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Language: en
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This is a short lecture on potentials. So
from the physical idea of potential energy,
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we outline another way to actually visualise
the dynamics of x dot =f(x). So you imagine
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a particle sliding down the walls of a potential
well, where the potential V(x) is defined
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by f(x)= -dv dx. A potential well is the region
that is surrounding a local minimum of potential
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energy. So let us make a simple minded plot
of V(x) versus x that is your potential well.
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We highlight the particle and the direction
which it is moving. So imagine the particle
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actually moving through the walls of the potential,
the negative sign in the definition of V actually
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comes from physics, essentially what this
shows is that the particle always moves downwards.
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Let us go ahead and develop some intuition
for ourselves, let x be a function of t and
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let us calculate the time derivative of V
is the function of x(t). So, invoking the
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good old chain rule from calculus yields dv
dt = dv dx times dx dt. So now x dot =f(x)
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= - dv dx and that is simply by the definition
of the potential. Thus, dv dt = - dv dx whole
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square which will less than or equal to zero.
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So V of t decreases along the trajectories,
that worth highlighting and thus the particle
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moves towards a lower potential. Now if the
particle is at an equilibrium when dv dx = 0
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and so V is simply a constant. Now note that
the local minima of V(x) gives us stable fixed
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points and the local maxima of V(x) gives
us unstable fixed points.
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Let us consider an example, graph the potential
for x dot = -x and identify all the equilibrium
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points. So, we need to find V(x) such that
- dv dx = - x this gives us V(x) =1/2 x square
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+ C1 where C1 is just an arbitrary constant.
So for now let C1 be 0, now let us plot V(x)
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versus x, this plot of V(x) versus x is rather
simple minded curve, which we can easily do
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by hand and were we go that is what the curve
look like.
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The only equilibrium point occurs at x = 0
and it is stable. The analytic solution for
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x dot = -x is just x = C1 e to the -t, where
C1 is a constant.
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Let us consider another example, graph the
potential for the system x dot = x - x cubed
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and identify all equilibrium points. So, we
set - dv dx = x - x cube and solving this
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we get V – 1/2 x square + 1/4 x to the 4
+ C1. Let C1=0. Now let us make the plot of
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V(x) versus x, so the plot of V(x) versus
x is little bit more involved highlight the
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local minima +1 and -1.
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So, the local minima is at x = plus minus
1 which implies stable equilibrium and the
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local maxima is at x = 0, which implies unstable
equilibrium. The system is bistable as it
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has 2 stable equilibrium. So here is an exercise,
can you find an analytical solution to x dot
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= x - x cubed. Let us consider another exercise,
let x dot =f(x) be a vector field on the line
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and use the existence of a potential function
V(x) to show that the solutions actually cannot
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oscillate.
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So, this second exercise is actually closely
related to the lecture where we talked about
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the impossibility of oscillations of x dot
=f(x). But here what I am saying is that can
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you use the existence of a potential function
V(x) to actually show that solutions of x
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dot =f(x) cannot oscillate.
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Now, this was a very short lecture, the intent
of the lecture was to introduce you to the
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notion of potentials and to highlight their
ability to analyse equations of the form x
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dot =f(x). Now you look at the definition
of a potential. So let us assume that we have
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potential function V(x) which is defined as
f(x) = - dv dx, then evaluating that, relationship
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allowed us to say something about the original
nonlinear system x dot =f(x).
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We offered one of two examples, but we left
you with an interesting exercise that I suggest
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that you actually try which was roughly as
follows, now can we actually use the notion
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of a potential as applied to an equation of
the form x dot =f(x) and prove using this
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notion that the solutions of x dot =f(x) will
actually not oscillate. They will actually
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not oscillate, this is something that we are
talked about earlier in the lectures in terms
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of impossibilities of oscillations, but now
can you use this notion of potential to make
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exactly the same point again.