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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
ZHAONING (NANCY) WANG
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
ZHAONING (NANCY) WANG
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
ZHAONING (NANCY) WANG
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
2 FIRST ORDER SYSTEMS
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
ZHAONING (NANCY) WANG
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
2 FIRST ORDER SYSTEMS
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
ZHAONING (NANCY) WANG
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
2 FIRST ORDER SYSTEMS
(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
ZHAONING (NANCY) WANG
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
2 FIRST ORDER SYSTEMS
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
ZHAONING (NANCY) WANG
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
2 FIRST ORDER SYSTEMS
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
ZHAONING (NANCY) WANG
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)
together with the initial conditions
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
ZHAONING (NANCY) WANG
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
3 SECOND ORDER SCALAR EQUATIONS
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
ZHAONING (NANCY) WANG
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
3 SECOND ORDER SCALAR EQUATIONS
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
ZHAONING (NANCY) WANG
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.
Example 3.7. Let us consider the equation
−(py′)′ + qy = λwy
on J = (a, b), where
J = (a, b) = (k, 1); k > 0;1
p(t)= t−2 sin
(1
t
),where k < t < 1; q = 0; w = 0.
24
3 SECOND ORDER SCALAR EQUATIONS
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).
Example 3.8. Let us consider the equation
−(py′)′ + qy = λwy
25
ZHAONING (NANCY) WANG
0.2 0.4 0.6 0.8 1.0
-1.0
-0.5
0.5
1.0
Figure 2: Example 3.7
on J = (a, b), where
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
3 SECOND ORDER SCALAR EQUATIONS
0.2 0.4 0.6 0.8 1.0
-0.4
-0.2
0.2
0.4
Figure 3: Example 3.8
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.
Example 3.9. Let us consider the equation
−(py′)′ + qy = λwy
27
ZHAONING (NANCY) WANG
on J = (a, b), where
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 <∞.
Example 3.10. Let us consider the equation
−(py′)′ + qy = λwy
on J = (a, b), where
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
3 SECOND ORDER SCALAR EQUATIONS
0.2 0.4 0.6 0.8 1.0
0.5
1.0
1.5
2.0
2.5
Figure 4: Example 3.9
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
ZHAONING (NANCY) WANG
0.2 0.4 0.6 0.8 1.0
-10
-8
-6
-4
-2
Figure 5: Example 3.10
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
3 SECOND ORDER SCALAR EQUATIONS
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
ZHAONING (NANCY) WANG
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
3 SECOND ORDER SCALAR EQUATIONS
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
ZHAONING (NANCY) WANG
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.
From the definition of Frechet derivative, we know that
[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 SECOND ORDER SCALAR EQUATIONS
(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.
Let us define the matrix X to be
X =
[x
px′
],
the matrix F to be
F =
[0
f
],
and the matrix Φ to be
Φ =
[φ1,1 φ2,1
φ2,1 φ2,2
].
From the variation of parameters formula it follows that
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
ZHAONING (NANCY) WANG
Hence, from equation (3.20), we know that
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
[prz′ − py′](u)r(u)du
)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
3 SECOND ORDER SCALAR EQUATIONS
|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
as r → 0 in L1(J,C). Thus,
‖∫ tcφ1,2(t, s)q(s)
(∫ sc
[prz′ − py′](u)r(u)du
)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
[prz′ − py′](u)du
)ds+
∫ t
c
[prz′ − py′]
∥∥∥∥=
∥∥∥∥∫ t
c
(φ1,2(t, s)q(s) + 1)
(∫ s
c
[prz′ − py′](u)du
)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.
From the definition of Frechet derivative, we know that
[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
ZHAONING (NANCY) WANG
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
[prz′ − py′](u)r(u)du
)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
[prz′ − py′](u)r(u)du
)ds
Note that φ2,2 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
|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
as r → 0 in L1(J,C). Thus,
‖∫ tcφ2,2(t, s)q(s)
(∫ sc
[prz′ − py′](u)r(u)du
)ds‖
‖r‖
≤‖r‖‖
∫ tcφ2,2(t, s)q(s)
(∫ sc
[prz′ − py′](u)du
)ds‖
‖r‖
≤∥∥∥∥∫ t
c
φ2,2(t, s)q(s)
(∫ s
c
[prz′ − py′](u)du
)ds
∥∥∥∥= M
∥∥∥∥∫ t
c
∫ t
c
ε
M(t− c)2du ds
∥∥∥∥ = Mε
M(t− c)2(t− c)2 = ε,
38
3 SECOND ORDER SCALAR EQUATIONS
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.
From the definition of Frechet derivative, we know that
[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.
Let us define the matrix X to be
X =
[x
px′
],
the matrix F to be
F =
[0
λr(t)z
],
and the matrix Φ to be
Φ =
[φ1,1 φ2,1
φ2,1 φ2,2
].
39
ZHAONING (NANCY) WANG
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 (3.23)
=
∫ 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)
Hence, from equation (3.25), we know that
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)
For equation (3.26), letting z = y + (z − y) we get
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
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.
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
3 SECOND ORDER SCALAR EQUATIONS
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
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 = 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.
From the definition of Frechet derivative, we know that
[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.
(6) For equation (3.27), letting z = y + (z − y) we get
(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
φ2,2(t, s, w)r(s)y(s, w)ds
+ λ
∫ t
c
φ2,2(t, s, w)[z(s, w)− y(s, w)]r(s)ds.
41
ZHAONING (NANCY) WANG
so that
(pz′)(t, w)− (py′)(t, w)− λ∫ t
c
φ2,2(t, s, w)r(s)y(s, w)ds
= 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)
as r → 0 in C. Thus,
‖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
[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
(pz′)(t, w)− (py′)(t, w)− λ∫ t
c
φ2,2(t, s, w)r(s)y(s, w)ds
= 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
3 SECOND ORDER SCALAR EQUATIONS
From the definition of Frechet derivative, we know that
[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.
Let us define the matrix X to be
X =
[x
px′
],
the matrix F to be
F =
[0
rwz
],
and the matrix Φ to be
Φ =
[φ1,1 φ2,1
φ2,1 φ2,2
].
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 (3.28)
=
∫ 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
ZHAONING (NANCY) WANG
Hence, from equation (3.30), we know that
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)
For equation (3.31), letting z = y + (z − y) we get
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
φ1,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds.
Note that φ1,2 is bounded on K × K, w ∈ L1loc(J,C), and z → y uniformly
on K by the continuous dependence of solutions in a compact interval. Suppose that
|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)
as r → 0 in C. Thus,
‖∫ 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
[z(s, λ)− y(s, λ)]ds
∥∥∥∥= M
∥∥∥∥∫ t
c
ε
M(t− c)ds
∥∥∥∥ = Mε
M(t− c)(t− c) = ε,
44
3 SECOND ORDER SCALAR EQUATIONS
which implies that
z(t, λ)− y(t, λ)−∫ 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 = 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.
From the definition of Frechet derivative, we know that
[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.
(8) For equation (3.32), letting z = y + (z − y) we get
(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
φ2,2(t, s, λ)rw(s)y(s, λ)ds
+
∫ t
c
φ2,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds.
so that
(pz′)(t, λ)− (py′)(t, λ)−∫ t
c
φ2,2(t, s, λ)rw(s)y(s, λ)ds
=
∫ t
c
φ2,2(t, s, λ)[z(s, λ)− y(s, λ)]rw(s)ds.
Note that φ2,2 is bounded on K × K, w ∈ L1loc(J,C), and z → y uniformly
on K by the continuous dependence of solutions in a compact interval. Suppose that
|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
ZHAONING (NANCY) WANG
as r → 0 in C. Thus,
‖∫ 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
[z(s, λ)− y(s, λ)]ds
∥∥∥∥= 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
φ2,2(t, s, λ)rw(s)y(s, λ)ds
=
∫ 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.
From the definition of Frechet derivative, we know that
[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 SECOND ORDER SCALAR EQUATIONS
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
ZHAONING (NANCY) WANG
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 SECOND ORDER SCALAR EQUATIONS
3.3.2 Examples
Example 3.13. Let us consider the equation
−(py′)′ + qy = λwy
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
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
ZHAONING (NANCY) WANG
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 SECOND ORDER SCALAR EQUATIONS
(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
ZHAONING (NANCY) WANG
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
3 SECOND ORDER SCALAR EQUATIONS
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
ZHAONING (NANCY) WANG
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
ZHAONING (NANCY) WANG
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
ZHAONING (NANCY) WANG
together with the initial conditions
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
5 CONNECTION BETWEEN THE TWO APPROACHES
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
ZHAONING (NANCY) WANG
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
5 CONNECTION BETWEEN THE TWO APPROACHES
10 20 30 40 50 60 70
-30
-20
-10
10
20
30
Figure 8: Example 5.2
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
on J = (a, b), where
J = (a, b) = (0, 1); k > 0; p(t) = sin(t),where 0 < t < 1; q = 0; f = 0.
61
ZHAONING (NANCY) WANG
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
5 CONNECTION BETWEEN THE TWO APPROACHES
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
on J = (a, b), where
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
ZHAONING (NANCY) WANG
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
ZHAONING (NANCY) WANG
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
ZHAONING (NANCY) WANG
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