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1、J. Differential Equations 248 (2010) 2149Contents lists available at ScienceDirectJournal of Differential E properties of solutions of linear second orderdifferential equations with random coefficientsSndor Csrg o, Lszl Hatvani,1University of Szeged, Bolyai Institute, Aradi vrtank tere 1, Szeged, H-
2、6720 Hungarya r t i c l ei n f oa b s t r a c tArticle history: Received 20 March 2009 Available online 28 August 2009MSC: primary 34F05, 34C11 secondary 70L05Keywords: Small solution Random oscillations Impulses Parametric resonance Stochastic parametric resonance Meissner s equation Problem of swi
3、ngingThe equationx?+a2(t)x=0,a(t) :=ak0if tk1?t0 (“parametric resonance” 4,6). By the use of Floquet s theory it can be shown 2 that the appropriate values are2T=j (j N)(1.7)(/is half of the period of the original unperturbed pendulum motion with=0). Following ourprobabilistic approach suggested in
4、the previous problem we now suppose that the coefficient a2is not a deterministic periodic function, but the lengths of the time intervals between the consecutive jump points are independent identically distributed random variables with values on interval0,2T.The problem of random swinging is again
5、to find the appropriate random variables and the values of Tfor which parametric resonance occurs (“stochastic parametric resonance”, see Definition 4.10).In 20 the authors studied the problem of small solutions for Eq. (1.1) with step function coeffi- cient a2tending to infinity monotonously as t .
6、The main theorem stated that it is almost sure that all solutions are small provided that the distances between the consecutive jump points are inde- pendent random variables of uniform distribution on0,1.This was generalized for non-monotonousstep function coefficients in 18.In this paper we give s
7、ufficient conditions for both stability and instability of the equilibrium position x=0 in the most general case: it is only assumed that the time intervals between theconsecutive jumps in the step function coefficient a2are independent, positive (not necessarily iden- tically distributed) random va
8、riables. The conditions will only contain the characteristic functions of the random variables. To illustrate stability results we formulate a corollary for the simplest case (see Corollary 2.6).Theorem A. Let the coefficient a2in (1.1) be a step function having the property (1.2), i.e.,a(t) =ak(tk1
9、?t0, 0.In the case ak? (k )(2.6) gives the evident estimate?(z0) ?0. Nevertheless, it can be shown 17 that there always exists at least one z0? = (0,0)Twith?(z0) =0; i.e., there exists at least one non-trivial small solution. Ifakisnot monotonous, then the case is much more difficult:?zk?is not mono
10、tonous either, and instead of (2.6) we have?k?n=1?1dn1?z0? ? ?zk? ?k?n=1?1+dn1+?z0?,(2.7)whered+:=maxd;0,d:=maxd;0(d R).26S. Csrg o, L. Hatvani / J. Differential Equations 248 (2010) 2149In general, (2.7) cannot be used to estimate?zk?by means ofak. (For example, in the case of (1.6) the estimate (2
11、.7) tells us nothing about the asymptotic behavior of?zk?.) Since the limit?(z0)does not exist any more, we have to work with the upper and lower limits? = ?(z0) :=liminf t?x2(t)+y2(t)?=liminf k?zk?2,? = ?(z0) :=limsup t?x2(t)+y2(t)?=limsup k?zk?2.(2.8)The global estimates (2.6), (2.7) cannot take i
12、nto account the fact that the effect of the contraction in Tkalso depends on y(tk0),i.e., on tk.For example, the first estimate in (2.6) is very rough, actually it tells us nothing if y(tk0) =0(k N); in this case the norm?zk?is constant and the solution is not small even if ak? (k ).However, this ca
13、se occurs if and only ifk=ak(tktk1) =mk(k N)with some mk N,so the situation seems to be rather “exceptional”. Now it is clear that in order to improve the estimates (2.6), (2.7) we have to incorporate the se- quencetninto the estimates. Buttnappears in solutions through the unknown sequencesy(tn).To
14、 overcome this difficulty and to justify the fact that the “bad”tn s are exceptional, we shall assume thatk=tktk1k=1is a random sequence and estimate the expected valuesE(?zk?2), k N.2.3. Estimate for the expected valuesE(?zk?2)Suppose that the lengthkof the intervaltk1,tk)is a random variable defin
15、ed on a probability space(k,Ak,Pk),k Nand1,2,.,k,.are totally independent. The limits?,?are functions of the random sequencek, so they are also random on the infinite product(,A,P)of the prob- ability spaces(k,Ak,Pk), k N(see 5). We are interested in the probability of the events? =0, perhaps? =0 (s
16、tability properties of the equilibrium x=0), or? 0,perhaps? 0 (instability prop- erties of x=0).The definitions of?,?use t ,so in the following we require that?k=1k= almost surely.(2.9)Let us state conditions which guarantee this property. Given a non-negative random variableand a constant c0,we define(c):=?
