Convergence and Accuracy of Euler's Method

It is easy to see that Euler's method converges for the special case of the equation $y^{\prime} = \lambda y$ with solution $y(T) = y_0 e^{\lambda T}$. For this example,  

$$y_N = y_0 (1 + h\lambda)^N = y_0 (1 + T\lambda/N)^N$$ (5)

Recalling that $\lim_{N\rightarrow \infty} (1 + 1/N)^N = e$, we see that $\lim_{N\rightarrow \infty} y_N = y_0 e^{\lambda T}$.

Error estimates show

• The global error at T is O(h) (first order accuracy).
• The error grows exponentially in time.
• The error increases with M. This suggests that one should take smaller steps where the solution is changing more rapidly. We will return to this below.

The error analysis above ignores round-off error. If one assumes that a fixed error is added at each time step, then the error estimate of is modified to $O(h) + O(\epsilon_{mach} h^{-1})$. That is, taking more steps reduces the discretization error, but increases the round-off error. Therefore, there is a point of diminishing returns where the total error increases as h decreases. Better results require not more effort, but more efficiency. The key is to take more terms of the Taylor series and reduce the discretization error to O(h^p), with p>1.