Resonant solutions in the presence of degeneracies for quasi-periodically perturbed systems

2012 1. Abstract 1.1 Statement of the results Let us consider the ordinary differential equation ?? ? ? = !0(B) + "F(!t, ,B), ?B = "G(!t, ,B), (1.1.1) where ( ,B) ? T à- B, with B an open subset of R, F,G : Td+1 à- B ? R and !0 : B ? R are real-analytic functions, ! ? Rd with d ? 2 and " is a (small) real parameter called the perturbation parameter ; hence the perturbation (F,G) is quasi-periodic in t with frequency vector !. Without loss of generality we can assume that ! has rationally independent components. Take the solution for the unperturbed system given by ( (t),B(t)) = ( 0 + !0(B0)t,B0), with B0 such that !0(B0) is resonant with !, i.e. such that there exists ( 0, ) ? Zd+1 for which !0(B0) 0 + ! · = 0. We want to study whether for some value of 0, that is for a suitable choice of the initial phase, such a solution can be continued under perturbation. The resonance condition between !0(B0) and ! yields a †œsimple resonance†� (or resonance of order 1) for the vector (!0(B0),!). The main assumptions on (1.1.1) are a Diophantine condition on the frequency vector of the perturbation and a non-degeneracy condition on the unperturbed system. More precisely we shall require that the vector (!0(B0),!) satisfies the condition X n?0 1 2n log inf ( 0, )?Zd+1 ( 0, )?( 0, ),0<|( 0, )|?2n |!0(B0) 0 + ! · | ?1 < ? (1.1.2) and that !? 0(B0) 6= 0. Up to a linear change of coordinates, we can (and shall) assume !0(B0) = 0, so that the vector , such that !0(B0) 0 +! · = 0, must be the null vector. Therefore it is not restrictive to formulate the assumptions on B0 and ! as follows. 1 Abstract Hypothesis 1. !0(B0) = 0 and ! satisfies the Bryuno condition B(!) < ?, where B(!) = X n?0 1 2n log 1 n(!) , n(!) = inf ?Zd 0<| |?2n |! · |. (1.1.3) Hypothesis 2. For B0 as in Hypothesis 1 one has !? 0 (B0) 6= 0. Note that if ! satisfies the standard Diophantine condition |! · | ? | |? for all ? Zd? , then it also satisfies the Bryuno condition, since m(!) ? 2?m in that case. Let us write F( , ,B) = X ?Zd ei · F ( ,B), G( , ,B) = X ?Zd ei · G ( ,B), (1.1.4) and note that, since F and G are real-valued functions, one has F? ( ,B) = F ( ,B)?, G? ( ,B) = G ( ,B)?. (1.1.5) By analogy with the periodic case, the function (1) 0 ( ) := G0( ,B0) will be called the first order Melnikov function. We look for a quasi-periodic solution to (1.1.1) with frequency vector !, that is a solution of the form ( (t),B(t)) = ( 0 + b(t),B0 + e B(t)), with b(t) = X ?Zd ei ·!tb , e B(t) = X ?Zd ei ·!tB . (1.1.6) Of course the existence of a quasi-periodic solution with frequency ! in the variables in which !0(B0) = 0 implies the existence of a quasi-periodic solution with frequency resonant with ! in terms of the original variables (that is, before performing the change of variables leading to !0(B0) = 0). If we set (t) := !0(B(t)) + "F(!t, (t),B(t)) and (t) = "G(!t, (t),B(t)), and write (t) = X ?Zd ei ·!t , (t) = X ?Zd ei ·!t , (1.1.7) in Fourier space (1.1.1) becomes (i! · )b = , 6= 0, (1.1.8a) (i! · )B = , 6= 0, (1.1.8b) 0 = 0, (1.1.8c) 0 = 0. (1.1.8d) 2 1.1 Statement of the results According to the usual terminology, we shall call (1.1.8a) and (1.1.8b) the range equations, while (1.1.8c) and (1.1.8d) will be referred to as the bifurcation equations. We start by looking for a formal solution ( (t),B(t)), with (t) = (t; ", 0) = 0 + X k?1 "kb(k)(t; 0) = 0 + X k?1 "k X ?Zd ei ·!tb(k) ( 0), B(t) = B(t; ", 0) = B0 + X k?1 "kB(k)(t; 0) = B0 + X k?1 "k X ?Zd ei ·!tB(k) ( 0) (1.1.9) and set U(t) := !0(B(t)) ? !? 0 (B0)(B(t) ? B0) and (t) = U(t) + "F(!t, (t),B(t)). Then define recursively for k ? 1 b(k) ( 0) = 1 (i! · ) (k) ( 0) + !? 0 (B0) (i! · )2 (k) ( 0), 6= 0 B(k) ( 0) = 1 (i! · ) (k) ( 0), 6= 0 B(k) 0 ( 0) = ? 1 !? 0(B0) (k) 0 ( 0), (1.1.10) where we denoted (k) ( 0) = [G(!t, (t),B(t))](k?1) and (k) ( 0) = [U(t)](k) + [F(!t, (t),B(t))](k?1) , with U(1) ( 0) = 0, so that (1) ( 0) = G ( 0,B0) and (1) ( 0) = F ( 0,B0), while, for k ? 2, [U(t)](k) = X s?2 1 s! @sB !0(B0) X 1+...+ s= i?Zd, i=1,...,s X k1+...+ks=k, ki?1 Ys i=1 B(ki) i ( 0), (1.1.11) and [P(!t, (t),B(t))](k?1) = X s?1 X p+q=s X 0+...+ s= 0, j?Zd j=p+1,...,s i?Zd , i=1,...,p 1 p!q! @p @q BP 0( 0,B0) à- à- X k1+...+ks=k?1, ki?1 Yp i=1 b(ki) i ( 0) Ys i=p+1 B(ki) i ( 0), P = F,G. (1.1.12) The series (1.1.9), with the coefficients defined as above and arbitrary 0, turn out to be a formal solution of (1.1.8a)-(1.1.8c): the coefficients b(k) ( 0), B(k) 0 ( 0) and B(k) ( 0) are well defined for all k ? 1 and all ? Zd? , and solve (1.1.8a)-(1.1.8c) order by order; moreover the functions b(k)(t; 0) and B(k)(t; 0) are analytic and quasi-periodic in t with frequency vector !. Note that if there exists k0 ? 1 such that (k) 0 ( 0) ? 0 for all k < k0, then the series (1.1.9) with the coefficients b(k) ,B(k) defined as in (1.1.10) solve the equations of motion up to order k0 ? 1 and moreover (k0) 0 is a well-defined function of 0. 3 Abstract Assume first that the system (1.1.1) is Hamiltonian, i.e. there exists a function H( , ,A,B) := ! ·A + h(B) + "f( , ,B), (1.1.13) where ( , ) ? Td+1 and (A,B) ? Rd à- B, with B an open subset of R, are canonically conjugate (action-angle) variables and the functions f : Td+1 à- B ? R and h : B ? R are real-analytic and such that !0(B) = @1h(B), @Bf( , ,B) = F( , ,B) and ?@ f( , ,B) = G( , ,B), so that the corresponding Hamilton equations for the variables ( ,B) are given by ?? ? ? = !0(B) + "@Bf(!t, ,B), ?B = ?"@ f(!t, ,B), (1.1.14) which are exactly of the form (1.1.1). Hypothesis 3. One has (k) 0 ( 0) := [?@ f( ,B)](k?1) 0 ? 0 for all k ? 1. Then we shall prove the following result. Theorem 1.1.1. Consider the system (1.1.14) and assume Hypotheses 1, 2 and 3. Then the series (1.1.9) are convergent for " small enough. Next we consider the more general system (1.1.1) and we assume that there exists k0 ? N such that all functions (k) 0 ( 0) are identically zero for 0 ? k ? k0 ?1, while (k0) 0 ( 0) is not identically vanishing. Again, by analogy with the periodic case, we shall call the function (k0) 0 ( 0) the k0-th order Melnikov function. Hypothesis 4. There exist k0 ? N and 0 such that (k) 0 ( 0) vanish identically for k < k0 and 0 is a zero of order ¯n for (k0) 0 ( 0), with ¯n odd. Moreover one has "k0!? 0(B0)@¯n 0 (k0) 0 ( 0) > 0. Then we shall prove the following result. Theorem 1.1.2. Consider the system (1.1.1) and assume Hypotheses 1, 2 and 4 to be satisfied. Then for " small enough there exists at least one quasi-periodic solution ( (t),B(t)) with frequency vector ! such that ( (t),B(t)) ? ( 0,B0) for " ? 0. 1.2 Remarks about the results and sketch of their proofs Quasi-periodic solutions to (1.1.14) with frequency vector ! describe lower-dimensional tori (ddimensional tori for a system with d + 1 degrees of freedom). Such tori are parabolic in the sense that the †œnormal frequency†� vanishes for " = 0. Theorems 1.1.1 and 1.1.2 imply the following result. 4 1.2 Remarks about the results and sketch of their proofs Theorem 1.2.1. Consider the system (1.1.14) and assume Hypotheses 1 and 2 to be satisfied. Then for " small enough there exists at least one quasi-periodic solution ( (t),B(t)) with frequency vector !. Proof. If all the coefficients (k) 0 = ?[@ f](k?1) 0 vanish identically for all k ? 1 we simply apply Theorem 1.1.1. Otherwise there exists k0 ? 1 such that all the coefficients (k) 0 ( 0) vanish identically for all k < k0 while (k0) 0 ( 0) is not identically zero and hence we can solve the equations of motion up to order k0 without fixing the parameter 0. Moreover one has (k0) 0 ( 0) = @ 0g(k0)( 0) with g(k0)( 0) := [B ?b](k0) 0 ? [h(B0 + B + B(k0))](k0) 0 ? [f(!t, 0 + b,B0 + B)](k0?1) 0 , because, if we denote b = kX0?1 k=1 b(k), B = kX0?1 k=1 B(k). one has @ 0 [f(!t, 0 + b,B0 + B)](k0?1) 0 = [@ f(!t, 0 + b,B0 + B)(1 + @ 0b)](k0?1) 0 + [@Bf(!t, 0 + b,B0 + B)@ 0B](k0?1) 0 = ? (k0) 0 ? [ ?B@ 0b](k0) 0 + [? b@ 0B](k0) 0 ? [!0(B0 + B + B(k0))@ 0(B + B(k0))](k0) 0 = ? (k0) 0 + @ 0[B ?b](k0) 0 ? @ 0 [h(B0 + B + B(k0))](k0) 0 . Since g(k0) is analytic and periodic, and then it has at least a maximum ? 0 and a minimum ?? 0 . Then Hypothesis 4 holds.