1 00:00:03,830 --> 00:00:08,990 When you put a cold drink on the kitchen counter the counter surface temperature will decrease. 2 00:00:08,990 --> 00:00:15,990 But, if the cold drink is removed, the counter will eventually return to room temperature. 3 00:00:16,710 --> 00:00:22,039 If instead we place a cup of tea on the counter, the counter temperature rises; but if we remove 4 00:00:22,039 --> 00:00:29,039 the cup of tea, the counter top eventually returns to room temperature. We say that the 5 00:00:29,109 --> 00:00:36,019 counter at room temperature is a stable equilibrium. In this video, we discuss the world from the 6 00:00:36,019 --> 00:00:42,440 perspective of equilibrium and stability, and in particular linear stability. 7 00:00:42,440 --> 00:00:48,030 This video is part of the Linearity video series. Many complex systems are modeled or 8 00:00:48,030 --> 00:00:53,540 approximated as linear because of the mathematical advantages. 9 00:00:53,540 --> 00:00:58,570 All the world is an initial value problem, and the matter merely state variables. However, 10 00:00:58,570 --> 00:01:04,260 and less poetically, there are alternative interpretations of physical, and indeed social, 11 00:01:04,260 --> 00:01:07,049 systems that can prove very enlightening. 12 00:01:07,049 --> 00:01:12,260 The purpose of this video is to introduce you to the framework of equilibrium and stability 13 00:01:12,260 --> 00:01:17,360 analysis. We hope this motivates you to study the topic in greater depth. 14 00:01:17,360 --> 00:01:22,310 To appreciate the material you should be familiar with elementary mechanics, ordinary differential 15 00:01:22,310 --> 00:01:29,310 equations, and eigenproblems. 16 00:01:30,890 --> 00:01:35,800 Let's look at the iced drink and hot teacup example from the perspective of equilibrium 17 00:01:35,800 --> 00:01:38,509 and stability. 18 00:01:38,509 --> 00:01:43,789 In this example, the governing partial differential equation is the heat equation shown here. 19 00:01:43,789 --> 00:01:49,320 At equilibrium, the solution of the governing equations is time-independent, that is, the 20 00:01:49,320 --> 00:01:52,070 partial time derivative is zero. 21 00:01:52,070 --> 00:01:57,310 This tells us that the del square temperature term must also be zero, which is only possible, 22 00:01:57,310 --> 00:02:03,690 given the boundary conditions, if the entire counter is at room temperature. 23 00:02:03,690 --> 00:02:08,149 Stability refers to the behavior of the system when perturbed from a particular equilibrium 24 00:02:08,149 --> 00:02:13,900 near the uniform counter temperature; a stable system returns to the equilibrium state; an 25 00:02:13,900 --> 00:02:19,390 unstable system departs from the equilibrium state. 26 00:02:19,390 --> 00:02:24,340 We say that this equilibrium is stable because if we perturb the temperature by increasing 27 00:02:24,340 --> 00:02:29,430 or decreasing the temperature slightly, it will return to room temperature, the equilibrium 28 00:02:29,430 --> 00:02:31,980 state, after enough time. 29 00:02:31,980 --> 00:02:37,440 Here we intuitively understand that the equilibrium is stable from our own experiences. But for 30 00:02:37,440 --> 00:02:42,480 other situations, we need mathematical methods for determining whether or not equilibria 31 00:02:42,480 --> 00:02:43,950 are stable. 32 00:02:43,950 --> 00:02:48,380 One way to determine if the equilibrium of this partial differential equation is stable 33 00:02:48,380 --> 00:02:51,020 is to apply an energy argument. 34 00:02:51,020 --> 00:02:55,240 Manipulation of this heat equation permits us to derive a relationship that describes 35 00:02:55,240 --> 00:02:59,810 how the "mean square departure" of the counter temperature from room temperature evolves 36 00:02:59,810 --> 00:03:02,290 over time, as shown here. 37 00:03:02,290 --> 00:03:07,599 Note u(x) is the deviation of the temperature in the counter from room temperature, Ω is 38 00:03:07,599 --> 00:03:14,599 the counter region, and Γ is the counter surface; d1 and d2 are positive constants—determined 39 00:03:14,709 --> 00:03:17,690 by the thermal properties of the counter. 40 00:03:17,690 --> 00:03:22,760 Because the right--hand side of this equation is negative, it drives the temperature fluctuations—the 41 00:03:22,760 --> 00:03:28,180 integral of u(x) squared over the counter region—to zero. 42 00:03:28,180 --> 00:03:35,180 This energy argument is the mathematical prediction of the behavior we observe physically. 43 00:03:39,120 --> 00:03:43,739 Linear stability theory refers to the case in which we limit our attention to initially 44 00:03:43,739 --> 00:03:50,410 small perturbations. This allows us to model the evolution with linear equation! Linearizing 45 00:03:50,410 --> 00:03:54,819 the governing equations has many mathematical advantages. 46 00:03:54,819 --> 00:03:59,700 Let's consider the following framework for linear stability analysis. 47 00:03:59,700 --> 00:04:05,870 choose a physical system of interest; develop a (typically nonlinear) mathematical 48 00:04:05,870 --> 00:04:10,819 model; identify equilibria; 49 00:04:10,819 --> 00:04:14,540 linearize the governing equations about these equilibria; 50 00:04:14,540 --> 00:04:21,540 convert the initial value problem to an eigenproblem; inspect the eigenvalues and associated eigenmodes 51 00:04:21,769 --> 00:04:23,860 (or eigenvectors) 52 00:04:23,860 --> 00:04:28,080 Let's see how to use this framework as we proceed through the example of the real, physical 53 00:04:28,080 --> 00:04:30,479 pendulum seen here. 54 00:04:30,479 --> 00:04:35,558 You see a large bob, the rod, and a flexural hinge, which is designed to reduce friction 55 00:04:35,558 --> 00:04:37,939 losses. 56 00:04:37,939 --> 00:04:42,990 Let's develop the mathematical model. We show here the simple pendulum consisting 57 00:04:42,990 --> 00:04:49,990 of a bob connected to a massless rod; we denote the (angular) position of the bob by theta(t), 58 00:04:50,360 --> 00:04:56,259 and the angular velocity of the bob by ω(t); g is the acceleration due to gravity; and 59 00:04:56,259 --> 00:05:00,180 L is the effective rod length for our simple pendulum. 60 00:05:00,180 --> 00:05:04,900 The effective length L is chosen such that the simple pendulum replicates the dynamics 61 00:05:04,900 --> 00:05:10,279 of the real, physical pendulum; L is calculated from the center of mass, the moment of inertia, 62 00:05:10,279 --> 00:05:13,789 and the mass of the compound pendulum. 63 00:05:13,789 --> 00:05:18,240 The dynamics may be expressed as a coupled system of ordinary differential equations 64 00:05:18,240 --> 00:05:23,219 that describe how the angular displacement and angular velocity evolve over time. 65 00:05:23,219 --> 00:05:30,080 These equations are nonlinear due to the presence of sin(θ) and the drag function fdrag(ω). 66 00:05:30,080 --> 00:05:32,409 The drag function is quite complicated. 67 00:05:32,409 --> 00:05:39,330 For large ω—it is equal to c|omega|omega, where c is a negative constant. 68 00:05:39,330 --> 00:05:43,270 But for very small angular velocities, near points on the trajectory where the pendulum 69 00:05:43,270 --> 00:05:48,669 isn't moving, or at least not moving fast, drag is given by to b omega, where b is a 70 00:05:48,669 --> 00:05:50,779 negative constant. 71 00:05:50,779 --> 00:05:57,150 Next, let's explore the validity of this model. Here you see a comparison between a numerical 72 00:05:57,150 --> 00:06:02,779 simulation, and an experiment, courtesy of Drs. Yano and Penn, respectively. 73 00:06:02,779 --> 00:06:08,550 The numerical simulation is created by calibrating the drag function to the experimental data. 74 00:06:08,550 --> 00:06:13,899 The agreement is quite good for both small and large initial displacement angles. 75 00:06:13,899 --> 00:06:19,689 However, because we are fitting the dissipation to the data, this comparison does not truly 76 00:06:19,689 --> 00:06:22,689 validate the mathematical model. 77 00:06:22,689 --> 00:06:27,419 To validate the mathematical model, we must focus on a system property that is largely 78 00:06:27,419 --> 00:06:33,499 independent of the here, small, dissipation, such as the natural frequency, or period, 79 00:06:33,499 --> 00:06:35,110 of the pendulum motion. 80 00:06:35,110 --> 00:06:40,020 And a comparison of the natural frequency of the physical pendulum to that predicted 81 00:06:40,020 --> 00:06:46,129 by the numerical model shows that the natural frequencies agree quite well, for any initial 82 00:06:46,129 --> 00:06:49,050 displacement angle, even large. 83 00:06:49,050 --> 00:06:55,058 We now look for equilibrium states, solutions that are independent of time. To find these 84 00:06:55,058 --> 00:07:00,779 equilibria we set the left--hand side of the equations to zero and solve for θ and ω. 85 00:07:00,779 --> 00:07:03,689 We can readily conclude that there are two equilibria: 86 00:07:03,689 --> 00:07:08,830 (θ, ω) = (0, 0), which we denote the "bottom"equilibrium; and 87 00:07:08,830 --> 00:07:13,110 (θ, ω) = (π, 0), which we denote the "top"equilibrium. 88 00:07:13,110 --> 00:07:17,960 The mathematical model actually has infinitely many equilibria corresponding to theta values 89 00:07:17,960 --> 00:07:20,099 that are integral multiples of pi. 90 00:07:20,099 --> 00:07:24,119 But for our stability analysis, two will suffice. 91 00:07:24,119 --> 00:07:29,469 Are these equilibrium solutions stable or unstable for the physical pendulum? Pause 92 00:07:29,469 --> 00:07:35,439 the video here and discuss. 93 00:07:35,439 --> 00:07:40,679 They are stable if a small nudge will result in commensurately small bob motion; They are 94 00:07:40,679 --> 00:07:46,580 unstable, if a small nudge will result in incommensurately large bob motion. So we can 95 00:07:46,580 --> 00:07:52,619 predict that the bottom equilibrium is stable; the top equilibrium, unstable. If our mathematical 96 00:07:52,619 --> 00:07:57,159 model is good, it should predict the same behavior as the physical system. 97 00:07:57,159 --> 00:08:01,869 But how do we mathematically analyze stability of our model? Let's work through the linear 98 00:08:01,869 --> 00:08:08,869 stability analysis framework for the bottom equilibrium, θ = 0 and ω = 0. First, we 99 00:08:09,459 --> 00:08:12,449 linearize the equations about the equilibrium. 100 00:08:12,449 --> 00:08:18,899 The linearized equations are only valid near the equilibrium, theta = 0 and omega =0, i.e. 101 00:08:18,899 --> 00:08:25,020 for small displacements, theta prime, with small angular velocities, omega prime. 102 00:08:25,020 --> 00:08:28,529 The first equation of our system is already linear. 103 00:08:28,529 --> 00:08:34,750 So we only need to worry about linearizing the sin(0+θ') term and the dissipation term 104 00:08:34,750 --> 00:08:39,549 of the second equation for small theta prime and omega prime. 105 00:08:39,549 --> 00:08:46,550 What is the linear approximation of sin(theta') about theta=0. Hint: use a Taylor series. 106 00:08:46,800 --> 00:08:53,080 Pause the video and write down your answer. 107 00:08:53,080 --> 00:08:59,060 If we assume that theta prime is small, we can approximate sin(θ') by theta' — the 108 00:08:59,060 --> 00:09:05,630 deviation of θ from 0— by ignoring the higher order terms in the Taylor series expansion 109 00:09:05,630 --> 00:09:09,020 of sin theta' about 0. 110 00:09:09,020 --> 00:09:13,450 Because we are linearizing near omega prime sufficiently close to zero, the dissipation 111 00:09:13,450 --> 00:09:20,330 term is asymptotic to b omega prime, as mentioned previously. We then substitute these expressions 112 00:09:20,330 --> 00:09:25,490 into our dynamical equations to obtain the, linear equations indicated. 113 00:09:25,490 --> 00:09:29,930 We must supply initial conditions, and the initial angle and angular velocity must be 114 00:09:29,930 --> 00:09:35,840 small in order for this linearized system of equations to be applicable. We now write 115 00:09:35,840 --> 00:09:41,080 the linear equations in matrix form, in order to prepare for the next step — formulation 116 00:09:41,080 --> 00:09:43,270 as an eigenproblem. 117 00:09:43,270 --> 00:09:49,520 To do this, we assume temporal behavior of the form eλt. This yields an eigenvalue problem 118 00:09:49,520 --> 00:09:50,840 for λ. 119 00:09:50,840 --> 00:09:55,630 Note that our matrix is 2×2 and hence there will be two eigenvalues and two associated 120 00:09:55,630 --> 00:09:58,690 eigenvectors, or eigenmodes. 121 00:09:58,690 --> 00:10:03,900 Once we obtain the eigenvalues and eigenvectors we may reconstruct the solution to the linearized 122 00:10:03,900 --> 00:10:10,690 equations, as shown here; the constants c1 and c2 are determined by the initial conditions. 123 00:10:10,690 --> 00:10:17,050 However, to determine stability, a simple inspection of the eigenvalues suffices: 124 00:10:17,050 --> 00:10:23,280 What happens to e to the lambda t when each of the eigenvalues has negative real part? 125 00:10:23,280 --> 00:10:29,430 Pause the video. 126 00:10:29,430 --> 00:10:36,430 We observe that eλt decays in this case, and the system is stable. 127 00:10:36,560 --> 00:10:43,250 What happens if any of the eigenvalues has positive real part? Pause the video. 128 00:10:43,250 --> 00:10:50,250 If an eigenvalue has positive real part — in which case eλt corresponds to exponential 129 00:10:52,680 --> 00:10:54,200 growth away from the equilibrium 130 00:10:54,200 --> 00:10:59,510 — the system is unstable; note that even one eigenvalue with a positive real part is 131 00:10:59,510 --> 00:11:05,840 sufficient to deem the system unstable, since sooner or later, no matter how small initially, 132 00:11:05,840 --> 00:11:08,140 the growing exponential term will dominate. 133 00:11:08,140 --> 00:11:13,630 Lastly, if the real part of the eigenvalue with largest real part is zero — in which 134 00:11:13,630 --> 00:11:19,900 case eλt is of constant magnitude — the system is marginally stable and requires further 135 00:11:19,900 --> 00:11:24,180 analysis. We now proceed for our particular system. 136 00:11:24,180 --> 00:11:28,240 For this problem it is easy to find the eigenvalues analytically. 137 00:11:28,240 --> 00:11:33,810 We observe that both λ1 and λ2 have a small negative real part - due to our negative damping 138 00:11:33,810 --> 00:11:34,840 coefficient, b. 139 00:11:34,840 --> 00:11:40,490 Thus, the system is stable as we predicted based on the physical pendulum. 140 00:11:40,490 --> 00:11:46,800 For our experimental system, b L over g is much less than 1. 141 00:11:46,800 --> 00:11:52,720 And thus the damping does indeed have little effect on the period of motion. If we recall 142 00:11:52,720 --> 00:11:58,280 the connection between the complex exponential and sine and cosine, we may conclude that 143 00:11:58,280 --> 00:12:05,150 the linearized system response is a very slowly decaying oscillation. This linear approximation 144 00:12:05,150 --> 00:12:09,790 to our governing equations can predict not just the stability of the system, but also 145 00:12:09,790 --> 00:12:15,500 because the system is stable, the evolution of the system—assuming of course small initial 146 00:12:15,500 --> 00:12:15,950 conditions 147 00:12:15,950 --> 00:12:22,040 Indeed, comparing our original non-linear numerical solution to the numerical solution 148 00:12:22,040 --> 00:12:26,600 obtained from the linear model, we find that the agreement between the two is very good 149 00:12:26,600 --> 00:12:29,470 for the low-amplitude case. 150 00:12:29,470 --> 00:12:33,860 As advertised, we obtain a solution to the linear model with which we can predict the 151 00:12:33,860 --> 00:12:39,680 motion resulting from small perturbations away from equilibrium. But as you see here, 152 00:12:39,680 --> 00:12:43,550 the accuracy of the linear theory is indeed limited to small perturbations 153 00:12:43,550 --> 00:12:48,380 — for even moderately larger angles, the linear model no longer adequately represents 154 00:12:48,380 --> 00:12:50,870 the behavior of the pendulum. 155 00:12:50,870 --> 00:12:55,820 But this is to be expected since the linear approximation of these governing equations 156 00:12:55,820 --> 00:13:01,750 is only valid when theta prime and omega prime are very close to zero. 157 00:13:01,750 --> 00:13:07,310 To Review We have thus linearized the governing equations 158 00:13:07,310 --> 00:13:12,370 for the pendulum near the equilibrium theta = 0, omega = 0. 159 00:13:12,370 --> 00:13:17,600 By solving an eigenvalue problem, we showed that the equilibrium was stable at this point. 160 00:13:17,600 --> 00:13:24,090 The linear equations even predict the behavior of the pendulum near this stable equilibrium. 161 00:13:24,090 --> 00:13:28,560 This video has focused primarily on the modal approach to stability analysis, which is the 162 00:13:28,560 --> 00:13:33,940 simplest and arguably the most relevant approach in many applications. However, there are other 163 00:13:33,940 --> 00:13:40,940 important approaches too, such as the "energy"approach, briefly mentioned at the outset of the video. 164 00:13:41,290 --> 00:13:48,290 We shall leave to you to analyze the case of the "top"equilibrium, which we predicted 165 00:13:48,560 --> 00:13:50,480 to be unstable. 166 00:13:50,480 --> 00:13:55,840 You'll find that there's an unstable mode AND a stable mode. The unstable mode will 167 00:13:55,840 --> 00:14:01,070 ultimately dominate, but the stable mode is still surprising. Our intuition suggests than 168 00:14:01,070 --> 00:14:07,800 any perturbation should grow. The stable mode requires a precise specification of both initial 169 00:14:07,800 --> 00:14:14,320 angular displacement AND initial angular velocity. This explains the apparent contradiction between 170 00:14:14,320 --> 00:14:18,070 the mathematics and our expectations.