Infinitely many solutions for a fourthorder differential equation on a nonlinear elastic foundation
Boundary Value Problems
Infinitely many solutions for a fourthorder differential equation on a nonlinear elastic foundation
Xiaodan Wang
In this paper, existence results of infinitely many solutions for a fourthorder 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 fourthorder twopoint 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 + bsγ
for s ∈ R.
(H) limu→+∞ 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) ≥ cuα,
f (x, u) ≤ cuβ .
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) ≤ bsγ
for s ∈ R.
(S) F(x, ) ≡ for any x ∈ [, ].
(S) There are constants k > and ζ ∈ [, ) with ζ < γ + such that
F(x, u) ≥ kuζ for any (x, u) ∈ [, ] × R.
(S) There exist constants k > and ζ ∈ [, ) such that
f (x, u) ≤ kuζ– 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 Cfunctional 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 finitedimensional 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
≥ ξξ >
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) ≥ Mu,
∀u ≥ L, ∀x ∈ [, ].
F(x, u) ≥ Mu ≥ 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 + cuβ+, (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 + Cuβ+, (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 + Cuβ+ 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) ≥ cuα – 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
= cλn
f (x, un)un – F(x, un) dx
α
cun – c dx
α
un dx – λnc.
This implies that
unα 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
≤ cuηβ + 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
cunηβ + 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( unα 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 – λkγkζ u ζ = ρk ≥
uζ dx ≥ u – λk u ζ∞
F(x, u) dx
for all u ∈ Zk with u = ρk := (λkγkζ )/(–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 + – λkc u ζ .
F(x, u) dx
uζ 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 + + kS¯ 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.
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