

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Solutions to exam 1 of math 310 for the spring 2009 semester. It includes the analysis of phase portraits and the harmonic oscillator equation. Students are expected to identify the type of each linear system based on the given phase portraits and find the general solution for the harmonic oscillator equation.
Typology: Exams
1 / 2
This page cannot be seen from the preview
Don't miss anything!
(a) x ′ = x(1 − x) − a
(b) x ′ = f (x) + a, where the graph of f (x) is shown below.
1
fHxL
(a) For what values of b does x exhibit oscillations?
Solution. We write the equation as a system X′^ =
− 1 −b
. This has eigenvalues λ± =
1 2
−b ±
b^2 − 4
. Oscillations occur when there is a nonzero imaginary component of the eigen- values, that is, when b 2 − 4 < 0, or when b < 2.
(b) Find the general solution in this case.
Solution. The general solution is a linear combination of eλt, i.e.
x(t) = e −bt
α cos
4 − b^2
2
t
4 − b^2
2
t
(c) Describe the motion of the mass in the case when b = 0.
Solution. When b = 0 the general solution consists of linear combinations of cosines and sines of t. Physically, this represents the case where the mass oscillates indefinitely with the same amplitude.
(a) X ′ =
X Eigenvalues ±
2; saddle: Portrait 2.
(b) X ′ =
X Eigenvalues 1 2 (5^ ±
5); source: Portrait 6.
(c) X ′ =
X Eigenvalues ± 2 i; center: Portrait 5.
(d) X
X One eigenvalue −1; degenerate source: Portrait 1.
(e) X ′ =
X Eigenvalues − 1 ± i; spiral sink: Portrait 4.
−2 0 2
−
0
2
x
y
−2 0 2
−
0
2
x
y
−2 0 2
−
0
2
x
y
−2 0 2
−
0
2
x
y
−2 0 2
−
0
2
x
y
−2 0 2
−
0
2
x
y
(f) X ′ =
X Eigenvalues 0, −5; stable line of equilibria: Portrait 3.
′ = aR + bJ, J ′ = −bR − aJ.
Solution. The eigenvalues are ±
a^2 − b^2. There are thus three possibilities, depending on whether a < b, a = b or a > b. If a < b, the eigenvalues are imaginary and there are indefinite oscillations. The two go around in cycles of love and hate, forever. If a > b the eigenvector corresponding to the positive eigenvalue is
−(a +
a^2 − b^2 )/b, 1
, which points in the 2nd quadrant. Thus as t → ∞, the solution tends toward (−∞, ∞) or (∞, −∞). That is, one ends up loving the other while the other hates the
former. If a = b there is only the zero eigenvalue. The canonical form is Y ′ =
, which has the
solution Y (t) = (αt + β, α), and the general solution is X(t) = (−αt − β + α, αt + β − 2 α), which again tends toward (−∞, ∞) or (∞, −∞). In either case, the lovers are doomed to end up in a situation where one loves the other but the other hates the former.
BONUS: For which values of α does the following initial value problems have a unique solution for all
0 ≤ t < ∞?
x ′ = x α , x(0) = 0.
Solution. First note that there can’t be any solution if α < 0, and that x(t) = 0 is always a solution for
α > 0. When α = 0 the equation says x′^ = 1, x(0) = 0, which has only the solution x(t) = t, and when
α = 1 we have proven that the only solution is x(t) = 0. Thus we must look to see for which positive values
of α there are solutions apart from the zero solution. We try to solve by separation of variables, supposing
α > 0 and α 6 = 1,
dx
xα^
= dt ⇒ x(t) = [(1 − α)t + C]
1 /(1−α)
We see that this can only satisfy the initial condition if α < 1. Thus there is a unique solution if α = 0 or
α ≥ 1.