Consider the equation of the linear oscillator u '' + u = h(theta), where the forcing term h : R -> R is 2 pi-periodic and positive. We show that the existence of a periodic solution implies the existence of a positive solution. To this aim we establish connections between this problem and some separation questions of convex analysis.Paris Observatory gran

Albouy, Alain

Ureña, Antonio

Ureña, Antonio J.

Communications in Contemporary Mathematics

English

arXiv

 Consider the equation of the linear oscillator [Formula: see text], where the forcing term [Formula: see text] is [Formula: see text]-periodic and positive. We show that the existence of a periodic solution implies the existence of a positive solution. To this aim we establish connections between this problem and some separation questions of convex analysis. </jats:p

Alain Albouy

Antonio J. Ureña

Crossref

An antimaximum principle for periodic solutions of a forced oscillator

Ureña Alcázar, Antonio Jesús

Repositorio Institucional Universidad de Granada

arXiv:2201.09272v1  [math.CA]  23 Jan 2022An antimaximum principle for periodic solutions of aforced oscillatorAlain Albouy1, Antonio J. Ureña21 IMCCE, CNRS-UMR8028, Observatoire de Paris77, avenue Denfert-Rochereau, 75014 Paris, FranceAlain.Albouy@obspm.fr2 Departamento de Matematica Aplicada, Facultad de CienciasCampus Universitario de FuentenuevaUniversidad de Granada, 18071, Granada, Spainajurena@ugr.esAbstract. Consider the equation of the linear oscillator u′′ + u = h(θ),where the forcing term h : R → R is 2π-periodic and positive. We show thatthe existence of a periodic solution implies the existence of a positive solution.To this aim we establish connections between this problem and some separationquestions of convex analysis.Key Words: Antimaximum principle, forced linear oscillator, positive so-lutions, separation of convex sets.1 IntroductionConsider the linear, second-order equationu′′ + ω2u = h(θ) , (1)where ω > 0 is a positive constant, the prime symbol stands for differentiationwith respect to the independent variable θ, the function h : R → R is continuousand 2π-periodic, and the twice continuously differentiable function u = u(θ) isthe unknown. It models the simplest oscillations that one can think of: those ofa forced harmonic spring in one spatial dimension. The well-known Fredholmalternative theorem (see, e.g., [4, Lemma 1.1, p. 146]) provides a precise answerto the question of existence and multiplicity of 2π-periodic solutions. Moreprecisely, if ω is not an integer then (1) has a unique 2π-periodic solution, and ifω ∈ Z then there are periodic solutions if and only if the nonresonance condition∫2π0h(θ) cosωθ dθ =∫2π0h(θ) sinωθ dθ = 0 (2)1holds. Morever, in this situation all solutions are 2π-periodic. Assuming nowthat h = h(θ) has a sign we consider the question:Does (2) imply the existence of solutions u of (1) having the sign of h?Under the label of antimaximum principles, this problem has been studied indepth when ω 6∈ Z . For instance, it follows from [7, Theorem 1.1] that theanswer is positive for 0 < ω ≤ 1/2 and negative for 1/2 < ω 6∈ Z. The mainresult of this note is the following:Theorem 1.1. Assume that ω = 1. If h = h(θ) satisfies (2), is nonnegative,and not identically zero, then (1) has a solution u = u(θ) with u(θ) > 0 for allθ ∈ R.Some remarks are in order:(i) Equivalently we may restate this theorem as follows: in the case ω = 1, if0 6≡ h ∈ C(R/2πZ,R) is nonnegative and equation (1) has a periodic solution,then it has a everywhere-positive solution.(ii) One may wonder whether Theorem 1.1 remains true for ω = 2, 3, 4... InProposition 4 we shall provide a counterexample showing that the answer isnegative for ω ≥ 3. The case ω = 2 is not well understood by the authors.(iii) We point out that (1) can be solved in a number of elementary ways thatare found in classical textbooks. For instance, for ω = 1 the formula of variationof constants givesu(θ) =(α +∫θ0h(τ) cos τ dτ)sin θ −(β +∫θ0h(τ) sin τ dτ)cos θwith α, β ∈ R. Alternatively one can use Fourier series: if h(θ) = a0 +∑+∞n=2(an cosnθ + bn sinnθ) then the solutions of (1) are given byu(θ) = a0 −+∞∑n=2(ann2 − 1cos nθ +bnn2 − 1sinnθ)+ α cos θ + β sin θ,with α, β ∈ R. Analyzing the sign of these expressions seems hard and so bothapproaches are apparently ill-suited to answer the question formulated above.(iv) A more sophisticated study would involve the method of lower and uppersolutions, for which we refer to the classical work [2]. Indeed, for nonnegative hequation (1) has the upper solution β(θ) ≡ 0 and the lower solution α(θ) ≡ Mwith M > 0 big. Notice however that they are in the reverse order. Eventhough a number of results of the lower and upper solutions literature considerthe reverse order case (see, e.g., [7]), we are unaware of any that might apply to(1).Our approach to the problem will be quite different and, to the best of ourknowledge, new. In broad terms, the key idea will consist in relating the differentperiodic solutions of equation (1) with the supporting linear forms of a givenconvex, positively homogeneous map of degree 1.22 Convexity and supporting linear formsThe following terminology is usual. A function ρ : RN → R is said to bepositively homogeneous of degree 1 provided thatρ(λz) = λρ(z), z ∈ RN , λ > 0.Assume now that ρ : RN → R is positively homogeneous of degree 1 and satisfiesρ(0) = 0. Given e∗ ∈ (RN)∗,• it is usually said that e∗ is a supporting linear form for ρ ifρ(z) ≥ 〈z, e∗〉 for every z ∈ RN ;• we shall say that e∗ is a strictly supporting linear form for ρ provided thatρ(z) > 〈z, e∗〉 for every z ∈ RN\{0}. (3)On the other hand, one easily checks that if ρ : RN → R is positivelyhomogeneous of degree 1, then it is also convex if and only ifρ(z1 + z2) ≤ ρ(z1) + ρ(z2), z1, z2 ∈ RN .Since every convex function defined on (an open convex subset of) a finite-dimensional vector space is continuous (see e.g. [3, Theorem 5.2.1]), in thissituation we recover the previous condition ρ(0) = 0.The classical version of the Hahn-Banach theorem (see, e.g. [1, p.1, ThéorèmeI.1]) states that every function ρ : RN → R which is positively homogeneous ofdegree 1 and convex admits supporting linear forms. The main result of thissection is the followingProposition 2.1. Let ρ : RN → R be positively homogeneous of degree 1, convexand differentiable at every point of RN\{0}. If in addition ρ 6∈ (RN)∗, then itadmits a strictly supporting linear form.This variation of the Hahn-Banach theorem can be deduced from knownresults (see [5, (9.12), p. 459]), but we shall give a hopefully simpler proof. Ourapproach will be divided in two steps. In the first one we shall see that underthe assumptions of Proposition 2.1 one has the inequalityρ(z) + ρ(−z) > 0 for every z ∈ RN\{0}. (4)This is indeed a first step towards the proof of Proposition 2.1 since condition(4) is clearly necessary for the existence of strictly supporting linear forms.Lemma 2.2. Let ρ : RN → R be positively homogeneous of degree 1 and convex.If moreover ρ 6∈ (RN )∗, then ρ(z) + ρ(−z) > 0 for every z ∈ RN\{0} where ρ isdifferentiable.3Proof. Convexity gives0 = ρ(0) ≤ρ(z) + ρ(−z)2,for every z ∈ RN . Thus, it will suffice to check that ρ(z) + ρ(−z) 6= 0 for everyz ∈ RN\{0} where ρ is differentiable. We use an inductive argument on thedimension N .If N = 1 the result is trivial: if ρ : R → R is positively homogeneous ofdegree 1, the equality ρ(−z0)+ρ(z0) = 0 for some z0 6= 0 implies that ρ is linear.Assuming the statement true for some N , let ρ : RN+1 → R be positivelyhomogeneous of degree 1, convex, and not linear, and suppose, by a contra-diction argument, that ρ(z0) + ρ(−z0) = 0 for some z0 ∈ RN+1\{0} where ρis differentiable. Choose some hyperplane H ⊂ RN+1 with z0 ∈ H . By theinductive assumption the restriction ρ∣∣H= e∗ must be linear. Fix some vectorz1 ∈ RN+1\H and let the function ρ1 : H → R be defined byρ1(h) := ρ(z1 + h)− 〈e∗, h〉, h ∈ H.Then ρ1 is convex. Moreover, denoting by | · | a fixed (arbitrary) norm in RNone haslim|h|→∞ρ1(h)|h|= lim|h|→∞(ρ(z1|h|+h|h|)−〈e∗,h|h|〉)== lim|h|→∞(ρ(z1|h|+h|h|)− ρ(h|h|))= 0,since ρ is continuous, thus uniformly continuous on compact sets, and z1/|h| → 0.Hence, we see that ρ1 : H → R is convex and has sublinear growth at infinity,and we deduce that ρ1(h) ≡ c is constant. Then,ρ(h + z1) = 〈e∗, h〉+ c, h ∈ H,and, by homogeneity,ρ(h+ λz1) = 〈e∗, h〉+ cλ, h ∈ H, λ ≥ 0.Similarly, there exists another constant d ∈ R such that ρ(h+λz1) = 〈e∗, h〉+dλfor every h ∈ H and λ ≤ 0. In particular, taking h = z0 ∈ H we see thatρ(z0 + λz1) ={〈e∗, z0〉+ cλ if λ ≥ 0,〈e∗, z0〉+ dλ if λ < 0.Remembering that ρ was assumed differentiable at z0 we deduce that c = d. Itmeans that ρ is linear, a contradiction. The result follows.To complete the proof of Proposition 2.1 we shall check the following lemma.It goes back to Minkowski [6, XXV, pp. 151-153], but we include a short prooffor completeness.4Lemma 2.3. Let ρ : RN → R be positively homogeneous of degree 1, convex andsatisfy (4). Then it admits a strictly supporting linear form.Proof. We argue by induction on the dimension N . If N = 1 then the resultis trivial. Assuming that it holds for some N , let ρ : RN+1 → R satisfy theassumptions above. The identification RN+1 ≡ RN × R allows us to write thepoints of RN+1 as pairs (z, λ) where z ∈ RN and λ ∈ R. Since the functionρ0 : RN → R given by z 7→ ρ(z, 0) is positively homogeneous of degree 1, convex,and satisfies (4), then it admits a strictly supporting linear form e∗0 ∈ (RN)∗.Let e∗ ∈ (RN+1)∗ be defined by〈e∗, (z, λ)〉 := 〈e∗0, z〉+ αλ for every (z, λ) ∈ RN+1,where α is a real number, to be fixed later. Homogeneity means that e∗ is astrictly supporting linear form for ρ if and only if the inequality in (3) holds forevery ordered pair of the form (z,±1) with z ∈ RN . Equivalently, if and only ifρ1(z) := 〈e∗0, z〉 − ρ(z,−1) < α < ρ2(z) := ρ(z, 1)− 〈e∗0, z〉, z ∈ RN . (5)All three functions ρ, ρ1, ρ2 are continuous. On the other hand homogeneitygiveslim sup|z|→+∞ρ1(z)|z|= max|z|=1(〈e∗0, z〉 − ρ0(z))< 0,and therefore, ρ1(z) → −∞ as |z| → ∞. Similarly, ρ2(z) → +∞ as |z| → ∞ andwe see that ρ1 attains its global maximum and ρ2 attains its global minimum.Therefore, it will be possible to find a constant α satisfying (5) if and only ifρ1(z1) < ρ2(z2) for every z1, z2 ∈ RN . Equivalently,ρ2(z2)− ρ1(z1) = ρ(z2, 1) + ρ(z1,−1)− 〈e∗0, z1 + z2〉 > 0, z1, z2 ∈ RN .When z2 = −z1 we obtain the inequality ρ(z2, 1) > ρ(−z2,−1), which holdstrue by assumption (4). When z2 6= −z1 one can use the inductive assumption toobtain ρ(z2, 1)+ρ(z1,−1) ≥ ρ(z1+z2, 0) > 〈e∗0, z1+z2〉. It proves the result.Proposition 2.1 will be needed in the next section in the case N = 2.3 Periodic functions of one variable vs. posi-tively homogeneous functions on the planeLet the linear spaces H, F be defined by:H :={ρ ∈ C(R2,R)∩C2(R2\{0},R) : ρ is positively homogeneous of degree 1},F := C2(R/2πZ,R).5One can construct a linear map Φ : H → F in the following way. Givenρ ∈ H we set Φ[ρ] := u, whereu(θ) = ρ(cos θ, sin θ), θ ∈ R.It is clear that Φ is bijective: for each u ∈ F there exists an unique ρ ∈ H suchthat Φ[ρ] = u. In fact, in polar coordinates ρ = Φ−1[u] is explicitly given by{ρ(reiθ) = ru(θ), r ≥ 0, θ ∈ R/2πZ,ρ(0) = 0,(6)where we use complex notation and write reiθ := (r cos θ, r sin θ). In the resultbelow we point out another property of this correspondence.Lemma 3.1. Let ρ ∈ H be given and set u := Φ[ρ]. Then ρ is convex if andonly if u′′(θ) + u(θ) ≥ 0 for every θ ∈ R.Proof. Differentiation in (6) shows that the matrix M of Hess ρ(reiθ) with re-spect to the orthonormal basis {eiθ, ieiθ} is given byM =(0 00 (u′′(θ) + u(θ))/r),for every r > 0 and θ ∈ R. Therefore, the inequality u′′(θ) + u(θ) ≥ 0 for everyθ ∈ R is equivalent to Hess ρ(z) being positive semidefinite for every z ∈ R2\{0}.The result follows.Proof of Theorem 1.1. Choose any solution f0 = f0(θ) of (1) and let the posi-tively homogeneous of degree 1 function ρ : R2 → R be defined as in (6). ByLemma 3.1, ρ is convex. Since it is further not linear (because h is not iden-tically zero), Proposition 2.1 states the existence of a supporting linear forme∗ ∈ (R2)∗, (x, y) 7→ ax + by. Thus, u(θ) := f0(θ) − a cos θ − b sin θ satisfies allthe required assumptions and completes the proof.4 Higher eigenvaluesThe main result of this section is the followingProposition 4.1. Let the integer ω ≥ 3 be fixed. Then, there exists a C∞function h : R/2πZ → R satisfying (2), with h(θ) > 0 for all θ ∈ R and suchthat (1) does not have positive solutions.Proof. We seth(θ) := u′′∗(θ) + ω2u∗(θ), θ ∈ R, (7)where u∗ : R/2πZ → R will be a suitable C∞ function satisfying(i) u′′∗(θ) + ω2u∗(θ) > 0 for all θ,6θu− 1232π−π......u = u∗(θ)πω2πω3πωFigure 1: The graph of u∗ for ω odd.(ii) u∗(0), u∗(3π/ω) < 0.In this way our function h will be 2π-periodic and positive. Moreover, condition(2) follows directly from the definition of h. In order to check that (1) does nothave positive solutions we notice that every solution u = u(θ) has the formu(θ) = u∗(θ) + α sin(ωθ) + β cos(ωθ), θ ∈ R,for some constants α, β ∈ R. The function θ 7→ α sin(ωθ) + β cos(ωθ) has adifferent sign at θ = 0 and θ = 3π/ω and therefore (ii) is not compatible withthe existence of solutions u with u(0), u(3π/ω) > 0.Using a convolution argument with a suitable smooth mollifier ϕ we see thatit actually suffices to construct the function u∗ : R/2πZ → R of class C1 andpiecewise C2. In this setting, condition (i) is assumed to hold only on the openintervals where u∗ is twice differentiable. The relevant property here is thatequation (1) is linear and therefore if u = u(θ) is a solution for a given forcingterm h, then u ∗ ϕ is a solution for the regularized forcing term h ∗ ϕ.With all this in mind we construct u∗ : R/2πZ → R as follows:u∗(θ) :=1/2− cosωθ if |θ| ≤ π/ω,3/2 if π/ω ≤ |θ| ≤ 2π/ω,1/2 + cosωθ if 2π/ω ≤ |θ| ≤ π,and extended by periodicity. It is C1 and piecewise C2; moreover, it satisfies(i)-(ii), thus concluding the proof.We conclude with some final remarks:• Concerning the possibility of avoiding subharmonic resonances we noticethat the function u∗ constructed above satisfies u∗(−θ) = u∗(θ) for everyθ ∈ R. In the case ω = 3 one further has that u∗(π/2 + θ) = u∗(π/2− θ)for every θ ∈ R. Assuming that the mollifier ϕ is chosen even, thesesymmetries are inherited by u∗ ◦ ϕ, and then by h := (u∗ ◦ ϕ)′′ + u∗ ◦ ϕ,and we deduce that∫2π0h(θ) cos θdθ =∫2π0h(θ) sin θdθ = 0. We do notknow whether it is possible to construct an improved example for the caseω = 3 satisfying also that∫2π0h(θ) cos 2θ dθ =∫2π0h(θ) sin 2θ dθ = 0.7• One can check that the trigonometric polynomial u∗(θ) = 1− 2 cos(2θ)−cos(4θ) also satisfies assumptions (i)-(ii) above in the case ω = 3. Thus,for ω = 3 the function h defined as in (7) lies under the framework ofProposition 4.1.• We do not know whether Theorem 1.1 (or Proposition 4.1) still holds forω = 2.References[1] Brezis, H., Analyse Fonctionnelle. Théorie et Applications. CollectionMathématiques Appliquées pour la Mâıtrise. Masson, Paris, 1983.[2] De Coster, C.; Habets, P., Two-point boundary value problems: lower andupper solutions. Mathematics in Science and Engineering, 205. Elsevier B. V.,Amsterdam, 2006.[3] Florenzano, M.; Le Van, C., Finite Dimensional Convexity and Optimization.Studies in Economic Theory; 13. Springer-Verlag, 2001.[4] Hale, J.K., Ordinary Differential Equations. Second edition. Robert E.Krieger Publishing Co., Inc., Huntington, N.Y., 1980.[5] Klee, V.L., Convex sets in linear spaces. Duke Math. J. 18 (1951), 443–466.[6] Minkowski, H., Gesammelte Abhandlungen, vol. II. B.G. Teubner, Leipzigand Berlin, 1911.[7] Zhang, M., Optimal conditions for maximum and antimaximum principlesof the periodic solution problem. Bound. Value Probl. 2010, Art. ID 410986.8

International audienceConsider the equation of the linear oscillator [Formula: see text], where the forcing term [Formula: see text] is [Formula: see text]-periodic and positive. We show that the existence of a periodic solution implies the existence of a positive solution. To this aim we establish connections between this problem and some separation questions of convex analysis

HAL-INSU

HAL: Hyper Article en Ligne

HAL-OBSPM

Consider the equation of the linear oscillator $u"+u=h(\theta)$, where the
forcing term $h:\mathbb R\to\mathbb R$ is $2\pi$-periodic and positive. We show
that the existence of a periodic solution implies the existence of a positive
solution. To this aim we establish connections between this problem and some
separation questions of convex analysis

arXiv.org e-Print Archive

https://digibug.ugr.es/bitstream/10481/77053/1/PRE%20PRINT%202201.09272.pdf

An antimaximum principle for periodic solutions of a forced oscillator

Abstract

Similar works

Full text

Available Versions