Thinking Graphically about Difference Equations

Preliminaries — A Brief Review

Recall the graph of a line:

DE-equilib-01.gif

And the "45 degree line" (equation Y=X — slope 1, y-intercept 0)

DE-equilib-02.gif

Step 1 : General form of a difference equation

(1)
\begin{equation} P_{n} = a P_{n-1} + b \end{equation}

The next value equals something times the previous value plus some increment.

In our compound interest example, a was 1 plus the interest rate and b was zero. In our population models, a was 1 plus the birth rate minus the death rate and b was the recruitment or immigration per time period. In our weasel examples a was 1 plus the reproduction rate and b was the number killed each year by hunters.

Step 2. Plot the change from step to step

(2)
\begin{equation} P_{n+1} = a P_{n} + b \end{equation}

This looks a lot like the equation for a line (if we think of pn+1 as y and pn as x). This makes sense since the very essence of difference equations is to express the next value as a function of the previous value (we might write next=f(previous) and this is the same as we do for a line: y=f(x)).

DE-equilib-03.gif

This is an odd little graph. How would we use it? Let's suppose some pn is some number C. We locate this on the horizontal axis. Then, to find pn+1 we go up to the line and across to the corresponding value on the vertical axis. Call this number D.

DE-equilib-04.gif

What comes next? Now D will be pn and we'll seek the next value. We locate D on the horizontal axis and repeat the process.

DE-equilib-05.gif

Now we know three points – three "states" of the system in succession. If we look up at the line and imagine how we have "moved" along it, we can depict how the system has moved.

DE-equilib-06.gif

Now

DE-equilib-07.gif

Step 3. Recall that at equilibrium, the system stays the same from one period to the next.

Call the value at which the system settles pe "p sub e" or the equilibrium value. It is still governed by the generic difference equation but it looks like this

(3)
\begin{equation} P_{e} = a P_{e} + b \end{equation}

This can be solved for pe:

(4)
\begin{align} P_e = \frac {b}{(1-a)} \end{align}

And, the equation can be written out in our usual terms, it looks like this

(5)
\begin{equation} p_{(n+1)} = p_{n} \end{equation}

But this is just the equation for a "45 degree line" – a line with slope 1 that goes through the origin (that is, the point 0,0).

DE-equilib-08.gif

Any time the system is at equilibrium it will be somewhere on this line – since, by definition, equilibrium is when pn=pn+1
Thus, if we draw a 45 degree line on the graph we drew above, we can locate the equilibrium.

DE-equilib-09.gif

Note that in the example above our line had a slope of less than 1. What happens if we have a line with a slope greater than 1?

DE-equilib-10.gif

For positions both above and below the equilibrium, the tendency is for the system to move AWAY from the equilibrium.
The difference we are recognizing here is between STABLE and UNSTABLE equilibria. In a stable equilibrium, a small change in the system results in a "self-correcting" move back to the equilibrium. In an unstable equilibrium, a small perturbation or bump results in a sharp and accelerating movement AWAY from the equilibrium point.
Consider these real world examples.

DE-equilib-11.gif
http://demo.physics.uiuc.edu/lectdemo/descript/801/comp.jpg
DE-equilib-12.gif
http://www.answers.com/topic/unstable-equilibrium
DE-equilib-13.gif
http://content.answers.com/main/content/img/oxford/Oxford_Sports/0199210896.stable-equilibrium.1.jpg
DE-equilib-14.gif
http://serc.carleton.edu/introgeo/models/EqStBOT.html
DE-equilib-15.gif
http://phys101.blogspot.com/2006/04/question-stable-and-unstable.html

What Can We Learn from the Slope of the Pn+1=f(Pn) Line?

Something we've seen graphically is very interesting. When the line describing our difference equation crosses the 45 degree line with slope less than one we get a stable equilibrium. When the line crosses with a slope greater than one we get an unstable equilibrium.
Let's think for a second whether there is any intuition in this. Recall that

(6)
\begin{align} P_e = \frac {b}{(1-a)} \end{align}

Consider a point one unit away from Pe. Since the slope of the line is a the next point would be Pe+a. If a<1 then our new point is closer to Pe than Pe+1 was. If a>1 then the new point is further away.
What is a is negative? If we move one unit away from equilibrium, what happens? Our next point is at Pe+a but this is on the other side of Pe since a is negative. A little thinking will get us to the fact that the point after this will again be to the right of Pe. With a negative sloping line our sequence oscillates. But does it converge or diverge. It turns out that the same rule holds as before. For absolute value greater than 1 we get divergence (an unstable equilibrium) and for absolute value less than 1 we get convergence (stable equilibrium).

Let's try our step by stepping with the following two diagrams

DE-equilib-17.gif

References and Resources

BLOSSOMS - Fabulous Fractals and Difference Equations