why does the local error increase in proportion to the square of the step size but the global error in proportion to the step size in euler's method for solving ODE’s? I am curious and have not been able to find a satisfactory answer
why does the local error increase in proportion to the square of the step size but the global error in proportion to the step size in euler's method for solving ODE’s? I am curious and have not been able to find a satisfactory answer
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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why does the local error increase in proportion to the square of the step size but the global error in proportion to the step size in euler's method for solving ODE’s? I am curious and have not been able to find a satisfactory answer
Expert Solution
Step 1: Explanation of Local errors
To understand why the local error increases in proportion to the square of the step size, we need to look at the Taylor series expansion of the exact solution around each point. Suppose we have an ODE of the form
where
On the other hand approximation by Euler method at
which is same as first two term of Taylor series ,therefore local error at
which is proportional to
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