Download Spring 2009 Math 310 Exam 1: Solving Phase Portraits and Harmonic Oscillators and more Exams Differential Equations in PDF only on Docsity!
Math 310 Exam 1 Spring 2009
- Each of the following equations depends on a parameter a. For each equation, (i) Plot the phase line when a = 0, and (ii) Plot the bifurcation diagram.
(a) x ′ = x(1 − x) − a
(b) x ′ = f (x) + a, where the graph of f (x) is shown below.
1
fHxL
- Consider the harmonic oscillator x ′′ + b x ′ + x = 0.
(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.
- Match each of the following phase portraits with the appropriate linear system below.
(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.
- Do opposites attract? Analyze the following system, assuming a and b are positive.
R
′ = 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.
