# Eigenvalue of Coupled Systems for Hammerstein Integral Equation with Two Parameters

Authors: Wanjun Li
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

By using the fixed-point index theory, we discuss the existence, multiplicity, and nonexistence of positive solutions for the coupled systems of Hammerstein integral equation with parameters.


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## Body

## 1. Introduction

In recent years, the study of solutions for Hammerstein integral equations has been an interesting topic, since the solution of some boundary value problems for differential equations are usually equivalent to solutions of Hammerstein integral equations [1–6]. And many results concerning the existence of solutions for Hammerstein integral equations have been obtained by many authors [1, 7–9]. For example, in [1] the Hammerstein integral equation:(1.1)ψ(x)=∫Gk(x,y)f(y,ψ(y))dy,
was considered, where G⊂ℝn is a bounded domain. When the nonlinear term is of the form f(u) or f(x,u)=∑i=1nai(x)uiαi, αi>0;i=1,2,…,n, some existence results of nonnegative solutions in C(G) for (1.1) were obtained; when the nonlinear term is a general f(x,u), some multiple results for (1.1) in space Lp(G)(p≥1) were derived. In [8], by means of the decomposition of the operator and the critical point theory, the existence of infinitely many solutions for (1.1) was considered. In [7] the integral equation(1.2)ψ(x)=λ∫[x,x+T]k(x,y)f(y,ψ(y-τ(y)))dy
was studied, where λ is a parameter. Using the Leggett-Williams fixed-point theorem, when λ belongs to some intervals, the existence of triple positive solutions for (1.2) was proved.More recently, in [9] the integral equation(1.3)u(x)=λ∫01k(x,y)f(y,u(y))dy,
was studied and obtained that there exists a λ*>0 such that (1.3) has at least two, one and no positive solutions for λ∈(0,λ*), λ=λ*, and λ>λ*, respectively.Motivated by the papers mentioned above, we consider the coupled systems of Hammerstein integral equation (1.4) in this paper,(1.4)u(x)=λ∫01k1(x,y)f1(y,u(y),v(y))dy,v(x)=μ∫01k2(x,y)f2(y,u(y),v(y))dy,
where λ, μ∈ℝ+2∖{(0,0)}, ℝ+=[0,+∞). Under some new assumptions, we show that there exists a continuous curve Γ separating ℝ+2∖{(0,0)} into two disjoint subsets 𝒪1 and 𝒪2 such that problem (1.4) has at least two, one and no positive solutions for 𝒪1, Γ and 𝒪2, respectively. Proofs of our results are mainly based on the fixed-point index theory, for this type of results see [10–12].The rest of the paper is organized as follows. In Section2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, we discuss the existence, multiplicity, and nonexistence of positive solution of the systems (1.4).The vector(u*,v*) is said to be a positive solution of problem (1.4) if and only if (u*,v*) satisfies problem (1.4) and u*(x)≥0,v*(x)>0, or u*(x)>0,v*(x)≥0 for any x∈[0,1].


## 2. Preliminaries and Lemmas

In the rest of the paper, we always suppose the following assumptions hold:(H1)
fi∈C[[0,1]×ℝ+×ℝ+,ℝ+],i=1,2;(H2)
ki∈C[[0,1]×[0,1],ℝ+],i=1,2;(H3)
there existσ, α, β∈(0,1), α<β such that ki(x,y)≥σki(y,y) for y∈(0,1) and x∈[α,β],i=1,2;(H4)
fi are nondecreasing on ℝ+2 for x∈[0,1], that is, fi(x,u1,v1)≤fi(x,u2,v2), i=1,2 whenever (u1,v1)≤(u2,v2), where the inequality on ℝ+2 can be understood componentwise and f1(x,0,0)>0 or f2(x,0,0)>0 for all x∈[0,1];(H5)
there exist constantscfi>0,i=1,2, such that fi(x,u,v)≥cfi(u+v) for all x∈[0,1];(H6)
lim∥(u,v)∥→∞(fi(x,u,v)/(u+v))=∞,i=1,2 uniformly for x∈[0,1].We will consider the Banach spaceE=C[0,1]×C[0,1] equipped with the standard norm (2.1)‖(u,v)‖=‖u‖+‖v‖=max⁡0≤x≤1⁡|u(x)|+max⁡0≤x≤1⁡|v(x)|,(u,v)∈C[0,1]×C[0,1].Define(2.2)P={(u,v)∈E:u(x)≥0,u(x)≥0,min⁡x∈[α,β]⁡(u(x)+v(x))≥σ‖(u,v)‖}.
It is easy to see that P is a cone in E.We define the operatorsTλ,Tμ:P→C[0,1] and T:P→C[0,1]×C[0,1] by(2.3)Tλ(u,v)(x)=λ∫01k1(x,y)f1(x,u(y),v(y))dy,Tμ(u,v)(x)=μ∫01k2(x,y)f2(y,u(y),v(y))dy,x∈[0,1],T(u,v)=(Tλ(u,v),Tμ(u,v)),∀(u,v)∈P.Obvious, the existence of a positive solution of problem (1.4) is equivalent to the existence of a nontrivial fixed point of T in P. It is easy to prove that the following lemma is true.Lemma 2.1.
Assume that(H1)–(H3) hold. Then T(P)⊂P and T:P→P is completely continuous.Finally we list two lemmas, which are crucial to prove our main results.Lemma 2.2 (see [13]).
LetE be a Banach space, K a cone and Ω an open bounded subset of E. Let θ∈Ω and T:K∩Ω¯→K be a completely continuous mapping. Suppose that Tx≠ϖx for all x∈K∩∂Ω and all ϖ≥1. Then i(T,K∩Ω,K)=1.Lemma 2.3 (see [13]).
LetE be a Banach space and K a cone in E. Assume that T:K¯r→K(Kr={x∈K,∥x∥<r,r>0}) is a compact map such that Tx≠x for x∈∂Kr. If ∥x∥≤∥Tx∥ for all x∈∂Kr, then i(T,Kr,K)=0.


## 3. Main Results

In this section, we consider the existence of positive solutions for (1.4) in E.Lemma 3.1.
Assume that(H1)–(H6) hold, and let Σ be a compact subset of ℝ+2∖{(0,0)}. Then there exists a constant CΣ>0 such that for all (λ,μ)∈Σ and all possible positive solutions (u(x),v(x)) of (1.4) at (λ,μ), one has ∥(u,v)∥≤CΣ.Proof.
Suppose by contradiction that there is a sequence(un,vn) of positive solutions of (1.4) at (λn,μn) such that (λn,μn)∈Σ for all n and ∥(un,vn)∥→∞. Then (un,vn)∈P and thus
(3.1)min⁡x∈[α,β](un(x)+vn(x))≥σ‖(un,vn)‖.
Since Σ is compact, the sequence {(λn,μn)}n=1∞ has a convergent subsequence which we denote without loss of generality still by {(λn,μn)}n=1∞ such that limn→∞λn=λ*, limn→∞μn=μ*, and at least one λ*>0 or μ*>0, hence for n sufficiently large, we have λn≥λ*/2>0.
Then by(H6), there exists Rf1>0 such that
(3.2)f1(t,u,v)≥Kf1(u+v),∀u+v≥Rf1,
where Kf1 satisfies
(3.3)λ*σ2Kf12∫αβk1(y,y)dy>1.
Thus forx∈[α,β] and ∀un+vn≥Rf1, by using (3.1) and (3.2), we get
(3.4)‖un‖≥un(x)=λn∫01k1(x,y)f1(y,un(y),vn(y))dy≥λ*2∫αβk1(x,y)f1(y,un(y),vn(y))dy≥λ*Kf12∫αβk1(x,y)(un(y)+vn(y))dy≥λ*σ2Kf12∫αβk1(y,y)(‖un‖+‖vn‖)dy>‖un‖+‖vn‖≥‖un‖
for all n sufficiently large. This is a contraction.Lemma 3.2.
Assume that(H1)–(H6) hold, and let (1.4) have a positive solution at (λ¯,μ¯). Then the problem also has a positive solution at (λ,μ) for all (λ,μ)≤(λ¯,μ¯).Proof.
Let(u¯,v¯) be a positive solution of (1.4) at (λ¯,μ¯), and let (λ,μ)∈ℝ+2∖{(0,0)} with (λ,μ)≤(λ¯,μ¯). First, we assume λ∈(0,λ¯), and μ∈(0,μ¯), then
(3.5)uλ¯(x)=λ¯∫01k1(x,y)f1(y,uλ¯(y),vμ¯(y))dy≥λ∫01k1(x,y)f1(y,uλ¯(y),vμ¯(y))dy,vμ¯(x)=μ¯∫01k2(x,y)f2(y,uλ¯(y),vμ¯(y))dy≥μ∫01k2(x,y)f2(y,uλ¯(y),vμ¯(y))dy.
Set
(3.6)Tλ(u,v)(x)=λ∫01k1(x,y)f1(y,u(y),v(y))dy,Tμ(u,v)(x)=μ∫01k2(x,y)f2(y,u(y),v(y))dy,
and let u0(x)=uλ¯(x), v0(x)=vμ¯(x), un(x)=Tλ(un-1(x),vn-1(x)), vn(x)=Tμ(un-1(x),vn-1(x)), T(un(x),vn(x))=(Tλ(un(x),vn(x)), Tμ(un(x),vn(x))), n=1,2,…. Then
(3.7)u0≥u1≥⋯≥un≥⋯≥σλcf1K1>θ,v0≥v1≥⋯≥vn≥⋯≥σμcf2K2>θ,
where Ki=∫αβki(y,y)dy, n=1,2. By the compactness of the operator Tλ(u,v)(x), Tμ(u,v)(x), Lemma 2.1, and the Lebesgue dominated convergence theorem, the sequence {un}n=0∞={Tλ(un-1,vn-1)}n=0∞, and {vn}n=0∞={Tμ(un-1,vn-1)}n=0∞ converges to u* and v*, respectively. It is clear that (u*,v*)∈P∖{(0,0)} and is a solution (1.4) at (λ,μ). Proof of the case μ=0, and λ∈(0,λ¯) or λ=0, and μ∈(0,λ¯) can be done similarly. The proof is complete.Lemma 3.3.
Assume that(H1)–(H4) hold. Then there exists (λ*,μ*)>(0,0) such that (1.4) has a positive solution for all (λ,μ)≤(λ*,μ*).Proof.
Letu0(x)=∫01k1(x,y)dy, v0(x)=∫01k2(x,y)dy, then (u0(x),v0(x))∈P∖{(0,0)}. Take Mfi=maxx∈[0,1]fi(x,u0(x),v0(x)), i=1,2. Then Mfi>0 and at (λ*,μ*)=(1/Mf1,1/Mf2), we get
(3.8)Tλ*(u,v)(x)=λ*∫01k1(x,y)f1(y,u(y),v(y))dy,Tμ*(u,v)(x)=μ*∫01k2(x,y)f2(y,u(y),v(y))dy.
This shows that (λ*,μ*)>(0,0) and
(3.9)u0(x)≥λ*∫01k1(x,y)f1(y,u0(y),v0(y))dy,v0(x)≥μ*∫01k2(x,y)f2(y,u0(y),v0(y))dy
for x∈[0,1]. Set
(3.10)un(x)=Tλ*(un-1(x),vn-1(x)),vn(x)=Tμ*(un-1(x),vn-1(x)),T(un(x),vn(x))=(Tλ*(un(x),vn(x)),Tμ*(un(x),vn(x))),n=1,2,…
for x∈[0,1]. Then
(3.11)u0≥u1≥⋯≥un≥⋯≥σλ*∫01k1(y,y)dy>θ,v0≥v1≥⋯≥vn≥⋯≥σμ*∫01k2(y,y)dy>θ.
We conclude the proof similarly to Lemma 3.2.Define𝒫={(λ,μ)∈ℝ+2∖{(0,0)}: (1.4) has a positive solution at (λ,μ)};  then by Lemma 3.3, 𝒫≠ϕ, and it is easy to see that (𝒫,≤) is a partially ordered set.Lemma 3.4.
Assume that(H1)–(H6) hold. Then (𝒫,≤) is bounded above.Proof.
Suppose to the contrary that there exists a fixed-point sequence{(un,vn)}(n=1,2,…) of T(u,v) at (λn,μn) such that limn→∞∥(λn,μn)∥=∞. Considering a subsequence if necessary, we assume limn→∞λn=∞. The proof for the case limn→∞μn=∞ can be shown by an analogous way. Then there are two cases to be considered:(i)
there exists a constantH>0 such that ∥(un,vn)∥≤H,n=0,1,2,…;(ii)
there exists a subsequence{(uni,vni)}i=1∞ such that limi→∞∥(uni,vni)∥=∞,which is impossible by Lemma 3.1. So we only consider (i). In view of (H4)–(H6) we can choose l0>0 such that f1(y,0,0)>l0H or f2(y,0,0)>l0H and further f1(y,un,vn)>l0H or f2(y,un,vn)>l0H for y∈[0,1]. Thus for x∈[α,β], we know
(3.12)un(x)+vn(x)=λn∫01k1(x,y)f1(y,un(y),vn(y))dy+μn∫01k2(x,y)f2(y,un(y),vn(y))dy≥λnσ∫αβk1(y,y)f1(y,un(y),vn(y))dy+μnσ∫αβk2(y,y)f2(y,un(y),vn(y))dy.
Now we will distinguish two cases.Case 1.
Iff1(y,un,vn)>l0H and f2(y,un,vn)>l0H, we have
(3.13)un(x)+vn(x)≥λnσl0HK1+μnσl0HK2≥(λn+μn)σl0HK0,
where K0=min(K1,K2), which implies that H≥(λn+μn)σl0HK1 or (λn+μn)≤1/σl0K1, which is a contradiction.Case 2.
If that one off1(y,un,vn)>l0H and f2(y,un,vn)>l0H is satisfied, without loss of generality, we assume f1(y,un,vn)>l0H. The proof for the case f2(y,un,vn)>l0H can be shown by an analogous way. By (H5) we have
(3.14)un(x)+vn(x)≥λnσl0HK1+μnσ2‖(un,vn)‖cf2K2≥λnσl0‖(un,vn)‖K1+μnσ2‖(un,vn)‖cf2K2≥(λn+μn)K3‖(un,vn)‖,
where K3=min(σl0K1,σ2cf2K2); which implies that ∥(un,vn)∥≥(λn+μn)K3∥(un,vn)∥ or (λn+μn)≤1/K3, which is a contradiction. The proof is complete.Lemma 3.5.
Assume that(H1)–(H6) hold. Then every chain in 𝒫 has a unique supremum in 𝒫.Proof.
Let𝒞 be a chain in 𝒫. Since 𝒫 is a partially ordered set, it is enough to show that 𝒞 has an upper bound in 𝒫. Without loss of generality, we may choose a distinct sequence {(λn,μn)}∈𝒞 such that (λn,μn)≤(λn+1,μn+1),n=1,2,…. By Lemma 3.4, two sequences {λn} and {μn} converge to, say, λ𝒞 and μ𝒞, respectively. If (λ𝒞,μ𝒞)∈𝒫, then the proof is done. Since the sequence {(λn,μn)} is bounded above, we may assume that the sequence belongs to a compact rectangle in ℝ+2∖{(0,0)} and Lemma 3.1 implies that the corresponding solutions {(un,vn)} are uniformly bounded in E. By the compactness of the integral operators Tλ and Tμ, the sequence {(un,vn)} has a subsequence converging to, say, {(un,vn)}∈E. We can easily show, by the Lebesgue convergence theorem, that (u𝒞,v𝒞) is a solution of (1.4) at (λ𝒞,μ𝒞). Thus (λ𝒞,μ𝒞)∈𝒫 and this completes the proof.Lemma 3.6.
Assume that(H1)–(H6) hold. Then there exists s*∈[μ*,μu] such that (1.4) has a positive solution at (0,s) for all 0<s≤s*, and no solution at (0,s) for all s>s*. Similarly, there exists r*∈[λ*,λu] such that (1.4) has a positive solution at (r,0) for all 0<s≤s*, and no solution at (r,0) for all r>r*, where (λu,μu) is upper bound of (𝒫,≤).Proof.
We know by Lemma3.3 that (1.4) has a positive solution at (0,s) for all 0<s≤μ*. Thus {(0,s):s>0}∩𝒫 is a nonempty chain in 𝒫 and by Lemma 3.5, it has a unique supremum of the form (0,s*) in 𝒫. The proof of the second part is similar.Lemma 3.7.
Assume that(H1)–(H6) hold. Then there exists a continuous curve Γ separating ℝ+2∖{(0,0)} into two disjoint subsets 𝒪1 and 𝒪2 such that problem (1.4) has at least one positive solution for (λ,μ)∈𝒪1∪Γ and no solution for (λ,μ)∈𝒪2.Proof.
We first construct the curveΓ on R+2∖{(0,0)}. Define
(3.15)Lx={(r,s)∈R+2∖{(0,0)}:s=r-x},x∈R.
At x=-s*, we know by Lemma 3.6, (0,s*)∈L-s*∩𝒫. Thus L-s*∩𝒫 is a nonempty chain in 𝒫 and Lemma 3.5 implies that the chain has a unique supremum. We show sup{L-s*∩𝒫}=(0,s*). Indeed, otherwise, we may choose (r̃,s̃)(≠(0,s*))∈L-s* such that (0,s*)<(r̃,s̃) and (1.4) has a solution at (r̃,s̃). Thus by Lemma 3.2, (1.4) has a solution at (0,s̃) and this contradicts Lemma 3.6. Similarly, we get sup{Lr*∩𝒫}=(r*,0).
For-s*<x<r*, we know by Lemmas 3.3, 3.5, and 3.6 that Lx∩𝒫 is a nonempty chain in 𝒫 and thus the chain also has a unique supremum.
We notice thatLx∩𝒫=∅, for x<-s* or x>r*. Now for x∈[-s*,r*], let us define
(3.16)Γ(x)=sup⁡(Lx∩P).
Then Γ:[-s*,r*]→R+2∖{(0,0)} is well defined and Γ(-s*)=(0,s*) and Γ(r*)=(r*,0). Similar to [10, Theorem  3.1], it is easy to prove that Γ is continuous on [-s*,r*].
Consequently, the curveΓ separates R+2∖{(0,0)} into two disjoint subsets 𝒪1 and 𝒪2, where 𝒪1 is bounded and 𝒪2 is unbounded. It is obvious that (1.4) has a positive solution at Γ(x) for all x∈[-s*,r*]. If (λ,μ)∈𝒪1 and so if (λ,μ)∈Lx0, then x0∈[-s*,r*] and (λ,μ)<Γ(x0). Thus by Lemma 3.2, (λ,μ)∈𝒫. On the other hand, if (λ,μ)∈𝒪2 and if (λ,μ)∈Lx0, then either Γ(x0) is not defined when x0∈[-s*,r*] or (λ,μ)>Γ(x0) when x0∉[-s*,r*]. We get (λ,μ)∉𝒫 for both cases and the proof is done.Now, we show the existence of the second positive solution for(λ,μ)∈𝒪1. Let (λ,μ)∈𝒪1, then we may choose x0∈[-s*,r*] such that (λ,μ)∈Lx0. We know by Lemma 3.7 that (1.4) has a positive solution at Γ(x0), so let (u*,v*) be the solution at Γ(x0) and let us denote Γ(x0)=(λ*,μ*). Then obviously (λ,μ)<(λ*,μ*) and we have the following theorem.Lemma 3.8.
Let(λ,μ)∈𝒪1. Then there exists ɛ0>0 and for all x∈[0,1], 0<ɛ≤ɛ0 such that
(3.17)uɛ*(x)>λ∫01k1(x,y)f1(y,uɛ*(y),vɛ*(y))dy,vɛ*(x)>μ∫01k2(x,y)f2(y,uɛ*(y),vɛ*(y))dy,
where uɛ*(x)=u*(x)+ɛ and vɛ*(x)=v*(x)+ɛ.Proof.
From(H4), there exists constant M>0 such that
(3.18)min⁡x∈[0,1]fi(x,u*(x),v*(x))≥M>0,i=1,2.
Then by the uniform continuity of fi on a compact set, there exists ɛ0>0 small enough such that for all x∈[0,1], 0<ɛ≤ɛ0,
(3.19)|f1(x,uɛ*(x),vɛ*(x))-f1(x,u*(x),v*(x))|<M(λ*λ-1),|f2(x,uɛ*(x),vɛ*(x))-f2(x,u*(x),v*(x))|<M(μ*μ-1),λf1(x,uɛ*(x),vɛ*(x))-λ*f1(x,u*(x),v*(x))<(λ*-λ)(M-f(x,u*(x),v*(x)))≤0,μf2(x,uɛ*(x),vɛ*(x))-μ*f2(x,u*(x),v*(x))<(μ*-μ)(M-f2(x,u*(x),v*(x)))≤0.
Furthermore, we have for all x∈[0,1](3.20)uɛ*(x)>λ*∫01k1(x,y)f1(y,u*(y),v*(y))dy>λ∫01k1(x,y)f1(y,uɛ*(y),vɛ*(y))dy.
The inequalities for vɛ* can be shown similarly and the proof is done.We now state and prove our main result in this section.Theorem 3.9.
Assume(H1)–(H6). Then there exists a continuous curve Γ separating ℝ+2∖{(0,0)} into two disjoint subsets 𝒪1 and 𝒪2 such that problem (1.4) has at least two positive solution on 𝒪1, at least one positive solution on Γ and no solution on 𝒪2.Proof.
By Lemma3.7, it is enough to show the existence of the second positive solution of (1.4) for (λ,μ)∈𝒪1. Let (λ,μ)∈𝒪1. Let (λ,μ)∈Lx0 and (u*,v*) be a positive solution of (1.4) at Γ(x0)=:(λ*,μ*) and let uɛ*(x)=u*+ɛ,vɛ*(x)=v*+ɛ, where ɛ is given in Lemma 3.8. Let
(3.21)Ω={(u,v)∈E:-ɛ<u(x)<uɛ*(x),-ɛ<v(x)<vɛ*(x),x∈[0,1]},
then Ω is bounded open in E, 0∈Ω and T:P∩Ω¯→P be a completely continuous mapping, since it is completely continuous. Let (u,v)∈P∩∂Ω. Then there exists xo∈[0,1] such that either u(xo)=uɛ*(xo) or v(xo)=vɛ*(xo). Suppose that u(xo)=uɛ*(xo). Then by (H4) and  Lemma 3.8,
(3.22)Tλ(u,v)(xo)=λ∫01k1(xo,y)f1(y,u(y),v(y))dy≤λ∫01k1(xo,y)f1(y,uɛ*(y),vɛ*(y))dy<uɛ*(xo)=u(xo)≤ϖu(xo),
for all ϖ≥1. Similarly, for the case v(xo)=vɛ*(xo), we can get Tμ(u,v)(xo)<ϖv(xo) for all ϖ≥1. Thus T(u,v)=(Tλ(u,v),Tμ(u,v))≠ϖ(u,v), for all (u,v)∈P∩Ω¯ and all ϖ≥1 and by Lemma 2.2,
(3.23)i(T,P∩Ω,P)=1.
Modifying (3.2) for λ*=λ, that is, there exists Rf1>0 such that
(3.24)f1(x,u,v)≥Kf1(u+v),∀u+v≥Rf1,
where Kf1 satisfies
(3.25)λσ2Kf12∫αβk1(y,y)dy>1.
Let R0=max{CΣ,(1/σ)Rf1,∥(uɛ*,vɛ*)∥}, where CΣ is given in Lemma 3.1 with Σ a compact rectangle in ℝ+2∖{(0,0)} containing (λ,μ). Let PR0={(u,v)∈P,∥(u,v)∥<R0}. Then (u,v)≠T(u,v) for (u,v)∈∂PR0, by Lemma 3.1. Furthermore, if (u,v)∈PR0, then
(3.26)u(t)+v(t)≥min⁡x∈[α,β](u(x)+v(x))≥σ∥(u,v)∥≥Rf1.
Thus by (3.24), f1(x,u(x),v(x))≥Kf1(u(x)+v(x)), for all x∈[α,β] and
(3.27)Tλ(u,v)(x)=λ∫01k1(x,y)f1(y,u(y),v(y))dy≥λσ∫αβk1(y,y)f1(y,u(y),v(y))dy≥λσKf1∫αβk1(y,y)(u(y)+v(y))dy≥λσ2Kf12∫αβk1(y,y)‖(u,v)‖ds>‖(u,v)‖.
Therefore ∥T(u,v)∥≥∥Tλ(u,v)∥>∥(u,v)∥ and by Lemma 2.3,
(3.28)i(T,PR0,P)=0.
Consequently by the additivity of the fixed-point index,
(3.29)0=i(T,PR0,P)=i(T,P∩Ω,P)+i(T,PR0∖P∩Ω¯,P).
Since i(T,P∩Ω,P)=1,i(T,PR0∖P∩Ω¯,P)=-1 and thus T has a fixed point on P∩Ω and another on PR0∖P∩Ω¯, and this completes the proof.



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*Source: 989401-2011-10-11.xml*