Numerical Methods - Euler Method Part 3

Numerical Methods for Solving Differential Equations

Euler's Method

Theoretical Introduction

(continued fromÌýlast page...)

Deriving the Euler's Method Formulas

Reminder: We're solving the initial value problem:

²â′Ìý=Ìýf(x, y)

y(xo)Ìý=Ìýyo

As we just saw in the graphical description of the method, the basic idea is to use a known point as a "starter," and then use the tangent line through this known point to jump to a new point. Rather than focus on a particular point in the sequence of points we're going to generate, let's be generic. Let's use the names:

  • (xn,Ìýyn) for the known point
  • (xn+1,Ìýyn+1) for the new point

Our picture, based on previous experience, should look something like this:

Plot

(Though the proximity of theÌýtrue solutionÌýto theÌýpoint (xn,Ìýyn)Ìýis, perhaps, a little optimistic.)

Our task here is to find formulas for the coordinates of the new point, the one on the right. Clearly it lies on theÌýtangent line, and thisÌýtangent lineÌýhas a known slope, namelyÌýf(xn,Ìýyn). Let's mark on our picture names for the sizes of theÌýx-jump, and theÌýy-jump as we move from the known point, (xn,Ìýyn), to the new point. Let's also write in the slope of theÌýtangent lineÌýthat we just mentioned. Doing so, we get:

Plot

The formula relatingÌýxnÌý²¹²Ô»åÌýxn+1Ìýis obvious:

xn+1Ìý=ÌýxnÌý+Ìýh

Also, we know from basic algebra thatÌýslopeÌý=Ìýrise / run, so applying this idea to the triangle in our picture, the formula becomes:

f(xn,Ìýyn)Ìý=ÌýΔy / h

which can be rearranged to solve for ΔyÌýgiving us:

ΔyÌý=ÌýhÌýf(xn,Ìýyn)

But, we're really after a formula forÌýyn+1. Looking at the picture, it's obvious that:

yn+1Ìý=ÌýynÌý+ÌýΔy

And, replacing ΔyÌýby our new formula, this becomes:

yn+1Ìý=ÌýynÌý+ÌýhÌýf(xn,Ìýyn)

And that's it! We've derived the formulas required to generate a numerical solution to an initial value problem using Euler's Method. Let'sÌýgo and lookÌýat a summary of the method.

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