Acta Mathematica Sinica, English Series Dec., 2007, Vol. 23, No. 12, pp. 2205–2212 Published online: May 5, 2006 DOI: 10.1007/s10114-005-0769-0 Http://www.ActaMath.com
Existence and Uniqueness of Fixed Point in Partially Ordered Sets and Applications to Ordinary Differential Equations Juan J. NIETO
´ Rosana RODR´ IGUEZ–LOPEZ
Departamento de An´ alisis Matem´ atico, Facultad de Matem´ aticas, Universidad de Santiago de Compostela, 15782, Spain E-mail:
[email protected] [email protected] Abstract We prove some fixed point theorems in partially ordered sets, providing an extension of the Banach contractive mapping theorem. Having studied previously the nondecreasing case, we consider in this paper nonincreasing mappings as well as non monotone mappings. We also present some applications to first-order ordinary differential equations with periodic boundary conditions, proving the existence of a unique solution admitting the existence of a lower solution. Keywords fixed point, partially ordered set, first-order differential equation, lower and upper solutions MR(2000) Subject Classification 47H10, 34B15
1
Preliminaries
In reference [1], we find some results on the existence of fixed points in partially ordered sets as well as some applications to matrix equations, and in [2] some results on the existence of a unique fixed point for nondecreasing mappings are applied to obtain a unique solution for a firstorder ordinary differential equation with periodic boundary conditions. Here we present some results concerning the existence of a unique fixed point for nonincreasing functions, showing their application in the field of ordinary differential equations and, finally, we study the nonmonotone case. The results we present here are similar to the theory exposed in [2] and are applicable in some cases where neither Tarski’s theorem [3, 4] (which requires X to be a complete lattice), nor theorems of Knaster–Tarski [5] or Amann [5, 6] (which require the existence of a supremum for every chain in X) are useful. Besides, since the base space has not necessarily a vectorial structure, the Schauder theorem can neither be applied to prove the existence of fixed points. As an application, we obtain conditions which guarantee the existence of a unique solution for periodic boundary value problems relative to ordinary differential equations. 2
Fixed Point Theorems
In [2], we prove some existence and uniqueness results for nondecreasing functions in a partially ordered set. Received June 8, 2005, Accepted July 28, 2005 Research partially supported by Ministerio de Educaci´ on y Ciencia and FEDER, Project MTM2004-06652-C0301, and by Xunta de Galicia and FEDER, Projects PGIDIT02PXIC20703PN and PGIDIT05PXIC20702PN
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Definition 1 Let (X, ≤) be a partially ordered set and f : X → X. We say that f is monotone nondecreasing if x, y ∈ X, x ≤ y =⇒ f (x) ≤ f (y). This concept coincides with the notion of nondecreasing function in the case where X = R and ≤ represents the usual total order in R. Theorem 1 ([2])
Let (X, ≤) be a partially ordered set and suppose that there exists a metric
d in X such that (X, d) is a complete metric space. Let f : X → X be a monotone nondecreasing mapping such that there exists k ∈ [0, 1) with d(f (x), f (y)) ≤ k d(x, y), ∀ x ≥ y. Assume that either f is continuous or X is such that if there is a nondecreasing sequence {xn } → x in X, then xn ≤ x, ∀ n.
(1)
If there exists x0 ∈ X with x0 ≤ f (x0 ), then f has a fixed point. Theorem 2 ([2]) Let (X, ≤) be a partially ordered set and suppose that there exists a metric d in X such that (X, d) is a complete metric space. Let f : X → X be a monotone nondecreasing mapping such that there exists k ∈ [0, 1) with d(f (x), f (y)) ≤ k d(x, y), ∀ x ≥ y. Assume that either f is continuous or X is such that if there is a nonincreasing sequence {xn } → x in X, then x ≤ xn , ∀ n.
(2)
If there exists x0 ∈ X with x0 ≥ f (x0 ), then f has a fixed point. The uniqueness of the fixed point and consequent global convergence of the method of successive approximations can be obtained by adding the hypothesis: Every pair of elements of X has a lower bound or an upper bound,
(3)
which is equivalent, as we have proved in [2], to For every x, y ∈ X, there exists z ∈ X which is comparable to x and y. With this assumption, for every x ∈ X, limn→+∞ f n (x) = y, where y is the fixed point of f such that y = limn→+∞ f n (x0 ), and hence f has a unique fixed point. Note that condition (3) is always valid if X is a lattice. In particular, we can take the set C([a, b], R) of continuous functions u : [a, b] → R equipped with the partial order x, y ∈ C([a, b], R),
x ≤ y ⇐⇒ x(t) ≤ y(t), for every t ∈ [a, b],
which is a lattice, since the elements max{x, y} and min{x, y}, defined by real variable operations, provide the least upper and greatest lower bounds of x and y, respectively. As a corollary of the previous results, we obtained in [2] an existence and uniqueness result for periodic boundary value problems relative to ordinary differential equations. In order to apply fixed point results, we make use of lower (or upper) solutions for some particular problem of the type ⎧ ⎨ u (t) = f (t, u(t)), t ∈ I = [0, T ], (4) ⎩ u(0) = u(T ),
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where T > 0, and f : I × R −→ R is a continuous function. Definition 2
A solution to (4) is a function u ∈ C 1 (I, R) satisfying conditions in (4).
Definition 3
A lower solution for (4) is a function α ∈ C 1 (I, R) such that α (t) ≤ f (t, α(t)), t ∈ I,
α(0) ≤ α(T ).
An upper solution for (4) verifies the reversed inequalities. Theorem 3 ([2]) Consider problem (4) with f continuous and suppose that there exist λ > 0, μ > 0, with μ < λ, such that 0 ≤ f (t, y) + λy − [f (t, x) + λx] ≤ μ(y − x), for all x, y ∈ R, y ≥ x.
(5)
Then the existence of a lower solution (or an upper solution) for (4) provides the existence of a unique solution of (4). 3
Nonincreasing functions
Definition 4 If (X, ≤) is a partially ordered set and f : X → X, we say that f is monotone nonincreasing if x, y ∈ X, x ≤ y =⇒ f (x) ≥ f (y). For nonincreasing functions, it is also possible to prove Fixed Point Theorems analogous to the results proved in [2]. Theorem 4 Let (X, ≤) be a partially ordered set verifying (3) and suppose that there exists a metric d in X such that (X, d) is a complete metric space. Let f : X → X be a nonincreasing function such that there exists k ∈ [0, 1) with d(f (x), f (y)) ≤ k d(x, y), ∀ x ≥ y. Suppose also that either f is continuous or X is such that if {xn } → x is a sequence in X whose consecutive terms are comparable, then there exists a subsequence {xnk }k∈N of {xn }n∈N such that every term is comparable to the limit x.
(6)
If there exists x0 ∈ X with x0 ≤ f (x0 ) or x0 ≥ f (x0 ), then f has a unique fixed point. Proof If f (x0 ) = x0 , then the proof is finished. Suppose that f (x0 ) = x0 . Following the lines of the proof of Theorem 2.1 in [1], we obtain that {f n (x0 )} is a convergent sequence in X. Indeed, by hypotheses, f n+1 (x0 ) and f n (x0 ) are comparable, for every n = 0, 1, 2, . . . and therefore, by induction, d(f n+1 (x0 ), f n (x0 )) ≤ kn d(f (x0 ), x0 ), n ∈ N, which allows us to prove, for m > n, that d(f m (x0 ), f n (x0 )) ≤ km−1 + km−2 + · · · + kn d(f (x0 ), x0 ) kn − km d(f (x0 ), x0 ) = 1−k n k d(f (x0 ), x0 ), ≤ 1−k
(7)
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so that {f n (x0 )}n∈N is a Cauchy sequence in X. The complete character of X implies the existence of x ¯ ∈ X such that limn→+∞ f n (x0 ) = x ¯. Next, we prove that x ¯ is a fixed point of f . Since the continuous case has been analyzed in [1, 2], suppose that condition (6) holds. Indeed, since f is nonincreasing, {f n (x0 )}n∈N is not necessarily monotone, but it is a convergent sequence with comparable consecutive terms, then, by hypotheses, there exists a subsequence {f nk (x0 )}k∈N consisting of terms which are comparable to the limit x ¯. Hence, for every k ∈ N, d(f (¯ x), x ¯) ≤ d(f (¯ x), f (f nk (x0 ))) + d(f nk +1 (x0 ), x ¯) ≤ kd(¯ x, f nk (x0 )) + d(f nk +1 (x0 ), x ¯) k→+∞
≤ d(¯ x, f nk (x0 )) + d(f nk +1 (x0 ), x ¯) −→ 0. This shows that d(f (¯ x), x ¯) = 0. Remark 1 Supposing that d(a, c) ≥ d(b, c), for a ≤ b ≤ c, the validity of (6) implies the validity of conditions (1) and (2), since, in the monotone case, the existence of a subsequence whose terms are comparable with the limit is equivalent to saying that all the terms in the sequence are also comparable with the limit. Taking into account this remark, the results previously discussed and the fact that, in conditions (1) and (2), the key is that the terms in the sequence (starting at a certain term) are comparable to the limit, we obtain the following result, which improves Theorem 2.1 in [1]: Theorem 5 Let (X, ≤) be a partially ordered set where (3) holds, and suppose that there exists a metric d in X such that (X, d) is a complete metric space. Let f : X → X be a monotone function (nondecreasing or nonincreasing) such that there exists k ∈ [0, 1) with d(f (x), f (y)) ≤ k d(x, y), ∀ x ≥ y. Suppose that either f is continuous or X is such that condition (6) holds. If there exists x0 ∈ X with x0 ≤ f (x0 ) or x0 ≥ f (x0 ), then f has a unique fixed point. 3.1 Application to Ordinary Differential Equations Theorem 6 Let f : I × R −→ R be continuous and suppose that there exist λ > 0, μ > 0, with μ < λ, such that −μ(y − x) ≤ f (t, y) + λy − [f (t, x) + λx] ≤ 0, for every x, y ∈ R, y ≥ x.
(8)
Then the existence of a lower solution (or an upper solution) for problem (4) provides the existence of a unique solution to (4). Proof Problem (4) can be written as ⎧ ⎨ u (t) + λu(t) = f (t, u(t)) + λu(t), t ∈ I = [0, T ], ⎩ u(0) = u(T ), and, equivalently, as the integral equation T G(t, s) [f (s, u(s)) + λu(s)] ds, t ∈ I, u(t) = 0
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⎧ λ(T +s−t) e ⎪ ⎪ , 0 ≤ s < t ≤ T, ⎨ λT e −1 G(t, s) = λ(s−t) ⎪ ⎪ e ⎩ , 0 ≤ t < s ≤ T. eλT − 1
where
We consider the complete metric space X = C(I, R) with the distance d(x, y) = sup |x(t) − y(t)|, x, y ∈ C(I, R), t∈I
and the following order relation: x, y ∈ C(I, R), x ≤ y if and only if x(t) ≤ y(t), ∀ t ∈ I. Define the operator A : C(I, R) −→ C(I, R), given by T G(t, s) [f (s, u(s)) + λu(s)] ds, t ∈ I, [A u](t) = 0
whose fixed points u ∈ C(I, R) are solutions u ∈ C 1 (I, R) of (4). Hypothesis (8) and the property G(t, s) > 0, for (t, s) ∈ I × I, imply, for u ≥ v, that T T [A u](t)= G(t, s) [f (s, u(s)) + λu(s)] ds≤ G(t, s) [f (s, v(s)) + λv(s)] ds = [A v](t), t ∈ I, 0
0
therefore A is nonincreasing. Besides, for u ≥ v, d(A u, A v) ≤
μ λ
d(u, v), and a lower (re-
spectively, upper) solution to (4) is such that α(t) ≤ [A (α)](t), t ∈ I (respectively, β(t) ≥ [A β](t), t ∈ I). The conclusion follows from applying Theorem 4, where uniqueness comes from the validity of condition (3). Condition (8) can be expressed as −(μ + λ)(y − x) ≤ f (t, y) − f (t, x) ≤ −λ(y − x), y ≥ x, that is, for every t, x fixed, the graph of the function of variable h (h ≥ 0), f (t, x + h) − f (t, x) stays in the region delimited by the lines through the origin with slopes −(μ + λ) and −λ, as shown in Figure 1, for different values of λ and μ.
Figure 1
μ = 1.7 < 2 = λ, μ = 0.8 < 1 = λ
Similarly, we can deal with the case where the graph of the above-mentioned function of variable h is inside a region of the type indicated in Figure 2.
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Figure 2
4
Other possible situations
Lack of Continuity and Monotonicity
As we affirmed in [2], to obtain the uniqueness of the fixed point for some function, it is not necessary for the function to be continuous. In this section, we show that the monotonicity of f is not essential for the existence of a unique fixed point in this context and we replace this condition by the preservation of comparable elements, that is, transformation of comparable elements into comparable elements, which is trivially verified if X (or f (X)) is totally ordered. Theorem 7 Let (X, ≤) be a partially ordered set and suppose that (3) holds and that there exists a metric d in X such that (X, d) is a complete metric space. Let f : X → X be such that f maps comparable elements into comparable elements, that is, ⎧ ⎪ ⎨ f (x) ≤ f (y) x, y ∈ X, x ≤ y =⇒ or ⎪ ⎩ f (x) ≥ f (y) and such that there exists k ∈ [0, 1) with d(f (x), f (y)) ≤ k d(x, y), ∀ x ≥ y. Suppose that either f is continuous or X is such that condition (6) holds. If there exists x0 ∈ X with x0 comparable to f (x0 ), then f has a unique fixed point x ¯. Moreover, ∀ x ∈ X, limn→+∞ f n (x) = x ¯. Proof Since x0 ∈ X is comparable to f (x0 ), then f n+1 (x0 ) and f n (x0 ) are comparable, for every n = 0, 1, 2, . . ., so that the argument exposed in the proof of Theorem 4 is valid. For a totally ordered complete metric space (for instance, in the real case), this theorem is the contractive mapping theorem. Next, we present some examples. Example 1 Take the complete metric space X = [0, +∞) with the distance d(x, y) = |x − y|, x, y ∈ X, and consider the usual total order relation. Let f be defined as ⎧ ⎨ x, x ∈ [n, n + 1), n even or n = 0, f (x) = ⎩ 2n + 1 − x, x ∈ [n, n + 1), n odd, which is not continuous at the positive integer numbers, since, for n even, f (n− ) = n − 1 = n = f (n+ ) and, for n odd, f (n− ) = n = n + 1 = f (n+ ). Function f maps comparable elements into comparable elements but there exists an infinite number of fixed points since the condition of contractivity over comparable elements is not valid: if x ≥ y, then x ∈ [n, n + 1),
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y ∈ [m, m + 1), for integer numbers n ≥ m ≥ 0 and, for instance, for even numbers n, m, we obtain that d(f (x), f (y)) = |x − y|. Example 2 Take the same complete metric space (X, d) of the previous example and consider the order relation ⎧ ⎪ x, y ∈ [0, 1], x ≤ y ⎪ ⎪ ⎨ x ≤ y ⇐⇒ or ⎪ ⎪ ⎪ ⎩ x, y ∈ (n, n + 1], for some n = 1, 2, . . . , x ≤ y. Let f be defined as
⎧ 1 ⎪ ⎪ x, x ∈ [0, 1], ⎪ ⎪ ⎪ ⎨ 2 1 f (x) = (n − 1) + (x − n), x ∈ (n, n + 1], n even, ⎪ 2 ⎪ ⎪ ⎪ 1 ⎪ ⎩ n − (x − n), x ∈ (n, n + 1], n odd, 2 which is not continuous at any positive integer number since, for an even number n, 3 f (n− ) = n − = n − 1 = f (n+ ) 2 and, for an odd number n, f (1− ) = 12 = 1 = f (1+ ), f (n− ) = n − 32 = n = f (n+ ). Function f maps comparable elements into comparable elements and the contractivity condition over this type of elements is true: if x ≥ y, then x, y ∈ [0, 1] or x, y ∈ (n, n + 1], for some n = 1, 2, . . . . If x, y ∈ [0, 1],
1 1 1 d(f (x), f (y)) = d x, y = d(x, y), 2 2 2 if x, y ∈ (n, n + 1] with n an even number,
1 1 1 d(f (x), f (y)) = d (n − 1) + (x − n), (n − 1) + (y − n) = d(x, y) 2 2 2 and, for x, y ∈ (n, n + 1] with n an odd number,
1 1 1 d(f (x), f (y)) = d n − (x − n), n − (y − n) = d(x, y). 2 2 2
1 }m is a sequence with comparable Condition (6) is not valid since, for n ∈ N fixed, {n + m consecutive terms and convergent towards n in X but no term in the sequence is comparable to the limit n. Also (3) fails, since given two positive integer numbers, there is neither an upper bound nor a lower bound. Nevertheless, the fixed point x ¯ = 0 is unique since, for the restriction f|[0,1] , conditions (3) and (6) hold in the interval [0, 1] and, for x > 1, f (x) is not comparable to x, since f ((n, n + 1]) ⊆ (n − 1, n], ∀ n ∈ N. The new results are more general than the fixed point theorem presented in [1], since we have proved that it is possible to remove the hypotheses of continuity and monotonicity of function f , providing alternative conditions, some of them relative to the space X.
Acknowledgment ments.
The authors thank the referee for useful remarks and interesting com-
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