I am so confused with how to determine the starting values for linear multistep methods.

I have searched the wiki page for linear multistep method. And it says that for Two-step Adams–Bashforth, the starting values are just got by Euler's method.

I have also found this page. And the case is that the exact solution is already known.

So my question is that: for common cases, how to determine the starting values?


In a nutshell, what this is saying is that you have a numerical method that requires certain initialization data and the given initial conditions do not provide that.

How do you get enough initial conditions when they are not given? The answer is to use another numerical method to jump start the process (is it clear why this is better than guessing and solves the problem of knowing a closed form solution). In the example you cite, they need two starting values and the BVP only provided one. They resort to using Euler's method (and any choice that gives you another point numerically would suffice) to find another value to jumpstart the two-step method.

In practice, things get a little a stranger still as you get to a point where you would do a fourth order method like the four - step Adams-Bashforth Method. For this:

  • We need four starting values, so we use a fourth order Runge-Kutta method to find those starting values (in problems, they choose a method where they know the exact solution for comparison purposes, but cheat and use it for starting the process as RK-4 is a lot of work).
  • The next step is to actually calculate an approximation using the fourth-order Adams-Bashforth explicit method and this is called a 'predictor'.
  • Lastly, we use an implicit Adams-Moulton method as a 'corrector'.

The implicit method is used to improve the approximations obtained by the explicit method. These are thusly known as 'predictor-corrector' methods and you can hopefully now do a bit of more research and see if comes together.

  • $\begingroup$ Nice write-up +1! $\endgroup$ – Namaste Jun 16 '13 at 0:26
  • $\begingroup$ my pleasure! Hope your day's been good! $\endgroup$ – Namaste Jun 16 '13 at 0:34

Using the exact solution to find the starting values is basically just cheating; what's the point in numerically integrating when you can already know those exact solutions? My guess is that the page that you linked is basically just demonstrating how you might use a multistep method, after you know the starting values use some external process (which in their case is just using the true values). However, in practice you need to use a one-step method (e.g. Euler's method) in order to calculate the starting values, as the wikipedia page correctly explains.

  • $\begingroup$ Yes, but is there some detailed information on finding the starting values? Or, just how much difference is there between Euler's method or some other one step methods? Thanks. $\endgroup$ – eccstartup Jun 15 '13 at 10:52

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