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ISRN Mathematical Analysis
VolumeΒ 2012Β (2012), Article IDΒ 869147, 11 pages
Research Article

Existence of Alternate Steady States in a Phosphorous Cycling Model

1Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762, USA
2Department of Mathematics and Statistics, University of North Carolina at Greensboro, Greensboro, NC 27412, USA

Received 20 January 2012; Accepted 8 February 2012

Academic Editors: J.-F.Β Colombeau, G. L.Β Karakostas, and P.Β Omari

Copyright Β© 2012 Dagny Butler et al. 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.


We analyze the positive solutions to the steady-state reaction diffusion equation with Dirichlet boundary conditions of the form: βˆ’Ξ”π‘’=πœ†[πΎβˆ’π‘’+𝑐(𝑒4/(1+𝑒4))],π‘₯∈Ω,𝑒=0,π‘₯βˆˆπœ•Ξ©. Here, Δ𝑒=div(βˆ‡π‘’) is the Laplacian of 𝑒, 1/πœ† is the diffusion coefficient, 𝐾 and 𝑐 are positive constants, and Ξ©βŠ‚β„π‘ is a smooth bounded region with πœ•Ξ© in 𝐢2. This model describes the steady states of phosphorus cycling in stratified lakes. Also, it describes the colonization of barren soils in drylands by vegetation. In this paper, we discuss the existence of multiple positive solutions leading to the occurrence of an S-shaped bifurcation curve. We prove our results by the method of subsuper solutions.

1. Introduction

The nonlinear boundary value problemξ‚Έπ‘’βˆ’Ξ”π‘’=πœ†πΎβˆ’π‘’+𝑐41+𝑒4ξ‚Ή=βˆΆπœ†π‘“(𝑒),π‘₯∈Ω,𝑒=0,π‘₯βˆˆπœ•Ξ©,(1.1) where Δ𝑒=div(βˆ‡π‘’) is the Laplacian of 𝑒,1/πœ† is the diffusion coefficient, 𝐾 and 𝑐 are positive constants, and Ξ©βŠ‚β„π‘ is a smooth bounded region with πœ•Ξ© in 𝐢2. This model describes phosphorus cycling in stratified lakes (see [1]). In particular, it illustrates the decrease in the amount of phosphorus in the epilimnion (upper layer) and the rapid recycling that occurs when the hypolimnion (lower layer) is depleted of oxygen. Here, 𝑒 is the mass or concentration of phosphorous (P) in the water column, and 𝐾 is the rate of P input from the watershed. The rate of recycling of P is given by 𝑐𝑒4/(1+𝑒4), where 𝑐 is the maximum recycling rate. The assumption here is that the recycling is primarily from the sediments. The same equation has also been used to describe plant colonization of barren soils in drylands (see [2]). In this case, 𝑒 is the amount of barren soil, and 𝑐𝑒4/(1+𝑒4) represents erosion by wind and runoff.

It is known that when 𝑓(𝑒)>0;[0,∞) with limπ‘’β†’βˆž(𝑓(𝑒)/𝑒)=0, a positive solution exists for all πœ†>0, and this solution is unique if 𝑒/𝑓(𝑒) is increasing. The existence of multiple positive solutions to such problems has also been studied extensively (see [3–6]). On the other hand, proving multiplicity results for nonlinearities with a falling zero (say at π‘Ÿ0>0) is very challenging and often remains an open problem (see [7] and example (iv) in [4]). For such problems, the solution space is restricted as β€–π‘’β€–βˆž<π‘Ÿ0 (see Figure 1). Our model falls in this category.

Figure 1: Restricted solution space for the falling zero problem.

Instead of working with the particular reaction term in (1.1), we will prove our results for a class of functions 𝑓 which satisfy the following hypothesis:(H1)π‘“βˆˆπΆ2([0,∞)),𝑓(𝑒)>0 on [0,π‘Ÿ0) and 𝑓(𝑒)<0 for 𝑒>π‘Ÿ0.

To state our multiplicity result, for 0<π‘Ž<𝑏, let𝑄(π‘Ž,𝑏,Ξ©)∢=(𝑏/𝑓(𝑏))((𝑁+1)/𝑁)𝑁+1𝑁2/𝑅2‖‖𝑒minπ‘Ž/Ξ©β€–β€–βˆžπ‘“βˆ—(π‘Ž),2𝑁𝑀/𝑓(𝑏)𝑅2ξ€Ύ,(1.2) where 𝐡𝑅=𝐡(0,𝑅) is the largest inscribed ball on Ξ©, π‘“βˆ—(𝑠)=maxπ‘‘βˆˆ[0,𝑠]𝑓(𝑑) (see Figure 2) and 𝑒Ω is the unique positive solution of βˆ’Ξ”π‘’=1 in Ξ©,𝑒=0 on πœ•Ξ©,.

Figure 2: Definition of 𝐡𝑅 and π‘“βˆ—.

Now, we establish the following result.

Theorem 1.1. Let π‘š,π‘€βˆˆ(0,π‘Ÿ0) be such that 𝑓 is nondecreasing in (π‘š,𝑀). Assume that there exists π‘βˆˆ[π‘š,𝑀] and π‘Žβˆˆ(0,𝑏) such that 𝑄(π‘Ž,𝑏,Ξ©)<1, then (1.1) has three positive solutions for πœ†βˆˆ((𝑏/𝑓(𝑏))((𝑁+1)/𝑁)𝑁+1(𝑁2/𝑅2),min{π‘Ž/β€–π‘’Ξ©β€–βˆžπ‘“βˆ—(π‘Ž),2𝑁𝑀/𝑓(𝑏)𝑅2}).

We will use the method of subsupersolutions to prove our results. By a subsolution (supersolution) of (1.1), we mean a function πœ“βˆˆπ‘Š1,2β‹‚(Ξ©)𝐢(Ξ©) such that πœ“=0 on πœ•Ξ© andξ€œΞ©ξ€œβˆ‡πœ“β‹…βˆ‡π‘žβ‰€(β‰₯)Ξ©πœ†π‘“(πœ“)π‘ž,(1.3) for every π‘žβˆˆ{πœ‚βˆˆπΆβˆž0(Ξ©)βˆΆπœ‚β‰₯0inΞ©}. Then, the following lemma holds.

Lemma 1.2. Let πœ“ be a subsolution of (1.1), and let πœ™ be a supersolution of (1.1) such that πœ“β‰€πœ™, then, (1.1) has a solution π‘’βˆˆπΆ2⋂𝐢(Ξ©)1(Ξ©) such that πœ“β‰€π‘’β‰€πœ™.

To establish our main multiplicity result (Theorem 1.1), we use the following very useful result discussed in [8, 9].

Lemma 1.3. Suppose that there exists a subsolution πœ“1, a strict supersolution 𝑍1, a strict subsolution πœ“2, and a supersolution 𝑍2 for (1.1) such that πœ“1<𝑍1<𝑍2, πœ“1<πœ“2<𝑍2, and let πœ“2≰𝑍1, then, (1.1) has at least three distinct solutions 𝑒1,𝑒2, and 𝑒3 such that πœ“1≀𝑒1<𝑒2<𝑒3≀𝑍2.

Note here that by πœ“1<πœ“2 we mean that πœ“1β‰€πœ“2 and πœ“1β‰ πœ“2.

We prove Theorem 1.1 in Section 2. The proof of Theorem 1.1 is motivated by the arguments in [7] where the authors establish a multiplicity result for a model used to describe a logistically growing species with grazing. In Section 3, we analyze in detail the phosphorus cycling model when 𝑓(𝑒)=πΎβˆ’π‘’+𝑐(𝑒4/(1+𝑒4)) has a unique positive zero π‘Ÿ0. This will be the case when 𝐾>𝐾0∢=(3/4)4√3/5βˆ’(1/4)4√3/55 and 𝑐≫1. We will prove that an 𝑆-shaped bifurcation curve occurs when 𝑐≫1 and 𝐾0<𝐾<9𝑐/16. This analysis turned out to be quite nontrivial and challenging. This study is motivated by the results in [10, 11] where such a multiplicity result for the case 𝑁=1 was discussed. Here, we extend this study for the higher dimension case. We also obtained more detailed analytical and computational results for the case 𝑁=1, which are presented in the appendix.

2. Proof of Theorem 1.1

To establish the multiplicity result, we have to construct a subsolution πœ“1, a strict supersolution 𝑍1, a strict subsolution πœ“2, and a supersolution 𝑍2 for (1.1) such that πœ“1<𝑍1<𝑍2, πœ“1<πœ“2<𝑍2, and πœ“2≰𝑍1. Clearly, πœ“1=0 is a strict subsolution since 𝑓(0)>0. For the large supersolution, choose 𝑍2=𝑀(πœ†)𝑒Ω where 𝑀(πœ†)>πœ†maxπ‘‘βˆˆ[0,π‘Ÿ0]𝑓(𝑑). Then, βˆ’Ξ”π‘2=𝑀(πœ†)β‰₯πœ†π‘“(𝑍2) making 𝑍2 a positive supersolution.

Now for the smaller strict supersolution, define 𝑍1=π‘Žπ‘’Ξ©/β€–π‘’Ξ©β€–βˆž. Since πœ†<π‘Ž/β€–π‘’Ξ©β€–βˆžπ‘“βˆ—(π‘Ž),βˆ’Ξ”π‘1=π‘Ž/β€–π‘’Ξ©β€–βˆž>πœ†π‘“βˆ—(π‘Ž)β‰₯πœ†π‘“βˆ—(π‘Žπ‘’Ξ©/β€–π‘’Ξ©β€–βˆž)β‰₯πœ†π‘“(π‘Žπ‘’Ξ©/β€–π‘’Ξ©β€–βˆž)=πœ†π‘“(𝑍1) in Ξ©. Here, π‘“βˆ—(𝑠)=maxπ‘‘βˆˆ[0,𝑠]𝑓(𝑑). Hence, 𝑍1 is a strict supersolution.

We will now construct the strict subsolution πœ“2. Let𝑓𝑓(𝑒)=(𝑒),𝑒<π‘š,𝑓(𝑒),𝑒β‰₯π‘š,(2.1) where 𝑓(𝑒) is defined so that the function 𝑓(𝑒) is strictly increasing on (0,𝑀) and 𝑓(𝑒)≀𝑓(𝑒) (see Figure 3).

Figure 3: Graph of 𝑓(𝑒).

Let𝜌⎧βŽͺ⎨βŽͺ⎩[],ξ‚Έξ‚€(π‘Ÿ)=1,π‘Ÿβˆˆ0,πœ–1βˆ’1βˆ’π‘…βˆ’π‘Ÿξ‚π‘…βˆ’πœ–π›½ξ‚Ήπ›Ό],π‘Ÿβˆˆ(πœ–,𝑅,𝛼,𝛽>1.(2.2) Note that⎧βŽͺ⎨βŽͺ⎩[],ξ‚Έξ‚€πœŒβ€²(π‘Ÿ)=0,π‘Ÿβˆˆ0,πœ–βˆ’π›Όπ›½1βˆ’π‘…βˆ’π‘Ÿξ‚π‘…βˆ’πœ–π›½ξ‚Ήπ›Όβˆ’1ξ‚€π‘…βˆ’π‘Ÿξ‚π‘…βˆ’πœ–π›½βˆ’1],π‘Ÿβˆˆ(πœ–,𝑅,𝛼,𝛽>1(2.3) and |πœŒξ…ž(π‘Ÿ)|<𝛼𝛽/(π‘…βˆ’πœ–). Now define 𝑀(π‘Ÿ)∢=π‘πœŒ(π‘Ÿ) andπœ“2ξ‚»(π‘₯)=ξ‚πœ“2,π‘₯βˆˆπ΅π‘…,0,π‘₯βˆˆΞ©βˆ’π΅π‘…,(2.4) where ξ‚πœ“2 is the solution ofβˆ’ξ‚πœ“2ξ…žξ…ž(π‘Ÿ)βˆ’π‘βˆ’1π‘Ÿξ‚πœ“ξ…ž2(π‘Ÿ)=πœ†π‘“(𝑀(π‘Ÿ)),π‘Ÿβˆˆ(0,𝑅),ξ‚πœ“ξ…ž2(0)=0=ξ‚πœ“2(𝑅),(2.5) and 𝐡𝑅 is the largest inscribed ball in Ξ©. Then, πœ“2βˆˆπ‘Š1,2β‹‚(Ξ©)𝐢(Ξ©) and πœ“2=0 on πœ•Ξ©. We will now establish that ξ‚πœ“2(π‘Ÿ)∈(𝑀(π‘Ÿ),𝑀] on [0,𝑅). Then, βˆ’Ξ”πœ“2=πœ†π‘“(𝑀(π‘Ÿ))<πœ†π‘“(ξ‚πœ“2(π‘Ÿ))β‰€πœ†π‘“(πœ“2(π‘Ÿ)) on [0,𝑅), while outside 𝐡𝑅, we have βˆ’Ξ”πœ“2=0=πœ†π‘“(0)=πœ†π‘“(πœ“2), and hence, πœ“2 will be a strict subsolution.

First, we will show that ξ‚πœ“2(π‘Ÿ)≀𝑀. Nowξ€·π‘Ÿπ‘βˆ’1ξ‚πœ“ξ…ž2ξ€Έ(π‘Ÿ)ξ…ž=βˆ’πœ†π‘Ÿπ‘βˆ’1𝑓(𝑀(π‘Ÿ)),ξ‚πœ“ξ…ž2(π‘Ÿ)=βˆ’πœ†π‘Ÿπ‘βˆ’1ξ€œπ‘Ÿ0π‘ π‘βˆ’1𝑓(𝑀(𝑠))𝑑𝑠,ξ‚πœ“2(𝑑)βˆ’ξ‚πœ“2ξ€œ(0)=βˆ’π‘‘0πœ†π‘Ÿπ‘βˆ’1ξ‚»ξ€œπ‘Ÿ0π‘ π‘βˆ’1𝑓(𝑀(𝑠))π‘‘π‘ π‘‘π‘Ÿ.(2.6) But ξ‚πœ“2(𝑅)=0. Hence, we getξ‚πœ“2ξ€œ(0)=𝑅0πœ†π‘Ÿπ‘βˆ’1ξ‚»ξ€œπ‘Ÿ0π‘ π‘βˆ’1ξ‚ξ‚Όβ‰€πœ†ξ‚π‘“π‘“(𝑀(𝑠))π‘‘π‘ π‘‘π‘Ÿ(𝑏)π‘ξ€œπ‘…0=π‘Ÿπ‘‘π‘ πœ†π‘“(𝑏)𝑅2𝑁2.since𝑏β‰₯π‘š,𝑓(𝑠)=𝑓(𝑠)for𝑠β‰₯π‘š(2.7) But πœ†<2𝑁𝑀/𝑓(𝑏)𝑅2. Hence, β€–ξ‚πœ“2β€–βˆž=ξ‚πœ“2(0)<𝑀.

Next, to establish ξ‚πœ“2>𝑀 on [0,𝑅], we will show that ξ‚πœ“ξ…ž2<π‘€ξ…žβ‰€0 on [0,𝑅]. This will be sufficient, since ξ‚πœ“2(𝑅)=𝑀(𝑅)=0. Now π‘€ξ…ž=0 and ξ‚πœ“ξ…ž2<0 in the interval [0,πœ–), and hence, ξ‚πœ“ξ…ž2<π‘€ξ…žβ‰€0 in that interval. For π‘Ÿ>πœ–, we haveβˆ’ξ‚πœ“ξ…ž2πœ†(π‘Ÿ)=π‘Ÿπ‘βˆ’1ξ€œπ‘Ÿ0π‘ π‘βˆ’1β‰₯πœ†π‘“(𝑀(𝑠))π‘‘π‘ π‘Ÿπ‘βˆ’1ξ€œπœ–0π‘ π‘βˆ’1=πœ†π‘“(𝑀(𝑠))π‘‘π‘ π‘Ÿπ‘βˆ’1ξ€œπœ–0π‘ π‘βˆ’1𝑓β‰₯πœ†ξ‚(𝑏)𝑑𝑠(since𝜌(𝑠)=1,𝑠<πœ–)𝑓(𝑏)π‘…π‘βˆ’1ξ€œπœ–0π‘ π‘βˆ’1=π‘‘π‘ πœ†π‘“(𝑏)π‘…π‘βˆ’1πœ–π‘π‘ξ‚€ξ‚ξ‚.since𝑏β‰₯π‘š,𝑓(𝑠)=𝑓(𝑠)for𝑠β‰₯π‘š(2.8) We also know that |π‘€ξ…ž(π‘Ÿ)|≀𝑏𝛼𝛽/(π‘…βˆ’πœ–). Hence, |ξ‚πœ“ξ…ž2(π‘Ÿ)|>|π‘€ξ…ž(π‘Ÿ)| if πœ†>𝛼𝛽(𝑏/𝑓(𝑏))(π‘π‘…π‘βˆ’1/(π‘…βˆ’πœ–)πœ–π‘). But min0<πœ–<𝑅(1/(π‘…βˆ’πœ–)πœ–π‘)=(𝑁+1)𝑁+1/𝑁𝑁𝑅𝑁+1, and this minimum is achieved at πœ–0=𝑁𝑅/(𝑁+1). Since πœ†>(𝑏/𝑓(𝑏))(𝑁2/𝑅2)((𝑁+1)/𝑁)𝑁+1=(𝑏/𝑓(𝑏))(π‘π‘…π‘βˆ’1/(π‘…βˆ’πœ–0)πœ–π‘0), we can choose πœ–=πœ–0 and 𝛼,𝛽>1 such that πœ†>𝛼𝛽(𝑏/𝑓(𝑏))(π‘π‘…π‘βˆ’1/(π‘…βˆ’πœ–0)πœ–π‘0). Hence, |ξ‚πœ“ξ…ž2(π‘Ÿ)|>|π‘€ξ…ž(π‘Ÿ)| on (0,𝑅). This implies 𝑀<ξ‚πœ“2. Thus, πœ“2 is a strict subsolution of (1.1) if (𝑏/𝑓(𝑏))(𝑁2/𝑅2)((𝑁+1)/𝑁)𝑁+1<πœ†<2𝑁𝑀/𝑓(𝑏)𝑅2. Furthermore, ξ‚πœ“2(0)>𝑀(0)=𝑏>π‘Ž=‖𝑍1β€–βˆž, that is, πœ“2≰𝑍1. Moreover, 𝑀(πœ†) can be chosen large enough so that πœ“2<𝑍2 and 𝑍1<𝑍2. Hence, by Lemma 1.3, Theorem 1.1 holds.

3. Results for the Example 𝑓(𝑒)=πΎβˆ’π‘’+𝑐(𝑒4/(1+𝑒4))

First, we will analyze some properties of this nonlinearity. We will show that for large 𝑐 we can find values of 𝐾 for which the function 𝑓(𝑒)=πΎβˆ’π‘’+𝑐(𝑒4/(1+𝑒4)) satisfies (H1), and we will also identify π‘š and 𝑀 such that 𝑓 is increasing in (π‘š,𝑀). Clearly, π‘“βˆˆπΆ2([0,∞)), 𝑓(0)=𝐾, and π‘“ξ…ž(0)=βˆ’1.

Proposition 3.1. If 𝑐>16/54√135, then there exists two points π‘š1π‘Žπ‘›π‘‘π‘š2 such that 0<π‘š1<π‘š2 and π‘“ξ…ž(π‘šπ‘–)=0 for 𝑖=1,2.

Proof. We have π‘“ξ…ž(𝑒)=βˆ’1+(4𝑐𝑒3/(1+𝑒4)2). So π‘“ξ…ž(𝑒)=0, when 1=4𝑐𝑒3/(1+𝑒4)2. Let 𝑔(𝑒)∢=4𝑐𝑒3/(1+𝑒4)2, and let β„Ž(𝑒)∢=1. We have 𝑔(𝑒)β‰₯0, 𝑔(0)=0, and limπ‘’β†’βˆžπ‘”(𝑒)=0. Since 𝑔′(𝑒)=4𝑐𝑒2(3βˆ’5𝑒4)/(1+𝑒4)3, we can see that 𝑔(𝑒) achieves a maximum of (5𝑐/16)4√135 at 𝑒=4√3/5. If maxπ‘₯∈(0,∞)𝑔(𝑒)=(5𝑐/16)4√135>1, then the line β„Ž(𝑒) will cut 𝑔(𝑒) at exactly two different points. Hence, if 𝑐>16/54√135, then there are exactly two positive points π‘š1<π‘š2 such that π‘“ξ…ž(π‘šπ‘–)=0 for 𝑖=1,2.

Proposition 3.2. If 𝐾>(3/4)4√3/5βˆ’(1/4)(4√3/5)5=∢𝐾0, then there exists a unique π‘Ÿ0>0 such that 𝑓(π‘Ÿ0)=0.

Proof. From Figure 5, we can see that if 𝑓(π‘š1)>0, then 𝑓(𝑒) has a unique positive zero. Since π‘“ξ…ž(π‘š1)=0, we obtain π‘π‘š31/(1+π‘š41)=(1+π‘š41)/4. So, 𝑓(π‘š1)=πΎβˆ’π‘š1+π‘π‘š41/(1+π‘š41)=πΎβˆ’π‘š1+(π‘š1(1+π‘š41)/4)=πΎβˆ’(3/4)π‘š1+π‘š51/4. Hence, 𝑓(π‘š1)>0 if 𝐾>(3/4)π‘š1+π‘š51/4. On analyzing π‘“ξ…žξ…ž(𝑒)=4𝑐𝑒2(3βˆ’5𝑒4)/(1+𝑒4)3, we see that the positive inflection of 𝑓(𝑒) occurs at 𝑒=4√3/5. Thus, π‘š1<4√3/5, and hence 𝐾>𝐾0 ensures that there exists a unique π‘Ÿ0 such that 𝑓(π‘Ÿ0)=0.

Choose π‘š=π‘š1 and 𝑀=π‘š2. Thus, given 𝐾>𝐾0, we can find 𝑐 large so that 𝑓(𝑒) is increasing on (π‘š,𝑀), and there exists a unique π‘Ÿ0>0 such that 𝑓(π‘Ÿ0)=0, that is, 𝑓(𝑒) satisfies (H1). Now, we will prove that the other assumptions in Theorem 1.1 hold in the given example.

We will select π‘βˆˆ[π‘š,𝑀] and π‘Žβˆˆ(0,𝑏) such that 𝑄(π‘Ž,𝑏,Ξ©)<1. The point at which the function 𝑒/𝑓(𝑒) has a minimum would be an ideal choice for 𝑏.

Proposition 3.3. If 𝐾<9𝑐/16, then 𝑒/𝑓(𝑒) has the shape given in Figure 6.

Proof. We have (𝑒/𝑓(𝑒))ξ…ž=(𝑓(𝑒)βˆ’π‘’π‘“β€²(𝑒))/(𝑓(𝑒))2. Hence, the critical points of 𝑒/𝑓(𝑒) are given by 𝑓(𝑒)βˆ’π‘’π‘“β€²(𝑒)=0, and in particular, the nonzero critical points are given by 𝐾+𝑐(𝑒8βˆ’3𝑒4)/(1+𝑒4)2=0. Solving for 𝑒, we get the positive critical points as 𝛼=4ξ”βˆš3π‘βˆ’2πΎβˆ’π‘(9π‘βˆ’16𝐾)/2(𝑐+𝐾) and 𝛽=4ξ”βˆš3π‘βˆ’2𝐾+𝑐(9π‘βˆ’16𝐾)/2(𝑐+𝐾). Note that if 𝐾<9𝑐/16, then 𝛼 and 𝛽 are positive real roots of (𝑒/𝑓(𝑒))β€² with 𝛼<𝛽. Hence, 𝑒/𝑓(𝑒) has a relative maximum at 𝛼 and a relative minimum at 𝛽.

Since 𝛽→4√3 as π‘β†’βˆž and π‘š<4√3/5, we have π‘š<𝛽. Furthermore, it is clear from Figure 4 that π‘€β†’βˆž as π‘β†’βˆž, so 𝛽<𝑀 for 𝑐≫1. Thus, we have π›½βˆˆ[π‘š,𝑀] for large 𝑐 and we choose 𝑏=𝛽. We also choose π‘Žβˆˆ(0,𝑀) such that 𝑓(π‘Ž)=π‘“βˆ—(π‘Ž)=𝑓(0) (see Figure 7).

Figure 4: 𝑔(𝑒) and β„Ž(𝑒).
Figure 5: Graph of 𝑓(𝑒) with π‘š and 𝑀.
Figure 6: Graph of 𝑒/𝑓(𝑒).
Figure 7: Graph of π‘“βˆ—(𝑒).

The following estimates hold for π‘Ž and 𝑀 for 𝑐≫1.

Proposition 3.4. For 𝑐≫1,(i)  5βˆšπ‘<𝑀<4βˆšπ‘,(ii)  1/3βˆšπ‘<π‘Ž<1/4βˆšπ‘.

Proof. (i) By the shape of the graph of 𝑓(𝑒) established in Propositions 3.1 and 3.3 (see Figure 5), it is enough if we prove that π‘“ξ…ž(5βˆšπ‘)>0 and π‘“ξ…ž(4βˆšπ‘)<0. We have π‘“ξ…ž(5βˆšπ‘)=βˆ’1+4𝑐8/5/(1+𝑐4/5)2=βˆ’1+4𝑐8/5/(1+2𝑐4/5+𝑐8/5)=βˆ’1+4/(1/𝑐8/5+2/𝑐4/5+1)>0, and π‘“ξ…ž(4βˆšπ‘)=βˆ’1+4𝑐7/4/(1+𝑐)2=βˆ’1+(4/(1/𝑐7/4+2/𝑐3/4+𝑐1/4))<0 for 𝑐≫1. Thus, 5βˆšπ‘<𝑀<4βˆšπ‘ for large 𝑐.
(ii) We have 𝑓(π‘Ž)=𝐾, simplifying which we get π‘Ž4βˆ’π‘π‘Ž3+1=0. Define 𝑗(𝑒)∢=𝑒4βˆ’π‘π‘’3+1. If 𝑒<π‘Ž, then 𝑗(𝑒)>0, and if π‘’βˆˆ(π‘Ž,𝑀), then 𝑗(𝑒)<0. We have 𝑗(1/3βˆšπ‘)=(1/3βˆšπ‘)4βˆ’π‘(1/3βˆšπ‘)3+1=1/𝑐4/3>0, and 𝑗(1/4βˆšπ‘)=(1/4βˆšπ‘)4βˆ’π‘(1/4βˆšπ‘)3+1=1/π‘βˆ’4βˆšπ‘+1<0 for 𝑐≫1. Hence, 1/3βˆšπ‘<π‘Ž<1/4βˆšπ‘ for large 𝑐. Note that since 𝑏=𝛽→4√3 and π‘Ž<1/4βˆšπ‘β†’0 as π‘β†’βˆž, π‘Žβˆˆ(0,𝑏) for 𝑐≫1.

Now we will discuss the existence of at least three positive solutions for a certain range of πœ† (see Theorem 1.1). Our aim is to prove that for 𝑐≫1 and 𝐾0<𝐾<9𝑐/16, 𝑄(π‘Ž,𝑏,Ξ©)<1. It is enough if we prove that𝑏𝑓(𝑏)𝑁+1𝑁𝑁+1𝑁2𝑅2ξ‚»π‘Ž<minβ€–β€–π‘’Ξ©β€–β€–βˆžπ‘“βˆ—,(π‘Ž)2𝑁𝑀𝑓(𝑏)𝑅2ξ‚Ό.(3.1) Note that for 𝑐≫1(𝑏/𝑓(𝑏))=4√3/(πΎβˆ’4√3+3𝑐/4) and 𝑀/𝑓(𝑏)=𝑀/(πΎβˆ’4√3+3𝑐/4)>5βˆšπ‘/(πΎβˆ’4√3+3𝑐/4). Also, π‘Ž/π‘“βˆ—(π‘Ž)=π‘Ž/π‘˜>1/𝐾3βˆšπ‘. Applying the estimates we obtained for π‘Ž/π‘“βˆ—(π‘Ž),𝑏/𝑓(𝑏), and 𝑀/𝑓(𝑏) to the above inequality, we get the following:4√3π‘˜βˆ’4√ξƒͺξ‚€3+3𝑐/4𝑁+1𝑁𝑁+1𝑁2𝑅2ξƒ―1<minβ€–β€–π‘’Ξ©β€–β€–βˆžπ‘˜3βˆšπ‘,2𝑁𝑅25βˆšπ‘π‘˜βˆ’4√3+3𝑐/4ξƒͺξƒ°.(3.2) Simplifying the above, we can see that 𝑄(π‘Ž,𝑏,Ξ©)<1 if4√3π‘˜/π‘βˆ’4√ξƒͺξ‚€3/𝑐+3/4𝑁+1𝑁𝑁+1𝑁2𝑅2𝑐<min2/3β€–β€–π‘’Ξ©β€–β€–βˆžπ‘˜,2𝑁𝑅2𝑐1/5π‘˜/π‘βˆ’4√3/𝑐+3/4ξƒͺξƒ°.(3.3) Clearly, this inequality is true for 𝑐≫1; hence, Theorem 1.1 holds.


Consider the two-point boundary value problemβˆ’π‘’ξ…žξ…ž=πœ†π‘“(𝑒),π‘₯∈(0,1),𝑒(0)=0=𝑒(1),(A.1) where 𝑓 satisfies the following hypotheses:(G1)π‘“βˆˆπΆ2([0,∞)),𝑓(𝑒)>0for0<𝑒<π‘Ÿ0 and 𝑓(𝑒)<0 for 𝑒>π‘Ÿ0 for some π‘Ÿ0>0,(G2) there exists π‘˜β‰₯0 such that 𝑓(𝑒)βˆ’π‘“(𝑣)β‰₯βˆ’π‘˜(π‘’βˆ’π‘£) for all 𝑒,π‘£βˆˆ[0,π‘Ÿ0) with 𝑒>𝑣.

Using the quadrature method (see [12]), it follows that (A.1) has a positive solution if and only ifβˆšβˆšπœ†=2ξ€œπœŒ0π‘‘π‘§βˆš[]𝐹(𝜌)βˆ’πΉ(𝑧)∢=𝐺(𝜌),(A.2) where ∫𝐹(𝑠)∢=𝑠0𝑓(𝑑)𝑑𝑑 and 𝜌=𝑒(1/2)=β€–π‘’β€–βˆž. Further, 𝑒(π‘₯) is symmetric about π‘₯=1/2 and is given byξ€œ0𝑒(π‘₯)π‘‘π‘§βˆš[]=√𝐹(𝜌)βˆ’πΉ(𝑧)12πœ†π‘₯,0<π‘₯<2.(A.3)

Equation (A.2) describes the bifurcation curve of positive solutions of (A.1), and it follows by results in [12] that limπœŒβ†’0+𝐺(𝜌)=0 and limπœŒβ†’π‘Ÿβˆ’0𝐺(𝜌)=∞. Furthermore, from [4], we haveπΊξ…žβˆš(𝜌)=2ξ€œ10𝐻(𝜌)βˆ’π»(πœŒπ‘ )[]𝐹(𝜌)βˆ’πΉ(πœŒπ‘ )3/2𝑑𝑠,(A.4) where 𝐻(𝑒)=𝐹(𝑒)βˆ’π‘’/2𝑓(𝑒). Note that 𝐻(0)=0 and π»ξ…ž(0)=1/2𝑓(0)>0. Hence, if there exists a point 𝜌0∈(0,π‘Ÿ0) such that 𝐻(𝜌0)<0, (see Figure 8) then πΊξ…ž(𝜌)<0 for a certain range of 𝜌; thus, the bifurcation diagram must be at least 𝑆-shaped. We will now prove that such a 𝜌0 exists when 𝐾>𝐾0 and 𝑐>5.20626𝐾.

Figure 8: Graph of 𝐻(𝑒).

Consider the case 𝑓(𝑒)=πΎβˆ’π‘’+𝑐(𝑒4/(1+𝑒4)). Clearly, given 𝐾>𝐾0, then for 𝑐≫1𝑓 satisfies (G1)-(G2) (see Proposition 3.3). Hence, 𝐺(𝜌) is defined for all πœŒβˆˆπ‘†=(0,π‘Ÿ0).

We have 𝐻(𝑒)=𝐹(𝑒)βˆ’π‘’/2𝑓(𝑒)=𝐾𝑒/2+π‘π‘’βˆ’π‘(𝑒5/2(1+𝑒4√))βˆ’π‘/42{βˆ’2tanβˆ’1√(1βˆ’2𝑒)+2tanβˆ’1√(1+√2𝑒)βˆ’ln(1+2𝑒+𝑒2√)/(1βˆ’2𝑒+𝑒2)}. Clearly, 4√3/5<π‘Ÿ0 (see Proposition 3.3); choose 𝜌0=4√3/5. Thus, 𝐻(𝜌0)=.440056πΎβˆ’.0845244𝑐, and hence, 𝐻(𝜌0)<0 if 𝑐>5.20626𝐾.

We finally used Mathematica to compute βˆšπœ†=𝐺(𝜌) in the case when 𝑓(𝑒)=πΎβˆ’π‘’+𝑐(𝑒4/(1+𝑒4)) and plotted the bifurcation diagrams. We found that the bifurcation diagrams are, in fact, exactly 𝑆-shaped when multiplicity occurred. Figures 9 and 10 describe the bifurcation diagrams for a certain value of 𝑐 and 𝐾.

Figure 9: Bifurcation diagram with 𝐾=1 and 𝑐=6. Here, π‘Ÿ0=6.9975.
Figure 10: Bifurcation diagram with 𝐾=1 and 𝑐=15. Here, π‘Ÿ0=15.9998.


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