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# EDO 9th Boyce Cap9 Resolução

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```the trajectories are can be written as
1
2
y2 + \u3c92(1\u2212 cos x) = c ,
in which c is an arbitrary constant.
(b) Multiplying by mL2 and reverting to the original physical variables, we obtain
1
2
mL2(
d\u3b8
dt
)2 +mL2\u3c92(1\u2212 cos \u3b8) = mL2c .
Since \u3c92 = g/L , the equation can be written as
1
2
mL2(
d\u3b8
dt
)2 +mgL(1\u2212 cos \u3b8) = E ,
in which E = mL2c .
9.3 469
(c) The absolute velocity of the point mass is given by v = Ld\u3b8/dt . The kinetic
energy of the mass is T = mv2/2 . Choosing the rest position as the datum, that
is, the level of zero potential energy, the gravitational potential energy of the point
mass is
V = mgL(1\u2212 cos \u3b8).
It follows that the total energy,T + V , is constant along the trajectories.
23.(a) A = 0.25
Since the system is undamped, and y(0) = 0 , the amplitude is 0.25 . The period is
estimated at \u3c4 \u2248 3.16 .
(b)
(a) A = 1/2 (b) A = 1
(c) A = 3/2 (d) A = 2
470 Chapter 9. Nonlinear Differential Equations and Stability
R \u3c4
A = 0.5 0.5 3.20
A = 1.0 1.0 3.35
A = 1.5 1.5 3.63
A = 2.0 2.0 4.17
(c) Since the system is conservative, the amplitude is equal to the initial amplitude.
On the other hand, the period of the pendulum is a monotone increasing function
of the initial position A .
It appears that as A \u2192 0 , the period approaches pi, the period of the corresponding
linear pendulum (2pi/\u3c9).
(d)
The pendulum is released from rest, at an inclination of 4\u2212 pi radians from the
vertical. Based on conservation of energy, the pendulum will swing past the lower
equilibrium position (\u3b8 = 2pi) and come to rest, momentarily, at a maximum rota-
tional displacement of \u3b8max = 3pi \u2212 (4\u2212 pi) = 4pi \u2212 4 . The transition between the
two dynamics occurs at A = pi , that is, once the pendulum is released beyond the
upright configuration.
26.(a) It is evident that the origin is a critical point of each system. Furthermore,
9.3 471
it is easy to see that the corresponding linear system, in each case, is given by
dx
dt
= y
dy
dt
= \u2212x .
The eigenvalues of the coefficient matrix are r1,2 = ± i . Hence the critical point of
the linearized system is a center.
(b) Using polar coordinates, it is also easy to show that
lim
r\u2192 0
\u2016g(x)\u2016
\u2016x\u2016 = 0 .
Alternatively, the nonlinear terms are analytic in the entire plane. Hence both
systems are almost linear near the origin.
(c) For system (ii), note that
x
dx
dt
+ y
dy
dt
= xy \u2212 x2(x2 + y2)\u2212 xy \u2212 y2(x2 + y2).
Converting to polar coordinates, and differentiating the equation r2 = x2 + y2 with
respect to t , we find that
r
dr
dt
= x
dx
dt
+ y
dy
dt
= \u2212(x2 + y2)2 = \u2212r4.
That is, r \u2032 = \u2212r3. It follows that r2 = 1/(2t+ c), where c = 1/r20. Since r \u2192 0
as t \u2192 0 , regardless of the value of r0 , the origin is an asymptotically stable
equilibrium point.
On the other hand, for system (i),
r
dr
dt
= x
dx
dt
+ y
dy
dt
= (x2 + y2)2 = r4.
That is, r \u2032 = r3. Solving the differential equation results in
r2 =
c\u2212 2t
(2t\u2212 c)2 .
Imposing the initial condition r(0) = r0 , we obtain a specific solution
r2 = \u2212 r
2
0
2 r20 t\u2212 1
.
Since the solution becomes unbounded as t \u2192 1/2r20 , the critical point is unstable.
27. The characteristic equation of the coefficient matrix is r2 + 1 = 0 , with complex
roots r1,2 = ± i . Hence the critical point at the origin is a center. The characteristic
equation of the perturbed matrix is r2 \u2212 2 ² r + 1 + ²2 = 0 , with complex conjugate
roots r1,2 = ² ± i . As long as ² 6= 0 , the critical point of the perturbed system is
a spiral point. Its stability depends on the sign of ² .
472 Chapter 9. Nonlinear Differential Equations and Stability
28. The characteristic equation of the coefficient matrix is (r + 1)2 = 0, with roots
r1 = r2 = \u22121. Hence the critical point is an asymptotically stable node. On the
other hand, the characteristic equation of the perturbed system is r2 + 2r + 1 + ² =
0, with roots r1,2 = \u22121 ±
\u221a\u2212² . If ² > 0 , then r1,2 = \u22121 ± i
\u221a
² are complex
roots. The critical point is a stable spiral. If ² < 0 , then r1,2 = \u22121 ±
\u221a|²| are
real and both negative (|²| ¿ 1). The critical point remains a stable node.
29.(d) Set k = sin(\u3b1/2) = sin(A/2) and g/L = 4 .
9.4
1.(a)
(b) The critical points are solutions of the system of equations
x(1.5\u2212 x\u2212 0.5 y) = 0
y(2\u2212 y \u2212 0.75x) = 0 .
The four critical points are (0 , 0), (0 , 2), (1.5 , 0) and (0.8 , 1.4).
(c) The Jacobian matrix of the vector field is
J =
(
3/2\u2212 2x\u2212 y/2 \u2212x/2
\u22123y/4 2\u2212 3x/4\u2212 2y
)
.
9.4 473
At the critical point (0 , 0), the coefficient matrix of the linearized system is
J(0 , 0) =
(
3/2 0
0 2
)
.
The eigenvalues and eigenvectors are
r1 = 3/2 , \u3be(1) =
(
1
0
)
; r2 = 2 , \u3be(2) =
(
0
1
)
.
The eigenvalues are positive, hence the origin is an unstable node.
At the critical point (0 , 2), the coefficient matrix of the linearized system is
J(0 , 2) =
(
1/2 0
\u22123/2 \u22122
)
.
The eigenvalues and eigenvectors are
r1 = 1/2 , \u3be(1) =
(
1
\u22120.6
)
; r2 = \u22122 , \u3be(2) =
(
0
1
)
.
The eigenvalues are of opposite sign. Hence the critical point is a saddle, which is
unstable.
At the critical point (1.5 , 0), the coefficient matrix of the linearized system is
J(1.5 , 0) =
(\u22121.5 \u22120.75
0 0.875
)
.
The eigenvalues and eigenvectors are
r1 = \u22121.5 , \u3be(1) =
(
1
0
)
; r2 = 0.875 , \u3be(2) =
(\u22120.31579
1
)
.
The eigenvalues are of opposite sign. Hence the critical point is also a saddle, which
is unstable.
At the critical point (0.8 , 1.4), the coefficient matrix of the linearized system is
J(0.8 , 1.4) =
( \u22120.8 \u22120.4
\u22121.05 \u22121.4
)
.
The eigenvalues and eigenvectors are
r1 = \u22121110 +
\u221a
51
10
, \u3be(1) =
(
1
3\u2212\u221a51
4
)
; r2 = \u22121110 \u2212
\u221a
51
10
, \u3be(2) =
(
1
3+
\u221a
51
4
)
.
The eigenvalues are both negative. Hence the critical point is a stable node, which
is asymptotically stable.
474 Chapter 9. Nonlinear Differential Equations and Stability
(d,e)
(f) Except for initial conditions lying on the coordinate axes, almost all trajectories
converge to the stable node at (0.8 , 1.4).
2.(a)
(b) The critical points are the solution set of the system of equations
x(1.5\u2212 x\u2212 0.5 y) = 0
y(2\u2212 0.5 y \u2212 1.5x) = 0 .
The four critical points are (0 , 0), (0 , 4), (1.5 , 0) and (1 , 1).
(c) The Jacobian matrix of the vector field is
J =
(
3/2\u2212 2x\u2212 y/2 \u2212x/2
\u22123y/2 2\u2212 3x/2\u2212 y
)
.
At the origin, the coefficient matrix of the linearized system is
J(0 , 0) =
(
3/2 0
0 2
)
.
The eigenvalues and eigenvectors are
r1 = 3/2 , \u3be(1) =
(
1
0
)
; r2 = 2 , \u3be(2) =
(
0
1
)
.
9.4 475
The eigenvalues are positive, hence the origin is an unstable node.
At the critical point (0 , 4), the coefficient matrix of the linearized system is
J(0 , 4) =
(\u22121/2 0
\u22126 \u22122
)
.
The eigenvalues and eigenvectors are
r1 = \u22121/2 , \u3be(1) =
(
1
\u22124
)
; r2 = \u22122 , \u3be(2) =
(
0
1
)
.
The eigenvalues are both negative, hence the critical point (0 , 4) is a stable node,
which is asymptotically stable.
At the critical point (3/2 , 0), the coefficient matrix of the linearized system is
J(3/2 , 0) =
(\u22123/2 \u22123/4
0 \u22121/4
)
.
The eigenvalues and eigenvectors are
r1 = \u22123/2 , \u3be(1) =
(
1
0
)
; r2 = \u22121/4 , \u3be(2) =
(
3
\u22125
)
.
The eigenvalues are both negative, hence the critical point is a stable node, which
is asymptotically stable.
At the critical point ( 1, 1), the coefficient matrix of the linearized system is
J(1 , 1) =
( \u22121 \u22121/2
\u22123/2 \u22121/2
)
.
The eigenvalues and eigenvectors are
r1 =
\u22123 +\u221a13
4
, \u3be(1) =
(
1
\u2212 1+
\u221a
13
2
)
; r2 = \u22123 +
\u221a
13
4
, \u3be(2) =
(
0
\u22121+\u221a13
2
)
.
The eigenvalues are of opposite sign, hence ( 1, 1) is a saddle, which is unstable.```