Exam (elaborations) TEST BANK FOR An introduction to Ordinary Differential Equations 1st Edition By Robinson J.C. (Solution Manual) An Introduction to
... [Show More] Ordinary Di®erential Equations Exercises and Solutions James C. Robinson 1 Radioactive decay and carbon dating Exercise 1.1 Radioactive isotopes decay at random, with a ¯xed probability of decay per unit time. Over a time interval ¢t, suppose that the probability of any one isotope decaying is k¢t. If there are N isotopes, how many will decay on average over a time interval ¢t? Deduce that N(t + ¢t) ¡ N(t) ¼ ¡Nk¢t; and hence that dN=dt = ¡kN is an appropriate model for radioactive decay. Over a time interval ¢t, Nk¢t isotopes will decay. We then have N(t + ¢t) ¡ N(t) = ¡Nk¢t: Dividing by ¢t gives N(t + ¢t) ¡ N(t) ¢t = ¡Nk; and letting ¢t ! 0 we obtain, using the de¯nition of the derivative, dN dt = ¡kN: Exercise 1.2 Plutonium 239, virtually non-existent in nature, is one of the radioactive materials used in the production of nuclear weapons, and is a by-product of the generation of power in a nuclear reactor. Its half-life is approximately 24 000 years. What is the value of k that should be used in (1.1) for this isotope? Since N(t) = N(s)e¡k(t¡s), half of the isotopes decay after a time T, where N(s + T) = 1 2N(s) = N(s)e¡kT ; 1 2 1 Radioactive decay and carbon dating i.e. when 1 2 = e¡kT . Thus the half-life T = ln 2=k (as derived in Section 1.1). If T = 24000 then k = ln 2=T ¼ 2:888 £ 10¡5. Exercise 1.3 In 1947 a large collection of papyrus scrolls, including the old- est known manuscript version of portions of the Old Testament, was found in a cave near the Dead Sea; they have come to be known as the `Dead Sea Scrolls'. The scroll containing the book of Isaiah was dated in 1994 using the radiocarbon technique1; it was found to contain between 75% and 77% of the initial level of carbon 14. Between which dates was the scroll written? We have N(1994) = pN(s) = N(s)e¡k(1994¡s); where 0:75 · p · 0:77. Taking logarithms gives log p = ¡k(1994 ¡ s); and so s = 1994 + log p k : With k = 1:216 £ 10¡4 this gives (approximately) ¡372 · s · ¡155; dating the scrolls between 372 BC and 155 BC. Exercise 1.4 A large round table hangs on the wall of the castle in Winch- ester. Many would like to believe that this is the Round Table of King Arthur, who (so legend would have it) was at the height of his powers in about AD 500. If the table dates from this time, what proportion of the original carbon 14 would remain? In 1976 the table was dated using the radiocarbon tech- nique, and 91.6% of the original quantity of carbon 14 was found2. From when does the table date? If the table dates from 500 AD then we would expect N(t) = e¡k(t¡500)N(500); and so in 2003 we have N(2003) = e¡1503kN(500): The proportion of 14C isotopes remaining should there be e¡1503k ¼ 83%. 1 A.J. Jull et al., `Radiocarbon Dating of the Scrolls and Linen Fragments from the Judean Desert', Radiocarbon (1995) 37, 11{19. 2 M. Biddle, King Arthur's Round Table (Boydell Press, 2001). Radioactive decay and carbon dating 3 However, we in fact have 91.6% remaining in 1976. Therefore N(1976) = 0:915N(s) = N(s)e¡k(1993¡s): Taking logarithms gives s = 1976 + log 0:916 k ¼ 1255; the table probably dates from during the reign of the English King Edward I, who took the throne in 1270 AD (once the wood was well seasoned) and had a passion for all things Arthurian. Exercise 1.5 Radiocarbon dating is an extremely delicate process. Suppose that the percentage of carbon 14 remaining is known to lie in the range 0:99p to 1:01p. What is the range of possible dates for the sample? Suppose that a proportion ®p of the original 14C isotopes remain. Then ®pN(s) = N(t) = e¡k(t¡s)N(s); and so log ® + log p = ¡k(t ¡ s): It follows that s = t + log p k + log ® k : (S1.1) Denote by S the value of this expression when ® = 1, i.e. S = t + (log p)=k. For a proportion 0:99p the expression (S1.1) gives s = S ¡ 82:65; while for a proportion 1:01p the expression gives s = S + 81:83 (both correct to two decimal places). Small errors can give a di®erence of over 160 years in the estimated date. 2 Integration variables There are no exercises for this chapter. 4 3 Classi¯cation of di®erential equations Exercise 3.1 Classify the following equations as ordinary or partial, give their order, and state whether they are linear or nonlinear. In each case identify the dependent and independent variables. (i) Bessel's equation (º is a parameter) x2y00 + xy0 + (x2 ¡ º2)y = 0; (ii) Burger's equation (º is a parameter) @u @t ¡ º @2u @x2 + u @u @x = 0; (iii) van der Pol's equation (m, k, a and b are parameters) mxÄ + kx = ax_ ¡ bx_ 3; (iv) dy=dt = t ¡ y2, (v) the wave equation (c is a parameter) @2y @t2 = c2 @2y @x2 ; (vi) Newton's law of cooling (k is a parameter and A(t) is a speci¯ed function) dT dt = ¡k(T ¡ A(t)); (vii) the logistic population model (k is a parameter) dp dt = kp(1 ¡ p); 5 6 3 Classi¯cation of di®erential equations (viii) Newton's second law for a particle of mass m moving in a potential V (x), mÄx = ¡V 0(x); (ix) the coupled equations in (3.9) x_ = x(4 ¡ 2x ¡ y) y_ = y(9 ¡ 3x ¡ 3y); and (x) dx dt = Ax; where x is an n-component vector and A is an n £ n matrix. (i) linear 2nd order ODE for y(x); (ii) nonlinear 2nd order PDE for u(x; t); (iii) nonlinear 2nd order ODE for x(t); (iv) nonlinear 1st order ODE for y(t); (v) linear 2nd order PDE for y(x; t); (vi) linear 1st order ODE for T(t); (vii) nonlinear 1st order ODE for p(t); (viii) 2nd order ODE for x(t), linear if V 0(x) = ax + b for some a; b 2 R, otherwise nonlinear; (ix) nonlinear 1st order ODE for the pair (x(t); y(t)); and (x) linear 1st order ODE for the vector x(t). 4 *Graphical representation of solutions using MATLAB Exercise 4.1 Plot the graphs of the following functions: (i) y(t) = sin 5t sin 50t for 0 · t · 3, (ii) x(t) = e¡t(cos 2t + sin 2t) for 0 · t · 5, (iii) T(t) = Z t 0 e¡(t¡s) sin s ds for 0 · t · 7; (iv) x(t) = t ln t for 0 · t · 5, (v) plot y against x, where x(t) = Be¡t + Ate¡t and y(t) = Ae¡t; for A and B taking integer values between ¡3 and 3. (i) >> t=linspace(0,3); >> y=sin(5*t).*sin(50*t); >> plot(t,y) The result is shown in Figure 4.1. (ii) >> t=linspace(0,5); >> x=exp(-t).*(cos(2*t)+sin(2*t)); >> plot(t,x) The result is shown in Figure 4.2. (iii) Use the short M-¯le exint.m: f=inline('exp(-(t-s)).*sin(s)','t','s'); for j=0:100; t(j+1)=7*j/100; T(j+1)=quad(f,0,t(j+1),[],[],t(j+1)); end 7 8 4 *Graphical representation of solutions using MATLAB 0 0.5 1 1.5 2 2.5 3 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Fig. 4.1. The graph of y(t) = sin 5t sin 50t against t. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 1.2 Fig. 4.2. The graph of x(t) = e¡t(cos 2t + sin 2t) against t. plot(t,T) The plot is shown in Figure 4.3. (iv) >> t=linspace(0,5); >> x=t.*log(t); >> plot(t,x) The resulting graph is shown in Figure 4.4. (v) Use the short M-¯le param.m: t=linspace(0,5); hold on for A=-3:3; *Graphical representation of solutions using MATLAB 9 0 1 2 3 4 5 6 7 −200 −100 0 100 200 300 400 500 600 700 800 Fig. 4.3. The graph of the integral in Exercise 4.1(iii). 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −1 0 1 2 3 4 5 6 7 8 9 Fig. 4.4. The graph of x(t) = t ln t against t. for B=-3:3; x=B*exp(-t)+A.*t.*exp(-t); y=A*exp(-t); plot(x,y) end end hold off See Figure 4.5. 10 4 *Graphical representation of solutions using MATLAB −3 −2 −1 0 1 2 3 −3 −2 −1 0 1 2 3 Fig. 4.5. A collection of curves de¯ned parametrically by x(t) = Be¡t +Ate¡t and y(t) = Ae¡t. Exercise 4.2 Draw contour plots of the following functions: (i) F(x; y) = x2 + y2 for ¡ 2 · x; y · 2; (ii) F(x; y) = xy2 for ¡ 1 · x; y · 1; with contour lines where F = §0:1, §0:2, §0:4, and §0:8; (iii) E(x; y) = y2 ¡ 2 cos x for ¡ 4 · x; y · 4; (iv) E(x; y) = x ¡ 1 3x3 + 1 2y2(x4 ¡ 2x2 + 2) for ¡2 · x · 4 and ¡2 · y · 2, showing the contour lines where E = 0, 0:5, 0:8, 1, 2, 3, and 4; (v) E(x; y) = y2 + x3 ¡ x for ¡ 2 · x; y · 2: (i) >> [x, y]=meshgrid(-2:.1:2, -2:.1:2); >> F=x.^2+y.^2; >> contour(x,y,F) These contours are shown in Figure 4.6. *Graphical representation of solutions using MATLAB 11 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Fig. 4.6. Contours of x2 + y2 (ii) >> [x, y]=meshgrid(-1:.1:1, -1:.1:1); >> F=x.*y.^2; >> contour(x,y,F,[.1 .2 .4 .8 -.1 -.2 -.4 -.8]) These contours are shown in Figure 4.7. −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Fig. 4.7. Contours of xy2 (iii) >> [x, y]=meshgrid(-4:.1:4, -4:.1:4); >> E=y.^2-2*cos(x); >> contour(x,y,E) These contours are shown in Figure 4.8. (iv) >> [x, y]=meshgrid(-2:.1:4, -2:.1:2); >> E=x-(x.^3)/3+(y.^2/2).*(x.^4-2*x.^2+2); >> contour(x,y,E,[0 .5 .8 1 2 3 4]) These contours are shown in Figure 4.9. (v) >> [x, y]=meshgrid(-2:.1:2, -2:.1:2); >> E=y.^2+x.^3-x; 12 4 *Graphical representation of solutions using MATLAB −4 −3 −2 −1 0 1 2 3 4 −4 −3 −2 −1 0 1 2 3 4 Fig. 4.8. Contours of y2 ¡ 2 cos x −2 −1 0 1 2 3 4 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Fig. 4.9. Contours of x ¡ 1 3x3 + 1 2y2(x4 ¡ 2x2 + 2) >> contour(x,y,E) These contours are shown in Figure 4.10. −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Fig. 4.10. Contours of y2 + x3 ¡ x 5 `Trivial' di®erential equations Exercise 5.1 Find the general solution of the following di®erential equa- tions, and in each case ¯nd also the particular solution that passes through the origin. (i) dµ dt = sin t + cos t; (ii) dy dx = 1 x2 ¡ 1 (use partial fractions) (iii) dU dt = 4t ln t; (iv) dz dx = xe¡2x; and (v) dT dt = e¡t sin 2t: (i) Integrating both sides of dµ dt = sin t + cos t with respect to t we get µ(t) = ¡cos t + sin t + c: 13 14 5 `Trivial' di®erential equations For the solution to pass through the origin we need µ(0) = 0, i.e. 0 = ¡1 + c or c = 1, and thus this solution is µ(t) = 1 ¡ cos t + sin t: (ii) We have dy dx = 1 x2 ¡ 1 = 1 2 µ 1 x ¡ 1 ¡ 1 x + 1 ¶ ; and so y(x) = 1 2 (log jx ¡ 1j ¡ log jx + 1j) + c = 1 2 log jx ¡ 1j jx + 1j + c: For this solution to pass through the origin we need 0 = y(0) = 1 2 log j ¡ 1j j1j + c; i.e. c = 0, and so we get y(x) = 1 2 log jx ¡ 1j jx + 1j : (iii) Integrating dU dt = 4t ln t we have U(t) = Z 4t ln t dt = 1 2t2 ln t ¡ 1 4t2 + c: To ensure that U(0) = 0 we need c = 0, and so U(t) = 1 2t2 ln t ¡ 1 4t2: (iv) Integrating the right-hand side of dz dx = xe¡2x by parts we have z(x) = Z xe¡2x dx = ¡ 1 2xe¡2x + 1 2 Z e¡2x dx = ¡ 1 2xe¡2x ¡ 1 4 e¡2x + c; `Trivial' di®erential equations 15 and for z(0) = 0 we need 0 = ¡(1=4) + c, i.e. c = 1=4 giving z(x) = 1 4 (1 ¡ e¡2x ¡ 2xe¡2x): (v) We need to integrate dT dt = e¡t sin 2t: The integral of e¡t sin 2t will be a linear combination of e¡t sin 2t and e¡t cos 2t. For z(t) = ®e¡t sin 2t + ¯e¡t cos 2t we have dz dt = ®(¡e¡t sin 2t + 2e¡t cos 2t + ¯(¡e¡t cos 2t ¡ 2e¡t sin 2t) = (¡® ¡ 2¯)e¡t sin 2t + (2® ¡ ¯)e¡t cos 2t; and so we need ¡® ¡ 2¯ = 1 and 2® ¡ ¯ = 0; i.e. ® = ¡1=5 and ¯ = ¡2=5. Therefore T(t) = ¡ e¡t sin 2t + 2e¡t cos 2t 5 + c: For T(0) = 0 we need 0 = ¡(2=5) + c, i.e. c = 2=5, and so T(t) = 2 ¡ e¡t(sin 2t + 2 cos 2t) 5 : Exercise 5.2 Find the function f(x) de¯ned for ¡¼=2 < x < ¼=2 whose graph passes through the point (0; 2) and has slope ¡tan x. We want to ¯nd a function f that satis¯es df dx = ¡tan x with f(0) = 2: So we integrate between the limits that correspond to x values 0 and x, f(x) ¡ f(0) = Z x 0 ¡tan ~x d~x = Z x 0 ¡sin ~x cos ~x d~x = [ln j cos ~xj]x 0 = ln j cos xj; 16 5 `Trivial' di®erential equations and so, since cos x > 0 for ¡¼=2 < x < ¼=2 f(x) = ln cos x + 2: Exercise 5.3 Find the function g(x) de¯ned for x > ¡1 that has slope ln(1 + x) and passes through the origin. The required function g(x) satis¯es dg dx = ln(1 + x) with g(0) = 0: Integrating both sides of the di®erential equation between 0 and x gives g(x) = g(0) + Z x 0 ln(1 + ~x) d~x = · (1 + ~x) ln(1 + ~x) ¡ ~x ¸x ~x=0 = (1 + x) ln(1 + x) ¡ x; since ln 1 = 0. Exercise 5.4 Find the solutions of the following equations satisfying the given initial conditions: (i) x_ = sec2 t with x(¼=4) = 0; (ii) y0 = x ¡ 1 3x3 with y(¡1) = 1; (iii) dµ dt = 2 sin2 t with µ(¼=4) = ¼=4; (iv) x dV dx = 1 + x2 with V (1) = 1; and (v) d dt £ x(t)e3t¤ = e¡t with x(0) = 3; `Trivial' di®erential equations 17 (i) Integrating x_ = sec2 t from ¼=4 to t gives x(t) = x(¼=4) + Z t ¼=4 sec2 ~t d~t = 0 + · tan ~t ¸t ~t=¼=4 = tan t ¡ 1 2 : (ii) Integrating y0 = x ¡ 1 3x3 from ¡1 to x gives y(x) = y(¡1) + Z x ¡1 ~x ¡ 1 3 ~x3 d~x = 1 + · ~x2 2 ¡ ~x4 12 ¸x ~x=¡1 = 1 + x2 2 ¡ x4 12 ¡ 1 2 + 1 12 = 7 12 + x2 2 ¡ x4 12: (iii) Integrating dµ dt = 2 sin2 t between ¼=4 and t we have µ(t) = µ(¼=4) + Z t ¼=4 2 sin2 ~t d~t = ¼=4 + Z t ¼=4 1 ¡ cos 2~t d~t = ¼=4 + · ~t ¡ 1 2 sin 2~t ¸t ~t=¼=4 = ¼=4 + t ¡ 1 2 sin 2t ¡ ¼=4 + 1 2 = 1 2 + t ¡ 1 2 sin 2t: (iv) Dividing xdV dx = 1 + x2 by x and then integrating between 1 and x we obtain V (x) = V (1) + Z x 1 1 ~x + ~x d~x = 1 + · ln ~x + 1 2 ~x2 ¸x ~x=1 = 1 + ln x + 1 2x2 ¡ ln 1 ¡ 1 2 = 1 2 + ln x + 1 2x2: 18 5 `Trivial' di®erential equations (v) Integrating both sides of d dt £ x(t)e3t¤ = e¡t between 0 and t gives x(t)e3t = x(0) + Z t 0 e¡~t d~t = 3 + · ¡e¡~t ¸t ~t=0 = 3 ¡ e¡t + 1 = 4 ¡ e¡t; and so x(t) = 4e¡3t ¡ e¡4t: Exercise 5.5 The Navier-Stokes equations that govern °uid °ow were given as an example in Chapter 3 (see equations (3.1) and (3.2)). It is not possible to ¯nd explicit solutions of these equations in general. However, in certain cases the equations reduce to something much simpler. Suppose that a °uid is °owing down a pipe that has a circular cross-section of radius a. Assuming that the velocity V of the °uid depends only on its distance from the centre of the pipe, the equation satis¯ed by V is 1 r d dr µ r dV dr ¶ = ¡P; where P is a positive constant. Multiply by r and integrate once to show that dV dr = ¡ Pr 2 + c r where c is an arbitrary constant. Integrate again to ¯nd an expression for the velocity, and then use the facts that (i) the velocity should be ¯nite at all points in the pipe and (ii) that °uids `stick' to boundaries (which means that V (a) = 0) to show that V (r) = P 4 (a2 ¡ r2); see Figure 5.1. (This is known as Poiseuille °ow.) `Trivial' di®erential equations 19 a V(r)=P(a2−r2)/4 Fig. 5.1. The quadratic velocity pro¯le in a circular pipe. Multiplying the equation by r we obtain d dr µ r dV dr ¶ = ¡Pr; and integrating both sides gives r dV dr = ¡ Pr2 2 + c; which implies that dV dr = ¡ Pr 2 + c r : Integrating this equation gives V (r) = ¡ Pr2 4 + c ln r + d: Since ln r ! ¡1 as r ! 0, for V (r) to be ¯nite when r = 0 we have to take c = 0. This then leaves V (r) = ¡ Pr2 4 + d; and to ensure that V (a) = 0 we take d = Pa2=4, so that V (r) = P 4 (a2 ¡ r2) as claimed. Exercise 5.6 An apple of mass m falls from a height h above the ground. Neglecting air resistance its velocity satis¯es m dv dt = ¡mg v(0) = 0; 20 5 `Trivial' di®erential equations where v = y_ and y is the height above ground level. Show that the apple hits the ground when t = s 2h g : The velocity at time t is given by v(t) = v(0) ¡ gt = ¡gt; and its height y above ground level satis¯es y_ = v(t) = ¡gt; and hence y(t) = y(0) ¡ 1 2gt2 = h ¡ 1 2gt2: It follows that y(t) = 0 when t = p 2h=g as claimed. Exercise 5.7 An artillery shell is ¯red from a gun, leaving the muzzle with velocity V . If the gun is at an angle µ to the horizontal then the initial horizontal velocity is V cos µ, and the initial vertical velocity is V sin µ (see Figure 5.2). The horizontal velocity remains constant, but the vertical ve- locity is a®ected by gravity, and obeys the equation v_ = ¡g. How far does the shell travel before it hits the ground? (Give your answer in terms of V and µ.) V q Fig. 5.2. Firing a shell at muzzle velocity V at an angle µ to the horizontal. The shell follows a parabolic path. The vertical velocity v satis¯es v_ = ¡g with v(0) = V sin µ; and so integrating both sides of the di®erential equation between times 0 and t we obtain v(t) = v(0) ¡ gt = V sin µ ¡ gt: The height y(t) of the shell at time t satis¯es y_ = v = V sin µ ¡ gt with y(0) = 0; `Trivial' di®erential equations 21 and so integrating both sides of this between zero and t we have y(t) = V t sin µ ¡ 1 2gt2: The shell strikes the ground at time t¤, where V t¤ sin µ ¡ 1 2gt2¤ = 0; i.e. when t¤ = (2V sin µ)=g. Since the horizontal velocity is constant and equal to V cos µ, the shell will have travelled a distance V t¤ cos µ = 2V 2 sin µ cos µ g = V 2 sin 2µ g : Exercise 5.8 In Dallas on 22 November 1963, President Kennedy was as- sassinated; by Lee Harvey Oswald if you do not believe any of the conspiracy theories. Oswald ¯red a Mannlicher-Carcano ri°e from approximately 90 m away. The sight on Oswald's ri°e was less than ideal: if the bullet travelled in a straight line after leaving the ri°e (at a velocity of roughly 700 m/s) then the sight aimed about 10cm too high at a target 90 m away. How much would the drop in the trajectory due to gravity compensate for this? (The initial vertical velocity v is zero, and satis¯es the equation v_ = ¡g, while the horizontal velocity is constant if we neglect air resistance.) There is nothing to slow down the horizontal velocity of the bullet if we neglect air resistance: so it takes the bullet 9/70 seconds to travel 90 m. In this time it will have dropped vertically, its height h satisfying d2h dt2 = ¡g: The solution of this, integrating twice, is h(t) = h(0) ¡ 1 2gt2; and so with t = 9=70 and h(0) = 0 this gives a drop of 0:081 m or 8:1 cm, compensating quite well for the dodgy sight. Exercise 5.9 This exercise ¯lls in the gaps in the proof of the Fundamental Theorem of Calculus. Suppose that f is continuous at x, i.e. given any ² > 0, there exists a ± = ±(²) such that j~x ¡ xj · ± ) jf(~x) ¡ f(x)j · ²: By writing f(x) = 1 ±x Z x+±x x f(x) d~x 22 5 `Trivial' di®erential equations show that for all ±x with j±xj · ±(²) ¯¯¯¯ f(x) ¡ 1 ±x Z x+±x x f(~x) d~x ¯¯¯¯ · ²; and hence that lim ±x!0 1 ±x Z x+±x x f(~x) d~x = f(x): You will need to use the fact that ¯¯¯¯ Z b a g(x) dx ¯¯¯¯ · Z b a jg(x)j dx · (b ¡ a) max x2[a;b] jg(x)j: We have ¯¯¯¯ f(x) ¡ 1 ±x Z x+±x x f(~x) d~x ¯¯¯¯ = ¯¯¯¯ 1 ±x Z x+±x x f(x) d~x ¡ 1 ±x Z x+±x x f(~x) d~x ¯¯¯¯ = 1 ±x ¯¯¯¯ Z x+±x x f(x) ¡ f(~x) d~x ¯¯¯¯ · 1 ±x Z x+±x x jf(x) ¡ f(~x)j d~x: Then, given any ² > 0, there exists a ±(²) > 0 such that if ±x < ±(²) then for every ~x 2 [x; x + ±x] we have jf(x) ¡ f(~x)j · ², and so ¯¯¯¯ f(x) ¡ 1 ±x Z x+±x x f(~x) d~x ¯¯¯¯ · 1 ±x Z x+±x x ² d~x = 1 ±x [² ±x] = ²: Therefore, using the de¯nition of a limit, lim ±x!0 1 ±x Z x+±x x f(~x) d~x = f(x); as claimed. 6 Existence and uniqueness of solutions Exercise 6.1 Which of the following di®erential equations have unique so- lutions (at least on some small time interval) for any non-negative initial condition (x(0) ¸ 0)? (i) x_ = x(1 ¡ x2) (ii) x_ = x3 (iii) x_ = x1=3 (iv) x_ = x1=2(1 + x)2 (v) x_ = (1 + x)3=2. In each of these questions we will denote by f(x) the right-hand side of the di®erential equation. We need to check whether or not f and f0 are continuous for x ¸ 0. (i) Here f(x) = x(1 ¡ x2) and f0(x) = 1 ¡ 3x2 are both continuous, so solutions are unique. [In fact for x(0) ¸ 0 solutions exist for all t 2 R, while for x(0) < 0 they `blow up' to x = ¡1 in a ¯nite time.] (ii) For this example f(x) = x3 and f0(x) = 3x2 so solutions are unique. [Solutions blow up in a ¯nite time unless x(0) = 0.] (iii) We have f(x) = x1=3 (which is continuous), but f0(x) = x¡2=3=3, so f0(x) ! 1 as x ! 0, and the solution of x_ = x1=3 with x(0) = 0 is not unique: for any choice of c ¸ 0, the function x(t) = 8< : 0 t < c ³ 2(t¡c) 3 ´3=2 t ¸ c solves this equation. [Note that unlike the example x_ = x1=2 this equation also makes sense for x < 0.] 23 24 6 Existence and uniqueness of solutions (iv) We have f(x) = x1=2(1 + x2) which is continuous, and f0(x) = 1 2x¡1=2(1 + x2) + 2x3=2: Near zero f0(x) ! 1, and so the solutions with x(0) = 0 are not unique. (v) The function f(x) = (1+x)3=2 is continuous, and f0(x) = 3 2(1+x)1=2 is also continuous, so solutions are unique. [Solutions blow up in a ¯nite time.] Exercise 6.2 The Mean Value Theorem says that if f is di®erentiable on an interval [a; b] then f(a)¡f(b) = (b¡a)f0(c) for some c 2 (a; b). Suppose that f(x) is di®erentiable with jf0(x)j · L for a · x · b. Use the Mean Value Theorem to show that for a · x; y · b we have jf(x) ¡ f(y)j · Ljx ¡ yj: The result is clearly true if x = y. Using the mean value theorem for x > y we have f(x) ¡ f(y) = f0(c)(x ¡ y) for some c 2 (x; y). It follows that jf(x) ¡ f(y)j · jf0(c)jjx ¡ yj; and since jf0(c)j · L we have jf(x) ¡ f(y)j · Ljx ¡ yj: (S6.1) If y > x then f(y)¡f(x) = f0(c)(y ¡x) and on taking the modulus of both sides we once again arrive at (S6.1). Exercise 6.3 This Exercise gives a simple proof of the uniqueness of solu- tions of x_ = f(x; t) x(t0) = x0; (S6.2) under the assumption that jf(x; t) ¡ f(y; t)j · Ljx ¡ yj: (S6.3) Suppose that x(t) and y(t) are two solutions of (S6.2). Write down the di®erential equation satis¯ed by z(t) = x(t) ¡ y(t), and hence show that d dt jzj2 = 2z[f(x(t); t) ¡ f(y(t); t)]: Existence and uniqueness of solutions 25 Now use (S6.3) to show that d dt jzj2 · 2Ljzj2: If dZ=dt · cZ it follows that Z(t) · Z(t0)ec(t¡t0) (see Exercise 9.7): use this to deduce that the solution of (S6.2) is unique. Hint: any two solutions of (S6.2) agree when t = t0. We have dx dt = f(x; t) and dy dt = f(y; t): It follows that dz dt = d dt (x ¡ y) = dx dt ¡ dy dt = f(x; t) ¡ f(y; t): Now, d dt jzj2 = dz dt 2 = 2z dz dt = 2z[f(x; t) ¡ f(y; t)]: Using the Lipschitz property (S6.3) we have f(x; t) ¡ f(y; t) · jf(x; t) ¡ f(y; t)j · Ljx ¡ yj = Ljzj; and so d dt jzj2 · 2Lzjzj · 2Ljzj2: It follows (using dZ=dt · cZ ) Z(t) · Z(t0)ec(t¡t0)) that jz(t)j2 · jz(t0)j2e2L(t¡t0): (S6.4) Since x(t0) = y(t0) we have z(t0) = 0, and so (S6.4) becomes jz(t)j2 = 0. It follows that z(t) = 0, and so x(t) = y(t), which shows that the two solutions must be identical. Exercise 6.4 (T) The proof of existence of solutions is much more involved than the proof of their uniqueness. We will consider here the slightly simpler case x_ = f(x) with x(0) = x0; (S6.5) assuming that jf(x) ¡ f(y)j · Ljx ¡ yj: (S6.6) 26 6 Existence and uniqueness of solutions The ¯rst step is to convert the di®erential equation into an integral equation that is easier to deal with: we integrate both sides of (S6.5) between times 0 and t to give x(t) = x0 + Z t 0 f(x(~t)) d~t: (S6.7) This integral equation is equivalent to the original di®erential equation: any solution of (S6.7) will solve (S6.5), and vice versa. The idea behind the method is to use the right-hand side of (S6.7) as a means of re¯ning any `guess' of the solution xn(t) by replacing it with xn+1(t) = x0 + Z t 0 f(xn(~t)) d~t: (S6.8) We start with x0(t) = x0 for all t, set x1(t) = x0 + Z t 0 f(x0) d~t; and continue in this way using (E6.6). The hope is that xn(t) will converge to the solution of the di®erential equation as n ! 1. (i) Use (S6.6) to show that jxn+1(t) ¡ xn(t)j · L Z t 0 jxn(~t) ¡ xn¡1(~t)j d~t; and deduce that max t2[0;1=2L] jxn+1(t) ¡ xn(t)j · 1 2 max t2[0;1=2L] jxn(t) ¡ xn¡1(t)j: (S6.9) (ii) Using (S6.9) show that max t2[0;1=2L] jxn+1(t) ¡ xn(t)j · 1 2n¡1 max t2[0;1=2L] jx1(t) ¡ x0(t)j: (S6.10) (iii) By writing xn(t) = [xn(t)¡xn¡1(t)]+[xn¡1(t)¡xn¡2(t)]+¢ ¢ ¢+[x1(t)¡x0(t)]+x0(t) deduce that max t2[0;1=2L] jxn(t) ¡ xm(t)j · 1 2N¡2 max t2[0;1=2L] jx1(t) ¡ x0(t)j for all n;m ¸ N. Existence and uniqueness of solutions 27 It follows that xn(t) converges to some function x1(t) as n ! 1, and therefore taking limits in both sides of (E6.6) implies that x1(t) = x0 + Z t 0 f(x1(~t)) d~t: Thus x1(t) satis¯es (S6.7), and so is a solution of the di®erential equation. The previous Exercise shows that this solution is unique. (i) We have xn+1(t) ¡ xn(t) = x0 + Z t 0 f(xn(~t)) d~t ¡ x0 ¡ Z t 0 f(xn¡1(~t)) d~t = Z t 0 f(xn(~t)) ¡ f(xn¡1(~t)) d~t; and so jxn+1(t) ¡ xn(t)j = ¯¯¯¯ Z t 0 f(xn(~t)) ¡ f(xn¡1(~t)) d~t ¯¯¯¯ · Z t 0 jf(xn(~t)) ¡ f(xn¡1(~t))j d~t · Z t 0 Ljxn(~t) ¡ xn¡1(~t)j d~t; = L Z t 0 jxn(~t) ¡ xn¡1(~t)j d~t; using (S6.6). Since, therefore, jxn+1(t) ¡ xn(t)j · Lt max ~t2[0;t] jxn(~t) ¡ xn¡1(~t)j · L 1 2L max ~t2[0;1=2L] jxn(~t) ¡ xn¡1(~t)j for any t 2 [0; 1=2L], it follows that max t2[0;1=2L] jxn+1(t) ¡ xn(t)j · 1 2 max t2[0;1=2L] jxn(t) ¡ xn¡1(t)j; (S6.11) as claimed. (ii) We will write Dn = max t2[0;1=2L] jxn(t) ¡ xn¡1(t)j; 28 6 Existence and uniqueness of solutions then (S6.11) reads Dn+1 · 1 2Dn. It follows easily that Dn+1 · 2¡(n¡1)D1 which is (S6.10). (iii) Taking (wlog) n > m we have xn(t) ¡ xm(t) = xn(t) ¡ xn¡1(t) + xn¡1(t) ¡ xn¡2(t) + : : : + xm+1(t) ¡ xm(t); it follows that max t2[0;1=2L] jxn(t) ¡ xm(t)j · Dn + Dn¡1 + : : : + Dm+1 · · 1 2n¡2 + 1 2n¡3 + : : : + 1 2m¡1 ¸ D1 · D1 2m¡2 : 7 Scalar autonomous ODEs Exercise 7.1 For each of the following di®erential equations draw the phase diagram, labelling the stationary points as stable or unstable. (i) x_ = ¡x + 1 (ii) x_ = x(2 ¡ x) (iii) x_ = (1 + x)(2 ¡ x) sin x (iv) x_ = ¡x(1 ¡ x)(2 ¡ x) (v) x_ = x2 ¡ x4 The Figures all show the phase diagram and the graph of the function f(x) on the right-hand side of the equation. (i) x_ = ¡x + 1: There is a single stationary point at x = 1, as shown in Figure 7.1. −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 S Fig. 7.1. The phase diagram for x_ = ¡x + 1. 29 30 7 Scalar autonomous ODEs (ii) x_ = x(2 ¡ x): There are two stationary points, x = 0 and x = 2, as shown in Figure 7.2. −1 −0.5 0 0.5 1 1.5 2 2.5 3 −1.5 −1 −0.5 0 0.5 1 U S Fig. 7.2. The phase diagram for x_ = x(2 ¡ x). (iii) x_ = (1 + x)(2 ¡ x) sin x: There are an in¯nite number of stationary points, x = ¡1, x = 2, and x = n¼ for any integer n, as shown in Figure 7.3. −8 −6 −4 −2 0 2 4 6 8 −30 −25 −20 −15 −10 −5 0 5 10 15 20 S U S U S U S Fig. 7.3. The phase diagram for x_ = (1+x)(2¡x) sin x for ¡7 · x · 7. There are other stationary points at x = n¼ for any integer n. (iv) x_ = ¡x(1 ¡ x)(2 ¡ x): There are three stationary points, x = 0, x = 1, and x = 2, as shown in Figure 7.4. (v) x_ = x2 ¡ x4: Scalar autonomous ODEs 31 −0.5 0 0.5 1 1.5 2 2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 S U S Fig. 7.4. The phase diagram for x_ = ¡x(1 ¡ x)(2 ¡ x). There three are stationary points, at x = 0, x = 1, and x = ¡1, as shown in Figure 7.5. −1.5 −1 −0.5 0 0.5 1 1.5 −1 −0.5 0 0.5 U U S Fig. 7.5. The phase diagram for x_ = x2 ¡ x4. Exercise 7.2 For the equations in Exercise 7.1 determine the stability of the stationary points analytically, by considering the sign of the derivative of the right-hand side. (i) When f(x) = ¡x + 1 we have f0(x) = ¡1, and the stationary point at x = 1 is stable. (ii) When f(x) = x(2¡x) we have f0(x) = 2¡2x: we have f0(0) = 2, so that the stationary point x = 0 is unstable; and we have f0(2) = ¡2, and this stationary point is stable. 32 7 Scalar autonomous ODEs (iii) For f(x) = (1 + x)(2 ¡ x) sin x we have f0(x) = (1 ¡ 2x) sin x + (1 + x)(2 ¡ x) cos x: So at x = ¡1 we have f0(¡1) = 3 sin(¡1) ¼ ¡2:52 < 0 and this point is stable; at x = 2 we have f0(2) = ¡3 sin 2 ¼ ¡2:73 < 0 and this point is also stable. At x = n¼ we have f0(n¼) = (1 + n¼)(2 ¡ n¼)(¡1)n; taking n = 0 gives f0(0) = 2, and this point is unstable. For integer n 6= 0, (1 + n¼)(2 ¡ n¼) · 0; and so f0(n¼) > 0 (these points are unstable) if n is odd and f0(n¼) < 0 if n is even (and these points are stable). (iv) We have f(x) = ¡x(1 ¡ x)(2 ¡ x), and so f0(x) = ¡3x2 + 6x ¡ 2: Therefore f0(0) = ¡2; f0(1) = 1; and f0(2) = ¡2; and so x = 0 and x = 2 are stable, while x = 1 is unstable. (v) Now we have f(x) = x2 ¡ x4, and so f0(x) = 2x ¡ 4x3. This gives f0(¡1) = 2; f0(0) = 0; and f0(1) = ¡2: We can only tell using this method that x = ¡1 is unstable and that x = 1 is stable. To ¯nd the stability of x = 0 we need to work from the phase diagram (it is unstable, or, if you prefer, `semi-stable', i.e. stable on one side and unstable on the other). Exercise 7.3 For all positive values of c ¯nd all the stationary points of dx dt = sin x + c; and determine analytically which are stable and unstable. Draw the portion of the phase diagram between ¡¼ and ¼. There are three di®erent cases, Scalar autonomous ODEs 33 0 · c < 1, c = 1, and c > 1. You will need to be more careful with the case c = 1. Stationary points occur whenever f(x) ´ sin x + c is zero, so whenever sin x = ¡c. If 0 · c < 1 then there are two solutions in (¡¼; ¼], x1 between ¡¼ and ¡¼=2, and x2 between ¡¼=2 and 0. Since f0(x) = cos x we have f0(x1) < 0 and f0(x2) > 0; so that x1 is stable and x2 is unstable, see Figure 7.6. −3 −2 −1 0 1 [Show Less]