# Boundary Value Problems for a Super-Sublinear Asymmetric Oscillator: The Exact Number of Solutions

Authors: Armands Gritsans; Felix Sadyrbaev
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

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## Abstract

Properties of asymmetric oscillator described by the equationx′′=−λ(x+)p+μ(x−)q (i), where p≥1 and 0<q≤1, are studied. A set of (λ,μ) such that the problem (i), x(0)=0=x(1) (ii), and |x′(0)|=α (iii) have a nontrivial solution, is called α-spectrum. We give full description of α-spectra in terms of solution sets and solution surfaces. The exact number of nontrivial solutions of the two-parameter Dirichlet boundary value problem (i), and (ii) is given.


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## Body

## 1. Introduction

Asymmetric oscillators were studied intensively starting from the works by Kufner and Fučík; see [1] and references therein. Simple equations like (2) given with the boundary conditions allow for complete investigation of spectra. It is known that the spectrum of the problem (2), (4) is a set of hyperbola looking curves in the (λ,μ)-plane. On the other hand, there is a plenty of works devoted to one-parameter case of equations x′′+λf(x)=0 given together with the two-point boundary conditions. Due to nonlinearity of f one should consider solutions x(t;α) with different α=x′(0). Bifurcation diagrams in terms of α and λ, or ∥x∥ and λ, can serve then to evaluate the number of solutions [2–4].In this paper we consider differential equations of the form(1)x′′=-λf(x+)+μg(x-),
where f=xp,p≥1, and g=xq,0<q≤1. Here λ and μ are nonnegative parameters, x+=max{x,0}, x-=max{-x,0}. This equation describes asymmetric oscillator with different nonlinear restoring forces on both sides of x=0. If f=g=x, then equation becomes famous Fučík equation
(2)x′′=-λx++μx-.
Properties of the Fučík spectrum are well known (the Fučík spectrum is a set of all pairs (λ,μ) where λ,μ≥0, such that the Dirichlet problem—(2) with boundary conditions x(0)=0=x(1)—has a non-trivial solution).The aim of our study in this paper is to describe properties of the spectrum of the problem(3)x′′=-λ(x+)p+μ(x-)q,0<q≤1,1≤p,(4)x(0)=0,x(1)=0.For this we study first the time maps for the related functions (Section2), then we give the analytical description of the spectrum (Section 3), and formulate the properties of the spectrum (Section 4), including the asymptotics. In Section 5 we consider the solution sets and solution surfaces which bear information on multiplicity of solutions to the problem. Analysis of properties of solution surfaces (Section 6) can give us estimations of the number of solutions to the problem (Section 7). These estimations contain also information of properties of solutions such as the number of zeros and evaluations of x′(0).This paper continues series of publications by the authors devoted to nonlinear asymmetric oscillations [5–8].


## 2. Time Maps

Consider the Cauchy problem(5)x′′=-λxr,x(0)=0,x′(0)=α,
where r>0. Solutions of this problem for α and λ positive have a zero. This zero will be denoted Tr(α,λ) and called time map function; more on time maps can be found in [9, 10].Proposition 1.
Supposer,α,λ>0, p>1 and 0<q<1. (1) For anyα,λ>0 the formula is valid:
(6)Tr(α,λ)=1λtr(αλ),wheretr(γ)=Tr(γ,1).(2) The functionTr(α,λ) for the problem (5) is
(7)Tr(α,λ)=2r/(r+1)(r+1)1/(r+1)A(r)α-(r-1)/(r+1)λ-1/(r+1),whereA(r)=∫01(ds/1-sr+1).(3) The functionTr(α,λ) for fixed r and α is strictly decreasing function of λ and possesses the properties
(8)limλ→0+Tr(α,λ)=+∞,limλ→+∞Tr(α,λ)=0.(4) The functionTp(α,λ) for fixed p and λ is decreasing function of α.(5) The functionTq(α,λ) for fixed q and λ is increasing function of α.Proof.
By standard computations.Remark 2.
T
r
(
α
,
λ
)
=
π
/
λ for r=1, irrespective of α.


## 3. Spectrum

Suppose that the problems (3) and (4) are considered with the additional condition
(9)|x′(0)|=α>0.Definition 3.
For givenα>0 a set of all nonnegative (λ,μ) for which the problems (3), (4), and (9) have a nontrivial solution is called α-spectrum.Remark 4.
A solution of (3) is a C2-function. Therefore x′(t) is continuous. If z1 and z2 are two consecutive zeros of x(t), then |x′(z1)|=|x′(z2)| since a solution in the interval (z1,z2) is symmetric with respect to the middle point. Therefore |x′(z)|=α for any zero point z and signs of x′(t) alternate.Theorem 5.
Theα-spectrum for the problem
(10)x′′=-λ(x+)p+μ(x-)q,x(0)=0,x(1)=0,|x′(0)|=α>0
consists of the following α-branches:
(11)F0+(α)={(λ,μ):Tp(α,λ)=1,μ≥0},F0-(α)={(λ,μ):λ≥0,Tq(α,μ)=1},F2i-1+(α)={(λ;μ):iTp(α,λ)+iTq(α,μ)=1},F2i-1-(α)={(λ;μ):iTq(α,λ)+iTp(α,μ)=1},F2i+(α)={(λ;μ):(i+1)Tp(α,λ)+iTq(α,μ)=1},F2i-(α)={(λ;μ):(i+1)Tq(α,μ)+iTp(α,λ)=1}.
The notation Fi+(α), respectively: Fi-(α), refers to solutions x which satisfy the initial conditions x(0)=0,x′(0)=α>0 (resp.: x′(0)=-α<0) and have exactly i zeros in the interval (0,1).Proof.
We prove the theorem only for solutions which have exactly one zero in(0,1) and satisfy the initial condition |x′(0)|=α>0.
Consider the casex′(0)=α>0. The first zero of x(t) appears at t=Tp and x′(Tp)=-α. The second zero is at t=Tp+Tq. One has that for solutions with exactly one zero in (0,1) the relation Tp+Tq=1 holds, which defines the branch F1+(α).
Supposex′(0)=-α. The first zero now is at t=Tq. The second one is at t=Tq+Tp and therefore Tq+Tp=1.
The branchesF1+(α) and F1-(α) are given by the equivalent relations Tp+Tq=1 and Tq+Tp=1, respectively, and therefore coincide.
Proof for solutions with different nodal structure is similar.Remark 6.
Ifp=q=1, then for any α>0 the α-spectrum is
(12)F0+(α)={(λ,μ):πλ=1,μ≥0},F0-(α)={(λ,μ):λ≥0,πμ=1},F2i-1+(α)={(λ;μ):iπλ+iπμ=1},F2i-1-(α)={(λ;μ):iπμ+iπλ=1},F2i+(α)={(λ;μ):(i+1)πλ+iπμ=1},F2i-(α)={(λ;μ):(i+1)πμ+iπλ=1}.
It is classical Fučík spectrum for the problems (2) and (4), see [1].


## 4. Properties of theα-Spectrum

Proposition 7.
(1)
The branchesF2i-1+(α) and F2i-1-(α) coincide.
(2)
The branchesF2i-1±(α) and F2i-1±(α) do not intersect unless i≠j.
(3)
The branchesF2i+(α) and F2j+(α) do not intersect unless i≠j.
(4)
The branchesF2i-(α) and F2j-(α) do not intersect unless i≠j.
(5)
Any branchFis(α) where i≥1 and s is either “+” or “−” is a graph of monotonically decreasing function μ=μ(λ).
(6)
The branchesF2i+(α) and F2i-(α) intersect once.Proof.
(
1
) It follows from (11) that F2i-1+(α) coincides with F2i-1-(α), since both sets (branches) are defined by symmetric relations:
(13)iTp(α,λ)+iTq(α,μ)=1,iTq(α,μ)+iTp(α,λ)=1.(2), (3) and (4) follows from (11), but (5) follows from the relations (11) and (7). (6) Indeed, any point of intersection satisfies the system
(14)(i+1)Tp(α,λ)+iTq(α,μ)=1,(i+1)Tq(α,μ)+iTp(α,λ)=1.
It follows that
(15)Tp(α,λ)=Tq(α,μ)
or
(16)2p/(p+1)(p+1)1/(p+1)A(p)α-(p-1)/(p+1)λ-1/(p+1)=2q/(q+1)(q+1)1/(q+1)A(q)α-(q-1)/(q+1)μ-1/(q+1).
The above relation defines a curve which is a graph of monotonically increasing function μ=C(p,q,α)λ(q+1)/(p+1), where C(p,q,α) is a constant computable from (16). This curve emanates from the origin and intersects the graph of monotonically decreasing from +∞ to 0 function μ=μ(λ) from (5) only once.Letλi(α,p) be a unique solution of the equation Tp(α,λ)=1/i; similarly, let μi(α,q) be a unique solution of the equation Tq(α,μ)=1/i.Proposition 8.
Supposeα,λ,μ>0.  (1) The branchF2i-1±(α)(i∈ℕ) is located in the sector
(17){(λ,μ):λ>λi(α,p),μ>μi(α,q)}and is a hyperbola looking curve in(λ,μ) plane with vertical asymptote λ=λi(α,p) and horizontal asymptote μ=μi(α,q).(2) The branchF2i+(α)(i∈ℕ) is located in the sector
(18){(λ,μ):λ>λi-1(α,p),μ>μi(α,q)}and is a hyperbola looking curve in(λ,μ) plane with vertical asymptote λ=λi-1(α,p) and horizontal asymptote μ=μi(α,q).(3) The branchF2i-(α)(i∈ℕ) is located in the sector
(19){(λ,μ):λ>λi(α,p),μ>μi-1(α,q)}and is a hyperbola looking curve in(λ,μ) plane with vertical asymptote λ=λi(α,p) and horizontal asymptote μ=μi-1(α,q).Proof.
Follows from the relations (11) and Proposition 7.Remark 9.
So the positive part of theα-spectrum
(20)F+(α)=⋃i=0+∞‍Fi+(α)
in the extended (λ,μ)-plane may be schematically described by the chain
(21)(λ1,0)→F0+(α)(λ1,+∞)→F1+(α)(+∞,μ1)→F2+(α)(λ2,+∞)→F3+(α)(+∞,μ3)→F4+(α)(λ3,+∞)⋯.
Similarly, the negative part of the α-spectrum
(22)F-(α)=⋃i=0+∞‍Fi-(α)
in the extended (λ,μ)-plane may be described as follows:
(23)(0,μ1)→F0-(α)(+∞,μ1)→F1-(α)(λ1,+∞)→F2-(α)(+∞,μ2)→F3-(α)(λ2,+∞)→F4-(α)(+∞,μ3)⋯.Proposition 10.
Letp>1 and 0<q<1 be fixed. It is true that
(24)limα→+0λ1(α,p)=+∞,limα→+∞λ1(α,p)=+0,limα→+0μ1(α,p)=+0,limα→+∞μ1(α,p)=+∞.Proof.
Follows from Propositions1 and 10.


## 5. Solution Sets and Solution Surfaces

Definition 11.
A solution set of the problems (3) and (4) is a set F of all triples (λ,μ,α)  (λ≥0, μ≥0,α>0) such that there exists a nontrivial solution of the problem.Let us distinguish between solutions of the problems (3) and (4) with different number of zeros in the interval (0,1). Let Fi+ be a set of all triples (λ,μ,α) such that there exists a nontrivial solution of the respective problems (3) and (4), x′(0)=α>0 which has exactly i zeros in (0,1),i=0,1,…, but Fi- be a set of all triples (λ,μ,α) such that there exists a nontrivial solution of the respective problems (3) and (4), x′(0)=-α<0 which has exactly i zeros in (0,1),i=0,1,…Definition 12.
F
i
+ will be called a positive i-solution surface, but Fi--a negative i-solution surface.


### 5.1. Description and Properties of a Solution Set

We will identify the cross section of a solution surfaceFis with the plane α=α0>0 (α0 is fixed) with its projection to the (λ,μ)-plane. This projection is in fact the respective branch Fis(α0) of the spectrum of the problem (3), (4), and (9).Theorem 13.
(1) Solution surfaces are unions of respectiveα-branches:
(25)Fi+=⋃α>0‍Fi+(α),Fi-=⋃α>0‍Fi-(α).
(2) A solution setF of the problems (3) and (4) is a union of all i-solution surfaces and is a union of all α-branches:
(26)F=⋃i=1∞‍Fi±=⋃α>0‍Fi±(α).
(3) Solution surfacesF2i-1+ and F2i-1- coincide.
(4) Solution surfacesF2i-1± and F2j-1± do not intersect unless i=j.
(5) For giveni≠j the solution surfaces F2i-1± and F2j-1± are centroaffine equivalent under the mapping Φi,j:ℝ3→ℝ3:
(27)(λ,μ,α)↦Φi,j(λ¯,μ¯,α¯),
whereλ¯=(j/i)2λ, μ¯=(j/i)2μ, α¯=(j/i)α.Proof.
(
1
), (2), (3), and (4) follow from definitions of F, Fi+, Fi-, Fi+(α), Fi(α)- and Proposition 7.
(
5
) First observe that for α,β,λ>0 we have by making use of the formula (6) that
(28)T(β,λ)=1λT(βλ,1)=1λT(αλβα,1)=1λT(αλα2/β2,1)=1λα2/β2αβT(αλα2/β2,1)=αβT(α,λα2β2).
The above formula is applicable to Tp and to Tq.
For giveni≠j and α>0 set α¯=(j/i)α. Suppose (λ,μ,α)∈F2i-1±:
(29)iTp(α,λ)+iTq(α,μ)=1.
Applying the rescaling formula (28), where α¯ replaces β, to the previous equation one has
(30)iα¯αTp(α¯,λα¯2α2)+iα¯αTq(α¯,μα¯2α2)=1,jTp(α¯,(ji)2λ)+jTq(α¯,(ji)2μ)=1,jTp(α¯,λ¯)+jTq(α¯,μ¯)=1,
therefore (λ¯,μ¯,α¯)∈F2j-1+. Since Φi,j-1=Φj,i, one has Φi,j(F2i-1±)=F2j-1±, and the surfaces F2i-1± and F2j-1± are centroaffine equivalent under the mapping Φi,j.Remark 14.
Since solution surfacesF2i-1± and F2j-1±(i≠j) are centro-affine equivalent, they have similar shape. Therefore it is enough to study properties of the solution surface F1±, in order to know properties of other odd-numbered solution surfaces. The same is true for F2i-1-.


### 5.2. Cross-Sections of Solution Surfaces with the Planesα=Const

A cross-section of any solution surfaceFi+ or Fi- by the plane α=const>0 locates in the sector Q1(α)={(λ,μ):λ>λ1(α),μ>μ1(α)}.Irrespective of the choice ofα no oscillatory (with at least one zero in (0,1)) solution of the problems (3), (4), and (9) exists for (λ,μ) in the “dead zone” below the envelope (see Figure 1).Figure 1
“Dead zone” beneath the envelope (in red) of branchesF1±(α) of the spectrum of the problems (3) and (4) for p=3, q=1/3 and α=7,9,12.The analytical description of envelopes, corresponding to branches of spectra, follows.


## 5.1. Description and Properties of a Solution Set

We will identify the cross section of a solution surfaceFis with the plane α=α0>0 (α0 is fixed) with its projection to the (λ,μ)-plane. This projection is in fact the respective branch Fis(α0) of the spectrum of the problem (3), (4), and (9).Theorem 13.
(1) Solution surfaces are unions of respectiveα-branches:
(25)Fi+=⋃α>0‍Fi+(α),Fi-=⋃α>0‍Fi-(α).
(2) A solution setF of the problems (3) and (4) is a union of all i-solution surfaces and is a union of all α-branches:
(26)F=⋃i=1∞‍Fi±=⋃α>0‍Fi±(α).
(3) Solution surfacesF2i-1+ and F2i-1- coincide.
(4) Solution surfacesF2i-1± and F2j-1± do not intersect unless i=j.
(5) For giveni≠j the solution surfaces F2i-1± and F2j-1± are centroaffine equivalent under the mapping Φi,j:ℝ3→ℝ3:
(27)(λ,μ,α)↦Φi,j(λ¯,μ¯,α¯),
whereλ¯=(j/i)2λ, μ¯=(j/i)2μ, α¯=(j/i)α.Proof.
(
1
), (2), (3), and (4) follow from definitions of F, Fi+, Fi-, Fi+(α), Fi(α)- and Proposition 7.
(
5
) First observe that for α,β,λ>0 we have by making use of the formula (6) that
(28)T(β,λ)=1λT(βλ,1)=1λT(αλβα,1)=1λT(αλα2/β2,1)=1λα2/β2αβT(αλα2/β2,1)=αβT(α,λα2β2).
The above formula is applicable to Tp and to Tq.
For giveni≠j and α>0 set α¯=(j/i)α. Suppose (λ,μ,α)∈F2i-1±:
(29)iTp(α,λ)+iTq(α,μ)=1.
Applying the rescaling formula (28), where α¯ replaces β, to the previous equation one has
(30)iα¯αTp(α¯,λα¯2α2)+iα¯αTq(α¯,μα¯2α2)=1,jTp(α¯,(ji)2λ)+jTq(α¯,(ji)2μ)=1,jTp(α¯,λ¯)+jTq(α¯,μ¯)=1,
therefore (λ¯,μ¯,α¯)∈F2j-1+. Since Φi,j-1=Φj,i, one has Φi,j(F2i-1±)=F2j-1±, and the surfaces F2i-1± and F2j-1± are centroaffine equivalent under the mapping Φi,j.Remark 14.
Since solution surfacesF2i-1± and F2j-1±(i≠j) are centro-affine equivalent, they have similar shape. Therefore it is enough to study properties of the solution surface F1±, in order to know properties of other odd-numbered solution surfaces. The same is true for F2i-1-.


## 5.2. Cross-Sections of Solution Surfaces with the Planesα=Const

A cross-section of any solution surfaceFi+ or Fi- by the plane α=const>0 locates in the sector Q1(α)={(λ,μ):λ>λ1(α),μ>μ1(α)}.Irrespective of the choice ofα no oscillatory (with at least one zero in (0,1)) solution of the problems (3), (4), and (9) exists for (λ,μ) in the “dead zone” below the envelope (see Figure 1).Figure 1
“Dead zone” beneath the envelope (in red) of branchesF1±(α) of the spectrum of the problems (3) and (4) for p=3, q=1/3 and α=7,9,12.The analytical description of envelopes, corresponding to branches of spectra, follows.


## 6. Envelopes of Solution Surfaces

Any solution surface for (3) is defined by one of the following relations:
(31)F2i-1±={(λ,μ,α):iTf(α,λ)+iTg(α,μ)=1}(i=1,2,…),F2i+={(λ,μ,α):(i+1)Tf(α,λ)+iTg(α,μ)=1}(i=0,1,…),F2i-={(λ,μ,α):iTf(α,λ)+(i+1)Tg(α,μ)=1}(i=0,1,…).
We use the unifying formula
(32)ℱ:kTf(α,λ)+mTg(α,μ)=1
and consider
(33)H(λ,μ,α)∶=kTf(α,λ)+mTg(α,μ)-1=0.Treatα as a parameter. The family of envelopes for H(λ,μ,α)=0 can be determined from the system
(34)H=0,∂H∂α=0.For the case of (3) one gets
(35)H(λ,μ,α)=kh(p)α-(p-1)/(p+1)λ-1/(p+1)+mh(q)α-(q-1)/(q+1)μ-1/(q+1)-1,
where
(36)h(p)=2p/(p+1)(p+1)1/(p+1)A(p),A(p)=∫01ds1-sp+1,∂H∂α=-kp-1p+1h(p)α-2p/(p+1)λ-1/(p+1)-mq-1q+1h(q)α-2q/(q+1)μ-1/(q+1).Consider the system(37)kh(p)α-(p-1)/(p+1)λ-1/(p+1)+mh(q)α-(q-1)/(q+1)μ-1/(q+1)=1,kp-1p+1h(p)α-2p/(p+1)λ-1/(p+1)+mq-1q+1h(q)α-2q/(q+1)μ-1/(q+1)=0
and exclude the parameter α. One obtains that
(38)μ=(ψ(p,q)m(q+1)(p-1)ψ(q,p)k(p+1)(q-1))1/(p-1)λ(q-1)/(p-1),
where
(39)ψ(p,q)=((q-1)(p+1)2(q-p)h(p))(p+1)(q-1).(1) For(40)F2i-1±:iTf(α,λ)+iTg(α,μ)=1,k=i,m=i, and the equation of the envelope is
(41)ℰ2i-1±∶μ=ω2i-1±(λ)∶=(ψ(p,q)i(q+1)(p-1)ψ(q,p)i(p+1)(q-1))1/(p-1)λ(q-1)/(p-1).(2) For(42)F2i+:(i+1)Tf(α,λ)+iTg(α,μ)=1,k=i+1, m=i, the equation of the envelope is
(43)ℰ2i+:μ=ω2i+(λ)∶=(ψ(p,q)i(q+1)(p-1)ψ(q,p)(i+1)(p+1)(q-1))1/(p-1)λ(q-1)/(p-1).(3) For(44)F2i-:iTf(α,λ)+(i+1)Tg(α,μ)=1,k=i, m=i+1, the equation of the envelope is
(45)ℰ2i-:μ=ω2i-(λ)∶=(ψ(p,q)(i+1)(q+1)(p-1)ψ(q,p)i(p+1)(q-1))1/(p-1)λ(q-1)/(p-1).Proposition 15.
For givenp>1 and 0<q<1 the location of the envelopes is as follows:  (1)ℰ2i-1±≺ℰ2i+,(2)ℰ2i-1±≺ℰ2i-,(3)
(a)  ifpq-1<0, then ℰ2i-≺ℰ2i+,(b)  ifpq-1=0, then ℰ2i-=ℰ2i+,(c)  ifpq-1>0, then ℰ2i+≺ℰ2i-,(4)ℰ2i+≺ℰ2(i+1)-1±,(5)ℰ2i-≺ℰ2(i+1)-1±,where ℰns≺ℰmt means that ωns(λ)<ωmt(λ) for any λ>0.Proof.
(
1
) First consider
(46)ℰ2i-1±:μ=ω2i-1±(λ)∶=(ψ(p,q)i(q+1)(p-1)ψ(q,p)i(p+1)(q-1))1/(p-1)λ(q-1)/(p-1),ℰ2i+:μ=ω2i+(λ):=(ψ(p,q)i(q+1)(p-1)ψ(q,p)(i+1)(p+1)(q-1))1/(p-1)λ(q-1)/(p-1).
For any λ>0,
(47)ω2i+(λ)ω2i-1±(λ)=i(q+1)(p-1)i(p+1)(q-1)(i+1)(p+1)(q-1)i(q+1)(p-1)=(ii+1)(p+1)(q-1)>1,
then ℰ2i-1±≺ℰ2i+.Remark 16.
Layout of envelopes depends onsign(pq-1) (see Figure 2).Layout of envelopes depends onsign(pq-1).
(a)
the casep=5 and q=1/3. Since pq-1>0, the envelopes are ordered as ℰ1±≺ℰ2+≺ℰ2-≺ℰ3±<⋯
(b)
the casep=3 and q=1/3. Since pq-1=0, the envelopes are ordered as ℰ1±≺ℰ2±≺ℰ3±<⋯
(c)
the casep=2 and q=1/3. Since pq-1<0, the envelopes are ordered as ℰ1±≺ℰ2-≺ℰ2+≺ℰ3±<⋯


## 7. The Number of Solutions by Geometrical Analysis of Solution Surfaces

We can detect the precise number of solutions to the problem for given positive(λ,μ). We can evaluate the initial values x′(0) for solutions on a basis of geometrical analysis of solution surfaces and the respective envelopes. The nodal structure of solutions can be described also.Letμ=c2λ be the family of rays, covering the first quadrant of the (λ,μ)-plane. Consider the cross-section of a solution surface (32) by the plane μ=c2λ in the (λ,μ,α)-space, for example, the curve ℱ|μ=c2λ, which is defined by the relations μ=c2λ and
(48)kTp(α,λ)+mTq(α,c2λ)=1.
This 3D curve ℱ|μ=c2λ can be regularly parameterized as
(49)λ=λ(r)∶=[cktf(cr)+mtg(r)c]2,μ=μ(r)∶=[cktf(cr)+mtg(r)]2,α=α(r)∶=r[cktf(cr)+mtg(r)],
where r>0. These formulas define homeomorphism ℝ+→ℱ|μ=c2λ, where ℝ+=(0,+∞).Let𝒢c be projection of ℱ|μ=c2λ to the (λ,α) plane, that is:
(50)𝒢c={(λ,α)∈ℝ+2:kTf(α,λ)+mTg(α,c2λ)=1}.
The curve 𝒢c can be regularly [11] parameterized as
(51)λ=λ(r):=[cktf(cr)+mtg(r)c]2,α=α(r)∶=r[cktf(cr)+mtg(r)],
where r>0. These formulas define homeomorphism ℝ+→𝒢c.Proposition 17.
There exists a unique parameterr0>0 such that the line λ=λ*>0(a)
does not intersect the curve𝒢c if 0<λ*<λ(r0),(b)
intersects the curve𝒢c only once if λ*=λ(r0),(c)
intersects the curve𝒢c exactly at two points if λ*>λ(r0).Proof.
(a) Consider equationλ′(r)=kc2tp′(cr)+mtq′(r)=0, which turns to
(52)kc2p-1p+1h(p)c-2p/(p+1)r-2p/(p+1)+mq-1q+1h(q)r-2q/(q+1)=0.
It can be found from the above that(53)r0=[-m(q-1)(p+1)h(q)k(p-1)(q+1)h(p)c2/(p+1)]-(p+1)(q+1)/2(p-q)
is the only root of the equation λ′(r)=0. Hence at the point (λ0,α0)=(λ(r0),α(r0)) the curve 𝒢c has a unique tangent line parallel to the Oα axis.
(b) Since the given parametrization of the curve𝒢c is regular, then α′(r0)≠0 and in some neighborhood of the point (λ0,α0) the curve 𝒢c can be represented as the graph of the function λ=φ(α). Now find
(54)φ′′(r0)=λ′′(r0)(α′(r0))2=2[cktf(cr)+mtg(r)][kc3tf′′(cr0)+mtg′′(r0)]c2[α′(r0)]2.
The next step is to detect the sign of the expression
(55)kc3tp′′(cr)+mtq′′(r)=kc32p(p-1)(p+1)2h(p)c-(3p+1)/(p+1)r-(3p+1)/(p+1)+m2q(q-1)(q+1)2h(q)r-(3q+1)/(q+1)=k2p(p-1)(p+1)2h(p)c3-(3p+1)/(p+1)r-(3p+1)/(p+1)+m2q(q-1)(q+1)2h(q)r-(3q+1)/(q+1)=r-(3q+1)/(q+1)×(k2p(p-1)(p+1)2h(p)c3-(3p+1)/(p+1)=r-(3q+1)(q+1)×r-(3(p+1)/(p+1))+((3q+1)/(q+1))r-(3q+1)/(q+1)+m2q(q-1)(q+1)2h(q)).
The routine calculations show that the expression in parentheses is equal to
(56)-mq-1q+1h(q)(2pp+1-2qq+1)=-mq-1q+1h(q)2(p-q)(p+1)(q+1)>0.
Hence the function λ=φ(α) has the strict local minimum at the point α=α0.
(c) It follows from the relations(57)limr→0+tp(r)=+∞,limr→+∞tp(r)=0+,limr→0+rtp(r)=0+,limr→+∞rtp(r)=+∞,limr→0+tq(r)=0+,limr→+∞tq(r)=+∞,limr→0+rtq(r)=0+,limr→+∞rtq(r)=+∞
that if a parameter r>0 goes from 0 to +∞, then a point (λ,α)∈𝒢c goes from the point (+∞,0+) to the point (+∞,+∞).
(d) It follows from the above argument that if the parameterr>0 goes from 0 to +∞ then a point (λ,α)∈𝒢c goes from (+∞,0+) to (λ0,α0), then turns to the right, and goes to (+∞,+∞). Since the parametrization of the curve 𝒢c is without self-intersection points, one can deduce that the curve 𝒢c is a union of two branches (i.e., graphs of the functions α=ψ1(λ) and α=ψ2(λ), where (λ0<λ<+∞) which do not intersect and are continuously “glued” at the point (λ0,α0).Before presenting “the exact number of solutions” result we make the following conventions:(1)
solution means a nontrivial solution of the problems (3) and (4),(2)
ℰns means the envelope, where n∈ℕ and s is either + or − or ±,(3)
we mean bys-solution a solution of the problem (3), (4): (a) a solution with x′(0)>0 if s=+; (b) a solution with x′(0)<0 if s=-; (c) two solutions with x′(0)=α>0 and x′(0)=-α<0 if s=±.Proposition 18.
Consider an envelope(58)ℰns={(λ,μ)∈ℝ+2∶μ=ωns(λ)},n>0.
The problems (3) and (4) have  (a)
nos-solutions with n zeroes in (0,1) if μ<ωns(λ),(b)
exactly ones-solution with even number n zeroes in (0,1) if s=+ or s=- and μ=ωns(λ),(c)
exactly twos-solutions with n zeroes in (0,1) if
n is odd, μ=ωns(λ) and s=±orn is even, μ>ωns(λ) and s=+ or s=-,(d)
exactly fours-solutions with odd number n of zeroes in (0,1) if s=± and μ>ωns(λ).Proof.
It can be verified analytically that(λ0,μ0)∈ℰns, where
(59)λ0=λ(r0)=1c2(ckh(p)(cr0)-(p-1)/(p+1)+mh(q)r0-(q-1)/(q+1))2.
and μ0=c2λ(r0).
Recall that the curve𝒢c is the projection of the 3D curve ℱ|μ=c2λ to the (λ,α)-plane. Taking in mind Proposition 17 we can assert that there exists a unique parameter r0>0, see (53), such that the line parallel to the Oα axis in the (λ,μ,α)-space going through the point (λ,μ), where λ>0 and μ=c2λ, and the curve ℱ|μ=c2λ (hence the solution surface ℱ also)  (i)
does not intersect ifμ<ωns(λ),(ii)
intersects only once ifμ=ωns(λ),(iii)
intersects exactly at two points ifμ>ωns(λ), (see Figure 3).
Depending on the value ofs one can detect the number of s-solutions as the theorem states. For example, in the case (d): if n is odd, μ>ωns(λ) and s=±, then the mentioned above line intersects the solution surface Fn+ exactly twice. Since Fn-=Fn+ the mentioned above line intersects the solution surface Fn-exactly twice also. Hence the problem (3), (4) has exactly four solutions with n zeros in (0;1).Figure 3
The casep=3, q=1/2, k=2, m=1 and c=1.2.Now we are able to prove the main theorem. In this theorem we suppose thatω0-(λ)=0 and ω0+(λ)=0 for all λ>0 and will use auxiliary envelopes ℰ0-:μ=ω0-(λ) and ℰ0+:μ=ω0+(λ).Theorem 19.
Suppose0<q<1<p.  (1)
Ifpq-1>0 and i∈ℕ, then the exact number of nontrivial solutions to the problems (3) and (4) is
(a)
8i-6 if ω2i-2-(λ)<μ<ω2i-1±(λ),(b)
8i-4 if μ=ω2i-1±(λ),(c)
8i-2 if ω2i-1±(λ)<μ<ω2i+(λ),(d)
8i-1 if μ=ω2i+(λ),(e)
8i if ω2i+(λ)<μ<ω2i-(λ),(f)
8i+1 if μ=ω2i-(λ).(2)
Ifpq-1=0 and i∈ℕ, then the exact number of nontrivial solutions to the problems (3) and (4) is
(a)
8i-6 if ω2i-2+(λ)<μ<ω2i-1±(λ),(b)
8i-4 if μ=ω2i-1±(λ),(c)
8i-2 if ω2i-1±(λ)<μ<ω2i+(λ),(d)
8iif μ=ω2i+(λ).(3)
Ifpq-1<0 and i∈ℕ then the exact number of nontrivial solutions to the problems (3) and (4) is
(a)
8i-6 if ω2i-2+(λ)<μ<ω2i-1±(λ),(b)
8i-4 if μ=ω2i-1±(λ),(c)
8i-2 if ω2i-1±(λ)<μ<ω2i-(λ),(d)
8i-1 if μ=ω2i-(λ),(e)
8i if ω2i-(λ)<μ<ω2i+(λ),(f)
8i+1 if μ=ω2i+(λ).Proof.
First notice that ifλ>0 and μ>0 then there exists exactly one ±-solution without zeros in (0,1), in fact, accordingly to conventions, two solutions without zeros in (0,1), and of opposite sign values of x′(0).
We consider only the casepq-1>0. Similarly two other possible cases pq-1<0 and pq-1=0 can be treated.
It follows from Proposition15 that the envelopes are ordered as
(60)ℰ0-≺ℰ1±≺ℰ2+≺ℰ2-≺ℰ3±≺⋯≺ℰ2i-1±≺ℰ2i+≺ℰ2i-≺ℰ2i+1±≺⋯.
Let us consider the case (1), whenpq-1>0, and the subcase (a): ω2i-2-(λ)<μ<ω2i-1±(λ) for i=1,2,….(i)
ifi=1 then there are 2=8·1-6 solutions without zeros;(ii)
ifi=2, then there are two solutions without zeros, two ±-solutions with 1 zero (that is, four solutions with 1 zero), two “+”-solution with 2 zeros and two “−”-solutions with 2 zeros; totally there are 10=8·2-6 (nontrivial) solutions;(iii)
ifi=3, then in addition to 10 solutions mentioned in the previous step, there are also two ±-solutions with 3 zeros (that is, four solutions with 3 zeros), two “+”-solutions with 4 zeros and two “−”-solution with 4 zeros; totally there are 18=8·3-6 (nontrivial) solutions and so on;(iv)
ifi∈ℕ, then totally there are 8i-6 (nontrivial) solutions.
The other cases can be considered analogously.



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*Source: 705984-2013-02-28.xml*