Diff. Eqns.
Problem Set 10
Solutions
1. Solve the initial-value problem
7
9
x0 =
x,
−6 −8
x (0) =
4
−3
(1)
Describe the behavior of the solution as t → ∞.
We start by assuming solutions of the form zert and solve the system
7−r
9
z1
0
=
−6 −8 − r
z2
0
(2)
for eigenvalues r and eigenvectors z. For (2) to hold, the matrix must be singular (zero
determinant).
7−r
9
(3)
−6 −8 − r = (7 − r) (−8 − r) − (−6) (9)
= r2 + r − 2
(4)
= (r + 2) (r − 1)
⇒ r = 1, −2.
(5)
r1 = 1:
6
9 0
−6 −9 0
⇒ 6z1 + 9z2 = 0 ⇒ z1 =
3
−2
1
−1
.
(6)
.
(7)
r2 = −2:
9
9 0
−6 −6 0
⇒ 9z1 + 9z2 = 0 ⇒ z2 =
Thus, the general solution is
x = c1
3
−2
t
e + c2
1
−1
e−2t .
Lastly we apply the initial condition.
4
3
1
3
1
c1
= x (0) = c1
+ c2
=
−3
−2
−1
−2 −1
c2
2 +
1 ·2/3 3
3
1 4
1
=⇒
−2 −1 −3
0 −1/3
4
−1/3 =⇒ c1 = c2 = 1.
So the solution for the initial value problem is
3
1
t
x=
e +
e−2t .
−2
−1
(8)
(9)
(10)
(11)
lim x = lim
t→∞
t→∞
3
−2
t
e +
1
−1
−2t
e
=
∞
−∞
(12)
2. Consider the system
0
x =
1 −1
5 −3
x.
(13)
Find the general solution of the system and express this solution in terms of
real-valued functions.
We start by assuming solutions of the form zert and solve the system
1−r
−1
z1
0
=
5
−3 − r
z2
0
(14)
for eigenvalues r and eigenvectors z. For (14) to hold, the matrix must be singular (zero
determinant).
1−r
−1 = (1 − r) (−3 − r) − (−1) (5)
(15)
5
−3 − r = r2 + 2r + 2 = 0
r1 = −1 + i:
2−i
−1
5
−2 − i
0 1 ∗(2+i) 5 −2 − i
=⇒
0
5 −2 − i
z1 =
1
2−i
⇒ r = −1 ± i
0
0 ⇒ 5z1 − (2 + i) z2 = 0
(16)
(17)
(18)
Note that we only need one eigenvalue/eigenvector pair, but if we did need z2 , it would be
z2 = z1 , since r2 = r1 . Since we have r1 and z1 , then one solution is
1
x1 =
e(−1+i)t
(19)
2−i
1
=
e−t (cos t + i sin t)
(20)
2−i
cos t
sin t
= e−t
+ ie−t
(21)
2 cos t + sin t
2 sin t − cos t
Thus our general solution is
−t
x = c1 e
cos t
2 cos t + sin t
−t
+ c2 e
sin t
2 sin t − cos t
(22)
3. Consider the mass-spring-damper equation
mu00 + γu0 + ku = 0
(23)
(a) Let x1 = u and x2 = u0 and write the second-order equation for u(t) as a
first-order system for x(t).
First we may note that x01 = x2 . Also, using these substitutions, (23) becomes
mx02 + γx2 + kx1 = 0
x02 = −
(24)
γ
k
x2 − x1 .
m
m
(25)
Using these two equations, we can form a system for x = [x1 , x2 ]T .
0
1
x0 =
x.
−k/m −γ/m
(26)
(b) Let m = 1 and γ = 4 and find the eigenvalues of the coefficient matrix in
part (a). How do the eigenvalues depend on the spring constant k > 0? Find a
critical value of k where the eigenvalues change type.
If m = 1 and γ = 4 then the coefficient matrix becomes
0
1
.
−k −4
(27)
This matrix has the following eigenvalues:
−λ
1
2
−k −4 − λ = −λ (−4 − λ) + k = λ + 4λ + k = 0
λ = −2 ±
(28)
√
4 − k.
(29)
If k < 4, the eigenvalues are real. If k > 4, the eigenvalues are imaginary.
(c) Sketch the behavior of solutions in the phase plane for the cases
(See Table 9.1.1, pg. 494 in the textbook).
(i) m = 1, γ = 4, k = 3
Eigenvalues are −1 and −3, both negative. The sketch should show an asymptotically stable node at (0, 0) (solution curves into the origin). Below is the solution with x(0) = (0, 2).
(ii) m = 1, γ = 4, k = 5
Eigenvalues are −2 ± i. These are complex with a negative real part, so there is an asymptotically stable spiral point at (0, 0) (solution spirals into the origin). Below is the solution
with x(0) = (0, 2).