Existence and Uniqueness Theorems of Ordered Contractive Map in Banach Lattices

Abstract

This paper presents some existence and uniqueness theorems of the fixed point for ordered contractive mapping in Banach lattices. Moreover, we prove the existence of a unique solution for first-order ordinary differential equations with initial value conditions by using the theoretical results with no need for using the condition of a lower solution or an upper solution.

1. Introduction and Preliminaries

Existence of fixed points in partial ordered complete metric spaces has been considered further recently in [1–6]. Many new fixed point theorems are proved in a metric space endowed with partial order by using monotone iterative technique, and their results are applied to problems of existence and uniqueness of solutions for some differential equation problems. In [6] the existence of a minimal and a maximal solution for a nonlinear problem is presented by constructing an iterative sequence with the condition of a lower solution or an upper solution.

In this paper, the theoretical results of fixed points are extended by using the theorem of cone and monotone iterative technique in Banach lattices. But the iterative sequences can be constructed with no need for using the condition of a lower solution or an upper solution. To demonstrate the applicability of our results, we apply them to study a problem of ordinary differential equations in the final section of the paper, and the existence and uniqueness of solution are obtained.

Let 𝐸 be a Banach space and 𝑃 a cone of 𝐸. We define a partial ordering ≤ with respect to 𝑃 by 𝑥≤𝑦 if and only if 𝑦−𝑥∈𝑃. A cone 𝑃⊂𝐸 is called normal if there is a constant 𝑁>0, such that 𝜃≤𝑥≤𝑦 implies ‖𝑥‖≤𝑁‖𝑦‖, for all 𝑥,𝑦∈𝐸. The least positive constant 𝑁 satisfying the above inequality is called the normal constant of 𝑃.

Let 𝐸 be a Riesz space equipped with a Riesz norm. We call 𝐸 a Banach lattice in the partial ordering ≤, if 𝐸 is norm complete. For arbitrary 𝑥,𝑦∈𝐸, sup{𝑥,𝑦} and inf{𝑥,𝑦} exist. One can see [7] for the definition and the properties about the lattice.

Let 𝐷⊂𝐸; the operator 𝐴∶𝐷→𝐸 is said to be an increasing operator if 𝑥,𝑦∈𝐷, 𝑥≤𝑦, implies 𝐴𝑥≤𝐴𝑦; the operator 𝐴∶𝐷→𝐸 is said to be a decreasing operator if 𝑥,𝑦∈𝐷, 𝑥≤𝑦, implies 𝐴𝑦≤𝐴𝑥.

Lemma 1.1 (see [8]). Let 𝑃 be a normal cone in a real Banach space 𝐸. Suppose that {𝑥_𝑛} is a monotone sequence which has a subsequence {𝑥_𝑛𝑖} converging to 𝑥^∗, then {𝑥_𝑛} also converges to 𝑥^∗. Moreover, if {𝑥_𝑛} is an increasing sequence, then 𝑥_𝑛≤𝑥^∗(𝑛=1,2,3,…); if {𝑥_𝑛} is a decreasing sequence, then 𝑥^∗≤𝑥_𝑛(𝑛=1,2,3,…).

Lemma 1.2 (see [9]). Let Ω be a bounded open set in a real Banach space 𝐸 such that 𝜃∈Ω; let 𝑃 be a cone of 𝐸. Let 𝐴∶𝑃∩Ω→𝑃 is completely continuous. Suppose that 𝑥≰𝐴𝑥,∀𝑥∈𝑃∩Ω. (1.1) Then 𝑖(𝐴,𝑃∩Ω,𝑃)=1.

Lemma 1.3 (see [9]). Let 𝐸 be a real Banach space, and let 𝑃⊂𝐸 be a cone. Assume Ω₁ and Ω₂ are two bounded open subsets of 𝐸 with 𝜃∈Ω₁⊂Ω₂ and Ω₁⊂Ω₂, and let 𝐴∶𝑃∩(Ω₂⧵Ω₁)→𝑃 is completely continuous. Suppose that eitherH₁𝑥≰𝐴𝑥,forall𝑥∈𝑃∩Ω₁ and 𝐴𝑥≰𝑥,forall𝑥∈𝑃∩Ω₂, orH₂𝐴𝑥≰𝑥,forall𝑥∈𝑃∩Ω₁ and 𝑥≰𝐴𝑥,forall𝑥∈𝑃∩Ω₂.Then 𝐴 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

2. Main Results

Theorem 2.1. Let 𝐸 be a real Banach lattice, and let 𝑃⊂𝐸 be a normal cone. Suppose that 𝐴∶𝐸→𝐸 is a decreasing operator such that there exists a linear operator 𝐿∶𝐸→𝐸 with spectral radius 𝑟(𝐿)<1 and 𝐴𝑣−𝐴𝑢≤𝐿(𝑢−𝑣),for𝑢,𝑣∈𝐸with𝑣≤𝑢. (2.1) Then the operator 𝐴 has a unique fixed point.

Proof. For any 𝑢₀∈𝐸, since 𝐴∶𝐸→𝐸, we have 𝐴𝑢₀∈𝐸. Now we suppose the following two cases.
Case (I). Suppose that 𝑢₀ is comparable to 𝐴𝑢₀. Firstly, without loss of generality, suppose that 𝑢₀≤𝐴𝑢₀. If 𝐴𝑢₀=𝑢₀, then the proof is finished. Suppose 𝐴𝑢₀≠𝑢₀. Since 𝐴 is decreasing together with 𝑢₀≤𝐴𝑢₀, we obtain by induction that {𝐴^𝑛+1(𝑢₀)} and {𝐴^𝑛(𝑢₀)} are comparable, for every 𝑛=0,1,2,…. Using the contractive condition (2.1), we can obtain by induction that 𝐴^𝑛+1𝑢₀−𝐴^𝑛𝑢₀≤𝑁𝐿^𝑛𝐴𝑢₀−𝑢₀,𝑛∈𝑁. (2.2) In fact, for 𝑛=1, using the fact that 𝑃 is normal, we have 𝐴𝑢₀−𝐴²𝑢₀≤𝑁𝐿𝐴𝑢₀−𝑢₀. (2.3) Suppose that (2.2) is true when 𝑛=𝑘 then when 𝑛=𝑘+1, we obtain 𝐴^𝑛+2𝑢₀−𝐴^𝑛+1𝑢₀=𝐴𝐴^𝑛+1𝑢₀−𝐴𝐴^𝑛𝑢₀≤𝑁𝐿𝐴^𝑛+1𝑢₀−𝐴^𝑛𝑢₀≤𝑁𝐿^𝑛+1𝐴𝑢₀−𝑢₀. (2.4) For any 𝑚,𝑛∈𝑁, 𝑚>𝑛, since 𝑃 is normal cone, we have 𝐴^𝑚𝑢₀−𝐴^𝑛𝑢₀=𝐴^𝑚𝑢₀−𝐴^𝑚−1𝑢₀+⋯+𝐴^𝑛+1𝑢₀−𝐴^𝑛𝑢₀≤𝑁𝐿^𝑚−1+𝐿^𝑚−2+⋯+𝐿^𝑛𝐴𝑢₀−𝑢₀≤𝑁𝑟𝐿^𝑚−1+𝐿^𝑚−2+⋯+𝐿^𝑛𝐴𝑢₀−𝑢₀≤𝑁𝑟𝐿^𝑚−1+𝑟𝐿^𝑚−2+⋯+𝑟𝐿^𝑛𝐴𝑢₀−𝑢₀. (2.5) Here 𝑁 is the normal constant.
Given a 𝛼 such that 𝑟(𝐿)<𝛼<1, since lim_𝑛→+∞‖𝐿^𝑛‖^1/𝑛=𝑟(𝐿)<𝛼<1, there exists a 𝑛₀∈𝑁 such that 𝐿^𝑛<𝛼^𝑛,𝑛≥𝑛₀. (2.6) For any 𝑚,𝑛∈𝑁, 𝑚>𝑛≥𝑛₀, since 𝑃 is normal cone, we have 𝐴^𝑚𝑢₀−𝐴^𝑛𝑢₀≤𝑁𝑟𝐿^𝑚−1+𝑟𝐿^𝑚−2+⋯+𝑟𝐿^𝑛𝐴𝑢₀−𝑢₀≤𝑁𝛼^𝑚−1+𝛼^𝑚−2+⋯+𝛼^𝑛𝐴𝑢₀−𝑢₀𝛼≤𝑁^𝑛−𝛼^𝑚1−𝛼𝐴𝑢₀−𝑢₀𝛼≤𝑁^𝑛1−𝛼𝐴𝑢₀−𝑢₀. (2.7) This implies that {𝐴^𝑛(𝑢₀)} is a Cauchy sequence in 𝐸. The complete character of 𝐸 implies the existence of 𝑥^∗∈𝑃 such that lim_𝑛→+∞𝐴^𝑛𝑢₀=𝑥^∗. (2.8) Next, we prove that 𝑥^∗ is a fixed point of 𝐴 in 𝐸. Since 𝐴 is decreasing and 𝑢₀≤𝐴𝑢₀, we can get 𝐴²𝑢₀≤𝐴𝑢₀.
So 𝐴𝑢₀−𝐴²𝑢₀≤𝐿𝐴𝑢₀−𝑢₀, (2.9) then 𝐴²𝑢₀−𝑢₀=𝐴𝑢₀−𝑢₀−𝐴𝑢₀−𝐴²𝑢₀≥(𝐼−𝐿)𝐴𝑢₀−𝑢₀≥𝜃. (2.10) It is easy to know that 𝐴² is increasing and 𝐴²𝑢₀≤𝐴⁴𝑢₀,𝐴³𝑢₀≤𝐴𝑢₀. (2.11) By induction, we obtain that 𝑢₀≤𝐴²𝑢₀≤⋯≤𝐴^2𝑛𝑢₀≤⋯≤𝐴^2𝑛+1𝑢₀≤⋯≤𝐴³𝑢₀≤𝐴𝑢₀. (2.12) Hence, the sequence {𝐴^𝑛(𝑢₀)} has an increasing Cauchy subsequence {𝐴^2𝑛(𝑢₀)} and a decreasing Cauchy subsequence {𝐴^2𝑛+1(𝑢₀)} such that lim_𝑛→+∞𝐴^2𝑛𝑢₀=𝑢^∗,lim_𝑛→+∞𝐴^2𝑛+1𝑢₀=𝑣^∗. (2.13) Thus Lemma 1.1 implies that 𝐴^2𝑛(𝑢₀)≤𝑢^∗, 𝑣^∗≤𝐴^2𝑛+1(𝑢₀).
Since {𝐴^𝑛(𝑢₀)} is a Cauchy sequence, we can get that 𝑢^∗=𝑣^∗=𝑥^∗.
Moreover 𝐴𝑥^∗−𝑥^∗≤𝐴𝑥^∗−𝐴𝐴^2𝑛𝑢₀+𝐴^2(𝑛+1)𝑢₀−𝑥^∗≤𝑁𝐿𝑥^∗−𝐴^2𝑛𝑢₀+𝐴^2(𝑛+1)𝑢₀−𝑥^∗≤𝑁𝛼𝑥^∗−𝐴^2𝑛𝑢₀+𝐴^2(𝑛+1)𝑢₀−𝑥^∗. (2.14) Thus ‖𝐴𝑥^∗−𝑥^∗‖=0. That is 𝐴𝑥^∗=𝑥^∗. Hence 𝑥^∗ is a fixed point of 𝐴 in 𝐸.
Case (II). On the contrary, suppose that 𝑢₀ is not comparable to 𝐴𝑢₀.
Now, since 𝐸 is a Banach lattice, there exists 𝑣₀ such that inf{𝐴𝑢₀,𝑢₀}=𝑣₀. That is 𝑣₀≤𝐴𝑢₀ and 𝑣₀≤𝑢₀. Since 𝐴 is a decreasing operator, we have 𝐴²𝑢₀≤𝐴𝑣₀,𝐴𝑢₀≤𝐴𝑣₀. (2.15) This shows that 𝑣₀≤𝐴𝑣₀. Similarly as the proof of case (I), we can get that 𝐴 has a fixed point 𝑥^∗ in 𝐸.
Finally, we prove that 𝐴 has a unique fixed point 𝑥^∗ in 𝐸. In fact, let 𝑢^∗ and 𝑣^∗ be two fixed points of 𝐴 in 𝐸.(1)If 𝑢^∗ is comparable to 𝑣^∗, 𝐴^𝑛(𝑢^∗)=𝑢^∗ is comparable to 𝐴^𝑛(𝑣^∗)=𝑣^∗ for every 𝑛=0,1,2,…, and 𝑢^∗−𝑣^∗=𝐴^𝑛𝑢^∗−𝐴^𝑛𝑣^∗≤𝑁𝛼^𝑛𝑢^∗−𝑣^∗, (2.16) which implies 𝑢^∗=𝑣^∗.(2)If 𝑢^∗ is not comparable to 𝑣^∗, there exists either an upper or a lower bound of 𝑢^∗ and 𝑣^∗ because 𝐸 is a Banach lattice, that is, there exists 𝑧^∗∈𝐸 such that 𝑧^∗≤𝑢^∗,𝑧^∗≤𝑣^∗ or 𝑢^∗≤𝑧^∗,𝑢^∗≤𝑧^∗. Monotonicity implies that 𝐴^𝑛(𝑧^∗) is comparable to 𝐴^𝑛(𝑢^∗) and 𝐴^𝑛(𝑣^∗), for all 𝑛=0,1,2,…, and 𝑢^∗−𝑣^∗=𝐴^𝑛𝑢^∗−𝐴^𝑛𝑣^∗≤𝐴^𝑛𝑧^∗−𝐴^𝑛𝑢^∗+𝐴^𝑛𝑧^∗−𝐴^𝑛𝑣^∗≤𝑁𝛼^𝑛𝑢^∗−𝑧^∗+𝑁𝛼^𝑛𝑧^∗−𝑣^∗. (2.17)This shows that ‖𝑢^∗−𝑣^∗‖→0 when 𝑛→+∞. Hence 𝐴 has a unique fixed point 𝑥^∗ in 𝐸.

Theorem 2.2. Let 𝐸 be a real Banach lattice, and let 𝑃⊂𝐸 be a normal cone. Suppose that 𝐴∶𝑃→𝑃 is a completely continuous and increasing operator such that there exists a linear operator 𝐿∶𝐸→𝐸 with spectral radius 𝑟(𝐿)<1 and 𝐴𝑢−𝐴𝑣≤𝐿(𝑢−𝑣),for𝑢,𝑣∈𝑃with𝑣≤𝑢. (2.18) Then the operator 𝐴 has a unique fixed point 𝑢^∗ in 𝑃.

Proof. For any 𝑟>0, let Ω={𝑥∈𝑃∶‖𝑥‖≤𝑟}. Now we suppose the following two cases.
Case (I). Firstly, suppose that there exists 𝑢₀∈𝜕Ω such that 𝑢₀≤𝐴𝑢₀. If 𝐴𝑢₀=𝑢₀, then the proof is finished. Suppose 𝐴𝑢₀≠𝑢₀. Since 𝑢₀≤𝐴𝑢₀ and 𝐴 is nondecreasing, we obtain by induction that 𝑢₀≤𝐴𝑢₀≤𝐴²𝑢₀≤𝐴³𝑢₀≤⋯≤𝐴^𝑛𝑢₀≤𝐴^𝑛+1𝑢₀≤⋯. (2.19) Similarly as the proof of Theorem 2.1, we can get that {𝐴^𝑛(𝑢₀)} is a Cauchy sequence in 𝐸. Since 𝐸 is complete, by Lemma 1.1, there exists 𝑢^∗∈𝐸,𝐴^𝑛(𝑢₀)≤𝑢^∗ such that lim_𝑛→+∞𝐴^𝑛𝑢₀=𝑢^∗. (2.20) Next, we prove that 𝑢^∗ is a fixed point of 𝐴, that is, 𝐴𝑢^∗=𝑢^∗. In fact 𝐴𝑢^∗−𝑢^∗≤𝐴𝑢^∗−𝐴𝐴^𝑛𝑢₀+𝐴^𝑛+1𝑢₀−𝑢^∗≤𝑁𝐿𝑢^∗−𝐴^𝑛𝑢₀+𝐴^𝑛+1𝑢₀−𝑢^∗≤𝑁𝛼𝑢^∗−𝐴^𝑛𝑢₀+𝐴^𝑛+1𝑢₀−𝑢^∗. (2.21) Now, by the convergence of {𝐴^𝑛(𝑢₀)} to 𝑢^∗, we can get ‖𝐴𝑢^∗−𝑢^∗‖=0. This proves that 𝑢^∗ is a fixed point of 𝐴.
Case (II). On the contrary, suppose that 𝑥≰𝐴𝑥 for all 𝑥∈𝜕Ω. Thus Lemma 1.2 implies the existence of a fixed point in this case also.
Finally, similarly as the proof of Theorem 2.1, we can get that 𝐴 has a unique fixed point 𝑥^∗ in 𝑃.

Theorem 2.3. Let 𝐸 be a real Banach lattice, and let 𝑃⊂𝐸 be a normal cone. Suppose that 𝐴∶𝑃→𝑃 is a completely continuous and increasing operator which satisfies the following assumptions:(i)there exists a linear operator 𝐿∶𝐸→𝐸 with spectral radius 𝑟(𝐿)<1 and 𝐴𝑢−𝐴𝑣≤𝐿(𝑢−𝑣),for𝑢,𝑣∈𝑃with𝑣≤𝑢; (2.22)(ii)𝑆={𝑥∈𝑃∶𝐴𝑥≤𝑥} is bounded.
Then the operator 𝐴 has a unique nonzero fixed point 𝑢^∗ in 𝑃.

Proof. Firstly, for any 𝑟>0, let Ω={𝑥∈𝑃∶‖𝑥‖≤𝑟}. Now we suppose the following two cases.
Case (I). Suppose that there exists 𝑢₀∈𝜕Ω such that 𝑢₀≤𝐴𝑢₀. Similarly as proof of Theorem 2.1, we get that 𝐴 has a nonzero fixed point 𝑢^∗ in 𝑃.
Case (II). On the contrary, suppose that 𝑥≰𝐴𝑥 for all 𝑥∈𝜕Ω. Now, since 𝑆 is bounded there exists 𝑅>𝑟 such that 𝐴𝑥≰𝑥 for all 𝑥∈𝑃 with ‖𝑥‖=𝑅. Thus Lemma 1.3 implies the existence of a nonzero fixed point in this case.
Finally, similarly as the proof of Theorem 2.1, we can get that 𝐴 has a unique non-zero fixed point 𝑢^∗ in 𝑃.

3. Applications

In this section, we use Theorem 2.1 to show the existence of unique solution for the first-order initial value problem 𝑢^′(𝑡)((𝑡[],𝑢(0)=𝑓𝑡,𝑢)),𝑡∈𝐼=0,𝑇=𝑢₀, (3.1) where 𝑇>0 and 𝑓∶𝐼×𝑅→𝑅 is a continuous function.

Theorem 3.1. Let 𝑓∶𝐼×𝑅→𝑅 be continuous, and suppose that there exists 0<𝜇<𝜆, such that −𝜇𝑦−𝑥≤𝑓𝑡,𝑦+𝜆𝑦−𝑓(𝑡,𝑥)+𝜆𝑥≤0,∀𝑦≥𝑥. (3.2) Then (3.1) has a unique solution 𝑢^∗.

Proof. It is easy to know that 𝐸=𝐶(𝐼) is a Banach space with maximum norm ‖⋅‖, and it is also a Banach lattice with maximum norm ‖⋅‖. Let 𝑃={𝑢∈𝐸|𝑢(𝑡)≥0,forall𝑡∈𝐼}, and 𝑃 is a normal cone in Banach lattice 𝐸. Equation (3.1) can be written as 𝑢^′(𝑡)(𝑡)((𝑡(𝑡)[],𝑢(0)+𝜆𝑢=𝑓𝑡,𝑢))+𝜆𝑢,𝑡∈𝐼=0,𝑇=𝑢₀. (3.3) This problem is equivalent to the integral equation 𝑢(𝑡)=𝑒^−𝜆𝑡𝑢₀+^𝑡₀𝑒^𝜆𝑠𝑓(𝑠,𝑢(𝑠))+𝜆𝑢(𝑠)𝑑𝑠. (3.4) Define operator 𝐴 as the following: (𝐴𝑢)(𝑡)=𝑒^−𝜆𝑡𝑢₀+^𝑡₀𝑒^𝜆𝑠𝑓(𝑠,𝑢(𝑠))+𝜆𝑢(𝑠)𝑑𝑠,𝑡∈𝐼. (3.5) Moreover, the mapping 𝐴 is decreasing in 𝑢. In fact, by hypotheses, for 𝑢≥𝑣, 𝑓(𝑡,𝑢(𝑡))+𝜆𝑢(𝑡)≤𝑓(𝑡,𝑣(𝑡))+𝜆𝑣(𝑡) (3.6) implies that (𝐴𝑢)(𝑡)=𝑒^−𝜆𝑡𝑢₀+^𝑡₀𝑒^𝜆𝑠𝑓(𝑠,𝑢(𝑠))+𝜆𝑢(𝑠)𝑑𝑠≤𝑒^−𝜆𝑡𝑢₀+^𝑡₀𝑒^𝜆𝑠𝑓(𝑠,𝑣(𝑠))+𝜆𝑣(𝑠)𝑑𝑠=(𝐴𝑣)(𝑡),𝑡∈𝐼, (3.7) so 𝐴 is decreasing. Besides, for 𝑢≥𝑣, 𝐴(𝑣)−𝐴(𝑢)=^𝑡₀𝑒^{𝜆(𝑠−𝑡)}𝑓(𝑠)(𝑠)((𝑠(𝑠≤𝑠,𝑣+𝜆𝑣−𝑓𝑠,𝑢))−𝜆𝑢)𝑑𝑠_𝑡0𝑒^{𝜆(𝑠−𝑡)}𝜇[𝑢(𝑠)−𝑣(𝑠)]𝑑𝑠=𝐿(𝑢−𝑣), (3.8) where 𝐿𝑢=∫^𝑡₀𝑒^{𝜆(𝑠−𝑡)}𝜇𝑢(𝑠)𝑑𝑠. Since 𝐴 is decreasing, then 𝐿 is positive linear operator.
Now, let us prove that the spectral radius 𝑟(𝐿)<1. For 𝑡∈𝐼, since 0<𝑒^{𝜆(𝑠−𝑡)}≤1, we have ‖𝐿𝑢‖=max_𝑡∈𝐼^𝑡₀𝑒^{𝜆(𝑠−𝑡)}𝜇𝑢(𝑠)𝑑𝑠≤𝜇^𝑡₀𝑒^{𝜆(𝑠−𝑡)}‖𝑢‖‖𝑢‖,𝐿𝑑𝑠≤𝜇𝑡₂𝑢=max_𝑡∈𝐼^𝑡₀𝑒^{𝜆(𝑠−𝑡)}𝜇𝐿(𝑢(𝑠))𝑑𝑠≤𝜇^2𝑡₀𝑒^{𝜆(𝑠−𝑡)}‖‖≤𝜇𝑠𝑑𝑠𝐿𝑢²𝑡2!²‖𝑢‖. (3.9) By mathematical induction, for any 𝑛∈𝑁, we have 𝐿^𝑛𝑢≤𝜇^𝑛𝑡𝑛!^𝑛‖𝑢‖,𝑡∈𝐼. (3.10) So 𝐿^𝑛≤𝜇^𝑛𝑇𝑛!^𝑛. (3.11) Since 0<𝜇<𝜆, we have 𝑟(𝐿)=lim_𝑛→+∞𝐿^𝑛1/𝑛=0<1. (3.12) So the condition of Theorem 2.1 holds, and Theorem 3.1 is proved.

Acknowledgments

The first author was supported financially by the NSFC (71240007), NSFSP (ZR2010AM005).