Nontrivial solutions for a boundary value problem with integral boundary conditions

  • Bingmei Liu1Email author,

    Affiliated with

    • Junling Li1 and

      Affiliated with

      • Lishan Liu2

        Affiliated with

        Boundary Value Problems20142014:15

        DOI: 10.1186/1687-2770-2014-15

        Received: 27 July 2013

        Accepted: 13 November 2013

        Published: 13 January 2014

        Abstract

        This paper concerns the existence of nontrivial solutions for a boundary value problem with integral boundary conditions by topological degree theory. Here the nonlinear term is a sign-changing continuous function and may be unbounded from below.

        1 Introduction

        Consider the following Sturm-Liouville problem with integral boundary conditions
        { ( L u ) ( t ) + h ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , ( cos γ 0 ) u ( 0 ) ( sin γ 0 ) u ( 0 ) = 0 1 u ( τ ) d α ( τ ) , ( cos γ 1 ) u ( 1 ) + ( sin γ 1 ) u ( 1 ) = 0 1 u ( τ ) d β ( τ ) ,
        (1.1)

        where ( L u ) ( t ) = ( p ˜ ( t ) u ( t ) ) + q ( t ) u ( t ) , p ˜ ( t ) C 1 [ 0 , 1 ] , p ˜ ( t ) > 0 , q ( t ) C [ 0 , 1 ] , q ( t ) < 0 , α and β are right continuous on [ 0 , 1 ) , left continuous at t = 1 and nondecreasing on [ 0 , 1 ] with α ( 0 ) = β ( 0 ) = 0 ; γ 0 , γ 1 [ 0 , π / 2 ] , 0 1 u ( τ ) d α ( τ ) and 0 1 u ( τ ) d β ( τ ) denote the Riemann-Stieltjes integral of u with respect to α and β, respectively. Here the nonlinear term f : [ 0 , 1 ] × ( , + ) ( , + ) is a continuous sign-changing function and f may be unbounded from below, h : ( 0 , 1 ) [ 0 , + ) with 0 < 0 1 h ( s ) d s < + is continuous and is allowed to be singular at t = 0 , 1 .

        Problems with integral boundary conditions arise naturally in thermal conduction problems [1], semiconductor problems [2], hydrodynamic problems [3]. Integral BCs (BCs denotes boundary conditions) cover multi-point BCs and nonlocal BCs as special cases and have attracted great attention, see [414] and the references therein. For more information about the general theory of integral equations and their relation with boundary value problems, we refer to the book of Corduneanu [4], Agarwal and O’Regan [5]. Yang [6], Boucherif [8], Chamberlain et al. [10], Feng [11], Jiang et al. [14] focused on the existence of positive solutions for the cases in which the nonlinear term is nonnegative. Although many papers investigated two-point and multi-point boundary value problems with sign-changing nonlinear terms, for example, [1520], results for boundary value problems with integral boundary conditions when the nonlinear term is sign-changing are rarely seen except for a few special cases [7, 12, 13].

        Inspired by the above papers, the aim of this paper is to establish the existence of nontrivial solutions to BVP (1.1) under weaker conditions. Our findings presented in this paper have the following new features. Firstly, the nonlinear term f of BVP (1.1) is allowed to be sign-changing and unbounded from below. Secondly, the boundary conditions in BVP (1.1) are the Riemann-Stieltjes integral, which includes multi-point boundary conditions in BVPs as special cases. Finally, the main technique used here is the topological degree theory, the first eigenvalue and its positive eigenfunction corresponding to a linear operator. This paper employs different conditions and different methods to solve the same BVP (1.1) as [7]; meanwhile, this paper generalizes the result in [17] to boundary value problems with integral boundary conditions. What we obtain here is different from [620].

        2 Preliminaries and lemmas

        Let E = C [ 0 , 1 ] be a Banach space with the maximum norm u = max 0 t 1 | u ( t ) | for u E . Define P = { u E u ( t ) 0 , t [ 0 , 1 ] } and B r = { u E u < r } . Then P is a total cone in E, that is, E = P P ¯ . P denotes the dual cone of P, namely, P = { g E g ( u ) 0 ,  for all  u P } . Let E denote the dual space of E, then by Riesz representation theorem, E is given by
        E = { v v  is right continuous on  [ 0 , 1 )  and is bounded variation on  [ 0 , 1 ] with  v ( 0 ) = 0 } .

        We assume that the following condition holds throughout this paper.

        (H1) u ( t ) 0 is the unique C 2 solution of the linear boundary value problem
        { ( L u ) ( t ) = 0 , 0 < t < 1 , ( cos γ 0 ) u ( 0 ) ( sin γ 0 ) u ( 0 ) = 0 , ( cos γ 1 ) u ( 1 ) + ( sin γ 1 ) u ( 1 ) = 0 .
        Let φ , ψ C 2 ( [ 0 , 1 ] , R + ) solve the following inhomogeneous boundary value problems, respectively:
        { ( L φ ) ( t ) = 0 , 0 < t < 1 , ( cos γ 0 ) φ ( 0 ) ( sin γ 0 ) φ ( 0 ) = 1 , ( cos γ 1 ) φ ( 1 ) + ( sin γ 1 ) φ ( 1 ) = 0 and { ( L ψ ) ( t ) = 0 , 0 < t < 1 , ( cos γ 0 ) ψ ( 0 ) ( sin γ 0 ) ψ ( 0 ) = 0 , ( cos γ 1 ) ψ ( 1 ) + ( sin γ 1 ) ψ ( 1 ) = 1 .

        Let κ 1 = 1 0 1 φ ( τ ) d α ( τ ) , κ 2 = 0 1 ψ ( τ ) d α ( τ ) , κ 3 = 0 1 φ ( τ ) d β ( τ ) , κ 4 = 1 0 1 ψ ( τ ) d β ( τ ) .

        (H2) κ 1 > 0 , κ 4 > 0 , k = κ 1 κ 4 κ 2 κ 3 > 0 .

        Lemma 2.1 ([7])

        If (H1) and (H2) hold, then BVP (1.1) is equivalent to
        u ( t ) = 0 1 G ( t , s ) h ( s ) f ( s , u ( s ) ) d s ,

        where G ( t , s ) C ( [ 0 , 1 ] × [ 0 , 1 ] , R + ) is the Green function for (1.1).

        Define an operator A : E E as follows:
        ( A u ) ( t ) = 0 1 G ( t , s ) h ( s ) f ( s , u ( s ) ) d s , u E .
        (2.1)

        It is easy to show that A : E E is a completely continuous nonlinear operator, and if u E is a fixed point of A, then u is a solution of BVP (1.1) by Lemma 2.1.

        For any u E , define a linear operator K : E E as follows:
        ( K u ) ( t ) = 0 1 G ( t , s ) h ( s ) u ( s ) d s , u E .
        (2.2)
        It is easy to show that K : E E is a completely continuous nonlinear operator and K ( P ) P holds. By [7], the spectral radius r ( K ) of K is positive. The Krein-Rutman theorem [21] asserts that there are ϕ P { 0 } and ω P { 0 } corresponding to the first eigenvalue λ 1 = 1 / r ( K ) of K such that
        λ 1 K ϕ = ϕ
        (2.3)
        and
        λ 1 K ω = ω , ω ( 1 ) = 1 .
        (2.4)
        Here K : E E is the dual operator of K given by:
        ( K v ) ( s ) = 0 s 0 1 G ( t , τ ) h ( τ ) d v ( t ) d τ , v E .
        The representation of K , the continuity of G and the integrability of h imply that ω C 1 [ 0 , 1 ] . Let e ( t ) : = ω ( t ) . Then e P { 0 } , and (2.4) can be rewritten equivalently as
        r ( K ) e ( s ) = 0 1 G ( t , s ) h ( s ) e ( t ) d t , 0 1 e ( t ) d t = 1 .
        (2.5)

        Lemma 2.2 ([7])

        If (H1) holds, then there is δ > 0 such that P 0 = { u P 0 1 u ( t ) e ( t ) d t δ u } is a subcone of P and K ( P ) P 0 .

        Lemma 2.3 ([22])

        Let E be a real Banach space and Ω E be a bounded open set with 0 Ω . Suppose that A : Ω ¯ E is a completely continuous operator. (1) If there is y 0 E with y 0 0 such that u A u + μ y 0 for all u Ω and μ 0 , then deg ( I A , Ω , 0 ) = 0 . (2) If A u μ u for all u Ω and μ 1 , then deg ( I A , Ω , 0 ) = 1 . Here deg stands for the Leray-Schauder topological degree in E.

        Lemma 2.4 Assume that (H1), (H2) and the following assumptions are satisfied:

        (C1) There exist ϕ P { 0 } , ω P { 0 } and δ > 0 such that (2.3), (2.4) hold and K maps P into P 0 .

        (C2) There exists a continuous operator H : E P such that
        lim u + H u u = 0 .

        (C3) There exist a bounded continuous operator F : E E and u 0 E such that F u + u 0 + H u P for all u E .

        (C4) There exist v 0 E and ζ > 0 such that K F u λ 1 ( 1 + ζ ) K u K H u v 0 for all u E .

        Let A = K F , then there exists R > 0 such that
        deg ( I A , B R , 0 ) = 0 ,

        where B R = { u E u < R } .

        Proof Choose a constant L 0 = ( δ λ 1 ) 1 ( 1 + ζ 1 ) + K > 0 . From (C2), for 0 < ε 0 < L 0 1 , there exists R 1 > 0 such that u > R 1 implies
        H u < ε 0 u .
        (2.6)
        Now we shall show
        u K F u + μ ϕ for any  u B R  and  μ 0 ,
        (2.7)

        provided that R is sufficiently large.

        In fact, if (2.7) is not true, then there exist u 1 B R and μ 1 0 satisfying
        u 1 = K F u 1 + μ 1 ϕ .
        (2.8)
        Since ϕ P { 0 } , e ( t ) P { 0 } , 0 1 ϕ ( t ) e ( t ) d t > 0 . Multiply (2.8) by e ( t ) on both sides and integrate on [ 0 , 1 ] . Then, by (C4), (2.5), we get
        0 1 u 1 ( t ) e ( t ) d t = 0 1 ( K F u 1 ) ( t ) e ( t ) d t + μ 1 0 1 ϕ ( t ) e ( t ) d t λ 1 ( 1 + ζ ) 0 1 0 1 G ( t , s ) h ( s ) u 1 ( s ) d s e ( t ) d t 0 1 ( K H u 1 ) ( t ) e ( t ) d t 0 1 v 0 ( t ) e ( t ) d t = λ 1 ( 1 + ζ ) 0 1 0 1 G ( t , s ) h ( s ) u 1 ( s ) e ( t ) d s d t 0 1 0 1 G ( t , s ) h ( s ) ( H u 1 ) ( s ) e ( t ) d s d t 0 1 v 0 ( t ) e ( t ) d t = λ 1 ( 1 + ζ ) 0 1 [ 0 1 G ( t , s ) h ( s ) e ( t ) d t ] u 1 ( s ) d s 0 1 [ 0 1 G ( t , s ) h ( s ) e ( t ) d t ] ( H u 1 ) ( s ) d s 0 1 v 0 ( t ) e ( t ) d t = λ 1 ( 1 + ζ ) r ( K ) 0 1 e ( s ) u 1 ( s ) d s r ( K ) 0 1 ( H u 1 ) ( s ) e ( s ) d s 0 1 v 0 ( t ) e ( t ) d t = ( 1 + ζ ) 0 1 u 1 ( t ) e ( t ) d t r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t 0 1 v 0 ( t ) e ( t ) d t .
        (2.9)
        Thus,
        0 1 u 1 ( t ) e ( t ) d t ζ 1 ( r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + 0 1 v 0 ( t ) e ( t ) d t ) .
        (2.10)
        By (2.9), 0 1 ( K H u 1 ) ( t ) e ( t ) d t = r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t holds. Then (2.3), (2.6) and (2.10) imply
        0 1 ( u 1 ( t ) + ( K H u 1 ) ( t ) + ( K u 0 ) ( t ) ) e ( t ) d t ζ 1 ( r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + 0 1 v 0 ( t ) e ( t ) d t ) + r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + 0 1 ( K u 0 ) ( t ) e ( t ) d t ζ 1 ( 1 + ζ ) r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + ζ 1 0 1 v 0 ( t ) e ( t ) d t + 0 1 ( K u 0 ) ( t ) e ( t ) d t ζ 1 ( 1 + ζ ) r ( K ) ε 0 u 1 + L 1 ,
        (2.11)

        where L 1 = ζ 1 0 1 v 0 ( t ) e ( t ) d t + 0 1 ( K u 0 ) ( t ) e ( t ) d t is a constant.

        (C3) shows F u 1 + u 0 + H u 1 P and (C1) implies μ 1 ϕ = μ 1 λ 1 K φ 1 P 0 . Then (C1), (2.8) and Lemma 2.2 tell us that
        u 1 + K H u 1 + K u 0 = K F u 1 + μ 1 ϕ + K H u 1 + K u 0 = K ( F u 1 + H u 1 + u 0 ) + μ 1 ϕ P 0 .
        The definition of P 0 yields
        0 1 ( u 1 + K H u 1 + K u 0 ) ( t ) e ( t ) d t δ u 1 + K H u 1 + K u 0 δ u 1 δ K H u 1 δ K u 0 .
        (2.12)
        It follows from (2.6), (2.11) and (2.12) that
        u 1 = δ 1 0 1 ( u 1 + K H u 1 + K u 0 ) ( t ) e ( t ) d t + K H u 1 + K u 0 ε 0 ( δ λ 1 ) 1 ( 1 + ζ 1 ) u 1 + L 1 δ 1 + ε 0 K u 1 + K u 0 = ε 0 L 0 u 1 + L 2 ,
        (2.13)

        where L 2 = K u 0 + L 1 δ 1 is a constant.

        Since 0 < ε 0 L 0 < 1 , then (2.13) deduces that (2.7) holds provided that R is sufficiently large such that R > max { L 2 / ( 1 ε 0 L 0 ) , R 1 } . By (2.13) and Lemma 2.3, we have
        deg ( I A , B R , 0 ) = 0 .

         □

        3 Main results

        Theorem 3.1 Assume that (H1), (H2) hold and the following conditions are satisfied:

        (A1) There exist two nonnegative functions b ( t ) , c ( t ) C [ 0 , 1 ] with c ( t ) 0 and one continuous even function B : R R + such that f ( t , x ) b ( t ) c ( t ) B ( x ) for all x R . Moreover, B is nondecreasing on R + and satisfies lim x + B ( x ) x = 0 .

        (A2) f : [ 0 , 1 ] × R R is continuous.

        (A3) lim inf x + f ( t , x ) x > λ 1 uniformly on t [ 0 , 1 ] .

        (A4) lim sup x 0 | f ( t , x ) x | < λ 1 uniformly on t [ 0 , 1 ] .

        Here λ 1 is the first eigenvalue of the operator K defined by (2.2).

        Then BVP (1.1) has at least one nontrivial solution.

        Proof We first show that all the conditions in Lemma 2.4 are satisfied. By Lemma 2.2, condition (C1) of Lemma 2.4 is satisfied. Obviously, B : E P is a continuous operator. By (A1), for any ε > 0 , there is L > 0 such that when x > L , B ( x ) < ε x holds. Thus, for u E with u > L , B ( u ) < ε u holds. The fact that B is nondecreasing on R + yields ( B u ) ( t ) B ( u ) for any u P , t [ 0 , 1 ] . Since B : R R + is an even function, for any u E and t [ 0 , 1 ] , ( B u ) ( t ) B ( u ) holds, which implies B u B ( u ) for u E . Therefore,
        B u B ( u ) < ε u , u E  with  u > L ,

        that is, lim u + B u u = 0 . Take H u = c 0 B u , for any u E , where c 0 = max t [ 0 , 1 ] c ( t ) > 0 . Obviously, lim u + H u u = 0 holds. Therefore H satisfies condition (C2) in Lemma 2.4.

        Take u 0 ( t ) b = max t [ 0 , 1 ] b ( t ) > 0 and ( F u ) ( t ) = f ( t , u ( t ) ) for t [ 0 , 1 ] , u E , then it follows from (A1) that
        F u + u 0 + H u P for all  u E ,

        which shows that condition (C3) in Lemma 2.4 holds.

        By (A3), there exist ε 1 > 0 and a sufficiently large number l 1 > 0 such that
        f ( t , x ) λ 1 ( 1 + ε 1 ) x , x l 1 .
        (3.1)
        Combining (3.1) with (A1), there exists b 1 0 such that
        f ( t , x ) λ 1 ( 1 + ε 1 ) x b 1 c 0 B ( x ) for all  x R ,
        and so
        F u λ 1 ( 1 + ε 1 ) u b 1 H u for all  u E .
        (3.2)
        Since K is a positive linear operator, from (3.2) we have
        ( K F u ) ( t ) λ 1 ( 1 + ε 1 ) ( K u ) ( t ) K b 1 ( K H u ) ( t ) , t [ 0 , 1 ] , u E .

        So condition (C4) in Lemma 2.4 is satisfied.

        According to Lemma 2.4, we derive that there exists a sufficiently large number R > 0 such that
        deg ( I A , B R , 0 ) = 0 .
        (3.3)
        From (A4) it follows that there exist 0 < ε 2 < 1 and 0 < r < R such that
        | f ( t , x ) | ( 1 ε 2 ) λ 1 | x | , t [ 0 , 1 ] , x R  with  | x | r .
        Thus
        | ( A u ) ( t ) | ( 1 ε 2 ) λ 1 ( K | u | ) ( t ) , t [ 0 , 1 ] , u E  with  u r .
        (3.4)
        Next we will prove that
        u μ A u for all  u B r  and  μ [ 0 , 1 ] .
        (3.5)

        If there exist u 1 B r and μ 1 [ 0 , 1 ] such that u 1 = μ 1 A u 1 . Let z ( t ) = | u 1 ( t ) | . Then z P and by (3.4), z ( 1 ε 2 ) λ 1 K z . The n th iteration of this inequality shows that z ( 1 ε 2 ) n λ 1 n K n z ( n = 1 , 2 , ), so z ( 1 ε 2 ) n λ 1 n K n z , that is, 1 ( 1 ε 2 ) n λ 1 n K n . This yields 1 ε 2 = ( 1 ε 2 ) λ 1 r ( K ) = ( 1 ε 2 ) λ 1 lim n K n n 1 , which is a contradictory inequality. Hence, (3.5) holds.

        It follows from (3.5) and Lemma 2.3 that
        deg ( I A , B r , 0 ) = 1 .
        (3.6)
        By (3.3), (3.6) and the additivity of Leray-Schauder degree, we obtain
        deg ( I A , B R B ¯ r , 0 ) = deg ( I A , B R , 0 ) deg ( I A , B r , 0 ) = 1 .

        So A has at least one fixed point on B R B ¯ r , namely, BVP (1.1) has at least one nontrivial solution. □

        Corollary 3.1 Using ( A 1 ) instead of (A1), the conclusion of Theorem  3.1 remains true.

        ( A 1 ) There exist three constants b > 0 , c > 0 and α ( 0 , 1 ) such that
        f ( x ) b c | x | α for any  x R .

        Declarations

        Acknowledgements

        The first two authors were supported financially by the National Natural Science Foundation of China (11201473, 11271364) and the Fundamental Research Funds for the Central Universities (2013QNA35, 2010LKSX09, 2010QNA42). The third author was supported financially by the National Natural Science Foundation of China (11371221, 11071141), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001) and the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province.

        Authors’ Affiliations

        (1)
        College of Sciences, China University of Mining and Technology
        (2)
        School of Mathematical Sciences, Qufu Normal University

        References

        1. Cannon JR: The solution of the heat equation subject to the specification of energy. Q. Appl. Math. 1963, 21: 155-160.MathSciNet
        2. Ionkin NI: Solution of a boundary value problem in heat conduction theory with nonlocal boundary conditions. Differ. Equ. 1977, 13: 294-304.MATHMathSciNet
        3. Chegis RY: Numerical solution of a heat conduction problem with an integral boundary condition. Litov. Mat. Sb. 1984, 24: 209-215.MATHMathSciNet
        4. Corduneanu C: Integral Equations and Applications. Cambridge University Press, Cambridge; 1991.MATHView Article
        5. Agarwal RP, O’Regan D: Infinite Interval Problems for Differential, Difference and Integral Equations. Kluwer Academic, Dordrecht; 2001.MATHView Article
        6. Yang ZL: Existence and nonexistence results for positive solutions of an integral boundary value problem. Nonlinear Anal. 2006, 65: 1489-1511. 10.1016/j.na.2005.10.025MATHMathSciNetView Article
        7. Yang ZL: Existence of nontrivial solutions for a nonlinear Sturm-Liouville problem with integral boundary value conditions. Nonlinear Anal. 2008, 68: 216-225. 10.1016/j.na.2006.10.044MATHMathSciNetView Article
        8. Boucherif A: Second-order boundary value problems with integral boundary conditions. Nonlinear Anal. 2009, 70: 364-371. 10.1016/j.na.2007.12.007MATHMathSciNetView Article
        9. Kong LJ: Second order singular boundary value problems with integral boundary conditions. Nonlinear Anal. 2010, 72: 2628-2638. 10.1016/j.na.2009.11.010MATHMathSciNetView Article
        10. Chamberlain J, Kong LJ, Kong QK: Nodal solutions of boundary value problems with boundary conditions involving Riemann-Stieltjes integrals. Nonlinear Anal. 2011, 74: 2380-2387. 10.1016/j.na.2010.11.040MATHMathSciNetView Article
        11. Feng MQ: Existence of symmetric positive solutions for a boundary value problem with integral boundary conditions. Appl. Math. Lett. 2011, 24: 1419-1427. 10.1016/j.aml.2011.03.023MATHMathSciNetView Article
        12. Li YH, Li FY: Sign-changing solutions to second-order integral boundary value problems. Nonlinear Anal. 2008, 69: 1179-1187. 10.1016/j.na.2007.06.024MATHMathSciNetView Article
        13. Li HT, Liu YS: On sign-changing solutions for a second-order integral boundary value problem. Comput. Math. Appl. 2011, 62: 651-656. 10.1016/j.camwa.2011.05.046MATHMathSciNetView Article
        14. Jiang JQ, Liu LS, Wu YH: Second-order nonlinear singular Sturm-Liouville problems with integral boundary conditions. Appl. Math. Comput. 2009, 215: 1573-1582. 10.1016/j.amc.2009.07.024MATHMathSciNetView Article
        15. Sun JX, Zhang GW: Nontrivial solutions of singular superlinear Sturm-Liouville problem. J. Math. Anal. Appl. 2006, 313: 518-536. 10.1016/j.jmaa.2005.06.087MATHMathSciNetView Article
        16. Han GD, Wu Y: Nontrivial solutions of singular two-point boundary value problems with sign-changing nonlinear terms. J. Math. Anal. Appl. 2007, 325: 1327-1338. 10.1016/j.jmaa.2006.02.076MATHMathSciNetView Article
        17. Liu LS, Liu BM, Wu YH: Nontrivial solutions of m -point boundary value problems for singular second-order differential equations with a sign-changing nonlinear term. J. Comput. Appl. Math. 2009, 224: 373-382. 10.1016/j.cam.2008.05.007MATHMathSciNetView Article
        18. Liu LS, Liu BM, Wu YH: Nontrivial solutions for higher-order m -point boundary value problem with a sign-changing nonlinear term. Appl. Math. Comput. 2010, 217: 3792-3800. 10.1016/j.amc.2010.09.038MATHMathSciNetView Article
        19. Graef JR, Kong LJ: Periodic solutions for functional differential equations with sign-changing nonlinearities. Proc. R. Soc. Edinb. A 2010, 140: 597-616. 10.1017/S0308210509000523MATHMathSciNetView Article
        20. Wang YQ, Liu LS, Wu YH: Positive solutions for a class of fractional boundary value problem with changing sign nonlinearity. Nonlinear Anal. 2011, 74: 6434-6441. 10.1016/j.na.2011.06.026MATHMathSciNetView Article
        21. Krein MG, Rutman MA: Linear operators leaving invariant a cone in a Banach space. Transl. Am. Math. Soc. 1962, 10: 199-325.
        22. Guo DJ, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, Orlando; 1988.MATH

        Copyright

        © Liu et al.; licensee Springer. 2014

        This article is published under license to BioMed Central Ltd. 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.