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Nonlinear Analysis 67 (2007) 1540–1549www.elsevier.com/locate/na
Multiple doubly periodic solutions of semi-linear damped beamequations
Peng Wang∗, Yukun An
Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing, 210016, PR China
Received 21 December 2005; accepted 14 July 2006
Abstract
The purpose of this work is to investigate the weakened Ambrosetti–Prodi type multiplicity results for weak doubly periodicsolutions of damped beam equations. By using the topological degree theory, the author obtains a result which is similar to theresult for damped wave equations in the literature.c© 2007 Published by Elsevier Ltd
Keywords: Damped beam equation; Double periodic solution; Multiplicity; Topological degree
1. Introduction
There is a large body of literature about Ambrosetti–Prodi type results for ordinary and partial differentialequations. For example see [5–10]. This type of result, the so-called Ambrosetti–Prodi type result, was initiated byAmbrosetti–Prodi [1] in 1972 in the study of a Dirichlet problem to elliptic equations.
Kim [3] has treated Ambrosetti–Prodi type multiplicity results for the Dirichlet periodic boundary value problemof semi-linear dissipative hyperbolic equations in n-dimensional space. Kim [2] has obtained Ambrosetti–Prodi typemultiplicity results for weak doubly periodic solutions of the nonlinear dissipative wave equation in one-dimensionalspace. More precisely, the author studied the equationsβut + ut t − uxx + g(t, x, u) = s + h(t, x) in Ω ,
u(t, 0) = u(t, 2π), ux (t, 0) = ux (t, 2π) t ∈ [0, 2π ],
u(0, x) = u(2π, x), ut (0, x) = ut (2π, x) x ∈ [0, 2π ].
(1)
Here β( 6=0) ∈ R, u = u(t, x),Ω = [0, 2π ] × [0, 2π ], h ∈ L2(Ω) with∫ ∫
Ω h(t, x)dtdx = 0. He obtained
Theorem 1.1. Let g : Ω × R → R be a continuous function satisfying
(g1) There exist a(t, x) ∈ L∞(Ω) and b(t, x) ∈ L2(Ω) such that
|g(t, x, u)| ≤ a(t, x)|u| + b(t, x) a.e. on Ω;
∗ Corresponding author.E-mail addresses: [email protected] (P. Wang), [email protected] (Y. An).
0362-546X/$ - see front matter c© 2007 Published by Elsevier Ltddoi:10.1016/j.na.2006.07.036
P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549 1541
(g2) g(t, x, u) ≥ 0;(g3) lim|u|→+∞ g(t, x, u) = +∞ uniformly on Ω .
Then there exist real numbers s0 ≤ s1 such that
(a) (1) has no weak doubly periodic solution for s < s0,(b) (1) has at least one weak doubly periodic solution for s = s1,(c) (1) has at least two weak doubly periodic solutions for s > s1.
Meanwhile, there are many papers about the existence of periodic solutions for the semi-linear wave equationswith or without damping. For example see [11,13–15]. In [12], Grossinho and Nkashama investigated a higher orderdissipative hyperbolic equation. In [4], An treated an Ambrosetti–Prodi type multiplicity result for the followingdamped beam equation:
ut t + δut + uxxxx − u + g(u) =1π
s sin x + h(t, x) in Ω ,
u(t, 0) = u(t, π) = uxx (t, 0) = uxx (t, π) = 0 t ∈ (0, 2π),
u(0, x) = u(2π, x), ut (0, x) = ut (2π, x) x ∈ (0, π).
(2)
Here Ω = (0, 2π) × (0, π), δ( 6=0) ∈ R, h ∈ L2(Ω) with∫ ∫
Ω h(t, x) sin xdtdx = 0, s is a real parameter andg ∈ C(R). The author obtained
Theorem 1.2. Let g : R → R be a continuous function. Moreover, we assume the following:
(H1) g(u) ≥ 0 for all u ∈ R,(H2) lim|u|→+∞ g(u) = +∞,(H3) there exist nonnegative constants a0, b0 and a, b with a ∈ (0, 7) such that
g(u) ≤ a0u + b0 for all u ≥ 0,
g(u) ≤ a|u| + b for all u ≤ 0,
then there exist constants s0 ≤ s1 such that
(1) (2) has no solution if s < s0,(2) (2) has at least one solutions if s = s1,(3) (2) has at least two solutions if s > s1.
Inspired by those mentioned papers, we will investigate in this note the Ambrosetti–Prodi type multiplicity resultsfor weak doubly periodic solutions of semi-linear damped beam equations:
ut t + δut + uxxxx + g(t, x, u) = s + h(t, x) in Ω ,
u(t, 0) = u(t, 2π), ux (t, 0) = ux (t, 2π),
uxx (t, 0) = uxx (t, 2π), uxxx (t, 0) = uxxx (t, 2π) t ∈ (0, 2π),
u(0, x) = u(2π, x), ut (0, x) = ut (2π, x) x ∈ (0, 2π).
(3)
Here δ(6=0) ∈ R, u = u(t, x),Ω = (0, 2π) × (0, 2π), h ∈ L2(Ω) with∫ ∫
Ω h(t, x)dtdx = 0, s is a real parameterand g : Ω × R → R is a continuous function.
2. Preliminaries
Let Z, R+ and R be the set of all integers, non-negative real numbers and real numbers, respectively, Λ = Z × Z.Let Ω = (0, 2π) × (0, 2π), and L1(Ω) be the space of measurable real-valued functions u : Ω → R which areLebesgue integrable over Ω with usual norm ‖ · ‖L1 , L2(Ω) be the usual space of square integrable functions withusual inner product (·, ·) and corresponding norm ‖ · ‖. For the Sobolev space H1(Ω), we denote the standard innerproduct by 〈u, v〉1 = (u, v) + (ux , vx ) + (ut , vt ), and norm by ‖u‖H1 .
1542 P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549
Denote φmn = ei(mt+nx), (m, n) ∈ Λ. Let H be the Hilbert space which is the closure for the L2-norm of the set ofu ∈ C2([0, 2π ] × [0, 2π ]) satisfying the boundary conditions, so that
H =
u =
∑Λ
umnφmn :
∑Λ
|umn|2 < ∞
.
Let us define the closed linear operator A, the abstract realization of the linear beam operator Au = ut t + δut +
uxxxx , in H by
Dom(A) =
u ∈ H | u(x, t) =
∑Λ
umnφmn and∑Λ
((n4− m2)2
+ m2δ2)|umn|2 < ∞
,
Au =
∑Λ
((n4− m2) + mδi)umnφmn
for all u(x, t) =∑
Λ umnφmn ∈ Dom(A).It is clear that the set of eigenvalues of A is
σ(A) = λmn = (n4− m2) + mδi | (m, n) ∈ Λ,
and
Ker A = R,
Im A =
h ∈ L2(Ω) :
∫ ∫Ω
h(t, x)dtdx = 0
.
Im A is closed and
Im A = (Ker A)⊥.
Hence,
H = Im A ⊕ Ker A.
A is not bijective but the restriction
A|Dom A∩Im A : Dom A ∩ Im A → Im A
is bijective.So we can define the continuous right inverse of A by
A−1= (A|Dom A∩Im A)−1
: Im A → Dom A ∩ Im A.
And
A−1u =
∑Λ0
1n4 − m2 + mδi
umnφmn
for all u =∑
Λ umnφmn , u ∈ Im A, where Λ0 = Λ \ (0, 0).In addition, we have the following result for operator A.
Lemma 2.1. There exist C1, C2 ∈ R+ such that the operator A satisfies
‖A−1u‖ ≤ C1‖u‖, ∀u ∈ Im A,
‖A−1u‖H1 ≤ C2‖u‖, ∀u ∈ Im A.
Proof. We know that
A−1u =
∑Λ0
1n4 − m2 + mδi
umnφmn
for all u =∑
Λ umnφmn , u ∈ Im A. For any (m, n) ∈ Λ0, we have
P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549 1543
(n4− m2)2
+ m2δ2≥ m2δ2
≥ δ2.
Denote C21 = δ2 (C1 ∈ R+); then
‖A−1u‖2
= 4π2∑Λ0
1(n4 − m2)2 + m2δ2 |umn|
2
≤ C214π2
∑Λ0
|umn|2
≤ C21‖u‖
2.
Hence
‖A−1u‖ ≤ C1‖u‖, ∀u ∈ Im A.
On the other hand,
‖A−1u‖2H1 = ‖A−1u‖
2+ ‖(A−1u)x‖
2+ ‖(A−1u)t‖
2
= 4π2∑Λ0
1(n4 − m2)2 + m2δ2 |umn|
2+ 4π2
∑Λ0
n2
(n4 − m2)2 + m2δ2 |umn|2
+ 4π2∑Λ0
m2
(n4 − m2)2 + m2δ2 |umn|2
= 4π2∑Λ0
1 + m2+ n2
(n4 − m2)2 + m2δ2 |umn|2.
Since for any (m, n) ∈ Λ0, we have
1 + m2+ n2
(n4 − m2)2 + m2δ2
=1 + m2
+ n2
(n2 + |m|)2(n2 − |m|)2 + m2δ2 .
If n2− |m| 6= 0, then
1 + m2+ n2
(n2 + |m|)2(n2 − |m|)2 + m2δ2 ≤1 + m2
+ n2
(n2 + |m|)2 ≤1 + m2
+ n2
(|n| + |m|)2
=1 + m2
+ n2
m2 + n2 + 2|mn|≤
1 + m2+ n2
n2 + m2 + 2< 1.
If n2− |m| = 0, then
1 + m2+ n2
(n2 + |m|)2(n2 − |m|)2 + m2δ2 =1 + n4
+ n2
n4δ2
≤3δ2 .
Let C22 = max1, 3
δ2 ; we obtain
‖A−1u‖2H1 ≤ C2
2 4π2∑Λ0
|umn|2
= C22‖u‖
2.
Hence
‖A−1u‖H1 ≤ C2‖u‖, ∀u ∈ Im A.
The proof is complete.
1544 P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549
By Lemma 2.1, we can find that the operator A−1: H → H is compact since the embedding H1 → L2 is compact.
Since
(A−1h)(t, x) =
∑(m,n)∈Λ0
1(n4 − m2) + mδi
hmnφmn,
we can represent A−1 as a convolution product
(A−1h)(t, x) = (K ∗ h)(t, x) =
∫ ∫Ω
K (t − s, x − y)h(s, y)dsdy.
Here
K (t, x) =1
4π2
∑(m,n)∈Λ0
1(n4 − m2) + mδi
φmn .
Now we can extend A−1 to L1(Ω) by defining
A−1: L1(Ω) → L2(Ω)
by the formula
(A−1h)(t, x) =
∫ ∫Ω
K (t − s, x − y)h(s, y)dsdy for h ∈ L1(Ω).
Then, by Holder’s inequality and Fubini’s theorem, we have the following lemma:
Lemma 2.2. ‖A−1h‖L2 ≤ ‖K‖L2‖h‖L1 .
Proof. Since
(A−1h)(t, x) =
∫ ∫Ω
K (t − s, x − y)h(s, y)dsdy,
we have, by extending h(t, x)2π -periodically in both variables to R × R and then changing variables,
(A−1h)(t, x) =
∫ t
t−2π
∫ x
x−2π
h(t − u, x − v)K (u, v)dudv.
By Holder’s inequality, we get
|(A−1h)(t, x)|2 ≤
[∫ ∫Ω
|h(t − u, x − v)||K (u, v)|dudv
]2
=
[∫ ∫Ω
(|h(t − u, x − v)|)12 (|h(t − u, x − v)|
12 |K (u, v)|)dudv
]2
≤
[∫ ∫Ω
|h(t − u, x − v)|dudv
] [∫ ∫Ω
|h(t − u, x − v)||K (u, v)|2dudv
]= ‖h‖L1
∫ ∫Ω
|h(t − u, x − v)||K (u, v)|2dudv.
By Fubini’s theorem∫ ∫Ω
∫ ∫Ω
|h(t − u, x − v)||K (u, v)|2dudvdtdx =
∫ ∫Ω
|K (u, v)|2
[∫ ∫Ω
|h(t − u, x − v)|dtdx]
dudv
= ‖h‖L1
∫ ∫Ω
|K (u, v)|2dudv
= ‖h‖L1‖K‖2L2 .
Therefore
‖A−1h‖L2 ≤ ‖K‖L2‖h‖L1 .
P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549 1545
3. Results and proofs
Let A : Dom(A) ⊆ H → H be defined as before. In addition, we define a substitution operator Gs : H → H by
(Gsu)(t, x) = g(t, x, u(t, x)) − s − h(t, x)
for (t, x) ∈ Ω , u ∈ H and s ∈ R.We consider the weak solutions of problem (3). A weaksolution u(x, t) of Eq. (3) means that u(x, t) ∈ H and
satisfies
(u, −δvt + vt t + vxxxx ) + (g(t, x, u) − s − h(t, x), v) = 0
for all v ∈ C2(Ω) and verifying the boundary conditions of (3). It is easy to check that u(t, x) is a weak solution ofEq. (3) if and only if u ∈ Dom(A) satisfying
Au + Gsu = 0. (4)
Let g : Ω × R → R be a continuous function satisfying the following:
(H1) g(t, x, u) ≥ 0 on Ω × R,(H2) lim|u|→+∞ g(t, x, u) = +∞ uniformly on Ω ,(H3) there exist a(t, x) ∈ L∞(Ω) and b(t, x) ∈ L2(Ω), such that
|g(t, x, u)| ≤ a(t, x)|u| + b(t, x) a.e. on Ω .
From the above conditions, it is clear that Gs is a bounded continuous map. By P we denote the continuousprojection P : H → Ker A. Then, for any open bounded set B in H, PGs : B → H is bounded andA−1(I − P)Gs : B → H is compact and continuous. Thus Gs is A-compact on B and the coincidence degreeDL(A + Gs, B) is well defined.
Lemma 3.1. If (H2) and (H3) are satisfied,∫ ∫
Ω h(t, x)dtdx = 0, then for any s∗∈ R+, there exists M(s∗) > 0
such that
‖u‖L2 ≤ M(s∗)
holds for each possible weak doubly periodic solution, u = u + u with u ∈ Ker A and u ∈ Im A, of (3) with s ≤ s∗.
Proof. Let u be any weak doubly periodic solution of (3). Then u is a solution of (4) where u = u + u with u ∈ Ker Aand u ∈ Im A. By applying A−1
on both sides of Eq. (4), we have, since
A−1|Im A = A−1
u = −A−1Gsu = A−1[−g(·, ·, u) + s + h(·, ·)].
Hence, by Lemma 2.2,
‖u‖L2 ≤ ‖K‖L2 [‖g(·, ·, u)‖L1 + 4π2|s| + ‖h‖L1 ].
By taking the inner product with 1 on both sides of (4), since 1 ∈ Ker A, we have∫ ∫Ω
g(t, x, u(t, x))dtdx = 4π2s.
Hence, ‖g(·, ·, u)‖L1 = 4π2s. Therefore, we have
‖u‖L2 ≤ ‖K‖L2 [8π2s + ‖h‖L1 ]
≤ ‖K‖L2 [8π2s∗+ ‖h‖L1 ] ≡ M(s∗).
The proof is complete.
1546 P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549
Lemma 3.2. If (H1)–(H3) are satisfied, then, for each s∗∈ R+, there exists γ (s∗) such that
|u| ≤ γ (s∗)
holds for each possible weak doubly periodic solution, u = u + u with u ∈ Ker A and u ∈ Im A, of (3) with s ≤ s∗.
Proof. Suppose there exists a constant s with s ≤ s∗ and that the corresponding weak doubly periodic solution un
of (3) with un is unbounded. Then un is a solution of (4) where un = un + un with u ∈ Ker A and u ∈ Im A. Wemay choose a subsequence, say again un, such that |un| → +∞ as n → +∞. Now suppose that un → +∞ asn → +∞. Let M0 > 2π M(s∗) where M(s∗) is given in Lemma 3.1 and let
Ωn =
(t, x) ∈ Ω : u(t, x) ≤ −
M0
4π2
.
Then
2π M(s∗) ≥
∫ ∫Ω
|un(t, x)|dtdx
≥
∫ ∫Ωn
|un(t, x)|dtdx
≥ |Ωn|
[M0
4π2
].
Therefore |Ωn| ≤ 4π2(2π M(s∗)/M0) and hence
|Ω \ Ωn| ≥ 4π2[1 − 2π M(s∗)/M0] > 0.
Since lim|u|→+∞ g(t, x, u) = +∞ uniformly on Ω , there exists C(s∗) > 0 such that
g(t, x, u) > 4π2s∗/|Ω \ Ωn|
for all n if |u| ≥ C(s∗).Since un → +∞, there exists N > 0 such that
un ≥M0
4π2 + C(s∗) if n ≥ N .
Hence, for (t, x) ∈ Ω \ Ωn and n ≥ N , we have
un(t, x) = un + un(t, x) ≥ C(s∗).
Thus, for n ≥ N , we have∫ ∫Ω\Ωn
g(t, x, un(t, x))dtdx > 4π2s∗.
On the other hand, by taking the inner product with 1 on both sides of (4), we have∫ ∫Ω
g(t, x, un(t, x))dtdx ≤ 4π2s∗.
Therefore, for n ≥ N , by (H1),
4π2s∗≥
∫ ∫Ω
g(t, x, un(t, x))dtdx
≥
∫ ∫Ω\Ωn
g(t, x, un(t, x))dtdx > 4π2s∗
which is impossible. Similarly, we can treat the case where un → −∞.The proof is completed.
P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549 1547
Lemma 3.3. If (H1)–(H3) are satisfied, then, for each s∗∈ R+, there exists an open bounded set B1(s∗) in H such
that, for any open bounded set B with H ⊃ B ⊇ B1(s∗),
DL(A + Gs, B) = 0
for all s ≤ s∗.
Proof. Since lim|u|→+∞ g(t, x, u) = +∞, for each s∗∈ R+ there exists γ ∗ > 0 such that g(t, x, u) > s∗ if
|u| ≥ γ ∗. Let B1(s∗) = u ∈ L2(Ω) : |u| < γ , ‖u‖L2 < M where u = u + u with u ∈ Ker A and u ∈ Im A, andγ > maxγ (s∗), γ ∗
, M > M(s∗), and M(s∗) and γ (s∗) are given in Lemmas 3.1 and 3.2. Let
s0 = min(t,x)∈Ω ,u∈R
g(t, x, u).
If (4) has a solution u for some s ∈ R, then, by taking the inner product with 1 on both sides of Eq. (4), we have
s0 ≤1
4π2
∫ ∫Ω
g(t, x, u(t, x))dtdx = s.
Thus (3) has no solution for s < s0. Hence for each open bounded set B ⊇ B1(s∗), we have
DL(A + Gs, B) = 0 for s < s0.
Chose s < s0 and define
F : (Dom A ∩ B) × [0, 1] → H
by
F(u, µ) = Au + G(1−µ)s+µsu, for s ≤ s∗.
From Lemmas 3.1 and 3.2, we have
0 6∈ F(Dom A ∩ ∂ B) × [0, 1] for s ≤ s∗.
By the homotopy invariance of degree, we get, for all s ≤ s∗,
DL(A + Gs, B) = DL(F(·, 1), B)
= DL(F(·, 0), B)
= 0,
and the proof is complete.
Lemma 3.4. If (H1)–(H3) are satisfied, then there exists s1 ≥ s0 such that, for each s∗ > s1, we can find an openbounded set B2(s∗) in H on which
|DL(A + Gs, B2(s∗))| = 1
for all s1 < s ≤ s∗.
Proof. Let (t0, x0, u0) ∈ Ω × R be such that
g(t0, x0, u0) = min(t,x)∈Ω ,u∈R
g(t, x, u)
and let
s1 = max(t,x)∈Ω ,u∈[u0−M,u0+M]
g(t, x, u).
Let
B2(s∗) = u ∈ L2(Ω) : u0 < u < γ , ‖u‖L2 < M
1548 P. Wang, Y. An / Nonlinear Analysis 67 (2007) 1540–1549
where γ and M are given by Lemma 3.3. If s > s1 and u is a possible solution of (4) such that u ∈ B2(s∗), then byLemmas 3.1 and 3.2, we have necessary u = u0 and
u0 − M < u(t, x) = u + u(t, x) < u0 + M
for all (t, x) ∈ Ω .By taking the inner product with 1 on both sides of (4), we have
14π2
∫ ∫Ω
g(t, x, u(t, x))dtdx = s.
But
s1 ≥1
4π2
∫ ∫Ω
g(t, x, u(t, x))dtdx = s
which is impossible. Thus, for s > s1,
DL(A + Gs, B2(s∗))
is well defined, and
DL(A + Gs, B2(s∗)) = DB(PGs, B2(s∗) ∩ Ker A, 0),
where PGs : H → Ker A is an operator defined by
(PGsu)(t, x) =1
4π2
∫ ∫Ω
g(t, x, u(t, x))dtdx − s
and DB denotes the Brouwer degree.Thus, for s1 < s ≤ s∗, we have PGs |KerL(u0) ≤ s1 − s < 0, and by definition of γ in Lemma 3.3,
PGs |KerL(γ ) > 0. Therefore
|DL(A + Gs, B2(s∗))| = |DB(PGs, B2(s∗) ∩ Ker A, 0)| = 1
and the proof is complete.
From the above lemmas, we get the main result of this paper and the proof is omitted since it is similar to [2].
Theorem 3.1. If (H1), (H2) and (H3) are satisfied, then there exist real numbers s0 ≤ s1 such that
(1) (3) has no weak doubly periodic solution if s < s0,(2) (3) has at least one weak doubly periodic solutions if s = s1,(3) (3) has at least two weak doubly periodic solutions if s > s1.
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