Second-Order Differential Equations

Transcription

1 WWW Problems and Soluions 3.1 Chaper 3 Second-Order Differenial Equaions Secion 3.1 Springs: Linear and Nonlinear Models www m Problem 3. (NonlinearSprings). A bod of mass m is aached o a wall b means of a spring and moves back and forh along is axis on a horizonal fricionless surface. Suppose ha measures he displacemen of he bod along he spring s axis from he equilibrium posiion; < 0 corresponds o a sreched spring, > 0oacompressed spring. The margin figure shows he bod wih viscous damping forces represened b a damper. The downward graviaional force on he bod is exacl canceled b an upward force exered b he supporing surface, so he mass moves onl horizonall, unaffeced b gravi. Under hese assumpions, Newon s Second Law (inerpreed for horizonal moion) implies ha m () = S(()) c () where c > 0 is he damping consan. Some possible formulas for he spring resoring force S() are given b equaions (2) (5). In pars (a) (c) below firs wrie he equivalen ssem as in IVP (11). Plo he orbis hrough he given iniial poins in he v-plane, and plo he corresponding soluion curves in he -plane. Inerpre our graphs in erms of he moion of a spring, and discuss he reali of each orbi. Esimae he periods of periodic soluions; does he period increase, decrease, or sa he same as he magniude of he iniial veloci increases? (a) (Undamped Hard Spring). = , 0 25; 0 = 0, v 0 = 0, 1, 3, 9. (b) (Undamped Sof Spring). = , 0 30; 0 = 0, v 0 = 0, 0.4, 0.9; 0 = 5, v 0 = 1.49, 1.51; 0 = 5, v 0 = 1.49, (c) (Damped Sof Spring). = , 0 40; 0 = 0, v 0 = 0, 2.44, (d) (Linearizaion). Linearize he ODE in par (c) a he saic equilibrium = 0, = 0. Plo orbis of he original and of he linearized ssem near his equilibrium sae, using he same iniial daa ses for he wo ssems. Explain he difference beween he wo sae porrais. Using he same daa ses, plo -soluion curves of he wo ssems, and explain wha ou see. Whenever we inerpre he resuls of our model in erms of he ssem being modeled, we mus be careful o resric our aenion o he regions where our model is valid. For springs, he model is valid onl when he displacemen is wihin he elasic limis of he spring. Tha is, he model is onl valid for displacemens ielding a resoring

2 3.2 Chaper 3/Second-Order Differenial Equaions spring force, posiive for < 0 and negaive for > 0. For he sof springs of (b) and (c), he elasic limi is no greaer han (10) 1/2 3.16, since for >(10) 1/2, S() will have he wrong sign for a resoring force. A cclical orbi [as in (a) and (b)] indicaes ha afer a cerain period of ime he ssem reurns o is original configuraion, and since ime does no appear explicil in he equaions of moion (i.e., he ODEs are auonomous) he ssem repeas is moion periodicall. Generall, he period will be differen for differen orbis, as a componen plo will show. In (c) he moion is damped, so energ is dissipaed and he ampliude decas wih ime. In ever case m = 1 and he corresponding ssem is = v, v = S() c. (a) Here we have = v, v = ,0 25. See Figs. 3(a). The equilibrium poin a he origin of Fig. 3(a), Graph 1 and he corresponding sraigh line soluion curve in Fig. 3(a), Graph 2 correspond o he spring a res a = 0 in is posiion of saic equilibrium. The oher hree soluions are periodic and correspond o he periodic oscillaions of he undamped spring. The periods of he hree soluions corresponding o 0 = 0, v 0 = 1, 3, 9 are approximael 12.3, 8.7, and 5.5, respecivel; he period appears o decrease as he ampliude increases. Apparenl he siffness of he spring is such ha he spring oscillaes faser and wih greaer ampliude if i is given a greaer iniial veloci. (c) Here, = v, v = v, The soluion wih iniial poin 0 = 0, v 0 = 2.44 is jus close enough o he res poin of saic equilibrium = 0, v = 0 [see Fig. 3(c), Graphs 1, 2], ha he damping forces he moion of he spring ino decaing oscillaions around he equilibrium. The soluion wih iniial daa 0 = 0, v 0 = 2.46 has jus ha exra amoun of iniial veloci (and so, energ) ha i appears o move beond he limis of validi of he model. See he orbi in Fig. 3(c), Graph 1 and he soluion curve in Fig. 3(c), Graph 2 ha escape o infini as increases. v = Problem 3.1.3(a), Graph 1. Problem 3.1.3(a), Graph 2.

3 3.2/ Second-Order ODEs and Their Properies 3.3 v = Problem 3.1.3(c), Graph 1. Problem 3.1.3(c), Graph 2. Secion 3.2 Second-Order ODEs and Their Properies www Problem 5. Find he soluion formula for he IVP = 2, (0) = 1, (0) = 1. Plo he soluion curve and orbi of his IVP. Wha is he larges -inerval on which he soluion of his IVP is defined? [Hin: Wrie 2 as ( 2 ).] To solve he IVP = 2, (0) = 1, (0) = 1 we use he hin o see ha = ( 2 ). Inegraing and appling he iniial condiions, we see ha = 2,so / 2 = 1. Inegraing again and appling he iniial condiions, we obain 1/ + 1 =. Solving for, wehave = 1/(1 ). Thelarges-inerval for his IVP is < 1. See Figs. 5, Graph 1 and Graph 2 for he soluion graph and orbi, respecivel. Problem 3.2.5, Graph 1. Problem 3.2.5, Graph 2. Secion 3.3 Undriven Consan Coefficien Linear ODEs, I www Problem 3. (Given a Soluion, Wha s he ODE?). Find an ODE of he form + a + b = 0, where a 2 > 4b for which he given funcion is a soluion, or else explain wh no such ODE exiss. (a) e e (b) e e (c) e + e 2 (d) 1 + e 3 (e) e e 3 (f) e 2 + e 2 (g) e π 3 (h) e 2 (i) + 2 (j) 2 e

4 3.4 Chaper 3/Second-Order Differenial Equaions Theorem shows us ha all possible soluions of + a + b = (D 2 + ad+ b)[] = P(D)[] = 0 are of he form C 1 e r1 + C 2 e r2 or C 1 e r1 + C 2 e r1 where P(r) = r 2 + ar + b = (r r 1 )(r r 2 ). Solving for hese cases is a maer of maching erms in he given soluion wih he appropriae roos. The values of C 1 and C 2 don affec our choice of differenial equaion (hese values are fixed when iniial condiions are chosen). (a) In he case of e e we see ha he roos are r 1 = 1andr 2 = 1 (we don reall care abou C 1 or C 2 ). So he characerisic polnomial P(r) mus have as roos 1 and 1. So P(r) = (r r 1 )(r r 2 ) = (r 1)(r + 1) = r 2 1. So we know ha P(D) = D 2 1andP(D)[] = (D 2 1)[] =. Our original ODE is = 0. (c) In he case of e + e 2 we see ha he roos are r 1 = 1andr 2 = 2. So P(r) = (r 1)(r + 2) = r 2 + r 2. So P(D) = D 2 + D 2andP(D)[] = (D 2 + D 2)[] = + 2. So he original ODE is + 2 = 0. (e) In he case of e e 3 we see ha he roos are r 1 = 2andr 2 = 3. So P(r) = (r 2)(r 3) = r 2 5r + 6. So P(D) = D 2 5D + 6andP(D)[] = (D 2 5D + 6)[] = So he original ODE is = 0. (g) In he case of e π 3 we see ha he roos are r 1 = π and r 2 = 0. So P(r) = (r π)r = r 2 rπ giving us P(D) = D 2 Dπ and P(D)[] = (D 2 Dπ)[] = π. So he original ODE is π = 0. (i) In he case of + 2 we see ha i can be wrien in he form 2e 0 + e 0 where boh roos are zero. So P(r) = r 2 giving P(D) = D 2 and P(D)[] = D 2 [] =.Sohe original ODE mus be = 0. Secion 3.4 Undriven Consan Coefficien Linear ODEs, II www Problem 3. Find he general real-valued soluion of each ODE. Then plo soluion curves in he -plane for 1 5, where (0) = 1, (0) = 6, 3, 0, 3, 6. Plo he corresponding orbis in a recangle in he -plane ha shows he main feaures of he orbis. Discuss our resuls. (a) + = 0 (b) = 0 (c) = 0 (d) + 10 = 0 (e) /9 = 0 (f) 3 /4 + /8 = 0 In each par, soluion curves are ploed in Graph 1, wih he corresponding orbis ploed in Graph 2; c 1 and c 2 are arbirar real consans. (a) The characerisic polnomial for + = (D 2 + D)[] = 0isr 2 + r, which facors o (r + 1)r. Thus, soluions of he ODE are given b e, e 0, and he linear combinaions of hese funcions. Tha is, he real-valued soluions are given b = c 1 e + c 2. See Figs. 3(a). From he figures and he soluion formulas we see ha as +, () c 2, while (if c 1 0), as, () ± depending on he sign of c 1. The orbis are slaned ras ha approach he -axis in he -plane as +.

5 3.4/ Undriven Consan Coefficien Linear ODEs, II 3.5 Problem 3.4.3(a), Graph 1. Problem 3.4.3(a), Graph 2. (c) The characerisic polnomial of he ODE = (D 2 + 3D + 2)[] = 0 is r 2 + 3r + 2 which facors o (r + 1)(r + 2). Soluions of he ODE are given b e, e 2, and he linear combinaions of hese funcions. So he real-valued soluions are given b = c 1 e + c 2 e 2. See Fig. 3(c), Graph 1. From he formula and he picures as + soluions deca o 0, bu as, soluions become unbounded. The orbis end o he origin of he -plane as + [Fig. 3(c), Graph 2]. Problem 3.4.3(c), Graph 1. Problem 3.4.3(c), Graph 2. (e) The characerisic polnomial of he ODE /9 = (D 2 1/9)[] = 0isr 2 1/9 which facors o (r + 1/3)(r 1/3). Soluions of he ODE are given b e /3, e /3, and he linear combinaions of hese funcions. The real-valued soluions are given b = c 1 e /3 + c 2 e /3. See Fig. 3(e), Graph 1. The soluion curves become unbounded as ± if c 1 and c 2 are nonzero. The ime span used here is fairl shor, so we onl see arcs of orbis (in some cases) [Fig. 3(e), Graph 2] ha would move ou of he recangle if he ime span, boh forward and backward, were longer. In Fig. 3(e), Graph 3 we have enlarged he x-recangle and lenghened he ime span, so we can reall see wha is going on.

6 3.6 Chaper 3/Second-Order Differenial Equaions Problem 3.4.3(e), Graph 1. Problem 3.4.3(e), Graph 2. Problem 3.4.3(e), Graph 3. Secion 3.5 Periodic Soluions and Simple Harmonic Moion www Problem 1. (Periodic Funcion Facs). (a) Show ha if f () has period T, hen f (ω) has period T/ω. (b) Suppose f () and g() have periods T and S, respecivel, wih T/S raional. Show ha h() = f () + g() is periodic wih period equal o he smalles value k such ha k = mt = ns for posiive inegers m and n. (c) Verif he claim in par (b) b graphing f () = sin(2), g() = sin(5),and f () + g() on he same plo. Wha is he period of each funcion? (d) Explain wh h() = f () + g() is no periodic if he raio T/S of he periods of f and g is irraional. (a) If f () has (fundamenal) period T, hent is he smalles posiive number such ha for an ime, f ( + T) = f (). Le g() = f (ω) and suppose ha g() has (fundamenal) period T 1 ;hais,t 1 is he smalles posiive number such ha g( + T 1 ) = g(). Then f (ω( + T 1 )) = f (ω) b he definiion of g(), so f (ω + ωt 1 ) = f (ω). Bu he smalles posiive value of T 1 for which his holds saisfies ωt 1 = T, since T is he period of f.so,t 1 = T/ω is he period of g() = f (ω). (c) See Fig. 1(c) for he graphs of sin 2 (long dashes) and sin 5 (shor dashes), wih periods T = π and S = 2π/5, respecivel. Using he noaion of par (b), wehave m = 2andn = 5sohak = 2(π) = 5(2π/5) = 2π. The solid curve is he graph of sin 2 + sin 5, which does indeed have period 2π.

7 3.6/ Driven Consan Coefficien Linear ODEs 3.7 sin 2,sin5,sin2 + sin 5 Problem 3.5.1(c). Secion 3.6 Driven Consan Coefficien Linear ODEs www Problem 9. (Hears and Ees). Find a soluion formula for + 25 = sin ω, where ω 5. Plo he soluion curve of he IVP wih (0) = (0) = 0, where ω = 4. Then plo he orbi for 0 20 in he recangle 0.1, Repea wih ω = 1. Tr o overla he wo graphs. [Hin: See Example ] To find a soluion for he equaion + 25 = sin ω where ω 5, firs we find he undriven soluion and hen we find a paricular soluion. The general real-valued soluion of he undriven equaion + 25 = 0is = C 1 cos5 + C 2 sin 5. To find a paricular soluion, firs noe ha sin ω = Im[e iω ]. So we firs solve (D )[z] = e iω where = Im[z]. Subsiuing z = Ae iω ino he ODE ields A = 1/(25 ω 2 ).Seing = Im[z] gives = sin ω/(25 ω 2 ). The general soluion of he driven ODE + 25 = sin ω is, for ω 5, = C 1 cos5 + C 2 sin sin ω 25 ω2 See Figs. 9(a), Graphs 1 and 2 for he orbis of he soluions where (0) = (0) = 0, ω = 4andhenω = 1. In Fig. 9(a), Graph 3, we have superimposed wo orbis b wriing a new ssem x = = 25x + sin(z) z = 0 and using iniial daa x = 0, = 0, z = 4, 1. You see wh we love o call his he hear and ees problem.

8 3.8 Chaper 3/Second-Order Differenial Equaions Problem 3.6.9, Graph 1. x Problem 3.6.9, Graph 2. x Problem 3.6.9, Graph 3. x Secion 3.7 The General Theor of Linear ODEs www Problem 7. (Inerlacing of Zeros of a Basic Soluion Pair). Suppose ha a() and b() are coninuous on < <, and suppose ha {u(), v()} is a basic soluion se for he ODE + a() + b() = 0. Show ha beween an wo consecuive zeros of one soluion here is precisel one zero of he oher soluion. Wh are he margin graphs no of a basic soluion pair? Le {u(), v()} be a basic soluion se for he ODE, + a() + b() = 0. Le c and d be consecuive zeros of u(). We will show firs ha v() has a leas one zero beween c and d. Le W() = W[u,v]() be he Wronskian of he basic soluion pair {u,v}. Since W() 0forall we ma as well assume ha W() >0[if W() <0, hen an analogous argumen works]. So, 0 < W(c) = v(c)u (c), and 0 < W(d) = v(d)u (d), where we have used u(c) = u(d) = 0. Now from he Vanishing Daa Theorem we see ha u (c) 0andu (d) 0sinceu() and v() are no idenicall 0. Since 0 < v(c)u (c) and 0 < v(d)u (d) and since c and d are consecuive zeros of u, i mus be ha u (c) and u (d) are of opposie sign. I follows ha v(c) and v(d) are of opposie sign, so v mus have a zero a some poin beween c and d. A similar argumen shows ha u() mus have a leas one zero beween an wo consecuive zeros of v(). Therefore, beween an wo consecuive zeros of one soluion here is precisel one zero of he oher soluion. Tha s wh he wo curves skeched in he margin canno form a basic soluion se for he ODE, + a() + b() = 0.