# Solutions series for some non-harmonic motion equations

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1 Fifth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 10, 2003, pp or ftp ejde.math.swt.edu (login: ftp) Solutions series for some non-harmonic motion equations A. Raouf Chouikha Abstract We consider the class of nonlinear oscillators of the form d 2 u + f(u) = ɛg(t) dt 2 u(0) = a 0, u (0) = 0, where g(t) is a 2T -periodic function, f is a function only dependent on u, and ɛ is a small parameter. We are interested in the periodic solutions with minimal period 2T, when the restoring term f is such that f(u) = ω 2 u+u 2 and g is a trigonometric polynomial with period 2T = π. By using method ω based on expanding the solution as a sine power series, we prove the existence of periodic solutions for this perturbed equation. 1 Introduction Consider the second order differential equation x + g(t, x, x, ɛ) = 0 u(0) = a 0, u (0) = 0, (1.1) where ɛ > 0 is a small parameter, g is a 2T -periodic function in t with g(t, x, 0, 0) = g(x) is independent of t. The existence of solutions for this equation in the case where g is independent on x and continuously differentiable has been studied by many authors. In the latter case, this proved the existence of solutions to x + g(t, x, ɛ) = 0. For a detailed review, we refer the reader to the book by Chow and Hale [1]. Furthermore, the example given by Hartman proved non existence cases of (1). So, we cannot expect to generalize their results. On the other hand, Loud, Mathematics Subject Classifications: 34A20. Key words: power series solution, trigonometric series. c 2003 Southwest Texas State University. Published February 28,

2 116 Solutions series EJDE/Conf/10 Willem and others considered the case where T is independent on ɛ. Concerning the case T = T (ɛ), Fonda and Zanolin studied the system u + f(u) = ɛe(t, u, ɛ) where f is semilinear with xf(x) > 0. Here e is a T ɛ -periodic function such that e(t, x, ɛ) < K, lim ɛ 0 T ɛ = + and lim ɛ 0 ɛt ɛ = 0. They proved the existence of periodic solutions. In the present paper we are interested in the class of non linear oscillators of the form d 2 u + f(u) = ɛg(t) dt2 (1.2) u(0) = a 0, u (0) = 0, where the restoring term f f(u) depends on u and the function g g(t) is 2T -periodic. We look for periodic solutions with minimal period 2T, when f and g satisfy suitable conditions. When f is a quadratic polynomial function, we are able to solve (1.1), in some situations and without ɛ to be small, or any restricted condition on the period T. We will prove for that the existence of a trigonometric expansions of (periodic) solutions in Sinus powers. Note that, some times under certain conditions, (1) cannot have a periodic solution, as described in the example below. However, under other suitable conditions of controllability of the period, Equation (1) has a periodic solution. A non existence result According to Hartman [5, p.39], Equation (1) in general does not have a non constant periodic solution, even if xg(t, x, x ) > 0. The following example, given by Moser, proves the non existence of a non constant periodic solution of Let x + φ(t, x, x ) = 0. φ(t, x, y) = x + x 3 + ɛf(t, x, y), ɛ > 0 satisfying the following conditions: φ C 1 (R 3 ); f(t + 1, x, y) = f(t, x, y) with f(0, 0, 0) = 0 and f(t, x, y) = 0 if xy = 0; uniformly in (t, y) R 2 ; φ x as x δf δy > 0 if xy > 0, and δf δy = 0 otherwise. and x, y satisfying x < ɛ, y < ɛ. In fact, we have xf(t, x, y) and yf(t, x, y)) > 0 if xy > 0, x < ɛ, y arbitrary and φ = 0 otherwise. We remark that δf δy is small. The function V = 2x2 + x 4 + 2x 2 satisfies V = 4ɛx f(t, x, x ), so that V < 0 if xx > 0, x < ɛ and V = 0 otherwise. Thus x cannot be periodic unless V = 0.

3 EJDE/Conf/10 A. Raouf Chouikha 117 A controllably periodic perturbation In this part, we give brief explanation of a method due to Farkas [4] inspired of the one of Poincaré. The determination of controllably periodic perturbed solution. This method appears to be efficient particularly for the perturbations of various autonomous systems. Since we know it for example a good application for perturbed Van der Pol equations type as well as Duffing equations type. The perturbation is supposed to be controllably periodic, i.e., it is periodic with a period which can be chosen appropriately. Let u 0 (t) be the periodic solution of the (unperturbed) equation. Under very mild conditions it is proved that to each small enough amplitude of the perturbation there belongs a one parameter family of periods such that the perturbed system has a unique periodic solution with this period. It is assumed the existence of the fundamental matrix solution of the first variational 2-dimensional system of ẋ = h(x) and the unique periodic solution p(t) corresponding to u 0 (t). For more details concerning applications of this method we may refer to [4]. 2 Periodic perturbations of Duffing non linear equations In this section, we will apply the methods of trigonometric series in sinus powers to resolve in an explicit way non linear oscillators of Duffing type. This approach already was used previously in an effective way, see [6] and [2]. One can also apply it for periodic perturbations of various differential equations. Our results generalize and improve those of [6] and [2]. Their also contribute in an interesting way to the general problem of the periodic perturbations roughly exposed in the introduction. Case of finite trigonometric polynomial Firstly, let us consider the restoring function f(u) = ω 2 u + u 2, where ω is a constant and g(t) a finite trigonometric polynomial with period 2T = π ω, which has a finite Fourier series expansion. So, it can be written g(t) = b 0 + b 2 sin 2 ( πt 2T ) + + b 2p 0 sin 2p0 ( πt 2T ) = n p 0 b 2n sin 2n ( πt 2T ), p 0 N. Consider the following system d 2 u dt 2 + ω2 u + u 2 = ɛg(t) u(0) = a 0, u (0) = 0. (2.1)

4 118 Solutions series EJDE/Conf/10 Theorem 2.1 The T -periodic solutions of (1.2) may be expressed in the form u(t) = c 0 + c 2 sin 2 ( πt 2T ) + c 4 sin 4 ( πt 2T ) + c 6 sin 6 ( πt 2T ) + = c 2n sin 2n ( πt 2T ) The coefficients c 2n satisfy a 0 < C, a 0 = c 0, 2ω 2 c 2 = ω 2 c 0 c ɛb 0, (ω = π 2T ) and the recursion formula (2n + 1)(2n + 2)c 2n+2 { (4n 2 1)c 2n 1 n = ω 2 r=0 c 2rc 2n 2r + ɛ ω b 2 2n, for n p 0, (4n 2 1)c 2n 1 n ω 2 r=0 c 2rc 2n 2r, for n > p 0. (2.2) Proof Suppose the solution u of equation (1.2) can be written in the form u(t) = c 2n sin 2n (ωt). We shall apply the previous method in [2] to: (i) obtain recursion formula of the coefficients c 2n, and (ii) prove the convergence of the series expansion of the solution. This method finds a solution to (1.2)in the form u(t) = c 0 + c 1 sin ωt + c 2 sin 2 ωt + c 3 sin 3 ωt +... (2.3) where c i, i = 0, 1, 2,..., are coefficients to be determined by the substitution of (2.2) in (1.2). In fact, we get ω = π 2T where T is the period of the solution. So, a trivial computation gives We get also 2ω 2 c 2 = ω 2 c 0 c ɛb 0. 12c 4 = 3c 2 2 ω 2 c 0c 2 + ɛ ω 2 b 2 By identification for n 1, we obtain the recursion formula of these coefficients (2n + 1)(2n + 2)c 2n+2 = (4n 2 1)c 2n 1 ω 2 n and r=0 c 2r c 2n 2r + ɛ ω 2 b 2n, for n p 0. (2.4) (2n + 1)(2n + 2)c 2n+2 = (4n 2 1)c 2n 1 n ω 2 c 2r c 2n 2r, for n > p 0. (2.5) Equation (1.2) implies c 1 = 0. Relations (2.3) yields c 3 = 0, c 5 = 0,.... However, the even order coefficients c 2n do not vanish. The solution to (1.2) can now be written as u(t) = c 0 [c 0 1 ω 2 c2 0 + ɛ ω 2 b 0] sin 2 ωt + [ 1 4 c 2 1 6ω 2 c 0c 2 + ɛ 12ω 2 b 2] sin 2 ωt +... (2.6) r=0

5 EJDE/Conf/10 A. Raouf Chouikha 119 Relations of the coefficients and further induction show that c 2i, i = 1, 2,... all vanish for a 0 = 0 and b 2i = 0. So, the trivial solution u 0 is included in (2.2) as a special case. The following lemma is strictly analogous of [2, Lemma 2], corresponding to the case ɛ = 0. It implies that c 2n < +, n 1 which implies the convergence of the expansion. Lemma 2.2 There exist a positive constants k and α verifying 1 < α < 3 2 such that the coefficients c 2n of the series expansion (2.2) solution of the differential equation (1.2) satisfy the inequality c 2n < k (2n) α. (2.7) Remark Following [6], it is interesting to write the power series solution for the system (1.2) in the form u(t) = v(sin ωt). Let g defined by g(t) = g(sin ωt). Then v is a solution of the differential equation (1 x 2 ) d2 v dx 2 x dv dx + v + 1 ω 2 v2 = ɛ ω 2 g(x) (2.8) dv v(0) = a 0, dx (0) = 0 Recall that the latter method permits to compare approximate solutions of the non-harmonic motion of the oscillator. A more general case Now, consider a more general situation where the function g(t) in (1.1) may be written as an infinite expansion in Sinus power g(t) = b 0 + b 2 sin 2 ( πt 2T ) + + b 2n sin 2n ( πt 2T ) + = b 2n sin 2n ( πt 2T ). It is the case in particular, when the function g(t) has a finite Fourier series expansion. When g has an (infinite) Fourier series expansion, g(t) may be expressed as an infinite expansion in Sinus power. But we have to prove its convergence. Now, we are interested in solving d 2 u dt 2 + ω2 u + u 2 = ɛ u(0) = a 0, u (0) = 0. b 2n sin 2n ( πt 2T ) We prove the following theorem which extends Theorem 2.1. (2.9)

6 120 Solutions series EJDE/Conf/10 Theorem 2.3 Suppose that the coefficients of the expansions of the function g(t) satisfies b 2n < 1, with β 1, (2n) β then the solutions of (2.8) may be expressed in the form u(t) = a 0 + c 2 sin 2 ( πt 2T ) + c 4 sin 4 ( πt 2T ) + c 6 sin 6 ( πt 2T ) + = c 2n sin 2n ( πt 2T ). The coefficients c 2n satisfy the conditions a 0 < C, a 0 = c 0, 2ω 2 c 2 = ω 2 c 0 c ɛb 0, (ω = π 2T ) and the recursion formula, for n > 0: (2n + 1)(2n + 2)c 2n+2 = (4n 2 1)c 2n 1 ω 2 n r=0 c 2r c 2n 2r + ɛ ω 2 b 2n. (2.10) Proof The proof starts in the same manner as that of Theorem 2.1. In order to establish the recursion relation between the coefficients we may proceed as previously. The difference is that one obtains a less good estimation of the c 2n. The crucial point is to state an analogous one of Lemma 2.4, ensuring thus the convergence of the series solution. It may be deduced from c 2n < + n 1 For that, we prove the following lemma. Lemma 2.4 For any positive number α such that 1 < α < 3/2, there exists a positive constant k satisfying k < 1 ω (3 2 α)41 α such that the coefficients c n of the series expansion solution of the differential equation (2.8) satisfies the inequality c n < k n α. (2.11) Proof We first notice that Lemma 2.4 gives an optimal result, because our method does not work for α = 3 2. The coefficients c n of the power series solution, satisfy the recursion formula (2.9). We shall prove there exist two positive constants k > 0, and α > 1, such that the following inequality holds c n < k n α for any integer n 1. Suppose for any n p, we get c n < k n α. In particular, it implies that 0<r<p c r c p r < 0<r<p k 2 r α (r p) α k 2 (p 1) α 1.

7 EJDE/Conf/10 A. Raouf Chouikha 121 Equality (2.9) gives c p+2 = p 2 1 (p + 1)(p + 2) c 1 p 1 p ω 2 (p + 1)(p + 2) Thus, if we prove the inequality k 2 r=0 c r c p r + ɛ ω 2 (p + 1)(p + 2) b p. p 2 1 k (p + 1)(p + 2) p α + 1 ω 2 (p + 1)(p + 2) (p 1) α 1 + ɛ b p ω 2 (p + 1)(p + 2) k (p + 2) α, we can conclude c p+2 < k (p + 2) α. (2.12) Since the coefficients b p satisfy the hypothesis b 2n < 1 (2n) β, with β 1, it is not difficult to exhibit an integer p 0 depending on β such that the following inequality holds for p p 0 : p 2 1 k (p + 1)(p + 2) p α + 1 ω 2 (p + 1)(p + 2) (p 1) α 1 k (p + 2) α (2.13) This inequality is equivalent to where k 2 k < β pf(p)g(p) (2.14) ω2 f(p) = p + 1 p (p 1 p + 2 )α 1 g(p) = 1 (p2 β ω 2 c 0 )(p + 2) α 1 (p + 1)p α Using MAPLE, we prove that f(p) is an increasing positive function in p. Moreover, for any p 1, f(p) is bounded below as The function g(p) is such that f(p) ( 3 2 )41 α. pg(p) = p (p2 β ω c 2 0 )( p+2 p )α 1 (p + 1) is a strictly decreasing and bounded function. More exactly, we may calculate the lower bound (3 2α) g(p) >. p Thus, if (3 2α) = ɛ > 0, it suffices to chose for inequality (17) to hold. k ( 3 2 )41 α (3 2α)

8 122 Solutions series EJDE/Conf/10 Remark for the case ɛ = 0: Note that the choice of k depends on α value. For α = 3/2, using MAPLE we can prove that pg(p) = p (p2 3 2ω 2 c 0 )( p+2 p )1/2 (p + 1) is positive and strictly decreasing to 0. While p 2 g(p) is a bounded function. Moreover, it appears that pf(p)g(p) is a decreasing function which tends to 0 as p tends to infinity. Thus, our method fails since it does not permit to determine a non negative constant k. References [1] S. N. Chow, J. Hale, Methods of bifurcation theory, Springer, Berlin (1982). [2] A. R. Chouikha, Remark on Periodic Solutions of Non-linear Oscillators, Appl. Math. Lett.,vol. 14 (8), p , (2001). [3] A. R. Chouikha, Fonctions elliptiques et bifurcations d equations differentielles, Canad. Math. Bull., vol. 40 (3), p , (1997). [4] M. Farkas, Estimates on the existence regions of perturbed periodic solutions, SIAM J. Math. Anal., vol 9, p , (1978). [5] P. Hartman, On boundary value problems for superlinear second order differential equation, J. of Diff Eq., vol 26, p , (1977). [6] A. Shidfar and A. Sadeghi, Some Series Solutions of the anharmonic Motion Equation, J. Math. Anal. Appl. 120, p (1986). [7] A. Shidfar and A. Sadeghi, The Periodic Solutions of Certain Non-linear Oscillators, Appl. Math. Lett., vol 3, n 4, p , (1990). A. Raouf Chouikha University of Paris-Nord, Laga, CNRS UMR 7539 Villetaneuse F-93430,

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