EDO 9th Boyce Cap9 Resolução
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EDO 9th Boyce Cap9 Resolução


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429
CH A P T E R
9
Nonlinear Differential Equations and
Stability
9.1
2.(a) Setting x= \u3be ert results in the algebraic equations(
5\u2212 r \u22121
3 1\u2212 r
)(
\u3be1
\u3be2
)
=
(
0
0
)
.
For a nonzero solution, we must have det(A\u2212 r I) = r2 \u2212 6 r + 8 = 0 . The roots of
the characteristic equation are r1 = 2 and r2 = 4 . For r = 2, the system of equa-
tions reduces to 3\u3be1 = \u3be2. The corresponding eigenvector is \u3be(1) = (1 , 3)T . Substi-
tution of r = 4 results in the single equation \u3be1 = \u3be2 . A corresponding eigenvector
is \u3be(2) = (1 , 1)T .
(b) The eigenvalues are real and positive, hence the critical point is an unstable
node.
430 Chapter 9. Nonlinear Differential Equations and Stability
(c,d)
(a) x1 vs t (b) x1 vs t
(c) x1 vs t (d) x1 vs t
3.(a) Solution of the ODE requires analysis of the algebraic equations(
2\u2212 r \u22121
3 \u22122\u2212 r
)(
\u3be1
\u3be2
)
=
(
0
0
)
.
For a nonzero solution, we must have det(A\u2212 r I) = r2 \u2212 1 = 0 . The roots of the
characteristic equation are r1 = 1 and r2 = \u22121. For r = 1, the system of equations
reduces to \u3be1 = \u3be2 . The corresponding eigenvector is \u3be(1) = (1 , 1)T . Substitution
of r = \u22121 results in the single equation 3 \u3be1 \u2212 \u3be2 = 0 . A corresponding eigenvector
is \u3be(2) = (1 , 3)T .
9.1 431
(b) The eigenvalues are real, with r1 r2 < 0 . Hence the critical point is a saddle.
(c,d)
(a) x1 vs t (b) x1 vs t
(c) x1 vs t (d) x1 vs t
5.(a) The characteristic equation is given by\u2223\u2223\u2223\u22231\u2212 r \u221251 \u22123\u2212 r
\u2223\u2223\u2223\u2223 = r2 + 2 r + 2 = 0 .
The equation has complex roots r1 = \u22121 + i and r2 = \u22121\u2212 i. For r = \u22121 + i,
the components of the solution vector must satisfy \u3be1 \u2212 (2 + i)\u3be2 = 0 . Thus the
corresponding eigenvector is \u3be(1) = (2 + i , 1)T . Substitution of r = \u22121\u2212 i results
432 Chapter 9. Nonlinear Differential Equations and Stability
in the single equation \u3be1 \u2212 (2\u2212 i)\u3be2 = 0 . A corresponding eigenvector is \u3be(2) =
(2\u2212 i , 1)T .
(b) The eigenvalues are complex conjugates, with negative real part. Hence the
origin is a stable spiral.
(c,d)
(a) x1 vs t (b) x1 vs t
(c) x1 vs t (d) x1 vs t
6.(a) Solution of the ODEs is based on the analysis of the algebraic equations(
2\u2212 r \u22125
1 \u22122\u2212 r
)(
\u3be1
\u3be2
)
=
(
0
0
)
.
9.1 433
For a nonzero solution, we require that det(A\u2212 r I) = r2 + 1 = 0. The roots of the
characteristic equation are r = ±i . Setting r = i , the equations are equivalent to
\u3be1 \u2212 (2 + i)\u3be2 = 0 . The eigenvectors are \u3be(1) = (2 + i , 1)T and \u3be(2) = (2\u2212 i , 1)T .
(b) The eigenvalues are purely imaginary. Hence the critical point is a center.
(c,d)
(a) x1 vs t (b) x1 vs t
7.(a) Setting x= \u3be ert results in the algebraic equations(
3\u2212 r \u22122
4 \u22121\u2212 r
)(
\u3be1
\u3be2
)
=
(
0
0
)
.
For a nonzero solution, we require that det(A\u2212 r I) = r2 \u2212 2r + 5 = 0. The roots
of the characteristic equation are r = 1 ± 2i . Substituting r = 1 \u2212 2i , the two
equations reduce to (1 + i)\u3be1 \u2212 \u3be2 = 0 . The two eigenvectors are \u3be(1) = (1 , 1 + i)T
and \u3be(2) = (1 , 1\u2212 i)T .
(b) The eigenvalues are complex conjugates, with positive real part. Hence the
origin is an unstable spiral.
434 Chapter 9. Nonlinear Differential Equations and Stability
(c,d)
(a) x1 vs t (b) x1 vs t
(c) x1 vs t (d) x1 vs t
8.(a) The characteristic equation is given by\u2223\u2223\u2223\u2223\u22121\u2212 r \u221210 \u22121/4\u2212 r
\u2223\u2223\u2223\u2223 = (r + 1)(r + 1/4) = 0 ,
with roots r1 = \u22121 and r2 = \u22121/4. For r = \u22121, the components of the solution
vector must satisfy \u3be2 = 0 . Thus the corresponding eigenvector is \u3be(1) = (1 , 0)T .
Substitution of r = \u22121/4 results in the single equation 3\u3be1/4 + \u3be2 = 0 . A corre-
sponding eigenvector is \u3be(2) = (4 ,\u22123)T .
9.1 435
(b) The eigenvalues are real and both negative. Hence the critical point is a stable
node.
(c,d)
(a) x1 vs t (b) x1 vs t
9.(a) Solution of the ODEs is based on the analysis of the algebraic equations(
3\u2212 r \u22124
1 \u22121\u2212 r
)(
\u3be1
\u3be2
)
=
(
0
0
)
.
For a nonzero solution, we require that det(A\u2212 r I) = r2 \u2212 2 r + 1 = 0. The single
root of the characteristic equation is r = 1 . Setting r = 1 , the components of
the solution vector must satisfy \u3be1 \u2212 2 \u3be2 = 0 . A corresponding eigenvector is \u3be =
(2 , 1)T .
(b) Since there is only one linearly independent eigenvector, the critical point is an
unstable, improper node.
436 Chapter 9. Nonlinear Differential Equations and Stability
(c,d)
(a) x1 vs t (b) x1 vs t
10.(a) The characteristic equation is given by\u2223\u2223\u2223\u22231\u2212 r 2\u22125 \u22121\u2212 r
\u2223\u2223\u2223\u2223 = r2 + 9 = 0 .
The equation has complex roots r1,2 = ± 3i. For r = \u22123i, the components of the
solution vector must satisfy 5 \u3be1 + (1\u2212 3i)\u3be2 = 0 . Thus the corresponding eigenvec-
tor is \u3be(1) = (1\u2212 3i ,\u22125)T . Substitution of r = 3i results in 5 \u3be1 + (1 + 3i)\u3be2 = 0 .
A corresponding eigenvector is \u3be(2) = (1 + 3i ,\u22125)T .
(b) The eigenvalues are purely imaginary, hence the critical point is a center.
9.1 437
(c,d)
(a) x1 vs t (b) x1 vs t
11.(a) The characteristic equation is (r + 1)2 = 0 , with double root r = \u22121 . It is
easy to see that the two linearly independent eigenvectors are \u3be(1) = (1 , 0)T and
\u3be(2) = (0 , 1)T .
(b) Since there are two linearly independent eigenvectors, the critical point is a
stable proper node.
(c,d)
438 Chapter 9. Nonlinear Differential Equations and Stability
(a) x1 vs t (b) x1 vs t
12.(a) Setting x= \u3be ert results in the algebraic equations(
2\u2212 r \u22125/2
9/5 \u22121\u2212 r
)(
\u3be1
\u3be2
)
=
(
0
0
)
.
For a nonzero solution, we require that det(A\u2212 r I) = r2 \u2212 r + 5/2 = 0. The roots
of the characteristic equation are r = 1/2 ± 3i/2 . Substituting r = 1/2 \u2212 3i/2 ,
the equations reduce to (3 + 3i)\u3be1 \u2212 5 \u3be2 = 0 . Therefore the two eigenvectors are
\u3be(1) = (5 , 3 + 3i)T and \u3be(2) = (5 , 3\u2212 3i)T .
(b) Since the eigenvalues are complex, with positive real part, the critical point is
an unstable spiral.
(c,d)
(a) x1 vs t (b) x1 vs t
9.1 439
14. Setting x \u2032 = 0 , that is, (\u22122 1
1 \u22122
)
x =
(
2
\u22121
)
,
we find that the critical point is x0 = (\u22121, 0)T . With the change of dependent
variable, x=x0+u , the differential equation can be written as
du
dt
=
(\u22122 1
1 \u22122
)
u.
The critical point for the transformed equation is the origin. Setting u= \u3be ert results
in the algebraic equations(\u22122\u2212 r 1
1 \u22122\u2212 r
)(
\u3be1
\u3be2
)
=
(
0
0
)
.
For a nonzero solution, we require that det(A\u2212 r I) = r2 + 4r + 3 = 0. The roots
of the characteristic equation are r = \u22123 , \u22121 . Hence the critical point is a stable
node.
15. Setting x \u2032 = 0 , that is, (\u22121 \u22121
2 \u22121
)
x =
(
1
\u22125
)
,
we find that the critical point is x0 = (\u22122, 1)T . With the change of dependent
variable, x=x0+u , the differential equation can be written as
du
dt
=
(\u22121 \u22121
2 \u22121
)
u.
The characteristic equation is det(A\u2212 r I) = r2 + 2r + 3 = 0, with complex conju-
gate roots r = \u22121 ± i\u221a2 . Since the real parts of the eigenvalues are negative, the
critical point is a stable spiral.
16. The critical point x0 satisfies the system of equations(
0 \u2212\u3b2
\u3b4 0
)
x =
(\u2212\u3b1
\u3b3
)
.
It follows that x0 = \u3b3/\u3b4 and y0 = \u3b1/\u3b2 . Using the transformation, x=x0+u , the
differential equation can be written as
du
dt
=
(
0 \u2212\u3b2
\u3b4 0
)
u.
The characteristic equation is det(A\u2212 r I) = r2 + \u3b2 \u3b4 = 0. Since \u3b2 \u3b4 > 0 , the roots
are purely imaginary, with r = ± i\u221a\u3b2\u3b4 . Hence the critical point is a center.
20. The system of ODEs can be written as
dx
dt
=
(
a11 a12
a21 a22
)
x.
440 Chapter 9. Nonlinear Differential Equations and Stability
The characteristic equation is r2 \u2212 p r + q = 0. The roots are given by
r1,2 =
p ±
\u221a
p2 \u2212 4q
2
=
p ± \u221a\u2206
2
.
The results can be verified using Table 9.1.1.
21.(a) If q > 0 and p < 0, then the roots are either complex conjugates with negative
real parts, or both real and negative.
(b) If q > 0 and p = 0, then the roots are purely imaginary.
(c) If