Steady-state solutions for a suspension bridge with intermediate supports

  • Claudio Giorgi1Email author and

    Affiliated with

    • Elena Vuk1

      Affiliated with

      Boundary Value Problems20132013:204

      DOI: 10.1186/1687-2770-2013-204

      Received: 5 March 2013

      Accepted: 19 July 2013

      Published: 9 September 2013

      Abstract

      This work is focused on a system of boundary value problems whose solutions represent the equilibria of a bridge suspended by continuously distributed cables and supported by M intermediate piers. The road bed is modeled as the junction of N = M + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq1_HTML.gif extensible elastic beams which are clamped to each other and pinned at their ends to each pier. The suspending cables are modeled as one-sided springs with stiffness k. Stationary solutions of these doubly nonlinear problems are explicitly and analytically derived for arbitrary k and a general axial load p applied at the ends of the bridge. In particular, we scrutinize the occurrence of buckled solutions in connection with the length of each sub-span of the bridge.

      MSC:35G30, 74G05, 74G60, 74K10.

      Keywords

      extensible elastic beam suspension bridge boundary value problems for nonlinear higher-order PDE bifurcation and buckling

      1 Introduction

      In this paper, we investigate the solutions of a system of one-dimensional nonlinear problems describing the steady-states of an extensible elastic suspension bridge with M intermediate supports (piers). In particular, we assume that the road bed of the bridge (deck) is composed of N = M + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq1_HTML.gif extensible elastic beams which are clamped to each other, pinned at their ends to each pier and suspended by a continuous distribution of flexible elastic cables. On account of the midplane stretching of the beams due to their elongations, a geometric nonlinearity appears into the bending equations of the deck.

      In the case of a bridge with a single span (no intermediate pier), the problem can be recast into a non-dimensional setting, where its length is supposed to be unitary, for simplicity. Let u = u ( x ) : [ 0 , 1 ] R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq2_HTML.gif be a non-dimensional field which accounts for the downward deflection of the bridge in the vertical plane with respect to its reference configuration, and let u + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq3_HTML.gif stand for its positive part, namely
      u + = { u if  u 0 , 0 if  u < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equa_HTML.gif
      Assuming that both ends of the bridge are pinned, the equation of the bending equilibrium looks like (see [1])
      { u ′′′′ + ( p u L 2 ( 0 , 1 ) 2 ) u + k u + = 0 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ1_HTML.gif
      (1.1)

      The term k u + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq4_HTML.gif accounts for the restoring force due to one-sided springs which models the supporting cables. Since we confine our attention to stationary conditions, we neglect the dynamical coupling between the deck and the main cable. The constant p represents a non-dimensional measure of the axial force acting at the ends of the span in the reference configuration. Accordingly, p is negative when the span is stretched, positive when compressed. The symbol ′ represents the derivative with respect to the argument.

      Assume now that the bridge span is composed of N extensible beams whose internal ends are hinged to M = N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq5_HTML.gif intermediate piers. Let x i ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq6_HTML.gif, i = 1 , , M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq7_HTML.gif, be the points at which these supports are located and let
      0 = x 0 < x 1 < < x N 1 < x N = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equb_HTML.gif
      We denote by I n = ( x n 1 , x n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq8_HTML.gif, n = 1 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq9_HTML.gif, the corresponding intervals between two subsequent piers and by | I n | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq10_HTML.gif the corresponding length, so that
      n = 1 N | I n | = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equc_HTML.gif
      According to the assumptions, the deflection u of the whole span obeys
      u ( x ) = u ( x ) = 0 , = 0 , 1 , , N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equd_HTML.gif
      Furthermore, we assume that consecutive sub-spans are mutually clamped at the common end, which in turn implies the continuity of u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq11_HTML.gif across the supported points x i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq12_HTML.gif, i = 1 , , M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq7_HTML.gif (see Figure 1).
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Fig1_HTML.jpg
      Figure 1

      The joint connecting two consecutive sub-spans and the intermediate pier.

      Let u n = u n ( x ) : I n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq13_HTML.gif, n = 1 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq9_HTML.gif, account for the downward deflection of the n th beam in the vertical plane, and let p n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq14_HTML.gif be the axial load acting on it. Of course, u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq15_HTML.gif can be viewed as the restriction of u on I n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq16_HTML.gif, namely u n = u | I n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq17_HTML.gif. Then, taking advantage of the nonlinear model (1.1) for a single supported span, each u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq15_HTML.gif solves the following boundary value problem:
      { u n ′′′′ + ( p n u n L 2 ( I n ) 2 ) u n + k u n + = 0 , in  I n u n ( x n 1 ) = u n ( x n ) = u n ( x n 1 ) = u n ( x n ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ2_HTML.gif
      (1.2)
      Such solutions must fulfill the mutual groove condition
      u i ( x i ) = u i + 1 ( x i ) , i = 1 , 2 , , M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ3_HTML.gif
      (1.3)
      We finally observe that the total elongation of the span equals the sum of the elongations of each beam, namely
      u L 2 ( 0 , 1 ) 2 = n = 1 N u n L 2 ( I n ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Eque_HTML.gif
      Throughout the paper, we assume a uniform distribution of the axial load, that is, p n = p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq18_HTML.gif, n = 1 , 2 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq19_HTML.gif. In addition, we let
      k = κ π 4 , κ R + , p = b π 2 , b R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equf_HTML.gif

      Our aim is to scrutinize the existence of suitable buckled solutions for u, which can be obtained by joining buckled solutions for u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq15_HTML.gif on each sub-span, n = 1 , 2 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq19_HTML.gif. For later convenience, we denote such solutions by u ( N ) = ( u 1 , , u N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq20_HTML.gif. We prove that they exist provided that the lengths of the sub-spans are properly chosen.

      1.1 Early contributions

      In recent years, an increasing attention has been payed to the analysis of buckling, vibrations and post-buckling dynamics of nonlinear beam models (see, for instance, [2, 3]). As far as we know, most of the papers in the literature deal with approximations and numerical simulations, and only few works are able to derive exact solutions (see [47]).

      The investigation of solutions to BVP (1.1), in dependence on p, represents a classical nonlinear buckling problem in the literature on structural mechanics. The notion of buckling, introduced by Euler more than two centuries ago, describes a static instability of structures due to in-plane loading. In this respect, the main concern is to find the critical buckling loads, at which a bifurcation of solutions occurs, and their associated mode shapes, called postbuckling configurations. In the case k = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq21_HTML.gif, a careful analysis of the corresponding buckled stationary states was performed in [7] for all values of p in the presence of a source with a general shape (see also [8]). By replacing u + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq3_HTML.gif with u in (1.1), we obtain a simpler model which was scrutinized in [1, 5].

      In addition, it is worth noting that (1.1) represents the static counterpart of quite many different evolution problems arising both from elastic, viscoelastic and thermo-elastic theories. An example is the following quasilinear equation:
      t t u + x x x x u + ( p x u L 2 ( 0 , 1 ) 2 ) x x u + t u + k u + = f , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ4_HTML.gif
      (1.4)
      which is obtained by matching the modeling of the extensible elastic beam (see [9, 10]) with the well-known equation describing the motion of a damped suspension bridge (see [11, 12])
      t t u + x x x x u + t u + k u + = f . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ5_HTML.gif
      (1.5)

      Free and forced vibrations of (1.5) were recently scrutinized in [13, 14], whereas the long-time behavior of (1.4) was described in [15] for all values of p.

      Obviously, solutions to BVP (1.1) represent the steady states of a lot of models more general than (1.4), for instance, when either the rotational inertia (as in the Kirchhoff theory) or some kind of damping are taken into account. In particular, (1.1) works either when external viscous forces are added or when some structural dissipation phenomena occur in the deck, as in thermoelastic and viscoelastic beams (see, for instance, [1618]).

      When the geometric nonlinear term into (1.1) is disregarded, the existence of nontrivial (positive) solutions to the corresponding system were established in [19] by the variational method. Therein, some nonlinearly perturbed versions were also scrutinized, but the set of assumptions made there no longer holds when the full model is considered.

      1.2 Outline of the paper

      To the best of our knowledge, this is the first paper in the literature dealing with exact solutions to the doubly nonlinear BVPs (1.2), even for N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq22_HTML.gif. As is well known, the analysis of the corresponding set S κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq23_HTML.gif of their stationary solutions takes a great importance in the longterm dynamics of the corresponding evolution system, especially when its structure is nontrivial [20]. The main results of this paper concern the steady states analysis of a bridge with N = 1 , 2 , 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq24_HTML.gif sub-spans and are stated in Section 2. In Section 2.1 we scrutinize the case of a single span without piers ( N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq22_HTML.gif) and we prove that increasing the value of the lateral load p, first a negative u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq25_HTML.gif, then a positive u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq26_HTML.gif buckled solution appear at equilibrium. In Section 2.2 a bridge with a single pier ( N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq27_HTML.gif) is considered. When the position of the pier is allowed to be asymmetric ( | I 1 | | I 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq28_HTML.gif), we establish a condition on ε = | I 1 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq29_HTML.gif in order that buckled static solutions exist. In particular, we prove that ε 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq30_HTML.gif as k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq31_HTML.gif. Taking advantage of these results, the analysis of a bridge with two symmetrically-placed piers ( N = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq32_HTML.gif) is performed in Section 2.3, where buckled static solutions are proved to exist provided that | I 1 | = | I 3 | | I 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq33_HTML.gif fulfills a suitable condition. In Section 3 we deal with the general problem of a bridge with M = N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq5_HTML.gif piers, and we discuss separately the cases when the number M of the piers is either odd or even. All buckled solutions are determined in a closed form and belong to C 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq34_HTML.gif. Each of them is constructed by rescaling and suitably collecting positive and negative solutions, u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq26_HTML.gif and u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq25_HTML.gif. For any given N, a general explicit formula is established to compute the bifurcation values as a function of k.

      2 Stationary states I

      2.1 A single span without piers ( N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq22_HTML.gif)

      The set S κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq23_HTML.gif of the bridge equilibria in this case consists of all (weak) solutions u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq35_HTML.gif to the following boundary value problem on the interval [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq36_HTML.gif (see Eq. (1.1))
      { u ′′′′ + ( b π 2 u L 2 ( 0 , 1 ) 2 ) u + κ π 4 u + = 0 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ6_HTML.gif
      (2.1)

      It is worth noting that S κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq23_HTML.gif is bounded in H 2 ( 0 , 1 ) H 0 1 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq37_HTML.gif for all b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq38_HTML.gif, κ R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq39_HTML.gif (see [20]).

      When κ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq40_HTML.gif, a general result was established in [7] for a class of non-vanishing sources. In [4], the same strategy with minor modifications was applied to a problem close to (2.1), where the one-sided springs are replaced by unyielding ties. We summarize here the results concerning stationary solutions in the case κ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq40_HTML.gif.

      Theorem 2.1 (see [7], Th. 4.1)

      When κ vanishes, the set S 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq41_HTML.gif is finite for all b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq38_HTML.gif. Letting S b = { j N : b > j 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq42_HTML.gif and j = | S b | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq43_HTML.gif, S 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq41_HTML.gif has exactly 2 j + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq44_HTML.gif elements:
      u 0 = 0 and u j ± ( x ) = C j ± sin j π x , j S b , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equg_HTML.gif
      where
      C j ± = ± 1 j 2 ( b j 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equh_HTML.gif

      The general case is much more complicated. Since the scheme devised in [5, 7] does not work in the present situation, we obtain here a limited result.

      Theorem 2.2 (Existence of buckled solutions)

      When κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, besides the null solution u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq46_HTML.gif, which exists for all b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq38_HTML.gif, problem (2.1) admits

      • a negative buckled solution, u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq47_HTML.gif, if b > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq48_HTML.gif;

      • a positive buckled solution, u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq49_HTML.gif, if b > 1 + κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq50_HTML.gif.

      Proof For all values b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq38_HTML.gif, problem (2.1) has at least the trivial solution, u 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq51_HTML.gif. For all b > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq48_HTML.gif, besides the null solution, we obtain also the negative buckled solution
      u ( 1 ) ( x ) = A 1 sin ( π x ) , with  A 1 ( = C 1 ) = 2 ( b 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equi_HTML.gif
      which solves the Woinowsky-Krieger problem (see [21])
      { u ′′′′ + ( b π 2 u L 2 ( 0 , 1 ) 2 ) u = 0 , in  [ 0 , 1 ] , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ7_HTML.gif
      (2.2)
      If b 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq52_HTML.gif, there exists only the null solution. Therefore, b = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq53_HTML.gif will be referred to as the smallest bifurcation value. By paralleling [5], we infer that when b > 1 + κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq54_HTML.gif, system (2.1) admits the positive buckled solution
      u ( 1 ) + ( x ) = A 1 + sin ( π x ) , with  A 1 + = 2 ( b 1 κ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equj_HTML.gif
      Since κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, we can easily check that u ( 1 ) + ( x ) < | u ( 1 ) ( x ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq55_HTML.gif for all x ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq56_HTML.gif (see Figure 2).  □
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Fig2_HTML.jpg
      Figure 2

      The buckled solutions u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq57_HTML.gif (a) and u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq58_HTML.gif (b) when κ = 0.1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq59_HTML.gif , b = 1.125 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq60_HTML.gif .

      2.2 A bridge with a single pier ( N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq27_HTML.gif)

      Now we consider a bridge with a single pier at x = x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq61_HTML.gif and two sub-spans. The whole solution u ( 2 ) = ( u 1 , u 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq62_HTML.gif is obtained by joining their deflections, u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq63_HTML.gif and u 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq64_HTML.gif, which solve problems (1.2) for n = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq65_HTML.gif and fulfill the mutual groove condition
      u 1 ( x 1 ) = u 2 ( x 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ8_HTML.gif
      (2.3)

      When | I 1 | = | I 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq66_HTML.gif, it is easy to check that this condition cannot be satisfied by any buckled solution. Then we choose | I 1 | | I 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq28_HTML.gif. Let x 1 = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq67_HTML.gif be the point at which the pier is located, so that | I 1 | = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq68_HTML.gif. For the sake of definiteness, let u 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq69_HTML.gif and u 2 < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq70_HTML.gif.

      Theorem 2.3 When κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, problems (1.2) for N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq27_HTML.gif admit two buckled solutions, called u ( 2 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq71_HTML.gif and u ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq72_HTML.gif, provided that ε = ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq73_HTML.gif and b > b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq74_HTML.gif, where the values of ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif and b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq76_HTML.gif are defined in (2.9) and (2.10), respectively.

      Proof In order to establish the form of u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq63_HTML.gif and u 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq64_HTML.gif, we rescale the domains [ 0 , ε ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq77_HTML.gif and [ ε , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq78_HTML.gif by virtue of the transformations x = ε y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq79_HTML.gif and x = ( 1 ε ) z + ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq80_HTML.gif, respectively. Hence, we get the problem
      { w ′′′′ + ε 2 ( b π 2 1 ε 2 w L 2 ( 0 , 1 ) 2 ) w + ε 4 κ π 4 w = 0 , in  [ 0 , 1 ] , v ′′′′ + ( 1 ε ) 2 ( b π 2 1 ( 1 ε ) 2 v L 2 ( 0 , 1 ) 2 ) v = 0 , in  [ 0 , 1 ] , w ( 0 ) = w ( 1 ) = w ( 0 ) = w ( 1 ) = 0 , v ( 0 ) = v ( 1 ) = v ( 0 ) = v ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ9_HTML.gif
      (2.4)

      where w ( y ) = u 1 ( ε y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq81_HTML.gif and v ( z ) = u 2 ( [ 1 ε ] z + ε ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq82_HTML.gif. Obviously, the null solution v = w = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq83_HTML.gif exists for all b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq38_HTML.gif. On the contrary, nontrivial solutions occur under special conditions.

      Arguing as in Section 2.1, we exploit the explicit expression of solutions u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq26_HTML.gif and u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq25_HTML.gif, respectively, and we obtain
      w ( y ) = A 2 + ( ε ) sin ( π y ) and v ( z ) = A 2 ( ε ) sin ( π z ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equk_HTML.gif
      where, for some ε κ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq84_HTML.gif,
      A 2 + ( ε κ ) = ε κ 2 [ b 1 ε κ 2 κ ε κ 2 ] , A 2 ( ε κ ) = ( ε κ 1 ) 2 [ b 1 ( 1 ε κ ) 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equl_HTML.gif
      Of course, such solutions both exist provided that
      b > b κ = max { 1 ( 1 ε κ ) 2 , 1 + κ ε κ 4 ε κ 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equm_HTML.gif
      and ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif satisfies the continuity condition (2.3), namely
      w ( 1 ) ε κ = v ( 0 ) 1 ε κ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ10_HTML.gif
      (2.5)
      Then, by joining w and v, we obtain the whole solution
      u ( 2 ) + ( x ) = { A 2 + ( ε κ ) sin ( π x / ε κ ) , x [ 0 , ε κ ) , 0 , x = ε κ , A 2 ( ε κ ) sin ( π ( x ε κ ) / ( 1 ε κ ) ) , x ( ε κ , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ11_HTML.gif
      (2.6)
      The function u ( 2 ) = ( u 2 , u 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq85_HTML.gif, which is obtained by symmetrizing u ( 2 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq86_HTML.gif with respect to the vertical line x = 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq87_HTML.gif, solves (1.2) under the same conditions (see Figure 3). Its expression is given by replacing ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif with 1 ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq88_HTML.gif and A 2 + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq89_HTML.gif with A 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq90_HTML.gif into (2.6),
      u ( 2 ) ( x ) = { A 2 ( ε κ ) sin ( π x / ( 1 ε κ ) ) , x [ 0 , 1 ε κ ) , 0 , x = 1 ε κ , A 2 + ( ε κ ) sin ( π ( x 1 + ε κ ) / ε κ ) , x ( 1 ε κ , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ12_HTML.gif
      (2.7)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Fig3_HTML.jpg
      Figure 3

      The buckled solutions u ( 2 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq91_HTML.gif (a) and u ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq92_HTML.gif (b) when κ = 5.9 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq93_HTML.gif , b = 5.75 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq94_HTML.gif .

      Computation of ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif and b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq76_HTML.gif. For each κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, in order to compute the unknown value ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif, which ensures the existence of buckled solutions, we exploit the continuity condition (2.5), namely
      ( 1 ε κ ) A 2 + ( ε κ ) + ε κ A 2 ( ε κ ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equn_HTML.gif
      Thus, the required value ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif has to satisfy the system
      { 1 ε κ 2 + κ ε κ 2 = 1 ( 1 ε κ ) 2 , 0 < ε κ < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ13_HTML.gif
      (2.8)
      which implies
      κ = 2 ε κ 1 ε κ 4 ( 1 ε κ ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ14_HTML.gif
      (2.9)
      It can be easily checked (see Figure 4) that ε κ ( 1 / 2 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq95_HTML.gif and
      lim κ 0 ε κ ε 0 = 1 2 , lim κ + ε κ = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equo_HTML.gif
      In order to compute the critical value of the axial force, b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq76_HTML.gif, we observe that by virtue of (2.9), we have
      b κ = 1 ( 1 ε κ ) 2 = 1 + κ ε κ 4 ε κ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ15_HTML.gif
      (2.10)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Fig4_HTML.jpg
      Figure 4

      The value of ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq96_HTML.gif as a function of κ .

      Summarizing, for any given κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, if b b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq97_HTML.gif no buckled solution exists, if b > b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq98_HTML.gif there exists a unique value of ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif, which produces two buckled solutions. As a consequence, b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq76_HTML.gif represents the bifurcation value. According to (2.10), we get
      A 2 + ( ε κ ) = ε κ 2 ( b b κ ) , A 2 ( ε κ ) = ( ε κ 1 ) 2 ( b b κ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equp_HTML.gif
      and A 2 < A 2 + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq99_HTML.gif. In addition, for all κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, we can easily obtain
      b κ > b 0 = 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equq_HTML.gif

      Then, when b 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq100_HTML.gif, the set S κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq23_HTML.gif contains only the trivial solution u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq46_HTML.gif. □

      In the limit κ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq101_HTML.gif, it follows that
      A 2 , A 2 + ( b 4 ) / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equr_HTML.gif

      This means that solutions u ( 2 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq86_HTML.gif and u ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq102_HTML.gif tend to coincide with the second bifurcation branch of problem (2.2), as expected.

      2.3 A bridge with two piers ( N = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq32_HTML.gif)

      When the bridge has two intermediate piers and three sub-spans, we shall construct solutions u ( 3 ) = ( u 1 , u 2 , u 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq103_HTML.gif, where u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq63_HTML.gif, u 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq64_HTML.gif and u 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq104_HTML.gif solve (1.2). It is easy to check that no buckled solution exists when | I 1 | = | I 2 | = | I 3 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq105_HTML.gif. Then we choose | I 1 | = | I 3 | | I 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq33_HTML.gif and construct a buckled solution by joining three (suitably rescaled) functions which have the form of either u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq26_HTML.gif or u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq25_HTML.gif.

      Theorem 2.4 When κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, problems (1.2) for N = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq32_HTML.gif admit two buckled solutions, u ( 3 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq106_HTML.gif and u ( 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq107_HTML.gif, provided that b > c κ + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq108_HTML.gif and b > c κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq109_HTML.gif, respectively, where the values of c κ + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq110_HTML.gif and c κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq111_HTML.gif are defined in (2.12) and (2.15).

      Proof The buckled solutions are constructed as follows. The former, u ( 3 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq112_HTML.gif, has two positive components with the same shape on I 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq113_HTML.gif and I 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq114_HTML.gif, and a negative one on I 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq115_HTML.gif. The latter, u ( 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq116_HTML.gif, is negative in I 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq113_HTML.gif and I 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq114_HTML.gif, but positive in I 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq115_HTML.gif. Accordingly, let
      w j = u ( 3 ) + | I 2 j 1 , j = 1 , 2 and v 0 = u ( 3 ) + | I 2 , v j = u ( 3 ) | I 2 j 1 , j = 1 , 2 and w 0 = u ( 3 ) | I 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equs_HTML.gif

      so that w k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq117_HTML.gif and v k < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq118_HTML.gif for k = 0 , 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq119_HTML.gif.

      In order to construct u ( 3 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq112_HTML.gif, we exploit the same argument as in Section 2.2. First, we argue that
      | I 1 | = | I 3 | = ξ , | I 2 | = 1 2 ξ , ξ ( 0 , 1 / 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equt_HTML.gif
      Then, we apply the rescaling procedure with the scale factor ξ / ( 1 ξ ) = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq120_HTML.gif and, by virtue of the transformations x = ξ y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq121_HTML.gif and x = ( 1 2 ξ ) z + ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq122_HTML.gif, we obtain
      { w j ′′′′ + ξ 2 ( b π 2 1 ξ 2 w j L 2 ( 0 , 1 ) 2 ) w j + κ ξ 4 π 4 w j = 0 , in  [ 0 , 1 ] , j = 1 , 2 , v 0 ′′′′ + ( 1 2 ξ ) 2 ( b π 2 1 ( 1 2 ξ ) 2 v 0 L 2 ( 0 , 1 ) 2 ) v 0 = 0 , in  [ 0 , 1 ] , w j ( 0 ) = w j ( 1 ) = w j ( 0 ) = w j ( 1 ) = 0 , j = 1 , 2 , v 0 ( 0 ) = v 0 ( 1 ) = v 0 ( 0 ) = v 0 ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ16_HTML.gif
      (2.11)
      Taking into account that ε = ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq73_HTML.gif is uniquely and implicitly given by (2.9), the corresponding value of ξ becomes
      ξ κ = ε κ 1 + ε κ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equu_HTML.gif
      so that
      lim κ 0 ξ κ = ξ 0 = 1 3 , lim κ + ξ κ = 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equv_HTML.gif
      Accordingly, we obtain
      w j ( y ) = A 3 + ( ξ κ ) sin ( π y ) , j = 1 , 2 , v 0 ( z ) = A 3 ( ξ κ ) sin ( π z ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equw_HTML.gif
      where
      A 3 + ( ξ κ ) = ξ κ 2 [ b 1 ξ κ 2 κ ξ κ 2 ] , A 3 ( ξ κ ) = ( 2 ξ κ 1 ) 2 [ b 1 ( 1 2 ξ κ ) 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equx_HTML.gif
      Of course, such solutions exist at the same time, provided that b > c κ + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq123_HTML.gif, where
      c κ + = max { 1 ( 1 2 ξ κ ) 2 , 1 + κ ξ κ 4 ξ κ 2 } = max { ( 1 + ε κ ) 2 ( 1 ε κ ) 2 , ( 1 + ε κ ) 4 + κ ε κ 4 ε κ 2 ( 1 + ε κ ) 2 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equy_HTML.gif
      where c κ + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq124_HTML.gif represents the bifurcation value. By virtue of (2.9), we get
      c κ + = ( 1 + ε κ ) 2 ( 1 ε κ ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ17_HTML.gif
      (2.12)
      If this is the case, we havethe whole solution (see Figure 5a)
      u ( 3 ) + ( x ) = { A 3 + ( ξ κ ) sin ( π x / ξ κ ) , x [ 0 , ξ κ ] , A 3 ( ξ κ ) sin ( π ( x ξ κ ) / ( 1 2 ξ κ ) ) , x [ ξ κ , 1 ξ κ ] , A 3 + ( ξ κ ) sin ( π ( x 1 + ξ κ ) / ξ κ ) , x [ 1 ξ κ , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ18_HTML.gif
      (2.13)
      In order to construct u ( 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq116_HTML.gif, we proceed as before by letting
      | I 1 | = | I 3 | = η , | I 2 | = 1 2 η , η ( 0 , 1 / 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equz_HTML.gif
      By means of a rescaling procedure with the scale factor η / ( 1 η ) = 1 ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq125_HTML.gif, the transformations x = η z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq126_HTML.gif and x = ( 1 2 η ) y + η http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq127_HTML.gif lead to
      { v j ′′′′ + η 2 ( b π 2 1 η 2 v j L 2 ( 0 , 1 ) 2 ) v j = 0 , in  [ 0 , 1 ] , j = 1 , 2 , w 0 ′′′′ + ( 1 2 η ) 2 ( b π 2 1 ( 1 2 η ) 2 w 0 L 2 ( 0 , 1 ) 2 ) w 0 + κ ( 1 2 η ) 4 π 4 w 0 = 0 , in  [ 0 , 1 ] , v j ( 0 ) = v j ( 1 ) = v j ( 0 ) = v j ( 1 ) = 0 , j = 1 , 2 , w 0 ( 0 ) = w 0 ( 1 ) = w 0 ( 0 ) = w 0 ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ19_HTML.gif
      (2.14)
      Accordingly, we find
      η κ = 1 ε κ 2 ε κ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equaa_HTML.gif
      where ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif is given by (2.9), and
      lim κ 0 η κ = η 0 = 1 3 , lim κ + η κ = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equab_HTML.gif
      So, we obtain
      v j ( z ) = A 3 ( η κ ) sin ( π z ) , j = 1 , 2 , w 0 ( y ) = A 3 + ( η κ ) sin ( π y ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equac_HTML.gif
      where
      A 3 ( η κ ) = η κ 2 [ b 1 η κ 2 ] , A 3 + ( η κ ) = ( 1 2 η κ ) 2 [ b 1 ( 1 2 η κ ) 2 κ ( 1 2 η κ ) 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equad_HTML.gif
      Of course, such solutions do exist together provided that
      b > c κ = max { 1 η κ 2 , 1 + κ ( 1 2 η κ ) 4 ( 1 2 η κ ) 2 } = max { ( 2 ε κ ) 2 ( 1 ε κ ) 2 , ( 2 ε κ ) 4 + κ ε κ 4 ε κ 2 ( 2 ε κ ) 2 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equae_HTML.gif
      where c κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq128_HTML.gif represents the bifurcation value. By virtue of (2.9), we get
      c κ = ( 2 ε κ ) 2 ( 1 ε κ ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ20_HTML.gif
      (2.15)
      By means of (2.9) it is easy to check that c κ < c κ + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq129_HTML.gif. The complete solution is given by (see Figure 5b)
      u ( 3 ) ( x ) = { A 3 ( η κ ) sin ( π x / η κ ) , x [ 0 , η κ ] , A 3 + ( η κ ) sin ( π ( x η κ ) / ( 1 2 η κ ) ) , x [ η κ , 1 η κ ] , A 3 ( η κ ) sin ( π ( x 1 + η κ ) / η κ ) , x [ 1 η κ , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ21_HTML.gif
      (2.16)
       □
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Fig5_HTML.jpg
      Figure 5

      The buckled solutions u ( 3 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq130_HTML.gif (a) and u ( 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq131_HTML.gif (b) when κ = 0.1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq59_HTML.gif and b = 1.225 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq132_HTML.gif .

      3 Stationary states II

      In this section we generalize the problem to a bridge with N sub-spans and M = N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq5_HTML.gif piers. The existence of buckled solutions is investigated in connection with the length of the sub-spans. Indeed, it is easy to check that no buckled solution exists when all of them are of the same length. Then a buckled solution may be obtained by collecting and joining N (suitably rescaled) functions of the same form as either u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq26_HTML.gif or u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq25_HTML.gif. To this end, we are forced to consider separately the cases when the number M of the piers is either odd or even. In the former case, indeed, we adopt a strategy which is close to that applied in Section 2.2. In the latter, we iterate the procedure devised in Section 2.3.

      Theorem 3.1 For any κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif, N N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq133_HTML.gif, (1.2) admits two buckled solutions.

      • In the odd case, M = 2 m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq134_HTML.gif, there exist u ( 2 m ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq135_HTML.gifand u ( 2 m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq136_HTML.gifprovided that b > b κ ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq137_HTML.gif, where the value of b κ ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq138_HTML.gifis characterized in (3.5).

      • In the even case, M = 2 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq139_HTML.gif, there exist u ( 2 m + 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq140_HTML.gifand u ( 2 m + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq141_HTML.gifprovided that b > c κ + ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq142_HTML.gifand b > c κ ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq143_HTML.gif, respectively, where the values of c κ + ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq144_HTML.gifand c κ ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq145_HTML.gifare characterized in (3.6).

      Proof The odd case. Solutions u ( 2 m ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq146_HTML.gif and u ( 2 m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq147_HTML.gif, m = ( M + 1 ) / 2 = N / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq148_HTML.gif, are assumed to change the sign alternately on the sub-spans. The superscript + (−) means that the solution is positive (negative) on I 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq113_HTML.gif. In order to construct them, we join 2m rescaled functions of the same form as either u ( 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq26_HTML.gif or u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq25_HTML.gif. Arguing as in the step N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq27_HTML.gif, we let
      | I 2 j 1 | = ε κ m , j = 1 , , m , | I 2 h + 2 | = 1 ε κ m , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equaf_HTML.gif

      where ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif is given by (2.9).

      Construction of u ( 2 m ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq146_HTML.gif. Let
      w j = u ( 2 m ) + | I 2 j 1 , j = 1 , , m , v h = u ( 2 m ) + | I 2 h + 2 , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equag_HTML.gif
      so that w j > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq149_HTML.gif and v h < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq150_HTML.gif. Since
      | I 2 j 1 I 2 j | = 1 m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equah_HTML.gif
      we stress that each restriction
      u ( 2 m ) + | I 2 j 1 I 2 j , j = 1 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equai_HTML.gif
      is similar to u ( 2 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq71_HTML.gif, rescaled by a factor 1 / m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq151_HTML.gif. In particular,
      I 2 j 1 = ( j 1 m , j 1 + ε κ m ) , j = 1 , , m , I 2 h + 2 = ( h + ε κ m , h + 1 m ) , h = 0 , , m 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equaj_HTML.gif
      Therefore, on I 2 j 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq152_HTML.gif we have
      w j ( x ) = A 2 m + sin ( π ( x j 1 m ) / ε κ m ) , j = 1 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equak_HTML.gif
      while on I 2 h + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq153_HTML.gif we obtain
      v h ( x ) = A 2 m sin ( π ( x h + ε κ m ) / 1 ε κ m ) , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equal_HTML.gif
      where
      A 2 m + = ε κ m 2 [ b m 2 ε κ 2 κ ε κ 2 m 2 ] , A 2 m = ε κ 1 m 2 [ b m 2 ( 1 ε κ ) 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ22_HTML.gif
      (3.1)
      Such solutions exist provided that
      b > b κ ( m ) = max { m 2 ε κ 2 + κ ε κ 2 m 2 , m 2 ( 1 ε κ ) 2 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ23_HTML.gif
      (3.2)
      By virtue of Lemma 3.2, which will be proved later, we get
      b κ ( m ) = m 2 b κ , m = N / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equam_HTML.gif
      Construction of u ( 2 m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq147_HTML.gif. Let
      v j = u ( 2 m ) | I 2 j 1 , j = 1 , , m , w h = u ( 2 m ) | I 2 h + 2 , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equan_HTML.gif
      so that w h > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq154_HTML.gif and v j < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq155_HTML.gif. Arguing as before, we obtain
      | I 2 j 1 | = 1 ε κ m , j = 1 , , m , | I 2 h + 2 | = ε κ m , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equao_HTML.gif
      where ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif is given by (2.9) and
      u ( 2 m ) | I 2 j 1 I 2 j , j = 1 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equap_HTML.gif
      is similar to u ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq72_HTML.gif, rescaled by a factor 1 / m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq151_HTML.gif. In particular,
      I 2 j 1 = ( j 1 m , j ε κ m ) , j = 1 , , m , I 2 h + 2 = ( h + 1 ε κ m , h + 1 m ) , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equaq_HTML.gif
      so on I 2 j 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq152_HTML.gif we have
      v j ( x ) = A 2 m sin ( π ( x j 1 m ) / 1 ε κ m ) , j = 1 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equar_HTML.gif
      while on I 2 h + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq153_HTML.gif we obtain
      w h ( x ) = A 2 m + sin ( π ( x h + 1 ε κ m ) / ε κ m ) , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equas_HTML.gif

      with A 2 m + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq156_HTML.gif, A 2 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq157_HTML.gif defined as in (3.1).

      The even case. In this case, we construct the solutions u ( 2 m + 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq158_HTML.gif and u ( 2 m + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq159_HTML.gif, where m = M / 2 = ( N 1 ) / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq160_HTML.gif.

      Construction of u ( 2 m + 1 ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq158_HTML.gif. Let
      w j = u ( 2 m + 1 ) + | I 2 j 1 , j = 1 , , m + 1 , v h = u ( 2 m + 1 ) + | I 2 h + 2 , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equat_HTML.gif
      so that w j > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq149_HTML.gif and v h < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq150_HTML.gif. Arguing as in the step N = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq32_HTML.gif, we obtain
      | I 2 j 1 | = ξ κ , j = 1 , , m + 1 , | I 2 h + 2 | = 1 ( 1 + m ) ξ κ m , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equau_HTML.gif
      where ξ κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq161_HTML.gif is given by ξ κ = ε κ m + ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq162_HTML.gif. Therefore
      | I 2 j 1 | = ε κ m + ε κ , | I 2 h + 2 | = 1 ε κ m + ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equav_HTML.gif
      and
      I 2 j 1 = ( j 1 m + ε κ , j 1 + ε κ m + ε κ ) , j = 1 , , m + 1 , I 2 h + 2 = ( h + ε κ m + ε κ , h + 1 m + ε κ ) , h = 0 , , m 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equaw_HTML.gif
      So, on I 2 j 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq152_HTML.gif we have
      w j ( x ) = A 2 m + 1 + sin ( π ( x j 1 m + ε κ ) / ε κ m + ε κ ) , j = 1 , , m + 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equax_HTML.gif
      while on I 2 h + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq153_HTML.gif we obtain
      v h ( x ) = A 2 m + 1 sin ( π ( x h + ε κ m + ε κ ) / 1 ε κ m + ε κ ) , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equay_HTML.gif
      where
      A 2 m + 1 + = ε κ m + ε κ 2 [ b ( m + ε κ ) 2 ε κ 2 κ ε κ 2 ( m + ε κ ) 2 ] , A 2 m + 1 = ε κ 1 m + ε κ 2 [ b ( m + ε κ ) 2 ( 1 ε κ ) 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equaz_HTML.gif
      Such solutions exist provided that
      b > c κ + ( m ) = max { ( m + ε κ ) 2 ε κ 2 + κ ε κ 2 ( m + ε κ ) 2 , ( m + ε κ ) 2 ( 1 ε κ ) 2 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ24_HTML.gif
      (3.3)
      By virtue of Lemma 3.2, we get
      c κ + ( m ) = ( m + ε κ ) 2 ( 1 ε κ ) 2 , m = ( N 1 ) / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equba_HTML.gif
      Construction of u ( 2 m + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq159_HTML.gif. Let
      v j = u ( 2 m + 1 ) | I 2 j 1 , j = 1 , , m + 1 , w h = u ( 2 m + 1 ) | I 2 h + 2 , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbb_HTML.gif
      so that w h > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq154_HTML.gif and v j < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq155_HTML.gif. Arguing as in the step N = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq32_HTML.gif, we obtain
      | I 2 j 1 | = η κ , j = 1 , , m + 1 , | I 2 h + 2 | = 1 ( 1 + m ) η κ m , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbc_HTML.gif
      where η κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq163_HTML.gif is given by η κ = 1 ε κ m + 1 ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq164_HTML.gif. Therefore
      | I 2 j 1 | = 1 ε κ m + 1 ε κ , | I 2 h + 2 | = ε κ m + 1 ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbd_HTML.gif
      and
      I 2 j 1 = ( j 1 m + 1 ε κ , j ε κ m + 1 ε κ ) , j = 1 , , m + 1 , I 2 h + 2 = ( h + 1 ε κ m + 1 ε κ , h + 1 m + 1 ε κ ) , h = 0 , , m 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Eqube_HTML.gif
      As a consequence, on I 2 j 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq152_HTML.gif we have
      v j ( x ) = B 2 m + 1 sin ( π ( x j 1 m + 1 ε κ ) / 1 ε κ m + 1 ε κ ) , j = 1 , , m + 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbf_HTML.gif
      while on I 2 h + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq153_HTML.gif we obtain
      w h ( x ) = B 2 m + 1 + sin ( π ( x h + 1 ε κ m + 1 ε κ ) / ε κ m + 1 ε κ ) , h = 0 , , m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbg_HTML.gif
      where
      B 2 m + 1 + = ε κ m + 1 ε κ 2 [ b ( m + 1 ε κ ) 2 ε κ 2 κ ε κ 2 ( m + 1 ε κ ) 2 ] , B 2 m + 1 = ε κ 1 m + 1 ε κ 2 [ b ( m + 1 ε κ ) 2 ( 1 ε κ ) 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbh_HTML.gif
      Such solutions exist provided that b > c κ ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq165_HTML.gif, where
      c κ ( m ) = max { ( m + 1 ε κ ) 2 ε κ 2 + κ ε κ 2 ( m + 1 ε κ ) 2 , ( m + 1 ε κ ) 2 ( 1 ε κ ) 2 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ25_HTML.gif
      (3.4)
      By virtue of Lemma 3.2, we get
      c κ ( m ) = ( m + 1 ε κ ) 2 ( 1 ε κ ) 2 , m = ( N 1 ) / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbi_HTML.gif

       □

      Lemma 3.2 (Characterization of the bifurcation values)

      For any given κ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq45_HTML.gif and m N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq166_HTML.gif,
      b κ ( m ) = m 2 ( 1 ε κ ) 2 = m 2 b κ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ26_HTML.gif
      (3.5)
      c κ ( m ) = ( m + 1 ε κ ) 2 ( 1 ε κ ) 2 < c κ + ( m ) = ( m + ε κ ) 2 ( 1 ε κ ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equ27_HTML.gif
      (3.6)

      where ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif and b κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq76_HTML.gif are computed by (2.9) and (2.10), respectively.

      Proof First, in view of (3.2), we have to prove that
      m 2 ( 1 ε κ ) 2 m 4 + κ ε κ 4 ε κ 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbj_HTML.gif
      which is equivalent to
      κ m 4 2 ε κ 1 ε κ 4 ( 1 ε κ ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbk_HTML.gif
      By replacing the expression of κ given by (2.9), we obtain
      1 m 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbl_HTML.gif
      which implies that (3.5) holds for all m N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq166_HTML.gif. Then, in view of (3.4), we have to prove
      ( m + 1 ε κ ) 2 ( 1 ε κ ) 2 ( m + 1 ε κ ) 4 + κ ε κ 4 ε κ ( m + 1 ε κ ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbm_HTML.gif
      which is equivalent to
      κ ( 2 ε κ 1 ) ( m + 1 ε κ ) 4 ε κ 4 ( 1 ε κ ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbn_HTML.gif
      By virtue of (2.9), we obtain
      1 ( m + 1 ε κ ) 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbo_HTML.gif
      which is identically satisfied for all admissible values ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif and for all m N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq166_HTML.gif. Hence, (3.6)1 holds. Finally, in view of (3.3), we need to prove
      ( m + ε κ ) 2 ( 1 ε κ ) 2 ( m + ε κ ) 4 + κ ε κ 4 ε κ ( m + ε κ ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbp_HTML.gif
      and this is equivalent to
      κ ( 2 ε κ 1 ) ( m + ε κ ) 4 ε κ 4 ( 1 ε κ ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbq_HTML.gif
      Applying (2.9) as in the previous cases, we obtain 1 ( m + ε κ ) 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq167_HTML.gif, which is identically satisfied for all admissible values ε κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq75_HTML.gif and for all m N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq166_HTML.gif so that (3.6)2 follows. Moreover, starting from (3.6) and taking into account that 1 / 2 < ε κ < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq168_HTML.gif, it is easy to check that
      c κ ( m ) < c κ + ( m ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbr_HTML.gif

       □

      Previous results can be summarized by means of a simple sketch which highlights the main bifurcation features (see Figure 6).
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Fig6_HTML.jpg
      Figure 6

      The bifurcation portrait: for N odd (on the left, m = N / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq169_HTML.gif ) and N even (on the right, m = ( N 1 ) / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq170_HTML.gif ).

      Finally, it is worth noting that by removing the coupling between the road-bed and the cable, we recover well-known results (see, for instance, [5, 8])

      Remark 3.3 In the limit κ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq101_HTML.gif, from (2.9) it follows ε κ 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_IEq171_HTML.gif, so that buckled solutions exist even if all sub-spans are equal. Moreover,
      b κ ( m ) ( 2 m ) 2 = N 2 , c κ ( m ) , c κ + ( m ) ( 2 m + 1 ) 2 = N 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-204/MediaObjects/13661_2013_Article_603_Equbs_HTML.gif

      Declarations

      Authors’ Affiliations

      (1)
      DICATAM, Università degli Studi di Brescia

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