Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2013, Article ID 705984, 9 pages http://dx.doi.org/10.1155/2013/705984 Research Article Boundary Value Problems for a Super-Sublinear Asymmetric Oscillator: The Exact Number of Solutions Armands Gritsans and Felix Sadyrbaev Daugavpils University, Department of Mathematics, Parades Street 1, 5400 Daugavpils, Latvia Correspondence should be addressed to Armands Gritsans; armands.gricans@du.lv Received 30 March 2012; Accepted 5 November 2012 Academic Editor: Paolo Ricci Copyright © 2013 A. Gritsans and F. Sadyrbaev. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. π π Properties of asymmetric oscillator described by the equation π₯σΈ σΈ = −π(π₯+ ) + π(π₯− ) (i), where π ≥ 1 and 0 < π ≤ 1, are studied. A set of (π, π) such that the problem (i), π₯(0) = 0 = π₯(1) (ii), and |π₯σΈ (0)| = πΌ (iii) have a nontrivial solution, is called πΌ-spectrum. We give full description of πΌ-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. 1. Introduction Asymmetric oscillators were studied intensively starting from Μ see [1] and references therein. the works by Kufner and FucΔ±Μk; 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 (π, π)-plane. On the other hand, there is a plenty of works devoted to one-parameter case of equations π₯σΈ σΈ + ππ(π₯) = 0 given together with the two-point boundary conditions. Due to nonlinearity of π one should consider solutions π₯(π‘; πΌ) with different πΌ = π₯σΈ (0). Bifurcation diagrams in terms of πΌ and π, or βπ₯β and π, can serve then to evaluate the number of solutions [2–4]. In this paper we consider differential equations of the form π₯σΈ σΈ = −ππ (π₯+ ) + ππ (π₯− ) , (1) where π = π₯π , π ≥ 1, and π = π₯π , 0 < π ≤ 1. Here π and π are nonnegative parameters, π₯+ = max{π₯, 0}, π₯− = max{−π₯, 0}. This equation describes asymmetric oscillator with different nonlinear restoring forces on both sides of π₯ = 0. If π = π = Μ equation π₯, then equation becomes famous FucΔ±Μk π₯σΈ σΈ = −ππ₯+ + ππ₯− . (2) Μ spectrum are well known (the FucΔ±Μk Μ Properties of the FucΔ±Μk spectrum is a set of all pairs (π, π) where π, π ≥ 0, such that the Dirichlet problem—(2) with boundary conditions π₯(0) = 0 = π₯(1)—has a non-trivial solution). The aim of our study in this paper is to describe properties of the spectrum of the problem π π π₯σΈ σΈ = −π(π₯+ ) + π(π₯− ) , π₯ (0) = 0, 0 < π ≤ 1, 1 ≤ π, π₯ (1) = 0. (3) (4) For this we study first the time maps for the related functions (Section 2), 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 π₯σΈ (0). This paper continues series of publications by the authors devoted to nonlinear asymmetric oscillations [5–8]. 2. Time Maps Consider the Cauchy problem π₯σΈ σΈ = −ππ₯π , π₯ (0) = 0, π₯σΈ (0) = πΌ, (5) 2 International Journal of Mathematics and Mathematical Sciences where π > 0. Solutions of this problem for πΌ and π positive have a zero. This zero will be denoted ππ (πΌ, π) and called time map function; more on time maps can be found in [9, 10]. Proposition 1. Suppose π, πΌ, π > 0, π > 1 and 0 < π < 1. πΌ 1 ), π‘ ( √π π √π πΉ0+ (πΌ) = {(π, π) : ππ (πΌ, π) = 1, π ≥ 0} , πΉ0− (πΌ) = {(π, π) : π ≥ 0, ππ (πΌ, π) = 1} , (1) For any πΌ, π > 0 the formula is valid: ππ (πΌ, π) = consists of the following πΌ-branches: + πΉ2π−1 (πΌ) = {(π; π) : πππ (πΌ, π) + πππ (πΌ, π) = 1} , (6) − πΉ2π−1 (πΌ) = {(π; π) : πππ (πΌ, π) + πππ (πΌ, π) = 1} , (11) πΉ2π+ (πΌ) = {(π; π) : (π + 1) ππ (πΌ, π) + πππ (πΌ, π) = 1} , where π‘π (πΎ) = ππ (πΎ, 1). πΉ2π− (πΌ) = {(π; π) : (π + 1) ππ (πΌ, π) + πππ (πΌ, π) = 1} . (2) The function ππ (πΌ, π) for the problem (5) is ππ (πΌ, π) = 2π/(π+1) (π + 1)1/(π+1) π΄ (π) πΌ−(π−1)/(π+1) π−1/(π+1) , (7) 1 where π΄(π) = ∫0 (ππ /√1 − π π+1 ). (3) The function ππ (πΌ, π) for fixed π and πΌ is strictly decreasing function of π and possesses the properties lim ππ (πΌ, π) = +∞, lim ππ (πΌ, π) = 0. π → 0+ π → +∞ (8) (4) The function ππ (πΌ, π) for fixed π and π is decreasing function of πΌ. (5) The function ππ (πΌ, π) for fixed π and π is increasing function of πΌ. Proof. By standard computations. Remark 2. ππ (πΌ, π) = π/√π for π = 1, irrespective of πΌ. The notation πΉπ+ (πΌ), respectively: πΉπ− (πΌ), refers to solutions π₯ which satisfy the initial conditions π₯(0) = 0, π₯σΈ (0) = πΌ > 0 (resp.: π₯σΈ (0) = −πΌ < 0) and have exactly π 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 |π₯σΈ (0)| = πΌ > 0. Consider the case π₯σΈ (0) = πΌ > 0. The first zero of π₯(π‘) appears at π‘ = ππ and π₯σΈ (ππ ) = −πΌ. The second zero is at π‘ = ππ + ππ . One has that for solutions with exactly one zero in (0, 1) the relation ππ + ππ = 1 holds, which defines the branch πΉ1+ (πΌ). Suppose π₯σΈ (0) = −πΌ. The first zero now is at π‘ = ππ . The second one is at π‘ = ππ + ππ and therefore ππ + ππ = 1. The branches πΉ1+ (πΌ) and πΉ1− (πΌ) are given by the equivalent relations ππ + ππ = 1 and ππ + ππ = 1, respectively, and therefore coincide. Proof for solutions with different nodal structure is similar. Remark 6. If π = π = 1, then for any πΌ > 0 the πΌ-spectrum is 3. Spectrum Suppose that the problems (3) and (4) are considered with the additional condition σ΅¨σ΅¨ σΈ σ΅¨σ΅¨ σ΅¨σ΅¨π₯ (0)σ΅¨σ΅¨ = πΌ > 0. σ΅¨ σ΅¨ (9) Definition 3. For given πΌ > 0 a set of all nonnegative (π, π) for which the problems (3), (4), and (9) have a nontrivial solution is called πΌ-spectrum. Remark 4. A solution of (3) is a πΆ2 -function. Therefore π₯σΈ (π‘) is continuous. If π§1 and π§2 are two consecutive zeros of π₯(π‘), then |π₯σΈ (π§1 )| = |π₯σΈ (π§2 )| since a solution in the interval (π§1 , π§2 ) is symmetric with respect to the middle point. Therefore |π₯σΈ (π§)| = πΌ for any zero point π§ and signs of π₯σΈ (π‘) alternate. Theorem 5. The πΌ-spectrum for the problem π π π₯σΈ σΈ = −π(π₯+ ) + π(π₯− ) , π₯ (1) = 0, π₯ (0) = 0, σ΅¨σ΅¨ σΈ σ΅¨σ΅¨ σ΅¨σ΅¨π₯ (0)σ΅¨σ΅¨ = πΌ > 0 σ΅¨ σ΅¨ (10) πΉ0+ (πΌ) = {(π, π) : π = 1, π ≥ 0} , √π πΉ0− (πΌ) = {(π, π) : π ≥ 0, + πΉ2π−1 (πΌ) = {(π; π) : π − πΉ2π−1 π = 1} , √π π π +π = 1} , √π √π π π = 1} , +π (πΌ) = {(π; π) : π √π √π πΉ2π+ (πΌ) = {(π; π) : (π + 1) π π +π = 1} , √π √π πΉ2π− (πΌ) = {(π; π) : (π + 1) π π = 1} . +π √π √π (12) Μ spectrum for the problems (2) and (4), see It is classical FucΔ±Μk [1]. International Journal of Mathematics and Mathematical Sciences 4. Properties of the πΌ-Spectrum 3 (2) The branch πΉ2π+ (πΌ) (π ∈ N) is located in the sector Proposition 7. {(π, π) : π > π π−1 (πΌ, π) , π > ππ (πΌ, π)} + − (1) The branches πΉ2π−1 (πΌ) and πΉ2π−1 (πΌ) coincide. ± ± (πΌ) and πΉ2π−1 (πΌ) do not intersect (2) The branches πΉ2π−1 unless π =ΜΈ π. + (3) The branches πΉ2π+ (πΌ) and πΉ2π (πΌ) do not intersect unless π =ΜΈ π. − (4) The branches πΉ2π− (πΌ) and πΉ2π (πΌ) do not intersect unless π =ΜΈ π. (5) Any branch πΉππ (πΌ) where π ≥ 1 and π is either “+” or “−” is a graph of monotonically decreasing function π = π(π). (6) The branches πΉ2π+ (πΌ) and πΉ2π− (πΌ) intersect once. πππ (πΌ, π) + πππ (πΌ, π) = 1. (13) (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 (π + 1) ππ (πΌ, π) + πππ (πΌ, π) = 1, =2 π΄ (π) πΌ Remark 9. So the positive part of the πΌ-spectrum +∞ πΉ+ (πΌ) = β πΉπ+ (πΌ) (20) π=0 in the extended (π, π)-plane may be schematically described by the chain πΉ0+ (πΌ) πΉ1+ (πΌ) πΉ3+ (πΌ) πΉ4+ (πΌ) πΉ2+ (πΌ) (π 1 , 0) σ³¨→ (π 1 , +∞) σ³¨→ (+∞, π1 ) σ³¨→ (π 2 , +∞) (21) σ³¨→ (+∞, π3 ) σ³¨→ (π 3 , +∞) ⋅ ⋅ ⋅ . πΉ− (πΌ) = β πΉπ− (πΌ) (22) π=0 in the extended (π, π)-plane may be described as follows: π΄ (π) πΌ−(π−1)/(π+1) π−1/(π+1) (π + 1) and is a hyperbola looking curve in (π, π) plane with vertical asymptote π = π π (πΌ, π) and horizontal asymptote π = ππ−1 (πΌ, π). (15) or 1/(π+1) (19) +∞ ππ (πΌ, π) = ππ (πΌ, π) π/(π+1) {(π, π) : π > π π (πΌ, π) , π > ππ−1 (πΌ, π)} Similarly, the negative part of the πΌ-spectrum It follows that 1/(π+1) (3) The branch πΉ2π− (πΌ) (π ∈ N) is located in the sector (14) (π + 1) ππ (πΌ, π) + πππ (πΌ, π) = 1. 2π/(π+1) (π + 1) and is a hyperbola looking curve in (π, π) plane with vertical asymptote π = π π−1 (πΌ, π) and horizontal asymptote π = ππ (πΌ, π). Proof. Follows from the relations (11) and Proposition 7. + Proof. (1) It follows from (11) that πΉ2π−1 (πΌ) coincides with − πΉ2π−1 (πΌ), since both sets (branches) are defined by symmetric relations: πππ (πΌ, π) + πππ (πΌ, π) = 1, (18) −(π−1)/(π+1) −1/(π+1) π (16) . The above relation defines a curve which is a graph of monotonically increasing function π = πΆ(π, π, πΌ)π(π+1)/(π+1) , where πΆ(π, π, πΌ) is a constant computable from (16). This curve emanates from the origin and intersects the graph of monotonically decreasing from +∞ to 0 function π = π(π) from (5) only once. Let π π (πΌ, π) be a unique solution of the equation ππ (πΌ, π) = 1/π; similarly, let ππ (πΌ, π) be a unique solution of the equation ππ (πΌ, π) = 1/π. Proposition 8. Suppose πΌ, π, π > 0. ± (1) The branch πΉ2π−1 (πΌ)(π ∈ N) is located in the sector {(π, π) : π > π π (πΌ, π) , π > ππ (πΌ, π)} (17) and is a hyperbola looking curve in (π, π) plane with vertical asymptote π = π π (πΌ, π) and horizontal asymptote π = ππ (πΌ, π). πΉ0− (πΌ) πΉ1− (πΌ) πΉ3− (πΌ) πΉ4− (πΌ) πΉ2− (πΌ) (0, π1 ) σ³¨→ (+∞, π1 ) σ³¨→ (π 1 , +∞) σ³¨→ (+∞, π2 ) (23) σ³¨→ (π 2 , +∞) σ³¨→ (+∞, π3 ) ⋅ ⋅ ⋅ . Proposition 10. Let π > 1 and 0 < π < 1 be fixed. It is true that lim π 1 (πΌ, π) = +∞, lim π πΌ → +∞ 1 lim π1 (πΌ, π) = +0, lim π1 (πΌ, π) = +∞. πΌ → +0 πΌ → +0 (πΌ, π) = +0, (24) πΌ → +∞ Proof. Follows from Propositions 1 and 10. 5. Solution Sets and Solution Surfaces Definition 11. A solution set of the problems (3) and (4) is a set πΉ of all triples (π, π, πΌ) (π ≥ 0, π ≥ 0, πΌ > 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). 4 International Journal of Mathematics and Mathematical Sciences Let πΉπ+ be a set of all triples (π, π, πΌ) such that there exists a nontrivial solution of the respective problems (3) and (4), π₯σΈ (0) = πΌ > 0 which has exactly π zeros in (0, 1), π = 0, 1, . . ., but πΉπ− be a set of all triples (π, π, πΌ) such that there exists a nontrivial solution of the respective problems (3) and (4), π₯σΈ (0) = −πΌ < 0 which has exactly π zeros in (0, 1), π = 0, 1, . . . Applying the rescaling formula (28), where πΌ replaces π½, to the previous equation one has Definition 12. πΉπ+ will be called a positive π-solution surface, but πΉπ− -a negative π-solution surface. πΌ πΌ2 πΌ πΌ2 π ππ (πΌ, π 2 ) + π ππ (πΌ, π 2 ) = 1, πΌ πΌ πΌ πΌ 5.1. Description and Properties of a Solution Set. We will identify the cross section of a solution surface πΉππ with the plane πΌ = πΌ0 > 0 (πΌ0 is fixed) with its projection to the (π, π)plane. This projection is in fact the respective branch πΉππ (πΌ0 ) of the spectrum of the problem (3), (4), and (9). π 2 π 2 πππ (πΌ, ( ) π) + πππ (πΌ, ( ) π) = 1, π π Theorem 13. (1) Solution surfaces are unions of respective πΌ-branches: πΉπ+ = β πΉπ+ (πΌ) , πΉπ− = β πΉπ− (πΌ) . πΌ>0 πΌ>0 (25) (2) A solution set πΉ of the problems (3) and (4) is a union of all π-solution surfaces and is a union of all πΌbranches: ∞ πΉ = βπΉπ± = β πΉπ± (πΌ) . (26) πΌ>0 π=1 + − (3) Solution surfaces πΉ2π−1 and πΉ2π−1 coincide. ± ± and πΉ2π−1 do not (4) Solution surfaces πΉ2π−1 intersect unless π = π. ± ± and πΉ2π−1 (5) For given π =ΜΈ π the solution surfaces πΉ2π−1 are centroaffine equivalent under the mapping Φπ,π : R3 → R 3 : Φπ,π (π, π, πΌ) σ³¨σ³¨→ (π, π, πΌ) , (27) where π = (π/π)2 π, π = (π/π)2 π, πΌ = (π/π)πΌ. Proof. (1), (2), (3), and (4) follow from definitions of πΉ, πΉπ+ , πΉπ− , πΉπ+ (πΌ), πΉπ (πΌ)− and Proposition 7. (5) First observe that for πΌ, π½, π > 0 we have by making use of the formula (6) that π (π½, π) = = For given π =ΜΈ π and πΌ > 0 set πΌ = (π/π)πΌ. Suppose ± : (π, π, πΌ) ∈ πΉ2π−1 πππ (πΌ, π) + πππ (πΌ, π) = 1. (30) πππ (πΌ, π) + πππ (πΌ, π) = 1, + therefore (π, π, πΌ) ∈ πΉ2π−1 . Since Φ−1 π,π = Φπ,π , one has ± ± ± ± Φπ,π (πΉ2π−1 ) = πΉ2π−1 , and the surfaces πΉ2π−1 and πΉ2π−1 are centroaffine equivalent under the mapping Φπ,π . ± ± Remark 14. Since solution surfaces πΉ2π−1 and πΉ2π−1 (π =ΜΈ π) are centro-affine equivalent, they have similar shape. Therefore it is enough to study properties of the solution surface πΉ1± , in order to know properties of other odd-numbered solution − . surfaces. The same is true for πΉ2π−1 5.2. Cross-Sections of Solution Surfaces with the Planes πΌ = Const. A cross-section of any solution surface πΉπ+ or πΉπ− by the plane πΌ = const > 0 locates in the sector π1 (πΌ) = {(π, π) : π > π 1 (πΌ), π > π1 (πΌ)}. Irrespective of the choice of πΌ no oscillatory (with at least one zero in (0, 1)) solution of the problems (3), (4), and (9) exists for (π, π) in the “dead zone” below the envelope (see Figure 1). 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: ± = {(π, π, πΌ) : πππ (πΌ, π) + πππ (πΌ, π) = 1} πΉ2π−1 (π = 1, 2, . . .) , π½ 1 π( , 1) √π √π πΉ2π+ = {(π, π, πΌ) : (π + 1) ππ (πΌ, π) + πππ (πΌ, π) = 1} 1 πΌ π½ πΌ 1 π( π( , 1) , 1) = √π √π πΌ √π √ππΌ2 /π½2 πΉ2π− = {(π, π, πΌ) : πππ (πΌ, π) + (π + 1) ππ (πΌ, π) = 1} (π = 0, 1, . . .) , (31) (π = 0, 1, . . .) . πΌ πΌ πΌ πΌ2 = , 1) = π (πΌ, π 2 ) . π( π½ π½ √ππΌ2 /π½2 π½ √ππΌ2 /π½2 1 (28) The above formula is applicable to ππ and to ππ . (29) We use the unifying formula F : πππ (πΌ, π) + πππ (πΌ, π) = 1 (32) and consider π» (π, π, πΌ) := πππ (πΌ, π) + πππ (πΌ, π) − 1 = 0. (33) International Journal of Mathematics and Mathematical Sciences 5 60 55 50 πΌ = 12 45 π 40 πΌ=9 35 πΌ=7 30 25 10 20 30 π 40 50 Figure 1: “Dead zone” beneath the envelope (in red) of branches πΉ1± (πΌ) of the spectrum of the problems (3) and (4) for π = 3, π = 1/3 and πΌ = 7, 9, 12. Treat πΌ as a parameter. The family of envelopes for π»(π, π, πΌ) = 0 can be determined from the system ππ» = 0. ππΌ π» = 0, and exclude the parameter πΌ. One obtains that π(π, π)π(π+1)(π−1) π=( ) π(π, π)π(π+1)(π−1) (34) 1/(π−1) π(π−1)/(π−1) , (38) where For the case of (3) one gets π» (π, π, πΌ) = πβ (π) πΌ−(π−1)/(π+1) π−1/(π+1) + πβ (π) πΌ−(π−1)/(π+1) π−1/(π+1) − 1, (35) where π (π, π) = ( (π − 1) (π + 1) ) 2 (π − π) β (π) (π+1)(π−1) . (39) (1) For 1/(π+1) β (π) = 2π/(π+1) (π + 1) π΄ (π) = ∫ 1 0 ± πΉ2π−1 : πππ (πΌ, π) + πππ (πΌ, π) = 1, π΄ (π) , ππ , √1 − π π+1 π−1 ππ» =− π β (π) πΌ−2π/(π+1) π−1/(π+1) ππΌ π+1 −π π = π, π = π, and the equation of the envelope is (36) 1/(π−1) π−1 β (π) πΌ−2π/(π+1) π−1/(π+1) . π+1 πΉ2π+ : (π + 1) ππ (πΌ, π) + πππ (πΌ, π) = 1, π π−1 β (π) πΌ−2π/(π+1) π−1/(π+1) = 0 π+1 (41) (42) π = π + 1, π = π, the equation of the envelope is + πβ (π) πΌ−(π−1)/(π+1) π−1/(π+1) = 1, +π π(π−1)/(π−1) . (2) For −(π−1)/(π+1) −1/(π+1) π−1 π β (π) πΌ−2π/(π+1) π−1/(π+1) π+1 ± E±2π−1 : π = π2π−1 (π) π (π, π) π(π+1)(π−1) := ( ) π (π, π) π(π+1)(π−1) Consider the system πβ (π) πΌ (40) (37) + E+2π : π = π2π (π) := ( π (π, π) π(π+1)(π−1) (π+1)(π−1) π (π, π) (π + 1) 1/(π−1) ) π(π−1)/(π−1) . (43) 6 International Journal of Mathematics and Mathematical Sciences (3) For πΉ2π− : πππ (πΌ, π) + (π + 1) ππ (πΌ, π) = 1, 350 (44) 300 250 π = π, π = π + 1, the equation of the envelope is π 200 − E−2π : π = π2π (π) (π+1)(π−1) π (π, π) (π + 1) := ( π (π, π) π(π+1)(π−1) 150 1/(π−1) (π−1)/(π−1) ) π . Proposition 15. For given π > 1 and 0 < π < 1 the location of the envelopes is as follows: (1)E±2π−1 βΊ E+2π , (2)E±2π−1 βΊ E−2π , 0 10 20 30 40 50 π (a) the case π = 5 and π = 1/3. Since ππ − 1 > 0, the envelopes are ordered as E±1 βΊ E+2 βΊ E−2 βΊ E±3 < ⋅ ⋅ ⋅ 300 (b) (c) βΊ E+2π , = E+2π , βΊ E−2π , 250 150 100 π‘ (π) for any π > 0. where Eπ π βΊ Eπ‘π means that πππ (π) < ππ : π= := ( 30 40 50 350 1/(π−1) π(π−1)/(π−1) , 300 (π) π π (π, π) π(π+1)(π−1) ) (π+1)(π−1) β°3± 250 200 β°2+ 150 1/(π−1) (π−1)/(π−1) π . (46) For any π > 0, + π2π (π) π(π+1)(π−1) π(π+1)(π−1) π (π+1)(π−1) = =( > 1, ) ± (π+1)(π−1) π2π−1 (π) (π + 1) π+1 π(π+1)(π−1) (47) βΊ 20 π (π) π (π, π) (π + 1) 10 (b) the case π = 3 and π = 1/3. Since ππ − 1 = 0, the envelopes are ordered as E±1 βΊ E±2 βΊ E±3 < ⋅ ⋅ ⋅ π (π, π) π(π+1)(π−1) := ( ) π (π, π) π(π+1)(π−1) + π2π β°1± 0 Proof. (1) First consider : π= β°2± 50 (5) E−2π βΊ E±2(π+1)−1 , ± π2π−1 β°3± π 200 (4) E+2π βΊ E±2(π+1)−1 , then β°1± 350 if ππ − 1 < 0, then E−2π if ππ − 1 = 0, then E−2π if ππ − 1 > 0, then E+2π (a) E±2π−1 β°2+ 50 (3) E+2π β°2− 100 (45) E±2π−1 β°3± β°2− 100 β°1± 50 0 10 20 30 40 50 π (c) the case π = 2 and π = 1/3. Since ππ − 1 < 0, the envelopes are ordered as E±1 βΊ E−2 βΊ E+2 βΊ E±3 < ⋅ ⋅ ⋅ Figure 2: Layout of envelopes depends on sign(ππ − 1). E+2π . Remark 16. Layout of envelopes depends on sign(ππ − 1) (see Figure 2). 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 (π, π). We can evaluate the initial values π₯σΈ (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. Let π = π2 π be the family of rays, covering the first quadrant of the (π, π)-plane. Consider the cross-section of a International Journal of Mathematics and Mathematical Sciences solution surface (32) by the plane π = π2 π in the (π, π, πΌ)space, for example, the curve F|π=π2 π , which is defined by the relations π = π2 π and 2 πππ (πΌ, π) + πππ (πΌ, π π) = 1. (48) This 3D curve F|π=π2 π can be regularly parameterized as π = π (π) := [ πππ‘π (ππ) + ππ‘π (π) π (49) where π > 0. These formulas define homeomorphism R+ → F|π=π2 π , where R+ = (0, +∞). Let Gπ be projection of F|π=π2 π to the (π, πΌ) plane, that is: 2 : πππ (πΌ, π) + πππ (πΌ, π π) = 1} . (50) The curve Gπ can be regularly [11] parameterized as πππ‘π (ππ) + ππ‘π (π) π 2 [πππ‘π (ππ) + ππ‘π (π)] [ππ3 π‘πσΈ σΈ (ππ0 ) + ππ‘πσΈ σΈ (π0 )] π2 [πΌσΈ (π0 )] 2 ], (51) where π > 0. These formulas define homeomorphism R+ → Gπ . Proposition 17. There exists a unique parameter π0 > 0 such that the line π = π ∗ > 0 = ππ3 2π (π − 1) +π 2π (π − 1) 2 (π + 1) 2π (π − 1) 2 (π + 1) +π (c) intersects the curve Gπ exactly at two points if π ∗ > π(π0 ). Proof. (a) Consider equation πσΈ (π) = ππ2 π‘πσΈ (ππ) + ππ‘πσΈ (π) = 0, which turns to (52) It can be found from the above that π (π − 1) (π + 1) β (π) ] π (π − 1) (π + 1) β (π) π2/(π+1) −(π+1)(π+1)/2(π−π) (53) is the only root of the equation πσΈ (π) = 0. Hence at the point (π 0 , πΌ0 ) = (π(π0 ), πΌ(π0 )) the curve Gπ has a unique tangent line parallel to the ππΌ axis. (b) Since the given parametrization of the curve Gπ is regular, then πΌσΈ (π0 ) =ΜΈ 0 and in some neighborhood of the β (π) π−(3π+1)/(π+1) 2 (π + 1) β (π) π−(3π+1)/(π+1) (55) = π−(3π+1)/(π+1) 2π (π − 1) × (π 2 (π + 1) β (π) π3−(3π+1)/(π+1) × π−(3(π+1)/(π+1))+((3π+1)/(π+1)) +π 2π (π − 1) 2 (π + 1) β (π)) . The routine calculations show that the expression in parentheses is equal to −π 2π π0 = [− (54) β (π) π3−(3π+1)/(π+1) π−(3π+1)/(π+1) 2π (π − 1) (b) intersects the curve Gπ only once if π ∗ = π(π0 ), π−1 β (π) π−2π/(π+1) = 0. π+1 . β (π) π−(3π+1)/(π+1) π−(3π+1)/(π+1) 2 (π + 1) (a) does not intersect the curve Gπ if 0 < π ∗ < π(π0 ), −1 ππ β (π) π−2π/(π+1) π−2π/(π+1) π+1 2 The next step is to detect the sign of the expression =π πΌ = πΌ (π) := π [πππ‘π (ππ) + ππ‘π (π)] , +π 2 (πΌσΈ (π0 )) ππ3 π‘πσΈ σΈ (ππ) + ππ‘πσΈ σΈ (π) πΌ = πΌ (π) := π [πππ‘π (ππ) + ππ‘π (π)] , π = π (π) := [ πσΈ σΈ (π0 ) πσΈ σΈ (π0 ) = ], 2 Gπ = {(π, πΌ) ∈ point (π 0 , πΌ0 ) the curve Gπ can be represented as the graph of the function π = π(πΌ). Now find = 2 π = π (π) := [πππ‘π (ππ) + ππ‘π (π)] , R2+ 7 π−1 2π 2π β (π) ( − ) π+1 π+1 π+1 2 (π − π) π−1 > 0. = −π β (π) π+1 (π + 1) (π + 1) (56) Hence the function π = π(πΌ) has the strict local minimum at the point πΌ = πΌ0 . (c) It follows from the relations lim π‘π (π) = +∞, π → 0+ lim ππ‘π (π) = 0+, π → 0+ lim π‘ π → +∞ π lim ππ‘π (π) = +∞, π → +∞ lim π‘π (π) = 0+, π → 0+ π → +∞ lim ππ‘π (π) = 0+, π → +∞ π → 0+ (π) = 0+, lim π‘π (π) = +∞, lim ππ‘π (π) = +∞ (57) 8 International Journal of Mathematics and Mathematical Sciences that if a parameter π > 0 goes from 0 to +∞, then a point (π, πΌ) ∈ Gπ goes from the point (+∞, 0+) to the point (+∞, +∞). (d) It follows from the above argument that if the parameter π > 0 goes from 0 to +∞ then a point (π, πΌ) ∈ Gπ goes from (+∞, 0+) to (π 0 , πΌ0 ), then turns to the right, and goes to (+∞, +∞). Since the parametrization of the curve Gπ is without self-intersection points, one can deduce that the curve Gπ is a union of two branches (i.e., graphs of the functions πΌ = π1 (π) and πΌ = π2 (π) , where (π 0 < π < +∞) which do not intersect and are continuously “glued” at the point (π 0 , πΌ0 ). 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) Eπ π means the envelope, where π ∈ N and π is either + or − or ±, (3) we mean by π -solution a solution of the problem (3), (4): (a) a solution with π₯σΈ (0) > 0 if π = +; (b) a solution with π₯σΈ (0) < 0 if π = −; (c) two solutions with π₯σΈ (0) = πΌ > 0 and π₯σΈ (0) = −πΌ < 0 if π = ±. Proposition 18. Consider an envelope Eπ π = {(π, π) ∈ R2+ : π = πππ (π)} , π > 0. (58) The problems (3) and (4) have (a) no π -solutions with π zeroes in (0, 1) if π < πππ (π), (b) exactly one π -solution with even number π zeroes in (0, 1) if π = + or π = − and π = πππ (π), (c) exactly two π -solutions with π zeroes in (0, 1) if π is odd, π = πππ (π) and π = ± or π is even, π > πππ (π) and π = + or π = −, (d) exactly four π -solutions with odd number π of zeroes in (0, 1) if π = ± and π > πππ (π). Proof. It can be verified analytically that (π 0 , π0 ) ∈ Eπ π , where π 0 = π (π0 ) = 1 −(π−1)/(π+1) −(π−1)/(π+1) 2 (ππβ (π) (ππ0 ) + πβ (π) π0 ). 2 π (59) and π0 = π2 π(π0 ). 200 π 100 0 200 β±β£π=π2 π π = π2 π 150 πΌ 100 50 0 0 π 100 β° 200 Figure 3: The case π = 3, π = 1/2, π = 2, π = 1 and π = 1.2. Recall that the curve Gπ is the projection of the 3D curve F|π=π2 π to the (π, πΌ)-plane. Taking in mind Proposition 17 we can assert that there exists a unique parameter π0 > 0, see (53), such that the line parallel to the ππΌ axis in the (π, π, πΌ)-space going through the point (π, π), where π > 0 and π = π2 π, and the curve F|π=π2 π (hence the solution surface F also) (i) does not intersect if π < πππ (π), (ii) intersects only once if π = πππ (π), (iii) intersects exactly at two points if π > πππ (π), (see Figure 3). Depending on the value of π one can detect the number of π -solutions as the theorem states. For example, in the case (d): if π is odd, π > πππ (π) and π = ±, then the mentioned above line intersects the solution surface πΉπ+ exactly twice. Since πΉπ− = πΉπ+ the mentioned above line intersects the solution surface πΉπ− exactly twice also. Hence the problem (3), (4) has exactly four solutions with π zeros in (0; 1). Now we are able to prove the main theorem. In this theorem we suppose that π0− (π) = 0 and π0+ (π) = 0 for all π > 0 and will use auxiliary envelopes E−0 : π = π0− (π) and E+0 : π = π0+ (π). Theorem 19. Suppose 0 < π < 1 < π. (1) If ππ − 1 > 0 and π ∈ N, then the exact number of nontrivial solutions to the problems (3) and (4) is − ± (π) < π < π2π−1 (π), (a) 8π − 6 if π2π−2 ± (b) 8π − 4 if π = π2π−1 (π), ± + (c) 8π − 2 if π2π−1 (π) < π < π2π (π), + (d) 8π − 1 if π = π2π (π), + − (e) 8π if π2π (π) < π < π2π (π), − (f) 8π + 1 if π = π2π (π). International Journal of Mathematics and Mathematical Sciences (2) If ππ − 1 = 0 and π ∈ N, then the exact number of nontrivial solutions to the problems (3) and (4) is + ± (π) < π < π2π−1 (π), (a) 8π − 6 if π2π−2 ± (b) 8π − 4 if π = π2π−1 (π), ± + (c) 8π − 2 if π2π−1 (π) < π < π2π (π), + (d) 8π if π = π2π (π). (3) If ππ − 1 < 0 and π ∈ N then the exact number of nontrivial solutions to the problems (3) and (4) is + ± (π) < π < π2π−1 (π), (a) 8π − 6 if π2π−2 ± (b) 8π − 4 if π = π2π−1 (π), ± − (c) 8π − 2 if π2π−1 (π) < π < π2π (π), − (d) 8π − 1 if π = π2π (π), − + (e) 8π if π2π (π) < π < π2π (π), + (f) 8π + 1 if π = π2π (π). Proof. First notice that if π > 0 and π > 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 π₯σΈ (0). We consider only the case ππ − 1 > 0. Similarly two other possible cases ππ − 1 < 0 and ππ − 1 = 0 can be treated. It follows from Proposition 15 that the envelopes are ordered as E−0 βΊ E±1 βΊ E+2 βΊ E−2 βΊ E±3 βΊ ⋅ ⋅ ⋅ βΊ E±2π−1 βΊ E+2π βΊ E−2π βΊ E±2π+1 βΊ ⋅ ⋅ ⋅ . (60) Let us consider the case (1), when ππ − 1 > 0, and the − ± subcase (a): π2π−2 (π) < π < π2π−1 (π) for π = 1, 2, . . .. (i) if π = 1 then there are 2 = 8 ⋅ 1 − 6 solutions without zeros; (ii) if π = 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) if π = 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) if π ∈ N, then totally there are 8π − 6 (nontrivial) solutions. 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Sadyrbaev, “On nonlinear FucΜΔ±Μk type spectra,” Mathematical Modelling and Analysis, vol. 13, no. 2, pp. 203–210, 2008. [6] A. Gritsans and F. Sadyrbaev, “Nonlinear spectra: the Neumann problem,” Mathematical Modelling and Analysis, vol. 14, no. 1, pp. 33–42, 2009. [7] A. Gritsans and F. Sadyrbaev, “Nonlinear spectra for parameter dependent ordinary differential equations,” Nonlinear Analysis: Modelling and Control, vol. 12, no. 2, pp. 253–267, 2007. [8] F. Sadyrbaev, “Multiplicity in parameter-dependent problems for ordinary differential equations,” Mathematical Modelling and Analysis, vol. 14, no. 4, pp. 503–514, 2009. [9] R. Schaaf, Global Solution Branches of Two Point Boundary Value Problems, vol. 1458 of Lecture Notes in Mathematics, Springer, Berlin, Germany, 1990. [10] A. Gritsans and F. Sadyrbaev, “Time map formulae and their applications,” in Proceedings LU MII “Mathematics, Differential Equations”, vol. 8, pp. 72–93, Riga, Latvia, 2008. [11] C. G. 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