a
1
(x)u(x)v(x)

b
a

_
b
a
(u(x)a
1
(x))

v(x) dx. (6.16)

Now, we consider the a
2
v

term. In this case it will take two integrations

by parts:
_
b
a
u(x)a
2
(x)v

(x) dx = a
2
(x)u(x)v

(x)

b
a

_
b
a
(u(x)a
2
(x))

v(x)

dx
= [a
2
(x)u(x)v

(x) (a
2
(x)u(x))

v(x)]

b
a
+
_
b
a
(u(x)a
2
(x))

v(x) dx. (6.17)

Combining these results, we obtain
< u, Lv > =
_
b
a
u(a
2
v

+a
1
v

+a
0
v) dx
= [a
1
(x)u(x)v(x) +a
2
(x)u(x)v

(x) (a
2
(x)u(x))

v(x)]

b
a
+
_
b
a
[(a
2
u)

(a
1
u)

+a
0
u] v dx. (6.18)
Inserting the boundary conditions for v, one has to determine boundary
conditions for u such that
[a
1
(x)u(x)v(x) +a
2
(x)u(x)v

(x) (a
2
(x)u(x))

v(x)]

b
a
= 0.
6.2 Properties of Sturm-Liouville Eigenvalue Problems 193
This leaves
< u, Lv >=
_
b
a
[(a
2
u)

(a
1
u)

+a
0
u] v dx < L

u, v > .
Therefore,
L

=
d
2
dx
2
a
2
(x)
d
dx
a
1
(x) +a
0
(x). (6.19)
When L

= L, the operator is called formally self-adjoint. When the do-

main of L is the same as the domain of L

, the term self-adjoint is used.

As the domain is important in establishing self-adjointness, we need to do a
complete example in which the domain of the adjoint is found.
Example 6.7. Determine L

and its domain for operator Lu =

du
dx
where u
satises the boundary conditions u(0) = 2u(1) on [0, 1].
We need to nd the adjoint operator satisfying < v, Lu >=< L

v, u > .
Therefore, we rewrite the integral
< v, Lu >=
_
1
0
v
du
dx
dx = uv|
1
0

_
1
0
u
dv
dx
dx =< L

v, u > .
From this we have the adjoint problem consisting of an adjoint operator and
the associated boundary condition:
1. L

=
d
dx
.
2. uv

1
0
= 0 0 = u(1)[v(1) 2v(0)] v(1) = 2v(0).
6.2.2 Lagranges and Greens Identities
Before turning to the proofs that the eigenvalues of a Sturm-Liouville problem
are real and the associated eigenfunctions orthogonal, we will rst need to
introduce two important identities. For the Sturm-Liouville operator,
L =
d
dx
_
p
d
dx
_
+q,
we have the two identities:
Lagranges Identity uLv vLu = [p(uv

vu

)]

.
Greens Identity
_
b
a
(uLv vLu) dx = [p(uv

vu

)]|
b
a
.
Proof. The proof of Lagranges identity follows by a simple manipulations of
the operator:
194 6 Sturm-Liouville Eigenvalue Problems
uLv vLu = u
_
d
dx
_
p
dv
dx
_
+qv
_
v
_
d
dx
_
p
du
dx
_
+qu
_
= u
d
dx
_
p
dv
dx
_
v
d
dx
_
p
du
dx
_
= u
d
dx
_
p
dv
dx
_
+p
du
dx
dv
dx
v
d
dx
_
p
du
dx
_
p
du
dx
dv
dx
=
d
dx
_
pu
dv
dx
pv
du
dx
_
. (6.20)
Greens identity is simply proven by integrating Lagranges identity.
6.2.3 Orthogonality and Reality
We are now ready to prove that the eigenvalues of a Sturm-Liouville problem
are real and the corresponding eigenfunctions are orthogonal. These are easily
established using Greens identity, which in turn is a statement about the
Sturm-Liouville operator being self-adjoint.
Theorem 6.8. The eigenvalues of the Sturm-Liouville problem are real.
Proof. Let
n
(x) be a solution of the eigenvalue problem associated with
n
:
L
n
=
n

n
.
The complex conjugate of this equation is
L
n
=
n

n
.
Now, multiply the rst equation by
n
and the second equation by
n
and
then subtract the results. We obtain

n
L
n

n
L
n
= (
n

n
)
n

n
.
Integrate both sides of this equation:
_
b
a
_

n
L
n

n
L
n
_
dx = (
n

n
)
_
b
a

n
dx.
Apply Greens identity to the left hand side to nd
[p(
n

n
)]|
b
a
= (
n

n
)
_
b
a

n
dx.
Using the homogeneous boundary conditions for a self-adjoint operator, the
left side vanishes to give
0 = (
n

n
)
_
b
a

2
dx.
The integral is nonnegative, so we must have
n
=
n
. Therefore, the eigen-
values are real.
6.2 Properties of Sturm-Liouville Eigenvalue Problems 195
Theorem 6.9. The eigenfunctions corresponding to dierent eigenvalues of
the Sturm-Liouville problem are orthogonal.
Proof. This is proven similar to the last theorem. Let
n
(x) be a solution of
the eigenvalue problem associated with
n
,
L
n
=
n

n
,
and let
m
(x) be a solution of the eigenvalue problem associated with
m
=

n
,
L
m
=
m

m
,
Now, multiply the rst equation by
m
and the second equation by
n
. Sub-
tracting the results, we obtain

m
L
n

n
L
m
= (
m

n
)
n

m
Similar to the previous prooof, we integrate both sides of the equation and
use Greens identity and the boundary conditions for a self-adjoint operator.
This leaves
0 = (
m

n
)
_
b
a

m
dx.
Since the eigenvalues are distinct, we can divide by
m

n
, leaving the
desired result,
_
b
a

m
dx = 0.
Therefore, the eigenfunctions are orthogonal with respect to the weight func-
tion (x).
6.2.4 The Rayleigh Quotient
The Rayleigh quotient is useful for getting estimates of eigenvalues and prov-
ing some of the other properties associated with Sturm-Liouville eigenvalue
problems. We begin by multiplying the eigenvalue problem
L
n
=
n
(x)
n
by
n
and integrating. This gives
_
b
a
_

n
d
dx
_
p
d
n
dx
_
+q
2
n
_
dx =
_
b
a

2
n
dx.
One can solve the last equation for to nd
=

_
b
a
_

n
d
dx
_
p
dn
dx
_
+q
2
n
_
dx
_
b
a

2
n
dx
.
196 6 Sturm-Liouville Eigenvalue Problems
It appears that we have solved for the eigenvalue and have not needed the
machinery we had developed in Chapter 4 for studying boundary value prob-
lems. However, we really cannot evaluate this expression because we do not
know the eigenfunctions,
n
(x) yet. Nevertheless, we will see what we can
determine.
One can rewrite this result by performing an integration by parts on the
rst term in the numerator. Namely, pick u =
n
and dv =
d
dx
_
p
dn
dx
_
dx for
the standard integration by parts formula. Then, we have
_
b
a

n
d
dx
_
p
d
n
dx
_
dx = p
n
d
n
dx

b
a

_
b
a
_
p
_
d
n
dx
_
2
q
2
n
_
dx.
Inserting the new formula into the expression for , leads to the Rayleigh
Quotient

n
=
p
n
dn
dx

b
a
+
_
b
a
_
p
_
dn
dx
_
2
q
2
n
_
dx
_
b
a

2
n
dx
. (6.21)
In many applications the sign of the eigenvalue is important. As we had
seen in the solution of the heat equation, T

+ kT = 0. Since we expect
the heat energy to diuse, the solutions should decay in time. Thus, we would
expect > 0. In studying the wave equation, one expects vibrations and these
are only possible with the correct sign of the eigenvalue (positive again). Thus,
in order to have nonnegative eigenvalues, we see from (6.21) that
a. q(x) 0, and
b. p
n
dn
dx
|
b
a
0.
Furthermore, if is a zero eigenvalue, then q(x) 0 and
1
=
2
= 0
in the homogeneous boundary conditions. This can be seen by setting the
numerator equal to zero. Then, q(x) = 0 and

n
(x) = 0. The second of these
conditions inserted into the boundary conditions forces the restriction on the
type of boundary conditions.
One of the (unproven here) properties of Sturm-Liouville eigenvalue prob-
lems with homogeneous boundary conditions is that the eigenvalues are or-
dered,
1
<
2
< . . . . Thus, there is a smallest eigenvalue. It turns out that
for any continuous function, y(x),

1
= min
y(x)
py
dy
dx
|
b
a
+
_
b
a
_
p
_
dy
dx
_
2
qy
2
_
dx
_
b
a
y
2
dx
(6.22)
and this minimum is obtained when y(x) =
1
(x). This result can be used to
get estimates of the minimum eigenvalue by using trial functions which are
continuous and satisfy the boundary conditions, but do not necessarily satisfy
the dierential equation.
6.3 The Eigenfunction Expansion Method 197
Example 6.10. We have already solved the eigenvalue problem

+ = 0,
(0) = 0, (1) = 0. In this case, the lowest eigenvalue is
1
=
2
. We can
pick a nice function satisfying the boundary conditions, say y(x) = x x
2
.
Inserting this into Equation (6.22), we nd

1

_
1
0
(1 2x)
2
dx
_
1
0
(x x
2
)
2
dx
= 10.
Indeed, 10
2
.
6.3 The Eigenfunction Expansion Method
In section 4.3.2 we saw generally how one can use the eigenfunctions of a
dierential operator to solve a nonhomogeneous boundary value problem. In
this chapter we have seen that Sturm-Liouville eigenvalue problems have the
requisite set of orthogonal eigenfunctions. In this section we will apply the
eigenfunction expansion method to solve a particular nonhomogenous bound-
ary value problem.
Recall that one starts with a nonhomogeneous dierential equation
Ly = f,
where y(x) is to satisfy given homogeneous boundary conditions. The method
makes use of the eigenfunctions satisfying the eigenvalue problem
L
n
=
n

n
subject to the given boundary conditions. Then, one assumes that y(x) can
be written as an expansion in the eigenfunctions,
y(x) =

n=1
c
n

n
(x),
and inserts the expansion into the nonhomogeneous equation. This gives
f(x) = L
_

n=1
c
n

n
(x)
_
=

n=1
c
n

n
(x)
n
(x).
The expansion coecients are then found by making use of the orthogo-
nality of the eigenfunctions. Namely, we multiply the last equation by
m
(x)
and integrate. We obtain
_
b
a
f(x)
m
(x) dx =

n=1
c
n

n
_
b
a

n
(x)
m
(x)(x) dx.
198 6 Sturm-Liouville Eigenvalue Problems
Orthogonality yields
_
b
a
f(x)
m
(x) dx = c
m

m
_
b
a

2
m
(x)(x) dx.
Solving for c
m
, we have
c
m
=
_
b
a
f(x)
m
(x) dx

m
_
b
a

2
m
(x)(x) dx
.
Example 6.11. As an example, we consider the solution of the boundary value
problem
(xy

+
y
x
=
1
x
, x [1, e], (6.23)
y(1) = 0 = y(e). (6.24)
This equation is already in self-adjoint form. So, we know that the associ-
ated Sturm-Liouville eigenvalue problem has an orthogonal set of eigenfunc-
tions. We rst determine this set. Namely, we need to solve
(x

+

x
= , (1) = 0 = (e). (6.25)
Rearranging the terms and multiplying by x, we have that
x
2

+x

+ (1 +x) = 0.
This is almost an equation of Cauchy-Euler type. Picking the weight function
(x) =
1
x
, we have
x
2

+x

+ (1 +) = 0.
This is easily solved. The characteristic equation is
r
2
+ (1 +) = 0.
One obtains nontrivial solutions of the eigenvalue problem satisfying the
boundary conditions when > 1. The solutions are

n
(x) = Asin(n ln x), n = 1, 2, . . . .
where
n
= n
2

2
1.
It is often useful to normalize the eigenfunctions. This means that one
chooses A so that the norm of each eigenfunction is one. Thus, we have
1 =
_
e
1

n
(x)
2
(x) dx
= A
2
_
e
1
sin(n ln x)
1
x
dx
= A
2
_
1
0
sin(ny) dy =
1
2
A
2
. (6.26)
6.4 The Fredholm Alternative Theorem 199
Thus, A =

2.
We now turn towards solving the nonhomogeneous problem, Ly =
1
x
. We
rst expand the unknown solution in terms of the eigenfunctions,
y(x) =

n=1
c
n

2 sin(n ln x).
Inserting this solution into the dierential equation, we have
1
x
= Ly =

n=1
c
n

2 sin(n ln x)
1
x
.
Next, we make use of orthogonality. Multiplying both sides by
m
(x) =

2 sin(m ln x) and integrating, gives

m
c
m
=
_
e
1

2 sin(m ln x)
1
x
dx =

2
m
[(1)
m
1].
Solving for c
m
, we have
c
m
=

2
m
[(1)
m
1]
m
2

2
1
.
Finally, we insert our coecients into the expansion for y(x). The solution
is then
y(x) =

n=1
2
n
[(1)
n
1]
n
2

2
1
sin(n ln(x)).
6.4 The Fredholm Alternative Theorem
Given that Ly = f, when can one expect to nd a solution? Is it unique? These
questions are answered by the Fredholm Alternative Theorem. This theorem
occurs in many forms from a statement about solutions to systems of algebraic
equations to solutions of boundary value problems and integral equations. The
theorem comes in two parts, thus the term alternative. Either the equation
has exactly one solution for all f, or the equation has many solutions for some
fs and none for the rest.
The reader is familiar with the statements of the Fredholm Alternative
for the solution of systems of algebraic equations. One seeks solutions of the
system Ax = b for A an nm matrix. Dening the matrix adjoint, A

through
< Ax, y >=< x, A

y > for all x, y, C

n
, then either
Theorem 6.12. First Alternative
The equation Ax = b has a solution if and only if < b, v >= 0 for all v
such that A

v = 0.
200 6 Sturm-Liouville Eigenvalue Problems
or
Theorem 6.13. Second Alternative
A solution of Ax = b, if it exists, is unique if and only if x = 0 is the only
solution of Ax = 0.
The second alternative is more familiar when given in the form: The solu-
tion of a nonhomogeneous system of n equations and n unknowns is unique
if the only solution to the homogeneous problem is the zero solution. Or,
equivalently, A is invertible, or has nonzero determinant.
Proof. We prove the second theorem rst. Assume that Ax = 0 for x = 0
and Ax
0
= b. Then A(x
0
+ x) = b for all . Therefore, the solution is not
unique. Conversely, if there are two dierent solutions, x
1
and x
2
, satisfying
Ax
1
= b and Ax
2
= b, then one has a nonzero solution x = x
1
x
2
such that
Ax = A(x
1
x
2
) = 0.
The proof of the rst part of the rst theorem is simple. Let A

v = 0 and
Ax
0
= b. Then we have
< b, v >=< Ax
0
, v >=< x
0
, A

v >= 0.
For the second part we assume that < b, v >= 0 for all v such that A

v = 0.
Write b as the sum of a part that is in the range of A and a part that in the
space orthogonal to the range of A, b = b
R
+ b
O
. Then, 0 =< b
O
, Ax >=<
A

b, x > for all x. Thus, A

b
O
. Since < b, v >= 0 for all v in the nullspace of
A

, then < b, b
O
>= 0. Therefore, < b, v >= 0 implies that 0 =< b,
O
>=<
b
R
+ b
O
, b
O
>=< b
O
, b
O
> . This means that b
O
= 0, giving b = b
R
is in the
range of A. So, Ax = b has a solution.
Example 6.14. Determine the allowed forms of b for a solution of Ax = b to
exist, where
A =
_
1 2
3 6
_
First note that A

= A
T
. This is seen by looking at
< Ax, y > = < x, A

y >
n

i=1
n

j=1
a
ij
x
j
y
i
=
n

j=1
x
j
n

j=1
a
ij
y
i
=
n

j=1
x
j
n

j=1
( a
T
)
ji
y
i
(6.27)
For this example,
A

=
_
1 3
2 6
_
6.4 The Fredholm Alternative Theorem 201
We next solve A

v = 0. This means, v
1
+ 3v
2
= 0. So, the nullspace of A

is
spanned by v = (3, 1)
T
. For a solution of Ax = b to exist, b would have to
be orthogonal to v. Therefore, a solution exists when
b =
_
1
3
_
.
So, what does this say about solutions of boundary value problems? There
is a more general theory for linear operators. The matrix formulations follows,
since matrices are simply representations of linear transformations. A more
general statement would be
Theorem 6.15. If L is a bounded linear operator on a Hilbert space, then
Ly = f has a solution if and only if < f, v >= 0 for every v such that
L

v = 0.
The statement for boundary value problems is similar. However, we need
to be careful to treat the boundary conditions in our statement. As we have
seen, after several integrations by parts we have that
< Lu, v >= S(u, v)+ < u, L

v >,
where S(u, v) involves the boundary conditions on u and v. Note that for
nonhomogeneous boundary conditions, this term may no longer vanish.
Theorem 6.16. The solution of the boundary value problem Lu = f with
boundary conditions Bu = g exists if and only if
< f, v > S(u, v) = 0
for all v satisfying L

v = 0 and B

v = 0.
Example 6.17. Consider the problem
u

+u = f(x), u(0) u(2) = , u

(0) u

(2) = .
Only certain values of and will lead to solutions. We rst note that L = L

=
d
2
dx
2
+ 1.
Solutions of
L

v = 0, v(0) v(2) = 0, v

(0) v

(2) = 0
are easily found to be linear combinations of v = sinx and v = cos x.
202 6 Sturm-Liouville Eigenvalue Problems
Next one computes
S(u, v) = [u

v uv

]
2
0
= u

(2)v(2) u(2)v

(2) u

(0)v(0) +u(0)v

(0). (6.28)
For v(x) = sin x, this yields
S(u, sinx) = u(2) +u(0) = .
Similarly,
S(u, cos x) = .
Using < f, v > S(u, v) = 0, this leads to the conditions
_
2
0
f(x) sin xdx = ,
_
2
0
f(x) cos xdx = .
Problems
6.1. Find the adjoint operator and its domain for Lu = u

+4u

3u, u

(0) +
4u(0) = 0, u

(1) + 4u(1) = 0.
6.2. Show that a Sturm-Liouville operator with periodic boundary conditions
on [a, b] is self-adjoint if and only if p(a) = p(b). [Recall, periodic boundary
conditions are given as u(a) = u(b) and u

(a) = u

(b).]
6.3. The Hermite dierential equation is given by y

2xy

+y = 0. Rewrite
this equation in self-adjoint form. From the Sturm-Liouville form obtained,
verify that the dierential operator is self adjoint on (, ). Give the inte-
gral form for the orthogonality of the eigenfunctions.
6.4. Find the eigenvalues and eigenfunctions of the given Sturm-Liouville
problems.
a. y

+y = 0, y

(0) = 0 = y

().
b. (xy

+

x
y = 0, y(1) = y(e
2
) = 0.
6.5. The eigenvalue problem x
2
y

xy

+ y = 0 with y(1) = y(2) = 0 is

not a Sturm-Liouville eigenvalue problem. Show that none of the eigenvalues
are real by solving this eigenvalue problem.
6.6. In Example 6.10 we found a bound on the lowest eigenvalue for the given
eigenvalue problem.
a. Verify the computation in the example.
6.4 The Fredholm Alternative Theorem 203
b. Apply the method using
y(x) =
_
x, 0 < x <
1
2
1 x,
1
2
< x < 1.
Is this an upper bound on
1
c. Use the Rayleigh quotient to obtain a good upper bound for the lowest
eigenvalue of the eigenvalue problem:

+ ( x
2
) = 0, (0) = 0,

(1) = 0.
6.7. Use the method of eigenfunction expansions to solve the problem:
y

+ 4y = x
2
, y(0) = y(1) = 0.
6.8. Determine the solvability conditions for the nonhomogeneous boundary
value problem: u

+ 4u = f(x), u(0) = , u

(1) = .