| |
Chapter 2: Stability of Two-Dimensional Maps
Section 4.4: Second-Order Difference Equations -
-
-
Example 4.4:
Second-order difference equations can be solved numerically
using Phaser. However, before they can be entered into Phaser,
they must be converted to a pair of first-order difference equations.
Consider, for example, the linear second-order difference equation
x(t+2) + p x(t+1) + q x(t) = 0. Using the new variables x(t) = x1 and
x(t+1) = x2, this second-order equation is equivalent to the
linear system of first-order equations:
(1)
where p and q are parameters.
Figure 4.4.1.
Solution of Eq.(1) satisfying the initial conditions x1 = 1, x2 = 0,
for the parameter values p = 6, q = 9.
Figure 4.4.2.
Graph of the solution in the (x1, t) plane from the previous figure.
Figure 4.4.3.
Phase portrait of Eq.(1) in the (x1, x2)-plane for p = -0.4, q = 0.995.
The origin is an asymptotically stable
fixed point.
Activities:
- Click on the first picture to load it into your local copy of Phaser.
Verify that the values of x1(t) agree with the formula x1(t) = (-3)t(1 - t).
- Click on the third picture to load it into your local copy of Phaser.
Change the parameter q = 1.
(PhaserTip:
Changing Parameters)
Clear and Go.
Do you notice any qualitative change in the picture?
- Change the parameter q = 1.01.
Clear and Go.
Do you notice any qualitative change in the picture?
-
[ Previous Section
| Next Section
| Main Index
| Phaser Tips
]
|