Boyce/DiPrima 9th ed, Ch 11.2: SturmLiouville Boundary Value Problems
Elementary Differential Equations and Boundary Value Problems, 9 th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
We now consider two-point boundary value problems of the
type obtained in Section 11.1 by separating the variables in a
heat conduction problem for a bar of variable material
properties and with a source term proportional to temperature.
This kind of problem also occurs in many other applications.
These boundary value problems are commonly known as
Sturm-Liouville problems.
They consist of a differential equation of the form
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
together with the boundary conditions
1 y(0)   2 y(0)  0, 1 y(1)  2 y(1)  0
Differential Operator Form
Consider the linear homogeneous differential operator L:
Ly  [ p( x) y]  q( x) y
The differential equation
[ p( x) y]  q( x) y  r ( x) y  0
can then be written as
Ly  r ( x) y
We assume that the functions p, p', q and r are continuous on
0  x  1, and further, that p(x) > 0 and r(x) > 0 on [0,1].
The boundary conditions are said to be separated, since each
one only involves one of the boundary points:
1 y(0)   2 y(0)  0, 1 y(1)  2 y(1)  0
Lagrange’s Identity
(1 of 3)
We next establish Lagrange’s identity, which is basic to the
study of linear boundary value problems.
Let u and v be functions having continuous second derivatives
on the interval 0  x  1. Then

1
0
1
L[u ]vdx   [( pu )v  quv]dx
0
Integrating the first term on the right twice by parts, we obtain
1
1
0
0
 L[u]vdx   p( x)u( x)v( x)
1
 p( x)u ( x)v( x) 0   [( pv)u  quv]dx
1
0
  p( x)u( x)v( x)  u ( x)v( x) 0   uL[v]dx
1
1
0
We thus obtain Lagrange’s identity:
1






L
[
u
]
v

uL
[
v
]
dx


p
(
x
)
u
(
x
)
v
(
x
)

u
(
x
)
v
(
x
)
0
0
1
Lagrange’s Identity Using
Boundary Conditions (2 of 3)
Suppose that u and v satisfy the boundary conditions
1 y(0)   2 y(0)  0, 1 y(1)  2 y(1)  0
If neither 2 nor 2 is zero, then
 p ( x)u ( x)v( x)  u ( x)v( x) 0
1
  p (1)u (1)v(1)  u (1)v(1)  p (0)u(0)v(0)  u (0)v(0)
 p (1)
1
u (1)v(1)  u (1)v(1)  p(0) 1 u (0)v(0)  u (0)v(0)
2
2
0
The result also holds if either 2 or 2 is zero.
Thus Lagrange’s identity reduces to
 L[u]v  uL[v] dx  0
1
0
Lagrange’s Identity: Inner Product Form
(3 of 3)
From the previous slide, we have Lagrange’s identity
 L[u]v  uL[v] dx  0
1
0
In Section 10.2 we introduced the inner product (u,v) with
1
(u, v)   u ( x)v( x)dx  0
0
Thus Lagrange’s identity can be rewritten as
L[u], v  u, L[v]  0,
provided the previously mentioned assumptions are satisfied.
Complex Inner Products
The inner product of two complex-valued functions on [0,1] is
defined as
1
(u , v)   u ( x)v ( x)dx  0
0
where is the complex conjugate of v.
Lagrange’s identity, as shown in the previous slides,
L[u], v  u, L[v]  0,
is still valid in this case since p(x), q(x), 1, 1, 2, and 2 are
all real quantities.
Sturm-Liouville Problems, Lagrange’s Identity
Consider again the Sturm-Liouville boundary value problem
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0
This problem always has eigenvalues and eigenvectors.
In Theorems 11.2.1 – 11.2.4 of this section, we will state
several of their important properties.
Each property is illustrated by the Sturm-Liouville problem
y  y  0, y (0)  0, y (1)  0,
whose eigenvalues are n = n2 2 and corresponding
eigenfunctions n(x) = sin n x.
These results rely on Lagrange’s identity.
Theorem 11.2.1
Consider the Sturm-Liouville boundary value problem
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0
All of the eigenvalues of this problem are real.
We give an outline of the proof on the next slide.
It can also be shown that all of the eigenfunctions are real.
Theorem 11.2.1: Proof Outline
Suppose that  is a (possibly complex) eigenvalue with
corresponding eigenvector .
Let  = u + iv, (x) = U(x) + iV(x), where u, v, U, V are real.
Then it follows from Lagrange’s identity that
L[ ],    , L[ ]  0
 r ,    , r   0
or
1
1
   ( x) ( x)r ( x)dx     ( x) ( x)r ( x)dx
0
0
Since r(x) is real, we have
     ( x) ( x)r ( x)dx  0
1
0

    U
1
0
2

( x)  V 2 ( x) r ( x)dx  0
Recalling r(x) > 0 on [0,1], it follows that  is real.
Theorem 11.2.2
Consider the Sturm-Liouville boundary value problem
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0
If 1(x) and 2(x) are two eigenfunctions corresponding to the
eigenvalues 1 and 2, respectively, and if 1  2, then
1
  ( x) ( x)r ( x)dx  0
0
1
2
Thus 1(x) and 2(x) are orthogonal with respect to the weight
function r(x).
Theorem 11.2.2: Proof Outline
Let 1, 2, 1(x) and 2(x) satisfy the hypotheses of theorem.
Then
L[1 ], 2   1, L[2 ]  0  r1,2   1, r2   0
or
1
1
1  1 ( x)2 ( x)r ( x)dx    1 ( x)2 ( x)r ( x)dx
0
0
Since r(x), 2 and 2(x) are all real, we have
1  2 0 1 ( x)2 ( x)r ( x)dx  0
1
Since 1  2, we have
1
  ( x) ( x)r ( x)dx  0
0
1
2
Theorem 11.2.3
Consider the Sturm-Liouville boundary value problem
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0
The eigenvalues are all simple. That is, to each eigenvalue
there corresponds one and only one linearly independent
eigenfunction.
Further, the eigenvalues form an infinite sequence and can be
ordered according to increasing magnitude so that
1  2  3    n  
Moreover,
lim n  
n 
Illustration of Sturm-Liouville Properties
The properties stated in Theorems 11.2.1 – 11.2.3 are again
illustrated by the Sturm-Liouville problem
y  y  0, y (0)  0, y (1)  0,
whose eigenvalues n = n2 2 are real, distinct and increasing,
such that n   as n  .
Also, the corresponding real eigenfunctions n(x) = sin n x are
orthogonal to each other with respect to r(x) = 1 on [0, 1].
Orthonormal Eigenfunctions
We now assume that the eigenvalues of the Sturm-Liouville
problem are ordered as indicated in Theorem 11.2.3.
Associated with the eigenvalue n is a corresponding real
eigenfunction n(x) determined up to a multiplicative constant.
It is often convenient to choose the arbitrary constant
multiplying each eigenfunction so as to satisfy the condition

1
0
n2 ( x)r ( x)dx  1, n  1, 2,
The eigenfunctions are said to be normalized, and form an
orthonormal set with respect to the weight function r, since
they are orthogonal and normalized.
Kronecker Delta
For orthonormal eigenfunctions, it is useful to combine the
orthogonal and normalization relations into one symbol.
To this end, the Kronecker delta mn is helpful:
 mn
0, if m  n

1, if m  n
Thus, for our orthonormal eigenfunctions, we have
1

0
m
( x)n ( x)r ( x)dx   mn
Example 1: Orthonormal Eigenfunctions
Consider the Sturm-Liouville problem
y  y  0, y (0)  0, y (1)  0,
In this case the weight function is r(x) = 1. The eigenvalues
and eigenfunctions are n = n2 2 and yn(x) = sin n x.
To find the orthonormal eigenfunctions, we choose kn so that

1
0
(k n sin n x) 2 dx  1, n  1, 2,
Now
1
1
2
2
2
1  k n  sin n xdx  k n / 2 1  cos 2n x dx  k n2 / 2,
0
0
and hence the orthonormal eigenfunctions are
n ( x)  2 sin n x, n  1, 2,
Example 2: Boundary Value Problem
(1 of 3)
Consider the Sturm-Liouville problem
y  y  0, y (0)  0, y (1)  y(1)  0
In this case is r(x) = 1. The eigenvalues n satisfy
sin    cos   0
and the corresponding eigenfunctions are
yn x   kn sin n x, n  1, 2,
To find the orthonormal eigenfunctions, we choose kn so that
 k
1
0
or
k
2
n

1
0

2
n
sin n x dx  1, n  1, 2, 
sin 2 n xdx  1, n  1, 2,
Example 2: Orthonormal Eigenfunctions
We have
1 k
2
n

1
0
sin 2
(2 of 2)
 x sin 2 n x 
1 1
cos 2 n x 
2
2
dx  k n  

n xdx  k n   
0 2

2
2
4 n 



1
0
2




2


sin
2



sin

cos

1

cos
n
n
n
n
n
n
2
2
2


 kn
 kn
 kn
,




2
4 n
2 n




where in the last step we have used
sin    cos   0
Thus the orthonormal eigenfunctions are
n ( x ) 
2 sin n x
(1  cos
2
n )
1/ 2
, n  1, 2, 
Eigenfunction Expansions
(1 of 2)
We now investigate expressing a given function f as a series of
eigenfunctions of the Sturm-Liouville boundary value problem
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0
For example, if f is continuous and has a piecewise continuous
derivative on [0, 1], and satisfies f (0) = f (1) = 0, then f can be
expanded in a Fourier sine series of the form

f ( x)   bn sin n x, bn 2 f ( x) sin n x dx
n 1
1
0
which converges for each x in [0, 1].
This expansion of f is given in terms of the eigenfunctions of
y  y  0, y (0)  0, y (1)  0
Eigenfunction Expansions
(2 of 2)
Suppose that a given function f, satisfying suitable conditions,
can be expressed in a series of orthonormal eigenfunctions
n(x) of the Sturm-Liouville boundary value problem.
Then

f ( x)   cnn ( x), cn   f ( x)n ( x)r ( x)dx
1
0
n 1
To obtain cm, we assume the series can be integrated term by
term, after multiplying both sides by m(x)r(x) and integrating:

1
0

f ( x)m ( x)r ( x)dx   cn  n ( x)m ( x)r ( x)dx  cm
n 1
1
0
In inner product form, we have
cm   f , m r , m  1, 2,
Theorem 11.2.4
Consider the Sturm-Liouville boundary value problem
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0
Let 1, 2, …, n,… be the normalized eigenfunctions for this
problem, and let f and f ' be piecewise continuous on 0  x 1.
Then the series

1
f ( x)   cnn ( x), cn   f ( x)n ( x)r ( x)dx
n 1
0
converges to [ f (x+) + f (x-)]/2 at each point x in the open
interval 0 < x < 1.
Theorem 11.2.4: Discussion
If f satisfies further conditions, then a stronger conclusion can
be established. Consider again the Sturm-Liouville problem
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0
Suppose f is continuous and f ' piecewise continuous on[0,1].
If 2 = 0, then assume that f(0) = 0. If 2 = 0, then assume that
f(1) = 0. Otherwise no boundary conditions need to be
prescribed for f. Then the series

f ( x)   cnn ( x), cn   f ( x)n ( x)r ( x)dx
n 1
1
0
converges to f (x) at each point in the closed interval [0, 1].
Example 3: Function f
(1 of 3)
Consider the function
f ( x )  x, 0  x  1
Recall from Example 2, for the Sturm-Liouville problem
y  y  0, y (0)  0, y (1)  y(1)  0,
the orthonormal eigenfunctions are
n ( x)  k n sin n x, kn 
2
(1  cos
2
n )
1/ 2
Then by Theorem 11.2.4, we have

f ( x)   cnn ( x), cn   f ( x)n ( x)r ( x)dx
n 1
1
0
, n  1, 2,
Example 3: Coefficients
(2 of 3)
Thus
1
1
0
0
cn   f ( x)n ( x)r ( x)dx  k n  x sin n x dx
Integrating by parts, we obtain
 sin n cos n
cn  k n 

 n
n


2 sin n
  kn

n

where in the last step we have used
sin    cos   0
Next, recall that
kn 
2
(1  cos
2
n )
1/ 2
, n  1, 2,
Example 3: Eigenfunction Expansion
We have
cn  k n
2 sin n
n

 2 sin 
2
n


 1  cos 2  
n
n


It follows that
cn 
2 2 sin n
n 1  cos
2
n
, n  1, 2, 
Thus


n 1
n 1
f ( x)   cn k n sin n x  4 
sin n sin n x

n 1  cos 2 n

(3 of 3)
Sturm-Liouville Problems and Algebraic
Eigenvalue Problems
Sturm-Liouville boundary value problems are of great
importance in their own right, but they can also be viewed as
belonging to a much more extensive class of problems that
have many of the same properties.
For example, there are many similarities between SturmLiouville problems and the algebraic system Ax = x, where
the n x n matrix A is real symmetric or Hermitian.
Comparing the results mentioned in Section 7.3 with those of
this section, in both cases the eigenvalues are real and the
eigenfunctions or eigenvectors form an orthogonal set, and can
be used as a basis for expressing an essentially arbitrary
function or vector, respectively, as a sum.
Linear Operator Theory
The most important difference is that a matrix has only a finite
number of eigenvalues and eigenvectors, while a SturmLiouville system has infinitely many.
It is of fundamental importance in mathematics that these
seemingly different problems – the matrix problem Ax = x
and the Sturm-Liouville problem,
[ p( x) y]  q( x) y  r ( x) y  0, 0  x  1
1 y (0)   2 y(0)  0, 1 y (1)   2 y(1)  0,
which arise in different ways, are actually parts of a single
underlying theory.
This theory is linear operator theory, and is part of the subject
of functional analysis.
Self-Adjoint Problems
(1 of 4)
Consider the boundary value problem consisting of the
differential equation L[y] = r(x)y, where
dny
dy
L y   Pn ( x) n    P1 ( x)  P0 ( x) y,
dx
dx
and n homogeneous boundary conditions at the endpoints.
If Lagrange’s identity
L[u], v  u, L[v]  0
is valid for every pair of sufficiently differentiable functions
that satisfy the boundary conditions, then the problem is said
to be self-adjoint.
Self-Adjoint Problems and
Structure of Differential Operator L
(2 of 4)
Lagrange’s identity involves restrictions on both the
differential equation and the boundary conditions.
The differential operator L must be such that the same operator
appears in both terms of Lagrange’s identity,
L[u], v  u, L[v]  0
This requires L to be of even order. Further, a second order
operator must have the form
Ly  [ p( x) y]  q( x) y
and a fourth order operator must have the form
Ly  [ p( x) y]  [q( x) y]  s( x) y
Higher order operators must have an analogous structure.
Self-Adjoint Problems and
Boundary Conditions (3 of 4)
In addition, the boundary conditions must be such as to
eliminate the boundary terms that arise during the integration
by parts used in deriving Lagrange’s identity.
For example, in a second order problem this is true for the
separated boundary conditions
1 y(0)   2 y(0)  0, 1 y(1)  2 y(1)  0
and also in other cases, one of which is given in Example 4, as
we will see.
Fourth Order Self-Adjoint Problems and
Eigenvalue, Eigenvector Properties (4 of 4)
Suppose we have self-adjoint boundary value problem for
L[y] = r(x)y, where L[y] is given by
Ly  [ p( x) y]  [q( x) y]  s( x) y
We assume that p(x), q(x), r(x), s(x) are continuous on [0,1]
and that the derivatives p', p'' and q' are also continuous.
Suppose also that p(x) > 0 and r(x) > 0 on [0,1].
Then there is an infinite sequence of real eigenvalues tending
to , the eigenfunctions are orthogonal with respect to the
weight function r, and an arbitrary function f can be expressed
as a series of eigenfunctions.
However, the eigenfunctions may not be simple in these more
general problems.
Sturm-Liouville Problems and Fourier Series
We have noted previously that Fourier sine (and cosine) series
can be obtained using the eigenfunctions of certain SturmLiouville problems involving the differential equation
y  y  0
This raises the question of whether we can obtain a full
Fourier series, including both sine and cosine terms, by
choosing a suitable set of boundary conditions.
The answer is yes, as we will see in the following example.
This example will also serve to illustrate the occurrence of
nonseparated boundary conditions.
Example 4: Boundary Value Problem
(1 of 3)
Consider the boundary value problem
y  y  0, y ( L)  y ( L)  0, y( L)  y( L)  0
This is not a Sturm-Liouville problem because the boundary
conditions are not separated.
The boundary conditions above are called periodic boundary
conditions since the require that y and y' assume the same
values at x = L as at x = -L.
It is straightforward to show that this problem is self-adjoint.
Example 4:
Eigenvalues and Eigenfunctions
(2 of 3)
Our boundary value problem is
y  y  0, y ( L)  y ( L)  0, y( L)  y( L)  0
It can be shown that 0 = 0 is an eigenvalue with
corresponding eigenfunction 0(x) = 1.
All other eigenvalues are given by
1   / L2 , 2  2 / L2 ,, n  n / L2 ,
To each of these eigenvalues there corresponds two linearly
independent eigenfunctions.
For example, the eigenfunctions corresponding to n are
n ( x)  cosn / L ,  n ( x)  sin n / L
This shows that the eigenfunctions may not be simple when
the boundary conditions are not separated.
Example 4: Eigenfunction Expansion (3 of 3)
Further, if we seek to expand a given function f of period 2L in
a series of eigenfunctions for this problem, we obtain
a0  
n x
n x 
f ( x) 
   an cos
 bn sin
2 n 1 
L
which is just the Fourier series of f.

L 