154 Chapter 9 Hints, Answers, and Solutions The particular trajectories are highlighted in the phase portraits below.

Transcription

1 54 Chapter 9 Hints, Answers, and Solutions 9. The Phase Plane The particular trajectories are highlighted in the phase portraits below Shown below is one possibility with x(t) and y(t) periodic. There are many others because the same parametric curve could be traced out from different initial points and with different speeds. x y t Shown below is one possibility with x(t) and y(t) approaching sinusoidal functions as t. There are many others because the same parametric curve could be traced out from different initial points and with different speeds.

2 9. Hints, Answers, and Solutions Shown below is one possibility with x(0) = y(0) = 0 and x(t) and y(t) each approaching a sinusoidal function with amplitude as t. x y t Shown below is one possibility with x(t) and y(t) each approaching a periodic function with amplitude as t Since x(t) and y(t) appear to be sinusoidal, the parametric curve is an ellipse. y 4 x

3 56 Chapter 9 Hints, Answers, and Solutions x y t t (0, 0), (, 0), and (3, 0) are unstable; (, 0) is neutrally stable (0, 0) and (3, 0) are asymptotically stable; (, 0) and (, 0) are unstable The first two plots in Figure 9 correspond to the two inner, periodic orbits in Figure 8, respectively. The third plot corresponds to the clockwise orbit comprising the upper portion of outer circle, and the fourth corresponds to the counterclockwise orbit comprising the lower portion of outer circle dy dx = x y = y dy = x dx = y = x = C = x + y = C. Orbits trace out circles centered at the origin. The orientation is clockwise, since (e.g.) x > 0 when y > 0. dy dx = x y = y dy = x dx = y = x = C = y x = C. These are hyperbolas. The phase portrait is shown below

4 9. Hints, Answers, and Solutions dy dx = xy xy dy = dx = = y = x + C. The phase portrait is shown below dy dx = xy y = dy dx = x = y = x + C. The phase portrait is shown below dy dx = y +x y xy = dy dx = y x + x = y + x y = x. Use the integrating factor x to obtain the solution y = 3 x + C x. The phase portrait is shown below dy dx = sin x y - = y dy = sin x dx = y = cos x + C = y = cos x + C Implicitly differentiate x + y = with respect to x to obtain dy dx = x y. Now using x + y = in the right-hand side of the slope equation, dy dx = (x+y) y ( x y ), we have dy (x + y) = dx y ( x y ) = (x + xy + y ) y ( (x + y )) = xy y = x y. Thus the slope field is tangent to the unit circle at each point on it. No single orbit traces out the entire unit circle, because (±, 0) are equilibrium points.

5 58 Chapter 9 Hints, Answers, and Solutions Rearrange the slope equation into (x + x )dx + ydy (xdy + ydx) = 0. Integrate to get x 3 + 3x + 3y 6xy = C d ϕ dx ϕ(x(t), y(t)) = dt x dt + ϕ y dy dt = ϕ ϕ x y + ϕ y ϕ = 0 = ϕ(x(t), y(t)) = C. x 9. Phase Portraits of Homogeneous Linear Systems 9... (a) unstable, saddle point; (b) real, opposite sign 9... (a) stable, spiral point; (b) nonreal, negative real part (a) stable, biaxial node; (b) real, negative (a) stable, coaxial node; (b) real, repeated, negative (a) unstable, biaxial node; (b) real, positive (a) unstable, spiral point; (b) nonreal, positive real part From the given eigenvectors, the straight line orbits trace out y = ±x. Since the eigenvalues are both negative, the origin is a stable node From the given eigenvectors, the straight line orbits trace out y = ±x. Since the eigenvalues are real with opposite signs, the origin is a stable node.

6 9. Hints, Answers, and Solutions From the given eigenvectors, the straight line orbits trace out y = x and y = x. Since the eigenvalues are both positive, the origin is an unstable node From the given eigenvectors, the straight line orbits trace out only y = x. Since the repeated eigenvalue is negative, the origin is a stable coaxial node. ( a b 9... Let A = c d ). From the given sign diagram, we know a, c > 0 while b, d < 0. Along the positive x axis, the system x = ax+by, y = cx+dy reduces to x = ax > 0 and y = cx > 0, i.e. x and y are both increasing. Along the negative x axis, we have precisely the opposite behavior. On the other hand, along the positive y axis, the system reduces to x = by < 0 and y = dy < 0, i.e. x and y are both decreasing. Along the negative y axis, we have precisely the opposite behavior. Since the real parts of the eigenvalues of A were negative by assumption, the orbits spiral in towards the origin as shown in the phase portrait below. Note that if the real parts of the eigenvalues of A were positive, the phase portrait would look the same except that the orbits would spiral outward from the origin. a b 9... Let A =. From the given sign diagram, we know a, c > 0 while b, d < 0. c d Along the positive x axis, the system x = ax+by, y = cx+dy reduces to x = ax > 0

7 60 Chapter 9 Hints, Answers, and Solutions and y = cx < 0, i.e. x is increasing while y is decreasing. Along the negative x axis, we have precisely the opposite behavior. On the other hand, along the positive y axis, the system reduces to x = by > 0 and y = dy < 0, i.e. x is increasing while y is decreasing. Along the negative y axis, we have precisely the opposite behavior. Since the real parts of the eigenvalues of A were negative by assumption, the orbits spiral in towards the origin as shown in the phase portrait below. Note that if the real parts of the eigenvalues of A were positive, the phase portrait would look the same except that the orbits would spiral outward from the origin. a b Let A =. From the given sign diagram, we know a, c, d < 0 while b > 0. c d Along the positive x axis, the system x = ax+by, y = cx+dy reduces to x = ax < 0 and y = cx < 0, i.e. x and y are both decreasing. Along the negative x axis, we have precisely the opposite behavior. On the other hand, along the positive y axis, the system reduces to x = by > 0 and y = dy < 0, i.e. x is increasing while y is decreasing. Along the negative y axis, we have precisely the opposite behavior. Since the real parts of the eigenvalues of A were negative by assumption, the orbits spiral in towards the origin as shown in the phase portrait below. Note that if the real parts of the eigenvalues of A were positive, the phase portrait would look the same except that the orbits would spiral outward from the origin. a b Let A =. From the given sign diagram, we know a, b, d < 0 while c > 0. c d Along the positive x axis, the system x = ax+by, y = cx+dy reduces to x = ax < 0 and y = cx > 0, i.e. x is decreasing while y is increasing. Along the negative x axis, we have precisely the opposite behavior. On the other hand, along the positive y

8 9. Hints, Answers, and Solutions 6 axis, the system reduces to x = by < 0 and y = dy < 0, i.e. x and y are both decreasing. Along the negative y axis, we have precisely the opposite behavior. Since the real parts of the eigenvalues of A were negative by assumption, the orbits spiral in towards the origin as shown in the phase portrait below (a, b) Note that if the real parts of the eigenvalues of A were positive, the phase portrait would look the same except that the orbits would spiral outward from the origin. (c, d) The eigenvalues of A are + k ± k 8k + 4. The real parts of the eigenvalues are plotted below as a function of the parameter k. 6 4 Re λ k -4-6 Notice the eigenvalues are nonreal when 4 3 < k < 4 + 3, corresponding to the straight part of the graph above. The following can be inferred about the type

9 6 Chapter 9 Hints, Answers, and Solutions of the equilibrium point as k changes: k < 0 : saddle point 0 < k < 4 3 : stable biaxial node k = 4 3 : stable coaxial node 4 3 < k < : stable spiral point k = : center < k < : unstable spiral point k = : unstable coaxial node k > : unstable biaxial node ( (a) The eigenvalues of A are k ± k 4 ). The real parts of the eigenvalues are plotted below as a function of the parameter k (b) Real and distinct when k > ; real and repeated when k = ±; imaginary when k = 0; nonreal complex when k <. (c) The following can be inferred about the type of the equilibrium point as k changes: k < : stable biaxial node k = : stable coaxial node < k < 0 : stable spiral point k = 0 : center 0 < k < : unstable spiral point k = : unstable coaxial node k > : unstable biaxial node ( (a) The eigenvalues of A are k ± k + k 7 ). The real parts of the eigenvalues are plotted below as a function of the parameter k

10 9. Hints, Answers, and Solutions 63 (b) Real and distinct when k < or k > + ; real and repeated when k = ± ; imaginary when k = ; nonreal complex when < k < +. (c) The following can be inferred about the type of the equilibrium point as k changes: k < : stable biaxial node k = : stable coaxial node < k < : stable spiral point k = : center < k < + : unstable spiral point k = + : unstable coaxial node + < k < : unstable biaxial node k > : saddle point A λi = λ (a + a )λ + A ; hence the eigenvalues are ( a + a ± ) (a + a ) 4 A. Since A has a repeated eigenvalue, it follows that (a +a ) = 4 A, and the repeated eigenvalue is (a + a ). Now, Ã λi = λ (a + a + ε)λ + A + a ε; hence its eigenvalues are λ, = ( a + a + ε ± ) (a + a + ε) 4( A + a ε). Using (a + a ) = 4 A, we simplify these as follows: λ, = ( a + a + ε ± ) (a + a ) + (a + a )ε + ε 4( A + a ε) = ( a + a + ε ± ) (a + a )ε + ε 4a ε = ( a + a + ε ± ε ( (a a ) + ε )). From this last form, the stated results follow by consideration of cases A reasonable sketch of the phase portrait can be made based on the eigenvalues, nullclines, and straight-line orbits. This information is summarized below. Eigenvalues: ( ± 5). Nullclines: y = 0, y = x. Straight-line orbits: m = ( ± 5) = y = ( ± 5)x. The phase portrait is shown below. Nullclines are dashed. Straight-line orbits are the thicker lines.

11 64 Chapter 9 Hints, Answers, and Solutions 9... A reasonable sketch of the phase portrait can be made based on the eigenvalues, nullclines, and straight-line orbits. This information is summarized below. Eigenvalues:,. Nullclines: y = 0, y = 3 x. Straight-line orbits: m =, = y = x, y = x. The phase portrait is shown below. Nullclines are dashed. Straight-line orbits are the thicker lines A reasonable sketch of the phase portrait can be made based on the eigenvalues, nullclines, and straight-line orbits. This information is summarized below. Eigenvalues: ±. Nullclines: y = 3x, y = x. Straight-line orbits: m = 3, = y = 3x, y = x. The phase portrait is shown below. Nullclines are dashed. Straight-line orbits are the thicker lines Eigenvalues: ± i. Nullclines: y = x, y = 4x. There are no straight-line orbits, since m = 4 m +m has no real solutions.

12 9.3 Hints, Answers, and Solutions Eigenvalues:,. Nullclines: y = 3 4x, y = x. Straight-line orbits: m = = y = x Eigenvalues: 0,. Nullclines: y = x, on which every point is an equilibrium point. Straight-line orbits: dy dx = everywhere except on y = x. So all nontrivial orbits trace out straight lines with slope. 9.3 Phase Portraits of Nonlinear Systems (a) We want to find all simultaneous solutions of x(y x ) = 0 and x y = 0. Begin by solving x y = 0 for y and then substituting the result into x(y x ) = 0. That gives x(x x ) = x(x + )(x ) = 0, whose solutions are x = 0,,. Now using y = x again, we get equilibrium points (0, 0), (, ), and (, 4). x + y x (b) J (x, y) =. Then x 0 J (0, 0) = has eigenvalues,, 0 J (, ) = has eigenvalues ± 3, J (, 4) = has eigenvalues 4 ( 3 ± 33). (c) (0, 0) is a stable node; (, ) and (, 4) are saddle points.

13 66 Chapter 9 Hints, Answers, and Solutions (e) (a) We want to find all simultaneous solutions of x( y) = 0 and y( x) = 0. First, x( y) = 0 if x = 0 or y =. If x = 0, then y( x) = y = 0 = y = 0. This gives the equilibrium point (0, 0). If y =, then y = y( x) = x = 0 = x =. This gives the equilibrium point (, ). y x (b) J (x, y) =. Then y + x 0 J (0, 0) = has eigenvalues ±, 0 0 J (, ) = has eigenvalues ± i. 0 (c) (0, 0) is a saddle point; (, ) is a possible center or spiral point. (e) (a) Solving x( y) = 0 and x y = 0 gives (0, 0) and (, ). y x (b) J (x, y) =. Then 0 J (0, 0) = has eigenvalues ±, 0 J (, ) = has eigenvalues ( ± i 3). (c) (0, 0) is a saddle point; (, ) is stable spiral point

14 9.3 Hints, Answers, and Solutions 67 (e) (a) Solving y = 0 and x(4 x) = 0 gives (0, 0) and (4, 0). (b) J (x, y) = 0. Then 4 x 0 0 J (0, 0) = has eigenvalues ±, J (4, 0) = has eigenvalues ± i. 4 0 (c) (0, 0) is a saddle point; (4, ) is possible center or spiral point. (e) (a) Solving x y = 0 and x y = 0 gives (, ) and (, ). (b) J (x, y) = x y. Then J (, ) = has eigenvalues ( ± 7), J (, ) = has eigenvalues (3 ± i 7). (c) (, ) is a saddle point; (, ) is stable spiral point.

15 68 Chapter 9 Hints, Answers, and Solutions (e) (a) Solving x y = 0 and y(x y) = 0 gives (, ), (, ), and (±, 0). x y (b) J (x, y) =. Then y x y J (, ) = has eigenvalues (3 ± i 7), J (, ) = has eigenvalues ( 3 ± i 7), J ( ( 8 ) 0, 0) = 0 has eigenvalues, 8, J ( ( ) 8 0, 0) = has eigenvalues 8,. 0 (c) (±, 0) are saddle points; (, ) is an unstable spiral point; (, ) is a stable spiral point. (e) (a) x + y = 0 and x y = 0 at (, ±) and (, ±). x y (b) J (x, y) =. Then x y J (, ) = and J (, ) = J (, ) have eigenvalues ± 8, J (, ) = and J (, ) = J (, ) have eigenvalues ± i.

16 9.3 Hints, Answers, and Solutions 69 (c) (, ) and (, ) are saddle points; (, ) is a stable spiral point; (, ) is an unstable spiral point. (e) (a) xy + 3 = 0 and x y + x + = 0 at (, 3). y x 3 (b) J (x, y) = xy x. J (, 3) = 4 (c) (, 3) is a saddle point. (e) 3 has eigenvalues ± (a) x y + y + = 0 and x y + x y = 0 at (, /). xy + x (b) J (x, y) = xy x. J (, /) = 0 (c) (, /) is a saddle point. has eigenvalues,. (e)

17 70 Chapter 9 Hints, Answers, and Solutions ( ) The Jacobian is J (x, y) =. For odd n, we get J (±nπ, 0) =, k cos x 0 k 0 which has eigenvalues ± k. Hence these equilibrium points are saddle points. Each odd n corresponds to the unstable position at the top of the pendulum s arc The x equation implies that y = 0 at every equilibrium point. Since ρ(0) = 0, it follows from the y equation( that the equilibrium ) points are (±nπ, 0), n = 0,,,... 0 The Jacobian is J (x, y) = k cos x ρ. For odd n, we get (y) 0 J (±nπ, 0) = k ρ which has eigenvalues ρ (0) ± ρ (0) + 4k. (0) Since the eigenvalues are real and of opposite sign, these equilibrium points are saddle points. For even n, we get 0 J (±nπ, 0) = k ρ which has eigenvalues ρ (0) ± ρ (0) 4k. (0) If ρ (0) is small relative to k, we get a pair of complex conjugate eigenvalues with negative real parts. Therefore the equilibrium points are stable spiral points. If ρ (0) is large, we get two negative real eigenvalues. Therefore the equilibrium points are stable nodes. As in the undamped case, each odd n corresponds to the unstable position at the top of the pendulum s arc, and each even n corresponds to the stable position at the bottom of the pendulum s arc. When ρ (0) is small relative to k, damped oscillations occur, corresponding to a stable spiral point. For larger values of ρ (0), damping prevents oscillation and causes the equilibrium point to be a stable node The x equation implies that y = 0 at every equilibrium point. The y equation then implies that x = 0 at every equilibrium point. Thus the sole equilibrium point is (0, 0). The Jacobian is ( J (x, y) = and so 0 kxy k(x ) 0 J (0, 0) =, which has eigenvalues ( k ± ) k k 4. A plot of the real parts of the eigenvalues (for k > 0) is shown below. 3 Re λ ), k 3 4

18 9.3 Hints, Answers, and Solutions 7 The eigenvalues are nonreal when 0 < k <, corresponding to the straight part of the graph above. The following can be inferred about the type of the equilibrium point as k changes: 0 < k < : unstable spiral point, k = : unstable coaxial node, k > : unstable biaxial node The equilibrium point is (k, /k). The Jacobian is y xy J (x, y) = y, xy k and so /k J (k k, /k) =, which has eigenvalues k3 ± k 6 6k 3 + /k k k. A plot of the real parts of the eigenvalues (for k > 0) is shown below. Re λ The eigenvalues are nonreal when 3 < k 3 < 3 +. The following can be inferred about the type of the equilibrium point as k changes: 0 < k 3 < 3 : stable node 3 < k 3 < : stable spiral point k = : possible center < k 3 < 3 + : unstable spiral point k 3 > 3 + : unstable node Phase portraits for each case are shown below. Stable node if 0 < k 3 < 3. Stable spiral point if 3 < k 3 <. k

19 7 Chapter 9 Hints, Answers, and Solutions Possible center if k =. Unstable spiral point if < k 3 < 3 +. Unstable node if k 3 > The nonnegative equilibrium points are (0, 0) and (, ). xy x J (x, y) =. y x y 0 J (0, 0) = has eigenvalues 0,. 0 0 J (, ) = has eigenvalues ± i. So (0, 0) is unstable, and (, ) is a stable spiral. The phase portrait is shown below

20 9.3 Hints, Answers, and Solutions The nonnegative equilibrium points are (0, 0), (, ), and (0, ). 3x y x 3 J (x, y) = 4xy. (+x +x y 0 J (0, 0) = has eigenvalues,. 0 J (, ) = has eigenvalues ( 3 ± 5). 0 J (0, ) = has eigenvalues,. 0 So (0, 0) is an unstable node; (, ) and (0, ) are saddle points The equilibrium points are (0, 0), (, 0), (, 0), and (3, 0). ( ) 0 J (x, y) = 5x 4 8x 3 + 5x. 34x J (0, 0) = has eigenvalues ± J (, 0) = has eigenvalues 0, J (, 0) = has eigenvalues ± i. 0 0 J (3, 0) = has eigenvalues ±. 0 So (0, 0) and (3, 0) are saddle points; (, 0) is a center; the nature of (, 0) is not determined (0, 0) and (3, 0) are stable spiral points, and (, 0) is a saddle point. (, 0) has characteristics of a stable node and a saddle point, consistent with the fact that the eigenvalues of J (, 0) are and 0.

21 74 Chapter 9 Hints, Answers, and Solutions The equilibrium points are (0, 0), (0, ), (5/7, 4/7), (6/7, /7), (, 0), and (3, 0). The Jacobian here is quite complicated, so a computer will be useful. It turns out that 6 0 J (0, 0) = has eigenvalues 6,, 0 6/3 0 J (0, ) = has eigenvalues, 6/3, /4 /7 J (5/7, 4/7) = has eigenvalues 4/7 /7 8 ( 5 ± 37), 6/35 4/35 J (6/7, /7) = has eigenvalues /7 /7 70 ( ± i 839), /5 8/5 J (, 0) = has eigenvalues, /5, 0 3/0 6/5 J (3, 0) = has eigenvalues, 3/0. 0 So (0, 0), (5/7, 4/7), and (, 0) are saddle points; (0, ) and (0, 3) are stable nodes; (6/7, /7) is an unstable spiral point (a) The equilibrium points are (0, 0) and (5/, 0). ( ) 0 6 4y J (x, y) = x + y + x + 8y 0 6 J (0, 0) = has eigenvalues 0 ( ± i 39). 0 6 J (5/, 0) = has eigenvalues 6, (b) So (0, 0) is a stable spiral point while (5/, 0) is a saddle point (c) First, (x /) + y = = (x /) + y dy dx = 0 = dy dx = x y. Also, the equation of the circle is equivalent to 4x + 4y = 3 + 4x. Use this in the right-hand side of the slope equation as follows: 4x + xy + 4y 0x y 6y y = = 3 6x + xy y (3 y)( x) 6y y = y(3 y) Therefore the slope field is tangent to the circle at each point on it. = x. y

22 9.3 Hints, Answers, and Solutions (a) The x equation implies that either x = 0 or y = 0 at every equilibrium point, and y equation implies that y = ±x at every equilibrium point. Thus the only equilibrium point is (0, 0). ( y x J (x, y) = x y ) ; thus J (0, 0) = (c) First, x + y ax = 0 = x + y dy dy dx a = 0 = dx = a x y. Also, on these circles we have y = ax x. So the right hand side of the slope equation becomes x y xy = x + ax x xy = x(x a) xy = a x. y (a) The orbits follow the level curves of the surface. (b) Nearby orbits trace out closed curves around (x 0, y 0 ) so this point is a center point (a) The equilibrium points are (0, 0) and (±/, 0). ( ) 0 J (x, y) = 4 4x. 0 0 J (0, 0) = has eigenvalues ±. 4 0 J (±/ 0, 0) = has eigenvalues ± i (b) Solve the separable equation to get the implicit family 4x 4 4x + y = C. (c) Solving x (4x4 4x +y ) = y (4x4 4x +y ) = 0 yields (0, 0) and (±/, 0). Now apply the second derivative test for functions of two variables: Let D(x, y) = ϕ xx ϕ yy (ϕ xy ). Since D(0, 0) < 0, we know ϕ(0, 0) is neither a max nor a min. On the other hand D(±/, 0) = 3 > 0 and ϕ xx (±/, 0) = 6 > 0 so a local min occurs at these points. A surface plot and a plot of the level curves are shown below.

23 76 Chapter 9 Hints, Answers, and Solutions 0.5 φ x y y x (d) With the help of the identity sin θ = sin θ cos θ, it is straightforward to check that (sin θ, sin θ) satisfies 4x 4 4x + y = 0. The phase portrait (including the Lissajous curve) is shown below (a) Solve the separable equation to get the implicit family y cos x = C. (b) Apply the second derivative test for functions of two variables: Let D(x, y) = ϕ xx ϕ yy (ϕ xy ). Here, D(x, y) = 4 cos x. When n is even, D(±nπ, 0) = 4 > 0 and ϕ xx (±nπ, 0) = > 0; so a local min occurs at each of these points. 9.4 Limit Cycles. (a) p = xx + yy = x(y + x( x y )) + y( x + y( x y )) = (x + y )( x y ) = p( p) (b) Suppose r p(0). Then by (a), p(t) is nondecreasing for t 0, and p(t) as t. On the other hand, if p(0) R, then p(t) is nonincreasing for all t 0, and p(t) as t. (c) To see that (0, 0) is the only equilibrium point of the system, look at yf(x, y) xg(x, y). As in (b), if 0 < p(0) < then p(t) is increasing for all t 0 and p(t) as t. Therefore any orbit initially near the origin will move away from the origin. (d) We can conclude that there exists a periodic orbit of the system in A r,r.

24 9.4 Hints, Answers, and Solutions Take M = {(x, y) x + y = r } and let p(t) = x(t) + y(t) as in Problem. Since p (t) 0 when x + y, the hypotheses of Theorem are satisfied. Therefore M is a forward invariant region. None ofthese disks contain a periodic orbit since (0, 0) is asymptotically stable. See Figure 3b in the section (a) p = xx + yy = x(ay + xϕ) + y( ax + yϕ) = (x + y )ϕ = pϕ (b) Notice that if R p(t) r, then p (t) < 0. Therefore every closed disk centered at the origin with radius r R is a forward invariant region. (c) Solve f(x, y) = 0 and g(x, y) = 0 simultaneously to see that (0, 0) is the only equilibrium point of the system. Moreover, p (t) > 0 when (x(t), y(t)) is near the origin. Thus any orbit initially near the origin will move away from the origin. (d) We can conclude that there exists a periodic orbit of the system in the disk x + y R Assume that a, b, c are positive. Let p = bx + ay. Then p = (bx + acy )ϕ. Let R R be such that x + y R whenever bx + acy R. Then p < 0 whenever x + y R. Thus every closed disk centered at (0, 0) with radius r R is a forward-invariant region. All of the same conclusions as in Problem 3 can now be reached (a) If x(t) 0 and y(t) = 0, then y (t) = bx(t) + β 0. If x(t) = 0 and y(t) 0, then x (t) = py(t) + α 0. Therefore, no orbit can leave the first quadrant. (b) First observe that x + y = (a + b)x + (p + q)y + α + β. Comparing the lines x + y = k and (a + b)x + (p + q)y + α + β = 0 and using the fact that a + b < 0, p + q < 0, α + β 0, and the assumption on k, we see that x + y = k lies above (a + b)x + (p + q)y + α + β = 0 in the first quadrant. Thus x + y < 0 when x + y = k. (c) By (a) and (b), the triangular region is a forward invariant region when k 4/3. Solving f(x, y) = 0 and g(x, y) = 0 simultaneously, we see that (/, /3) is the only equilibrium point, and it is a stable spiral point. Therefore every first quadrant orbit approaches (/, /3) as t. The phase portrait is shown below

25 78 Chapter 9 Hints, Answers, and Solutions (a) Solve f(x, y) = g(x, y) = 0 to see that (0, 0) and (3/, 6/5) are the only equilibrium points of the system with nonnegative entries. Now + xy /4 + x J (x, y) =, xy x /4 J (0, 0) = has eigenvalues ( ± 0), 8/5 5/ J (3/, 6/5) = has eigenvalues 3/5 5/4 40 (7 ± i 75). (b) (x + y) = x y + 9y/4 = (9/4 x )y (c) For each fixed y > 0, the minimum value of x y x+y/4 occurs where xy = 0, i.e., at x = /y. Therefore, x y x + y/4 /y /y + y/4 = y 4 4y for all y > 0 and all x. Therefore, if y >, then dy dx = x y + x + y x y x + y/4 ( x y + x + y) 4y y 4y + 4 y 4, where the last inequality comes from maximizing x y + x + y (for fixed y > 0). (d) Observe that 4y + y 4 is decreasing for y > with limit 4 as y. Thus, for all y > 5 its value is less than 0/, which is less than 5. (e, f) The only equilibrium point is (, 3), which is an unstable spiral point. A forwardinvariant region containing (, 3) is { (x, y) x 0, y 0, y 9 } 33 (x + 5), x + y. 4 4 This follows from: (i) x = when x = 0, and y = 3x 0 when y = 0 and x 0. (ii) (x + y) = x 0 if x. (iii) For fixed y > 0, x y + 3x is maximized by x = 3 y ; so y 9 4y for all y > 0. (iv) For fixed y > 0, x y 4x + is minimized by x = y ; so x 4/y for all y > 0. (v) By (iii) and (iv), dy dx 9/(4y) 4/y = 9 dy 4y 6 ; thus dx 9 4 if y 5. (vi) The line x + y = c intersects the line y = 9 4 x + 5 at x =, if c = 33/ (a) (i) x = a > 0 when x = 0, and y = b > 0 when y = 0. (ii) (x + y) = a + b x 0 if x a + b.

26 9.4 Hints, Answers, and Solutions 79 (iii) y b and x a 4y = 4ay 4y for all (x, y) in the first quadrant; thus dy dx 4by 4ay for all (x, y) in the first quadrant. Therefore, dy dx b/a = b/a for all x > 0 and y > a. (iv) The line x + y = c intersects the line y = b a x + a at x = a + b, if c = (a+b)(a+b)+ a. From (i) (iv) it follows that the region in the first quadrant bounded by the lines y = b a x + (a+b)(a+b)+ a and x + y = a is forward invariant There is no equilibrium point, and hence no periodic orbit Every periodic orbit must enclose the origin. Therefore, since the first quadrant is a forward-invariant region (by Theorem ), no periodic orbit can exist The only equilibrium point is (, ); so if there is a periodic orbit, it must enclose (, ) Every periodic orbit must enclose a point on the line y = x, which consists entirely of equilibrium points. This implies that a periodic orbit must pass through an equilibrium point, which is impossible. Therefore, no periodic orbit can exist The system has no periodic orbit. One example is x = xy, y = ( x) ( y) (a) Let p be the period of the orbit. By Green s theorem, Ω (f x + g y ) dx dy = p 0 ( g x + f y )dt = p ( g f + f g)dt = 0. 0 (b) f x + g y = + cos y + sin x 0 for all x, y and not identically zero on any region; therefore, for any region Ω, Ω (f x + g y ) dx dy 0. (c) With y = x, the equation becomes the system x = y, y = ρ(x)y µ(x). Thus f x + g y = ρ(x). The result now follows (a) Clearly (0, 0), (±, 0) are equilibrium points of this system. x( 6x + 3x 4 + y ) (x )y J (x, y) = 3x + xy x 3 y x x4 3y. 0 J (0, 0) = has eigenvalues ±. 0 J (, 0) = has eigenvalues 4 (3 ± i 3 ). J (, 0) = has eigenvalues 4 ( ± i 3 ). Therefore, (0, 0) is a saddle point, (, 0) is an unstable spiral point, and (, 0) is a stable spiral point. (b) Let us consider the behavior of orbits crossing the x axis near (±, 0). First note that on the x axis, we have y = x( x ). Now, x > implies y < 0 while 0 < x < implies y > 0. Therefore, the flow around the spiral point (, 0) is clockwise. Similarly, < x < 0 implies y < 0 while x < implies y > 0. Therefore the flow around the spiral point (, 0) is clockwise.

27 80 Chapter 9 Hints, Answers, and Solutions (d) (a) The equilibrium points are (0, 0), (0, ), (, ), (, ). The x-nullcline is the circle x + (y ) = ( ), and the y-nullclines are the lines x = 0 and y =. (b) Suppose that y(0) =. Then y(t) = for all t, and so x = x 4 for all t. Therefore, x (t) 0 for all t if x(0) and x (t) > 0 for all t if x(0) >. (c) Differentiate the ellipse equation implicitly with respect to x. (d) d dt ϕ = yy (xyx + x y ) y y + xx = (x xy)x + (y x y )y = x( y)(x + y y) + (y x y )x( y) = 0. Observe that ϕ(x, ) is constant and that implicit differentiation of ϕ(x, y) = C (with respect to x) yields the slope equation in (c). (e) ϕ x = ϕ y = 0 and ϕ xx ϕ yy ϕ xy = y(y 6) 4x > 0 at each of (0, 0) and (0, ). Therefore, those equilibrium points are surrounded by periodic orbits. (f) Consider a closed level curve Γ of ϕ that surrounds (x, y ). The gradient ϕ = (ϕ x, ϕ y ) is orthogonal to Γ and points out of the region enclosed by Γ (since ϕ increases in the direction of ϕ). The vector (f, g) is tangent to any orbit, and the cosine of the angle θ between (f, g) and ϕ at any point has the same sign as ϕ (f, g) = ϕ x f + ϕ y g, which we assume is nonpositive on Γ. Thus θ π/, which implies that (f, g) is either tangent to Γ or points into the region enclosed by Γ ϕ x f +ϕ y g = xy yx = 0. Therefore, disks centered at (0, 0) are forward-invariant. (In fact, orbits are circular.) ϕ x f + ϕ y g = x(y x) + y( x y) = (x + y ) 0. Therefore, disks centered at (0, 0) are forward-invariant ϕ x f + ϕ y g = x(y x 3 ) xy = x 4 0. Therefore, disks centered at (0, 0) are forward-invariant. Also, (0, 0) is asymptotically stable.

28 9.5 Hints, Answers, and Solutions Beyond the Plane Solving f(x, y, z) = g(x, y, z) = h(x, y, z) = 0 yields only (0, 0, 0). Now a a 0 a a 0 J (x, y, z) = b z x = J (0, 0, 0) = b 0, y x c 0 0 c which has eigenvalues c, ( (a + ) ± (a ) + 4ab). Since a > 0, 0 b <, and c > 0, all the eigenvalues are real and negative. Hence the equilibrium point is asymptotically stable Solving f(x, y, z) = g(x, y, z) = h(x, y, z) = 0 yields (0, 0, 0), ( 9.30, 9.30, 7), and (9.30, 9.30, 7). The Jacobian is Then J (x, y, z) = 8 z x. y x 3. J (0, 0, 0) has eigenvalues λ =.83,.83, 3., J ( 9.30, 9.30, 7) has λ = 4.0,.00 ±.03i, J (9.30, 9.30, 7) has λ = 4.0,.00 ±.03i. At each equilibrium point, J has an eigenvalue with positive real part. Therefore each equilibrium point is unstable Solving f(x, y, z) = g(x, y, z) = h(x, y, z) = 0 yields (0, 0, 0), ( 9.35, 9.35, 7), and (9.35, 9.35, 7). The Jacobian is Then J (x, y, z) = 8 z x. y x 3.4 J (0, 0, 0) has eigenvalues.83,.83, 3.4, J ( 9.35, 9.35, 7) has eigenvalues 4.3,.006 ±.09i, J (9.35, 9.35, 7) has eigenvalues 4.3,.006 ±.09i. At each of the two nonzero equilibrium points, J has eigenvalues with negative real part. Therefore each such equilibrium point is asymptotically stable.

29 8 Chapter 9 Hints, Answers, and Solutions (a) dl dt = 0; so l(x, y, z) is a conserved quantity. (b) dl dt 0 always and the origin is a minimum of l. Therefore l is a Lyapunov function. 5. (a) dl dt = (x + x z + y z ); so l(x, y, z) is not a conserved quantity. (b) dl dt 0 always and the origin is a minimum of l. Therefore l is a Lyapunov function (a) dl dt = x z y z + xyz ; so l(x, y, z) is not a conserved quantity. (b) dl dt = x z y z + xyz = z(x + y) > 0 whenever z < 0. Therefore l is not a Lyapunov function (a) dl dt = x xy + x z xyz; so l(x, y, z) is not a conserved quantity. (b) dl dt = x xy + x z xyz = x(x y)( + z), which is positive whenever 0 < y < x and z >. Therefore l is not a Lyapunov function dl dt = l xx + l y y = g(x)h(x, y)f(y) + f(y)h(x, y)g(x) = Take h(x, y) = x + y 3, f(y) = y, and g(x) = x. Then use (a) to get l(x, y) = x + y. Then (0, 0) is a minimum of l and dl dt = xx + yy = 0. Therefore, l is a Lyapunov function. Now apply the Lyapunov stability theorem to conclude that (0, 0) is a stable equilibrium point Write the system as x = k xy ( a y y apply (a) to get h(x, y) = xy, f(y) = k (a y y ), y = k xy b x x. With ) (b x), and g(x) = k, x l(x, y) = k x k y + bk ln x + ak ln y. Now we show that l is a Lyapunov function. First, l x (x, y) = k + bk x l y (x, y) = k + ak y, which are each 0 at (b, a). Also, and l xx l yy (l xy ) = ( bk x )( ak y ) 0 = abk k x y, which is positive at (b, a), while l xx (b, a) < 0. Therefore, l has a local minimum at (b, a). Finally, ( ) ( ) dl dt = k + bk x + k + ak y x y ( ) ( ) = k + bk k x(a y) + k + ak ( k y(b x)) x y = k k x(a y) + b(a y) + y(b x) a(b x) = 0. Therefore, l is a Lyapunov function at (b, a), which implies that (b, a) is stable. dϕ dt = ϕ xx + ϕ y y + ϕ z z. Replace x, y, z with the values from the given system. Simplification gives dϕ dt = 0.