University of Iowa
Iowa Research Online
Theses and Dissertations
Spring 2011
Bifurcation theory for a class of second order
differential equations
Alvaro Correa
University of Iowa
Copyright 2011 Alvaro Ramon Correa
This dissertation is available at Iowa Research Online: http://ir.uiowa.edu/etd/940
Recommended Citation
Correa, Alvaro. "Bifurcation theory for a class of second order differential equations." PhD (Doctor of Philosophy) thesis, University of
Iowa, 2011.
http://ir.uiowa.edu/etd/940.
Follow this and additional works at: http://ir.uiowa.edu/etd
Part of the Mathematics Commons
BIFURCATION THEORY FOR A CLASS OF SECOND ORDER
DIFFERENTIAL EQUATIONS
by
Alvaro Correa
An Abstract
Of a thesis submitted in partial fulfillment of the
requirements for the Doctor of Philosophy
degree in Mathematics
in the Graduate College of
The University of Iowa
May 2011
Thesis Supervisor: Professor Yi Li
1
ABSTRACT
We consider the existence of positive solutions of the nonlinear two point
boundary value problem u00 + λf (u) = 0, u(−1) = u(1) = 0, where f (u) = u(u −
a)(u − b)(u − c)(1 − u), 0 < a < b < c < 1, as the parameter λ varies through positive
values. Every solution u(x) is an even function, and when it exists, it is uniquely
identified by α = u(0). We study how the number of solutions changes when the
parameter varies, i.e. we will be focusing on the locations of bifurcation points.
The authors P. Korman, Y. Li and T. Ouyang ( ”Computing the location and
the direction of bifurcation”, Mathematical Research Letters, accepted ), prove that a
necessary and sufficient condition for α to be a bifurcation point is
1/2
Z
α
G(α) ≡ F (α)
0
where F (α) =
Rα
0
f (α) − f (τ )
dτ − 2 = 0,
[F (α) − F (τ )]3/2
f (u) du. We will prove that G(α) has vertical asymptotes at α = b,
α = 1 and at any point α ∈ (0, 1) for which
Rα
0
f (u) du = 0. We will use the
asymptotic behavior of G to estimate intervals where G(α) 6= 0, that is, intervals
where there is no bifurcation point.
2
Abstract Approved:
Thesis Supervisor
Title and Department
Date
BIFURCATION THEORY FOR A CLASS OF SECOND ORDER
DIFFERENTIAL EQUATIONS
by
Alvaro Correa
A thesis submitted in partial fulfillment of the
requirements for the Doctor of Philosophy
degree in Mathematics
in the Graduate College of
The University of Iowa
May 2011
Thesis Supervisor: Professor Yi Li
Graduate College
The University of Iowa
Iowa City, Iowa
CERTIFICATE OF APPROVAL
PH.D. THESIS
This is to certify that the Ph.D. thesis of
Alvaro Correa
has been approved by the Examining Committee for the
thesis requirement for the Doctor of Philosophy degree
in Mathematics at the May 2011 graduation.
Thesis Committee:
Yi Li, Thesis Supervisor
Juan Gatica
Weimin Han
Laurent Jay
Tong Li
ACKNOWLEDGEMENTS
I would like to thank my family for their love and support. My parents have
always encouraged me to continue my studies and I would like to dedicate this dissertation to them.
I want to thank, from the bottom of my heart, my supervisor, Yi Li, for
his continuous support during my work on my Ph.D thesis, as well as his patience,
motivation, enthusiasm and accessibility. His guidance and exemplary professionalism
helped me during my studies at The University of Iowa. I would also like to say a
special thank you to my thesis committee: Juan Gatica, Weimin Han, Tong Li,
Laurent Jay
ii
ABSTRACT
We consider the existence of positive solutions of the nonlinear two point
boundary value problem u00 + λf (u) = 0, u(−1) = u(1) = 0, where f (u) = u(u −
a)(u − b)(u − c)(1 − u), 0 < a < b < c < 1, as the parameter λ varies through positive
values. Every solution u(x) is an even function, and when it exists, it is uniquely
identified by α = u(0). We study how the number of solutions changes when the
parameter varies, i.e. we will be focusing on the locations of bifurcation points.
The authors P. Korman, Y. Li and T. Ouyang ( ”Computing the location and
the direction of bifurcation”, Mathematical Research Letters, accepted ), prove that a
necessary and sufficient condition for α to be a bifurcation point is
1/2
Z
α
G(α) ≡ F (α)
0
where F (α) =
Rα
0
f (α) − f (τ )
dτ − 2 = 0,
[F (α) − F (τ )]3/2
f (u) du. We will prove that G(α) has vertical asymptotes at α = b,
α = 1 and at any point α ∈ (0, 1) for which
Rα
0
f (u) du = 0. We will use the
asymptotic behavior of G to estimate intervals where G(α) 6= 0, that is, intervals
where there is no bifurcation point.
iii
TABLE OF CONTENTS
CHAPTER
1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
1
Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2 DEFINITIONS AND BACKGROUND . . . . . . . . . . . . . . . . .
4
2.1
2.2
Bifurcation approach . . . . . . . . . . . . . . . . . . . . . . . .
General properties . . . . . . . . . . . . . . . . . . . . . . . . . .
5
6
3 GENERAL STUDY OF THE PARAMETRIC SPACE . . . . . . . . .
11
3.1
3.2
Region that satisfies the sufficient condition . . . . . . . . . . . .
Basic Behaviour of Solutions . . . . . . . . . . . . . . . . . . . .
11
15
4 ASYMPTOTIC BEHAVIOR OF G . . . . . . . . . . . . . . . . . . .
21
4.1
4.2
4.3
On the Asymptotic Behavior of G(α) . . . . . . . . . . . . . . .
Further Analysis of Asymptotic Behavior, Part I. . . . . . . . . .
Further Analysis of Asymptotic Behavior, Part II. . . . . . . . .
21
30
39
5 LOCATION OF BIFURCATIONS . . . . . . . . . . . . . . . . . . . .
45
5.1
5.2
5.3
Applications of Theorem 2.2.8 . . . . . . . . . . . . . . . . . . .
General Behaviour of Solutions . . . . . . . . . . . . . . . . . . .
Some generalizations . . . . . . . . . . . . . . . . . . . . . . . . .
45
45
54
REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
iv
1
CHAPTER 1
INTRODUCTION
In this thesis, we study exact bifurcation diagrams and the existence of multiple positive solutions to the Dirichlet problem
u00 (x) + λf (u(x)) = 0 on (−1, 1), u(−1) = u(1) = 0,
(1.1)
depending on a positive parameter λ. We recall that solutions of (1.1) are even
functions, with u0 (x) < 0 for x > 0, and hence any solution is uniquely identified
by α = u(0), see [12]. A bifurcation occurs where the number of solutions change
as the parameter λ changes. Since the value of u(0) = α uniquely identifies u(x),
then u(0) = α stands for u. In order to draw solution curves, one puts the λ on the
horizontal axis and the parameter α on the vertical axis. It is customary to refer to
these curves as bifurcation diagrams.
In applications, information about the shape of solution branches is needed.
Branches of (1.1) are smooth. We are then interested in whether or not these curves
have turns with respect to the λ-direction, and if so, how many bifurcations there are
and where they are located.
The problem (1.1) arises in many different physical situations, for example, in
the theory of thermal ignition of gases, in quantum field theory and in population
dynamics (see e.g. [18]). The nonlinearity of f in combustion problems, describes
intermediate steady states of the temperature distribution u, λ then measures the
amount of unburnt substance. Here the bifurcation points have significant physical
2
implications, and that is important whether or not these exist (see e.g. [22]).
We study the open problem where f (u) = u(u − a)(u − b)(u − c)(d − u), and f
does not depend explicitly on x and whose roots satisfy 0 < a < b < c < d < +∞. A
necessary condition for a positive solution of (1.1) to exist is that
Rc
b
Rb
0
f (t) dt > 0, and
f (t) dt > 0. Roughly speaking, these hypotheses imply that the graph of f has two
positive humps and two negative bumps, where each positive bump has a bigger area
than the previous negative hump. We want to show how bumps of the function f may
affect the ”bifurcation diagrams” or the multiplicity results. The convexity properties
of f (u) will be important to determine the location of the bifurcation points.
This problem, when f (u) = (u−a)(u−b)(c−u), i.e. a cubic polynomial whose
roots are three distinct constants a ≤ b < c, was studied by J. Smoller and A. Wasserman [19], who attempted to solve the problem in general, and succeeded in solving
it for a = 0. Later S.-H. Wang [20] solved the problem under some restrictions on a.
Both papers used phase-plane analysis. Then P. Korman, Y. Li and T. Ouyang [12],
[13] used bifurcation theory to attack the problem. Their technique was a careful use
of the Crandall and Rabinowitz theorem but again some restrictions were necessary
(all of the above mentioned papers covered more general types of f (u), behaving like
cubic functions). Finally in [14] the problem is completely understood for all such
cubic polynomials.
In order to study the number of solutions of (1.1) and bifurcations, we must
study the qualitative shape of the bifurcation diagram. The shape of such bifurcation
diagram is determined by its turning points.
3
Our main tool comes from [14] where P. Korman, Y. Li and T. Ouyang present
a formula, where the relevant point is that it allows us to compute all α0 s where a
turn may occur, and another formula, which allows to compute the direction of the
turn.
1.1
Overview
In Chapter 2, we provide definitions and background information. Necessary
lemmas will be announced and some will be proved. All of the results cited in this
chapter are based on the papers [7], [8], [9], [12], [13], [14], [15].
In Chapter 3, we derive a region that satisfies necessary and sufficient conditions for the existence of positive solution. We end the chapter with an application
of Contraction Mapping Theorem.
Chapter 4 is the heart of this thesis. Using delicate integral estimates we will
study the asymptotic behavior of G, where G(α) ≡ F (α)1/2
and F (α) =
Rα
0
Rα
f (α)−f (τ )
0 [F (α)−F (τ )]3/2
dτ − 2
f (u) du. Such asymptotic analysis enable us to narrow the range of
possible bifurcation points.
In Chapter 5, we explore and prove a series of properties which restrict the
location of a bifurcation point by studying the change of concavity of the graph of f
and the points where the rays from 0 and b touche the graph of f .
4
CHAPTER 2
DEFINITIONS AND BACKGROUND
We present exact multiplicity results for the boundary value problem of the
type
u00 (x) + λf (u) = 0 on (−1, 1), u(−1) = u(1) = 0,
(2.1)
with the nonlinearity f (u) = −u(u − a)(u − b)(u − c)(u − d), 0 < a < b < c < d < +∞
and λ a positive parameter. It is convenient to consider the problem on the interval
(−1, 1), because positive solutions are even in this interval. By shifting and scaling,
we can replace the interval (−1, 1) for any interval (a, b). Most results in the literature
are obtained when f = f (u), i.e. when the problem is autonomous. Changes in the
sign of f (u) lead to multiple positive solutions of (2.1). All of the results cited in this
chapter are based on the papers [7], [8], [9], [12], [13], [14], [15].
For the quintic nonlinearity in problem (2.1), by letting u = dν, we may
assume that d = 1, so that our nonlinearity is f (u) = u(u − a)(u − b)(u − c)(1 − u),
with new a, b, c, i.e. we consider
u00 (x) + λu(u − a)(u − b)(u − c)(1 − u) = 0 on (−1, 1), u(−1) = u(1) = 0. (2.2)
This substitution allows us to “compactify” the parameter set, since now 0 <
a < b < c < 1. Note that u ≡ 0 is a trivial solution of (2.2).
5
2.1
Bifurcation approach
The implicit function theorem for Banach spaces is the basic tool for continuation of solutions.The following version for Banach spaces is given by M.G. Crandall
and P.H. Rabinowitz [2].
Theorem 2.1.1. Let X, Λ and Z be Banach spaces,and f (x, λ) a continuous mapping
of an open set U ⊂ X × Λ → Z. Assume that f has a Frechet derivative with
respect to x, fx (x, λ) which is continuous on U. Assume that f (x0 , λ0 ) = 0 for some
(x0 , λ0 ) ∈ U. If fx (x0 , λ0 ) is an isomorphism (i.e. 1:1 and onto) of X onto Z, then
there is a ball Br = {λ :k λ−λ0 k} < r and unique continuous map x(λ) : Br (λ0 ) → X,
such that f (x(λ), λ) ≡ 0, x(λ0 ) = x0 . If f is the class C p , so is x(λ), p ≥ 1.
We call (x0 , λ0 ) a regular solution if it satisfies the conditions of Theorem
2.1.1, otherwise we call it a singular solution. The main tool for Philip Korman, Yi
Li, Tiancheng Ouyang’s papers has been the M.G. Crandall and P.H. Rabinowitz
Theorem. It gives us conditions to have continuation of solutions through a critical
point.
Next we state a bifurcation theorem of Crandall-Rabinowitz [3].
Theorem 2.1.2. Let X and Y be Banach spaces. Let (λ, x) ∈ R × X and let F be a
continuously differentiable mapping of an open neighborhood of (λ, x) into Y . Let the
null-space N (Fx (λ, x)) = span x0 be one-dimensional and codim R(Fx (λ, x)) = 1.
Let Fλ (λ, x) 6∈ R(Fx (λ, x)). If Z is a complement of span x0 in X, then the solutions
of F (λ, x) = F (λ, x) near (λ, x) form a curve (λ(s), x(s)) = (λ + τ (s), x + sx0 + z(s)),
6
where s → (τ (s), z(s)) ∈ R × Z is a continuously differentiable function near s = 0
and τ (0) = τ 0 (0) = 0, z(0) = z 0 (0) = 0.
I will denote derivatives of u(x) by either u0 (x) or ux and mix both notations
to make our proofs more transparent (u0x will denote the second derivative of u(x),
when convenient.)
2.2
General properties
It is well known that the method of supersolution and subsolution yields not
only existence of a solution but it also locates the solution between given bounds. To
use this method, one must be able to construct a supersolution γ and an sub-solution
ψ of (2.1) (with the appropriate boundary conditions) so that 0 ≤ ψ ≤ γ. Remember
that a function γ(x) ∈ C 2 (−1, 1) ∩ C 0 [−1, 1] is called a supersolution of (2.1) if
γ 00 + λf (γ) ≤ 0 on (−1, 1), γ(−1) ≥ 0, γ(1) ≥ 0.
(2.3)
A subsolution ψ(x) is defined by reversing the inequalities in (2.3). The following result is standard.
Lemma 2.2.1. Let γ(x) and ψ(x) be respectively super- and subsolutions of (2.1),
and γ(x) ≥ ψ(x) on (−1, 1) with γ(x) 6≡ ψ(x), then γ(x) > ψ(x) on (−1, 1).
We shall often use this lemma with either γ(x) or ψ(x) or both being solution
of (2.1). The following lemma is a consequence of the first.
Lemma 2.2.2. Let u(x) be a nontrivial solution of (2.1) with f (u) ≡ 0. If u(x) ≥ 0
on (−1, 1) then u > 0 on (−1, 1).
7
Lemma 2.2.3. Let ξ ∈ (−1, 1) be any critical point of u(x), i.e. u0 (ξ) ≡ 0. Then
u(x) is symmetric with respect to ξ.
Proof. Let ν(x) = u(2ξ − x). Then ν 00 (x) + λf (ν) ≡ 0 on the interval (−1, 1) ∩ (2ξ −
1, 2ξ + 1). Then both u(x) and ν(x) satisfy the equation (2.1) and ν(ξ) = u(ξ) and
ν 0 (ξ) = u0 (ξ) = 0. By uniqueness of initial value problems, u(x) ≡ ν(x), and the
proof follows.
From the lemmas above, it follows that any positive solution of (2.1) is an even
function, moreover u0 (x) > 0 on (−1, 0), and u0 (x) < 0 on (0, 1). Thus, α ≡ u(0) is
the maximal value of solution.
The next lemma shows that it is impossible that two solutions of (2.1) share
the same α.
Lemma 2.2.4. The value of u(0) = α uniquely identifies the solutions pair (λ, u(x))
(i.e. there is at most one λ, with at most one solution u(x), so that u(0) = α).
Proof. Assume on the contrary that we have two solution pairs (λ, u(x)) and (µ, ν(x)),
with u(0) = ν(0) = α. Clearly, λ 6= µ, since otherwise we have a contradiction with
uniqueness of initial value problems. (Recall that u0 (0) = ν 0 (0) = 0). Then u( √1λ x)
and ν( √1λ x) are both solutions of the same initial value problem
u00 + f (u) = 0,
u(0) = α,
u0 (0) = 0,
and hence u( √1λ x) ≡ ν( √1λ x), but this is impossible, since the first function vanishes
at x =
√
√
λ, while the second at x = µ.
8
To use the bifurcation theory approach, we need to linearize (2.1)
ω 00 (x) + λf 0 (u(x))ω = 0 on (−1, 1), ω(−1) = ω(1) = 0,
(2.4)
where u(x) is a solution of (2.1). If (2.4) has a nontrivial solution, then we call u(x)
a singular solution of (2.1). If ω(x) ≡ 0 is the only solution of (2.4), we say that the
solution u(x) is nonsingular.
Lemma 2.2.5. Let u(x) be a positive solution of (2.1), with
u0 (1) < 0.
(2.5)
If the problem (2.4) admits a nontrivial solution, then it does not change sign, i.e.
we may assume that ω(x) > 0 on (−1, 1).
Proof. The function u0 (x) also satisfies the linear equation (2.4). By the condition
(2.5), u0 (x) is not a multiple of ω(x). Hence its roots are interlaced with those of
ω(x). If ω(x) had a root ξ inside say (−1, 0), then u0 (x) would have to vanish on
(−1, ξ), which is impossible.
Condition (2.5) will hold for any positive solution, provided that
f (0) ≥ 0,
(2.6)
see, e.g., p.107 in M. Renardi and R.C. Rogers [17]. If f (u) < 0, then we can have
u0 (1) = 0.
Lemma 2.2.6. If the problem (2.4) admits nontrivial solutions, then the solution set
is one dimensional. If moreover u(x) is a positive solution, satisfying (2.5), then ω
is an even function.
9
Proof. By uniqueness of the initial value problem the value of w0 (1) uniquely determines ω(x), and hence the null space is one dimensional. Turning to the second claim,
u(x) is positive, then it is even. Hence ω(−x) also solve (2.4). Since the null space
is one dimensional, ω(−x) = cω(x) for some constant c. Evaluating this relation at
x = 0, we conclude that c = 1 (since ω(0) > 0 by the previous lemma), which is the
desired symmetry.
Lemma 2.2.7. Any two positive solutions of (2.1) do not intersect inside (−1, 1)
(i.e., they are strictly ordered on (−1, 1)).
Proof. Let u(x) and ν(x) be two intersecting solutions. Since both of them are even
functions, they intersect on the half-interval (0, 1) as well. Let 0 < ξ < η < 1 be two
consecutive intersection points. If ν(x) > u(x) on (ξ, η), then | u0 (ξ) |>| ν 0 (ξ) |, while
| u0 (η) |<| ν 0 (η) |. The energy E(x) + 21 u0 (x)2 + λF (u(x)) is constant for any solution
u(x). But at ξ, u(x) has higher energy than ν(x), and at η the order is reversed, a
contradiction.
The next theorem was proved in [12] where f satisfies
f (u) = u(u − b)(c − u) for − 1 < x < 1, u(−1) = u(1) = 0,
(2.7)
with 0 < b < c < +∞ and
Z
c
f (u)du > 0.
(2.8)
0
Theorem 2.2.8. Under the conditions (2.7) , (2.8) there is a critical λ0 > 0 such that
for λ < λ0 the problem (2.1) has no nontrivial solutions, it has exactly one nontrivial
10
solution for λ = λ0 , and exactly two nontrivial solutions for λ > λ0 . Moreover, all
solutions lie on a single solution curve, which for λ > λ0 has two branches denoted
by u− (x, λ) < u+ (x, λ), with u+ (x, λ) strictly monotone increasing in λ, u− (0, λ)
strictly monotone decreasing in λ, and limλ→∞ u+ (x, λ) = c, limλ→∞ u− (x, λ) = 0 for
x ∈ (−1, 1)\{0}, while u− (0, λ) > b for all λ > λ0 .
11
CHAPTER 3
GENERAL STUDY OF THE PARAMETRIC SPACE
This chapter studies necessary and sufficient conditions for the existence of
positive solutions of the quintic nonlinearity problem
u00 (x) + λu(u − a)(u − b)(u − c)(1 − u) = 0 on (−1, 1), u(−1) = u(1) = 0, (3.1)
as well as some basic behaviours of positive solutions.
P. Hess showed that
Rb
0
f (s) ds > 0 and
R1
b
f (s) ds > 0 was a sufficient condition
for the existence of positive solution of the problem (3.1), i.e. he showed how bumps
of the function f affect the existence of positive solutions. We are able to prove
this using the argument due by A. Ambrosetti and P.H.Rabiniwitz [8, p.12] that was
outlined in the Theorem 2.2.8.
3.1
Region that satisfies the sufficient condition
In the next theorem, we will derive a region that satisfies sufficient condition
for the existence of positive solution. This region will be very important for numerical
computation.
Theorem 3.1.1. Let f (s) = s(s − a)(s − b)(s − c)(1 − s) with 0 < a < b < c < 1,
then
Rb
0
f (s) ds > 0 and
R1
b
f (s) ds > 0 is equivalent to
√
1 −2a + 3a2
< b ≤ 1 + a 1 − 3a + a2
0<a< ,
3
−1 + 2a
and
−5ab + 3b2 + 3ab2 − 2b3
−10a + 5b + 5ab − 3b2
<c<
−2 + 3a − 3b + 4ab − 3b2 + ab2 − 2b3 ,
−3 + 5a − 4b + 5ab − 3b2
12
or
√
1
2
0<a<
, (1 + a − 1 − 3a + a < b < 1
3
and
Proof. Since
Rb
0
Rb
0
b<c<
−2 + 3a − 3b + 4ab − 3b2 + 3ab2 − 2b3 .
−3 + 5a − 4b + 5ab − 3b2
f (s) ds =
1 3
b (b(2b2 + 5c − 3b(1 + c)) + a(−3b2 − 10c + 5b(1 + c)).
60
f (s) ds > 0 is equivalent to
Then
1 3
b (b(2b2 +5c−3b(1+c))+a(−3b2 −10c+5b(1+c)
60
Multiplying the last inequality by
60
,
b3
> 0.
and grouping in a and b we obtain
b(2b2 + 5c − 3b(1 + c)) + a(−3b2 − 10c + 5b(1 + c)) > 0,
so a(−3b2 − 10c + 5b(1 + c)) > −b(2b2 + 5c − 3b(1 + c)), and also notice that (−3b2 −
10c + 5b(1 + c)) < 0, hence
a<
Denoting L(b, c) =
−b(2b2 + 5c − 3b(1 + c)
.
−3b2 − 10c + 5b(1 + c)
−3b2 −3b2 c+2b3 +5bc
,
3b2 +10c−5b(1+c)
(3.2)
we have
a < L(b, c).
(3.3)
We will prove that L is increasing in (b, c) and L(c, c) < 31 . Taking derivative, we
obtain
∂L
6b4 + 50c2 − 20b3 (1 + c) − 60bc(1 + c) + 15b2 (1 + 5c + c2 )
=
.
∂b
(3b2 + 10c − 5b − 5bc)2
Define N (b) = 6b4 + 50c2 − 20b3 (1 + c) − 60bc(1 + c) + 15b2 (1 + 5c + c2 ), and b = c,
13
0 < < 1, then
N (c) = 50c2 − 60c2 (1 + c) + 15c2 (1 + 5c + c2 )2 − 20c3 (1 + c)3 + 60c4 4
= 50c2 − 60c2 − 60c3 + 15c2 2 + 75c3 2 + 15c4 2 − 20c3 3 − 20c4 3 + 6c4 4
= (152 − 203 + 64 )c4 + (−60 + 752 − 203 )c3 + (50 − 60 + 152 )c2
= c2 [(152 − 203 + 64 )c2 + (−60 + 752 − 203 )c + (50 − 60 + 152 )].
Consider
M (c) = (152 − 203 + 64 )c2 + (−60 + 752 − 203 )c + (50 − 60 + 152 ).
Let h() = 15 − 20 + 62 , (3.4)then h0 (0 ) = 0 ⇔ 0 =
5
3
(3.4)
> 1.
Since 0 ∈
/ (0, 1) and h is a quadratic function in , to know the sign of h in (0, 1) is
necessary to check the values of h in the extreme of the interval (0, 1). But h(0) = 15
and h(1) = 1, then
h() = 15 − 20 + 62 > 0 in (0, 1).
(3.5)
From (3.4), M 0 (c) = 2(152 − 203 + 64 )c + (−60 + 752 − 203 ), so M 0 (c0 ) = 0 if
and only if c0 =
60−75+202
.
2h()
Claim 1. 60 − 75 + 202 > 2(15 − 20 + 62 ).
P roof . Define k() = −123 + 602 − 105 + 60, ∈ (0, 1), then k 0 () = −362 +
120 − 105 < 0, so k() is decreasing and 3 = k(1) ≤ k() for all ∈ (0, 1). Hence
k() > 0. Then we may conclude the claim.
Using the claim and (3.5), we obtain that c0 > 1.
Claim 2. t() = 152 − 60 + 50 > 0.
14
P roof . t0 (0 ) = 0 ⇔ 0 = 2 ∈
/ (0, 1). But t(0) = 50 and t(1) = 5, so t() > 0 in (0, 1).
Using (3.4) and (3.5), M is a quadratic function in c, where the principal coefficient
152 − 203 + 64 > 0, and its critical point c0 ∈
/ (0, 1). To determine the sign of M in
(0, 1) is enough to evaluate M at c = 0 and c = 1. Note that M (0) = 152 −60+50 =
l(), so l0 (0 ) = 0 if and only if 0 = 2, but l(0) = 50 and l(1) = 5, therefore M (0) > 0.
Also, it is easy to verify that M (1) = 64 − 403 + 1052 − 120 + 50 = H() > 0.
This follows because H 0 () = 243 − 1202 + 210 − 120, H 00 () = 722 − 240 + 210
and H 000 (0 ) = 0 if and only if 0 =
240
144
> 1, so H 00 (0) = 210 > 0 and H 00 (1) = 42 > 0
therefore H 00 () is increasing, then H 0 (0) < H 0 () < H 0 (1) for all ∈ (0, 1), but
H 0 (1) = −6, consequently H 0 () < 0, so H is decreasing, then H(1) < H() < H(0).
Hence M (1) = H() > 0. Consequently M (c) > 0.
But N (c) = c2 M (c) > 0, so N (b) > 0. Then
N (b)
∂L
=
> 0.
2
∂b
(3b + 10c − 5b − 5bc)2
(3.6)
We conclude that L is increasing in b, so
L(b, c) < L(c, c) =
Claim 3. L(c, c) ≤
2c2 − c3
.
5c − 2c2
(3.7)
1
3
P roof . Define w(c) = 3c2 − 8c + 5 with c ∈ (0, 1). Then w0 (c0 ) = 0 ⇔ c0 =
8
6
>
1. Since w(1) = 0, we conclude that w(c) ≥ 0. Therefore 6c2 − 3c3 ≤ 5c − 2c2 .
Consequently 3(2c2 − c3 ) ≤ (5c − 2c2 ). So L(c, c) < 13 .
From (3.3), (3.7) and Claim 3, we have
1
0<a< .
3
15
The rest of the proof of this theorem is similar.
3.2
Basic Behaviour of Solutions
The structure of the set of solution of the problem (3.1) is richer in properties.
For a first simple approach, we notice that solutions of (3.1) are fixed points of the
operator T : L2 [−1, 1] →
− L2 [−1, 1] by T u = h, where h(x) =
R1
−1
Z(x, y)f (u(y)) dy
and Z is the Green’s function corresponding to the problem (3.1).
We want to find out properties of the Dirichlet-branches. The next properties are
very useful to understand the behavior of these branches.
Lemma 3.2.1. Any nontrivial positive solution of (3.1) satisfies u0 (1) 6= 0.
Proof. It follows from the fact that f (0) = 0.
Lemma 3.2.2. If u is a nontrivial positive solution of (3.1) with maximum value
α = u(0), then α ∈ (0, 1).
Proof. Suppose that α = 0, then u(0) = 0 and u0 (0) = 0. By uniqueness of initial
value problems, u ≡ 0, a contradiction.
Suppose that α = 1, then u(0) = 1 and u0 (0) = 0. By uniqueness of initial value
problems, u ≡ 1, a contradiction.
Consider α > 1, therefore u00 (0)+λf (u(0)) = 0, so u00 (0) = −λf (u(0)) = −λf (α) > 0,
a contradiction.
Lemma 3.2.3. Let u a nontrivial positive solution of (3.1), then
Rα
0
f (s) ds > 0.
16
Proof. Suppose that u is a nontrivial positive solution of (3.1), then
1 0
(u (x))2 = λ[F (α) − F (u(x))].
2
(3.8)
Let x = 1, substituting this in(3.8), we obtain
1 0
(u (1))2 = λ[F (α) − F (u(1))]
2
= λF (α).
(3.9)
From Lemma (3.2.1) F (α) > 0.
Lemma 3.2.4. Let u a nontrivial positive solution of (3.1) then α ∈ (a, b) ∪ (c, 1).
Proof. From Lemma (3.2.2) α ∈ (0, a] ∪ (a, b) ∪ [b, c] ∪ (c, 1). If α = a then u ≡ a, a
contradiction.
If α ∈ (0, a) then u00 (0) > 0, a contradiction.
The same proof applies if α ∈ [b, c].
Lemma 3.2.5. Let α = u(0), then f (α) > 0.
Proof. It is clear that the proof of the lemma follows from the Lemma (3.2.4).
Lemma 3.2.6. If α ∈ (a, b) then
Proof.
Rb
0
f (s) ds =
Rα
0
f (s) ds +
Rb
0
Rb
Lemma 3.2.7. If α ∈ (c, 1) then
α
f (s) ds > 0.
f (s) ds.
R1
0
f (s) ds > 0 and
R1
b
f (s) ds > 0.
Proof. Note that b ∈ [u(1), u(0)], so there is x0 ∈ (0, 1) such that u(x0 ) = b. Therefore
2
u0 (x0 ) = λ[F (α) − F (b)] > 0,
17
so F (α) > F (b). Suppose that
R1
b
f (s) ds ≤ 0, then F (1) ≤ F (b), so F (α) > F (1), a
contradiction.
Lemma 3.2.8. If α ∈ (a, b) then α ∈ (γ0 , b) where γ0 is such that
Proof.
Rα
0
f (s) ds =
R γ0
0
f (s) ds +
Rb
γ0
R γ0
0
f (s) ds = 0.
f (s) ds > 0.
Lemma 3.2.9. If α ∈ (c, 1) then α ∈ (γ1 , 1) where γ1 is such that
R γ1
b
f (s) ds = 0.
Proof. Because F (α) > F (b), α ∈ (γ1 , 1).
We can use Green’s function to turn (3.1) into an equivalent integral equation.
Applying the Green’s function to the equation
−u00 (x) = λf1 (u)
(3.10)
gives
Z
1
Z(x, y)f1 (u(y)) dy,
u(x) = λ
(3.11)
−1
where Z is the Green’s function for (3.10), i.e.,




 12 (x + 1)(1 − y), −1 ≤ x ≤ y ≤ 1,
Z(x, y) =



 1 (y + 1)(1 − x), −1 ≤ y ≤ x ≤ 1,
2
(3.12)
and





f 0 (0)u,
u ≤ 0,





f1 (u) = u(u − a)(u − b)(u − c)(1 − u), 0 ≤ u ≤ 1,








f 0 (1)(u − 1),
u > 0.
(3.13)
18
Define σf = max{|f 0 (u(x))| : −1 ≤ x ≤ 1}, P 2 =
λ¯0 =
1
.
P σf
Elementary computations show that P 2 =
4
.
45
R1 R1
−1
−1
Z 2 (x, y) dy dx and
The next two theorems say
that there is a unique solution if λ < λ¯0 , but the equation (3.1) always has the trivial
solution.
Theorem 3.2.10. Assume α = u(0) ∈ (γ0 , b), then the integral equation (3.11)
possesses a unique solution u(x) ∈ L2 [−1, 1] for every small value of the parameter λ
that satisfies λ < λ¯0 , i.e. u ≡ 0.
Proof. Define the mapping T : L2 [−1, 1] →
− L2 [−1, 1] by T u = h, where h(x) =
λ
R1
−1
Z(x, y)f1 (u(y)) dy.
This map is well defined for each u ∈ L2 [−1, 1], h ∈
R1
L2 [−1, 1], because Z is bounded. Let t(x) =
−1
Z(x, y)f1 (u(y)) dy. By Cauchy-
Schwarz inequality
Z 1
|Z(x, y)f1 (u(y))| dy
Z(x, y)f1 (u(y)) dy ≤
−1
−1
1/2 Z
Z 1
2
|Z(x, y)| dy
≤
Z
1
1/2
1
2
|f1 (u(y))| dy
−1
−1
then
Z
2
1
2
|t(x)| ≤
Z
1
2
|Z(x, y)| dy
|f1 (u(y))| dy
−1
−1
therefore
Z
1
Z
2
1
Z
1
2
|t(x)| dx ≤
−1
Z
1
|Z(x, y)| dy
−1
Z
1
Z
−1
1
=
−1
< ∞,
−1
2
|f1 (u(y))| dy
dx
−1
Z
|Z(x, y)| dy dx
1
2
−1
2
|f1 (u(y))| dy
19
so
t(x) ∈ L2 [−1, 1].
(3.14)
We know that L2 [−1, 1] is a complete metric space with metric
Z
1/2
1
2
|u1 (x) − u2 (x)| dx
d(u1 , u2 ) =
.
−1
Note that d(T u1 , T u2 ) = d(h1 , h2 ), where T ui = hi and hi (x) = λ
R1
−1
Z(x, y)f1 (ui (y)) dy
for i = 1, 2. We will show that T is a contraction mapping for small λ.
This can be seen as follows,
1
−1
Z
1
Z
1
Z
1
d(T u1 , T u2 ) = λ
Z
−1
2 !1/2
Z(x, y)[f1 (u1 (y)) − f1 (u2 (y))] dy dx
Z(x, y) |f1 (u1 (y)) − f1 (u2 (y))| dy
≤ λ
−1
Z
1
Z
1
2
Z
Z(x, y) dy
−1
1
−1
Z
1
2
≤ λ
1
|f1 (u1 (y))
−1
1/2 Z 1
2
1/2
− f1 (u2 (y))| dy
2
dx
1/2
|f1 (u1 (y)) − f1 (u2 (y))| dy
Z(x, y) dy dx
−1
Z
dx
−1
≤ λ
Z
!1/2
2
−1
−1
1/2
1
2
|f1 (u1 (y)) − f1 (u2 (y))| dy
≤ λP
−1
Z
1
≤ λP
2
1/2
|σf (u1 (y) − u2 (y))| dy
,
−1
so
Z
1
d(T u1 , T u2 ) ≤ λP σf
2
1/2
|u1 (y) − u2 (y)| dy
−1
≤ λP σf |d(u1 , u2 )| .
If λ < λ¯0 , then d(T u1 , T u2 ) < d(u1 , u2 ), therefore T is a contraction mapping, so
there is a unique u ∈ L2 [−1, 1] such that T u = u.
20
Theorem 3.2.11. Assume α = u(0) ∈ (γ1 , 1), then the integral equation (3.11)
possesses a unique solution u(x) ∈ L2 [−1, 1] for every small value of the parameter λ
that satisfies λ < λ¯0 .
Proof. The same proof works.
21
CHAPTER 4
ASYMPTOTIC BEHAVIOR OF G
The shape of any bifurcation diagram is determined by its turning points.
In this chapter, I will use a theorem derived in [14] that provides a necessary and
sufficient condition on α for be a bifurcation point. The theorem will be used to find
places where there are no bifurcation points. Also, the formula given in [14] can be
used to compute numerically all turning points.
We will assume throughout this chapter that
Rb
0
f (t) dt > 0, and
It is clear that there exists γ0 ∈ (a, b) and γ1 ∈ (c, 1) such that
R γ1
b
R γ0
0
R1
b
f (t) dt > 0.
f (t) dt = 0, and
f (t) dt = 0.
We will adopt the following notation
Jk = max{|f (k) (u)| : u ∈ [0, 1]} > 0, k = 0, 1, 2, 3.
j0 = min{f (u) : u ∈ [0, 1]} < 0.
In addition, let x0 ∈ (a, b) such that f 0 (x0 ) = 0.
4.1
On the Asymptotic Behavior of G(α)
The following result was proved by P. Korman, Y. Li, T. Ouyang in [14].
Theorem 4.1.1. A solution of problem (2.1) with the maximal value α = u(0) is a
bifurcation point if and only if
1/2
Z
G(α) ≡ F (α)
0
with F (u) =
Ru
0
f (t) dt.
α
f (α) − f (τ )
[F (α) − F (τ )]3/2
dτ − 2 = 0,
(4.1)
22
An intuitive study of (4.1) tells us, for example, that G(α) approaches to +∞
when α → γ0+ . We want to explore and provide a rigorous proof of the asymptotic
behavior of G when α is close to γ0 , b, γ1 and 1. The next eight theorems are the
main results of this thesis.
Theorem 4.1.2. limα→b− G(α) = −∞, i.e., to the left of b but near b there is no
bifurcation point.
Proof. For τ close to α, f and F can be expressed
f (τ ) = f (α) + (τ − α)f 0 (α) +
(τ − α)2 00
f (α) + O((τ − α)3 ),
2
(4.2)
F (τ ) = F (α) + (τ − α)F 0 (α) +
(τ − α)2 00
F (α) + O((τ − α)3 ).
2
(4.3)
From (4.2), (4.3) and using the fact that F 0 = f and F 00 = f 0 , we obtain
f (α) − f (τ ) = (α − τ )f 0 (α) −
(α − τ )2 00
f (α) + O((α − τ )3 ),
2
(4.4)
F (α) − F (τ ) = (α − τ )f (α) −
(α − τ )2 0
f (α) + O((α − τ )3 ).
2
(4.5)
and
Then
f (α) − f (τ )
[F (α) − F (τ )]3/2
=
=
=
(α−τ ) 00
f (α) + O((α − τ )2 )]
2
) 0
τ )3/2 [f (α) − (α−τ
f (α) + O((α − τ )2 )]3/2
2
) 00
[f 0 (α) − (α−τ
f (α) + O((α − τ )2 )]
2
) 0
τ )1/2 [f (α) − (α−τ
f (α) + O((α − τ )2 )]3/2
2
) 00
[f 0 (α) − (α−τ
f (α) + f1 ]
2
,
(α−τ
τ )1/2 [f (α) − 2 ) f 0 (α) + f2 ]3/2
(α − τ )[f 0 (α) −
(α −
(α −
(α −
(4.6)
23
where f1 , f2 = O((α − τ )2 ). Therefore, there are constants C1 , C2 , δ1 , δ2 > 0 such that
|f1 | ≤ C1 |(α − τ )2 | for |α − τ | < δ1
and
|f2 | ≤ C2 |(α − τ )2 | for |α − τ | < δ2 .
Fix > 0 such that
0
0
< min{
−f (b) −f (b)
,
,
2J2
2C2
s
−f 0 (b) b − x0 b − γ0
,
,
, δ1 , δ2 }.
4C1
2
2
(4.7)
Let α ∈ (b − , b). Then
f 0 (α) = f 0 (b) + (α − b)f 00 (z),
z ∈ (α, b)
= f 0 (b) + (b − α)(−f 00 (z))
≤ f 0 (b) + J2 ,
(4.8)
from (4.7), f 0 (α) + 21 (−f 0 (b)) < 0 for all α ∈ (b − , b). Also
α − τ 00 1
0
2 f (α) < 4 (−f (b)) ,
(4.9)
1
(−f 0 (b)) ,
4C1
(4.10)
(α − τ )2 <
|α − τ | <
1
(−f 0 (b)),
2C2
(4.11)
for any τ ∈ (α − , α).
Define
Z
I1 (α) =
0
α−
f (α) − f (τ )
[F (α) − F (τ )]3/2
dτ,
(4.12)
24
α
f (α) − f (τ )
α−
[F (α) − F (τ )]3/2
Z
I2 (α) =
2f 0 (α) − f 0 (b)
K(α) =
2
α
Z
(4.13)
dτ
α−
S(α) =
dτ,
(α −
τ )1/2 [f (α)
+
(α−τ )
(−f 0 (α)
2
− f 0 (b))]3/2
J0 − j0
(α − ),
[F (α) − F (α − )]3/2
, (4.14)
(4.15)
and
T (α) = F (α)1/2 [S(α) + K(α)] − 2.
(4.16)
Therefore from (4.1), (4.12) and (4.13) we obtain
G(α) = F (α)1/2 I1 (α) + I2 (α) − 2.
(4.17)
From (4.10) and (4.11)
1
|f1 | ≤ (−f 0 (b)),
4
(4.18)
and
|f2 | ≤
α−τ
(−f 0 (b)).
2
We will prove that I1 (α) ≤ S(α). For every τ ∈ (0, α − ) we have that
F (τ ) ≤ F (α − ),
and
f (α) − f (τ ) ≤ J0 − j0 .
(4.19)
25
Therefore
f (α) − f (τ )
J0 − j0
≤
.
3/2
[F (α) − F (τ )]
[F (α) − F (α − )]3/2
Hence
Z
α−
f (α) − f (τ )
dτ
[F (α) − F (τ )]3/2
Z α−
J0 − j0
≤
dτ
[F (α) − F (α − )]3/2
0
J0 − j0
=
(α − ).
[F (α) − F (α − )]3/2
I1 (α) =
0
It follows from (4.15) that I1 (α) ≤ S(α). Note that
lim− S(α) =
α→b
J0 − j0
(b − ) is finite.
[F (b) − F (b − )]3/2
(4.20)
Next, we will be working with I2 (α). Observe that from (4.6), (4.13), (4.18) and
(4.19) follows that
(α−τ ) 00
f (α) + f1 ]
2
dτ
(α−τ
) 0
3/2
f
(α)
+
f
]
α− (α − τ )1/2 [f (α) −
2
2
Z α
f 0 (α) − 21 f 0 (b)
3 dτ
1/2 [f (α) − α−τ f 0 (α) − α−τ f 0 (b)] 2
α− (α − τ )
2
2
Z α
dτ
1
f 0 (α) − f 0 (b)
(α−τ
)
2
(−f 0 (α)
α− (α − τ )1/2 [f (α) +
2
Z
I2 (α) =
≤
=
[f 0 (α) −
α
− f 0 (b))]3/2
.
From (4.14) we obtain I2 (α) ≤ K(α). Consequently G(α) ≤ T (α). Let A = f (α) and L =
−f 0 (α) − f 0 (b), then
K(α) =
1 0 f (α) − f (b)
2
0
Z
α
α−
dτ
(α −
τ )1/2 [A
+
(α−τ )
L]3/2
2
,
26
√
dτ
Take u = 2 α − τ , then du = − √α−τ
. therefore
K(α) =
=
=
=
Z
1 0 0
−du
f (α) − f (b)
L 2 3/2
√
2
2 [A + 8 u ]
√
Z
du
1 0 2 0
.
f (α) − f (b)
2
[A + L8 u2 ]3/2
0
√
1 0
0
f (α) − 2 f (b) 8
√
2A3 + LA2
√
f 0 (α) − 21 f 0 (b) 8
r
.
3
2
0
0
2f (α) + f (α) − f (α) − f (b)
0
Hence limα→b− T (α) = −∞. Consequently limα→b− G(α) = −∞.
Theorem 4.1.3. limα→1− G(α) = −∞, i.e., near 1, there is no bifurcation point.
Proof. The proof is similar to the previous theorem.
Theorem 4.1.4. limα→γ0+ G(α) = +∞, i.e., to the right of γ0 but near γ0 , there is
no bifurcation point.
Proof. Fix > 0 such that
a f 00 (0) b − γ0
,
< min{ ,
, γ0 − a}.
2 J3
4
(4.21)
From (4.21) it follows that f 00 (x) > 0 for any x ∈ (0, ). Let α ∈ (γ0 , γ0 + ). Define
the following integrals
Z
I1 (α) =
[F (α) − F (τ )]3/2
0
Z
I2 (α) =
f (α) − f (τ )
a
f (α) − f (τ )
[F (α) − F (τ )]3/2
dτ,
(4.22)
dτ,
(4.23)
27
γ0 −
Z
f (α) − f (τ )
I3 (α) =
[F (α) − F (τ )]3/2
a
dτ,
(4.24)
dτ,
(4.25)
and
α
f (α) − f (τ )
γ0 −
[F (α) − F (τ )]3/2
Z
I4 (α) =
then
G(α) = F (α)1/2 I1 (α) + I2 (α) + I3 (α) + I4 (α) − 2.
(4.26)
Next, we will study every of these integrals separately. For every τ ∈ [, a], we have
that f (τ ) ≤ 0 and F (a) ≤ F (τ ). Therefore
Z
a
I2 (α) =
Z
a
≥
Z
≥
a
f (α) − f (τ )
[F (α) − F (τ )]3/2
f (α)
[F (α) − F (τ )]3/2
f (α)
dτ
dτ
dτ
[F (α) − F (a)]3/2
f (α)
≥
(a − ).
[F (α) − F (a)]3/2
(4.27)
So
I2 (α) ≥
f (α)
[F (α) − F (a)]3/2
(a − )
(4.28)
Note that
F (α) > 0,
for every γ0 < α.
(4.29)
28
Also, for all τ ∈ [a, γ0 − ], f (τ ) ≤ J0 and F (τ ) ≤ F (γ0 − ). Consequently
γ0 −
Z
I3 (α) =
a
γ0 −
Z
≥
a
γ0 −
Z
≥
f (α) − f (τ )
[F (α) − F (τ )]3/2
−J0
[F (α) − F (τ )]3/2
−J0
dτ
dτ
dτ
[F (α) − F (γ0 − )]3/2
−J0
≥
(γ0 − a − ).
[F (α) − F (γ0 − )]3/2
a
(4.30)
Using Mean Value Theorem, there are c0 , c1 ∈ (τ, α) such that
f (α) − f (τ ) = (α − τ )f 0 (c0 )
(4.31)
F (α) − F (τ ) = (α − τ )f (c1 ).
(4.32)
and
Hence
Z
α
I4 (α) =
f (α) − f (τ )
dτ
[F (α) − F (τ )]3/2
(α − τ )f 0 (c0 )
=
dτ
3/2
γ0 − [(α − τ )f (c1 )]
Z α
f 0 (c0 )
=
dτ
1/2 [f (c )]3/2
1
γ0 − (α − τ )
Z α
f 0 (c0 )
1
=
dτ.
3/2
1/2
f (c1 )
γ0 − (α − τ )
γ0 −
Z α
Let u = (α − τ )1/2 , then
f 0 (c0 )
I4 (α) =
f (c1 )3/2
Z
0
√
−2 du
α−γ0 +
√
α−γ0 +
Z
2f 0 (c0 )
=
du
f (c1 )3/2 0
2f 0 (c0 ) √
=
α − γ0 + .
f (c1 )3/2
(4.33)
29
The Taylor expansion of F around 0 is
F (τ ) = F (0) + τ F 0 (0) +
τ 2 00
τ3
F (0) + F 000 (c3 ),
2
6
(4.34)
where c3 ∈ (0, τ ). Note that F (0) = F 0 (0) = 0. Since τ ∈ [0, ], then f 00 (c3 ) > 0. We
have
F (τ ) =
τ2 0
τ3
f (0) + f 00 (c3 ),
2
6
(4.35)
then
F (α) − F (τ ) = F (α) −
τ2 0
τ3
f (0) − f 00 (c3 )
2
6
< F (α) + τ 2 (−f 0 (0)).
(4.36)
Therefore
Z
I1 (α) =
0
f (α) − f (τ )
3/2
[F (α) − F (τ )]
Z
dτ ≥
0
f (α)
dτ.
[F (α) + τ 2 (−f 0 (0))]3/2
(4.37)
Define
H(α) ≡ F (α)1/2 [
f (α)
(a − )
[F (α) − F (a)]3/2
f (α) − fmax
+
(γ0 − a − )
[F (α) − F (γ0 − )]3/2
2f 0 (c0 )
+
(α − γ0 + )
f (c1 )3/2
Z f (α)
+
dτ ] − 2.
2
0
3/2
0 [F (α) + τ (−f (0))]
From (4.26), (4.27), (4.30), (4.33) and (4.37) we have
G(α) ≥ H(α).
(4.38)
30
On the other hand
lim+ H(α) =
α→γ0
F (γ0 )1/2 f (γ0 )
(a − )
[F (γ0 ) − F (a)]3/2
F (γ0 )1/2 (f (γ0 ) − fmax ) √
γ0 − a − [F (γ0 ) − F (γ0 − )]3/2
2F (γ0 )1/2 f 0 (c0 )
+
f (c1 )3/2
Z dτ
1/2
+ lim+ F (α) f (α)
− 2.
2
0
3/2
α→γ0
0 [F (α) + τ (−f (0))]
+
Define A0 = (−f 0 (0)). Hence
1/2
lim+ F (α)
f (α)
α→γ0
=
0
lim+ F (α)1/2 f (α)
lim+
α→γ0
Let
A0
τ
F (α)
= t, then dτ =
f (α)
lim+
α→γ0 F (α)
Z
0
q
f (α)
F (α)
F (α)
dt.
A0
1
[1 +
τ 2 A0 3/2
]
F (α)
0
Z
0
Z
α→γ0
=
Z
1
dτ
[F (α) + τ 2 A0 ]3/2
1
dτ
2A
0 3/2
F (α)3/2 [1 + τF (α)
]
1
[1 +
τ 2 A0 3/2
]
F (α)
dτ.
Therefore
dτ =
lim+
α→γ0
f (α)
1/2
A0 F (α)1/2
Z
0
q
A0
F (α)
dt
(1 + t2 )3/2
= +∞.
Hence limα→γ0+ H(α) = +∞. Consequently from (4.38) limα→γ0+ G(α) = +∞
Theorem 4.1.5. limα→γ1+ G(α) = +∞, i.e., near γ1+ , there is no bifurcation point.
Proof. The proof is similar to previous theorem.
4.2
Further Analysis of Asymptotic Behavior, Part I.
From the last section, we knew that close to γ0 , b, γ1 and 1, bifurcations can
not occur. In this section, we want to describe four intervals that do not contain
31
bifurcation points. For numerical studies we want to avoid them. The following five
lemmas are preparation for the main result of this section. We will assume through
this section is fixed and it satisfies
3
9J2
+
0 < < min{−
2J3 J3
r
9J22 −J3 f 0 (b) −3f 0 (b) b − x0
+
,
,
}.
4
3
4J2
2
(4.39)
In addition, we define
h
δ = min{
i3/2
F (b) − F (b − )
2 −f 0 (b)
,
, √
11
J2
4 2(1 + a)(1 + b)(1 + c)(b − )(−f 0 (b) +
J2 1/2
)
6
}. (4.40)
We assume throughout this section that α ∈ (b − δ, b).
Lemma 4.2.1. J0 − j0 ≤ 2(1 + a)(1 + b)(1 + c).
Proof. Since f (u) = −u5 + (a + b + c + 1)u4 + (−a − b − c − ab − ac − bc)u3 + (ab +
ac + bc + abc)u2 − abcu, a direct calculation reveals
f (u) ≤ (a + b + c + 1) + (ab + ac + bc + abc)
= (1 + a) + (b + c) + a(b + c) + bc(1 + a)
= (1 + a)(1 + b)(1 + c).
Also
f (u) ≥ −1 − (a + b + c + ab + ac + bc) − abc
= −(1 + a)(1 + b)(1 + c),
so J0 − j0 ≤ 2(1 + a)(1 + b)(1 + c).
Lemma 4.2.2. F (α) − F (α − ) ≥ F (b) − F (b − ).
32
Proof. Define g(x) = F (x) − F (x − ). Then g 0 (x) = f (x) − f (x − ) < 0 for all
x ∈ [b − , b], so g(α) ≥ g(b), therefore F (α) − F (α − ) ≥ F (b) − F (b − ).
0
Lemma 4.2.3. f (α) −
(α−τ ) 00
f (α)
2
+
(α−τ )2 000
f (ᾱ)
6
0
<
f (b)
,
2
for every τ, ᾱ ∈
(α − , α).
¯ between a and b such that
Proof. There exists a number ᾱ
0
0
00
0
00
¯ ) = f (b) − (b − α)f (ᾱ
¯)
f (α) = f (b) + (α − b)f (ᾱ
0
≤ f (b) + (b − α)J2
0
≤ f (b) + J2 .
(4.41)
On the other hand, for any τ, ᾱ ∈ (α − , α)
0
f (α) −
(α − τ ) 00
(α − τ )2 000
(α − τ )
(α − τ )2
0
f (α) +
f (ᾱ) ≤ f (α) +
J2 +
J3
2
6
2
6
2
0
≤ f (α) + J2 + J3 .
(4.42)
2
6
From (4.41) and (4.42), we obtain
(α − τ ) 00
(α − τ )2 000
3J2
J3
0
f (α) +
f (ᾱ) ≤ f (b) +
+ 2 .
2
6
2
6
q
0
9J22
−J3 f 0 (b)
9J2
3
+
, then J3 2 + 9J2 + 3f (b) < 0, therefore
Since 0 < < − 2J3 + J3
4
3
0
f (α) −
0
3J2
J3
f (b)
f (b) +
+ 2 <
.
2
6
2
0
0
So f (α) −
(α−τ ) 00
f (α)
2
+
(α−τ )2 000
f (ᾱ)
6
(4.43)
0
<
f (b)
.
2
0
Lemma 4.2.4. Let A = f (α) and B =
−f (α)
2
+
J2
,
6
then
B
A
> 1.
33
0
Proof. We can write f in the following form
0
0
00
f (α) = f (b) + (α − b)f (b̄),
where b̄ ∈ (α, b). Hence
0
0
f (α) ≤ f (b) + (b − α)J2 .
So
0
f (α) ≤ f 0 (b) + J2 .
From our hypothesis, <
−3f 0 (b)
,
4J2
(4.44)
therefore f 0 (α) < 0. Using (4.44) we obtain
0
0
−f (b) J2
−f (α) J2
+ ) ≥ (
− ).
(
2
6
2
3
(4.45)
Similarly
0
f (α) = f (b) + (α − b)f (b) +
(α − b)2 00 ¯
f (b̄),
2
where ¯b̄ ∈ (α, b). So
(b − α)2 00 ¯
f (b̄)
2
(b − α) 00 ¯
0
= [−f (b) +
f (b̄)](b − α),
2
0
f (α) = −f (b)(b − α) +
consequently
0
f (α) < [−f (b) +
J2
](b − α).
2
(4.46)
0
On other hand, since <
−3f (b)
,
4J2
it implies that
0
f (b)
−J2
<
,
4
3
(4.47)
34
0
f (b)
−J2
<
2
3
(4.48)
and
0
−3f (b)
J2
> .
8
2
(4.49)
Using (4.49), we obtain
0
J2
3f (b)
0
−f (b) + < −f (b) −
.
2
8
0
(4.50)
From (4.47), (4.48) and (4.50) we get
0
0
( −f2(b) − J32 )
( −f2(b) − J32 )
>
(−f 0 (b) − 38 f 0 (b))
(−f 0 (b) + J22 )
2
>
.
11
Therefore
0
−f (b) J2
J2
0
(
− ) > (−f (b) + )(b − α).
2
3
2
(4.51)
From (4.46) and (4.51)
0
−f (b) J2
(
− ) > f (α).
2
3
Using (4.45) and (4.52), we obtain
0
−f (α) J2
(
+ ) > f (α).
2
6
So
B
A
> 1.
0
Lemma 4.2.5. f (α) ≤ −2(b − α)f (b).
(4.52)
35
Proof.
(α − b)2 00 ¯
f ( b0 )
2
(α − b)2 00 ¯
0
= −(b − α)f (b) +
f (b0 )
2
0
f (α) = f (b) + (α − b)f (b) +
(4.53)
where b¯0 ∈ (α, b). Since
0
b−α≤−
2f (b)
,
J2
we get
(α − b)2 00 ¯
0
f (b0 ) ≤ −(b − α)f (b).
2
0
It follows from (4.53) f (α) ≤ −2(b − α)f (b).
Theorem 4.2.6. G(α) < 0 for every α ∈ (b − δ, b).
Proof. Let α ∈ (b − δ, b). Consider, using as in (4.39), the following integrals
α−
Z
I1 (α) =
f (α) − f (τ )
[F (α) − F (τ )]3/2
0
dτ,
(4.54)
and
α
Z
I2 (α) =
f (α) − f (τ )
dτ.
(4.55)
[F (α) − F (τ )]3/2
It follows from (4.1) that G(α) = F (α)1/2 I1 (α) + I2 (α) − 2. For any τ ∈ (0, α − ),
α−
F (τ ) ≤ F (α − ), so from (4.54) we get
Z
α−
I1 (α) =
0
Z
α−
≤
0
Z
≤
0
α−
f (α) − f (τ )
[F (α) − F (τ )]3/2
J0 − j0
[F (α) − F (τ )]3/2
J0 − j0
dτ
dτ
[F (α) − F (α − )]3/2
dτ
36
So
I1 (α) ≤
J0 − j0
[F (α) − F (α − )]3/2
(α − ).
(4.56)
From Lemma 4.2.1, Lemma 4.2.2 and (4.56) we obtain
Z
α−
f (α) − f (τ )
(4.57)
(τ − α)2 00
f (ᾱ)
f (τ ) = f (α) + (τ − α)f (α) +
f (α) +
(τ − α)3 ,
2
6
(4.58)
0
3/2
[F (α) − F (τ )]
dτ ≤
J0 − j0
(b − ).
I1 (α) =
[F (b) − F (b − )]3/2
From Taylor’s formula we get
000
0
and
000
F (τ ) = F (α) + (τ − α)F 0 (α) +
(τ − α)2 00
F (α¯0 )
F (α) +
(τ − α)3 , (4.59)
2
6
where ᾱ, α¯0 ∈ (τ, α). From (4.58), (4.59) and using the fact that F 0 = f and F 00 = f 0 ,
we obtain
000
f (ᾱ)
(α − τ ) 00
0
f (α) +
(τ − α)2
f (α) − f (τ ) = (α − τ ) f (α) −
2
6
(4.60)
00
(α − τ ) 0
f (α¯0 )
2
F (α) − F (τ ) = (α − τ ) f (α) −
f (α) +
(τ − α) .
2
6
(4.61)
and
It follows from Lemma 4.2.3 and (4.60) that
0
(α − τ )f (b)
.
f (α) − f (τ ) ≤
2
(4.62)
Hence, by (4.60), (4.61) and (4.62)
0
f (α) − f (τ )
f (b)
≤
0
[F (α) − F (τ )]3/2
2(α − τ )1/2 f (α) + (α − τ )( −f 2(α) +
f 00 (α¯0 )
(α
6
3/2 .
− τ ))
37
Since
0
00
0
−f (α) f (α¯0 )
−f (α) J2
f (α) + (α − τ )(
+
(α − τ )) ≤ f (α) + (α − τ )(
+ ).
2
6
2
6
then
0
f (α) − f (τ )
f (b)
≤
0
3/2
[F (α) − F (τ )]
2(α − τ )1/2 f (α) + (α − τ )( −f 2(α) +
J2
)]3/2
6
(4.63)
From (4.55) and (4.63) and using the notation given in Lemma (4.2.4), it follows
1 0
I2 (α) ≤
f (b)
2
Z
α
α−
(α −
τ )1/2 [A
dτ
.
+ (α − τ )B]3/2
Let
u=
√
dτ
α − τ , then du = − √
,
2 α−τ
(4.64)
so
Z
0
I2 (α) ≤ f (b)
0
√
√
= f 0 (b)
Z
0
Let u =
q
A
s
B
2
−du
[A + Bu2 ]3/2
du
[A + Bu2 ]3/2
2
then A + Bu = A(1 + s ) and du =
0
f (b)
I2 (α) ≤
AB 1/2
Z √ B
A
0
q
A
ds.
B
ds
[1 + s2 ]3/2
From Lemma 4.2.4 it follows that
Z 1
0
f (b)
ds
I2 (α) ≤
1/2
2 3/2
AB
0 [1 + s ]
0
f (b)
s 1
=
.
AB 1/2 1 + s2 0
Therefore
(4.65)
38
Hence
0
f (b)
I2 (α) ≤ √
.
2AB 1/2
(4.66)
Using Mean Value Theorem, we obtain that
0
0
00
−f (α) = −f (b) − (α − b)f (b̄), where b̄ ∈ (α, b).
0
≤ −f (b) + J2 ,
and consequently
0
0
−f (b) J2
−f (α)
≤
+ .
2
2
2
(4.67)
0
Since <
−3f (b)
,
4J2
we get
0
−f (b) J2
0
+ < −f (b).
2
2
(4.68)
Hence, by (4.67) and (4.68)
J2 1/2
0
B 1/2 < − f (b) + .
6
(4.69)
By Lemma 4.2.5, (4.66) and (4.69) it follows that
1
I2 (α) < − √
2 2(b − α)(−f 0 (b) +
J2 1/2
)
6
.
(4.70)
From Lemma 4.2.1, (4.57) and (4.70) we obtain
I1 (α) + I2 (α) <
2(1 + a)(1 + b)(1 + c)
3/2
[F (b) − F (b − )]
1
(b − ) − √
2 2(b − α)(−f 0 (b) +
Then
G(α) < 0 holds for all α ∈ (b − δ, b).
J2 1/2
)
6
.
39
Theorem 4.2.7. There exists δ > 0 such that G(α) < 0 for every α ∈ (1 − δ, 1).
Proof. The proof is similar to previous theorem.
4.3
Further Analysis of Asymptotic Behavior, Part II.
The next Theorem will study the asymptotic behavior of G(α) as α gets close
to γ0 from the right. Fix such that
0 < < min{
f 00 (0) f (γ0 ) γ0 − a a b − a
,
,
, ,
}.
J3
4J1
2
4 2
(4.71)
We will assume through this section, as before, that A0 = −f 0 (0). In addition, we
define
n A 2
0
δ = min ,
,
J0
16A0 J0
f (γ0 )2
√
2F (b)J0
(b
(f (γ0 ))3/2
o
√
− a − ) +
8
F (b)J1
f (γ0 )3/2
+1
2
.
(4.72)
We start with the following lemma that is needed for the proof of the Theorem of
this section.
Lemma 4.3.1. For all x ∈ (0, ), f 00 (x) > 0.
Proof. It follows from (4.71).
Theorem 4.3.2. G(α) > 0 for all α ∈ (γ0 , γ0 + δ).
Proof. Let α ∈ (γ0 , γ0 + δ), therefore α − γ0 <
Z
I1 (α) =
I2 (α) =
and α − γ0 < . Define
f (α) − f (τ )
[F (α) − F (τ )]3/2
0
Z
A0 2
J0
a
f (α) − f (τ )
[F (α) − F (τ )]3/2
dτ,
(4.73)
dτ,
(4.74)
40
α−
Z
I3 (α) =
f (α) − f (τ )
α
f (α) − f (τ )
α−
[F (α) − F (τ )]3/2
Z
dτ,
(4.75)
dτ.
(4.76)
[F (α) − F (τ )]3/2
a
I4 (α) =
There exist y1 , y2 ∈ (γ0 , α), and y3 ∈ (α − , α) such that
f (α) = f (γ0 ) + f 0 (y2 )(α − γ0 ),
(4.77)
F (α − ) = F (α) − F 0 (y3 ),
(4.78)
F (α) = F (γ0 ) + (α − γ0 )F 0 (y1 ),
(4.79)
and in addition for τ ∈ (0, ), there is y ∈ (0, τ ) ⊂ (0, ) such that
F (τ ) = F (0) + τ F 0 (0) +
τ 2 00
τ3
F (0) + F 000 (y).
2
6
(4.80)
We also note that F (γ0 ) = 0, F 0 (y1 ) = f (y1 ) and (α − γ0 )J0 < A0 2 .
From (4.79) follows
F (α) = (α − γ0 )f (y1 )
≤ (α − γ0 )J0
≤ A0 2
(4.81)
Since F (0) = F 0 (0) = 0, F 00 (0) = f 0 (0) and F 000 (y) = f 00 (y), it follows from (4.80)
F (α) − F (τ ) = F (α) −
τ2 0
τ3
f (0) − f 00 (y).
2
6
(4.82)
From Lemma 4.3.1 and (4.82) we get that for any τ ∈ (0, )
F (α) − F (τ ) ≤ F (α) − τ 2 f 0 (0),
(4.83)
41
so
1
1
≥
.
F (α) − F (τ )
F (α) − τ 2 f 0 (0)
(4.84)
From (4.84) we obtain
Z
I1 (α) =
0
Z
≥
f (α) − f (τ )
[F (α) − F (τ )]3/2
f (α)
dτ
dτ
[F (α) − τ 2 f 0 (0)]3/2
Z f (α)
dτ
=
h
i3/2 .
F (α)3/2 0
A0
2
1 + F (α) τ
0
Let
q
A0
τ
F (α)
= t, then dτ =
q
F (α)
dt.
A0
Therefore
q
I1 (α) ≥
(4.85)
f (α)
Z
1/2
A0 F (α)
A0
F (α)
0
dt
.
(1 + t2 )3/2
(4.86)
From (4.81) and (4.86) we get
I1 (α) ≥
Z
f (α)
1/2
A0 F (α)
0
1
dt
,
(1 + t2 )3/2
(4.87)
from which follows
f (α)
I1 (α) ≥ √
.
2A0 F (α)
Also, from (4.77), f (α) ≥ f (γ0 ) − J1 . Since <
f (α) >
f (γ0 )
,
2J1
f (γ0 )
.
2
(4.88)
this implies
(4.89)
Using (4.81), we have
F (α) ≤ (α − γ0 )J0 .
(4.90)
42
Consequently
f (γ0 )
F (α)1/2 I1 (α) > √
.
2 2A0 J0 (α − γ0 )1/2
(4.91)
Note that on [, a], f (τ ) ≤ 0 and F (a) ≤ F (τ ). Therefore we have
a
Z
I2 (α) =
a
Z
≥
a
Z
≥
f (α) − f (τ )
[F (α) − F (τ )]3/2
f (α)
[F (α) − F (τ )]3/2
f (α)
dτ
dτ
dτ
[F (α) − F (a)]3/2
f (α)
≥
(a − ).
[F (α) − F (a)]3/2
(4.92)
So
F (α)1/2 I2 (α) ≥ 0.
(4.93)
F (α − ) = F (α) − f (y3 ).
(4.94)
By (4.78)
Similarly, f (y3 ) = f (γ0 ) + f 0 (y4 )(y3 − γ0 ), where y4 ∈ (min{y3 , γ0 }, max{y3 , γ0 }). So
f (y3 ) ≥ f (γ0 ) − J1 >
f (γ0 )
,
2
(4.95)
therefore
F (α) − F (α − ) >
f (γ0 )
.
2
(4.96)
But
Z
I3 (α) =
a
α−
f (α) − f (τ )
[F (α) − F (τ )]3/2
dτ,
43
then
−J0
(α − − a)
[F (α) − F (α − )]3/2
√
−2 2J0
(b − a − ).
≥
f (γ0 )3/2 3/2
I3 (α) ≥
(4.97)
Since F (α)1/2 ≤ F (b)1/2 , (4.97) implies that
1/2
F (α)
√
−2 2J0 F (b)1/2
I3 (α) ≥
(b − a − ).
f (γ0 )3/2 3/2
(4.98)
On the other hand, by definition of I4 (α)
α
f (α) − f (τ )
α−
[F (α) − F (τ )]3/2
Z
I4 (α) =
dτ.
Now
f (α) − f (τ ) = (α − τ )f 0 (y5 )
(4.99)
F (α) − F (τ ) = (α − τ )f (y6 )
(4.100)
and
where y5 , y6 ∈ (τ, α) ⊂ (α − , α). Note that for any x ∈ (α − , α)
f (α) ≤ f (x) + J1 .
(4.101)
From (4.71), (4.89) and (4.101) we get
f (x) >
f (γ0 )
.
4
(4.102)
44
Hence
α
f 0 (y5 )
dτ
1/2 f (y )3/2
6
α− (α − τ )
√
2 f 0 (y5 )
=
f (y6 )3/2
√
−2 J1
≥
f (y6 )3/2
√
−16 J1
.
>
f (γ0 )3/2
Z
I4 (α) =
Then we have
1/2
F (α)
√
−16 F (b)1/2 J1
I4 (α) >
.
f (γ0 )3/2
(4.103)
We conclude from (4.91), (4.98) and (4.103) that
p
p
2 2F (b)J0
16 F (b)J1
f (γ0 )
G(α) > √
−
(b − a − ) −
−2
(f (γ0 ))3/2
f (γ0 )3/2
2 2A0 J0 (α − γ0 )1/2
From (4.72) G(α) > 0.
Theorem 4.3.3. There exists δ > 0 such that G(α) > 0 for every α ∈ (γ1 , γ1 + δ).
Proof. The proof is similar to previous theorem.
45
CHAPTER 5
LOCATION OF BIFURCATIONS
5.1
Applications of Theorem 2.2.8
Inspired by Theorem 2.2.8, we conjecture that there are two values λ0 , λ1 , of
the parameter λ for which one of the following holds:
Case 1: If λ0 6= λ1 , then problem (3.1) has exactly one positive solution for 0 <
λ = min{λ0 , λ1 }, exactly two positive solutions for min{λ0 , λ1 } < λ < max{λ0 , λ1 },
exactly three positive solutions for λ = max{λ0 , λ1 }, and exactly four nontrivial
positive solutions for λ > max{λ0 , λ1 }.
Case 2: If λ0 = λ1 , the problem (3.1) has exactly one positive solution for 0 < λ =
λ0 = λ1 , and exactly four nontrivial positive solutions for λ > λ0 = λ1 .
Moreover, we conjecture as well that all solutions lie on two smooth curves which
are parabola-like, having exactly one turn to the right on such curve. The first curve
Γ0 , the lower curve, has a turning point at λ0 with limλ→∞,Γ+0 u(0, λ) = b while
limλ→∞,Γ−0 u(0, λ) = γ0 . The second curve Γ1 , the upper curve, has a turning point at
λ1 with limλ→∞,Γ+1 u(0, λ) = 1 while limλ→∞,Γ−1 u(0, λ) = γ1 , where γ0 is the unique
root of
R γ0
0
f (s) ds = 0 in (a, b) and γ1 is the unique root of
5.2
R γ1
b
f (s) ds = 0 in (c, 1).
General Behaviour of Solutions
We begin by deriving some lemmas. By our assumptions on the function f , it
is clear that there are exactly three points where a ray starting at the origin touches
the graph of f (u). We denote the first point by β1 , i.e. β1 ∈ (a, b) is the first solution
46
of equation
f 0 (β) =
f (β)
.
β
(5.1)
We recall, from the analysis in [10], that turning (or singular) points of (2.1) can
occur only when (2.4) has a nontrivial solution w(x). Furthermore in such a case we
can choose w(x) to be strictly positive on (−1, 1).
The next lemma was proved in [12] for the cubic case, f (u) = u(u − a)(u − b).
Fortunately, the same proof applies for (3.1), which we present for completeness.
Lemma 5.2.1. Let u(x) be any critical point of (3.1), 0 < u(x) < b. Then
u(0) > β1 .
(5.2)
Proof. We will show that if u(0) ≤ β1 , then the only solution of (2.4) is w ≡ 0. First,
we claim that
f 0 (u) >
f (u)
for 0 < u < β1 .
u
(5.3)
Indeed, denote p(u) = uf 0 (u) − f (u). Then p(0) = p(β1 ) = 0, and p0 (u) = uf 00 (u).
It follows that p0 (u) > 0 near u = 0, and p0 (u) < 0 near u = β1 . Since p(u) has no
roots in (0, β1 ) (since solution of (5.1) is unique) it follows that p(u) > 0 on (0, β1 ),
establishing (5.3). We now rewrite (2.1) in the form
u00 + λ
f (u)
u = 0.
u
Using the Sturm comparison theorem and (5.3), we conclude that (2.4) cannot have
a positive solution w(x). (By (5.3) any solution of (2.4) would have to vanish on
47
(−1, 1)). Since any nontrivial solution of (2.4) has to be positive and even, the proof
is complete.
Suppose that the function f (u) is concave up at b, that is f 00 (b) ≥ 0. It is clear
that there is exactly one point where a ray out of the point b touches to the right the
graph of f (u). We denote this point by β2 , i.e. β2 is the only solution of equation
f 0 (β) =
f (β)
, for β > b.
β−b
Theorem 5.2.2. Suppose that f 00 (b) ≥ 0, and let u(x) be any singular solution of
(3.1), and that γ1 < u(0) < 1. Then u(0) > β2 .
Proof. Suppose that u(0) ≤ β2 . We shall reach a contradiction. Note that f 0 (u) >
f (u)
u−b
for all b < u ≤ β2 . Also,
0 = u00 + λf (u)
λf (u)
(u − b)
u−b
λf (u)
λf (u)
= u00 +
u−
b.
u−b
u−b
= u00 +
So,
u00 +
λf (u)
λf (u)
u=
b.
u−b
u−b
(5.4)
We define,
r1 (x) = λf 0 (u),
r2 (x) =
λf (u)
,
u−b
(5.5)
(5.6)
48
and
g(x) = λ
f (u)
b.
u−b
(5.7)
By (5.4), (5.6) and (5.7) we have
u00 + r2 (x)u = g(x)
(5.8)
and, since the linearized must have positive solution w, we have that
w00 + r1 (x)w = 0.
(5.9)
We define the new dependent variable
ν=
u−b
.
w
(5.10)
So, u − b = νw. A simple computation shows that
u00 = ν 00 w + 2ν 0 w0 + νw00 .
(5.11)
Remember that by Lemma 2.8, if the problem (2.4) admits a nontrivial solution w,
then it does not change sign, i.e. we may assume that w(x) > 0 on (−1, 1). We get
ν 00 +
2w0 0 w00
u00
ν +
ν= .
w
w
w
(5.12)
on (−1, 1).
From (5.8)
u00 = g(x) − r2 (x)u.
(5.13)
Combining (5.12) and (5.13), we get
ν 00 +
2w0 0 w00
g(x) − r2 (x)u
ν +
ν=
.
w
w
w
(5.14)
49
We find then that
(u)
− λfu−b
u
w
λf (u) (u − b)
= −
u−b w
g(x) − r2 (x)u
=
w
λf (u)
b
u−b
so we can write
g(x) − r2 (x)u
= −r2 (x)ν.
w
(5.15)
As before, dividing (5.9) by the positive quantity w, we get
w00
= −r1 (x).
w
(5.16)
Replacing (5.15) and (5.16) into (5.14), we obtain
2w0 0
ν − r1 (x)ν = −r2 (x)ν
w
(5.17)
2w0 0
ν + (r2 (x) − r1 (x))ν = 0.
w
(5.18)
ν 00 +
therefore,
ν 00 +
Note that r2 (x) < r1 (x), since f 0 (u) >
f (u)
.
u−b
Choose x0 such that u(x0 ) = b. If x0 = 1, then u(x0 ) = u(1) = 0 = b, a contradiction.
Similarly x0 6= −1, so x0 ∈ (−1, 1).
Consider the equation
ν 00 +
2w0 0
ν + (r2 (x) − r1 (x))ν = 0 on [−x0 , x0 ].
w
ν is a even function, since ν(−x0 ) =
u(−x0 )−b
w(−x0 )
(5.19)
= ν(x0 ).
Since ν is continuous on [−x0 , x0 ], ν attains a maximum value at some number d in
50
[−x0 , x0 ]. Let ν(d) = max{ν(x) : x ∈ [−x0 , x0 ]}. Then there are two possibilities for
d:
1. if d = x0 , in such case ν(d) = 0, a contradiction, since ν(0) =
u(0)−b
w(0)
> 0.
2. if d ∈ (−x0 , x0 ), then ν 00 (d) ≤ 0 and ν 0 (d) = 0, so
0 = ν 00 (d) +
2w0 (d) 0
ν (d) + (r2 (d) − r1 (d))ν
w(d)
= ν 00 (d) + (r2 (d) − r1 (d))ν
< 0,
a contradiction. This finishes the proof of the theorem.
Also, note that the function f (u) could be, for example, concave up at b0 with
b0 < b. Let b∗ = max{b, b0 }. It is clear that there is exactly one point where a ray
out of the point b∗ touches to the right the graph of f (u). We denote this point by
β3 , i.e., β3 ∈ (c, 1) is the only solution of equation
f 0 (β) =
f (β) − f (b∗ )
for β > b∗ .
β − b∗
Theorem 5.2.3. Let u(x) be any singular solution of (3.1) with 0 < u(0) < 1.
Suppose that f 00 (b∗ ) ≥ 0, then
u(0) > β3 .
(5.20)
Proof. Suppose that u(0) ≤ β3 . We shall reach a contradiction. Note that f 0 (u) >
f (u)−f (b∗ )
u−b∗
for all b∗ < u ≤ β3 . Each step that follows has a straightforward analog in
the setting of Theorem ??.
51
Theorem 5.2.4. Suppose that α = u(0) ∈ (γ0 , b). Let λ0 =
γ02
Rb
a
f (s) ds
. If 0 ≤ λ ≤ λ0 ,
then the problem (3.1) does not possess any positive solutions.
Proof. Suppose that (3.1) has a positive solution for some λ, 0 < λ ≤ λ0 , then
1 0
[u (s)]2 + λ
2
Z
u(s)
f (x) dx = 0.
u(0)
Thus,
1 0
[u (s)]2 = −
2λ
Define =
Rb
a
Z
u(s)
Z
u(0)
f (x) dx ≤
f (x) dx =
u(0)
Z
u(s)
b
f (x) dx.
a
1
f (s) ds. Using the last equation, we obtain [u0 (s)] ≤ (2λ) 2 . Solving
this last inequality and using the fact that u0 (s) < 0 on (0, 1), we obtain
1
u0 (s) ≥ −(2λ) 2 .
(5.21)
Integrating (5.21),
Z
1
1
u0 (s) ds ≥ −(2λ) 2 .
(5.22)
0
Consequently,
1
u(1) − u(0) ≥ −(2λ) 2 .
(5.23)
So,
λ≥
u(0)2
γ2
> 0 = λ0 ,
2
2
(5.24)
and therefore λ > λ0 . This contradicts the assumption 0 ≤ λ ≤ λ0 .
Theorem 5.2.5. Assume that α = u(0) ∈ (γ1 , 1). Let λ1 =
Rb
a
γ12
R
.
f (s) ds+ c1 f (s) ds
0 ≤ λ ≤ λ1 , then problem (3.1) does not possess any positive solutions.
If
52
Proof. Suppose that (3.1) has a positive solution for some λ, 0 < λ ≤ λ1 , then
1 0
[u (s)]2 + λ
2
Z
u(s)
f (x) dx = 0.
u(0)
Thus,
1 0
[u (s)]2 = −
2λ
Define =
Rb
a
Z
u(s)
Z
u(0)
f (x) dx ≤
f (x) dx =
u(0)
f (s) ds+
Z
u(s)
R1
c
b
Z
f (x) dx +
a
1
f (s) ds.
c
1
f (s) ds. Using the last equation, we obtain [u0 (s)] ≤ (2λ) 2 .
Solving this last inequality and using the fact that u0 (s) < 0 on (0, 1), we obtain
1
u0 (s) ≥ −(2λ) 2 .
(5.25)
Integrating (5.25),
Z
1
1
u0 (s) ds ≥ −(2λ) 2 .
(5.26)
0
Consequently,
1
u(1) − u(0) ≥ −(2λ) 2 .
(5.27)
So,
λ≥
u(0)2
γ2
> 1 = λ1 ,
2
2
(5.28)
therefore λ > λ1 . This contradicts the assumption 0 ≤ λ ≤ λ1 .
Lemma 5.2.6. Consider α = u(0) ∈ (γ0 , b), and let λ0 =
γ0
M
and M = max{|f (u)| :
0 < u < b}. If 0 ≤ λ ≤ λ0 , then problem (3.1) does not possess any positive solutions.
53
Proof. u satisfies (3.1) if and only if u satisfies the integral equation
Z
1
G(x, y)f (u(y)) dy
u(x) = λ
(5.29)
−1
where G is the Green’s function for (3.1), i.e.,




 21 (x + 1)(1 − y), −1 ≤ x ≤ y ≤ 1,
G(x, y) =



 1 (y + 1)(1 − x), −1 ≤ y ≤ x ≤ 1,
2
from which G(x, y) ≥ 0 on [−1, 1] × [−1, 1] and
R1
−1
(5.30)
G(x, y) dy ≤ 1.
Hence
Z
1
|u(x)| ≤ λ
G(x, y)|f (u(y))| dy
−1
Z
1
≤ λ max{|f (u)| : 0 ≤ u ≤ b}
G(x, y) dy
−1
≤ λM.
Hence, if γ0 < u(0) < b, then
γ0 ≤ |u(x)|
≤ λM.
so,
λ>
γ0
.
M
Lemma 5.2.7. Suppose that α = u(0) ∈ (γ1 , 1), and let λ1 =
γ1
,
k
and M =
max{|f (u)| : 0 < u < 1}. If 0 ≤ λ ≤ λ1 , then problem (3.1) does not possess
any positive solutions.
54
Under our assumptions on f , there is a constant µ > 0, such that f (u) ≤ µu
for all u > 0.
Lemma 5.2.8. If u is a non-trivial solution of (3.1), then λ ≥
π2
4µ
Proof.
1
1
1
π2
λµ
u dx ≥ λ
f (u)udu =
u dx ≥
4
−1
−1
−1
Z
so λ ≥
Z
2
Z
02
Z
1
u2 dx,
−1
π2
.
4µ
Lemma 5.2.9. If α = u(0) ∈ (γ0 , b), then problem (3.1) has no non-trivial solutions
γ2
2
γ0
π
0
,M
, R b f (s)
for 0 ≤ λ < max{ 4µ
}.
ds
a
Lemma 5.2.10. If α = u(0) ∈ (γ1 , 1), then problem (3.1) has no non-trivial solutions
γ2
2
γ1
π
,M
, R b f (s) ds+1R 1 f (s) ds }.
for 0 ≤ λ < max{ 4µ
a
c
5.3
Some generalizations
We shall be interested in the following Dirichlet boundary problem,
u00 + u(u − b1 )(u − b2 )...(u − b2n−1 )(1 − u) = 0,
x ∈ (−1, 1),
u(−1) = u(1) = 0,
(5.31)
with constants 0 = b0 < b1 < b2 < ...... < b2n−1 < b2n = 1, a positive parameter
λ, and n ≥ 1. The polynomial f (u) = u(u − b1 )(u − b2 ).....(u − b2n−1 )(1 − u) has
a negative hump over (0, b1 ) followed by a positive hump (b1 , b2 ). We shall refer to
them as a pair of humps. This polynomial has n pairs of humps. It is well known that
each solution branch has its maximum value inside a single positive hump, and that
55
it is necessary to have
R b2
0
f (u) du > 0 in order for solutions with maximum value in
(b1 , b2 ) to exist.We shall assume this condition to hold, and similarly for the other
humps, otherwise we can combine pairs of humps.
To use the bifurcation theory approach, we need to linearize (5.31). Consider
the linearized problem for (5.31),
w00 (x) + λf 0 (u(x))w = 0 on (−1, 1), w(−1) = w(1) = 0,
(5.32)
where u(x) is a solution of (5.31). If w(x) is a nontrivial solution of (5.32), then we
call u(x) a singular solution of (5.31). If w(x) ≡ 0 is the only solution of (5.32), we
say that the solution u(x) is nonsingular. In this section, we generalize some results
from the previous section for the problem (5.31). We may repeat each proof with
some modifications to conclude each of the followings results.
Lemma 5.3.1. Any nontrivial positive solution of (3.1) satisfies u(0) ∈
Fn
k=1 (γk−1 , b2k ),
n ≥ 1.
Lemma 5.3.2. if u(0) ∈ (b2k−1 , b2k ), then γk−1 < u(0) < b2k , k = 1, 2, ..., n, n ≥ 1.
Suppose that the function f (u) is concave up at b2k−2 , that is f 00 (b2k−2 ) = 0,
for some k = 1, 2, ...., n, n ≥ 1 It is clear that there is exactly one point where a ray
out of the point b2k−2 touches to the right the graph of f (u). We denote this point
by βk , i.e. βk is the only solution of equation
f 0 (β) =
f (β)
.
β − b2k−2
56
Theorem 5.3.3. Let u(x) be any critical point of (5.32) with u(x) ∈ (0, b2k ), and
suppose that f 00 (b2k−2 ) ≤ 0. Then
u(0) > βk .
From our assumptions there is γk−1 ∈ (b2k−1 , b2k ) that satisfies
R γk−1
b2k−2
f (s) ds =
0, n ≥ 1, k = 1, 2, ..., n.
Theorem 5.3.4. limα→b− G(α) = −∞, i.e., near b2k , there is no bifurcation point,
2k
for all n ≥ 1, k = 1, 2, ..., n.
+
Theorem 5.3.5. limα→γ + G(α) = +∞, i.e., near γk−1
, there is no bifurcation point,
k−1
for all n ≥ 1, k = 1, 2, ..., n.
Lemma 5.3.6. Suppose that α = u(0) ∈ (b2k−1 , b2k ) and γk−1 ∈ (b2k−1 , b2k ). Define
λk−1 =
Pk
2
γk−1
R b2j
j=1 b2j−1
f (s) ds
. If 0 ≤ λ ≤ λk−1 , then problem (3.1) does not possess any
positive solutions.
Lemma 5.3.7. Assume α = u(0) ∈ (γk , b2k ) and define λk =
γk
,
M
where M =
max{|f (u)| : 0 < u < b2k }, n ≥ 1, k = 1, 2, ..., n. If 0 ≤ λ ≤ λk , then problem
(5.31) does not possess any positive solutions.
Lemma 5.3.8. Suppose α = u(0) ∈ (γk , b2k ), then the problem (5.31) possesses
2
γk
π
,M
, Pk
no non-trivial solutions for 0 ≤ λ < max{ 4µ
γk2
R b2j
j=1 b2j−1
max{|f (u)| : 0 < u < b2k }, n ≥ 1, k = 1, 2, ..., n.
f (s) ds
}, where M =
57
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