# Forced Oscillations Of Elastic Strings With Nonlinear Damping-Books Pdf

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58 M COUNTRYMAN and R KANNAN, here to the interesting article 8 in which approximate equations to the full problem. were obtained by using a perturbation argument We also mention the work of 4. where the stability theory for these problems is discussed. 2 Equations of motion We first derive for the sake of completeness the equations. of motion as in 6 Thus let TQ be the tension in the string in equilibrium position. A the area of cross section m CA the mass per unit length and E the Young s. modulus Before we derive the equations of 6 let us first recall Carrier s general. equations of motion 2 Thus if the element of length dx in the equilibrium position. has a vertical displacement v v x t at end A and displacements u uxdx and. v vvdx at B then the local tension is, T x T0 EA 53 2 1. where ds is the new length of the element dx Thus, T x TQ EA 1 1 vx 2 u2x i 2 2. and the equations of motion are, CAu T sinfl 2 3, CAvtt TcosO 2 4. Assuming v 0 and ux small but allowing for tension variation we would have. T x T0 ux and sin 6 ux, substituting these expressions for T x and sin 6 in 2 3 we get.
Uu UXX 2 TUxUxx, which was the equation studied in 2. In 6 the authors assume as in 2 above that v 0 and tan 6 is small However. considering a deformation of the string into a space curve we proceed as follows. Let dx be an element that is deformed into the space element ds Assuming. longitudinal displacements of the ends to be negligible. ds dx2 dv2 dz2 X 2 dx 1 y2x z2 y2 z2 2, Thus ds dx yl z2 y2 z2 2 dx The local tension T is given by. T To EA r T EM, T0 EA y2x z2x z2x 2, FORCED OSCILLATIONSOF ELASTICSTRINGS 59. The potential energy of the element dx is thus, rQ EAk kdx T0 y2x z2x EAg T y z 2. Usually T0 is very small compared to EA and is thus dropped from the second. term on the right Let f x cos cot be the external force per unit length acting in the. y direction Then the potential energy of the string for an arbitrary deformation is. u Jo Uyl zl zl 2 yf x coscot dx, The kinetic energy is.
t 1 f y z dx, By Hamilton s principle the first variation d Jq T U dt 0 and one can see that. y x t and z x t must satisfy the Euler equations given by. y clyxx c yxxy c yxzl cos wt 2 6, 3 2 2 1 2 9, z cozxx C Z 2C z z 2C iz y l 0 2 7. Finally it must be remarked as noted in 6 that near resonance yx and zx. are appreciable and the nonlinear terms cannot be ignored Further the longitudinal. velocity c is much larger than the transverse velocity cQ. 3 Existence of periodic solutions to the coupled equations Following the ideas in. 3 we now demonstrate the existence of a unique periodic solution to the nonlinearly. coupled system of equations analogous to 2 6 2 7 in the presence of damping. Thus we have the system of equations, U Uxx UlUXX UxVl x P X 0 3 1. vt vxx vlvxx vxul x y u x o 3 2, with the boundary conditions. m 0 t u n t 0 u x t u x t 2n 3 3, v 0 t v n t 0 v x t v x t 2n 3 4.
The hypotheses on p q and y are as follows, i p x t and q x t are defined for 0 x n oc t oo and are. 2n periodic in t p q L 0 n x 0 2n, ii fl and y are strictly monotone increasing continuous functions. iii there exist constants a and b such that, a Z m l y b t m l 3 5. 60 M COUNTRYMAN ANDR KANNAN, for some m 1 and g R, iv there exist constants ci and d such that. m cl d r y Z c2 d2 i m 3 6, Let D denote the set of all functions u x t 0 x n oo t oo which.
are of class C and satisfy u x t 2n u x t w 0 t u n t 0 Further. let A denote the completion of D by means of the norm. Wu x ll H2, where denotes the norm in L2 0 n x 0 2n we will use L2 for the rest of. this paper, The space L has a complete orthonormal system given by. 4 kl 2 1 127i1 cos kt sin Ix ykl 2 Xl n sin kt sin Ix. Xt x 2 1 27t 2sinIx, for k 1 2 and 1 2, Let S0 be the subspace generated by x 1 2 and the orthog. onal complement of S0 in L2 generated by f kIand y kl k 1 2 and. Any pair of elements u v e L2 has a Fourier series given by. u x t M x w x v x t V x W2 x 3 7, where w w e S and w w2 e S0. We can now define P L S0 to be the projection operator defined by Pu w. for u G L2 and w given by 3 7, Finally let Rn L2 L2 be the truncation operator which truncates the Fourier.
series for any u e L2 by deleting all the terms with k n or I n or both If we. now consider the truncated problems, U x UlUxx 0 P. Rn v vxx vlvxx vxul x y vt o, we get the corresponding equivalent system of equations. I P Rn un uxx u uxx u v x P ut p 0 3 8, PRnK Uxx U1UXX. I P Rn vtt vxx v vxx vxu x y vt q 0 3 10, PRnlVU VXX VlVXX VxUl x y V 9 0 3 1 1. We now outline the various steps in proving the existence of a periodic solution. to 3 1 3 2, FORCED OSCILLATIONSOF ELASTICSTRINGS 61.
Step I Using the notations of 3 7 and dropping the subscript n for the sake of. simplicity we first consider the system of auxiliary equations in order to apply the. Leray Schauder principle given by, U t Ulxx 0 x uu A 7 p R Piuu UlUxx uxvl x p 0 3 12. wlxx APR 0 uu uxuxx uxv2x x p 0 3 13, vut v xx Vvu P R y v 1 v2xvxx uxv2x x q 0 3 14. w2xx WR V Vlt v2xvxx u2xvx x q 0 3 15, Here ux ut are 27r periodic in t and all of u wt w and w2 are zero at x 0. For A 0 1 we first show that all possible solutions of 3 12 3 15 are. bounded in v4n norm independent of X, Step II We note that when X 0 3 12 3 15 has only the trivial solution. ul vl w 1 w2 0, Step III By the Leray Schauder principle we conclude that 3 12 3 15 has a.
solution for X 1 and in this case 3 12 3 15 coincides with 3 8 3 11 which. depends on n, Step IV Having solved 3 8 3 11 for each n we now show that the correspond. ing solutions un uln win vn v n w2n have a subsequence that converges to. a solution u v of 3 1 3 2, Step V We now show that if 3 1 3 2 has a solution it must be unique. We now outline the proofs of Steps I IV and V Multiplying 3 12 and 3 14. by u t and vu respectively and integrating over 0 n x 0 In we get using the. notation for the inner product, 1 X uu 2 X I P R 0 uu uxuxx uxv2x x p uu 0 3 16. 1 l vu 2 X I P R y vu v2xvxx vxux x q vu 0 3 17, It is easy to see that uUt uu ulxx uu 0 and similarly for v Using 3 5. in 3 16 and 3 17 we get, a Uu C UlUxx Uu Uxv2x x Uu P MU 3 18.
bWVu C v2xvxx vu VxVl x Vu V Vu 3 19, Adding two equations and using Dirichlet boundary conditions in x and periodic. boundary conditions in t it follows that, IKrllm land llvull i 3 2. are bounded independent of n Note that we dropped the subscript n at the. beginning of Step I This implies that, IK II and f lt 3 21. are bounded independent of n, 62 M COUNTRYMAN and R KANNAN. For every x ux x t and vx x t have mean value zero as a function of t. ux x t J uu x x dx 0 x 7i 0 t 2n 3 22, and a similar equality for vx.
Using 3 22 in conjunction with 3 20 we get, r2n rji r2n rn r. II C V u x t m xdxdt f r uu X T m ldTdxdt, Jo Jo Jo Jo IJo. This implies, ll ill i and similarly 3 23, are bounded independent of n. Proceeding as we did from 3 21 to 3 22 we have, Mj and v are bounded independent of n 3 24. Note that in 3 20 and hence in all subsequent estimates the bound is dependent. only on p x t and q x t, We now obtain bounds on w v and vx To this end we multiply 3 12 by uy.
and 3 13 by w and integrate both of the equations over 0 2n x 0 n Recall. h m w 1 2 ulxx ux u 2, Wlxx Wl II Lvll2 Wlxx Wl H lll2. Also from 3 21 and 3 24 w uj m and v are bounded Then. from 3 12 and 3 13 we get, llWl ll 11W, Ijc11 1 U U MI P. X R fi uu uj iuxuxx uxv2x x p M 0 3 25, lltojl2 XPR p uu wx u2xuxx wx uxv2x x wx p tu 0 3 26. with similar equations resulting from 3 14 and 3 15 when multiplied by vx and. w2 and integrated, Adding 3 25 and 3 26 we obtain, IKJ2 H iii2 ihJi2 mi uluxx i. where u ux wx, uu u ux ux 0, uxvl x xvl Ux, Using that w1 and u are bounded and the growth condition 3 6 on ft we.
can conclude from 3 27 that, m v and io are bounded independent of n 3 28. FORCED OSCILLATIONS OF ELASTIC STRINGS 63, Also from 3 13 and wlxx w to1JC we could have obtained instead of. wlx 2 PR P uu wx uxuxx wx uxvl x wl p wj 0 3 29, Adding 3 25 and 3 29 and using that w1 M and tu are bounded we. conclude that, io1 is bounded independent of n 3 30. From 3 21 3 24 3 28 and 3 30 it follows that, IML IKIU ll ilL H JI2 IKJI2 2 IK II2 IK II2 1 2.
IImIL is bounded independent of n, and similarly, w is bounded independent of n. This proves Step I i e all possible solutions of 3 12 3 15 are bounded indepen. By the Leray Schauder principle 3 8 3 11 has solutions and all such solutions. are bounded independent of n and vn, A passage to the limit argument now ensures that there exist periodic solutions. to 3 1 and 3 2 which are limit points of the sequences un and vn This. step follows standard arguments of passing through subsequences and closure type. properties The key assumption is that ft is strictly monotone and continuous. Finally we establish the uniqueness of the solution to 3 1 3 2 Thus let u v. i 1 2 be two sets of solutions to 3 1 3 2 i e, u tt u xx u xu xx u xvx P u t P x 3J1. v vL v x vL v xu x x y v o 3 32, Adding these two equations and setting A u vl we get. A t A xx x U xx V xV xx V x x X U U y v t P 1, A t Kx 4x4 P u y v p 3 33.
Similarly subtracting 3 32 from 3 31 and setting Bl ul v we get. K Kx B B xx fi ui y vf p g 3 34, We now write 3 33 for i 1 and i 2 subtract these two equations multiplying. the difference by a A2f and integrate over 0 n x 0 2n Using the periodicity. of A with respect to t and the property that the A satisfy Dirichlet boundary. conditions in x we get, 4x4 4x4 4 2, 64 M COUNTRYMAN and R KANNAN. Simplifying, y v p u y v2t u u v v 0 3 35, Similarly from 3 34 we get. y v p u y v2t u u v v 0 3 36, Adding 3 35 and 3 36 we have. p u p u u u 0 3 37, y v y v2t v v2t 0 3 38, The strict increasing nature of and y as assumed at the beginning of this section.
implies that u u2 and vj vf Now repeating the above steps with 3 33 and. 3 34 excepting that we take inner products with A x B x and so on we conclude. that the solution to 3 1 3 2 is unique, 4 Discrete approximations In this section we develop discrete approximations. of the system, u cluxx cWxuxx t 4 1, Vn ClVxx C VlVxx hc ulxvx y Ut fl X 0 4 2. subject to, u 0 t u 7i t 0 u x t u x t 2n 4 3, v 0 v n t 0 v x t v x t 2n 4 4. We begin our presentation of the numerical method by discussing finite difference. approximations for the derivatives which occur We employ the standard second. order centered differences in t Thus at the point xj t we have. utl LJzi 0 k2, u U J XMiJ X fe2, where utj w x tj u ih j 1 k For the derivatives in we employ. averaged centered differences to create an implicit system at each time level To fix. the ideas consider the term, 2 3 2 2 d r 2 1 2 s3, C0Uxx ICl vWv dx C0Ux 2CMx.
We may approximate ux x t by, FORCED OSCILLATIONSOF ELASTICSTRINGS 65. d 2 1 2 j Ui 2 j 1 2 Ui 2 j Ui l 2J, The outer derivative in 4 5 is approximated in a similar way We see that. d 2Ui l 2 j, Ui l 2 j o 9, dx h dx Ui l 2 j 1 2 j, o i l Uij Ui lJ. 2 Ui j Uij Ui j, C0 0 h2 4 6, Note that the final expression in 4 6 is just the standard second order centered. difference clu, cl4 u Similarly, d ic2 w 2 A31 c ra.
dx 2 1 V h 2h3 dx U l 2 t, K 1 2J 3 I 3 0 h2, 1 2 i K I J. Finally we approximate j uxv2 by, d 2 d 1 2 Ui 2 j Vi 2 j Vi 2 j. h3 dX W 2 Ui l 2 j Vi l 2 j 1 2 0 z, i i Vi o 2, 1 V 2 K K Vi 0 2. If we replace the derivatives in 4 1 by the differences given above and with dif. ferences in 4 6 4 7 and 4 8 averaged between time lines j 1 and j 1 we. 66 M COUNTRYMAN and R KANNAN, ui J i 2uij uij i, o Mi l y l 2uiJ ui l j l Ui l j 1 2ui j ui J l. 4 WI 1 1 Ui j 0 Ui j M l 1, 4 T4 M 1 7 1 Mi 7 l V l 7 1 Vi j.
7 K 7 Vl 7 l 2, After multiplication of 4 9 by h2 rearrangement yields. 2 i y i 0 7 i l j l fif1 J L, k2 J 1 2 i y i V 2k U J. wri M l 7 1 W 7 l I 7 1 M l 7 l, 2 1 y 1 1 7 1 7 l 2. 7 1 7 1 7 1 l 7 l 2, j h2 2 72 C2, 2p 0 c0 2 j, 4 2 V UiJ 1 1 7 l. 2 l 1 7 l 7 l K 1 7 l U 7 2, l l f 7 1 l y l 2 AVl 7 4 10.
To obtain the discrete analog of 4 2 modify 4 10 by interchanging u and v. FORCED OSCILLATIONSOF ELASTICSTRINGS 67, replacing ft with y and replacing with f2lj Thus we obtain. vi lJ l vij xf Vij l l 1 3, 2 l Vi l j l Vi j l ui l j i Vi J l 2. Vij l J 1 1 2, 2 K i y 1 y 3 Kj i V i 3, K 1 J I J l l l 2. Vij i K l i i 2 247 4 n, We may now write 0 and g2jj 0 where gUj is defined to be the left hand. side of 4 10 and g2lJ is defined to be the left hand side of 4 11 We assume that. u and v are known at the grid points x and x 1 2 N. Therefore the unknowns at each time step are the values ut x and vi J x i. 1 2 N and all terms inside the curly brackets in 4 10 and 4 11 are con. sidered known We may solve for the w 1 and vt at each time step by the. generalized Newton method That is the system can be solved iteratively by. iJ l iJ l dgUj duiJ i 4 U, dg2ij dvij i, where w is a relaxation parameter 0 a 2.
a k2J 2k 2k, 2 l M y l M l y l, 2 K i w i 2 Vi i i 2. 68 M COUNTRYMAN and R KANNAN, 8 J I k2J 2k V 2 r, f i 1 2 K Vl i l 2. 4 M 1 J 1 U J Ui j i Ui l j 0, Note that dgUJ dui J x 0 and dg2ij dvj j 0 since 0 and 0. Also we have, dgUj c2 3cf 2 c 2, dui 1 7 1 2 W 1, 1 1 Mi l Vi 1. FORCED OSCILLATIONS OF ELASTIC STRINGS WITH NONLINEAR DAMPING By M COUNTRYMAN Louisiana Tech University Ruston Louisiana AND R KANNAN University of Texas Arlington Texas 1 Introduction In this paper we discuss the question of existence and numerical approximation of periodic solutions of nonlinear elastic string vibrations in the pres ence of damping Starting with an analysis of

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