1. No category

Sample Solutions of Assignment 3 for MAT3270B:
2.8,2.3,2.5,2.7
1. Transform the given initial problem into an equivalent problem with
the initial point at the origin
(a).
(b).
dy
dt
dy
dt
= t2 + y 2 , y(1) = 2,
= 1 − y 3 , y(−1) = 3
Answer: (a)Let t = s + 1, y = w + 2, then
dy
=1
dw
dt
=1
ds
the original problem can be written as
dw
= (s + 1)2 + (w + 2), w(0) = 0.
ds
(b)Let t = s − 1, y = w + 3, then
dy
=1
dw
dt
=1
ds
the original problem can be written as
dw
= 1 + (w + 3)2 , w(0) = 0.
ds
2. Use the method of successive approximations to solve the given
initial value problem:(1) Determine φn (t); (2) Find the limit of φn
1
2
0
(a). y = 2(y + 1), y(0) = 0
0
(b). y = y + 1 − t, y(0) = 0
Answer:
(a) If y = φ(t), then the corresponding integral equation is
Z t
φ(t) =
2(φ(s) + 1)ds
0
If the initial approximation is φ0 (t) = 0, then
Z t
φ1 (t) =
2ds = 2t
0
Z
t
φ2 (t) =
2(2t + 1)ds = 2t + 2t2
0
and
Z
t
φn (t) =
2(φn−1 (s) + 1)ds =
0
n
X
2 k tk
k!
1
lim φn (t) = e2t − 1
n→∞
(b) If y = φ(t), then the corresponding integral equation is
Z t
φ(t) =
(φ(s) + 1 − s)ds
0
If the initial approximation is φ0 (t) = 0, then
Z t
1
(1 − s)ds = t − t
φ1 (t) =
2!
0
Z t
1
1
φ2 (t) =
(1 − s)ds = t − t3
2
3!
0
and
Z
t
φn (t) =
(φn−1 (s) + 1 − s)ds = t −
0
tn+1
(n + 1)!
lim φn (t) = t
n→∞
3. A mass of 0.25 kg is dropped from rest in a medium offering a
resistance of 0.2|v|, where v is measured in m/sec.
3
(a)If the mass is dropped from a height of 30m, find its velocity when
it hits the ground.
(b)If the mass is to attain a velocity of no more than 10m/sec, find the
maximum height from which it can dropped.
(c)Suppose that the resistance force is k|v|, where v is measured in
m/sec and k is a constant. If the mass is dropped from a height of
30m and must hit the ground with velocity of no more than 10m/sec,
determine the coefficient of the resistance k that is required.
Answer: Denote the mass m, the height of the mass x(t),the velocity
v = x0 , the resistance force k|v|, , the initial height h, then, x0 (0) =
0, x(0) = h. By Newton’s 2nd Law, we have
mx00 = −mg − kx0
x0 (0)
=0
x(0)
=h
Hence
k 0
x
m
x00 +
⇒
k
= −g
k 0
x)
m
e m t (x00 +
k
= −ge m t
k
k
⇒
(e m t x0 )0 = −ge m t
⇒
e m t x0 = − m
g(e m t − 1)
k
⇒
x0 = − mg
(1 − e− m t )
k
k
k
k
⇒ x=h−
mg
(t
k
+
k
m −m
(e t
k
− 1))
Here g ∼
= 9.8m/s2 .
(a). From h = 30, i.e.
mg
(t
k
k
m −m
(e t
k
k
−m
t
+
3.6305s.Then v = x0 = − mg
(1 − e
k
− 1)) = 30, we have t ∼
=
) = −11.5789m/s.
4
(b).
|x0 | ≤ 10
⇒
mg
(1
k
k
− e− m t ) ≤ 10
⇒
t ≤ 2.11824s
⇒
h ≤ 13.4485m
(c). According to the problem,


x =



 v =
x =



h =


v =
k
h − mg
(t + m
(e− m t − 1))
k
k
k
x0 = − mg
(1 − e− m t )
k
0
30
−10
which gives t = 3.9523s and k = 0.239438kg/s. Hence the coefficient
of the resistance k must be 0.239438 kg/s.
4. Find the escape velocity for a body projected upward with an initial velocity v0 from a point x0 = ξR above the surface of the earth,
where R is the radius of the earth and ξ is a constant. Neglect the
air resistance, find the initial altitude from which the body must be
launched in order to reduce the escape velocity to 85/100 of its value
at the earth’s surface.
Answer: Denote the mass of the body m, by Newton 2nd Law, we
have

m
 mx00 = GM
x2
0
x (0) = v0

x(0) = ξR
5
Here
GM
R2
= g, i.e. GM = gR2 .Then
mx00 =
GM m
x2
⇒
x00 = −GM x−2
⇒
2x0 x00 = −2GM x−2 x0
⇒ (x0 )2 − v02 = 2GM ( x1 −
1
)
ξR
Let x → ∞ which means the body can escape, we have
(x0 )2 − v02 = − 2GM
ξR
⇒ v02 = (x0 )2 +
2GM
ξR
≥
2GM
ξR
q
Hence the escape velocity on the earth surface is
2GM
.
R
Assume the
escape velocity reduce 85% of its value on the earth surface, i.e.
2GM
ξR
⇒
85 2 2GM
= ( 100
) R
ξ=
400
R
289
400
i.e. the initial altitude must be ( 289
− 1)R =
111
R.
289
5. Suppose that a certain population has a growth rate that varies with
time and that this population satisfies the differential equation
dy
= (0.5 + sin (kt))y/N.
dt
If y(0) = y0 , find the time τ at which the population has doubled.
Suppose y0 = 1, k = 2π, N = 5 estimate τ .
Answer: The original equation can be written as
1
dy
= (0.5 + sin (kt))dt
y
N
then we get
1
1
y = ce N (0.5t− k cos (kt))
6
1
and c = y0 e N k by y(0) = y0 .
Assuming y(τ ) = 2y0 , and substituting y0 = 1, k = 2π, N = 5 to the
above result, then
−1
πτ − cos 2πτ = 10π log (2e 10π )
Hence the τ ≈ 7.2500
6. In the following problems, sketch the graph of f (y) versus y, determine the critical points and classify each one as asymptotically stable
or unstable.
(a).
(b).
(c).
dy
dt
dy
dt
dy
dt
= y(y − 1)(y − 2), y0 ≥ 0
= (y − 1)(ey − 1), −∞ < y0 < +∞
√
= ay − b y, a > 0, b > 0, y0 ≥ 0
Answer:
(To be continued)
7
1.5
1
0.5
-1
1
3
2
-0.5
-1
t
-1.5
Figure 1. for problem 6
f (y) = y(y − 1)(y − 2)
2.5
2
1.5
1
0.5
-1
-0.5
0.5
1
1.5
2
b
Figure 2. for problem 6
f (y) = (y − 1)(ey − 1);
(a). From the figure of f (y) = y(y − 1)(y − 2), there are 3 critical
points y = 0, y = 1, y = 2 and y = 1 is stable, y = 0, y = 2 are
unstable.
(b). From the figure of f (y) = (y − 1)(ey − 1), there are 2 critical
points y = 0, y = 1 and y = 0 is stable, y = 1 are unstable.
8
4
3.5
3
2.5
2
1.5
1
0.5
0
−0.5
−1
0
1
2
3
4
5
6
7
8
9
10
Figure 3. for problem 6
√
f (y) = ay − b y.
√
(c). From the figure of f (y) = ay − b y, there are 2 critical points
y = 0, y =
b2
a2
and y = 0 is stable, y =
b2
a2
is unstable.
7. Solve the Gompertz equation
dy
K
= ry log ( ), y(0) = y0 > 0
dt
y
Answer: The Original ODE can be rewritten as
y0
= r ln K − r ln y
y
9
Let w = ln y, we have
w0 + rw = r ln K
⇒
(ert w)0 = rert ln K
⇒
ert w − w0 = (ert − 1) ln K
⇒ w = (1 − e−rt ) ln K + w0 e−rt
y = y0e
⇒
−rt
K 1−e
−rt
8. Consider the following Schaefer’s model in population dynamics:
dy
y
= r(1 − )y − Ey
dt
K
Suppose E < r. Find the equilibrium points and state if they are stable
or unstable.
Answer: Let f (y) = r(1 −
0
y
)y
K
− Ey, then
f (y) = r(1 −
y
y
)−r −E
K
K
Setting f (y) = 0, then y = 0, or y = k(1 − Er )
0
At the point y = 0, f (0) = r − E > 0, this is an unstable equilibrium
point.
0
At the point y = k(1− Er ), f (0) = E −r < 0, this is a stable equilibrium
point.
9. Consider the following bifurcation equation
dy
= ²x − x3
dt
10
Show that for ² < 0, there exists only one critical point which is asymptotically stable; while for ² > 0, there are three critical points, of which
one is unstable and the other two are stable.
Answer: (a).If ² < 0, then ²x − x3 = −x(x2 + |²|) = 0 has only one
root,so there exist only one critical point x = 0. Since x > 0, ²x−x3 < 0
and x < 0, ²x − x3 > 0, the critical point x = 0 is stable.
√
√
(b). If ² > 0, then ²x − x3 = −x(x − ²)(x − ²) = 0 has 3 root,so
√
√
there exist 3 critical points − ², 0, ².
√
√
Since x < − ², ²x − x3 > 0 and − ² < x < 0, ²x − x3 < 0, the critical
√
point x = − ² is stable.
√
√
Since − ² < x < 0, ²x − x3 < 0 and 0 < x < ², ²x − x3 > 0, the
critical point x = 0 is unstable.
√
√
Since x > ², ²x − x3 < 0 and 0 < x < ², ²x − x3 > 0, the critical
√
point x = ² is stable.
√
i.e. the critical point x = 0 is unstable and the critical points x = ± ²
is stable.
10. Solve the following Chemical Reactions equations
dx
= α(p − x)(q − x), x(0) = x0
dt
Answer: By separate variable method,
dx
= αdt
(p − x)(q − x)
Case 1: p = q
dx
(x−p)2
⇒
1
x0 −p
−
= αdt
1
x−p
⇒ x=p+
= αdt
x0 −p
α(x0 −p)t+1
11
Case 2: p 6= q
dx
(p−x)(q−x)
⇒
1
x−p
−
1
x−q
∗
⇒ ln( x−p
x−q
x−p
x−q
⇒
⇒
x=
= (p − q)αdt
x0 −q
)
x0 −p
=
= αdt
= (p − q)αt
x0 −p (p−q)αt
e
x0 −q
qαe(p−q)αt (p−x0 )−p(q−x0 )
αe(p−q)αt (p−x0 )−(q−x0 )
11. Use Euler’s method to find approximate values of the solution of
the given initial value problem at t = 0.5, 1, 1.5, 2, 2.5, 3 with h = 0.1
√
0
(a). y = 5 − 3 y, y(0) = 2
0
(b). y = −ty + 0.1y 3 , y(0) = 1
Answer:
(a) Setting
√
f (t, y) = 5 − 3 y
t0 = 0, y0 = 2, f0 = f (0, 2) = 0.757
y1 = y0 + f0 h = 2 + 0.1 ∗ 0.757 = 2.0757
Setting fn = f (tn , yn ), yn+1 = yn + 0.1fn . Hence, we get the following
results:
y(0.5) = y5 = 2.30800
y(1) = y10 = 2.49006
y(1.5) = y15 = 2.60023
y(2) = y20 = 2.66773
y(2.5) = y25 = 2.70939
y(3) = y30 = 2.73521
12
(b) Setting
f (t, y) = −ty + 0.1y 3
t0 = 0, y0 = 1, f0 = f (0, 2) = 0.1
y1 = y0 + f0 h = 1 + 0.1 ∗ 0.1 = 1.01
Setting fn = f (tn , yn ), yn+1 = yn + 0.1fn . Hence, we get the following
results:
y(0.5) = y5 = 0.950517
y(1) = y10 = 0.687550
y(1.5) = y15 = 0.369188
y(2) = y20 = 0.145990
y(2.5) = y25 = 0.0421429
y(3) = y30 = 0.00872877