Download Homework #9 - UC Davis Mathematics

Prof. Hunter 22B Homework 8 Solutions
Steffen Docken and Paige Buck-Moyer
December 10, 2014
7.4
1. Prove the generalization of Theorem 7.4.1 as expressed in the sentence that includes Eq. (8), for
any arbitrary value of the integer k.
Ans: We are proving that if x(1) , . . . , x(k) are solutions of Eq. (3), then
x = c1 x(1) (t) + · · · + ck x(k) (t)
is also a solution for any constants c1 , . . . , ck . Theorem 7.471 gives us the base case for k = 2.
Now, assume that for a fixed j < k,
y = c1 x(1) (t) + · · · + cj x(j) (t)
is also a solution for any constants c1 , . . . , cj . Then by Theorem 7.4.1,
z = cy y(t) + cj+1 x(j+1) (t) = cy c1 x(1) (t) + · · · + cy cj x(j) (t) + cj+1 x(j+1) (t)
is also a solution. Therefore, by induction, if x(1) , . . . , x(k) are solutions of Eq. (3), then
x = c1 x(1) (t) + · · · + ck x(k) (t)
is also a solution for any constants c1 , . . . , ck .
4. If x1 = y and x2 = y 0 , then the second order equation
y 00 + p(t)y 0 + q(t)y = 0
corresponds to the system
x01 = x2
x02 = −q(t)x1 − p(t)x2
Show that if x(1) and x(2) are a fundamental set of solutions of the system of first order ODEs, and
if y (1) and y (2) are a fundamental set of solutions of the second order ODE, then W [y (1) , y (2) ] =
W [x(1) , x(2) ], where c is a nonzero constant.
Ans: If y (1) and y (2) are a fundamental set of solutions of the second order ODE, then
dy (1) (2)
dy (2)
−
y
dt
dt
and x(2) are a fundamental set of solutions of the system of first order ODEs, then
x (t) x12 (t) = x11 (t)x22 (t) − x12 (t)x21 (t)
W [x(1) , x(2) ] = 11
x21 (t) x22 (t) W [y (1) , y (2) ] = y (1)
And if x(1)
Now, since y = x1 and y 0 = x2 , y (1) and y (2) are linear combinations of x11 and x12 and similar
(1)
(2)
for dydt , dydt . So, we can write
y (1) = c1 x11 +c2 x12 ,
y (2) = c3 x11 +c4 x12 ,
dy (1)
= c1 x21 +c2 x22 ,
dt
and
dy (2)
= c3 x21 +c4 x22
dt
Plugging this in gives
W [y (1) , y (2) ] = (c1 x11 + c2 x12 )(c3 x21 + c4 x22 ) − (c1 x21 + c2 x22 )(c3 x11 + c4 x12 )
= c1 x11 c3 x21 + c1 x11 c4 x22 + c2 x12 c3 x21 + c2 x12 c4 x22 −
(c1 x21 c3 x11 + c1 x21 c4 x12 + c2 x22 c3 x11 + c2 x22 c4 x12 )
= (c1 c4 − c2 c3 )x11 x22 + (c2 c3 − c1 c4 )x12 x21
= (c1 c4 − c2 c3 )(x11 x22 − x12 x21 )
= (c1 c4 − c2 c3 )W [x(1) , x(2) ]
= cW [x(1) , x(2) ]
1
7.5 27. The eigenvalues and eigenvectors of a matrix A are
1
r1 = 1, ξ (1) =
;
r2 = 2,
2
ξ (2) =
1
−2
Consider the corresponding system x0 = Ax.
(a) Sketch a phase portrait of the system.
(b) Sketch the trajectory passing through the initial point (2, 3).
(c) For the trajectory in part (b), sketch the graphs of x1 versus t and of x2 versus t on the same
set of axes.
Ans: All graphs are on the next page.
2
3
29. Consider the equation
ay 00 + by 0 + cy = 0,
where a, b, and c are constants with a 6= 0. In Chapter 3 it was shown that the general solution
depended on the roots of the characteristic equation
ar2 + br + c = 0.
(a) Transform the second order ODE into a system of first order equations
by letting x1 = y,
x
1
x2 = y 0 . Find the system of equations x0 = Ax satisfied by x =
.
x2
(b) Find the equation that determines the eigenvalues of the coefficient matrix A in part (a).
Note that this equation is just the characteristic equation of the second order ODE.
Ans:
(a) First we set x1 = y and x2 = y 0 then we manipulate the equation ay 00 + by 0 + cy = 0 with
a 6= 0 to get y 00 = − ab y 0 − ac y replacing y with x1 , y 0 with x2 and y 00 with x02 gives
x1 = x2
c
b
x002 = − x2 − x1
a
a
Putting this in matrix form gives
0 x1
0
=
x2
− ac
1
− ac
x1
x2
(b)
A − λI =
−λ
− ac
1 − ab − λ
c
b
c
b
det(A − λI) = −λ(− − λ) + = λ2 + λ +
a
a
a
a
Using the quadratic formula to find the values for λ results in
q b
b 2
−a ±
− 4(1) ac
a
λ=
2
q
b2
c
− ab ±
a2 − 4 a
λ=
2
q
b2
4ac
− ab ±
a2 − a2
λ=
2
q
b2 −4ac
− ab ±
a2
λ=
2
λ=
− ab ±
√
b2 −4ac
a
2
√
−b ± b2 − 4ac
λ=
2a
This is exactly the solution to the characteristic equation.
4
7.6
9. Find the solution to
x0 =
−5
−3
1
1
x,
x(0) =
1
1
Describe the behavior of the solution as t → ∞
Ans: To solve the initial value problem, we must first find the eigenvalues of the matrix. So, we
have
1−λ
−5 = (1 − λ)(−3 − λ) + 5 = λ2 + 2λ + 2 = 0
1
−3 − λ Therefore,
√
Which gives (2−i)x1 −5x2 = 0 or x2 =
Therefore, let x1 = 1, so x2 = 2−i
5 and
4−8
√
−4
= −1 ± i
2
2
For the eigenvalue corresponding to λ1 = −1 + i, we need to solve for x1 and x2 , such that
2−i
−5
x1
0
=
1
−2 − i
x2
0
λ=
−2 ±
2−i
5 x1
x
= −1 ±
since the two equations in the system are equivalent.
(1)
=
1
2−i
5
is our first eigenvector. For the eigenvalue corresponding to λ1 = −1 − i, we need to solve for x1
and x2 , such that
2+i
−5
x1
0
=
1
−2 + i
x2
0
Which gives (2+1)x1 −5x2 = 0 or x2 =
Therefore, let x1 = 1, so x2 = 2+i
5 and
2+i
5 x1
since the two equations in the system are equivalent.
x(2) =
1
2+i
5
is our second eigenvector. So, the general solution is
1
(−1+i)t
(−1−i)t
x = c1 e
+ c2 e
2−i
5
Using the initial conditions, we have
1
= c1
1
1
2−i
5
+ c2
1
1
2+i
5
2+i
5
Therefore, c1 = 1 − c2 , and plugging this into the second equation gives
2+i
2−i
+ c2
5
5
2−i
2+i 2−i
1=
+ c2
−
5
5
5
3+i
2i
= c2
5
5
3+i
= c2
2i
1 − 3i
= c2
2
1 = (1 − c2 )
5
and c1 =
1+3i
2
So,
x=
1 + 3i (−1+i)t
e
2
1
2−i
5
+
1 − 3i (−1−i)t
e
2
1
2+i
5
And, since the real component of both eigenvalues is negative, the solution will spiral in towards
0 as t → ∞.
28. A mass m on a spring with constant k satisfies the differential equation
mu00 + ku = 0,
where u(t) is the displacement at time t of the mass from its equilibrium position.
(a) Let x1 = u, x2 = u0 , and show that the resulting system is
0
1
0
x =
x.
−k/m 0
(b) Find the eigenvalues of the matrix for the system in part (a).
(c) Sketch several trajectories of the system. Choose one of your trajectories, and sketch the
corresponding graphs of x1 versus t and x2 versus t. Sketch both graphs on one set of axes.
(d) What is the relation between the eigenvalues of the coefficient matrix and the natural frequency of the spring-mass system?
Ans:
(a) First we solve the given differential equation for the value of u00 .
mu00 + ku = 0
mu00 = −ku
k
u00 = − u
m
k
Now we substitute x1 = u and x2 = u0 to get the equations x01 = x2 and x02 = − m
Putting
this in vector form gives
0 0
1
x1
x1
=
k
x2
x2
−m
0
(b)
1
A − λI =
−λ
r !
r !
k
k
k
2
det(A − λI) = λ +
= λ−i
λ+i
m
m
m
r
k
λ = ±i
m
−λ
k
−m
(c) See figures on the next page.
(d) The natural frequency of an oscillator, ω is defined as
so λ = ±iω
6
q
k
m.
q
k
Our eigenvalues are λ = ±i m
,
7