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Hello, good morning I hope you are comfortable
by now with two point boundary value
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problems. And we have discussed a finite difference
methods and how to solve two point
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boundary value problems using finite difference
methods. So, but there are other certain
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methods for two point boundary value problems,
you see we started with initial value
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problems, and then we have come to boundary
value problems. Now, having the
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knowledge of initial value of problems can
we use it to solve boundary value problems
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as well.
So, this is slightly a different technique
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compare to finite difference methods called
shooting method. So, as the word suggests
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a sometimes to reach the target, sometimes
we may under shoot, and sometimes we may over
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shoot, in the sense we may reach here
above the target, sometimes below the target.
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So, the main idea here is with what angle
we should fire, so that we reach the target,
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this is a kind of a intuition. So, let us
a start
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discussing shooting method.
.
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.Shooting methods for boundary value problems,
so let us start with some physics. For
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example, when we have autonomous systems say
d x by d t equals to some f of x y t and
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d y by d t is g of x y t. So, this is called
autonomous system. For example, the book by
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G
F Siemens and ODE discusses about this. So,
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accordingly we must have x of some initial
condition t naught is x naught y of t naught
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is y naught. So, that means we are looking
for x of t y of t at any given time, so this
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is our solution right now what a shooting
method does. So, we are trying to solve for
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so this is our let us say target, so then
what is
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the corresponding initial angle. So, that
we hit the target so let us switch over to
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boundary value problems.
.
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So, in the context of boundary value problem
shooting, so
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we have say y double dash
plus y dash y is r of x then we have the boundary
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conditions. Now, what we are trying to
do is convert, so this is b v p now we know
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how to solve IVP. So, can we convert star
convert star as a initial value problem, yes
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it is possible how do we supply
initial
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condition, why we have to supply initial condition
on y dash because already we have y
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of a, right?
So, if you supply a initial condition y dash
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then this boundary value problem can be
converted to initial value problem. So, how
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do so y double plus p y prime q y is r and
y
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of a is gamma 1 and y of y dashed of a is
some alpha. Now, this is our IVP now what
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is
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.our idea suppose if you solve we can get
the solution at any grid point. So, what is
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the
idea?
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.
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The idea is solve IVP to get the solution
at the boundary point y of b, now what may
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happen so we know the curve let us assume
the end points. So, this is y of a this is
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y of b,
now we have started with some y dashed of
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a equals to alpha let us say with this we
hit
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here. That means, something it raised and
then we reach here so that means this is y
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at b
with alpha. Suppose we slightly at just the
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slope, suppose next time we reach here so
this
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is y of at b and let us say here it is y dashed
of a equals to some delta. So, this is at
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delta,
so what is happening?
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If you adjust your slope you are tracing a
different path, but what is our target so
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this is y
of b, so this is y of b. So, this a and this
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is b so this is y of b, so we need to guess
or
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rather we need to find out what could be the
slope with which if you fire, we can reach
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the target. And how do we reach the target?
As close as possible, so that means the
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problem reduces to finding suitable slope,
so this is the target. So, is that clear with
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a
shooting method now let us formulize.
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..
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So, we have
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so these are boundary value problem now B
v p. As a system y dash is z z
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dashed is r minus p z minus q y and y of a
is gamma one and z of we need z of a is y
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dashed of a, so we would like to convert b
v p as a system and if you convert we need
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this missing z of a. So, this is the slope,
right? Now let us say alpha so then we have
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equivalent IVP, so we have given b v p we
have converted into equivalent IVP.
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Then IVP can be solved using a favorite method,
so your favorite method say R-k
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method or Taylor series extra. So, when we
solve we have a b v p we converted into
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equivalent IVP. However we have something
to do with alpha so what we say we are
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saying some alpha, so we do not know what
is this alpha? So, then once we convert to
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equivalent IVP.
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..
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We can x say a then x 1 x 2 then x n equals
to b and
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here we should mention our y of a is
gamma one and y dashed of a is alpha. Now,
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if you compute so we can get the value y,
y
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of a is given here y of x 1. Of course, with
this alpha, so one can compute y of b with
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this
alpha. Suppose, we change delta so that means
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this corresponds to now whatever may be
the slope one chooses, what is our target?
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Target is y of b equals gamma 2, hence
compare y of b obtained with some alpha minus
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gamma 2 we have to compare, if is less
than epsilon then we have solved the b v p,
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do you get it? See we have some initial
guesses alpha that is our slope so then you
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solve the IVP using some favorite method,
we
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get the values at every grid point.
So, we get solution at the boundary point
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as well with the slope chosen, now suppose
somebody has solved with alpha as a slope
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one of your friend's have solved with delta
as
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a slope. Then we check the difference because
our target is gamma two so the difference
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must be less than epsilon where epsilon is
some prior signed. So, this is a desired
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accuracy, now with some alpha you get a let
us say up to 10 power minus 2 that means
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up to two decimals. Suppose your friend with
some other that is let us say delta gets a
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more than definitely your friend solution
is more closer to the target. So, what is
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our
aim? See this, what we are comparing and the
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condition we are looking for looking for
alpha, such that
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so this is what we are looking for. So, that
means you need to find.
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..
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So, we are looking for this that means find
the roots of this equation phi of alpha, so
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this
we treat it as some non-linear equation, we
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know several methods to find the roots of
this
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equation. So, how do we find the roots of
phi alpha equals to 0, we can try we have
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learnt can you name some of the methods. So
for example, you have Newton Raphson
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method any other, yes Secant method so let
us define very popular Newton Raphson
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method. Of course, here the prime denotes
derivative with respect to alpha, but the
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question is how do we find the derivative.
So, then let us say we switch over to Secant
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method, so this is secant method and this
is
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Newton Raphson method, so to solve phi of
alpha equals to 0 using secant method one
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need two initial approximations say alpha
0, alpha one so we need two initial
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approximations alpha 0, alpha 1. So, then
what we do with this, so what we do we
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compute phi of alpha and phi of alpha and
minus 1. So, when we have these initial
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approximations, that means these two are two
different slopes.
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..
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So, the two initial slopes alpha 0 alpha 1
define 2 IVP s, so IVP 1 y double plus p y
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prime q y is r y of a is gamma 1 and y dashed
of a is alpha 0 IVP 2 IVP 2. So, we have
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two different IVP's then what we do solve
IVP 1 and IVP 2. Solve IVP 1 and IVP 2 to
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get y at b using alpha 0 and y at b using
alpha 1. So, then when we know this what will
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be phi of alpha 0 and phi of alpha 1.
So, that means with this initial slope if
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you start what, what would be the value at
the
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boundary point. So, this is what we get then
how far we are from the actual target
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because gamma 2 is our actual target. Then
similarly, with this slope how far we are
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so
once we know this we can compute alpha 2 using
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Secant method using Secant method.
So, then this will be our new slope that could
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be a refinement so then we obtain alpha
two what do we do?
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..
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Solve IVP 3 alpha 2 then solve for what y
of b alpha 2 then check
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if this is less than if
yes then the b v p is solved if no refine
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alpha via Secant method. So, this is we do
it so
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once we get some initial slopes, we start
solving the IVP's then we get the value at
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the
boundary point then compare with the target.
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So, let us they are far away then we refine
to refine, we use secant method or we can
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use Newton Raphson method, but right now
we are discussing secant method. So, once
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you refine then we get a new slope again we
have to solve the IVP with the new slope to
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get the value at the boundary point.
So, once you get the value at the boundary
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point again you compare with the target that
is the actual value which we are looking for
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and check how far we are. Suppose, we are
slightly close, but we are not so happy then
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again we take that value and then refine
using Secant method and we do this process
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until we are happy. That means, until the
difference between the actual gamma 2, which
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we are looking for and the value at the
end point which we have obtained via the initial
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value problem they differ up to decide
accuracy. So, let us start with a problem
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shooting method via secant method.
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..
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Example, suppose this is our example y of
0 is 1 and y of 1 is y of 1 is say half then
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h is
one-third. Now, the 2 IVP is so choose, choose
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alpha 0 and alpha 1, alpha 0 is 1.2 and
alpha one is 1.5. So, these are definitely
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so these initial choices etc may be some from
the physical data or some experimental data.
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So, this is a kind of a guess, so that means
we define now IVP's, IVP 1 then IVP 2, so
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this is IVP 2 so then
y of 1, let us take 5, now
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we have to do? We have to solve, since for
the sake of simplicity I would like to try
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just
a Euler method please excuse me because Taylor
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series R-k method there little tedious,
but you can try with the Taylor series R-k
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method.
.
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.So, we define using Euler method for the
IVP's, so we reduce
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to the system then IVP 1 y
of 0 is one then y dashed which we have chosen
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is 1.2. Now, for this x 0 is 0. So, y n
plus 1 is y n plus h y n prime that is z n
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z n plus 1 is z n plus h 6 y n square minus
x n.
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Now, using this let us compute y 1 which is
y of one third so y 0 so y 0 is 1 h and z
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0 is
1.2. So, this will be 1.4 z 1 is z 0 plus
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h 6 y 0 square. So, z 0 is 1.2 plus one-third
6 y 0
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square so that is 6 minus x 0 so this will
be 3.2.
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.
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Then we proceed further y 2 this is y 1 plus
h z 1, so this will be and z 2 z 1 h 6 y 1
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square minus x 1. So, this have computed so
z 2 then y 3, y 2, z 2, so we get y 3 so this
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is
y at 1 using so this is y at b alpha 0. Similarly,
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for IVP 2 so we have y dash z, z dash y of
0 is 1 z of 0 is 1.5. So, we can obtain y
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one is same all the method y 1 is 1.5 z 1
is 3.5
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then y 2 is you want me to write down.
So, y 1 is 1 plus 1 by 3, so this is 1.5 z
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1 is z 0 plus h 6 y 0 1 minus. So, this will
be then
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y 2 is y 1 plus h z 1 and z 2 z 2 if you compute
z 1 1 by 6 6 y 1 square minus x 1. So, this
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is then y 3 is 5.29 and what is this y of
1 using 1.5, so either of them so this is
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4.79 and
what is our target so our target, target was
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so assuming either of them is a not close
enough then we have to go for second method.
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..
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So y of 1 using 1.2 minus 5 minus 5, so this
is 4.7966 minus 5 and y of 1 using 1.5 minus
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5 this was 5.1, so assuming both are not less
than epsilon, let us say. Now, let us define
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second method, so alpha 2 is alpha 1 minus,
so these are essentially our 5 of alpha 0
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and
5 of alpha. So, alpha 1, 1.5 then
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so this will be so gamma 2 get cancelled it
is essentially
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y of 1 using 1.5 y of 1, 1.2 into 5 of alpha
1. So, this is y of 1 using 1.5 minus 5. So,
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this
we have this 0.3 by this difference, so the
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difference of 1 at 1.5 this 5.29 difference
this
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one and this difference we have. So, this
is equals one point so this alpha 2. So having
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computed alpha two what we have to do?
.
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.Solve z of 0 is alpha 2 which is 1.32 again
Euler method we solve let us say on solving
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we get. So, I am not giving the details
y 3, 4.965 and what is this y 3 y at 1 using
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1.32
and is it close to five depends, so you can
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check so mod minus gamma 2. So, this is mod
four point. So, this is
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so suppose you are not happy so this is not
epsilon because so then
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what we do? Compute alpha 3, so alpha 3, alpha
2. So, this 1.28, so we are adjusting so
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then what we do?
.
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Solve IVP using z 0 equals 2 alpha 3 equals
1.28, so then we get let us say
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something
like this then it is not close so then we
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have to again compute alpha 4. So, we have
to
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continue the process like this, so when we
are happy with that desired accuracy then
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we
say we have reached the slope approximate
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slopes with which if we reach the target.
That means, the entire path has been so this
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is y of b say you have computed using initial
value problems some error. So, this is computed
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with some alpha k. So, this distance is
epsilon right so this is shooting method using
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Secant method, now we have discussed
already while shooting method is applied all
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that we have to do is we have to refine our
slope. Sometimes we are all shooting sometimes
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00:40:01,480 --> 00:40:08,630
we are under shooting and all that, so
can be refined slope using any other method
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00:40:08,630 --> 00:40:15,190
so we have ah standard Newton Raphson
method as well. So let us look into this using
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Newton Raphson method.
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..
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So, shooting and Newton Raphson so this is
slightly tricky, so consider y double prime
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00:40:37,869 --> 00:40:48,470
is
so I have considered this type, complete non-linear
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so y dash of a is alpha so this is our
alpha. Now, what is our phi of alpha, now
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the idea is solve for alpha using N-R method.
Now, if you want to define Newton Raphson
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method, so we have to compute so the issue
is how to compute pie dash of alpha n.
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.
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So let us phi dash of alpha equals see what
was our phi let us start with phi of alpha
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is y
of b alpha minus gamma two, then if I differentiate
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with alpha treating this as two
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.variable function of this argument as well
as the parameter. So, this let us call some
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eta
of alpha b, say then we have y double is f
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of x, y of b s or alpha and y dash b alpha
differentiate with respect to alpha both sides
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see this prime is x this is what?
Now, if I differentiate with respect to alpha,
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alpha goes in I can take of so this is eta
alpha, so this is remember at b. So, we get
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eta alpha double, so prime is I am denoting,
so
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this is taken there equals, now right hand
we had differentiated with the alpha. So,
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this
we use the chain rule x is independent dou
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f by dou y dou y by dou alpha plus dou f dou
y prime dou y dou y.
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So, please try to understand so this is right
so this is the definition so dou y by dou
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alpha
we are denoting by eta alpha, then we have
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this original given ode that written at b
differentiating both sides with respect to
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alpha. So, I have shown the working, so this
is
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the transmission so this reduces two of differentiate
with respect to alpha, left had side
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become this and right hand becomes using chain
rule this.
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.
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Then we have y of a gamma 1 if you differentiate
with alpha and right hand side is 0 and
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we have initial condition, if you differentiate
with the alpha this will be this is not
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derivative just to know that it depends on
alpha and right hand side is one. Therefore,
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what did we achieve, we have converted and
one more, this is what? This is eta alpha
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and what about this prime can be given to
eta right, how see for example.
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..
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We are looking for so
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this will be d by d x of eta alpha so that
will be prime.
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.
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So, with this notation this becomes eta alpha
and eta alpha a is 0 1. So, what did we
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achieve in order to find phi of alpha we have
obtained this. So, essentially phi dash of
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alpha if we solve see we are looking for phi
dash of alpha in order to use in our Newton
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Raphson method now what did we obtain? So
this is say some on
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solving T, we obtain
eta alpha which is phi dash that of alpha.
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Of course, b so this gives a big tool to compute
phi dash that of alpha, so let us see for
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example.
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..
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If y double is y simplest case we are taking
say y 0 is 1 and y of 1 is 0 is our boundary
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value problem. Now, what is our f our f was
just y therefore, 1 is 0 therefore, the IVP
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for
eta the IVP for eta alpha reduces to what
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was our IVP? So, this reduces to dou of dou
y is
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1 and this is 0 and boundary conditions so
we have to find. So, this implies and y dash
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of
0 is alpha this implies 1.
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.
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So, accordingly the IVP is eta alpha prime
is eta alpha 0, now how do we solve this again
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system let eta alpha is some mu then mu prime
is eta alpha. Accordingly then mu of so
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.this is the system so let h is 1 by 3. So,
then we have to define Euler method of course,
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equals eta alpha n plus h z n sorry h mu n
then mu n plus 1 mu n plus h eta alpha n.
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So,
from this let us solve eta 1, I am dropping
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alpha because too many suffixes so eta 1 is
eta
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0 plus h mu 0 and mu 1 is mu 0 plus h eta
1 eta 1. So, that is eta mu 1, mu 1 is mu
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0, plus
mu 0 plus h eta 0, so this is 0 so this is
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one. Similarly, eta 2 is eta 1 plus h to mu
1, so
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this will be
and mu 2 mu 1 h eta 1, so this will be
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see essentially we are solving this
using Euler method.
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.
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So, ultimately we need a eta 3 so eta 2 plus
h mu 2, so this will be 9, so this is our
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value
mu 2 is
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so this will be
so we have obtained eta 3. So, this is nothing
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but phi dash of alpha
of course, with this b right, therefore we
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can obtain phi of alpha n. So, this is alpha
n
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minus y of b at using alpha n minus gamma
2 and this is eta, so this is alpha 1. So,
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let us
choose say alpha 0 is 1 and this we have to
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solve with alpha 0 obtained y at 1 using alpha
0. Say this is point suppose same problem
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we have to solve so if we solve, let us say
this
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is the value. So, this the value we have obtained
minus gamma 2 so gamma 2 was the
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given gamma 2 was 0 and eta alpha just now
we obtained. So, this is so we get defined
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value.
So, this is our alpha one then again we have
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to obtain, so let us say this is this is some
one point right, so we get some value let
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us say this is some 0.92 check. So, then obtain
y
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at 1 using alpha 1 again refine to get alpha
2. So, we continue until we get the ah desired
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.accuracy, so shooting method the idea is
same either, we use secant method or we use
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Newton Rahpson method, the idea remain the
same. Time and again we have to solve
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several initial value problems for each mu
slope we solve the mu initial value problem
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with that slope and then try to obtain the
solution. And check how far we are from the
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target then if you are not satisfied refine
your new slope to get a new slope using secant
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method. Only difference is in case of Newton
Raphson to solve the phi dash of alpha
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again we have to solve a system of equations
for finding phi dash of alpha.
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So, that means one set of IVP for the original
problem and another set of ivp is for eta,
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which is our slope five dash of alpha. So,
this is a slightly complicated, otherwise
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the
underline principle remain the same. So, with
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this we have reviewed more or less
different methods for two point boundary value
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problems. So, in the next class we will
discuss further more topics related to 2.1
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value problems and do some problems bye.
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.