Math Problem Set #3 Solution 19 February 2001

Transcription

1 Math Problem Set #3 Solution 19 February 2001 Exercises: 1. B & D, Section 2.3, problem #3. In your answer, give both exact values and decimal approximations for the amount of salt in the tank at the end of ten minutes and at the end of twenty minutes. Solution: Let S 1 (t denote the amount of salt in the tank at time t for t = 0 to 10 minutes. Then S 1 (t obeys a differential equation S (t = (flux in (flux out. The flux in is equal to the concentration of salt in the water flowing in times the rate of flow. That is, flux in = 1 2 lb gal 2 gal min = 1 lb min. The flux out is equal to the concentration of salt in the tank times the flow rate. That is, flux out = S(t 100 lb gal 2 gal min = S(t 50 lb min. The initial condition is S 1 (0 = 0, and so our initial value problem for S 1 reads S 1(t = 1 S 1(t This is a first-order linear differential equation; we can easily solve it using the method of integrating factors. Moving the S 1 term to the left side and multiplying through by e t 50, we get e t 50 S 1 (t + e t S 1 (t = e t We recognize the left side as the derivative of e t 50 S1 (t, so we may integrate both sides to get e t 50 S1 (t = 50e t 50 + C1.

2 Now using the initial condition S 1 (0 = 0, we deduce that C 1 = 50. We make this subsititution and multiply through by e t 50 to find the solution S 1 (t = 50 50e t After all that, the only thing we need from this equation is the amount of salt in the tank at t = 10, that is S 1 (10 = 50 50e 1 5 = 50 ( 1 e lb. Let S 2 (t denote the amount of salt in the tank during the next ten minutes. The function S 2 (t obeys a differential equation similar to the equation for S 1, but with the inflow equal to zero. Thus S 2 (t = S(2 We can solve this differential equation by inspection. It is clear that S 2 (t = C 2 e t We now use the results from the earlier part of the problem as an initial condition for S 2. We have Thus from which we get Thus we have S 2 (10 = S 1 (10 = 50 ( 1 e ( 1 e 1 5 = S2 (10 = C 2 e 1 5 C 2 = 50 ( e S 2 (t = 50 ( e e t To answer the question, we need only evaluate S 2 at t = 20. We get S 2 (20 = 50 ( e e lb.

3 2. Consider the first-order difference equation y n+1 = f(y n. A number y is an equilibrium solution or fixed point of this difference equation if y = f(y. (a Determine (algebraically the equilibrium solution to the difference equation y n+1 = y n (b Use a calculator or computer to approximate the first ten terms in the solution to the initial value problem y n+1 = y n /2 + 2; y 0 = 1. Based on your results, make a conjecture about the behavior of the terms y n as n goes to infinity. (c Draw a stair-step diagram to illustrate the solution to the initial value problem in part (2b. (d Repeat parts (2a through (2c for the difference equation y n+1 = y n /2 + 6 with the same initial condition. Solution: (a We need to solve the equation y = y This is simple enough. We get y 2 = 2, so y = 4 is the fixed point. (b I used a TI-85 to run this difference equation. I set y1 to 0.5x + 2, then stored 1 in x, entered y1 [STO] x, and pressed the enter key repeatedly. The first ten terms of the solution (to five significant figures are 1, 2.5, 3.25, 3.625, , , , , It appears that the y i are increasing toward the equilibrium solution of 4. (c Here is the stair-step diagram for this function. It shows the solution increasing toward the equilibrium point.

4 x (d We find the equilibrium solution by solving y = y y = 6 2 y = 4. Running the difference equation on the TI-85 as above, we get the sequence 1, 5.5, 3.25, 4.375, , , , , , It appears that the terms of the sequence are again approaching 4, but this time they are alternately greater than and less than 4. Here is the stair-step diagram for this problem. It shows the solution zeroing in on the equilibrium from alternating sides.

5 x 3. Consider the difference equation y n+1 = y n y3 n y n 3y 2 n 1. (a Determine (algebraically the equilibrium solutions for this differential equation. (b Use a calculator or computer to guess the values of lim n y n for y 0 = 0.4, y 0 = 0.5, and y 0 = 0.6. Notice anything strange? (c (Optional, for extra credit, brownie points, and so on Define a function L(α by L(α = lim n y n when y 0 = α. In part (3b we found L(0.4, L(0.5, and L(0.6. Produce a graph of L and zoom in on the the number between α = 0.4 and α = 0.5 where L changes values. Solution: (a The equation we need to solve is y = y y3 y 3y 2 1. Clearly this implies that y ± 1/3 and that y 3 y = 0. We can factor y 3 y as y(y 2 1. The roots are thus 0 and ±1. These are the three fixed points of the given difference equation.

6 (b I used a calculator to run the difference equation. For y 0 = 0.4, the terms approach 0 rapidly. In fact, y 4 is on the order of 10 12, and the calculator reports y 5 as exactly 0. (This just means the calculator can t distinguish y 5 from 0. Thus we conjecture that lim n y n = 0. For y 0 = 0.5, we find y 1 = 1, and since 1 is a fixed point, it follows that lim n y n = 1 in this case. For y 0 = 0.6, after ten iterations the calculator reports that all terms are equal to 1. So we conjecture that lim n y n = 1. What s strange about this? Why does the solution starting at 0.5 converge to a fixed point that is less than the solution starting at 0.4? (c I ve put this solution in a separate Maple worksheet. 4. Consider the logistic difference equation y n+1 = ρy n (1 y n. (a When ρ = 3.2, the logistic difference equation has a stable 2-cycle. That is, there are two numbers x 1 and x 2 such that the sequence x 1, x 2, x 1, x 2, x 1, x 2,... is a solution to the difference equation. Use a calculator or computer to approximate a and b. (HINT: start the difference equation running at almost any initial condition, and the solution will approach the stable 2-cycle. (b The numbers x 1 and x 2 in part (4a satisfy the equations x 2 = ρx 1 (1 x 1 x 1 = ρx 2 (1 x 2. Use one of these equations to make a substitution in the other and thus find a quartic equation (with coefficients in terms of ρ that is satisfied by x 1. (c Set ρ = 3.2 in the equation from part (4b, and try to solve the equation. One of the roots will be zero; use a computer or calculator to approximate the other three. Verify that x 1 and x 2 from part (4a are both roots of the equation. Can you find exact expressions for x 1 and x 2? (d When ρ = 3.48, the logistic difference equation has a stable 4-cycle. Use a calculator or computer to approximate the four numbers in this cycle. Solution: (a I used the TI-85 to do this. First I set y1 equal to r x (1-x, then set r to 3.2. Then I stored 0.2 in x, entered y1 [STO] x and pressed the enter key repeatedly.

7 After twenty iterations or so, the process settled down to the numbers x and x (b Making the suggested substitution, we get x 1 = ρ [ρx 1 (1 x 1 ] [1 [ρx 1 (1 x 1 ]] = ρ 2 x 1 (1 x 1 [1 ρx 1 + ρx 2 1]. We cancel a factor of x 1 from each side (this eliminates the root x 1 = 0, and subtract 1 from each side to get 0 = (ρ 2 ρ 2 x 1 (1 ρx 1 + ρx = (ρ 2 1 (ρ 2 + ρ 3 x 1 + 2ρ 3 x 2 1 ρ 3 x 3 1. The quartic mentioned in the problem is simply x 1 times this cubic expression. (c Let r 1, r 2, and r 3 denote the roots of this cubic. Using the root feature of the TI-85, we estimate the roots to be r r r It appears that r 1 and r 3 are the numbers in our 2-cycle. To find exact expressions for r 1 and r 3, we appeal to Maple, Derive, Mathematica or any other computer algebra program that knows the cubic formula. (We might also choose to look up the cubic formula and apply it by hand, but we would find that very difficult and time-consuming. Maple tells me that the three roots are ρ 1 ρ, ρ + 1 ± ρ2 2ρ 3. 2ρ Evaluating these roots numerically (with ρ = 3.2 confirms that the numbers ρ + 1 ± ρ 2 2ρ 3 2ρ = 21 ± agree with the numbers we found using the iterative process above. (d I followed the same procedure as in the first part of this problem. After about sixty iterations, the solution settles down to the repeating sequence , , ,