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ISRN Mathematical Analysis
VolumeΒ 2012Β (2012), Article IDΒ 869147, 11 pages
doi:10.5402/2012/869147
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.

Abstract

We analyze the positive solutions to the steady-state reaction diffusion equation with Dirichlet boundary conditions of the form: βˆ’ Ξ” 𝑒 = πœ† [ 𝐾 βˆ’ 𝑒 + 𝑐 ( 𝑒 4 / ( 1 + 𝑒 4 ) ) ] , π‘₯ ∈ Ξ© , 𝑒 = 0 , π‘₯ ∈ πœ• Ξ© . Here, Ξ” 𝑒 = d i v ( βˆ‡ 𝑒 ) is the Laplacian of 𝑒 , 1 / πœ† is the diffusion coefficient, 𝐾 and 𝑐 are positive constants, and Ξ© βŠ‚ ℝ 𝑁 is a smooth bounded region with πœ• Ξ© in 𝐢 2 . 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 ξ‚Έ 𝑒 βˆ’ Ξ” 𝑒 = πœ† 𝐾 βˆ’ 𝑒 + 𝑐 4 1 + 𝑒 4 ξ‚Ή = ∢ πœ† 𝑓 ( 𝑒 ) , π‘₯ ∈ Ξ© , 𝑒 = 0 , π‘₯ ∈ πœ• Ξ© , ( 1 . 1 ) where Ξ” 𝑒 = d i v ( βˆ‡ 𝑒 ) is the Laplacian of 𝑒 , 1 / πœ† is the diffusion coefficient, 𝐾 and 𝑐 are positive constants, and Ξ© βŠ‚ ℝ 𝑁 is a smooth bounded region with πœ• Ξ© in 𝐢 2 . 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 𝑐 𝑒 4 / ( 1 + 𝑒 4 ) , 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 𝑐 𝑒 4 / ( 1 + 𝑒 4 ) represents erosion by wind and runoff.

It is known that when 𝑓 ( 𝑒 ) > 0 ; [ 0 , ∞ ) with l i m 𝑒 β†’ ∞ ( 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 [36]). On the other hand, proving multiplicity results for nonlinearities with a falling zero (say at π‘Ÿ 0 > 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€– 𝑒 β€– ∞ < π‘Ÿ 0 (see Figure 1). Our model falls in this category.

fig1
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: ( H 1 ) 𝑓 ∈ 𝐢 2 ( [ 0 , ∞ ) ) , 𝑓 ( 𝑒 ) > 0 on [ 0 , π‘Ÿ 0 ) and 𝑓 ( 𝑒 ) < 0 for 𝑒 > π‘Ÿ 0 .

To state our multiplicity result, for 0 < π‘Ž < 𝑏 , let 𝑄 ( π‘Ž , 𝑏 , Ξ© ) ∢ = ( 𝑏 / 𝑓 ( 𝑏 ) ) ( ( 𝑁 + 1 ) / 𝑁 ) 𝑁 + 1 ξ€· 𝑁 2 / 𝑅 2 ξ€Έ ξ€½ β€– β€– 𝑒 m i n π‘Ž / Ξ© β€– β€– ∞ 𝑓 βˆ— ( π‘Ž ) , 2 𝑁 𝑀 / 𝑓 ( 𝑏 ) 𝑅 2 ξ€Ύ , ( 1 . 2 ) where 𝐡 𝑅 = 𝐡 ( 0 , 𝑅 ) is the largest inscribed ball on Ξ© , 𝑓 βˆ— ( 𝑠 ) = m a x 𝑑 ∈ [ 0 , 𝑠 ] 𝑓 ( 𝑑 ) (see Figure 2) and 𝑒 Ξ© is the unique positive solution of βˆ’ Ξ” 𝑒 = 1 in Ξ© , 𝑒 = 0 on πœ• Ξ© ,.

fig2
Figure 2: Definition of 𝐡 𝑅 and 𝑓 βˆ— .

Now, we establish the following result.

Theorem 1.1. Let π‘š , 𝑀 ∈ ( 0 , π‘Ÿ 0 ) be such that 𝑓 is nondecreasing in ( π‘š , 𝑀 ) . Assume that there exists 𝑏 ∈ [ π‘š , 𝑀 ] and π‘Ž ∈ ( 0 , 𝑏 ) such that 𝑄 ( π‘Ž , 𝑏 , Ξ© ) < 1 , then (1.1) has three positive solutions for πœ† ∈ ( ( 𝑏 / 𝑓 ( 𝑏 ) ) ( ( 𝑁 + 1 ) / 𝑁 ) 𝑁 + 1 ( 𝑁 2 / 𝑅 2 ) , m i n { π‘Ž / β€– 𝑒 Ξ© β€– ∞ 𝑓 βˆ— ( π‘Ž ) , 2 𝑁 𝑀 / 𝑓 ( 𝑏 ) 𝑅 2 } ) .

We will use the method of subsupersolutions to prove our results. By a subsolution (supersolution) of (1.1), we mean a function πœ“ ∈ π‘Š 1 , 2 β‹‚ ( Ξ© ) 𝐢 ( Ξ© ) such that πœ“ = 0 on πœ• Ξ© and ξ€œ Ξ© ξ€œ βˆ‡ πœ“ β‹… βˆ‡ π‘ž ≀ ( β‰₯ ) Ξ© πœ† 𝑓 ( πœ“ ) π‘ž , ( 1 . 3 ) for every π‘ž ∈ { πœ‚ ∈ 𝐢 ∞ 0 ( Ξ© ) ∢ πœ‚ β‰₯ 0 i n Ξ© } . 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“ ≀ πœ™ , then, (1.1) has a solution 𝑒 ∈ 𝐢 2 β‹‚ 𝐢 ( Ξ© ) 1 ( Ξ© ) such that πœ“ ≀ 𝑒 ≀ 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 𝑍 1 , a strict subsolution πœ“ 2 , and a supersolution 𝑍 2 for (1.1) such that πœ“ 1 < 𝑍 1 < 𝑍 2 , πœ“ 1 < πœ“ 2 < 𝑍 2 , and let πœ“ 2 β‰° 𝑍 1 , then, (1.1) has at least three distinct solutions 𝑒 1 , 𝑒 2 , and 𝑒 3 such that πœ“ 1 ≀ 𝑒 1 < 𝑒 2 < 𝑒 3 ≀ 𝑍 2 .

Note here that by πœ“ 1 < πœ“ 2 we mean that πœ“ 1 ≀ πœ“ 2 and πœ“ 1 β‰  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 ) = 𝐾 βˆ’ 𝑒 + 𝑐 ( 𝑒 4 / ( 1 + 𝑒 4 ) ) has a unique positive zero π‘Ÿ 0 . This will be the case when 𝐾 > 𝐾 0 ∢ = ( 3 / 4 ) 4 √ 3 / 5 βˆ’ ( 1 / 4 ) 4 √ 3 / 5 5 and 𝑐 ≫ 1 . We will prove that an 𝑆 -shaped bifurcation curve occurs when 𝑐 ≫ 1 and 𝐾 0 < 𝐾 < 9 𝑐 / 1 6 . 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 𝑍 1 , a strict subsolution πœ“ 2 , and a supersolution 𝑍 2 for (1.1) such that πœ“ 1 < 𝑍 1 < 𝑍 2 , πœ“ 1 < πœ“ 2 < 𝑍 2 , and πœ“ 2 β‰° 𝑍 1 . Clearly, πœ“ 1 = 0 is a strict subsolution since 𝑓 ( 0 ) > 0 . For the large supersolution, choose 𝑍 2 = 𝑀 ( πœ† ) 𝑒 Ξ© where 𝑀 ( πœ† ) > πœ† m a x 𝑑 ∈ [ 0 , π‘Ÿ 0 ] 𝑓 ( 𝑑 ) . Then, βˆ’ Ξ” 𝑍 2 = 𝑀 ( πœ† ) β‰₯ πœ† 𝑓 ( 𝑍 2 ) making 𝑍 2 a positive supersolution.

Now for the smaller strict supersolution, define 𝑍 1 = π‘Ž 𝑒 Ξ© / β€– 𝑒 Ξ© β€– ∞ . Since πœ† < π‘Ž / β€– 𝑒 Ξ© β€– ∞ 𝑓 βˆ— ( π‘Ž ) , βˆ’ Ξ” 𝑍 1 = π‘Ž / β€– 𝑒 Ξ© β€– ∞ > πœ† 𝑓 βˆ— ( π‘Ž ) β‰₯ πœ† 𝑓 βˆ— ( π‘Ž 𝑒 Ξ© / β€– 𝑒 Ξ© β€– ∞ ) β‰₯ πœ† 𝑓 ( π‘Ž 𝑒 Ξ© / β€– 𝑒 Ξ© β€– ∞ ) = πœ† 𝑓 ( 𝑍 1 ) in Ξ© . Here, 𝑓 βˆ— ( 𝑠 ) = m a x 𝑑 ∈ [ 0 , 𝑠 ] 𝑓 ( 𝑑 ) . Hence, 𝑍 1 is a strict supersolution.

We will now construct the strict subsolution πœ“ 2 . Let  𝑓 ξ‚»  𝑓 ( 𝑒 ) = ( 𝑒 ) , 𝑒 < π‘š , 𝑓 ( 𝑒 ) , 𝑒 β‰₯ 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).

869147.fig.003
Figure 3: Graph of  𝑓 ( 𝑒 ) .

Let 𝜌 ⎧ βŽͺ ⎨ βŽͺ ⎩ [ ] , ξ‚Έ ξ‚€ ( π‘Ÿ ) = 1 , π‘Ÿ ∈ 0 , πœ– 1 βˆ’ 1 βˆ’ 𝑅 βˆ’ π‘Ÿ  𝑅 βˆ’ πœ– 𝛽 ξ‚Ή 𝛼 ] , π‘Ÿ ∈ ( πœ– , 𝑅 , 𝛼 , 𝛽 > 1 . ( 2 . 2 ) Note that ⎧ βŽͺ ⎨ βŽͺ ⎩ [ ] , ξ‚Έ ξ‚€ 𝜌 β€² ( π‘Ÿ ) = 0 , π‘Ÿ ∈ 0 , πœ– βˆ’ 𝛼 𝛽 1 βˆ’ 𝑅 βˆ’ π‘Ÿ  𝑅 βˆ’ πœ– 𝛽 ξ‚Ή 𝛼 βˆ’ 1 ξ‚€ 𝑅 βˆ’ π‘Ÿ  𝑅 βˆ’ πœ– 𝛽 βˆ’ 1 ] , π‘Ÿ ∈ ( πœ– , 𝑅 , 𝛼 , 𝛽 > 1 ( 2 . 3 ) and | 𝜌 ξ…ž ( π‘Ÿ ) | < 𝛼 𝛽 / ( 𝑅 βˆ’ πœ– ) . Now define 𝑀 ( π‘Ÿ ) ∢ = 𝑏 𝜌 ( π‘Ÿ ) and πœ“ 2 ξ‚» ( π‘₯ ) =  πœ“ 2 , π‘₯ ∈ 𝐡 𝑅 , 0 , π‘₯ ∈ Ξ© βˆ’ 𝐡 𝑅 , ( 2 . 4 ) where  πœ“ 2 is the solution of βˆ’  πœ“ 2 ξ…ž ξ…ž ( π‘Ÿ ) βˆ’ 𝑁 βˆ’ 1 π‘Ÿ  πœ“ ξ…ž 2  ( π‘Ÿ ) = πœ† 𝑓 ( 𝑀 ( π‘Ÿ ) ) , π‘Ÿ ∈ ( 0 , 𝑅 ) ,  πœ“ ξ…ž 2 ( 0 ) = 0 =  πœ“ 2 ( 𝑅 ) , ( 2 . 5 ) and 𝐡 𝑅 is the largest inscribed ball in Ξ© . Then, πœ“ 2 ∈ π‘Š 1 , 2 β‹‚ ( Ξ© ) 𝐢 ( Ξ© ) and πœ“ 2 = 0 on πœ• Ξ© . We will now establish that  πœ“ 2 ( π‘Ÿ ) ∈ ( 𝑀 ( π‘Ÿ ) , 𝑀 ] on [ 0 , 𝑅 ) . Then, βˆ’ Ξ” πœ“ 2   = πœ† 𝑓 ( 𝑀 ( π‘Ÿ ) ) < πœ† 𝑓 (  πœ“ 2 ( π‘Ÿ ) ) ≀ πœ† 𝑓 ( πœ“ 2 ( π‘Ÿ ) ) on [ 0 , 𝑅 ) , while outside 𝐡 𝑅 , we have βˆ’ Ξ” πœ“ 2 = 0 = πœ† 𝑓 ( 0 ) = πœ† 𝑓 ( πœ“ 2 ) , and hence, πœ“ 2 will be a strict subsolution.

First, we will show that  πœ“ 2 ( π‘Ÿ ) ≀ 𝑀 . Now ξ€· π‘Ÿ 𝑁 βˆ’ 1  πœ“ ξ…ž 2 ξ€Έ ( π‘Ÿ ) ξ…ž = βˆ’ πœ† π‘Ÿ 𝑁 βˆ’ 1  𝑓 ( 𝑀 ( π‘Ÿ ) ) ,  πœ“ ξ…ž 2 ( π‘Ÿ ) = βˆ’ πœ† π‘Ÿ 𝑁 βˆ’ 1 ξ€œ π‘Ÿ 0 𝑠 𝑁 βˆ’ 1  𝑓 ( 𝑀 ( 𝑠 ) ) 𝑑 𝑠 ,  πœ“ 2 ( 𝑑 ) βˆ’  πœ“ 2 ξ€œ ( 0 ) = βˆ’ 𝑑 0 πœ† π‘Ÿ 𝑁 βˆ’ 1 ξ‚» ξ€œ π‘Ÿ 0 𝑠 𝑁 βˆ’ 1  ξ‚Ό 𝑓 ( 𝑀 ( 𝑠 ) ) 𝑑 𝑠 𝑑 π‘Ÿ . ( 2 . 6 ) But  πœ“ 2 ( 𝑅 ) = 0 . Hence, we get  πœ“ 2 ξ€œ ( 0 ) = 𝑅 0 πœ† π‘Ÿ 𝑁 βˆ’ 1 ξ‚» ξ€œ π‘Ÿ 0 𝑠 𝑁 βˆ’ 1  ξ‚Ό ≀ πœ†  𝑓 𝑓 ( 𝑀 ( 𝑠 ) ) 𝑑 𝑠 𝑑 π‘Ÿ ( 𝑏 ) 𝑁 ξ€œ 𝑅 0 = π‘Ÿ 𝑑 𝑠 πœ† 𝑓 ( 𝑏 ) 𝑅 2 𝑁 2 ξ‚€   . s i n c e 𝑏 β‰₯ π‘š , 𝑓 ( 𝑠 ) = 𝑓 ( 𝑠 ) f o r 𝑠 β‰₯ π‘š ( 2 . 7 ) But πœ† < 2 𝑁 𝑀 / 𝑓 ( 𝑏 ) 𝑅 2 . Hence, β€–  πœ“ 2 β€– ∞ =  πœ“ 2 ( 0 ) < 𝑀 .

Next, to establish  πœ“ 2 > 𝑀 on [ 0 , 𝑅 ] , we will show that  πœ“ ξ…ž 2 < 𝑀 ξ…ž ≀ 0 on [ 0 , 𝑅 ] . This will be sufficient, since  πœ“ 2 ( 𝑅 ) = 𝑀 ( 𝑅 ) = 0 . Now 𝑀 ξ…ž = 0 and  πœ“ ξ…ž 2 < 0 in the interval [ 0 , πœ– ) , and hence,  πœ“ ξ…ž 2 < 𝑀 ξ…ž ≀ 0 in that interval. For π‘Ÿ > πœ– , we have βˆ’  πœ“ ξ…ž 2 πœ† ( π‘Ÿ ) = π‘Ÿ 𝑁 βˆ’ 1 ξ€œ π‘Ÿ 0 𝑠 𝑁 βˆ’ 1  β‰₯ πœ† 𝑓 ( 𝑀 ( 𝑠 ) ) 𝑑 𝑠 π‘Ÿ 𝑁 βˆ’ 1 ξ€œ πœ– 0 𝑠 𝑁 βˆ’ 1  = πœ† 𝑓 ( 𝑀 ( 𝑠 ) ) 𝑑 𝑠 π‘Ÿ 𝑁 βˆ’ 1 ξ€œ πœ– 0 𝑠 𝑁 βˆ’ 1  𝑓 β‰₯ πœ†  ( 𝑏 ) 𝑑 𝑠 ( s i n c e 𝜌 ( 𝑠 ) = 1 , 𝑠 < πœ– ) 𝑓 ( 𝑏 ) 𝑅 𝑁 βˆ’ 1 ξ€œ πœ– 0 𝑠 𝑁 βˆ’ 1 = 𝑑 𝑠 πœ† 𝑓 ( 𝑏 ) 𝑅 𝑁 βˆ’ 1 πœ– 𝑁 𝑁 ξ‚€   . s i n c e 𝑏 β‰₯ π‘š , 𝑓 ( 𝑠 ) = 𝑓 ( 𝑠 ) f o r 𝑠 β‰₯ π‘š ( 2 . 8 ) We also know that | 𝑀 ξ…ž ( π‘Ÿ ) | ≀ 𝑏 𝛼 𝛽 / ( 𝑅 βˆ’ πœ– ) . Hence, |  πœ“ ξ…ž 2 ( π‘Ÿ ) | > | 𝑀 ξ…ž ( π‘Ÿ ) | if πœ† > 𝛼 𝛽 ( 𝑏 / 𝑓 ( 𝑏 ) ) ( 𝑁 𝑅 𝑁 βˆ’ 1 / ( 𝑅 βˆ’ πœ– ) πœ– 𝑁 ) . But m i n 0 < πœ– < 𝑅 ( 1 / ( 𝑅 βˆ’ πœ– ) πœ– 𝑁 ) = ( 𝑁 + 1 ) 𝑁 + 1 / 𝑁 𝑁 𝑅 𝑁 + 1 , and this minimum is achieved at πœ– 0 = 𝑁 𝑅 / ( 𝑁 + 1 ) . Since πœ† > ( 𝑏 / 𝑓 ( 𝑏 ) ) ( 𝑁 2 / 𝑅 2 ) ( ( 𝑁 + 1 ) / 𝑁 ) 𝑁 + 1 = ( 𝑏 / 𝑓 ( 𝑏 ) ) ( 𝑁 𝑅 𝑁 βˆ’ 1 / ( 𝑅 βˆ’ πœ– 0 ) πœ– 𝑁 0 ) , we can choose πœ– = πœ– 0 and 𝛼 , 𝛽 > 1 such that πœ† > 𝛼 𝛽 ( 𝑏 / 𝑓 ( 𝑏 ) ) ( 𝑁 𝑅 𝑁 βˆ’ 1 / ( 𝑅 βˆ’ πœ– 0 ) πœ– 𝑁 0 ) . Hence, |  πœ“ ξ…ž 2 ( π‘Ÿ ) | > | 𝑀 ξ…ž ( π‘Ÿ ) | on ( 0 , 𝑅 ) . This implies 𝑀 <  πœ“ 2 . Thus, πœ“ 2 is a strict subsolution of (1.1) if ( 𝑏 / 𝑓 ( 𝑏 ) ) ( 𝑁 2 / 𝑅 2 ) ( ( 𝑁 + 1 ) / 𝑁 ) 𝑁 + 1 < πœ† < 2 𝑁 𝑀 / 𝑓 ( 𝑏 ) 𝑅 2 . Furthermore,  πœ“ 2 ( 0 ) > 𝑀 ( 0 ) = 𝑏 > π‘Ž = β€– 𝑍 1 β€– ∞ , that is, πœ“ 2 β‰° 𝑍 1 . Moreover, 𝑀 ( πœ† ) can be chosen large enough so that πœ“ 2 < 𝑍 2 and 𝑍 1 < 𝑍 2 . Hence, by Lemma 1.3, Theorem 1.1 holds.

3. Results for the Example 𝑓 ( 𝑒 ) = 𝐾 βˆ’ 𝑒 + 𝑐 ( 𝑒 4 / ( 1 + 𝑒 4 ) )

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 ) = 𝐾 βˆ’ 𝑒 + 𝑐 ( 𝑒 4 / ( 1 + 𝑒 4 ) ) satisfies ( H 1 ) , and we will also identify π‘š and 𝑀 such that 𝑓 is increasing in ( π‘š , 𝑀 ) . Clearly, 𝑓 ∈ 𝐢 2 ( [ 0 , ∞ ) ) , 𝑓 ( 0 ) = 𝐾 , and 𝑓 ξ…ž ( 0 ) = βˆ’ 1 .

Proposition 3.1. If 𝑐 > 1 6 / 5 4 √ 1 3 5 , then there exists two points π‘š 1 π‘Ž 𝑛 𝑑 π‘š 2 such that 0 < π‘š 1 < π‘š 2 and 𝑓 ξ…ž ( π‘š 𝑖 ) = 0 for 𝑖 = 1 , 2 .

Proof. We have 𝑓 ξ…ž ( 𝑒 ) = βˆ’ 1 + ( 4 𝑐 𝑒 3 / ( 1 + 𝑒 4 ) 2 ) . So 𝑓 ξ…ž ( 𝑒 ) = 0 , when 1 = 4 𝑐 𝑒 3 / ( 1 + 𝑒 4 ) 2 . Let 𝑔 ( 𝑒 ) ∢ = 4 𝑐 𝑒 3 / ( 1 + 𝑒 4 ) 2 , and let β„Ž ( 𝑒 ) ∢ = 1 . We have 𝑔 ( 𝑒 ) β‰₯ 0 , 𝑔 ( 0 ) = 0 , and l i m 𝑒 β†’ ∞ 𝑔 ( 𝑒 ) = 0 . Since 𝑔 β€² ( 𝑒 ) = 4 𝑐 𝑒 2 ( 3 βˆ’ 5 𝑒 4 ) / ( 1 + 𝑒 4 ) 3 , we can see that 𝑔 ( 𝑒 ) achieves a maximum of ( 5 𝑐 / 1 6 ) 4 √ 1 3 5 at 𝑒 = 4 √ 3 / 5 . If m a x π‘₯ ∈ ( 0 , ∞ ) 𝑔 ( 𝑒 ) = ( 5 𝑐 / 1 6 ) 4 √ 1 3 5 > 1 , then the line β„Ž ( 𝑒 ) will cut 𝑔 ( 𝑒 ) at exactly two different points. Hence, if 𝑐 > 1 6 / 5 4 √ 1 3 5 , then there are exactly two positive points π‘š 1 < π‘š 2 such that 𝑓 ξ…ž ( π‘š 𝑖 ) = 0 for 𝑖 = 1 , 2 .

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

Proof. From Figure 5, we can see that if 𝑓 ( π‘š 1 ) > 0 , then 𝑓 ( 𝑒 ) has a unique positive zero. Since 𝑓 ξ…ž ( π‘š 1 ) = 0 , we obtain 𝑐 π‘š 3 1 / ( 1 + π‘š 4 1 ) = ( 1 + π‘š 4 1 ) / 4 . So, 𝑓 ( π‘š 1 ) = 𝐾 βˆ’ π‘š 1 + 𝑐 π‘š 4 1 / ( 1 + π‘š 4 1 ) = 𝐾 βˆ’ π‘š 1 + ( π‘š 1 ( 1 + π‘š 4 1 ) / 4 ) = 𝐾 βˆ’ ( 3 / 4 ) π‘š 1 + π‘š 5 1 / 4 . Hence, 𝑓 ( π‘š 1 ) > 0 if 𝐾 > ( 3 / 4 ) π‘š 1 + π‘š 5 1 / 4 . On analyzing 𝑓 ξ…ž ξ…ž ( 𝑒 ) = 4 𝑐 𝑒 2 ( 3 βˆ’ 5 𝑒 4 ) / ( 1 + 𝑒 4 ) 3 , we see that the positive inflection of 𝑓 ( 𝑒 ) occurs at 𝑒 = 4 √ 3 / 5 . Thus, π‘š 1 < 4 √ 3 / 5 , and hence 𝐾 > 𝐾 0 ensures that there exists a unique π‘Ÿ 0 such that 𝑓 ( π‘Ÿ 0 ) = 0 .

Choose π‘š = π‘š 1 and 𝑀 = π‘š 2 . Thus, given 𝐾 > 𝐾 0 , we can find 𝑐 large so that 𝑓 ( 𝑒 ) is increasing on ( π‘š , 𝑀 ) , and there exists a unique π‘Ÿ 0 > 0 such that 𝑓 ( π‘Ÿ 0 ) = 0 , that is, 𝑓 ( 𝑒 ) satisfies ( H 1 ) . Now, we will prove that the other assumptions in Theorem 1.1 hold in the given example.

We will select 𝑏 ∈ [ π‘š , 𝑀 ] and π‘Ž ∈ ( 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 < 9 𝑐 / 1 6 , then 𝑒 / 𝑓 ( 𝑒 ) has the shape given in Figure 6.

Proof. We have ( 𝑒 / 𝑓 ( 𝑒 ) ) ξ…ž = ( 𝑓 ( 𝑒 ) βˆ’ 𝑒 𝑓 β€² ( 𝑒 ) ) / ( 𝑓 ( 𝑒 ) ) 2 . Hence, the critical points of 𝑒 / 𝑓 ( 𝑒 ) are given by 𝑓 ( 𝑒 ) βˆ’ 𝑒 𝑓 β€² ( 𝑒 ) = 0 , and in particular, the nonzero critical points are given by 𝐾 + 𝑐 ( 𝑒 8 βˆ’ 3 𝑒 4 ) / ( 1 + 𝑒 4 ) 2 = 0 . Solving for 𝑒 , we get the positive critical points as 𝛼 = 4  √ 3 𝑐 βˆ’ 2 𝐾 βˆ’ 𝑐 ( 9 𝑐 βˆ’ 1 6 𝐾 ) / 2 ( 𝑐 + 𝐾 ) and 𝛽 = 4  √ 3 𝑐 βˆ’ 2 𝐾 + 𝑐 ( 9 𝑐 βˆ’ 1 6 𝐾 ) / 2 ( 𝑐 + 𝐾 ) . Note that if 𝐾 < 9 𝑐 / 1 6 , 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 β†’ 4 √ 3 as 𝑐 β†’ ∞ and π‘š < 4 √ 3 / 5 , we have π‘š < 𝛽 . Furthermore, it is clear from Figure 4 that 𝑀 β†’ ∞ as 𝑐 β†’ ∞ , so 𝛽 < 𝑀 for 𝑐 ≫ 1 . Thus, we have 𝛽 ∈ [ π‘š , 𝑀 ] for large 𝑐 and we choose 𝑏 = 𝛽 . We also choose π‘Ž ∈ ( 0 , 𝑀 ) such that 𝑓 ( π‘Ž ) = 𝑓 βˆ— ( π‘Ž ) = 𝑓 ( 0 ) (see Figure 7).

869147.fig.004
Figure 4: 𝑔 ( 𝑒 ) and β„Ž ( 𝑒 ) .
869147.fig.005
Figure 5: Graph of 𝑓 ( 𝑒 ) with π‘š and 𝑀 .
869147.fig.006
Figure 6: Graph of 𝑒 / 𝑓 ( 𝑒 ) .
869147.fig.007
Figure 7: Graph of 𝑓 βˆ— ( 𝑒 ) .

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

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

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 𝑓 ξ…ž ( 5 √ 𝑐 ) > 0 and 𝑓 ξ…ž ( 4 √ 𝑐 ) < 0 . We have 𝑓 ξ…ž ( 5 √ 𝑐 ) = βˆ’ 1 + 4 𝑐 8 / 5 / ( 1 + 𝑐 4 / 5 ) 2 = βˆ’ 1 + 4 𝑐 8 / 5 / ( 1 + 2 𝑐 4 / 5 + 𝑐 8 / 5 ) = βˆ’ 1 + 4 / ( 1 / 𝑐 8 / 5 + 2 / 𝑐 4 / 5 + 1 ) > 0 , and 𝑓 ξ…ž ( 4 √ 𝑐 ) = βˆ’ 1 + 4 𝑐 7 / 4 / ( 1 + 𝑐 ) 2 = βˆ’ 1 + ( 4 / ( 1 / 𝑐 7 / 4 + 2 / 𝑐 3 / 4 + 𝑐 1 / 4 ) ) < 0 for 𝑐 ≫ 1 . Thus, 5 √ 𝑐 < 𝑀 < 4 √ 𝑐 for large 𝑐 .
(ii) We have 𝑓 ( π‘Ž ) = 𝐾 , simplifying which we get π‘Ž 4 βˆ’ 𝑐 π‘Ž 3 + 1 = 0 . Define 𝑗 ( 𝑒 ) ∢ = 𝑒 4 βˆ’ 𝑐 𝑒 3 + 1 . If 𝑒 < π‘Ž , then 𝑗 ( 𝑒 ) > 0 , and if 𝑒 ∈ ( π‘Ž , 𝑀 ) , then 𝑗 ( 𝑒 ) < 0 . We have 𝑗 ( 1 / 3 √ 𝑐 ) = ( 1 / 3 √ 𝑐 ) 4 βˆ’ 𝑐 ( 1 / 3 √ 𝑐 ) 3 + 1 = 1 / 𝑐 4 / 3 > 0 , and 𝑗 ( 1 / 4 √ 𝑐 ) = ( 1 / 4 √ 𝑐 ) 4 βˆ’ 𝑐 ( 1 / 4 √ 𝑐 ) 3 + 1 = 1 / 𝑐 βˆ’ 4 √ 𝑐 + 1 < 0 for 𝑐 ≫ 1 . Hence, 1 / 3 √ 𝑐 < π‘Ž < 1 / 4 √ 𝑐 for large 𝑐 . Note that since 𝑏 = 𝛽 β†’ 4 √ 3 and π‘Ž < 1 / 4 √ 𝑐 β†’ 0 as 𝑐 β†’ ∞ , π‘Ž ∈ ( 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 𝐾 0 < 𝐾 < 9 𝑐 / 1 6 , 𝑄 ( π‘Ž , 𝑏 , Ξ© ) < 1 . It is enough if we prove that 𝑏 ξ‚€ 𝑓 ( 𝑏 ) 𝑁 + 1 𝑁  𝑁 + 1 𝑁 2 𝑅 2 ξ‚» π‘Ž < m i n β€– β€– 𝑒 Ξ© β€– β€– ∞ 𝑓 βˆ— , ( π‘Ž ) 2 𝑁 𝑀 𝑓 ( 𝑏 ) 𝑅 2 ξ‚Ό . ( 3 . 1 ) Note that for 𝑐 ≫ 1 ( 𝑏 / 𝑓 ( 𝑏 ) ) = 4 √ 3 / ( 𝐾 βˆ’ 4 √ 3 + 3 𝑐 / 4 ) and 𝑀 / 𝑓 ( 𝑏 ) = 𝑀 / ( 𝐾 βˆ’ 4 √ 3 + 3 𝑐 / 4 ) > 5 √ 𝑐 / ( 𝐾 βˆ’ 4 √ 3 + 3 𝑐 / 4 ) . Also, π‘Ž / 𝑓 βˆ— ( π‘Ž ) = π‘Ž / π‘˜ > 1 / 𝐾 3 √ 𝑐 . Applying the estimates we obtained for π‘Ž / 𝑓 βˆ— ( π‘Ž ) , 𝑏 / 𝑓 ( 𝑏 ) , and 𝑀 / 𝑓 ( 𝑏 ) to the above inequality, we get the following:  4 √ 3 π‘˜ βˆ’ 4 √ ξƒͺ ξ‚€ 3 + 3 𝑐 / 4 𝑁 + 1 𝑁  𝑁 + 1 𝑁 2 𝑅 2 ξƒ― 1 < m i n β€– β€– 𝑒 Ξ© β€– β€– ∞ π‘˜ 3 √ 𝑐 , 2 𝑁 𝑅 2  5 √ 𝑐 π‘˜ βˆ’ 4 √ 3 + 3 𝑐 / 4 ξƒͺ ξƒ° . ( 3 . 2 ) Simplifying the above, we can see that 𝑄 ( π‘Ž , 𝑏 , Ξ© ) < 1 if  4 √ 3 π‘˜ / 𝑐 βˆ’ 4 √ ξƒͺ ξ‚€ 3 / 𝑐 + 3 / 4 𝑁 + 1 𝑁  𝑁 + 1 𝑁 2 𝑅 2 ξƒ― 𝑐 < m i n 2 / 3 β€– β€– 𝑒 Ξ© β€– β€– ∞ π‘˜ , 2 𝑁 𝑅 2  𝑐 1 / 5 π‘˜ / 𝑐 βˆ’ 4 √ 3 / 𝑐 + 3 / 4 ξƒͺ ξƒ° . ( 3 . 3 ) Clearly, this inequality is true for 𝑐 ≫ 1 ; hence, Theorem 1.1 holds.

Appendix

Consider the two-point boundary value problem βˆ’ 𝑒 ξ…ž ξ…ž = πœ† 𝑓 ( 𝑒 ) , π‘₯ ∈ ( 0 , 1 ) , 𝑒 ( 0 ) = 0 = 𝑒 ( 1 ) , ( A . 1 ) where 𝑓 satisfies the following hypotheses:(G1) 𝑓 ∈ 𝐢 2 ( [ 0 , ∞ ) ) , 𝑓 ( 𝑒 ) > 0 f o r 0 < 𝑒 < π‘Ÿ 0 and 𝑓 ( 𝑒 ) < 0 for 𝑒 > π‘Ÿ 0 for some π‘Ÿ 0 > 0 ,(G2) there exists π‘˜ β‰₯ 0 such that 𝑓 ( 𝑒 ) βˆ’ 𝑓 ( 𝑣 ) β‰₯ βˆ’ π‘˜ ( 𝑒 βˆ’ 𝑣 ) for all 𝑒 , 𝑣 ∈ [ 0 , π‘Ÿ 0 ) with 𝑒 > 𝑣 .

Using the quadrature method (see [12]), it follows that (A.1) has a positive solution if and only if √ √ πœ† = 2 ξ€œ 𝜌 0 𝑑 𝑧 √ [ ] 𝐹 ( 𝜌 ) βˆ’ 𝐹 ( 𝑧 ) ∢ = 𝐺 ( 𝜌 ) , ( A . 2 ) where ∫ 𝐹 ( 𝑠 ) ∢ = 𝑠 0 𝑓 ( 𝑑 ) 𝑑 𝑑 and 𝜌 = 𝑒 ( 1 / 2 ) = β€– 𝑒 β€– ∞ . Further, 𝑒 ( π‘₯ ) is symmetric about π‘₯ = 1 / 2 and is given by ξ€œ 0 𝑒 ( π‘₯ ) 𝑑 𝑧 √ [ ] = √ 𝐹 ( 𝜌 ) βˆ’ 𝐹 ( 𝑧 ) 1 2 π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 l i m 𝜌 β†’ 0 + 𝐺 ( 𝜌 ) = 0 and l i m 𝜌 β†’ π‘Ÿ βˆ’ 0 𝐺 ( 𝜌 ) = ∞ . Furthermore, from [4], we have 𝐺 ξ…ž √ ( 𝜌 ) = 2 ξ€œ 1 0 𝐻 ( 𝜌 ) βˆ’ 𝐻 ( 𝜌 𝑠 ) [ ] 𝐹 ( 𝜌 ) βˆ’ 𝐹 ( 𝜌 𝑠 ) 3 / 2 𝑑 𝑠 , ( A . 4 ) where 𝐻 ( 𝑒 ) = 𝐹 ( 𝑒 ) βˆ’ 𝑒 / 2 𝑓 ( 𝑒 ) . Note that 𝐻 ( 0 ) = 0 and 𝐻 ξ…ž ( 0 ) = 1 / 2 𝑓 ( 0 ) > 0 . Hence, if there exists a point 𝜌 0 ∈ ( 0 , π‘Ÿ 0 ) such that 𝐻 ( 𝜌 0 ) < 0 , (see Figure 8) then 𝐺 ξ…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 𝜌 0 exists when 𝐾 > 𝐾 0 and 𝑐 > 5 . 2 0 6 2 6 𝐾 .

869147.fig.008
Figure 8: Graph of 𝐻 ( 𝑒 ) .

Consider the case 𝑓 ( 𝑒 ) = 𝐾 βˆ’ 𝑒 + 𝑐 ( 𝑒 4 / ( 1 + 𝑒 4 ) ) . Clearly, given 𝐾 > 𝐾 0 , then for 𝑐 ≫ 1 𝑓 satisfies (G1)-(G2) (see Proposition 3.3). Hence, 𝐺 ( 𝜌 ) is defined for all 𝜌 ∈ 𝑆 = ( 0 , π‘Ÿ 0 ) .

We have 𝐻 ( 𝑒 ) = 𝐹 ( 𝑒 ) βˆ’ 𝑒 / 2 𝑓 ( 𝑒 ) = 𝐾 𝑒 / 2 + 𝑐 𝑒 βˆ’ 𝑐 ( 𝑒 5 / 2 ( 1 + 𝑒 4 √ ) ) βˆ’ 𝑐 / 4 2 { βˆ’ 2 t a n βˆ’ 1 √ ( 1 βˆ’ 2 𝑒 ) + 2 t a n βˆ’ 1 √ ( 1 + √ 2 𝑒 ) βˆ’ l n ( 1 + 2 𝑒 + 𝑒 2 √ ) / ( 1 βˆ’ 2 𝑒 + 𝑒 2 ) } . Clearly, 4 √ 3 / 5 < π‘Ÿ 0 (see Proposition 3.3); choose 𝜌 0 = 4 √ 3 / 5 . Thus, 𝐻 ( 𝜌 0 ) = . 4 4 0 0 5 6 𝐾 βˆ’ . 0 8 4 5 2 4 4 𝑐 , and hence, 𝐻 ( 𝜌 0 ) < 0 if 𝑐 > 5 . 2 0 6 2 6 𝐾 .

We finally used Mathematica to compute √ πœ† = 𝐺 ( 𝜌 ) in the case when 𝑓 ( 𝑒 ) = 𝐾 βˆ’ 𝑒 + 𝑐 ( 𝑒 4 / ( 1 + 𝑒 4 ) ) 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 .

869147.fig.009
Figure 9: Bifurcation diagram with 𝐾 = 1 and 𝑐 = 6 . Here, π‘Ÿ 0 = 6 . 9 9 7 5 .
869147.fig.0010
Figure 10: Bifurcation diagram with 𝐾 = 1 and 𝑐 = 1 5 . Here, π‘Ÿ 0 = 1 5 . 9 9 9 8 .

References

  1. S. R. Carpenter, D. Ludwig, and W. A. Brock, β€œManagement of eutrophication for lakes subject to potentially irreversible change,” Ecological Applications, vol. 9, no. 3, pp. 751–771, 1999. View at Publisher Β· View at Google Scholar
  2. M. Scheffer, W. Brock, and F. Westley, β€œSocioeconomic mechanisms preventing optimum use of ecosystem services: an interdisciplinary theoretical analysis,” Ecosystems, vol. 3, no. 5, pp. 451–471, 2000. View at Publisher Β· View at Google Scholar
  3. R. Aris, β€œOn stability criteria of chemical reaction engineering,” Chemical Engineering Science, vol. 24, no. 1, pp. 149–169, 1969. View at Publisher Β· View at Google Scholar
  4. K. J. Brown, M. M. A. Ibrahim, and R. Shivaji, β€œS-shaped bifurcation curves,” Nonlinear Analysis, vol. 5, no. 5, pp. 475–486, 1981. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  5. J.-P. Kernevez, G. Joly, M.-C. Duban, B. Bunow, and D. Thomas, β€œHysteresis, oscillations, and pattern formation in realistic immobilized enzyme systems,” Journal of Mathematical Biology, vol. 7, no. 1, pp. 41–56, 1979. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  6. M. Ramaswamy and R. Shivaji, β€œMultiple positive solutions for classes of p-Laplacian equations,” Differential and Integral Equations, vol. 17, no. 11-12, pp. 1255–1261, 2004.
  7. E. Lee, S. Sasi, and R. Shivaji, β€œS-shaped bifurcation curves in ecosystems,” Journal of Mathematical Analysis and Applications, vol. 381, no. 2, pp. 732–741, 2011. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  8. H. Amann, β€œFixed point equations and nonlinear eigenvalue problems in ordered Banach spaces,” SIAM Review, vol. 18, no. 4, pp. 620–709, 1976. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  9. R. Shivaji, β€œA remark on the existence of three solutions via sub-super solutions,” in Nonlinear Analysis and Applications, vol. 109 of Lecture Notes in Pure and Applied Mathematics, pp. 561–566, Dekker, New York, NY, USA, 1987. View at Zentralblatt MATH
  10. J. Jiang and J. Shi, β€œBistability dynamics in some structured ecological models,” in Spatial Ecology, R. S. Cantrell, C. Cosner, and S. Ruan, Eds., Mathematical and Computational Biology, pp. 33–62, Chapman & Hall/CRC, 2009.
  11. R. Shivaji, β€œA remark on the existence of three solutions via sub-super solutions, Nonlinear Analysis and Applications- Ed V,” Lakshmikantham, pp. 561–566, 1987.
  12. T. Laetsch, β€œThe number of solutions of a nonlinear two point boundary value problem,” Indiana University Mathematics Journal, vol. 20, pp. 1–13, 1970. View at Publisher Β· View at Google Scholar