Numerical ODEs · Definition

Local and global error

numerical methodsnumerical odes

Before you start you need to know

When approximating a differential equation step by step, each step introduces a small mistake. Measuring how these mistakes accumulate tells you whether a method is reliable.

Definition

The local error is the difference between the exact ODE solution over one step and the method's approximation starting from the exact value.

\[e_{n+1}=y(t_{n+1})-y_{n+1}\]

If Euler's method has local error of order \(O(h^2)\), the global error is usually of order \(O(h)\), one power less, because the local error accumulates over about \(1/h\) steps. This relation is the basis of convergence analysis.

Consider \(y'=y\), \(y(0)=1\), with \(h=0.1\). After the first Euler step we get \(y_1=1.1\), while \(y(0.1)=e^{0.1}\approx1.10517\), so the local error is about \(0.00517\).

A typical mistake is confusing local with global error and assuming that a method with small local error must have an equally small global error. The global error also depends on the stability of the method.

Think about it

Why can a method with a small local error still have a large global error?

Because local errors accumulate and propagate at each step, and the method's stability can amplify them.

Exercises 10

  1. What does the local error measure in a numerical ODE method?
  2. What does the global error measure in a numerical ODE method?
  3. If a method's local error is \(O(h^2)\), what order does the global error typically have?
  4. With Euler's method, what is the local error when computing \(y_{n+1}\) from \(y(t_n)\)?
  5. What best describes how local errors build up into the global error?
  6. For \(y'=y\), \(y(0)=1\), and \(h=0.1\), what is approximately the local error of the first Euler step?
  7. If halving h roughly halves the global error of Euler's method, what is the order of the global error?
  8. If the local error of one step is \(C h^3\), roughly how many steps accumulate over the interval \([0,1]\)?
  9. A method has local error \(O(h^2)\). If h goes from 0.1 to 0.05, how does the global error change approximately if its global order is \(O(h)\)?
  10. For \(y'=-2y\), \(y(0)=1\) and \(h=0.1\), the first Euler step gives \(y_1=0.8\). If the local error of that step is \(0.0187\), what is the approximate relative local error with respect to \(y(0.1)=e^{-0.2}\approx0.81873\)?

References

  • Richard L. Burden, J. Douglas Faires, Annette M. Burden, Numerical Analysis, 10th edition, Cengage Learning, 2016.
  • Lloyd N. Trefethen, David Bau III, Numerical Linear Algebra, 1st edition, SIAM, 1997.