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# EDO 9th Boyce Cap9 Resolução

DisciplinaEquações Diferenciais Ordinárias2.005 materiais14.770 seguidores
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```0
x
g(s)ds > 0 for x < 0 .
Therefore V (x, y) is positive definite. The rate of change of V along any trajectory
is
V\u2d9 = Vx
dx
dt
+ Vy
dy
dt
= g(x) · (y) + y [\u2212g(x)\u2212 y] = \u2212y2 .
It follows that V\u2d9 (x, y) is only negative semidefinite. Hence the origin is a stable
critical point.
(b) Given
V (x, y) =
1
2
y2 +
1
2
y sin(x) +
\u222b x
0
sin(s)ds ,
It is easy to see that V (0 , 0) = 0 . The rate of change of V along any trajectory is
V\u2d9 = Vx
dx
dt
+ Vy
dy
dt
=
[
sin x+
y
2
cos x
]
(y) +
[
y +
1
2
sin x
]
[\u2212 sin x\u2212 y] =
=
1
2
y2 cos x\u2212 1
2
sin2 x\u2212 y
2
sin x\u2212 y2 .
For \u2212pi/2 < x < pi/2 , we can write sin x = x\u2212 \u3b1x3/6 and cos x = 1\u2212 \u3b2 x2/2 , in
which \u3b1 = \u3b1(x) , \u3b2 = \u3b2(x) . Note that 0 < \u3b1 , \u3b2 < 1 . Then
V\u2d9 (x , y) =
y2
2
(1\u2212 \u3b2 x
2
2
)\u2212 1
2
(x\u2212 \u3b1x
3
6
)2 \u2212 y
2
(x\u2212 \u3b1x
3
6
)\u2212 y2 .
9.6 507
Using polar coordinates,
V\u2d9 (r , \u3b8) = \u2212r
2
2
[1 + sin \u3b8 cos \u3b8 + h(r , \u3b8)] = \u2212r
2
2
[
1 +
1
2
sin 2\u3b8 + h(r , \u3b8)
]
.
It is easy to show that
|h(r , \u3b8)| \u2264 1
2
r2 +
1
72
r4.
So if r is sufficiently small, then |h(r , \u3b8)| < 1/2 and \u2223\u2223 12 sin 2\u3b8 + h(r , \u3b8)\u2223\u2223 < 1 . Hence
V\u2d9 (x , y) is negative definite. Now we show that V (x, y) is positive definite. Since
g(u) = sin u ,
V (x, y) =
1
2
y2 +
1
2
y sin(x) + 1\u2212 cos x .
This time we set
cos x = 1\u2212 x
2
2
+ \u3b3
x4
24
.
Note that 0 < \u3b3 < 1 for \u2212pi/2 < x < pi/2 . Converting to polar coordinates,
V (r, \u3b8) =
r2
2
[
1 + sin \u3b8 cos \u3b8 \u2212 r
2
12
sin \u3b8 cos3 \u3b8 \u2212 \u3b3 r
2
24
cos4 \u3b8
]
=
r2
2
[
1 +
1
2
sin 2\u3b8 \u2212 r
2
12
sin \u3b8 cos3 \u3b8 \u2212 \u3b3 r
2
24
cos4 \u3b8
]
.
Now
\u2212 r
2
12
sin \u3b8 cos3 \u3b8 \u2212 \u3b3 r
2
24
cos4 \u3b8 > \u22121
8
for r < 1 .
It follows that when r > 0 ,
V (r, \u3b8) >
r2
2
[
7
8
+
1
2
sin 2\u3b8
]
\u2265 3 r
2
16
> 0 .
Therefore V (x, y) is indeed positive definite, and by Theorem 9.6.1 , the origin is
an asymptotically stable critical point.
12.(a) We consider the linear system(
x
y
)\u2032
=
(
a11 a12
a21 a22
)(
x
y
)
.
Let V (x, y) = Ax2 +Bxy + Cy2, in which
A = \u2212a
2
21 + a
2
22 + (a11a22 \u2212 a12a21)
2\u2206
B =
a12a22 + a11a21
\u2206
C = \u2212a
2
11 + a
2
12 + (a11a22 \u2212 a12a21)
2\u2206
,
508 Chapter 9. Nonlinear Differential Equations and Stability
and \u2206 = (a11 + a22)(a11a22 \u2212 a12a21). Based on the hypothesis, the coefficients A
and B are negative. Therefore, except for the origin, V (x, y) is negative on each of
the coordinate axes. Along each trajectory,
V\u2d9 = (2Ax+By)(a11 x+ a12 y) + (2Cy +Bx)(a21 x+ a22 y) = \u2212x2 \u2212 y2.
Hence V\u2d9 (x, y) is negative definite. Theorem 9.6.2 asserts that the origin is an
unstable critical point.
(b) We now consider the system(
x
y
)\u2032
=
(
a11 a12
a21 a22
)(
x
y
)
+
(
F1(x , y)
G1(x , y)
)
,
in which F1(x , y)/r \u2192 0 and G1(x , y)/r \u2192 0 as r \u2192 0 . Let
V (x, y) = Ax2 +Bxy + Cy2,
in which
A =
a221 + a
2
22 + (a11a22 \u2212 a12a21)
2\u2206
B = \u2212a12a22 + a11a21
\u2206
C =
a211 + a
2
12 + (a11a22 \u2212 a12a21)
2\u2206
,
and \u2206 = (a11 + a22)(a11a22 \u2212 a12a21). Based on the hypothesis, A ,B > 0 . Ex-
cept for the origin, V (x, y) is positive on each of the coordinate axes. Along each
trajectory,
V\u2d9 = x2 + y2 + (2Ax+By)F1(x , y) + (2Cy +Bx)G1(x , y) .
Converting to polar coordinates, for r 6= 0 ,
V\u2d9 = = r2 + r(2A cos \u3b8 +B sin \u3b8)F1 + r(2C sin \u3b8 +B cos \u3b8)G1
= r2 + r2
[
(2A cos \u3b8 +B sin \u3b8)
F1
r
+ (2C sin \u3b8 +B cos \u3b8)
G1
r
]
.
Since the system is almost linear, there is an R such that\u2223\u2223\u2223\u2223(2A cos \u3b8 +B sin \u3b8) F1r + (2C sin \u3b8 +B cos \u3b8) G1r
\u2223\u2223\u2223\u2223 < 12 ,
and hence
(2A cos \u3b8 +B sin \u3b8)
F1
r
+ (2C sin \u3b8 +B cos \u3b8)
G1
r
> \u22121
2
for r < R . It follows that
V\u2d9 >
1
2
r2
as long as 0 < r < R . Hence V\u2d9 is positive definite on the domain
D =
{
(x , y) |x2 + y2 < R2} .
By Theorem 9.6.2, the origin is an unstable critical point.
9.7 509
9.7
3. The equilibrium solutions of the ODE
dr
dt
= r(r \u2212 1)(r \u2212 3)
are given by r1 = 0 , r2 = 1 and r3 = 3 . Note that
dr
dt
> 0 for 0 < r < 1 and r > 3 ;
dr
dt
< 0 for 1 < r < 3 .
r = 0 corresponds to an unstable critical point. The equilibrium solution r2 = 1 is
asymptotically stable, whereas the equilibrium solution r3 = 3 is unstable. Since
the critical values are isolated, a limit cycle is given by
r = 1 , \u3b8 = t+ t0
which is asymptotically stable. Another periodic solution is found to be
r = 3 , \u3b8 = t+ t0
which is unstable.
5. The equilibrium solutions of the ODE
dr
dt
= sin pir
are given by r = n , n = 0 , 1 , 2 , . . . . Based on the sign of r \u2032 in the neighborhood
of each critical value, the equilibrium solutions r = 2k , k = 1 , 2 , . . . correspond to
unstable periodic solutions, with \u3b8 = t+ t0 . The equilibrium solutions r = 2k + 1 ,
k = 0 , 1 , 2 , . . . correspond to stable limit cycles, with \u3b8 = t+ t0 . The solution
r = 0 represents an unstable critical point.
6. The equilibrium solutions of the ODE
dr
dt
= r|r \u2212 2|(r \u2212 3)
are given by r1 = 0 , r2 = 2 and r3 = 3 . Note that
dr
dt
< 0 for 0 < r < 3 ;
dr
dt
> 0 for r > 3 .
r = 0 corresponds to an asymptotically stable critical point. The equilibrium so-
lution r2 = 2 is semistable, whereas the equilibrium solution r3 = 3 is unstable.
Since the critical values are isolated, a semistable limit cycle is given by
r = 2 , \u3b8 = \u2212t+ t0.
Another periodic solution is found to be
r = 3 , \u3b8 = \u2212t+ t0
which is unstable.
510 Chapter 9. Nonlinear Differential Equations and Stability
10. Given F (x , y) = a11 x+ a12 y and G(x , y) = a21 x+ a22 y , it follows that
Fx +Gy = a11 + a22 .
Based on the hypothesis, Fx +Gy is either positive or negative on the entire plane.
By Theorem 9.7.2, the system cannot have a nontrivial periodic solution.
12. Given that F (x , y) = \u22122x\u2212 3y \u2212 xy2 and G(x , y) = y + x3 \u2212 x2y ,
Fx +Gy = \u22121\u2212 x2 \u2212 y2.
Since Fx +Gy < 0 on the entire plane, Theorem 9.7.2 asserts that the system
cannot have a nontrivial periodic solution.
14.(a) Based on the given graphs, the following table shows the estimated values:
µ = 0.2 T \u2248 6.29
µ = 1.0 T \u2248 6.66
µ = 5.0 T \u2248 11.60
(b) The initial conditions were chosen as x(0) = 2 , y(0) = 0 .
T \u2248 6.38 .
T \u2248 7.65 .
9.7 511
T \u2248 8.86 .
T \u2248 10.25 .
(c) The period, T , appears to be a quadratic function of µ .
15.(a) Setting x = u and y = u \u2032, we obtain the system of equations
dx
dt
= y
dy
dt
= \u2212x+ µ(1\u2212 1
3
y2)y .
512 Chapter 9. Nonlinear Differential Equations and Stability
(b) Evidently, y = 0 . It follows that x = 0 . Hence the only critical point of the
system is at (0 , 0). The components of the vector field are infinitely differentiable
everywhere. Therefore the system is almost linear. The Jacobian matrix of the
vector field is
J =
(
0 1
\u22121 µ\u2212 µy2
)
.
At the critical point (0 , 0), the coefficient matrix of the linearized system is
J(0 , 0) =
(
0 1
\u22121 µ
)
,
with eigenvalues
r1,2 =
µ
2
± 1
2
\u221a
µ2 \u2212 4 .
If µ = 0 , the equation reduces to the ODE for a simple harmonic oscillator. For the
case 0 < µ < 2 , the eigenvalues are complex, and the critical point is an unstable
spiral. For µ \u2265 2 , the eigenvalues are real, and the origin is an unstable node.
(c) The initial conditions were chosen as x(0) = 2 , y(0) = 0 .
A \u2248 2.16 and T \u2248 6.65 .
(d)
A \u2248 2.00 and T \u2248 6.30 .
9.7 513
A \u2248 2.04 and T \u2248 6.38 .
A \u2248 2.6 and T \u2248 7.62 .
A \u2248 4.37 and T \u2248 11.61 .
(e)
A T
µ = 0.2 2.00 6.30
µ = 0.5 2.04 6.38
µ = 1.0 2.16 6.65
µ = 2.0 2.6 7.62
µ = 5.0 4.37 11.61
514 Chapter 9. Nonlinear Differential Equations and Stability
16.(a) The critical points are solutions of the algebraic system
µx+ y \u2212 x(x2 + y2) = 0
\u2212x+ µ y \u2212 y(x2 + y2)```