Region of Stability for the Model Problem

The Trapezoidal Method is an A-stable Method; Comparison to Euler

Let's consider the model problem
dy/dt = -20*y
y(0) = 1
0 <= t <= 2
We know that the solution of this problem decays, and that in fact
y(t) = exp(-20*t)

Let's compare the solutions obtained with the Trapezoidal Method, and with a simple explicit method (in this case: Euler's Method). We will do this for different timesteps. A simple script that can be downloaded here . Again, it uses the functions deriv.m and exact.m .

Euler's Method


h=0.025
euler_h=0.025.gif
h=0.09
euler_h=0.09.gif
h=0.11
euler_h=0.11.gif

Clearly, the solution is unstable in the last case; in particular, it diverges. In the second case (h=0.09) the solution still oscillates (not very good !!), but at least the asymptotic behaviour (decay) is correct. We get a "pretty good" result for h=0.025.

Now, let's look at the same graphs if we employ the Trapezoidal method instead:



h=0.025
trap_h=0.025.gif
h=0.09
trap_h=0.09.gif
h=0.11
trap_h=0.11.gif

In all cases, the solution behaves correctly qualitatively, i.e., it decays. In fact, we can use arbitrarily large timesteps to get the correct qualitative behaviour. This is why the method is called A-stable. The region of stability is the entire left half plane.

h=0.3
trap_h=0.3.gif

The Model Problem dy/dt = +20y


h=0.01
model+20_0.01.gif

h=0.11
model+20_0.11.gif


What do we learn: With both methods, and the timesteps chosen, the results are not particularly good. But: the qualitative behaviour is correct, in the sense that the solution grows.

Let's look at the same problem using Backward Euler

h=0.002
model+20_0.01.gif

h=0.11
model+20_0.11.gif

Clearly, the qualitative behaviour is wrong for h=0.11. This is so because 20*h is in the region of absolute stability!!