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On resonances in transversally vibrating strings induced by an external force and a time- dependent coefficient in a Robin boundary condition

Wang, Jing; van Horssen, Wim T.; Wang, Jun-Min DOI 10.1016/j.jsv.2021.116356 Publication date 2021 Document Version Final published version Published in Journal of Sound and Vibration

Citation (APA) Wang, J., van Horssen, W. T., & Wang, J-M. (2021). On resonances in transversally vibrating strings induced by an external force and a time-dependent coefficient in a Robin boundary condition. Journal of Sound and Vibration, 512, 1-16. [116356]. https://doi.org/10.1016/j.jsv.2021.116356 Important note To cite this publication, please use the final published version (if applicable). Please check the document version above.

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Journal of Sound and Vibration

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On resonances in transversally vibrating strings induced by an external force and a time-dependent coefficient in a Robin boundary condition Jing Wang a,b,∗, Wim T. van Horssen b, Jun-Min Wang a a School of Mathematics and Statistics, Beijing Institute of Technology, Beijing, 100081, PR China b Department of Mathematical Physics, Delft Institute of Applied Mathematics, Delft University of Technology, Mekelweg 4, Delft, 2628CD, Netherlands

ARTICLEINFO ABSTRACT

Keywords: In this paper an initial– on a bounded, fixed interval is considered for Robin boundary condition a one-dimensional and forced string equation subjected to a Dirichlet boundary condition at Time-dependent coefficient one end of the string and a Robin boundary condition with a slowly varying time-dependent Interior layer analysis coefficient at the other end of the string. This problem may serve as a simplified model Multiple-timescales perturbation method describing transverse or longitudinal vibrations as well as resonances in axially moving cables Resonance zone for which the length changes in time. By introducing an adapted version of the method of separation of variables, by using averaging and singular perturbation techniques, and by finally using a three time-scales perturbation method, resonances in the problem are detected and accurate, analytical approximations of the solutions of the problem are constructed. It will turn √ out that small order 휀 excitations can lead to order 휀 responses when the frequency of the external force satisfies certain conditions. Finally, numerical simulations are presented, which are in full agreement with the obtained analytical results.

1. Introduction

In the last few decades, high-rise buildings have become increasingly popular. The higher buildings rise, the more vulnerable they become to wind and earthquake influences. At the same time, it poses particular design challenges to the vertical transportation systems such as high-rise elevators. Not only can external wind and earthquake forces cause building sway, it can also damage the elevator cables. If a frequency of the external wind force coincides with one of the natural frequencies of the elevator cable, large oscillations can occur and damage can be caused. This phenomenon is called resonance. In most cases resonance finally leads to failures. In order to prevent such failures, it is important to understand the nature of these vibrations. There is a lot of research on these types of problems. Kaczmarczyk [1] analysed resonance in a catenary-vertical cable with slowly varying length under a periodic excitation. Fajans et al. [2] introduced auto-resonant (nonstationary) excitation of the diocotron mode in non-neutral plasmas. Chen and Yang [3] considered the stability in parametric resonance of axially moving viscoelastic beams with time-dependent speed. Friedland et al. [4] proposed auto-resonant phase-space holes in plasmas. Kimura et al. [5] proposed forced vibration analysis of an elevator rope with both ends excited by wind-induced displacement sway of the building. Sandilo and van Horssen [6] studied auto-resonance phenomena in a space–time-varying mechanical system. Gaiko and van Horssen [7] considered lateral vibrations of a

∗ Corresponding author at: Department of Mathematical Physics, Delft Institute of Applied Mathematics, Delft University of Technology, Mekelweg 4, Delft, 2628CD, Netherlands. E-mail address: [email protected] (J. Wang). https://doi.org/10.1016/j.jsv.2021.116356 Received 16 February 2021; Received in revised form 24 June 2021; Accepted 11 July 2021 Available online 11 August 2021 0022-460X/© 2021 The Authors. Published by Elsevier Ltd. This is an open access article under the CC BY license

(http://creativecommons.org/licenses/by/4.0/). J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

Fig. 1. The transverse vibrating string with a time-varying spring-stiffness support at 푥 = 퐿, and an external force 휀퐴 cos(휔푡).

vertically moving string with in time harmonically varying length, and in [8] the authors further discussed resonances and vibrations in an elevator cable system due to boundary sway. Kaczmarczyk [9] studied how simulation models and control strategies can be deployed to mitigate the effects of resonance conditions induced by wind loads and long-period seismic excitations. Zhu and Wu [10] studied the transverse vibration of the translating string with sinusoidally varying velocities. In this paper we are motivated by resonance phenomena occurring in a transversally vibrating string (see Fig.1 ), where one end of the string is fixed, and the other one is attached to a spring for which the stiffness properties change in time (due to fatigue, temperature change, and so on). Mathematically, we will show how to (approximately) solve an initial–boundary value problem for a nonhomogeneous wave equation on a bounded, fixed interval with a Dirichlet type of boundary condition at one endpoint, and a Robin type of boundary condition with a time-dependent coefficient at the other end. Actually, the Robin boundary condition is an interesting one to study from the application and mathematical point of view. The wave equations involving a Robin type of boundary condition with a time-varying coefficient can be regarded as simple models for vibrations of elevator or mining cables in the study of axially moving strings with time-varying lengths. Chen et al. [11] considered an analytical vibration response in the time domain for an axially translating and laterally vibrating string with mixed boundary conditions. Further Chen et al. [12] investigated the exchange of vibrational energy of a finite length translating tensioned string model with mixed boundary conditions applying d’Alembert’s principle and the reflection properties. Wang et al. [13] designed an output feedback controller to regulate the state of a wave equation on a time-varying spatial interval with an unknown boundary disturbance. Wang et al. [14] studied a typical flexible hoisting system and proposed an absorber with artificial intelligence optimization to reduce system vibrations. For more information on initial–boundary value problems for axially moving continua, the reader is referred to [15–21]. Also in other fields, Robin boundary conditions play an important role, and are sometimes called impedance boundary conditions in electromagnetic problems or convective boundary conditions in heat transfer problems. Usually the method of separation of variables (SOV), or the (equivalent) Laplace transform method is used to solve initial value problem for a wave equation on a bounded interval for various types of boundary conditions with constant coefficients. However, when a Robin boundary condition with a time-dependent coefficient is involved in the problem, the afore-mentioned methods are not applicable. For this reason, van Horssen and Wang in [21] employed the method of d’Alembert to solve a homogeneous wave equation involving Robin type of boundary conditions with time-dependent coefficients. In this method, the time domain can be divided into finite intervals of length 2, so that the initial conditions extension procedure for each interval coincides with the previous ones. Accordingly, one can obtain an analytical expression in a rather straightforward way for the solution on the time-interval [0, 2푛] with 푛 = 1, 2, 3, … , 푁, and 푁 not too large. But, one will encounter computational issues for large N. When 푡 is relatively large, the amount of calculations is very large and the calculation process is complicated. Therefore, in this paper we want to construct accurate approximations of the solutions of these problems on long timescales by using a different procedure. This paper is organized as follows. In Section2 , the problem is formulated and a short motivation is given for the methods which are used in this paper. In Section3 , an adapted version of the method of separation of variables is introduced. This approach allows us to define a new slow variable 휏 = 휀푡, and to separate 푢(푥, 푡) as 푇 (푡, 휏)푋(푥, 휏). In Section4 , averaging and singular perturbation techniques are used to detect resonance zones, and to determine the scalings which are presented in the problem. By using these scalings, a three time-scales perturbation method is used in Section5 to construct accurate, analytical approximations of the solutions of the initial–boundary value problem. In Section6 numerical simulations are presented, which are in full agreement with the obtained, analytical approximations. As numerical method a standard finite difference method is used in this paper for simplicity, but of course also more advanced methods such as the finite element method (as has been used in [22]) can be applied. Finally, in Section7 some concluding remarks will be made.

2. Formulation of the problem

By using Hamilton’s principle, the governing equation of motion to describe the transversal vibration of a string as shown in Fig.1 can be derived, and is given by:

휌푢푡푡(푥, 푡) − 푃 푢푥푥(푥, 푡) = 휀퐴 cos(휔푡), 휔 > 0, 0 < 푥 < 퐿, 푡 > 0, (1) where 휌 is the mass density, 푃 is the axial tension (which is assumed to be constant), 퐿 is the distance between the supports, and 푢 describes the lateral displacement of the string. The term 휀퐴 cos(휔푡) in( 1) is a small external force acting on the whole string, where 휀, 휔 and 퐴 are constants with 0 < 휀 ≪ 1, 휔 > 0 and 퐴 ∈ R. The boundary conditions and initial conditions are given by:

푢(0, 푡) = 0, 푃 푢푥(퐿, 푡) + 푘(푡)푢(퐿, 푡) = 0, 푘(푡) = 1 + 휀푡, 푡 ≥ 0, (2)

2 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

푢(푥, 0) = 휀푢0(푥), 푢푡(푥, 0) = 휀푢1(푥), 0 < 푥 < 퐿, (3) where 푘(푡) is the time-varying stiffness of the spring at 푥 = 퐿. The boundary condition at 푥 = 0 is a Dirichlet type of boundary condition, and the boundary condition at 푥 = 퐿 is a Robin type of boundary condition with a time-dependent coefficient 푘(푡). In this example, the choice of 푘(푡) leads to a spring which becomes stiffer and stiffer in time. For simplicity, based on the Buckingham Pi theorem, the following dimensionless parameters are used: √ √ 푥 푢 푡 푃 퐿 휌 푥 = , 푢 = , 푡 = , 푘 = 푘, 휀 = 휀퐿 , 퐿 퐿 퐿 휌 푃 푃 √ √ 퐴 휌 푃 푢0 푢1 퐴 = √ , 휔 = 퐿휔 , 푢0 = , 푢1 = , 휌푃 푃 휌 퐿2 퐿 by which, the governing equation(1), the boundary conditions(2), and the initial conditions(3) can be rewritten into the following non-dimensional form:

푢푡푡(푥, 푡) − 푢푥푥(푥, 푡) = 휀퐴 cos(휔푡), 휔 > 0, 0 < 푥 < 1, 푡 > 0, (4) 푢(0, 푡) = 0, 푡 ≥ 0,

푢푥(1, 푡) + 푘(푡)푢(1, 푡) = 0, 푘(푡) = 1 + 휀푡, 푡 ≥ 0, (5)

푢(푥, 0) = 휀푢0(푥), 0 < 푥 < 1, 푢푡(푥, 0) = 휀푢1(푥), 0 < 푥 < 1, (6) respectively, where the overbar notations are omitted for convenience. Due to the Robin boundary condition with the time-dependent coefficient 푘(푡), traditional, analytical methods, such as the method of separation of variables (SOV), and the (equivalent) Laplace transform method, can usually not be used. By putting 푘(푡) = 1+휀푡 an additional difficulty is introduced: for 푡 < 푂( 1 ), 휀푡 is a small term, while for 푡 = 푂( 1 ), 휀푡 is not a small term. So, we need to analyse 휀 휀 this problem from a new view-point. Firstly, since the coefficient changes slowly in time, we study the problem in Section3 by an adapted version of the method of separation of variables, in which an extra independent slow time variable 휏 = 휀푡 is defined, and 푢(푥, 푡) can be separated as 푇 (푡, 휏)푋(푥, 휏). Then, by using the boundary conditions, the original partial differential equation can be transformed into linear ordinary differential equations with slowly varying (prescribed) frequencies. Unexpectedly (or not), the slow variation leads to a singular perturbation problem. By applying an interior layer analysis in the averaging procedure in Section4 a resonance manifold is found. Three different scalings turn out to be present in the problem. For that reason, a three-timescales perturbation method is used in Section5 to construct explicit approximations of the solutions of the initial–boundary value problem (4)–(6).

3. An adapted version of the method of separation of variables

First of all, in the method of separation of variables we consider the homogeneous part of Eq.(4) subject to the homogeneous boundary conditions(5): { 푢 (푥, 푡) − 푢 (푥, 푡) = 0, 0 < 푥 < 1, 푡 0, 푡푡 푥푥 ≥ (7) 푢(0, 푡) = 0, 푢푥(1, 푡) + 푘(푡)푢(1, 푡) = 0, 푘(푡) = 1 + 휀푡, 푡 ≥ 0. Note that the coefficient 푘(푡) in the Robin boundary condition is slowly varying in time. So, in order to derive a solution of problem (7), we define an extra slow time variable 휏 = 휀푡, which will be treated independently from the variable 푡. Hence 푢(푥, 푡) becomes a new function ̄푢(푥, 푡, 휏) and further problem(7) becomes

2 ̄푢푡푡(푥, 푡, 휏) + 2휀̄푢푡휏 (푥, 푡, 휏) + 휀 ̄푢휏휏 (푥, 푡, 휏) − ̄푢푥푥(푥, 푡, 휏) = 0, 0 < 푥 < 1, 푡 > 0, 휏 > 0,

̄푢(0, 푡, 휏) = 0, ̄푢푥(1, 푡, 휏) + (1 + 휏)̄푢(1, 푡, 휏) = 0, 푡 ≥ 0, 휏 ≥ 0. (8) By looking for a nontrivial solution ̄푢(푥, 푡, 휏) in the form 푇 (푡, 휏)푋(푥, 휏), the governing equations of(8) can be approximately written as

2 푋(푥, 휏)푇푡푡(푡, 휏) + 2휀푋(푥, 휏)푇푡휏 (푡, 휏) + 2휀푋휏 (푥, 휏)푇푡(푡, 휏) − 푋푥푥(푥, 휏)푇 (푡, 휏) + 푂(휀 ) = 0, or equivalently as 푇 (푡, 휏) 푋 (푥, 휏) 푡푡 + 푂(휀) = 푥푥 , 0 < 푥 < 1, 푡 0, 휏 0. (9) 푇 (푡, 휏) 푋(푥, 휏) ≥ ≥ The 푂(1) part of the left-hand side of Eq.(9) is a function of 푡 and 휏, and the right-hand side is a function of 푥 and 휏. To be equal, both sides need to be equal to a function of 휏. Let this function be −휆2(휏) (which will be defined later), so we get 푇 (푡, 휏) 푋 (푥, 휏) 푡푡 = 푥푥 = −휆2(휏), 0 < 푥 < 1, 푡 0, 휏 0, 푇 (푡, 휏) 푋(푥, 휏) ≥ ≥ implying:

2 푋푥푥(푥, 휏) + 휆 (휏)푋(푥, 휏) = 0, 0 < 푥 < 1, 휏 ≥ 0,

3 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

Fig. 2. For 푡 = 0, intersection points of 푦 = tan 휆 and 푦 = −휆 are giving 휆푛(0).

2 푇푡푡(푡, 휏) + 휆 (휏)푇 (푡, 휏) = 0, 푡 ≥ 0, 휏 ≥ 0. (10) From the boundary condition( 8), we obtain

푇 (푡, 휏)푋(0, 휏) = 0 ⇒ 푋(0, 휏) = 0,

푇 (푡, 휏)푋푥(1, 휏) + (1 + 휏)푇 (푡, 휏)푋(1, 휏) = 0 ⇒ 푋푥(1, 휏) + (1 + 휏)푋(1, 휏) = 0. (11)

In accordance with the first equation for 푋(푥, 휏) in( 10), a nontrivial solution 푋푛(푥, 휏) (satisfying( 11)) is

푋푛(푥, 휏) = 퐵푛(휏) sin(휆푛(휏)푥), (12) where 퐵푛(휏) is a function of 휏 only, and 휆푛(휏) is the 푛th positive root of 휆 (휏) tan(휆 (휏)) = − 푛 . (13) 푛 1 + 휏

For 휏 = 0 it is indicated in Fig.2 how 휆푛(0) can be obtained. In should be observed that the eigenfunctions 푋푛(푥, 휏) are orthogonal on 0 < 푥 < 1. And so, the general solution of( 4)–(6) can be expanded in the following form: ∞ ∑ 푢(푥, 푡) = ̄푢(푥, 푡, 휏) = 푇푛(푡, 휏) sin(휆푛(휏)푥), (14) 푛=1 where the boundary conditions( 5) are automatically satisfied. Substituting Eq.( 14) into Eq.( 4), and into Eq.( 6) yields ∞ ∑ 푑휆 (휏) [(푇 + 2휀푇 + 휆2(휏)푇 ) sin(휆 (휏)푥) + 2휀푥 푛 푇 cos(휆 (휏)푥)] = 휀퐴 cos(휔푡) + 푂(휀2), 푛,푡푡 푛,푡휏 푛 푛 푛 푑휏 푛,푡 푛 푛=1 ∞ ∑ [푇푛(0, 0) sin(휆푛(0)푥)] = 휀푢0(푥), 푛=1 ∞ ∑ 푑휆 (0) [(푇 (0, 0) + 휀푇 (0, 0)) sin(휆 (0)푥) + 휀푇 (0, 0) 푛 푥 cos(휆 (0)푥)] 푛,푡 푛,휏 푛 푛 푑휏 푛 푛=1

= 휀푢1(푥). (15)

Now, by multiplying the first equation in( 15) with sin(휆푘(휏)푥), and the second and third equations in( 15) with sin(휆푘(0)푥), by integrating the so-obtained equation from 푥 = 0 to 푥 = 1, and by using the orthogonality properties of the sin-functions on 0 < 푥 < 1, we obtain the following differential equations for 푘 = 1, 2, 3, …, and 푡 > 0, 휏 > 0:

푇 + 휆2(휏)푇 = 휀[−2푇 − 2 ∑∞ 푑휆푛(휏) 푐 (휏)푇 + 퐴푑 (휏) cos(휔푡)] + 푂(휀2), 푡 0, 휏 0, ⎧ 푘,푡푡 푘 푘 푘,푡휏 푛=1 푑휏 푛,푘 푛,푡 푘 ≥ ≥ ⎪ 1 ∫0 푢0(휉) sin(휆푘(0)휉)푑휉 ⎪푇푘(0, 0) = 휀 = 휀퐹푘, 1 휆 휉 휆 휉 푑휉 ⎪ ∫0 sin( 푘(0) ) sin( 푘(0) ) 1 푢 (휉) sin(휆 (0)휉)푑휉 ⎪푇 (0, 0) + 휀푇 (0, 0) = 휀 ∫0 1 푘 (16) ⎨ 푘,푡 푘,휏 1 ∫0 sin(휆푘(0)휉) sin(휆푘(0)휉)푑휉 ⎪ 1 휉 cos(휆 (0)휉) sin(휆 (0)휉)푑휉 ⎪ − 휀 ∑∞ 푇 (0, 0) 푑휆푛(0) ∫0 푛 푘 푛=1 푛 푑휏 1 ⎪ ∫0 sin(휆푘(0)휉) sin(휆푘(0)휉)푑휉 ⎪ ⎩ = 휀퐺푘.

4 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356 where 푐푛,푘(휏) and 푑푘(휏) are functions of 휏, and are given by: 1 ∫0 푥 cos(휆푛(휏)푥) sin(휆푘(휏)푥)푑푥 푐푛,푘(휏) = , 1 2 ∫0 sin (휆푘(휏)푥)푑푥 1 ∫0 푥 cos(휆푘(휏)푥) sin(휆푘(휏)푥)푑푥 푐푘,푘(휏) = , 1 2 ∫0 sin (휆푘(휏)푥)푑푥 1 ∫ sin(휆푘(휏)푥)푑푥 4(1 − cos(휆 (휏))) 푑 (휏) = 0 = 푘 . (17) 푘 1 2 2휆푘(휏) − sin(2휆푘(휏)) ∫0 sin (휆푘(휏)푥)푑푥 ̃ To simplify the formula, we define a new dependent variable: 푇푘(푡) = 푇푘(푡, 휏), yielding 푇̃ + 휆2(휏)푇̃ = 휀[−2훴∞ 푑휆푛(휏) 푐 (휏)푇̃ + 퐴푑 (휏) cos(휔푡)] + 푂(휀2), ⎧ 푘,푡푡 푘 푘 푛=1 푑휏 푛,푘 푛,푡 푘 ⎪ ̃ ⎨푇푘(0) = 휀퐹푘, (18) ⎪ ̃ ⎩푇푘,푡(0) = 휀퐺푘, where 휏 = 휀푡, 푡 ≥ 0. In the next section we will use the averaging method to detect resonance zones in problem(18), and to determine time-scales which describe the solution of(18) accurately.

4. Averaging and resonance zones

The linear ordinary differential equation(18) with the slowly varying frequency 휆푘(휏) as given by(13), can be analysed by making use of the averaging method. In this section it will be shown that an interior layer analysis (including a rescaling and balancing procedure) leads to a description of an (un-)expected resonance manifold and leads to time-scales which describe the solution of (18) sufficiently accurately. To apply the method of averaging to(18) the following standard transformations are introduced: 푡 휙푘(푡) = 휆푘(휀푠)푑푠 and 훷 = 휔푡, (19) ∫0 ̃ ̃ and 푇푘(푡), 푇푘,푡(푡) are described by 퐴푘(푡), 퐵푘(푡) in the following way: ̃ 푇푘(푡) = 퐴푘(푡) sin(휙푘(푡)) + 퐵푘(푡) cos(휙푘(푡)), ̃ 푇푘,푡(푡) = 휆푘(휏)퐴푘(푡) cos(휙푘(푡)) − 휆푘(휏)퐵푘(푡) sin(휙푘(푡)). (20) Problem(18) can now be rewritten in the following problem:

̇ 푑휆푘(휏) 1 2 푑휆푘(휏) 1 ⎧퐴푘 = 휀[− ( + 2푐푘,푘(휏))퐴푘 cos (휙푘(푡)) + ( + 푐푘,푘(휏))퐵푘 sin(2휙푘(푡)) ⎪ 푑휏 휆푘(휏) 푑휏 2휆푘(휏) 푑휆푛(휏) 푑휆푛(휏) ⎪ − 2훴푛 푘 푐푛,푘(휏)퐴푛 cos(휙푛(푡)) cos(휙푘(푡)) + 2훴푛 푘 푐푛,푘(휏)퐵푛 sin(휙푛(푡)) cos(휙푘(푡)) ⎪ ≠ 푑휏 ≠ 푑휏 + 퐴푑푘(휏) (cos(훷 + 휙 (푡)) + cos(훷 − 휙 (푡)))], ⎪ 2휆 (휏) 푘 푘 ⎪ 푘 퐵̇ = 휀[ 푑휆푘(휏) ( 1 + 푐 (휏))퐴 sin(2휙 (푡)) − 푑휆푘(휏) ( 1 + 2푐 (휏))퐵 sin2(휙 (푡)) ⎪ 푘 푑휏 2휆 (휏) 푘,푘 푘 푘 푑휏 휆 (휏) 푘,푘 푘 푘 ⎪ 푘 푘 + 2훴 푑휆푛(휏) 푐 (휏)퐴 cos(휙 (푡)) sin(휙 (푡)) − 2훴 푑휆푛(휏) 푐 (휏)퐵 sin(휙 (푡)푡) sin(휙 (푡)) (21) ⎨ 푛≠푘 푑휏 푛,푘 푛 푛 푘 푛≠푘 푑휏 푛,푘 푛 푛 푘 ⎪ 퐴푑푘(휏) − (sin(훷 + 휙푘(푡)) − sin(훷 − 휙푘(푡)))], ⎪ 2휆푘(휏) ⎪ ̇휏 휀, ⎪ = ⎪훷̇ = 휔, ⎪ ⎪휙̇ = 휆 (휏). ⎩ 푘 푘 ̇ ̇ ̇ ̇ Resonance in(21), due to the external forcing with frequency 휔, can be expected when 훷 − 휙푘 ≈ 0, or 훷 + 휙푘 ≈ 0. But since 휔 > 0 and 휆푘(휏) > 0, resonance only will occur when 휔 1 휔 휔 ≈ 휆 (휏) ⇔ 휏 ≈ − − 1 ⇔ 푡 ≈ (− − 1). (22) 푘 tan 휔 휀 tan 휔 휆 휏 Since 휆 (휏) satisfies(13), that is, tan(휆 (휏)) = − 푘( ) , it follows (see also Fig.2) that when t increases, then the value of 휆 (휏) 푘 푘 1+휏 푘 increases. Besides, for t tending to infinity, 휆푘(휏) tends to 푘휋 (with 푘 = 1, 2, …). Therefore, 휆푘(휏) is increasing in time and

0 < 휆푘(0) ≤ 휆푘(휏) < 푘휋, tan(휆푘(0)) = −휆푘(0). (23) From(23) we can then conclude that:

1. When the external force frequency 휔 is bounded away from 휆푘(0) by a constant and satisfies (푘 − 1)휋 ≤ 휔 < 휆푘(0), then no resonance will occur; 2. When the external force frequency 휔 satisfies 0 < 휆 (0) 휔 < 푘휋, then resonance will occur around 푡 = 1 (− 휔 −1). Moreover, 푘 ≤ 휀 tan 휔 for − 휔 − 1 > 0, and − 휔 − 1 = 푂(휀), the resonance time zone is around 푡̃ with 푡̃ = 푂(1); and when − 휔 − 1 = 푂(1), the tan 휔 tan 휔 tan 휔 resonance time zone is around 푡̃ with 푡̃ = 푂( 1 ). 휀

5 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

As long as we stay out of the resonance time zone (or equivalently, the resonance manifold), the variables 퐴푘 and 퐵푘 are slowly varying in time. For that reason we can average the right-hand side of the equations in(21) over 휙푘 and 훷 while keeping 퐴푘 and 퐵푘 constant. The averaged equation for 퐴푘 and 퐵푘 now become

⎧ ̇푎 푑휆푘(휏) 1 푎 퐴 = −휀 ( + 푐푘,푘(휏))퐴 , ⎪ 푘 푑휏 2휆푘(휏) 푘 (24) ⎨ ̇ 푎 푑휆푘(휏) 1 푎 퐵 = −휀 ( + 푐푘,푘(휏))퐵 , ⎪ 푘 푑휏 2휆푘(휏) 푘 ⎩ where the upper index 푎 indicates that this is the averaged function. From the expression for 푐푘,푘 in(17), we then obtain 1 푑휆 (휏) 1 ∫ 푥 cos(휆푘(휏)푥) sin(휆푘(휏)푥)푑푥 퐴̇푎 = −휀 푘 ( + 0 )퐴푎 푘 1 푘 푑휏 2휆푘(휏) 2 ∫0 sin (휆푘(휏)푥)푑푥 푑 1 2 휆 휏 푥 푑푥 휀 푑휆푘(휏) 푑(ln(휆푘(휏))) (ln(∫0 sin ( 푘( ) ) )) 푎 = − ( + )퐴푘 2 푑휏 푑휆푘(휏) 푑휆푘(휏) 푑(ln( 1 − sin(2휆푘(휏)) )) 휀 푑휆푘(휏) 푑(ln(휆푘(휏))) 2 4휆푘(휏) 푎 = − ( + )퐴푘 2 푑휏 푑휆푘(휏) 푑휆푘(휏) 휆푘(휏) sin(2휆푘(휏)) 1 푑(ln( − )) = − ( 2 4 )퐴푎, (25) 2 푑푡 푘 which implies that 퐶 퐶 퐴푎 1 , 퐵푎 2 , 푘 = √ 푘 = √ (26) 2휆푘(휏) − sin(2휆푘(휏)) 2휆푘(휏) − sin(2휆푘(휏)) with √ 휀퐺푘 2휆푘(0) − sin(2휆푘(0)) √ 퐶1 = , 퐶2 = 휀퐹푘 2휆푘(0) − sin(2휆푘(0)), (27) 휆푘(0) where 퐺푘 and 퐹푘 are given in(16). Hence, outside the resonance manifold the solution of system(18) is given by √ 휀퐺 2휆 (0) − sin(2휆 (0)) ̃ 푘 푘 푘 푇푘(푡) = √ sin(휙푘(푡)) 휆푘(0) 2휆푘(휏) − sin(2휆푘(휏)) √ 휀퐹푘 2휆푘(0) − sin(2휆푘(0)) + √ cos(휙푘(푡)), (28) 2휆푘(휏) − sin(2휆푘(휏)) where 휙푘(푡) is defined in(19). When 휔 satisfies 휆푘(0) ≤ 휔 < 푘휋 for a certain 푘 (with 푘 = 1, 2, …) then a resonance zone will occur. We introduce

휓 = 훷(푡) − 휙푘(푡), (29) and rescale 휏 − 휏 = 훿(휀) ̄휏 with ̄휏 = 푂(1) and 휏 = − 휔 − 1. System(21) then becomes: 푘 푘 tan 휔

⎧ ̇ 푑휆푘(휏) 1 2 푑휆푘(휏) 1 퐴푘 = 휀[− ( + 2푐푘,푘(휏))퐴푘 cos (휙푘(푡)) + ( + 푐푘,푘(휏))퐵푘 sin(2휙푘(푡)) ⎪ 푑휏 휆푘(휏) 푑휏 2휆푘(휏) ⎪ − 2훴 푑휆푛(휏) 푐 (휏)퐴 cos(휙 (푡)) cos(휙 (푡)) + 2훴 푑휆푛(휏) 푐 (휏)퐵 sin(휙 (푡)) cos(휙 (푡)) ⎪ 푛≠푘 푑휏 푛,푘 푛 푛 푘 푛≠푘 푑휏 푛,푘 푛 푛 푘 퐴푑푘(휏) ⎪ + cos(훷 + 휙푘(푡)) + cos(휓)], ⎪ 2휆푘(휏) 퐵̇ = 휀[ 푑휆푘(휏) ( 1 + 푐 (휏))퐴 sin(2휙 (푡)) − 푑휆푘(휏) ( 1 + 2푐 (휏))퐵 sin2(휙 (푡)) ⎪ 푘 푑휏 2휆 (휏) 푘,푘 푘 푘 푑휏 휆 (휏) 푘,푘 푘 푘 ⎪ 푘 푘 + 2훴 푑휆푛(휏) 푐 (휏)퐴 cos(휙 (푡)푡) sin(휙 (푡)) − 2훴 푑휆푛(휏) 푐 (휏)퐵 sin(휙 (푡)) sin(휙 (푡)) ⎪ 푛≠푘 푑휏 푛,푘 푛 푛 푘 푛≠푘 푑휏 푛,푘 푛 푛 푘 (30) ⎨ 퐴푑푘(휏) − sin(훷 + 휙푘(푡)) − sin(휓)], ⎪ 2휆푘(휏) ⎪ ⎪ ̇휏 = 휀, ⎪훷̇ = 휔, ⎪ ⎪ ̄휏̇ = 휀 , ⎪ 훿(휀) ⎪ ̇휓 = 휔 − 휆푘(휏푘 + 훿(휀) ̄휏). ⎩

To simplify(30) it should be observed that for 휏 = 휏푘 + 훿(휀) ̄휏 we have

̇휓 = 휔 − 휆푘(휏푘 + 훿(휀) ̄휏) = 휆푘(휏푘) − 휆푘(휏푘 + 훿(휀) ̄휏) 푑휆 = 휆 (휏 ) − (휆 (휏 ) + 훿(휀) ̄휏 푘 + 푂(훿2(휀))) 푘 푘 푘 푘 푑휏 |휏=휏푘 푑휆 = −훿(휀) ̄휏 푘 + 푂(훿2(휀)). (31) 푑휏 |휏=휏푘

6 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

휆 휏 By differentiating(13), that is, tan(휆 (휏)) = − 푘( ) with respect to 휏, we obtain 푘 1+휏 1 푑휆 −1 푑휆 휆 푘 = 푘 + 푘 . (32) 2 2 cos 휆푘 푑휏 1 + 휏 푑휏 (1 + 휏) And so, it follows from(32) that 푑휆 휆 cos2 휆 푘 = 푘 ⋅ 푘 |휏=휏푘 2 |휏=휏푘 푑휏 (1 + 휏) 1 + 휏 + cos 휆푘 sin 2휔 sin2 휔 = − = , (33) 2 2(1 + 휏푘 + cos 휔) 휔 − sin 휔 cos 휔 which implies for ̇휓 (see(31)): sin2 휔 ̇휓 = − 훿(휀) ̄휏 + 푂(훿2(휀)). (34) 휔 − sin 휔 cos 휔 2 From(33) and from 휆 (0) 휔 < 푘휋, we obtain sin 휔 0. 푘 ≤ 휔−sin 휔 cos 휔 ≠ Based on(34) it now follows from(34) that a balance in system(30) occurs when 휀 = 훿(휀), and this implies for the averaging 훿 휀 √ ( ) procedure in the resonance zone that 훿(휀) = 휀. So, together with 휏 − 휏푘 = 훿(휀) ̄휏, it follows from(30) that √ 휏 ̄휏 = 휀(푡 − 푡 ), 푡 = 푘 . (35) 푘 푘 휀 Further, from(34), we obtain 1 sin2 휔 휓(푡) = 휓(푡 ) − 훼휀(푡 − 푡 )2, 훼 = . (36) 푘 2 푘 휔 − sin 휔 cos 휔 Hence, in the resonance zone, we can write 1 1 휔 cos(휓(푡)) = cos(− 훼휀(푡 − 푡 )2 + 휔푡 − 휙 (푡 )), 푡 = − ( + 1). (37) 2 푘 푘 푘 푘 푘 휀 tan 휔

So taking into account(35), let us average system(30) over the fast variables. Then, the averaged equations for 퐴푘 and 퐵푘 become

⎧퐴̇푎 = −휀 푑휆푘(휏) ( 1 + 푐 (휏))퐴푎 + 휀 퐴푑푘(휏) cos(휓), 푘 푑휏 2휆 (휏) 푘,푘 푘 2휆 (휏) ⎪ 푘 푘 (38) ⎨퐵̇ 푎 = −휀 푑휆푘(휏) ( 1 + 푐 (휏))퐵푎 + 휀 퐴푑푘(휏) sin(휓), ⎪ 푘 푑휏 2휆 (휏) 푘,푘 푘 2휆 (휏) ⎩ 푘 푘 where the upper index 푎 indicates that this is the averaged function. 푎 It follows from(37) and(38) that 퐴푘 can be written as 퐶 푡 푎 1 퐴휀 ̄ 1 ̄ 2 ̄ 퐴푘 = + ℎ푘(휀푡) cos[− 훼휀(푡 − 푡푘) + 휔푡푘 − 휙푘(푡푘)]푑푡, (39) 푙푘(휀푡) 푙푘(휀푡) ∫0 2 where 퐶1 is given by(27) and √ 푙푘(휀푡) = 2휆푘(휀푡) − sin(2휆푘(휀푡)), (40) 푙 (휀푡̄) 2(1 − cos(휆 (휀푡̄))) ℎ (휀푡̄) = 푘 푑 (휀푡̄) = 푘 . (41) 푘 2휆 (휀푡̄) 푘 √ 푘 휆푘(휀푡̄) 2휆푘(휀푡̄) − sin(2휆푘(휀푡̄)) 1 For 푡̄ = 푡 + 푂( √ ), 휏 = 휀푡 , 푘 휀 푘 푘 √ 휔 √ ℎ (휀푡̄) = ℎ (휀푡 + 푂( 휀)) = ℎ (− − 1 + 푂( 휀)) 푘 푘 푘 푘 tan 휔 휔 √ 푑ℎ (푎) = ℎ (− − 1) + 푂( 휀) ⋅ 푘 + h.o.t, (42) 푘 tan 휔 푑푎 |푎=휏푘 푑ℎ 푎 where 푘( ) is bounded due to(33). Then, 푑푎 |푎=휏푘 휔 푡 퐶 휀퐴ℎ푘(− − 1) 푎 1 tan 휔 1 ̄ 2 ̄ 퐴푘 = + cos[− 훼휀(푡 − 푡푘) + 휔푡푘 − 휙푘(푡푘)]푑푡 푙푘(휀푡) 푙푘(휀푡) ∫0 2 + h.o.t, (43) 퐴ℎ (− 휔 −1) 퐶1 푘 tan 휔 where 퐶1 is given by(27). We know that = 푂(휀) and = 푂(1), and so it is important to consider the order of 푙푘(휀푡) 푙푘(휀푡) 푡 1 2 휀 cos[− 훼휀(푡 − 푡푘) + 휔푡푘 − 휙푘(푡푘)]푑푡.̄ (44) ∫0 2 √ By setting 푢 = 1 훼휀(푡 − 푡 ), we obtain 2 푘 푡 1 2 휀 cos[− 훼휀(푡 − 푡푘) + 휔푡푘 − 휙푘(푡푘)]푑푡̄ ∫0 2

7 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

√ Fig. 3. 퐶퐹 푟(푡) (a) and 푆퐹 푟(푡) (b) have a resonance jump from 푂( 휀) to 푂(1) around 푡 = 100.

√ 훼휀 √ (푡−푡 ) 2휀 2 푘 = cos(−푢2 + 휔푡 − 휙 (푡 ))푑푢 √ 훼휀 푘 푘 푘 훼 ∫ (−푡 ) 2 푘 √ √ 2휀 2휀 = cos(휔푡 − 휙 (푡 ))퐶 (푡) + sin(휔푡 − 휙 (푡 ))푆 (푡), (45) 훼 푘 푘 푘 퐹 푟 훼 푘 푘 푘 퐹 푟 where √ 훼휀 √ 훼휀 (푡−푡 ) (푡−푡 ) 2 푘 2 푘 휏 퐶 (푡) = cos(푢2)푑푢, 푆 (푡) = sin(푢2)푑푢, 푡 = 푘 . (46) 퐹 푟 √ 훼휀 퐹 푟 √ 훼휀 푘 ∫ (−푡 ) ∫ (−푡 ) 휀 2 푘 2 푘 1 When 0 푡 푂( √ ), ≤ 푘 ≤ 휀

⎧ 1 0 ≤ 퐶퐹 푟(푡) ≤ 푂(1), 0 ≤ 푡 ≤ 푡푘 + 푂( √ ); ⎪ 휀 (47) ⎨ 1 퐶 (푡) = 푂(1), 푡 > 푡 + 푂( √ ). ⎪ 퐹 푟 푘 휀 ⎩ 1 When 푡 > 푂( √ ), 푘 휀

√ 1 ⎧ 퐶 (푡) = 푂( 휀), 0 푡 < 푡 − 푂( √ ); ⎪ 퐹 푟 ≤ 푘 휀 ⎪ √ 1 1 푂( 휀) 퐶 (푡) 푂(1), 푡 − 푂( √ ) 푡 푡 + 푂( √ ); (48) ⎨ ≤ 퐹 푟 ≤ 푘 휀 ≤ ≤ 푘 휀 ⎪ 1 퐶 (푡) = 푂(1), 푡 > 푡 + 푂( √ ). ⎪ 퐹 푟 푘 휀 ⎩

In the same way, 푆퐹 푟(푡, ̂훼) also satisfies( 47) and( 48). So, the presence of the functions 퐶퐹 푟(푡) and 퐶퐹 푟(푡) causes resonance jumps in the system. 퐶퐹 푟(푡) and 푆퐹 푟(푡) are plotted for 휀 = 0.01, 휔 = 2.2889, and so 푡푘 = 100 in Fig.3 . Above all, it follows from( 43) that √ 2 2휀퐴(1 − cos(휆푘(휀푡))) 퐴 = 휔 1 푘 √ |푡=− − 훼휆푘(휀푡)(2휆푘(휀푡) − sin(2휆푘(휀푡))) 휀푡푎푛휔 휀

⋅ [cos(휔푡푘 − 휙푘(푡푘))퐶퐹 푟(푡) + sin(휔푡푘 − 휙푘(푡푘))푆퐹 푟(푡)] + 푂(휀), (49) and

1 ⎧퐴 = 푂(휀), 0 푡 < 푡 − 푂( √ ); ⎪ 푘 ≤ 푘 휀 ⎪ √ 1 1 푂(휀) 퐴 푂( 휀), 푡 − 푂( √ ) 푡 푡 + 푂( √ ); (50) ⎨ ≤ 푘 ≤ 푘 휀 ≤ ≤ 푘 휀 ⎪ √ 1 퐴 = 푂( 휀), 푡 > 푡 + 푂( √ ). ⎪ 푘 푘 휀 ⎩

Similarly, 퐵푘 also satisfies( 50). So, in the resonance zone, ̃ 푇푘(푡) √ = 휀푀1[(cos(휔푡푘 − 휙푘(푡푘))퐶퐹 푟(푡) + sin(휔푡푘 − 휙푘(푡푘))푆퐹 푟(푡)) sin(휙푘(푡))

+ (sin(휔푡푘 − 휙푘(푡푘))퐶퐹 푟(푡) − cos(휔푡푘 − 휙푘(푡푘))푆퐹 푟(푡)) cos(휙푘(푡))] + 푂(휀)

8 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

√ 푡 = 휀푀1[(cos(휔푡푘)퐶퐹 푟(푡) + sin(휔푡푘)푆퐹 푟(푡)) sin( 휆푘(휀푡̄)푑푡̄) ∫푡푘 푡 + (sin(휔푡푘)퐶퐹 푟(푡) − cos(휔푡푘)푆퐹 푟(푡)) cos( 휆푘(휀푡̄)푑푡̄)] + 푂(휀), (51) ∫푡푘 where 퐶퐹 푟(푡) and 푆퐹 푟(푡) are given in(46), and √ 2 2퐴(1 − cos(휆푘(휀푡))) 푀 = 휔 1 . (52) 1 √ |푡=− − 훼휆푘(휀푡)(2휆푘(휀푡) − sin(2휆푘(휀푡))) 휀 tan 휔 휀

Hence, when the external force frequency 휔 is bounded away from 휆푘(0) by a constant and satisfies (푘 − 1)휋 ≤ 휔 < 휆푘(0), for all 푘 = 1, 2, … , (where 휆푘(0) satisfies(23)), then no resonance will occur, and for an 푂(휀) external excitation there is only an 푂(휀) response, which is described in detail in(28). When the external force frequency 휔 satisfies 휆푘(0) ≤ 휔 < 푘휋, for a certain 푘 (with 휔 √ 푘 = 1, 2, …) and 휆 (0) satisfies(23), then a resonance will occur for 푡 near − − 1 , and for the 푂(휀) external excitation an 푂( 휀) 푘 휀 tan(휔) 휀 amplitude response will occur, which is described in detail in(51). In the next section, the occurrence of the (un)expected timescales will be used to construct accurate approximation of the solution for problem(15) and for the original problem(7) when a resonance zone exists. When a resonance zone does not exist then the solution of problem(7) will remain 푂(휀) for 푡 = 푂(휀−1).

5. Three-timescales perturbation method

In the previous section, it was shown that (under certain condition on the external frequency 휔) resonance can occur around time 푡 = 1 (− 휔 − 1). For this reason, we rescale 푡 by defining 푡 = 푡̃ + 1 (− 휔 − 1), and 휏 = 휀푡̃ − 휔 − 1. Thus, problem(18) can 휀 tan 휔 휀 tan 휔 tan 휔 be rewritten in 푡̃ as follows 푇̃ 휆2 휏 푇̃ 휀 훴푛=∞ 푑휆푛(휏) 푐 휏 푇̃ ⎧ 푘,푡̃푡̃ + 푘( ) 푘 = [−2 푛=1 푛,푘( ) 푛,푡̃ ⎪ 푑휏 + 퐴푑 (휏) cos(휔푡̃ + 휔 (− 휔 − 1))] + 푂(휀2), ⎪ 푘 휀 tan 휔 (53) ⎨푇̃ ( 1 ( 휔 + 1)) = 휀퐹 , ⎪ 푘 휀 tan 휔 푘 ⎪푇̃ ( 1 ( 휔 + 1)) = 휀퐺 . ⎩ 푘,푡̃ 휀 tan 휔 푘 In this section, we study problem(53) in detail under the assumption that 휔 is such that a resonance zone exits for the 푘th oscillation mode. The application of the straightforward expansion method to solve(53) will result in the occurrence of so-called secular terms which cause the approximations of the solutions to become unbounded on long timescales. For this reason, to remove secular terms, √ √ we introduce three timescales 푡0 = 푡̃, 푡1 = 휀푡̃, 푡2 = 휀푡̃. The time-scale 푡1 = 휀푡̃ is introduced because of the size of the resonance zone which has been found in the previous section, and the other two time-scales are the natural scalings for weakly nonlinear ̃ √ equations such as(53). By using the three timescales perturbation method, the function 푇푘(푡̃; 휀) is supposed to be a function of 푡0, 푡1 and 푡2, ̃ √ √ 푇푘(푡̃; 휀) = 푤푘(푡0, 푡1, 푡2; 휀). (54) √ By substituting(54) into(53), we obtain the following equations up to 푂(휀 휀):

⎧ 휕2푤 √ 휕2푤 휕2푤 휕2푤 √ 휕2푤 푘 + 휆2(푡 )푤 + 2 휀 푘 + 휀(2 푘 + 푘 ) + 2휀 휀 푘 ⎪ 휕푡2 푘 2 푘 휕푡 휕푡 휕푡 휕푡 휕푡2 휕푡 휕푡 0 0 1 0 2 1 1 2 ⎪ 푛=∞ 푑휆푛(푡2) 휕푤푛 ⎪= 휀[−2훴 푐푛,푘(푡2) + 퐴푑푘(푡2) cos(휔(푡0 − 푎))] 푛=1 푑푡2 휕푡0 ⎪ √ 푑휆 (푡 ) 휕푤 −2휀 휀[훴푛=∞ 푛 2 푐 (푡 ) 푛 ], (55) ⎨ 푛=1 푑푡 푛,푘 2 휕푡 ⎪ √ 2 1 ⎪푤푘(푎, 푏, 푐; 휀) = 휀퐹푘, 휕푤 √ √ 휕푤 √ 휕푤 √ ⎪ 푘 (푎, 푏, 푐; 휀) + 휀 푘 (푎, 푏, 푐; 휀) + 휀 푘 (푎, 푏, 푐; 휀) = 휀퐺 , ⎪ 휕푡 휕푡 휕푡 푘 ⎩ 0 1 2 where 1 휔 1 휔 휔 푎 = ( + 1), 푏 = √ ( + 1), 푐 = + 1. (56) 휀 tan(휔) 휀 tan(휔) tan(휔) √ By using a three-timescales perturbation method, 푤푘(푡0, 푡1, 푡2; 휀) will be approximated by the formal asymptotic expansion √ √ √ √ 푤푘(푡0, 푡1, 푡2; 휀) = 휀푤푘,0(푡0, 푡1, 푡2; 휀) + 휀푤푘,1(푡0, 푡1, 푡2; 휀) √ √ 2 + 휀 휀푤푘,2(푡0, 푡1, 푡2; 휀) + 푂(휀 ). (57) √ √ It is reasonable to assume this solution form since the function 푤푘(푡0, 푡1, 푡2; 휀) analytically depends on 휀, and we are interested in approximations of the solution of Eqs.(4)–(6), when the initial conditions and the external excitation are of 푂(휀). By substituting √ (57) into problem(55), and after equating the coefficients of like powers in 휀, we obtain as: √ the 푂( 휀)-problem: 휕2푤 휕푤 푘,0 2 푘,0 + 휆 (푡2)푤푘,0 = 0, 푤푘,0(푎, 푏, 푐) = 0, (푎, 푏, 푐) = 0, (58) 휕푡2 푘 휕푡 0 0

9 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356 the 푂(휀)-problem: 휕2푤 휕2푤 푘,1 2 푘,0 + 휆 (푡2)푤푘,1 = −2 + 퐴푑푘(푡2) cos(휔(푡0 − 푎)), 휕푡2 푘 휕푡 휕푡 0 0 1 휕푤푘,1 휕푤푘,0 푤푘,1(푎, 푏, 푐) = 퐹푘, (푎, 푏, 푐) = − (푎, 푏, 푐) + 퐺푘, (59) 휕푡0 휕푡1 √ and the 푂(휀 휀)-problem: 휕2푤 휕2푤 휕2푤 휕2푤 푘,2 2 푘,1 푘,0 푘,0 + 휆 (푡2)푤푘,2 = −2 − 2 − 휕푡2 푘 휕푡 휕푡 휕푡 휕푡 휕푡2 0 0 1 0 2 1 휕푤 ∞ 푑휆푛(푡2) 푛,0 − 2훴푛=1 푐푛,푘(푡2) , 푑휏 휕푡0 휕푤푘,2 휕푤푘,0 휕푤푘,1 푤푘,2(푎, 푏, 푐) = 0, (푎, 푏, 푐) = − (푎, 푏, 푐) − (푎, 푏, 푐). (60) 휕푡0 휕푡2 휕푡1 √ The 푂( 휀)− problem has as solution √ 푤푘,0(푡0, 푡1, 푡2; 휀) = 퐶푘,1(푡1, 푡2) sin(휆푘(푡2)푡0) + 퐶푘,2(푡1, 푡2) cos(휆푘(푡2)푡0), (61) where 퐶 and 퐶 are still unknown functions of the slow variables 푡 and 푡 , and they can be determined by avoiding secular terms 푘,1 푘,2 √ 1 2 in the solutions of the 푂(휀)− and 푂(휀 휀)− problems. By using the initial conditions(58), it follows that 퐶푘,1(푏, 푐) = 퐶푘,2(푏, 푐) = 0. Now, we shall solve the 푂(휀)− problem(59). This problem (outside as well as inside the resonance manifold) can be written as 휕2푤 휕퐶 휕퐶 푘,1 2 푘,1 푘,2 + 휆 (푡2)푤푘,1 = −2휆푘(푡2)[ cos(휆푘(푡2)푡0) − sin(휆푘(푡2)푡0)] 휕푡2 푘 휕푡 휕푡 0 1 1 + 퐴푑푘(푡2) cos(휔(푡0 − 푎)), 휕푤푘,1 휕푤푘,0 푤푘,1(푎, 푏, 푐) = 퐹푘, (푎, 푏, 푐) = − (푎, 푏, 푐) + 퐺푘. (62) 휕푡0 휕푡1 By introducing the transformation (푤 , 푤 ) → (퐷 , 퐷 ) with 푘,1 푘,1,푡0 푘,1 푘,2 √ 푤푘,1(푡0, 푡1, 푡2; 휀) √ √ = 퐷푘,1(푡0, 푡1, 푡2; 휀) sin(휆푘(푡2)푡0) + 퐷푘,2(푡0, 푡1, 푡2; 휀) cos(휆푘(푡2)푡0), √ 푤 (푡 , 푡 , 푡 ; 휀) 푘,1,푡0 0 1 2 √ √ = 휆푘(푡2)[퐷푘,1(푡0, 푡1, 푡2; 휀) cos(휆푘(푡2)푡0) − 퐷푘,2(푡0, 푡1, 푡2; 휀) sin(휆푘(푡2)푡0)], it follows that the partial differential equation in(62) is equivalent with the system

⎧ ̇ 휕퐶푘,1 휕퐶푘,2 퐷푘,1 = − [cos(2휆푘(푡2)푡0) + 1] + sin(2휆푘(푡2)푡0) ⎪ 휕푡1 휕푡1 ⎪ 퐴푑푘(푡2) + [cos(휔푡0 − 휔푎 + 휆푘(푡2)푡0) + cos(휔푡0 − 휔푎 − 휆푘(푡2)푡0)], ⎪ 2휆푘(푡2) (63) ⎨ ̇ 휕퐶푘,1 휕퐶푘,2 퐷푘,2 = sin(2휆푘(푡2)푡0) − [1 − cos(2휆푘(푡2)푡0)] ⎪ 휕푡1 휕푡1 ⎪ 퐴푑 (푡 ) − 푘 2 [sin(휔푡 − 휔푎 + 휆 (푡 )푡 ) − sin(휔푡 − 휔푎 − 휆 (푡 )푡 )], ⎪ 휆 푡 0 푘 2 0 0 푘 2 0 ⎩ 2 푘( 2) 휕 where the overdot represents differentiation with respect to 푡0, that is, ̇ = . 휕푡0 Outside of the resonance zone, whether it exists or not it should be observed that the last terms in both equations in(63) do not give rise to secular terms in 퐷푘,1 and 퐷푘,2. To avoid secular terms, 퐶푘,1 and 퐶푘,2 have to satisfy the following conditions

휕퐶푘,1 휕퐶푘,2 = 0, = 0. (64) 휕푡1 휕푡1 ̄ ̄ Then, 퐶푘,1(푡1, 푡2) = 퐶푘,1(푡2), and 퐶푘,2(푡1, 푡2) = 퐶푘,2(푡2). Inside the resonance zone, we observe that cos(휔푡0 − 휔푎 − 휆푘(푡2)푡0) and sin(휔푡0 − 휔푎 − 휆푘(푡2)푡0) might cause secular terms. In 1 accordance with the resonance timescale of 푂( √ ), see(34), it is convenient to rewrite the arguments of the cos − and sin − function 휀 as 훼 휔푡 − 휔푎 − 휆 (푡 )푡 = − 푡2 − 휔푎, 0 푘 2 0 2 1 where 훼 is given by(36). Accordingly, the solution of 푤푘,1 in Eq.(62) has unbounded terms in 푡0 unless 휕퐶 푘,1 훼 2 −2휆푘(푡2) + 퐴푑푘(푡2) cos[− 푡1 − 휔푎] = 0, 휕푡1 2

10 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356 휕퐶 푘,2 훼 2 2휆푘(푡2) − 퐴푑푘(푡2) sin[− 푡1 − 휔푎] = 0, (65) 휕푡1 2 which implies that 퐴 휔푎 푑 푡 퐴 휔푎 푑 푡 ̄ cos( ) 푘( 2) ̄ sin( ) 푘( 2) ̄ 퐶푘,1(푡1, 푡2) = 퐶푘,1(푡2) + √ 퐶퐹 푟(푡1) − √ 푆퐹 푟(푡1), 2훼휆푘(푡2) 2훼휆푘(푡2) 퐴 휔푎 푑 푡 퐴 휔푎 푑 푡 ̄ sin( ) 푘( 2) ̄ cos( ) 푘( 2) ̄ 퐶푘,2(푡1, 푡2) = 퐶푘,2(푡2) − √ 퐶퐹 푟(푡1) − √ 푆퐹 푟(푡1), (66) 2훼휆푘(푡2) 2훼휆푘(푡2) where √ 훼 √ 훼 푡 푡 2 2 퐶̄ (푡) = cos(푥2)푑푥, and 푆̄ (푡) = sin(푥2)푑푥, (67) 퐹 푟 √ 훼 퐹 푟 √ 훼 ∫ 푏 ∫ 푏 2 2 and which are the well-known Fresnel integrals. Thus, it follows from(63) that 퐴푑 푡 ̄ 푘( 2) 퐷푘,1(푡0, 푡1, 푡2) = 퐷푘,1(푡1, 푡2) + sin(휔푡0 − 휔푎 + 휆푘(푡2)푡0) 2휆푘(푡2)(휔 + 휆푘(푡2)) 퐴푑푘(푡2) + sin(휔푡0 − 휔푎 − 휆푘(푡2)푡0), 2휆푘(푡2)(휔 − 휆푘(푡2)) 퐴푑 푡 ̄ 푘( 2) 퐷푘,2(푡0, 푡1, 푡2) = 퐷푘,2(푡1, 푡2) + cos(휔푡0 − 휔푎 + 휆푘(푡2)푡0) 2휆푘(푡2)(휔 + 휆푘(푡2)) 퐴푑푘(푡2) − cos(휔푡0 − 휔푎 − 휆푘(푡2)푡0). (68) 2휆푘(푡2)(휔 − 휆푘(푡2)) where 퐷̄ and 퐷̄ are still unknown functions of the slow variables 푡 and 푡 . The undetermined behaviour with respect to 푡 and 푘,1 푘,2 √ 1 2 1 푡2 can be used to avoid secular terms in the 푂(휀 휀)− problem(60), and in the high order problems. Taking into account the secularity conditions, the general solution of the 푂(휀) equation is given by √ 푤푘,1(푡0, 푡1, 푡2; 휀) = 퐷푘,1(푡0, 푡1, 푡2) sin(휆푘(푡2)푡0) + 퐷푘,2(푡0, 푡1, 푡2) cos(휆푘(푡2)푡0), (69) where 퐷푘,1(푡0, 푡1, 푡2) and 퐷푘,1(푡0, 푡1, 푡2) are given by(68) and

휕푤푘,0 퐷푘,2(푎, 푏, 푐) = 퐹푘, 퐷푘,1(푎, 푏, 푐) = − (푎, 푏, 푐) + 퐺푘. (70) 휕푡1 √ The 푂(휀 휀)− problem(60) outside and inside the resonance manifold can be written as 휕2푤 휕퐷 휕퐷 푘,2 2 푘,1 푘,2 + 휆 (푡2)푤푘,2 = −2휆푘(푡2)[ cos(휆푘(푡2)푡0) − sin(휆푘(푡2)푡0)] 휕푡2 푘 휕푡 휕푡 0 1 1 휕퐶푘,1 휕퐶푘,2 − 2휆푘(푡2)[ cos(휆푘(푡2)푡0) − sin(휆푘(푡2)푡0)] 휕푡2 휕푡2 2 2 휕 퐶푘,1 휕 퐶푘,2 + [ sin(휆푘(푡2)푡0) + cos(휆푘(푡2)푡0)] 휕푡2 휕푡2 1 1 푛=∞ 푑휆푛(푡2) − 2훴푛=1 푐푛,푘(푡2)휆푘(푡2)[퐶푛,1 cos(휆푘(푡2)푡0) 푑푡2 − 퐶푛,2 sin(휆푘(푡2)푡0)], 휕푤푘,2 휕푤푘,0 휕푤푘,1 푤푘,2(푎, 푏, 푐) = 0, (푎, 푏, 푐) = − (푎, 푏, 푐) − (푎, 푏, 푐). (71) 휕푡0 휕푡2 휕푡1

To avoid secular terms in the solution 푤푘,2 in Eq.(71), the following conditions have to be imposed 2 휕퐷푘,1 휕퐶푘,1 휕 퐶푘,2 푑휆푘(푡2) −2휆푘(푡2) − 2휆푘(푡2) + − 2푐푘,푘(푡2)휆푘(푡2) 퐶푘,1 = 0, 휕푡 휕푡 휕푡2 푑푡 1 2 1 2 2 휕퐷푘,2 휕퐶푘,2 휕 퐶푘,1 푑휆푘(푡2) 2휆푘(푡2) + 2휆푘(푡2) + + 2푐푘,푘(푡2)휆푘(푡2) 퐶푘,2 = 0. (72) 휕푡 휕푡 휕푡2 푑푡 1 2 1 2 Next, we analyse this Eq.(72) inside and outside the resonance manifold. Inside the resonance zone, substituting(66) and(68) into(72), we obtain the following secularity conditions: 휕퐷̄ 푑퐶̄ 푑휆 푡 푘,1 푘,1 푘( 2) ̄ − 2 − 2 − 2푐푘,푘(푡2) 퐶푘,1 휕푡1 푑푡2 푑푡2 푑( 퐴 cos(휔푎)푑푘(푡2) ) 휆 푡 퐴푑 (푡 ) 푘( 2) ̄ ̄ 푘 2 2 − (퐶퐹 푟(푡1) − 푆퐹 푟(푡1)) + 훼푡1 sin[−훼푡1 − 휔푎] 푑푡2 휆푘(푡2)

11 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356 푑휆 푡 퐴 휔푎 푑 푡 푘( 2) cos( ) 푘( 2) ̄ ̄ + 푐푘,푘(푡2) (푆퐹 푟(푡1) − 퐶퐹 푟(푡1)) = 0, (73) 푑푡2 휆푘(푡2) and 휕퐷̄ 푑퐶̄ 푑휆 푡 푘,2 푘,2 푘( 2) ̄ 2 + 2 + 2푐푘,푘(푡2) 퐶푘,2 휕푡1 푑푡2 푑푡2 푑( 퐴 cos(휔푎)푑푘(푡2) ) 휆 푡 퐴푑 (푡 ) 푘( 2) ̄ ̄ 푘 2 2 − (퐶퐹 푟(푡1) + 푆퐹 푟(푡1)) − 훼푡1 cos[−훼푡1 − 휔푎] 푑푡2 휆푘(푡2) 푑휆 푡 퐴 휔푎 푑 푡 푘( 2) cos( ) 푘( 2) ̄ ̄ − 푐푘,푘(푡2) (푆퐹 푟(푡1) + 퐶퐹 푟(푡1)) = 0. (74) 푑푡2 휆푘(푡2) ̄ ̄ Solving(73) and(74) for 퐷푘,1 and 퐷푘,2, we observe that the solution will be unbounded in 푡1, due to terms which are ̄ ̄ only depending on 푡2. Therefore, to have secular-free solutions for 퐷푘,1 and 퐷푘,2, the following conditions have to be imposed independently 휕퐶̄ 푑휆 푡 휕퐶̄ 푑휆 푡 푘,1 푘( 2) ̄ 푘,2 푘( 2) ̄ + 푐푘,푘(푡2) 퐶푘,1 = 0, + 푐푘,푘(푡2) 퐶푘,2 = 0, (75) 휕푡2 푑푡2 휕푡2 푑푡2 together with 푑( 퐴 cos(휔푎)푑푘(푡2) ) 휕퐷̄ 휆 푡 퐴푑 (푡 ) 푘,1 푘( 2) ̄ ̄ 푘 2 2 −2 − (퐶퐹 푟(푡1) − 푆퐹 푟(푡1)) + 훼푡1 sin[−훼푡1 − 휔푎] 휕푡1 푑푡2 휆푘(푡2) 푑휆 푡 퐴 휔푎 푑 푡 푘( 2) cos( ) 푘( 2) ̄ ̄ + 푐푘,푘(푡2) (푆퐹 푟(푡1) − 퐶퐹 푟(푡1)) = 0, (76) 푑푡2 휆푘(푡2) and 푑( 퐴 cos(휔푎)푑푘(푡2) ) 휕퐷̄ 휆 푡 퐴푑 (푡 ) 푘,2 푘( 2) ̄ ̄ 푘 2 2 2 + (퐶퐹 푟(푡1) − 푆퐹 푟(푡1)) − 훼푡1 cos[−훼푡1 − 휔푎] 휕푡1 푑푡2 휆푘(푡2) 푑휆 푡 퐴 휔푎 푑 푡 푘( 2) cos( ) 푘( 2) ̄ ̄ − 푐푘,푘(푡2) (푆퐹 푟(푡1) + 퐶퐹 푟(푡1)) = 0. (77) 푑푡2 휆푘(푡2) ̄ ̄ where 퐷푘,1 and 퐷푘,2 can be found by integration of(76) and(77), but we omit the details because of cumbersome expressions. ̄ ̄ Next, from(75) we obtain the functions 퐶푘,1(푡2) and 퐶푘,2(푡2):

푡2 푑휆푘(푠) 푡2 푑휆푘(푠) ̄ − ∫푐 푐푘푘(푠) 푑푠 ̄ − ∫푐 푐푘푘(푠) 푑푠 퐶푘,1(푡2) = 푚1푒 푑푠 , 퐶푘,2(푡2) = 푚2푒 푑푠 , (78) ̄ ̄ where 푚1 and 푚2 are constants. Since 퐶푘,1(푏, 푐) = 0 and 퐶푘,2(푏, 푐) = 0, together with(66), this implies that 퐶푘,1(푡2), 퐶푘,2(푡2) are both identically equal to zero. So, in the resonance zone, √ 푤푘,0(푡0, 푡1, 푡2; 휀) 퐴 휔푎 푑 푡 퐴 휔푎 푑 푡 cos( ) 푘( 2) ̄ sin( ) 푘( 2) ̄ = [ √ 퐶퐹 푟(푡1) − √ 푆퐹 푟(푡1)] sin(휆푘(푡2)푡0) 2훼휆푘(푡2) 2훼휆푘(푡2) 퐴 휔푎 푑 푡 퐴 휔푎 푑 푡 cos( ) 푘( 2) sin( ) 푘( 2) ̄ − [ √ 푆퐹 푟(푡1) + √ 퐶퐹 푟(푡1)] cos(휆푘(푡2)푡0) 2훼휆푘(푡2) 2훼휆푘(푡2) ̄ ̄ = 푀1[(cos(휔푎)퐶퐹 푟(푡1) − sin(휔푎)푆퐹 푟(푡1)) sin(휆푘(푡2)푡0) ̄ ̄ − (cos(휔푎)푆퐹 푟(푡1) + sin(휔푎)퐶퐹 푟(푡1)) cos(휆푘(푡2)푡0)] + ℎ.표.푡, (79) where 푀1 is given by(52). Outside the resonance zone, it follows from(64) and(72) that 휕퐶̄ 푑휆 푡 휕퐶̄ 푑휆 푡 푘,1 푘( 2) ̄ 푘,2 푘( 2) ̄ + 푐푘,푘(푡2) 퐶푘,1 = 0, + 푐푘,푘(푡2) 퐶푘,2 = 0, 휕푡2 푑푡2 휕푡2 푑푡2 which implies that

푡2 푑휆푘(푠) 푡2 푑휆푘(푠) ̄ − ∫푐 푐푘푘(푠) 푑푠 ̄ − ∫푐 푐푘푘(푠) 푑푠 퐶푘,1(푡2) = 푚1푒 푑푠 퐶푘,2(푡2) = 푚2푒 푑푠 , (80) ̄ ̄ where 푚1 and 푚2 are constants. Since 퐶푘,1(푏, 푐) = 0, 퐶푘,2(푏, 푐) = 0, this implies that 퐶푘,1(푡2), and 퐶푘,2(푡2) are identically equal to zero outside the resonance zone. √ Now we can solve the 푂(휀 휀)− problem, where √ 푤푘,2(푡0, 푡1, 푡2; 휀) = 퐸푘,1(푡0, 푡1, 푡2) sin(휆푘(푡2)푡0) + 퐸푘,2(푡0, 푡1, 푡2) cos(휆푘(푡2)푡0), (81) where 퐸푘,1 and 퐸푘,2 are still unknown functions of the variable 푡0, and the slow variables 푡1, 푡2, and they can be obtained by avoiding secular terms from higher order problems. Moreover,

퐸푘,2(푎, 푏, 푐) = 0,

12 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

Fig. 4. (a) The solution 푢(0.5, 푡) of the system with 휔 = 2.0917, and the resonance time 푡 ≈ 20, 휆1(휀 ⋅ 20) = 휔. (b) The solution 푢(0.5, 푡) of the system with 휔 = 2.4556, and the resonance time 푡 ≈ 200, 휆1(휀 ⋅ 200) = 휔.

휕푤푘,0 휕푤푘,1 퐸푘,1(푎, 푏, 푐) = − (푎, 푏, 푐) − (푎, 푏, 푐). (82) 휕푡2 휕푡1 Note that in Eqs.( 69) and( 81), 퐷 , 퐷 , 퐸 , 퐸 are yet also undetermined functions. All these unknown functions can be 푘,1 푘,2 푘,√1 푘,2 determined from the 푂(휀2)− problem and 푂(휀2 휀)− problem. At this moment, only the first term in the expansion of the solution for the string problem is important from the physical point of view. We are not interested in high-order approximations; that is why we take ̄ ̄ ̄ ̄ 퐷푘,1(푡1, 푡2) = 퐷푘,1(푏, 푐), 퐷푘,2(푡1, 푡2) = 퐷푘,2(푏, 푐),

퐸푘,1(푡0, 푡1, 푡2) = 퐸푘,1(푎, 푏, 푐), 퐸푘,2(푡0, 푡1, 푡2) = 퐸푘,2(푎, 푏, 푐). (83) Thus, an approximation of the solution of Eqs.( 4)–(6) is given by

푢(푥, 푡) ∞ ∑ √ √ √ = [ 휀푤푛,0(푡0, 푡1, 푡2; 휀) + 휀푤푛,1(푡0, 푡1, 푡2; 휀) 푛=1 √ √ 2 + 휀 휀푤푛,2(푡0, 푡1, 푡2; 휀)] sin(휆푛(휏)푥) + 푂(휀 ), (84) where 푤푘,0, 푤푘,1 and 푤푘,2 are given by( 61),( 69) and( 81).

• When the external force frequency 휔 is bounded away from 휆푘(0) by a constant and satisfies (푘 − 1)휋 ≤ 휔 < 휆푘(0), with 휆푘(0) √ given by( 23), for all 푘 = 1, 2, …, it follows from( 64) and( 80) that 푤푘,0(푡0, 푡1, 푡2; 휀) = 0. Further from( 54) and( 57), we obtain ̃ 푇푘(푡) = 푂(휀), i.e., for the 푂(휀) external excitation, there is an 푂(휀) response. This case can be referred to as the non-resonant case. • When the external force frequency 휔 satisfies 휆푘(0) ≤ 휔 < 푘휋, with 휆푘(0) given by( 23), for a fixed 푘 = 1, 2, …, it follows from √ 푤 푡 , 푡 , 푡 휀 (79) that 푛,0( 0 1 2; )푛≠푘 = 0 and √ 푤푘,0(푡0, 푡1, 푡2; 휀) ̄ ̄ = 푀1[(cos(휔푎)퐶퐹 푟(푡1) − sin(휔푎)푆퐹 푟(푡1)) sin(휆푘(푡2)푡0) ̄ ̄ − (cos(휔푎)푆퐹 푟(푡1) + sin(휔푎)퐶퐹 푟(푡1)) cos(휆푘(푡2)푡0)], (85) √ where 푀1 satisfies( 52). For the solution 푢(푥, 푡) of the original problem( 69) this implies a resonance jump from 푂(휀) to 푂( 휀) around 푡 = 1 (− 휔 − 1) in the 푘th oscillation mode. 휀 tan 휔 It should be observed that 푇̃ (푡) as calculated by using a three-timescales perturbation method agrees well with the approximation 푘 √ √ as calculated by using the averaging method. Further, by using the first term 푤푘,0(푡0, 푡1, 푡2; 휀) as 푢(푥, 푡) = [ 휀푤푘,0 + 푂(휀)] sin(휆푘(휏)푥) with 휆푘(0) ≤ 휔 < 푘휋 and 휆푘(0) given by( 23), 푢(0.5, 푡) is plotted for 휀 = 0.01, 퐴 = 1, 휔 = 2.0917, and so 푡푘 = 20 in Fig.4 (a) and 푢(0.5, 푡) is plotted for 휀 = 0.01, 퐴 = 1, 휔 = 2.4556, and so 푡푘 = 200 in Fig.4 (b).

6. Numerical results

In this section the finite difference method is used to present numerical approximations of the vibration response and energy of the string. The computations are performed by using the parameters 휀 = 0.01, 퐴 = 1. Let us assume that the initial displacement is prescribed by 푓 = sin(1.7155푥), and the initial velocity 푔 = 0 for 0 ≤ 푥 ≤ 1. There will be different behaviour in the amplitude

13 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

Fig. 5. The solution 푢(0.5, 푡) (a) and the energy 퐸(푡) (b) of the system with 휔 = 2.0917, and the resonance time 푡 ≈ 20, 휆1(휀 ⋅ 20) = 휔.

Fig. 6. The solution 푢(0.5, 푡) (a) and the energy 퐸(푡) (b) of the system with 휔 = 2.4556, and the resonance time 푡 ≈ 200, 휆1(휀 ⋅ 200) = 휔.

Fig. 7. The solution 푢(0.5, 푡) (a) and the energy 퐸(푡) (b) of the system with 휔 = 1.5, there is no resonance.

response of the solution for different choices of the parameter 휔. Note that the following numerical results are computed based on 푂(휀) approximations of the equations. Higher order terms in the equations are neglected due to their unimportant contribution in the solution. By using( 22), and according to our analytical results, resonance occurs around times 휔 1 푡 = − − . (86) 휀 ⋅ tan 휔 휀

14 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

When the external force frequency 휔 is bounded away from 휆푘(0) by a constant and satisfies (푘−1)휋 ≤ 휔 < 휆푘(0), with 휆푘(0) given by(23), for all 푘 = 1, 2, … Fig.5 shows the displacement at 푥 = 0.5 and the vibratory energy of the system, respectively, for times √ up to푡 = 200 for 휔 = 2.0917. We observe that for 푡 ≈ 20 the response amplitudes of the vibration become of order 휀 from order 휀 at 푡 = 0. Similarly, Fig.6 shows the displacement and vibratory energy of the system for times up to 푡 = 400 for 휔 = 2.4556. Again we √ observe that the response amplitudes of the vibration become of order 휀 but now for 푡 ≈ 200. When the external force frequency 휔 is bounded away from 휆푘(0) by a constant and satisfies (푘 − 1)휋 ≤ 휔 < 휆푘(0), with 휆푘(0) given by(23), for all 푘 = 1, 2, …, Fig.7 shows the displacement and vibratory energy of the system for times up to푡 = 400 for 휔 = 1.58, but now there is no resonance and the response amplitudes of the vibration are still of order 휀. These numerical results are in agreement with our results as presented in Section4 and in Section5. Moreover, in these figures, the shadowed bands represent the resonance zones, which have the size 1 of (푂( √ )) as was also obtained analytically. Therefore, from Figs.5–7, we can conclude that the general dynamic behaviour of the 휀 solution as approximated numerically is in complete agreement with the analytic approximations as obtained in Section4 and in Section5.

7. Conclusions

In this paper resonances in a transversally vibrating string are studied. A small, externally applied and harmonic force with frequency 휔 is acting on the whole string. The string is fixed at one end, and at the other end a spring is attached for which the stiffness slowly varies in time. In this paper one choice is made for the slowly varying stiffness, but also other choices can be made and a similar analysis as presented in this paper can be given. By assuming that the small external force is of order 휀 and by assuming that the initial values are also small and of order 휀, it is shown in this paper that resonances can occur for certain values of 휔, and √ that the amplitude response (in case of resonance) is of order 휀. To obtain these results an adapted version of the method of separation of variables is introduced, and perturbation methods, (such as averaging methods, singular perturbation techniques, and multiple timescales perturbation methods) are used. Not only the interval for 휔 for which resonance occurs is determined, but also the times and time-intervals are found for which this resonance occurs. Furthermore, explicit, and accurate approximations of the solution of the initial–boundary value problem are constructed. All approximations are valid on time-scales of order 휀−1. Also a finite difference method is applied to construct numerical approximations of the solution of the initial–boundary value problem. These numerical approximations are in full agreement with the analytically obtained approximations.

CRediT authorship contribution statement

Jing Wang: Methodology, Implementation of the computer code and supporting algorithms, Writing – original draft. Wim T. van Horssen: Ideas, Formulation of overarching research goals and aims, Writing – review & editing. Jun-Min Wang: Writing – review & editing.

Declaration of competing interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Acknowledgements

This research is supported by the National Natural Science Foundation of China under Grant 62073037. The first author Jing Wang would like to thank Delft Instituted of Applied Mathematics (DIAM), Delft University of Technology, for providing such an excellent research environment, and would like to thank the financial support from the China Scholarship Council (CSC).

Appendix. Numerical approximation

Firstly, we introduce a uniform mesh 훥푥, a constant discretization time 훥푡, and a rectangular mesh consisting of points (푥푖, 푡푗 ) with 푥푖 = 푖훥푥 and 푡푗 = 푗훥푡, where 푖 = 1, 2, 3, … , 푁, 푗 = 1, 2, …, with 푁훥푥 = 1. Following the finite difference method and by using the Taylor series expansion, the second order space and time derivatives can be discretized by

휕2푢 푢(푥푖+1, 푡푗 ) − 2푢(푥푖, 푡푗 ) + 푢(푥푖−1, 푡푗 ) = + 푂((훥푥)2), (87) 휕푥2 (훥푥)2 휕2푢 푢(푥푖, 푡푗+1) − 2푢(푥푖, 푡푗 ) + 푢(푥푖, 푡푗−1) = + 푂((훥푡)2). (88) 휕푡2 (훥푥)2 Substituting the finite difference formulae into Eqs.(4)–(6), and rearranging the terms, we end up with the linear iterative system

2 2 2 푢푖,푗+1 = 휎 푢푖+1,푗 + 2(1 − 휎 )푢푖,푗 + 휎 푢푖−1,푗 − 푢푖,푗−1 + 휀퐴 cos(휔푡푗 ), (89) where 휎 = 훥푡 . From the boundary condition(5) it follows that 훥푥 푢푛−1,푗 푢0,푗 = 0, 푢푛,푗 = . (90) 1 + 푘(푡푗 )훥푥

15 J. Wang et al. Journal of Sound and Vibration 512 (2021) 116356

(푗) 푇 (푗) 푇 Let us introduce the following vector: 푈 = [푢1,푗 , 푢2,푗 , … , 푢푛−2,푗 , 푢푛−1,푗 ] , 푆 = [̄푠푗 , ̄푠푗 , … , ̄푠푗 ] , ̄푠푗 = 휀퐴 cos(휔푡푗 ), ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ 푛−1 times

⎛2(1 − 휎2) 휎2 0 ⋯ ⋯ 0 ⎞ ⎜ 휎2 2(1 − 휎2) 휎2 ⋯ ⋯ 0 ⎟ ⎜ ⎟ 퐵 = ⎜ ⋮ ⋱ ⋱ ⋱ ⋱ ⋮ ⎟ , ⎜ 0 ⋯ ⋯ 휎2 2(1 − 휎2) 휎2 ⎟ ⎜ 휎2 ⎟ ⎜ 0 ⋯ ⋯ 0 휎2 2(1 − 휎2) + ⎟ ⎝ 1+푘(푡푗 )훥푥 ⎠ then the iteration process can be rewritten in the following matrix form

푈 (푗+1) = 퐵푈 (푗) − 푈 (푗−1) + 푆(푗), (91) where the initial conditions imply: 1 1 푢 = 푓 = 푓(푥 ), 푢 = 휎2푓 + (1 − 휎2)푓 + 휎2푓 + 훥푡푔 . (92) 푖,0 푖 푖 푖,1 2 푖+1 푖 2 푖−1 푖

References

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