资源预览内容
第1页 / 共59页
第2页 / 共59页
第3页 / 共59页
第4页 / 共59页
第5页 / 共59页
第6页 / 共59页
第7页 / 共59页
第8页 / 共59页
第9页 / 共59页
第10页 / 共59页
亲,该文档总共59页,到这儿已超出免费预览范围,如果喜欢就下载吧!
资源描述
1.3 Solutions to Dynamical System A dynamical system: an+1 = r an, a0 = an initial value. In some cases, the behavior predicted by the dynamical systems is characterized by the mathematical structure of the system. In some cases, there are wild variations in the behavior caused by only small changes in the initial values of the dynamical system. In some cases, small changes in the proportionality constants cause wildly different predictions.1The Method of Conjecture The method of conjecture is a powerful mathematical technique to hypothesize the form of a solution to a dynamical system and then to accept or reject the hypothesis. Example 1 A Savings Certificate Revisited A savings certificate initially worth $1000 accumulated interest paid each month at 1% of the balance. No deposits or withdrawals occurred in the account, determining the dynamical system an+1 = 1.01 an, (1.5) a0 = 1000.2Conjecture For k = 1, 2, 3, , the term ak in the dynamical system (1.5) is ak = (1.01)k a0. (1.6)Test the Conjecture We test the conjecture by examining if the formula for ak satisfies the system of Equation (1.5) upon substitution.(1.6) is true for k = 0, suppose it is true for k = n, then for k = n + 1 we have an+1 = 1.01an = 1.01(1.01)n1000 = (1.01)n+11000, which make (1.6) true for every positive integer n, so the conjecture (1.6) is accepted.3Conclusion The solution for the term ak in dynamical system (1.5) is ak = (1.01)k 1000, or ak = (1.01)k a0, k = 0, 1, 2,This solution allows us to compute the balance ak in the account after k months directly, for example, after 10 years, a120 = (1.01)120 1000 $3303.90, and after 30 years, a360 = (1.01)360 1000 $35949.64.4THEOREM 1 The solution of the linear dynamical systeman+1 = r an with r any nonzero constant is ak = rk a0, where a0 is a given initial value.5Example 2 Sewage Treatment A sewage treatment plant processes raw sewage to produce usable fertilizer and clean water by removing all other contaminants. The process is such that each hour 12% of remaining contaminants in a processing tank are removed. What percentage of the sewage would remain after 1 day? How long would it take to lower the amount of sewage by half? How long until the level of sewage is down to 10% of the original level? 6Solution Let the initial amount of sewage contaminants be a0 and let an denote the amount after n hour, we have an+1 = an 0.12an = 0.88an, so from Theorem 1 the solution is ak = (0.88)k a0.After 1 day, k = 24, a24 = (0.88)24 a0 = 0.0465 a0. 7Half the original contaminants remain when ak = 0.5 a0, 0.5 a0 = (0.88)k a0, k = log 0.5/log 0.88 = 5.42.The level of sewage being down to 10% of the original one gives (0.88)k a0 = 0.1 a0, k = log 0.1/log 0.88 = 18.01.8Long-Term Behavior of an+1 = r an for r ConstantValues for which a dynamical system remains constant at those values, once reached, are called equilibrium values of the system.r = 0 Constant solution and equilibrium value at 0 r = 1 All initial values are constant solutions r 1 Growth without bound (Examples and Figures are omitted) 9Dynamical System of the Form an+1 = r an + b, Where r and b Are Constants Definition A number a is called an equilibrium value or fixed point of a dynamical system an+1 = f(an) if ak = a for all k = 1, 2, 3,when a0 = a. That is, ak = a is a constant solution to the dynamical system.A consequence of the definition is that a is an equilibrium value for an+1 = f(an) if and only if a = f(a) when a0 = a.10Example 3 Prescription for Digoxin Recall that digoxin is used in the treatment of heart patients. Suppose we prescribe a daily drug dosage of 0.1 mg and know that half the digoxin remains in the bloodstream at the end of each dosage period. This results in the dynamical system an+1 = 0.5an + 0.1.11Now consider three starting values, or initial doses: A: a0 = 0.1 B: a0 = 0.2 C: a0 = 0.3The numerical solutions for each case are in following table and figure:12ABC nananan 00.100000000000000.20.30000000000000 10.150000000000000.20.25000000000000 20.175000000000000.20.22500000000000 30.187500000000000.20.21250000000000 40.193750000000000.20.20625000000000 50.196875000000000.20.20312500000000 60.198437500000000.20.20156250000000 70.199218750000000.20.20078125000000 80.199609375000000.20.20039062500000 90.19980468750000.20.20019531250000 100.199902343750000.20.20009765625000 110.199951171875000.20.20004882812500 120.199975585937500.20.20002441406250 130.199987792968750.20.20001220703125 140.199993896484380.20.2000061035156313Figure 1.18 Three initial digoxin doses Note that the value 0.2 is an equilibrium value, and when we start below the equilibrium (Case A) or above the equilibrium (Case C), we approach the equilibrium value as a limit.
收藏 下载该资源
网站客服QQ:2055934822
金锄头文库版权所有
经营许可证:蜀ICP备13022795号 | 川公网安备 51140202000112号