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Spring 2009 Math 310 Exam 1: Solving Phase Portraits and Harmonic Oscillators, Exams of Differential Equations

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

Pre 2010

Uploaded on 08/18/2009

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Math 310 Exam 1 Spring 2009
1. 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
1
x
-1
1
fHxL
2. Consider the harmonic oscillator x′′ +b x+x= 0.
(a) For what values of bdoes xexhibit oscillations?
Solution. We write the equation as a system X=0 1
1b. This has eigenvalues λ±=
1
2b±b24. Oscillations occur when there is a nonzero imaginary component of the eigen-
values, that is, when b24<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) = ebt "αcos 4b2
2t!+βsin 4b2
2t!#.
(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.
3. Match each of the following phase portraits with the appropriate linear system below.
(a) X=1 1
11XEigenvalues ±2; saddle: Portrait 2.
(b) X=3 1
1 2XEigenvalues 1
2(5 ±5); source: Portrait 6.
(c) X=0 2
2 0XEigenvalues ±2i; center: Portrait 5.
(d) X=1 2
01XOne eigenvalue 1; degenerate source: Portrait 1.
(e) X=1 1
11XEigenvalues 1±i; spiral sink: Portrait 4.
pf2

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Math 310 Exam 1 Spring 2009

  1. 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 1 x
    • 1

1

fHxL

  1. 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

  • β sin

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. 

  1. 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.

  1. 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.