1 00:00:00,880 --> 00:00:10,710 This is a short lecture on potentials. So from the physical idea of potential energy, 2 00:00:10,710 --> 00:00:21,630 we outline another way to actually visualise the dynamics of x dot =f(x). So you imagine 3 00:00:21,630 --> 00:00:33,720 a particle sliding down the walls of a potential well, where the potential V(x) is defined 4 00:00:33,720 --> 00:00:49,530 by f(x)= -dv dx. A potential well is the region that is surrounding a local minimum of potential 5 00:00:49,530 --> 00:00:59,420 energy. So let us make a simple minded plot of V(x) versus x that is your potential well. 6 00:00:59,420 --> 00:01:07,390 We highlight the particle and the direction which it is moving. So imagine the particle 7 00:01:07,390 --> 00:01:18,420 actually moving through the walls of the potential, the negative sign in the definition of V actually 8 00:01:18,420 --> 00:01:29,259 comes from physics, essentially what this shows is that the particle always moves downwards. 9 00:01:29,259 --> 00:01:41,709 Let us go ahead and develop some intuition for ourselves, let x be a function of t and 10 00:01:41,709 --> 00:01:48,670 let us calculate the time derivative of V is the function of x(t). So, invoking the 11 00:01:48,670 --> 00:02:03,149 good old chain rule from calculus yields dv dt = dv dx times dx dt. So now x dot =f(x) 12 00:02:03,149 --> 00:02:14,410 = - dv dx and that is simply by the definition of the potential. Thus, dv dt = - dv dx whole 13 00:02:14,410 --> 00:02:18,490 square which will less than or equal to zero. 14 00:02:18,490 --> 00:02:28,890 So V of t decreases along the trajectories, that worth highlighting and thus the particle 15 00:02:28,890 --> 00:02:44,780 moves towards a lower potential. Now if the particle is at an equilibrium when dv dx = 0 16 00:02:44,780 --> 00:02:57,130 and so V is simply a constant. Now note that the local minima of V(x) gives us stable fixed 17 00:02:57,130 --> 00:03:08,310 points and the local maxima of V(x) gives us unstable fixed points. 18 00:03:08,310 --> 00:03:19,150 Let us consider an example, graph the potential for x dot = -x and identify all the equilibrium 19 00:03:19,150 --> 00:03:32,070 points. So, we need to find V(x) such that - dv dx = - x this gives us V(x) =1/2 x square 20 00:03:32,070 --> 00:03:47,680 + C1 where C1 is just an arbitrary constant. So for now let C1 be 0, now let us plot V(x) 21 00:03:47,680 --> 00:03:55,650 versus x, this plot of V(x) versus x is rather simple minded curve, which we can easily do 22 00:03:55,650 --> 00:04:00,883 by hand and were we go that is what the curve look like. 23 00:04:00,883 --> 00:04:13,950 The only equilibrium point occurs at x = 0 and it is stable. The analytic solution for 24 00:04:13,950 --> 00:04:24,700 x dot = -x is just x = C1 e to the -t, where C1 is a constant. 25 00:04:24,700 --> 00:04:35,120 Let us consider another example, graph the potential for the system x dot = x - x cubed 26 00:04:35,120 --> 00:04:46,100 and identify all equilibrium points. So, we set - dv dx = x - x cube and solving this 27 00:04:46,100 --> 00:05:00,790 we get V – 1/2 x square + 1/4 x to the 4 + C1. Let C1=0. Now let us make the plot of 28 00:05:00,790 --> 00:05:12,160 V(x) versus x, so the plot of V(x) versus x is little bit more involved highlight the 29 00:05:12,160 --> 00:05:14,660 local minima +1 and -1. 30 00:05:14,660 --> 00:05:24,521 So, the local minima is at x = plus minus 1 which implies stable equilibrium and the 31 00:05:24,521 --> 00:05:37,160 local maxima is at x = 0, which implies unstable equilibrium. The system is bistable as it 32 00:05:37,160 --> 00:05:49,449 has 2 stable equilibrium. So here is an exercise, can you find an analytical solution to x dot 33 00:05:49,449 --> 00:06:02,560 = x - x cubed. Let us consider another exercise, let x dot =f(x) be a vector field on the line 34 00:06:02,560 --> 00:06:14,290 and use the existence of a potential function V(x) to show that the solutions actually cannot 35 00:06:14,290 --> 00:06:15,290 oscillate. 36 00:06:15,290 --> 00:06:22,630 So, this second exercise is actually closely related to the lecture where we talked about 37 00:06:22,630 --> 00:06:30,280 the impossibility of oscillations of x dot =f(x). But here what I am saying is that can 38 00:06:30,280 --> 00:06:38,050 you use the existence of a potential function V(x) to actually show that solutions of x 39 00:06:38,050 --> 00:06:41,680 dot =f(x) cannot oscillate. 40 00:06:41,680 --> 00:06:48,590 Now, this was a very short lecture, the intent of the lecture was to introduce you to the 41 00:06:48,590 --> 00:06:54,910 notion of potentials and to highlight their ability to analyse equations of the form x 42 00:06:54,910 --> 00:07:01,250 dot =f(x). Now you look at the definition of a potential. So let us assume that we have 43 00:07:01,250 --> 00:07:12,300 potential function V(x) which is defined as f(x) = - dv dx, then evaluating that, relationship 44 00:07:12,300 --> 00:07:20,230 allowed us to say something about the original nonlinear system x dot =f(x). 45 00:07:20,230 --> 00:07:26,530 We offered one of two examples, but we left you with an interesting exercise that I suggest 46 00:07:26,530 --> 00:07:33,130 that you actually try which was roughly as follows, now can we actually use the notion 47 00:07:33,130 --> 00:07:43,320 of a potential as applied to an equation of the form x dot =f(x) and prove using this 48 00:07:43,320 --> 00:07:53,830 notion that the solutions of x dot =f(x) will actually not oscillate. They will actually 49 00:07:53,830 --> 00:07:57,550 not oscillate, this is something that we are talked about earlier in the lectures in terms 50 00:07:57,550 --> 00:08:03,550 of impossibilities of oscillations, but now can you use this notion of potential to make 51 00:08:03,550 --> 00:08:05,210 exactly the same point again.