Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 RESEARCH Open Access Contraction mapping principle with generalized altering distance function in ordered metric spaces and applications to ordinary differential equations Yongfu Su* * Correspondence: [email protected] Department of Mathematics, Tianjin Polytechnic University, Tianjin, 300387, P.R. China Abstract The aim of this paper is to present the definition of a generalized altering distance function and to extend the results of Yan et al. (Fixed Point Theory Appl. 2012:152, 2012) and some others, and to prove a new fixed point theorem of generalized contraction mappings in a complete metric space endowed with a partial order by using generalized altering distance functions. The results of this paper can be used to investigate a large class of nonlinear problems. As an application, we discuss the existence of a solution for a periodic boundary value problem. Keywords: contraction mapping principle; partially ordered metric spaces; fixed point; generalized altering distance function; ordinary differential equations 1 Introduction The Banach contraction mapping principle is a classical and powerful tool in nonlinear analysis. Weak contractions are generalizations of the Banach contraction mapping, which have been studied by several authors. In [–], the authors prove some types of weak contractions in complete metric spaces, respectively. In particular, the existence of a fixed point for weak contractions and generalized contractions was extended to partially ordered metric spaces in [, –]. Among them, some involve altering distance functions. Such functions were introduced by Khan et al. in [], where they present some fixed point theorems with the help of such functions. First, we recall the definition of an altering distance function. Definition . An altering distance function is a function ψ : [, ∞) → [, ∞) which satisfies: (a) ψ is continuous and non-decreasing. (b) ψ = if and only if t = . Recently, Harjani and Sadarangani proved some fixed point theorems for weak contractions and generalized contractions in partially ordered metric spaces by using the altering distance function in [, ], respectively. Their results improve the theorems of [, ]. © 2014 Su; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 2 of 15 Theorem . ([]) Let (X, ≤) be a partially ordered set and suppose that there exists a metric d ∈ X such that (X, d) is a complete metric space. Let f : X → X be a continuous and non-decreasing mapping such that d f (x), f (y) ≤ d(x, y) – ψ d(x, y) , for x ≥ y, where ψ : [, ∞) → [, ∞) is continuous and non-decreasing function such that ψ is positive in (, ∞), ψ() = and limt→∞ ψ(t) = ∞. If there exists x ∈ X with x ≤ f (x ), then f has a fixed point. Theorem . ([]) Let (X, ≤) be a partially ordered set and suppose that there exists a metric d ∈ X such that (X, d) is a complete metric space. Let f : X → X be a continuous and non-decreasing mapping such that ψd f (x), f (y) ≤ ψ d(x, y) – φ d(x, y) , for x ≥ y, where ψ and φ are altering distance functions. If there exists x ∈ X with x ≤ f (x ), then f has a fixed point. Subsequently, Amini-Harandi and Emami proved another fixed point theorem for contraction type maps in partially ordered metric spaces in []. The following class of functions is used in []. Let denote the class of those functions β : [, ∞) → [, ) which satisfy the condition: β(tn ) → ⇒ tn → . Theorem . ([]) Let (X, ≤) be a partially ordered set and suppose that there exists a metric d such that (X, d) is a complete metric space. Let f : X → X be an increasing mapping such that there exists an element x ∈ X with x ≤ f (x ). Suppose that there exists β ∈ such that d f (x), f (y) ≤ β d(x, y) d(x, y) for each x, y ∈ X with x ≥ y. Assume that either f is continuous or M is such that if an increasing sequence xn → x ∈ X, then xn ≤ x, ∀n. Besides, if for each x, y ∈ X there exists z ∈ m which is comparable to x and y, then f has a unique fixed point. In , Yan et al. proved the following result. Theorem . ([]) 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 T : X → X be a continuous and non-decreasing mapping such that ψ d(Tx, Ty) ≤ φ d(x, y) , ∀x ≥ y, where ψ is an altering distance function and φ : [, ∞) → [, ∞) is a continuous function with the condition ψ(t) > φ(t) for all t > . If there exists x ∈ X such that x ≤ Tx , then T has a fixed point. Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 3 of 15 The aim of this paper is to present the definition of generalized altering distance function and to extend the results of Yan et al. [] and some others, and to prove a new fixed point theorem of generalized contraction mappings in a complete metric space endowed with a partial order by using generalized altering distance functions. The results of this paper can be used to investigate a large class of nonlinear problems. As an application, we discuss the existence of a solution for a periodic boundary value problem. 2 Main results We first give the definition of generalized altering distance function as follows. Definition . A generalized altering distance function is a function ψ : [, ∞) → [, ∞) which satisfies: (a) ψ is non-decreasing; (b) ψ = if and only if t = . We first recall the following notion of a monotone non-decreasing function in a partially ordered set. Definition . If (X, ≤) is a partially ordered set and T : X → X, we say that T is monotone non-decreasing if x, y ∈ X, x ≤ y ⇒ T(x) ≤ T(y). This definition coincides with the notion of a non-decreasing function in the case where X = R and ≤ represents the usual total order in R. In what follows, we prove the following theorem, which is the generalized type of Theorems .-.. Theorem . 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 T : X → X be a continuous and nondecreasing mapping such that ψ d(Tx, Ty) ≤ φ d(x, y) , ∀x ≥ y, where ψ is a generalized altering distance function and φ : [, ∞) → [, ∞) is a right upper semi-continuous function with the condition: ψ(t) > φ(t) for all t > . If there exists x ∈ X such that x ≤ Tx , then T has a fixed point. Proof Since T is a non-decreasing function, we obtain by induction that x ≤ Tx ≤ T x ≤ T x ≤ · · · ≤ T n x ≤ T n+ x ≤ · · · . () Put xn+ = Txn . Then, for each integer n ≥ , from () and, as the elements xn+ and xn are comparable, we get ψ d(xn+ , xn ) = ψ d(Txn , Txn– ) ≤ φ d(xn , xn– ) . () Using the condition of Theorem . we have d(xn+ , xn ) < d(xn , xn– ). () Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 4 of 15 Hence the sequence {d(xn+ , xn )} is decreasing and, consequently, there exists r ≥ such that d(xn+ , xn ) → r+ , as n → ∞. Consider the properties of ψ and φ, letting n → ∞ in () we get ψ(r) ≤ lim ψ d(xn+ , xn ) ≤ lim φ d(xn , xn– ) ≤ φ(r). n→∞ n→∞ By using the condition: ψ(t) > φ(t) for all t > , we have r = , and hence d(xn+ , xn ) → , () as n → ∞. In what follows, we will show that {xn } is a Cauchy sequence. Suppose that {xn } is not a Cauchy sequence. Then there exists ε > for which we can find subsequences {xnk } with nk > mk > k such that d(xnk , xmk ) ≥ ε () for all k ≥ . Further, corresponding to mk we can choose nk in such a way that it is the smallest integer with nk > mk and satisfying (). Then d(xnk– , xmk– ) < ε. () From () and (), we have ε ≤ d(xnk , xmk ) ≤ d(xnk ), xnk– + d(xnk– , xmk ) < d(xnk , xnk– ) + ε. Letting k → ∞ and using (), we get lim d(xnk , xmk ) = ε. k→∞ () By using the triangular inequality we have d(xnk , xmk ) ≤ d(xnk , xnk– ) + d(xnk– , xmk– ) + d(xmk– , xmk ), d(xnk– , xmk– ) ≤ d(xnk– , xnk ) + d(xnk , xmk ) + d(xmk , xmk– ). Letting k → ∞ in the above two inequalities and using () and (), we have lim d(xnk– , xmk– ) = ε. k→∞ As nk > mk and xnk– and xmk– are comparable, using () we have ψ d(xnk , xmk ) ≤ φ d(xnk– , xmk– ) . () Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 5 of 15 Consider the properties of ψ and φ, letting k → ∞ and taking into account () and (), we have ψ(ε) ≤ φ(ε). From the condition ψ(t) > φ(t) for all t > , we get ε = , which is a contradiction. This shows that {xn } is a Cauchy sequence and, since X is a complete metric space, there exists z ∈ X such that xn → z as n → ∞. Moreover, the continuity of T implies that z = lim xn+ = lim Txn = Tz, n→∞ n→∞ and this proves that z is a fixed point. This completes the proof. In what follows, we prove that Theorem . is still valid for T being not necessarily continuous, assuming the following hypothesis in X: If (xn ) is a non-decreasing sequence in X such that xn → x then xn ≤ x for all n ∈ N. () Theorem . 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. Assume that X satisfies (). Let T : X → X be a non-decreasing mapping such that ψ d(Tx, Ty) ≤ φ d(x, y) , ∀x ≥ y, where ψ is a generalized altering distance functions and φ: [, ∞) → [, ∞) is a right upper semi-continuous function with the condition ψ(t) > φ(t) for all t > . If there exists x ∈ X such that x ≤ Tx , then T has a fixed point. Proof Following the proof of Theorem . we only have to check that T(z) = z. As (xn ) is a non-decreasing sequence in X and limn→∞ xn = z the condition () gives us that xn ≤ z for every n ∈ N and consequently, ψ d xn+ , T(z) = ψ d T(xn ), T(z) ≤ φ d(xn , z) . Letting n → ∞ and taking into account that ψ is an altering distance function, we have ψ d z, T(z) ≤ φ(). Using condition of theorem we have φ() = , this implies (d(z, T(z))) = . Thus, d(z, T(z)) = or equivalently, T(z) = z. Now, we present an example where it can be appreciated that the hypotheses in Theorems . and Theorems . do not guarantee uniqueness of the fixed point. An example appears in []. Let X = {(, ), (, )} ⊂ R and consider the usual order (x, y) ≤ (z, t) ⇔ x ≤ z, y ≤ t. Thus, (x, y) is a partially ordered set whose different elements are not comparable. Besides Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 6 of 15 (X, d ) is a complete metric space considering d the Euclidean distance. The identity map T(x, y) = (x, y) is trivially continuous and non-decreasing and condition () of Theorem . is satisfied, since the elements in X are only comparable to themselves. Moreover, (, ) ≤ T(, ) = (, ) and T has two fixed points in X. In what follows, we give a sufficient condition for the uniqueness of the point in Theorems . and .. This condition is: for x, y ∈ X there exists a lower bound or an upper bound. () In [] it is proved that condition () is equivalent to: for x, y ∈ X there exists z ∈ X which is comparable to x and y. () Theorem . Adding condition () to the hypotheses of Theorem . (resp. Theorem .) we obtain the uniqueness of the fixed point of T. Proof Suppose that there exist z, y ∈ X which are fixed points. We distinguish two cases. Case . If y is comparable to z then T n (y) = y is comparable to T n (z) = z for n = , , , . . . and ψ d(z, y) = ψ d T n (z), T n (y) ≤ φ d T n– (z), T n– (y) ≤ φ d(z, y) . As we have the condition ψ(t) > φ(t) for t > we obtain d(z, y) = and this implies z = y. Case . If y is not comparable to z then there exists x ∈ X comparable to y and z. Monotonicity of T implies that T n (x) is comparable to T n (y) and to T n (z) = z, for n = , , , . . . Moreover, ψ d z, T n (x) = ψ d T n (z), T n (x) ≤ φ d T n– (z), T n– (x) = φ d z, T n– (x) . () Hence, ψ is a generalized altering distance function and we have the condition ψ(t) > φ(t) for t > , this gives us that {d(z, f n (x))} is a non-negative decreasing sequence and, consequently, there exists γ such that lim d z, T n (x) = γ . n→∞ Letting n → ∞ in () and, taking into account the properties of ψ and φ, we obtain ψ(γ ) ≤ φ(γ ). This and the condition ψ(t) > φ(t) for t > imply γ = . Analogously, it can be proved that lim d y, T n (x) = . n→∞ Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 7 of 15 Finally, as lim d z, T n (x) = lim d y, T n (x) = n→∞ n→∞ the uniqueness of the limit gives us y = z. This finishes the proof. Remark . Under the assumption of Theorem ., it can be proved that for every x ∈ X, limn→∞ T n (x) = z, where z is the fixed point (i.e. the operator f is Picard). Remark . Theorem . is a particular case of Theorem . for ψ being the identity function, and φ(t) = t – ψ(t). Theorem . is a particular case of our Theorem . for φ(t) being replaced by ψ(t) – φ(t). Theorem . is a particular case of Theorem . for ψ being the identity function, and φ(t) = β(t)t. Theorem . is also a particular case of Theorem . for ψ and φ being continuous. Example . The following are some generalized altering distance functions: ⎧ ⎨, t = , ⎩[t] + , t > , ⎧ ⎨, t = , ψ (t) = ⎩λ([t] + ), t > , ψ (t) = where α > is a constant. ψ (t) = ⎧ ⎨t, ≤ t < , ⎩αt , t ≥ , where α ≥ is a constant. We choose ψ(t) = ψ (t) and ⎧ ⎨t , ≤ t < , φ(t) = ⎩βt, t ≥ , where < β < α is a constant. By using Theorem ., we can get the following result. Theorem . 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 T : X → X be a continuous and nondecreasing mapping such that ≤ d(Tx, Ty) < d(Tx, Ty) ≥ ⇒ d(Tx, Ty) ≤ d(x, y) , α d(Tx, Ty) ≤ βd(x, y) ⇒ for any x, y ∈ X. If there exists x ∈ X such that x ≤ Tx , then T has a fixed point. Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 8 of 15 3 Application to ordinary differential equations In this section we present two examples where our Theorems . and . can be applied. The first example is inspired by []. We study the existence of a solution for the following first-order periodic problem: ⎧ ⎨u (t) = f (t, u(t)), t ∈ [, T], ⎩u() = u(T), () where T > and f : I × R → R is a continuous function. Previously, we considered the space C(I) (I = [, T]) of continuous functions defined on I. Obviously, this space with the metric given by d(x, y) = sup x(t) – y(t) : t ∈ I , for x, y ∈ C(I), is a complete metric space. C(I) can also be equipped with a partial order given by x, y ∈ C(I), x≤y ⇔ x(t) ≤ y(t)m, for t ∈ I. Clearly, (C(I), ≤) satisfies condition (), since for x, y ∈ C(I) the functions max{x, y} and min{x, y} are least upper and greatest lower bounds of x and y, respectively. Moreover, in [] it is proved that (C(I), ≤) with the above mentioned metric satisfies condition (). Now we give the following definition. Definition . A lower solution for () is a function α ∈ C () (I) such that ⎧ ⎨α (t) ≤ f (t, α(t)), for t ∈ I, ⎩α() ≤ α(T). Theorem . Consider problem () with f : I × R → R continuous and suppose that there exist λ, α > with α≤ λ(eλT – ) T(eλT + ) such that for x, y ∈ R with x ≥ y ≤ f (t, x) + λx – f (t, y) + λy ≤ α g(x – y), where g(t) : [, +∞) → [, +∞) is a light upper semi-continuous function with g() = , g(t) < t , ∀t > . Then the existence of a lower solution for () provides the existence of an unique solution of (). Proof Problem () can be written as ⎧ ⎨u (t) + λu(t) = f (t, u(t)) + λu(t), for t ∈ I = [, T], ⎩u() = u(T). Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 9 of 15 This problem is equivalent to the integral equation T u(t) = G(t, s) f s, u(s) + λu(s) ds, where G(t, s) is the Green function given by ⎧ ⎨ eλ(T+s–t) , ≤ s < t ≤ T, eλT – G(t, s) = λ(s–t) ⎩e , ≤ t < s ≤ T. eλT – Define F : C(I) → C(I) by T (Fu)(t) = G(t, s) f s, u(s) + λu(s) ds. Note that if u ∈ C(I) is a fixed point of F then u ∈ C (I) is a solution of (). In what follows, we check that the hypotheses in Theorems . and . are satisfied. The mapping F is non-decreasing, since we have u ≥ v, and using our assumption. We can obtain f (t, u) + λu ≥ f (t, v) + λv which implies, since G(t, s) > , that for t ∈ I T G(t, s) f s, u(s) + λu(s) ds T G(t, s) f s, v(s) + λv(s) ds = (Fv)(t). (Fu)(t) = ≥ Besides, for u ≥ v, we have d(Fu, Fv) = sup(Fu)(t) – (Fv)(t) t∈I = sup (Fu)(t) – (Fv)(t) t∈I T = sup t∈I T ≤ sup t∈I G(t, s) f s, u(s) + λu(s) – f s, v(s) – λv(s) ds G(t, s)α g u(s) – v(s) ds. () Using the Cauchy-Schwarz inequality in the last integral we get T G(t, s)α g u(s) – v(s) ds ≤ T α g u(s) – v(s) ds G(t, s) ds T . () Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 10 of 15 The first integral gives us T t G(t, s) ds = T G(t, s) ds + G(t, s) ds t t = eλ(T+s–t) ds + (eλT – ) = e(λT–) λ(eλT – ) = eλT + . λ(eλT – ) t T eλ(s–t) ds (eλT – ) () The second integral in () gives the following estimate: T α g u(s) – v(s) ds ≤ α g u – v · T = α g d(u, v) · T. () Taking into account ()-() we have λT e + · α g d(u, v) · T d(Fu, Fv) ≤ sup λT – ) λ(e t∈I λT √ e + = · α · T · g d(u, v) λT λ(e – ) and from the last inequality we obtain d(Fu, Fv) ≤ eλT + · α · T · g d(u, v) λT λ(e – ) or, equivalently. λ eλT – d(Fu, Fv) ≤ eλT + · α · T · g d(u, v) . By our assumption, as α≤ λ(eλT – ) T(eλT + ) , the last inequality gives us λ eλT – d(Fu, Fv) ≤ λ eλT – · g d(u, v) , and, hence, d(Fu, Fv) ≤ g d(u, v) . () Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 11 of 15 Put ψ(t) = t and φ(t) = g(t). Obviously, ψ is a generalized altering distance function, ψ(t) and φ(t) satisfy the condition of ψ(t) > φ(t) for t > . From (), we obtain for u ≥ v ψ d(Fu, Fv) ≤ φ d(u, v) . Finally, let α(t) be a lower solution for (); we claim that α ≤ F(α). In fact α (t) + λα(t) ≤ f t, α(t) + λα(t), for t ∈ I. We multiply by eλt , α(t)eλt ≤ f t, α(t) + λα(t) eλt , for t ∈ I, and this gives us t f s, α(s) + λα(s) eλs ds, α(t)e ≤ α() + λt for t ∈ I. () As α() ≤ α(T), the last inequality gives us α()eλt ≤ α(T)eλT ≤ α() + T f s, α(s) + λα(s) eλs ds, and so T α() ≤ eλs f s, α(s) + λα(s) ds. λT e – This and () give us t α(t)e ≤ λt eλ(T+s) f s, α(s) + λα(s) ds + eλT – T t eλs f s, α(s) + λα(s) ds eλT – and, consequently, T G(t, s) f s, α(s) + λα(s) ds = t eλ(s–t) f s, α(s) + λα(s) ds eλT – eλ(T+s–t) ds + eλT – α(t) ≤ t = (Fα)(t), for t ∈ I. Finally, Theorems . and . show that F has an unique fixed point. Example . In Theorem ., we can choose the function g(t) as follows: () g (t) = ln(t + ); () ⎧ ⎪ ⎪ ⎨t , ≤ t < , g (t) = , t = , ⎪ ⎪ ⎩ t, < t < +∞. Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 12 of 15 () g (t) = ⎧ ⎨t , ⎩t – , ≤ t ≤ , < t < +∞. The functions g (t), g (t) are continuous and non-decreasing. The function g (t) is right upper semi-continuous. If we choose g(t) = g (t) in Theorem ., we obtain the result of []. Example . Consider the following first-order periodic problem: ⎧ ⎨u (t) = sin t et – βx, t ∈ [, T], () ⎩u() = u(T). Let f (t, x) = sin t – βx, et x ∈ [, ∞), t ∈ [, ], then f (t, x) is continuous. Further, for x ≥ y, we have x–y . f (t, x) + λx – f (t, y) + λy = (λ – β) We chose β ∈ [, λ] such that (λ – β) ≤ λ(eλT – ) T(eλT + ) . Taking g(t) = ( t ) for all t ∈ [, +∞), we have f (t, x) + λx – f (t, y) + λy = (λ – β) g(x – y). By using Theorem ., we know that the first-order periodic problem () has a unique solution. A second example where our results can be applied is the following two-point boundary value problem of the second order differential equation: ⎧ ⎨– d x = f (t, x), x ∈ [, ∞), t ∈ [, ], dt ⎩x() = x() = . () It is well known that x ∈ C [, ] is a solution of () that is equivalent to x ∈ C[, ] being a solution of the integral equation x(t) = G(t, s)f s, x(s) ds, for t ∈ [, ], Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 13 of 15 where G(t, s) is the Green function given by ⎧ ⎨t( – s), ≤ t ≤ s ≤ , G(t, s) = ⎩s( – t), ≤ s ≤ t ≤ . () Theorem . Consider problem () with f : I × R → [, ∞) continuous and nondecreasing with respect to the second variable and suppose that there exists ≤ α ≤ such that for x, y ∈ R with x ≥ y f (t, x) – f (t, y) ≤ α g(x – y), () where g(t) : [, +∞) → [, +∞) is a light upper semi-continuous function with g() = , g(t) < t , ∀t > . Then our problem () has a unique non-negative solution. Proof Consider the cone P = x ∈ C[, ] : x(t) ≥ . Obviously, (P, d) with d(x, y) = sup{|x(t) – y(t)| : t ∈ [, ]} is a complete metric space. Consider the operator given by (Tx)(t) = G(t, s)f s, x(s) ds, for x ∈ P, where G(t, s) is the Green function appearing in (). As f is non-decreasing with respect to the second variable, for x, y ∈ P with y ≥ x and t ∈ [, ], we have (Ty)(t) = G(t, s)f s, y(s) ds ≥ G(t, s)f s, x(s) ds ≥ (Tx)(t), and this proves that T is a non-decreasing operator. Besides, for y ≥ x and taking into account (), we obtain d(Ty, Tx) = sup (Tx)(t) – (Ty)(t) t∈[,] = sup (Tx)(t) – (Ty)(t) t∈[,] = sup G(t, s) f s, x(s) – f s, y(s) ds t∈[,] ≤ sup G(t, s)α g x(s) – y(s) G(t, s)α g x(s) – y(s) ds t∈[,] ≤ sup t∈[,] = α ln y – x + sup G(t, s) ds. t∈[,] () Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Page 14 of 15 It is easy to verify that G(t, s) ds = –t t + and that G(t, s) ds = . sup t∈[,] These facts, the inequality (), and the hypothesis < α ≤ give us α g(x – y) ≤ g x – y = g d(x, y) . d(Tx, Ty) ≤ Hence d(Ty, Tx) ≤ g d(x, y) . Put ψ(t) = t , φ(t) = g(t), obviously ψ is an altering distance function, ψ and φ satisfy the condition of ψ(t) > φ(t), for t > . From the last inequality, we have ψ d(Tx, Ty) ≤ φ d(x, y) . Finally, as f and G are non-negative functions T = G(t, s)f (s, ) ds ≥ and Theorems . and . tell us that F has a unique non-negative solution. Remark . In Theorem ., we can choose g(t) as g (), g (t), and g (t) as well as in Theorem .. Example . Consider the following two-point boundary value problem of the second order differential equation: ⎧ ⎨– d x = sin t + x , x ∈ [, ∞), t ∈ [, ], +cos tπ et dt ⎩x() = x() = . () Let f (t, x) = sin t x + , t e + cos tπ x ∈ [, ∞), t ∈ [, ], then f (t, x) is continuous and non-decreasing with respect to the second variable. Further, for x ≥ y, we have y x – ≤ f (t, x) – f (t, y) = + cos tπ + cos tπ x–y . Su Fixed Point Theory and Applications 2014, 2014:227 http://www.fixedpointtheoryandapplications.com/content/2014/1/227 Taking g(t) = t for all t ∈ [, +∞). By using Theorem ., we know that the two-point boundary value problem () has a unique non-negative solution. Competing interests The author declares that they have no competing interests. Received: 29 May 2014 Accepted: 8 October 2014 Published: 3 November 2014 References 1. Khan, MS, Swaleh, M, Sessa, S: Fixed point theorems by altering distances between the points. Bull. Aust. Math. Soc. 30(1), 1-9 (1984) 2. Dhutta, P, Choudhury, B: A generalization of contractions in partially ordered metric spaces. Appl. Anal. 87, 109-116 (2008) 3. Rhoades, BE: Some theorems on weakly contractive maps. Nonlinear Anal. 47, 2683-2693 (2001) 4. Aydi, H, Karapınar, E, Bessem, S: Fixed point theorems in ordered abstract spaces. Fixed Point Theory Appl. 2012, Article ID 76 (2012) 5. Nieto, JJ, Pouso, RL, Rodríguez-López, R: A fixed point theorem for contraction type maps in partially ordered metric spaces and application to ordinary differential equations. Proc. Am. Math. Soc. 135, 2505-2517 (2007) 6. 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