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Integrating with Modified Euler's Method

Date: 08/13/2007 at 01:35:12
From: Saji
Subject: modified Eulerís method

Using modified Eulerís method find the value of y at x = 0.2 with 
h = 0.1 where dy/dx = 1 - y with y(0) = 0.



Date: 08/15/2007 at 08:06:47
From: Doctor Luis
Subject: Re: modified Eulerís method

Hi Saji,

To get you started, first look at this link

  Euler Method
    http://en.wikipedia.org/wiki/Euler_integration 

Basically, the Euler integration method gives you a way to obtain the
values of y(x) at specific points x, based on the differential equation

  dy/dx = f(x,y)

subject to an appropriate initial condition y_0 (i.e. the value of
y(x) at x_0).

As a numerical approximation, you can approximate the derivative
by a forward difference as follows:

  y'(x_n) ~ (y_{n+1} - y_n)/h

Additionally, from the assumption that y satisfies the differential
equation

  y'(x_n) = f(x_n,y_n)

you can obtain the following relation between successive y's:

  y_{n+1} = y_n + h * f(x_n,y_n)

This formula makes sense if you think of y as distance, x as time, 
and dy/dx = f as speed.  So, in that sense, this formula says that 
your new position (y_{n+1}) is equal to the old position (y_n) plus
whatever distance you traveled in a time interval equal to h.

Successive values y_0, y_1, y_2, ... may be obtained in that manner,
in time increment values of h.

In your case, of course, x_0 = 0, y_0 = 0, f(x,y) = 1 - y, but I'll
keep the discussion general so you'll be able to solve similar
problems in the future.

With a spacing h, you can see that

  x_n = x_0 + n * h,

so that

  x_1 = h  = 0.1
  x_2 = 2h = 0.2

The y_n's are defined such that

  y_0 = y(x_0)
  y_1 = y(x_1)
  y_2 = y(x_2)

Performing the Euler integration is done as follows:

  y_1 = y_0 + h * f(x_0, y_0)
  y_2 = y_1 + h * f(x_1, y_1)

Note that, just as y_0 is needed for calculating y_1, you'll need to
find y_1 in order to find y_2.

However, that's not the answer to your problem just yet.  The problem
asks you to use the *modified* Euler method.  To explain what that is,
let's go back to our speed analogy

  y_{n+1} = y_n + h * f(x_n,y_n)

You'll note that the time interval covered is from x_n to x_{n+1}, and
that the speed is evaluated at the *start* of this time interval. 

Well, that doesn't sound very accurate, right?  After all, the speed
dy/dx is changing throughout the time interval.  Using the speed at
the beginning of the interval might overshoot or undershoot the
correct value of the distance.

Wouldn't it be better if we used the value of the speed evaluated at
the midpoint of the time interval?  That sounds like a better estimate
of the average speed during the interval, right?  Well, that's exactly
what the modified Euler integration method uses.

In other words, the modified equation we want is:

  y_{n+1} = y_n + h * f(x_{n+1/2}, y_{n+1/2})

Applying this to your problem will give you

  y_1 = y_0 + h * f(x_{1/2}, y_{1/2})
  y_2 = y_1 + h * f(x_{3/2}, y_{3/2})

Oops.  To find y_1 from y_0 we need the value of y_{1/2} first.  Well,
we only need an estimate, so we can apply the regular Euler method
first, but with a half-step of h/2 instead:

  y_{1/2} = y_0 + (h/2) * f(x_0, y_0)

And that gives us y_{1/2}!

But what about y_{3/2}, you ask?  Well, that's a full-step away from
y_{1/2}, so now you can use the more accurate modified method to find
an estimate for y_{3/2}, where the "velocity" term is evaluated at the
midpoint "time" x_1.

To summarize the modified Euler integration method:

First take a half-step of h/2 with the regular method

  y_{1/2} = y_0 + (h/2) * f(x_0, y_0)

Next, take full steps by using the modified method making sure to
evaluate the "velocity" terms at the appropriate midpoint of each
interval.

  y_{1}   = y_0     +   h * f(x_{1/2}, y_{1/2})
  y_{3/2} = y_{1/2} +   h * f(x_{1},   y_{1})
  y_{2}   = y_{1}   +   h * f(x_{3/2}, y_{3/2})

And that's it!

I hope it all makes sense.  If it looks complicated, just take it a
step at a time.  Just don't forget the following basic definitions:

First, we have the "times"

  x_0     = 0
  x_{1/2} = (1/2)h
  x_1     = h
  x_{3/2} = (3/2)h
  x_2     = 2h

and, one at a time, find the corresponding "distances"

  y_0     = y(x_0)
  y_{1/2} = y(x_{1/2})
  y_1     = y(x_1)
  y_{3/2} = y(x_{3/2})
  y_2     = y(x_2)

Let us know if you have any more questions!

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Calculus

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