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Very good morning, you know the course title
is numerical solutions of ordinary and
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partial differential equations. So, for the
first lecture is I would like to give little
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bit of
motivation with respect to some examples,
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so as you know numerical when we say
numerical solutions of ordinary and partial
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deferral equations. Yeah, many times we try
to take some examples and then try to solve
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analytically, but sometimes we may not get
analytical solutions.
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So, then we look for numerical solutions,
but then unless we have regress analysis of
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what kind of methods we work for, what kind
of problems we cannot really go for a trial
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and error. So, the main aim of this course
is to learn different methods numerically
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how
to solve ordinary and partial equations.
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.
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So, the first lecture is motivation. So, let
us see as I mentioned why do we require
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numerical solution? Definitely one does not
possess analytical solution in hand, so the
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simplest example is interpolation. So, what
is interpolation? Method of constructing new
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data points within the range of a discrete
set of known data points? So, for example,
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you
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.are given x and the corresponding function
and we would like to get let us say x in
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between say 2.5 or 4.7 or 6.8, so whatever.
.
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The aim of interpolation is approximate, what
is that interpolation? It supports a method
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of estimating the function at interned points,
so f of x will be approximated by a
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polynomial. Once you get an approximate polynomial,
you can get the function value at
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any point, so the given data points have been
approximated by a suitable polynomial. So,
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then where ever you want we substitute the
corresponding value and get the approximate
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value functional value at that point. Now,
more or less the story remains the same even
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in case of solutions of ordinary differential
equations. So, when I say the story remains
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the same, I mean definitely we also try to
approximate the given ordinary differential
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equations and get the approximate solution.
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..
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So, many problems come from real life applications,
they have to be handled numerically
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because as you know real life situations are
really complex. So, we have to handle them
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numerically, maybe there are few examples
which can be handled analytically. So, I
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would like to explain with some examples how
we arrive at an ordinary or partial
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differential equations, so then how we take
it further to solve numerically.
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.
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So, the first example I will consider is very
mean in physics, it is very simplest kind
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of
thing, it is conservation of mass. So, what
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do you mean by conservation of mass? So,
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.you take any flow quantity which we represent
by a vector field, so any flow quantity
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which we represent by vector field v conservation
of mass states that the divergence of
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the vector correspond vector is 0 provided
there are no sources and sinks. So, in case
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when you have source and sinks present, so
then the divergence v is equal to the
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corresponding source or sinks down.
So, I have written here divergence of v equals
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to q, so where v is the vector field, so
which represents let us say a flow velocity
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is flux of magnetic field etcetera and q is
the
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corresponding source or sinks down. It depends
on the special coordinates, so we are
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considering the study case, the simplest case.
Hence, this is the form, so when there is
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no
source and sinks q will be 0.
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.
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Now, let us consider another physical concept
that is, when a flow is irrotational, we
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represent generally curl of v is 0, as you
know curl of vector represent rotation. So,
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when
curl of v is 0, the flow is called irrotational,
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of course I am not describing the conditions
under which an irrotational field. It can
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be express as a gradient of scalar; there
are some
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restrictions on the domain, so I have not
coded those restrictions. Your velocity field
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v
bar can be expressed as gradient of a scalar
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where phi is a scalar.
Now, when phi is a scalar function this equation
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is this, so you substitute v equals to
grade phi in divergence v equals to q. So,
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then the left hand side v is Laplacean, so
Laplacean of phi is equals to q, so this is
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a PDE which is popularly known as position
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.equation. So, this is Poisson equation, so
when q is 0, so then the corresponding
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homogeneous equation is called Laplace equation.
So, this is the Laplace equation, now
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whether it is PDE for example, this particular
equation three whether it is an ODE or it
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is
a PDE depends on the coordinates.
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.
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For example, let us see the Poisson equation
is the Poisson equation is tau square phi
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equals q. Now, let us say we are talking about
three dimensional, complete three
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dimensional coordinate system x, y, z. So,
then your re proclaim becomes tau square phi
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by tau x square tau square phi by tau y square
tau square phi by tau z square equals to q.
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So, this is a PDE and what kind of PDE it
is a second order PDE right, so suppose your
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quantities depends only on one coordinates,
so that means any quantity f just depends
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on
say function of x.
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Then, your PDE reduces to simply d square
phi by d x square phi, so this is a PDE
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whereas, this is a ODE, so these are few examples
that means the same physical
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principles that is conservation of mass which
is a represented by Poisson equation here.
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Depending on the coordinate system for example,
when it depends on three coordinates,
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we arrive at a PDE when it depends just one
of the coordinates only then we arrive at
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ODE.
Now, if q is 0, then we have correspondingly
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the simplest first or second order ODE
where d square phi by d x square equals to
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0. So, this can be obviously solved
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.analytically, but depending on the complexity
of the physical situation, we arrive at some
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time corresponding equations which cannot
be solved analytically.
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.
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So, let us come back to the Laplace equation,
so this is a Laplace equation, now as I
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mentioned that the flow quantities depend
only on one of the coordinates. Then this
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reduces to an ordinary differential equation
as I mentioned, when I solve this analytically
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in simple situations, but sometimes it is
very difficult depending on the coordinates.
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Sometimes, depending on the geometry, the
analytical solution cannot be obtained, so
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then we have to look for numerical solutions.
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..
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So, the main aim of the course is to learn
numerical solution, now let us consider a
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second example, so the second example is have
considered which is very popularly
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known as Fourier’s law. So, this comes on
heat conduction problems, so what is that,
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let
us see, so Fourier’s law is law of heat
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conduction the time rate of heat transfer
through a
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material is proportional to a negative gradient
in that temperature and to the area. So, q
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bar vector field which is heat flux and this
is related to the heat temperature in this
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form.
So, this is k, the thermal conductivity of
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the medium and that is the temperature gradient,
so this is very popularly known as Fourier’s
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law now one dimension form.
.
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.So, one dimensional form, for example in
x direction we consider, so then q x is minus
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k
dt by x q x is minus k d t by x. So, for example,
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if we consider a rod of length l, then the
gradient can be represented like this T 2
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minus T 1, this should be suffix 1. So, T
2
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minus t 1 by l which is minus 1 over k q x,
now if you consider only this combination,
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T
2 is expressed like this. So, you just make
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a note of it, T t 2 T 1 minus l by k q x,
so we
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will come back to this very soon, so one can
define an initial value problem as follows,
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so same Fourier’s law d t by d x is some
constant time of T x.
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.
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So, I have considered the simple case one
dimension, so if it is more than one
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dimensions accordingly, then this would be
radiant and this also depends on more than
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one dimension. Then we need at any coordinate
what would be the initial temperature, so
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why do we require only at one point, see this
is as long as q. We consider q as only
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function of T, then d t by d x equals to something.
Assuming, this is a linear this is a first
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order ODE, now how many arbitrary constant
exist in a solution, it is 1 because it is
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a
first order ODE.
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Therefore, one requires initial condition
to solve this, so if we define an initial
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value
problem like this, then the star exactly suggest
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solution of a kind of more or less T 2 is
T
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1 plus, so what is this T 1 at a known x naught.
So, correspondingly T 2 at some other
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solution is given by this, so this is a kind
of solution of the initial value problem.
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So, that
means, if you generalize one would express
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T 3 is T 2. So, similarly, a general T n plus
1
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.is t n plus some incremental. So, this is
general motivation how an initial value problem
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may occur and then correspond what will be
the solution.
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.
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This is a kind of an approximation when we
say T 2 plus T 1excreta, because the
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gradient has been approximated by this. Now,
we are in a position to define IVP more
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formally, so let us look at few examples before
we define more formally.
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.
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So, for example, d y by d x equals x plus
y, so this is a first order ODE as I mentioned
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we need how many initial condition we need
one, why we need one because this is a first
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.order ODE. So, let us define your condition
as y of may be 1 equals some quantity, so
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this can be considered as initial value problem.
So, similarly, one can consider examples
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say minus x d y by d x plus x square y equals
say cos x. So, what kind of defines like
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ordinary definition is this second order ordinary
differential equation. Now, to solve it
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we need two conditions, so let us say y of
0 is 0 then y of 3 is 6.
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So, this is also initial value problem, so
we would like to consider such initial value
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problems and device numerical methods to handle
them. So, before we define more
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formally when we say numerical method how
we arrive at a numerical method what is a
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motivation I would like to explain with reference
to simple method, which is a kind of
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semi analytical. So, why do I say semi analytical
is not really numerical, but it is
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numerical, so it is definitely an interesting
because any numerical method we expect a
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kind of iteration.
So, what do you mean by iteration, you plug
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in something and then get an improved
version, then again you plug in that process
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and get improved version so on. So, for it
let
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us consider one such semi analytical method
which suggests what a numerical solution
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is, let us see that.
.
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So, I will explain with respect to an example,
so the example I consider is d y by d x is
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1
plus y square, so as I mentioned, this is
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a first order ODE. So, I would need one
condition, so I have taken these conditions
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very simple. So, this is initial value problem
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.and first order, now when I said semi numerical,
which means to some extent we proceed
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analytically and then switch over to numerical.
So, that is the motivation you integrate,
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say this is star, integrating star from x
naught to x, so x naught will be x d y by
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d x, d x
equals integral x naught to x 1 plus y square
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d x. so this is the step which is definitely
analytical.
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.
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Now, let us expand this, we integrate d y
by d x d x so that we get it as y of x minus
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y of
x naught equals integral x naught x 1 plus
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y square d x. So, if you perform integration
on
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left hand side we get this, so this can be
written as y of x equals y of x naught plus,
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so
now at this stage if you look at it the corresponding
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differently equation has been
reduced. An equation which involves an integral,
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hence you may call this is an integral
equation so that the next question is how
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do we solve it? Remember we have not yet
made use of the initial condition, now how
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do you solve it if you look at it right hand
side also contains why we are looking for
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y where as right side also contains y.
Therefore, how do we proceed further, so this
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is where I defined this method as a semi
numerical, because now we cannot proceed further
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in some analytically. Therefore, we
would like to bring in the numerical concept
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in some in the sense a kind of iterative
process, so in order to solve say this is
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a an iterative method. So, what is the iterative
method, we have our initial condition in hand
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which we have not yet made use of...
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.So, we would like to pick up that initial
condition and see left hand side, we are trying
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to
obtain so that should be our answer. So, whereas
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right hand side contains a same quantity
y, therefore what is a iteration as I mention
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whatever approximate in hand you put it in
your process and get slightly refined. So,
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here we have right hand side y, now we plug
in
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our initial value as a first approximation.
So, let y 0 of x equals y of x 0, so this
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is our
initial approximation, now for this problem
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y 0 is defined as 0, so let us see what will
happen.
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.
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So, y of x which is y of x 0 and x 0 is 0,
0 to x 1 plus y square will be now y 0
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approximation d x. Now, y of x 0 is 0 1 plus
and this is 0. So, this is simply x, so we
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get
a first approximation, therefore we denoted
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this, so we denote this y 1 x equals to x.
Now, we expect that this is a refinement over
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the initially get; now we would like to
make use of this and get the next approximation.
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..
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So, let us see how do we do it, so we defined
y 2 x as y x 0 plus 0 to x 1 plus y 1 square,
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so this is what we defined. Now, what is our
y 1, now we should plug in that the right
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hand side, so let us see y 2, x is y of x
0, x 1 plus y 1. So, this is again this is
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0, 1 plus y 1
is x, so this is 1 plus, so this is x cube
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by 3, so this is our y 2, this is our y 2,
now we
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would like to improve up on this.
.
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So, how do we do it, defined y 3 process
to x 1 plus y 2 square, so this is 0, 1 plus
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y 2 is
x plus, so this is 7 into 9 plus, so this
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what we get y 3. Now, what is story next step,
so
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.we got the expression, so why I said it is
a kind of semi numerical. It is looks as if
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an
analytical method because you are seeing only
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the expressions of r known numeric, but
what is the motivation behind it? You get
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some initial approximate in hand, you plug
in
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that improve and you get a refined version
plug in get a refined version.
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So, this is a kind of numerical where as the
expression in hand is looking like a
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analytical, so what we would like to do, now
suppose we need solution at a particular
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point. Since, we have the expression you definitely
plug in whatever we have obtain and
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00:29:18,960 --> 00:29:27,179
you get the solution, but how do we know that
this is really exact solution of the given
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00:29:27,179 --> 00:29:34,860
initial value of problem. So, there is an
issue, what is that issue we have done up
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00:29:34,860 --> 00:29:38,840
to three
terms that means we find and we have obtained
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00:29:38,840 --> 00:29:49,110
y 3, but one of your friend let us say thus
it ten times. So, then what do you expect
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00:29:49,110 --> 00:29:58,590
the number of terms will be more, so naturally
when you compute the value at a particular
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00:29:58,590 --> 00:30:02,690
point using whatever just now we have
obtained.
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00:30:02,690 --> 00:30:10,120
Let us say your friend used ten terms minus
ten refinements and then the person gets a
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00:30:10,120 --> 00:30:16,840
huge expression and get a value using that
expression, then if at all you have analytical
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00:30:16,840 --> 00:30:24,380
solution in hand and try to compare. So, the
solution applied by your friend appears to
215
00:30:24,380 --> 00:30:29,540
be
more closer to the analytical solution, why
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00:30:29,540 --> 00:30:40,010
is that because your friend used more number
of terms. So, that means the moral is whatever
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00:30:40,010 --> 00:30:46,940
we have obtained is not the complete
analytical, so there is a kind of numerical
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00:30:46,940 --> 00:30:57,020
process involved which is bringing in some
kind of error. So, I would like to use this
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00:30:57,020 --> 00:31:01,750
word for the first time, so why do we say
error,
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00:31:01,750 --> 00:31:08,990
it is obvious we have used only three terms.
So, then we got some value, but your friend
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00:31:08,990 --> 00:31:13,090
used ten terms and got little refinement,
so
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00:31:13,090 --> 00:31:18,910
that means you can compare, if use three terms,
the solution and then use ten terms.
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00:31:18,910 --> 00:31:29,679
There is some difference, so that is contribution
of the error if you stop after three terms,
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00:31:29,679 --> 00:31:40,700
so you can refine and then improve. Now, is
there any particular name given to this
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00:31:40,700 --> 00:32:01,300
method, of course yes, so it is popularly
known as Picard’s method of successive
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00:32:01,300 --> 00:32:15,980
approximation. Now, the next question comes
definitely who would ensure that the
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00:32:15,980 --> 00:32:21,630
method gives solutions which converge to the
exact solution.
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00:32:21,630 --> 00:32:22,630
..
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00:32:22,630 --> 00:32:49,049
So, that is the Picard’s existence and uniqueness
theorem, who would ensure the
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convergence solution that is Picard’s existence
and uniqueness theorem, so I am not
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00:32:55,400 --> 00:33:03,970
stating the theorem how are you can refer
the standard books given in this course. So,
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00:33:03,970 --> 00:33:10,520
you can refer to know what is Picard’s existence
and uniqueness theorem, so this
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00:33:10,520 --> 00:33:24,059
theorem ensures that the solution converges.
Now, what y 3 we obtained is x plus by 3
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00:33:24,059 --> 00:33:43,010
plus 2 x 5 by 15 plus x 7 by 63, so an intelligent
person would try to say can we close it
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like this appears to be a nice series. So,
can we get a close form expression at we
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00:33:49,250 --> 00:33:54,730
immediately conclude that so this could be
our exact solution.
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00:33:54,730 --> 00:34:09,319
Fortunately, the example which we have taken
it is supporting this is tan x, so situations
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00:34:09,319 --> 00:34:16,840
where the series supports a close form expression.
Then we will be tempted to say this is
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00:34:16,840 --> 00:34:25,389
a solution, however numerically 10 x 1, it
is expanded in any computer core. Definitely,
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00:34:25,389 --> 00:34:31,450
one has to use terms up to certain number
and then rest will be through that is where
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00:34:31,450 --> 00:34:34,840
I am
defining those chances ever.
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We will discuss in more detail about what
do you mean by error, now in situations where
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00:34:43,639 --> 00:34:52,451
we cannot get such close form obviously we
have to say that this is the solution. For
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example, now somebody would like to get at
6. So, we substitute in the corresponding
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00:35:02,580 --> 00:35:22,140
expression excreta, now are we in a position
to define general yes of course we are in
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00:35:22,140 --> 00:35:28,240
a
position to define the general process.
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00:35:28,240 --> 00:35:29,240
..
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00:35:29,240 --> 00:35:55,210
This term is always y of x 0, so this is any
f of x y, you consider what is this corresponds
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00:35:55,210 --> 00:36:18,540
to IVP y dash equals to f of x y with initial
condition. So, what is n there n is 1, 2 so
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00:36:18,540 --> 00:36:24,180
on.
So, this is the Picard’s method of successive
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00:36:24,180 --> 00:36:33,619
approximation and we ensure the
convergence under Picard’s existence and
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00:36:33,619 --> 00:36:35,750
uniqueness theorem.
.
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So, let us try another example, so let us
say your problem is y dash equals to y dash
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00:36:46,770 --> 00:37:04,260
equals to say 1 plus x y and let us say that
cautions y of 0 equals to 1. So, we defined
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00:37:04,260 --> 00:37:25,190
first approximation y of x is y of x naught
plus integral 0 to x, so y 1 of x would be
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00:37:25,190 --> 00:37:26,190
y of x
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00:37:26,190 --> 00:37:52,870
.0 is 1, 0 to x 1 plus x y 0 x d x 1 plus
0 to x 1 plus x. So, it our y 0 x, we have
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to consider
the initial condition as first approximation,
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00:37:57,190 --> 00:38:09,970
therefore this is 1, so it is so plus
so this is
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00:38:09,970 --> 00:38:25,600
our y 1. Now, we would improve up on this
what do we improve up on this, now y 2 is
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00:38:25,600 --> 00:38:41,839
y
0 1 plus x y 1 x.
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00:38:41,839 --> 00:38:42,839
.
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00:38:42,839 --> 00:39:01,619
So, this is 1 plus x, so just we have plained
y 1, so that is 1 plus x plus x square by
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00:39:01,619 --> 00:39:27,370
2, so
this is x cube by 3, so this is y 2 x. Now,
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00:39:27,370 --> 00:39:38,070
if one would get the simple substitutions,
now
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00:39:38,070 --> 00:39:56,020
the question is how long we continue, so we
can refine, so as I mentioned what are the
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conditions under which convergence. Then when
do you stop the conditions under which
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00:40:00,850 --> 00:40:10,450
is converges there are restrictions on f,
so I would request you to refer to the book
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00:40:10,450 --> 00:40:13,730
given
in the references for Picard’s existence
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and uniqueness. Now, with reference to this
example, I would like to say when we stop,
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so when do we stop if we look at this
contains five terms, whereas earlier one contains
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three terms. So, y 2 is this plus some
additional terms, so naturally if you further
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refine if you get some additional terms so
keeping this you when we stop.
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00:40:53,750 --> 00:40:54,750
..
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00:40:54,750 --> 00:41:08,300
You consider at some k plus 1 stage, then
you compare with the value the previous step
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if this is less than or equals to some excel
on where excel on is to be assigned. So, then
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00:41:25,599 --> 00:41:43,000
we conclude that
the method converges up to over excel on pre
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00:41:43,000 --> 00:41:54,731
assigned. So, for example
let us say mod of y 5 of 0.3 minus y 4 of
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00:41:54,731 --> 00:42:06,859
0.3 is less than or equals to
say for example,
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00:42:06,859 --> 00:42:14,880
now the question is whether we stop our alteration
here or we proceed. Further,
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00:42:14,880 --> 00:42:20,060
somebody expect this is not a fair enough
accuracy, what do you mean by this? The
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00:42:20,060 --> 00:42:24,880
difference between the solution obtained fourth
step and the solution obtained at fifth
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00:42:24,880 --> 00:42:29,380
step.
So, the difference suppose we are not happy,
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so then we have to proceed further and let
us say we proceeded further less than or equals
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00:42:42,420 --> 00:42:56,060
to say 0.35, that means the solution
obtains seventh step, eighth step. They are
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00:42:56,060 --> 00:43:04,960
agreed up to two decimal places, therefore
now it depends what is our requirement, so
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00:43:04,960 --> 00:43:09,830
if somebody expects that I need solution up
to two decimals.
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00:43:09,830 --> 00:43:19,010
So, then we stop suppose somebody says we
have proceed further, so definitely we will
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00:43:19,010 --> 00:43:27,420
compute further, so this is Picard’s method
of successive approximations. I hope you got
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00:43:27,420 --> 00:43:35,490
a clear idea of what is an iterative process
and then how do we stop at the particular
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00:43:35,490 --> 00:43:43,510
nitration level. So, that is sorely depends
on our requirement when I said it is on this
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00:43:43,510 --> 00:43:58,990
quantity, now having learnt the semi analytical
method we most proceed for more
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00:43:58,990 --> 00:44:08,810
numerical methods, so called numerical methods.
So, before we proceed further defined
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00:44:08,810 --> 00:44:17,030
.formally what is a initial value problem,
let us define more formally what is an initial
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00:44:17,030 --> 00:44:25,680
value problem, so the simplest is y dash equals
to f of x y.
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00:44:25,680 --> 00:44:26,680
.
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00:44:26,680 --> 00:44:46,119
So, here f can be linear, non-linear, so example
y dash equals to x plus y dash equals to x
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00:44:46,119 --> 00:44:58,950
y square plus sin x, as this kind as I mentioned
this is a first order equation, therefore
299
00:44:58,950 --> 00:45:12,040
we
defined more general. So, for we had been
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00:45:12,040 --> 00:45:26,160
discussing only first order IVP, so can we
generalize it, yes of course we generalized.
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00:45:26,160 --> 00:45:27,160
.
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00:45:27,160 --> 00:46:00,900
.For example, you defined x, y, y 1, y 2,
n x, so that means nth derivative IVP, it
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00:46:00,900 --> 00:46:08,240
is
expecting up to n minus 1, so right hand side
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00:46:08,240 --> 00:46:19,940
is processor which expect these are the
values. So, first consider y, y at x naught
305
00:46:19,940 --> 00:46:41,080
is given than y 1 at x naught still y 1 next
y 2
306
00:46:41,080 --> 00:47:04,230
still at x naught
this is like this. So, what is the remark,
307
00:47:04,230 --> 00:47:11,349
all the values are defined at only
one point, all the values if you see there
308
00:47:11,349 --> 00:47:25,369
been defined at only one point. So, this problem
is called initial value problem, the reason
309
00:47:25,369 --> 00:47:28,720
is all the values have been defined at only
one
310
00:47:28,720 --> 00:47:29,720
point.
.
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00:47:29,720 --> 00:47:40,510
So, I must explain this notation little bit,
so let us understand that the notation is
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00:47:40,510 --> 00:47:57,980
x n plus
1 is x n plus h x n plus h, so y of x n is
313
00:47:57,980 --> 00:48:11,570
y n y of x n plus h is y of x n plus 1. This
is y n
314
00:48:11,570 --> 00:48:34,040
plus 1, and then y dash the x n is y n prime,
therefore y 3 of x n is this is at x n. Now,
315
00:48:34,040 --> 00:48:41,500
when we solve a numerical solutions, when
we solve when we IVP numerically, we try
316
00:48:41,500 --> 00:48:48,790
to get the solution at particular points,
therefore more formal I will give you more
317
00:48:48,790 --> 00:48:52,650
formal
definition little later. So, we try to get
318
00:48:52,650 --> 00:49:02,099
at a particular point, therefore our accesses,
suppose this is x 0 like interpolation we
319
00:49:02,099 --> 00:49:12,600
divide into parts. So, this is x 1, this is
x 2, x 3
320
00:49:12,600 --> 00:49:32,930
and x n, now each this is our h step size
this is the step size.
321
00:49:32,930 --> 00:49:33,930
..
322
00:49:33,930 --> 00:50:03,220
So, for example we are given interval is 0
to 2 and say h is 0, 0.2, 0.4, 1, 1.2 so on,
323
00:50:03,220 --> 00:50:12,050
so
this is our x 0 x 1 x 2 x 3 excreta. So, in
324
00:50:12,050 --> 00:50:17,940
general if a, b is an interval, so then if
you make
325
00:50:17,940 --> 00:50:32,980
it into n points, so then b minus a by n is
h. So, with this modulation, we try to solve
326
00:50:32,980 --> 00:50:45,000
numerical methods, so I would like to define
more formulae IVP in my next lecture. So,
327
00:50:45,000 --> 00:50:54,040
let us see some more physical examples where
we come across ODE or p s. So, one is
328
00:50:54,040 --> 00:50:58,130
conservational mass and then law we have discussed.
.
329
00:50:58,130 --> 00:51:12,859
.Now, I would like to mention another example,
Fick’s law
330
00:51:12,859 --> 00:51:37,500
of diffusion, so this is, if v is
any flow quantity say velocity, then v dot
331
00:51:37,500 --> 00:51:47,119
grad c where c is the n mass transfer
concentration where c is the concentration
332
00:51:47,119 --> 00:51:58,589
and d is the diffusivity. So, v dot grad c
is d
333
00:51:58,589 --> 00:52:11,270
trance square c, so this is conservation of
mass, so for example if you consider in two
334
00:52:11,270 --> 00:52:26,150
dimensions, so say v bar is u v, then we have
u tau c by tau x plus v tau c by tau y equals
335
00:52:26,150 --> 00:52:49,170
d. So, this is PDE second order PDE for c,
however one must know v for given u and v
336
00:52:49,170 --> 00:52:54,640
this defines a second order PDE and one can
solve.
337
00:52:54,640 --> 00:53:03,660
So, we discuss various methods to solve these
second order methods as well, so first we
338
00:53:03,660 --> 00:53:11,820
concentrate on ordinary differential equations.
Then we switch over to partial differential
339
00:53:11,820 --> 00:53:18,170
equations, so we have learnt couple of physical
examples, conservation mass when
340
00:53:18,170 --> 00:53:29,170
conservation of heat conduction equation.
Then conservation of any concentration, so
341
00:53:29,170 --> 00:53:32,850
in
next lecture, we defined mode formulae what
342
00:53:32,850 --> 00:53:37,850
is an initial value problem and what is a
boundary value problem and then start with
343
00:53:37,850 --> 00:53:41,880
at least one or two methods.
Thank you.
344
00:53:41,880 --> 00:53:41,880
.