Midterm 2 solutions

MATH 4428 M IDTERM E XAM 2 S OLUTIONS
Problem 1.
Consider the following dynamical system.
dx1
= x1 + 3x2
dt
dx2
= x1 − x2
dt
(a) [10 points]
The phase portrait of this dynamical system is figure B below.
(A)
(B)
(C)
(D)
Solution.
There are many ways to solve this. The easiest is to pick a point and calculate what the
derivative vector should be there. You could pick any point, but for the sake of simplicity,
let’s look at the point (0, 2). The derivative vector there should be
dx1
d~x
(0, 2) = 0 + 3 · 2 = 6
6
dt
=
=⇒
dx2
−2
(0, 2) = 0 − 2
= −2
dt
dt
That vector points mostly right and a little bit downward. Now look at the vectors in
each phase portrait at the point (0, 2). The only one that has a vector starting from (0, 2)
and going right and down is figure (B), so that’s the answer (keep in mind that there are
different versions of the exam with these plots in a different order).
Problem 2.
Consider the following dynamical system.
dx1
= −x1 + x2
dt
dx2
= x2 − x22
dt
(a) [5 points] Find all equilibrium points of the system.
Solution.
We want to find all points (x1 , x2 ) that satisfy the equations
0 = −x1 + x2
0 = x2 − x22
From the first equation we get x1 = x2 and from the second equation, x2 (1 − x2 ) = 0,
which means x2 = 0 or 1, so the two equilibrium points are (0, 0) and (1, 1).
(b) [10 points] Show if each equilibrium point is asymptotically stable or not.
Solution.
To check if a point ~x0 is asymptotically stable, we check if all the eigenvalues of
∇F (~x0 ) are negative.
−1
1
∇F =
0 1 − 2x2
So for the point (0, 0), that matrix is
−1 1
∇F (0, 0) =
0 1
which is a triangular matrix, so the eigenvalues are on the diagonal: −1 and 1. Since
one is positive, (0, 0) is an unstable equilibrium point.
−1 1
∇F (1, 1) =
0 −1
Again, since it’s a triangular matrix, the eigenvalues are on the diagonal: −1 and −1.
Since they’re both negative, (1, 1) is an asymptotically stable equilibrium point.
Problem 3.
Consider the following discrete dynamical system.
∆x1 = 12 (−x1 + x2 )
∆x2 = x21 − x2
(a) [5 points] Find all equilibrium points of the system.
Solution.
The equilibrium points satisfy ∆x1 = 0 and ∆x2 = 0, which give us the equations
0 = 12 (−x1 + x2 )
0 = x21 − x2
The first equation means x1 = x2 . The second one means x21 = x2 . Combining them
gives us x21 = x1 , which we can change to x21 − x1 = 0, and thus x1 (x1 − 1) = 0, so
x1 = 0 or 1. Therefore the equilibrium points are (0, 0) and (1, 1).
(b) [10 points] Show if each equilibrium point is asymptotically stable or not.
Solution.
An equilibrium point of a discrete dynamical system is asymptotically stable if the
eigenvalues of ∇G(~x0 ) have complex absolute values less than 1, where G(~x) =
~x + F (~x).
1
1
(−x1 + x2 )
(x1 + x2 )
x1
2
2
=
G(~x) = ~x + F (~x) =
+
x21 − x2
x21
x2
1 1
1 1
1 1
∇G = 2 2
=⇒
∇G(0, 0) = 2 2
and ∇G(1, 1) = 2 2
2x1 0
0 0
2 0
1 1
2
2 is a triangular matrix, so the eigenvalues are on the diagonal: 0 and 1 . There2
0 0
fore (0, 0) is asymptotically stable. For the other matrix ∇G(1, 1), which we’ll label
as A, we have to compute the eigenvalues the usual way.
λ − 1 − 1 2
2
=0
det(λI − A) = 0
=⇒
=⇒
(λ − 12 )λ − 1 = 0
−2
λ
So we need to solve the equation λ2 − 12 λ − 1 = 0. We use the quadratic formula.
λ=
1
2
±
q
2
1
4
s
+4
=
1
4
±
q
17
16
=⇒
|λ| =
1 2
4
+
q 2
17
16
=
q
1
16
+
17
16
=
q
18
.
16
We want to know if |λ| > 1. This is the same thing as checking if |λ|2 > 1. So, since
18
> 1, we see that (1, 1) is an unstable equilibrium point.
|λ|2 = 16
Problem 4.
Consider the following dynamical system.
dx1
= −3x1 + x2
dt
dx2
= x1 − 3x2
dt
(a) [5 points] Write out the equations for applying (forward) Euler’s Method to this
dynamical system, with time step size h.
Solution.
= xn1 + h(−3xn1 + xn2 )
xn+1
1
xn+1
= xn2 + h(xn1 − 3xn2 )
2
(b) [5 points]
The exact solution of the above dynamical system is of the form
1 −2t
1
~x(t) = c1
e + c2
e−4t ,
1
−1
which goes to ~0 as t → ∞ for any choice of initial value ~x(0).
Let ~xhn be the value given by (forward) Euler’s Method at the nth time step, with
time step size h. What is the largest we can take h so that ~xhn goes to ~0 for any choice
of initial value?
Solution.
Look at the answer to part (a). That can be written as a discrete dynamical system.
∆xn1 = h(−3xn1 + xn2 )
∆xn2 = h(xn1 − 3xn2 )
0
~
This system has one equilibrium point: 0 =
. If that equilibrium point is stable, then
0
~xnh will go to ~0 for any choice of initial value. So we just repeat the same procedure that
we used in problem 3. We define F (~xn ) to be
h(−3xn1 + xn2 )
n
F (~x ) =
h(xn1 − 3xn2 )
and we use that to define G(~xn ),
xn1
h(−3xn1 + xn2 )
(1 − 3h)xn1 + hxn2
G(~x ) = ~x + F (~x ) = n +
=
x2
h(xn1 − 3xn2 )
hxn1 + (1 − 3h)xn2
n
n
n
If the eigenvalues of ∇G(0, 0) have eigenvalues with complex absolute values less than 1,
the discrete system is stable.
1 − 3h
h
1 − 3h
h
∇G =
=⇒
∇G(0, 0) =
.
h
1 − 3h
h
1 − 3h
Call this matrix A. Now we compute the eigenvalues of A.
λ − (1 − 3h)
−h
= (λ−(1−3h))2 −h2 = λ2 −2(1−3h)λ+(1−3h)2 −h2 .
0 = det(λI−A) = −h
λ − (1 − 3h)
We use the quadratic formula to find the roots.
p
√
2(1 − 3h) ± 4(1 − 3h)2 − 4((1 − 3h)2 − h2 )
2(1 − 3h) ± 4h2
=
= (1 − 3h) ± h
λ=
2
2
So λ1 = 1 − 4h and λ2 = 1 − 2h are the two eigenvalues.
|1 − 4h| < 1
|1 − 2h| < 1
⇐⇒
⇐⇒
1 − 4h < 1 and
1 − 2h < 1 and
− (1 − 4h) < 1
− (1 − 2h) < 1
⇐⇒
⇐⇒
0 < h < 12
0<h<1
Since our time step has to be positive, we know h > 0, so ~xnh → ~0 when h <
both inequalities are satisfied.
1
2
because