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Solutions to a some exercises in Boyce DiPrima
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9.1.1 Investigate the critical point. r 1 < 0 < r 2
a) The eigenvalues are gotten from the characteristic equation
(3 − r)(− 2 − r) + 4 = r^2 − r − 2 = (r + 1)(r − 2) = 0.
So r 1 = −1 and r 2 = 2. The eigenvector ξ solves (A − rI)ξ = 0.
r 1 = −1:
(A − rI)ξ(1)^ =
[
]
ξ(1)^ = 0 ⇒ ξ(1)^ =
[
]
r 2 = 2:
(A − rI)ξ(1)^ =
[
]
ξ(2)^ = 0 ⇒ ξ(2)^ =
[
]
b) We have the case that r 1 < 0 < r 2. Table 9.1.1 gives that the critical point is an unstable saddle point.
c) See d).
d) The following Maple code will produce the phase portrait in figure 1.
with(DEtools): phaseportrait([D(x1)(t) = 3x1(t)-2x2(t), D(x2)(t) = 2x1(t)-2x2(t)], [x1(t), x2(t)], t=-1.5..0.5, [[x1(0)=0.5, x2(0)=1],[x1(0)=1,x2(0)=0.5], [x1(0)=-0.5, x2(0)=-1],[x1(0)=-1,x2(0)=-0.5]]);
The solutions in figure 2 are also gotten from Maple.
plot([ exp(-t) + exp(2t), 2exp(-t) + exp(2t), exp(-t) + 0exp(2*t)], t=0..0.5, color=blue);
0
2
4
x
–2 –1 1 2 x
Figur 1: Phase portrait of 9.1.
1
2
3
0 0.1 0.2 0.3 0.4 0. t
Figur 2: Solutions of x 1 for (c 1 , c 2 ) = (1, 1), (2, 1) and (1, 0)
0
2
4
x
–4 –2 2 4 x
Figur 3: Phase portrait of 9.1.
0
10
20
30
40
50
60
70
0.2 0.4 0.6 0.8 1 t
Figur 4: Solutions of x 1 for (c 1 , c 2 ) = (1, 1), (2, 1) and (1, 0).
9.1.5 Investigate the critical point. r 1 , r 2 are complex.
a) The eigenvalues are gotten from the characteristic equation
(1 − r)(− 3 − r) + 5 = r^2 + 2r + 2 = (r − (−1 + i))(r − (− 1 − i)) = 0.
So r 1 = −1 + i and r 2 = − 1 − i. The eigenvector ξ solves (A − rI)ξ = 0.
r 1 = −1 + i:
(A − rI)ξ(1)^ =
[
2 − i − 5 1 − 2 − i
]
ξ(1)^ = 0
⇒ ξ 1 (1) − (− 2 − i)ξ 2 (1) = 0
⇒ ξ(1)^ =
[
2 − i 1
]
r 2 = − 1 − i: ξ(2)^ = ξ(1)^ =
[
2 + i 1
]
b) We have the case that r 1 and r 2 are complex with negative real part. Table 9.1.1 gives that the critical point is an asymptotically stable spiral point.
c) See d).
d) The following Maple code will produce the phase portrait in figure 5.
with(DEtools): phaseportrait([D(x1)(t) = x1(t) - 5x2(t), D(x2)(t) = x1(t) - 3x2(t)], [x1(t), x2(t)], t=-2.5..4, [[x1(0)=0.5, x2(0)=0.5],[x1(0)=-0.5,x2(0)=-0.5]], stepsize=0.05, linecolor=blue);
The solutions in figure 6 are also gotten from Maple.
plot([ exp(2t) + exp(4t), 2exp(2t) + exp(4t), exp(2t) + 0exp(4t)], t=0..1, color=blue);
9.2.5a Critical points of non-linear systems.
A critical point x^0 = (x 0 , y 0 ) is characterized by
[
x′(x 0 , y 0 ) y′(x 0 , y 0 )
]
[
]
. We
need to solve
0 = x − xy = x(1 − y) ⇒ x = 0 or y = 1 (1) 0 = y + 2xy = y(1 + 2x). ⇒ x = − 0 .5 or y = 0. (2)
Consider the relations satisfying (1).
x = 0: Now (2) is satisfied for y = 0, so we have the critical point x^01 = (0, 0).
y = 1: Now (2) is satisfied for x = − 0 .5, so we have the critical point x^02 = (− 0. 5 , 1).
9.2.7a Critical points of non-linear systems.
A critical point x^0 = (x 0 , y 0 ) is characterized by
[
x′(x 0 , y 0 ) y′(x 0 , y 0 )
]
[
]
. We
need to solve
0 = x − x^2 − xy = x(1 − x − y) ⇒ x = 0 or y = 1 − x (3)
0 =
y −
y^2 −
xy =
y(2 − y − 3 x) ⇒ y = 0 or y = 2 − 3 x. (4)
Consider the relations satisfying (3).
x = 0: In order for (4) to be satisfied we must have y = 0 or y = 2 − 3 · 0 = 2, so we have the critical points x^01 = (0, 0) and x^02 = (0, 2).
y = 1 − x: Now (4) gives y = 0 ⇒ x = 1 or
y = 2 − 3 x ⇒ 1 − x = 2 − 3 x ⇒ x = 0.5 and y = 0. 5.
Thus we have the critical points x^03 = (1, 0) and x^04 = (0. 5 , 0 .5).n
9.2.21 Non-linear, autonomous system
a) Since the derivatives have no explicit t dependence we can write
dy dx
dy/dt dx/dt
− sin x y
We learned how to solve this kind of equations in chapter 2.
ydy = − sin xdx 1 2
y^2 = cos x + c
H(x, y) =
y^2 − cos x = c.
b) We consult Maple for plotting of trajectories.
with(plots): H := (x,y) -> y^2/2 - cos(x); implicitplot({H(x,y) = 0, H(x,y) = 1, H(x,y) = 2, H(x,y) = 3}, x=-5..10, y=-10..10);
0
1
2
y
–4 –2 2 4 6 8 10 x
Figur 7: Plot of H(x, y) = c for c = 0, 1 , 2 and 3.
9.3.5 Almost linear systems.
a) We want to find x^0 = (x 0 , y 0 ) s.t. dxdt (x^0 ) = dydt (x^0 ) = 0. Note that dxdt = 0 for x = −2 and x = y.
x = −2: dy dt = (4 + 2)(y − 2) = 6y − 12 = 0 ⇒ y = 2.
x = y:
dy dt
= (4 − y)(y + y) = 2y(4 − y) = 0 ⇒ y = 0 or y = 4.
Thus we have three critical points x^01 = (− 2 , 2), x^02 = (0, 0) and x^03 = (4, 4).
d) Maple helps us draw the phase portrait.
with(DEtools): DEplot([D(x)(t) = (2 + x(t))(y(t) - x(t)), D(y)(t) = (4 - x(t))(y(t) + x(t))], [x(t), y(t)], t=0..2, x=-2.5..4.5, y=-1..5, [[x(0)=3, y(0)=1], [x(0)=-2.1, y(0)=2.1], [x(0)=0.1, y(0)=0.1]], stepsize=0.05, linecolor=blue);
0
1
2
3
4
5
y
–2 –1 1 2 3 4 x
Figur 8: The critical points (0,0) and (-2, 2) are unstable while (4,4) is stable.
9.3.8 Almost linear systems.
a) We want to find x^0 = (x, y) s.t. dxdt (x^0 ) = dydt (x^0 ) = 0. Note that dxdt = x(1 − x − y) = 0 for x = 0 and x = 1 − y.
x = 0: dy dt
(2y − y^2 − 3 · 0) =
y(2 − y) = 0 ⇒ y = 0 or y = 2.
x = 1 − y:
dy dt
(2y − y^2 − 3(1 − y)y) =
y(2y − 1) = 0 ⇒ y = 0 or y = 0. 5.
Thus we have three critical points (0, 0), (0, 2), (1, 0) and (0. 5 , 0 .5).
b) The system can be linearized around the critical point (x 0 , y 0 ) by
d dt
[
x y
]
[
Fx(x 0 , y 0 ) Fy (x 0 , y 0 ) Gx(x 0 , y 0 ) Gy (x 0 , y 0 )
] [
x − x 0 y − y 0
]
where dxdt = F (x, y) and dydt = G(x, y). Here F (x, y) = x − x^2 − xy and G(x, y) = 14 (2y − y^2 − 3 xy) so we get the derivatives
Fx = 1 − 2 x − y Fy = −x
Gx = −
y
Gy =
(2 − 2 y − 3 x)
Now let us linearize the system around each critical point
(x 0 , y 0 ) = (0, 0) : Inserting these values in (6) gives
d dt
[
x y
]
[
] [
x y
]
(x 0 , y 0 ) = (0, 2) : Inserting these values in (6) gives
d dt
[
x y
]
[
] [
x y − 2
]
(x 0 , y 0 ) = (1, 0) : Inserting these values in (6) gives
d dt
[
x y
]
[
] [
x − 1 y
]
(x 0 , y 0 ) = (0. 5 , 0 .5) : Inserting these values in (6) gives
d dt
[
x y
]
[
] [
x − 1 y
]
c) Calculate the eigenvalue for each of the critical points
(x 0 , y 0 ) = (0, 0) : (1 − r)(
− r) = 0.
r 1 = 12 and r 2 = 1 are two positive eigenvalues. According to table 9.3. this critical point is an unstable node for the non-linear system. (x 0 , y 0 ) = (0, 2) : (− 1 − r)(−
− r) = 0.
r 1 = −1 and r 2 = − 12 are two negative eigenvalues. According to table 9.3.1 this critical point is an asymptotically stable node for the non-linear system.
