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5. Recycling Force Evaluations

5.1. One Force Evaluation per Step

%\subsubsubsection{General Form} {\bf General Form}

3rd order not possible: in Eq. (
199) we see that starting the first force calculation at time zero implies that the coefficient for in should be , and not as we insist upon above. [EXPAND THIS]

This means that we only have to expand up to powers in .


we get

This has to be equal to the Taylor series expansions:

This implies:



From the last two, we get . Two parameter freedom, with :

%\subsubsubsection{Second Order Recycle Conditions}

{\bf Second Order Recycle Conditions}

Now insist that :

Already okay. So we are left with a two-parameter freedom.

For and , simplest choice: leapfrog.

[check for which values time symmetry; presumably only for and ]

5.2. Two Force Evaluations per Step

%\subsubsubsection{General Form} {\bf General Form}

Let us expand up to powers in .

we get

This has to be equal to the Taylor series expansions:

which implies

This has to be equal to the Taylor series expansions:

which implies

%\subsubsubsection{Third Order Recycle Conditions}

{\bf Third Order Recycle Conditions}

Now insist that :

seven equations for ten variables.

From the last two:

Combining that with the equation above the two we just used:

Two possibilities: or . In the first case, and from the next to last equation and the one above that. Contradiction. Hence:

Introduce and , and use as the third parameter. Then

So in general:

Example: for , :

Like Simpon's rule for velocity integration.

Another example: for , :

%\subsubsubsection{Exact Recycling}

{\bf Exact Recycling}

So far, only pseudo-FSAL, or better: pseudo Runge Kutta!

Can we make it really FSAL Runge Kutta, to all orders?

With the first two:

combining this with the third:

leading to

but alas:

So this doesn't work: the coefficients for the and terms in the expression for in Eq. (
237) blow up.

%\subsubsubsection{A Search for a Fourth Order Scheme}

{\bf A Search for a Fourth Order Scheme}

Given that we have three free parameters left in our construction of a recycling scheme that is third order correct, it is tempting to search for a fourth-order scheme, based on only two new force calculations per time step.

Repeating the previous analysis to one order higher in , we get

we get

This has to be equal to

In addition to the previous conditions, we get the following two additional requirements:


This has to be equal to

We get the additional equations:

We have thus eleven conditions gathered so far for the ten unknown parameters . A priori we would expect to find no solutions in such an overdetermined system. However, let's see how far we get when we try. Let us list the conditions here together:

Subtracting the 6th and 7th equation, we find

and subtracting the 7th and 8th equation, we find

Together, these two expressions imply

There are three solutions: , , and . The first two solutions can be discarded, because they would imply that the left-hand side of the 6th, 7th, and 8th equations above would all have the same value, contradicting the fact that their right-hand sides have different values. We thus find

With this result, we can use the remaining information in the 6th, 7th, and 8th equations above to determine the other two values:

The 5th equation gives us

Subtracting the 9th and 10th equation, we find

and plugging in the values we have found so far gives us

The 9th and 10th equation then give us:

Since the 11th equation gives us

we conclude that

We can now write the 2nd and 3rd equations as

Subtraction those expressions gives us


and plugging this back in the expressions above gives

The 1st equation then gives

Remarkably, we have been able to solve the eleven equations for the ten unknowns and found a consistent solution! To summarize:

Not only that, it turns out that we get an additional bonus: these solutions solve the previous relation for demanding the , which was

We have thus found a consistent set of solutions for ten variables satisfying twelve equations. Could we be really lucky? Could it be that in fact ? If that were true, our fourth-order scheme would allow us to recycle the last force calculation, and we would really have obtained a fourth-order scheme with an effective costs of only two new force calculations per step. This does sound too good to be true, but let's just check.

It was too good to be true!

We thus have:

Looking at the equations this way, we can in fact see directly that a fourth-order scheme doesn't work ('directly' once you have become sufficiently familiar with all these expressions). For the scheme to be fourth order, the position where the last force calculation is computed should agree to third order with the new position at the end of the time step. However, the latter has a term which the former lacks, and since the difference is of first order in , there is a real third-order difference between the two positions, hence between the forces computed in these two positions. The upshot is that this will introduce a fourth-order error in the velocity in the next step, when we recycle the last force calculation. Our scheme is thus only third-order accurate when we recycle, even though it is fourth-order accurate if we decide to compute all three new forces anew at each step.

By the way, as a fourth-order scheme, it is listed in Abramowitz and Stegun's welknown Handbook of Mathematical Functions as eq. 25.5.22, but with a typo: the error in the position is listed as being , while it really should be ; in addition no error is listed for the velocity. As we have seen, for the velocity, too, the error is .
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