time-averaged canonical perturbation of nonlinear vibrations of a spinning disk

Nonlinear Analysis: Real World Applications 7 (2006) 319–340www.elsevier.com/locate/na

Time-averaged canonical perturbation of nonlinearvibrations of a spinning disk

Natalie Baddour∗, Jean W. ZuDepartment of Mechanical and Industrial Engineering, University of Toronto, 5 King’s College Road, Toronto,

Ontario, Canada, M5S 3G8

Received 17 January 2003; accepted 24 March 2005

Abstract

This paper considers the dynamics of a spinning disk with the inclusion of in-plane inertia. Theinclusion of the in-plane inertia couples the in-plane and transverse vibrations of the spinning disk.By considering one mode of each vibration, the problem can be reduced to a non-linear 2 DOFmodel. An analytical solution of this 2 DOF model is then developed using a time-averaged canonicalperturbation approach and is compared with that of a numerical procedure. Through the use of theseanalytical and numerical tools, it becomes apparent that the inclusion of the in-plane inertia leads tothe possibility of internal resonance between the oscillators, depending on the relationship betweenthe natural frequencies of the two oscillators.� 2005 Elsevier Ltd. All rights reserved.

Keywords: Nonlinear spinning disk coupled vibrations

1. Introduction

Spinning disks can be found in many engineering applications. Common industrial ap-plications include circular sawblades, turbine rotors, brake systems, fans, flywheels, gears,grinding wheels, precision gyroscopes and computer storage devices. Spinning disks mayexperience severe vibrations which could lead to fatigue failure of the system. Thus, thedynamics of spinning disks has attracted much research interest over the years.

∗ Corresponding author. Tel./fax: +1 416 978 1287.E-mail address: [email protected] (N. Baddour).

1468-1218/$ - see front matter � 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.nonrwa.2005.03.004

http://www.elsevier.com/locate/na

mailto:[email protected]

320 N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319–340

Many authors have investigated the vibrations of spinning disks using linear theory.The original papers are by Lamb and Southwell [19] and by Southwell [26] where thedisk was modelled as a spinning membrane with added bending stiffness. Another popularapproach in the literature is to model the spinning disks as a pure membrane with nobending stiffness [24,13,25,16]. The incorporation of both the bending stiffness of the diskand the effect of rotation leads to a fourth order PDE that is difficult to solve. As a result,various researchers have applied different solution techniques to the linear analysis of thefree transverse vibrations of spinning disks [7,14,11,20]. In addition to considering thetransverse vibrations of a spinning disk, the problem of free planar vibrations has also beeninvestigated using linear theory [27,18,8,12,10].

Although Linear models have been extensively used in the literature for many years,nonlinear models have also been employed since it is known from the theory of stationarydisks that the linear theory breaks down when the transverse displacement is on the order ofthe thickness of the disk [21,23,1,2]. However, in the nonlinear case the in-plane inertia ofthe disk has always been neglected [21,22,15] and the authors could not find any papers thatconsidered its inclusion in the nonlinear dynamics of the rotating disk. With the inclusion ofthe in-plane inertia, the simplest possible representation of the nonlinear dynamics result ina nonlinear 2 DOF system. In this paper, the process by which this simplified 2 DOF systemis obtained is outlined. The main thrust of the paper is to develop an analytical solutionto this simplified 2 DOF system and to verify whether this analytical solution succeeds incapturing the various dynamics of the system.

2. Equations of motion

Since all previous models [21,22,15] in the literature ignore the in-plane inertia of thedisk in order that a stress function may be used, it is necessary to rederive the equations ofmotion so that the in-plane inertia of the disk may be kept. This results in three equationsof motion for the dynamics instead of the usual two—a considerable complication. Thisderivation was performed in detail in [5].

To consider the effect of the inclusion of in-plane inertia, Galerkin’s procedure is usedin this paper to produce a simplified problem. The result will be a two degree of freedommodel—one for each of the time dependences of the in-plane and transverse vibrations.When the in-plane inertia is included, the time dependence of the in-plane vibrations is anextra independent quantity that must be taken into account.

To this purpose, let us choose the displacements as

u(r, �, t) = ueq(r) + U(r) cos(m�)c(t), (1)

v(r, �, t) = −V (r) sin(m�)c(t), (2)

w(r, �, t) = W(r) cos(n�)�(t), (3)

where ueq, U, V and W are assumed to be known and will be discussed momentarily. Theonly unknowns here are the time dependences, c(t) and �(t). The ueq portion is the in-planeequilibrium displacement due to the spin of the disk. That is, ueq is a symmetrical equi-librium in-plane displacement that occurs in the plane of the disk as a result of its rotation

N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319–340 321

(that is, due to the centrifugal force). The form for U and V are chosen from the solutionof the in-plane linear problem [4,9,8]. Similarly, W is taken to be one of the mode shapesof the linear strain model of the spinning disk, derived by using purely linear strains [6].

Once a choice for the space dependent part of the solution has been made, it still remainsto use this choice to derive the equations of motion for the temporal part of the solution. Oneoption is to substitute the assumed forms of the solution into the equations of motion andthen apply the method of Galerkin. A completely equivalent calculation is to substitute theassumed form of the solution into the expressions for kinetic and potential energies. Sincethe spatial form of the solution is known, the integration can be performed explicitly overspace. The kinetic and potential energies are thus reduced to functions of c(t) and �(t) givenin Eqs. (1)–(3), that is we have a 2 DOF system. Lagrange’s equations are then applied toyield the two equations of motion that are sought. The advantage of actually calculatingthe potential and kinetic energies of the system, as opposed to directly using Galerkin’smethod is that conserved quantities are more readily apparent. Such conservation laws canbe used to simplify the equations of motion and could aid in the finding of an analyticsolution. Furthermore, the machinery of Hamilton’s equations and canonical perturbationtheory is also available in this formulation. For these reasons, the simplified 2 DOF ordinarydifferential equations of motion are calculated by first constructing the kinetic and potentialenergies, integrating over space and then using Lagrange’s equations.

The potential and kinetic energies can be found from

PE = 1

2

∫ h

−h

∫ 2�

0

∫ a

0(�rr�rr + �� + 2�r��r�)r dr d� dz, (4)

KE =∫ 2�

0

∫ a

0(hKE1 + h3KE3)r dr d�, (5)

where the integration over z has already been incorporated into the expression for the kineticenergy, and KE1 and KE3 are given by [5]

KE1 = ��2(v2 + u2 + r2 + 2ru) + 2��

[u

�v

�t− v

�u

�t+ r

�v

�t

]

+ �

[(�u

�t

)2

+(

�v

�t

)2

+(

�w

�t

)2]

, (6)

KE3 = ��2

3r2

[(�w

��

)2

+ r2(

�w

�r

)2]

+ �

3r2

[(�2w

��t

)2

+ r2(

�2w

�r�t

)2]

+ 2��

3r

[�2w

��t

(�w

�r

)− �2w

�r�t

(�w

��

)]. (7)

Linear stress–strain relations along with nonlinear von Karman strains are used in theexpression for potential energy to complete the derivation. Once the integrations over z, �and r are performed, the resulting expressions for the kinetic and potential energies becomesimple functions of c and � and resemble the expressions of a two degree of freedom system.Indeed, by taking only one modeshape for each of the in-plane and out of plane vibrations,we have restricted the system to having only two degrees of freedom.


After having obtained the expressions for kinetic and potential energy, it is a simplematter to form the Lagrangian and to derive the equations of motion of the system usingLagrange’s equations. The coupled equations of motion can be shown to be

c + �2cc + k1�

2 = 0, (8)

� + �2�� + 2k1c� + k3�

3 = 0. (9)

Clearly, �c and �� are the natural frequencies of the linear in-plane and transverse modesthat were chosen. The expressions for k1 and k3 are complicated [3], but it suffices to notethat they will depend on the amplitudes of the chosen modes. For certain choices of modes,it is possible that k1 = 0, which indicates that the in-plane and transverse modes in questionare uncoupled and will vibrate independently. However, for other choices of modes, k1 willnot be zero and the in-plane and transverse vibrations will be coupled. Eqs. (8) and (9)will be referred to as the two degrees of freedom (2 DOF) model since the equations ofmotion are essentially those of a nonlinear two degree of freedom system. Thus, currentwork considers Eqs. (8) and (9) to be the fundamental equations describing the dynamicsof the system.

3. Time-averaged canonical perturbation approach

Eqs. (8) and (9) can also be arrived at from the point of view of Hamilton’s equations.That is, Eqs. (8) and (9) are equivalent to

c = �H

�pc

,

� = �H

�p�,

pc = − �H

�c,

p� = − �H

��, (10)

where pc and p� are the conjugate momenta to c and �, and the Hamiltonian is given by

H = p2c

2+ p2

�

2+ �2

c

2c2 + �2

�

2�2 + k1c�

2 + k3

4�4. (11)

Eqs. (10) represent a general Hamiltonian system. Since the equations of motion of the 2DOF model can be derived from a Hamiltonian, they are a Hamiltonian system. The generalobjective here is to transform the given Hamiltonian system into a simpler Hamiltoniansystem and to use the result to construct an approximate solution.

Generally, a transformation of variables from generalized coordinates q, p to a new setof coordinates q�, p� can be of the form:

q� = q�(q, p, t),

p� = p�(q, p, t), (12)


with the corresponding transformed differential equations

q� = q�(q, p, t),

p� = p�(q, p, t). (13)

Suppose that Eqs. (13) have the same symmetry as the original equations. That is, supposethat there exists a function H�(q�, p�, t) such that

q� = �H�

�p�

p� = −�H�

�q�. (14)

Then the transformation (12) is known as a canonical transformation [17]. That is, canon-ical transformations transform Hamiltonian differential equations into another system ofdifferential equations of the same type.

One possible approach to solving the equations of motion is to find a canonical transfor-mation such that the transformed Hamiltonian H� is a constant function, that is a functionthat does not depend on time or any of the generalized coordinates or momenta. Then Eqs.(14) reduce to the simple form of q� = p� = 0 and can be easily integrated. In reality,finding such a transformation can be just as difficult as solving the original equations ofmotion. However, it is possible to use a canonical transformation that effectively transformsa simple Hamiltonian to aid in the construction of an approximate solution for a problemwith a more complicated Hamiltonian. That is, take a transformation that essentially solvesa simple related problem and apply it to the more complicated problem. Even though such atransformation will not solve the more complicated problem, it will help in the constructionof an approximate solution. This is the approach that will be undertaken here.

Consider the Hamiltonian of the 2DOF model:

H = p2c

2+ p2

�

2+ �2

c

2c2 + �2

�

2�2 + k1c�

2 + k3

4�4. (15)

This Hamiltonian can be expressed as

H = H(0) + H(1), (16)

where

H(0) = p2c

2+ p2

�

2+ �2

c

2c2 + �2

�

2�2, (17)

H(1) = k1c�2 + k3

4�4. (18)

The motivation for doing this is that H(0) can be recognized as the Hamiltonian of twouncoupled linear harmonic oscillators. H(1) brings in coupling by way of the third order k1term. The fourth order term in H(1) does not couple the two oscillators. Rather, it has theeffect of turning the linear � oscillator into a nonlinear Duffing oscillator. The k1 term is theonly term that couples the two oscillators.


Thus the first step in our approximate solution will be to find a canonical transformationthat transforms H(0) into a constant. This can easily be done by solving the Hamilton–Jacobiequation associated with H(0) [17]. Now this canonical transformation is equally validno matter which Hamiltonian is employed. It just has the advantage of simplifying thisparticular Hamiltonian H(0), but it can be used with the entire original Hamiltonian H.After thus transforming the Hamiltonian, approximate solutions for the ensuing Hamiltonianequations will be sought.

As previously mentioned, the first step in the procedure is to seek the complete solu-tion S(c, �, pc, p�, t) of the Hamilton–Jacobi equation associated with H(0)(c, �, pc, p�).Namely, solve

H(0)

(c, �,

�S

�c,�S

��, t

)+ �S

�t= 0. (19)

That is, solve

1

2

(�S

��

)2

+ 1

2

(�S

�c

)2

+ �2c

2c2 + �2

�

2�2 + �S

�t= 0. (20)

To achieve this, write

S = −ht + W1(c, 1) + W2(�, 2) (21)

where h = 1 + 2 is a constant. If Eq. (21) is substituted into Eq. (20), then Eq. (20) canbe satisfied if

1

2

(�W1

�c

)2

+ �2c

2c2 = 1, (22)

1

2

(�W2

��

)2

+ �2�

2�2 = 2. (23)

The preceding equations can be integrated directly to yield

S = −1t − 2t +∫ √

21 − �2cc

2 dc +∫ √

22 − �2��

2 d�. (24)

The required transformation can now be found from

1 = �S

�1= −t +

∫dc√

21 − �2cc

2(25)

= − t + 1

�c

arcsin

(�cc√21

), (26)

or equivalently

c =√

21

�c

sin �c(t − 1). (27)


Similarly, we obtain

� =√

22

��sin ��(t − 2). (28)

Furthermore, the generalized momenta can also be transformed

pc = �S

�c=√

21 − �2cc

2 =√21 cos �c(t − 1), (29)

p� = �S

��=√

22 − �2��

2 =√22 cos ��(t − 2). (30)

Note that Eqs. (27)–(30) are essentially a canonical transformation from the old variablesc, �, pc, p� to the new variables 1, 2, 1, 2. It can be verified that if (27)–(30) are substi-tuted into H(0), the result is the constant 1 +2. This is no coincidence: the transformationwas specifically constructed to do so. In fact, Eqs. (27)–(30) are precisely the solution tothe uncoupled two oscillator problem where 1, 2, 1, 2 are to be taken as constants ofintegration. The transformed Hamiltonian for the uncoupled two oscillator problem is givenby H(0) + �S/�t , which is precisely zero (by construction). With the transformed Hamil-tonian being zero, Hamilton’s equations assume a particularly simple, easily-integrableform. Namely, they give the result that 1, 2, 1, 2 must be constants to satisfy the newHamilton’s equations.

While the canonical transformation given by Eqs. (27)–(30) transforms H(0) into zero,it does not do so for the entire Hamiltonian H = H(0) + H(1). The transformed totalHamiltonian becomes H� =H(0) +�S/�t +H(1) =H(1)(1, 2, 1, 2, t). The ‘perturbingHamiltonian’ H(1) can be expressed as

H(1)(1, 2, 1, 2, t) = k1c(1, 1, t)�2(2, 2, t) + k3

4�4(2, 2, t)

= 2k1

√2√

1 sin(�c(t − 1))2 sin(�� (t − 2))2

�c�2�

+ k322 sin(�c(t − 2))

4

�4�

,

H(1) = − 2

8�c�4�(−8k1

√21�

2� sin(�ct − �c1)

+ 4k1

√21�

2� sin((�c + 2��)t − �c1 − 2��2)

+ 4k1

√21�

2� sin((�c − 2��)t − �c1 + 2��2) − 3k32�c

− k32�c cos(4��t − 4��2) + 4k32�c cos(2��t − 2��2)).

Note from the above definition of H(1) that there is a difference in its form depending onwhether or not �c = 2��. Although it may not be obvious yet, if �c = 2��, the system willdisplay internal resonance and H(1) will take on a different form.


From the above, using the new variables 1, 2, 1 and 2, Hamilton’s equations (10)become

1 = �H(1)

�1= 1

22k1

√21

�2�

(−2 cos(−�ct + �c1)

+ cos(−�ct + �c1 − 2��t + 2��2)

+ cos(�ct − �c1 − 2��t + 2��2)), (31)

2 = �H(1)

�2= 1

2

2

�1�32

(2k1

√21�

2� cos(−�ct + �c1 − 2��t + 2��2)

− 2k1

√21�

2� cos(�ct − �c1 − 2��t + 2��2)

− k32�c sin(−4��t + 4��2)

+ 2k32�c sin(−2��t + 2��2)), (32)

1 = − �H(1)

�1= 1

42k1

√2

�2��c

√1

(2 sin(−�ct + �c1)

− sin(−�ct + �c1 − 2��t + 2��2)

+ sin(�ct − �c1 − 2��t + 2��2)), (33)

2 = − �H(1)

�2= 1

4�c�4�(−2k1

√21�

2� sin(−�ct + �c1 − 2��t + 2��2)

+ 4k1√

2√

1�2� sin(−�ct + �c1)

+ 2k1

√21�

2� sin(�ct − �c1 − 2��t + 2��2)

− 3k32�c − k32�c cos(−4��t + 4��2)

+ 4k32�c cos(−2��t + 2��2)). (34)

If the initial conditions for (8) and (9) are given by

�(0) = �0, (35)

d�(0)

dt= 0, (36)

c(0) = c0, (37)

dc(0)

dt= 0, (38)

the corresponding initial values for 1, 2, 1, 2 can be worked out from the transformationequations (27 )–(30) to be

01 = 1

2�2

cc20, (39)

01 = − �

2�c

, (40)


02 = 1

2�2

��20, (41)

02 = − �

2��. (42)

Now, Eqs. (31)–(34) are exact—no approximations have been made at this point. We haveused a valid transformation to change the variables and obtained the equations of motionfor the transformed variables. If these equations could be solved, they would provide thesolution to the original problem. In practice though, these equations are no easier to solvethan the original equations of motion. However, Eqs. (31)–(34) offer a good starting pointfor approximate solutions and make certain features of the solutions easy to spot.

To construct an approximate solution, approximations will be made fairly early in thegame, namely at H(1) itself. Here the perturbing Hamiltonian can be further separated andreplaced with its temporal mean. Hamilton’s equations for 1, 2, 1, 2 are then foundusing this time-averaged Hamiltonian instead of the original Hamiltonian. The benefit ofthis approach is that for our problem the ensuing Hamilton’s equations can be solved exactly.It should be noted that this approach is referred to in the literature as a canonical perturbationapproach, although it is not the only approach that is referred to as such.

Consider again the expression for H(1) from which Eqs. (31)–(34) are derived. Thetemporal mean of H(1) can easily be calculated. Note that the temporal mean of H(1)

will have two different values depending on whether or not �c = 2��. If �c = 2�� thensin((�c − 2��)t + (�c1 − 2��2)) is not a function of time, and will not average out tozero. Thus, for the no-resonance case where �c �= 2� we obtain

H(1)av = 3k32

2

8�4�

. (43)

Similarly, for the internal resonance case where �c = 2��, we obtain

H(1)av,IR = 3k32

2

8�4�

− k12√

21

2�c�2�

sin(2��(2 − 1)). (44)

To construct an approximate solution to our problem, H(1)av and H

(1)av,IR in Eqs. (43) and (44)

are used to derive the Hamilton equations of motion instead of the exact value of H(1).There are thus two possible sets of equations of motion depending on whether or not thecondition for internal resonance is met. However, the advantage of this formulation is thatin both cases the Hamilton equations of motion can be solved exactly.

4. Solution of the time-averaged approximation

4.1. Case 1: No internal resonance

For the case when there is no internal resonance, that is where �c �= 2��, then theHamilton equations of motion are given by

1 = �H(1)av

�1= 0, (45)


2 = �H(1)av

�2= 0, (46)

1 = −�H(1)av

�1= 0, (47)

2 = −�H(1)av

�2= −3k32

8�4�

. (48)

Integrating these four equations yields

1 = 01, (49)

2 = 02, (50)

1 = 01, (51)

2 = −3k302

4�4�

t + 02. (52)

Thus the solution to the original problem becomes

c =√

201

�c

sin(�c(t − 01)) (53)

� =√

202

��sin

(��

((1 + 3k30

2

4�4�

)t − 0

2

)). (54)

Note that for this case, since 01,

02,

01,

02 are all constants, the solution for c and � have

constant amplitudes. Furthermore, note that the frequency of � depends on k3 so that we havethe familiar result of frequency depending on amplitude. Substituting from Eqs. (39)–(42)for 0

1, 02,

01,

02 yields

c = c0 cos(�ct) (55)

� = �0 cos

[��t

(1 + 3k3�2

0

8�2�

)]. (56)

Note that the solution for c in this case is exactly what it would be if the oscillators wereuncoupled—it is not affected by the coupling.

4.2. Case 2: Internal resonance

For the case when there is internal resonance, that is where �c = 2��, then the Hamiltonequations of motion are given by

1 = �H(1)av,IR

�1= k12

√21

2�2�

cos(2��(1 − 2)), (57)


2 = �H(1)av,IR

�2= −k12

√21

2�2�

cos(2��(1 − 2)), (58)

1 = −�H(1)av,IR

�1= − k12

√2

8√

1�3�

sin(2��(1 − 2)), (59)

2 = −�H(1)av,IR

�2= −3k32

8�4�

− k1√

21

4�3�

sin(2��(1 − 2)). (60)

Since H(1)av,IR does not contain the time t explicitly, it must be a conserved quantity, that is,

a constant. Let us denote this constant as g1, so that H(1)av,IR = g1. Adding Eqs. (57) and (58)

yields 1 + 2 = 0, or 1 + 2 = g2, where g2 is a constant of integration. Both constantsg1 and g2 can be found from the initial values of 1, 2, 1 and 2. This yields

g1 = 3k3

32+ k1

4, (61)

g2 = 5

2�2

� . (62)

Solving Eq. (44) for the sine term, using the fact that 1=g2−2, and using the trigonometricidentity sin2(x) = 1 − cos2(x), it can be shown that 2 must satisfy

22 = − 9k2

3

16�6�4

2 − k21

2�2�

32 + (320k2

1 + 192k3k1 + 72k23)

256�2�

22

− (9k23 + 48k3k1 + 64k2

1)

256�2

� . (63)

Note that Eq. (63) is an equation purely in terms of 2. That is, this ODE can be integratedfor 2. Integrating this ODE will be discussed subsequently. Once 2 is known, the othervariables can be found by integration from

1 = g2 − 2, (64)

1 = −12k322 + 3k3�4

� + 8k1�4�

32(22 − 5�2�)�

4�

, (65)

2 = −12k322 + 3k3�4

� + 8k1�4�

32�4�

. (66)

Thus, the crux of the problem is to solve Eq. (63) for 2.Recall that 2 is the amplitude �. Before even solving for 2, it is worthwhile to remark

that in the non-resonance case 2 was a constant and in this (resonance) case, it will clearlynot be a constant. This is the most important observation that can be made here since eventhough we can solve for 2 exactly, the calculations are not very intuitive.


It can be verified that (2a2 − �2�) is a factor of the quartic in Eq. (63). Thus, to solve Eq.

(63), consider a change of variables by letting

2 = �2�

(1

2+ 1

y

), (67)

or equivalently

y = 2�2�

22 − �2�

. (68)

Then Eq. (63) can be written as

− 4��√14k2

1 + 12k1k3

dy

dt= ±√(y − y1)(y − y2)(y − y3), (69)

where y1,y2 and y3 satisfy y3 > y2 > y1 and are the roots of the cubic

y3 + (8k21 + 12k1k3 − 9k2

3)

14k21 + 12k1k3

y2 − (8k21 + 18k2

3)

14k21 + 12k1k3

y − 9k23

14k21 + 12k1k3

. (70)

Note that this assumes that the roots to the above cubic are real. Should two of the roots becomplex, a different procedure must be considered and will be discussed after this one.

The solution of Eq. (69) is given by

y = (y2 − y1)sn2(M(t − t0), k) + y1, (71)

where

M =√

(y3 − y1)(14k21 + 12k1k3)

8��, (72)

k =√

(y2 − y1)

(y3 − y1), (73)

and sn denotes the Jacobian elliptic function with parameter k. Note that sn is a generalizationof the trigonometric sine function. In fact, sn reduces to the sine function for k = 0. Thejacobian elliptic function can be defined from

u =∫ x

0

dx√(1 − x2)(1 − k2x2)

= sn−1(x), (74)

where sn−1(x) denotes the inverse Jacobian elliptic function. That is, x = sn(u, k). Bydifferentiating, it can be shown that

du

dx= 1√

(1 − x2)(1 − k2x2). (75)

Note the similarity between Eqs. (75) and (69). In fact, Eq. (75) and Eq. (69) are related bya simple change of variable of the form x = √

A + By.


It still remains to find t0. This can be found from the initial conditions. Recall that2(t = 0) = �2

�/2 if �0 = 1. This implies that y → ∞ at t = 0. Thus sn(−Mt0, k) → ∞,implies that −Mt0 = iK ′ where

K ′ =∫ 1

0

dx√(1 − x2)(1 − (k′)2x2)

(76)

=∫ �

2

0

d�√1 − (k′)2sin2 �

= F(k′, �

2

), (77)

and (k′)2 + k2 = 1. Note that F denotes the complete elliptic integral of the first kind withparameter k′. It follows that

sn(M(t − t0), k) = sn(Mt + iK ′, k) (78)

= 1

k

1

sn(Mt, k)(79)

= 1

kns(Mt, k). (80)

Thus finally we have that

2 = �2�

(1

2+ 1

y

), (81)

y = (y3 − y1)ns2(Mt, k) + y1. (82)

For the more general initial condition where �0 �= 1, it follows that 2(t = 0) = �2��

20/2.

Thus y(t = 0) = 2/�20 − 1. This yields

sn(−Mt0, k) =√

2 − y1(�20 − 1)

(y2 − y1)(�20 − 1)

. (83)

Hence the solution for y becomes

y = (y2 − y1)sn2(M(t − t0), k) + y1 (84)

t0 = 1

MF

[k, arcsin

(√2 − y1(�2

0 − 1)

(y2 − y1)(�20 − 1)

)]. (85)

The above solution is only valid if y1,y2, and y3 are real. For imaginary roots the solutionproceeds as follows. The equation for 2 is written as[

−16

9k23

](d2

dt

)2

=(

2 − �2�

2

)(2 − x1)[(2 − x4)

2 + x25 ], (86)


where x1,x4 and x5 are real and

(2 − x1)[(2 − x4)2 + x2

5 ] = 32 + (16k2

1 + 9k23)

18k23

�2�

22

− (3k3 + 8k1)2

36k23

�4�2 − (3k3 + 8k1)

2

72k23

�6� . (87)

Let Z1 = (2 − �2�

2 )(2 − x1) and Z2 = (2 − x4)2 + x2

5 . Now consider Z1 − �Z2:

Z1 − �Z2 = (1 − �)22 +

(2�x4 − �2

�

2− x1

)2 + �2

�x1

2− �(x2

4 + x25 ). (88)

Clearly, Z1 − �Z2 is a quadratic in 2. This quadratic is a perfect square if we choose �such that the discriminant of the quadratic is zero. Thus, let �1 and �2 denote the roots ofthe discriminant:

−4x25�2 +

[4x2

4 −(

�2�

2+ x1

)x4 + 4x2

5 + 2�2�x1

]� +

(�2

�

2− x1

)2

= 0. (89)

Eq. (89) is itself a quadratic in � and as such has two possible solutions. Further, it is not aquadratic if x5 = 0, as this case would reduce to the previously discussed case of real yi..Solving for �i yields

�1,2 = 1

4x25

[− �2

�x4 − 2x1x4 + 2x24 + 2x2

5 + �2�x1

±√

(x25 + (x1 − x4)

2)(4x25 + (2x4 − �2

�)2)

]. (90)

The expression under the square root sign is always positive so �1 and �2 are both realquantities. Take �2 to be the quantity associated with the positive sign in Eq. (90) while �1is associated with the negative sign. Since the coefficient of �2 in Eq. (89) is negative whilethe constant term is positive, it is easily seen that �1 < 0. Solving the expression for �2 inEq. (90) for �2

� , it can easily be shown that for �2� to be a real quantity that �2 �1. It can be

shown that �2 = 1 results in Z1 and Z2 being proportional, clearly an impossibility sinceone has real roots while the other has imaginary roots. Thus, it follows that �2 > 1.

It follows that

Z1 − �1Z2 = P 21

4(1 − �1), (91)

Z1 − �2Z2 = P 22

4(1 − �1), (92)

where

P1 = 2(1 − �1)2 − �2�

2− x1 + 2�1x4, (93)


P2 = 2(1 − �2)2 − �2�

2− x1 + 2�2x4. (94)

Solving for Z1 and Z2 gives

Z1 = 1

4(�2 − �1)

[�2P

21

(1 − �1)− �1P

22

(1 − �2)

], (95)

Z2 = 1

4(�2 − �1)

[P 2

1

(1 − �1)− P 2

2

(1 − �2)

]. (96)

Now consider the change of variables y = P1/P2. Then after some algebra, Eq. (86) can bewritten as(

dy

dt

)2

= N2�2(y2 + g2)(h2 − y2), (97)

where

N2 = 9k23(�2

� + 2x1 − 4x4)2

256(1 − �1)2 , (98)

g2 = 1 − �1

�2 − 1, (99)

h2 = �1(1 − �1)

�2(1 − �2). (100)

In its present form, Eq. (97) can be solved to give

y =√

�1(1 − �1)

�2(1 − �2)cn

⎡⎣N

√(�2 − �1)(1 − �1)

(�2 − 1)(t − t0),

√�1

�1 − �2

⎤⎦ . (101)

Since

y = P1

P2=

2(1 − �1)2 − �2�

2− x1 + 2�1x4

2(1 − �2)2 − �2�

2− x1 + 2�2x4

, (102)

this can be solved for 2 to give

2 =2x4(�2y − �1) +

(�2

�

2+ x1

)(1 − y)

2(�2y − �1) + 2(1 − y). (103)

It remains to find t0. Using the initial condition 2(t = 0) = �2��0/2, define y0 as the value

of y at t = 0 such that

y0 = y(0) = �2��

20(2�1 − 1) + 2x1 − 4�1x4

�2��

20(2�2 − 1) + 2x1 − 4�2x4

. (104)


It then follows that

t0 = − 1

N

√(�2 − 1)

(�2 − �1)(1 − �1)F

⎡⎣arccos

(y0

h

),

√�1

�1 − �2

⎤⎦ . (105)

5. Numerical simulations

In this section, numerical simulations of the 2 DOF models are considered to show theeffectiveness of the approximate analytical solution. All numerical results were obtained byusing MATLAB. Numerical solutions to the nonlinearly coupled equations are included forcomparison and were obtained using built-in fourth and fifth order Runge–Kutta proceduresin MATLAB.

5.1. Resonance case

For the first case, the following values for the constants are taken: k1 = 2, k3 = 2, �c =4, �� =2, c0 =0.2, �0 =0.2. This is clearly a case where �c =2��, thus internal resonanceis expected.

Let us now consider how the approximate solution fares in capturing the expected periodicvariation of amplitude. In Figs. 1 and 2, the numerical solution for � and c are comparedto the approximate solution generated by the time-averaged canonical perturbation theory.It can be seen that the solution produced by the time-averaged perturbation solution doestrack the actual amplitude variation of the solution. In fact, in this case the time-averagedperturbation solution closely matches the numerical solution.

As a second example, consider the case k1 =4, k3 =6, �c =2, �� =1, c0 =0.1, �0 =0.1.This is also a case where �c = 2��, thus internal resonance is expected.

The graphs of � and c for the numerical solution of the 2 DOF model and the time-averagedHamiltonian perturbation solution are shown in Figs. 3 and 4, respectively. Again, for thiscase, the time-averaged Hamiltonian canonical perturbation solution is a good approximatesolution to the 2 DOF model.

5.2. Non-resonance case

As another example, consider the case k1 = 2, k3 = 2, �c = 2, �� = 4, c0 = 0.1, �0 = 0.1.This is not a case where �c = 2��, thus no resonance is expected. The graphs of � and c forthe numerical solution of the 2 DOF model and the time-averaged Hamiltonian perturbationsolution are shown in Figs. 5 and 6, respectively. It can be seen from the graphs that there isgood agreement between the numerical and time-averaged canonical perturbation solutions.

Consider a similar example with k1 = 2, k3 = 2, �c = 2, �� = 4, c0 = 0.4, �0 = 0.4. Thisis the same as the previous case, with a slight change in the initial conditions. The graphs of� and c for the numerical solution of the 2 DOF model and the time-averaged Hamiltonianperturbation solution are shown in Figs. 7 and 8, respectively.

Once again, there is good agreement between the approximate solution and the exactsolution. Note also how the frequency of the solutions depends on the initial conditions.


NumericalTime-Av Hamiltonian

0 5 10 15 20 25 30 35 40 45 50 -0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5tau for numerical and Time-Averaged Hamiltonian Solutions

Time

tau

Fig. 1. Comparison of � with numerical solution and time-averaged canonical perturbation solution.


0 5 10 15 20 25 30 35 40 45 50 -0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2c for numerical and Time-Averaged Hamiltonian Solutions

Time

c

Fig. 2. Comparison of c with numerical solution and time-averaged canonical perturbation solution.



0 5 10 15 20 25 30 35 40 45 50 -0.25

-0.2

- 0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2


Time

tau



0 5 10 15 20 25 30 35 40 45 50 -0.15

-0.1

-0.05

0

0.05

0.1


Time

c




0 5 10 15 20 25 30 35 40 45 50 -0.15

-0.1

-0.05

0

0.05

0.1


Time

tau



0 5 10 15 20 25 30 35 40 45 50- 0.15

-0.1

-0.05

0

0.05

0.1


Time

c



0 5 10 15 20 25 30 35 40 45 50 -0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4


Time

tau


0 5 10 15 20 25 30 35 40 45 50 -0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3


Time

c



The only change in this case from the previous case is in the initial conditions. However,note that in this case although the time-averaged canonical perturbation solution correctlycaptures the frequency of the solution, it slightly underpredicts the minimum of the solutionfor c.

5.3. Conclusions

In conclusion, for the case of coupled oscillators that we examined, the possibility ofinternal resonance between the oscillators arises, depending on the relationship betweenthe natural frequencies of the two oscillators. In the case of internal resonance, there isa sort of ‘energy sharing’ between the two oscillators. The amplitudes of vibrations ofboth are not constant but rather vary periodically. This is reminiscent of beats between twolinear harmonic oscillators. For the linear harmonic oscillator, the beat frequency is easilyfound, whereas for the nonlinearly coupled oscillators it is not. For the non-resonant casenumerical simulations showed that the amplitude of the system would not show the sameperiodic variation.

We can see from the simulations of the numerical and time-averaged canonical analyticalsolution that the analytical solution does a good job of matching the numerical solution forboth possible cases of internal resonance or no internal resonance.

6. Summary

In this paper, an analytical solution to the nonlinear spinning disk problem is given. Themethod of time-averaged canonical perturbation solution is developed for the 2 DOF non-linear spinning disk model. Numerical simulations of this solution along with simulatingthe 2 DOF system directly show that the analytical solution agrees quite well with thepredictions of the numerical solution for both resonance and non-resonance cases. Thus,the time-averaged canonical perturbation solution does a good job of capturing the variousdynamics of the 2 DOF model.

References

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time-averaged canonical perturbation of nonlinear vibrations of a spinning disk

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