Sturm -Lioville Problems

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45

Sturm-Liouville Problems

45.1 PreliminariesOn the Study of Sturm-Liouville Problems

Sturm-Liouville problems are boundary-value problems that naturally arise when solving cer-

tain partial differential equation problems using the separation of variables method.1 It is the

theory behind Sturm-Liouville problems that, ultimately, justifies the separation of variables

method for these partial differential equation problems. The simplest applications lead to the var-

ious Fourier series, and less simple applications lead to generalizations of Fourier series involving

Bessel functions, Hermite polynomials, etc.

Unfortunately, there are several difficulties with our studying Sturm-Liouville problems:

1. Motivation: Had we the time, we would first discuss partial differential equation problems

and develop the separation of variables method for solving certain important types of

problems involving partial differential equations. We would then see how these Sturm-Liouville problems arise and why they are so important. But we dont have time. Instead,

Ill briefly remind you of some results from linear algebra that are analogous to the results

we will, eventually, obtain.

2. Another difficulty is that the simplest examples (which are very important since they lead

to the Fourier series) are too simple to really illustrate certain elements of the theory,

while the other standard examples tend to get complicated and require additional tricks

which distract from illustrating the theory. Well deal with this as well as we can.

3. The material getstheoretical. Sorry, there is no way around this. The end results, however,

are very useful in computations, especially now that we have computers to do the tedious

computations. I hope we get to that point.

4. Finally, I must warn you that, in most texts, the presentation of the Sturm-Liouville

theory stinks. In differential equation texts, this material is usually near the end, which

means that the author just wants to finish the damn book, get it published. In texts on

partial differential equations, the results are usually quoted with some indication that the

proofs can be found in a good text on differential equations.

1 not to be confused with the separation of variables method discussed in chapter 4 for solving separable first-order

ordinary differential equations.

4/22/2013


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Chapter & Page: 452 Sturm-Liouville

Linear Algebraic Antecedents

Let us briefly review some elements from linear algebra which we will be recreating using

functions instead of finite-dimensional vectors. But first, let me remind you of a little complex

variable algebra. This is because our column vectors and matrices may have complex values.

Complex Conjugates and Magnitudes

Weve certainly used complex numbers before in this text, but I should remind you that a complex

number z is something that can be written as

z = x + i y

where x and y are real numbers the real and imaginary parts, respectively of z. The

corresponding complex conjugate z and magnitude|z| of z are then given by

z = x i y and |z| =

x 2 +y 2 .

Now if x is a real number positive, negative or zero then x2

=|x |2

. However, you caneasily verify that z2 =|z|2 . Instead, we have

|z|2 = x 2 + y2 = x 2 (i y)2 = (x i y)(x+ i y) = zz .

Recollections From Linear Algebra

When we treat N as a vector space, we are discussing the vector space of all N1 matriceswith real components. That is, saying that v and w are vectors in N will simply mean

v = v1

v2

...

vN

and w = w1

w2

...

wN

where the vks and wks are real numbers.

Recall that the norm (or length) of the above v ,v , is computed by

v=

|v1|2 + |v2|2 + |vN|2

and that, computationally, the classic dot product of the above v and w is

v w = v1w1 + v2w2 + + vNwN

= [v1, v2, . . . , vN]

w1

w2...

wN

= vTw

where vTw is the matrix product with vT being the transpose of matrix v (i.e., the matrix

constructed from v by switching rows with columns).


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Preliminaries Chapter & Page: 453

Observe that, since v2 =|v|2 when v is real,v v = v1v1 + v2v2 + + vNvN

=|v1|2 +|v2|2 + +|vN|2 =v2 .Also recall that a set of vectors

{v1, v2, v2, . . .

} is said to be orthogonal if and only if

vm vn = 0 whenever m=n .And finally, recall that a matrix A is symmetric if and only if AT = A , and that, for

symmetric matrices, we have the following theorem from linear algebra (and briefly mentioned

in chapter 39):

Theorem 45.1

Let A be a symmetric NN matrix (with real-valued components). Then both of the followinghold:

1. All the eigenvalues of A are real.

2. There is an orthogonal basis for N consisting of eigenvectors for A .

We, ultimately, will obtain an analog to this theorem involving a linear differential operator

instead of a matrix A .

Vectors and Matrices with Complex-Valued Components

Everything mentioned above can be generalized to N , the vector space of all N1 matriceswith complex components. In this case, saying that v and w are vectors in N simply means

v = v1

v2

...

vN

and w = w1

w2

...

wN

where the vks and wks are complex numbers. Since the components are complex, we define

the complex conjugate of v in the obvious way,

v =

v1v2...

vN

.

The natural norm (or length) of v is still given by

v=

|v1|2 + |v2|2 + |vN|2

However, when we try to relate this to the classic dot product v v , we getv2 =|v1|2 +|v2|2 + +|vN|2

= v1v1 + v2v2 + + vNvN= (v) v = v v .


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This suggests that, instead of using the classic dot product, we use the (standard vector) inner

productof v with w , which is denoted by v | w and defined by by

v| w= v w = v1w1 + v2w2 + + vNwN

= [v1, v2, . . . , vN]

w1

w2...

wN

= (v)Tw .

Then

v| v= v v =v2 .

We also adjust our notion of orthogonality by saying that any set{v1, v2, . . .} of vectors in

N isorthogonalif and only if

vm vn = 0 whenever m=n .The inner product for vectors with complex components is the mathematically natural ex-

tension of the standard dot product for vectors with real components. Some easily verified (and

useful) properties of this inner product are given in the next theorem. Verifying it will be left as

an exercise (see exercise 45.4).

Theorem 45.2 (properties of the inner product)

Suppose and are two (possibly complex) constants, and v , w , and u are vectors in N .

Then

1.

v

| w

=w

| v

,

2. u| v + w= u| v+ u| w ,

3. v + w| u= v| u+ w| u ,and

4. v| v=v .

Later, well define other inner products for functions. These inner products will have very

similar properties to those in given in the last theorem.

Adjoints

Theadjointof any matrix A denoted A is the transpose of the complex conjugate of A ,

A = (A)T (equivalently, AT) .That is, A is the matrix obtained from matrix A by replacing each entry in A with its complex

conjugate, and then switching the rows and columns (or first switch the rows and columns and

then replace the entries with their complex conjugates you get the same result either way).


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!Example 45.1: If

A =

1 + 2i 3 4i 5i6i 7 8i

,

then

A =1 + 2i 3 4i 5i

6i 7 8iT

=1 2i 3 + 4i 5i

6i 7 8i

T=

1 2i 6i3 + 4i 75i 8i

.This adjoint turns out to be more useful than the transpose when we allow vectors to have

complex components.

A matrix A isself adjoint2 if and only if A =A . Note that:1. A self-adjoint matrix is automatically square.

2. IfA is a square matrix with just real components, thenA =AT , and A is self adjointmeans the same as A is symmetric.

If you take the proof of theorem 45.1 and modify it to take into account the possibility of

complex-valued components, you get

Theorem 45.3

Let A be a self-adjoint NN matrix. Then both of the following hold:1. All the eigenvalues of A are real.

2. There is an orthogonal basis for N consisting of eigenvectors for A .

This theorem is noteworthy because it will help explain thesource of some of theterminology

that we will later be using.

What is more noteworthy is what the above theorem says about computing Av when A is

self adjoint. To see this, let b1 , b2 , b3 , . . . , bN

be any orthogonal basis for N consisting of eigenvectors for A (remember, the theorem says

there is such a basis), and let

{ 1, 2, 3, . . . , N}be the corresponding set of eigenvectors (so Abk = kbk for k= 1, 2, . . . ,N ). Since the setof bks is a basis, we can express v as a linear combination of these basis vectors.

v

= v1b

1

+ v2b

2

+ v3b

3

+ + vNb

N

=

N

k=1

vkbk . (45.1)

Av can then be computed via

Av = A v1b1 + v2b2 + v3b3 + + vNbN= v1Ab1 + v2Ab2 + v3Ab3 + + vNAbN

= v11b1 + v22b2 + v33b3 + + vNNbN .2 the termHermitianis also used


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If N is large, this could be a lot faster than doing the basic component-by-component matrix

multiplication. (And in our analog with functions, Nwill be infinite.)

Finding Vector Components

One issue is finding the components (v1, v2, v3, . . . , vN) in formula (45.1) for v . But a useful

formula is easily derived. First, take the inner product of both sides of (45.1) with one of thebks , say, b1 and then repeatedly use the linearity property described in theorem 45.2,

b1 v = b1 v1b1 + v2b2 + v3b3 + + vNbN

= v1b1 b1+ v2 b1 b2+ v3 b1 b3+ + vN b1 bN .

Remember, this set of bks is orthogonal. That meansb1 bk = 0 if 1=k .

On the other hand, b1 b1 = b12 .Taking this into account, we continue our computation of b1 v :

b1 v = v1 b1 b1+ v2 b1 b2+ v3 b1 b3+ + vN b1 bN

= v1b12 + v2 0+ v3 0+ + vN 0 .

So, b1 v = v1 b12 .

Solving for v1 gives us

v1

= b1

v

b12 .Repeating this using each bk yields the next theorem:

Theorem 45.4

Let b1 , b2 , b3 , . . . , bN

be an orthogonal basis for N . Then, for any vector v in N ,

v = v1b1 + v2b2 + v3b3 + + vNbN =N

k1vkb

k (45.2a)

where

vk=bk v bk2 for k=1, 2, 3, . . . , N . (45.2b)

If you know a little about Fourier series, then you may recognize formula set (45.2) as a

finite-dimensional analog of the Fourier series formulas. If you know nothing about Fourier

series, then Ill tell you that a Fourier series for a function f is an infinite-dimensional analog

of formula set (45.2). We will eventually see this.


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About Normalizing a Basis

In the above, we assumed that b1 , b2 , b3 , . . . , bN

is an orthogonal basis for N . If it had been orthonormal, then we would also have had

bk = 1 for k= 1, 2, 3, . . . , N ,and the formulas in our last theorem would have simplified somewhat. In fact, given any orthog-

onal set b1 , b2 , b3 , . . . , bN

,

we can construct a corresponding orthonormal setn1 , n2 , n3 , . . . , nN

by just letting

nk = bk

bk

for k=1, 2, 3, . . . , N .

Moreover, if bk

is an eigenvector for a matrix A with corresponding eigenvalue k, so is nk

.When we so compute the nks from the bks , we are said to be normalizing our bks .

Some authors like to normalize their orthogonal bases because it does yield simpler formulas for

computing with such a basis. These authors, typically, are only deriving pretty results and are not

really using them in applications. Those that really use the results rarely normalize, especially

when the dimension is infinite (as it will be for us), because normalizing leads to artificial

formulas for the basis vectors, and dealing with these artificial formulas for basis vectors usually

complicates matters enough to completely negate the advantages of having the simpler formulas

for computing with these basis vectors.

We wont normalize.

Inner Products and Adjoints

You should recall (or be able to quickly confirm) that

(AB)T = BTAT ,

ATT = A , (AB) = AB and A = A .

With these and the definition of the adjoint, you can easily verify that

(AB) = BA and A = A .Also observe that, if

v=

v1

v2

...

vN

and w =

w1

w2

...

wN

,

then

vw = v1 , v2 , . . . , vN

w1

w2...

wN

= v1w1 + v2w2+ + vNwN =v| w .


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So, for any N N matrix A and any two vectors v and u in N ,

v

Au

= v

Au

= vAu = (Av)u = Av| u .

Consequently, if A is self adjoint (i.e., A

=A ), thenv| Au=Av| u for every v, u in N

Its a simple exercise in linear algebra to show that the above completely characterizes

adjointness and self adjointness for matrices. That is, you should be able to finish proving

the next theorem. Well use the results to extend the these concepts to things other than matrices.

Theorem 45.5 (characterization of adjointness)

Let A be an NN matrix. Then,

v Au =Av| u for every v and u in N .

Moreover, both of the following hold:

1. B=A if and only if

v| Bu= Av| u for every v and u in N .

2. A is self adjoint if and only if

v| Au=Av| u for every v and u in N .

Comments

To be honest, we are not going to directly use the material weve developed over the past several

pages. The reason we went over the theory of self-adjoint matrices and related material

concerning vectors in N is that the Sturm-Liouville theory well be developing is a functional

analog of what we just discussed, using differential operators and functions instead of matrices

and vectors in N . Understanding the theory and computations weve just developed should

expedite learning the theory and computations we will be developing.

45.2 Boundary-Value Problems with Parameters

Our interest will be in homogeneous boundary-value problems in which the differential equations

involve both the unknown function which we will start denoting by or (x ) and a

parameter (basically, is just some yet undetermined constant). The equations that will be

of interest can all be written as

A(x )d2

dx 2 + B(x) d

d x+ [C(x ) + ] = 0 (45.3a)


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Boundary-Value Problems with Parameters Chapter & Page: 459

or, equivalently, as

A(x )d2

d x 2 + B(x ) d

d x+ C(x ) = (45.3b)

where A , B and Care known functions. (We use instead of in the last equation forfuture convenience.)

A solution to such a problem is a pair (,) that satisfies the given problem (both the

differential equation and the boundary conditions. To avoid triviality, we will insist that the

function be nontrivial (i.e., not always zero). Traditionally, the is called an eigenvalue, and

the corresponding is called an eigenfuction. Well often refer to the two together, (,) , as

an eigen-pair. The entire problem can be called an eigen-problem, a boundary-value problem

or a Sturm-Liouville problem (though we wont officially define just what a Sturm-Liouville

problem is until later.

In our problems, we will need to find the general solution = to the given differentialequation for each possible value of , and then apply the boundary conditions to find all possible

eigen-pairs. Technically, at this point, can be any complex number. However, thanks to the

foreknowledge of the author, we can assume is real. Why this is a safe assumption will be one

of the things we will later need to show (its analogous to the fact that self-adjoint matrices havejust real eigenvalues). What we cannot yet do, though, is assume the s all come from some

particular subinterval of the real line. This means you must consider all possible real values for

, and take into account that the form of the solution may be quite different for different

ranges of these s .

!Example 45.2: The simplest (and possibly most important) example of such an boundary-

value problem with parameter is

+ = 0 (45.4a)with boundary conditions

(0) = 0 and (L) = 0 (45.4b)where L is some positive number (think of it as a given length).

The above differential equation is a simple second-order homogeneous differential equa-

tion with constant coefficients. Its characteristic equation is

r2 + = 0 ,with solution

r =

.

In this example, the precise formula for (x ) , the equations general solution corre-

sponding to a particular value of , depends on whether >0 , =

0 or 0 . For convenience, let = . Then

r= ,

and

(x ) = c1ex + c2ex

where c1 and c2 are arbitrary constants, and, as already stated, = .


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Applying the first boundary condition:

0 = (0) = c1e0 + c2e0 = c1 + c2 ,

which tells us that

c2=

c1 ,

and thus,

(x ) = c1ex c1ex = c1

ex ex .Combining this with the boundary condition at x= L yields

0 = (L) = c1

eL eL .Now eL >e 0 =1 and eL 0 ,

and the only way we can have

0 = c1

eL eLis for c1=0 . Thus, the only solution to this problem with 0 ). Thus, the only solution to the differentialequation that satisfies the boundary conditions when =0 is

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

again, just the trivial solution.


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Boundary-Value Problems with Parameters Chapter & Page: 4511

> 0 : This time, it is convenient to let =

, so that the solution to the characteristic

equation becomes

r =

= i .While the general formula for can be written in terms of complex exponentials, it

is better to recall that these complex exponentials can be written in terms of sines and

cosines, and that the general formula for can then be given as

(x ) = c1cos(x )+ c2sin(x)

where, again, c1 and c2 are arbitrary constants, and, as already stated, =

.

Applying the first boundary condition:

0 = (0)= c1cos( 0)+ c2sin( 0)= c1 1+ c2 0 = c1 ,

which tells us that

(x ) = c2sin(x ) .With the boundary condition at x= L, this gives

0 = (L) = c2sin(L) .

To avoid triviality, we want c2 to be nonzero. So, for the boundary condition at x= Lto hold, we must have

sin(L) = 0 ,

which means thatL = an integral multiple of ,

Moreover, since =

>0 , L must be a positive integral multiple of . Thus, we

have a list of allowed values of ,

k= kL

with k= 1, 2, 3, . . . ,

a corresponding list of allowed values for ,

k = (k)2 = k

L 2

with k= 1, 2, 3, . . .

(these are the eigenvalues), and a corresponding list of (x )s (the corresponding eigen-

functions),

k(x ) = cksin(kx ) = cksin

k

Lx

with k= 1, 2, 3, . . .

where the cks are arbitrary constants. For our example, these are the only nontrivial

eigenfunctions.


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In summary, we have a list of solutions to our Sturm-Liouville problem, namely,

(k, k) for k= 1, 2, 3, . . .where, for each k,

k= kL

2

and

k(x ) = cksin(kx ) = cksin

k

Lx

(with the cks being arbitrary constants).

In particular, if L= 1 , then the solutions to our Sturm-Liouville problem" are given by(k, k(x)) =

k2 2 , cksin(kx )

for k= 1, 2, 3, . . . .

45.3 The Sturm-Liouville Form for a DifferentialEquation

To solve these differential equations with parameters, you will probably want to use form (45.3a),

as we did in our example. However, to develop and use the theory that we will be developing

and using, we will want to rewrite each differential equation in itsSturm-Liouville form

d

d x

p(x )

d

d x

+ q(x ) = w(x)

where p , q and w are known functions.

!Example 45.3: Technically, the equation

+ = 0is not in Sturm-Liouville form, but the equivalent equation

= is, with p=1 , q= 0 and w=1 .

!Example 45.4: The equation

ddx

sin(x ) dd x+ cos(x ) = x 2

is in Sturm-Liouville form, with

p(x ) = sin(x ) , q(x ) = cos(x ) and w(x ) = x 2 .On the other hand,

xd2

d x 2 + 2 d

d x+ [sin(x ) + ] = 0

is not in Sturm-Liouville form.


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The Sturm-Liouville Form for a Differential Equation Chapter & Page: 4513

Fortunately, just about any differential equation in form (45.3a) or (45.3a) can converted to

Sturm-Liouville form using a procedure similar to that used to solve first-order linear equations.

To describe the procedure in general, lets assume we have at least gotten our equation to the

form

A(x)d2

d x 2 + B(x ) d

d x+ C(x ) = .

To illustrate the procedure, well use the equation

xd2

d x 2 + 2 d

d x+ sin(x ) =

(with (0, ) being our interval of interest).Here is what you do:

1. Divide through by A(x ) , obtaining

d2

d x 2 + B(x )

A(x) + C(x)

A(x) = 1

A(x ) .

Doing that with our example yields

d2

dx 2 + 2

x

d

d x+ sin(x)

x = 1

x .

2. Compute the integrating factor

p(x ) = e

B(x)A(x)

d x,

ignoring any arbitrary constants.

In our example,

B(x)A(x ) d x = 2x d x = 2 lnx + c .

So (ignoring c ),

p(x ) = e

B(x)A(x)

d x = e2 lnx = elnx 2 = x 2 .

(As an exercise, you should verify that

d p

d x= pB

A.

In a moment, well use this fact.)

3. Using the p(x ) just found:(a) Multiply the differential equation resulting from step 1 by p(x ) , obtaining

pd2

d x 2 + pB

A

d

d x+ p C

A = p

A ,

(b) observe that (via the product rule),

d

d x

p

d

d x

= p d

2

dx 2 + pB

A

d

dx,


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(c) and rewrite the differential equation according to this observation,

d

d x

p

d

d x

+ p C

A = p

A .

In our case, we have

x 2

d2

dx 2 + 2

x

d

d x+ sin(x )

x

= x 2

1

x

x 2 d2dx 2

+ 2x ddx

+ xsin(x ) = x .

Oh look! By the product rule,

d

d x

p

d

d x

= d

dx

x 2

d

d x

= x 2 d

2

d x 2 + 2x d

d x.

Using this to simplify the first two terms of our last differential equation

above, we get

d

d x

x 2

d

d x

+ xsin(x ) = x .

It is now in the desired form, with

p(x ) = x 2 , q(x ) = xsin(x ) and w(x ) = x .

45.4 Boundary Conditions for Sturm-LiouvilleProblems

The boundary conditions for a Sturm-Liouville problem will have to be homogeneous as defined

in the previous chapter. There is an additional condition that well derive in this section that will

help allow us to view a certain operator as self adjoint. That operator is given by the left side of

the differential equation of interest when that differential equation is written in Sturm-Liouville

form,d

dx

p(x )

d

dx

+ q(x) = w(x) .

For the rest of this chapter, this operator will be denoted by L . That is, given any suitably

differentiable function ,

L[] = dd x

p(x )

d

d x

+ q(x )

where p and q are presumably known real-valued functions on some interval (a, b) . Note that

our differential equation can be written as

L[] = w .


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Boundary Conditions for Sturm-Liouville Problems Chapter & Page: 4515

Greens Formula, and Sturm-Liouville AppropriateBoundary Conditions

To describe the additional boundary conditions needed, we need to derive an identity of Green.

You should start this derivation by doing exercise 45.10 on page 4536, in which you verify the

following lemma :

Lemma 45.6 (the preliminary Greens formula)

Let (a, b) be some finite interval, and let L be the operator given by

L[] = dd x

p(x )

d

d x

+ q(x )

where p and q are any suitably smooth and integrable functions on (a, b) . Then ba

fL[g] d x = p(x)f(x ) dgd x

b

a

b

a

pd f

d x

dg

d xd x +

ba

q f g d x (45.5)

whenever f and g are suitably smooth and integrable functions on (a, b) .

Now suppose we have two functions u and v on (a, b) (assumed suitably smooth and

integrable, but, possibly, complex valued), and suppose we want to compare ba

uL[v] d x and b

a

L[u]vd x .

(Why? Because this will lead to our extending the notion of self adjointness as characterized

in theorem 45.5 on page 458)

If p and q are real-valued functions, then it is trivial to verify that

L[u] = LuUsing this and the above preliminary Greens formula, we see that b

a

uL[v] d x b

a

L[u]vd x = b

a

uL[v] d x b

a

v L

u

d x

=

pudv

dx

ba

b

a

pdu

dx

dv

dxd x +

ba

quvd x

pvdu

d x

b

a

b

a

pdv

d x

du

d xd x +

b

a

qvud x

.

Nicely enough, most of the terms on the right cancel out, leaving us with:

Theorem 45.7 (Greens formula)

Let (a, b) be some interval, p and q any suitably smooth and integrable real-valued functions

on (a, b) , and L the operator given by

L[] = dd x

p(x )

d

d x

+ q(x ) .


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Then ba

uL[v] d x b

a

L[u]vd x = p

udv

d x v du

dx

ba

(45.6)

for any two suitably smooth and differentiable functions u and v.

Equation(45.6)is known as Greens formula. (Strictly speaking, the right sideof the equationis the Greens formula for the left side).

Now we can state what sort of boundary conditions are appropriate for our discussions.

We will refer to a pair of homogeneous boundary conditions at x= a and x= b as beingSturm-Liouville appropriateif and only if

p

u

dv

dx v du

d x

ba

= 0 . (45.7)

whenever both u and v satisfy these boundary conditions.

For brevity, we may say appropriate when we mean Sturm-Liouville appropriate. Do

note that the function p in equation (45.7) comes from the differential equation in the Sturm-

Liouville problem. Consequently, it is possible that a particular pair of boundary conditions isSturm-Liouville appropriate when using one differential equation, and not Sturm-Liouville

appropriate when using a different differential equation. This is particularly true when the

boundary conditions are of the periodic or boundedness types.

!Example 45.5: Consider the boundary conditions

(a) = 0 and (b) = 0 .If u and v satisfy these conditions; that is,

u(a) = 0 and u(b) = 0and

v(a) = 0 and v(b) = 0 ,then, assuming p(a) and p(b) are finite numbers,

p

u

dv

d x v du

d x

ba

= p(b)

(u(b)) dv

dx

x=b

v(b) d u

d x

x=b

p(a)

(u(a)) dv

d x

x=a

v(a) d u

d x

x=a

= p(b)

0 dv

d x

x=b

0 d u

d x

x=b

p(a)

0 dvdx

x=a

0 d ud x

x=a

= 0 .

So

(a) = 0 and (b) = 0 .are Sturm-Liouville appropriate boundary conditions, at least whenever p(a) and p(b) are

finite.


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Boundary Conditions for Sturm-Liouville Problems Chapter & Page: 4517

A really significant observation is that, whenever u and v satisfy appropriate boundary

conditions, then Greens formula reduces to ba

uL[v] d x b

a

L[u]vd x = 0 ,

which, in turn, gives us the following lemma, which, in turn, suggests just what we will be usingfor inner products and self adjointness in the near future.

Lemma 45.8

Let (a, b) be some interval, p and q any suitably smooth and integrable real-valued functions

on (a, b) , and L the operator given by

L[] = dd x

p(x )

d

d x

+ q(x ) .

Assume, further, that u(x ) and v(x ) satisfy Sturm-Liouville appropriate boundary conditions

at a and b . Then ba

uL[v] d x = b

a

L[u]vd x .

The Rayleigh Quotient

Before we stray too far from the preliminary Greens formula (lemma 45.6), lets take a look

at the integral ba

L[] d x

when is an eigenfunction corresponding to eigenvalue for some Sturm-Liouville problemwith differential equation

d

d x

p(x)

d

d x

+ q(x ) = w for a < x < b .

In terms of the operator L , this equation is

L[] = w ,

and, so,

ba

L[] d x = ba

[w] d x = ba

( (x )) (x )w(x ) d x = ba

||2 w d x .

On the other hand, using the preliminary Greens formula gives us

ba

L[] d x = p ddx

ba

b

a

pd

d x

d

d xd x +

ba

qd x

= p ddx

ba

b

a

p

dd x2 d x +

ba

q ||2 d x


18/44


= p ddx

ba

b

a

p

dd x2 q ||2

d x .

Together, the above tells us that

b

a ||2

w d x = b

a L[] d x = pd

d x b

a

b

ap dd x

2

q ||2 d x .

Cutting out the middle and solving for gives us the next lemma.

Lemma 45.9 (Rayleighs Quotient)

Let (,) be an eigen-pair for a Sturm-Liouville problem with differential equation

d

d x

p(x)

d

d x

+ q(x ) = w for a < x < b .

Then

= p

d

d xb

a+

b

ap dd x

2

q ||2 dx b

a

||2 w d x. (45.8)

Equation (45.8) is called the Rayleigh quotient. It relates the value of an eigenvalue to any

corresponding eigenfunction. It is of particular interest in the Sturm-Liouville problems in which

the boundary conditions ensure that

pd

dx

b

a

= 0 .

Then the Rayleigh quotient reduces to

=

ba

p

dd x2 q ||2

d x b

a

||2 w d x.

Now look at the right side of this equation. Remember, p , q and w are all real-valued functions

on (a, b) . It should then be clear that the integrals on the right are all real valued. Thus, must

be real valued.

If, in addition,

q(x ) 0 for a < x


19/44

Sturm-Liouville Problems Chapter & Page: 4519

must always be positive. On the other hand, if q= 0 on (a, b) (which is quite common), thenthe Rayleigh quotient further reduces to

=

ba

p

d

d x

2

d x

ba

||2 w d x 0 .

With a little thought, you can then show that the only possible eigenfunctions corresponding to a

zero eigenvalue, =0 , are nonzero constant functions, and, consequently, if nonzero constantfunctions do not satisfy the boundary conditions, then cannot be zero; all the eigenvalues are

positive.

45.5 Sturm-Liouville ProblemsFull Definition

Finally, we can fully define the Sturm-Liouville problem: A Sturm-Liouville problem is a

boundary-value problem consisting of both of the following:

1. A differential equation that can be written in the form

d

d x

p(x )

d

d x

+ q(x ) = w(x ) for a < x


20/44


A Important Class of Sturm-Liouville Problems

There are several classes of Sturm-Liouville problems. One particularly important class goes

by the name of regular Sturm-Liouville problems. A Sturm-Liouville problem is said to be

regularif and only if all the following hold:

1. The functions p , q and w are all real valued and continuous on the closed interval[a, b] , with p being differentiable on (a, b) , and both p and w being positive on theclosed interval[a, b] .

2. We have homogeneous regular boundary conditions at both x= a and x= b . That is

a (a)+ a (a) = 0

where a and a are constants, with at least one being nonzero, and

b (b)+ b(b) = 0

where a and b are constants, with at least one being nonzero.

Many, but not all, of the Sturm-Liouville problems generated in solving partial differential

equation problems are regular. For example, the problem considered in example 45.2 on page

459 is a regular Sturm-Liouville problem.

So What?

It turns out that the eigenfunctions from a Sturm-Liouville problem on an interval (a, b) can

often be used in much the same way as the eigenvectors from a self-adjoint matrix to form an

orthogonal basis for a large set of functions. Consider, for example, the eigenfunctions from

the example 45.2

k(x ) = cksin

k

Lx

for k= 1, 2, 3, . . . .

If f is any reasonable function on (0,L ) (say, any continuous function on this interval), then

we will discover that there are constants c1, c2, c3, . . . such that

f(x ) =

k=1cksin

k

Lx

for all x in (0,L ) .

This infinite series, called the (Fourier) sine series for f on (0,L ) , turns out the be extremely

useful in many applications. For one thing, it expresses any function on (0,L ) in terms of the

well-understood sine functions.Our next goal is to develop enough of the necessary theory to discover what I just said we

will discover. In particular we want to learn how to compute the cks in the above expression

for f(x) . It turns out to be remarkable similar to the formula for computing the vks in theorem

45.4 on page 456. Take a look at it right now. Of course, before we can verify this claim, we

will have to find out just what we are using for an inner product.

All this, alas, will take some time.


21/44


22/44


1. There is not an independent pair of eigenvectors corresponding to . This means the

eigenspace corresponding to eigenvalue is one dimensional (i.e., is a simple

eigenvalue), and every eigenfunction is a constant multiple of 1.

2. There is an independent pair of eigenvectors corresponding to . This means the

eigenspace corresponding to eigenvalue is two dimensional (i.e., is a double

eigenvalue), and every solution to the differential equation (with the given ) is an

eigenfunction.

45.7 Inner Products, Orthogonality and GeneralizedFourier Series

We must, briefly, forget about Sturm-Liouville problems, and develop the basic analog of the innerproduct described for finite dimensional vectors at the beginning of this chapter. Throughout this

discussion, (a, b) is an interval, and w=w(x ) is a function on (a, b) satisfying

w(x ) > 0 whenever a < x < b .

This function, w , will be called theweight functionfor our inner product. Often, it is simply the

constant 1 function (i.e., w(x )=1 for every x).Also, throughout thisdiscussion,let us assume thatour functions areall piecewise continuous

and sufficiently integrable for the computations being described. Well discuss what suitably

integrable means if I decide it is relevant. Otherwise, well slop over issues of integrability.

Inner Products for Functions

Let f and g be two functions on the interval (a, b) . We define theinner product (with weight

function w )of f with g denoted f| g by

f | g = b

a

f(x)g(x)w(x ) d x .

For convenience, lets say that the standard inner productis simply the inner product with weight

function w=1 ,

f | g = b

af(x )g(x ) d x .

For now, you can consider the inner product to be simply shorthand for some integrals that will

often arise in our work.

!Example 45.6: Let (a, b)=(0, 2) and w(x )=x 2 . Then

f | g = 2

0

f(x )g(x )x 2 d x .


23/44

Inner Products, Orthogonality and Generalized Fourier Series Chapter & Page: 4523

In particular 6x+ 2i

1x=

20

(6x+ 2i ) 1x

x 2 d x

=

2

0

(6x 2i )x d x

= 2

0

6x 2 i 2x d x

= 2x 3 i x 22

0= 16 4i .

The inner product of functions just defined is, in many ways, analogous to the inner product

defined for finite dimensional vectors at the beginning of this chapter. To see this, well verify

the following theorem, which is very similar to theorem 45.2 on page 454.

Theorem 45.11 (properties of the inner product)

Let

| be an inner product as just defined above. Suppose and are two (possibly

complex) constants, and f , g, and h are functions on (a, b) . Then

1. f | g = g | f,

2. h | f+ g = h | f+ h | g ,

3. f+ g| h = f | h+ g | h ,and

4. f | f 0 with f | f = 0 if and only if f=0 on (a, b) .

PROOF: To Be Written (Someday) --- Take Notes in Class

Norms

Recall that the norm of any vector v in N is related to the inner product of v with itself by

v=

v| v .

In turn, for each inner product| , we define the correspondingnormof a function f by

f

= f | f .In general

f2 = f | f = b

a

f(x )f(x)w(x ) d x = b

a

|f(x )|2 w(x ) d x .

Do note that, in a loose sense,

f is generally small over (a, b) f is small .


24/44


!Example 45.7: Let (a, b)=(0, 2) and w(x )=x 2 . Then

f=

f | f = 2

0

(f(x))f(x )x 2 d x = 2

0

|f(x )|2 x 2 d x .

In particular,

5x+ 6i2 = 2

0

(5x+ 6i )(5x+ 6i )x 2 d x

= 2

0

(5x 6i )(5x+ 6i )x 2 d x

= 2

0

25x 2 + 36

x 2 d x

= 2

0

25x 4 + 36x 2 d x

= 5x 5

+12x 320

= 256 .

So

5x+ 6i=

256 = 16 .

Orthogonality

Recall that any two vectors v and w in N are orthogonal if and only if

v| w= 0 .

Analogously, we say that any pair of functions f and g isorthogonal(over the interval) (with

respect to the inner product, or with respect to the weight function) if and only if

f | g = 0 .

More generally, we will refer to any indexed set of nonzero functions

{ 1, 2, 3, . . .}

as beingorthogonalif and only if

k| n = 0 whenever k=n .

If, in addition, we have

k= 1 for each k ,

then we say the set is orthonormal. For our work, orthogonality will be important, but we wont

spend time or effort making the sets orthonormal.


25/44


!Example 45.8: Consider the setsin

k

Lx

:k= 1, 2, 3, . . .

which is the set of sine functions (without the arbitrary constants) obtained as eigenfunctions

in example 45.2 on page 459. The interval is (0,L ) . For the weight function, well use

w(x )= 1 . Observe that, if k and n are two different positive integers, then, using thetrigonometric identity

2sin(A) sin(B) = cos(A B ) cos(A +B) ,

we havesin

k

Lx

sinnL

x

=

L0

sin

k

Lx

sin

n

Lx

d x

= L

0

sin

k

Lx

sin

n

Lx

d x = = 0 .

So sin

k

Lx:k= 1, 2, 3, . . .

is an orthogonal set of functions on (0,L ) with respect to the weight function w(x )=1 .

Generalized Fourier Series

Now suppose{1, 2, 3, . . . } is some orthogonal set of nonzero functions on (a, b) , and fis some function that can be written as a (possibly infinite) linear combination of the ks ,

f(x) = k

ckk(x ) for a < x < b .

To find each constant ck , first observe what happens when we take the inner product of both

sides of the above with one of the ks , say, 3. Using the linearity of the inner product and the

orthogonallity of our functions, we get

3| f =

3

k

ckk

=

k

ck 3| k

= k

ck32 if k=30 if k=3

= c3 32 .So

c3 = 3| f32 .

Since there is nothing special about k=3 , we clearly have

ck= k| fk2 for all k .


26/44


More generally, whether or not f can be expressed as a linear combination of the ks ,

we define the generalized Fourier series for f (with respect to the given inner product and

orthogonal set{1, . . . ,} ) to be

G.F.S.[f]|x =k

ckk(x )

where, for each k

ck= k| fk2 .

The cks are called the correspondinggeneralized Fourier coefficientsof f .3 Dont forget:

k| f = b

a

k(x )f(x )w(x ) d x

and

k2 = b

a

|k(x )|2 w(x) d x .

We will also refer to G.F.S.[

f]

as the expansion of f in terms of the k

s . If the k

s

just happen to be eigenfunctions from some Sturm-Liouville problem, we will even refer to

G.F.S.[f] as theeigenfunction expansionof f .

!Example 45.9: From previous exercises, we know

{1, 2, 3, . . . } =

sin

k

Lx:k=1, 2, 3, . . .

is an orthogonal set of functions on (0,L ) with respect to the weight function w(x )= 1 .(Recall that this is a set of eigenfunctions for the Sturm-Liouville problem from example 45.2

on page 459.) Using this set,

G.F.S.[f]|x = k

ckk(x ) with ck= k

| f

k2

becomes

G.F.S.[f]|x =

k=1ck sin

k

Lx

with

ck= k| fk2 =

L0

sin

k

Lx

f(x) dx L

0

sin

k

Lx

2

d x

=

L0

f(x) sin

k

Lx

dx L

0

sin2

k

Lx

d x

.

Since L0

sin2

k

Lx

d x = = L2

.

the above reduces to

G.F.S.[f]|x =

k=1ck sin

k

Lx

with ck= 2

L

L0

f(x) sin

k

Lx

d x .

3 Compare the formula for the generalized Fourier coefficients with formula in theorem (45.4) on page 456 for the

components of a vector with respect to any orthogonal basis. They are virtually the same!


27/44


In particular, suppose f(x )=x for 0< x < L. Then the above formula for ck yields

ck= 2L

L0

xsin

k

Lx

d x

= 2

L 2xk coskL x L

0 + 2

k L

0 coskL x d x

= 2L

0 + 2L

kcos

k

LL

+

2

k

2sin

k

Lx

L0

= (1)k 4

k.

So, using the given interval, weight function and orthogonal set, the generalized Fourier series

for f(x )=x , is

k=1(1)k 4

ksin

k

Lx

.

(In fact, this is the classic Fourier sine series for f(x )=x on (0,L ) .)

Approximations and Completeness

Lets say we have an orthogonal set of functions{1, 2, 3, . . . } and some function f on(a, b) . Let

G.F.S.[f]|x =

k=1ckk(x )

In practice, to avoid using the entire series, we may simply wish to approximate fusing theNth

partial sum,

f(x ) N

k=1ckk(x ) .

The error in using this is

EN(x) = f(x )N

k=1ckk(x) ,

and the square of its norm

EN2 =

f(x )N

k=1ckk(x )

2

= ba

f(x )N

k=1ckk(x)

2

w(x ) d x

give a convenient measure of how good this approximation is. The above integral is sometimes

known as the (weighted) mean square error in usingN

k=1ckk(x ) for f(x ) on the interval(a, b) .

We, of course, hope the error shrinks to zero (as measured by EN ) as N . If wecan be sure this happens no matter what piecewise continuous function f we start with, then we

say the orthogonal set{1, 2, 3, . . . } iscomplete.


28/44


Now if{1, 2, 3, . . . } is complete, then taking the limits above yieldf G.F.S.[f]= 0 .

Equivalently

baf(x)

k=1 c

kk(x )2

w(x) d x = 0 .In practice, all of this usually means that the infinite series

kckk(x) converges to f(x) at

every x in (a, b) at which f is continuous. In any case, if theset{1, 2, 3, . . . } is complete,then we can view the corresponding generalized Fourier series for a function f as being the

same as that function, and can write

f(x ) =

k

ckk(x ) for a < x < b

where

ck= k| fk2 .

In other words, a complete orthogonal set of (nonzero) functions can be viewed as a basis forthe vector space of all functions of interest.

45.8 Sturm-Liouville Problems and EigenfunctionExpansions

Suppose we have some Sturm-Liouville problem with differential equation

dd x

p(x) d

d x

+ q(x ) = w for a < x < band Sturm-Liouville appropriate boundary conditions. As usual, well let L be the operator

given by

L[] = dd x

p(x )

d

d x

+ q(x ) .

Remember, by definition, p and q are real-valued functions and w is a positive function on

(a, b) .

Self-Adjointness and Some Immediate Results

Glance back at lemma 45.8 on page 4517. It tells us that, whenever u and v are sufficiently

differentiable functions satisfying the boundary conditions, ba

uL[v] d x = b

a

L[u]vd x .

Using the standard inner product for functions on (a, b) ,

f | g = b

a

f(x )g(x ) d x ,


29/44

Sturm-Liouville Problems and Eigenfunction Expansions Chapter & Page: 4529

we can write this equality as

u | L[v]= L[u]| v ,which looks very similar to the equation characterizing self-adjointness for a matrixA in theorem

45.5 on page 458. Because of this we often say either that Sturm-Liouville problems areself

adjoint, or that the operator L is self adjoint.4

The terminology is important for communication,but what is even more important is that functional analogs to the results obtained for self-adjoint

matrices also hold for here.

To derive two important results, let (1, 1) and (2, 2) be two solutions to our Sturm-

Liouville problem (i.e., 1 and 2 are two eigenvalues, and 1 and 2 are corresponding

eigenfunctions). From a corollary to Greens formula (lemma 45.8, noted just above), we know ba

1L[2] d x =

ba

L[1]2d x .

But, from the differential equation in the problem, we also have

L[2]

= 2w2 ,

L[1] = 1w1and thus,

L[1] = (1w1) = 1w1

(remember w is a positive function). So, ba

1L[2] d x =

ba

L[1]2d x

b

a

1(2w2) d x =

ba

(1w1)2d x

2 b

a

12w d x = 1

ba

12w d x .

Since the integrals on both sides of the last equation are the same, we must have either

2 = 1 or b

a

12w d x = 0 . (45.10)

Now, we did not necessarily assume the solutions were different. If they are the same,

(1, 1) = (2, 2) = (,)and the above reduces to

= or b

a

w d x = 0 .

But, since w is a positive function and is necessarily nontrivial, ba

w d x = b

a

| (x )|2 w(x ) d x > 0 (=0) .

4 The correct terminology is that L is a self-adjoint operatoron the vector space of twice-differentiable functions

satisfying the given homogeneous boundary conditions.


30/44


So we must have

= ,which is only possible if is a real number. Thus

FACT: The eigenvalues are all real numbers.

Now suppose 1 and 2 are not the same. Then, since they are differentrealnumbers, wecertainly do not have

2 = 1 .Line (45.10) then tells us that we must have b

a

12w d x = 0 ,

which we can also write as

1| 2 = 0using the inner product with weight function w ,

f | g = ba

f(x)g(x)w(x ) d x .

Thus,

FACT: Eigenfunctions corresponding to different eigenvalues are orthogonal with

respect to the inner product with weight function w(x ) .

By the way, since the eigenvalues are real, it is fairly easy to show that the real part and the

imaginary part of each eigenfunction is also an eigenfunction. From this it follows that we can

always choose real-valued functions as our basis for each eigenspace.

What all the above means, at least in part, is that we will be constructing generalized Fourier

series using eigenfunctions from Sturm-Liouville problems, and that the inner product used will

be based on the weight function w(x ) from the differential equation

d

d x

p(x)

d

d x

+ q(x ) = w for a < x < b .

in the Sturm-Liouville problem. Accordingly, we may refer to w(x ) and the corresponding inner

product on (a, b) as thenatural weight functionand thenatural inner productcorresponding to

the given Sturm-Liouville problem.

Other Results Concerning Eigenvalues

Using the Rayleigh quotient it can be shown that there is a smallesteigenvalue 0 for each

Sturm-Liouville problem normally encountered in practice. The rest of the eigenvalues are

larger. Unfortunately, verifying this is beyond our ability (unless the Sturm-Liouville problem

is sufficiently simple). Another fact you will just have to accept without proof is that, for each

Sturm-Liouville problem normally encountered in practice, the eigenvalues form an infinite

increasing sequence

0 < 1 < 2 < 3 < with

limk

k= .


31/44

Sturm-Liouville Problems and Eigenfunction Expansions Chapter & Page: 4531

The Eigenfunctions

Let us assume that

E= { 0, 1, 2, 3, . . . }is the set of all distinct eigenvalues for our Sturm-Liouville problem (indexed so that 0 is the

smallest and k < k+1 in general). Remember, each eigenvalue will be either a simple or adouble eigenvalue. Next, choose a set of eigenfunctions

B = { 0, 1, 2, 3, . . . }

as follows:

1. For each simple eigenvalue, choose exactly one corresponding (real-valued) eigenfunc-

tion for B .

2. For each double eigenvalue, choose exactly one orthogonal pair of corresponding (real-

valued) eigenfunctions for B .

Remember, this set of functions will be orthogonal with respect to the weight function w fromthe differential equation in the Sturm-Liouville problem. (Note: Each k is an eigenfunction

corresponding to eigenvalue k only if all the eigenvalues are simple.)

Now let f be a function on (a, b) . Since B is an orthogonal set, we can construct the

corresponding generalized Fourier series for f

G.F.S.[f(x )] =

k=0ckk(x )

with

ck= k

| f

k2 = b

a

k(x)f(x )w(x) dx b

a

|k(x)|2 w(x) d x.

The obvious question to now ask is Is this orthogonal set of eigenfunctions complete? That is,

can we assume

f =

k=0ckk on (a, b) ?

The answer is yes, at least for the Sturm-Liouville problems normally encountered in practice.

But you will have to trust me on this.5 There are some issues regarding the convergence of

k=0ckk(x ) when x is a point at which f is discontinuous or when x is an endpoint of

(a, b) and f does not satisfy the same boundary conditions as in the Sturm-Liouville problem,

but we will gloss over those issues for now.Finally (assuming reasonable assumptions concerning the functions in the differential

equation), it can be shown that the graphs of the eigenfunctions corresponding to higher values

of the eigenvalues wiggle more than those corresponding to the lower-valued eigenvalues. To

5 One approach to proving this is to consider the problem of minimizing the Rayleigh quotient over the vector space

U of functions orthogonal to the vector space of all the eigenfunctions. You can then (I think) show that, if the

orthogonal set of eigenfunctions in not orthogonal, then U is not empty, and this minimization problem has a

solution (,) and that this solution must also satisfy the Sturm-Liouville problem. Hence is an eigenfunction

in U , contraryto the definition ofU . So U must be emptyand the given set of eigenfunctions must be complete.


32/44


be precise, eigenfunctions corresponding to higher values of the eigenvalues must cross the X

axis (i.e., be zero) more oftern that do those corresponding to the lower-valued eigenvalues. To

see this (sort of), suppose 0 is never zero on (a, b) (so, it hardly wiggles this is typically the

case with 0 ). So 0 is either always positive or always negative on (a, b) . Since0 willalso be an eigenfunction, we can assume weve chosen 0 to always be positive on the interval.

Now let k be an eigenfunction corresponding to another eigenvalue. If it, too, is never zero on(a, b) , then, as with 0, we can assume weve chosen k to always be positive on (a, b) . But

then,

0| k = b

a

0(x)k(x)w(x ) >0

d x > 0 ,

contrary to the known orthogonality of eigenfunctions corresponding to different eigenvalues.

Thus, each k other than 0 must be zero at least at one point in (a, b) . This idea can be

extended, showing that eigenfunctions corresponding to high-valued eigenvalues cross the X

axis more often than do those corresponding to lower-valued eigenvalues, but requires developing

much more differential equation theory than we have time (or patience) for.

45.9 The Main Results Summarized (Sort of)A Mega-Theorem

We have just gone through a general discussion of the general things that can (often) be derived

regarding the solutions to Sturm-Liouville problems. Precisely what can be proven depends

somewhat on the problem. Here is one standard theorem that can be found (usually unproven)

in many texts on partial differential equations and mathematical physics. It concerns the regular

Sturm-Liouville problems (see page 4520).

Theorem 45.12 (Mega-Theorem on Regular Sturm-Liouville problems)

Consider a regular Sturm-Liouville problem6 with differential equation

d

d x

p(x )

d

d x

+ q(x ) = w(x) for a < x < b

and homogeneous regular boundary conditions at the endpoints of the finite interval (a, b) .

Then, all of the following hold:

1. All the eigenvalues are real.

2. The eigenvalues form an ordered sequence

0 < 1 < 2 < 3 < ,

with a smallest eigenvalue (usually denoted by 0 or 1) and no largest eigenvalue. In

fact,

k as k .6 i.e., p , q and w are real valued and continuous on the closed interval[a, b] , with p being differentiable on

(a, b) , and both p and w being positive on the closed interval[a, b] .


33/44

The Main Results Summarized (Sort of) Chapter & Page: 4533

3. All the eigenvalues are simple.

4. If, for each eigenvalue k, we choose a corresponding eigenfunction k, then the set

B = { 0, 1, 2, 3, . . . }

is a complete, orthonormal set of functions relative to the inner product

u | v = b

a

u(x )v(x )w(x ) d x .

Moreover, if f is any piecewise smooth function on (a, b) , then the corresponding

generalized Fourier series of f ,

G.F.S.[f] =

k=0ckk(x ) with ck= k| fk2

,

converges for each xin (a, b) , and it converges to

(a) f(x ) if f is continuous at x.

(b) the midpoint of the jump in f ,

lim0+

1

2[f(x+ ) + f(x )] ,

if f is discontinuous at x.

5. 0, the eigenfunction in B corresponding to the smallest eigenvalue, is never zero on

(a, b) . Moreover, for each k, k+1 has exactly one more zero in (a, b) than does k.

6. Each eigenvalue is related to any corresponding eigenfunction via the Rayleigh

quotient

=p d

d x

ba

+ b

a

p

dd x2 q ||2 d x

2 .

Similar mega-theorems can be proven for other Sturm-Liouville problems. The main dif-

ference occurs when we have periodic boundary conditions. Then most of the eigenvalues are

doubleeigenvalues, and our complete set of eigenfunctions looks like

{ . . . , k, k, . . . }

where{

k, k}

is an orthogonal pair of eigenfunctions corresponding to eigenvalue k . (Typi-

cally, though, the smallest eigenvalue is still simple.)


34/44


Additional Exercises

45.1. Let 1

=3 , 2

= 2 ,

b1 =

2

3

and b2 =

3

2

,

and assume (1, b1) and (2, b

2) are eigenpairs for a matrix A .

a. Verify that{b1, b2} is an orthogonal set.b. Compute

b1 and b2 .c. Express each of the following vectors in terms of b1 and b2 using the formulas in

theorem 45.4 on page 456.

i. u = 46 ii. v = 125 iii. w = 01d. Compute the following using the vectors from the previous part. Leave your answers

in terms of b1 and b2 .

i. Au ii. Av iii. Aw

45.2. Let 1=2 , 2=4 , 3=6 ,

b1 =

1

2

1

, b2 =

2

10

and b3 =

1

2

5

.

and assume (1, b1) , (2, b

2) and (3, b3) are eigenpairs for a matrix A .

a. Verify that{b1, b2, b3} is an orthogonal set.b. Compute

b1 , b2 and b3 .c. Express each of the following vectors in terms of b1 , b2 and b3 using the formulas

in theorem 45.4 on page 456.

i. u =

47

22

ii. v =

12

3

iii. w =

01

0

d. Compute the following using the vectors from the previous part. Leave your answersin terms of b1 , b2 and b3 .


45.3. Let 1= 1 , 2=0 , 3=1 ,

b1 =

11

1

, b2 =

12

1

and b3 =

10

1

.


35/44

Additional Exercises Chapter & Page: 4535

and assume (1, b1) , (2, b

2) and (3, b3) are eigenpairs for a matrix A .

a. Verify that{b1, b2, b3} is an orthogonal set.b. Compute

b1

,

b2

and

b3

.

c. Express each of the following vectors in terms of b1 , b2 and b3 using the formulas

in theorem 45.4 on page 456.

i. u =

58

3

ii. v =

03

6

iii. w =

10

0

d. Compute the following using the vectors from the previous part. Leave your answers

in terms of b1 , b2 and b3 .


45.4. Verify each of the claims in theorem 45.2 on page 454.

45.5. Finish verifying the claims in theorem 45.5 on page 458. In particular show that, if B

and A are N N matrices such that v| Bu= Av| u for every v, u in N ,

then B=A .45.6. Find the general solutions for each of the following corresponding to each real value .

Be sure the state the values of for which each general solution is valid. (Note: Some

of these are Euler equations see chapter 19.)

a. + 4 = b. + 2 =

c. x 2 +x = for 0 < xd. x 2 + 3x = for 0 < x

45.7. Rewrite each of the following equations in Sturm-Liouville form.

a. + 4 + = 0b. + 2 = c. x 2 +x = for 0 < xd. x 2 + 3x = for 0 < xe.

2x

= (Hermites Equation)

f.

1 x 2 x = for 1 < x


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a. (0) = 0 and (L) = 0b. (0) = 0 and (L) = 0

45.9. Solve the following Sturm-Liouville problems:

a. =

for 0 < x < 2 with (0)=

(2 ) and (0)

=(2 )b. x 2 +x = for 1 < x < e with (1)=0 and (e )=0c. x 2 +3x = for 1 < x < e with (1)=0 and (e )=0

45.10. (Preliminary Greens formula) Assume (a, b) is some finite interval, and let L be the

operator given by

L[] = ddx

p(x )

d

d x

+ q(x) .

where p and q are any suitably smooth and integrable functions on (a, b) . Using

integration by parts, show that ba

fL[g] d x = p(x )f(x ) dgd x

ba

b

a

pd f

dx

dg

d xd x +

ba

q f g d x

whenever f and g are suitably smooth and integrable functions on (a, b) .

45.11. Let L be the differential operator

L[] = dd x

p(x )

d

d x

+ q(x )

where p and q are continuous functions on the closed interval

[a, b

].

a. Verify that the boundary conditions

(a) = 0 and (b) = 0

is a Sturm-Liouville appropriate set of boundary conditions.

b. Show that any pair of homogeneous regular boundary conditions ata andb is Sturm-

Liouville appropriate.

c. Suppose our boundary conditions are periodic; that is,

(a) = (b) and (a) = (b) .

What must p satisfy for these boundary conditions to be Sturm-Liouville appropriate?

45.12. Compute the following, assuming the interval is (0, 3) and the weight function is

w(x )=1 .a. x| sin(2x ) b. x 2 9 + i 8x c.

9 + i 8x

x 2 d. e i 2x x e. x f. 9 + i 8x


37/44


g. sin(2x ) h.ei 2x

45.13. Compute the following, assuming the interval is (0, 1) and the weight function is

w(x )=x .a. x| sin(2x ) b. x

2

9 + i 8x c. 9 + i 8x x 2 d. e i 2x x e. x f. 4 + i 8xg. sin(2x ) h.

ei 2x45.14. Determine all the values for so that

ei 2x2

, ei 2x2

is an orthogonal set on (0, 1) with weight function w(x )=x .45.15. LetL be a positive value. Verify thateach of the following setsof functions is orthogonal

on (0,L ) with respect to the weight function w(x )=1 .a.

cos

k

Lx:k= 1, 2, 3, . . .

b.

ei 2kx /L :k= 0, 1, 2, 3, . . .

45.16. Consider

{1, 2, 3, . . . } =

cos

1

Lx

, cos

2

Lx

, cos

3

Lx

, . . .

,

which, in the exercise above, you showed is an orthogonal set of functions on (0,L )

with respect to the weight function w(x )=1 . Using this interval, weight function andorthogonal set, do the following:

a. Show that, in this case, the generalized Fourier series

G.F.S.[f] = k=1

ckk(x ) with ck= k| fk2

is given by

G.F.S.[f] =

k=1ckcos

k

Lx

with ck= 2L

L0

f(x ) cos

k

Lx

d x .

b. Find the generalized Fourier series for

i. f(x ) = x ii. f(x ) =

1 if 0 < x < L/2

0 if L/2< x < L

iii. f(x ) =

x if 0 < x < L/2

Lx if L/2< x < Lc. Show that this set of cosines is not a complete set by showing

f(x ) =

k=1ckcos

k

Lx

with ck= 2L

L0

f(x ) cos

k

Lx

d x

when f is the constant function f(x )=1 on (0,L ) .


38/44


45.17. In several exercises above, you considered either the Sturm-Liouville problem

+ = 0 for 0< x < L with (0)=0 and (L)=0

or, at least, the above differential equation. You may use the results from those exercises

to help answer the ones below:

a. What is the natural weight function w(x ) and inner product f| g correspondingto this Sturm-Liouville problem?

b. We know that

{0, 1, 2, . . . , k, . . .} =

0,

1

L

2,

2

L

2, . . . ,

k

L

2, . . .

is the complete set of eigenvalues for this Sturm-Liouville problem. Now write out a

corresponding orthogonal set of eigenfunctions (without arbitrary constants),

{0, 1, 2, . . . , k, . . .} .

c. Compute the norm of each of your ks from your answer to the last part.

d. To what does each coefficient in the generalized Fourier series

G.F.S.[f] =

k=0ckk(x ) with ck= k| fk2

,

reduce to using the eigenfunctions and inner products from previous parts of this

exercise?

e. Using the results from the last part, find the generalized Fourier series for the follow-

ing:

i. f(x )=1 ii. f(x )=x

iii. f(x )=

1 if 0 < x < L/2

0 if L/2< x < L


x 2 +x = for 1< x < e with (1)=0 and (e )=0


to answer the ones below:

a. What is the natural weight function w(x ) and inner product

f

|g

corresponding

to this Sturm-Liouville problem?

b. We know that

{1, 2, 2, . . . , k, . . .} =

1, 4, 9, . . . , k2, . . .



{1, 2, 3, . . . , k, . . .} .


39/44


c. Compute the norm of each of your ks from your answer to the last part. (A simple

substitution may help.)


G.F.S.[f] =

k=0ckk(x ) with ck=

k|

f

k2 ,


exercise?


ing:

i. f(x )=1 ii. f(x )=ln |x | iii. f(x )=x


x 2 + 3x = for 1 < x < e with (1)=0 and (e )=0


to answer the ones below:

a. What is the natural weight function w(x ) and inner product f| g correspondingto this Sturm-Liouville problem?

b. We know that

{1, 2, 2, . . . , k, . . .} = 2, 5, 10, . . . , k2 + 1, . . .



{1, 2, 3, . . . , k, . . .}

c. Compute the norm of each of your ks from your answer to the last part. (A simple

substitution may help.)


G.F.S.[f] =

k=0ckk(x ) with ck=

k

| f

k2 ,


exercise?


ing:

i. f(x )=1 ii. f(x )=x


40/44


45.20. Consider the Sturm-Liouville problem1 x 2x + = 0 for 1< x


41/44



42/44


Some Answers to Some of the Exercises

WARNING! Most of the following answers were prepared hastily and late at night. They

have not been properly proofread! Errors are likely!

1b.

b1

=

13 and

b2

=

13

1c i. u=2b1

1c ii. v=3b1 2b21c iii. w= 3

13b1 + 2

13b2

1d i. Au=6b11d ii. Av=9b1 + 4b21d iii. Aw= 9

13b1 4

13b2

2b.b1= 6 , b2= 5 and b3= 30

2c i. u=2b1 + 3b2 4b32c ii. v= 4

3b1 1

3b3

2c iii. w= 13

b1 15

b2 + 115

b3

2d i. Au=4b1

+ 12b2

24b3

2d ii. Av= 83

b1 2b3

2d iii. Aw= 23

b1 45

b2 + 25

b3

3b.b1= 3 , b2= 6 and b3= 2

3c i. u=2b1 3b2 + 4b33c ii. v=b1 + 2b2 + 3b33c iii. w= 1

3b1 + 1

6b2 1

2b3

3d i. Au= 2b1 + 4b33d ii. Av= b1 + 3b33d iii. Aw= 1

3b1 1

2b3

6a. If 4 , (x )=a cos(x ) + b sin(x ) with=

4 +

6b. If < 1 , (x ) = ae (1+)x + be(1)x with =

1 ; If = 1 , 1(x ) =aex + bx ex ; If >1 ,(x )=a ex cos(x ) + bex sin(x ) with=

1

6c. If < 0 ,(x )= ax + b x with =

; If = 0 ,0(x )= aln |x |+b ;If >0 ,(x )=a cos(ln |x |) + b sin(ln |x |) with=

6d. If 1 ,(x )=a x1 cos(ln |x |) + bx 1 sin(ln |x|) with=

1

7a. + 4 = 7b.

d

d x

e2x

d

d x

= e2x

7c. d

d xx dd x = 1x

7d. d

d x

x 3

d

d x

= x

7e. d

d x

ex

2d

dx

= ex 2

7f. d

d x

1 x 2 d

d x

= 1

1 x 2


43/44


7g. d

d x

x ex

d

d x

= ex

8a. (0, 0(x ))=(0, c0)and (k, k(x ))=

k

L

2, ckcos

k

Lx

fork=1, 2, 3, . . .

8b. (k, k(x ))= (2k+ 1)

2L 2

, cksin(2k+ 1)

2Lx fork=1, 2, 3, . . .

9a. (0, 0(x ))=(0, c0)and (k, k(x ))=

k2 , ck,1cos(k x ) + ck,2sin(kx )

fork=1, 2, 3, . . .9b. (k, k(x ))=

k2 , cksin(kln |x |)

fork=1, 2, 3, . . .

9c. (k, k(x ))=

k2 + 1 , ckx1 sin(kln |x |)

fork=1, 2, 3, . . .11c. p(b)= p(a)12a. 3

212b. 81 + 162i12c. 81 162i12d.

3

2i

12e. 3

12f.

819

12g. 3212h.

3

13a. 12

13b. 9

4+ i8

5

13c. 9

4 i8

5

13d. 1

2 2+ 1

2i

13e. 1

2

13f. 2

6

13g. 1

21 + 12

13h. 1

214. can be any integer except 1 .

16b i.

k=1

2L

(k )2

(1)k 1 cos k

Lx

16b ii.

k=1

2

ksin

k

2

cos

k

Lx

17a. w(x )=1 , f | g=L

0 (f(x ))g(x ) d x

17b. 1, cos1

Lx , cos

2

Lx , . . . , cos

k

Lx , . . .

17c. 1=L , coskL

x= L

2

17d. c0= 1L

L0

f(x ) d x , ck= 2L

L0

f(x ) cos

k

Lx

d x fork>0

17e i. 1

17e ii. L

2+

k=1

2L

(k )2

(1)k 1 cos k

Lx


44/44


17e iii. 1

2+

k=1

2

ksin

k

2

cos

k

Lx

18a. w(x )= 1x

, f | g= e1

(f(x ))g(x )1

xd x

18b. {sin(1 ln |x |) , sin(2 ln |x |) , sin(3 ln |x|) , . . . , sin(kln |x |) , . . .}18c.

sin(kln

|x

|)

=

2

18d. ck= 2

e1

f(x ) sin(kln |x |) 1x

d x

18e i.

k=1

2

k

1 (1)k sin(kln |x |)

18e ii.

k=1

(1)k+1 2k

sin(kln |x |)

18e iii.

k=1

2k

( 2 + k2)

1 (1)ke sin(kln |x |)19a. w(x )

=x , f | g

= e

1 (f(x ))g(x)x d x

19b. x1 sin(1 ln |x |) , x1 sin(2 ln |x|) , x1 sin(3 ln |x|) , . . . , x1 sin(kln |x|) , . . .19c.

x1 sin(kln |x |)=2

19d. ck= 2

e1

f(x ) sin(kln |x |) d x

19e i.

k=1

2k

1 + k2

1 (1)ke sin(kln |x |)x

19e ii.

k=1

2k

(+ 1)2 + k2 1 (1)ke(+1) sin(kln |x |)

x

20a. 0=0 , 1=1 , 2=4 , 3=9

20b.

d

d x 1 x 2 dd x = 11 x 2 20c. f | g=

11

(f(x ))g(x )

1 x 21/2 d x20d i.

20d ii.

2

20d iii.

3

8

20d iv.

220e i. 0

20e ii. 4

3x

20e iii. 43

x

20e iv. 4

3x 4

5x 3

Sturm -Lioville Problems

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