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Good morning, in the last class we have learnt
Range Kutta method for solving initial
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value problems, which is a single step method
and we have learnt second order method.
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Now, let us see how higher order Range Kutta
methods can be developed, and further we
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will see how to solve higher order equations
as well.
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.
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So, let us talk about, last time we talked
about second order, so now let us go 1 step
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ahead, that is third order. So, that is 3
stage Range Kutta method, so that means we
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are
using 3 slopes, so we define the method as
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follows
W 1 K 1 W 2 K 2 S 3 K 3, where K 1
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is f of x y K 2 a 2 h y plus a 2 1 h K 1 K
3 a 3 h y plus a 3 1 h K 1 plus a 3 2 h K
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2. So, I
would like to make remark here, sometimes
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in some books K 1, K 2, K 3 will be defined
with h in front, in which case this h will
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be taken out. So, that means if you multiply
h
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inside so h K 1 will be new K 1, h K 2 will
be new K 2, h K 3 will be new K 3.
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Now, how do we determine the coefficients
W 1, W 2, W 3, a 2, a 3, a 2 1, a 3 1, a 3
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2 so
total how many 8 so one has to compare with
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the Taylor series method and then try to
determine the coefficient. Since we are using
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three stage method, we expect that we
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.should equate terms of 2 h cube. So, for
the second order method I have shown you the
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detailed calculation, so I expect that if
1 would follow those detailed calculations,
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then
we arrive at a system. Now, we have to make
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some arbitrary choice, so I make the
following choice say a 2 1 is a 2, a 3 1 is
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a 3 minus a 3 2, so that means 1 2 3.
.
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So, with this we obtain system of equations
as follows, so the system of equations the
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weight coefficients they satisfy this. So,
I repeat again by choosing the above arbitrary
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coefficients, we get system and that system
has been reduced using this arbitrary, then
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this will be the reduced system so this should
be an exercise for you to obtain this
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system. So, once we get the system what are
the standard methods? One can have
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arbitrary choice like in second order method,
we had different choices for example, alpha
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is something and beta is something half and
half, so things like that.
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..
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So, one can have arbitrary try choices, but
unless you show that the method is really
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sensible, we do not make any arbitrary try
choice. So, there are couple of standards
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methods, so what are they? So, the first one
is Heun’s method, we make the following
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choice and a 2 is one-third a 3 is two-third.
So, you may see this is a 2, we made a choice
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then a 3 a 3 2, so that will determine a 2
1 a 3 1 and in turn W 1 W 2 W 3, so this is
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the
set.
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So, this is the following method, since W
2 is 0 K 2 goes off, but then we need K 2
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to
compute K 3. So, this is K 2 and K 3, this
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is Heun’s method. So, we can see define
K 2
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we are using K 1 to define K 3 we should have
used K 1 K 2, but since W 2 is 0, so K 2
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does not explicitly appear.
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..
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So, this would have been K 1 K 2, but we have
other choice as well, so that is more
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standard R- K 3, so in short notation I am
writing. So, for what is the choice, so this
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is
called standard and this yields the following
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method
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correspondingly K 1. So, a 2 h a 2 is
half therefore, half h K 3 observe the way
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we have defined that let to this kind of.
So,
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this is the first time we have coming across
minus sign in this, this is because we made
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an arbitrary choice.
So, what was our arbitrary choice, our arbitrary
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try choice was a 3 1 equals a 3 minus a 3
2, looking at these values a 3 1 is a 3 minus
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a 3 2, so this is 1 minus 2, which is minus
1
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therefore, this we get a minus sign. So, how
did we arrive at these methods, we have
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defined an approximation then we can Telesis
expansion of the exact and then we
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consider the h expansion of the approximate
formula, and try to match the terms and
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since we are using 3 slopes, so we have gone
up to that. So, how to obtain the
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coefficients, I left it as an exercise, so
I am sure making the second order method,
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one
can obtain the system.
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..
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Now, let us proceed for the fourth order method.
So, this is a fourth order R K method,
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so we define K 1, K 2, K 3, for K 3 K 1 and
K 2 are used K 4, a 4 1, a 4 2, a 4 3. So,
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this
is our definition, so this is aligned with
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the general method with l slopes. Now, again
our
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duty is to compute the arbitrary coefficients
up to some accuracy. So how do we do?
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Again we do this expand Taylor series and
h expansion and compare. So, again I am
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living this exercise, so when we do this we
obtain the following, it is a lengthy system
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then getting nasty as b plus c.
.
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.So, really very lengthy, but it is worth
understanding because fourth order R K method
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is
very popular and most of the calculations
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are done using fourth order R K method. And
we have an additional condition, which we
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get this standard condition sigma W i is 1,
this is also one of the equations out of that
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and further we have more equations. So, we
I
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will try to write it here, so it is a very
big, so this is we get and again one has to
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look for
the arbitrary coefficients and then try to
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get specific method.
.
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So, what are the standard choosing a 2 equals
to a 3 equals to half we get, then W 1 W 4
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once a 4 1 a 4 2, so this will define the
method as follows. So, this is the coefficients
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and
then the method
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is our fourth order R K method. Of course,
the algebra is very tedious
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and we have done second order, so if we try
to follow, we get the corresponding third
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order and fourth order methods.
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..
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Now, we should try to see what happens to
the error in each case because we did not
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talk
it, we will talk about this little later.
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So, before we go further, let us try to solve
problem.
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Let us say calculate y f 0.1 that will y f
0.2 because h is 0.2 y f 0.4, see what is
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our so
what is x 0, so this will be h 1, so essentially
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this is y 1, this is y 2. So, let us try to
compute, so f is minus 2 x y square, so we
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should try to compute K 1. Obviously, our
n
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is 0 we are trying to compute y 1 therefore
n is 0, so the given from here y 0 is 1, so
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what
is this K 1 is minus 2, x 0 is 0 and y 0 is
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1 square.
So, this is 0 K 2 is minus 2 x 0 plus h by
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2 y 0 plus h K 1 by 2 square because K is
defined as x n plus h by 2 comma y n plus
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h K 1 by 2, so this is what we get. So, this
is
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equals minus 2 x 0 is 0 h is 0.2 so we get
0.1 there and y 0 is 1 and K 1 is 0, so this
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is 0,
so we get 1 there, so this will be minus 0.2,
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then K 3 minus 2 x 0 plus h by 2 y 0 plus
h K
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2 by 2 square so minus 2, this already we
computed 0.1 and y 0 1 plus h by 2 K, so K
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3.
So, this will be
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minus 0.2 1 minus, so this will be 0.1, so
this will be 1 minus 0.02, so
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this is 0.2 times 0.98.
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..
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So, K 3 we have the value, so this is our
K 3 then K 4, so K 4 is defined as minus 2
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x 0
plus h y 0 plus h K 3 square, so K 4 x n plus
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h, h should be there. So, this is minus 2
x 0
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is 0 h is 0.2 y 0 is a 1 h is 0.2 and K 3
is minus 0.096 square. Make a pardon; this
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is
square missing here, so this value is incorrect,
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so we get a new value. So, we can use it
accordingly, this is incorrect, so we get
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K 3 is some x then K 4 is some y, so then
we get
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y 1 equals to y of 0.2 equals to y 0 that
is 1 plus h is 0.2 times K 1, K 1 is 0 2 K
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2 plus 2
K 3, K 3 was x plus K 4 some y, so we get
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the answer.
Make a pardon; we have to simplify and get
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x and y, this is just to explain the method
so
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that is how we compute the solution. Now,
having learnt third order and fourth order
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R K
method, we can try some more examples, but
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this is higher order method. So, far we
have solved an only first order initial value
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problem, which is y dash equals to f of x
y y f
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x z equals to y 0. Now, let us go 1 step ahead.
So, can we use these methods learnt to
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solve higher order initial value problems,
so what are the methods we have learnt?
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..
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Let us says Euler method, Taylor series method
and R K methods, so can we use these
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methods to solve higher order equations. So,
let us try to do that, so higher order
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equations, we have
first order I v p y dash equals to f of x
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y and y of x 0 is y 0, then
second order I v p, so I will try to write
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standard method y double plus p x y dash q
x y is
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some r of x. And since this is second order
to define second order initial value problems,
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so this is our initial condition, but here
we need two conditions because this is a second
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order equation y of x 0 is y 0 y dash of x
0 is y 1.
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So, how do we solve this? And we have learnt
how to solve first order initial value
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problem, with that knowledge can we solve
second order I v p. So, the answer is yes,
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certainly we can. So, how do we do this? So,
we tried to do this by reducing the given
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second order 2, a couple systems of first
order equations, so how do we do it? Let us
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consider an example.
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..
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Suppose, here the equation is y double plus
2 x y dash equals to 0 and y of 0 is 2 y dash
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0
is minus 1, so this is our star. Let us say
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now can we reduce this to couple system of
equations? Yes of course let y dash equals
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to z then y double will be of course, x is
our
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independent variable, y is dependent variable.
Now, we have defined new dependent
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variable y dash equals to z and hence y double
is z prime. Therefore, using this notation
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star reduces to y double becomes z and y double
becomes z prime and y dash becomes z
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therefore, we obtain a couple systems as follows.
Now the dependent variables are y and z, so
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we need the corresponding conditions, we
have y 0 is 2 and y dash of 0 is minus 1,
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but our y dash is z therefore, this will be
supported by z of 0 is minus 1. So, our given
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second order initial value problem has been
reduced to couple first order system and where
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is the coupling? You can see y dash is z
and z dash is related, so this involves y
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so that means one cannot solve explicitly
this
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unless we solve this and vice versa. So, given
system independently one can solve or
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given second order by reducing to couple first
order one can solve. So, let us try to learn
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how to solve system using Eulers method and
if possible R K method.
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..
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So, solving a system of equations using Euler’s
method, Consider the following system,
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so here x is independent variable, y and z
are the dependant variables supported by the
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following initial conditions. Say h is 0.2,
so x 0 is 0, y 0 is 1, z 0 is minus 1 h is
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0.2, so
these are all parameters. Now, we have to
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define Euler’s method for this system, so
if
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you recall we have y n plus 1 equals to y
n plus h times f, this was our Euler. So,
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00:33:48,090 --> 00:33:51,649
similar
thing for this system, we have to write, but
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treating y dash equals to f of x y z, z dash
equals to g of x y z and y of x 0 is y 0,
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z of x 0 is z 0. So, this is a general system,
so we
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know f and g, for this we have to define Euler
method, so how do we do?
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.
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00:34:36,700 --> 00:35:05,670
.So, we define y n plus h
and for Z g, now our h was 0.2, so for the
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given example f is
this, g is this. So, let us write down our
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f is x plus y z, g is all that we need is
this. Now,
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let to compute y 1, y 1 equals to y of 0.2
equals to y 0, so let us repeat all that x
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00:35:46,440 --> 00:35:59,570
0 is 0, y
0 is 1, z 0 is minus 1 h 0.2. So, y 0 is 1
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h 0.2 and we have compute f of x 0 plus y
0 z 0 so
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that is x 0 is 0, y 0 is 1, z 0 is minus 1.
So this is 1 minus 0.2, this is 0.8 then z
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00:36:36,240 --> 00:36:52,350
1, which is
z of 0.2 is z 0 h, so we have to have g of
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x 0, y 0, z 0 so g is y 0 so 1 x 0 is 0 z
0 is minus
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00:37:02,340 --> 00:37:15,530
1. So, this whole thing is this is 0, so this
is minus 0.8, so that means we got y 1 and
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z 1.
.
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Now, let us proceed one step
y 2, 0.4 is y 1, so y 1 just computed, so
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let us have the data
here, so y 1 that was computed
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and x 1 we need
y 1 plus h times f of x 1, y 1, z 1. So, y
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00:38:05,380 --> 00:38:15,530
1
plus h times, what was our f, this x 1 plus
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00:38:15,530 --> 00:38:55,260
y 1, z 1, so x 1, y 1, z 1 this recalls. So,
we can
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00:38:55,260 --> 00:39:18,380
simplify we get, and z 2 z of 0.4, so z 1
that is minus 0.8 h then g of x 1, y 1, z
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00:39:18,380 --> 00:39:56,900
1 that is
our g was this, so y 1 plus x 1 z 1. So, we
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get y 2 and z 2, so this is Euler method,
so one
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can also try R K method. So, let us try to
use may be points method or may be R K
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fourth order, we can try. So, another example
because there is a slight change involved
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in
solving system for R K method, so what is
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that trick, we will see.
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..
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For example we talk about Heun’s method,
so what was the method y n plus 1 is y n plus
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00:41:00,210 --> 00:41:45,060
h by 4 K 1 is f of x y K 2 is x plus y plus
Heun’s method, but we have a system y dashed
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is x y z, z dashed is g of x y z. So, that
means we need to define similar set one set
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00:41:57,120 --> 00:41:59,770
for
this, one set for this.
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.
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So, how do we do it, so we define y n plus
1 is y n plus h by 4 K 1 plus and K 1 is f
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of x
y z. So, before we write for z, z n plus h
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00:42:37,370 --> 00:42:49,490
by 4 some other notation and l 1 is g of x
y z and
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K 2 f of x plus 1 by 3 h, y plus 1 by 3 h
K 1 and z plus 1 by 3 hl 1. So, this is the
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00:43:09,530 --> 00:43:16,640
.difference while defining R K method coefficients,
this is the difference because this is 2
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00:43:16,640 --> 00:43:22,580
dependant variables, so for y we are computing
K 1, K 2, K 3 and for z we are
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computing l 1, l 2, l 3.
Therefore, in z the increment will be in the
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terms of l's and l 2 g of y plus 1 by 3 h
K 1 z
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00:43:41,620 --> 00:44:10,960
plus 1 by 3 h l 1 similarly, K 3 is f of x
plus 2 by 3 h, y plus 2 by 3 K 2 and h h l
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00:44:10,960 --> 00:44:35,870
2 and l 3
2 by 3 h K 2 z plus 2 by 3 h l 2. So, you
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try to understand this because this is very
important while solving system, since we have
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two dependent variables y and z. And
since the system is defined as follows y dash
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is f of x y z, z dash is g of x y z, where
x is
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independent variable and y and z are dependent
variables. We are defining one set for y
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and another set for z and this is the set.
So, you can see unless we compute one, we
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cannot compute the other. So, this time instead
of taking a system, let us take y double
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plus with some conditions.
.
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So, let y dashed is z then this becomes z
dashed equals x y minus 2 z from here, and
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y 1
is 1 z 1 is minus 1 and say h is 0.6. So,
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we have to compute say K 1, K 1 equals 2,
so
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before we compute, we can write what is our
f, f is simply z and g is x y minus 2 z. So,
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K
1 is now n equals 0, K 1 is z 0, which is
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00:46:30,040 --> 00:46:44,170
we can write x 0 is 1, y 0 is 1, z 0 is minus
1. So,
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this is minus 1 then l 1 we do is parallel.
Why we have to do this because to compute
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00:46:57,210 --> 00:47:04,490
K
2, you need K 2 if it is function of z, you
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00:47:04,490 --> 00:47:19,900
need l 1. So, unless you compute l 1, you
cannot compute K 2, l 1 is g of x y z, g of
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00:47:19,900 --> 00:47:22,910
x 0 y 0 z 0.
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00:47:22,910 --> 00:47:34,510
.So, our g was this therefore, x 0 y 0, x
0 is 1, y 0 is 1 minus 2, z 0 is minus 1.
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00:47:34,510 --> 00:47:59,200
So, this is 3
then K 2, K 2 is defined as f of x y z dependency
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like this, but for the given case, since f
is only z, so we get z plus 1 by 3. So, z
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00:48:09,370 --> 00:48:17,600
0 plus 1 by 3 h l 1, so z 0 is minus 1 and
h is 0.6
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00:48:17,600 --> 00:48:37,510
by 3 and l 1 is 3. So, this is minus 0.4 then
l 2 is g of corresponding increments. What
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00:48:37,510 --> 00:48:41,800
are
the increments, x plus one-third h y plus
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00:48:41,800 --> 00:48:46,270
one-third h K 1 and z plus one-third h l 1,
so we
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00:48:46,270 --> 00:49:03,820
have to have x 0 plus one-third h y 0 plus
one-third h K 1 minus 2 times z 0 plus onethird
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00:49:03,820 --> 00:49:20,610
h l 1.
So, this is l 2, we can simplify x 0 is 1
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00:49:20,610 --> 00:49:26,890
and one-third h, h is this so one-third is
0.2, so 1.2
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00:49:26,890 --> 00:49:39,110
then y 0 is 1 and one-third h is 0.2 and K
1 is minus 1. So, 1 minus 1 minus one-third
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00:49:39,110 --> 00:49:45,220
h is
0.2, so let us write one-third h is 0.2 and
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K 1 is minus 1, so minus 0.2 minus 2 times
z 0
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00:49:51,990 --> 00:50:07,980
is minus 1, one-third h is 0.2 l 1 is 3. So,
this can be simplified 1.2 times 0.8 there
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00:50:07,980 --> 00:50:19,200
minus
2 times so this becomes plus 0.4, so you get
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00:50:19,200 --> 00:50:45,590
0.96, so this will be 6 so this is l 2.
.
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00:50:45,590 --> 00:50:59,120
So, now having completed l 2, the next step
is K 3. So, K 3 is z 0 plus two-third h l
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00:50:59,120 --> 00:51:08,740
2 so
that is z 0 is minus 1 plus two-third h l
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00:51:08,740 --> 00:51:49,300
2 is 1.76. So, this minus 1 plus l 3, l 3
is little
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00:51:49,300 --> 00:52:10,240
lengthy x 0 plus two-third h y 0 plus two-third
h K 2 minus 2 times z 0 plus two-third h l
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00:52:10,240 --> 00:52:21,490
2, so we can get the value. So, for example
having computed K 3, we get y of 1.6 equals
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00:52:21,490 --> 00:52:37,730
to this is our y 1 and the formula defined
is y 1 y 0 plus h by 4 K 1 plus 3 K 3. So,
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00:52:37,730 --> 00:52:46,160
y 0 is
1 h 0.6 by 4 and K 1 minus 1 plus 3 K 3 plus
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00:52:46,160 --> 00:53:20,210
3, so this
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00:53:20,210 --> 00:53:30,110
is minus sign. So, we simplified
we get the answer.
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00:53:30,110 --> 00:53:42,250
.So, this is the r K method for solving system
of equation, the important of R K is how to
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00:53:42,250 --> 00:53:52,310
define the coupled system like this, the coefficients.
This is very important in R K
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00:53:52,310 --> 00:53:59,990
method, so you can see here two dependant
variables y and z. Therefore, one system for
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00:53:59,990 --> 00:54:04,130
one dependant variable, other system for other
dependant variable, but they are coupled
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00:54:04,130 --> 00:54:12,810
and the coupling forces that you compute K
1 then l 1 will be computed, unless l 1 is
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00:54:12,810 --> 00:54:18,420
computed K 2 cannot be computed, unless l
2 is computed K 3 cannot be computed.
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00:54:18,420 --> 00:54:28,950
So, in this sense you may try the R K fourth
order for a system of equation, which is
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00:54:28,950 --> 00:54:40,930
definitely a little bit of algebra is involved.
Now, in the next lecture we try to learn for
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00:54:40,930 --> 00:54:48,370
example, somebody gives this is an approximate
method say R K method only, but how
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00:54:48,370 --> 00:54:54,080
do you determine what will be the error corresponding
to this approximation, so that is
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00:54:54,080 --> 00:55:00,960
an important task. So, we should try to estimate
what should be the order for a given
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00:55:00,960 --> 00:55:08,170
approximate method. So, then we try to learn
little bit of as I mentioned for computing
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00:55:08,170 --> 00:55:10,150
y
1, some error is involved.
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00:55:10,150 --> 00:55:15,780
Now, we are using y 1 to compute y 2 so the
error will be propagated. Now, if small
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00:55:15,780 --> 00:55:21,670
disturbances are creating a small error at
early stage creates large disturbances then
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00:55:21,670 --> 00:55:26,800
definitely the system is not good for us because
such an approximation is not good, that
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00:55:26,800 --> 00:55:32,770
means this is the concept of stability. So,
we try to discuss all this things in the next
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00:55:32,770 --> 00:55:33,770
lecture.
Thank you.
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.