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In this segment, we're going to derive Runge-Kutta second-order method.
So Runge-Kutta second-order method, which is the way
to solve first-order ordinary differential equations.
We're going to look at how Runge and Kutta derived this
formula.|So again,
what does Runge-Kutta second-order method say?|It is that if you have the differential equation
of this particular form, then you can write down
the numerical method to solve for y with respect . . .
as a function of x by using this particular general formula, and it'll be second-order
accurate, where
these, a1 plus a2 is equal to
1,
and then a2 p2 is equal to . . . a2 p1
is equal to 1/2, and a2 q11
is equal to 1/2, these are the three equations which they obtained
in order to be able to plug them into this particular equation, where k1
is equal to the value of the function at xi, comma, yi, and k2 is equal
to the value of the function at xi plus p1
times h, yi plus q11
times k1 times h.|So what we
have in the Runge-Kutta second-order method is this is the formula which is used, where
k1 and k2 in the formula are given by these expressions of calculating the value
of the function at certain points, and you are going to satisfy
these three equations to find out what the values of a1, a2, p1, and q11 you're going to use.
So what we want to do in this segment is to figure out how did they get this?
How were they able to obtain these three equations so as to write this as the general
form for Runge-Kutta second-order method?|So this involves use of Taylor
series, so we do need to pay some attention here.|So let's go ahead and see that how . . .
how this was done.|So let's suppose if you look at the Taylor series for
y as a function of x will be given by this.
So we're writing the Taylor series
of one variable, and we're going to write the first three terms
of the Taylor series, because that's what is used in the Runge-Kutta second-order method,
and that's why it's called the Runge-Kutta second-order method, because you're using the first three terms of the
Taylor series, times h squared, and plus the
terms which you are leaving are of the order of h cubed, so that's how we've got to look at it.
This is the Taylor series for y as a function of x.|Give me the value of the function of y at a certain point,
its derivative at that particular point, its second derivative at that particular point, and so on and so forth,
and if I'm going to forget about the terms after the third term of this Taylor series, then
those terms will be the order of h cubed.|Now, we already know what h is, h is nothing but
the difference between the point where I want to and the point where I am.
Now, since we know that the differential equation itself is dy by dx is
equal to f, x, comma, y, this is what we are trying to solve, what I'm going to do is I'm going to
substitute this back in here, into the Taylor series expression, and this is what I'm going to get,
I'm going to get y-sub-i-plus-1 is equal to yi, plus the value of the function
at xi, comma, yi, times h, plus 1
by factorial 2, f prime, xi, comma, yi, times h
squared, plus the order of the terms of h cubed.|So because,
since dy by dx is this function, f, I can just substitute it back there at calculated
points, these two points, and this one is the derivative of the function, f, because I have the second derivative
of y, so if I take the second derivative of y, I get the first derivative of f, and that's why you see only an
f prime there.|So having said that, in order to
be able to calculate f prime, what we need to do is we need to apply the chain rule, and the chain rule
goes as follows, the f prime of x, comma, y, the derivative of f with
respect to x is given by the chain rule.|The reason why it's given by the chain rule is because of the fact that . . .
that y is a function of x, so
you don't have two independent variables, you have y as a dependent variable . . . y as a dependent variable
and x is the independent variable, so the chain rule looks like this, it's the partial derivative of
f with respect to x, plus the partial derivative with respect to y,
times the dy by dx, that's what you get there.
So if I'm going to substitute f prime, xi,
comma, yi, it will be nothing but del f by del x being
calculated at that point, and del f by del y
being calculated at that point, and the dy by dx is
nothing but the value of the function, f, being calculated at that particular point.
So I'm going to substitute this expression now back into this Taylor series expansion
which I had, and see what I get.|And this is what I'm going to get, I'm going to
get y-sub-i-plus-1 is equal to yi, plus
the value of the function at xi, comma, yi, times h, so that's from the
second term of the Taylor series, and then the last term
which I will get is 1/2 del f
by del x, at xi, yi, times h
squared, plus
1/2 del f by del y, at xi, comma,
yi, times the value of the function at xi, comma, yi, because
that's dy by dx at that particular point, times h squared, plus the order of
the h cubed terms.|So once I have
that, what I want to be able to do is that I want to equate it to something so that I can develop
those three equations, three unknowns.|So this is what I know, that, hey, in order to be able to get second-order
accuracy, I need to take the first three terms of the series, of the Taylor series, and now I'm getting
four terms here, I'm not adding one more term of the Taylor series, it's just that the
second term has now two parts to it, because I used the chain rule for the derivative
of the function of f.
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