Existence of solutions for a class of nonlinear boundary value problems on half-line

  • Türker Ertem1Email author and

    Affiliated with

    • Ağacık Zafer1

      Affiliated with

      Boundary Value Problems20122012:43

      DOI: 10.1186/1687-2770-2012-43

      Received: 28 January 2012

      Accepted: 16 April 2012

      Published: 16 April 2012

      Abstract

      Consider the infinite interval nonlinear boundary value problem

      ( p ( t ) x ) + q ( t ) x = f ( t , x ) , t t 0 0 , x ( t 0 ) = x 0 , x ( t ) = a v ( t ) + b u ( t ) + o ( r i ( t ) ) , t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equa_HTML.gif

      where u and v are principal and nonprincipal solutions of (p(t)x')' + q(t)x = 0, r1(t) = o(u(t)(v(t)) μ ) and r2(t) = o(v(t)(u(t)) μ ) for some μ ∈ (0, 1), and a and b are arbitrary but fixed real numbers.

      Sufficient conditions are given for the existence of a unique solution of the above problem for i = 1, 2. An example is given to illustrate one of the main results.

      Mathematics Subject Classication 2011: 34D05.

      Keywords

      Boundary value problem singular half-line principal nonprincipal

      1. Introduction

      Boundary value problems on half-line occur in various applications such as in the study of the unsteady flow of a gas through semi-infinite porous medium, in analyzing the heat transfer in radial flow between circular disks, in the study of plasma physics, in an analysis of the mass transfer on a rotating disk in a non-Newtonian fluid, etc. More examples and a collection of works on the existence of solutions of boundary value problems on half-line for differential, difference and integral equations may be found in the monographs [1, 2] For some works and various techniques dealing with such boundary value problems (we may refer to [36] and the references cited therein).

      In this article by employing principal and nonprincipal solutions we introduce a new approach to study nonlinear boundary problems on half-line of the form
      ( p ( t ) x ) + q ( t ) x = f ( t , x ) , t t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ1_HTML.gif
      (1.1)
      x ( t 0 ) = x 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ2_HTML.gif
      (1.2)
      x ( t ) = a v ( t ) + b u ( t ) + o ( r ( t ) ) , t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ3_HTML.gif
      (1.3)
      where a and b are any given real numbers, u and v are principal and nonprincipal solutions of
      ( p ( t ) x ) + q ( t ) x = 0 , t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ4_HTML.gif
      (1.4)

      and pC([0, ∞), (0, ∞)), qC([0, ∞), ℝ) and fC([0, ∞) × ℝ, ℝ).

      We will show that the problem (1.1)-(1.3) has a unique solution in the case when
      r ( t ) = o ( u ( t ) ( v ( t ) ) μ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ5_HTML.gif
      (1.5)
      and
      r ( t ) = o ( v ( t ) ( u ( t ) ) μ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ6_HTML.gif
      (1.6)

      where μ ∈ (0, 1) is arbitrary but fixed real numbers.

      The nonlinear boundary value problem (1.1)-(1.3) is also closely related to asymptotic integration of second order differential equations. Indeed, there are several important works in the literature, see [716], dealing with mostly the asymptotic integration of solutions of second order nonlinear equations of the form
      x = f ( t , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equb_HTML.gif
      The authors are usually interested in finding conditions on the function f(t, x) which guarantee the existence of a solution asymptotic to linear function
      x ( t ) = a t + b , t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ7_HTML.gif
      (1.7)
      We should point out that u(t) = 1 and v(t) = t are principal and nonprincipal solutions of the corresponding unperturbed equation
      x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equc_HTML.gif
      and the function x(t) in (1.7) can be written as
      x = a v ( t ) + b u ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equd_HTML.gif

      Note that v(t) → ∞ as t → ∞ but u(t) is bounded in this special case. It turns out such information is crucial in investigating the general case. Our results will be applicable whether or not u(t) → ∞ (v(t) → ∞) as t → ∞.

      2. Main results

      It is well-known that [17, 18] if the second order linear Equation (1.4) has a positive solution or nonoscillatory at , then there exist two linearly independent solutions u(t) and v(t), called principal and nonprincipal solutions of the equation. The principal solution u is unique up to a constant multiple. Moreover, the following useful properties are satisfied:
      lim t u ( t ) v ( t ) = 0 , t * 1 p ( t ) u 2 ( t ) d t = , t * 1 p ( t ) v 2 ( t ) d t < , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Eque_HTML.gif

      where t* ≥ 0 is a sufficiently large real number.

      Let v be a principal solution of (1.4). Without loss of generality we may assume that v (t) > 0 if tt1 for some t1 ≥ 0. It is easy to see that
      v ( t ) = u ( t ) t 1 t 1 p ( s ) u 2 ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ8_HTML.gif
      (2.1)

      is a nonprincipal solution of (1.4), which is strictly positive for t > t1.

      Theorem 2.1. Let t0 > t1. Assume that the function f satisfies
      f ( t , x ) h 1 ( t ) g ( x ) + h 2 ( t ) , t t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ9_HTML.gif
      (2.2)
      and
      f ( t , x 1 ) - f ( t , x 2 ) k ( t ) v ( t ) x 1 - x 2 , t t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ10_HTML.gif
      (2.3)
      where gC([0, ∞), [0, ∞)) is bounded; h1, h2, kC([t0, ∞), [0, ∞)). Suppose further that
      t 0 u ( s ) k ( s ) d s μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ11_HTML.gif
      (2.4)
      and
      1 p ( t ) u 2 ( t ) t u ( s ) h i ( s ) d s β ( t ) , t t 0 , i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ12_HTML.gif
      (2.5)
      for some βC([t0, ∞), [0, ∞)) such that
      t 0 t β ( s ) d s = o ( ( v ( t ) ) μ ) , t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ13_HTML.gif
      (2.6)
      If either
      v ( t ) , t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ14_HTML.gif
      (2.7)
      or else
      b = x 0 u ( t 0 ) - a t 1 t 0 1 p ( s ) u 2 ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ15_HTML.gif
      (2.8)

      then there is a unique solution x(t) of (1.1)-(1.3), where r is given by (1.5).

      Proof. Denote by M the supremum of the function g over [0, ∞). Let X be a space of functions defined by
      X = x C ( t 0 , , ) | x ( t ) l 1 v ( t ) + l 2 u ( t ) , t t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equf_HTML.gif
      where
      l 1 = ( M + 1 ) p ( t 0 ) u 2 ( t 0 ) β ( t 0 ) + a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equg_HTML.gif
      and
      l 2 = x 0 u ( t 0 ) + a t 1 t 0 1 p ( s ) u 2 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equh_HTML.gif
      Note that X is a complete metric space with the metric d defined by
      d ( x 1 , x 2 ) = sup t t 0 1 v ( t ) x 1 ( t ) - x 2 ( t ) , x 1 , x 2 X . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equi_HTML.gif
      Define an operator F on X by
      ( F x ) ( t ) = - u ( t ) t 0 t 1 p ( s ) u 2 ( s ) s u ( τ ) f ( τ , x ( τ ) ) d τ d s + a v ( t ) + x 0 u ( t 0 ) - a t 1 t 0 1 p ( s ) u 2 ( s ) d s u ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equj_HTML.gif
      In view of conditions (2.2) and (2.5) we see that F is well defined. Next we show that F XX. Indeed, let xX, then
      ( F x ) ( t ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) s u ( τ ) f ( τ , x ( τ ) ) d τ d s + a v ( t ) + l 2 u ( t ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) t 0 u ( τ ) f ( τ , x ( τ ) ) d τ d s + a υ ( t ) + l 2 u ( t ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) t 0 u ( τ ) ( h 1 ( τ ) g ( x ( τ ) ) + h 2 ( τ ) ) d τ d s + a v ( t ) + l 2 u ( t ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) t 0 u ( τ ) ( M h 1 ( τ ) + h 2 ( τ ) ) d τ d s + a υ ( t ) + l 2 u ( t ) ( M + 1 ) p ( t 0 ) u 2 ( t 0 ) β ( t 0 ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) d s + a v ( t ) + l 2 u ( t ) l 1 v ( t ) + l 2 u ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equk_HTML.gif

      which means that F xX.

      Using (2.1), (2.3) and (2.4) we also see that
      ( F x 1 ) ( t ) - ( F x 2 ) ( t ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) s u ( τ ) f ( τ , x 1 ( τ ) ) - f ( τ , x 2 ( τ ) ) d τ d s u ( t ) t 0 t 1 p ( s ) u 2 ( s ) s u ( τ ) k ( τ ) v ( τ ) x 1 ( τ ) - x 2 ( τ ) d τ d s d ( x 1 , x 2 ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) s u ( τ ) k ( τ ) d τ d s d ( x 1 , x 2 ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) t 0 u ( τ ) k ( τ ) d τ d s d ( x 1 , x 2 ) v ( t ) t 0 u ( τ ) k ( τ ) d τ μ d ( x 1 , x 2 ) v ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equl_HTML.gif

      where x1, x2X arbitrary. This implies that F is a contracting mapping.

      Thus according to Banach contraction principle F has a unique fixed point x. It is not difficult to see that the fixed point solves (1.1) and (1.2). It remains to show that x(t) satisfies (1.3) as well. It is not difficult to show that
      x ( t ) - a v ( t ) - b u ( t ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) s u ( τ ) f ( τ , x ( τ ) ) d τ d s + c u ( t ) u ( t ) t 0 t 1 p ( s ) u 2 ( s ) s u ( τ ) ( M h 1 ( τ ) + h 2 ( τ ) ) d τ d s + c u ( t ) ( M + 1 ) u ( t ) t 0 t β ( s ) d s + c u ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equm_HTML.gif
      where
      c = x 0 u ( t 0 ) - a t 1 t 0 1 p ( s ) u 2 ( s ) d s - b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equn_HTML.gif

      If (2.7) is satisfied, then in view (2.6) and the above inequality we easily obtain (1.3). In case (2.8) holds, then c = 0 and hence we still have (1.3).

      From Theorem 2.1 we deduce the following Corollary.

      Corollary 2.2. Assume that the function f satisfies (2.2) and
      f ( t , x 1 ) - f ( t , x 2 ) k ( t ) t x 1 - x 2 , t t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equo_HTML.gif
      where kC([t0, ∞), [0, ∞)). Suppose further that
      t 0 k ( s ) d s μ ; t h i ( s ) d s β ( t ) , t t 0 , i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equp_HTML.gif
      for some μ ∈ (0, 1) and βC([t0, ∞), [0, ∞)), where
      t 0 t β ( s ) d s = o ( t μ ) , t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equq_HTML.gif
      Then for each a, b ∈ ℝ the boundary value problem
      x = f ( t , x ) , t t 0 , x ( t 0 ) = x 0 , x ( t ) = a t + b + o ( t μ ) , t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equr_HTML.gif

      has a unique solution.

      Let υ be a nonprincipal solution of (1.4). Without loss of generality we may assume that v(t) > 0, if tt2 for some t2 ≥ 0. It is easy to see that [17, 18]
      u ( t ) = v ( t ) t 1 p ( s ) v 2 ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ16_HTML.gif
      (2.9)
      is a principal solution of (1.4) which is strictly positive. Take t2 large enough so that
      t 1 p ( s ) v 2 ( s ) d s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equs_HTML.gif

      Then from (2.9), we have v (t) ≥ u(t) for tt2, which is needed in the proof of the next theorem.

      Theorem 2.3. Let t0t2. Assume that the function f satisfies (2.2) and (2.3). Suppose further that
      t 0 v ( s ) k ( s ) d s μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ17_HTML.gif
      (2.10)
      and
      1 p ( t ) v 2 ( t ) t v ( s ) h i ( s ) d s β ( t ) , t t 0 , i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ18_HTML.gif
      (2.11)
      for some βC([t0, ∞), [0, ∞)) such that
      t 0 t β ( s ) d s = o ( ( u ( t ) ) μ ) , t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ19_HTML.gif
      (2.12)
      If either
      u ( t ) , t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ20_HTML.gif
      (2.13)
      or else
      a = x 0 v ( t 0 ) - b t 0 1 p ( s ) v 2 ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ21_HTML.gif
      (2.14)

      then there is a unique solution x(t) of (1.1) - (1.3), where r is given by (1.6).

      Proof. Let X be a space of functions defined by
      X = x C ( t 0 , , ) | x ( t ) l 1 v ( t ) + l 2 u ( t ) , t t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equf_HTML.gif
      where
      l 1 = ( M + 1 ) p ( t 0 ) u ( t 0 ) v ( t 0 ) β ( t 0 ) + x 0 v ( t 0 ) + b t 0 1 p ( s ) v 2 ( s ) d s and l 2 = b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equt_HTML.gif

      Again, X is a complete metric space with the metric d defined in the proof of the previous theorem.

      We define an operator F on X by
      ( F x ) ( t ) = - v ( t ) t 0 t 1 p ( s ) v 2 ( s ) s v ( τ ) f ( τ , x ( τ ) ) d τ d s + x 0 v ( t 0 ) - b t 0 1 p ( s ) v 2 ( s ) d s v ( t ) + b u ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equu_HTML.gif

      The remainder of the proof proceeds similarly as in that of Theorem 2.1 by using (2.2), (2.3), (2.9)-(2.14).

      Corollary 2.4. Assume that the function f satisfies (2.2) and (2.3). Suppose further that
      t 0 s k ( s ) d s μ ; 1 t 2 t s h i ( s ) d s β ( t ) , t t 0 , i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equv_HTML.gif
      for some μ ∈ (0, 1) and βC([t0, ∞), [0, ∞)), where
      1 t β ( s ) d s = o ( 1 ) , t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equw_HTML.gif
      If for any given a, b ∈ ℝ the condition (2.14) holds then the boundary value problem
      x = f ( t , x ) , t t 0 , x ( t 0 ) = x 0 , x ( t ) = a t + b + o ( t ) , t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equx_HTML.gif

      has a unique solution.

      3. An example

      Consider the boundary value problem
      ( t x ) = 1 t 2 arctan x + t ν , t t 0 , ν < - 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ22_HTML.gif
      (3.1)
      x ( t 0 ) = x 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ23_HTML.gif
      (3.2)
      x ( t ) = a ln t + b + o ( ( ln t ) μ ) , t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ24_HTML.gif
      (3.3)
      where t0 > t1 = 1 and μ ∈ (0, 1) are chosen to satisfy
      1 + ln t 0 t 0 μ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equ25_HTML.gif
      (3.4)
      Note that since
      lim t 0 1 + ln t 0 t 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equy_HTML.gif

      for any given μ ∈ (0, 1) there is a t0 such that (3.4) holds.

      Comparing with the boundary value problem (1.1)-(1.3) we see that p(t) = t, q(t) = 0, and f(t, x) = (1/t2) arctan x + t υ . The corresponding linear equation becomes
      ( t x ) = 0 , t t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equz_HTML.gif
      Clearly, we may take
      u ( t ) = 1 and v ( t ) = ln t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equaa_HTML.gif
      Let
      h 1 ( t ) = 1 t 2 , h 2 ( t ) = t ν , g ( x ) = arctan x , k ( t ) = ln t t 2 , β ( t ) = 1 t 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equab_HTML.gif
      then it is easy to see that
      f ( t , x ) 1 t 2 arctan x + t ν = h 1 ( t ) g x + h 2 ( t ) , f ( t , x 1 ) - f ( t , x 2 ) 1 t 2 x 1 - x 2 = k ( t ) v ( t ) x 1 - x 2 , t 0 k ( s ) d s = t 0 ln s s 2 d s = 1 + ln t 0 t 0 μ by (3 .4) 1 t t h 1 ( s ) d s 1 t t 1 s 2 d s = 1 t 2 = β ( t ) , t t 0 , 1 t t h 2 ( s ) d s = - t ν ν + 1 β ( t ) , t t 0 , t 0 t β ( s ) d s = t 0 t 1 s 2 d s = 1 t 0 - 1 t = o ( ( ln t ) μ ) , t , μ ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equac_HTML.gif
      and
      v ( t ) = ln t , t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equad_HTML.gif

      i.e., all the conditions of Theorem 2.1 are satisfied. Therefore we may conclude that if (3.4) holds, then the boundary value problem (3.1)-(3.3) has a unique solution.

      Furthermore, we may also deduce that there exist solutions x1(t) and x2(t) such that
      x 1 ( t ) = 1 + o ( ( ln t ) μ ) , t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equae_HTML.gif
      and
      x 2 ( t ) = ln t + o ( ( ln t ) μ ) , t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-43/MediaObjects/13661_2012_Article_149_Equaf_HTML.gif

      by taking (a, b) = (0, 1) and (a, b) = (1, 0), respectively.

      Declarations

      Acknowledgements

      This work was supported by the Scientific and Technological Research Council of Turkey (TUBITAK) under project number 108T688.

      Authors’ Affiliations

      (1)
      Department of Mathematics, Middle East Technical University

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