University:
University of California, Berkeley
Course:
COMPSCI 164 | Programming Languages and Compilers
Academic year:

2020
Views:

259

Pages:

3
Author:

larbiwxcvbn

Cobwebbing As an alternative to algebraically solving for a closed-form solution, we can solve graphically using cobwebbing. To do so, we plot the updating function and add the line y = x to the graph, which makes the process work. To cobweb, begin with the initial condition, M₀, and locate that point on the function. Draw a horizontal line from this point until it intersects the line y = x. From this point of intersection, draw a vertical line until it intersects the function again to find M₁. Repeat this process. Ex 1.6.1 Our tree-growth model was governed by the rule ht+1 = ht + 1. Use cobwebbing to find subsequent tree heights, starting with h₀ = 1. Sol'n: The updating function is f(ht) = ht + 1. This is a line with slope 1 and h-intercept 1 (point-slope form). Thus, graph with a line y = x and a line f(ht) = ht + 1, showing the cobwebbing process Cobwebbing provides us with a qualitative answer, as it is incumbent on one's ability to accurately graph by hand, limiting its accuracy. In the previous example, our lines were parallel. What if the lines cross? Such points are called equilibria. As we will see, the change between successive intervals decreases near such points. At such a point, the output is exactly equal to the input. Definition: A point M* is an equilibrium point of Mt+1 = f(Mt) if f(M*) = M*. **1.6.2** The concentration of medicine in a patient's blood is halved every day, and then increased by 1 mg/L. Thus, Mt+1 = 0.5Mt + 1.0. Let M₀ = 1 mg/L. Solve graphically. **Sol'n:** We have a slope of 0.5 and vertical intercept of 1. [Image of a graph with a line y = x and a line f(Mt) = 0.5Mt + 1, showing the cobwebbing process] If we had chosen M₀ = 2, the concentration would have decreased to reach 2. To find the equilibrium points algebraically, replace Mt+1 with Mt on both sides of the dynamical rule, set them equal, and solve for M*. Ex 1.6.7: Suppose bt+1 = rbₜ, r ∈ ℝ. Find the equilibrium. Sol'n: b* = rb* b*(1-r) = 0 ⇒ b* = 0 is the equilibrium point. Note that in the special case r = 1, all b* are in equilibrium. Ex 1.6.8: Mt+1 = 0.5Mt + S, S is a dosage parameter. Find the equilibria. Sol'n: Set M* = 0.5M* + S 0.5M* = S M* = 2S Equilibrium is proportional to daily dosage. Ex 1.6.10: Mt+1 = (1-α)Mt + S, α is % absorbed daily. Find the equilibrium points. Sol'n: M* = (1-α)M* + S αM* = S M* = S/α The more absorbed, the lower the equilibrium. The more injected, the higher the equilibrium.