Math 215
Spring 2010
Assignment 9
You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples.
§9.1: 8(ab), 9(ab), 15, §9.2: 4, 22(a), §9.3: 8(abc), 9(abc)
µ
¶
−1 −1
• (§9.1: 8(ab)) For x0 =
x, (a) find the eigenvalues and eigenvectors,
0 −1/4
(b) classify the critical point (0, 0) as to type and determine whether it is stable,
asymptotically stable, or unstable.
Answer.
The characteristic equation is det(A − rI) = (r + 1)(r + 1/4) = 0, with
roots r1 = −1 and r2 = −1/4. For r = −1, the eigenvector equation reduces to
ξ2 = 0. Thus the corresponding eigenvector is ξ (1) = (1, 0)T . For r = −1/4, the
eigenvector equation reduces to 3ξ1 /4 + ξ2 = 0. Thus the corresponding eigenvector is
ξ (2) = (4, −3)T . The eigenvalues are real and both negative. Hence the critical point
(0, 0) is a stable node.
µ
¶
3 −4
0
• (§9.1: 9(ab)) For x =
x, (a) find the eigenvalues and eigenvectors, (b)
1 −1
classify the critical point (0, 0) as to type and determine whether it is stable, asymptotically stable, or unstable.
Answer. The characteristic equation is det(A − rI) = r2 − 2r + 1 = 0, with single
root r = 1. For r = 1, the eigenvector equation reduces to ξ1 − 2ξ2 = 0. Thus the
corresponding eigenvector is ξ (1) = (2, 1)T . Since r = 1 > 0, the critical point (0, 0) is
an unstable improper node.
µ
¶
µ ¶
−1 −1
−1
0
• (§9.1: 15) For x =
x+
, determine the critical point x = x0 ,
2 −1
5
and then classify its type and examine its stability by making the transformation
x = x0 + u.
µ
¶
µ ¶
−1 −1
1
Answer.
Setting x0 = 0, i.e.,
x+
, we find the critical point is
2 −1
−5
x0 = (−2, 1)T . With the change
variable x = x0 + u, the differential
µ of dependent
¶
−1 −1
system can be written as u0 =
u. The characteristic equation is det(A −
2 −1
√
rI) = r2 + 2r + 3 = 0, with complex conjugate roots r = −1 ± 2i. Since the real
parts of the eigenvalues are negative, the critical point x0 is a stable spiral.
• (§9.2: 4) Sketch the trajectory corresponding to the solution satisfying the specified
initial conditions, and indicate the direction of motion for increasing t:
√
dx/dt = ay, dy/dt = −bx, a > 0, b > 0;
x(0) = a, y(0) = 0.
Answer. The trajectories of the system satisfy the ODE dy/dx =
is separable, with
aydy = −bxdx.
bx
ay .
The equation
Hence the trajectories are given by bx2 + ay 2 = C 2 in which C is arbitrary. These are
ellipses. For the given initial condition, we have ba + 0 = C 2 , and the trajectory is
√
y2
x2
a + b = 1. At t = 0 we have (dx/dt, dy/dt) = (0, −b a) which is pointing downward.
Thus the solution is moving along the ellipse clockwise.
[The sketch is omitted]
• (§9.2: 22(a)) Find an equation of the form H(x, y) = c satisfied by the trajectory of
dx/dt = 2x2 y − 3x2 − 4y,
Answer.
dy/dt = −2xy 2 + 6xy.
The trajectories are solutions of the DE
−2xy 2 + 6xy
dy
= 2
,
dx
2x y − 3x2 − 4y
which can be written as (2xy 2 − 6xy)dx + (2x2 y − 3x2 − 4y)dy = 0. The resulting
ODE is exact, with
∂H
= 2xy 2 − 6xy,
∂x
∂H
= 2x2 y − 3x2 − 4y.
∂y
Integrating the first term in x, we find that H(x, y) = x2 y 2 − 3x2 y + h(y). It follows
2
2
0
that ∂H
∂y = 2x y − 3x + h (y). Comparing the two partial derivatives, we obtain
h0 (y) = −4y and thus h(y) = −2y 2 + c. Hence
H(x, y) = x2 y 2 − 3x2 y − 2y 2 .
• (§9.3: 8(abc)) Consider
1
dy/dt = (2y − y 2 − 3xy).
4
dx/dt = x − x2 − xy,
(a) Determine all critical points.
(b) Find the corresponding linear system near each critical point.
(c) Find the eigenvalues of each linear system. What conclusions can you then draw
about the nonlinear system?
Answer.
(a) The critical points are given by the solution set of
x(1 − x − y) = 0,
2
y(2 − y − 3x) = 0.
If x = 0, then either y = 0 or y = 2. If x 6= 0 then x = 1 − y and y(2 − y − 3 + 3y) = 0.
Thus y = 0 (then x = 1) or y = 1/2 (then x = 1/2). The critical points are (0, 0),
(0, 2), (1, 0) and (1/2, 1/2).
(b,c) Note that F (x, y) = x − x2 − xy and G(x, y) = 14 (2y − y 2 − 3xy). The Jacobian
matrix of the vector field is
µ
¶ µ
¶
Fx Fy
1 − 2x − y
−x
J=
=
.
Gx Gy
−3y/4
1/2 − y/2 − 3x/4
µ
¶
1 0
At (0, 0),
= Jx+ error, where J =
, with eigenvalues 1, 1/2. The
0 1/2
eigenvalues are real and both positive. Hence the critical point is an unstable node.
µ
¶
−1
0
d
, with eigenvalues −1, −1/2.
At (0, 2), dt x = Jx+ error, where J =
−3/2 −1/2
The eigenvalues are both negative. Hence the critical point is a stable node.
µ
¶
−1 −1
d
At (1, 0), dt x = Jx+ error, where J =
, with eigenvalues −1, −1/4.
0 −1/4
The eigenvalues are both negative. Hence the critical point is a stable node.
µ
¶
−1/2 −1/2
d
At (1/2, 1/2), dt x = Jx+ error, where J =
, with eigenvalues
−3/8 −1/8
√
1
16 (−5 ± 57). The eigenvalues are real with opposite sign. Hence the critical point
is an unstable saddle.
d
dt x
• (§9.3: 9(abc)) Repeat the above problem for
1
dx/dt = (2 + y)(y − x),
2
Answer.
1
dy/dt = (2 − x)(y + x).
2
(a) The critical points are given by the solution set of
1
(2 + y)(y − x) = 0,
2
1
(2 − x)(y + x) = 0.
2
If y = −2, then either x = 2 or x = −2y = 4. If y 6= −2 then y = 2x and
(2 − x)(2x + 12 x) = 0. Thus x = 0 (then y = 0) or x = 2 (then y = 1). The critical
points are (0, 0), (2, −2), (4, −2) and (2, 1).
(b,c) Note that F (x, y) = 2y + y 2 − x − xy/2 and G(x, y) = 2y − xy + x − x2 /2. The
Jacobian matrix of the vector field is
µ
¶ µ
¶
Fx Fy
−1 − y/2 2 + 2y − x/2
J=
=
.
Gx Gy
−y + 1 − x
2−x
µ
¶
√
−1 2
At (0, 0),
= Jx+ error, where J =
, with eigenvalues 12 (1 ± 17). The
1 2
eigenvalues are real with opposite sign. Hence the critical point is an unstable saddle.
d
dt x
3
µ
¶
√
0 −3
= Jx+ error, where J =
, with eigenvalues ± 3i. The
At (2, −2),
1 0
eigenvalues have zero real parts. The stability property of the critical point cannot
be decided from the linear approximation.
µ
¶
√
0 −4
d
At (4, −2), dt x = Jx+ error, where J =
, with eigenvalues −1 ± 5.
−1 −2
The eigenvalues are real with opposite sign. Hence the critical point is an unstable
saddle.
µ
¶
√
−3/2 3
d
At (2, 1), dt x = Jx+ error, where J =
, with eigenvalues 14 (−3 ± 87i).
−2 0
The eigenvalues are complex conjugate with negative real part. Hence the critical
point is a stable spiral.
d
dt x
4