Infinitely many solutions for a fourth-order differential equation on a nonlinear elastic foundation

Boundary Value Problems, Nov 2013

In this paper, existence results of infinitely many solutions for a fourth-order differential equation with nonlinear boundary conditions are established. The proof is based on variational methods. Some recent results are improved and extended.

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Infinitely many solutions for a fourth-order differential equation on a nonlinear elastic foundation

Boundary Value Problems Infinitely many solutions for a fourth-order differential equation on a nonlinear elastic foundation Xiaodan Wang In this paper, existence results of infinitely many solutions for a fourth-order differential equation with nonlinear boundary conditions are established. The proof is based on variational methods. Some recent results are improved and extended. ⎨⎧⎪ u() = f (x; u);  < x < ; u() = u () = ; ⎪⎩ u () = ; u () = g(u()) - 1 Introduction In this paper, we consider a beam equation with nonlinear boundary conditions of the type where f ∈ C([, ], R) and g ∈ C(R) are real functions. This kind of problem arises in the study of deflections of elastic beams on nonlinear elastic foundations. The problem has the following physical description: a thin flexible elastic beam of length  is clamped at its left end x =  and resting on an elastic device at its right end x = , which is given by g. Then, the problem models the static equilibrium of the beam under a load, along its length, characterized by f . The derivation of the model can be found in [, ]. Owing to the importance of fourth-order two-point boundary value problems in describing a large class of elastic deflection, there is a wide literature that deals with the existence and multiplicity results for such a problem with different boundary conditions (see, for instance, [–] and the references therein). Motivated by the above works, in the present paper we study the existence of infinitely many solutions for problem (.) when the nonlinear term f (x, u) satisfies the superlinear condition and sublinear condition at the infinity on u, respectively. As far as we know, this case has never before been considered. Now we state our main results. 1.1 The superlinear case We give the following assumptions. (H) g is odd and satisfies  s  g(t) dt – g(s)s ≥ , g(t) dt ≥  for all s ∈ R.  s (H) There exist constants a, b ≥  and γ ∈ [, ) such that g(s) ≤ a + b|s|γ for s ∈ R. (H) lim|u|→+∞ F(ux,u) = +∞ uniformly for x ∈ [, ], where F(x, u) = u f (x, t) dt. (H) F(x, ) ≡ ,  ≤ F(x, u) = o(|u|) as |u| →  uniformly for x ∈ [, ]. (H) There exist constants α > ,  < β <  + αα– , c, c >  and L >  such that for every x ∈ [, ] and u ∈ R with |u| ≥ L, f (x, u)u – F(x, u) ≥ c|u|α, f (x, u) ≤ c|u|β . Theorem . Assume that (H)-(H) hold and F is even in u. Then problem (.) has infinitely many solutions. Remark . There exist some functions satisfying (H)-(H), but not satisfying the wellknown (AR)-condition,  < θ F(x, u) ≤ f (x, u)u, ∀u > , x ∈ [, ], for some θ > . For example, take f (x, u) = u ln( + u) + +uu . Then F(x, u) = |u| ln( + |u|). Obviously, (H)-(H) are satisfied. Note that f (x, u)u – F(x, u) = |u| ln  + |u| |u|  + |u| ≥ |u| ln , ∀|u| ≥ , and f (x, u) ≤  ln  + |u| |u| + +|u| u||  ln  + |u| |u| ≤ |u|  , ∀|u| ≥ L, for L being large enough, which implies (H). However, it is easy to see that f does not satisfy (AR)-condition. 1.2 The sublinear case We make the following assumptions. (S) g is odd and satisfies g(s)s ≥  for any s ∈ R. (S) There exist constants b >  and γ ∈ [, ) such that g(s) ≤ b|s|γ for s ∈ R. (S) F(x, ) ≡  for any x ∈ [, ]. (S) There are constants k >  and ζ ∈ [, ) with ζ < γ +  such that F(x, u) ≥ k|u|ζ for any (x, u) ∈ [, ] × R. (S) There exist constants k >  and ζ ∈ [, ) such that f (x, u) ≤ k|u|ζ– for any (x, u) ∈ [, ] × R. Theorem . Assume that (S)-(S) hold and F is even in u. Then problem (.) has infinitely many solutions. Remark . The condition (S) implies that s g(t) dt ≥ . The remainder of this paper is organized as follows. In Section , some preliminary results are presented. In Section , we give the proofs of our main results. 2 Variational setting and preliminaries In this section, the following two theorems will be needed in our argument. Let E be a Banach space with the norm · and E = j∈N Xj with dim Xj < ∞ for any j ∈ N. Set Yk = jk= Xj, Zk = j∞=k Xj and Bk = {u ∈ Yk : u ≤ ρk}, Nk = {u ∈ Zk : u = rk} for ρk > rk > . Consider the C-functional λ : E → R defined by λ(u) = A(u) – λB(u), λ ∈ [, ]. Assume that: (C) λ maps bounded sets to bounded sets uniformly for λ ∈ [, ]. Furthermore, λ(–u) = λ(u) for all (λ, u) ∈ [, ] × E. (C) B(u) ≥  for all u ∈ E; A(u) → ∞ or B(u) → ∞ as u → ∞; or (C) B(u) ≤  for all u ∈ E; B(u) → –∞ as u → ∞. For k ≥ , define k := {γ ∈ C(Bk, E) : γ is odd; γ |∂Bk = id}, ck(λ) := infγ ∈ k maxu∈Bk λ(γ (u)), bk(λ) := infu∈Zk, u =rk λ(u), ak(λ) := maxu∈Yk, u =ρk λ(u). Theorem . ([, Theorem .]) Assume that (C) and (C) (or (C) ) hold. If bk(λ) > ak(λ) for all λ ∈ [, ], then ck(λ) ≥ bk(λ) for all λ ∈ [, ]. Moreover, for a.e. λ ∈ [, ], there exists a sequence {ukn(λ)}n∞= such that supn ukn(λ) < ∞, λ(ukn(λ)) →  and λ(ukn(λ)) → ck(λ) as n → ∞. Theorem . ([, Theorem .]) Suppose that (C) holds. Furthermore, we assume that the following conditions hold: (D) B(u) ≥ ; B(u) → ∞ as u → ∞ on any finite dimensional subspace of E. (D) There exist ρk > rk >  such that ak(λ) := infu∈Zk, u =ρk λ(u) ≥  > bk(λ) := maxu∈Yk, u =rk λ(u) for all λ ∈ [, ] and dk(λ) := infu∈Zk, u ≤ρk λ(u) →  as k → ∞ uniformly for λ ∈ [, ]. Then there exist λn → , u(λn) ∈ Yn such that λn |Yn (u(λn)) = , λn (u(λn)) → ck ∈ [dk(), bk()] as n → ∞. In particular, if {u(λn)} has a convergent subsequence for every k, then  has infinitely many nontrivial critical points {uk} ⊂ E \ {} satisfying (uk) → – as k → ∞. Now we begin describing the variational formulation of problem (.), which is based on the function space E = u ∈ H(, ); u() = u () =  , where H(, ) is the Sobolev space of all functions u : [, ] → R such that u and its distributional derivative u are absolutely continuous and u belongs to L(, ). Then E is a Hilbert space equipped with the inner product and the norm u, v = u (x)v (x) dx, u = u , where · p denotes the standard Lp norm. In addition, E is compactly embedded in the spaces L(, ) and C[, ], and therefore, there exist immersion constants S, S¯ >  such that (.) (.) (.) (.) (.) (.)    u      u u  ≤ S u , and u ∞ ≤ S¯ u . Next, we consider the functional J : E → R defined by J(u) =  u  –   where F , G are the primitives F x, u(x) dx + G u() , F(x, u) = f (x, t) dt, and G(u) = g(t) dt. Since f , g are continuous, we deduce that J is of class C and its derivative is given by J (u), ϕ = u (x)ϕ (x) dx – f x, u(x) ϕ(x) dx + g u() ϕ() for all u, ϕ ∈ E. Then we can infer that u ∈ E is a critical point of J if and only if it is a (classical) solution of problem (.). Now we define a class of functionals on E by Jλ(u) =  u  + G u() – λ = A(u) – λB(u),  λ ∈ [, ].  F x, u(x) dx It is easy to know that Jλ ∈ C(E; R) for all λ ∈ [, ] and the critical points of J = J correspond to the weak solutions of problem (.). We choose a completely orthonormal basis {ej} of E and define Xj := Rej. Then Zk , Yk can be defined as that at the beginning of Section . (.) (.) (.) (.) Lemma . Under the assumptions of Theorem ., there exists ρk >  large enough such that ak(λ) := maxu∈Yk, u =ρk Jλ(u) ≤  for all λ ∈ [, ]. Proof Let u ∈ Yk , then there exists  >  such that meas x ∈ [, ] : u(x) ≥  u ≥ , ∀u ∈ Yk \ {}. Otherwise, for any positive integer n, there exists un ∈ Yk \ {} such that  meas x ∈ [, ] : un(x) ≥ n un < n for all k. Set vn(x) := unu(nx) ∈ Yk \ {}, then vn =  and meas x ∈ [, ] : vn(x) ≥ n < n for all k. Since dim Yk < ∞, it follows from the compactness of the unit sphere of Yk that there exists a subsequence, say {vn}, such that vn converges to some v in Yk . Hence, we have v = . By the equivalence of the norms on the finite-dimensional space Yk , we have vn → v in L[, ], i.e.,   |vn – v| dx →  as n → ∞. Thus there exist ξ, ξ >  such that meas x ∈ [, ] : v(x) ≥ ξ ≥ ξ. In fact, if not, we have  meas x ∈ [, ] : v(x) ≥ n = , i.e., meas x ∈ [, ] : v(x) < n = , for all positive integer n. This implies that  <    v(x)  dx < n →  as n → ∞, which gives a contradiction. Therefore, (.) holds. Now let  = x ∈ [, ] : v(x) ≥ ξ ,  n = x ∈ [, ] : vn(x) < n and n⊥ = [, ] \ n. By (.) and (.), we have meas( n ∩ ) = meas  \ n⊥ ∩  ≥ meas( ) – meas n⊥ ∩   ≥ ξ – n for all positive integer n. Let n be large enough such that ξ – n ≥  ξ and ξ – n ≥  ξ. Then we have vn(x) – v(x)  ≥ ξ –   n  ξ , ≥   ∀x ∈ n ∩ . This implies that   |vn – v| dx ≥ |vn – v| dx n∩  ≥  ξ meas( n ∩ ) ≥  ξ ξ –  n ≥  ξξ >  for all large n, which is a contradiction with (.). Therefore, (.) holds. For any u ∈ Yk , let u = {x ∈ [, ] : |u(x)| ≥  u }. By condition (H), for M = λ ≥  > , there exists L >  such that Hence one has F(x, u) ≥ M|u|, ∀|u| ≥ L, ∀x ∈ [, ]. F(x, u) ≥ M|u| ≥ M  u , ∀x ∈ u, for all u ∈ Yk with u ≥ L . It follows from (H)-(H) and (.) that Jλ(u) =  u  + u() g(x) dx – λ  ≤  u  + aS¯ u + bS¯ γ + u γ + – λ F(x, u) dx u F(x, u) dx ≤  u  + aS¯ u + bS¯ γ + u γ + – λM  u  = –  u  + aS¯ u + bS¯ γ + u γ +,  for all u ∈ Yk with u ≥ L . Since γ < , for u = ρk large enough, we have Jλ(u) ≤ . Lemma . Under the assumptions of Theorem ., there exist rk > , bk → ∞ such that bk(λ) := infu∈Zk, u =rk Jλ(u) ≥ bk for all λ ∈ [, ]. Proof Set γk := supu∈Zk, u = u ∞. Then γk →  as k → ∞. Indeed, it is clear that  < γk+ ≤ γk , so that γk → γ¯ ≥ , as k → ∞. For every k ≥ , there exists uk ∈ Zk such that uk =  and uk ∞ > γk/. By the definition of Zk , uk  in E. Then it implies that uk →  in C[, ]. Thus we have proved that γ¯ = . By (H), we have F(x, u) ≤ c + c|u|β+, (x, u) ∈ [, ] × R. By (H), for any > , there exists δ >  such that F(x, u) ≤ |u|, ∀x ∈ [, ], |u| ≤ δ. Therefore, there exists C = C( ) >  such that F(x, u) ≤ |u| + C|u|β+, (x, u) ∈ [, ] × R. Hence, for any u ∈ Zk , choose = (λp)–, by (H) and (.), we have Jλ(u) =  u  + u() g(x) dx – λ  ≥  u  – λ  ≥  u  – λ S u  – λC u β∞+ ≥  u  – λCγkβ+ u β+. |u| + C|u|β+ dx F(x, u) dx (.)  Let rk := (λCγkβ+) –β . Then, for any u ∈ Zk with u = rk , we have Jλ(u) ≥  λCT γkβ+ –β := bk → ∞ uniformly for λ as k → ∞. Lemma . Under the assumptions of Theorem ., there exist λn →  as n → ∞, {un(k)}n∞= ⊂ E such that Jλn (un(k)) → , Jλn (un(k)) ∈ [bk, ck], where ck = supu∈Bk (u). Proof It is easy to verify that (C) and (C) of Theorem . hold. By Lemmas ., . and Theorem ., we can obtain the result. Proof of Theorem . For the sake of notational simplicity, in what follows we always set un = un(k) for all n ∈ N. By Lemma ., it suffices to prove that {un}n∞= is bounded and possesses a strong convergent subsequence in E. If not, passing to a subsequence if necessary, we assume that un → ∞ as n → ∞. In view of (H), there exists c >  such that f (x, u)u – F(x, u) ≥ c|u|α – c for all (x, u) ∈ [, ] × R, and combining (H), we have Jλn (un) – Jλn (un)un =  g(x) dx – g un() un() un()  + λn ≥ λn = cλn       f (x, un)un – F(x, un) dx α c|un| – c dx α |un| dx – λnc. This implies that  |un|α dx un →  as n → ∞. Note that from (H),  < β <  + αα– . Let η = α(αβ––) , then η > ,  ηβ –  = η – . α By (H), there exists c >  such that f (x, u) η η ≤ c|u|ηβ + c, ∀(x, u) ∈ [, ] × R. By (.), (H) and the Hölder inequality, one has Jλn (un)un = un  + g un() un() – λn f (x, un)un dx   (.) (.) (.) ≥ un  – a + b un() γ un() – λn η f (x, un) dx Cη un ≥ un  – aS¯ un – bS¯ γ + un γ + – λn η f (x, un) dx Cη un , (.)  η  η     where Cη >  is a constant independent of n. By (.) we obtain   η f (x, un) dx ≤ η c|un|ηβ + c dx   ≤ c ≤ c     α |un| dx α |un| dx /α /α   α(ηβ–) |un| α– dx – α + c un (ηβ–) + c, combining this inequality with (.) and (.) yields that  (  |f (x, un)|η dx) η un ≤ c(  |un|α dx)/α un /α un (ηβ–) un η– α + c un η  η →  as n → ∞. Combining this with (.), we have  = un  un  ≤ Jλn (un)un un  + aS¯ + un bS¯ γ + un –γ +  λn(  |f (x, un)|η dx) η Cη un →  as n → ∞, since  ≤ γ < . This is a contradiction. Therefore, {un}n∞= is bounded in E. Without loss of generality, we may assume un wk in E. Then un → wk in C[, ]. Note that un – wk  = Jλn (un) – Jλn (wk) (un – wk) – g un() – g wk()  un() – wk() + λn  f (x, un) – f (x, wk) (un – wk) dx. Taking n → ∞, we have limn→∞ un – wk = , which means that un → wk in E and J(wk) = . Hence, J has a critical point wk with J(wk) ∈ [bk, ck]. Consequently, we obtain infinitely many solutions since bk → ∞. Lemma . Under the assumptions of Theorem ., there exists ρk small enough such that ak(λ) := infu∈Zk, u =ρk Jλ(u) ≥  and dk(λ) := infu∈Zk, u ≤ρk Jλ(u) →  as k → ∞ uniformly for λ ∈ [, ]. Proof For any u ∈ Zk , by using γk := supu∈Zk, u = u ∞ defined in Lemma ., together with (S) and (S), we have Jλ(u) =  u  + u() g(x) dx – λ  ≥  u  – λk  ≥  u  – λkγkζ u ζ =  ρk ≥   |u|ζ dx ≥  u  – λk u ζ∞ F(x, u) dx for all u ∈ Zk with u = ρk := (λkγkζ )/(–ζ). Obviously, ρk →  as k → ∞. So ak(λ) := infu∈Zk, u =ρk Jλ(u) ≥  and dk(λ) := infu∈Zk, u ≤ρk Jλ(u) →  as k → ∞ uniformly for λ ∈ [, ]. Lemma . Under the assumptions of Theorem ., there exists rk small enough such that bk(λ) := maxu∈Yk, u =rk Jλ(u) <  for all λ ∈ [, ]. Proof For any u ∈ Yk , by (S)-(S) and the equivalence of the norms on the finitedimensional space Yk , we have Jλ(u) =  u  + u() g(x) dx – λ  ≤  u  + bS¯ γ + u γ + – λk  ≤  u  + bS¯ γ + u γ + – λkc u ζ . F(x, u) dx |u|ζ dx Since ζ < γ +  < , for u = rk < ρk small enough, we can get Jλ(u) <  for all λ ∈ [, ]. that Proof of Theorem . It is easy to verify that (C) and (D) hold under the assumptions of Theorem .. By Lemmas . and ., the condition (D) is also satisfied. Therefore, by Theorem . there exist λn → , u(λn) := un ∈ Yn such that Jλn |Yn (un) = , Jλn (un) → ck ∈ [dk (), bk ()] as n → ∞. In the following we show that {un}n∞= is bounded. Indeed, note un  = Jλn (un) –  g(x) dx + λn F(x, un) dx ≤ M + bS¯ γ + un γ + + k ≤ M + bS¯ γ + un γ + + kS¯ ζ un ζ , ∀n ∈ N,  un()     (.) for some M > . Since  < γ +  < , (.) yields that {un} is bounded in E. By a standard argument, this yields a critical point uk of J such that J(uk ) ∈ [dk (), ck ()]. Since dk () → – as k → ∞, we can obtain infinitely many critical points. Competing interests The author declares that she has no competing interests. Authors’ contributions The author read and approved the final manuscript. Acknowledgements The author would like to express her sincere thanks to the referees for their helpful comments. 1. Ma, TF: Positive solutions for a beam equation on a nonlinear elastic foundation . Math. Comput. Model . 39 , 1195 - 1201 ( 2004 ) 2. Ma, TF, da Silva, J: Iterative solutions for a beam equation with nonlinear boundary conditions of third order . Appl. Math. Comput . 159 , 11 - 18 ( 2004 ) 3. Bonanno , G , Di Bella , B: A boundary value problem for fourth-order elastic beam equations . J. Math. Anal. Appl . 343 , 1166 - 1176 ( 2008 ) 4. Bonanno , G , Di Bella , B: Infinitely many solutions for a fourth-order elastic beam equation . Nonlinear Differ. Equ. Appl . 18 , 357 - 368 ( 2011 ) 5. Han, G , Xu, Z: Multiple solutions of some nonlinear fourth-order beam equations . Nonlinear Anal . 68 , 3646 - 3656 ( 2008 ) 6. 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Xiaodan Wang. Infinitely many solutions for a fourth-order differential equation on a nonlinear elastic foundation, Boundary Value Problems, 2013, 258,