Questions:
1) For which integer q can you conclude that the error of the
 explicit Euler method,
 method of Heun,
 implicit Euler method,
 Adams Moulton method,
 Adams Moulton method - linearisation 1,
 Adams Moulton method - linearisation 2
behaves like 𝑂(𝛿𝑡 𝑞 )?
We will assume here that we are talking about the global error. We know that the global error is decreased by
a factor of roughly 2𝑝 for a method of order p and each halving of the step size. Looking at the error reduction
rates from our results, we can say that:






explicit Euler method: q = 1
method of Heun: q = 2
implicit Euler method, q=1
Adams Moulton method, q=2
Adams Moulton method - linearisation 1, q=1
Adams Moulton method - linearisation 2, q=1
2) In the lecture, we saw that both the implicit Euler and the Adams Moulton method are unconditionally
stable und, thus, give us stable solutions for every choice of 𝛿𝑡 . For the example of this worksheet we stated in
c) and d) that the resulting equation for each timestep is sometimes not solvable and, thus, the method cannot
be applied for certain 𝛿𝑡 . Can you explain this apparent discrepancy?
The main issue here is that the problem is stiff. Consider the general form of the logistic equation:
A method which satisfies the stability condition may still be numerically unstable1. In the case of the logistic
differential equation 𝑝̇ = 7𝑝 (1 −
𝑝
10
) = 7𝑝 −
7
10
𝑝2 , we know the solution is of the that the problem is
relatively stiff because the exponential power.
How do we define stability? What happens exactly at this point? Is Adams-Moulton really not stable, or does
something else happen? Actually, we always get a solution to Euler…so the only question is the first
Thoughts:



Could it be that it IS stable, but just not consistent?
How does one define stable? In fact, Adams-Moulton is in this case stable…something else is
happening here. It doesn’t converge because it’s not consistent?
The methods are stable, it’s the starting point that is the problem. In Worksheet 1, we started with
y(0) = 1, which is above the unstable y=0 equilibrium. This time we start above the stable y=10
equilibrium, and we overshoot past the unstable equilibrium if the time step is too large. There is in
fact an unstable region for implicit Euler, it’s just much smaller than that for explicit Euler. See:
http://books.google.com/books?id=opd2NkBmMxsC&pg=PA76&dq=implicit+euler+stability+
region&hl=en&ei=PFneTPHSKJL24gbWhY0y&sa=X&oi=book_result&ct=result&resnum=8&ve
d=0CE8Q6AEwBw#v=onepage&q=implicit%20euler%20stability%20region&f=false

The discrepancy lies in the nature of the problem that is being solved. This is an example of a logistic
population model. Thus, the differential equation describing the population dynamics is non-linear.
Implicit Euler and Adams-Moulton methods are only unconditionally stable for every choice of 𝛿𝑡 if
the differential equation is linear (is this correct?!).
1
Linz, P 1985, Analytical and numerical methods for Volterra equations, SIAM, p. 185.
3) Can you give a reason why the linearisation 1 of the Adams Moulton method works better?
Tutor said: Stability is not the question. The question is: “How big is the influence from each term, and how
much error does each term (y(n) and y(n+1))”. We need to look at the way that y(n) and y(n+1) move, and how
much each one is weighted.
For the first method, there is less influence from the next term. For the calculation of y(1), the magnitude of
the next term is very large due to the strong negative gradient at y(0). The first linearization method weights
this next term much less than the second linearization method, hence it has a smaller error at the next point.
This then snowballs for all further y-value predictions.
4) Which type of methods (explicit/implicit) would you choose for:
 The initial value problem from worksheet 1,
The problem from worksheet 1 is much less stiff than that of worksheet 2. Also, the initial value (p(0)
= 1) is above the unstable equilibrium p=0, and below the attractive equilibrium p=10. Hence an
explicit method would be more than suitable here, because the solution will always converge to p=10
for reasonable values of the step size.
Probably an explicit method would be fine. The problem has an initial value at y=1, and y=10 is an
attractive equilibrium. Hence, for reasonable values of the time step size, the solution will converge to
the equilibrium. This of course also has the benefit of faster running times and simpler
implementation.Perhaps here check the stability region for explicit methods on the worksheet 1
problem. If indeed the explicit methods are unconditionally stable for the first worksheet problem,
then choosing an explicit method would be better because we wouldn’t have to choose too small a
time step, and hence it wins over implicit methods on the grounds of computation cost.

The initial value problem (1,2) from this worksheet?
The problem from this worksheet is much stiffer than that of the first worksheet. In order to avoid
instability resulting from the strong negative gradient at the beginning, an implicit method should be
chosen. Perhaps implicit Euler is the best option.

For the last part, where exactly can we put an X? The explicit Euler gives us values to the end
but doesn’t oscillate. For the cases where it oscillates a little bit, do NOT write an X! Only do
it for the solutions that look natural.