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# EDO 9th Boyce Cap9 Resolução

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```(\u22121 \u22121
0 1\u2212 \u3b1
)
,
which means that this equilibrium is a saddle when 0 < \u3b1 < 1 and an asymptotically
stable node when \u3b1 > 1.
At the critical point (0, \u3b1) the Jacobian is
J(0, \u3b1)) =
(
1\u2212 \u3b1 0
\u2212\u3b1(2\u3b1\u2212 1) \u2212\u3b1
)
,
which implies that this equilibrium is a saddle when 0 < \u3b1 < 1 and an asymptoti-
cally stable node when \u3b1 > 1.
At the critical point (1/2, 1/2) the Jacobian is given by
J(1/2, 1/2) =
( \u22121/2 \u22121/2
1/2\u2212 \u3b1 \u22121/2
)
.
This is an asymptotically stable node when 0 < \u3b1 < 1 and a saddle point when
\u3b1 > 1.
(f) Phase portraits:
(a) \u3b1 = 3/4 (b) \u3b1 = 1 (c) \u3b1 = 5/4
9.5 493
9.5
1.(a)
(b) The critical points are solutions of the system of equations
x(1.5\u2212 0.5 y) = 0
y(\u22120.5 + x) = 0 .
The two critical points are (0 , 0) and (0.5 , 3).
(c) The Jacobian matrix of the vector field is
J =
(
3/2\u2212 y/2 \u2212x/2
y \u22121/2 + x
)
.
At the critical point (0 , 0), the coefficient matrix of the linearized system is
J(0 , 0) =
(
3/2 0
0 \u22121/2
)
.
The eigenvalues and eigenvectors are
r1 = 3/2 , \u3be(1) =
(
1
0
)
; r2 = \u22121/2 , \u3be(2) =
(
0
1
)
.
The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable.
At the critical point (0.5 , 3), the coefficient matrix of the linearized system is
J(0.5 , 3) =
(
0 \u22121/4
3 0
)
.
The eigenvalues and eigenvectors are
r1 = i
\u221a
3
2
, \u3be(1) =
(
1
\u22122 i\u221a3
)
; r2 = \u2212i
\u221a
3
2
, \u3be(2) =
(
1
2 i
\u221a
3
)
.
The eigenvalues are purely imaginary. Hence the critical point is a center, which is
stable.
494 Chapter 9. Nonlinear Differential Equations and Stability
(d,e)
(f) Except for solutions along the coordinate axes, almost all trajectories are closed
curves about the critical point (0.5 , 3).
2.(a)
(b) The critical points are the solution set of the system of equations
x(1\u2212 0.5 y) = 0
y(\u22120.25 + 0.5x) = 0 .
The two critical points are (0 , 0) and (0.5 , 2).
(c) The Jacobian matrix of the vector field is
J =
(
1\u2212 y/2 \u2212x/2
y/2 \u22121/4 + x/2
)
.
At the critical point (0 , 0), the coefficient matrix of the linearized system is
J(0 , 0) =
(
1 0
0 \u22121/4
)
.
The eigenvalues and eigenvectors are
r1 = 1 , \u3be(1) =
(
1
0
)
; r2 = \u22121/4 , \u3be(2) =
(
0
1
)
.
9.5 495
The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable.
At the critical point (0.5 , 2), the coefficient matrix of the linearized system is
J(0.5 , 2) =
(
0 \u22121/4
1 0
)
.
The eigenvalues and eigenvectors are
r1 = i/2 , \u3be(1) =
(
1
\u22122 i
)
; r2 = \u2212i/2 , \u3be(2) =
(
1
2 i
)
.
The eigenvalues are purely imaginary. Hence the critical point is a center, which is
stable.
(d,e)
(f) Except for solutions along the coordinate axes, almost all trajectories are closed
curves about the critical point (0.5 , 2).
4.(a)
(b) The critical points are the solution set of the system of equations
x(9/8\u2212 x\u2212 y/2) = 0
y(\u22121 + x) = 0 .
The three critical points are (0 , 0), (9/8 , 0) and (1 , 1/4).
496 Chapter 9. Nonlinear Differential Equations and Stability
(c) The Jacobian matrix of the vector field is
J =
(
9/8\u2212 2x\u2212 y/2 \u2212x/2
y \u22121 + x
)
.
At the critical point (0 , 0), the coefficient matrix of the linearized system is
J(0 , 0) =
(
9/8 0
0 \u22121
)
.
The eigenvalues and eigenvectors are
r1 = 9/8 , \u3be(1) =
(
1
0
)
; r2 = \u22121 , \u3be(2) =
(
0
1
)
.
The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable.
At the critical point (9/8 , 0), the coefficient matrix of the linearized system is
J(9/8 , 0) =
(\u22129/8 \u22129/16
0 1/8
)
.
The eigenvalues and eigenvectors are
r1 = \u221298 , \u3be
(1) =
(
1
0
)
; r2 =
1
8
, \u3be(2) =
(
9
\u221220
)
.
The eigenvalues are of opposite sign, hence the critical point (9/8 , 0) is a saddle,
which is unstable.
At the critical point (1 , 1/4), the coefficient matrix of the linearized system is
J(1 , 1/4) =
(\u22121 \u22121/2
1/4 0
)
.
The eigenvalues and eigenvectors are
r1 =
\u22122 +\u221a2
4
, \u3be(1) =
(\u22122 +\u221a2
1
)
; r2 =
\u22122\u2212\u221a2
4
, \u3be(2) =
(\u22122\u2212\u221a2
1
)
.
The eigenvalues are both negative. Hence the critical point is a stable node, which
is asymptotically stable.
(d,e)
9.5 497
(f) Except for solutions along the coordinate axes, all solutions converge to the
critical point (1 , 1/4).
5.(a)
(b) The critical points are solutions of the system of equations
x(\u22121 + 2.5x\u2212 0.3 y \u2212 x2) = 0
y(\u22121.5 + x) = 0 .
The four critical points are (0 , 0), (1/2 , 0), (2 , 0) and (3/2 , 5/3).
(c) The Jacobian matrix of the vector field is
J =
(\u22121 + 5x\u2212 3x2 \u2212 3y/10 \u22123x/10
y \u22123/2 + x
)
.
At the critical point (0 , 0), the coefficient matrix of the linearized system is
J(0 , 0) =
(\u22121 0
0 \u22123/2
)
.
The eigenvalues and eigenvectors are
r1 = \u22121 , \u3be(1) =
(
1
0
)
; r2 = \u22123/2 , \u3be(2) =
(
0
1
)
.
The eigenvalues are both negative, hence the critical point (0 , 0) is a stable node,
which is asymptotically stable.
At the critical point (1/2 , 0), the coefficient matrix of the linearized system is
J(1/2 , 0) =
(
3/4 \u22123/20
0 \u22121
)
.
The eigenvalues and eigenvectors are
r1 =
3
4
, \u3be(1) =
(
1
0
)
; r2 = \u22121 , \u3be(2) =
(
3
35
)
.
The eigenvalues are of opposite sign, hence the critical point (1/2 , 0) is a saddle,
which is unstable.
498 Chapter 9. Nonlinear Differential Equations and Stability
At the critical point (2 , 0), the coefficient matrix of the linearized system is
J(2 , 0) =
(\u22123 \u22123/5
0 1/2
)
.
The eigenvalues and eigenvectors are
r1 = \u22123 , \u3be(1) =
(
1
0
)
; r2 = 1/2 , \u3be(2) =
(
6
\u221235
)
.
The eigenvalues are of opposite sign, hence the critical point (2 , 0) is a saddle,
which is unstable.
At the critical point (3/2 , 5/3), the coefficient matrix of the linearized system is
J(3/2 , 5/3) =
(\u22123/4 \u22129/20
5/3 0
)
.
The eigenvalues and eigenvectors are
r1 =
\u22123 + i\u221a39
8
, \u3be(1) =
(\u22129+i 3\u221a39
40
1
)
; r2 =
\u22123\u2212 i\u221a39
8
, \u3be(2) =
(\u22129\u2212i 3\u221a39
40
1
)
.
The eigenvalues are complex conjugates. Hence the critical point (3/2 , 5/3) is a
stable spiral, which is asymptotically stable.
(d,e)
(f) The single solution curve that converges to the node at (1/2 , 0) is a separatrix.
Except for initial conditions on the coordinate axes, trajectories on either side of
the separatrix converge to the node at (0 , 0) or the stable spiral at (3/2 , 5/3).
6. Given that t is measured from the time that x is a maximum, we have
x =
c
\u3b3
+
cK
\u3b3
cos(
\u221a
ac t)
y =
a
\u3b1
+K
a
\u3b1
\u221a
c
\u3b1
sin(
\u221a
ac t) .
The period of oscillation is evidently T = 2pi/
\u221a
ac . Both populations oscillate
about a mean value. The following is based on the properties of the cosine and sine
functions. The prey population x is maximum at t = 0 and t = T . It is a minimum
9.5 499
at t = T/2 . Its rate of increase is greatest at t = 3T/4. The rate of decrease of the
prey population is greatest at t = T/4 . The predator population y is maximum
at t = T/4 . It is a minimum at t = 3T/4 . The rate of increase of the predator
population is greatest at t = 0 and t = T . The rate of decrease of the predator
population is greatest at t = T/2 . In the following example, the system in Problem
2 is solved numerically with the initial conditions x(0) = 0.7 and y(0) = 2 . The
critical point of interest is at (0.5 , 2). Since a = 1 and c = 1/4 , it follows that the
period of oscillation is T = 4pi .
(a) Prey (b) Predator (c) Predator vs Prey
8.(a) The period of oscillation for the linear system is T = 2pi/
\u221a
ac . In system (2),
a = 1 and c = 0.75 . Hence the period is estimated as T = 2pi/
\u221a
0.75 \u2248 7.2552 .
(b) The estimated period appears to agree with the graphic in Figure 9.5.3.
(c) The critical point of interest```