Boyce/DiPrima 9th ed, Ch 11.2: Sturm-Liouville Boundary Value ProblemsElementary Differential Equations and Boundary Value Problems, 9th 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

together with the boundary conditions

10,0)()(])([ xyxryxqyxp

0)1()1(,0)0()0( 2121 yyyy

Differential Operator Form

Consider the linear homogeneous differential operator L:

The differential equation

can then be written as

We assume that the functions p, p', q and r are continuous on0 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:

yxqyxpyL )(])([

0)()(])([ yxryxqyxp

yxryL )(

0)1()1(,0)0()0( 2121 yyyy

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

Integrating the first term on the right twice by parts, we obtain

We thus obtain Lagrange’s identity:

1

0

1

0])([][ dxquvvupvdxuL

1

0

1

0

1

0

1

0

1

0

1

0

][)()()()()(

])([)()()()()()(][

dxvuLxvxuxvxuxp

dxquvuvpxvxuxpxvxuxpvdxuL

1

0

1

0)()()()()(][][ xvxuxvxuxpdxvuLvuL

Lagrange’s Identity Using Boundary Conditions (2 of 3)

Suppose that u and v satisfy the boundary conditions

If neither 2 nor 2 is zero, then

The result also holds if either 2 or 2 is zero.

Thus Lagrange’s identity reduces to

0

)0()0()0()0()0()1()1()1()1()1(

)0()0()0()0()0()1()1()1()1()1(

)()()()()(

2

1

2

1

1

0

vuvupvuvup

vuvupvuvup

xvxuxvxuxp

0)1()1(,0)0()0( 2121 yyyy

0][][1

0 dxvuLvuL

Lagrange’s Identity: Inner Product Form (3 of 3)

From the previous slide, we have Lagrange’s identity

In Section 10.2 we introduced the inner product (u,v) with

Thus Lagrange’s identity can be rewritten as

provided the previously mentioned assumptions are satisfied.

0][][1

0 dxvuLvuL

0)()(),(1

0 dxxvxuvu

,0][,],[ vLuvuL

Complex Inner Products

The inner product of two complex-valued functions on [0,1] is defined as

where is the complex conjugate of v.

Lagrange’s identity, as shown in the previous slides,

is still valid in this case since p(x), q(x), 1, 1, 2, and 2 are all real quantities.

0)()(),(1

0 dxxvxuvu

,0][,],[ vLuvuL

Sturm-Liouville Problems, Lagrange’s Identity

Consider again the Sturm-Liouville boundary value problem

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

whose eigenvalues are n = n2 2 and corresponding eigenfunctions n(x) = sin n x.

These results rely on Lagrange’s identity.

,0)1(,0)0(,0 yyyy

0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

Theorem 11.2.1

Consider the Sturm-Liouville boundary value problem

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.

0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

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

or

Since r(x) is real, we have

Recalling r(x) > 0 on [0,1], it follows that is real.

0,,0][,],[ rrLL

1

0

1

0)()()()()()( dxxrxxdxxrxx

0)()()(0)()()(1

0

221

0 dxxrxVxUdxxrxx

Theorem 11.2.2

Consider the Sturm-Liouville boundary value problem

If 1(x) and 2(x) are two eigenfunctions corresponding to the eigenvalues 1 and 2, respectively, and if 1 2, then

Thus 1(x) and 2(x) are orthogonal with respect to the weight function r(x).

0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

0)()()(1

0 21 dxxrxx

Theorem 11.2.2: Proof Outline

Let 1, 2, 1(x) and 2(x) satisfy the hypotheses of theorem.

Then

or

Since r(x), 2 and 2(x) are all real, we have

Since 1 2, we have

0,,0][,],[ 21212121 rrLL

1

0 21

1

0 211 )()()()()()( dxxrxxdxxrxx

0)()()(1

0 2121 dxxrxx

0)()()(1

0 21 dxxrxx

Theorem 11.2.3

Consider the Sturm-Liouville boundary value problem

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

Moreover,

0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

n 321

n

nlim

Illustration of Sturm-Liouville Properties

The properties stated in Theorems 11.2.1 – 11.2.3 are again illustrated by the Sturm-Liouville problem

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].

,0)1(,0)0(,0 yyyy

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

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.

,2,1,1)()(1

0

2 ndxxrxn

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:

Thus, for our orthonormal eigenfunctions, we have

nm

nmmn if,1

if,0

mnnm dxxrxx 1

0)()()(

Example 1: Orthonormal Eigenfunctions

Consider the Sturm-Liouville problem

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

Now

and hence the orthonormal eigenfunctions are

,0)1(,0)0(,0 yyyy

,2,1,1)sin(1

0

2 ndxxnkn

,2/2cos12/sin11

0

221

0

22 nnn kdxxnkxdxnk

,2,1,sin2)( nxnxn

Example 2: Boundary Value Problem (1 of 3)

Consider the Sturm-Liouville problem

In this case is r(x) = 1. The eigenvalues n satisfy

and the corresponding eigenfunctions are

To find the orthonormal eigenfunctions, we choose kn so that

or

0)1()1(,0)0(,0 yyyyy

,2,1,1sin1

0

2 ndxxk nn

,2,1,sin nxkxy nnn

,2,1,1sin1

0

22 nxdxk nn

0cossin

Example 2: Orthonormal Eigenfunctions (2 of 2)

We have

where in the last step we have used

Thus the orthonormal eigenfunctions are

,2

cos1

2

cossin

4

2sin2

4

2sin

22

2cos

2

1sin1

2222

1

0

21

0

21

0

22

nn

n

nnnn

n

nnn

n

nn

nnnn

kkk

xxkdx

xkxdxk

0cossin

,2,1,)cos1(

sin2)(

2/12

n

xx

n

nn

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

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

which converges for each x in [0, 1].

This expansion of f is given in terms of the eigenfunctions of

dxxnxfbxnbxfn

nn

1

01

sin)(2,sin)(

0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

0)1(,0)0(,0 yyyy

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

To obtain cm, we assume the series can be integrated term by term, after multiplying both sides by m(x)r(x) and integrating:

In inner product form, we have

dxxrxxfcxcxf nn

nnn

1

01

)()()(),()(

mn

mnnm cdxxrxxcdxxrxxf

1

1

0

1

0)()()()()()(

,2,1,, mrfc mm

Theorem 11.2.4

Consider the Sturm-Liouville boundary value problem

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

converges to [ f (x+) + f (x-)]/2 at each point x in the open interval 0 < x < 1.

0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

dxxrxxfcxcxf nn

nnn

1

01

)()()(),()(

Theorem 11.2.4: Discussion

If f satisfies further conditions, then a stronger conclusion can be established. Consider again the Sturm-Liouville problem

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

converges to f (x) at each point in the closed interval [0, 1].

dxxrxxfcxcxf nn

nnn

1

01

)()()(),()(

0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

Example 3: Function f (1 of 3)

Consider the function

Recall from Example 2, for the Sturm-Liouville problem

the orthonormal eigenfunctions are

Then by Theorem 11.2.4, we have

,0)1()1(,0)0(,0 yyyyy

10,)( xxxf

,2,1,)cos1(

2,sin)(

2/12

nkxkx

n

nnnn

dxxrxxfcxcxf nn

nnn

1

01

)()()(),()(

Example 3: Coefficients (2 of 3)

Thus

Integrating by parts, we obtain

where in the last step we have used

Next, recall that

dxxxkdxxrxxfc nnnn 1

0

1

0sin)()()(

n

nn

n

n

n

nnn kkc

sin2cossin

0cossin

,2,1,)cos1(

22/12

nkn

n

Example 3: Eigenfunction Expansion (3 of 3)

We have

It follows that

Thus

,2,1,cos1

sin222

ncnn

nn

12

1 cos1

sinsin4sin)(

n nn

nn

nnnn

xxkcxf

n

n

nn

nnn kc

sin2

cos1

2sin22

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 Sturm-Liouville 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 Sturm-Liouville 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,

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.

,0)1()1(,0)0()0(

10,0)()(])([

2121

yyyy

xyxryxqyxp

Self-Adjoint Problems (1 of 4)

Consider the boundary value problem consisting of the differential equation L[y] = r(x)y, where

and n homogeneous boundary conditions at the endpoints.

If Lagrange’s identity

is valid for every pair of sufficiently differentiable functions that satisfy the boundary conditions, then the problem is said to be self-adjoint.

,)()()( 01 yxPdx

dyxP

dx

ydxPyL

n

n

n

0][,],[ vLuvuL

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,

This requires L to be of even order. Further, a second order operator must have the form

and a fourth order operator must have the form

Higher order operators must have an analogous structure.

0][,],[ vLuvuL

yxqyxpyL )(])([

yxsyxqyxpyL )(])([])([

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

and also in other cases, one of which is given in Example 4, as we will see.

0)1()1(,0)0()0( 2121 yyyy

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

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.

yxsyxqyxpyL )(])([])([

Sturm-Liouville Problems and Fourier Series

We have noted previously that Fourier sine (and cosine) series can be obtained using the eigenfunctions of certain Sturm-Liouville problems involving the differential equation

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.

0 yy

Example 4: Boundary Value Problem (1 of 3)

Consider the boundary value problem

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.

0)()(,0)()(,0 LyLyLyLyyy

Example 4: Eigenvalues and Eigenfunctions (2 of 3)

Our boundary value problem is

It can be shown that 0 = 0 is an eigenvalue with corresponding eigenfunction 0(x) = 1.All other eigenvalues are given by

To each of these eigenvalues there corresponds two linearly independent eigenfunctions.

For example, the eigenfunctions corresponding to n are

This shows that the eigenfunctions may not be simple when the boundary conditions are not separated.

0)()(,0)()(,0 LyLyLyLyyy

,/,,/2,/ 222

21 LnLL n

LnxLnx nn /sin)(,/cos)(

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

which is just the Fourier series of f.

1

0 sincos2

)(n

nn L

xnb

L

xna

axf