How to estimate the local error and the global error for Runge-Kutta method used for solve a system of differential equations in practice?

I use Richardson extrapolation for select a adaptive step [Solving Ordinary Differential Equations I - Nonstiff Problems 167-168p].

I seen the following formula for estimate a global error for one differential equation: $\delta_{i+1} = \delta_{i}\cdot e^{\int^{t_{i+1}}_{t_i} \mu ds} + r_k$, where $r_k$ local error at step, $\mu$ - maximal eigen value of $\frac{J + J^T}{2}$ ($J$ is jacobian of system). But I didn't find its explanation and application to system of ODE.


1 Answer 1


The local error is typically estimated using something like an embedded pair of Runge-Kutta methods. The classic example would be the 4(5) pair (for instance, Dormand-Prince), where the error over a single step would be estimated by comparing the fourth- and fifth-order solutions. Other methods include approaches like halving the time step, or Richardson extrapolation.

The global estimate you're referring to is one by Dahlquist, using the logarithmic norm, which is a bound on the growth of the norm of the solution to an ODE. (See reference 3 of this review paper on the logarithmic norm by Soderlind. A similar, weaker result was proven by Gronwall using Lipschitz constants; see Gronwall's inequality.)

There are a variety of methods for bounding and estimating the global error in solutions of ODEs, and I'm sure I'll fail to list all of them.

For estimates, the most recent paper I'm aware of is by Constantinescu, which has a large number of references worth looking at; other notable papers include the now-dated review by Skeel, the adjoint approach by Estep (he has a whole series of adjoint-based a posteriori error estimation papers), and the adjoint approach by Cao & Petzold.

Bounds could also be computed using interval arithmetic and Taylor methods. The thesis by Joseph Scott should have a number of references discussing this topic.

  • $\begingroup$ Thank you very much! I very glad to receive so nice answer. $\endgroup$ Commented Oct 18, 2014 at 9:53

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