First Order Difference Equations

Differential equation are great for modeling situations where there is a continually changing population or value.  If the change happens incrementally rather than continuously then differential equations have their shortcomings.  Instead we will use difference equations which are recursively defined sequences.  Examples of incrementally changes include salmon population where the salmon spawn once a year, interest that is compound monthly, and seasonal businesses such as ski resorts. 

What makes this first order is that we only need to know the most recent previous value to find the next value.  It also comes from the differential equation

        y'  =  g(n,y(n)) 

Recalling the limit definition of the derivative this can be written as 

 if we think of h and n as integers, then the smallest that h can become without being 0 is 1.  The differential equation becomes

        y(n+1) - y(n)  =  g(n,y(n))

        y(n+1)  = y(n) +g(n,y(n))

Now letting 

        f(n,y(n))  =  y(n) +g(n,y(n))

and putting into sequence notation gives

        y_n+1  =  f(n,y_n)

If the first order difference depends only on y_n (autonomous in Diff EQ language), then we can write

        y₁  =  f(y₀),    y₂  =  f(y₁)  =  f(f(y₀)),     y₃  =  f(y₂)  =  f(f(f(y₀)))  =  f ³(y₀)

In general, 

       y_n  =  f ⁿ(y₀)

Solutions to a finite difference equation with 

       y_n+1  =  y_n

Are called equilibrium solutions.  We find them by setting 

        y_n  =  f(n,y_n)

An finite difference equation is called linear if f(n,y_n) is a linear function of y_n.  

Example

Each year, 1000 salmon are stocked in a creak and the salmon have a 30% chance of surviving and returning to the creak the next year.  How many salmon will be in the creak each year and what will be population in the very far future?

Solution

This is a linear finite difference equation with 

        y_n+1  =  .3y_n + 1000

We have 

        y₀  =  1000,    y₁  =  .3y₀ + 1000,    y₂  =  .3y₁ + 1000  =  .3(.3y₀ + 1000) + 1000

        y₃  =  .3y₂ + 1000  =  .3(.3(.3y₀ + 1000) + 1000) + 1000  =  1000 + .3(1000) + .3²(1000) + .3³y₀

In general, 

        y_n  =  1000(1 + .3 + .3² + .3³ + ... + .3^n-1) + .3ⁿy₀

The first term is a geometric series, so the equation can be written as

                    1000(1 - .3ⁿ)
         y_n  =                             + .3ⁿy₀
                        1 - .3

Notice that the limiting population will be 1000/.7  =  1429 salmon.

More generally for the linear first order difference equation

        y_n+1  =  ry_n + b

The solution is 

                    b(1 - rⁿ)
         y_n  =                             + rⁿy₀
                        1 - r

Recall the logistics equation 

        y'  =  ry(1 - y/K)

 After some work, it can be modeled by the finite difference logistics equation 

        u_n+1  =  ru_n(1 - u_n)

The equilibrium can be found by solving 

        u_n  =  ru_n(1 - u_n)

A quadratic that has solution 

        u_n  =  0    or    u_n  =  (r - 1)/r  

To determine the stability of the equilibrium points, look at values of u_n very close to the equilibrium value.  For the first point, u_n is much larger than (u_n)²,  so the logistics equation can be approximated by 

        u_n+1  =  ru_n(1 - u_n)  =  ru_n - ru_n²  @ ru_n 

For |r| < 1, this converges to 0, thus the equilibrium point is stable.

For the other equilibrium value, write

        u_n  =  (r - 1)/r  + v_n  

So that the equilibrium value becomes 

        v_n  =  0

We can now substitute into the difference equation and chop off the nonlinear term to get

         v_n+1  =  (2 - r)v_n 

This converges to 0 for 

          |2 - r| < 1

or

        1 < r < 3

So the equilibrium point is stable in this range.

At r  =  1, we say that there is an exchange of stability.  For r > 3, the sequence exhibits strange behavior.  In particular for 3 < r < 3.57 the sequence is periodic, but past this value there is chaos.

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