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这是一篇加拿大作业案例分享是应用数学作业限时测试
Week 1
The idea of discrete models in applied mathematics and science which are formulated in terms
of dierence equations (as opposed to continuous models which are formulated in terms of dierential equations).
Some simple examples: Radioactive decay, population growth.
Iteration dynamics for the simple dierence equation/discrete dynamical system, xn+1 = cxn,
for which x = 0 is the unique xed point, with solution xn = cnx0. For |c| < 1, x = 0 is an attractive xed point, for |c| > 1, x = 0 is a repulsive xed point. Phase portraits associated
with this dynamical system.
You are not required to memorize any discrete models. Instead, you should be able to use any
given information to construct a discrete model and analyze it. (This includes the sampling of a continuous solution x(t) of a dierential equation and nding the dierence equation satised by its samples xn = x(nT) for some T > 0.)
Week 2
Additional examples of dierence equations, e.g., simple interest vs. compound interest.
The simple Euler dierence scheme to provide approximations to the exact solution of the
rst order DE,
dy/dx= f(x,y), y(x0) = y0 . (1)
(Nothing more complicated than this, e.g., the quadratic Euler scheme which was the subject of a Bonus Problem in a later Problem Set.)
The general dierence equation xn = axn1 + b. It helps to know that this involves the iterationof the function f(x) = ax+b with xed point ˉx =b1a for a 6= 1. The dynamics of this iteration scheme depends on a, e.g., if |a| < 1, then the xed point ˉx is attractive, etc.. Finite dierences (Lecture 4) is intended to be supplementary. Theory of dierence equations, starting on Page 44, Lecture 4. There is not much on Page 44 – just the idea of a solution of a d.e.. Linear dierence equations. This is the main subject of this section. You should know the general form for an nth order linear dierence equation (Page 45) and the idea of the existence and uniqueness of solutions to it (Page 46). Homogeneous solutions, general solutions Second order linear dierence equations (starting in Lecture 5, Page 49). Fundamental set of solutions and the condition involving their determinants (Wronskians) in Theorem 5 (Page 50). Linear second order dierence equations with constant coecients (Page 55), in the general form of Eq. (79), i.e.,yk+2 + pyk+1 + qyk = 0, k ≥ 0. (2) Assuming a solution of the form yk = mk yields the associated characteristic (quadratic) equation, Eq. (82), m2 + pm + q = 0. (3) The three cases to consider: (i) two distinct real roots, (ii) equal real roots and (iii) complex conjugate roots. Week 3 Characteristic equation associated with second order linear homogeneous d.e. with constant coecients (cont’d): (iii) complex conjugate roots. In this case, extract real and imaginary parts to produce two, linearly-independent real-valued solutions. Asymptotic behaviour of solutions to linear second-order homogeneous d.e.’s with constant coecients. Determined by roots of characteristic equation. The two most important cases: – Both roots mi have magnitudes less than one (Page 64, top), summarized in Theorem 1. – Both roots mi have magnitudes greater than one (Pages 65-66). Another important case: distinct real roots m1 6= m2, with |m1| > 1 and |m2| < 1 (Page 67). Then all nonzero solutions behave as Yk C1mk 1 as k →∞. Analysis of a discrete model for the propagation of annual plants – you don’t have to know the details of this model. The most important part was the analysis, i.e., to determine the conditions – in terms of the parameters used in the model – for which the plant species will grow and not decay to zero. You should be able to analyze a given, simple model in terms of its parameters. Solutions to inhomogeneous linear second-order dierence equations with constant coecients – the Method of Undetermined Coecients. It is sucient to know Case 1 (rk = a), Case 2 (rk = ak) and Case 3 (rk = kn), as well as complications that result when rk coincides with a root of the characteristic equation. Don’t worry about Cases 4-6.