Electronic Journal of Qualitative Theory of Differential Equations Proc. 7th Coll. QTDE, 2004, No. 9 1-20; http://www.math.u-szeged.hu/ejqtde/

Forced Oscillations of Beams on Elastic Bearings

Michal Fečkan ^{∗}

Department of Mathematical Analysis, Comenius University Mlynská dolina, 842 48 Bratislava - Slovakia and Mathematical Institute, Slovak Academy of Sciences Štefánikova 49, 814 73 Bratislava, Slovakia email: Michal.Feckan@fmph.uniba.sk

Abstract We study the existence of weak periodic solutions for certain damped and forced linear beam equations resting on semi-linear elastic bearings. Conditions for the periodic forcing term and semi-linear elastic bearings are derived which ensure either the existence or nonexistence of periodic solutions of the beam equation. Topological degree arguments are used to achieve these results.

Keywords. beam equations, periodic solutions, topological degree AMS Subject Classification. 35B10, 35B20, 74K10

1 Introduction

In this note, we consider a periodically forced and damped beam resting on two different bearings with purely elastic responses. The length of the beam is π/4. The equation of vibrations is as follows

\begin{matrix}u_{tt} + u_{xxxx} + δu_{t} + h(x, t) = 0 , \\ u_{xx}(0, ·) = u_{xx}(π/4, ·) = 0 , \\ u_{xxx}(0, ·) = −ku(0, ·) − f(u(0, ·)) , \\ u_{xxx}(π/4, ·) = ru(π/4, ·) + g(u(π/4, ·)) , \end{matrix}  (1)

where δ > 0, r ≥ 0, k ≥ 0 are constants, h ∈ C([0, π/4] × S^{T}), and f, g ∈ C(ℝ) have at most linear growth at infinity. Here S^{T} is the circle S^{T} = ℝ/{T ℤ}.

The undamped and unforced case of the form

\begin{matrix}u_{tt} + u_{xxxx} = 0 , \\ u_{xx}(0, ·) = u_{xx}(π/4, ·) = 0 , \\ u_{xxx}(0, ·) = −f(u(0, ·)) , \\ u_{xxx}(π/4, ·) = f(u(π/4, ·)) \end{matrix}  (2)

∗partially supported by Grant GA-SAV 2/1140/21.

This paper is in final form and no version of it will be submitted for publication elsewhere.

is studied in [7] and [9] by using variational methods, where among others the following results are proved.

Theorem. ([9]) If the function f(u) satisfies the following assumptions:

(i) f ∈ C^{1}(ℝ), f(−u) = −f(u) for all u ∈ ℝ.

(ii) For any C > 0 there is a K(C) such that f(u) ≥ Cu − K(C) for all u ≥ 0.

(iii) \frac{1}{2}f(u)u − F (u) ≥ c_{1}|f(u)| − c_{2} for all u ∈ ℝ, where F (u) = ∫_{0}^{u} f(s) ds and

c_{1}, c_{2} are positive constants.

(iv) f(0) = f′(0) = 0.

Then there is a sufficiently large positive integer M such that equation (2) possesses at least one nonzero time periodic solution with the period 2πM^{2}.

Theorem. ([7]) If the continuous function f(u) is odd on ℝ, C^{1}-smooth near u = 0 with f′(0) > 0 and lim_{|u|→∞} f(u)/u = 0. Then equation (2) possesses

infinitely many odd time periodic solutions with periods densely distributed in an interval (a_{1}, 2a_{1}) for a constant a_{1} > 0.

It is pointed out in [9] that equation (2) is a simple analogue of a more complicated shaft dynamics model introduced in the works [5] and [6].

When the nonlinearities and parameters are small, i.e. (1) is of the form

\begin{matrix}u_{tt} + u_{xxxx} + εδu_{t} + εμh(x, \sqrt{ε}t) = 0 , \\ u_{xx}(0, ·) = u_{xx}(π/4, ·) = 0 , \\ u_{xxx}(0, ·) = −εf(u(0, ·)), u_{xxx}(π/4, ·) = εf(u(π/4, ·)) , \end{matrix}  (3)

where ε > 0 and μ are small parameters, δ > 0 is a constant, f ∈ C^{2}(ℝ), h ∈ C^{2}([0, π/4] × ℝ) and h(x, t) is 1-periodic in t. Then by using analytic methods, the following result is proved in [2] and [3].

Theorem. ([2], [3]) Let the following assumptions hold:

(I) f(0) = 0, f′(0) < 0 and the equation ẍ+f(x) = 0 has a homoclinic solution γ(t) ≠ 0 that is a non trivial bounded solution such that lim_{t→±∞} γ(t) = 0.

(II) The homoclinic solution γ_{1}(t) := \frac{\sqrt{π}}{2} γ(2\sqrt{\frac{2}{π}} t) is non-degenerate, that is the linear equation

v̈ + \frac{24}{π}f′(\frac{2}{\sqrt{π}} γ_{1}(t))v = 0

has no nontrivial bounded solutions.

(III) 10.5705675493 · |f′(0)| < δ.

If η ≠ 0 can be chosen in such a way that the equation

δ ∫_{−∞}^{∞} γ̇_{1}(s)^{2} ds + \frac{2}{\sqrt{π}} η ∫_{−∞}^{∞} ∫_{0}^{π/4}γ̇_{1}(s)h(x, s + α) dx ds = 0

has a simple root α, then there exists ε̄ > 0 such that for any ε ∈ (0, ε̄] and μ =

\sqrt{ε}η, equation (3) has a unique bounded solution on ℝ near γ(2\sqrt{\frac{2}{π}} (\sqrt{ε}t − α)) which is exponentially homoclinic to a unique small periodic solution of (3). Moreover, the Smale horseshoe (see [12]) can be embedded into the dynamics of (3).

Finally, a damped case is studied in [8] of the form

\begin{matrix}u_{tt} + u_{xxxx} + δu_{t} + h(x, t) = 0 , \\ u_{xx}(0, ·) = u_{xx}(π/4, ·) = 0 , \\ u_{xxx}(0, ·) = −f(u(0, ·)) , \\ u_{xxx}(π/4, ·) = g(u(π/4, ·)) , \end{matrix}  (4)

where δ is a positive constant, f and g are analytic, the function h ∈ C([0, π/4]× S^{T}) is splitted as follows

h(x, t) = 8\frac{θ_{2} − 2θ_{1}}{T π} + 96\frac{θ_{1} − θ_{2}}{T π^{2}} x + p(x, t)

for θ_{1,2} ∈ ℝ and

∫_{0}^{T} ∫_{0}^{π/4} p(x, t) dx dt = ∫_{0}^{T} ∫_{0}^{π/4} xp(x, t) dx dt = 0 .

Conditions are found in [8] between the numbers θ_{1,2}, the function p(x, t) and the nonlinearities f, g under which (4) has a T -periodic solution. It is also shown that under certain assumptions, constants θ_{1,2} are functions of p(x, t) in order to get a T -periodic solution of (4).

In this note, we are interested in T -periodic vibrations of (1) by using topological degree arguments. We show the existence of T -periodic vibrations of (1) for r > 0, k > 0 and f, g sublinear at infinity. Also a generic result is derived for this case when in addition f, g ∈ C^{1}(ℝ). If either r = 0 or k = 0, then we derive Landesman-Lazer type conditions on h, f, g for showing either existence or nonexistence results of T -periodic vibrations of (1).

2 Setting of the Problem

By a weak T -periodic solution of (1), we mean any u(x, t) ∈ C([0, π/4] × S^{T}) satisfying the identity

\begin{matrix}∫_{0}^{T} ∫_{0}^{π/4}[u(x, t)\{v_{tt}(x, t) + v_{xxxx}(x, t) − δv_{t}(x, t)\} + h(x, t)v(x, t)] dx dt \\ + ∫_{0}^{T} \{(ku(0, t) + f(u(0, t)))v(0, t) \\ +(ru(π/4, t) + g(u(π/4, t)))v(π/4, t)\} dt = 0 \end{matrix} (5)

for any v(x, t) ∈ C^{∞}([0, π/4] × S^{T}) such that the following boundary value conditions hold

v_{xx}(0, ·) = v_{xx}(π/4, ·) = v_{xxx}(0, ·) = v_{xxx}(π/4, ·) = 0 .  (6)

The eigenvalue problem

\begin{matrix}w_{xxxx}(x) = μ^{4}w(x) , \\ w_{xx}(0) = w_{xx}(π/4) = 0, w_{xxx}(0) = w_{xxx}(π/4) = 0 \end{matrix}

is known [9] to possesses a sequence of eigenvalues μ_{k}, k = −1, 0, 1, ⋯ with

μ_{−1} = μ_{0} = 0

and

cos(μ_{k}π/4) cosh(μ_{k}π/4) = 1, k = 1, 2, · · · .  (7)

The corresponding orthonormal in L^{2}(0, π/4) system of eigenvectors is

w_{−1}(x) = \frac{2}{\sqrt{π}} , w_{0}(x) = \frac{16}{π} (x − \frac{π}{8} ) \sqrt{\frac{3}{π}}
w_{k}(x) = \frac{4}{\sqrt{π}W_{k}} [ cosh(μ_{k}x) + cos(μ_{k}x)

−\frac{cosh ξ_{k}−cos ξ_{k}}{sinh ξ_{k}−sin ξ_{k}} ( sinh(μ_{k}x) + sin(μ_{k}x))]

where the constants W_{k} are given by the formulas

W_{k} = cosh(ξ_{k}) + cos ξ_{k} − \frac{cosh ξ_{k}−cos ξ_{k}}{sinh ξ_{k}−sin ξ_{k}} ( sinh ξ_{k} + sin ξ_{k})

for ξ_{k} = μ_{k}π/4. From (7) we get the asymptotic formulas

1 < μ_{k} = 2(2k + 1) + r(k) ∀k ≥ 1

along with

|r(k)| ≤ c̄_{1}e^{−c̄_{2}k} ∀k ≥ 1 ,

where c̄_{1}, c̄_{2} are positive constants. Moreover, the eigenfunctions {w_{i}}_{i=−1}^{∞} are uniformly bounded in C([0, π/4]).

3 Preliminary Results

Let H_{1}(x, t) ∈ C([0, π/4] × S^{T}), H_{2}(t), H_{3}(t) ∈ C(S^{T}) be continuous T -periodic functions and consider the equation

\begin{matrix}∫_{0}^{T} ∫_{0}^{π/4}[z(x, t)\{v_{tt}(x, t) + v_{xxxx}(x, t) − δv_{t}(x, t)\} + H_{1}(x, t)v(x, t)] dx dt \\ + ∫_{0}^{T} \{H_{2}(t)v(0, t) + H_{3}(t)v(π/4, t)\} dt = 0 \end{matrix} (8)

for any v(x, t) ∈ C^{∞}([0, π/4] × ℝ) satisfying the boundary conditions (6) along with

∫_{0}^{π/4} v(x, t)dx = ∫_{0}^{π/4} xv(x, t)dx = 0 ∀t ∈ S^{T} .  (9)

Note that conditions (9) correspond to the orthogonality of v(x, t) to w_{−1}(x) and w_{0}(x), for any t ∈ S^{T}. We look for z(x, t) in the form

z(x, t) = ∑_{i=1}^{∞} z_{i}(t)w_{i}(x) .  (10)

We formally put (10) into (8) to get a system of ordinary differential equations

z̈_{i}(t) + δż_{i}(t) + μ_{i}^{4}z_{i}(t) = h_{i}(t) ,  (11)

where

h_{i}(t) = −( ∫_{0}^{π/4}H_{1}(x, t)w_{i}(x) dx + H_{2}(t)w_{i}(0) + H_{3}(t)w_{i}(π/4)) .  (12)

Let us put

M_{1} = sup_{i≥1,x} |w_{i}(x)|, M_{2} = 4M_{1} ∑_{i=1}^{∞} 1/μ_{i}^{2} < ∞ M_{3} = sup_{i≥1} i^{2}/μ_{i}^{2} < ∞ . (13)

Since μ_{i} > 0 for i ≥ 1, equation (11) has a unique T -periodic solution z_{i}(t), for 2μ_{i}^{2} > δ given by

z_{i}(t) = \frac{2}{ω̄_{i}} ∫_{−∞}^{t} e^{−δ(t−s)/2} sin (\frac{ω̄_{i}}{2} (t − s)) × h_{i}(s) ds ,  (14)

where ω̄_{i} = \sqrt{4μ_{i}^{4} − δ^{2}}, for 2μ_{i}^{2} = δ given by

z_{i}(t) = ∫_{−∞}^{t} e^{−δ(t−s)/2}(t − s) × h_{i}(s) ds ,  (15)

and for 2μ_{i}^{2} < δ given by

z_{i}(t) = ∫_{−∞}^{t} \frac{1}{ω̃_{i}} (e^{(−δ+ω̃_{i})(t−s)/2} − e^{(−δ−ω̃_{i})(t−s)/2}) × h_{i}(s) ds ,  (16)

where ω̃_{i} = \sqrt{δ^{2} − 4μ_{i}^{4}}. Let ∥ · ∥_{∞} denote the maximum norm on [0, T]. From (14) for 3μ_{i}^{4} > δ^{2} we get

\begin{matrix}∥z_{i}∥_{∞} ≤ \frac{4}{ω̄_{i}δ} ∥h_{i}∥_{∞} ≤ \frac{4}{μ_{i}^{2}δ} ∥h_{i}∥_{∞} \\ ∥ż_{i}∥_{∞} ≤ \frac{4}{δ}∥h_{i}∥_{∞} , \end{matrix}  (17)

and for 3μ_{i}^{4} < δ^{2} < 4μ_{i}^{4} from (14) we get

\begin{matrix}∥z_{i}∥_{∞} ≤ ∥h_{i}∥_{∞} ∫_{−∞}^{t} e^{−δ(t−s)/2}(t − s) ds = \frac{4}{δ^{2}} ∥h_{i}∥_{∞} \\ ≤ _{\frac{4}{\sqrt{3}μ_{i}^{2}δ}} ∥h_{i}∥_{∞} ≤ \frac{4}{μ_{i}^{2}δ} ∥h_{i}∥_{∞} , \\ ∥ż_{i}∥_{∞} ≤ \frac{4}{δ}∥h_{i}∥_{∞} , \end{matrix}  (18)

where we use in derivation of (18) the inequality | sin x| ≤ |x| ∀x ∈ ℝ. From (15) for 2μ_{i}^{2} = δ we get

\begin{matrix}∥z_{i}∥_{∞} ≤ ∥h_{i}∥_{∞} ∫_{−∞}^{t} e^{−δ(t−s)/2}(t − s) ds = \frac{4}{δ^{2}} ∥h_{i}∥_{∞} \\ = \frac{2}{μ_{i}^{2}δ} ∥h_{i}∥_{∞} ≤ \frac{4}{μ_{i}^{2}δ} ∥h_{i}∥_{∞} , \\ ∥ż_{i}∥_{∞} ≤ \frac{4}{δ}∥h_{i}∥_{∞} . \end{matrix}  (19)

From (16) for 2μ_{i}^{2} < δ we get

\begin{matrix}∥z_{i}∥_{∞} ≤ \frac{1}{μ_{i}^{4}} ∥h_{i}∥_{∞} \\ ∥ż_{i}∥_{∞} ≤ \frac{δ}{μ_{i}^{4}} ∥h_{i}∥_{∞} ≤ δ∥h_{i}∥_{∞} . \end{matrix}  (20)

From (12) we get

∥h_{i}∥_{∞} ≤ M_{1}(\frac{π}{4} ∥H_{1}∥_{∞} + ∥H_{2}∥_{∞} + ∥H_{3}∥_{∞}) .  (21)

We consider the Banach space C([0, π/4] × S^{T}) with the usual maximum norm ∥ · ∥_{∞}. We need the following result.

Proposition 1. A sequence {z^{n}(x, t)}_{n=1}^{∞} ⊂ C([0, π/4] × S^{T}) is precompact if there is a constant M > 0 such that

sup_{i≥1,n≥1} ∥z_{i}^{n}∥_{∞}i^{2} < M, sup_{i≥1,n≥1} ∥ż_{i}^{n}∥_{∞} < M ,  (22)

where z^{n}(x, t) = ∑_{i=1}^{∞} z_{i}^{n}(t)w_{i}(x).

Proof. From (22) we get

|z_{i}^{n}(t)|i^{2} ≤ M, |ż_{i}^{n}(t)| ≤ M ∀t ∈ S^{T} .

By the Arzela-Ascoli theorem, there is a subsequence {z_{1}^{n_{k}}}_{k=1}^{∞} of {z_{1}^{n}}_{n=1}^{∞} such that z_{1}^{n_{k}}(t) → z_{1}^{0}(t) uniformly on S^{T}. Similarly we have a subsequence {z_{2}^{n_{ks}} }_{s=1}^{∞} of {z_{2}^{n_{k}}}_{k=1}^{∞} such that z_{2}^{n_{ks}} (t) → z_{2}^{0}(t) uniformly on S^{T}. Then we follow this construction. By using the Cantor diagonal procedure, we find an increasing sequence {m_{k}}_{k=1}^{∞} ⊂ ℕ such that z_{i}^{m_{k}}(t) → z_{i}^{0}(t) uniformly on S^{T} for any i ≥ 1. Clearly we have

sup_{i≥1} ∥z_{i}^{0}∥_{∞}i^{2} ≤ M .

Hence z^{0}(x, t) = ∑_{i=1}^{∞} z_{i}^{0}(t)w_{i}(x) ∈ C([0, π/4] × S^{T}). Let ε > 0 be given. Then

we choose i_{1} ∈ ℕ so large that 2M_{1}M ∑_{i=i_{1}}^{∞} i^{−2} < ε/2. We estimate

\begin{matrix}∥z^{m_{k}} − z^{0}∥_{∞} ≤ M_{1} ∑_{i=1}^{∞} ∥z_{i}^{m_{k}} − z_{i}^{0}∥_{∞} \\ ≤ 2M_{1}M ∑_{i=i_{1}}^{∞} i^{−2} + M_{1} ∑_{i=1}^{i_{1}} ∥z_{i}^{m_{k}} − z_{i}^{0}∥_{∞} < \frac{ε}{2} + M_{1} ∑_{i=1}^{i_{1}} ∥z_{i}^{m_{k}} − z_{i}^{0}∥_{∞} . \end{matrix}

For 1 ≤ i < i_{1}, we have

∥z_{i}^{m_{k}} − z_{i}^{0}∥_{∞} → 0

as k → ∞. Hence ∥z^{m_{k}} − z^{0}∥_{∞} < ε/2 for k large . This implies z^{m_{k}} → z^{0} in C([0, π/4] × S^{T}). The proof is finished.

Now if h_{i}(t), i ≥ 1 is given by (12) and T -periodic z_{i}(t) are defined by (11), then z(x, t) given by (10) satisfies z(x, t) ∈ C([0, π/4] × S^{T}). Indeed, from

∑_{i=1}^{∞} μ_{i}^{−2} < ∞ and (17) - (20) we have that the series (10) is uniformly convergent.

Hence z(x, t) ∈ C([0, π/4] × S^{T}) and (17) - (20) also imply

∥z∥_{∞} ≤ M_{1} ∑_{2μ_{i}<δ} \frac{1}{μ_{i}^{4}} ∥h_{i}∥_{∞} + M_{1} ∑_{2μ_{i}≥δ} \frac{4}{μ_{i}^{2}δ} ∥h_{i}∥_{∞}

≤ M_{1} ∑_{i=1}^{∞} (\frac{1}{μ_{i}^{2}} + \frac{4}{μ_{i}^{2}δ} )∥h_{i}∥_{∞}

≤ M_{2}(\frac{1}{δ} + \frac{1}{4})(\frac{π}{4} ∥H_{1}∥_{∞} + ∥H_{2}∥_{∞} + ∥H_{3}∥_{∞}) .

Moreover, we derive

sup_{i≥1} ∥z_{i}∥_{∞}i^{2} ≤ M_{3}(\frac{4}{δ} + 1) (\frac{π}{4} ∥H_{1}∥_{∞} + ∥H_{2}∥_{∞} + ∥H_{3}∥_{∞})  (23)

and

sup_{i≥1} ∥ż_{i}∥_{∞} ≤ (\frac{4}{δ} + δ) (\frac{π}{4} ∥H_{1}∥_{∞} + ∥H_{2}∥_{∞} + ∥H_{3}∥_{∞}) .  (24)

We also know from [3] that such z(x, t) satisfies (8). On the other hand, if

z(x, t) ∈ C([0, π/4]×S^{T}) satisfies (8), then z(x, t) = ∑_{i=1}^{∞} z_{i}(t)w_{i}(x) in L^{2}(0, π/4)

for a.e. t ∈ S^{T}. By inserting v(x, t) = ϕ(t)w_{i}(x) in (8) with ϕ ∈ C^{∞}(S^{T}), we get (11) with (12). So z(x, t) has the above properties.

Finally, let us define the following Banach space

\begin{matrix}C_{0}([0, π/4] × S^{T}) := \{z(x, t) ∈ C([0, π/4] × S^{T}) | ∫_{0}^{π/4}z(x, t) dx = \\ ∫_{0}^{π/4}z(x, t)x dx = 0 ∀ t ∈ S^{T} \} \end{matrix}

with the maximum norm ∥ · ∥_{∞} on [0, π/4] × S^{T}. Summarizing, we arrive at the following result.

Proposition 2. For any given functions H_{1}(x, t) ∈ C([0, π/4] × S^{T}), H_{2}(t), H_{3}(t) ∈ C(S^{T}), equation (8) has a unique solution z(x, t) ∈ C_{0}([0, π/4] × S^{T}) of the form

z(x, t) = ∑_{i=1}^{∞} z_{i}(t)w_{i}(x) .

Such a solution satisfies the condition (9) along with:

(a) z(x, t) ∈ X for the Banach space

\begin{matrix}X = \{z(x, t) ∈ C([0, π/4] × S^{T}) | z(x, t) = ∑_{i=1}^{∞} z_{i}(t)w_{i}(x), \\ sup_{i≥1} ∥z_{i}∥_{∞}i^{2} < ∞\} . \end{matrix}

(b) ∥z∥_{∞} ≤ M_{2}(\frac{1}{δ} + \frac{1}{4}) (\frac{π}{4} ∥H_{1}∥_{∞} + ∥H_{2}∥_{∞} + ∥H_{3}∥_{∞}).

(c) The mapping L : C([0, π/4] × S^{T}) × C(S^{T}) × C(S^{T}) → C_{0}([0, π/4] × S^{T}) defined by L(H_{1}, H_{2}, H_{3}) := z(x, t) is compact.

Proof. Properties (a) and (b) are proved above. Property (c) follows from Lemma 1 and inequalities (23) and (24). The proof is finished.

4 Nonhomogeneous Linear Problems

In this section, we consider the linear problem

\begin{matrix}u_{tt} + u_{xxxx} + δu_{t} + h(x, t) = 0 , \\ u_{xx}(0, ·) = u_{xx}(π/4, ·) = 0 , \\ u_{xxx}(0, ·) = −ku(0, ·) − f_{1}(t) , \\ u_{xxx}(π/4, ·) = ru(u(π/4, ·) + f_{2}(t) , \end{matrix}  (25)

where h(x, t) ∈ C([0, π/4] × S^{T}), f_{1}(t), f_{2}(t) ∈ C(S^{T}). Of course, we consider (25) in the sense of (5). Now we split u(x, t) as follows

u(x, t) = y_{1}(t)w_{−1}(x) + y_{2}(t)w_{0}(x) + z(x, t)

with z(x, t) ∈ C_{0}([0, π/4] × S^{T}). Then (25) is equivalent to the system

\begin{matrix}ÿ_{1}(t) + δẏ_{1}(t) + \frac{2}{\sqrt{π}} ∫_{0}^{π/4}h(x, t) dx \\ +\frac{4}{π} (k + r)y_{1}(t) + \frac{4\sqrt{3}}{π}(r − k)y_{2}(t) \\ +\frac{2}{\sqrt{π}} (kz(0, t) + rz(π/4, t) + f_{1}(t) + f_{2}(t)) = 0 , \end{matrix}  (26)

\begin{matrix}ÿ_{2}(t) + δẏ_{2}(t) + \frac{16}{π} \sqrt{\frac{3}{π}} ∫_{0}^{π/4}h(x, t)(x − \frac{π}{8} ) dx \\ +\frac{4\sqrt{3}}{π}(r − k)y_{1}(t) + \frac{12}{π}(k + r)y_{2}(t) \\ +2\sqrt{\frac{3}{π}} (rz(π/4, t) − kz(0, t) + f_{2}(t) − f_{1}(t)) = 0 , \end{matrix}  (27)

\begin{matrix}∫_{0}^{T} ∫_{0}^{π/4}[z(x, t)\{v_{tt}(x, t) + v_{xxxx}(x, t) − δv_{t}(x, t)\} + h(x, t)v(x, t)] dx dt \\ + ∫_{0}^{T} \{(k (\frac{2}{\sqrt{π}} y_{1}(t) − 2\sqrt{\frac{3}{π}} y_{2}(t) + z(0, t)) + f_{1}(t)) v(0, t) \\ +(r(\frac{2}{\sqrt{π}} y_{1}(t) + 2\sqrt{\frac{3}{π}} y_{2}(t) + z(π/4, t)) + f_{2}(t))v(π/4, t)\} dt = 0 \end{matrix} (28)

where v(x, t) ∈ C^{∞}([0, \frac{π}{4} ] × S^{T}) satisfies the conditions (6), (9). Let us define

L_{1} : C(S^{T}) × C(S^{T}) → C_{0}([0, π/4] × S^{T}) ,
L_{2} : C_{0}([0, π/4] × S^{T}) → C_{0}([0, π/4] × S^{T}) ,
H_{4} ∈ C_{0}([0, π/4] × S^{T})

by

L_{1}(y_{1}, y_{2}) := L(0, k(\frac{2}{\sqrt{π}} y_{1}(t) − 2\sqrt{\frac{3}{π}} y_{2}(t)), r(\frac{2}{\sqrt{π}} y_{1}(t) + 2\sqrt{\frac{3}{π}} y_{2}(t))) ,
L_{2}(z) := L(0, kz(0, t), rz(π/4, t)) ,
H_{4}(t) := L(h(x, t), f_{1}(t), f_{2}(t)) .

Then according to Proposition 2, equation (28) has the form

z = L_{2}(z) + L_{1}(y_{1}, y_{2}) + H_{4} .  (29)

Moreover, operators L_{1} and L_{2} are compact. Furthermore, since for r > 0, k > 0 the matrix

A = \frac{4}{π} ( \begin{matrix}(k + r) \sqrt{3}(r − k) \\ \sqrt{3}(r − k) 3(r + k) \end{matrix} )

is invertible, the system

ÿ + δẏ + Ay = h̄(t) = (h_{1}(t), h_{2}(t)) ∈ C(S^{T})^{2}  (30)

has a unique T -periodic solution y = (y_{1}, y_{2}) := L_{3}(h_{1}, h_{2}). Let us define

L_{4} : C_{0}([0, π/4] × S^{T}) → C(S^{T})^{2}, H_{5} ∈ C(S^{T})

given by

L_{4}(z) := L_{3}(\frac{2}{\sqrt{π}} (kz(0, t) + rz(π/4, t)) , 2\sqrt{\frac{3}{π}} (rz(π/4, t) − kz(0, t)) ) ,

H_{5}(t) := L_{3}(\frac{2}{\sqrt{π}} ∫_{0}^{π/4}h(x, t) dx + \frac{2}{\sqrt{π}} (f_{1}(t) + f_{2}(t)),

\frac{16}{π} \sqrt{\frac{3}{π}} ∫_{0}^{π/4}h(x, t)(x − \frac{π}{8} ) dx + 2\sqrt{\frac{3}{π}} (f_{2}(t) − f_{1}(t))) .

Then (26) and (27) are equivalent to

y = L_{4}(z) + H_{5}(t)  (31)

and L_{4} is compact. Consequently, in order to solve uniquely equations (29) and (31), we must show that if

\begin{matrix}z = L_{2}(z) + L_{1}(y_{1}, y_{2}), z ∈ C_{0}([0, π/4] × S^{T}) \\ y = L_{4}(z), y = (y_{1}, y_{2}) ∈ C(S^{T})^{2} , \end{matrix}  (32)

then z = 0 and y = 0. Equation (32) is equivalent to the system

\begin{matrix}z̈_{j}(t) + δż_{j}(t) + μ_{j}^{4}z_{j}(t) \\ + ∑_{s=−1}^{∞} (kz_{s}(t)w_{s}(0)w_{j}(0) + rz_{s}(t)w_{s}(π/4)w_{j}(π/4)) = 0 \end{matrix}  (33)

for z_{j} ∈ C^{2}(S^{T}), j ≥ −1 with sup_{j≥−1} ∥z_{j}∥_{∞}(j^{2} + 1) < ∞. Let us expand

z_{j}(t) = ∑_{m∈ℤ} e^{ı2πmt/T} c_{mj} .

Note that c_{mj} ∼ j^{−2} as j → ∞ uniformly for m ∈ ℤ. Then by (33) we derive

\begin{matrix}c_{mj} (μ_{j}^{4} − \frac{4π^{2}m^{2}}{T^{2}} + ı\frac{2δπ}{T} m) \\ + ∑_{s=−1}^{∞} (kc_{ms}w_{s}(0)w_{j}(0) + rc_{ms}w_{s}(π/4)w_{j}(π/4)) = 0 . \end{matrix}  (34)

By taking c_{mj} = a_{mj} + ıb_{mj}, from (34) we derive

a_{mj} (μ_{j}^{4} − \frac{4π^{2}m^{2}}{T^{2}} ) − \frac{2δπ}{T} mb_{mj}

+ ∑_{s=−1}^{∞} (kw_{s}(0)w_{j}(0) + rw_{s}(π/4)w_{j}(π/4))a_{ms} = 0 ,

a_{mj} \frac{2δπ}{T} m + (μ_{j}^{4} − \frac{4π^{2}m^{2}}{T^{2}} ) b_{mj}

+ ∑_{s=−1}^{∞} (kw_{s}(0)w_{j}(0) + rw_{s}(π/4)w_{j}(π/4))b_{ms} = 0 .

Since a_{mj}, b_{mj} ∼ j^{−2} as j → ∞, we get

∑_{j=−1}^{∞} (a_{mj}^{2} + b_{mj}^{2}) \frac{2πδ}{T} m = 0 ,

hence a_{mj} = b_{mj} = 0 for any m ≠ 0 and j. For m = 0 we get

a_{0j}μ_{j}^{4} + ∑_{s=−1}^{∞} (kw_{s}(0)w_{j}(0) + rw_{s}(π/4)w_{j}(π/4))a_{0s} = 0 ,  (35)

b_{0j}μ_{j}^{4} + ∑_{s=−1}^{∞} (kw_{s}(0)w_{j}(0) + rw_{s}(π/4)w_{j}(π/4))b_{0s} = 0 .  (36)

We put a_{0j}(μ_{j}^{4} + 1) = e_{j} and from (35) we get

∑_{j=−1}^{∞} \frac{e_{j}^{2}}{μ_{j}^{4}+1} \frac{μ_{j}^{4}}{μ_{j}^{4}+1} + k( ∑_{s=−1}^{∞} w_{s}(0)\frac{e_{s}}{μ_{s}^{4}+1})^{2} + r( ∑_{s=−1}^{∞} w_{s}(π/4)\frac{e_{s}}{μ_{s}^{4}+1})^{2} = 0 . (37)

From (37) for r > 0, k > 0 we immediately get e_{j} = 0 for j ≥ 1 and

\frac{2}{\sqrt{π}} e_{−1} − 2\sqrt{\frac{3}{π}} e_{0} = 0, \frac{2}{\sqrt{π}} e_{−1} + 2\sqrt{\frac{3}{π}} e_{0} = 0 ,

which imply also e_{−1} = e_{0} = 0. Similar results hold for (36). Hence, (32) has the only solution z = 0 and y = 0. Consequently, (29) and (31) are uniquely solvable in z, y for r > 0, k > 0. Summarizing, we arrive at the following result.

Proposition 3. If r > 0, k > 0 then for any given functions h(x, t) ∈ C([0, π/4] × S^{T}) and f_{1}(t), f_{2}(t) ∈ C(S^{T}), equation (25) has a unique solution u(x, t) ∈ C([0, π/4] × S^{T}) of the form

u(x, t) = ∑_{i=−1}^{∞} z_{i}(t)w_{i}(x) .

Such a solution satisfies:

(a) u(x, t) ∈ Y for the Banach space

\begin{matrix}Y = \{u(x, t) ∈ C([0, π/4] × S^{T}) | u(x, t) = ∑_{i=−1}^{∞} z_{i}(t)w_{i}(x), \\ ∥u∥ := sup_{i≥−1} ∥z_{i}∥_{∞}(|i| + 1)^{2} < ∞\} . \end{matrix}

(b) ∥u∥, ∥u∥_{∞} ≤ c (∥h∥_{∞} + ∥f_{1}∥_{∞} + ∥f_{2}∥_{∞}) for a constant c > 0.

(c) The mapping L̃ : C([0, π/4] × S^{T}) × C(S^{T}) × C(S^{T}) → C([0, π/4] × S^{T}) defined by L̃(h, f_{1}, f_{2}) := u(x, t) is compact.

We also define a compact mapping L̄ : C(S^{T}) × C(S^{T}) → C([0, π/4] × S^{T}) by L̄(f_{1}, f_{2}) := L̃(0, f_{1}, f_{2}). We denote by ∥L̄∥ the norm of L̄.

Now we study the case when r = 0 and k > 0 in (25). Then equation (29) remains, but the matrix A is no more invertible. Equation (30) has a T -periodic solution if and only if

∫_{0}^{T}(\sqrt{3}h_{1}(t) + h_{2}(t)) dt = 0

and the linear equation

ÿ + δẏ + Ay = 0

has the only T -periodic solutions y(t) = c(\sqrt{3}, 1), c ∈ ℝ. Consequently, we are still working with Fredholm operators of index 0 possessing forms of compact perturbations of identity operators [11]. Hence, in order to study (25) we consider like in (33) the equations

z̈_{j}(t) + δż_{j}(t) + μ_{j}^{4}z_{j}(t) + k ∑_{s=−1}^{∞} z_{s}(t)w_{s}(0)w_{j}(0) = 0  (38)

z̈_{j}(t) + δż_{j}(t) + μ_{j}^{4}z_{j}(t) + k ∑_{s=−1}^{∞} z_{s}(t)w_{s}(0)w_{j}(0) + h_{j}(t) = 0  (39)

for h_{j} ∈ C(S^{T}), z_{j} ∈ C^{2}(S^{T}), j ≥ −1 with sup_{j≥−1} ∥z_{j}∥_{∞}(j^{2} + 1) < ∞. Like for

(33), we get that z_{−1}(t) = c\sqrt{3}, z_{0}(t) = c, c ∈ ℝ, z_{j}(t) = 0, j ≥ 1 for (38). According to the above comments, the set of all {h_{j}(t)}_{j≥−1} for which (39) is solvable must have a codimension 1. For this reason, we consider an adjoint equation to (38) of the form

v̈_{j}(t) − δv̇_{j}(t) + μ_{j}^{4}v_{j}(t) + k ∑_{s=−1}^{∞} v_{s}(t)w_{s}(0)w_{j}(0) = 0  (40)

for v_{j} ∈ C^{2}(S^{T}), j ≥ −1 with sup_{j≥−1} ∥v_{j}∥_{∞}(j^{2} + 1) < ∞. Like above we get

v_{−1}(t) = c\sqrt{3}, v_{0}(t) = c, c ∈ ℝ, v_{j}(t) = 0, j ≥ 1 for (40). By multiplying (39) with v_{j}(t) and using integration by parts, we get

\begin{matrix}∫_{0}^{T} z_{j}(t)(v̈_{j}(t) − δv̇_{j}(t) + μ_{j}^{4}v_{j}(t)) dt \\ +k ∫_{0}^{T} ( ∑_{s=−1}^{∞} z_{s}(t)w_{s}(0)w_{j}(0)v_{j}(t)) dt + ∫_{0}^{T} h_{j}(t)v_{j}(t) dt = 0 . \end{matrix}  (41)

Inserting (40) to (41) we obtain

\begin{matrix}k ∫_{0}^{T} ∑_{s=−1}^{∞} (z_{s}(t)w_{s}(0)w_{j}(0)v_{j}(t) − z_{j}(t)w_{s}(0)w_{j}(0)v_{s}(t)) dt \\ + ∫_{0}^{T} h_{j}(t)v_{j}(t) dt = 0 . \end{matrix}  (42)

Since v_{j}(t) ∼ j^{−2}, z_{j}(t) ∼ j^{−2} uniformly on S^{T}, (42) implies

0 = ∑_{s=−1}^{∞} ∫_{0}^{T} h_{j}(t)v_{j}(t) dt = ∫_{0}^{T}(\sqrt{3}h_{−1}(t) + h_{0}(t)) dt .  (43)

We recall that the set of all {h_{j}(t)}_{j≥−1} for which (39) is solvable has a codimension 1. Then condition (43) is necessary and also sufficient for solvability of (39). We note

\begin{matrix}h_{−1}(t) = \frac{2}{\sqrt{π}} ∫_{0}^{π/4}h(x, t) dx + \frac{2}{\sqrt{π}} (f_{1}(t) + f_{2}(t)) , \\ h_{0}(t) = \frac{16}{π} \sqrt{\frac{3}{π}} ∫_{0}^{π/4}h(x, t)(x − \frac{π}{8} ) dx + 2\sqrt{\frac{3}{π}} (f_{2}(t) − f_{1}(t)) . \end{matrix}

Then condition (43) has the form

\frac{4}{π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)x dx dt + ∫_{0}^{T} f_{2}(t) dt = 0 .  (44)

Finally, the corresponding kernel to (38) is spanned by the function

z_{−1}(t)w_{−1}(x) + z_{0}(t)w_{0}(x) = \frac{16}{π}\sqrt{\frac{3}{π}} x .  (45)

Summarizing we get the next result.

Proposition 4. If r = 0, k > 0 then for any given functions h(x, t) ∈ C([0, π/4]×S^{T}) and f_{1}(t), f_{2}(t) ∈ C(S^{T}), equation (25) has a solution u(x, t) ∈ C([0, π/4] × S^{T}) if and only if condition (44) holds. Such a solution is unique if

∫_{0}^{T} ∫_{0}^{π/4}u(x, t)x dx dt = 0 .  (46)

Moreover, the mapping K : C_{1} → C_{2} is compact where

C_{1} := {(h, f_{1}, f_{2}) ∈ C([0, π/4] × S^{T}) × C(S^{T})^{2} | condition (44) holds} ,

C_{2} := {u ∈ C([0, π/4] × S^{T}) | condition (46) holds}

are Banach spaces endowed with the maximum norms and the mapping K is defined by K(h, f_{1}, f_{2}) := u(x, t).

Similarly we derive the next results.

Proposition 5. If r > 0, k = 0 then for any given functions h(x, t) ∈ C([0, π/4]×S^{T}) and f_{1}(t), f_{2}(t) ∈ C(S^{T}), equation (25) has a solution u(x, t) ∈ C([0, π/4] × S^{T}) if and only if

\frac{4}{π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)(\frac{π}{4} − x) dx dt + ∫_{0}^{T} f_{1}(t) dt = 0 .  (47)

Such a solution is unique if

∫_{0}^{T} ∫_{0}^{π/4}u(x, t)(\frac{π}{4} − x) dx dt = 0 .  (48)

Moreover, the mapping K̃ : C̃_{1} → C̃_{2} is compact where

C̃_{1} := {(h, f_{1}, f_{2}) ∈ C([0, π/4] × S^{T}) × C(S^{T})^{2} | condition (47) holds} ,

C̃_{2} := {u ∈ C([0, π/4] × S^{T}) | condition (48) holds}

are Banach spaces endowed with the maximum norms and the mapping K is defined by K̃(h, f_{1}, f_{2}) := u(x, t).

Proposition 6. If r = k = 0 then for any given functions h(x, t) ∈ C([0, π/4]×S^{T}) and f_{1}(t), f_{2}(t) ∈ C(S^{T}), equation (25) has a solution u(x, t) ∈ C([0, π/4] × S^{T}) if and only if the both conditions (44) and (47) hold. Such a solution is unique if the both conditions (46) and (48) hold.

Moreover, the mapping K̄ : C̄_{1} → C̄_{2} is compact where

C̄_{1} := {(h, f_{1}, f_{2}) ∈ C([0, π/4] × S^{T}) × C(S^{T})^{2} | conditions (44), (47) hold} ,

C̄_{2} := {u ∈ C([0, π/4] × S^{T}) | conditions (46), (48) hold}

are Banach spaces endowed with the maximum norms and the mapping K̄ is defined by K̄(h, f_{1}, f_{2}) := u(x, t).

5 Nonlinear Problems

In this section, we present the main results concerning equation (1).

Theorem 1. If r > 0, k > 0 and there are positive constants c_{11}, c_{12}, c_{21}, c_{22} along with

c_{12} + c_{22} < 1/∥L̄∥

and such that

\begin{matrix}|f(u)| ≤ c_{11} + c_{12}|u|, ∀u ∈ ℝ \\ |g(u)| ≤ c_{21} + c_{22}|u|, ∀u ∈ ℝ , \end{matrix}

then for any given function h(x, t) ∈ C([0, π/4] × S^{T}), equation (1) possesses a weak T -periodic solution u(x, t) ∈ C([0, π/4] × S^{T}).

Proof. By using the above results, the proof is standard. According to Proposition 3, equation (1) is equivalent to

u = F (u) := L̃(h, 0, 0) + L̄(f(u(0, ·)), g(u(π/4, ·))) .  (49)

Proposition 3 also implies the compactness of the mapping

F : C([0, π/4] × S^{T}) → C([0, π/4] × S^{T}) .

From the assumptions of Theorem 1 and (b) of Proposition 3, we get

\begin{matrix}∥F (u)∥_{∞} ≤ c∥h∥_{∞} + ∥L̄∥(∥f(u(0, ·))∥_{∞} + ∥g(u(π/4, ·))∥_{∞}) \\ ≤ c∥h∥_{∞} + ∥L̄∥(c_{11} + c_{21} + (c_{12} + c_{22})∥u∥_{∞}) . \end{matrix}  (50)

Since ∥L̄∥(c_{12} + c_{22}) < 1, there is a unique τ > 0 such that

τ = c∥h∥_{∞} + ∥L̄∥(c_{11} + c_{21} + (c_{12} + c_{22})τ) .

Consequently, (50) implies that the ball

B_{τ} = {u ∈ C([0, π/4] × S^{T}) | ∥u∥ ≤ τ}

is mapped to itself by the mapping F. The Schauder fixed point theorem ensures the existence of a fixed point u ∈ B_{τ} of F. This gives a weak T -periodic solution of (1). The proof is finished.

Of course, when f, g have sublinear growth at infinity:

lim_{|u|→∞} f(u)/u = 0, lim_{|u|→∞} g(u)/u = 0

and r > 0, k > 0, then the assumptions of Theorem 1 hold and equation (1) possesses a weak T -periodic solution u(x, t) ∈ C([0, π/4] × S^{T}) for any h(x, t) ∈ C([0, π/4] × S^{T}).

The implicit function theorem together with Proposition 3 gives the next result.

Theorem 2. If r > 0, k > 0, f(0) = f′(0) = g(0) = g′(0) = 0 and f, g ∈ C^{1}(S^{T}), then there are positive constants K_{1}, ε_{0} such that for any given function h(x, t) ∈ C([0, π/4] × S^{T}) with ∥h∥_{∞} < ε_{0}, equation (1) possesses a unique small weak T -periodic solution u(x, t) ∈ C([0, π/4] × S^{T}) satisfying ∥u∥_{∞} ≤ K_{1}∥h∥_{∞}.

Now we suppose that r = 0 and k > 0 in equation (1). Let

P : C([0, π/4] × S^{T}) × C(S^{T})^{2} → C_{1}

be a continuous projection and let

C_{2} ⊕ ℝx = C([0, π/4] × S^{T})

be a continuous splitting u(x, t) = v(x, t) + c\frac{4}{π} x with v(x, t) ∈ C_{2} and c ∈ ℝ. Then according to Proposition 4, equation (1) is equivalent to the system

v = λK(P (h, f(v(0, ·)), g(c + v(π/4, ·)))) ,  (51)

\frac{4}{π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)x dx dt + ∫_{0}^{T} g(c + λv(π/4, t)) dt = 0  (52)

for λ = 1. Now we can prove the next result.

Theorem 3. Let r = 0 and k > 0. If sup_{u∈ℝ} |f(u)| < ∞, finite limits

lim_{u→±∞} g(u) := g_{±}

exist and it holds

\frac{4}{T π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)x dx dt ∈ (−g_{−}, −g_{+}) .  (53)

Then equation (1) possesses a weak T -periodic solution u(x, t) ∈ C([0, π/4] × S^{T}). On the other hand, if

|\frac{4}{T π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)x dx dt| > sup_{u∈ℝ} |g(u)|  (54)

then equation (1) has no weak T -periodic solutions.

Proof. This is a Landesman-Lazer type result [4], [10]. We consider (51) and (52) for 0 ≤ λ ≤ 1 on C_{2} ⊕ ℝ. Since h, f, g are bounded, Proposition 4 implies that any solution of (51) must satisfy ∥v∥_{∞} ≤ K_{1}, for a constant K_{1} > 0. We take the set

B = {(v, c) ∈ C_{2} ⊕ ℝ | ∥v∥_{∞} < K_{1} + 1, |c| < K_{2}}

for a fixed large K_{2} > 0. If (v, c) ∈ ∂B then either ∥v∥_{∞} = K_{1} + 1 and then (51) does not hold, or ∥v∥_{∞} ≤ K_{1} + 1 and c = ±K_{2}, and then (52) does not hold according to (53). Hence we can apply Leray-Schauder degree to (51) and (52) on B [4], [10]. For λ = 0 we get a function

c → \frac{4}{T π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)x dx dt + g(c) ,

which according to (53) changes the sign on [−K_{2}, K_{2}]. Consequently, (51) and (52) are solvable on B. On the other hand, if (52) holds then clearly (54) can not be satisfied. The proof is finished.

Similarly we get the next result.

Theorem 4. Let r = 0 and k > 0. If sup_{u∈ℝ} |f(u)| < ∞, finite limits

lim_{u→±∞} g(u) := g_{±}

exist and g is monotonic on ℝ. Then equation (1) possesses a weak T -periodic solution if (53) holds and it has no weak T -periodic solutions if

\frac{4}{T π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)x dx dt ∉ [−g_{−}, −g_{+}] .

If in addition, g is strictly monotonic on ℝ, then equation (1) possesses a weak T -periodic solution if and only if (53) holds.

By using Proposition 5, similar arguments hold for the case r > 0 and k = 0. We state this result for the reader convenience.

Theorem 5. Let r > 0 and k = 0. If sup_{u∈ℝ} |g(u)| < ∞, finite limits

lim_{u→±∞} f(u) := f_{±}

exist and it holds

\frac{4}{T π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)(\frac{π}{4} − x) dx dt ∈ (−f_{−}, −f_{+}) .  (55)

Then equation (1) possesses a weak T -periodic solution u(x, t) ∈ C([0, π/4] × S^{T}). On the other hand, if

|\frac{4}{T π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)(\frac{π}{4} − x) dx dt| > sup_{u∈ℝ} |f(u)|  (56)

then equation (1) has no weak T -periodic solutions.

Theorem 4 can be also modified for the case r > 0 and k = 0. Now we study the case that r = k = 0 in equation (1). This is a codimension two problem. The above approach to Theorem 3 can be used with the following modifications. Let

P̄ : C([0, π/4] × S^{T}) × C(S^{T})^{2} → C̄_{1}

be a continuous projection and let

C̄_{2} ⊕ ℝ(1 − \frac{4}{π} x) ⊕ ℝx = C([0, π/4] × S^{T})

be a continuous splitting u(x, t) = v(x, t) + c_{1}(1 − \frac{4}{π} x) + c_{2} \frac{4}{π} x with v(x, t) ∈ C̄_{2} and c_{1}, c_{2} ∈ ℝ. Then according to Proposition 6, equation (1) is equivalent to the system

v = λK̄(P̄(h, f(c_{1} + v(0, ·)), g(c_{2} + v(π/4, ·)))) ,  (57)

\frac{4}{π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)(\frac{π}{4} − x) dx dt + ∫_{0}^{T} f(c_{1} + λv(0, t)) dt = 0  (58)

\frac{4}{π} ∫_{0}^{T} ∫_{0}^{π/4}h(x, t)x dx dt + ∫_{0}^{T} g(c_{2} + λv(π/4, t)) dt = 0  (59)

for λ = 1. By repeating the proof of Theorem 3 to (57)-(59), we get the next result.

Theorem 6. Let r = k = 0. If finite limits

lim_{u→±∞} f(u) := f_{±}, lim_{u→±∞} g(u) := g_{±}

exist and the both conditions (53) and (55) hold, then equation (1) possesses a weak T -periodic solution u(x, t) ∈ C([0, π/4] × S^{T}). On the other hand, if one of the conditions (54) and (56) is satisfied, then equation (1) has no weak T -periodic solutions.

Now let us suppose that f, g ∈ C^{1}(ℝ). If we consider equation (8) for any v(x, t) ∈ C^{∞}([0, π/4] × ℝ) satisfying the boundary conditions (6) and also orthogonal to each w_{i}(x), i = −1, 0, ⋯, i_{1} for i_{1} ∈ ℕ large and fixed, like in (9). Then we look for z(x, t) in the form

z(x, t) = ∑_{i=i_{1}+1}^{∞} z_{i}(t)w_{i}(x) ,

and we get a result similar to Proposition 2 with an estimate as (b) for M_{2} → 0 as i_{1} → ∞. Consequently, we can locally reduce by means of the Ljapunov-Schmidt method the solvability of (1) to finite-dimensional mappings. In this way, we can repeat the proof of the Sard-Smale theorem [4], [11] for (1). Moreover, by following a method of [11], we can prove the next result.

Theorem 7. Let the assumptions of Theorem 1 hold along with that f, g ∈ C^{1}(ℝ). Then there is an open and dense subset C_{3} ⊂ C([0, π/4]×S^{T}) such that for any given h(x, t) ∈ C_{3}, equation (1) possesses a finite nonzero number of weak T -periodic solutions u(x, t) ∈ C([0, π/4] × S^{T}). This number of solutions is constant on each connected components of C_{3}.

Finally, we note that the question on the existence of a global bounded weak solution of (1) remains open when h(x, t) is only bounded on [0, π/4] × S^{T}. A combination of methods of [1] and this paper would be hopeful.

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(Received July 21, 2003)