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.