The Impact of Multistability on Hysteresis Arising in Linear and Nonlinear Systems
NNoname manuscript No. (will be inserted by the editor)
The Impact of Multistability on Hysteresis Arising inLinear and Nonlinear Systems
Gina Faraj Rabbah · Amenda Chow
Received: date / Accepted: date
Abstract
Hysteresis is typically depicted as a looping behaviour in a systemsinput-output graph. This looping behaviour relates to multiple stable equilib-ria (that is, multistability) in the system. This work examines some necessarystability conditions for linear and nonlinear ordinary differential equations toexhibit hysteresis. Examples and simulations are presented supporting this.Additionally, the shape of hysteresis loops due to different types of multista-bility (e.g. continuum of equilibria or isolated equilibria) are described.
Keywords
Hysteresis loops · Dynamic systems · Differential equations · Equilibrium · Stable
Hysteresis is a common phenomenon found in real-world systems such as inbiological fields [2,11,21,26], magnetism [4,9,27], ferroelectric materials [7,30,38], circuits [14,35] and economics [12]. The following discuss hysteresis in-depth [8,17,18,23,26,27,36].The study of hysteresis is important for understanding the behaviour ofa system given changes in the input; however, its appearance in a dynamicalsystem can make modelling the system more difficult. Consequently, the morehysteresis is understood, the easier it can be to control the hysteresis arisingin a given dynamical system. Take, for example, a population of insects whichis sensitive to temperature changes. At certain temperatures there is an out-break of the population and at other temperatures a dramatic decline. Thesedynamics can lead to hysteretic behaviour, and hence understanding this canhelp control the population [11].
Department of Mathematics and Statistics, York University, Toronto, CanadaE-mail: [email protected] or [email protected] a r X i v : . [ m a t h . D S ] A ug Gina Faraj Rabbah, Amenda Chow
To define hysteresis, consider the following definition.
Definition 1 [23, Definition 3]A hysteretic system is one which has(i) multiple stable equilibrium points and(ii) system dynamics that are faster than the time scale at which inputs arevaried.The first part of definition 1 is known as the principle of multistability [29]. It is a necessary condition for a dynamical system to exhibit hysteresis;however it is not a sufficient condition. Consider a very simple dynamicalsystem, ˙ x ( t ) = 0, which has a continuum of stable equilibrium points. Thesolution to ˙ x ( t ) = 0 is a constant function, which does not exhibit hysteresis.Additional references that consider the relationship between equilibria andhysteresis are [1,2,3,9,11,16,17,29,34,39,37].An equilibrium solution is necessarily constant for all time t ; however,consider steady state solutions which only require convergence to a constantsolution after some t . An example of a steady state solution could be e − t cos( t )since this goes to zero as t approaches infinity. In this work, we relax the firstpart of definition 1 and demonstrate steady state solutions can also lead tothe existence of hysteresis.The second part of definition 1 is related to the transition speed betweenstable equilibria. When the input is near a stable equilibrium point, the systemdynamics tend to want to stay there because of the stability. But the inputwill vary enough that system dynamics shifts to the next available stableequilibrium and due to the fast dynamics of the system, this transition betweenequilibria occurs virtually instantaneously compared to the rate of change ofthe input. If these dynamics were displayed in an input-output graph, theywould be represented by a “jump” (ie. steep portion of graph) as depicted inFigure 1. However, this jump is not always so obvious, especially in systemswith a continuum of stable equilibria. In this paper, we explore the variousshapes of input-output graphs that may arise.The second part of definition 1 is not easy to compute. Fortunately, thereis a test for hysteresis that can be used. Test 1 [27] A system exhibits hysteresis if a nontrivial closed curve withperiodic input in the input-output graph of the system persists as the frequencyof the input approaches zero.
This nontrivial closed curve is called a hysteresis loop . If the loop degen-erates such that the graph can be described by a function, this is a trivialclosed curve, so in test 1 these are excluded. Figure 2 depicts an example of anontrivial closed curve and a trivial closed curve.It is important to note the presence of looping is not sufficient to concludehysteresis but rather, as test 1 indicates, the loops must persist as the frequencyof the input goes to zero. In other words, persists means the input-output graphalways exhibits a nontrivial closed curve even as the frequency of the inputapproaches zero. Likewise, since the loop must be closed, it must therefore itle Suppressed Due to Excessive Length 3 -10 -5 0 5 10 15-2-1012
Fig. 1:
Suppose -1 and 1 are two stable equilibria of a given dynamical system. Whenthe input is such that the output is at -1, the system dynamics tend to want to stay at -1because of the stability. But the input will vary enough that system dynamics shifts to thenext available stable equilibrium, namely 1. Due to the fast dynamics of the system, thistransition from -1 and 1 occurs virtually instantaneously compared to the rate of change ofthe input. This is represented by a “jump” (ie. steep portion of graph) from -1 to 1 whenthe value of the input is 5. -1 -0.5 0 0.5 1-2-1012 -1 -0.5 0 0.5 1-2-1012
Fig. 2:
Two input-output graphs with one displaying looping behaviour (nontrivial closedcurve, left) and the other not displaying looping (trivial closed curve, right). always be bounded. Hence, the size of the loop is finite as the frequency of theinput goes to zero; that is, the loop cannot grow without bounds.Definition 1 also suggests looping behaviour. This is because for a systemwith fast dynamics, the periodic input that steers the system dynamics fromequilibrium A to equilibrium B differs when the input steers the system fromequilibrium B to equilibrium A. This creates a loop, and consequently, hys-teresis is often said to be path-dependent. See Figure 3 for an illustration. Thispath-dependence can be thought of as a lag that the system undergoes as ittransitions between stable equilibria due to changes in the input. Interestingly,hysteresis derives from ancient Greek to mean “lag behind”[15].Hysteresis is also related to bifurcations [6, page 18]. In particular, we areinterested in bifurcation in the context of changes in the stability of the systemdue to changes in the input.To establish hysteresis in the dynamical systems presented in this paper,the following procedure is applied. The equilibria or steady state solutions ofeach system is first determined and then their stability is established. Thisis a necessary condition of hysteresis as indicated in definition 1. To establishstability, concepts such as phase portraits, linearization, bifurcation values and
Gina Faraj Rabbah, Amenda Chow -1 -0.5 0 0.5 1-2-1012
Fig. 3:
A loop is formed when the output follows different paths as the input varies. Thediffering paths has been emphasized through the use of two different colours in the hysteresisloop. eigenvalues are employed. Corresponding figures are constructed using Maple.A constant input is considered when establishing equilibria and its stability.Constructing the input-output graph of each system is also necessary inorder to apply test 1. The output is the solution to the dynamical system. Inthis paper, we consider ordinary differential equations (ODE). Let t be timeand u ( t ) be the input, which is introduced into the ODE to construct the input-output graphs. These input-output graphs are generated using MATLAB. Inthis paper, the periodic input u ( t ) is typically sin( ωt ) so that the frequencyof the input is ω . However, in the analysis of equilibria, the input is constantwith respect to time to fit with the definition of an equilibrium point. This isalso the approach taken in [23]. What affects the shape of hysteresis loops?
When the shape of the hysteresisloop is unchanged as the frequency of the periodic input changes, the systemis said to exhibit rate-independent hysteresis; otherwise, the hysteresis is saidto be rate-dependent [27]. Some authors define rate independence as necessaryfor hysteresis to occur [8,12,22,36]. In rate-dependent systems, the frequencyof the input has been shown to affect the area of the hysteresis loop [1,5,32]. In [25], the authors examine how different porous materials affect theshape of hysteresis loops. In [40], the authors explore pinched hysteresis loopswhich are closed curves that are not simple; that is, the loop crosses itself.Figure 4 depicts an example of a pinched hysteresis loop. In [13], the authorsstudy systems that exhibit hysteresis loops shaped like butterfly wings, someof which could also be classified as pinched loops.Operators are commonly used to describe hysteresis [8,36]. In particular,operators are found such that they best fit the shape of the hysteresis loop.For example, the bottom plot in Figure 5 shows a hysteresis loop with twooutputs, -1 and 1. The operator F ( u ( t )) = (cid:26) − , − < u ( t ) < , < u ( t ) < (cid:27) can be used to describe the shape of the hysteresis loop in Figure 5. This styleof hysteresis operator has been used to reduce frequent ‘on’ and ‘off’ switching itle Suppressed Due to Excessive Length 5 -1 -0.5 0 0.5 1-2-1012 Fig. 4:
A closed curve that is not simple; that is, the loop formed crosses or “pinches” itself. -10 -5 0 5 10 15-2-1012 -10 -5 0 5 10 15-2-1012 -10 -5 0 5 10 15-2-1012
Fig. 5:
In the top left, the system dynamics are depicted as “jumping” from -1 to 1 as theperiodic input u ( t ) varies. In the top right, as u ( t ) continues to vary, the system dynamicsfollows a different path from 1 to -1. Together, this creates a hysteresis loop with two outputs,-1 and 1, as shown in the bottom plot. from devices like thermostats [23]. However, hysteresis operators do not lookat the intrinsic properties of the dynamical system that are causing hysteresisin the first place, whereas definition 1 does.In Figure 5, when the input is between 0 and 5, it is not clear what theoutput is unless the previous output is known. For this reason, hysteresis isoften said to have a memory, and consequently, the memory effect in hysteresisis related to its looping behaviour.Since multiple stable equilibria are essential to a hysteretic system as notedin definition 1, there is likely a correlation between the equilibria of a systemand the shape of the hysteresis loop. This relationship is explored in greaterdepth in this paper for ODEs. In particular, ODEs can have different types ofequilibria such as a continuum of stable equilibria, a finite number of discretestable equilibria and the presence or absence of unstable equilibria. A set ofdiscrete equilibria is also called isolated equilibria in this context. Hysteresis Gina Faraj Rabbah, Amenda Chow arising from a continuum of equilibria is known as traversal-type hysteresis andhysteresis arising from isolated equilibria is known as bifurcation-type hystere-sis [28].Hysteresis loops arising in both linear and nonlinear systems are also dis-cussed. Conditions in which hysteresis can never occur, due to the principleof multistability not being satisfied, are also presented. Observations of theseexamples and summary discussions are reviewed in the last section. x ( t ) = a ( t ) x ( t ) + b ( t ) , (1a) x (0) = x , (1b)where a ( t ) is a continuous function of t with t ≥ x is an arbitrary constant,and b ( t ) is some continuous function of the input, u ( t ) . For example, b ( t ) = u ( t )or b ( t ) = ˙ u ( t ) | u ( t ) | . Proposition 1
When a ( t ) is a nonzero constant, (1) cannot exhibit hystere-sis. Proof: Let a ( t ) be a nonzero constant, a ( t ) = ¯ a . The equilibria are foundwhen ˙ x = 0. When the input is constant, say U , the equilibrium point is¯ x = − U ¯ a . Notice that this means that there is only one equilibrium point, so,this system cannot have multiple stable equilibria. Therefore, by the principleof multistability , (1) cannot exhibit hysteresis when a ( t ) is a nonzero constant. (cid:117)(cid:116) Figure 6 depicts the input-output graph of (1) with a = − .
1, and depictsloops that degenerate to the graph of a function as the frequency of the inputgoes to zero. By test 1, it follows that (1) does not exhibit hysteresis as ex-pected by Proposition 1.Now, suppose for (1), a ( t ) = 0. Then, we have ˙ x = b ( t ) and the solutionis x ( t ) = (cid:82) t b ( t ) dt . Depending on b ( t ), this system may exhibit hysteresis.Figure 7 shows a system with a ( t ) = 0 and b ( t ) = sin( ωt ) (that is, b ( t ) = u ( t )where u ( t ) = sin( ωt )) which does not exhibit hysteresis because, while thereis persistent looping, the size of the loops are growing without bounds as thefrequency of the input goes to zero. The system ˙ x ( t ) = sin( ωt ) , x (0) = − x ( t ) = 1 − cos( ωt ) ω − . As ω →
0, the output x ( t ) = − cos( ωt ) ω − ∞ leading to loops thatare expanding to infinity. Hence, the loop is not bounded as the frequencyapproaches 0. Therefore, by test 1, the system is not hysteretic for this input. itle Suppressed Due to Excessive Length 7 -1 -0.5 0 0.5 1-2-1012 =1 -1 -0.5 0 0.5 1-2-1012 =0.1-1 -0.5 0 0.5 1-2-1012 =0.01 -1 -0.5 0 0.5 1-2-1012 =0.001 Fig. 6:
The input-output graph of (1) with a ( t ) = − . , x = 1 and b ( t ) = | ˙ u ( t ) | u ( t ), where u ( t ) is the input, sin( ωt ) , and ω is the frequency of the input. Since the loops degenerate as ω goes to zero, by test 1, this system is not hysteretic. On the other hand, Figure 8 shows a system with a ( t ) = 0 , b ( t ) = | ˙ u ( t ) | u ( t )and x = 0 . x ( t ) = | ˙ u ( t ) | u ( t ). For u ( t ) = sin( ωt ), the solutionto ˙ x = | ˙ u ( t ) | u ( t ) is x ( t ) = (cid:40) cos( ωt ) + 0 . ωt ) ≤ − cos( ωt ) + 0 . ωt ) > (cid:41) . Notice ω does not appear in the denominator unlike in the case of ˙ x ( t ) = u ( t )and hence this leads to loops that are bounded. Also notice when the input isconstant for ˙ x ( t ) = | ˙ u ( t ) | u ( t ), the derivative of the input is 0. Hence, there area continuum of equilibria for constant inputs, which is a necessary conditionfor hysteresis as noted in definition 1.In addition, notice that a ( t ) from (1) could be some function of t . Figures9 and 10 have a ( t ) = − e − t and a ( t ) = − t +1 , respectively, which both showpersistent finitely-sized closed loops similar in shape to Figure 8. This may bedue to − e − t and − t +1 approaching zero as t → ∞ , which for large t , mimicsthe case when a ( t ) = 0. On the other hand, when a ( t ) = − t or a ( t ) = − t +1 +1,which do not approach 0 as t → ∞ , no hysteresis is suggested since persistentlooping does not occur in the corresponding input-output maps, see Figures 11and 12, respectively.Observe that the systems in Figures 9 and 10 are considered to have steadystate solutions since after some t , we have − e − t and − t +1 approaching 0. Thatis, rather than in the case of an equilibrium solution which is constant for alltime; we have a constant solution after some time, t . Gina Faraj Rabbah, Amenda Chow -1 -0.5 0 0.5 1-20246810 =1 -1 -0.5 0 0.5 10102030 =0.1-1 -0.5 0 0.5 1050100150200250 =0.01 -1 -0.5 0 0.5 10500100015002000 =0.001
Fig. 7:
The input-output graph of (1) with a ( t ) = 0 , x = − b ( t ) = u ( t ) = sin( ωt ),where ω is the frequency of the input. The loop expands to infinity as ω →
0, so the systemdoes not exhibit hysteresis. -1 -0.5 0 0.5 1-0.500.511.52 =1 -1 -0.5 0 0.5 1-0.500.511.52 =0.1-1 -0.5 0 0.5 1-0.500.511.52 =0.01 -1 -0.5 0 0.5 1-0.500.511.52 =0.001=0.0001=0.00001
Fig. 8:
The input-output graph of (1) with a ( t ) = 0 , x = 0 . b ( t ) = | ˙ u ( t ) | u ( t ), where ω is the frequency of the input u ( t ) = sin( ωt ). By test 1, this system appears to exhibitrate-independent hysteresis. itle Suppressed Due to Excessive Length 9 -1 -0.5 0 0.5 1-0.500.511.52 =1 -1 -0.5 0 0.5 1-0.500.511.52 =0.1-1 -0.5 0 0.5 1-0.500.511.52 =0.01 -1 -0.5 0 0.5 1-0.500.511.52 =0.001=0.0001=0.00001 Fig. 9:
The input-output graph of (1) with a ( t ) = − e − t , x = 0 . b ( t ) = | ˙ u ( t ) | u ( t ),where ω is the frequency of the input u ( t ) = sin( ωt ). By test 1, this system appears toexhibit hysteresis and rate-independence is observed as the frequency of the input goes tozero. -1 -0.5 0 0.5 1-0.500.511.52 =1 -1 -0.5 0 0.5 1-0.500.511.52 =0.1-1 -0.5 0 0.5 1-0.500.511.52 =0.1 -1 -0.5 0 0.5 1-0.500.511.52 =0.001=0.0001=0.00001 Fig. 10:
The input-output graph of (1) with a ( t ) = − t +1 , x = 0 . b ( t ) = | ˙ u ( t ) | u ( t ),where ω is the frequency of the input u ( t ) = sin( ωt ). By test 1, this system appears toexhibit hysteresis. Duhem model ˙ x ( t ) = α | ˙ u ( t ) | ( βu ( t ) − x ( t )) + γ ˙ u ( t ) , (2a) x (0) = x , (2b) -1 -0.5 0 0.5 1-1-0.500.51 =1 -1 -0.5 0 0.5 1-1-0.500.51 =0.1-1 -0.5 0 0.5 1-1-0.500.51 =0.01 -1 -0.5 0 0.5 1-1-0.500.51 =0.001 Fig. 11:
The input-output graph of (1) with a ( t ) = − t, x = 0 . b ( t ) = | ˙ u ( t ) | u ( t ),where ω is the frequency of the input u ( t ) = sin( ωt ). The loops degenerate as the frequencyof the input goes to zero. -1 -0.5 0 0.5 1-1-0.500.51 =1 -1 -0.5 0 0.5 1-1-0.500.51 =0.1-1 -0.5 0 0.5 1-1-0.500.51 =0.01 -1 -0.5 0 0.5 1-1-0.500.51 =0.001 Fig. 12:
The input-output graph of (1) with a ( t ) = − t +1 +1 , x = 0 . b ( t ) = | ˙ u ( t ) | u ( t ),where ω is the frequency of the input u ( t ) = sin( ωt ). The loops degenerate as the frequencyof the input goes to zero. is considered. The input u ( t ) is the magnetic field strength and the output, x ( t ), is the magnetic flux density [10,27]. Hysteresis is often associated tononlinear systems, see for instance [27]; however, in this context, equation (2) isclassified as linear based on (1) with a ( t ) = − α | ˙ u ( t ) | and b ( t ) = α | ˙ u ( t ) | βu ( t ) + γ ˙ u ( t ). The parameters α, β, γ are positive constants. Proposition 2
The Duhem Model in (2) cannot exhibit hysteresis for α < . itle Suppressed Due to Excessive Length 11 Proof: The equilibria of (2) are determined by setting the input u ( t ) to beconstant, meaning ˙ u ( t ) = 0. This leads to ˙ x ( t ) = 0 for all t . Therefore, (2) hasa continuum of equilibria.Let f ( x ) = α | ˙ u ( t ) | ( βu ( t ) − x ( t )) + γ ˙ u ( t ). It follows that dfdx = − α | ˙ u ( t ) | . (3)The sign of (3) determines the stability of the equilibrium points of (2). Notice dfdx does not depend on x and hence its sign is not affected by a particularequilibrium point. If α < dfdx >
0, which implies the continuum ofequilibria is unstable and hence (2) cannot exhibit hysteresis. (cid:117)(cid:116)
Figure 13 shows the input-output graphs of (2) with α <
0, which dis-plays no looping in the input-output graph of (2), and hence confirms thereis no hysteresis. On the other hand, Figure 14 depicts a scenario when α > -1 -0.5 0 0.5 1-20246 =1 -1 -0.5 0 0.5 1-20246 =0.1-1 -0.5 0 0.5 1-20246 =0.01 -1 -0.5 0 0.5 1-20246 =0.001
Fig. 13:
The input-output graph of (2) with α = − , β = 0 . , γ = 1 , x = 0 and u ( t ) =sin( ωt ), where ω is the frequency of the input. No closed curves appear, and therefore thesystem does not exhibit hysteresis. y ( t ) + p ( t ) ˙ y ( t ) + q ( t ) y ( t ) = b ( t ) , (4a) y (0) = y , ˙ y (0) = y , (4b) -1 -0.5 0 0.5 1-101 =1 -1 -0.5 0 0.5 1-101 =0.1-1 -0.5 0 0.5 1-101 =0.01 -1 -0.5 0 0.5 1-101 =0.001 Fig. 14:
The input-output graph of (2) with α = 0 . , β = 0 . , γ = 1 , x = 0 and u ( t ) =sin( ωt ), where ω is the frequency of the input. This system is exhibits rate-independenthysteresis. where p ( t ) , q ( t ) are continuous functions of t ≥ y , y are arbitrary constantsand b ( t ) is a continuous function of the input, u ( t ). Proposition 3
Equation (4) cannot exhibit hysteresis when q ( t ) is a nonzeroconstant function. Proof: Let q ( t ) be a nonzero constant, ¯ q . Let x = y ( t ) and x = ˙ y ( t ),writing (4) into a system of first order ODEs leads to˙ x = x . (5a)˙ x = − p ( t ) x − ¯ qx + b ( t ) . (5b) x (0) = y , x (0) = y . (5c)For constant input b ( t ) = U , there is only one equilibrium point, (¯ x , ¯ x ) =( U ¯ q , principle of multistability is not satisfied and this system cannotexhibit hysteresis. (cid:117)(cid:116) Figure 15 depicts the input-output graphs of (4) satisfying the conditionsin Proposition 3 and the absence of hysteresis is demonstrated. On the otherhand, Figure 21 depicts the input-input graphs of (4) when q ( t ) is not a con-stant function; that is, not satisfying the condition in Proposition 3, and thepresence of hysteresis is shown. The same function for p ( t ) and b ( t ) are usedin both instances. Proposition 4
Equation (4) cannot exhibit hysteresis when p ( t ) < for some t . itle Suppressed Due to Excessive Length 13 -1 -0.5 0 0.5 1-2-1012 =1 -1 -0.5 0 0.5 1-2-1012 =0.1-1 -0.5 0 0.5 1-2-1012 =0.01 -1 -0.5 0 0.5 1-2-1012 =0.001 Fig. 15:
The input-output graph of (6) with p ( t ) = e − t + 1, q ( t ) = 1 , y = 1 , y = 0 and b ( t ) = | ˙ u ( t ) | u ( t ) where u ( t ) = sin( ωt ). The system is not hysteretic because the loops donot persist as the frequency of the input goes to zero. Proof: Let x = y ( t ) and x = ˙ y ( t ), write (4) as a system of first orderlinear ODEs as follows ˙ x = x , (6a)˙ x = − p ( t ) x − q ( t ) x + b ( t ) , (6b) x (0) = y , x (0) = y . (6c)The Jacobian matrix of equation (6) is J = (cid:20) − q ( t ) − p ( t ) (cid:21) . Notice J does not depend on x and x , and hence is not influenced by specificequilibrium points or steady state solutions. The characteristic equation is λ + p ( t ) λ + q ( t ) = 0and the eigenvalues are λ = − p ( t ) + (cid:112) ( p ( t )) − ∗ q ( t )2 , (7) λ = − p ( t ) − (cid:112) ( p ( t )) − ∗ q ( t )2 . (8)Since the square root function is always positive or imaginary, the real partof λ and λ can be positive if p ( t ) < t , which means the equilibria or the steady state solutions will not be stable and hence (6) cannot exhibithysteresis. (cid:117)(cid:116) Figures 16 and 17 display the input-output graphs of (4) for examplesin which p ( t ) is not strictly positive; that is, satisfying the assumptions inProposition 4. In both, hysteresis is not present. Figure 18 provides an exampleof (4) in which p ( t ) is a non-negative function; however, hysteresis is also notpresent. This illustrate the assumption in Proposition 4 is necessary but notsufficient to establish the absence of hysteresis. These three examples have acontinuum of equilibria for constant input u ( t ). That is, the equilibria are ofthe form ( c, , where c is a constant. The next example demonstrates thepresence of hysteresis for systems of the form in (4). -1 -0.5 0 0.5 1-10-50510 =1 -1 -0.5 0 0.5 1-10-50510 =0.1-1 -0.5 0 0.5 1-10-50510 =0.01 -1 -0.5 0 0.5 1-1500-1000-500050010001500 =0.001 Fig. 16:
The input-output graph of (4) with p ( t ) = − . q ( t ) = | ˙ u ( t ) | , y = 1 , y = 0and b ( t ) = | ˙ u ( t ) | u ( t ) where u ( t ) = sin( ωt ). Looping behaviour does not appear as ω goes tozero.Therefore, this system is not hysteretic. Consider ¨ y ( t ) + p ( t ) ˙ y ( t ) + α | ˙ u ( t ) | y ( t ) = αβ | ˙ u ( t ) | u ( t ) + γ ˙ u ( t ) , (9a) y (0) = y , ˙ y (0) = y . (9b)This is a second order system that was constructed to be similar to the firstorder Duhem model in (2). Equation (9) is a specific example of (4), where q ( t ) = α | ˙ u ( t ) | and b ( t ) = αβ | ˙ u ( t ) | u ( t ) + γ ˙ u ( t ) . Figures 19 and 20 display the input-output graphs of (9) where p ( t ) is apositive constant function and u ( t ) = sin( ωt ). Figure 21 provides an examplewhere p ( t ) is not a constant function. In all three figures, persistent looping isshown indicating the presence of hysteresis. itle Suppressed Due to Excessive Length 15 -1 -0.5 0 0.5 1-10-50510 =1 -1 -0.5 0 0.5 1-10-50510 =0.1-1 -0.5 0 0.5 1-10-50510 =0.01 -1 -0.5 0 0.5 1-10-50510 =0.001 Fig. 17:
The input-output graph of (4) with p ( t ) = sin( t ), q ( t ) = | ˙ u ( t ) | , y = y = 0 and b ( t ) = | ˙ u ( t ) | u ( t ) where u ( t ) = sin( ωt ), which indicates the system is not hysteretic. -1 -0.5 0 0.5 1-1-0.500.51 =1 -1 -0.5 0 0.5 1-1-0.500.51 =0.1-1 -0.5 0 0.5 1-1-0.500.51 =0.01 -1 -0.5 0 0.5 1-1-0.500.51 =0.01 Fig. 18:
The input-output graph of (4) with p ( t ) = | sin( t ) | , q ( t ) = | ˙ u ( t ) | , y = y = 0 and b ( t ) = | ˙ u ( t ) | u ( t ), where u ( t ) = sin( ωt ). This system is not hysteretic because the loops donot persist as the frequency of the input goes to zero. To determine the equilibria of (9), let x = y ( t ) and x = ˙ y ( t ), which leadsto ˙ x = x , ˙ x = − p ( t ) x − α | ˙ u ( t ) | x + αβ | ˙ u ( t ) | u ( t ) + γ ˙ u ( t ) . To solve for the equilibria, once again u ( t ) is treated as a constant and thus˙ u ( t ) = 0. Meaning, the equilibria are of the form ( c, c is an arbitraryconstant since any constant x ( t ) will lead to ˙ x = 0. Therefore, (9) has acontinuum of equilibria. Using the eigenvalues from (7) and (8), the equilibriaare stable if the real part of λ and λ are negative. This can occur if p ( t ) ispositive for all t . -2 -1 0 1 2-1-0.500.51 =1 -2 -1 0 1 2-1-0.500.51 =0.1-2 -1 0 1 2-1-0.500.51 =0.01 -2 -1 0 1 2-1-0.500.51 =0.001=0.0001=0.00001 Fig. 19:
The input-output graph of (9) with u ( t ) = sin( ωt ), p ( t ) = 0 . β = α = 1, γ = 0and y = − . , y = 0. This system is hysteretic. Also, notice the appearance of pinchedhysteresis loops and rate-independence emerges as ω approaches zero. x ( t ) = f ( x ( t )) (10a) x (0) = x , (10b)where f is a nonlinear, differentiable and continuous function of x , x is aconstant and t ≥
0. Note that f can be a function of the input u ( t ) , which isknown. Recall that when exploring equilibria and stability, u ( t ) is a constantfunction and when constructing hysteresis loops, u ( t ) is a periodic function. itle Suppressed Due to Excessive Length 17 -1 -0.5 0 0.5 1-505 =1 -1 -0.5 0 0.5 1-505 =0.1-1 -0.5 0 0.5 1-505 =0.01 -1 -0.5 0 0.5 1-505 =0.001=0.0001=0.00001 Fig. 20:
The input-output graph of (9) with u ( t ) = sin( ωt ), p ( t ) = 0 . , β = α = γ = 1 and y = 0 . , y = 0. This system is hysteretic and rate-independence emerges as ω approacheszero. -1 -0.5 0 0.5 1-1-0.500.51 =1 -1 -0.5 0 0.5 1-1-0.500.51 =0.1-1 -0.5 0 0.5 1-1-0.500.51 =0.01 -1 -0.5 0 0.5 1-1-0.500.51 =0.01 Fig. 21:
The input-output graph of (9) with p ( t ) = e − t + 1, α = β = 1, γ = 0 , y = 0 . y = 0. The system is hysteretic and rate-independence appears as ω goes to zero. This system can have either a continuum of equilibria or a set of isolatedequilibria. Here is a continuum example,˙ x ( t ) = | ˙ u ( t ) | ( x ( t ) − ( x ( t )) + u ( t )) . (11)To determine the equilibria, set the input u ( t ) to be a constant, which means˙ u ( t ) = 0, and hence ˙ x = 0. This means any constant is an equilibrium pointand they are stable. For the periodic input u ( t ) = sin( ωt ), the input-output graph of (11) isdepicted in Figure 22. Persistent looping is demonstrated, which indicatesthe presence of hysteresis. Furthermore, the loops exhibit rate-independence.The first order Duhem model also lead to rate-independent hysteresis, seeequation (2) and Figure 14. -1 -0.5 0 0.5 1-2-1012 =1 -1 -0.5 0 0.5 1-2-1012 =0.1-1 -0.5 0 0.5 1-2-1012 =0.01 -1 -0.5 0 0.5 1-2-1012 =0.001=0.0001=0.00001 Fig. 22:
The input-output graph of (11) with u ( t ) = sin( ωt ) and x = 1. This system ishysteretic and rate-independence is observed. In the case of isolated equilibria, let the corresponding set of isolated equi-libria of (10), for a constant input, be denoted S = { ¯ x , ¯ x , ..., ¯ x n } for n being some natural number. Proposition 5
Equation (10) cannot exhibit hysteresis when dfdx (¯ x j ) > ,where ¯ x j ∈ S for all j ∈ [0 , n ] . Proof: To determine the stability of an equilibrium point, consider firstan illustration; that is, a phase portrait as depicted in Figure 23. Solutiontrajectories need to be approaching the equilibrium point in order to be stable.Let G be a region around x ( t ) such that there is only one equilibrium point,¯ x j ∈ S , in G . Let G ∈ G such that x ( t ) < ¯ x j and G ∈ G such that x ( t ) > ¯ x j .As can be seen in Figure 23, solution trajectories approach the equilibriumpoint when f ( x ( t )) > x ( t ) ∈ G (12a) f ( x ( t )) < x ( t ) ∈ G . (12b)A similar analysis can be found in [41, Page 119]. Since f (¯ x j ) = 0 and (12) itle Suppressed Due to Excessive Length 19 Fig. 23:
The phase portrait of (10) depicting a stable equilibrium point. For illustrationpurposes, the equilibrium point has been chosen as zero. For stability, solution trajectoriesgreater than the equilibrium must decrease towards the equilibrium (hence, ˙ x <
0) andsolution trajectories less than the equilibrium must increase towards the equilibrium (hence,˙ x >
Fig. 24:
The phase portrait of (10) depicting an unstable equilibrium point. The red curveis an unstable equilibrium solution. must hold for ¯ x j to be stable this must mean that ˙ x is decreasing in terms of x at ¯ x j because f changes sign from positive to negative at ¯ x j . The other wayaround would be unstable as shown in Figure 24. This is equivalent to stating: An isolated equilibrium point of (10) , ¯ x j ∈ S , is stable when dfdx (¯ x j ) < . (13)Hence, when all ¯ x j are such that dfdx (¯ x j ) >
0, all the equilibria in S areunstable and there are no stable equilibria. Therefore, by the principle of multi-stability, (10) would not have hysteresis. (cid:117)(cid:116) Note the statement in Proposition 5 is for all j but in fact, all but one j is also sufficient. This is because at least two stable equilibria are needed for hysteresis as per definition 1. The proof showed dfdx (¯ x j ) > x j to be unstable. Hence, if all j we have dfdx (¯ x j ) >
0, then there are no stableequilibria. If all but one j are such that dfdx (¯ x j ) >
0, then there is only onestable equilibrium point.For isolated equilibria, we can have the following proposition. This propo-sition is mentioned in [33, Page 10] but without a proof.
Proposition 6 If (10) has two or more isolated stable equilibria then inbetween each stable equilibrium point there must be an unstable equilibriumpoint. Proof: Suppose ¯ x = a and ¯ x = b are two stable isolated equilibria. Thenthere must be a third equilibrium point that is unstable in between them sothat solution trajectories less than a are increasing towards a and solutiontrajectories greater than b are decreasing towards b . Otherwise, if there is nounstable point between a and b , trajectories less than a would be increasingtowards a and therefore moving away from b thus making b unstable. Figure25 provides a geometric description. (cid:117)(cid:116) Fig. 25:
Illustration of the need for an unstable equilibrium point in between two isolatedstable equilibria, ‘a’ and ‘b’. The blue curves are particular solutions that converge to stableequilibria and the red curve is an unstable equilibrium solution.
From the proof of Proposition 5, in order for (10) to possibly exhibit hys-teresis f ( x ( t )) cannot be strictly positive or negative. And, with the consider-ation of Proposition 6, f ( x ( t )) must have at least three isolated roots so as tohave at least two isolated stable equilibria. If (10) has isolated equilibria andexhibits hysteresis, this is known as bifurcation-type hysteresis. Bifurcationin this context means a change in stability that occurs due to changes of theinput. itle Suppressed Due to Excessive Length 21 Recall that in (10), f is a function of the input u ( t ). Suppose u ( t ) = U is aconstant input. To determine the bifurcation values of (10), find U satisfying f ( x ( t ) , U ) = 0 , (14a) dfdx ( x ( t ) , U ) = 0 . (14b)The values of U are the bifurcation values. This is because f ( x ( t ) , U ) = 0will provide equilibrium points and from Proposition 5, the sign of dfdx ( x ( t ) , U )indicates when there is a change in stability.Consider the following example˙ x ( t ) = x ( t )(1 − ( x ( t )) ) + u ( t ) , (15a) x (0) = − , (15b)which is of the form (10) where f ( x ( t ) , u ( t )) = x ( t ) − x ( t ) + u ( t ). In discussionsregarding hysteresis, equation (15) can be found in such works as [31, Page336], [28], [5, Page 329].For different values of the constant input, there are different scenarios for x satisfying f ( x ( t ) , U ) = x ( t ) − x ( t ) + U = 0 . For instance, when the constant input is set to U = 1, equation (15) has oneequilibrium point. When the constant input is zero, equation (15) has threeequilibria. The time derivative in (15) is zero when x ( t ) equals 1, 0 and − u ( t ) = 0. Therefore, these are the equilibria of (15). Using the conditionin (13), the equilibrium points 1 and − U in (14) where f ( x ( t ) , U ) = x ( t ) − x ( t ) + U , it followsthat U ≈ − . U ≈ . u ( t ) = sin( ωt ) with loops persisting as ω goes to zero. Further-more, for smaller values of ω , rate-independence is observed. Figure 28 is amagnified version of one of the loops of equation (15) and shows the jumpsof the hysteresis loop occur when the periodic input takes values around thebifurcation values -0.3849001797 and 0.3849001797. This is where the systemrapidly moves between the stable equilibria 1 and -1.Consider another first order nonlinear example˙ x ( t ) = | u ( t ) | x ( t ) (cid:18) − x ( t ) q (cid:19) − ( x ( t )) x ( t )) , (16a) x (0) = 10 . (16b)This example can be found in [20], [24, Page 7-8] and represents the classicSpruce Budworm Model. In this model, u ( t ) depends on the birth rate ofthe spruce budworms and the constant q depends on the carrying capacity of Fig. 26:
A bifurcation diagram for (15), where the dashed line indicates an unstable portionof the graph. The dots are the bifurcation values which correspond to U ≈ − . U ≈ . -1 -0.5 0 0.5 1-2-101 =1 (a) -1 -0.5 0 0.5 1-2-101 =0.1 (b) -1 -0.5 0 0.5 1-2-101 =0.01 (c) -1 -0.5 0 0.5 1-2-101 =0.001=0.0001=0.00001 (d) Fig. 27:
The input-output graph of (15) with u ( t ) = sin( ωt ) for various values of ω .Equation (15) has two stable equilibria and one unstable equilibrium point and exhibitsrate-independent hysteresis as ω → the environment. The carrying capacity is how much of the population theenvironment can support. Equation (16) is of the form of (10) when f ( x ( t )) = | u ( t ) | x ( t ) (cid:18) − x ( t ) q (cid:19) − ( x ( t )) x ( t )) . When u ( t ) is set to be the constant input 0.52 and q = 25, equation (16)exhibits multiple stable equilibria. The authors in [20, Page 327] indicate 0.52 itle Suppressed Due to Excessive Length 23 -0.4 -0.35 -0.3 -0.25 -0.2 -0.150.20.40.60.8 =0.001=0.0001=0.00001 Fig. 28:
A magnified image of Figure 27d shows the vertical jumps of the loop converge toits bifurcation values as ω → is a physically realistic value for the input and the value of q was chosen toensure the existence of multiple equilibria. Using these values to solve for x satisfying 0 . (cid:18) − x ( t )25 (cid:19) − x ( t )1 + ( x ( t )) = 0 , it follows that the equilibria of (16) are¯ x = 0 , ¯ x = 0 . , ¯ x = 1 . , and¯ x = 22 . x , ¯ x are un-stable equilibria and ¯ x , ¯ x are stable. The phase portrait in Figure 29 supportsthis.Fig. 29: A phase portrait of (16) illustrating ¯ x = 0 , ¯ x = 1 . x = 0 . , ¯ x = 22 . When the birth rate is treated as the input of the system, the model exhibitshysteresis. Physically, the birth rate is always positive, therefore the input, u ( t ) = sin( ωt ) is placed in absolute value. The corresponding hysteresis loopsof (16) are shown in Figure 30 and indicates persistent looping as ω goes tozero. (a) (b) (c) (d) Fig. 30:
The input-output graph of (16) with u ( t ) = sin( ωt ) and q = 25 for various valuesof ω . Equation (16) has two stable equilibria and two unstable equilibria. Using (14) where f ( x ( t ) , u ( t )) = | u ( t ) | x ( t ) (cid:18) − x ( t ) q (cid:19) − ( x ( t )) x ( t )) . and Maple, the bifurcation values of (16) are approximately 0 , . . y ( t ) = f ( y ( t ) , ˙ y ( t )) , (17a) y (0) = y , ˙ y (0) = y , (17b) itle Suppressed Due to Excessive Length 25 Fig. 31:
A bifurcation diagram for (16), where the dashed line indicates an unsta-ble portion of the graph. The dots are the bifurcation values which correspond to U =0 , . , . U ,there is always an equilibrium point at 0. Fig. 32:
A magnified image of Figure 30d shows the vertical jumps of the hysteresis loopsconverge to its bifurcation values as the frequency of the input ω approaches 0. where f is a continuous function of t ≥ y , y are constants. As in (10), f can be a function of the input u ( t ) , which is known. Turning (17) into asystem of first order equations by setting x ( t ) = y ( t ) and x = ˙ y ( t ) leads to˙ x ( t ) = x ( t ) , (18a)˙ x ( t ) = f ( x , x ) , (18b) x (0) = y , x (0) = y . (18c)Let n be some natural number, and S = { ¯ X , ¯ X , ... ¯ X n } , where ¯ X j = ( x j , x j ) ∈ S for j ∈ [0 , n ] is an equilibrium point of (18). The capitalization of the letter“x” is used to specify equilibrium points in a second order system. Proposition 7
Equation (17) cannot exhibit hysteresis if for all j ∈ [0 , n ] ,the equilibrium ¯ X j ∈ S satisfies ∂f∂x ( ¯ X j ) > or ∂f∂x ( ¯ X j ) > . As discussed at the end of the proof of Proposition 5, there are actually twocases in which the system cannot exhibit hysteresis: for all j or for all but one j . Proof: From (18), an equilibrium point ¯ X j ∈ S has ¯ x j , ¯ x j such that¯ x j = 0 and f (¯ x j , ¯ x j ) = 0. Substituting ¯ x j = 0 into f leads to f (¯ x j ,
0) = 0.Thus the equilibria of (18) can be written in the form¯ X j = (¯ x j , , where ¯ x j is a root of f .Consider the Jacobian matrix of equation (17) evaluated at ¯ X j J = (cid:34) ∂f∂x (cid:12)(cid:12)(cid:12) ¯ X j ∂f∂x (cid:12)(cid:12)(cid:12) ¯ X j (cid:35) , whose corresponding characteristic equation is λ − λ ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12) ¯ X j − ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12) ¯ X j = 0 . For simplicity, let a = ∂f∂x (cid:12)(cid:12)(cid:12) ¯ X j and b = ∂f∂x (cid:12)(cid:12)(cid:12) ¯ X j . Solving the characteristicequation leads to λ = a + √ a + 4 b ,λ = a − √ a + 4 b . In order to be stable the real part of both λ and λ must be ≤
0, which means a has to be ≤
0. If b >
0, then √ a + 4 b > a , meaning λ >
0. Hence, in orderfor ¯ x j to be stable a = ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12) ¯ X j ≤ , and b = ∂f∂x (cid:12)(cid:12)(cid:12)(cid:12) ¯ X j ≤ . (19)Therefore, if all ¯ x j (or all but one) have ∂f∂x (cid:12)(cid:12)(cid:12) ¯ X j >
0, or ∂f∂x (cid:12)(cid:12)(cid:12) ¯ X j >
0, thenequation (17) cannot have multiple stable equilibria and therefore, cannotexhibit hysteresis. (cid:117)(cid:116)
Consider the following second order nonlinear equation¨ y ( t ) = − ˙ y ( t ) −
10 sin( y ( t )) + u ( t ) (20a) y (0) = 0 , ˙ y (0) = 0 . (20b)When u ( t ) = 0, this equation has two equilibria, (0 ,
0) and ( π, x = y ( t ) and x = ˙ y ( t ), so f ( x , x ) = − x −
10 sin( x ) + u ( t ). Then,using (19), we see that only (0 ,
0) is stable (i.e. there is only one j such that ∂f∂x ( x j ) ≤ . Hence, by Proposition 7, this system cannot exhibit hysteresis.Figure 33 provides the input-output graphs of (20) and persistent looping isnot demonstrated. itle Suppressed Due to Excessive Length 27 -1 -0.5 0 0.5 1-0.200.2 =1 (a) -1 -0.5 0 0.5 1-0.200.2 =0.1 (b) -1 -0.5 0 0.5 1-0.200.2 =0.01 (c) -1 -0.5 0 0.5 1-0.200.2 =0.001 (d)
Fig. 33:
The input-output graph of (20) with u ( t ) = sin( ωt ) for various values of ω . Thissystem only has two equilibria and of the two, only one is stable. Persistent looping doesnot occur. Consider ¨ y ( t ) = − y ( t ) − ( y ( t )) ( y ( t ) + 1) + u ( t ) , (21a) y (0) = − , ˙ y (0) = 0 , (21b)which also fits into the form of (17). By re-writing into a system of first orderODEs, this becomes˙ x ( t ) = x ( t ) , ˙ x ( t ) = − x ( t ) − ( x ( t )) ( x ( t ) + 1) + u ( t ) ,x (0) = − , x (0) = 0 . Also consider¨ y ( t ) = − ˙ y ( t ) − y ( t ) (( y ( t )) − . y ( t ) + 0 .
5) + u ( t ) , (22a) y (0) = − . , ˙ y (0) = 0 . (22b)By re-writing into a system of first order ODEs, this becomes˙ x ( t ) = x ( t ) , ˙ x ( t ) = − x ( t ) − x ( t ) (( x ( t )) − . x ( t ) + 0 .
5) + u ( t ) ,x (0) = − . , x (0) = 0 . For a constant input of 0, the equilibria are { (-1,0) and (0,0) } and { ( − . , ,
0) and (0 . , } for (21) and (22), respectively. All of these equilibria arestable according to (19) where f ( x , x ) = − x ( t ) − ( x ( t )) ( x ( t ) + 1) + u ( t )and f ( x , x ) = − x ( t ) − x ( t )) (( x ( t )) − . x ( t ) + 0 .
5) + u ( t ) for (21)and (22), respectively. The input-output graphs of (21) and (22) can be foundin Figures 34 and for 35, respectively. They indicate the presence of hysteresis. -1 -0.5 0 0.5 1-2-1012 =1 (a) -1 -0.5 0 0.5 1-2-1012 =0.1 (b) -1 -0.5 0 0.5 1-2-1012 =0.01=0.001 (c) -1 -0.5 0 0.5 1-2-1012 =0.0001=0.00001=0.000001 (d) Fig. 34:
The input-output graph of (21) with input u ( t ) = sin( ωt ) for various values of ω .Persistent looping indicates the presence of hysteresis. The bifurcation values of (21) and (22) are calculated by mimicking thefirst order case in (14). In the second order case, f depends on x and x and U ; however, we have x = 0 in the stability analysis. It follows by solving for U in f ( x , , U ) = 0 ,∂f∂x ( x , , U ) = 0 ,∂f∂x ( x , , U ) = 0 , that the bifurcation values of (21) and (22) can be found. The reason for f ( x , , U ) = 0 is because that is the solution for the equilibrium points.The reason for ∂f∂x ( x , , U ) = 0 and ∂f∂x ( x , , U ) = 0 is by considering the itle Suppressed Due to Excessive Length 29 -1 -0.5 0 0.5 1-2-1012 =1 -1 -0.5 0 0.5 1-2-1012 =0.1-1 -0.5 0 0.5 1-2-1012 =0.01=0.001 -1 -0.5 0 0.5 1-2-1012 =0.0001=0.00001=0.000001 Fig. 35:
The input-output graph of (22) with input u ( t ) = sin( ωt ) for various values of ω .Persistent looping indicates the presence of hysteresis. condition in (19), which indicates that the system changes stability when thederivative changes sign. Notice for (21) and (22), ∂f∂x = − ∂f∂x = − x that will change the sign of ∂f∂x . Using Maple to do the computations, for (21), the bifurcation values are0 and , which in decimal form is approximately 0 . , − . . In this section, observations and comparisons of the hysteresis arising in thepreceding linear and nonlinear examples are discussed.1. Conditions for multistability in which linear first and second order systemscan never exhibit hysteresis are presented. See Propositions 1, 2, 3 and4. Similar results were shown for nonlinear systems in Propositions 5, 6and 7. However, when these conditions are not satisfied, both the absence orpresence of hysteresis is possible as illustrated in various examples provided.2. For first and second linear systems to possibly exhibit hysteresis, the equi-libria are never isolated; that is, a continuum of stable equilibria is re-quired. This is known as traversal-type hysteresis . Furthermore, the shape
Fig. 36:
The bifurcation diagram of (21), on the left and of (22) on the right. The dashedlines are unstable portions of the graph. The dots indicate the bifurcation values of eachsystem: U = 0 and on the left and U = 0 , − . . U = 0, ∂f∂x has a double root at x = 0. This means even though(0 ,
0) is a stable equilibrium point of (22), it is very close to an unstable region and hencein the right plot, (
U, y ) = (0 ,
0) is located in an unstable region where y = x . -0.02 0 0.02 0.04 0.06 0.08-0.3-0.2-0.100.1 =0.0001=0.00001=0.000001 0.1 0.12 0.14 0.16 0.18 0.2-0.8-0.7-0.6-0.5-0.4 =0.0001=0.00001=0.000001 Fig. 37:
A magnified image of Figure 34d shows the vertical jumps of the loop approachingits bifurcation values 0 and as the frequency of the input ω goes to zero. -0.05 0 0.0500.10.20.3 =0.0001=0.00001=0.000001 0.05 0.1 0.15 0.2-0.5-0.4-0.3-0.2 =0.0001=0.00001=0.000001 Fig. 38:
A magnified image of Figure 35d to show the vertical jumps of the loops ap-proaching to its bifurcation values of -0.01243505829 and 0.09004734629 as ω approacheszero. of the corresponding hysteresis loops lack obvious vertical “jumps.” SeeFigures 8, 14, 19, 20, 21 and 22. itle Suppressed Due to Excessive Length 31
3. Figures 27, 30, 34 and 35 correspond to systems with two or three sta-ble isolated equilibria and depict bifurcation-type hysteresis . Their hystere-sis loops are distinguished by two “horizontal” portions, which representthe stable equilibria, and “jumps” between them. Simulations indicate thejumps converge towards the bifurcation values as the frequency approaches0.4. The requirement in definition 1 of stable equilibrium points is relaxed andreplaced with steady state solutions. In particular, examples of dynamicalsystems with steady state solutions (not equilibrium points) are shown toexhibit hysteresis. See Figures 9 and 10.5. Loops that appear to grow without bound as the frequency of the inputgoes to zero are excluded as hysteresis loops. See for example Figure 7.6. Figures 8, 14 and 22 depict rate-independent hysteresis loops. In all othercases, rate-dependent hysteresis loops are shown; however, for Figures 9,10, 19, 20, 27 30, 34 35 rate-independence of the hysteresis loops emergedfor small values of ω .In regards to observation 6, a continuation could be to show all rate de-pendent hysteresis loops are approximately rate independent as the frequencyof the input approaches to zero. This would explain why some definitions ofhysteresis require rate-independence [8,12,22,36]. In [19], the authors quantifyrate-dependence using the solution to the dynamical system α ( t ) and compar-ing it to the solution of the same system, β ( t ) , with a time transformation φ ( t ). For instance, let ε = || β ( t ) − α ( φ ( t )) || under the L -norm. If the errorfunction, ε , is approximately 0 as the frequency of the input approaches 0, thenthe dynamical system is defined as rate-independent at low frequencies. Thisapproach may be further developed to replace the second part of Definition 1.As well, the observations made about hysteresis loops here and in literaturemay in the future provide a way to predict or even standardize hysteresis loopshapes. This can also lead to future work on controller design to modify theshape of hysteresis loops by adjusting the types of equilibria that arise indynamical systems. Acknowledgements
This research was partially funded by a “York University FacultyAssociation Minor Research Grant”.
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