Equilibria Analysis in Dynamic Systems: Unstable Equilibria in Non-Linear Systems - Prof. , Assignments of Mathematics

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Math 170 Assignment #8 Solutions

EX-1). Discuss the equilibria of the following three systems:

Rn+1 = R^3 n − 3 Rn (1) Sn+1 = Sn + 1 (2) Tn+1 = Tn (3)

SOLUTION: In system (1), we find that the graph of y = x^3 − 3 x intersects y = x at − 2 , 0, and 2, so these are our equilibrium points. However, starting at any nonequilibrium point and plotting the first few steps of our spider diagram, it seems that the system never begins to settle down toward any particular equilibrium. Thus we suspect all three equilibria are unstable. Indeed they are. To see this, we will recall what we saw in Exercise 2 of Homework 7. In that case, we considered general linear systems Rn+1 = a + bRn and we saw that when − 1 < b < 1 the equilibrium point was stable and was unstable otherwise. Although this system is not linear, we can approximate it by a linear system near each of the equilibrium points (notice that although the graph is a curve, it does not curve very much if we look at it up close, hence close up we can think of it being straight). This idea of straightening the curve out a little bit is an application of derivatives; namely we approximate the curve near a given point by its tangent line at that point. In particular, taking the derivative of x^3 − 3 x we have 3x^2 − 3. This is the slope of the tangent line at a given point x, so we evaluate this at the equilibrium points to find equations for the tangent lines to the curve at the equilibria. At −2 and 2 our tangent line has slope 9, while at 0 the tangent line has slope −3. Thus we can calculate equations describing the tangent lines at these points: at x = −2 the tangent line is y = 16 +9x, at x = 0 the tangent line is y = − 3 x, and at x = 2 the tangent line is y = −16+9x. Since each of these equations is linear, it is in the form y = a + bx. Since the three b-values of our tangent lines at the equilibria all lie outside the critical range of − 1 < b < 1, we see that these equilibria are all unstable as we suspected. Note that this does not imply that NO point converges to a given equilibrium; rather if we choose a starting position randomly, it will almost certainly not converge to any equilibrium point but will instead bounce around the graph forever.

We can view system (2) as modeling the amount of money a gradeschool bully acquires over time. Each day he bullies the hapless nerd in the class into giving him his lunch money. So each day his money increases by $1 (school lunches are apparently very cheap at his school). Since the hapless nerd is vig- ilant about coming to class every day, the bully always makes a buck and thus there is never a point where he stops bullying. Equivalently there is no equi- librium in this system, since there is no amount of money which will cause the bully will stop taking the hapless nerd’s money. Indeed if we graph this system, its graph runs parallel to our y = x line, hence never intersects y = x and in particular we verify visually that the system has no equilibrium points.

System (3) is just as easy to analyze. Think of this system as representing the number of rocks in a pile. Left to itself, the rocks will not reproduce to make more rocks for the pile, nor will they spontaneously disappear. Thus, in the absence of outside influence, any size pile will remain the same size forever. It follows that EVERY size pile is an equilibrium in this system. Are these equilibria stable or unstable? Well if we add a rock to our pile, the pile will stay at the new amount of rocks instead of moving back toward the original number of rocks. Thus every point is an unstable equilibrium in this system.

6.3-5). Draw another stage of the argyle process. SOLUTION: If done correctly, you will have 24 little diamonds in your pic- ture, and 4 line segments at the corner of the picture.

6.3-13). Describe the locations of infinitely many points in the Sierpinski Dust fractal. SOLUTION: As is often the case in math, we will reduce this to something we have already done. Look at the bottom edge of our original triangle. When we remove the hexagon, we also remove the middle third of this edge. Sound familiar? It should be. If we keep removing hexagons, we will be removing mid- dle thirds from the segments of this bottom edge and hence the bottom edge will become a copy of the Cantor Set. In Homework 7, we already described the points in the Cantor set, so in the last homework we already described an infinite set of points in the Sierpinski Dust fractal without knowing it at the time.

6.3-20). Draw the third stage of replacing (1, 2 ,

√

5)-triangles by 5 smaller triangles. SOLUTION: See page 255 in the book for an idea of what this process will look like.

6.3-21). Draw the first few steps of a Koch Stool. SOLUTION: If I have to explain to you how to draw, you should probably think about repeating kindergarten.

6.3-34). Draw the first three stages of a Sierpinski Carpet which removes the middle square of a 5 × 5 grid at each step. SOLUTION: See solution to previous problem.

6.3-35). Draw the first three stages of a Sierpinski Carpet which removes the middle four squares of a 4 × 4 grid at each step. SOLUTION: I’ll give you two guesses as to which previous solution I will refer you to. If you use both guesses, you might seriously consider retaking preschool as well.

6.3-39). In the last two problems, how much of the final pattern is gold after infinitely many steps. SOLUTION: Instead of recommending you repeat kindergarten as in the last few problems, I will instead give a solution to this one. In both these cases instead of adding up the areas of the gold regions at each step, it is easier to

add up the area that remains purple and subtract this from the total area of the quilt (this is exactly like what we did in the previous homework when we calculated the final length of the Cantor Set). For convenience we will take the quilt to have area 1. Now in (6.3-34) we removed the middle square of a 5 × 5 grid, so we are left with 2425 of the original area as purple. At each subsequent step, each purple square is again subdivided in this way, so the amount of purple left after n steps is just

( 24

25

)n

. Since 2425 < 1, this quantity goes to 0 as n goes to infinity. Thus after infinitely many steps the gold will have total area 1 and the purple will have total area 0. Similarly in (6.3-35) we replace each purple square by 12 smaller purple squares and 4 smaller gold squares, so that the amount of purple left after n stages is

( 12

16

)n ; again since 1216 < 1 this quantity must go to 0 as n goes to infinity, hence after infinitely many steps the gold will have total area 1 as in the previous case.