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
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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

                y(x)=f(x,y(x))  with initial condition  y(x0)=y0


Then the exact solution at x0+h,where h is the step size given by 

                         y(x0+h)=y(x0)+hy(x0)+h22y(x0)+O(h3)

               where    o(h3) represents the higher order term that are negligible for small h.


On the other hand approximation by Euler method at  x0+h  is given by

                                      y1=y0+hf(x0,y0)

which is same as first two term of Taylor series ,therefore local error at x0+h  is

                                                   y(x0+h)y1=h22y(x0)+O(h3)

   which is proportional to h2  for small.h.

                            

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