1

An upper bound for the global error in Euler's method to solve a first order ODE numerically is given by the equation \begin{equation} \frac{Mh}{2L} ( e^{L(t_{i+1}-a)} - 1), \end{equation} where $t_{i+1}$ is the $i+1$ cell in the abcsisa and $a$ is the first abcisa. M is a bound for the second derivative of the ODE unknown $y$ and $L$ is the Lipschitz continuity property bound. This is derived in most of numerical analysis books.

My question is the following: Having $L$ small is good for convergence and stability (right?) in the sense that $L$ small implies little change. If $L=0$ then the function $f(t,y)$ (on the right hand side of the ODE) is constant on $y$. If $L <1$ the function is contracting. So why is $L$ in the denominator of the bound? My intuition (which is not working good) says that the error bound should diminish as $L$ diminshes.

What is wrong with my intuition here?

Thanks.

Herman Jaramillo
  • 2,994

1 Answers1

2

First this bound is not a good estimation. It is useful to say the error is bounded, but in practice it is completely inaccurate. Then the error does not explode when $L\rightarrow 0$. You can develop $\frac{1}{L}(e^{L(t_{i+1}−a)}−1)=\frac{1}{L}(1+L(t_{i+1}−a)+O(L^2)−1)=t_{i+1}−a+O(L)$. So you see that the error diminishes as $L$ tends to $0$, which is logical.

Adrien Laurent
  • 352