# Multiple Positive Solutions for the Dirichlet Boundary Value Problems by Phase Plane Analysis

Authors: A. Kirichuka; F. Sadyrbaev
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We consider boundary value problems for scalar differential equationx′′+λfx=0, x(0)=0, x(1)=0, where f(x) is a seventh-degree polynomial and λ is a parameter. We use the phase plane method combined with evaluations of time-map functions and make conclusions on the number of positive solutions. Bifurcation diagrams are constructed and examples are considered illustrating the bifurcation processes.


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## Body

## 1. Introduction

Nonlinear boundary value problems for ordinary differential equations still form rapidly developed branch of classical analysis. The traditional issues like existence of solutions, uniqueness, and continuous dependence on boundary data are discussed in a number of classical and modern sources [1, 2]. Less studied are complicated problems of the number of solutions as well as of their dependence on parameters. Of special value are results on existence of positive solutions due to multiple applications. We mention here the works [3–7], where two-point boundary value problems with parameters were considered for the second-order ordinary differential equations. The problem of finding multiple positive solutions was treated in [8]. Nonlinearities of polynomial type were considered in [9]. The time-map technique was applied for investigation of similar problems in [10].Our goal in this note is to demonstrate how elementary phase plane analysis combined with evaluations of time-map functions can provide the researchers with significant information on the number and properties of solutions. We have chosen problem(1)x′′+λfx=0,x0=0,x1=0,where f(x) is a seventh-degree polynomial. Our technique is based on a phase plane analysis. To find positive solutions we will use the first zero function (the so called time-map function). By the first zero function we mean the mapping γ→t1(γ), where t1(γ) is the first zero on the right of a solution of the Cauchy problem(2)x0=0,x′0=γ.The paper [11] discusses the cases of f(x) being third- and fifth-degree polynomials. It was observed that problem (1) may have, respectively, three or five positive solutions. We focus on the case of f(x) being seventh-degree polynomial.In Section2, we provide basic facts about first zero functions. In Section 3, we formulate proposition about number of positive solutions for the Dirichlet boundary value problem. The example in Section 4 provides the detailed description of the respective time-map functions, solutions, and bifurcation diagrams. In the final section, we summarize the results and make conclusions.


## 2. Basic Facts about Time-Map Function

Consider differential equation(3)x′′+fx=0.If x(t,γ) is a solution of (3) with the initial conditions(4)x0=0,x′0=γ, then we denote by t1(γ) the first zero function (time-map) for Cauchy problem (3), (4).Consider the problem with a parameter(5)X′′+λfX=0,X0=0,X′0=αand denote the first zero function U(α,λ).The relation between these two time-map functions was established previously [11, 12]:(6)Uα,λ=1λt1αλ.If x(t;γ) is a solution of the Cauchy problem (3), (4), then(7)Xt;α=xλt;αλsolves the initial value problem (5). For details consult [12].


## 3. Nonlinearity with Seventh-Degree Polynomial

In the sequel, we consider the problem withf(x) the seventh-degree polynomial as follows:(8)x′′=λfx,λ>0,where(9)fx=x-ax-bx-cx-dx-ex-gx-h,0<a<b<c<d<e<g<hwith the Dirichlet conditions(10)x0=0,x1=0.Let functionF(x) be the primitive function of f(x)(11)Fx=∫0xfsdsand let the conditions(12)Fa<Fc<Fe<Fhbe fulfilled.Let us consider phase plane for (8). The value of the first zero function at α is the time needed to move from a point (0,α) to the first intersection point with the X′-axis. Denote by X(t,α) a solution of the Cauchy problem (5). A set of (λ,α) such that (λ,X(t,α)) solves the Dirichlet problem (8), (10) will be calleda solution curve. All such (λ,α) satisfy the equality(13)Uα,λ=1.The phase portrait for (8) has 7 critical points; 3 of them are the points of type “center" and 4 are points of type “saddle”: (a,0), (c,0), (e,0), and (h,0).Consider equation(14)x′′=fx=x-ax-bx-cx-dx-ex-gx-h.Suppose thatγ1, γ2, γ3, and γ4 are initial values such that trajectories Xt,γi,X′t,γi enter the saddle points (as in Figure 1).Figure 1
The phase portrait of (14).Part of the phase portrait for (14) has the form shown in Figure 1.We will look now for solutions of the Dirichlet problem (14), (10) choosing the initial conditions in one of the four intervals separately γ∈(0,γ1), (γ1,γ2), (γ2,γ3), (γ3,γ4).We note that phase portraits of (8) and (14) are equivalent (critical points are the same and trajectories are in one-to-one correspondence due to formula (7)). If we change λ then the initial values for trajectories entering saddle points change also as αi=λγi(i=1,2,3,4).Proposition 1.
For anyλ>0, there exists γ∈(0,γ1(λ)) such that T(γ)=1, where T(γ) is the time-map function for (14).Proof.
Letx(t,γ1) be a solution which goes to the saddle point at x=a. Consider solutions x(t,γ) for 0<γ<γ1 where T(γ) is the time needed for a point to move along phase trajectory from point (0,γ) to point (0,-γ). T(γ) tends to zero as γ goes to zero (since (0,0) is not a critical point) and T(γ) tends to +∞ as γ goes to γ1. By continuity of T(γ) there exists γ∈(0,γ1) such that T(γ)=1.Proposition 2.
For the first zero functionT(γ) of (14) in each of the intervals (γ1,γ2), (γ2,γ3), and (γ3,γ4) there exist Timin=minγi<γ<γi+1T(γ), i=1,2,3, respectively.Proof.
Consider solutionsx(t,γ) for γ1<γ<γ2 where T(γ) is the time needed for a point to move along phase trajectory from point (0,γ) to point (0,-γ). Then T(γ) tends to +∞ as γ goes to γ1 and T(γ) tends to +∞ as γ goes to γ2. Function T(γ) is continuous since the right side of (14) is a polynomial and there is continuous dependence of solutions on initial data. Therefore, there is a minimum of T1min in (γ1,γ2) at some γ1min and, in this case also, the time function has a “U” shaped graph, similarly to T2min and T3min.Remark 3.
The graph of the time-map function has U-shaped segments and the number of positive solutions may be one, two, or more solutions depending on whether the graph ofT(γ) has one, two, or more intersections with the unity level (the unity refers to the length of the interval).Proposition 4.
There existsΛ>0 such that λ>Λ the Dirichlet problem (8), (10) has at least seven solutions.Proof.
It follows from Propositions1 and 2 and formula (6), where t1(α) is the time-map function for equation x′′=fx and f(x) is given in (9).Remark 5.
The relative positions ofTimin, i=1,2,3, do not change if λ tends to +∞. This follows from (6).Remark 6.
Precise number of solutions of problem (8), (10) depends on convexity of time-map function U(α,λ) with respect to α and this can be checked using criteria for ∂U/∂α and ∂2U/∂α2 as in [13].


## 4. Examples

### 4.1. Example 1

Consider equation(15)x′′=λx-1x-2x-4x-5x-9x-10x-12.The phase portrait of (15) has 7 critical points; 3 of them are points of type “center” and 4 are points of type “saddle”: (1,0), (4,0), (9,0), and (12,0) (see Figure 2).Figure 2
The phase portrait of (15), λ=0.0343.Then(16)fx=x-1x-2x-4x-5x-9x-10x-12.The condition (17)F1<F4<F9<F12is fulfilled (see Figure 3).Figure 3
Graphs of the functionf(x) and F(x) (dashed).Next we bring all the time-maps together and as a result get the bifurcation diagram in Figure4.Figure 4
The bifurcation diagram for Example1.The graphs ofU(α,λ) for seven values of λ are depicted in Figures 5–18.Figure 5
The graph of the functionU(α,0.01).Figure 6
The solution of problem (15), (10), α≈17.0132.Figure 7
The graph of the functionU(α,0.01824).Figure 8
Two solutions of problem (15), (10), α≈22.9775, α≈29.8788.Figure 9
The graph of the functionU(α,0.02).Figure 10
Three solutions of problem (15), (10), α≈24.0606, α≈29.4663, and α≈32.5832.Figure 11
The graph of the functionU(α,0.02222).Figure 12
Four solutions of problem (15), (10), α≈25.3608, α≈30.0365, α≈34.7552, and α≈41.4674.Figure 13
The graph of the functionU(α,0.0241).Figure 14
Five solutions of problem (15), (10), α≈26.4119, α≈30.6763, α≈36.3664, α≈40.8286, and α≈42.7074.Figure 15
The graph of the functionU(α,0.02711).Figure 16
Six solutions of problem (15), (10), α≈28.0128, α≈28.4601, α≈31.8026, α≈38.7197, α≈42.1239, and α≈45.4026.Figure 17
The graph of the functionU(α,0.0343).Figure 18
Seven solutions of problem (15), (10), α≈31.5093, α≈31.6248, α≈32.3915, α≈34.6074, α≈43.6897, α≈45.7099, and α≈51.0893.


### 4.2. Example 2

Consider equation(18)x′′=λx-1x-2x-5x-6x-8x-10x-12.The 012B phase portrait of (18) has 7 critical points; 3 of them are the points of type “center" and 4 are points of type “saddle”: (1,0), (5,0), (8,0), and (12,0). The structure of the phase portrait for (18) is the same as that for (15).Then(19)fx=x-1x-2x-5x-6x-8x-10x-12.The condition for the saddle points(20)F1<F5<F8<F12is fulfilled also.If we look at the bifurcation diagram in Figure19, we see that there is no such value λ where at some α the Dirichlet problem (18), (10) has seven solution because the difference between minimum values of the time-map function is large enough.Figure 19
The bifurcation diagram for Example2.


## 4.1. Example 1

Consider equation(15)x′′=λx-1x-2x-4x-5x-9x-10x-12.The phase portrait of (15) has 7 critical points; 3 of them are points of type “center” and 4 are points of type “saddle”: (1,0), (4,0), (9,0), and (12,0) (see Figure 2).Figure 2
The phase portrait of (15), λ=0.0343.Then(16)fx=x-1x-2x-4x-5x-9x-10x-12.The condition (17)F1<F4<F9<F12is fulfilled (see Figure 3).Figure 3
Graphs of the functionf(x) and F(x) (dashed).Next we bring all the time-maps together and as a result get the bifurcation diagram in Figure4.Figure 4
The bifurcation diagram for Example1.The graphs ofU(α,λ) for seven values of λ are depicted in Figures 5–18.Figure 5
The graph of the functionU(α,0.01).Figure 6
The solution of problem (15), (10), α≈17.0132.Figure 7
The graph of the functionU(α,0.01824).Figure 8
Two solutions of problem (15), (10), α≈22.9775, α≈29.8788.Figure 9
The graph of the functionU(α,0.02).Figure 10
Three solutions of problem (15), (10), α≈24.0606, α≈29.4663, and α≈32.5832.Figure 11
The graph of the functionU(α,0.02222).Figure 12
Four solutions of problem (15), (10), α≈25.3608, α≈30.0365, α≈34.7552, and α≈41.4674.Figure 13
The graph of the functionU(α,0.0241).Figure 14
Five solutions of problem (15), (10), α≈26.4119, α≈30.6763, α≈36.3664, α≈40.8286, and α≈42.7074.Figure 15
The graph of the functionU(α,0.02711).Figure 16
Six solutions of problem (15), (10), α≈28.0128, α≈28.4601, α≈31.8026, α≈38.7197, α≈42.1239, and α≈45.4026.Figure 17
The graph of the functionU(α,0.0343).Figure 18
Seven solutions of problem (15), (10), α≈31.5093, α≈31.6248, α≈32.3915, α≈34.6074, α≈43.6897, α≈45.7099, and α≈51.0893.


## 4.2. Example 2

Consider equation(18)x′′=λx-1x-2x-5x-6x-8x-10x-12.The 012B phase portrait of (18) has 7 critical points; 3 of them are the points of type “center" and 4 are points of type “saddle”: (1,0), (5,0), (8,0), and (12,0). The structure of the phase portrait for (18) is the same as that for (15).Then(19)fx=x-1x-2x-5x-6x-8x-10x-12.The condition for the saddle points(20)F1<F5<F8<F12is fulfilled also.If we look at the bifurcation diagram in Figure19, we see that there is no such value λ where at some α the Dirichlet problem (18), (10) has seven solution because the difference between minimum values of the time-map function is large enough.Figure 19
The bifurcation diagram for Example2.


## 5. Conclusions

For a polynomial of the type (9) but with arbitrary large odd number of zeros k in presence of the condition F(a1)<F(a3)<⋯<F(ak) (ai are the zeros) there always exists large enough λ such that the respective boundary value problem (1) has at least k positive solutions.Ifλ is not large then in presence of the condition F(a1)<F(a3)<⋯<F(ak) the number of solutions of the boundary value problem (1) depends on the relative positions of minima of the first zero function (time-map) T(γ). These relative positions may be regulated by the choice of parameters a1,a2,…,ak.



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