Dynamical Models and Optimal Control

Transcription

1 . Dynamical Models and Optimal Control A friendly introduction Solutions of esercizes Sandro Salsa Annamaria Squellati

2 Chapter. Solutions. Salsa-Squellati, Dynamical systems. The (horizontal) straight line x =is a constant solution. Then, it solves the problem with initial condition x () =. Assume x 6=,wefind Z dx (x ) = x Z te t dt, = te t e t + c, c R. The Cauchy problem x () = is satisfied if c =, and the corresponding solution is x (t) = te t e t +.. (a) This is a separable equation and x =is a particular solution. Separation of the variables brings to and, integrating, we find dx x = t t dt, x 6=, t 6= ±, ln x = ln t + c, c R. x = k t (k = e c hence, k is a positive constant). In order to interpolate the solutions along the intervals (, ), (, ) and (, + ), and to get rid of the absolute value appearing in the last formula, we let k be also negative or, and solutions can be extended to R. Then, the general solution is x = k(t ), k R. (b) If t 6= the equation is equivalent to x = Separation of the variables, and an integration yield to Z Z ( + x )dx = t dt, x + x x + x /9 k ex/+x t( + x ). = log t + c assuming c =logk as k> = logk t = t. Let k have any sign; then, the general solution can be written as t = k ex/+x /9, k R \{}. The image of the solutions t = t(x) is (, ), ifk< or (, + ), ifk>.

3 . Salsa-Squellati, Dynamical systems (c) This is a linear equation, with continuous coefficients in intervals such as I k =(kπ, (k +)π) (k Z). Using the formula (.) in page 5, we find µ x = e Z cot tdt c + cos te cot tdt dt = Z = e µc log sin t + cos te log sin t dt = sin t in any interval where sin t is positive, we get: x = sin t µ Z c + µ Z c + while, in any interval where sin t is negative, we get: hence, the general solution is in any I k. x = sin t (d) If t 6= ±, the equation is equivalent to µ c cos t sin t dt ; Z cos t sin tdt ; cos t sin tdt ; x = µc sin t cos t, c R x = tx t + tx t (.) and x =is a particular solution. Furthermore, the function P (t) =Q(t) = t is continuous in (, ) (, ) (, + ); thus,the t local existence theorem holds. As x 6=,wedividebyx both sides of equation, obtaining x x = t t x + t t. Substituting z = x, the bernoulli equation is transformed into a linear one; we have z = x x and (.) becomes z = t t z t t. Applying the formula in page 7 in each intervals (, ), (, ), (, + ), weobtain z(t) = e Z t t µc dt t t e t t dt dt = = p à Z! t t c p t = t = p à Z! t t c p t sign( t )dt = = p Ã! t c p = c p t. t

4 . Salsa-Squellati, Dynamical systems On account of the initial substitution x = z, finally, we find x(t) = c p t. (e) Denote f(t, x) = t + x tx ; f is a continuous function in any quadrant of the plane (axis t and x are excluded). The equation can be rewritten as + x x = t x. The substitution z = x t gives t and it can be integrated as a separable equation z z + tz = +z z, z dz = t dt = log t + c, c R z = p log t + k, k =c. The general solution is x = t p log t + k, k R. (f) Let A = {(t, x) R : t 6= and +tx 6=} namely, the subset of the plane (t, x) without the point belonging to the x axis and the points of the line described by the equation +tx =.InAthe differential equation can be equivalently rewritten as This is an exact equation, since (x +tx )dt +(t +t x )dx =. (x +tx ) x = (t +t x ) t =+6tx. The general solution is found using formula (.4) in the page 4. For instance, we choose (t,x )=(, ) (check that the denominator is not zero), and have Z x t +t s ds = c, c R, tx + t x = c.. (a) This is a linear differential equation; we solve it using the formula (.): y (x) = e x s/(+s )ds = Z x +x Z x log( x) dr = r +x. r ( + r ) e r s/(+s )ds dr =

5 4. Salsa-Squellati, Dynamical systems (b) This is a separable equation. Separation of the variables leads to e y dy = xe x dx; using formula (.9), we find Z y e t dt = Z x te t dt e y = xe x + e x, y = x +ln(x +). 4. The general solution of the associated homogeneous equation is x (t) =ke t, k R (a) Given that f is a second order polynomial, we use the similarity method to search for a particular solution such as u (t) =at + bt + c. Computing u, and substituting into the equation, we obtain that is at + b at + bt + c = t +t at +(a b) t + b c = t +t Thechoiceoftheparametersa, b, c is performed solving the system a = a b = b c =, leading to a = b = c =. The particular solution is u (t) = t t. (b) We search for a particular solution such as u (t) =a sin t + b cos t. Computing u and substituting, we are yield to a cos t b sin t (a sin t + b cos t) =cost that is (a b)cost +( b a)sint =cost Hence, the parameters a and b solve the system ½ a b = a b =. We find a = 5 b = 5 and the particular solution is u (t) = 5 sin t cos t. 5

6 . Salsa-Squellati, Dynamical systems 5 (c) As a first candidate, we consider the particular solution u (t) =ae t.computingu and substituting into the equation, we find ae t ae t e t which is false independently on the choice of the parameter a. Indeed, the function x = e t is a solution of the homogeneous equation x x =. Then, we consider the particular solution u (t) =ate t. Computing u, and substituting into the equation, we deduce ae t +ate t ate t e t which is true if a =. A particular solution is u (t) =te t. 5. The elasticity of a function f is E f (x) = xf (x) f (x). Assuming x> and f(x) >, inthefirstcasewehave f (x) f(x) = a + b x, ln f(x) = ax + b ln x + c, f (x) = kx b e ax (k>). In the second case, we have f (x) f(x) = ax + b + c x, ln f (x) = ax + bx + c ln x, f (x) = kx c e ax /+bx (k >). 6. Any continuous function on a compact set is bounded; hence, since f x (t, x) is continuous on K, thereexists M such that f x (t, x) M for every (t, x) K. Fix t, for every (x,t), (x,t) K we apply the Lagrange theorem to the function f (t, x) in the interval [x,x ]:wefind thus f (t, x ) f (t, x ) x x = f x (t, c), c (x,x ) f (t, x ) f (t, x ) = f x (t, c)(x x ) M x x. Then, f is lipschitz with constant M with respect to x, uniformly with respect to t. 7. Consider s b, weapplytof in the interval [a, s], the fundamental theorem of integral calculus: We deduce: f(s) f(a) = f(s) f(a) = Z s a f (u) du. Z s a Z s a k(s a) f (u) du f (u) du k s + k a

7 6. Salsa-Squellati, Dynamical systems Figura.. hence, f(s) f(a) + k s + k a. Consider now any (t, x ) S. We apply the previous formula in the x direction. We have: f(t, x) f(t, x ) + k x + k x c + k x since f(t, a) is a continuous function over the compact [a, b], and then it is bounded. 8. We have x (t) =a [M x (t)], whose solution is x (t) =M ( e at ). 9. As shown in the phase diagram Figure., equilibria are x =(semistable) and x =(unstable).. Let ϕ (t) be a T -periodic solution of the equation x = f (x), then ϕ (T )=ϕ(). Applying the Rolle theorem, there exists a point t where ϕ (t )=. If x = ϕ (t ), then substituting into the equation, we obtain f (x )= and then the solution of the problem ½ x = f (x) x (t )=x is the constant ψ (t) x. The local existence and uniqueness theorem requires that ϕ and ψ coincide, thus ϕ is constant.. (a) The function f (x) = x is derivable on R\{}. The local existence and uniqueness theorem hypothesis are then satisfied for k 6=. (b) If k =, x =is a solution. We search for the general solution of the equation. Consider x 6=, separation of the variables gives dx x = dt x = t + c, c R s µ x = ± t + c

8 The functions and s t µ x = t t> s t µ x = t t>. Salsa-Squellati, Dynamical systems 7 both solve the problem with the initial condition x () = but they are locally different functions. We argue that the (local) uniqueness of the solution does not hold, not even locally.. We denote and deduce f(t, x) = t x +t x, f x (t, x) = t + t x ; then, f is continuous for (t, x) (, + ) [, + ), andf x is continuous for (t, x) (, + ) (, + ). The continuity of f ensures the existence of at least one solution. For x =, thesufficient hypothesis guaranteeing uniqueness does not hold. It is immediate to see that x =is a solution. On the other hand, integration of the Bernoulli equation, when you set z = x,gives z = t z + t, z= t(t + c), c R, and we deduce the general solution x = t (t + c). These solutions are defined for t(t + c). In particular, for c = we find a solution whose domain is t (and it can be connected with x =for t<). This problem has, then, multiple local solutions. The particular solution x =is also known as boundary solution, since it coincide with the boundary of the set where the existence and uniqueness theorem holds.. The hypothesis (a) and (b) bring to the equation d I(t) =αis βi dt recalling that S(t)+I(t) =N, we deduce that the illness evolution model is described by the differential equation d dt I(t) =(αn β)i(t) αi(t). (.) Denote by N ( <N <N) the number of individuals infected at time t =. Integrating the separable equation (.), for αn β =,wehave: I(t) = N αt + and then lim I(t) =; t +

9 8. Salsa-Squellati, Dynamical systems meaning that the whole population is due to become healthy. As αn β 6=, the equation (.) can be integrated using the Bernoulli technique or, again, considering it as a separable equation. Its solution is (αn β)n I(t) = (αn αn β)e (β αn)t + αn and we deduce that lim I(t) = t + <N< β α αn β N> β α α. Then, two kind of longtime behavior might be expected: if <N β, the infection will vanish; instead, α if N> β αn β the number of infected people will reach the endemic level. α α 4. We assume that F C (R ).Thedifferential equation of the family is found eliminating c between the equations ½ y +x 4cx + c = y +4x 4c =, (the second equation is obtained by differentiation of the first one). Solve the latter equation with respect to c, and substitute into the former to obtain µ y y +x (y +x +4x)x + = that is y 4xy 4x +8y =. 5. Equilibria solve the equation f (x) =x x + mx = x x + m x =.Accordingtothevalueofm, we have: X if m>, x =; X if m =, x =and x =; X if <m<, x =and x =± m (both of them are positive). Since f (x) =x 4x + m, weobtain f () = m>. Then, for every m > the point x = is an unstable equilibrium. The phase diagram representing the cases m =and m> is in Figure.. If m =, the point x =is a semistable equilibrium; if <m< the two equilibria x = m and x =+ m are respectively asymptotically stable and unstable equilibrium. Chapter. Solutions. The general solution of the first problem is µ s n = c n, c R; as a consequence, for every c R we have s n.

10 . Salsa-Squellati, Dynamical systems 9 Figura.. The second problem is solved by s n = c ( ) n +, c R. If c =, we deduce s n = (wehaveaconstantsolution).otherwise,wefind an oscillating solution and s n.. As n +, wehave: (a) s n +, (b) s n =4 n. (c) s n does not converge; indeed {s n } = {4, 4, 4, 4, 4, 4,...}.. In the first case, we search for a particular solution such as u n = an + b where a and b are real parameters. Substituting u n = an + b, and u n+ = a (n +)+b in the equation, we find a (n +)+b = ³ a (an + b)+n n + a + b = n. We deduce a =,b= 4. Then, a particular solution is u n =n 4. In the second case, we search for a particular solution such as u n = a n where ar to be determined. Substituting in the equation u n = a n, and u n+ = a n+ =a n,wehave a n = a n + n n a =. Then, a particular solution is u n = n. 4. We consider the equation and rewriting it for t +,t+,...,wehave Y t = λy t + X t, t (.) Y t+ = λy t + X t+ Y t+ = λy t+ + X t+ Y t+ = λy t+ + X t+...

11 . Salsa-Squellati, Dynamical systems Divide the first by λ, the second by λ, and so on, to deduce λ Y t+ = Y t + λ X t+ λ Y t+ = λ Y t+ + λ X t+ λ Y t+ = λ Y t+ + λ X t+.. Add side by side; cancelation of the terms like λ Y t+, λ Y t+,... yields to =Y t + + X i= µ i X t+i. λ Since {X t } is bounded and λ>, we point out that the series formula gives that Y t = + X i= µ i X t+i λ + X i= µ i X t+i is convergent. The last λ is a particular solution of the equation (.). Thus, the general solution of the equation is Y t = cλ t they diverge all, but the particular solution. 5. We consider the function g is continuous in [a, b]. Furthermore, + X i= µ i X t+i, c R; λ g (x) =f (x) x; g (a) =f (a) a, g(b) =f (b) b. If f (a) =a, then a is a fixed point for f; iff (b) =b, then b is a fixed point for f. If neither f (a) =a, nor f (b) =b, we have g (a) > and g (b) <, and hence the continuous function g has at least one zero in (a, b), i.e. there exists x (a, b) and g (x )=f (x ) x =. As a consequence x solves the equation f (x) =x, namely, it is a fixed point for f. 6. On account of the continuity of f, wehave lim n + f (x n)=f lim x n n + µ = f (l) Using the recurrence equation, the uniqueness of the limit and the fact that x n+ l, we deduce thus, l is a fixed point for f x n+ = f (x n ) l = f (l),

12 . Salsa-Squellati, Dynamical systems 7. First, we consider x >x. We want to prove that for every n>, p (n) : x n >x n. We shall use mathematical induction on the statement p(n). First, we prove that the statement is true for n =; then we prove that if the statement is true for n, thenitistrueforn +. By hypothesis, p (n) is true for n =. We prove that p (n) p (n +). Assuming p (n) is true, since f is increasing, we deduce f (x n ) >f(x n ), that is x n+ >x n then, p (n +)is true. Induction concludes that p (n) is true for every n ; thatis,x n is increasing. We can repeat the reasoning for x <x, deducing that x n is decreasing. In any case, x n is monotonic, hence, it is regular. 8. The generatrix function f (x) = +x has the fixed point x =; this is an asymptotically stable equilibrium point, in fact f () = 4. Given that f is increasing, the sequence s n is regular for every a (> ) ; as the Figure. shows, we see that for <a< it is increasing; for a =it is constant (s n ) and for a> it is decreasing. In any case, s n. 9. (a) We explicitly write the sequence x n : Figura.. x = f (x )=f (f (s )) = f (s )=s, x = f (x )=f (f (s )) = f (s )=s 4,. x n+ = f (x n )=f (f (s n )) = f (s n+ )=s n+.

13 . Salsa-Squellati, Dynamical systems The proof for y n is analogous. (b) Since f is increasing, x n and y n are monotonic. If x n = s n is increasing, namely if s n <s n+,we deduce that s n+ = f (s n ) >f(s n+ )=s n+ since f is decreasing; thus, y n = s n+ is decreasing. The other cases are analogous. (c) If x n = s n x (x is a fixed point for f ), then the continuity of f implies that y n = s n+ = f (s n ) f (x ).Ify n = s n+ y (y is a fixed point for f ), then, the continuity of f implies that x n = s n = f (s n ) f (y ). Thus, the uniqueness of the limit gives x = f (y ) and y =(x ). In particular, if f has the same fixed points of f, we deduce x = y. (d) If x = y = l then s n l, s n+ l and hence s n l. Viceversa, if s n l then we have also x n = s n l and y n = s n+ l, thus, x = y = l.. Introducing x n+ = f (x n ), the generatrix function is f (x) = µ x + A. The fixed points of f solve x µ x + A ; x x = µ Ax they are x = ± A.Giventhatf (x) = and f ³± A asymptotically stable equilibria. Consider, for instance, the recurrence sequence defined by the equation x n+ = µ x n + 7 x n x =4. =, both of them are locally As the phase diagram in Figure.4 shows, the solution of the problem (.4) converges to 7. We find (.4) Figura.4. Phase diagram for the solution of the problem (.4) We recall that, if f is decreasing, then f f is increasing. We remind that any fixed point for f is a fixed point for every iterate f k. The converse is not true.

14 . Salsa-Squellati, Dynamical systems x = µ 4+ 7 = 4 8 =.875, x = µ ' x = µ ' x 4 = µ ' After three iteracions, we have three exact decimal digits.. The generatrix function is f (x) = +x. In order to determine the fixed points of f, we solve the equation x = +x, that is x + x =.Thefixed points are x = and x =.Sincef (x) =, we deduce that ( + x) Given that f ( ) = and f () =. f ( ) > and f () <, the point x = is an unstable equilibrium, and the point x =is an asymptotically stable equilibrium. For a =, the solution is constant. For every a>(a 6= ), the corresponding solution s n converges to with oscillating behavior. We verify this fact considering, for instance, a (, ). Using the results of exercise 9, the sequences s n and s n+ are monotone (and regular). In particular, if s (, ),s n is increasing and s n+ is decreasing. Furthermore, we find that f (x) =f (f (x)) = has the same fixed points of f, since the equation +/ ( + x) = +x +x +x +x = x is equivalent to x + x =.Thus,s n and s n+ and, as a consequence s n. The case a> canbeprovedusingthesameargument. The behavior of s n is sketched in Figure.5.. The equilibrium equation is Anditissolvedby p t+ = p t + σ (a bp t + c dp t )=[ σ (b + d)] p t + σ (a + c) p t =[ σ (b + d)] t (p p )+p, Where p = a + c b + d. Then p t converges to p, if and only if σ (b + d) <, namely if σ< b + d.. (a) The generatrix function is f (x) =xe x /. We draw its graph in the first quardrant, where the function is non negative. We deduce that f () = and lim x + f (x) =+.

15 4. Salsa-Squellati, Dynamical systems Figura Figura.6. Graphoff (x) =xe x / Since f (x) = x e x /, f (x) = as x =; henceforth, x =is a maximum. (b) The fixed points of f solve the equation xe x / = x. They are Given that x = and x =. ³ f () = e> and f = < both of them are unstable equilibria.

16 4. (a) The equilibrium points solve the equation x = f (x) = they are x =and, when α 6=, x = α β.since we have f (x) = αx +βx ; α ( + βx), µ α f () = α, f = β α.. Salsa-Squellati, Dynamical systems 5 Therefore, if <α<, x=is a stable equilibrium, and x = α (< ) is an unstable equilibrium; if β α>, x=is an unstable equilibrium and x = α is a stable equilibrium. β In the case α =, the only fixed point is x =and since f () =, the linearization criterion cannot be applied. A deeper analysis would be necessary to determine the stability nature of this point. The dynamics in a neighborhood of the fixed points is monotonic since f is positive for every x, and thus f is increasing. (b) When α = β =, the phase diagram is represented in the Figure.7, and we deduce that, for every x > x n (indeed, this is the behavior for every β) Figura.7. (c) Consider We have x = x +x, x = x +x + x +x = ( xn+ = x n +x n x >. x +x, x = x +x + x +x = x +x,...,

17 6. Salsa-Squellati, Dynamical systems and we deduce converging to zero for every x >. x n = x +nx, Chapter 4. Solutions. The linear differential equations are: (a) x (t) =; (b) x (t)+x (t) =, (b) x (t)+5x (t)+6x (t) =.. (a) The characteristic equation of the associated homogeneous equation is λ = whose solutions are λ = ±. A candidate particular solution is u (t) =a sin t, and we deduce a =. The general solution is x (t) =c e t + c e t sin t c,c R. (b) The characteristic equation of the associated homogeneous equation is λ += whose solutions are λ = ±i. A particular solution of the complete equation is u (t) =t. Thegeneral solution is x (t) =c cos t + c sin t +t, c,c R. (c) The characteristic equation of the associated homogeneous equation is λ λ += whose solutions are λ =± i. As a particular solution of the complete equation we use u (t) =at + b, deriving and substituting into the differential equation we deduce a =,b= 7. The general solution is x (t) =e t (c sin t + c cos t)+t + 7, c,c R.. (a) The general solution is x (t) =c cos t + c sin t, c,c R. The constants c and c are found solving the system ³ π x = c = ³ π x = c = π. The solution of the Cauchy problem is x (t) = π cos t. (b) The general solution of the differential equation is x (t) =c e t + c e t et.

18 . Salsa-Squellati, Dynamical systems 7 In order to solve the Cauchy problem, we compute x x (t) = c e t +c e t et. On account of the initial conditions, we deduce the system c + c = c +c =; hence, c = c = 4. The solution of the Cauchy problem is x (t) = e t + 4 et et. 4. The solution of the characteristic equation is (with multiplicity ). (a) Since is not a solution of the characteristic equation, as a candidate particular solution we choose u = At + Bt + Ct + D. Substituting into the equation, we find A, B, C, D, andhave u = t 6t. (b) Since is not a solution of the characteristic equation, while is (with double multiplicity), as a candidate particular solution we choose u = At e t + Be t. Substituting into the equation, we obtain u = t e t + e t. (c) Since ± i are not solutions of the characteristic equation, as a candidate particular solution we choose u = e t (A sin t + B cos t). Substituting into the equation, we have u = e t sin t. 5. (a) Therootsofthecharacteristicequationare and ± i. The general solution is Ã! Ã!! y = c e t + e Ãc t/ cos t + c sin t, where c,c,c are arbitrary real constants. The positive root of the characteristic equation implies that the zero solution is an unstable equilibrium. (b) The characteristic equation s root are and ± i, both with multiplicity.

19 8. Salsa-Squellati, Dynamical systems The general solution is ³ ³ y = c + c t +(c + c 4 t)cos t +(c 5 + c 6 t)sin t, where c i,i=,...,6 are arbitrary constants. The characteristic equation has roots with zero real part; given that they are not simple, the zero solution is an unstable equilibrium. 6. (a) Therootsofthecharacteristicequationare±i, with multiplicity. Since is not a solution of the characteristic equation, as a particular solution, we choose u = At + Bt + C. Substituting into the equation, we are yield to find A =,B=, C = 4. The general solution is then y =(c + c t)sint +(c + c 4 t)cost + t 4, where c i R, i=,...,4. (b) Therootsofthecharacteristicequationare, with multiplicity and, simple. Since is a double root of the characteristic equation, as a particular solution we choose u = e t (At + Bt + C). We find A = 6, B = 7,C= 7. The general solution is µ y = c e t + e t c + c t + t 7 t 7 + t4, 6 where c,c,c R. (c) The roots of the characteristic equation are and. Since and ±i don t solve the characteristic equation, as a particular solution we choose u = Ae t + B sin t + C cos t. We are yield to find A =, B= 65,C= The general solution is y = c e t + c e t/ e t + 65 sin t 8 cos t, 65 with c,c R. 7. The characteristic equation is λ λ + λ =, whose solutions are and (double root), corresponding to the following solutions of the differential equation:, e t,andte t. Given that is a double root of the characteristic equation, as a particular solution we choose We obtain that u(t) =At e t. u (t) =A(t +t)e t, u (t) =A(t +4t +)e t, u (t) =A(t +6t +6)e t

20 . Salsa-Squellati, Dynamical systems 9 and, substituting into the differential equation, we obtain: A(t +6t +6)e t A(t +4t +)e t + A(t +t)e t = e t, and hence Thus, the general solution is A =. y = c + c e t + c te t + t e t. (.5) The constants c,c,c are to be determined in order to satisfy the initial conditions. Deriving (.5), we obtain: y = c e t + c (t +)e t + t +t e t y = c e t + c (t +)e t + t +4t + e t. On account of the initial conditions, we are yield to the system c + c = c + c = c +c +=, solved by c =, c =, c =. The solution of the Cauchy problem is y = + µ t + t e t. 8. The zero solution is stable, thanks to the stability criterion for third order equations (we have positive coefficients and 6 9 > ). 9. The characteristic equation is λ +δλ + ω =. We analyze three cases. If δ>ωthe characteristic equation has two (negative) distinct real roots λ = δ p δ ω and λ = δ + p δ ω and the general solution is y (t) =C e λt + C e λt, C,C R. If δ = ω the characteristic equation has one double root λ = δ and the general solution is y (t) =(C + C t) e δt, C,C R. p If δ<ωthe characteristic equation has complex roots λ = δ ± i ω δ and the general solution is y (t) =e δt (C sin νt + C cos νt), C,C R, p denoting ν = ω δ. Anyhow,wehavethatlim t + y (t) =.

21 . Salsa-Squellati, Dynamical systems Chapter 5. Solutions. The characteristic equation is λ 5λ +6= whose solutions are λ =and λ =. Therefore, the general solution is x n = C n + C n, C,C R. In order to satisfy the Cauchy problem x =and x =,wesolvethesystem ½ C + C = C +C =, and deduce Then, the particular solution is. (a) Thecharacteristicequationis C =, C =. x n = n + n. λ λ +=, thus, λ =, and λ =. The general solution is x n = C µ n + C, C,C R. The zero solution is neutrally stable. (b) The characteristic equation is whose roots are ±i. The general solution is λ += x n = C sin n π + C cos n π, where C and C are real, arbitrary constants. The zero solution is neutrally stable. (c) The characteristic equation is λ 4λ +4=, then, is a -multiplicity root. A particular (constant) solution is u n =. Therefore, the general solution is x n = n (C + C n)+, C,C R. The constant solution is unstable.. (a) Thecharacteristicequationis λ λ += whose roots are λ =and λ =. We search for a particular solution of the complete equation such as x n = A n. Substituting into the equation, we find A =/. Then, the general solution is x n = c + c n + n, c,c R. (b) The characteristic equation is λ λ + λ =.

22 . Salsa-Squellati, Dynamical systems Its roots are λ =, and λ = ±i. We seek for a particular solution such as u n = An; weareyieldtofind A (n +) A (n +)+A (n +) An = and deduce A =. The general solution of the equation is x n = c + c cos n π + c sin n π, c,c,c R. 4. The characteristic equation is λ += with roots λ =, and λ, = ± i =cosπ ± i sin π. The general solution of the equation is x t = c ( ) t + c sin πt + c cos πt c,c,c R. Considering the initial value problem x =,x =,x =, we obtain the system c + c = c + c + c = c + c c =; solved by c =,c =, and c =. The solution of the Cauchy problem is We point out that then, x t is 6-periodic. x t =( ) t + x t+6 =( ) t+6 + sin πt. sin π (t +6) = x t 5. We tell two cases: λ,λ R and λ,λ complex conjugate. Case. Considering real roots, from λ < and λ <, we deduce λ λ <. If λ and λ have the same sign λ λ = λ λ, hence λ λ <. (.6) On the other hand, (.6) holds if the roots λ and λ have opposite sign or one of them is zero, too. From <λ < and <λ <, we deduce ( λ )( λ ) >, ( + λ )(+λ ) > (.7) namely and thus +λ λ >λ + λ, +λ λ > λ λ λ + λ < +λ λ. (.8) Viceversa, if (.8) holds, then (.7) and (.6) imply λ < and λ <.

23 . Salsa-Squellati, Dynamical systems Actually, (.7) hold if both roots are less than or if both roots are grater than, too;butinthese situations their product would be grater than. Case. If the roots are complex and conjugate, i.e. λ = α ± iβ, we have λ = λ = p α + β, λ λ = α + β and we immediately deduce that λ < and λ < λ λ <. The condition (.8) is true for every pair of complex and conjugate numbers; in fact λ + λ =α and (.8) becomes α < +α + β that is ( α ) + β >. 6. (a) The zero solution is asymptotically stable if and only if the roots of the characteristic equation are inside the circumference of radius (i.e. their modulus is less than ). The characteristic equation is its roots have a less than modulus if λ λ a = a <, < a, therefore, the zero solution is asymptotically stable if and only if <a<. (b) The characteristic equation is with roots λ λ 8 = λ = ± r 8. The general solution is x t = c à + r! t à + c 8 r! t, c,c R. 8 In order to find the solution satisfying the initial conditions x =, and x =, we determine c and c solving the system c à + c = r! à r! c + + c 8 =. 8 We are led to The solution is r à x t = r c = r c =. + r! t à 8 r! t. 8

24 . Salsa-Squellati, Dynamical systems 7. The model is described by the second order equation Y t ( s v) Y t + vy t = A ( + g) t. (.9) We search for a particular solution such as Y t = km t where m =+g. Substituting into (.9), we find km t ( s v) km t + vkm t = A m t m t km km ( s v)+kv A m = thus The particular solution is k = Y t = A m m ( s v) m + v. A m +t m ( s v) m + v. If the solutions of the homogeneous equation tend to as t +, namely if the roots of the characteristic equation λ ( s v) λ + v =are, in modulus, less than (i.e. when v<), then all the solutions of the complete equation tend to Y t. On the other hand, if the solutions of the homogeneous equation don t tend to, the solutions of the complete equation will diverge or oscillate, according to the sign of the discriminant of the characteristic equation. Chapter 6. Solutions. The eigenvalues of the coefficients matrix are λ = 4, λ =, corresponding, for instance, to the eigenvectors µ µ u = and u =. The general solution is µ x y = c µ e 4t + c µ where c and c are arbitrary constants. The zero solution is not stable.. The eigenvalues of the coefficients matrix are λ =,λ = corresponding, for instance, to the eigenvectors µ µ u = and u 4 =. The general solution is µ x y µ = c 4 + c µ where c and c are arbitrary constants. The initial condition gives e t e t ½ =c + c =4c +c, = ½ c = c = and, therefore, the solution if the Cauchy problem is µ µ µ x = + y 4 e t.

25 4. Salsa-Squellati, Dynamical systems. (a) From the first equation we deduce y = x + x, derivation gives y = x + x. Substitution into the second equation gives x + x = x x + x, namely The characteristic equation has complex roots λ =± i. Thus, x x +x =. λ λ += x = e t (c cos t + c sin t) where c and c are real constants. Now, we compute x = e t (c cos t + c sin t)+e t ( c sin t + c cos t) and y = x + x = e t (c sin t c cos t). The general solution of the system is The solution of the system can also be written as µ x y ½ x = e t (c cos t + c sin t) y = e t (c sin t c cos t). = c µ cos t sin t e t + c µ sin t cos t e t. (b) We solve the first equation, obtaining x = c e t. A substitution into the second equation gives y =y + c e t whose general solution is The system s general solution is y = c e t + c te t. ½ x = c e t y = c e t + c te t. 4. (a) The coefficients matrix is we find A = µ tr A = < and det A => therefore, the zero solution of the system is asymptotically stable. (b) The coefficients matrix has positive trace (thus, at least one eigenvalue is positive). We argue that the zero solution is unstable.

26 . Salsa-Squellati, Dynamical systems 5 5. The coefficients matrix has eigenvalues λ =, λ =,λ =, corresponding, for instance, to the eigenvectors u =, u = and u =. The general solution is x y z = c e t + c e t + c where c,c and c are arbitrary constants. The initial condition gives =c + c + c =c c, 4=c + c +c = c = c = c = e t then, the solution of the Cauchy problem is x y = e t + z e t + e t. 6. The coefficients matrix is triangular; then, the elimination method may be suggested. We find x = c e t y = c e t + c e t z = c e t + c te t + c e t. The initial condition gives =c =c + c, = c + c = and, then, the solution of the Cauchy problem is x = e t y = e t e t z = e t te t +e t. 7. Both systems have an unstable zero solution. c = c = c = 8. (a) The coefficients matrix is µ A = a a The origin is asymptotically stable if and only if. det A > and tra < ; that is, if and only if hence, for (b) The coefficients matrix is a > and +a<, A = a<. µ 4

27 6. Salsa-Squellati, Dynamical systems whose eigenvalues are corresponding, for instance, to the eigenvectors µ λ = and λ = and µ 4. The general solution is µ x u = c µ e t + c µ 4 e t, c,c R. Chapter 7. Solutions. (a) The eigenvalues of the coefficients matrix are ± i. They are complex and their real part is positive, hence the origin is an unstable focus. (b) The coefficients matrix has the -multiplicity (non regular) eigenvalue. Therefore, the origin is an improper node.. We have d dt E (x, y) =(x + y) x +(x +y) y =(x + y)( x y)+(x +y)(x + y) =. The origin is a center.. (a) The origin is an isolated critic point. The eigenvalues of the coefficients matrix µ A = µ µ are and, and two corresponding eigenvectors are, for instance, and. The steady state (, ) is an unstable proper node; the linear manifolds are y =and y = x. The straight line y =is a particular solution. Since ẏ =if and only if y =, the trajectories have no points with horizontal slope. The vertical-slope isocline is y = x. The (homogeneous) equation of the trajectories is y = dy dx = y x + y for x + y 6=. Introducing y = zx, we obtain: z + xz = z z +, xz = z + z z +. We deduce that z =and z = are particular solutions ( corresponding to y =and y = x). As y 6=,andy6= x, wehave: z + z + z dz = z + dx, ln x z =ln x + k, k R.

28 . Salsa-Squellati, Dynamical systems 7 The equation of the family of trajectories is y + x = cy, c R, with the straight line y =. Then, parabolae with horizontal axis, with vertex on the straight line y = x are union of trajectories. (b) The coefficients matrix µ A = µ µ has the eigenvalues and, corresponding, for instance, to the eigenvectors and. The straight line y = x is a stable linear manifold, the straight line y = x is an unstable linear manifold. The straight lines y = x and y = x are horizontal- and vertical-slope isoclines, respectively. The (homogeneous) trajectories equation is y = dy dx = x + y x +y for x +y 6=. Introducing y = zx, we obtain: z + xz = +z z +, xz = z z +. The straight lines z =and z = are particular solutions corresponding to y = x and y = x. As y 6= x, andy 6= x, wehave: Z z + z dz = (z x dx, ln +)(z ) =ln x + k, k R. The orbits are represented by the equation (y x) (y + x) =c, c R. 4. The equivalent system is with the coefficient matrix We have We analyze the cases: ½ ẋ = y ẏ = x ky µ A = k. tr A = k <, det A =>, = k 4. <, =, >. If <k<, the origin is a stable focus: the solutions tend to the steady state with oscillations. If k =the origin is an improper node. If k> the origin is a stable proper node. For every k, the solutions are definitively monotone.

29 8. Salsa-Squellati, Dynamical systems 5. (a) The steady states solve the system ½ x + y = x + y =, they are (, ), (, ). In order to determine their nature, we use the linearization criterion. The jacobian matrix associated to the system is µ y A(x, y) =. Considering the point (, ), we have det A =, tr A =, =(tr A) 4detA = 8. Since A has complex eigenvalues (with positive real part), the point (, ) is an unstable focus for the linearized system. Considering the point (, ), wefind det A =. Since the eigenvalues of A are real and have opposite sign, the point (, ) is a saddle for the linearized system. In view of the theorem., these results can be extended to the original system. (b) The steady states solving the system ½ x + x y = y = are (, ) and (, ). The jacobian matrix associated to the system is µ +xy x A(x, y) =. At (, ), wededuce det A =, tr A =, =; therefore, (, ) is a (stable) proper node for the linearized system. At (, ) we deduce det A = ; hence, (, ) is a saddle point for the linearized system. According to the theorem., theseresultscanbe extended to the original system. 6. The origin is the only steady state. We prove that it is a stable star. Indeed, on account of the theorem., thecoefficient matrix of the linearized system is µ A = and given that f(x, y) = x + o ρ +ε, x y = f(ρ, θ) =ρ 5 (cos θ) (sin θ) ρ 5. Then, for every m, there exists a trajectory crossing the origin with slope m. Moreprecisely,theaxisx and y are union of trajectories (each straight line is made of three trajectories: two half straight line and the steady state).

30 In order to find the vertical-slope isoclines, we solve ẋ =,thatis x = and x y =;. Salsa-Squellati, Dynamical systems 9 we deduce that the line x y =is a genuine vertical-slope isocline, while x =is a trajectory, indeed. Instead, ẏ is zero if and only if y =, that is, the straight line y =is a horizontal-slope isocline and a trajectory of the system. We find no genuine horizontal-slope isoclines. The figure.8 describes the phase diagram in the first quadrant. Figura (a) The equilibrium, solving the system x + y = x +x + y = according to the conditions x>, y>,, is(, ). The linearization criterion proves that it is a saddle. In fact, the jacobian matrix of the system, at (, ) is µ / / / and it has real eigenvalues, with opposite sign. (b) The vertical-slope isocline is y = x. The horizontal-slope isocline is y = x +x. (c) We find ẋ> as y>x; and we have ẏ> as y> x. The phase portrait is in figure.9. +x (d) No. In fact, the origin is a center for the linearized system, since the jacobian matrix at (, ) is µ / / /

31 . Salsa-Squellati, Dynamical systems Figura.9. and it has pure imaginary eigenvalues. Hence, the conclusion of the linarization criterion cannot be applied to the original system. 8. (a) The equilibrium points solve the system ½ x +y = x y = and are (, ) and (, ). Using the linearization technique we find that (, ) is a stable proper node while (, ) is a saddle. (b) The vertical-slope isocline is y = x; the horizontal-slope isocline is y = x. (c) We have ẋ> as x<y, and ẏ> as y<x. The phase portrait is sketched in the Figure.. 9. The funcion y = f (x) is strictly decreasing; on the other hand, the function x = g (y) is strictly increasing. We deduce that there exists one only equilibrium point. We have ẋ> as x<g(y), and ẏ> as y>f(x). The phase portrait is represented in the Figure.. We observe that the steady state is a saddle.. The jacobian matrix of the system is namely, at (, ), Given that H (x, y) = µ y xy k y µ H = H (, ) = k. det H = and tr H = k

32 . Salsa-Squellati, Dynamical systems as k 6= we use the linearization criterion. Figura.. If k<, the zero solution is asymptotically stable for the linearized system and for the original system as well. As k>, the zero solution is unstable both for the linearized and the original system. As k =, we use the Liapunov technique. Consider the function this is (evidently) positive definite in R. We have V (x, y) =x + y, V (x, y) =x y xy +y x y = x y y 4 in R. Thus, V is a Liapunov function for the system, and therefore the zero solution is asymptotically stable. Chapter 8. Solutions. (a) The eigenvalues µ of the µ coefficients matrix are λ =and λ =; two corresponding eigenvectors are, for instance, and, respectively. The general solution is µ µ µ xt = c y + c t t,c,c R. (b) In order to solve the initial value problem we have to solve the system ½ =c + c = c +c. We obtain c =, c =, then, the particular solution is µ xt y t = µ.

33 . Salsa-Squellati, Dynamical systems Figura... The coefficients matrix is A = 4 and the associated characteristic equation is h i det (A λi) =( λ) ( λ) 4 =. Its eigenvalues are λ =,λ =and λ =4, corresponding for instance, to the eigenvectors v = 8, v =, v = 4 6, respectively, solutions of the systems x +y z = y +4z = y +z =, The general solution of the system is, then, x t y t = c ϕ + c z t y z = y +4z = y + z =, + c 4 t 4 6 x +y z = y +4z = y z =. c,c,c R. where ϕ = v,,,,... ª. In order to determine the solution starting from x =(,, ) T, we search for c,c,c solving =8c + c +4c = c +6c =c +c. We are yield to c =/4, c =, c =/4.

34 . (a) The coefficients matrix of the system is whose characteristic equation is A = µ 4 det (A λi) =(4 λ)( λ)+=λ 6λ +9=. We find the eigenvalue λ =with -multiplicity. The associated eigenspace solves ½ x + y = x y =. It has dimension, and is generated, for instance, by the eigenvector λ =. We search for a generalized eigenvector, solving the system ½ u + v =. Salsa-Squellati, Dynamical systems µ, associated to the eigenvalue µ We find, for instance,. u v =. The general solution of the system is µ µ µ xt = y t c t The zero solution is unstable. (b) The coefficients matrix of the system is A = + c µ +t t µ,c,c R. with characteristic equation det (A λi) =( λ) += with roots λ =± ³ i = cos π ± i sin π. An eigenvector associated to the eigenvalue + i solves ½ iu + v = µ For instance, we consider µ xt = y t c t + µ u iv =. i. The general solution of the system is πt πt cos sin sin πt + c πt, c,c R. cos The zero solution is unstable, since the modulus of the eigenvectors is grater than. 4. The coefficients matrix of the system is and we deduce µ..7 A =.6. tr (A) =.5, det (A) =.6, tr (A) =.5 < + det(a) =.64 therefore, using the stability criterion for -dimensional system, we deduce that the zero solution is asymptotically stable.

35 4. Salsa-Squellati, Dynamical systems 5. (a) From the first equation, we obtain y t = x t+ + x t and then y t+ = x t+ + x t+ ; a substitution into the second equation gives x t+ x t+ +x t =. The characteristic equation is λ λ += with roots ± i = ³ cos π 4 ± i sin π ; then, 4 µ x t = t/ c cos πt 4 + c sin πt, c,c R. 4 We deduce that µ y t = (t+)/ π (t +) c cos + c sin 4 µ π (t +) + t/ 4 µ = t/ c cos πt 4 c sin πt 4 + c sin πt 4 + c cos πt 4 µ = t/ c cos πt 4 + c sin πt 4 ; c cos πt 4 + c sin πt = 4 µ + t/ c cos πt 4 + c sin πt 4 = thus, the general solution of the system is µ x t = t/ c cos πt 4 + c sin πt 4 µ y t = t/ c cos πt 4 + c sin πt 4, c,c R. (b) From the second equation, we obtain x t = y t+ 4y t and x t+ = y t+ 4y t+ and substituting into the first equation, we have y t+ 6y t+ +9y t =; therefore, We deduce y t =(c + c t) t, c,c R. x t =(c + c (t +)) t+ 4(c + c t) t =(( c +c ) c t) t and, summing up, we find the general solution of the system ½ xt =(( c +c ) c t) t y t =(c + c t) t, c,c R. 6. We know that ½ It = K t K t Y t = C t + I t. Then, we deduce that is Y t = C t + I t = C t + K t K t =.6Y t +.Y t + a (Y t Y t ) Considering the characteristic equation Y t (.6+a) Y t +(a.) Y t =. λ (.6+a) λ + a. = we find =(a a )(a a ) with a '.77 and a '.. Weanalyzethreecases.

36 . Salsa-Squellati, Dynamical systems 5 If a<a,wehavecomplexroots:λ = ρ (cos θ ± sin θ),where ρ = a.. The solution is Y t = ρ t (c cos θt ± c sin θt), c,c R. If a. <, namely if a<., Y t has dumped oscillations and tends to zero;, if a =., Y t has bounded oscillations (constant in modulus); if a>., Y t has explosive oscillations. If a = a,wehavetherootλ ', with double multiplicity. The (diverging) solution is Y t = λ t (c t + c ), c,c R. If a <a, wefind two real roots λ and λ, where λ λ = a. >. The (diverging) solution is Y t = c λ t + c λ t, c,c R. 7. The number of graduated students forecasted in years is about Denoting the solution of the problem is x = x 5 = L 5 x The population grows exponentially, since the dominant eigenvalue of L is.474 >. 9. The solution of the system is x t = T t c + x where c has to be determined using initial number of working, under repair or testing cars, and x =(I T) b. The constant solution x t = x is asymptotically stable, since the eigenvalues of the matrix T have 4 modulus less than. For any initial condition, the number of the cars in running order, under repair or under testing, respectively, tends to x =(I T) b = b = b When T is given, we can govern the system with an exogenous choice. For instance, if the expected number of working cars is f = perweek,wehavetobuyb (working) cars, that is / per week... Denote T = , and i = γ,..85 We point out that years correspond to 5 time units used in the Leslie model. 4 With MatLab we easily compute them and find.5,.4786, and.9889.

37 6. Salsa-Squellati, Dynamical systems the difference equation describing the population is x t+ = Tx t + i. The equilibrium population x is found solving the system x = Tx + i, and we deduce x =(I T) i, that is x = γ = γ γ γ The equilibrium is asymptotically stable, since the eigenvalues of T are λ =.85, λ =.9, λ =.95, and they have modulus less than.. The steady states solve the system ½ x =(+a) x xy x y =.5y + xy. Equilibria are: (, ) and p =(.5,a.5), having positive coordinates if and only if a>.5. The jacobian matrix of the system is µ +a y x x J (x, y) = y.5+x In the origin, we find µ +a J (, ) =.5 with eigenvalues +a> and.5. Since an eigenvalue is greater than, the origin is unstable. In p we have µ.5.5 J (.5,a.5) =. a.5 The eigenvalues of J solve the characteristic equation They have modulus less than if the conditions λ.5λ (a.5) =..5 (a.5) < and.5 < (a.5) are satisfied; that is, if.5 <a<.5. Therefore, p is an asymptotically stable equilibrium if and only of.5 <a<.5.

38 Chapter 9. Solutions. Salsa-Squellati, An Introduction to Dynamic Optimization 7. (a) The extremals of the functional solve the Euler equation Here, we have and, then, The Euler equation is whose solutions are (b) Given that we deduce The Euler equation is whose solutions are fẋ =ẋ, J (x) = Z b a f (t, x (t), ẋ (t)) dt d dt f ẋ (t, x, ẋ) f x (t, x, ẋ) =. f (t, x, ẋ) =x + ẋ x sin t d dt f ẋ =ẍ, ẍ x = sin t f x =x sint. x = c e t + c e t + sin t, c,c R. fẋ =ẋ, f (t, x, ẋ) =ẋ x + xe t d dt f ẋ =ẍ, f x = x + e t. ẍ +x = e t x = c sin t + c cos t + 4 et, c,c R. (c) Since =ẋ f (t, x, ẋ) t, the function f does not depend on x; then, the Euler equation becomes fẋ (t, ẋ) =c, c R, namely and we deduce ẋ = c t ; the extremals are ẋ t = c x = c t 4 + c, c,c R.. We have and, consequently, we obtain f (t, x, ẋ) =ẋ +x (ẋ + t)+4x ; fẋ (t, x, ẋ) =ẋ +x, d dt f ẋ (t, x, ẋ) =ẍ +ẋ, f x (t, x, ẋ) =ẋ +t +8x.

39 8. Salsa-Squellati, An Introduction to Dynamic Optimization The Euler equation is ẍ 4x = t, whose general solution is The boundary conditions x () = x () = yield to bx (t) = x (t) =c e t + c e t 4 t, c,c R. 4(e e ) et 4(e e ) e t 4 t. The extremal bx is a minimizer since f (t, x (t), ẋ (t)) = ẋ +xẋ +4x +xt is convex with respect to x and ẋ. Infact, f is a quadratic function (plus a linear one) and the associated coefficients matrix µ A = 4 is positive definite, since det A >, and tra >.. Given that f (t, x, ẋ) =ẋ +t ẋ does not depend on x, the Euler equation becomes namely fẋ (t, ẋ) =+t ẋ = c, c R. ẋ = c t whose solution is represented by the family of functions x = c + c, c,c R. t We determine c and c using the initial condition x () =, and the transversality condition fẋ (,x(), ẋ ()) =. We obtain x (t) t +. This is actually a minimizer, since f is convex in x and ẋ (indeed, fẋẋ =t,f xẋ = f xx =, hance the hessian matrix associated to f computed with respect to x and ẋ is positive semidefinite). 4. Since does not depend on x, the Euler equation is f (t, x, ẋ) =g (t) p +ẋ fẋ (t, ẋ) = g (t) ẋ +ẋ = c, c R, thus, ẋ = c qg (t) c. If g (t) = t, we deduce Z x (t) = c t c dt + c =c p t c + c, c,c R.

40 . Salsa-Squellati, An Introduction to Dynamic Optimization 9 If g (t) = t, we deduce x (t) =c, c R, when c =and Z x (t) = ct c t dt + c = c p c t + c, c 6=,c R. The condition x (a) =A gives the minimizer bx (t) =A (constant), in fact, the transversality condition implies c =. 5. From we obtain fẋ (t, x, ẋ) =e t ẋ, and we deduce the Euler equation Its general solution is The condition x () = gives The transversality condition gives implying c =. We obtain fẋ (b, ẋ (b)) = f (t, x, ẋ) =e t ẋ +x, d dt f ẋ (t, x, ẋ) =4e t ẋ +e t ẍ, f x (t, x, ẋ) =6e t x, ẍ +ẋ x =. x (t) =c e t + c e t, c,c R. c + c =. lim t + f ẋ (t, x (t), ẋ (t)) = lim t + et c e t + c e t = x (t) =e t which is a minimizer given that f is convex in x and ẋ. 6. Consider we have f (t, x, ẋ) =e rt ẋ +xẋ x, fẋ (t, x, ẋ) = e rt ( ẋ +x), f x (t, x, ẋ) =e rt (ẋ x), d dt f ẋ (t, x, ẋ) = e rt [ r ( ẋ +x) ẍ +ẋ]. The Euler equation is ẍ rẋ +(r ) x =. If r 6=, the general solution is x (t) =c e t + c e (r )t, c,c R. The condition x () = x requires using the transversality condition c + c = x ; lim t + f ẋ (t, x (t), ẋ (t)) =,

41 4. Salsa-Squellati, An Introduction to Dynamic Optimization we deduce namely which is satisfied for every c. Then, the functions represent the extremals. Since we deduce that ³ (r )t lim t + e rt c e t (r ) c e (r )t +c e t +c e =, lim c e t ( r +4)= t + bx c (t) =c e t +(x c ) e (r )t, J (bx c )= J (x) = Z + Z + e rt (ẋ x) dt c R e (r )t (x c ) (r ) dt and that J (bx )=. Then, the optimal choice is c = x and the maximizer is If r =, the general solution of the Euler equation is bx (t) =x e t. x (t) =e t (c + c t), c,c R. The condition x () = x gives c = x. The transversality condition c e t = lim t + is satisfied for every c. As extremals, we deduce the functions Nevertheless, bx c (t) =e t (x + c t), c R. J (bx c )= Z + 4c dt is convergent only with for c =. Thus, we find again the solution 7. Given that bx (t) =x e t. the Euler equation is fẋ = a (t) ẋ +b (t) x, f x =b (t) ẋ +c (t) x, d dt f ẋ = a (t)ẍ +a (t) ẋ +b (t) x +b (t) ẋ, a (t)ẍ + a (t) ẋ +(b (t) c (t)) x =. (a) With the conditions x (t )=x,x(t )=x,witht fixed and t free, the transversality condition is f (t,x(t ), ẋ (t )) ẋ (t ) fẋ (t,x(t ), ẋ (t )) = namely a (t ) ẋ (t ) +b (t ) ẋ (t ) x (t )+c(t ) x (t ) ẋ (t )(a(t ) ẋ (t )+b(t ) x (t )) =

42 . Salsa-Squellati, An Introduction to Dynamic Optimization 4 that is c (t ) x a (t ) ẋ (t ) =. (b) With the conditions x (t )=x,x(t ) free, with t and t fixed, the transversality condition is fẋ (t,x(t ), ẋ (t )) = that is a (t ) ẋ (t )+b (t ) x (t )=. Asufficient condition which guarantees that the extremal is a minimizer is the convexity of f. f is convex if the matrix µ a (t) b (t) H (t) = b (t) c (t) is positive semidefinite for every t [t,t ]. If H (t) is positive definite the minimizer is unique. 8. Let us introduce the multiplier λ and the functional L (x) = Z ẋ λx dt. The Euler equation is then whose solutions are ẍ + λ = x (t) = λ t 4 + c t + c, c,c R. The constants c,c and λ have to be determined using the integral constraint with the final and initial conditions: Z Z xdt = µ λ t4 + c t + c dt = B, x () = c =, x () = λ 4 + c + c =. We obtain c =6B 4, c =, λ =4(B ). 9. We consider the maximization case. On account of the theorem 5., the transversality condition is If R (t) is a differentiable function, we have ˆf x (t ) δx +[ˆf (t ) x (t ) ˆf x (t )]δt. δx = R (t ) δt and then n ˆf (t )+[R (t ) ˆx (t )] ˆf x (t )o δt. Given that δt is arbitrary, we deduce ˆf (t )+[R (t ) ˆx (t )] ˆf x (t )=.

43 4. Salsa-Squellati, An Introduction to Dynamic Optimization. We write J (x) = f (t, x, x ) dt + f (t, x, x ) dt. t t and denote by δx the variation of ˆx (t ). Using (9.9) in page 8, we have δj (ˆx)[v, δx,δt ]= Z t t Z t Z t ˆf d dt ˆf x vdt + ˆf x (t ) δx +[ˆf (t ) ˆx (t ) ˆf x (t )]δt + Z t ˆf d t dt ˆf x vdt ˆf x (t +) δx [ ˆf (t +) ˆx (t +) ˆf x (t +)]δt = Z t = ˆf d dt ˆf h x vdt + ˆfx (t ) ˆf i x (t +) δx + t h ˆf (t ) ˆx (t ) ˆf x (t ) ˆf (t +) + ˆx (t +) ˆf i x (t +) δt. The Erdmann-Weierstrass conditions are obtained equating to zero each term of the previous formula and recalling that v, δx + and δt are arbitrary.. (a) Since fẋ =ẋ the first Erdmann-Weierstrass condition at the possible angle t = a is namely lim f ẋ (a) = lim f ẋ (a) t a t a + lim t a ẋ (a) = lim t a + ẋ (a) hence, the derivative is continuous at a. Thus, extremals with angles cannot exist. (b) The Erdmann-Weierstrass conditions at the possible angle in t = a require lim ẋ (a)( ẋ (a)) ( ẋ (a)) ẋ (a)( ẋ (a)) ( ẋ (a)) t a = lim t a + lim ( ẋ (a)) ( ẋ (a)) t a = lim ( ẋ (a)) ( ẋ (a)) t a + They are verified in the following situations:. lim t a ẋ (a) =lim t a + ẋ (a) (there is no angle);. lim t a ẋ (a) =, lim t a + ẋ (a) =;. lim t a ẋ (a) =, lim t a + ẋ (a) =. Then,someextremalwithanglescanexist.Since the functional explicitly only depends on ẋ, theeuler equation is fẋẋ ẍ = whose solution is x (t) =c t + c, c,c R. The possible extremals with angles are straight lines segments such as x (t) =k and x (t) =x + k.