# Existence of Alternate Steady States in a Phosphorous Cycling Model

Authors: Dagny Butler; Sarath Sasi; R. Shivaji
Journal: ISRN Mathematical Analysis
 (2012)

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## 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⊂ℝN 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.


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## Body

## 1. Introduction

The nonlinear boundary value problem(1.1)-Δu=λ[K-u+cu41+u4]=:λf(u),x∈Ω,u=0,x∈∂Ω,
where Δu=div(∇u) is the Laplacian of u,1/λ is the diffusion coefficient, K and c are positive constants, and Ω⊂ℝN 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 whenf(u)>0;[0,∞) with limu→∞(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.Restricted solution space for the falling zero problem.
(a)
Whenf(u)>0:[0,∞)(b)
A falling zero problemInstead 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, for0<a<b, let(1.2)Q(a,b,Ω)∶=(b/f(b))((N+1)/N)N+1(N2/R2)min⁡{a/‖eΩ‖∞f*(a),2NM/f(b)R2},
where BR=B(0,R) is the largest inscribed ball on Ω, f*(s)=maxt∈[0,s]f(t) (see Figure 2) and eΩ is the unique positive solution of -Δe=1 in Ω,e=0 on ∂Ω,.Definition ofBR and f*.
(a)
BR in Ω(b)
f*(s)=maxt∈[0,s]f(t)Now, we establish the following result.Theorem 1.1.
Letm,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/∥eΩ∥∞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∈W1,2(Ω)⋂C(Ω̅) such that ψ=0 on ∂Ω and(1.3)∫Ω∇ψ⋅∇q≤(≥)∫Ωλf(ψ)q,
for every q∈{η∈C0∞(Ω):η≥0inΩ}. Then, the following lemma holds.Lemma 1.2.
Letψ 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 (Theorem1.1), we use the following very useful result discussed in [8, 9].Lemma 1.3.
Suppose that there exists a subsolutionψ1, 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 byψ1<ψ2 we mean that ψ1≤ψ2 and ψ1≠ψ2.We prove Theorem1.1 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)3/54-(1/4)3/545 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 Theorem1.1

To establish the multiplicity result, we have to construct a subsolutionψ1, 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)eΩ where M(λ)>λmaxt∈[0,r0]f(t). Then, -ΔZ2=M(λ)≥λf(Z2) making Z2 a positive supersolution.Now for the smaller strict supersolution, defineZ1=aeΩ/∥eΩ∥∞. Since λ<a/∥eΩ∥∞f*(a),-ΔZ1=a/∥eΩ∥∞>λf*(a)≥λf*(aeΩ/∥eΩ∥∞)≥λf(aeΩ/∥eΩ∥∞)=λf(Z1) in Ω. Here, f*(s)=maxt∈[0,s]f(t). Hence, Z1 is a strict supersolution.We will now construct the strict subsolutionψ2. Let(2.1)f̃(u)={f̂(u),u<m,f(u),u≥m,
where f̂(u) is defined so that the function f̃(u) is strictly increasing on (0,M) and f̃(u)≤f(u) (see Figure 3).Figure 3
Graph off̃(u).Let(2.2)ρ(r)={1,r∈[0,ϵ],1-[1-(R-rR-ϵ)β]α,r∈(ϵ,R],α,β>1.
Note that(2.3)ρ′(r)={0,r∈[0,ϵ],-αβ[1-(R-rR-ϵ)β]α-1(R-rR-ϵ)β-1,r∈(ϵ,R],α,β>1
and |ρ′(r)|<αβ/(R-ϵ). Now define w(r):=bρ(r) and(2.4)ψ2(x)={ψ̃2,x∈BR,0,x∈Ω-BR,
where ψ̃2 is the solution of(2.5)-ψ̃2′′(r)-N-1rψ̃2′(r)=λf̃(w(r)),r∈(0,R),ψ̃2′(0)=0=ψ̃2(R),
and BR is the largest inscribed ball in Ω. Then, ψ2∈W1,2(Ω)⋂C(Ω̅) and ψ2=0 on ∂Ω. We will now establish that ψ̃2(r)∈(w(r),M] on [0,R). Then, -Δψ2=λf̃(w(r))<λf̃(ψ̃2(r))≤λf(ψ2(r)) on [0,R), while outside BR, we have -Δψ2=0=λf(0)=λf(ψ2), and hence, ψ2 will be a strict subsolution.First, we will show thatψ̃2(r)≤M. Now(2.6)(rN-1ψ̃2′(r))′=-λrN-1f̃(w(r)),ψ̃2′(r)=-λrN-1∫0rsN-1f̃(w(s))ds,ψ̃2(t)-ψ̃2(0)=-∫0tλrN-1{∫0rsN-1f̃(w(s))ds}dr.
But ψ̃2(R)=0. Hence, we get(2.7)ψ̃2(0)=∫0RλrN-1{∫0rsN-1f̃(w(s))ds}dr≤λf̃(b)N∫0Rrds=λf(b)2NR2(sinceb≥m,f̃(s)=f(s)fors≥m).
But λ<2NM/f(b)R2. Hence, ∥ψ̃2∥∞=ψ̃2(0)<M.Next, to establishψ̃2>w on [0,R], we will show that ψ̃2′<w′≤0 on [0,R]. This will be sufficient, since ψ̃2(R)=w(R)=0. Now w′=0 and ψ̃2′<0 in the interval [0,ϵ), and hence, ψ̃2′<w′≤0 in that interval. For r>ϵ, we have(2.8)-ψ̃2′(r)=λrN-1∫0rsN-1f̃(w(s))ds≥λrN-1∫0ϵsN-1f̃(w(s))ds=λrN-1∫0ϵsN-1f̃(b)ds(sinceρ(s)=1,s<ϵ)≥λf̃(b)RN-1∫0ϵsN-1ds=λf(b)RN-1ϵNN(sinceb≥m,f̃(s)=f(s)fors≥m).
We also know that |w′(r)|≤bαβ/(R-ϵ). Hence, |ψ̃2′(r)|>|w′(r)| if λ>αβ(b/f(b))(NRN-1/(R-ϵ)ϵN). But min0<ϵ<R(1/(R-ϵ)ϵN)=(N+1)N+1/NNRN+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))(NRN-1/(R-ϵ0)ϵ0N), we can choose ϵ=ϵ0 and α,β>1 such that λ>αβ(b/f(b))(NRN-1/(R-ϵ0)ϵ0N). Hence, |ψ̃2′(r)|>|w′(r)| on (0,R). This implies w<ψ̃2. 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, ψ̃2(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 Examplef(u)=K-u+c(u4/(1+u4))

First, we will analyze some properties of this nonlinearity. We will show that for largec 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.
Ifc>16/51354, then there exists two points  m1andm2 such that  0<m1<m2 and  f′(mi)=0 for  i=1,2.Proof.
We havef′(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 limu→∞g(u)=0. Since g′(u)=4cu2(3-5u4)/(1+u4)3, we can see that g(u) achieves a maximum of (5c/16)1354 at u=3/54. If maxx∈(0,∞)g(u)=(5c/16)1354>1, then the line h(u) will cut g(u) at exactly two different points. Hence, if c>16/51354, then there are exactly two positive points m1<m2 such that f′(mi)=0 for i=1,2.Proposition 3.2.
IfK>(3/4)3/54-(1/4)(3/54)5=:K0, then there exists a unique  r0>0 such that  f(r0)=0.Proof.
From Figure5, we can see that if f(m1)>0, then f(u) has a unique positive zero. Since f′(m1)=0, we obtain cm13/(1+m14)=(1+m14)/4. So, f(m1)=K-m1+cm14/(1+m14)=K-m1+(m1(1+m14)/4)=K-(3/4)m1+m15/4. Hence, f(m1)>0 if K>(3/4)m1+m15/4. On analyzing f′′(u)=4cu2(3-5u4)/(1+u4)3, we see that the positive inflection of f(u) occurs at u=3/54. Thus, m1<3/54, and hence K>K0 ensures that there exists a unique r0 such that f(r0)=0.Choosem=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 selectb∈[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.
IfK<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=3c-2K-c(9c-16K)/2(c+K)4 and β=3c-2K+c(9c-16K)/2(c+K)4. 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.Sinceβ→34 as c→∞ and m<3/54, 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 off(u) with m and M.Figure 6
Graph ofu/f(u).Figure 7
Graph off*(u).The following estimates hold fora and M for c≫1.Proposition 3.4.
Forc≫1,(i)
c5<M<c4,(ii)
1/c3<a<1/c4.Proof.
(i) By the shape of the graph off(u) established in Propositions 3.1 and 3.3 (see Figure 5), it is enough if we prove that f′(c5)>0 and f′(c4)<0. We have f′(c5)=-1+4c8/5/(1+c4/5)2=-1+4c8/5/(1+2c4/5+c8/5)=-1+4/(1/c8/5+2/c4/5+1)>0, and f′(c4)=-1+4c7/4/(1+c)2=-1+(4/(1/c7/4+2/c3/4+c1/4))<0 for c≫1. Thus, c5<M<c4 for large c.
(ii) We havef(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/c3)=(1/c3)4-c(1/c3)3+1=1/c4/3>0, and j(1/c4)=(1/c4)4-c(1/c4)3+1=1/c-c4+1<0 for c≫1. Hence, 1/c3<a<1/c4 for large c. Note that since b=β→34 and a<1/c4→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 ofλ (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 that(3.1)bf(b)(N+1N)N+1N2R2<min⁡{a‖eΩ‖∞f*(a),2NMf(b)R2}.
Note that for c≫1(b/f(b))=34/(K-34+3c/4) and M/f(b)=M/(K-34+3c/4)>c5/(K-34+3c/4). Also, a/f*(a)=a/k>1/Kc3. Applying the estimates we obtained for a/f*(a),b/f(b), and M/f(b) to the above inequality, we get the following:(3.2)(34k-34+3c/4)(N+1N)N+1N2R2<min⁡{1‖eΩ‖∞kc3,2NR2(c5k-34+3c/4)}.
Simplifying the above, we can see that Q(a,b,Ω)<1 if(3.3)(34k/c-34/c+3/4)(N+1N)N+1N2R2<min⁡{c2/3‖eΩ‖∞k,2NR2(c1/5k/c-34/c+3/4)}.
Clearly, this inequality is true for c≫1; hence, Theorem 1.1 holds.



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*Source: 869147-2012-05-14.xml*