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Sample Solutions of Assignment 9 for MATH3270A for ODE

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Sample Solutions of Assignment 9 for MATH3270A for ODE
November 11,2013
Contact Mr. XIAO Yao(yxiao a math.cuhk.edu.hk) directly if you have any questions on the
solutions
1. For the following ODEs, find the eigenvalues and eigenvectors, and classify the critical point
(0, 0) as to type and determine whether it is stable, asymptotically stable, or unstable
a.
x =
3 в€’2
2 в€’2
x,
b.
x =
1 в€’5
1 в€’3
x,
c.
x =
2 в€’5
0 2
x,
d.
x =
1 в€’2
1 в€’1
x
Answer: (a). The eigen-system of the matrix is
r1 = в€’1, Оѕ 1 =
1
2
r2 = 2, Оѕ 2 =
2
1
Since one of the eigenvalues is positive, (0, 0) is unstable.
(b). The eigen-system of the matrix is
r1 = в€’1 в€’ i, Оѕ 1 =
2в€’i
1
r2 = в€’1 + i, Оѕ 2 =
2+i
1
Since the real part of the eigenvalues is negative, (0, 0) is stable and asymptotically stable.
(c). The eigen-system of the matrix is
r1 = r2 = 2, Оѕ 1 =
1
0
Since the eigenvalues are positive, (0, 0) is unstable.
(d). The eigen-system of the matrix is
r1 = в€’i, Оѕ 1 =
1в€’i
1
r2 = i, Оѕ 2 =
1+i
1
Since the eigenvalues are pure imaginative, (0, 0) is stable but not asymptotically stable.
1
2
О» Вµ
в€’Вµ О»
2. Transform the following problems into the form x =
a.
x =
3 в€’2
4 в€’2
x,
b.
Answer: The eigen-system of matrix is
в€љ
1
7
r1 = в€’ i
, Оѕ1 =
2
2
в€љ
1
7
r2 = + i
, Оѕ2 =
2
2
x =
x:
2 в€’2
3 в€’2
x.
в€љ
5
8
в€’i
1
5
8
+i
1
7
8
в€љ
в€љ
5
8
The real part and imaginative part of eigenvector is
7
8
and
1
7
8
0
. Let T =
5
8
1
в€љ
7
8
0
,
then
T AT =
в€љ
7
2
1
2
1
2в€љ
в€’1
7
2
в€’
Let y = Tв€’1 x, we get
y = Tв€’1 ATy =
в€љ
7
2
1
2
1
2в€љ
в€’
7
2
y.
(b) The eigen-system of matrix is
в€љ
в€љ
r1 = в€’i 2, Оѕ 1 =
2
3
в€’i
1
в€љ
r2 = i 2, Оѕ 2 =
2
3
+i
1
в€’1
T AT =
0
в€љ
в€’ 2
y = T ATy =
2
3
1
в€љ
and
в€љ
2
0
Let y = Tв€’1 x, we get
в€’1
в€љ
2
3
The real part and imaginative part of eigenvector is
then
2
3
0
в€љ
в€’ 2
в€љ
2
0
y.
3. Determine the critical points for each of the following systems
dx
dy
a.
= x(2x в€’ 3y),
= y(1 в€’ x в€’ y)
dt
dt
dx
dy
1
1
3
b.
= x в€’ x2 в€’ xy,
= y в€’ y 2 в€’ xy
dt
dt
2
4
4
dx
dy
2
c.
= y,
= Вµ(1 в€’ x )y в€’ x, Вµ > 0
dt
dt
2
3
0
. Let T =
2
3
1
в€љ
2
3
0
,
3
dx
dy
= y + x(1 в€’ x2 в€’ y 2 ),
= в€’x + y(1 в€’ x2 в€’ y 2 )
dt
dt
Answer: (a). From
d.
F =
x(2x в€’ 3y)
=0
G = y(1 в€’ x в€’ y) = 0
we know that the critical points are (0, 0), (0, 1), ( 35 , 52 ).
(b). From
x в€’ x2 в€’ xy
F =
1
y
2
G=
=0
в€’ 14 y 2 в€’ 34 xy = 0
we know that the critical points are (0, 0), (0, 2), (1, 0), ( 12 , 21 ).
(c). From
F =
y
=0
G = Вµ(1 в€’ x2 )y в€’ x = 0
we know that the critical point is (0, 0).
(d). From
F =
y + x(1 в€’ x2 в€’ y 2 )
=0
G = в€’x + y(1 в€’ x2 в€’ y 2 ) = 0
we know that the critical point is (0, 0).
4. Show that the trajectories of the nonlinear undamped pendulum equation
d2 Оё g
+ sin Оё = 0
dt2
l
are given by
g
y2
(1 в€’ cos x) +
=c
l
2
dОё
.
dt
Answer: Let x = Оё, y =
where x = Оё, y =
dОё
,
dt
the nonlinear undamped pendulum equation can be rewritten as
x
y
=
в€’ gl
y
sin x
Hence the trajectory equation is
в€’ gl sin x
dy
=
dx
y
Solve this equation by separation of variable method, we get
1 2 g
y + (1 в€’ cos x) = c
2
l
(1)
4
5. Prove that if a trajectory starts at a noncritical point of the system
dx
= f (x)
dt
then it can not reach a critical point in a finite length of time.
Answer: If not, i.e. assume that the solution x = x(t) satisfies x(0) = a and x(t0 ) = x0 , where
x0 is a critical point and a is not. Then, we deduce that x(t) is also the solution of the following
initial value problem:
dx
= f (x), x(t = t0 ) = x0
(1)
dt
with x(t = 0) = a. Since x0 is a critical point, x(t) в‰Ў x0 is a trivial solution of (1) and hence the
unique solution by uniqueness. This is a contradiction with the fact x(t = 0) = a.
6. Assuming that the trajectory corresponding to a solution x = П†(t), y = П€(t), в€’в€ћ < t < +в€ћ,
of an autonomous system is closed, show that the solution must be periodic.
Answer: Since the trajectory is closed, there exist at least one point (x0 , y0 ) such that П†(t0 ) =
x0 , П€(t0 ) = y0 and a number T > 0 such that П†(t0 + T ) = x0 , П€(t0 + T ) = y0 . Let О¦(t) =
П†(t + T ), ОЁ(t) = П€(t + T ), then it is easy to know that (О¦(t), ОЁ(t)) is also the solution of the
equation with the same initial data: О¦(t0 ) = П†(t0 + T ) = x0 , ОЁ(t0 ) = П€(t0 + T ) = y0 . Then, by
the uniqueness theorem, we know that П†(t) = О¦(t) = П†(t + T ), П€(t) = ОЁ(t) = П€(t + T ), в€Ђt.
That is, the solution is periodic.
7. Write the following spring-mass system as a system of equations by introducing x = u, y =
d2 u
du
+ c + ku = 0
2
dt
dt
Find out the critical point and analyze the stability of the critical point.
m
Answer: Let x = u, y =
du
dt
and by direct computation, we have
x
y
=
0
1
k
в€’ m в€’ mc
x
y
From
F =
y
k
G = в€’m
xв€’
=0
c
y
m
=0
we know that the critical point is (0, 0). The eigenfunction of the matrix is
r2 +
c
k
r+
=0
m
m
du
dt
5
For
c
m
< 0, we know that there is at least one eigenvalue which real part is positive, The critical
point (0, 0) is unstable.
For
For
c
m
c
m
k
> 0 and m
> 0, by Problem 15 in Section 7.8, (0, 0) is asymptotically stable.
k
> 0 and m < 0, then there must be have two real eigenvalues. One of them is negative
and the other is positive. The critical point (0, 0) is saddle point and unstable.
For
c
m
> 0 and
k
m
= 0, the eigenvalues are 0, в€’ mc < 0, The critical point (0, 0) is stable but not
asymptotically stable.
k
= 0, the equation is reduced to x = y, y = 0. Hence The critical point (0, 0)
For mc = 0 and m
is unstable.
For
c
m
= 0 and
k
m
> 0, there is two pure imaginative eigenvalue. (0, 0) is a center and is stable
k
< 0, there is two real eigenvalue, One of them
but not asymptotically stable. For mc = 0 and m
is negative and the other is positive. The critical point (0, 0) is saddle point and unstable.
8. Show that the system is almost linear and (0, 0) is a stable critical pint of the system
dx
= в€’x в€’ xy 2 ,
dt
dy
= в€’y в€’ x2 y .
dt
Answer: From
F = в€’x в€’ xy 2 = 0
G = в€’y в€’ x2 y = 0
we know that the critical point is (0, 0). Since F, G are continuous differential, the system is
almost linear.
Let V (x, y) = x2 + y 2 is positive definite, we have
W =
V В· (F, G) = в€’2(x2 + y 2 + 2x2 y 2 )
then W is negative definite. Hence, critical point (0, 0) is stable.
9. Show that the system is almost linear and (0, 0) is an asymptotically stable critical point of
the system
dx
1
dy
= в€’ x3 + 2xy 2 ,
= в€’y 3
dt
2
dt
Answer: From
F = в€’ 12 x3 + 2xy 2 = 0
G=
в€’y 3
=0
we know that the critical point is (0, 0). Since F, G are continuous differential, the system is
almost linear. Let V (x, y) = ax2 + by 2 , then
W =
V В· (F, G) = в€’ax4 + 4ax2 y 2 в€’ 2by 4
6
Then if we choose 0 < 2a < b, eg. a = 1, b = 3, then V is positive definite and W is negative
definite. Hence the critical point (0, 0) is asymptotically stable.
10. Determine all real critical points of the following system of equations
dy
dx
= x + y2,
=x+y
a.
dt
dt
b.
dx
= 1 в€’ y,
dt
dy
= x2 в€’ y 2
dt
Answer: (a) From
F = x + y2 = 0
G=
x+y
=0
we know that the critical points are (0, 0), (в€’1, 1).
(b) From
F =
1в€’y
=0
G = x2 в€’ y 2 = 0
we know that the critical points are (в€’1, 1), (1, 1).
11. Consider the following problem. Determine the eigenvalues and critical points, classify the
type of critical point and determine whether it is stable, asymptotically stable, or unstable.
x =
Here
1
в€’1
x
is a real number.
Answer: The eigen-system of matrix is
r1 = Оµ в€’ i,
Оѕ1 =
r2 = Оµ + i, Оѕ 2 =
i
1
в€’i
1
0
1
and
1
0
of the matrix is Оµ2 + 1, (0,0) is the only critical point of the system.
The real part and imaginative part of eigenvector is
. Since the determinant
For Оµ < 0, the real part of the eigenvalues are negative, hence (0, 0) is asymptotically stable.
For Оµ = 0, the two eigenvalues are pure imaginative number, hence (0, 0) is center and it is stable
but not asymptotically stable.
stable.
For Оµ > 0, the real part of the eigenvalues are positive, hence (0, 0) is unstable.
7
Section 9.1
For each of the following systems:
(a) Find the eigenvalues and eigenvectors.
(b) Classify the critival point (0, 0) as to type and determine wether it is stable, asymptotic
stable or unstable.
(c) Sketch several trajectories in the phase plane and also sketch some typical graphs of x1
versus t.
(d) Use a
dx
1.
=
dt
dx
=
3.
dt
dx
=
7.
dt
dx
=
9.
dt
dx
12.
=
dt
Answer:
computer to plot accurately the curves requested in part (c).
3
2
x
в€’2 в€’2
2
3
x
в€’1 в€’2
3 в€’2
x
4 в€’1
3 в€’4
x
1 в€’1
2 в€’ 25
x
9
в€’1
5
1. О»1 = 2,
О»2 = в€’1,
Оѕ1 =
3. О»1 = 1,
О»2 = в€’1,
Оѕ1 =
2
в€’1
3
в€’1
,
Оѕ2 =
,
Оѕ2 =
1
в€’2
1
в€’1
. So (0, 0) is an unstable saddle point.
. So (0, 0) is an unstable saddle point.
1
1
, Оѕ2 =
. Since the eigenvalues are
1+i
1в€’i
complex conjugates with positive real part, (0, 0) is a unstable spiral point.
2
9. О»1 = 1, О»2 = 1, Оѕ =
Since О»1 = О»2 = 1 > 0 and have only one linearly
1
independent eigenvectors, (0, 0) is unstable improper node.
5
5
12. О»1 = 1/2 + 3i/2, О»2 = 1/2 в€’ 3i/2, Оѕ1 =
, Оѕ2 =
. Since the
3 + 3i
3 в€’ 3i
eigenvalues are complex conjugates with positive real part, (0, 0) is a unstable spiral point.
7. О»1 = 1 + 2i,
О»2 = 1 в€’ 2i,
Оѕ1 =
Graphs of Part (c), (d) are omitted here.
In the following problem, determine the critical point x = x0 , and then classify its type and
examine its stability by making transformation x = x0 + u.
dx
в€’2 1
в€’4
14.
=
x+
.
1 в€’2
2
dt
dx
0 в€’ОІ
О±
16.
=
x+
; О±, ОІ, Оі, Оґ > 0.
Оґ
0
в€’Оі
dt
8
Answer:
14. Solve
в€’2 1
1 в€’2
x0 +
в€’4
2
0
0
=
.
The critical point is x0 = (в€’2, 0)t . Take the transformation x = x0 + u,
du
=
dt
в€’2 1
1 в€’2
u.
The eigenvalues are О»1 = в€’1, О»2 = в€’3, so x0 is node and asymptotically stable.
16. Solve
0 в€’ОІ
Оґ 0
x0 +
О±
в€’Оі
=
0
0
.
The critical point is x0 = ( ОіОґ , О±ОІ )t . Take the transformation x = x0 + u,
du
=
dt
0 в€’ОІ
Оґ 0
u.
в€љ
The eigenvalues are О»1,2 = В±i ОІОґ, so x0 is center and stable.
18. Consider the system x = Ax, and suppose that A has one zero eigenvalue.
(a). Show that x = 0 is a critical point and that, in addition, every point on a certain straight
line through the origin is also a critical point.
(b). Let r1 = 0 and r2 = 0, and let Оѕ 1 and Оѕ 2 be corresponding eigenvectors. Show that the
trajectories are as indicated in Figure 9.1.8.
Answer:
(a) Suppose A =
a b
c d
. From Ax = 0, we know that x = 0 and x1 = в€’ ab x2 .
(b). The solution is
x = c1 Оѕ 1 + c2 Оѕ 2 er2 t .
19. In this problem we indicate how to show that the trajectories are ellipses when the eigenvalues
are pure imaginary. Consider the system
x
y
=
a11 a12
a21 a22
x
y
(i)
(a) Show that the eigenvalues of the coefficient matrix are pure imaginary if and only if
a11 + a22 = 0,
a11 a22 в€’ a12 a21 > 0
(ii)
9
(b) The trajectories of the system (i) can be found by converting Eqs. (i) into the single
equation
dy
dy/dt
a21 x + a22 y
=
=
dx
dx/dt
a11 x + a12 y
Use the first of Eqs. (ii) to show that Eq. (iii) is exact.
(c) By integrating Eq. (iii) show that
(iii)
a21 x2 + 2a22 xy в€’ a12 y 2 = k.
(iv)
where k is a constant. Use Eqs. (ii) to conclude that the graph of Eq. (iv) is always an ellipse.
Hint: What is the discriminant of the quadratic form in Eq. (iv)?
Answer: (a) The
characteristic equation is
О»2 в€’ (a11 + a22 )О» + (a11 a22 в€’ a12 a21 ) = 0,
Clearly, its roots are pure imaginary if and only if Eq. (ii).
(b) Eq. (iii) can be rewritten as
(a21 x + a22 y)dx в€’ (a11 x + a12 y)dy = 0.
Then
∂N
∂M
в€’
= a22 + a11 = 0.
∂y
∂x
Therefore, Eq. (iii) is exact.
(c) Integral (iii) one has
1
1
a21 x2 + a22 xy в€’ a12 y 2 = c or a21 x2 + 2a22 xy в€’ a12 y 2 = k.
2
2
(d) Since a12 a21 < a11 a22 ≤ 0, a12 = 0, a21 = 0. WLOG, assume that a21 > 0, then a12 < 0.
The corresponding matrix to the quadratic form in Eq. (iv) is
A=
a21 a22
a22 в€’a12
with
a21 > 0,
det A = в€’a21 a12 в€’ a22 a22 = в€’a21 a12 + a22 a11 > 0.
So Eq. (iv) is always an ellipse by the theory of the linear algebra.
10
Section 9.2
In the following problems sketch the trajectory corresponding to the solution satisfying the
specified initial conditions, and indicate the direction of motion for increasing t.
1. dx/dt = в€’x, dy/dt = в€’3y; x(0) = 4, y(0) = 2.
3.dx/dt = в€’y, dy/dt = x; x(0) = 3, y(0) = 0 and x(0) = 0, y(0) = 3.
Answer:
1. Clearly, (0, 0) is a critical point. The characteristic equation and the eigenvalues, eigenvectors are
(О» + 1)(О» + 3) = 0. О»1 = в€’1, О»2 = в€’3
Therefore, (0, 0) is a node and asymptotically stable. It is easy to show that the two solutions
corresponding the given initial conditions are
x
y
=
4eв€’t
2eв€’2t
Then The trajectories are
y = x2 /8.
3. Clearly, (0, 0) is a critical point. The characteristic equation and the eigenvalues, eigenvectors are
(О» + i)(О» в€’ i) = 0. О»1 = i, О»2 = в€’i
Therefore, (0, 0) is a center and stable. It is easy to show that the two solutions corresponding
the given initial conditions are
x
y
=
3 cos t
sin t
and
x
y
=
в€’3sint
3 cos t
The trajectories are
x2 + y 2 = 9.
The pictures of the trajectories are omitted.
For each of the systems in the following problems:
(a) Find all critical points (equilibrium solutions).
(b) Use a computer to draw a direction field and phase portrait for the system.
(c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable,
stable, or unstable, and classify it as to type.
6. dx/dt = 1 + 2y, dy/dt = 1 в€’ 5x2 .
9. dx/dt = y(2 в€’ x в€’ y), dy/dt = в€’x в€’ y в€’ 2xy.
15.dx/dt = x(2 в€’ x в€’ y), dy/dt = в€’x + 3y в€’ 2xy Answer:
11
в€љ
в€љ
6. F = 1 + 2y = 0, G = 1 в€’ 5x2 = 0. The critical points are (В± 5/5, в€’1/2). (в€’ 5/5, в€’1/2)
в€љ
is saddle point and unstable. ( 5/5, в€’1/2) is center and stable.
в€љ
в€љ
9. F = y(2 − x − y) = 0, G = −x − y − 2xy = 0. The critical points are (0, 0), (1 ∓ 2, 1 ± 2).
в€љ
в€љ
в€љ
в€љ
(0, 0) is spiral and asymptotically stable. (1 в€’ 2, 1 + 2) and (1 + 2, 1 в€’ 2) are saddle points
and unstable.
15.F = x(2 в€’ x в€’ y), G = в€’x + 3y в€’ 2xy.The critical points are (0, 0),which is node and
unstable;(1, 1),which is a saddle point and unstable,(3, в€’1) which is a spiral point and asymptotically stable. Part (b) is omitted here.
In each of the following problems:
(a) Find an equation of the form H(x, y) = c satisfied by the trajectories.
(b) Plot several level curves of the function H. These are trajectories of the given system.
Indicate the direction of motion on each trajectory.
19.dx/dt = y, dy/dt = 2x + y.
24. Duffing’s equations :dx/dt = y, dy/dt = −x + (x3 /6).
Answer:
19.By observation,y = 2x + y = y + 2y,thus y = c1 e2t + c2 eв€’t .And substitute it into the
former equation,we get: x = 2c1 e2t в€’ c2 eв€’t.Thus we have H(x, y) = (x + y)(x в€’ 2y)2 = c
24.
в€’x + x3 /6
dy
=
dx
y
ydy = (в€’x + x3 /6)dx
1 2
1
1
y = в€’ x 2 + x4 + c
2
2
24
Thus
1
1
1
H(x, y) = x2 + y 2 в€’ x4 = c
2
2
24
25.Given that x = П†(t), y = П€(t) is a solution of the autonomous system
dx/dt = F (x, y), dy/dt = G(x, y)
for О± < t < ОІ,show that x = О¦(t) = П†(tв€’s), y = ОЁ(t) = П€(tв€’s) is a solution for О±+s < t < ОІ +s
for any real number s.
Proof: By the condition:
dφ(t)/dt = F (φ(t), ψ(t)), dψ(t) = G(φ(t), ψ(t)) on (α, β)
thus
dφ(t − s)/dt = F (φ(t − s), ψ(t − s)), dψ(t − s) = G(φ(t − s), ψ(t − s)) on (α + s, β + s)
12
i.e.
dΦ(t)/dt = F (Φ(t), Ψ(t)), dΨ(t) = G(Φ(t), Ψ(t)) on (α + s, β + s)
26. Prove that for the system
dx
dy
= F (x, y),
= G(x, y)
dt
dt
there is at most one trajectory passing through a given point (x0 , y0 ).
Hint: Let C0 be the trajectory generated by the solution x = П†0 (t), y = П€0 (t), with П†0 (t0 ) =
x0 , П€0 (t0 ) = y0 . and let C1 be the trajectory generated by the solution x = П†1 (t), y = П€1 (t),
with П†1 (t0 ) = x0 , П€1 (t0 ) = y0 . Use the fact that the system is autonomous and also the existence
and uniqueness theorem to show that C0 and C1 are the same.
Proof: Let C0 and C1 as in hint. WLOG, suppose that C0 is represented by y = H0 (x) and C1
is represented by y = H1 (x). Then
П€0 = H0 (П†0 ), П€1 = H1 (П†1 ), H0 (x0 ) = H1 (x0 ) = y0 .
Clearly, П€0 = H0 (П€0 ) and П€1 = H1 (П†1 ) satisfy the equations
G(φ0 , ψ0 )dφ0 − F (φ0 , ψ0 )dψ0 = 0,
G(φ1 , ψ1 )dφ1 − F (φ1 , ψ1 )dψ1 = 0,
and the initial conditions
П€0 = y0 as П†0 = x0 ,
П€1 = y0 as П†1 = x0 .
That is, H0 (x) and H1 (x) satisfy the same ODE
dH
F (x, H)
=
,
dx
G(x, H)
and the same initial condition H1 (x0 ) = H2 (x0 ) = y0 . According the existence and uniqueness
G(x, H)dx в€’ F (x, H)dH = 0,
theorem, H1 (x) = H2 (x). Therefore C1 = C2 .
or
13
Section 9.3
In each problems,verify that (0, 0) is a critical point,show that the system is locally linear,and
discuss the type and stability of the critical point (0, 0) by examing the corresponding linear
system.
1.dx/dt = x в€’ y 2 , dy/dt = x в€’ 2y + 2x2
3.dx/dt = (1 + x) sin y, dy/dt = 1 в€’ 4x в€’ cos y
Answer:
1.F (x, y) = x в€’ y 2 , G(x, y) = x в€’ 2y + 2x2 ,hence (0, 0) is a critical point.And F (x, y) and
G(x, y) have continuous partial derivatives at the origin,thus it’s locally linear,and we have:
d
dt
x
y
=
Fx (0, 0) Fy (0, 0)
Gx (0, 0) Gy (0, 0)
x
y
+
x1
y1
Where the coefficient matrix is
Fx (0, 0) Fy (0, 0)
Gx (0, 0) Gy (0, 0)
=
1 0
1 в€’2
Thus the eigenvalues are r1 = в€’2, r2 = 1,For both the linear and locally linear system ,(0, 0) is
an saddle point and unstable.
3.F (x, y) = (1 + x) sin y, G(x, y) = 1 в€’ 4x в€’ cos y,hence (0, 0) is a critical point.And F (x, y)
and G(x, y) have continuous partial derivatives at the origin,thus it’s locally linear,and we have:
d
dt
x
y
=
Fx (0, 0) Fy (0, 0)
Gx (0, 0) Gy (0, 0)
x
y
+
x1
y1
Where the coefficient matrix is
Fx (0, 0) Fy (0, 0)
Gx (0, 0) Gy (0, 0)
=
0 1
в€’4 0
Thus the eigenvalues are r1 = в€’2i, r2 = 2i,(0, 0) is a center and stable for the linear system and
center or spiral point and indeterminate for the locally linear system.
In each of problems below
(a)Determine all critical point of the given system of equations.
(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?
(d)Draw a phase portrait of the nonlinear system to confirm your conlusions,or to extend them
in those cases whter the linear system does not provide difinite information about the nonlinear
14
system.
5.dx/dt = (3 + x)(y в€’ x), dy/dt = (4 в€’ x)(y + x)
7.dx/dt = 1 в€’ 2y, dy/dt = x2 в€’ y 2
9.dx/dt = (3 + y)(y в€’ 0.5x), dy/dt = (2 в€’ x)(y + 0.5x)
11.dx/dt = 2x + y + xy 3 , dy/dt = x в€’ 2y в€’ xy
13.dx/dt = x в€’ 2y 2 , dy/dt = y в€’ 2x2
15.dx/dt = в€’2x в€’ y в€’ x(x2 + y 2 ), dy/dt = x в€’ y + y(x2 + y 2 )
17.dx/dt = 4 в€’ y 2 , dy/dt = (1 + x)(y в€’ x)
Answer:
5.F (x, y) = (3 + x)(y в€’ x), G(x, y) = (4 в€’ x)(y + x),they are all continuously differentiable.
Fx = в€’3 + y в€’ 2x,
Fy = 3 + xGx = 4 в€’ y в€’ 2x,
Gy = 4 в€’ x
Thus the system is locally nonlinear. For the criticla points:
F (x, y) = 0
в‡’
Gx, y = 0
x = в€’3
or
y=3
x=4
or
y=4
x=0
y=0
At (в€’3, 3),linear system is
d
u
6 0
u
=
v
7
7
v
dt
Thus eigenvalues are r1 = 6, r2 = 7,so for both linear and locally linear system,(в€’3, 3) is a node
and unstable.
At (4, 4),linear system is
d
u
в€’7 7
u
=
в€’8 0
v
dt v
Thus eigenvalues are r1 , r2 < 0,so for both linear and locally linear system,(4, 4) is a node and
asymptotically stable.
At (0, 0),linear system is
d
u
в€’7 7
u
=
в€’8 0
v
dt v
Thus eigenvalues are r1 < 0 < r2 ,so for both linear and locally linear system,(0, 0) is a saddle
point and unstable.
7.F (x, y) = 1 в€’ 2y, G(x, y) = x2 в€’ y 2 ,they are all continuously differentiable.
Fx = 0,
Fy = в€’2Gx = 2x,
Gy = в€’2y
Thus the system is locally nonlinear. For the criticla points:
F (x, y) = 0
в‡’
Gx, y = 0
x = в€’ 21
or
y = 12
x=
y=
At (в€’ 21 , 12 ),linear system is
d
dt
u
v
=
0 в€’2
в€’1 в€’1
u
v
1
2
1
2
15
Thus eigenvalues are r1 = в€’2, r2 = 1,so for both linear and locally linear system,(в€’ 12 , 12 ) is a
saddle point and unstable.
At ( 12 , 12 ),linear system is
d
dt
u
v
в€љ
7
i,so
2
Thus eigenvalues are r = в€’ 12 В±
=
0 в€’2
1 в€’1
u
v
for both linear and locally linear system,( 12 , 12 ) is a spiral
point and asymptotically stable.
9.F (x, y) = (3 + y)(y в€’ 0.5x), G(x, y) = (2 в€’ x)(y + 0.5x),they are all continuously differentiable.
Fx = в€’1.5 в€’ 0.5y,
Fy = 3 + 2y в€’ 0.5xGx = 1 в€’ y в€’ x,
Gy = 2 в€’ x
Thus the system is locally nonlinear. For the criticla points:
F (x, y) = 0
в‡’
Gx, y = 0
x=2
or
y = в€’3
x=6
or
y = в€’3
x=0
or
y=0
x=2
y=1
At (0, 0),linear system is
d
dt
Thus eigenvalues are r1 =
point and unstable.
1
4
u
v
=
в€’ 32 3
1 2
u
v
в€љ
97
,so
4
В±
for both linear and locally linear system,(0, 0) is a saddle
At (2, 1),linear system is
d
u
в€’2 4
u
=
в€’2 0
v
dt v
в€љ
Thus eigenvalues are r = в€’1 В± 7i < 0,so for both linear and locally linear system,(2, 1) is a
spiral point and asymptotically stable.
At (2, в€’3),linear system is
d
u
0 в€’4
u
=
2 0
v
dt v
в€љ
Thus eigenvalues are r = В±2 2i,so for liner system,(0, 0) is a center point and stable and center
or spiral point and indeterminate for locally linear system.
At (6, в€’3),linear system is
d
dt
u
v
=
0 в€’6
в€’2 в€’4
u
v
Thus eigenvalues are r = в€’6, 2,so for both linear and locally linear system,(6, в€’3) is a saddle
point and unstable.
11.F (x, y) = 2x + y + xy 3 , G(x, y) = x в€’ 2y в€’ xy,they are all continuously differentiable.
Fx = 2 + y 3 ,
Fy = 1 + 3xy 2 Gx = 1 в€’ y,
Gy = в€’2 в€’ x
16
Thus the system is locally nonlinear. For the criticla points:
F (x, y) = 0
в‡’
Gx, y = 0
x=0
or
y=0
x = в€’1.1935
y = в€’4797
At (0, 0),linear system is
d
dt
u
v
=
2 1
1 в€’2
u
v
в€љ
Thus eigenvalues are r1 = В± 5,so for both linear and locally linear system,(0, 0) is a saddle
point and unstable.
At (в€’1.1935, в€’1.4797),linear system is
d
dt
u
v
в€’1.2399 в€’6.8393
2.4797 в€’0.80655
=
u
v
Thus eigenvalues are r = в€’1.0232В±4.1125i,so for both linear and locally linear system,(в€’1.1935, в€’1.4797)
is a spiral point and asymptotically stable.
13.F (x, y) = x в€’ 2y 2 , G(x, y) = y в€’ 2x2 ,they are all continuously differentiable.
Fx = 1,
Fy = в€’4yGx = в€’4x,
Gy = 1
Thus the system is locally nonlinear. For the criticla points:
F (x, y) = 0
в‡’
Gx, y = 0
x=0
or
y=0
x=
y=
1
2
1
2
At ( 21 , 12 ),linear system is
d
dt
u
v
=
1 в€’2
в€’2 1
u
v
Thus eigenvalues are r = в€’1, 3,so for both linear and locally linear system,( 12 , 12 ) is a saddle point
and unstable.
At (0, 0),linear system is
d
u
1 0
u
=
0 1
v
dt v
Thus eigenvalues are r = 1, 1,so for both linear and locally linear system,(0, 0) is a node or spiral
point and unstable.
15.F (x, y) = в€’2x в€’ y в€’ x(x2 + y 2 ), G(x, y) = x в€’ y + y(x2 + y 2 ),they are all continuously
differentiable.
Fx = в€’2 в€’ x2 + y 2 ,
Fy = в€’1 в€’ 2xyGx = 1 + 2xy,
Gy = в€’1 + x2 + 3y 2
Thus the system is locally nonlinear. For the criticla points:
F (x, y) = 0
в‡’
Gx, y = 0
x=0
or
y=0
x = в€’0.33076
or
y = 1.0924
x = 0.33076
y = в€’1.0924
17
At (0, 0),linear system is
d
dt
Thus eigenvalues are r = (в€’3 В±
u
v
=
в€’2 в€’1
1 в€’1
u
v
в€љ
3i)/2,so for both linear and locally linear system,(0, 0) is a
saddle point and asymptotically stable.
At (в€’0.33076, 1.0924) and (0.33076, в€’1.0924),linear system is
d
dt
u
v
в€’3.5216 в€’0.27735
0.27735
2.6895
=
u
v
Thus eigenvalues are r = в€’3.5092, 2.6771,so for both linear and locally linear system,they are
saddle points and unstable.
17.F (x, y) = 4 в€’ y 2 , G(x, y) = (1 + x)(y в€’ x),they are all continuously differentiable.
Fx = 0,
Fy = в€’2yGx = 1 + y в€’ 2x,
Gy = 1 + x
Thus the system is locally nonlinear. For the criticla points:
F (x, y) = 0
в‡’
Gx, y = 0
x=2
or
y=2
x = в€’2
or
y = в€’2
x = в€’1
or
y=2
0 в€’4
в€’3 3
u
v
x = в€’1
y = в€’2
At (2, 2),linear system is
d
dt
Thus eigenvalues are r1 =
point and unstable.
3
2
u
v
=
в€љ
В±
57
,so
2
for both linear and locally linear system,(2, 2) is a saddle
d
dt
u
v
At (в€’2, в€’2),linear system is
Thus eigenvalues are r = в€’ 12 В±
=
0 4
1 в€’1
u
v
в€љ
7
2
< 0,so for both linear and locally linear system,(в€’2, в€’2) is a
saddle point and unstable.
At (в€’1, 2),linear system is
d
dt
u
v
=
d
dt
u
v
=
0 в€’4
3 0
u
v
0 4
в€’1 0
u
v
в€љ
Thus eigenvalues are r = В±2 3i,so for liner system,(в€’1, 2) is a center point and stable and
center or spiral point and indeterminate for locally linear system.
At (в€’1, в€’2),linear system is
Thus eigenvalues are r = В±2i,so for both linear and locally linear system,(в€’1, в€’2) is a center
point and stable and center or spiral point and indeterminate for locally linear system.
18
20.Consider the autonomous system
dx/dt = x, dy/dt = в€’2y + x3 .
(a)Show that the critical point (0, 0) is a saddle point.
(b)Sketch the trajectories for the corresponding linear system, and show that the trajectory for
which x в†’ 0, y в†’ 0 as t в†’ в€ћ is given by x = 0.
(c)Determine the trajectories for the nonlinear system x = 0 by integrating the equation for
dy/dx.Show that the trajectory corresponding to x = 0 for the linear system is unaltered,but
that the one corresponding to y = 0 is y = x3 /5.Sketch several of the trajectories for the nonlinear
system.
Answer:
(a) F (x, y) = x, G(x, y) = в€’2y + x3 ,continuous and differentiable,and Fx = 1, Fy = 0, Gx =
1 0
3x2 , Gy = в€’2,thus the coefficient matrix at (0, 0) is
and the eigenvalues are r =
0 в€’2
в€’2, 1,thus (0, 0) is a saddle point and unstable.
1
0
(b)The general solution for the linear sytem is x = c1
e t + c2
eв€’2t ,thus x(t) =
0
1
c1 et , y(t) = c2 eв€’2t thus the only trajectory for which x в†’ 0, y в†’ 0 as t в†’ в€ћ is given by x = 0.
(c)For x = 0,x = c1 et where c1 = 0,substitute it into the equation of y,
y=
c21 c2 x3
+
x2
5
And for the linear system,x = c1 et , y = c2 eв€’2t ,thus the trajectory is x2 y = C,thus when x = 0,the
3
trajectory is unaltered. For y = 0,c2 = 0 thus y = x5
25.This problem extends Problem 24 to a damped pendulu.The equations of motion are
dx/dt,
dy/dt = в€’4 sin x в€’ Оіy,
where Оі is the damping coefficient,with the initial conditions x(0) = 0, y(0) = v.
(a)For Оі = 1/4,plot x versus t for v = 2 and for v = 5.Explain these plots in terms of the motions
of the pendulum that they represent.Also explain how they are related to the corresponding
graphs in Problem 24(a).
(b)Estimate the critical value vc of the initial velocity where the transition from one type of
motion to the other occurs.
(c)Represent part (b) for other values of Оі and determine how vc depends on Оі.
Answer:
It involves the coding process of MATLAB or other computer assistance software.Solve it
yourselves if you are interested.
19
26.Theorem 9.3.2 provides no information about the stability of a critical point of a locally linear
system if that point is a center of the corresponding linear system.That this must be the case is
illustrated by the systems
dx/dt = y + x(x2 + y 2 )
dy/dt = в€’x + y(x2 + y 2 )
(i)
dx/dt = y в€’ x(x2 + y 2 )
dy/dt = в€’x в€’ y(x2 + y 2 )
(ii)
(a)Show that (0, 0) is a critical point of each system and,furthermore,is a center of the corresponding linear system.
(b)Show that each system.
(c)Let r2 = x2 + y 2 ,and note that xdx/dt + ydy/dt = rdr/dt.For system (ii),show that dt/dt < 0
and that r в†’ 0 as t в†’;hence the critical point is asymptotically stable.For system (i),show
that the solution fo the initial value problem for r with r = r0 at t = 0 becomes unbounded as
t в†’ 1/2r02 ,and hence the critical point is unstable.
Answer:
Let F1 (x, y) = y + x(x2 + y 2 ), G1 (x, y) = в€’x + y(x2 + y 2 ), F2 (x, y) = y в€’ x(x2 + y 2 ), G2 (x, y) =
в€’x в€’ y(x2 + y 2 )
(a)Let F1 (x, y) = 0, G1 (x, y) = 0,thus (0, 0) is the only critical point of the first system.Let
F2 (x, y) = 0, G2 (x, y) = 0,thus (0, 0) is the only critical point of the second system.
A1 =
∂x F1 ∂y F1
∂x G1 ∂y G1
3x2 + y 2 1 + 2xy
в€’1 + 2xy x2 + 3y 2
=
=
0 1
в€’1 0
The eigenvalues of A1 are r = В±i.Hence (0, 0) is a center of the corresponding linear system.
A2 =
∂x F2 ∂y F2
∂x G2 ∂y G2
=
в€’3x2 в€’ y 2 1 в€’ 2xy
в€’1 в€’ 2xy в€’x2 в€’ 3y 2
=
0 1
в€’1 0
The eigenvalues of A2 are r = В±i.Hence (0, 0) is a center of the corresponding linear system.
(b)Since (0, 0) is isolated critical point of each system and F1 , F2 , G1 , G2 are all at least twice
differentialble,each system is locally linear.
dt
(c)Let r2 = x2 + y 2 ,then r dt
= x dx
+ y dy
.For system (i),
dt
dt
rdr = x[y + x(x2 + y 2 )] + y[в€’x + y(x2 + y 2 )]
= (x2 + y 2 )2
= r4
Thus
dr
dt
= r3 ,together with the initial condition:r = в€љ
r0
.If
1в€’2r02 t
t→
1
,r
2r02
For system (ii)
r
dr
= x[y в€’ x(x62 + y 2 )] + y[в€’x в€’ y(x2 + y 2 )]
dt
= в€’(x2 + y 2 )2
= в€’r4
Thus
dr
dt
= в€’r3 ,so the system is asymptotically stable.
в†’ +в€ћ,it is unstable.
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