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Hello, in the last class we have reviewed
some exercises. I mean while solving exercises
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we have reviewed some of the single step methods.
So, let us continue to do that and we
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missed may be solving system of equation and
then... Of course, we have learnt R K
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methods, but only explicit. So, may be with
reference to some implicit nature. So, let
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us
try to attempt some problems.
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.
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So, this is tutorial two still continuing
single step methods. Solve for y of point
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3 point 6
when y x is the solution of
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second order equation. So, we need two mutual
conditions
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sorry. Choose h is point 3. So, solve for
y of point 3 and the y of point 6 when y x
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is the
solution of the second order equation this
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initial value problem and choose h point 3.
So,
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that means to compute y of point 3 with this.
So, x 0 is 0, so y of point 3 is y 1 y of
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point
6 is y 2.
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Now as we discussed so earlier so this is
a second order. So, what do we have to do
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convert it to couple system of fist order
equation and then try to adopt one of the
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methods. The story is not over, which method
we have to mention? Solve using runge -
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.kutta of order four right. If we solve runge
- kutta order four then lot of general idea
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we
may get.
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.
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So what was our equation? Y double minus x
y dashed plus y equals 0 and y of 0 is 1 and
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y dash of 0 is minus 1 right? Now we have
to convert this to system. Let y dash is z.
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So,
then this becomes this implies z dashed minus
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x z plus y equals to 0. So, therefore, we
have the following system y dash equals to
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z, z dash d is minus y plus x z, y of zero
is 1.
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And y dash of 0 becomes z of 0 is minus 1.
So, this is our system, so this our system
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that
we have to solve.
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Now since it is a remark x is independent
variable y and z dependent variables. So,
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therefore, when we use R K fourth order we
have to be careful and we have to define two
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different functions. So, how do we define?
So, the general with reference to general
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system y dash equals to f of x y z, which
is z it is z dash these g of x y z, which
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is minus
y plus x z. So, corresponding to y we define
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a specific functions and correspond z we
define different functions. So how do we do
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so how do we do?
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..
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y n plus 1 equals y n plus h by 6 k 1 plus
2 k 2 plus k 3 plus k 4 right? Then so we
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define. So, let us have simultaneously z n
plus 1 z n plus h by 6. So, we have to use
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a
different notation. So, let us use a different
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notation l 1 plus 2, l 2 plus l 3, l 4. So
what is
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k 1? K 1 is f of x, y, z and what is l 1?
G of x, y, z then k 2, therefore l 2 and k
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3. l 3, k 4,
l 4. So, these are the two different coupled
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so unless you compute k 1 one cannot
compute l 2. So, unless we compute l 1 one
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cannot compute k 2. So, they are coupled
right. So, let us proceed further.
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.
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.Now n is 0 so we have f of x, y, z is, f
of x, y, z is z. g of x, y, z is minus y plus
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x z so k 1
is. So, let us write down the given that as
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well y 0 is 0 z 0 is minus 1 x 0 is 0. So
z 0 is
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minus 1. And what is l 1? So, this is 0 0.
So, l 1 is y 0 sorry y zero is 1. So this
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will be 1
minus 1. So this is minus 1. Then k 2 is z
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0 plus h l 1 by 2 minus 1 plus l 1 is minus
1.
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So, this will be
this l 2, l 2 will be minus. So, this on we
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have defined l 2 is… so we have
a all three involved so the increments are
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h by 2, h k 1 by 2, h l 1 by 2. So from here
minus y 0 plus this plus x 0 plus h by 2 z
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0 plus h l 1 by 2. So this will be minus k
1 is
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minus 1 plus x 0 is 0, z 0 is minus 1, l 1
is minus 1. So this will be minus 1 minus
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so this
will be so this is our l 2.
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.
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Now k 3 so l 2 so this will be, then l 3,
k 2, l 2. So this will be, then k 4, l 3.
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So, this is
then l 4, k 3 value minus y 0. This is z 0
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so this is z 0. So, this can be simplified
and we
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get minus point.
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..
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So, having obtained these values we have
y 0, h by 6, then k 1 plus 2, k 2 plus 2,
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k 3 plus
k 4, so this l 1 plus 2, l 2 plus 2. So, this
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can be. So, y of point 6 left as an exercise
so
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similar method. So, you can try to do it.
So, with a with R K method we could solve
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system. So that we get both simultaneously
that is how to use R K method and
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simultaneously how to solve system of equations.
And if you observe the R K method,
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which we have discussed both second order
and then three stage and forth order etcetera.
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So, these are all explicit in fact tell a
series explicit method and the all that. So,
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there are
some methods which are implicit. So, what
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do you mean by implicit? So, we will discuss
in detail when we go to multi step methods,
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but just a brief what is implicit method and.
Then we quickly look into implicit Runge - Kutta
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method. So let us do that.
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..
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So, suppose a method is defined as y n plus
1 equals to y n minus 2 plus 3 f and plus
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1.
And this f n plus 1 is f of. So, that means
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we are asking for computing y n plus 1, but
right hand side is also asking. So such method
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is implicit right?
.
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So, let us look at it is a kind of problem
only, but in terms of implicit R K method
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derive
an implicit Runge - Kutta method of the form.
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So, you may observe why we are calling
this implicit. So, you have left hand side
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you have k 1, but right hand side as well
we
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have. So, derive an implicit Runge - Kutta
method of the form this
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that is second order.
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.So, how do you proceed? So, we in general
in explicit we have learnt we have expanded
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this and tell us series and then we have expanded
this in powers of h. And then we
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expanded the y n, y of x n plus 1 in tell
a series. And then match the coefficient of
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equal
parts of h.
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Similar sought of thing we need to do here,
but you may observe right hand side you
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have k 1. And left hand side we have k 1.
So that means if you keep on substituting
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k 1
this same expression recursively k 1 will
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be sitting on the right hand side. So, let
us see
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how do we proceed further? So, k 1 expanding
alpha h dou f by dou x plus beta k 1. So,
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the higer term etcetera. So, this is h f n
alpha h square plus h beta dou f by dou y
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k 1. So
k 1 is sitting. So now again we need to substitute
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k 1.
.
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And the other terms. So, therefore y n plus
1 is y n plus w 1 h f n plus. If you multiply
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these h square beta, h square beta h f n dou
y plus h q terms. So, plus so tell a series.
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Now comparing these two because the method
is second order we can pair up to h
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square. So, this approximate, this and for
f w 1 must be 1. Then for f x coefficient
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is half
there and here alpha w 1 is half. Then for
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here
w 1 is half. So, this implies alpha equals
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to beta equals to half.
So, therefore, y n plus 1 is y n plus h x
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n plus h by 2 y n plus k 1 by 2, where k 1
is. So it
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is a recursive because if you substitute we
keep on getting k 1 recursively k 1 there
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k 1.
Again if you substitute again you get k 1,
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but now the less up to second order by
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.comparing by we get the coefficient and then
this is explicit. Why it is explicit? You
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cannot recall. K 1 we are expecting to compute,
but it is given the right hand side. So,
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this is a very important observation in implicit
R K method.
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.
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So, let us solve some problems, but one remark
before we proceed is generally
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expanding in a tell a series is very difficult
because recursively how long we do it. So
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it
is. So, let us see through this example slightly.
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Different story with h equals to point 3
using
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second order implicit Runge - Kutta method
right. So just now we have derived by
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expansion now we would like to solve a problem.
So, this was our implicit R K method. For
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the given problem f of x y is this x 0 0 y
0 is 1
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right. Now let us try to compute k 1 h f of
this so therefore minus 2 x n plus h by 2
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y n
plus square. So, minus h 2 x n plus h, which
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is an implicit equation for k 1. And one may
use any iterative method because we have to
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solve for k 1. See this we have to solve the
problem if you keep on substituting the cursively
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so that is not end. The method has been
obtained in that fashion, but now we would
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like to compute the solution. So this is an
implicit equation. So we have to solved using
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any itorative method.
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..
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So to this extent let us define f of k 1 as
k 1 plus h. So, this is k 1, which is point
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3. Now
since it is any itorative method we try to
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use Newton - Rophson method we propose to
use Newton - Rophson method. So, therefore
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k 1 s plus 1 say l plus 1 minus. So let us
propose to have Newton - Rophson method. So,
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accordingly let us compute. Now we
need an initial guess. Assume k 1 0 equals
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h. So, this is f or f s minus 2 x 0 square
y 0
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square this is minus 2 point 3 x 0 square
0. So, k 1 is 0.
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.
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.Now let us compute f dashed. Say f dashed
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k 1 is 1 plus 2 times half. So, this is 1
plus
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0.3 2 x n 2, 2 get cancelled, Y n plus this.
Now so this will be x 0, y 0 so 1 plus k 1
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0, 0.
So, this equals, then f of k 1 0. This will
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be… so this will be.
.
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So, therefore... So, we have f of this is
f dashed of is. So, therefore k 1 1 is. So,
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this is
minus. So, this k 1 prime now we have to compute
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f of k 1 sorry not k 1 prime k 1 1.
Now we have to compute this so we may do using
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calculator and also we need this. And
hence k 1 2 we get. So, one may stop here
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or proceed until. So, we proceed until two
consecutive. The difference the difference
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was the two consecutive values is less then
x
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large some pre assigned we will proceed. So,
let us say here we stop. So, one we stop we
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get. So this is y 1 y 0 plus k 1. So, this
is the answer. So, this is how to sole using
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implicit R K method.
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..
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Now
we have solve lot of stability problems for
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the express it so let us do it for implicit.
Find the reason of stability for the implicit
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Runge - Kutta method given by this for the
I
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V P? Please make a note. So, this is a case
of f of x y equals. So, that is no explicit
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dependency on x. So, this is a kind of a special
case. Now for this we would like to have
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stability interval, stability analysis. So,
now let us proceed with our reference equation.
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.
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y dashed equals to lambda y this implies k
1 is h and k 2 is h. so you can see k 2 is
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implicit. So, k 1 lambda h y n, so since k
2 is explicit we try to collect the co efficient
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.and solve as an algebraic equation. So this
implies. So, this is kind of algebraic equation
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let.
.
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So, then we have k 2 equals. So, this our
k 2. So, therefore y n plus 1 is y n plus.
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So,
what was our method? Y n plus 1 by 4, k 1
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plus. So, this is y n plus 1 by 4. So, this
is our
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difference equation, is the difference equation
whose characteristic equation is. So, this
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is
a characteristic equation. So, now once we
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have the characterise equation. So, the
procedure is for absolute stability we have
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to put the condition of on the roots and then
try to determine. So, let us do that.
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..
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For absolute stability, where lamda is negative.
So, this implies minus 1, so this implies
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right. So, from from this we get, we get look
at it so if you bring it this side so we get
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h
bar. So, h bar by 6 h bar 6 plus h bar less
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than or equals to 0 and since lamda h s d
0 h or
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h bar is equal to the minus 6. So, this is
one side. So, then we have to see while the
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left.
So, this is the right inequality now we have
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to see the left.
.
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So, the left inequality. So this this gives
so minus 2. So, this gives
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and which is true for.
So, this holds and true for h bar greater
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than 2 minus 6, so therefore stability interval.
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.From the left we, from the right way got
h bar greater than equal to 6. And the left
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inequality this and this holds for this, so
the stability interval minus 6 to 0. So, for
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implicit, the problem with implicit is for
deriving a method we have use to tell a series.
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But then for a specific method we have to
solve an algebraic equation. Sometimes it
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is
nor linear and that we have solved using using
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Newton - Rephson method. Now let us
have some exercises for your for benefit of
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practice.
.
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Obtain solution of the system y dashed equals
to z. z dashed equals to minus 4 y minus 2
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z y of 0 is 1 z of 0 is 1 by Euler’s method.
Two Runge - Kutta fourth order
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using
step
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size h equals to 0.2. Obtain the solution
of the system using step size h is equal 0.2
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compute y of 0.2, y of 0.4.
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..
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Right two use implicit Runge - Kutta method
of second order to solve y dashed equals to
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x plus y y of point 4 is this with a step
size h is equals to 0.2. Compute y of 0.6.
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So,
These are some exercises for you you can try
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and then feel more confident on both
explicit and implicit R K methods. However
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as I mention we discuss in detail what is
an
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explicit method, when we proceed to multi
step methods. So, let us wait for the lectures
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on multi step methods to hear more on implicit
methods.
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Thank you until then bye.
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.