SOLUTIONS Chapter 1 Problem 1.1. 2,t-i) = —#2,t-i and Problem 1.2. Because |/0^0. Problem 1.3. a) No, because the deviations do not converge to zero, b) Yes, because the deviations remain bounded. Problem 1.4. a) P = 4 and b) yt = -j/t-i, 2/o = -1: P = 2. Problem 1.5. Prom (1.13) 1 1 + a (7-M)" =« X = i where 5 = 1/(1-a/3). Problem 1.6. A characteristic vector 5 and the corresponding characteristic value A are determined by Xs = Ms, that is, coordi- natewise XjSij = ctS2j and XjS2j — Psij. After multiplication: X'JSIJS2J = a(3sijS2,j- No characteristic vector may have coordinate zero, because then the other coordinate would also be zero. Therefore A^ = a(3. Thus we may normalize as siy3; = 1, that is, AjS2,j = A that is, 52,i = ^/®/f3 and 52,2 = -y/a//3. From (1.20) x0 = £isi + 6^2, whence the given initial state x0 = (#i,o>£2,o) yields ^-. (1.21) pro- vides the solution. 283 284 Solutions Problem 1.7. First we solve the scalar equation #1>t = rriiiXi^-i +wi. We substitute the solution obtained into the scalar equation X2,t = ™>22%i,t-i + ^2i^i,t-i + ^2, and so on. The only complication arises from the exponential terms on the R.H.S., which can be tackled, see (1.36). Problem 1.8. Column j of matrix M shows the image of the unit vector e^, thus magnification and rotation are combined. The characteristic equation is P(X) = A2 - 2p(coscp)A + p2 = 0 [(1.27)- (1.28)], yielding the foregoing characteristic roots. The geometric meaning also suggests this result. Problem 1.9. According to Problem 1.6, Mi has two character- istic values: 1 and -1. According to Problem 1.7, both characteristic values of Mi are equal to 1, however, there exists only a single inde- pendent characteristic vector: s = (0,1). Problem 1.10. Xt = Xxt-i 4- w where #*, A and w are scalars. Theorem 1.1: if A ^ 1, then x° = w/(l - A). Theorem 1.2: trivial. Theorem 1.3: empty. Theorem 1.4: stability — 1 < A < 1. Theorem 1.5: 0 < A < 1. Theorem 1.6: $ = 1/|A|. Theorem 1.7: xt = xt-i is meaningless. Theorem 1.8 reduces to Theorem 1.4. Problem 1.11. a) Ai = A2 > 0. b) at least one sign-change, at least two sign-changes. Problem 1.12. a) 0 l/a 0 **•( b) 6n = 0. c) Example 1.15. d) Similarly to Problem 1.6, the characteristic equation of matrix / — B(k) is P(X) = (A — I)2 — af3kik2 = 0, whence A = 1 ± y/aflkik2, Ai > 1. Chapter 2 Problem 2.1. Programming. Problem 2.2. Insert the stationary values into (2.18)-(2.19). Equations y° = Y°l+c and Y° = A{y°) yield the traditional equation l y° = Ay° + c, whence y° = (/ - A)~ c, Y° = A{y°). Considering the stationary conditions (2.20)-(2.21): system y° + (d)z° = y*,Y° + Dx Z° = Y* has a positive solution in (z°,Z°), assuming that y° < y* and Y° < Y* hold. Solutions 285 Problem 2.3. For |A| = 1, expression + e2 - is denned for all -1 < SftA < 1. The hyperbolic function |TT(A)|2 reaches its maximum at one end point. Since TT(1) = (/? — «)/(l — e) and TT(—1) = ((3 + a)/(l + s), under our assumptions, 0 < TT(1) < TT(—1) hold. Problem 2.4. (See Lovell, 1962 and Martos, 1990.) For uniform norms and reactions, b = (1 + 7e)l = /Jl holds, that is, matrix (2.43) is a rational function of AT, that is, A is a likewise function of v. A = 1 - (1 - u)e - (1 - i/)/?(l - vP)~lve. Problem 2.5. Let now fa = bi + fc^, a; = 6; and si = 1 - &;, 7rt = pi, z = 1,... ,n. On the basis of Problem 2.3 one can prove again that, due to 7Ti(—1) < 0, p[iV(—p(—1))] < 1 is sufficient but generally not necessary. If, however, N is 2-cyclic, then —p(N) is also a dominant characteristic value, that is, p[N(—p(—l))] < 1 is also necessary. Chapter 3 Problem 3.1. a) f(x) > x holds, though 0 < f'(x) = 1 - (1 + ex)~2ex < 1. b) Interval (—00,00) is not compact. Problem 3.2. a) The map is indeed not a contraction, since for (3 = 1 and x0 = 0.01, x\ = 50.005, while for y0 = 1, yx = 1. b) Comparing the arithmetic and geometric means yields xt > y//5 (t > 1). Then 0 < f'(x) = 1/2 - p/(2x2) < 1, that is, contrac- tion. Note that /'(>//?) = 0, that is, the convergence is very fast. Problem 3.3.* Draw the diagram known from the cobweb -cycle. After some trials one can discover the following case by case approach. a) Figure 3.1:Ifl<a<2, then the fixed point f(x°) - x° lies in (0;l/2). Due to symmetry, for x* = 1 - x°, f(x*) = x°. If x* < x0 < 1, then x± < x°; if 1/2 <x0 < x*, then x° < xi < 1/2. We have demonstrated that the system can stay to the right from 1/2 for at most one time-period. If x° < xt < 1/2, then x° < xt+i < xt\ 286 Solutions if 0 < xt < x°, then Xt < Xt+i < 1/2. Because of monotonicity, the limit exists, which is naturally a fixed point, that is, is equal to x°. b) Figure 3.2: If 2 < a < 3, then there exists such a number 0 < x\ < 1/2, for which f{xl) = 1/2. Because of symmetry, x\ = 1 — xj, f\x\) = 1/2. If xl < x0 < 1, then xx < x\. If 1/2 < x0 < z$, then x\ < xi < 1/2. If xt < x\, then xt < xt+i < x\. If x\ < xt < 1/2, then xt < #t+i < x2; that is, sooner or later the system arrives to the open interval (xJjXj) and stays there. Then the contraction map theorem is applicable, because * 1 =*= c) For a = 2, there is a borderline case between a) and b). Problem 3.4.* If matrix A (or B) is invertible, then the proof of the Lemma is elementary: ABx = Ax, y = A~xx => BAy = Ay. Hence via a limit-argument the Lemma applies for any matrix P A. Thus p(AB--C) = p(B--CA), that is, matrix D/ (x2) = D/(xi)D/(xP) • • -D/(x2) is stable. Problem 3.5. a) x° = 2 - 2x° => x° = 2/3. b) x2 = 2xi and xi = 2 - 2x2 = 2 - 4xi => xi = 2/5 => x2 = 4/5. c) Out of three, either two points are less than 1/2 or two points are greater than 1/2. Then continue as in b). d) Unstable, because |/'(-)| > 2, or 4,8 > 1. Problem 3.6. sin2<p = 2 sin </? cos </? implies <pt = 2<pt-i, and so on. Problems 3.7-3.9. Programming. Chapter 4 Problems 4.1—4.3. Programming. Problem 4.4. In fact, we used (4.28') rather than the time- invariance of At at (4.29). Problem 4.5.* Introducing the general control-theoretic frame- work with bounds, we have the state space equation xt = xt-i + But, and the planned control equation u\ = — {k)xt and the actual control equation u\ \iv?it<u\\ u?t \iu\<v?it<v%', i = l,...,n. { a ll a a i i,t — i' where u\ and uf are lower and upper control bounds, respectively. Solutions 287 Similarly to Section 2.2, we can generalize the analysis not only to the extended stock signalling model but to any Metzlerian control model. Chapter 5 Problem 5.1. Because of f(t,x) — \x, with notation h — t/k, the xk{ti,k) = Xk(U-i,k) + \xk(U-i,k)h = (1 + \h)xk(U-i,k). According to the formula of the geometric series, the solution is Xk(t) = #fc(0)(l + Xt/k)k, therefore for k —• oo, xu{t) converges to ext. Problem 5.2. Due to f[r,x(r)] = XX(T), the fcth integral equa- tion is xk+1{t) = x(0) + X / xk{r)dT. Jo For approximation 0, xo(t) = 1, the formula holds. Mathematical induction: Plug in approximation k and integrate by terms: the inte- gral of term j is equal to XXH^ 1/[(j + l)j!], which is just term j + 1, j = 0,1,..., fc, and term 0 remains 1. Problem 5.3. For function f(x) = x2/3, the difference-quotient [f(x) - f(0)]/(x - 0) = rc"1/3 is unbounded around x « 0. Problem 5.4. a) dx/dt = Xx =>• x~ldx = Xdt => logx = A£ -f xt+c c Xt c => x(t) = e ^ x(0) = e => x(t) = xoe . b) dx/dt = x2lz => x~2lzdx = dt => 3x^3 = t => x(0 = (^/3)3. c) dx/dt = x2 => x~2dx = dt => -x~l = t + c => x(t) = -l/(t H- c) => x(0) = -1/c and so on. Problem 5.5. Using (5.17*)-(5.18*) and Theorem 5.7, we have A = i and s = (1, —i)/2. Then 5Rs = (l,0)/2 and Ss = -(0,l)/2. Problem 5.6. We shall apply Theorem 5.8. According to the relation among the roots and coefficients, Ai + A2 = —a and Ai A2 = /?. a) If a system is stable, then the real parts of both roots are negative, that is, a > 0 and (3 > 0 (either real or complex), b) Assume a > 0 and 0 > 0.