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Mathematics MATC46, Assignment 2
Solutions to Selected Problems
First question:
dx
dy
= x(y − 1)
= − (2y + (x − 1)(x − 4))
dt
dt
Critical points: x = 0 or y = 1, and 2y + (x − 1)(x − 4) = 0 implies if
x = 0 then y = −2). If y = 1 then
(x − 1)(x − 4) + 2 = 0
or
x2 − 5x + 6 = 0
or (x − 2)(x − 3) = 0, so x = 2 or x = 3. So there are three critical points
(0, −2), (2, 1), (3, 1)
The matrix is
A=
"
∂F
∂x
∂G
∂x
∂F
∂y
∂G
∂y
#
"
=
"
−3 0
5 −2
y−1
x
−2x + 5 −2
#
At the critical point (0, −2),
A=
#
The eigenvalues are −2 and −3, so this is a stable node.
At the critical point (3, 1),
A=
"
0
6
−7 −2
#
The eigenvalues λ satisfy
λ2 + 2λ + 3 = 0
√
√
so λ = 12 (−2 ± 4 − 4 · 3) = −1 ± 2i
This is a stable spiral.
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At the critical point (2, 1), the matrix is
A=
"
0 2
1 −2
#
The eigenvalues λ satisfy λ2 + 2λ − 2 = 0 so λ = −1 ±
point.
Sect. 9.2 #5(a)
dx
= x(1 − y), dy
= y(1 + 2x)
dt
dt
Critical points:
x = 0 OR y = 1
√
3. This is a saddle
(1)
AND
y = 0 OR x = −1/2
(2)
From( 1), if x = 0 then the possibility x = −1/2 in (2) is impossible, so
for (2) to be satisfied, we require y = 0. Thus one critical point is (0, 0).
On the other hand, if y = 1 is satisfied (in (1)) then y = 0 is impossible
in (2), so we must have x = −1/2. Thus there is one more critical point
(−1/2, 1).
Sect. 9.2 #7(a)
The critical points are:
x(1 − x − y) = 0 → x = 0 OR x + y = 1
(3)
1 y 3x
y( − − ) = 0 → y = 0 OR 3x + y = 2
(4)
2 4
4
First consider x = 0 in (3).
This forces y = 0 or y = 2 in (4). So there are two critical points (0, 0)
and (0, 2).
If on the other hand x + y = 1 in (3), then y = 0 in (4) yields the critical
point (1, 0). We can also solve (4) by 3x+y = 2 which gives x = 1/2, y = 1/2.
So the critical points are (0, 0), (0, 2), (1, 0) and (1/2, 1/2).
Sect. 9.2 #24(a)
dx
= y,
dt
dy
− x + x3 /6
dt
This leads to
2
dy
−x + x3 /6
=
dx
y
or
ydy = (−x + x3 /6)dx
or
y 2 /2 = −x2 /2 + x4 /24 + C
or
x2 + y 2 − x4 /12 = C.
Sect. 9.3 #3
dx
= (1 + x) sin(y)
(5)
dt
dy
= 1 − x − cos(y)
(6)
dt
(0, 0) is a critical point because the right hand sides of equations (5) and
(6) vanish at this value.
Using the Taylor expansion of sin(y),
sin(y) = y − y 3 /3! + . . .
so the linear part of the right hand side of (5) is y.
The Taylor expansion of cos(y) is 1 − y 2 /2 + . . . so the linear part of the
right hand side of (6) is −x.
Note that |sin(y) − y| ≤ |y|3 (by the Taylor expansion) Also
|1 − cos(y)| ≤ |y|2
√
Since |y| ≤ x2 + y 2 it follows that the system is almost linear.
The corresponding linear system is
d
dt
x
y
where
A=
!
"
=A
0 1
−1 0
x
y
#
!
.
The eigenvalues are ±i, so (0, 0) is a center.
Sect. 9.3 #26
3
(a)(0, 0) is a critical point because the right hand side of both equations
(i) and (ii) is 0 when x = y = 0.
The linear part is
!′
!
x
x
=A
y
y
where
A=
"
0 1
−1 0
#
.
So the eigenvalues are ±i and the critical point is a center.
(b)The system is almost linear because the nonlinear part is
~v =
x
y
!
r2
(equation (i)) and
x
y
~v = −
!
r2
(equation (ii)) the norm of this is r3 so |~v |/|
x
y
!
| = r2 → 0 as r → 0.
(c) Equation (i):
r
so computing
r
dr
dx
dy
=x +y
dt
dt
dt
dx
dy
dr
=x +y
= r4
dt
dt
dt
so
1
=t+C
2r2
and since r = r0 at t = 0 we find C = − 2r12 so
−
0
r=q
1
2(C − t)
which becomes unbounded as t → C. Hence the critical point is unstable.
Equation (ii):
rdr = −r4
4
so
dr
= −dt
r3
and
1
=t+C
2r2
or
which tends to 0 as t → ∞.
Sect. 9.3 #30
Substituting y =
dx
dt
1
r=q
2(t + C)
dx
d2 x
+ c(x) + g(x) = 0
2
dt
dt
this becomes the system
dx
=y
dt
dy
= −c(x)y − g(x)
dt
(b) The right hand side is 0 when x = 0 and y = 0 since g(0) = 0.
Subtract
!
!
x
x
A
=
−c(0)y − g ′ (0)x
y
y
we obtain the linear system
dx
=y
dt
dy
= −c(0)y − g ′ (0)x
dt
The nonlinear part is
(−c(x) + c(0))y − (g(x) − g ′ (0)x)
The first term tends to 0 as x → 0 because c(x) is continuously differentiable and |y|/r is bounded. The second term tends to 0 as x → 0 because
the Taylor expansion of g(x) is
|g(x) − g ′ (0)x| ≤ |g ′′ (0)/2|x2
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so |g(x) − g ′ (0)x|/r → 0 as r → 0.
(c) The linear part is
A=
"
0
1
′
−g (0) −c(0)
#
The eigenvalues are therefore
λ± =
−c(0) ±
q
c(0)2 − 4g ′ (0)
2
q
If g ′ (0) > 0, and c(0) > 0, then c(0)2 − 4g ′ (0) < c(0) so both λ± are
real < 0, and hence the critical point is asymptotically stable.
If c(0) < 0 or g ′ (0) < 0 then either λ+ > 0 and λ2 < 0 or both λ+ and
λ− are > 0 and hence the critical point is unstable.
Sect. 9.6 #2 x′ = F (x, y), y ′ = G(x, y) where
F (x, y) = −(1/2)x3 + 2xy 2
G(x, y) = −y 3 . The associated vector field is
W = (F (x, y), G(x, y))
We study Liapunov functions of the form V (x, y) = ax2 + cy 2 . Then
▽V · W = (2ax, 2cy) · (−1/2x3 + 2xy 2 , −y 3 )
= a(−x4 + 4x2 y 2 ) − 2cy 4
Set u = x2 , v = y 2 and define α = −a, β = 4a, γ = −2c. Then this
expression becomes αu2 + βuv + γv 2 .
If we choose a > 0 and c > 0, then V is positive definite and α < 0..
According to Theorem 9.6.4, if 4αγ − β 2 > 0 then αu2 + βuv + γv 2 is negative
definite. If we choose c > 2a, ▽V · W is negative definite. For example, we
could choose V (x, y) = x2 + 3y 2 . Then (0, 0) is asymptotically stable, by
Theorem 1.
Sect. 9.6 #5
The vector field is
(F (x, y), G(x, y))
where
F (x, y) = y − xf (x, y),
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G(x, y) = −x − yf (x, y).
As above, choose V (x, y) with ▽V = (2ax, 2cy) so
▽V · W = (2ax, 2cy) · (y − xf (x, y), −x − yf (x, y))
(7)
= 2(a − c)xy − 2(ax2 + cy 2 )f (x, y)
. If we choose a = c > 0, then if f > 0 ⇒ the inner product (7) is negative
definite (asymptotically stable) while if f < 0 then the inner product is
positive definite (unstable).
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