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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,  
 \begin{displaymath}
y_N = y_0 (1 + h\lambda)^N = y_0 (1 + T\lambda/N)^N\end{displaymath} (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 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(hp), with p>1.


next up previous
Next: Higher Order Methods: Runge-Kutta Up: Euler's Method Previous: Euler's Method
G. Bard Ermentrout
1/9/1998