31.5 Systems of Differential Equations 1041 PROBLEMS FOR SECTION 31.5 1. Let x = x(t) be the number of hundreds of animals of species A at time t . Let y = y(t) be the number of hundreds of animals of species B at time t. For each system of differential equations, describe the nature of the interaction between the two species. What happens to each species in the absence of the other? (a) dx dt = 0.02x − 0.001x 2 − 0.002xy dy dt = 0.008y − 0.004y 2 − 0.001xy (b) dx dt = 0.02x − 0.01xy dy dt =−0.01y + 0.08xy (c) dx dt = 0.02x − 0.001x 2 + 0.002xy dy dt = 0.03y − 0.006y 2 + 0.001xy For each system of differential equations in Problems 2 through 4, find the nullclines and identify the equilibrium solutions. 2. dx dt = 0.02x − 0.001x 2 − 0.002xy dy dx = 0.008y − 0.004y 2 − 0.001xy 3. dx dt = 0.02x − 0.01xy dy dt =−0.01y + 0.08xy 4. dx dt = 0.02x − 0.001x 2 + 0.002xy dy dt = 0.03y − 0.006y 2 + 0.001xy 5. Let x = x(t) be the number of thousands of animals of species A at time t . Let y = y(t) be the number of thousands of animals of species B at time t. Suppose dx dt = x − 0.5xy dy dt = y − 0.5xy. (a) Is the interaction between species A and B symbiotic, competitive, or a predator- prey relationship? (b) What are the equilibrium populations? (c) Find the nullclines and draw directed horizontal and vertical tangent lines in the phase-plane (as in Figures 31.28 and 31.30). (d) The nullclines divide the first quadrant of the phase-plane into four regions. In each region determine the general direction of the trajectories. (e) If x = 0, what happens to y(t)? How is this indicated in the phase-plane? If y = 0, what happens to x(t)? How is this indicated in the phase-plane? (f) Use the information gathered in parts (b) through (e) to sketch representative solution trajectories in the phase-plane. Include arrows indicating the direction the trajectories are traveled. (g) For each of the initial conditions given below, describe how the number of species of A and B change with time and what the situation will look like in the long run. i. x(0) = 2 y(0) = 1.8 ii. x(0) = 2 y(0) = 2.3 iii. x(0) = 2.2 y(0) = 2 1042 CHAPTER 31 Differential Equations (h) Does this particular model support or challenge Charles Darwin’s principle of competitive exclusion? Problems 6 through 8 give systems of differential equations modeling competition between two species. In each problem find the nullclines. The nullclines will divide the phase-plane into regions; find the direction of the trajectories in each region. Use this information to sketch a phase-plane portrait. Then interpret the implications of your portrait for the long-term outcome of the competition. x(t) and y(t) give the number of thousands of animals of species A and B, respectively. 6. dx dt = 0.03x − 0.01x 2 − 0.01xy dy dt = 0.05y − 0.01y 2 − 0.01xy 7. dx dt = 0.04x − 0.02x 2 − 0.01xy dy dt = 0.04y − 0.01y 2 − 0.01xy 8. dx dt = 0.01x(2 − 2x − y) dy dt = 0.01y(1 − y − 0.25x) 9. Suppose we modify the Volterra predator-prey equations to reflect competition among prey for limited resources and competition among predators for limited resources. The equations would be of the form dx dt = k 1 x − k 2 x 2 − k 3 xy dy dt =−k 4 y−k 5 y 2 +k 6 xy where k 1 , k 2 , ,k 6 are positive constants. Consider the system dx dt = x(1 − 0.5x − y) dy dt = y(−1 − 0.5y + x). (a) Find the equilibrium points. (b) Do a qualitative phase-plane analysis. (In fact, solution trajectories will spiral in toward the non-trivial equilibrium point.) For Problems 10 and 11, suppose that for a given system of differential equations dy dt = f(x,y) dx dt = g(x, y) a trajectory with initial conditions y(0) = y 0 and x(0) = x 0 is as drawn. On the same set of axes, sketch possible graphs of x(t) and y(t) corresponding to the trajectory shown. (There are infinitely many correct answers. What characteristics must they all share?) 31.5 Systems of Differential Equations 1043 10. y 2 2 x 11. y x 1 1 2 2 12. y x 2 1 12 1044 CHAPTER 31 Differential Equations 13. Consider the following systems of differential equations. (a) dx dt =−2 (b) dx dt =+3y (c) dx dt =+10x dy dt =−4x dy dt =−3x dy dt =+10y Match each of the pairs of differential equations with the correct figure below. Explain your rationale briefly. y x (ii) y x (viii) y x (iii) y x (i) y x (xii) y x (v) y x (ix) (x) y x (vi) y x (vii) y x (iv) y x y x (xi) 31.6 Second Order Homogeneous Differential Equations with Constant Coefficients 1045 14. Match the phase-plane diagram with one of the systems of differential equations. Explain your reasoning. (a) dx dt =−4x+y (b) dx dt = x − 4y (c) dx dt =−x+4y (d) dx dt = x − 4y dy dt = x − 4y dy dt = 4x − y dy dt =−4x−y dy dt =−4x−y y x 31.6 SECOND ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS In the last section we looked at a frictionless model of a vibrating spring from the perspective of a system of differential equations. We sketched solution curves in the position-velocity phase-plane, but we did not find position as a function of time. In this section we will revisit this problem from the perspective of a second order differential equation, solve explicitly for position in terms of time, and tackle the issue of friction. By the end of the section we will be able to solve any second order homogeneous differential equation with constant coefficients and will have uncovered a surprising and beautiful mathematical result as a byproduct of our efforts. This result ties together differential equations, series, exponential functions and trigonometric functions. Consider the block-spring system analyzed after Example 31.26 in Section 31.5. We let x(t) be the position of the block at time t, where x = 0 corresponds to the equilibrium position of the system. Newton’s second law says force = mass · acceleration. x 0 x = 0 Figure 31.32 The forces on the system are the force of the spring, given by −kx in accordance with Hooke’s law, and the force of friction. The force of friction acts opposite the direction of 1046 CHAPTER 31 Differential Equations motion; it is reasonable to model friction as proportional to velocity. Velocity is given by dx dt and acceleration by d 2 x dt 2 . Therefore, we can deduce from Newton’s law that −kx − λ dx dt = m d 2 x dt 2 where k and λ are positive constants and m is the mass of the block. m d 2 x dt 2 + λ dx dt = 0 Equivalently, d 2 x dt 2 + λ m dx dt + k m x = 0. This is a second order differential equation of the form d 2 x dt 2 + b dx dt + cx = 0. If x(0) = x 0 > 0and x (0) = 0 then we might expect solutions as shown in Figure 31.33, with part (a) corresponding to the frictionless case and part (b) corresponding to some friction. x 0 x t x (a) (b) x 0 t Figure 31.33 Definition A differential equation of the form y + by + cy = 0 is called a second order homogeneous differential equation with constant coefficients. “Second order” refers to the second derivative, “constant coefficients” means the coefficients of y , y , and y must be constants, and “homogeneous” refers to the zero on the right-hand side of the equation. EXERCISE 31.9 Show that if y 1 and y 2 are solutions to the differential equation y + by + cy = 0 then y = C 1 y 1 + C 2 y 2 is also a solution, for any constants C 1 and C 2 . FACT If y 1 and y 2 are solutions to y + by + cy = 0 and y 1 and y 2 are not constant multiples of one another, then the general solution can be written in the form y = C 1 y 1 + C 2 y 2 where C 1 and C 2 are constants. (If y 1 = Cy 2 , C constant, we say y 1 and y 2 are linearly independent.) In light of the fact presented, we would like to guess two solutions to y + by + cy = 0. For inspiration, we’ll look at the simpler case with b = 0. As a warm-up, do Exercise 31.10. 31.6 Second Order Homogeneous Differential Equations with Constant Coefficients 1047 EXERCISE 31.10 For each differential equation, find two linearly independent solutions. i. y = y ii. y = 0 iii. y =−y ◆ EXAMPLE 31.30 Find the general solution to each of the following differential equations. (a) y = k 2 y (b) y = 0 (c) y =−k 2 y SOLUTION (a) y 1 = e kt and y 2 = e −kt are solutions, so the general solution is y = C 1 e kt + C 2 e −kt . (b) y 1 = 1 and y 2 = t are solutions, so the general solution is y = C 1 + C 2 t. (c) y 1 = sin(kt ) and y 2 = cos(kt) are solutions, so the general solution is y = C 1 sin(kt ) + C 2 cos(kt). Recall that in Chapter 30 we looked at the differential equation y =−yand showed that if the solution had a power series expansion it must be of the form y = C 1 sin t + C 2 cos t. ◆ Now let’s return to the general case y + by + cy = 0. Is it possible for a solution to be of the form e rt ? Suppose y = e rt . Then y = re rt and y = r 2 e rt . y + by + cy ? = 0 r 2 e rt + bre rt + ce rt ? = 0 e rt (r 2 + br + c) ? = 0 r is a solution to r 2 + br + c = 0. y = e rt is a solution. Given the differential equation y + by + cy = 0, we call the quadratic equation r 2 + br + c = 0 the characteristic equation.We’ll break our analysis into three cases. Case 1. Suppose the characteristic equation has two distinct real roots, r 1 and r 2 . Then y 1 = e r 1 t and y 2 = e r 2 t are solutions to y + by + cy = 0. The general solution is y = c 1 e r 1 t + c 2 e r 2 t . Case 2. Suppose the characteristic equation has a double root: r 1 = −b 2 . Then y 1 = e r 1 t is a solution. We need another solution. Claim: y 2 = te r 1 t is a solution. If y 2 = te r 1 t then y 1 2 = tr 1 e r 1 t + e r 1 t and y 2 = r 1 e r 1 t + tr 2 1 e r 1 t + r 1 e r 1 t . 1048 CHAPTER 31 Differential Equations y 2 + by 2 + cy 2 ? = 0 e r 1 t (r 1 + tr 2 1 +r 1 ) +b(e r 1 t )(tr 1 + 1) + cte r 1 t ? = 0 e r 1 t [2r 1 + tr 2 1 +btr 1 +b +tc] ? = 0 t(r 2 1 + br 1 +c 0 ) + 2r 1 + b ? = 0 2r 1 + b ? = 0 We know r 1 = −b 2 . 2 −b 2 + b ? = 0 −b + b =0 √ y 2 = te r 1 t is a solution. Therefore, the general solution is of the form y =C 1 e r 1 t + C 2 te r 1 t . Case 3. Suppose the characteristic equation has complex roots ∝±βi where i = √ −1. We might first be tempted to turn away from this case, but a look back at Figure 31.33 should push us forward. Both curves in Figure 31.33 cross the t -axis multiple times whereas the functions y = C 1 e r 1 t + C 2 e r 2 t and y = (C 1 t + C 2 )e r 1 t have at most one zero. If ∝±βi are roots of the characteristic equation then y 1 = e (∝+βi)t and y 2 = e (∝−βi)t are solutions to y +by + cy = 0. e (∝+βi)) t =e ∝t ·e βit . How can we make sense of e βit ? Let’s turn to Taylor series. We know e u = 1 + u + u 2 2! + u 3 3! + u 4 4! +···. Let u = βit. Then e βit =1 +βit + (βit) 2 2! + (βit) 3 3! + (βit) 4 4! + (βit) 5 5! +··· i= √ −1, so i 2 =−1, i 3 =−i,i 4 =1, i 5 = i, etc. e βit =1 +iβt − (βt) 2 2! − i (βt) 3 3! + (βt) 4 4! + i (βt) 5 5! − (βt) 6 6! +··· Gather together all terms without factors of i appearing. e βit = 1 − (βt) 2 2! + (βt) 4 4! − (βt) 6 6! +··· cos(βt ) +i βt − (βt) 3 3! + (βt) 5 5! −··· sin(βt) We obtain what is known as Euler’s Formula: e iβt =cos(βt) + i sin(βt).Wedefine e (∝+βi)t to be e (∝+βi)t =e ∝t (cos(βt) + i sin(βt)). The general solution to y + by + cy is therefore given by y = C 1 e ∝t · e βit +C 2 e ∝t ·e −βit =e ∝t C 1 cos(βi) + C 1 i sin(βt ) + C 2 cos(−βt) +C 2 i sin(−βt) . Cosine is an even function and sine is an odd function, so 31.6 Second Order Homogeneous Differential Equations with Constant Coefficients 1049 y = e ∝t (C 1 + C 2 ) cos(βt) + (C 1 − C 2 )i sin(βt ) or y = e ∝t C 3 cos(βt) + C 4 sin(βt ) . Let’s return to the vibrating spring example. x + λ m x + k m x = 0. The characteristic equa- tion r 2 + λ m r + k m = 0, has roots given by r = −λ m ± λ 2 m 2 − 4 k m 2 = −λ 2m ± 1 2m λ 2 − 4km. EXERCISE 31.11 Argue that if the characteristic equation r 2 + λ m r + k m = 0 has two real roots then they are both negative. What physical situation does this case correspond to? It is referred to as “overdamped”. EXERCISE 31.12 The situation corresponding to one real root, r = −λ 2m , is called “critically damped.” Find the solution given that the spring is compressed 0.5 inches and released and is critically damped. Your answer will be in terms of λ and m. Sketch this solution. EXERCISE 31.13 The situation corresponding to complex roots is called “underdamped”. Show that in all three situations, overdamped, critically damped, and underdamped, lim t→∞ x(t) = 0 provided λ>0.Explain why this makes sense. EXERCISE 31.14 Use Euler’s Formula to compute e πi . PROBLEMS FOR SECTION 31.6 Solve the differential equations in Problems 1 through 8. 1. y − 4y =−3 2. y − 2y + y = 0 3. y + 25y = 0 4. y + 5y = 0 5. y + 5y = 0 6. y = y + 2y 7. 3y + 3y + 3y = 0 8. d 2 x dt 2 + 25 = 3 dx dt In Problems 9 through 11, find the particular solution corresponding to the initial conditions given. 9. d 2 x dt 2 + dx dt = 2x, x(0) =−1, x (0) = 0 1050 CHAPTER 31 Differential Equations 10. 2x + 6x = 0, x(0) = 0, x (0) = 4 11. 4 d 2 x dt 2 − 4 dx dt =−x, x(0) = 1, x (0) = 2 Problems 12 through 15 refer to the equation x + bx + cx = 0. 12. What condition(s) must be satisfied to have a periodic solution? If the solution is periodic, what will its period be? 13. Find constants b and c such that if x(0) = 5 and x (0) = 0 then x(n) = 5 for any integer n. Is this pair of constants unique? 14. Suppose b>0and c>0.Is lim t→∞ x(t) necessarily zero? Explain. Interpret your response in terms of vibrating springs. 15. Suppose b<0 and c>0. For x(0) and x (0) not both zero, is it possible that lim t→∞ |x(t)|=L,where L is finite? Explain. 16. A spring with a 2-kg mass has a natural length of 0.6 m. A 10 N force is required to compress it to a length of 0.5 m. If the spring is compressed to 0.4 m and released, find the position of the mass at time t. Assume a frictionless system. 17. Write a second order homogeneous differential equation that is satisfied by y(t) = e t sin t. (The answer is not unique.) 18. Compute the following. (a) e 2πi (b) e −πi . coefficients” means the coefficients of y , y , and y must be constants, and “homogeneous” refers to the zero on the right-hand side of the equation. EXERCISE 31.9 Show that if y 1 and y 2 are. 0.006y 2 + 0.001xy 5. Let x = x(t) be the number of thousands of animals of species A at time t . Let y = y(t) be the number of thousands of animals of species B at time t. Suppose dx dt = x −. implications of your portrait for the long-term outcome of the competition. x(t) and y(t) give the number of thousands of animals of species A and B, respectively. 6. dx dt = 0.03x − 0.01x 2 −