Mathematics 345 Homework Assignment 4 Due Thursday 11 February 2016

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Mathematics 345
Homework Assignment 4
Due Thursday 11 February 2016
1. Consider the following differential equation derived from the Kepler problem in physics.
ẍ = −x−2 + x−3 .
and determine the potential function
(a) Write the equation in the form ẍ = − dV
dx
V (x).
(b) Introduce a new variable ẋ = y. Rewrite the equation into a system of two
1st-order ODEs of the form
ẋ = f (x, y),
ẏ = g(x, y).
(c) Show that the total energy
1
E = y 2 + V (x),
2
where V (x) is the potential found in (a) is a conserved quantity. (Hint: Show
= ∂E
ẋ + ∂E
ẏ = 0.)
that dE
dt
∂x
∂y
(d) Sketch the phase portrait of the system obtained in (b). (Clearly mark the nullclines, fixed point(s), vector field directions, and some representative phase trajectories).
(e) Show that the steady state found in (d), correspond to the local minimum of the
potential function V (x). Therefore, it is consistent with predictions of Theorem
6.5.1 and with the phase portrait obtained in (d).
(f) Generate an ode file for the system found in (b) (using the ode files from Homework
#3 as a starting template). Use XPPAUT to generate a phase portrait, print a
hard copy of it and attach it to this assignment.
Answer:
(a)
1
1
V (x) = − + 2 .
x 2x
(b)
ẋ = y,
ẏ = −x−2 + x−3 = −x−3 (x − 1).
(c)
ẋ = 0 nullcline :
y = 0;
ẏ = 0 nullcline :
x = 1.
There exists a unique fixed point (xs , ys ) = (1, 0).
The Jacobian
J(x, y) =
0
1
−4
x (2x − 3) 0
⇒
J(1, 0) =
0 1
−1 0
.
Thus, T r = 0, Det = 1.
λ2 + Det = 0
λ2 + 1 = 0
⇒
⇒
λ1,2 = ±i.
Therefore, (xs , ys ) = (1, 0) is a centre. The directions of the vector field is shown
in the sketch of the phase portrait.
(d) The hand-sketched phase portrait is shown in the left panel graph below.
y
y’=0
1.5
1
0.5
x’=0
0
x
-0.5
-1
-1.5
0
0.5
1
1.5
2
2.5
3
3.5
4
(e)
dV
= x−2 − x−3 = x−3 (x − 1) = 0
dx
⇒
xc = 1.
Also,
d2 V
= −2x−3 + 3x−4 = −x−4 (2x − 3)
⇒
2
dx
d2 V
|x=1 = 1 > 0
⇒
xc = 1 is a local min.
dx2
(f) The phase portrait generated by XPP is shown in the right panel of the graph
above.
2. Consider a dimensionless version of the Lotka-Volterra model that describes a predatorprey type interaction.
ẋ = αx − xy,
ẏ = xy − y,
where α > 0 is a parameter.
(a) Sketch all the nulllclines.
(b) Find all fixed points.
(c) Classify the fixed points and sketch a local phase portrait near each fixed point.
(d) Sketch the global phase portrait using the results obtained in (a)-(c).
(e) Show that E = x + y − ln(x) − α ln(y) is a conserved quantity.
(f) Based on Theorem 6.5.1, which one of the two steady states correspond to the
local minimum of the conserved quantity E?
(g) Generate an ode file for the system. Use XPPAUT to generate a phase portrait,
print a hard copy of it and attach it to this assignment.
2
Answer:
(a) Nullclines are determined by: x = 0 and y = α for x′ = 0; y = 0 and x = 1 for
y ′ = 0. A sketch of the nullclines and the direction of trajectories is given below.
y
x’=0
α
y’=0
0
1
x’=0
y’=0
x
Figure 1: A sketch of the nullclines and vector directions.
(b) There are 2 critical points: (xs , ys )=(0, 0) and (1, α).
α − y −x
.
(c) The above Lotka-Volterra model: (1) The Jacobian is: J(x) =
x−
y
1
α 0
0 −1
Therefore, at (0,0): J(0, 0) =
; and at (1,1): J(1, α) =
.
0 −1
α 0
The eigenvalues forJ(0, 0) are λ1 =α > 0, λ2 = −1 < 1. The respective eigenvec1
0
tors are: v1 =
and v2 =
. Therefore, (0,0) is an unstable saddle node,
0
1
the outward direction (v1 ) is the horizontal direction and the inward direction
(v2 ) is the vertical direction.
The eigenvalues for J(1, α) are λ1 =
neutrally stable centre.
√
√
αi, λ2 = − αi. Therefore, (1,α) is a
For ( α , 1)
For (0, 0)
Figure 2: A sketch of the local phase portrait at the two steady states
3
y
x’=0
α
y’=0
0
1
x
x’=0
y’=0
Figure 3: Phase portrait of the Lotka-Volterra system.
(d) The global phase portrait is given the figure.
(e)
∂E
∂E
1
α
dE
=
ẋ+
ẏ = (1− )(αx−xy)+(1− )(xy−y) = (x−1)(α−y)+(y−α)(x−1) = 0.
dt
∂x
∂y
x
y
(f) (1,α) is the one that correspond to the local minimum since it is surrounded by
closed orbits and is a nonlinear centre.
(g)
2
1.5
1
0.5
0
-0.5
-1
-1
-0.5
0
0.5
1
1.5
2
Figure 4: Phase portrait of the Lotka-Volterra system.
4
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