Existence of Alternate Steady States in a Phosphorous Cycling Model

Abstract

We analyze the positive solutions to the steady-state reaction diffusion equation with Dirichlet boundary conditions of the form: −Δ𝑢=𝜆[𝐾−𝑢+𝑐(𝑢⁴/(1+𝑢⁴))],𝑥∈Ω,𝑢=0,𝑥∈𝜕Ω. Here, Δ𝑢=div(∇𝑢) is the Laplacian of 𝑢, 1/𝜆 is the diffusion coefficient, 𝐾 and 𝑐 are positive constants, and Ω⊂ℝ^𝑁 is a smooth bounded region with 𝜕Ω in 𝐶². 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 problem−Δ𝑢=𝜆𝐾−𝑢+𝑐𝑢⁴𝑢=01,+𝑥𝑢⁴∈𝜕=Ω∶,𝜆𝑓(𝑢),𝑥∈Ω,(1.1) where Δ𝑢=div(∇𝑢) is the Laplacian of 𝑢,1/𝜆 is the diffusion coefficient, 𝐾 and 𝑐 are positive constants, and Ω⊂ℝ^𝑁 is a smooth bounded region with 𝜕Ω in 𝐶². 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, 𝑢 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 𝑐𝑢⁴/(1+𝑢⁴), 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, 𝑢 is the amount of barren soil, and 𝑐𝑢⁴/(1+𝑢⁴) represents erosion by wind and runoff.

It is known that when 𝑓(𝑢)>0;[0,∞) with lim_𝑢→∞(𝑓(𝑢)/𝑢)=0, a positive solution exists for all 𝜆>0, and this solution is unique if 𝑢/𝑓(𝑢) 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 𝑟₀>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 ‖𝑢‖_∞<𝑟₀ (see Figure 1). Our model falls in this category.

(a) When 𝑓(𝑢)>0∶[0,∞)

(b) A falling zero problem

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)𝑓∈𝐶²([0,∞)),𝑓(𝑢)>0 on [0,𝑟₀) and 𝑓(𝑢)<0 for 𝑢>𝑟₀.

To state our multiplicity result, for 0<𝑎<𝑏, let𝑄(𝑎,𝑏,Ω)∶=(𝑏/𝑓(𝑏))((𝑁+1)/𝑁)^𝑁+1𝑁²/𝑅²min𝑎/‖‖𝑒_Ω‖‖_∞𝑓^∗(𝑎),2𝑁𝑀/𝑓(𝑏)𝑅²,(1.2) where 𝐵_𝑅=𝐵(0,𝑅) is the largest inscribed ball on Ω, 𝑓^∗(𝑠)=max_{𝑡∈[0,𝑠]}𝑓(𝑡) (see Figure 2) and 𝑒_Ω is the unique positive solution of −Δ𝑒=1 in Ω,𝑒=0 on 𝜕Ω,.

(a) 𝐵𝑅 in Ω

(b) 𝑓∗(𝑠)=max𝑡∈[0,𝑠]𝑓(𝑡)

Figure 2 

Definition of 𝐵𝑅 and 𝑓∗.

Now, we establish the following result.

Theorem 1.1. Let 𝑚,𝑀∈(0,𝑟₀) be such that 𝑓 is nondecreasing in (𝑚,𝑀). Assume that there exists 𝑏∈[𝑚,𝑀] and 𝑎∈(0,𝑏) such that 𝑄(𝑎,𝑏,Ω)<1, then (1.1) has three positive solutions for 𝜆∈((𝑏/𝑓(𝑏))((𝑁+1)/𝑁)^𝑁+1(𝑁²/𝑅²),min{𝑎/‖𝑒_Ω‖_∞𝑓^∗(𝑎),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 𝜓∈𝑊^1,2(Ω)⋂𝐶(Ω) such that 𝜓=0 on 𝜕Ω and_Ω∇𝜓⋅∇𝑞≤(≥)_Ω𝜆𝑓(𝜓)𝑞,(1.3) for every 𝑞∈{𝜂∈𝐶^∞0(Ω)∶𝜂≥0inΩ}. Then, the following lemma holds.

Lemma 1.2. Let 𝜓 be a subsolution of (1.1), and let 𝜙 be a supersolution of (1.1) such that 𝜓≤𝜙, then, (1.1) has a solution 𝑢∈𝐶²(Ω)⋂𝐶¹(Ω) such that 𝜓≤𝑢≤𝜙.

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 𝜓₁, a strict supersolution 𝑍₁, a strict subsolution 𝜓₂, and a supersolution 𝑍₂ for (1.1) such that 𝜓₁<𝑍₁<𝑍₂, 𝜓₁<𝜓₂<𝑍₂, and let 𝜓₂≰𝑍₁, then, (1.1) has at least three distinct solutions 𝑢₁,𝑢₂, and 𝑢₃ such that 𝜓₁≤𝑢₁<𝑢₂<𝑢₃≤𝑍₂.

Note here that by 𝜓₁<𝜓₂ we mean that 𝜓₁≤𝜓₂ and 𝜓₁≠𝜓₂.

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(𝑢)=𝐾−𝑢+𝑐(𝑢⁴/(1+𝑢⁴)) has a unique positive zero 𝑟₀. This will be the case when 𝐾>𝐾₀∶=(3/4)√⁴3/5−(1/4)√⁴3/5⁵ 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 𝜓₁, a strict supersolution 𝑍₁, a strict subsolution 𝜓₂, and a supersolution 𝑍₂ for (1.1) such that 𝜓₁<𝑍₁<𝑍₂, 𝜓₁<𝜓₂<𝑍₂, and 𝜓₂≰𝑍₁. Clearly, 𝜓₁=0 is a strict subsolution since 𝑓(0)>0. For the large supersolution, choose 𝑍₂=𝑀(𝜆)𝑒_Ω where 𝑀(𝜆)>𝜆max_{𝑡∈[0,𝑟0]}𝑓(𝑡). Then, −Δ𝑍₂=𝑀(𝜆)≥𝜆𝑓(𝑍₂) making 𝑍₂ a positive supersolution.

Now for the smaller strict supersolution, define 𝑍₁=𝑎𝑒_Ω/‖𝑒_Ω‖_∞. Since 𝜆<𝑎/‖𝑒_Ω‖_∞𝑓^∗(𝑎),−Δ𝑍₁=𝑎/‖𝑒_Ω‖_∞>𝜆𝑓^∗(𝑎)≥𝜆𝑓^∗(𝑎𝑒_Ω/‖𝑒_Ω‖_∞)≥𝜆𝑓(𝑎𝑒_Ω/‖𝑒_Ω‖_∞)=𝜆𝑓(𝑍₁) in Ω. Here, 𝑓^∗(𝑠)=max_{𝑡∈[0,𝑠]}𝑓(𝑡). Hence, 𝑍₁ is a strict supersolution.

We will now construct the strict subsolution 𝜓₂. Let𝑓(𝑢)=𝑓𝑓((𝑢𝑢)),,𝑢𝑢<≥𝑚𝑚,,(2.1) where 𝑓(𝑢) is defined so that the function 𝑓(𝑢) is strictly increasing on (0,𝑀) and 𝑓(𝑢)≤𝑓(𝑢) (see Figure 3).

Figure 3 

Graph of 𝑓(𝑢).

Let𝜌(𝑟)=⎧⎪⎨⎪⎩11,−1−𝑅−𝑟𝑟∈[0,𝜖],𝑅−𝜖^𝛽^𝛼,𝑟∈(𝜖,𝑅],𝛼,𝛽>1.(2.2) Note that𝜌′(𝑟)=⎧⎪⎨⎪⎩0−,𝛼𝛽1−𝑅−𝑟𝑟∈[0,𝜖],𝑅−𝜖^𝛽^𝛼−1𝑅𝑅−−𝜖𝑟^𝛽−1,𝑟∈(𝜖,𝑅],𝛼,𝛽>1(2.3) and |𝜌^(𝑟)|<𝛼𝛽/(𝑅−𝜖). Now define 𝑤(𝑟)∶=𝑏𝜌(𝑟) and𝜓₂(𝑥)=𝜓²0,,𝑥𝑥∈∈𝐵Ω_𝑅−,𝐵_𝑅,(2.4) where 𝜓₂ is the solution of−𝜓^2(𝑟)−𝑁−1𝑟𝜓^2𝜓_2(𝑟)=𝜆𝑓(𝑤(𝑟)),𝑟∈(0,𝑅),(0)=0=𝜓₂(𝑅),(2.5) and 𝐵_𝑅 is the largest inscribed ball in Ω. Then, 𝜓₂∈𝑊^1,2(Ω)⋂𝐶(Ω) and 𝜓₂=0 on 𝜕Ω. We will now establish that 𝜓₂(𝑟)∈(𝑤(𝑟),𝑀] on [0,𝑅). Then, −Δ𝜓₂=𝜆𝑓(𝑤(𝑟))<𝜆𝑓(𝜓₂(𝑟))≤𝜆𝑓(𝜓₂(𝑟)) on [0,𝑅), while outside 𝐵_𝑅, we have −Δ𝜓₂=0=𝜆𝑓(0)=𝜆𝑓(𝜓₂), and hence, 𝜓₂ will be a strict subsolution.

First, we will show that 𝜓₂(𝑟)≤𝑀. Now𝜓𝑟^𝑁−1_2𝜓^2(𝑟)^=−𝜆𝑟^𝑁−1𝑓(𝑤(𝑟)),(𝑟)=−𝜆𝑟^𝑁−1^𝑟0𝜓₂𝑠^𝑁−1𝑓(𝑤(𝑠))𝑑𝑠,(𝑡)−𝜓₂(0)=−^𝑡0𝜆𝑟^𝑁−1^𝑟0𝑠^𝑁−1𝑓(𝑤(𝑠))𝑑𝑠𝑑𝑟.(2.6) But 𝜓₂(𝑅)=0. Hence, we get𝜓₂(0)=^𝑅0𝜆𝑟^𝑁−1^𝑟0≤𝜆𝑓(𝑏)𝑠^𝑁−1𝑓(𝑤(𝑠))𝑑𝑠𝑑𝑟𝑁^𝑅0=𝜆𝑓(𝑏)𝑟𝑑𝑠2𝑁𝑅²since𝑏≥𝑚,𝑓(𝑠)=𝑓(𝑠)for𝑠≥𝑚.(2.7) But 𝜆<2𝑁𝑀/𝑓(𝑏)𝑅². Hence, ‖𝜓₂‖_∞=𝜓₂(0)<𝑀.

Next, to establish 𝜓₂>𝑤 on [0,𝑅], we will show that 𝜓^2<𝑤^≤0 on [0,𝑅]. This will be sufficient, since 𝜓₂(𝑅)=𝑤(𝑅)=0. Now 𝑤^=0 and 𝜓^2<0 in the interval [0,𝜖), and hence, 𝜓^2<𝑤^≤0 in that interval. For 𝑟>𝜖, we have−𝜓^2(𝑟)=𝜆𝑟^𝑁−1^𝑟0≥𝜆𝑠^𝑁−1𝑓(𝑤(𝑠))𝑑𝑠𝑟^𝑁−1^𝜖0=𝜆𝑠^𝑁−1𝑓(𝑤(𝑠))𝑑𝑠𝑟^𝑁−1^𝜖0≥𝜆𝑓(𝑏)𝑠^𝑁−1𝑓(𝑏)𝑑𝑠(since𝜌(𝑠)=1,𝑠<𝜖)𝑅^𝑁−1^𝜖0=𝜆𝑓(𝑏)𝑠^𝑁−1𝑑𝑠𝑅^𝑁−1𝜖^𝑁𝑁since𝑏≥𝑚,𝑓(𝑠)=𝑓(𝑠)for𝑠≥𝑚.(2.8) We also know that |𝑤^(𝑟)|≤𝑏𝛼𝛽/(𝑅−𝜖). Hence, |𝜓^2(𝑟)|>|𝑤^(𝑟)| if 𝜆>𝛼𝛽(𝑏/𝑓(𝑏))(𝑁𝑅^𝑁−1/(𝑅−𝜖)𝜖^𝑁). But min_0<𝜖<𝑅(1/(𝑅−𝜖)𝜖^𝑁)=(𝑁+1)^𝑁+1/𝑁^𝑁𝑅^𝑁+1, and this minimum is achieved at 𝜖₀=𝑁𝑅/(𝑁+1). Since 𝜆>(𝑏/𝑓(𝑏))(𝑁²/𝑅²)((𝑁+1)/𝑁)^𝑁+1=(𝑏/𝑓(𝑏))(𝑁𝑅^𝑁−1/(𝑅−𝜖₀)𝜖^𝑁0), we can choose 𝜖=𝜖₀ and 𝛼,𝛽>1 such that 𝜆>𝛼𝛽(𝑏/𝑓(𝑏))(𝑁𝑅^𝑁−1/(𝑅−𝜖₀)𝜖^𝑁0). Hence, |𝜓^2(𝑟)|>|𝑤^(𝑟)| on (0,𝑅). This implies 𝑤<𝜓₂. Thus, 𝜓₂ is a strict subsolution of (1.1) if (𝑏/𝑓(𝑏))(𝑁²/𝑅²)((𝑁+1)/𝑁)^𝑁+1<𝜆<2𝑁𝑀/𝑓(𝑏)𝑅². Furthermore, 𝜓₂(0)>𝑤(0)=𝑏>𝑎=‖𝑍₁‖_∞, that is, 𝜓₂≰𝑍₁. Moreover, 𝑀(𝜆) can be chosen large enough so that 𝜓₂<𝑍₂ and 𝑍₁<𝑍₂. Hence, by Lemma 1.3, Theorem 1.1 holds.

3. Results for the Example 𝑓(𝑢)=𝐾−𝑢+𝑐(𝑢⁴/(1+𝑢⁴))

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(𝑢)=𝐾−𝑢+𝑐(𝑢⁴/(1+𝑢⁴)) satisfies (H1), and we will also identify 𝑚 and 𝑀 such that 𝑓 is increasing in (𝑚,𝑀). Clearly, 𝑓∈𝐶²([0,∞)), 𝑓(0)=𝐾, and 𝑓^(0)=−1.

Proposition 3.1. If 𝑐>16/5√⁴135, then there exists two points 𝑚₁𝑎𝑛𝑑𝑚₂ such that 0<𝑚₁<𝑚₂ and 𝑓^(𝑚_𝑖)=0 for 𝑖=1,2.

Proof. We have 𝑓^(𝑢)=−1+(4𝑐𝑢³/(1+𝑢⁴)²). So 𝑓^(𝑢)=0, when 1=4𝑐𝑢³/(1+𝑢⁴)². Let 𝑔(𝑢)∶=4𝑐𝑢³/(1+𝑢⁴)², and let ℎ(𝑢)∶=1. We have 𝑔(𝑢)≥0, 𝑔(0)=0, and lim_𝑢→∞𝑔(𝑢)=0. Since 𝑔′(𝑢)=4𝑐𝑢²(3−5𝑢⁴)/(1+𝑢⁴)³, we can see that 𝑔(𝑢) achieves a maximum of (5𝑐/16)√⁴135 at 𝑢=√⁴3/5. If max_{𝑥∈(0,∞)}𝑔(𝑢)=(5𝑐/16)√⁴135>1, then the line ℎ(𝑢) will cut 𝑔(𝑢) at exactly two different points. Hence, if 𝑐>16/5√⁴135, then there are exactly two positive points 𝑚₁<𝑚₂ such that 𝑓^(𝑚_𝑖)=0 for 𝑖=1,2.

Proposition 3.2. If 𝐾>(3/4)√⁴3/5−(1/4)(√⁴3/5)⁵=∶𝐾₀, then there exists a unique 𝑟₀>0 such that 𝑓(𝑟₀)=0.

Proof. From Figure 5, we can see that if 𝑓(𝑚₁)>0, then 𝑓(𝑢) has a unique positive zero. Since 𝑓^(𝑚₁)=0, we obtain 𝑐𝑚³¹/(1+𝑚⁴¹)=(1+𝑚⁴¹)/4. So, 𝑓(𝑚₁)=𝐾−𝑚₁+𝑐𝑚⁴¹/(1+𝑚⁴¹)=𝐾−𝑚₁+(𝑚₁(1+𝑚⁴¹)/4)=𝐾−(3/4)𝑚₁+𝑚⁵¹/4. Hence, 𝑓(𝑚₁)>0 if 𝐾>(3/4)𝑚₁+𝑚⁵¹/4. On analyzing 𝑓^(𝑢)=4𝑐𝑢²(3−5𝑢⁴)/(1+𝑢⁴)³, we see that the positive inflection of 𝑓(𝑢) occurs at 𝑢=√⁴3/5. Thus, 𝑚₁<√⁴3/5, and hence 𝐾>𝐾₀ ensures that there exists a unique 𝑟₀ such that 𝑓(𝑟₀)=0.

Choose 𝑚=𝑚₁ and 𝑀=𝑚₂. Thus, given 𝐾>𝐾₀, we can find 𝑐 large so that 𝑓(𝑢) is increasing on (𝑚,𝑀), and there exists a unique 𝑟₀>0 such that 𝑓(𝑟₀)=0, that is, 𝑓(𝑢) satisfies (H1). Now, we will prove that the other assumptions in Theorem 1.1 hold in the given example.

We will select 𝑏∈[𝑚,𝑀] and 𝑎∈(0,𝑏) such that 𝑄(𝑎,𝑏,Ω)<1. The point at which the function 𝑢/𝑓(𝑢) has a minimum would be an ideal choice for 𝑏.

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

Proof. We have (𝑢/𝑓(𝑢))^=(𝑓(𝑢)−𝑢𝑓′(𝑢))/(𝑓(𝑢))². Hence, the critical points of 𝑢/𝑓(𝑢) are given by 𝑓(𝑢)−𝑢𝑓′(𝑢)=0, and in particular, the nonzero critical points are given by 𝐾+𝑐(𝑢⁸−3𝑢⁴)/(1+𝑢⁴)²=0. Solving for 𝑢, we get the positive critical points as 𝛼=⁴3𝑐−2𝐾−√𝑐(9𝑐−16𝐾)/2(𝑐+𝐾) and 𝛽=⁴3𝑐−2𝐾+√𝑐(9𝑐−16𝐾)/2(𝑐+𝐾). Note that if 𝐾<9𝑐/16, then 𝛼 and 𝛽 are positive real roots of (𝑢/𝑓(𝑢))′ with 𝛼<𝛽. Hence, 𝑢/𝑓(𝑢) has a relative maximum at 𝛼 and a relative minimum at 𝛽.

Since 𝛽→√⁴3 as 𝑐→∞ and 𝑚<√⁴3/5, we have 𝑚<𝛽. Furthermore, it is clear from Figure 4 that 𝑀→∞ as 𝑐→∞, so 𝛽<𝑀 for 𝑐≫1. Thus, we have 𝛽∈[𝑚,𝑀] for large 𝑐 and we choose 𝑏=𝛽. We also choose 𝑎∈(0,𝑀) such that 𝑓(𝑎)=𝑓^∗(𝑎)=𝑓(0) (see Figure 7).

Figure 4 

𝑔(𝑢) and ℎ(𝑢).

Figure 5 

Graph of 𝑓(𝑢) with 𝑚 and 𝑀.

Figure 6 

Graph of 𝑢/𝑓(𝑢).

Figure 7 

Graph of 𝑓∗(𝑢).

The following estimates hold for 𝑎 and 𝑀 for 𝑐≫1.

Proposition 3.4. For 𝑐≫1,(i)  √⁵𝑐<𝑀<√⁴𝑐,(ii)  1/√³𝑐<𝑎<1/√⁴𝑐.

Proof. (i) By the shape of the graph of 𝑓(𝑢) established in Propositions 3.1 and 3.3 (see Figure 5), it is enough if we prove that 𝑓^(√⁵𝑐)>0 and 𝑓^(√⁴𝑐)<0. We have 𝑓^(√⁵𝑐)=−1+4𝑐^8/5/(1+𝑐^4/5)²=−1+4𝑐^8/5/(1+2𝑐^4/5+𝑐^8/5)=−1+4/(1/𝑐^8/5+2/𝑐^4/5+1)>0, and 𝑓^(√⁴𝑐)=−1+4𝑐^7/4/(1+𝑐)²=−1+(4/(1/𝑐^7/4+2/𝑐^3/4+𝑐^1/4))<0 for 𝑐≫1. Thus, √⁵𝑐<𝑀<√⁴𝑐 for large 𝑐.
(ii) We have 𝑓(𝑎)=𝐾, simplifying which we get 𝑎⁴−𝑐𝑎³+1=0. Define 𝑗(𝑢)∶=𝑢⁴−𝑐𝑢³+1. If 𝑢<𝑎, then 𝑗(𝑢)>0, and if 𝑢∈(𝑎,𝑀), then 𝑗(𝑢)<0. We have 𝑗(1/√³𝑐)=(1/√³𝑐)⁴−𝑐(1/√³𝑐)³+1=1/𝑐^4/3>0, and 𝑗(1/√⁴𝑐)=(1/√⁴𝑐)⁴−𝑐(1/√⁴𝑐)³+1=1/𝑐−√⁴𝑐+1<0 for 𝑐≫1. Hence, 1/√³𝑐<𝑎<1/√⁴𝑐 for large 𝑐. Note that since 𝑏=𝛽→√⁴3 and 𝑎<1/√⁴𝑐→0 as 𝑐→∞, 𝑎∈(0,𝑏) for 𝑐≫1.

Now we will discuss the existence of at least three positive solutions for a certain range of 𝜆 (see Theorem 1.1). Our aim is to prove that for 𝑐≫1 and 𝐾₀<𝐾<9𝑐/16, 𝑄(𝑎,𝑏,Ω)<1. It is enough if we prove that𝑏𝑓(𝑏)𝑁𝑁+1^𝑁+1𝑁²𝑅²<min𝑎‖‖𝑒_Ω‖‖_∞𝑓^∗(𝑎),𝑓2(𝑁𝑏)𝑀𝑅².(3.1) Note that for 𝑐≫1(𝑏/𝑓(𝑏))=√⁴3/(𝐾−√⁴3+3𝑐/4) and 𝑀/𝑓(𝑏)=𝑀/(𝐾−√⁴3+3𝑐/4)>√⁵𝑐/(𝐾−√⁴3+3𝑐/4). Also, 𝑎/𝑓^∗(𝑎)=𝑎/𝑘>1/𝐾√³𝑐. Applying the estimates we obtained for 𝑎/𝑓^∗(𝑎),𝑏/𝑓(𝑏), and 𝑀/𝑓(𝑏) to the above inequality, we get the following:⁴√3𝑘−√⁴3+3𝑐/4𝑁𝑁+1^𝑁+1𝑁²𝑅²<min1‖‖𝑒_Ω‖‖_∞𝑘√³𝑐,2𝑅𝑁²⁵√𝑐𝑘−√⁴3+3𝑐/4.(3.2) Simplifying the above, we can see that 𝑄(𝑎,𝑏,Ω)<1 if⁴√3𝑘/𝑐−√⁴3/𝑐+3/4𝑁𝑁+1^𝑁+1𝑁²𝑅²<min𝑐^2/3‖‖𝑒_Ω‖‖_∞𝑘,2𝑅𝑁²𝑘/𝑐−√⁴𝑐3^1/5/𝑐+3/4.(3.3) Clearly, this inequality is true for 𝑐≫1; hence, Theorem 1.1 holds.

Appendix

Consider the two-point boundary value problem−𝑢^=𝑢(𝜆0𝑓)=(𝑢)0,=𝑥𝑢(∈1)(,0,1),(A.1) where 𝑓 satisfies the following hypotheses:(G1)𝑓∈𝐶²([0,∞)),𝑓(𝑢)>0for0<𝑢<𝑟₀ and 𝑓(𝑢)<0 for 𝑢>𝑟₀ for some 𝑟₀>0,(G2) there exists 𝑘≥0 such that 𝑓(𝑢)−𝑓(𝑣)≥−𝑘(𝑢−𝑣) for all 𝑢,𝑣∈[0,𝑟₀) with 𝑢>𝑣.

Using the quadrature method (see [12]), it follows that (A.1) has a positive solution if and only if√𝜆=√2^𝜌0𝑑𝑧√[𝐹(𝜌)−𝐹(𝑧)]∶=𝐺(𝜌),(A.2) where 𝐹(𝑠)∶=∫^𝑠0𝑓(𝑡)𝑑𝑡 and 𝜌=𝑢(1/2)=‖𝑢‖_∞. Further, 𝑢(𝑥) is symmetric about 𝑥=1/2 and is given by^0𝑢(𝑥)𝑑𝑧√[𝐹(𝜌)−𝐹(𝑧)]=√2𝜆𝑥,0<𝑥<12.(A.3)

Equation (A.2) describes the bifurcation curve of positive solutions of (A.1), and it follows by results in [12] that lim_𝜌→0⁺𝐺(𝜌)=0 and lim_𝜌→𝑟⁻⁰𝐺(𝜌)=∞. Furthermore, from [4], we have𝐺^(𝜌)=√2¹⁰𝐻(𝜌)−𝐻(𝜌𝑠)[𝐹(𝜌)−𝐹(𝜌𝑠)]^3/2𝑑𝑠,(A.4) where 𝐻(𝑢)=𝐹(𝑢)−𝑢/2𝑓(𝑢). Note that 𝐻(0)=0 and 𝐻^(0)=1/2𝑓(0)>0. Hence, if there exists a point 𝜌₀∈(0,𝑟₀) such that 𝐻(𝜌₀)<0, (see Figure 8) then 𝐺^(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(𝑢).

Consider the case 𝑓(𝑢)=𝐾−𝑢+𝑐(𝑢⁴/(1+𝑢⁴)). Clearly, given 𝐾>𝐾₀, then for 𝑐≫1𝑓 satisfies (G1)-(G2) (see Proposition 3.3). Hence, 𝐺(𝜌) is defined for all 𝜌∈𝑆=(0,𝑟₀).

We have 𝐻√(𝑢)=𝐹(𝑢)−𝑢/2𝑓(𝑢)=𝐾𝑢/2+𝑐𝑢−𝑐(𝑢⁵/2(1+𝑢⁴))−𝑐/4√2{−2tan⁻¹(1−√2𝑢)+2tan⁻¹(1+√2𝑢)−ln(1+2𝑢+𝑢²)/(1−√2𝑢+𝑢²)}. Clearly, √⁴3/5<𝑟₀ (see Proposition 3.3); choose 𝜌₀=√⁴3/5. Thus, 𝐻(𝜌₀)=.440056𝐾−.0845244𝑐, and hence, 𝐻(𝜌₀)<0 if 𝑐>5.20626𝐾.

We finally used Mathematica to compute √𝜆=𝐺(𝜌) in the case when 𝑓(𝑢)=𝐾−𝑢+𝑐(𝑢⁴/(1+𝑢⁴)) 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, 𝑟0=6.9975.

Figure 10 

Bifurcation diagram with 𝐾=1 and 𝑐=15. Here, 𝑟0=15.9998.