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On nonlocal three-point boundary value problems of Duffing equation with mixed nonlinear forcing terms

Ahmed Alsaedi* and Mohammed HA Aqlan

Author Affiliations

Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box. 80203, Jeddah 21589, Saudi Arabia

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Boundary Value Problems 2011, 2011:47  doi:10.1186/1687-2770-2011-47

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2011/1/47


Received:15 June 2011
Accepted:25 November 2011
Published:25 November 2011

© 2011 Alsaedi and Aqlan; 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.

Abstract

In this paper, we investigate the existence and approximation of the solutions of a nonlinear nonlocal three-point boundary value problem involving the forced Duffing equation with mixed nonlinearities. Our main tool of the study is the generalized quasilinearization method due to Lakshmikantham. Some illustrative examples are also presented.

Mathematics Subject Classification (2000): 34B10, 34B15.

Keywords:
Duffing equation; nonlocal boundary value problem; quasilinearization; quadratic convergence

1 Introduction

The Duffing equation plays an important role in the study of mechanical systems. There are multiple forms of the Duffing equation, ranging from dampening to forcing terms. This equation possesses the qualities of a simple harmonic oscillator, a nonlinear oscillator, and has indeed an ability to exhibit chaotic behavior. Chaos can be defined as disorder and confusion. In physics, chaos is defined as behavior so unpredictable as to appear random, allowing great sensitivity to small initial conditions. The chaotic behavior can emerge in a system as simple as the logistic map. In that case, the "route to chaos" is called period-doubling. In practice, one would like to understand the route to chaos in systems described by partial differential equations such as flow in a randomly stirred fluid. This is, however, very complicated and difficult to treat either analytically or numerically. The Duffing equation is found to be an appropriate candidate for describing chaos in dynamic systems. The advantage of a pseudo-chaotic equation like the Duffing equation is that it allows control of the amount of chaos it exhibits. Chaotic oscillators are important tools for creating and testing models that are more realistic. This is why the Duffing equation is of great interest. The use of the Duffing equation aids in the dynamic behavior of chaos and bifurcation, which studies how small changes in a function can cause a sudden change in behavior [1]. Another important application of the Duffing equation is in the field of the prediction of diseases. A careful measurement and analysis of a strongly chaotic voice has the potential to serve as an early warning system for more serious chaos and possible onset of disease. This chaos is with the help of the Duffing equation. In fact, the success at analyzing and predicting the onset of chaos in speech and its simulation by equations such as the Duffing equation has enhanced the hope that we might be able to predict the onset of arrhythmia and heart attacks someday [2].

The Duffing equation is a mathematical representation of the oscillator. Both the equation and oscillator are prone to many output waveforms. One of the simplest waveforms includes simple harmonic motion like a pendulum. Other waveforms are considerably more complex and can quickly be described as shear oscillatory chaos. The Duffing equation can be a forced or unforced damped chaotic harmonic oscillator. Exact solutions of second-order nonlinear differential equations like the forced Duffing equation are rarely possible due to the possible chaotic output. There do exist a number of powerful procedures for obtaining approximate solutions of nonlinear problems such as Galerkin's method, expansion methods, dynamic programming, iterative techniques, the method of upper and lower bounds, and Chapligin method to name a few. The monotone iterative technique coupled with the method of upper and lower solutions [3] manifests itself as an effective and flexible mechanism that offers theoretical as well as constructive existence results in a closed set, generated by the lower and upper solutions. In general, the convergence of the sequence of approximate solutions given by the monotone iterative technique is at most linear. To obtain a sequence of approximate solutions converging quadratically, we use the method of quasilinearization. The origin of the quasilinearization lies in the theory of dynamic programming [4,5]. Agarwal [6] discussed quasilinearization and approximate quasilinearization for multipoint boundary value problems. In fact, the quasilinearization technique is a variant of Newton's method. This method applies to semilinear equations with convex (concave) nonlinearities and generates a monotone scheme whose iterates converge quadratically to a solution of the problem at hand. The nineties brought new dimensions to this technique when Lakshmikantham [7,8] generalized the method of quasilinearization by relaxing the convexity assumption. This development was so significant that it attracted the attention of many researchers, and the method was extensively developed and applied to a wide range of initial and boundary value problems for different types of differential equations. A detailed description of the quasilinearization method and its applications can be found in the monograph [9] and the papers [10-26] and the references therein.

In this paper, we study a nonlinear nonlocal three-point boundary value problem of the forced Duffing equation with mixed nonlinearities given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M1">View MathML</a>

(1.1)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M2">View MathML</a>

(1.2)

where N(t, x) ∈ C[J × ℝ, ℝ] is such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M3">View MathML</a>

(1.3)

and gi: ℝ → ℝ (i = 1,2) are given continuous functions. The details of such a decomposition can be found in Section 1.5 of the text [9]. In (1.3), it is assumed that f(t,x) is nonconvex, k(t,x) is nonconcave, and H(t,x) is a Lipschitz function:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M4">View MathML</a>

A quasilinearization technique due to Lakshmikantham [9] is applied to obtain an analytic approximation of the solution of the problem (1.1-1.2). In fact, we obtain sequences of upper and lower solutions converging monotonically and quadratically to a unique solution of the problem at hand. It is worth mentioning that the forced Duffing equation with mixed nonlinearities has not been studied so far.

2 Preliminaries

As argued in [12], the solution x(t) of the problem (1.1-1.2) can be written in terms of the Green's function as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M5">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M6">View MathML</a>

Observe that G(t,s) < 0 on [0,1] × [0,1].

Definition 2.1. We say that α C2[J, ℝ] is a lower solution of the problem (1.1-1.2) if

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M7">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M8">View MathML</a>

and β C2[J, ℝ] will be an upper solution of the problem (1.1-1.2) if the inequalities are reversed in the definition of lower solution.

Now we state some basic results that play a pivotal role in the proof of the main result. We do not provide the proof as the method of proof is similar to the one described in the text [9].

Theorem 2.1. Let α and β be lower and upper solutions of (1.1-1.2), respectively. Assume that

(i) fx(t,x) + kx(t,x) - L > 0 for every (t,x) ∈ J × ℝ.

(ii) g1 and g2 are continuous on ℝ satisfying the one-sided Lipschitz condition:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M9">View MathML</a>

Then α(t) ≤ β(t), t J.

Theorem 2.2. Let α and β be lower and upper solutions of (1.1-1.2), respectively, such that α(t) ≤ β(t), t J. Then, there exists a solution x(t) of (1.1-1.2) such that α(t) ≤ x(t) ≤ β(t), t J.

3 Main result

Theorem 3.1. Assume that

(A1) α0, β0 C2[J, ℝ] are lower and upper solutions of (1.1-1.2), respectively.

(A2) N C[J × ℝ, ℝ] be such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M10">View MathML</a>

where fx(t, x), kx(t, x), fxx(t, x), kxx(t, x) exist and are continuous, and for continuous functions ϕ, χ,(fxx(t, x) + ϕxx(t, x)) ≥ 0, (kxx(t, x) + χxx(t, x)) ≤ 0 with ϕxx ≥ 0, χxx ≤ 0 for every (t, x) ∈ S, where S = {(t, x) ∈ J × ℝ: α0(t) ≤ x(t) ≤ β0(t)}. H(t, x) satisfies the one-sided Lipschitz condition:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M11">View MathML</a>

where L > 0 is a Lipschitz constant and fx(t, x) + kx(t, x) - L > 0 for every (t, x) ∈ S.

(A3) For <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M82">View MathML</a> are continuous on ℝ satisfying <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M83">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M84">View MathML</a> with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M85">View MathML</a> on ℝ for some continuous functions ψi(x).

Then, there exist monotone sequences {αn} and {βn} that converge in the space of continuous functions on J quadratically to a unique solution x(t) of the problem (1.1-1.2).

Proof. Let us define F: J × ℝ → ℝ by F(t, x) = f(t, x) + ϕ(t, x), K: J × ℝ → ℝ by K(t, x) = k(t, x) + χ(t, x), Gi: ℝ → ℝ by Gi(x) = gi(x) + ψi(x), i = 1, 2. By the assumption (A2) and the generalized mean value theorem, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M12">View MathML</a>

(3.1)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M13">View MathML</a>

(3.2)

Interchanging x and y, (3.1) and (3.2) take the form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M14">View MathML</a>

(3.3)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M15">View MathML</a>

(3.4)

By the assumption (A3), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M16">View MathML</a>

(3.5)

which, on interchanging x and y yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M17">View MathML</a>

(3.6)

We set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M18">View MathML</a>

and for i = 1,2,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M19">View MathML</a>

Observe that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M20">View MathML</a>

(3.7)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M21">View MathML</a>

(3.8)

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M22">View MathML</a>

(3.9)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M23">View MathML</a>

(3.10)

Now, we consider the problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M24">View MathML</a>

(3.11)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M25">View MathML</a>

(3.12)

Using (A1), (3.7) and (3.8), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M26">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M27">View MathML</a>

which imply that α0 and β0 are, respectively, lower and upper solutions of (3.11-3.12). Thus, by Theorems 2.1 and 2.2, there exists a solution α1 for the problem (3.11-3.12) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M28">View MathML</a>

(3.13)

Next, consider the problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M29">View MathML</a>

(3.14)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M30">View MathML</a>

(3.15)

Using (A1), (3.9) and (3.10), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M31">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M32">View MathML</a>

which imply that α0 and β0 are, respectively, lower and upper solutions of (3.14-3.15). Again, by Theorems 2.1 and 2.2, there exists a solution β1 of (3.14-3.15) satisfying

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M33">View MathML</a>

(3.16)

Now we show that α1(t) ≤ β1(t). For that, we prove that α1(t) is a lower solution and β1(t) is an upper solution of (1.1-1.2). Using the fact that α1(t) is a solution of (3.11-3.12) satisfying α0(t) ≤ α1(t) ≤ β0(t) and (3.7-3.8), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M34">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M35">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M36">View MathML</a>

By the above inequalities, it follows that α1 is a lower solution of (1.1-1.2).

In view of the fact that β1(t) is a solution of (3.14-3.15) together with (3.9), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M37">View MathML</a>

and by virtue of (3.10), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M38">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M39">View MathML</a>

Thus, β1 is an upper solution of (1.1-1.2). Hence, by Theorem 2.1, it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M40">View MathML</a>

(3.17)

Combining (3.13, 3.16) and (3.17) yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M41">View MathML</a>

Now, by induction, we prove that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M42">View MathML</a>

For that, we consider the boundary value problems

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M43">View MathML</a>

(3.18)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M44">View MathML</a>

(3.19)

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M45">View MathML</a>

(3.20)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M46">View MathML</a>

(3.21)

Assume that for some n > 1, α0(t) ≤ αn(t) ≤ βn(t) ≤ β0(t) and we will show that αn+1(t) ≤ βn+1(t).

Using (3.7), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M47">View MathML</a>

By (3.8), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M48">View MathML</a>

which yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M49">View MathML</a>

Thus, αn is a lower solution of (3.18-3.19). In a similar manner, we find that βn is an upper solution of (3.18-3.19). Thus, by Theorems 2.1 and 2.2, there exists a solution αn+1(t) of (3.18-3.19) such that αn(t) ≤ αn+1(t) ≤ βn(t), t J. Similarly, it can be proved that αn(t) ≤ βn+1(t) ≤ βn(t), t J, where βn+1(t) is a solution of (3.20-3.21) and αn(t), βn(t) are lower and upper solutions of (3.20-3.21), respectively. Next, we show that αn+1(t) ≤ βn+1(t).

For that, we have to show that αn+1(t) and βn+1(t) are lower and upper solutions of (1.1-1.2), respectively. Using (3.7, 3.8) together with the fact that αn+1(t) is a solution of (3.18-3.19), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M50">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M51">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M52">View MathML</a>

which implies that αn+1 is a lower solution of (1.1-1.2). Employing a similar procedure, it can be proved that βn+1 is an upper solution of (1.1-1.2). Hence, by Theorem 2.1, it follows that αn+1(t) ≤ βn+1(t). Therefore, by induction, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M53">View MathML</a>

Since [0,1] is compact and the monotone convergence is pointwise, it follows that {αn} and {βn} are uniformly convergent with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M54">View MathML</a>

such that α0(t) ≤ x(t) ≤ y(t) ≤ β0(t), where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M55">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M56">View MathML</a>

By the uniqueness of the solution (which follows by the hypotheses of Theorem 2.1), we conclude that x(t) = y(t). This proves that the problem (1.1-1.2) has a unique solution x(t) given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M57">View MathML</a>

In order to prove that each of the sequences {αn}, {βn} converges quadratically, we set zn(t) = βn(t) - x(t) and rn(t) = x(t) - αn(t), and note that zn ≥ 0, rn ≥ 0. We will only prove the quadratic convergence of the sequence {rn} as that of {zn} is similar. By the mean value theorem, we find that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M59">View MathML</a>

where αn ζ5, ζ8 βn, αn ζ6, ζ7 x, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M61">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M62">View MathML</a>.

Now we define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M63">View MathML</a>

and obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M64">View MathML</a>

where αn γ1, δ1, ρ1, σ1 x, αn γ2 x, and αn δ2, ρ2, σ2 αn+1. Letting <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M66">View MathML</a> and M0 as an upper bound on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M67">View MathML</a>, we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M68">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M69">View MathML</a>, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M70">View MathML</a>. This completes the proof.

4 Examples

Example 4.1. Consider the problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M71">View MathML</a>

(4.1)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M72">View MathML</a>

(4.2)

Here f(t, x) = 2x - tcos(πx/2), k(t, x) ≡ 0, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M73">View MathML</a>. Let α0 = 0 and β0 = 1 be lower and upper solutions of (4.1-4.2), respectively. We note that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M74">View MathML</a>. Further, we choose <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M75">View MathML</a>. We note that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M76">View MathML</a>. Thus, all the conditions of Theorem (3.1) are satisfied. Hence, the conclusion of Theorem 3.1 applies to the problem (4.1-4.2).

Example 4.2. Consider the nonlinear boundary value problem given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M77">View MathML</a>

(4.3)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M78">View MathML</a>

(4.4)

where f(t, x) = 7x + sin(πxt/2), k(t, x) = -tcos(πx/2), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M79">View MathML</a>. Let α0 = 0 and β0 = 1 be lower and upper solutions of (4.1-4.2), respectively. Observe that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M80">View MathML</a>

and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M86">View MathML</a>. For positive constants M1, M2, N1, N2, we choose

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M81">View MathML</a>

such that fxx(t, x) + ϕxx(t, x) = π2t2[2M1 - cos(πtx/2)]/4 ≥ 0, kxx + χxx = -π2[8M2 - tcos(πx/2)]/4 ≤ 0. Clearly, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/47/mathml/M87">View MathML</a>. Thus, all the conditions of Theorem 3.1 are satisfied. Hence, the conclusion of theorem (3.1) applies to the problem (4.3-4.4).

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

Both authors, AA and MHA, contributed to each part of this work equally and read and approved the final version of the manuscript.

Acknowledgements

The authors thank the referees for their useful comments. This research was partially supported by Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia.

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