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 y_{1} = f(y_{0}), y_{2} = f(y_{1}) = f(f(y_{0})), y_{3} = f(y_{2}) = f(f(f(y_{0}))) = f ^{3}(y_{0}) In general, y_{n} = f^{ n}(y_{0}) 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 y_{0} = 1000, y_{1} = .3y_{0} + 1000, y_{2} = .3y_{1} + 1000 = .3(.3y_{0} + 1000) + 1000 y_{3} = .3y_{2} + 1000 = .3(.3(.3y_{0} + 1000) + 1000) + 1000 = 1000 + .3(1000) + .3^{2}(1000) + .3^{3}y_{0} In general, y_{n} = 1000(1 + .3 + .3^{2} + .3^{3} + ... + .3^{n1}) + .3^{n}y_{0} The first term is a geometric series, so the equation can be written as
1000(1  .3^{n}) 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  r^{n}) 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_{n}^{2} @ 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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