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# EDO 9th Boyce Cap9 Resolução

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```(d,e)
476 Chapter 9. Nonlinear Differential Equations and Stability
(f) Trajectories approaching the critical point (1 , 1) form a separatrix. Solutions
on either side of the separatrix approach either (0 , 4) or (1.5 , 0).
4.(a)
(b) The critical points are solutions of the system of equations
x(1.5\u2212 0.5x\u2212 y) = 0
y(0.75\u2212 y \u2212 0.125x) = 0 .
The four critical points are (0 , 0), (0 , 3/4), (3 , 0) and (2 , 1/2).
(c) The Jacobian matrix of the vector field is
J =
(
3/2\u2212 x\u2212 y \u2212x
\u2212y/8 3/4\u2212 x/8\u2212 2y
)
.
At the origin, the coefficient matrix of the linearized system is
J(0 , 0) =
(
3/2 0
0 3/4
)
.
The eigenvalues and eigenvectors are
r1 = 3/2 , \u3be(1) =
(
1
0
)
; r2 = 3/4 , \u3be(2) =
(
0
1
)
.
The eigenvalues are positive, hence the origin is an unstable node.
At the critical point (0 , 3/4), the coefficient matrix of the linearized system is
J(0 , 3/4) =
(
3/4 0
\u22123/32 \u22123/4
)
.
The eigenvalues and eigenvectors are
r1 = 3/4 , \u3be(1) =
(\u221216
1
)
; r2 = \u22123/4 , \u3be(2) =
(
0
1
)
.
The eigenvalues are of opposite sign, hence the critical point (0 , 3/4) is a saddle,
which is unstable.
9.4 477
At the critical point (3 , 0), the coefficient matrix of the linearized system is
J(3 , 0) =
(\u22123/2 \u22123
0 3/8
)
.
The eigenvalues and eigenvectors are
r1 = \u22123/2 , \u3be(1) =
(
1
0
)
; r2 = 3/8 , \u3be(2) =
(\u22128
5
)
.
The eigenvalues are of opposite sign, hence the critical point (0 , 3/4) is a saddle,
which is unstable.
At the critical point (2 , 1/2), the coefficient matrix of the linearized system is
J(2 , 1/2) =
( \u22121 \u22122
\u22121/16 \u22121/2
)
.
The eigenvalues and eigenvectors are
r1 =
\u22123 +\u221a3
4
, \u3be(1) =
(
1
\u2212 1+
\u221a
3
8
)
; r2 = \u22123 +
\u221a
3
4
, \u3be(2) =
(
0
\u22121+\u221a3
8
)
.
The eigenvalues are negative, hence the critical point (2 , 1/2) is a stable node,
which is asymptotically stable.
(d,e)
(f) Except for initial conditions along the coordinate axes, almost all solutions
converge to the stable node (2 , 1/2).
7. It follows immediately that
(\u3c31X + \u3c32Y )2 \u2212 4\u3c31\u3c32XY = \u3c321X2 + 2\u3c31\u3c32XY + \u3c322Y 2 \u2212 4\u3c31\u3c32XY
= (\u3c31X \u2212 \u3c32Y )2.
Since all parameters and variables are positive, it follows that
(\u3c31X + \u3c32Y )2 \u2212 4(\u3c31\u3c32 \u2212 \u3b11\u3b12)XY \u2265 0 .
Hence the radicand in Eq.(39) is nonnegative.
478 Chapter 9. Nonlinear Differential Equations and Stability
10.(a) The critical points consist of the solution set of the equations
x(²1 \u2212 \u3c31x\u2212 \u3b11y) = 0
y(²2 \u2212 \u3c32y \u2212 \u3b12x) = 0 .
If x = 0 , then either y = 0 or y = ²2/\u3c32 . If ²1 \u2212 \u3c31x\u2212 \u3b11y = 0 , then solving for
x results in x = (²1 \u2212 \u3b11y)/\u3c31. Substitution into the second equation yields
(\u3c31\u3c32 \u2212 \u3b11\u3b12)y2 \u2212 (\u3c31² 2 \u2212 ² 1\u3b12)y = 0 .
Based on the hypothesis, it follows that (\u3c31²2 \u2212 ²1\u3b12)y = 0. So if \u3c31²2 \u2212 ²1\u3b12 6= 0,
then y = 0, and the critical points are located at (0 , 0), (0 , ²2/\u3c32) and (²1/\u3c31, 0).
For the case \u3c31² 2 \u2212 ² 1\u3b12 = 0, y can be arbitrary. From the relation
x = (²1 \u2212 \u3b11y)/\u3c31,
we conclude that all points on the line \u3c31x+ \u3b11y = ²1 are critical points, in addi-
tion to the point (0 , 0).
(b) The Jacobian matrix of the vector field is
J =
(
²1 \u2212 2\u3c31x\u2212 \u3b11y \u2212\u3b11x
\u2212\u3b12y ²2 \u2212 2\u3c32y \u2212 \u3b12x
)
.
At the origin, the coefficient matrix of the linearized system is
J(0 , 0) =
(
²1 0
0 ²2
)
,
with eigenvalues r1 = ²1 and r2 = ²2 . Since both eigenvalues are positive, the
origin is an unstable node.
At the point (0 , ²2/\u3c32), the coefficient matrix of the linearized system is
J(0 , ²2/\u3c32) =
(
(²1\u3b12 \u2212 \u3c31²2)/\u3b12 0
²2\u3b12/\u3c32 \u2212²2
)
,
with eigenvalues r1 = (²1\u3b12 \u2212 \u3c31²2)/\u3b12 and r2 = \u2212²2 . If \u3c31²2 \u2212 ²1\u3b12 > 0 , then
both eigenvalues are negative. Hence the point (0 , ²2/\u3c32) is a stable node, which
is asymptotically stable. If \u3c31²2 \u2212 ²1\u3b12 < 0 , then the eigenvalues are of opposite
sign. Hence the point (0 , ²2/\u3c32) is a saddle, which is unstable.
At the point (²1/\u3c31, 0), the coefficient matrix of the linearized system is
J(²1/\u3c31, 0) =
(\u2212²1 \u2212²1\u3b11/\u3c31
0 (\u3c31²2 \u2212 ²1\u3b12)/\u3c31
)
,
with eigenvalues r1 = (\u3c31²2 \u2212 ²1\u3b12)/\u3c31 and r2 = \u2212²1 . If \u3c31²2 \u2212 ²1\u3b12 > 0 , then
the eigenvalues are of opposite sign. Hence the point (²1/\u3c31, 0) is a saddle, which
is unstable. If \u3c31²2 \u2212 ²1\u3b12 < 0 , then both eigenvalues are negative. In that case
the point (²1/\u3c31, 0) is a stable node, which is asymptotically stable.
(c) As shown in part (a), when \u3c31²2 \u2212 ²1\u3b12 = 0, the set of critical points consists of
(0, 0) and all the points on the straight line \u3c31x+ \u3b11y = ²1. Based on part (b), the
9.4 479
origin is still an unstable node. Setting y = (²1 \u2212 \u3c31x)/\u3b11 , the Jacobian matrix of
the vector field, along the given straight line, is
J =
( \u2212\u3c31x \u2212\u3b11x
\u2212\u3b12(²1 \u2212 \u3c31x)/\u3b11 \u3b12x\u2212 ²1\u3b12/\u3c31
)
.
The characteristic equation of the matrix is
r2 +
[
²1\u3b12 \u2212 \u3b12\u3c31x+ \u3c321x
\u3c31
]
r = 0 .
Using the given hypothesis, (²1\u3b12 \u2212 \u3b12\u3c31x+ \u3c321x)/\u3c31 = ² 2 \u2212 \u3b12x+ \u3c31x. Hence the
characteristic equation can be written as
r2 + [² 2 \u2212 \u3b12x+ \u3c31x] r = 0 .
First note that 0 \u2264 x \u2264 ²1/\u3c31. Since the coefficient in the quadratic equation is
linear, and
² 2 \u2212 \u3b12x+ \u3c31x =
{
² 2 at x = 0
² 1 at x = ²1/\u3c31 ,
it follows that the coefficient is positive for 0 \u2264 x \u2264 ²1/\u3c31. Therefore, along the
straight line \u3c31x+ \u3b11y = ²1, one eigenvalue is zero and the other one is negative.
Hence the continuum of critical points consists of stable nodes, which are asymp-
totically stable.
11.(a) The critical points are solutions of the system of equations
x(1\u2212 x\u2212 y) + \u3b4a = 0
y(0.75\u2212 y \u2212 0.5x) + \u3b4b = 0 .
Assume solutions of the form
x = x0 + x1\u3b4 + x2\u3b42 + . . .
y = y0 + y1\u3b4 + y2\u3b42 + . . . .
Substitution of the series expansions results in
x0(1\u2212 x0 \u2212 y0) + (x1 \u2212 2x1x0 \u2212 x0y1 \u2212 x1y0 + a)\u3b4 + . . . = 0
y0(0.75\u2212 y0 \u2212 0.5x0 ) + (0.75 y1 \u2212 2y0y1 \u2212 x1y0/2\u2212 x0y1/2 + b)\u3b4 + . . . = 0 .
(b) Taking a limit as \u3b4\u2192 0 , the equations reduce to the original system of equations.
It follows that x0 = y0 = 0.5 .
(c) Setting the coefficients of the linear terms equal to zero, we find that
\u2212y1/2\u2212 x1/2 + a = 0
\u2212x1/4\u2212 y1/2 + b = 0 ,
with solution x1 = 4a\u2212 4b and y1 = \u22122a+ 4b .
(d) Consider the ab -parameter space. The collection of points for which b < a
represents an increase in the level of species 1. At points where b > a , x1\u3b4 < 0 .
480 Chapter 9. Nonlinear Differential Equations and Stability
Likewise, the collection of points for which 2b > a represents an increase in the
level of species 2 . At points where 2b < a , y1\u3b4 < 0 .
It follows that if b < a < 2b , the level of both species will increase. This condition
is represented by the wedge-shaped region on the graph. Otherwise, the level of
one species will increase, whereas the level of the other species will simultaneously
decrease. Only for a = b = 0 will both populations remain the same.
12.(a) The critical points consist of the solution set of the equations
\u2212y = 0
\u2212\u3b3 y \u2212 x(x\u2212 0.15)(x\u2212 2) = 0 .
Setting y = 0 , the second equation becomes x(x\u2212 0.15)(x\u2212 2) = 0 , with roots
x = 0 , 0.15 and 2 . Hence the critical points are located at (0 , 0), (0.15 , 0) and
(2 , 0). The Jacobian matrix of the vector field is
J =
(
0 \u22121
\u22123x2 + 4.3x\u2212 0.3 \u2212\u3b3
)
.
At the origin, the coefficient matrix of the linearized system is
J(0 , 0) =
(
0 \u22121
\u22120.3 \u2212\u3b3
)
,
with eigenvalues
r1,2 = \u2212\u3b32 ±
1
10
\u221a
25 \u3b32 + 30 .
Regardless of the value of \u3b3 , the eigenvalues are real and of opposite sign. Hence
(0 , 0) is a saddle, which is unstable.
At the critical point (0.15 , 0), the coefficient matrix of the linearized system is
J(0.15 , 0) =
(
0 \u22121
0.2775 \u2212\u3b3
)
,
with eigenvalues
r1,2 = \u2212\u3b32 ±
1
20
\u221a
100 \u3b32 \u2212 111 .
If 100 \u3b32 \u2212 111 \u2265 0 , then the eigenvalues are real. Furthermore, since r1r2 =
0.2775 , both eigenvalues will have the same sign. Therefore the critical point
is a node, with its stability dependent on the sign of \u3b3 . If 100 \u3b32 \u2212 111 < 0 , the
9.4```