sturm-liouville oscillation theory for di ... - williams...

Sturm-Liouville Oscillation Theory for

Differential Equations and Applications to

Functional Analysis

by

ZHAONING WANG

Mihai Stoiciu, Advisor

A thesis submitted in partial fulfillment

of the requirements for the

Degree of Bachelor of Arts with Honors

in Mathematics

WILLIAMS COLLEGE

Williamstown, Massachusetts USA

May 11, 2011

ZHAONING (NANCY) WANG

Abstract

We study the connection between second-order differential equations and their corre-

sponding difference equations. With this connection in mind, we investigate quanti-

tative and qualitative properties of the zeros of the solutions of differential/difference

equations and of the eigenvalues of the associated Jacobi matrices. In particular, we

study various applications of the Sturm-Liouville Oscillation Theory to differential

equations and spectral theory.

Zhaoning (Nancy) Wang

Williams College

Williamstown, MA 01267

2

Acknowledgements

For months of guidance, inspiration, and insight, I am most grateful to my advisor,

Prof. Mihai Stoiciu for his excellent advising and confidence in me. I am also very

appreciative of the helpful feedback I received from my second reader, Prof. Stewart

D. Johnson. I would also like to thank the professors and students of the Williams

College Mathematics Department, who have provided me with the challenges and the

support that have made my four years at Williams meaningful and rewarding. Finally,

I would like to thank my parents for their unconditional love and support, which have

made it possible for me to accomplishment my goals.

3


Contents

1 Introduction 5

2 First Order Systems 7

2.1 Existence and Uniqueness of Solutions . . . . . . . . . . . . . . . . . . 7

2.2 Properties of the First Order Systems . . . . . . . . . . . . . . . . . . . 12

3 Second Order Scalar Equations 19

3.1 Existence and Uniqueness of Solutions . . . . . . . . . . . . . . . . . . 19

3.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Differentiable Dependence . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.3 The Prufer Transformations . . . . . . . . . . . . . . . . . . . . . . . . 47

3.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.4 Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.4.2 Oscillation Criteria . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.5 Sturm Comparison Theorem . . . . . . . . . . . . . . . . . . . . . . . . 54

4 Sturm Oscillation and Comparison Theorems 55

4.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.1.1 Construction of Difference Equations from a Jacobi Matrix . . . 55

4.2 Sturm Oscillation and Comparison Theorems . . . . . . . . . . . . . . . 57

5 Connection between the Two Approaches 57

5.1 Construction of the Jacobi Matrix from Differential Equations . . . . . 57

5.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

6 References 65

Appendices 66

A. Calculations for the Prufer Transformation Example . . . . . . . . . . . . 66

B. Calculations for the Difference/Differential Equations . . . . . . . . . . . 67

C. Calculations for the Difference/Differential Equations . . . . . . . . . . . 68

4

1 INTRODUCTION

1 Introduction

In 1836 and 1837, Jacques Charles Francois Sturm (1803–1855) and Joseph Liouville

(1809–1882) published a series of papers on second order linear ordinary differential

equations of the form

− (py′)′ + qy = f on J (1.1)

together with the initial conditions

y(c) = h, (py′)(c) = k, c ∈ J, h, k ∈ C (1.2)

where

J = (a, b), −∞ ≤ a < b ≤ +∞, 1

p, q, f : J → C. (1.3)

This marked the beginning of the investigation of the “Sturm-Liouville Problems” and

the establishment of the Sturm-Liouville Theory.

The discrete analog of the differential equation above is the difference equation

defined by the following recurrence relations

(hu)n = anun+1 + bnun + an−1un−1 = xu (1.4)

for n = 1, 2, · · · with u0 = 0, where h is some operator on u, x is a parameter, u, un

are functions of x for every n.

Among all the Ordinary Differential Equations (ODE), there are some equations

that are explicitly solvable, meaning that for those equations, we are able to find

a closed-form expression of specific functions that satisfy the differential equations.

However, the vast majority of the ordinary differential equations, unfortunately, are

not explicitly solvable so that we are unable to find specific expressions that solve the

differential equations.

In the latter case, thanks to Sturm and Liouville’s contributions, the Sturm-

Liouville Oscillation Theory can be utilized to investigate qualitative properties of

the solutions to some of the differential equations that we cannot solve explicitly. In

particular, we are well-informed of the number and the location of the zeros of the

solutions.

5


The Sturm-Liouville Oscillation Theory has applications in functional analysis

and the analysis of operators. For certain matrices M such as the finite n× n Jacobi

matrices JN defined by

Jn =

b1 a1 0 · · · · · · 0

a1 b2 a2...

0 a2 b3. . .

......

. . . . . . . . ....

.... . . bn−1 an−1

0 · · · · · · · · · an−1 bn

,

one can construct differential equations which have, as solutions, the characteristic

polynomial of M . The zeros of the solutions of the differential equations are precisely

the eigenvalues of M . Therefore, the Sturm-Liouville Oscillation Theory also provides

information about the eigenvalues of the matrix M .

Now, we state the Sturm Oscillation Theorem using the difference equations

approach:

Theorem 1.1. (Sturm Oscillation Theorem). Let P1, P2, · · · , Pn be a sequence of

monic orthogonal polynomials associated with the Jacobi matrix J . If P`(x0) 6= 0,

` = 0, 1, · · · , n, then the number of eigenvalues of Jn above x0 is

#{j | 0 ≤ j ≤ n− 1 so that sgn(Pj+1(x0)) 6= sgn(Pj(x0))}.

Using the same approach, the Sturm Comparison Theorem can be stated as:

Theorem 1.2. (Sturm Comparison Theorem) If u and v are linearly independent

solutions of

x0un = an+1un+1 + bn+1un + anun−1,

v has a sign change between every pair of sign changes of u.

Finally, we state the Sturm Comparison Theorem using the differential equa-

tions approach:

Theorem 1.3. Consider a singular endpoint of equation

(py′)′ + qy = 0 on J, 1/p, q ∈ L1loc(J,R), p > 0 a.e. on J = (a, b). (1.5)

6

2 FIRST ORDER SYSTEMS

Now, consider a comparison equation

(Pz′)′ +Qz = 0 on J. (1.6)

Assume 1/P, Q ∈ L1loc(J,R) and satisfy

Q ≥ q, 0 < P ≤ p on J. (1.7)

Suppose y is a non-trivial solution of (1.5) satisfying y(c) = 0 = y(d) for some

c, d ∈ J, c < d. Then every solution of (1.6) has a zero in the closed interval [c, d].

2 First Order Systems

2.1 Existence and Uniqueness of Solutions

Definition 2.1. (Contraction Mapping). A contraction mapping, or contraction, on

a metric space (M,d) is a function f from M to itself, with the property that there is

some real number k < 1 such that for all x and y in M ,

d(f(x), f(y)) ≤ k d(x, y).

The smallest such value of k is called the Lipschitz constant of f . Contractive maps

are sometimes called Lipschitzian maps. If the above condition is satisfied for k ≤ 1,

then the mapping is said to be non-expansive.

The following theorem is a classic result for first-order systems stated in Zettl

(2005). Zettl also provides the outline for two methods of proving the theorem.

Theorem 2.2. (Existence and Uniqueness). Let J be any integral, open, closed, half

open, bounded or unbounded; let n,m ∈ N. If

P ∈Mn(L1loc(J,C)) (2.1)

and

F ∈Mn,M(L1loc(J,C)) (2.2)

7


then every initial value problem (IVP)

Y ′ = PY + F (2.3)

Y (u) = C, u ∈ J, C ∈Mn,m(C) (2.4)

has a unique solution defined on all of J . Furthermore, if C,P, F are all real-valued,

then there is a unique real valued solution.

Proof. (Method 1: Defining a contraction mapping on a Banach space)

First, we note that if Y is a solution of the IVP (2.3), (2.4) then we integrate

both sides of (2.3), which yields

Y (t) = C +

∫ t

u

(PY + F ), t ∈ J. (2.5)

Conversely, every solution of the integral equation (2.5) has a unique solution

on [u, c] if c > u and on [c, u] if c < u. Assume c > u. Let

B = {Y : [u, c]→Mn,m(C), Y continuous}.

Now, we need to define a contraction mapping T on the Banach space B such

that ‖Tx‖ ≤ k‖x‖, where k < 1. Then, we are able to apply the contraction mapping

principle later in order to show that T has only one unique fixed point.

Following Bielecki (1956), we define the norm of any function Y ∈ B to be

‖Y ‖ = sup{exp

(−K

∫ t

u

|P (s)|ds)|Y (t)|, t ∈ [u, c]}, (2.6)

where K is a fixed positive constant K > 1. It is easy to see that with this norm B is

a Banach space since the Banach space is a metric space with norm. Let the operator

T : B → B be defined by

(TY )(t) = C +

∫ t

u

(PY + F )(s)ds, t ∈ [u, c], Y ∈ B. (2.7)

Then for Y, Z ∈ B we have

TY (t)− (TZ)(t) =

∫ t

u

P (s)(Y (s)− Z(s))ds

8


so that

|TY (t)− (TZ)(t)| ≤∫ t

u

|P (s)||Y (s)− Z(s)|ds

Now, let us look at exp

(−K

∫ t

u

|P (s)|ds)|(TY )(t) − (TZ)(t)|. By the above

inequality, we have

exp

(−K

∫ t

u

|P (s)|ds)|(TY )(t)− (TZ)(t)|

≤ exp

(−K

∫ t

u

|P (s)|ds)∫ t

u

|P (s)||Y (s)− Z(s)|ds

=

∫ t

u

exp

(−K

∫ t

u

|P (r)|dr)|P (s)||Y (s)− Z(s)|ds

=

∫ t

u

exp

(−K

∫ s

u

|P (r)|dr)

exp

(−K

∫ t

s

|P (r)|dr)|P (s)||Y (s)− Z(s)|ds

=

∫ t

u

exp

(−K

∫ t

s

|P (r)|dr)|P (s)| exp

(−K

∫ s

u

|P (r)|dr)|Y (s)− Z(s)|ds

=

∫ t

u

exp

(−K

∫ t

s

|P (r)|dr)|P (s)|‖Y − Z‖ds

= ‖Y − Z‖∫ t

u

|P (s)| exp

(−K

∫ t

s

|P (r)|dr)ds.

Let w(s) = −∫ t

s

|P (r)|dr and thus by the Fundamental Theorem of Calculus,

dw(s) = |P (s)|ds. Then by substitution we have∫ t

u

|P (s)| exp

(−K

∫ t

u

|P (r)|dr)ds

=

∫ 0

−R u

s |P (r)|drexp(−Kw(s))dw(s)

=

∫ R us |P (r)|dr

0

exp(−Kw(s))dw(s)

=

[−exp(−Kw)

K

]w=R u

s |P (r)|dr

w=0

=1

K− 1

Kexp

(−K

∫ u

s

|P (r)|dr)

≤ 1

K.

9


Hence, we have

exp

(−K

∫ t

u

|P (s)|ds)|(TY )(t)− (TZ)(t)|

≤ ‖Y − Z‖∫ t

u

|P (s)| exp

(−K

∫ t

s

|P (r)|dr)ds

≤ 1

K‖Y − Z‖.

Therefore, since by definition,

‖TY − TZ‖ = sup{exp

(−K

∫ t

u

|P (s)|ds)|(TY )(t)− (TZ)(t)|, t ∈ [u, c]},

and1

K‖Y − Z‖ is also an upper bound, we have

‖TY − TZ‖ ≤ 1

K‖Y − Z‖.

From the contraction mapping principle in Banach space it follows that the map

T has a unique fixed point and therefore the IVP (2.3), (2.4) has a unique solution on

[u, c]. The proof for the case c < u is similar; in this case the norm of B is modified to

‖Y ‖ = sup{exp

(K

∫ t

u

|P (s)|ds)|Y (t)|, t ∈ [c, u]},

Since there is a unique solution on every compact subinterval [u, c] and [c, u] for c ∈ J ,

c 6= u it follows that there is a unique solution on J . To establish the furthermore part

take the Banach space of real-valued functions, that is,

B = {Y : [u, c]→Mn,m(R), Y continuous}.

and proceed similarly. This completes the first proof.

Proof. (Method 2: Standard successive approximations proof)

For the second proof we construct a solution of (2.5) by successive approxima-

tions. Define

Y0(t) = C, Yn+1(t) = C +

∫ t

u

(PYn + F ), t ∈ J, n = 0, 1, 2, · · · (2.8)

10


Then Yn is a continuous function on J for each n ∈ N. We show that the sequence

{Yn : n ∈ N} converges to a function Y uniformly on each compact subinterval of J

and that the limit function Y is the unique solution of the integral equation (2.5) and

hence also of the IVP (2.3), (2.4). Choose b ∈ J, b > u and define

p(t) =

∫ t

u

|P (s)|ds, t ∈ J ;Bn(t) = maxu≤s≤t

|Yn+1(s)− Yn(s)|, u ≤ t ≤ b. (2.9)

Then,

Yn+1(t)− Yn(t) =

∫ t

u

P (s)[Yn(s)− Yn−1(s)]ds, t ∈ J, n ∈ N. (2.10)

From this we get

|Y2(t)− Y1(t)| ≤ B0(t)

∫ t

u

|P (s)|ds = B0(t)p(t) ≤ B0(t)p(b), u ≤ t ≤ b. (2.11)

|Y3(t)− Y2(t)| =

∫ t

u

|P (s)||Y2(t)− Y1(t)|ds ≤∫ t

u

|P (s)|B0(s)p(s)ds

≤ B0(t)

∫ t

u

|P (s)|p(s)ds ≤ B0(b)p2(t)

2!

≤ B0(b)p2(b)

2!, u ≤ t ≤ b.

From this and mathematical induction we get

|Yn+1(t)− Yn(t)| ≤ B0(b)pn(b)

n!, u ≤ t ≤ b.

Hence for any k ∈ N, by the triangle inequality,

|Yn+k+1(t)− Yn(t)| ≤ |Yn+k+1(t)− Yn+k(t)|+ |Yn+k(t)− Yn+k−1(t)|+

· · ·+ |Yn+1(t)− Yn(t)|

≤ B0(b)pn(b)

n![1 +

p(b)

n+ 1+

p2(b)

(n+ 2)(n+ 1) + · · ·].

Choose m large enough so that p(b)/(n+1) ≤ 1/2, then p2(b)/((n+2)(n+1)) ≤1/4, etc. when n > m and the term in brackets is bounded above by 2. It follows that

the sequence {Yn : n ∈ N0} converges uniformly, say to Y , on [u, b]. From this it

11


follows that Y satisfies the integral equation (2.5) and hence also the IVP (2.3) and

(2.4) on [u, b].

To show that Y is the unique solution assume Z is another one; then Z is

continuous and therefore |Y − Z| is bounded, say by M > 0 on [u, b]. Then

|Y (t)− Z(t)| =∣∣∣∣∫ t

u

P (s)[Y (s)− Z(s)]ds

∣∣∣∣ ≤M

∫ t

u

|P (s)|ds ≤Mp(t), u ≤ t ≤ b.

Now proceeding as above we get

|Y (t)− Z(t)| ≤Mpn(t)

n!≤M

pn(b)

n!, u ≤ t ≤ b, n ∈ N.

Note that as n approaches ∞, M pn(b)

n!approaches 0 so that therefore, Y = Z

on [u, b]. There is a similar proof for the case when b < u. This completes the second

proof.

2.2 Properties of the First Order Systems

Theorem 2.3. (The Gronwall Inequality)

(i) (The “right” Gronwall inequality) Let J = [a, b]. Assume g in L1(J,R) with

g ≥ 0 a.e., f real valued and continuous on J. If y is continuous, real valued, and

satisfies

y(t) ≤ f(t) +

∫ t

a

g(s)y(s)ds, a ≤ t ≤ b, (2.12)

then

y(t) ≤ f(t) +

(∫ t

a

f(s)g(s) exp

(∫ t

s

g(u)du

)ds

), a ≤ t ≤ b, (2.13)

For the special case when f(t) = c, a constant, we get

y(t) ≤ c exp

(∫ t

a

g(s)ds

), a ≤ t ≤ b. (2.14)

For the special case when f is nondecreasing on [a, b] we get

y(t) ≤ f(t) exp

(∫ t

a

g(s)ds

), a ≤ t ≤ b. (2.15)

12


(ii) (The “left” Gronwall inequality) Let J = [a, b]. Assume g in L1(J,R) with g ≥ 0

a.e., f real valued and continuous on J. If y is continuous, real valued, and

satisfies

y(t) ≤ f(t) +

∫ b

t

g(s)y(s)ds, a ≤ t ≤ b, (2.16)

then

y(t) ≤ f(t) +

(∫ b

t

f(s)g(s) exp

(∫ s

t

g(u)du

)ds

), a ≤ t ≤ b, (2.17)

For the special case when f(t) = c, a constant, we get

y(t) ≤ c exp

(∫ b

t

g(s)ds

), a ≤ t ≤ b. (2.18)

For the special case when f is nondecreasing on [a, b] we get

y(t) ≤ f(t) exp

(∫ b

t

g(s)ds

), a ≤ t ≤ b. (2.19)

Proof. For part (i) let z(t) =

∫ t

a

gy, t ∈ J and note that

z′ = gy ≤ g

(f +

∫ t

a

gy

)= gf + gz =⇒ z′ − gz ≤ gf a.e.

Hence, we have

exp

(−∫ s

a

g(u)du

)[z′(s)− g(s)z(s)]

= exp

(−∫ s

a

g(u)du

)z′(s)− exp

(−∫ s

a

g(u)du

)g(s)z(s)

=

[exp

(−∫ s

a

g(u)du

)z(s)

]′≤ g(s)f(s) exp

(−∫ s

a

g(u)du

), a ≤ s ≤ b.

Integrating both sides of the inequality above from a to t, we get

exp

(−∫ t

a

g(u)du

)z(t) ≤

∫ t

a

g(s)f(s) exp

(−∫ s

a

g(u)du

)ds, a ≤ t ≤ b.

13


From (2.12) and the above line we obtain

y(t) ≤ f(t) + z(t) ≤ f(t) + exp

(∫ t

a

g(u)du

)exp

(−∫ t

a

g(u)du

)z(t)

≤ f(t) + exp

(∫ t

a

g(u)du

)∫ t

a

g(s)f(s) exp

(−∫ s

a

g(u)du

)ds

= f(t) +

∫ t

a

g(s)f(s) exp

(∫ t

a

g(u)du

)exp

(−∫ s

a

g(u)du

)ds

= f(t) +

(∫ t

a

f(s)g(s) exp

(∫ t

s

g(u)du

)ds

), a ≤ t ≤ b.

This concludes the proof of (2.13). The two special cases follow from (2.13).

If f(t) = c, where c is a constant, then by (2.13), we have

y(t) ≤ f(t) +

(∫ t

a

f(s)g(s) exp

(∫ t

s

g(u)du

)ds

)= c+

(∫ t

a

cg(s) exp

(∫ t

s

g(u)du

)ds

)= c− c exp

(∫ t

s

g(s)ds

)∣∣∣∣s=ts=a

= c exp

(∫ t

a

g(s)ds

), a ≤ t ≤ b.

If f is a nondecreasing on [a, b], then by (2.13), we have

y(t) ≤ f(t) +

(∫ t

a

f(s)g(s) exp

(∫ t

s

g(u)du

)ds

)≤ f(t) + f(t)

(∫ t

a

g(s) exp

(∫ t

s

g(u)du

)ds

)= f(t)− f(t) exp

(∫ t

s

g(s)ds

)∣∣∣∣s=ts=a

= f(t) exp

(∫ t

a

g(s)ds

), a ≤ t ≤ b.

For part (ii) let z(t) =

∫ b

t

gy, t ∈ J and note that

z′ = −gy ≥ −g(f +

∫ b

t

gy

)= −gf − gz =⇒ z′ + gz ≥ −gf a.e.

14


Hence, we have

exp

(−∫ b

s

g(u)du

)[z′(s) + g(s)z(s)]

= exp

(−∫ b

s

g(u)du

)z′(s) + exp

(−∫ b

s

g(u)du

)g(s)z(s)

=

[exp

(−∫ b

s

g(u)du

)z(s)

]′≥ −g(s)f(s) exp

(−∫ b

s

g(u)du

), a ≤ s ≤ b.

Integrating both sides of the inequality above from t to b, we get

− exp

(−∫ b

t

g(u)du

)z(t) ≥ −

∫ b

t

g(s)f(s) exp

(−∫ b

s

g(u)du

)ds, a ≤ t ≤ b

=⇒ exp

(−∫ b

t

g(u)du

)z(t) ≤

∫ b

t

g(s)f(s) exp

(−∫ b

s

g(u)du

)ds, a ≤ t ≤ b.

From (2.16) and the above line we obtain

y(t) ≤ f(t) + z(t) ≤ f(t) + exp

(∫ b

t

g(u)du

)exp

(−∫ b

t

g(u)du

)z(t)

≤ f(t) + exp

(∫ b

t

g(u)du

)∫ b

t

g(s)f(s) exp

(−∫ b

s

g(u)du

)ds

= f(t) +

∫ b

t

g(s)f(s) exp

(∫ b

t

g(u)du

)exp

(−∫ b

s

g(u)du

)ds

= f(t) +

(∫ b

t

f(s)g(s) exp

(∫ s

t

g(u)du

)ds

), a ≤ t ≤ b.

This concludes the proof of (2.17). The two special cases follow from (2.17).

If f(t) = c, where c is a constant, then by (2.17), we have

y(t) ≤ f(t) +

(∫ b

t

f(s)g(s) exp

(∫ s

t

g(u)du

)ds

)= c+

(∫ b

t

cg(s) exp

(∫ s

t

g(u)du

)ds

)= c− c exp

(∫ s

t

g(s)ds

)∣∣∣∣s=bs=t

= c exp

(∫ b

t

g(s)ds

), a ≤ t ≤ b.

15


If f is a nondecreasing on [a, b], then by (2.17), we have

y(t) ≤ f(t) +

(∫ b

t

f(s)g(s) exp

(∫ s

t

g(u)du

)ds

)≤ f(t) + f(t)

(∫ b

t

g(s) exp

(∫ s

t

g(u)du

)ds

)= f(t)− f(t) exp

(∫ s

t

g(s)ds

)∣∣∣∣s=bs=t

= f(t) exp

(∫ b

t

g(s)ds

), a ≤ t ≤ b

and this completes the proof.

Theorem 2.4. (Abel’s Formula) Let J = (a, b), and assume that P ∈Mn(L1loc(J,C)).

If Y is an n× n matrix solution of

Y ′ = PY on J,

then for any u, t ∈ J, we have

(detY )(t) = (detY )(u) exp

(∫ t

u

traceP (s)ds

).

Proof. First, we want to show that y = detY satisfies that first order scalar equation

y′ − py = 0,

where p = traceP.

Let yij denote the ijth entry of Y (t). Note that by the definition of the deter-

minant,

y = detY =∑σ∈Sn

ε(σ)y1σ(1) · · · ynσ(n),

where σ(i) is in the symmetry group for every i, and hence

y′ =n∑i=1

∑σ∈Sn

ε(σ)y1σ(1) · · · y′iσ(i) · · · ynσ(n)

by the product rule.

16


By the definition of the determinant again, it follows that

y′ =dy

dt=

d

dt(detY (t)) =

n∑i=1

det

y11(t) · · · y1n(t)...

. . ....

y′i1(t) · · · y′in(t)...

. . ....

yn1(t) · · · ynn(t)

=

n∑i=1

n∑j=1

y′ij(t)Mij, (2.20)

where Mij is the corresponding matrix when the ith row and jth column of the previous

matrix removed.

Since Y ′ = PY, by the multiplication of matrices, we know that

y′ij(t) =n∑k=1

pik(t)ykj(t).

Thus, the ith row of the equation (2.20) can be replaced by

(y′i1(t), · · · , y′in(t)) =n∑k=1

pik(yk1(t), · · · , ykn(t)).

Recall the property of the determinant that if matrix B is obtained by multiply-

ing one row or column of matrix A by some number c, then detB = c detA. Plugging

the equality above into equation (2.20) and by the above property of the determinant,

we have

dy

dt=

d

dt(detY (t))

=n∑i=1

det

y11(t) · · · y1n(t)...

. . ....

y′i1(t) · · · y′in(t)...

. . ....

yn1(t) · · · ynn(t)

=

n∑i=1

n∑j=1

y′ij(t)Mij

=n∑i=1

n∑j=1

n∑k=1

pik(t)ykj(t)Mij

=n∑i=1

n∑k=1

pik(t)n∑j=1

ykj(t)Mij,

17


where the last term

n∑i=1

n∑k=1

pik(t)n∑j=1

ykj(t)Mij =n∑i=1

n∑k=1

pik(t) det

y11(t) · · · y1n(t)...

. . ....

yk1(t) · · · ykn(t)...

. . ....

yn1(t) · · · ynn(t)

.

Note that the row written in the middle denotes the generic kth row.

When k 6= i, then the above matrix will have two rows that are exactly the

same and will also miss the ith row. Instead, when i = k, the above matrix will be

exactly Y (t).

Note that if two rows of a matrix are the same, then the determinant of that

matrix will be 0. Here is the explanatin: the determinant should be the additive inverse

of itself in this case, since interchanging the rows (or columns) does not change the

absolute value of the determinant, but reverses the sign of the determinant. The only

number for which this is possible is when it is equal to 0.

Hence, the last term in the last equation can be simplified to

n∑i=1

n∑k=1

pik(t) det

y11(t) · · · y1n(t)...

. . ....

yk1(t) · · · ykn(t)...

. . ....

yn1(t) · · · ynn(t)

=

n∑i=1

pii(t) detY (t) = detY (t)n∑i=1

pii(t)

= y · traceP = py.

On the left-hand side, we have y′, and we have shown above that the right-hand

side can be simplified as py. Hence, we have y′ = py, where y′ = detY and p = traceP.

With y′ = py, for any u, t ∈ J, we have

1

ydy = p =⇒

∫ t

u

1

ydy =

∫ t

u

p =⇒ ln y(t)− ln y(u) =

∫ t

u

p(s)ds.

Exponentiating both sides, we have

y(t) = y(u) exp

(∫ t

u

p(s)ds

),

18

3 SECOND ORDER SCALAR EQUATIONS

from which it follows that

(detY )(t) = (detY )(u) exp

(∫ t

u

traceP (s)ds

),

which completes the proof.

3 Second Order Scalar Equations

In this section, we study the initial value problems (IVP) consisting of the second order

scalar equation

− (py′)′ + qy = f on J (3.1)


y(c) = h, (py′)(c) = k, c ∈ J, h, k ∈ C (3.2)

where

J = (a, b), −∞ ≤ a < b ≤ +∞, 1

p, q, f : J → C. (3.3)

3.1 Existence and Uniqueness of Solutions

Definition 3.1. (Solution). By a solution of equation (5.1) we mean a function y :

J → C such that y and y[1] = py′ are absolutely continuous on each compact subinterval

of J and the equation is satisfied a.e. on J . Given a solution y we refer to y[1] as its

quasi-derivative to distinguish it from the classical derivative y′.

Note that the classical derivative of a solution y, in general, exists only almost

everywhere but the quasi-derivative y[1] = (py′) is absolutely continuous on all compact

subintervals of J and thus exists and is continuous at each point of the open interval

J .

Theorem 3.2. Assume that

1/p, q, f ∈ L1loc(J,C). (3.4)

19


Then every initial value problem (IVP) (5.1), (5.2), (5.3) has a solution defined on all

of J and this solution is unique. Moreover, if all the data p, q, f, h, k is real, then there

is a unique real-valued solution on J .

Proof. First, let us construct the following matrices:

P =

[0 1/p

q 0

], F =

[0

−f

], Y =

[y

py′

]. (3.5)

Then, we observe that

PY + F =

[0 1/p

q 0

][y

py′

]+

[0

−f

]

=

[y′

qy − f

]

=

[y′

(py′)′

]= Y ′.

Hence, the equation (5.1) is equivalent to the first order system

Y ′ = PY + F on J, (3.6)

in the sense that, given any scalar solution y of (5.1) the vector Y defined by (3.1) is

a solution of the system (3.6) and conversely, given any vector solution Y of system

(3.6) its top component y is a solution of (5.1). Theorem 3.2 follows from this system

representation and Theorem 2.2 of the previous section.

In the theory of boundary value problems the spectral parameter λ and the

weight function w play important roles; thus we also study the equation

− (py′)′ + qy = λwy on J (3.7)

where

J = (a, b), −∞ ≤ a < b ≤ +∞, 1

p, q, w : L1

loc(J,C). (3.8)

20


In this case, we use the following notation

y = y(·, c, h, k, 1/p, q, w, λ), y[1] = y[1](·, c, h, k, 1/p, q, w, λ). (3.9)

Definition 3.3. (Regular and Singular Endpoints). Let J = (a, b),−∞ ≤ a < b ≤ ∞,

and consider the equation (5.1) with conditions (3.4).

The endpoint a is said to be regular (or equation (5.1) is regular at a) if

1/p, q, f ∈ L1((a, d),C) (3.10)

for some d ∈ J ; otherwise it is called singular. Similarly, the endpoint b is said to be

regular if

1/p, q, f ∈ L1((d, b),C) (3.11)

for some d ∈ J ; otherwise it is called singular. Note that, given condition (3.4), if

(3.10) and (3.11) hold for some d ∈ J, then they hold for any d ∈ J.

The following theorem is stated in Zettl (2005) without correct proof.

Theorem 3.4. Let (3.7),(3.8) hold in R and assume that p > 0 a.e. on J and λ is

real. Then the zeros of every nontrivial solution y of (3.7) are isolated in the interior

of J and also at regular endpoint of J . If a nontrivial solution y has a zero at a regular

endpoint of J then there is an appropriate one sided neighborhood of this endpoint in

which y has no other zero. Thus only a singular endpoint of J can be an accumulation

point of zeros of any nontrivial solution y of (3.7).

Proof. Suppose that a and b are both regular endpoints of J . By definition, for any

k ∈ J , we know that

1/p, q, f ∈ L1((a, k),C)

and that

1/p, q, f ∈ L1((k, b),C)

so that

1/p, q, f ∈ L1((a, b),C),

which implies that 1/p, q, f are all locally integrable on the interval (a, b).

21


First, we show that if y has zeros at c, d ∈ (a, b) with c < d, then (py′)(h) = 0

for some h ∈ (c, d). Then, we have

0 = y(d)− y(c) =

∫ d

c

y′ =

∫ d

c

1

p(py′) = (py′)(h)

∫ d

c

1

p

by the Mean Value Theorem for the Lebesgue integral (recall that (py′) is continuous

on J). Moreover, we are allowed to conduct the operations above because 1/p ∈L1((a, b),C).

Hence, either

(py′)(h) = 0

or ∫ d

c

1

p= 0

for the above equality to hold. However, we claim that it is impossible to have the

latter: by our hypothesis that p > 0 a.e. in (a, b), we know that 1/p > 0 a.e. in (a, b)

so that ∫ d

c

1

p> 0

since the integral of a positive function on some interval (c, d) is always positive. Hence,

we know that it must be true that (py′)(h) = 0.

Now, in order to prove the Theorem, let us suppose that there exists a sequence

{tn ∈ (a, b) : n ∈ N0} such that tn → t0 and y(tn) = 0 for all n ∈ N0. That is, the

sequence above is composed of zeros of y converging to t0. Then, since y is continuous,

we know that y(t0) = 0.

Then, suppose that there exists another sequence {hn ∈ (a, b) : n ∈ N0} such

that hn → t0 and (py′)(hn) = 0 for all n ∈ N0. Such a sequence must exist since we have

shown above that there exists an h between any two zeros of y such that (py′)(h) = 0.

Then, since y[1] is continuous in (a, b), we know that (py′)(t0) = 0. Hence, we have

y(t0) = 0

and

y[1](t0) = 0 =⇒ y′(t0) = 0.

However, y(t0) = 0 and y[1](t0) implies that y is identically zero on J . Thus,

y = 0 is the only solution by the uniqueness of initial value problem. Recall that we

22


have assumed y to be a nontrivial solution. This contradiction completes the proof.

Theorem 3.5. (Sturm Separation Theorem). Let (3.7),(3.8) hold in R and assume

that p > 0 a.e. on J and λ is real. Suppose that y and z are linearly independent

solutions of (3.7). Then z has a zero strictly between any two zeros of y.

Proof. Suppose

y(c) = 0 = y(d)

with c 6= d. Since y and z are linearly independent, neither is the trivial solution

and they have no common zero. By Theorem 3.4, we may assume that c and d are

consecutive zeros of y and that c < d. Further, replacing y by −y if necessary, we may

assume that y > 0 on the open interval (c, d). Then

0 < y(t)− y(c) =

∫ t

c

1

p(py′).

This imples that (py′)(c) > 0. Otherwise, since py′ is continuous and (py′)(c)

is not zero, the assumption that (py′)(c) < 0 implies that it is negative in some right

neighborhood of c and this contradicts the above equation. Similarly we get that

(py′)(d) < 0.

Now, multiplying the equation for y by z and the equation for z by y and

subtracting we get

0 = z(py′)′ − y(pz′)′ = [z(py′)− y(pz′)]′.

An integration yields

z(d)(py′)(d)− y(c)(pz′)(c) = 0.

Assume that z has no zero in (c, d). If z > 0 on (c, d) then the left hand side of the

above equation is positive giving a contradiction. A similar contradiction is reached if

z < 0, which completes the proof.

3.1.1 Examples

Example 3.6. Let us consider the equation

−(py′)′ + qy = λwy

23


on J = (a, b), where

J = (a, b) = (0, 1);1

p(t)= t−2 sin

(1

t

),where 0 < t < 1; q = 0; w = 0.

Then, we have

−(py′)′ = 0

First, we need to solve the differential equation. Since −(py′)′ = 0, we know

that (py′)′ = 0. Thus, py′ = C, where C is a constant. If C = 0, then y′ = 0 so that

one possible solution is u(t) = 1. If C 6= 0, then y′ =C

pso that

y =

∫C

p= C

∫1

p= C

∫t−2 sin

(1

t

)dt = C cos

(1

t

).

Hence, the other possible solution is

v(t) = cos

(1

t

).

Let us observe the solution v(t). As t → 0,1

t→ ∞ so that the solution has

infinitely many zeros in any right neighborhood of 0, as shown below.

We claim that 0 must be a singular endpoint in this case. Here is the reason:

Let d be any point in (0, 1), then

∫ d

0

1

p(t)dt =

∫ d

0

t−2 sin

(1

t

)dt = cos

(1

t

)∣∣∣∣t=dt=0

= cos

(1

d

)− lim

t→0cos

(1

t

),

which does not exist since the limit of cos

(1

t

)as t→ 0 does not exist.


−(py′)′ + qy = λwy


J = (a, b) = (k, 1); k > 0;1

p(t)= t−2 sin

(1

t

),where k < t < 1; q = 0; w = 0.

24


0.2 0.4 0.6 0.8 1.0

-1.0

-0.5

0.5

1.0

Figure 1: Example 3.6

Then, we have

−(py′)′ = 0

As shown in the previous example, we still have two solutions:

u(t) = 1, v(t) = cos

(1

t

).

Let us observe the solution v(t). As t → k,1

t→ 1

k< ∞ and As t → 1,

1

t→ 1 < ∞. Hence, the solution has finitely many zeros in the interval (k, 1), as

shown below.

We claim that k must be a regular endpoint in this case. Here is the reason:

Let d be any point in (k, 1), then

∫ d

k

1

p(t)dt =

∫ d

k

t−2 sin

(1

t

)dt = cos

(1

t

)∣∣∣∣t=dt=k

= cos

(1

d

)− cos

(1

k

)<∞,

so that 1/p ∈ L1((k, d),C).


−(py′)′ + qy = λwy

25


0.2 0.4 0.6 0.8 1.0

-1.0

-0.5

0.5

1.0



J = (a, b) = (0, 1); k > 0;1

p(t)= t−3/2 sin

(1

t

)+ (1/2)t−1/2 cos

(1

t

); q = 0; w = 0.

Then, we have

−(py′)′ = 0

Using the same method as the above, the solutions to this differential equation

are:

u(t) = 1, v(t) =√t cos

(1

t

).

Let us observe the solution v(t). As t → 0,1

t→ ∞ so that the solution has

infinitely many zeros in any right neighborhood of 0, as shown below. Thus, 0 is an

accumulation point and must be a singular endpoint.

To check that 0 is a singular endpoint is somewhat trickier than in the previous

cases. Here, we must distinguish between functions that are Riemann integrable and

Lebesgue integrable.

We claim that 1/p(t) is Riemann integrable on the interval (0, d), where d ∈(0, 1), but not Lebesgue integrable. To calculate the Riemann integral, we must use

improper Riemann integral in this case because the function is not defined at 0.

26


0.2 0.4 0.6 0.8 1.0

-0.4

-0.2

0.2

0.4


Let d be any point in (0, 1), then

∫ d

0

1

p(t)dt =

∫ d

0

t−3/2 sin

(1

t

)+ (1/2)t−1/2 cos

(1

t

)dt =

√t cos

(1

t

)∣∣∣∣t=dt=0

=√d cos

(1

d

)− lim

t→0

√t cos

(1

t

)=√d cos

(1

d

)− 0

=√d cos

(1

d

).

Hence, 1/p(t) is indeed improperly Riemann integrable on (0, d).

However, note that ∫ d

0

∣∣∣∣ 1

p(t)

∣∣∣∣ dt→∞since there is no longer the cancelation between the positive values and negative values.

This implies that the function 1/p(t) is not Lebesgue integrable on (0, d). By definition,

0 must be a singular endpoint.


−(py′)′ + qy = λwy

27



J = (0, a); a <∞;1

p(t)=

1

tα,where k < t < 1 and α ∈ [−∞, 1); q = 0; w = 0.

Then, we have

−(py′)′ = 0

First, we need to solve the differential equation. Since −(py′)′ = 0, we know

that (py′)′ = 0. Thus, py′ = C, where C is a constant. If C = 0, then y′ = 0 so that

one possible solution is u(t) = 1. If C 6= 0, then y′ =C

pso that

y =

∫C

p= C

∫1

p= C

∫1

tαdt =

C

−α + 1

1

tα−1.

Hence, the other possible solution is

v(t) =1

(−α + 1)tα−1=

t1−α

(−α + 1).

Let us observe the solution v(t). Recall that α < 1 so that 1−α > 0. As t→ 0,

t1−α → 0. Since v(t) is strictly increasing, we know that v does not have any zeros on

the interval (0, a) with a <∞.


−(py′)′ + qy = λwy


J = (0, a); a <∞;1

p(t)=

1

tα,where k < t < 1 and α ∈ [1,∞); q = 0; w = 0.

Then, we have

−(py′)′ = 0

As we have shown in the previous example, one possible solution is u(t) = 1

and the other possible solution is

v(t) =1

(−α + 1)tα−1=

t1−α

(−α + 1).

28


0.2 0.4 0.6 0.8 1.0

0.5

1.0

1.5

2.0

2.5


Let us observe the solution v(t). Recall that α ≥ 1 so that α− 1 ≥ 0. As t→ 0,

tα−1 → 0 so that v(t) → −∞. Since v(t) is strictly increasing and v(t) < 0, we know

that v does not have any zeros on the interval (0, a) with a <∞.However, we claim that 0 is be a singular endpoint in this case. Here is the

reason:

Let d be any point in (0, a), then

∫ d

0

1

p(t)dt =

∫ d

0

1

tαdt =

t1−α

(−α + 1)

∣∣∣∣t=dt=0

=d1−α

(−α + 1)− lim

t→0

t1−α

(−α + 1),

which does not exist since the limit oft1−α

(−α + 1)as t→ 0 does not exist when α ≥ 1.

3.2 Differentiable Dependence

Now, we will introduce the concept of the Frechet derivative, which is a derivative de-

fined on Banach spaces. Named after Maurice Frechet, it is commonly used to formalize

the concept of the functional derivative used widely in the calculus of variations. That

is, each derivate is regarded as a bounded linear operator T . Intuitively, it generalizes

the idea of linear approximation from functions of one variable to functions on Banach

29


0.2 0.4 0.6 0.8 1.0

-10

-8

-6

-4

-2


spaces. The Frechet derivative should be contrasted to the more general Gateaux

derivative which is a generalization of the classical directional derivative.

Definition 3.11. (Frechet derivative in Banach spaces) Let V and W be Banach

spaces, and U ⊆ V be an open subset of V. A function f : U → W is called Frechet

differentiable at x ∈ U if there exists a bounded linear operator T : V → W such that

limh→0

‖f(x+ h)− f(x)− T (h)‖W‖h‖V

= 0.

Theorem 3.12. (Differentiable dependence on the equation) Let conditions (3.7), (3.8)

hold, and let K be a compact subinterval of J. Fix t, c ∈ K, h, k ∈ C. Then the maps

q → y(t, q), q → y[1](t, q), w → y(t, w), w → y[1](t, w), 1/p → y(t, 1/p), 1/p →y[1](t, 1/p) from L1(K,C) to C as well as the maps λ → y(t, λ), λ → y[1](t, λ) from Cto C are Frechet differentiable and their derivatives are given by

y′(t, q)(r) = −∫ t

c

φ1,2(t, s)y(t, s)r(s)ds, r ∈ L1(K,C),

(py′)′(t, q)(r) = −∫ t

c

φ2,2(t, s)y(t, s)r(s)ds, r ∈ L1(K,C),

30


y′(t, 1/p)(r) =

∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(py′)(x)r(x)dx

)ds+

∫ t

c

(py′)(x)r(x)dx, r ∈ L1(K,C),

(py′)′(t, 1/p)(r) =

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(py′)(x)r(x)dx

)ds, r ∈ L1(K,C),

y′(t, w)(r) = λ

∫ t

c

φ1,2(t, s, w)y(s, w)r(s)ds, r ∈ L1(K,C),

(py′)′(t, w)(r) = λ

∫ t

c

φ2,2(t, s, w)y(s, w)r(s)ds, r ∈ L1(K,C),

y′(t, λ)(r) =

∫ t

c

φ1,2(t, s, λ)w(s)y(s, λ)ds, λ ∈ C,

(py′)′(t, λ)(r) =

∫ t

c

φ2,2(t, s, λ)w(s)y(s, λ)ds, λ ∈ C.

Proof. (1) We will prove the first formula.

Let

−(py′)′ + qy = 0, y(c) = h, (py′)′(c) = k;

−(pz′)′ + (q + r)y = 0, z(c) = h, (pz′)′(c) = k.

Let

x = z − y = y(t, q + r)− y(t, q).

Then we have

−(px′)′ + qx = −rz, x(c) = 0, (px′)(c) = 0.

Let us define the matrix X to be

X =

[x

px′

],

the matrix F to be

F =

[0

−rz

],

and the matrix Φ to be

Φ =

[φ1,1 φ2,1

φ2,1 φ2,2

].

31


From the variation of parameters formula it follows that

X(t) =

∫ t

u

Φ(t, s)F (s)ds =

∫ t

u

Φ(t, s)F (s)ds (3.12)

=

∫ t

u

[φ1,1(t, s) φ2,1(t, s)

φ2,1(t, s) φ2,2(t, s)

][0

(−r(t))z(t)

]ds (3.13)

=

∫ t

u

[φ1,2(−r(t))z(t)

φ2,2(−r(t))z(t)

]ds. (3.14)

Hence, from equation (3.14), we know that

x(t) =

∫ t

c

φ1,2(t, s)(−r(s))z(s)ds (3.15)

px′(t) =

∫ t

c

φ2,2(t, s)(−r(s))z(s)ds. (3.16)

For equation (3.15), letting z = y + (z − y) we get

z(t)− y(t) =

∫ t

c

φ1,2(t, s)(−r(s))[y(s) + (z(s)− y(s))]ds

=

∫ t

c

φ1,2(t, s)(−r(s))y(s)ds+

∫ t

c

φ1,2(t, s)[z(s)− y(s)](−r(s))ds

= −∫ t

c

φ1,2(t, s)r(s)y(s)ds−∫ t

c

φ1,2(t, s)[z(s)− y(s)]r(s)ds

so that

z(t)− y(t) +

∫ t

c

φ1,2(t, s)(r(s))y(s)ds

= −∫ t

c

φ1,2(t, s)[z(s)− y(s)]r(s)ds.

Note that φ1,2 is bounded on K×K and z → y uniformly on K by the continuous

dependence of solutions in a compact interval. Suppose that |φ1,2| ≤ M for some

positive real number M . Let ε > 0 be given and let

z(s)− y(s) =ε

M(t− c)

32


as r → 0 in L1(J,C). Thus,

‖ −∫ tcφ1,2(t, s)[z(s)− y(s)]r(s)ds‖

‖r‖

≤‖‖r‖

∫ tcφ1,2(t, s)[z(s)− y(s)]ds‖

‖r‖≤∥∥∥∥∫ t

c

φ1,2(t, s)[z(s)− y(s)]ds

∥∥∥∥≤ M

∥∥∥∥∫ t

c

[z(s)− y(s)]ds

∥∥∥∥= M

∥∥∥∥∫ t

c

ε

M(t− c)ds

∥∥∥∥= M

ε

M(t− c)(t− c) = ε,

which implies that

z(t)− y(t) +

∫ t

c

φ1,2(t, s)(r(s))y(s)ds = −∫ t

c

φ1,2(t, s)[z(s)− y(s)]r(s)ds

= o(r) as r → 0 in L1(J,C).

Note that

limr→0

‖y(t, q + r)− y(t, q)− T (t, q)(r)‖C‖r‖C

=‖z(t)− y(t)− [D2(y)](t, q)(r)‖C

‖r‖C= 0.

From the definition of Frechet derivative, we know that

[D2(y)](t, q)(r) = −∫ t

c

φ1,2(t, s)y(t, s)r(s)ds

is indeed the derivative of y(t, q) with respect to q.

(2) For equation (3.16), letting z = y + (z − y) we get

pz′(t)− py′(t) =

∫ t

c

φ2,2(t, s)(−r(s))[y(s) + (z(s)− y(s))]ds

=

∫ t

c

φ2,2(t, s)(−r(s))y(s)ds+

∫ t

c

φ2,2(t, s)[z(s)− y(s)](−r(s))ds

= −∫ t

c

φ2,2(t, s)r(s)y(s)ds−∫ t

c

φ2,2(t, s)[z(s)− y(s)]r(s)ds

33


so that

z(t)− y(t) +

∫ t

c

φ2,2(t, s)(r(s))y(s)ds = −∫ t

c

φ2,2(t, s)[z(s)− y(s)]r(s)ds.

Note that φ2,2 is bounded on K×K and z → y uniformly on K by the continuous

dependence of solutions in a compact interval. Suppose that |φ2,2| ≤ M for some

positive real number M . Let ε > 0 be given and let

z(s)− y(s) =ε

M(t− c)

as r → 0 in L1(J,C). Thus, following the same steps as in the last part, we have

‖ −∫ tcφ2,2(t, s)[z(s)− y(s)]r(s)ds‖

‖r‖

≤‖‖r‖

∫ tcφ2,2(t, s)[z(s)− y(s)]ds‖

‖r‖

≤ M

∥∥∥∥∫ t

c

[z(s)− y(s)]ds

∥∥∥∥= M

ε

M(t− c)(t− c)

= ε,

which implies that

pz′(t)− py′(t) +

∫ t

c

φ2,2(t, s)(r(s))y(s)ds = −∫ t

c

φ2,2(t, s)[z(s)− y(s)]r(s)ds

= o(r) as r → 0 in L1(J,C).

Note that

limr→0

‖py′(t, q + r)− py′(t, q)− T (t, q)(r)‖C‖r‖C

=‖pz(t)− py(t)− [D2(py

′)](t, q)(r)‖C

‖r‖C= 0.


[D2(py′)](t, q)(r) = −

∫ t

c

φ2,2(t, s)y(t, s)r(s)ds

is indeed the derivative of py′(t, q) with respect to q.

34


(3) In order to derive the formulas for the derivatives with respect to 1/p we proceed

as follows.

Let1

pr=

1

p+ r, r ∈ L1(J,C)

and let y = y(t, 1/p), z = z(t, 1/pr). Set

x(t) =

∫ t

c

1

p(py′ − prz′)ds.

Then, we have

−(px′)′ + qx = f, x(c) = 0, (px′)(c) = 0, f(t) = −q(t)∫ t

c

(prz′)rds.


X =

[x

px′

],

the matrix F to be

F =

[0

f

],


Φ =

[φ1,1 φ2,1

φ2,1 φ2,2

].


X(t) =

∫ t

u

Φ(t, s)F (s)ds (3.17)

=

∫ t

u

Φ(t, s)F (s)ds (3.18)

=

∫ t

u

[φ1,1(t, s) φ2,1(t, s)

φ2,1(t, s) φ2,2(t, s)

][0

f(t)

]ds (3.19)

=

∫ t

u

[φ1,2f(t)

φ2,2f(t)

]ds (3.20)

35



x(t) =

∫ t

c

φ1,2(t, s)f(s)ds = −∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(prz′)(u)r(u)du

)ds (3.21)

px′(t) =

∫ t

c

φ2,2(t, s)f(s)ds = −∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(prz′)(u)r(u)du

)ds.(3.22)

For equation (3.21), note that

z(t)− y(t) =

∫ t

c

[1

pr(prz

′)− 1

p(py′)

]ds

=

∫ t

c

[(1

p+ r

)(prz

′)− 1

p(py′)

]ds

=

∫ t

c

[1

p(prz

′)− 1

p(py′)

]+

∫ t

c

(prz′)rds

= −x(t) +

∫ t

c

(prz′)rds

=

∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(prz′)(u)r(u)du

)ds+

∫ t

c

(prz′)rds.

Letting prz′ = py′ + [prz

′ − py′], we get∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(prz′)(u)r(u)du

)ds+

∫ t

c

(prz′)rds

=

∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(py′)(u)r(u)du

)ds+

∫ t

c

(py′)rds

+

∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

[prz′ − py′](u)r(u)du

)ds+

∫ t

c

[prz′ − py′]rds.

Hence, we know that

z(t)− y(t)−(∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(py′)(u)r(u)du

)ds+

∫ t

c

(py′)rds

)=

∫ t

c

φ1,2(t, s)q(s)

(∫ s

c


)ds+

∫ t

c

[prz′ − py′]rds.

Note that φ1,2 is bounded on K × K, q ∈ L1(K,C) and prz′ → py′ uniformly

on K by the continuous dependence of solutions in a compact interval. Suppose that

36


|φ1,2q + 1| ≤M for some positive real number M . Let ε > 0 be given and let

z(s)− y(s) =ε

M(t− c)2


‖∫ tcφ1,2(t, s)q(s)

(∫ sc


)ds+

∫ tc[prz

′ − py′]r‖‖r‖

≤‖r‖‖

∫ tcφ1,2(t, s)q(s)

(∫ sc

[prz′ − py′](u)du

)ds+

∫ tc[prz

′ − py′]‖‖r‖

≤∥∥∥∥∫ t

c

φ1,2(t, s)q(s)

(∫ s

c


)ds+

∫ t

c

[prz′ − py′]

∥∥∥∥=

∥∥∥∥∫ t

c

(φ1,2(t, s)q(s) + 1)

(∫ s

c


)ds

∥∥∥∥= M

∥∥∥∥∫ t

c

∫ t

c

ε

M(t− c)2du ds

∥∥∥∥= M

ε

M(t− c)2(t− c)2 = ε,

which implies that

z(t)− y(t)−(∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(py′)(u)r(u)du

)ds+

∫ t

c

(py′)r

)= o(r) as r → 0 in L1(J,C).

Note that

limr→0

‖y(t, 1/p+ r)− y(t, 1/p)− T (t, 1/p)(r)‖C

‖r‖C

=‖z(t)− y(t)− [D2(y)](t, 1/p)(r)‖C

‖r‖C= 0.


[D2(y)](t, 1/p)(r) =

∫ t

c

φ1,2(t, s)q(s)

(∫ s

c

(py′)(x)r(x)dx

)ds+

∫ t

c

(py′)(x)r(x)dx

is indeed the derivative of y(t, 1/p) with respect to 1/p.

(4) Now, we want to derive the Frechet derivative of py′(t, 1/p) with respect to 1/p.

37


Note that

px′ = p

(1

p(py′ − prz′)

)= py′ − prz′.

From equation (3.22), we have

prz′ − py′ = −(px′) =

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(prz′)(u)r(u)du

)ds.

Letting prz′ = py′ + [prz

′ − py′], we get∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(prz′)(u)r(u)du

)ds

=

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(py′)(u)r(u)du

)ds+

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c


)ds.

Hence, we know that

prz′ − py′ =

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(py′)(u)r(u)du

)ds

+

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c


)ds

Note that φ2,2 is bounded on K × K, q ∈ L1(K,C) and prz′ → py′ uniformly


|φ2,2q| ≤M for some positive real number M . Let ε > 0 be given and let

z(s)− y(s) =ε

M(t− c)2


‖∫ tcφ2,2(t, s)q(s)

(∫ sc


)ds‖

‖r‖

≤‖r‖‖

∫ tcφ2,2(t, s)q(s)

(∫ sc


)ds‖

‖r‖

≤∥∥∥∥∫ t

c

φ2,2(t, s)q(s)

(∫ s

c


)ds

∥∥∥∥= M

∥∥∥∥∫ t

c

∫ t

c

ε

M(t− c)2du ds

∥∥∥∥ = Mε

M(t− c)2(t− c)2 = ε,

38


which implies that

(prz′)(t)− (py′)(t)−

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(py′)(u)r(u)du

)ds

= o(r) as r → 0 in L1(J,C).

Note that

limr→0

‖(py′)(t, 1/p+ r)− (py′)(t, 1/p)− T (t, 1/p)(r)‖C

‖r‖C

=‖(prz′)(t)− (py′)(t)− [D2(py

′)](t, 1/p)(r)‖C‖r‖C

= 0.


[D2(py′)](t, 1/p)(r) =

∫ t

c

φ2,2(t, s)q(s)

(∫ s

c

(py′)(x)r(x)dx

)ds

is indeed the derivative of (py′)(t, 1/p) with respect to 1/p.

(5) We will prove the next two formulas. Let

−(py′)′ + qy = λwy, y(c) = h, (py′)′(c) = k;

−(pz′)′ + qy = λ(w + r)z, z(c) = h, (pz′)′(c) = k.

Let x = z − y = y(t, w + r)− y(t, w). Then

−(px′)′ + (q − λw)x = λr(t)z, x(c) = 0, (px′)(c) = 0.


X =

[x

px′

],

the matrix F to be

F =

[0

λr(t)z

],


Φ =

[φ1,1 φ2,1

φ2,1 φ2,2

].

39



X(t) =

∫ t

u

Φ(t, s)F (s)ds =

∫ t

u

Φ(t, s)F (s)ds (3.23)

=

∫ t

u

[φ1,1(t, s, w) φ2,1(t, s, w)

φ2,1(t, s, w) φ2,2(t, s, w)

][0

λr(t)z(t, w)

]ds (3.24)

=

∫ t

u

[φ1,2λr(t)z(t, w)

φ2,2λr(t)z(t, w)

]ds (3.25)


x(t) = λ

∫ t

c

φ1,2(t, s, w)r(s)z(s, w)ds (3.26)

px′(t) = λ

∫ t

c

φ2,2(t, s, w)r(s)z(s, w)ds. (3.27)


z(t, w)− y(t, w) = λ

∫ t

c

φ1,2(t, s, w)r(s)[y(s, λ) + (z(s, w)− y(s, w))]ds

= λ

∫ t

c

φ1,2(t, s, w)r(s)y(s, w)ds

+ λ

∫ t

c

φ1,2(t, s, w)[z(s, w)− y(s, w)]r(s)ds.

so that

z(t, w)− y(t, w)− λ∫ t

c


= λ

∫ t

c

φ1,2(t, s, w)[z(s, w)− y(s, w)]r(s)ds.

Note that φ1,2 is bounded on K ×K, λ ∈ C, and z → y uniformly on K by the

continuous dependence of solutions in a compact interval. Suppose that |φ1,2| ≤M for

some positive real number M . Let ε > 0 be given and let

z(s, λ)− y(t, w) =ε

λM(t− c)

40


as r → 0 in C. Thus,

‖λ∫ tcφ1,2(t, s, w)[z(s, λ)− y(s, λ)]r(s)ds‖

‖r‖

≤‖‖r‖λ

∫ tcφ1,2(t, s, w)[z(s, λ)− y(s, λ)]ds‖

‖r‖

≤∥∥∥∥λ∫ t

c

φ1,2(t, s, w)[z(s, λ)− y(s, λ)]ds

∥∥∥∥≤ M

∥∥∥∥λ∫ t

c

[z(s, λ)− y(s, λ)]ds

∥∥∥∥ = M

∥∥∥∥λ∫ t

c

ε

λM(t− c)ds

∥∥∥∥= M λ

ε

λM(t− c)(t− c) = ε,

which implies that

z(t, w)− y(t, w)− λ∫ t

c


= λ

∫ t

c

φ1,2(t, s, w)[z(s, w)− y(s, w)]r(s)ds = o(r) as r → 0 in C.

Note that

limr→0

‖y(t, w + r)− y(t, w)− T (t, w)(r)‖C‖r‖C

=‖z(t, w)− y(t, w)− [D2(y)](t, w)(r)‖C

‖r‖C= 0.


[D2(y)](t, w) = λ

∫ t

c

φ1,2(t, s, w)y(s, w)r(s)ds

is indeed the derivative of y(t, w) with respect to w.


(pz′)(t, w)− (py′)(t, w) = λ

∫ t

c

φ2,2(t, s, w)r(s)[y(s, w) + (z(s, w)− y(s, w))]ds

= λ

∫ t

c


+ λ

∫ t

c

φ2,2(t, s, w)[z(s, w)− y(s, w)]r(s)ds.

41


so that

(pz′)(t, w)− (py′)(t, w)− λ∫ t

c


= λ

∫ t

c

φ2,2(t, s, w)[z(s, w)− y(s, w)]r(s)ds.

Note that φ2,2 is bounded on K ×K, λ ∈ C, and z → y uniformly on K by the

continuous dependence of solutions in a compact interval. Suppose that |φ2,2| ≤M for

some positive real number M . Let ε > 0 be given and let

z(s, λ)− y(t, w) =ε

λM(t− c)


‖λ∫ tcφ2,2(t, s, w)[z(s, λ)− y(s, λ)]r(s)ds‖

‖r‖

≤‖‖r‖λ

∫ tcφ2,2(t, s, w)[z(s, λ)− y(s, λ)]w(s)ds‖

‖r‖

≤∥∥∥∥λ∫ t

c

φ2,2(t, s, w)[z(s, λ)− y(s, λ)]ds

∥∥∥∥≤ M

∥∥∥∥λ∫ t

c


∥∥∥∥ = M

∥∥∥∥λ∫ t

c

ε

λM(t− c)ds

∥∥∥∥= M λ

ε

λM(t− c)(t− c) = ε,

which implies that

(pz′)(t, w)− (py′)(t, w)− λ∫ t

c


= λ

∫ t

c

φ2,2(t, s, w)[z(s, w)− y(s, w)]r(s)ds

= o(r) as r → 0 in C.

Note that

limr→0

‖(py′)(t, w + r)− (py′)(t, w)− T (t, w)(r)‖C‖r‖C

=‖z(t, w)− y(t, w)− [D2(py

′)](t, w)(r)‖C

‖r‖C= 0.

42



[D2(py′)](t, w) = λ

∫ t

c

φ2,2(t, s, w)y(s, w)r(s)ds

is indeed the derivative of (py′)(t, w) with respect to w.

(7) We will prove the last two formulas.

Let

−(py′)′ + qy = λwy, y(c) = h, (py′)′(c) = k;

−(pz′)′ + qy = (λ+ r)wz, z(c) = h, (pz′)′(c) = k.

Let

x = z − y = y(t, λ+ r)− y(t, λ).

Then

−(px′)′ + (q − λw)x = rwz, x(c) = 0, (px′)(c) = 0.


X =

[x

px′

],

the matrix F to be

F =

[0

rwz

],


Φ =

[φ1,1 φ2,1

φ2,1 φ2,2

].


X(t) =

∫ t

u

Φ(t, s)F (s)ds =

∫ t

u

Φ(t, s)F (s)ds (3.28)

=

∫ t

u

[φ1,1(t, s, λ) φ2,1(t, s, λ)

φ2,1(t, s, λ) φ2,2(t, s, λ)

][0

rw(t)z(t, λ)

]ds (3.29)

=

∫ t

u

[φ1,2rw(t)z(t, λ)

φ2,2rw(t)z(t, λ)

]ds (3.30)

43



x(t) =

∫ t

c

φ1,2(t, s, λ)rw(s)z(s, λ)ds (3.31)

px′(t) =

∫ t

c

φ2,2(t, s, λ)rw(s)z(s, λ)ds. (3.32)


z(t, λ)− y(t, λ) =

∫ t

c

φ1,2(t, s, λ)rw(s)[y(s, λ) + (z(s, λ)− y(s, λ))]ds

=

∫ t

c

φ1,2(t, s, λ)rw(s)y(s, λ)ds

+

∫ t

c

φ1,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds.

so that

z(t, λ)− y(t, λ)−∫ t

c

φ1,2(t, s, λ)rw(s)y(s, λ)ds =

∫ t

c


Note that φ1,2 is bounded on K × K, w ∈ L1loc(J,C), and z → y uniformly


|φ1,2w| ≤M for some positive real number M . Let ε > 0 be given and let

z(s, λ)− y(s, λ) =ε

M(t− c)


‖∫ tcφ1,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds‖

‖r‖

≤‖‖r‖

∫ tcφ1,2(t, s, λ)[z(s, λ)− y(s, λ)]w(s)ds‖

‖r‖

≤∥∥∥∥∫ t

c

φ1,2(t, s, λ)[z(s, λ)− y(s, λ)]w(s)ds

∥∥∥∥≤ M

∥∥∥∥∫ t

c


∥∥∥∥= M

∥∥∥∥∫ t

c

ε

M(t− c)ds

∥∥∥∥ = Mε

M(t− c)(t− c) = ε,

44


which implies that

z(t, λ)− y(t, λ)−∫ t

c


=

∫ t

c

φ1,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds = o(r) as r → 0 in C.

Note that

limr→0

‖y(t, λ+ r)− y(t, λ)− T (t, λ)(r)‖C

‖r‖C=‖z(t, λ)− y(t, λ)− [D2(y)](t, λ)(r)‖C

‖r‖C= 0.


[D2(y)](t, λ) =

∫ t

c

φ1,2(t, s, λ)y(s, λ)w(s)ds

is indeed the derivative of y(t, λ) with respect to λ.


(pz′)(t, λ)− (py′)(t, λ) =

∫ t

c

φ2,2(t, s, λ)rw(s)[y(s, λ) + (z(s, λ)− y(s, λ))]ds

=

∫ t

c


+

∫ t

c


so that

(pz′)(t, λ)− (py′)(t, λ)−∫ t

c


=

∫ t

c


Note that φ2,2 is bounded on K × K, w ∈ L1loc(J,C), and z → y uniformly


|φ2,2w| ≤M for some positive real number M . Let ε > 0 be given and let

z(s, λ)− y(s, λ) =ε

M(t− c)

45



‖∫ tcφ2,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds‖

‖r‖

≤‖‖r‖

∫ tcφ2,2(t, s, λ)[z(s, λ)− y(s, λ)]w(s)ds‖

‖r‖

≤∥∥∥∥∫ t

c

φ2,2(t, s, λ)[z(s, λ)− y(s, λ)]w(s)ds

∥∥∥∥≤ M

∥∥∥∥∫ t

c


∥∥∥∥= M

∥∥∥∥∫ t

c

ε

M(t− c)ds

∥∥∥∥= M

ε

M(t− c)(t− c)

= ε,

which implies that

(pz′)(t, λ)− (py′)(t, λ)−∫ t

c


=

∫ t

c

φ2,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds

= o(r) as r → 0 in C.

Note that

limr→0

‖(py′)(t, λ+ r)− (py′)(t, λ)− T (t, λ)(r)‖C

‖r‖C

=‖z(t, λ)− y(t, λ)− [D2(py

′)](t, λ)(r)‖C‖r‖C

= 0.


[D2(py′)](t, λ) =

∫ t

c

φ2,2(t, s, λ)y(s, λ)w(s)ds

is indeed the derivative of (py′)(t, λ) with respect to λ.

46


3.3 The Prufer Transformations

3.3.1 Definition

Recall that we have been focusing on second-order differential equations of the form

−(py′)′ + qy = λwy,

where

J = (a, b), −∞ ≤ a < b ≤ +∞, 1

p, q, w : L1

loc(J,C).

In this section, we will consider the substitution y = ρ sin θ and py′ = ρ cos θ, where

θ =

[θ1

θ2

].

We claim that the differential equation above is equivalent to

θ′j = rj cos2 θj + gj sin2 θj

on J = [a, b], where ∞ ≤ a < b ≤ ∞ with j = 1, 2. The initial conditions are

θj(cj) = αj, cj ∈ J, αj ∈ R, j = 1, 2.

Moreover, we have

rj = 1/pj, gj = λw − q ∈ L1(J,R).

Recall that ρ, θ, y, py′ are all functions of t. Let x = py′, and we have

ρ2 = x2 + y2.

Taking the derivative of both sides with respect to t, we have

2ρ · ρ′ = 2x · x′ + 2y · y′,

which implies that

ρ =x · x′ + y · y′

ρ′.

47


Now, let us consider x′ and y′.

x′ = ρ′ cos θ − ρθ′ sin θ =ρ′

ρx− yθ′ (3.33)

y′ = ρ′ sin θ + ρθ′ cos θ =ρ′

ρy + xθ′. (3.34)

Multiplying (3.33) by y and (3.34) by x and taking the difference, we have

xy′ − yx′ = (x2 + y2)θ′ = ρ2θ′.

Hence, we have

θ′ =xy′ − yx′

ρ2.

By our original equation and the substitution,

x = ρ cos θ, y′ =x

p,

y = ρ sin θ, x′ = qy − λwy = (q − λw)y.

Thus, we have

θ′ =xy′ − yx′

ρ2

=ρ cos θ x

p− ρ sin θ(q − λw)y

ρ2

=ρ cos θ ρ cos θ

p− ρ sin θ(q − λw)ρ sin θ

ρ2

= cos θcos θ

p− sin θ(q − λw) sin θ

=1

pcos2 θ + (λw − q) sin2 θ

= r cos2 θ + g sin2 θ,

where r = 1/p, g = λw − q.Now, we want to construct specific examples to show that the Prufer transfor-

mation is valid.

48


3.3.2 Examples


−(py′)′ + qy = λwy


J = (a, b) = (0, 1);1

p(t)= t−2 sin

(1

t

),where 0 < t < 1; q = 0; w = 0.

Then, we have

−(py′)′ = 0

As solved above, the possible solutions are

u(t) = 1, v(t) = cos

(1

t

).

Now, we want to check if the differential equation

θ′ = r cos2 θ + g sin2 θ

gives the same results. By Mathematica (codes in the appendix), it is easy to show that

the solutions above satisfies the differential equation after the Prufer transformation.

3.4 Oscillation

In this subsection, we will study the oscillatory properties of the equation

My = (py′)′ + qy = 0 on J = (a, b), −∞ ≤ a < b ≤ ∞, (3.35)

where the coefficients are assumed to satisfy

1/p, q ∈ L1loc(J,R), p > 0 a.e. on J. (3.36)

We can think of M as an operator on the function y.

49


3.4.1 Definitions

Definition 3.14. (Principal and Non-principal Solutions). Let u, v be real solutions

of (3.35). Then

• u is called a principal solution at a if

(1) u(t) 6= 0 for t ∈ (a, d] and some d ∈ J,

(2) every solution y of (3.35) which is not a multiple of u satisfies u(t)/y(t)→ 0

as t ∈ a+.

Note that y(t) 6= 0 for t in some right neighborhood of a by the Sturm

Separation Theorem.

• v is called a non-principal solution at a if

(1) v(t) 6= 0 for t ∈ (a, d] and some d ∈ J,

(2) v is not a principal solution at a.

Principal and non-principal solutions at b are defined similarly.

Definition 3.15. (Oscillation and Non-Oscillation). Let (3.35) and (3.36) hold. Then

endpoint a is called oscillatory (O) if there is a nontrivial solution y which has a

zero in the interval (a, c) for every c in J . Similarly b is oscillatory if there is nontrivial

solution which has a zero in the interval (c, b) for every c in J . By the Sturm Separation

Theorem, if one nontrivial solution has this property at an endpoint then all nontrivial

solutions have this property. Hence oscillation is a property of the equation (3.35) and

does not depend on any particular solution. An endpoint is non-oscillatory (NO) if ti

is not oscillatory.

3.4.2 Oscillation Criteria

In this section, we will discuss various criteria for equation (3.35) to be oscillatory or

non-oscillatory at a particular endpoint.

Theorem 3.16. The equation (3.35) is non-oscillatory at a if and only if there exists

a principal solution at a.

Proof. Since by definition a principal solution at a is non-oscillatory at a, we need only

show the converse. Let y1 and y2 be linearly independent real solutions of (3.35). Since

50


(3.35) is non-oscillatory, it follows from the Sturm Separation Theorem that there is

an α in I such that y1(t), y2(t) 6= 0 for a < t < α. Since y1, y2 are linearly independent

real solutions, it follows that

limt→a

y1(t)

y2(t)= L, where −∞ ≤ L ≤ ∞,

exists.

If L is finite, define u = y1 − Ly2, v = y2; otherwise, define u = y2, v = y1. In

any case, u and v are real linearly independent solutions of (3.35) such that

limt→a

u(t)

v(t)= 0.

Let y be any solution of (3.35) which is not a multiple of u. Then there exist constants

c, d such that d 6= 0 and y = cu+ dv. Therefore,

limt→a

∥∥∥∥y(t)

u(t)

∥∥∥∥ = limt→a

∥∥∥∥c+ dv(t)

u(t)

∥∥∥∥ =∞.

Hence, by definition, u(t) = o(y(t)) as t→ a. This shows that u is a principal solution

at a.

Theorem 3.17. Let (3.35), (3.36) hold.

• If ∫ b

c

1

p=∞ and

∫ b

c

q =∞

for some c ∈ J, then the equation (3.35) is oscillatory at b.

• If ∫ c

a

1

p=∞ and

∫ c

a

q =∞

for some c ∈ J, then the equation (3.35) is oscillatory at a.

Proof. We prove the result for the endpoint b only since the proof for a is similar.

Assume non-oscillation at b and let u, v be positive principal and non-principal solu-

tions, respectively, on [d, b) for some d ∈ J. Noting that q = −(pv′)′/v on [d, b) and

51


integrating by parts we get

−∫ t

d

q =

∫ t

d

(pv′)′

v=

(pv′)

v(t)− (pv′)

v(d)−

∫ t

d

pv′−v′

v2

=(pv′)

v(t)− (pv′)

v(d) +

∫ t

d

1

p

(pv′)2

v2.

Note that

−∫ t

d

q = −∞

as t → b. Hence,(pv′)(t)

v(t)→ −∞ as t → b and pv′ is negative near b. Therefore, we

know that v must be decreasing in a left neighborhood of b and hence has a limit at b,

say,

v(t)→ L as t→ b, , 0 ≤ L <∞.

From this it follows that, for some h ∈ [d, b), and some ε > 0, we have

v2(t) > L2 + ε,1

v2(t)>

1

L2 + ε, d ≤ h ≤ t < b,

which implies that ∫ t

h

1

pv2≥ 1

L2 + ε

∫ t

h

1

p, d ≤ h ≤ t < b.

This and the hypothesis imply that∫ b

h

1

pv2=∞,

contradicting the fact that ∫ b

h

1

pv2<∞

since v is a non-principal solution at b.

Theorem 3.18. Let J = (1,∞), p = 1 on J , q ∈ L1loc([1,∞),R) and assume that q

is piecewise continuous on J . Set q+(t) = max{q(t), 0}, q− = max{−q(t), 0}, t ∈ J.Assume that for each ε > 0 there is a δ > 0 such that for any subinterval I of J of

length δ

either

∫I

q+ < ε or

∫I

q− < ε.

52


If

−∞ ≤ lim inft→∞

∫ t

1

q < lim supt→∞

∫ t

1

q ≤ ∞,

then equation (3.35) is oscillatory at ∞.

Example 3.19. It follows readily from Theorem 3.18 that the equation

y′′ + qy = 0

is oscillatory at ∞ when q(t) = k sin(t) or q(t) = k cos(t) for any k ∈ R, k 6= 0. More

generally, for q(t) = tck sin(t) or q(t) = tck cos(t), with −1 < c ∈ R, k 6= 0.

20 40 60 80 100

-10

-5

5

10

Figure 6: Example 3.19: q(t) = t0.5 sin(t)

20 40 60 80 100

-400 000

-200 000

200 000

400 000

Figure 7: Example 3.19: q(t) = t3 cos(t).

53


3.5 Sturm Comparison Theorem

Theorem 3.20. Consider a singular endpoint of equation

(py′)′ + qy = 0 on J, 1/p, q ∈ L1loc(J,R), p > 0 a.e. on J = (a, b). (3.37)

Now, consider a comparison equation

(Pz′)′ +Qz = 0 on J. (3.38)

Assume 1/P, Q ∈ L1loc(J,R) and satisfy

Q ≥ q, 0 < P ≤ p on J. (3.39)

Suppose y is a non-trivial solution of (3.37) satisfying y(c) = 0 = y(d) for some

c, d ∈ J, c < d. Then every solution of (3.38) has a zero in the closed interval [c, d].

Proof. Since the zeros of y are isolated by Theorem (3.4), we may assume that c and

d are consecutive zeros of y. Replacing y by −y, if necessary, we may also assume that

y > 0 on (c, d). Suppose that z is a nontrivial solution of (3.38) which has no zero in

[c, d]. Then a direct, albeit tedious, computation yields the Picone identity

[y

z(py′z − Pz′y)]′ = (Q− q)y2 + (p− P )y′2 + P

(yz′ − zy′)2

z2. (3.40)

An integration yields∫ d

c

(Q− q)y2 +

∫ d

c

(p− P )y′2 = −∫ d

c

P(yz′ − zy′)2

z2. (3.41)

The hypothesis (3.39) implies that yz′− zy′ = 0 a.e. on [c, d]. Hence, f = y/z = 0, and

consequently, y = 0 on [c, d]. By the existence-uniqueness theorem, this implies that y

is the trivial solution on J and this contradiction completes the proof.

54

4 STURM OSCILLATION AND COMPARISON THEOREMS

4 Sturm Oscillation and Comparison Theorems

4.1 Basics

4.1.1 Construction of Difference Equations from a Jacobi Matrix

In this case, given the Jacobi matrix

J =

b1 a1 0 · · · · · · · · ·a1 b2 a2

0 a2 b3. . .

.... . . . . . . . .

.... . . . . . . . .

.... . . . . .

,

we consider the difference equation discussed in the very beginning, that is,

(hu)n = anun+1 + bnun + an−1un−1 = xu (4.1)

for n = 1, 2, · · · with u0 = 0.

Given a sequence of parameters a1, a2, · · · and b1, b2, · · · for the difference equa-

tion (4.1), we look at the fundamental solution, un(x), defined recursively by u1(x) = 1

and

anun+1(x) + (bn − x)un(x) + an−1un−1(x) = 0 (4.2)

with u0 ≡ 0. Hence,

un+1(x) = a−1n (x− bn)un(x)− a−1

n an−1un−1(x) = 0 (4.3)

Clearly, (4.3) implies, by induction, that un+1 is a polynomial of degree n with leading

term (an · · · a1)−1xn. Thus, we define for n = 0, 1, 2, · · ·

pn(x) = un+1(x), Pn(x) = (a1 · · · an)pn(x). (4.4)

Then (4.2) becomes

an+1pn+1(x) + (bn+1 − x)pn(x) + anpn−1(x) = 0 (4.5)

55


for n = 0, 1, 2, · · · . One also sees that

xPn(x) = Pn+1(x) + bn+1Pn(x) + a2nPn−1(x), (4.6)

where pn are orthonormal polynomials for a suitable measure on R and the Pn are what

are known as monic orthogonal polynomials.

Now, let Jn be the finite n× n matrix defined by

Jn =

b1 a1 0 · · · · · · 0

a1 b2 a2...

0 a2 b3. . .

......

. . . . . . . . . 0...

. . . bn−1 an−1

0 · · · · · · 0 an−1 bn

.

Proposition 4.1. The eigenvalues of Jn are precisely the zeros of Pn(x). We have

Pn(x) = det(xI − Jn). (4.7)

Proof. Let ϕ(x) be the vector where ϕj(x) = pj−1(x), j = 1, · · · , n. Then (4.2) implies

that

(Jn − x)ϕ(x) = −anpn(x)δn, (4.8)

where δn is the vector (0, 0, · · · , 0, 1). Thus every zero of pn is an eigenvalue of Jn.

Conversely, if ϕ is an eigenvector of Jn, then both ϕj and ϕj solve (4.3), so ϕj =

ϕ1ϕj(x). This implies that x is an eigenvalue only if pn(x) is zero and that eigenvalues

are simple.

Since Jn is real symmetric and eigenvalues are simple, pn(x) has n distinct

eigenvalues x(n)j , j = 1, · · · , n with x

(n)j−1 < x

(n)j . Thus, since pn and Pn have the same

zeros, we have

Pn(x) =n∏i=1

(x− xnj ) = det(x− Jn),

which completes the proof.

56

5 CONNECTION BETWEEN THE TWO APPROACHES

As proved above, from the Jacobi matrices representation of a polynomial Pn(x),

we have

Pn(x) = det(x− Jn). (4.9)

The determinant relation (4.9) has an important consequence concerning eigen-

value counting for Jn with carryover to the infinite J. As already noted, the eigenvalues

of Jn are exactly the zeros {x(n)j (dµ)}n−1

j=0 of Pn(x). Thus, the eigenvalues of Jn−1 and

Jn interlace since that is true of the zeros (this also follows from the fact that Jn−1 is

the restriction of Jn to a codimension 1 subspace.)

4.2 Sturm Oscillation and Comparison Theorems

Theorem 4.2. (Sturm Oscillation Theorem). If P`(x0) 6= 0, ` = 0, 1, · · · , n, then the

number of eigenvalues of Jn;F above x0 is

#{j|0 ≤ j ≤ n− 1 so that sgn(Pj+1(x0)) 6= sgn(Pj(x0))}.

Proof. See Simon (2005), page 4–5.

Theorem 4.3. (Sturm Comparison Theorem) If u and v are linearly independent

solutions of

x0un = an+1un+1 + bn+1un + anun−1,

v has a sign change between every pair of sign changes of u.

Proof. The proof immediately follows from the proof for the Sturm Comparison The-

orem using the differential equations approach.

5 Connection between the Two Approaches

5.1 Construction of the Jacobi Matrix from Differential Equa-

tions

In this section, we will again look at the initial value problems consisting of the second

order scalar equation

− (py′)′ + qy = f on J (5.1)

57



y(c) = 0, y′(c) = 1/2, c ∈ J, h, k ∈ C (5.2)

where

J = (a, b), −∞ ≤ a < b ≤ +∞, 1

p, q, f : J → C. (5.3)

In order to verify that the Sturm-Liouville problem defined using the differential

equation is essentially the same as its discrete analog, that is, the difference equations,

we need to construct an appropriate Jacobi matrix from the differential equation above,

and the difference equations will immediately follow from the Jacobi matrix.

Let (tn)∞n=0 be a sequence of complex numbers. Let (pn) be a sequence of points

on the function p defined by pn = p(tn+1) for all n ∈ {0, 1, · · · }, (qn) be a sequence of

points on the function q defined by qn = q(tn+1) for all n ∈ {0, 1, · · · }, (fn) be a sequence

of points on the function f defined by fn = f(tn+1) for all n ∈ {0, 1, · · · }. Finally, define

a sequence of points yn on the solution y by yn = y(tn+1). With sufficiently large n,

define

yn+1/2 =1

2(yn+1 + yn).

Moreover, define the first derivatives as

p′ = pn+1/2 − pn−1/2

y′ = yn+1/2 − yn−1/2

and the second derivatives as

p′′ = p′n+1/2 − p′n−1/2 = pn+1 − 2pn + pn−1

y′′ = y′n+1/2 − y′n−1/2 = yn+1 − 2yn + yn−1.

Now, we are ready to state a theorem in the special case when p = 1 using the

definition above:

Theorem 5.1. For the Schrodinger equation of the form

−y′′ + qy = f,

58


where q, f ∈ L1(J,C), using the definitions above, we can write its corresponding

difference equation as

−yn+1 + (2 + qn)yn − yn−1 = fn.

Moreover, its corresponding Jacobi matrix is given by

Jn =

2 + q1 −1 0 · · · · · · 0

−1 2 + q2 −1...

0 −1 2 + q3. . .

......

. . . . . . . . . 0...

. . . 2 + qn−1 −1

0 · · · · · · 0 −1 2 + qn

.

Proof. The proof immediately follows from the definition.

The general case is more complicated. Note that by the product rule,

(py′)′ = py′′ + p′y′.

Hence, we have

−(py′)′ + qy = f

⇐⇒ −py′′ − p′y′ + qy = f

⇐⇒ −pn(yn+1 − 2yn + yn−1)− (pn+1/2 − pn−1/2)(yn+1/2 − yn−1/2) + qnyn = fn

⇐⇒ −pnyn+1 + (2pn + qn)yn − pnyn−1 − pn+1/2yn+1/2 + pn−1/2yn+1/2

+pn+1/2yn−1/2 − pn−1/2yn−1/2 = fn

⇐⇒ −((1/2)pn+1/2 + pn − (1/2)pn−1/2)yn+1 + (2pn + qn)yn

−((1/2)pn+1/2 − pn − (1/2)pn−1/2)yn−1 = fn.

In this case, there is no obvious sequence {an} such that

an+1 = −((1/2)pn+1/2 + pn − (1/2)pn−1/2)

and

an = −((1/2)pn+1/2 − pn − (1/2)pn−1/2).

59


If such sequence does exist, then letting bn = 2pn + qn, and we will have our corre-

sponding Jacobi matrix

Jn =

b1 a1 0 · · · · · · 0

a1 b2 a2...

0 a2 b3. . .

......

. . . . . . . . ....

.... . . bn−1 an−1

0 · · · · · · · · · an−1 bn

.

Now, let c = t1. Note that the initial conditions in this case becomes

y(c) = y(t1) = y0 = 0

and

y′(c) = y′(t1) = y3/2 − y1/2 =y2 + y1

2− y1 + y0

2=y2 − y0

2≈ y1 − y0

2=y1

2=

1

2

so that y1 = 1.

5.2 Examples

Example 5.2. First, we will give an example of Theorem 5.1.

Let us consider the differential equation

y′′(x) + y(x) = sin(x)

with boundary conditions y(0) = 0 and y′(0) = 1. By the Theorem above, we can

write its corresponding difference equation as

yn+2 − yn+1 + yn = sin(n)

with initial conditions y0 = 0 and y1 = 1.

Here, choosing tn = n, we have the following graphs of the solution to the dif-

ference equation (the one with smaller amplitude) and that to the differential equation

(the one with larger amplitude).

60


10 20 30 40 50 60 70

-30

-20

-10

10

20

30


Zeros of the solution to the differential equation:

{0, 4.0782, 7.4722, 10.7228, 13.925, 17.1051, 20.2734, 23.4346, 26.5912, 29.7446,

32.8958, 36.0453, 39.1935, 42.3408, 45.4872, 48.6331, 51.7784, 54.9233, 58.0678, 61.2121,

64.3561, 67.4998}.

Zeros of the solution to the difference equation:

{0, 3, 5.028, 8.4039, 11.5854, 14.7121, 17.8162, 20, 9089, 23.9948, 27.0712,

30.1465, 33.2213, 36.2959, 39.3703, 42.4445, 45.5185, 48.5922, 51.6657, 54.7389, 57.8116,

60.884, 63.9559, 67.0272}.

Example 5.3. Now, we will give an example connecting the zeros of the solutions

to the differential/difference equation and the eignvalues of the corresponding Jacobi

matrix.

Let us consider the equation

−(py′)′ + qy = f


J = (a, b) = (0, 1); k > 0; p(t) = sin(t),where 0 < t < 1; q = 0; f = 0.

61


In this case, we have

−(py′)′ = 0.

Let pn = sin(n/6), where n ∈ [1, 5]. Then,

b1 = −2p1 ≈ −2 · 0.1659 = −0.3318

b2 = −2p2 ≈ −2 · 0.7401 = −1.4802

b3 = −2p3 ≈ −2 · 0.4794 = −0.9588

b4 = −2p4 ≈ −2 · 0.6184 = −1.2368

b5 = −2p5 ≈ −2 · 0.7401 = −1.4802

a1 = −((1/2)p3/2 + p1 − (1/2)p1/2) ≈ −1.1005

a2 = −((1/2)p5/2 + p2 − (1/2)p3/2) ≈ −0.7098

a3 = −((1/2)p7/2 + p3 − (1/2)p5/2) ≈ 0.3335

a4 = −((1/2)p9/2 + p4 − (1/2)p7/2) ≈ 1.0702

and

J5 =

b1 a1 0 0 0

a1 b2 a2 0 0

0 a2 b3 a3 0

0 0 a3 b4 a4

0 0 0 a4 b5

.

In this case, the eigenvalues of Jn are

−2.1799,−1.3264,−0.7394, −0.3344,−0.08197.

Now, we can calculate the corresponding P5(x). Using Mathematica and the

recurrence relation derived above, we have

P5(x) ≈ (2.1799 + x)(1.3264 + x)(0.7394 + x)(0.3344 + x)(0.0820 + x).

Solving for P5(x) = 0, we have the following solutions:

−2.1799,−1.3264,−0.7394,−0.3344,−0.0820.

Note that these solutions are very close to the eigenvalues for J , which verifies

62


our construction above. They are slightly different due to the rounding errors in the

calculations, especially the calculation of P5(x) using the recurrence relations, during

which small rounding errors are multiplied and thus become more significant.

Example 5.4. Below is a more complicated example connecting the zeros of the so-

lutions to the differential/difference equation and the eignvalues of the corresponding

Jacobi matrix. This example was also discussed in the previous section.

Let us consider the equation

−(py′)′ + qy = f


J = (a, b) = (0, 1); k > 0; p(t) = t2/ sin(1/t),where 0 < t < 1, ; q = 0; f = 0.

Let pn = (n/6)2/ sin(6/n), where n ∈ [1, 5]. Then,

b1 = −2p1 ≈ 0.1988

b2 = −2p2 ≈ −1.5747

b3 = −2p3 ≈ −0.5499

b4 = −2p4 ≈ −0.8911

b5 = −2p5 ≈ −1.4902

a1 = −((1/2)p3/2 + p1 − (1/2)p1/2) ≈ 0.1342

a2 = −((1/2)p5/2 + p2 − (1/2)p3/2) ≈ −0.9572

a3 = −((1/2)p7/2 + p3 − (1/2)p5/2) ≈ −0.3183

a4 = −((1/2)p9/2 + p4 − (1/2)p7/2) ≈ −0.5630

and

J5 =

b1 a1 0 0 0

a1 b2 a2 0 0

0 a2 b3 a3 0

0 0 a3 b4 a4

0 0 0 a4 b5

.

63


In this case, the eigenvalues of Jn are

−2.1866,−1.8184,−0.6245, 0.0849, 0.2375.

Now, we can calculate the corresponding P5(x). Using Mathematica and the

recurrence relation derived above, we have

P5(x) ≈ (−0.2375 + x)(−0.0849 + x)(0.6245 + x)(1.8184 + x)(2.1865 + x).

Solving for P5(x) = 0, we have the following solutions:

−2.1866,−1.8184,−0.6245, 0.0849, 0.2375.

Note that these solutions are exactly the same as the eigenvalues for J , which

verifies our construction above.

64

6 References

[1] A. Bielecki, Une remarque sur la methode de Banach-Cacciopoli-Tikhonov dans

la theorie des equations differentielles ordinaires, Bull. Acad. Polon. Sci. Cl. III

4(1956), 261-264.

[2] S. Elaydi, An Introduction to Difference Equations, Second Edition, Springer-

Verlag, New York, 1999.

[3] H.D. Niessen, A. Zettl, Singular Sturm-Liouville problems: the Friedrichs

extension and comparison of eigenvalues, Proc. London Math. Soc. 64 (1992) 545-

578.

[4] Hermann Schulz-Baldes, Sturm intersection theory for periodic Jacobi matrices

and linear Hamiltonian systems, unpublished.

[5] B. Simon, Orthogonal Polynomials on the Unit Circle, Part 1: Classical Theory,

AMS Colloquium Series, American Mathematical Society, Providence, RI, 2005.

[6] B. Simon, Sturm oscillation and comparison theorems, Sturm-Liouville Theory:

Past and Present (W. Amrein, A. Hinz, and D. Pearson, eds.), pp. 29-43,

Birkhauuser, Basel, 2005.

[7] G. Szego. Orthogonal Polynomials, 4th ed. Providence, RI: Amer. Math. Soc., pp.

44-47 and 54-55, 1975.

[8] G. Teschl, Jacobi Operators and Completely Integrable Nonlinear Lattices,

Mathematical Surveys and Monographs 72, Amer. Math. Soc., Providence, RI,

2000.

[9] G. Teschl, Topics in Real and Functional Analysis, unpublished, available online

at http://www.mat.univie.ac.at/~gerald/.

[10] W. Walter, Ordinary differential equations, Graduate Texts in Mathematics,

Springer, New York, 1998.

[11] A. Zettl, Sturm-Liouville Theory, Amer. Math. Soc., Mathematical Surveys and

Monographs 121, 2005.

65

http://www.mat.univie.ac.at/~gerald/


Appendix

Calculations for the Prufer Transformation Example

In[1]:= Clear!y"In[2]:= p :! 1#$t^$"2%#Sin!1#t"%In[3]:= q :! 0

In[36]:= $ :! 0

In[5]:= w :! 0

In[6]:= DSolve!D!"$p#y'!t"%, t" % q#y!t" & w #y!t", y!t", t"Out[6]= !!y"t# ! C"2# " C"1# Cos$1

t%&&

In[7]:= r :! 1#pIn[37]:= g :! $# w " q

In[12]:= v :! Cos&1t'

In[14]:= v1 :! FullSimplify!p#D!v, t""In[15]:= v1

Out[15]= 1

In[38]:= Solve!Tan!'" & v, '"Out[38]= !!# ! ArcTan$Cos$1

t%%&&

In[40]:= ' :! ArcTan&Cos&1t''

In[42]:= D!', t"Out[42]=

Sin' 1t(

t2 )1 " Cos' 1t(2*

r#$Cos!'"%^2Out[19]=

Sin' 1t(

t2 )1 " Cos' 1t(2*

In[43]:= Clear!'"In[44]:= DSolve!y'!t" & r#$Cos!y!t""%^2 % g#$Sin!y!t""%^2, y!t", t"

Power::infy! : !Infinite expression1

0encountered. "

Power::infy! : !Infinite expression1

0encountered. "

#::indet! : !Indeterminate expression ComplexInfinity$ComplexInfinity encountered. "

Out[44]= ++,,

66

Calculations for the Difference/Differential Equations

In[522]:= f@n_D := Sin@nê6D;b1 = N@H-2L*f@1DD;b2 = N@H-2L*f@2DD;b3 = N@H-2L*f@3DD;b4 = N@H-2L*f@4DD;b5 = N@H-2L*f@5DD;a1 = N@-H1ê2*f@3ê2D + f@1D - 1ê2*f@1ê2DLD;a2 = N@-H1ê2*f@5ê2D + f@2D - 1ê2*f@3ê2DLD;a3 = N@-H1ê2*f@7ê2D + f@3D - 1ê2*f@5ê2DLD;a4 = N@-H1ê2*f@9ê2D + f@4D - 1ê2*f@7ê2DLD;a5 = N@-H1ê2*f@11ê2D + f@5D - 1ê2*f@9ê2DLD;

In[533]:= u0 = 0;u1 = 1;u2 = a1^H-1L*Hx - b1L*u1;u3 = a2^H-1L*Hx - b2L*u2 - a2^H-1L*a1*u1;u4 = a3^H-1L*Hx - b3L*u3 - a3^H-1L*a2*u2;u5 = a4^H-1L*Hx - b4L*u4 - a4^H-1L*a3*u3;u6 = a5^H-1L*Hx - b5L*u5 - a5^H-1L*a4*u4;

In[550]:= p0 = u1;P0 = p0;p1 = u2;P1 = a1*p1;P2 = x*P1 - b2*P1 - a1^2*P0;P3 = x*P2 - b3*P2 - a2^2*P1;P4 = x*P3 - b4*P3 - a3^2*P2;P5 = x*P4 - b5*P4 - a4^2*P3;

In[558]:= Simplify@P5D

0.9999999999999999` H0.08196965357438073` + xLH0.33441959901099777` + xL H0.7394167814645523` + xLH1.3264061277550918` + xL H2.179913886914066` + xL

In[559]:= Solve@P5 ã 0, xD

88x Ø -2.1799138869140684`<,8x Ø -1.32640612775509`<, 8x Ø -0.7394167814645526`<,8x Ø -0.3344195990109978`<, 8x Ø -0.08196965357438073`<<

In[549]:= Eigenvalues@88b1, a1, 0, 0, 0<, 8a1, b2, a2, 0, 0<, 80, a2, b3, a3, 0<, 80, 0, a3, b4, a4<, 80, 0, 0, a4, b5<<D

8-2.1799138869140666`, -1.32640612775509`,-0.739416781464554`, -0.3344195990109972`, -0.08196965357438074`<

67


Calculations for the Difference/Differential Equations

In[380]:= f@n_D := Hnê6L^2êSin@6ênD;b1 = N@H-2L*f@1DD;b2 = N@H-2L*f@2DD;b3 = N@H-2L*f@3DD;b4 = N@H-2L*f@4DD;b5 = N@H-2L*f@5DD;a1 = N@-H1ê2*f@3ê2D + f@1D - 1ê2*f@1ê2DLD;a2 = N@-H1ê2*f@5ê2D + f@2D - 1ê2*f@3ê2DLD;a3 = N@-H1ê2*f@7ê2D + f@3D - 1ê2*f@5ê2DLD;a4 = N@-H1ê2*f@9ê2D + f@4D - 1ê2*f@7ê2DLD;a5 = N@-H1ê2*f@11ê2D + f@5D - 1ê2*f@9ê2DLD;

In[391]:= u0 = 0;u1 = 1;u2 = a1^H-1L*Hx - b1L*u1;u3 = a2^H-1L*Hx - b2L*u2 - a2^H-1L*a1*u1;u4 = a3^H-1L*Hx - b3L*u3 - a3^H-1L*a2*u2;u5 = a4^H-1L*Hx - b4L*u4 - a4^H-1L*a3*u3;u6 = a5^H-1L*Hx - b5L*u5 - a5^H-1L*a4*u4;

In[398]:= p0 = u1;P0 = p0;P1 = a1*p1;P2 = x*P1 - b2*P1 - a1^2*P0;P3 = x*P2 - b3*P2 - a2^2*P1;P4 = x*P3 - b4*P3 - a3^2*P2;P5 = x*P4 - b5*P4 - a4^2*P3;

In[405]:= Simplify@P5D

1.` H-0.2375459596746414` + xL H-0.08490891578562153` + xLH0.6245388095387586` + xL H1.818400552959919` + xL H2.186549506241636` + xL

In[415]:= Solve@P5 ã 0, xD

88x Ø -2.1865495062416342`<,8x Ø -1.8184005529599214`<, 8x Ø -0.6245388095387583`<,8x Ø 0.08490891578562156`<, 8x Ø 0.2375459596746413`<<

In[419]:= Eigenvalues@88b1, a1, 0, 0, 0<, 8a1, b2, a2, 0, 0<, 80, a2, b3, a3, 0<, 80, 0, a3, b4, a4<, 80, 0, 0, a4, b5<<D

8-2.1865495062416325`, -1.818400552959922`,-0.624538809538759`, 0.23754595967464148`, 0.08490891578562179`<

68

sturm-liouville oscillation theory for di ... - williams...

Documents