Matlab Lab 1 Solutions Problem 1 a. Solve f(y) = 0, i.e. 1 ‐ sin y = 0, or sin y = 1. The solutions are (π/2) + 2nπ, n∈Ζ. b. Note that the right‐hand‐side function of the ODE, f(y) = 1‐ sin y is continuous and differentiable for all y so that the existence and uniqueness theorem applies to any initial condition. Therefore, the graphs of different solutions may not cross each other and we see that the solution with initial condition y(0) = 4 approaches, as t→∞, the equilibrium value π/2+ 2π and, as t→‐∞, the equilibrium value π/2. c. The slopes of solutions are y’ = 1 – sin y, where ‐1 ≤ sin y ≤ 1, that is ‐1 ≤ ‐sin y ≤ 1, which (adding 1 to all three members of the inequalities) is equivalent to 1+(‐1) ≤ 1 ‐ sin y ≤ 1+1, that is 0 ≤ 1 – sin y ≤ 2. In particular, 0 ≤ y’ = 1 – sin y. d. Let v be the solution of the IVP y’ = 1 – sin y, y(0) = 4, and let w be the solution of the IVP y’ = 1 – sin y, y(0) = 4 ‐ 2π. It looks from the graph that the shapes of the graphs are identical and that the graph of w is actually that of v displaced by 2π units down. This means that we postulate w(t) = v(t) ‐ 2π. This is far from automatic: we must consider the function x(t) = v(t) ‐ 2π and prove that x and w are the same function. This can be accomplished by checking that x is a solution of the IVP y’ = 1 – sin y, y(0) = 4 ‐ 2π, since then, uniqueness of solutions would tell us that x and w are the same function (because w is a solution of this IVP by definition). First note that the initial condition is satisfied by x since x(0) = v(0) ‐ 2π = 4 ‐ 2π. Finally, note that x satisfies the ODE y’ = 1 – sin y since x’ = [v ‐ 2π]’ = v’ – 0 = 1 – sin v = 1 – sin(v ‐ 2π) = 1 – sin x, as needed. e. Problem 2 a. Solve f(y) = 0, i.e. y(y+1)(y‐2) = 0. The solutions are y = 0, y = ‐1, and y = 2. b. The direction field for the given differential equation and window is: The modified Matlab file LAB01_a.m used to generate this figure is: function LAB01_a
%driver for dirfield7.m
%window=[0,5,-2,4];
figure(1)
set(gcf,'DefaultAxesFontSize',16)
dirfield7(@f,[0,5],[-2,4],'arrow');
title('Direction field of y''= y(y+1)(y-2)'); % note the use of ''
%-------------------------------------------------------------function dydt=f(t,y)
% defines right-hand side of ODE
dydt= y.*(y+1).*(y-2);
We see from the figure that, as t→∞, the solution corresponding to the initial condition y(0) = 1.9 approaches the equilibrium value 0, while that corresponding to the initial condition y(0) = 2.1 grows unboundedly to ∞. c. No two curves cross each other. The reason is that, if there was a point of intersection for 2 curves, (τ, ξ), say, then the initial value problem corresponding to the initial condition y(τ) = ξ would have two different solutions corresponding to the 2 graphs that pass through this point. This contradicts the uniqueness theorem, which for this ODE holds with any initial condition since the right‐hand‐side f(y) = y (y+1)(y‐2) is continuous and differentiable for all values of t and y. d. The solution corresponding to the initial condition y(0) = ½ cannot eventually (i.e. for any t) become ‐1 because the constant function y = ‐1 is the solution of the ODE corresponding to the initial condition y(0) = ‐1 and solutions corresponding to different initial conditions at the same initial time (here t = 0) cannot intersect because of uniqueness. Problem 4.i We can solve the ODE explicitly since it is linear: y’ – 2y = 1/t has integrating factor µ(t) = exp(‐2t) and we obtain d/dt [µy] = µ(t)/t, whereby y = exp(2t) ∫ [µ(s)/s] ds. Unfortunately, there is not closed form for an antiderivative F of µ(t)/t, so that we cannot “see” the graphs of the solutions from the explicit formula y(t) = exp[2(t‐ t0)] y0 + exp(2t) [F(t)‐F(t0)] corresponding to the initial condition y(t0) = y0. On the other hand, if t0 , y0 > 0, we see that the limit of y(t) as t→∞ is ∞, since then µ(s)/s > 0 for s ∈ [t0 , ∞) says F is increasing on that interval. a. We see in the figure that as t approaches ∞ solutions approach either ∞ or ‐∞. b. Yes, in some cases. It seems from the figure that the curve y = 1/(2t) separates solutions that go to ∞ from those that go to ‐∞ as t approaches ∞. Two initial conditions that are arbitrarily close, one on each side of the curve, lead to substantially different solutions (large changes). c. No periodic solutions exist here since they all approach ∞ or ‐∞. d. For t approaching ‐∞ all solutions seems to approach y = 0 asymptotically. Problem 4.ii We can solve the ODE explicitly since it is linear: y’ + ½ y = sin t has integrating factor µ(t) = exp(½ t) and we obtain d/dt [µy] = µ(t) sin t, whereby y(t)= C exp(‐½t) + 2(sin t ‐2cos t)/5, where C is an arbitrary real number. It follows immediately that for any initial condition, as t approaches ∞, the solution of the IVP approaches the periodic function p(t) = 2(sin t ‐2cos t)/5 = 2/√5 sin(t+α), where tan α = ‐2. This periodic function can be expressed in the standard form p(t) = A sin(ωt+α), where the amplitude is A = 2/√5 ≅ 0.8944, the angular frequency is ω = 1, and the angle of phase is α ≅ ‐1.1071. Naturally, the behavior of solutions for t approaching ‐∞ depends on the sign of C: they approach ∞ if C is positive and ‐∞ if C is negative. a. We see in the figure that as t approaches ∞ all solutions seem to approach a periodic sine‐like function (that is, of course, p(t) = 2/√5 sin(t+α), where tan α = ‐2). b. No, any variation in initial condition disappears in the long run (i.e. for large enough t). The reason is explained in a. c. The figure seems to suggest a unique periodic solution. Counting the horizontal distance between a maximum and the fifth after it (approximately 32, from ‐17 to 15) we estimate a period of 6.4; we know the periodic solution analytically and it has period 2π/ω ≅ 6.3. d. No solutions approach any asymptotes, since they all become infinite as t approaches ‐∞ and all approach the periodic function p as t approaches ∞.