Chap 42 Exercises
Exercise 1 Describe the stability of each of the following flows.
Exercise 2 Consider this two-dimensional flow field:
There are three fixed points visible. The next plots zoom in on each of the fixed points.
- Is the fixed point in (A) stable or not?
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- Is the fixed point in (B) stable or not?
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- The fixed point in (C) is called a βsaddle.β It is stable in one direction and unstable in another. Which of these is correct?
Stable in y-direction and unstable in the x-direction
Stable in the x-direction and unstble in y
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Hereβs a system which has 4 fixed points in the region shown.
Plots (D) and (E) zoom in on two regions.
- Which of the following is the best description of the behavior near the fixed point in (D)?
Stable and rotating clockwise
Stable and rotating counter-clockwise
Unstable and rotating clockwise
Unstable and rotating counter-clockwise
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- Which of the following is the best description of the behavior near the upper left fixed point in (E)? (Neutal stability means neither stable nor unstable; the trajectory just orbits around the fixed point.)
Unstable and rotating clockwise
Stable and rotating counter-clockwise
βNeutral stabilityβ and rotating clockwise
βNeutral stabilityβ and rotating counter-clockwise
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Letβs look a little more closely at the upper-left fixed point in graph (E):
The pattern in figure (F) is clockwise rotation around the fixed point. This kind of pattern is of fundamental importance in physics and engineering.
Exercise 3
In plot (A), how many fixed points are visible?
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- In plot (B), how many fixed points are visible?
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- In plot (C), how many fixed points are visible?
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- In plot (D), how many fixed points are visible?
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Exercise 4 In each of the following graphs, a flow field is annotated with a red contour and a blue contour. Your task is to determine whether a contour corresponds to a zero of the horizontal component of flow, a zero of the vertical component of flow, or neither. (Remember, if the contour is at a zero of the horizontal flow, the flow on the contour will be entirely vertical. And vice versa.)
Plot A
:- Zero of horizontal flow
- Zero of vertical flow
- Neither
Plot A
:- Zero of horizontal flow
- Zero of vertical flow
- Neither
Plot B
:- Zero of horizontal flow
- Zero of vertical flow
- Neither
Plot B
:- Zero of horizontal flow
- Zero of vertical flow
- Neither
Activities
Exercise 5 XREF not implemented yet shows a flow field of the pendulum system and three pairs of trajectories, one pair for each of three initial conditions. Each trajectory starts at
A. Read the three different initial conditions from the graph.
B. The dark gray trajectories are for the original (nonlinear) pendulum system while the
C. The flow field corresponds to either the nonlinear (gray) or linearized (
Exercise 6 The rabbit-fox system is nonlinear:
Write down, in English, the names of the Greek letters in the above formula.
Using algebra, find the fixed point
of the rabbit-fox system.
- Linearize the rabbit-fox system around the fixed point and write down the linearized dynamics in this form:
Exercise 7 In this exercise, you will work with a particular function f()
of two variables. Construct the function this way:
The goal of this exercise is to explore the connections between the optimization method of gradient ascent or descent and dynamical systems.
Hereβs a plot of the function and its gradient field.
You can construct the f()
this way:
Use these two functions to define a dynamical system:
Use
integrateODE()
to integrate the equations numerically from the initial condition . Plot the resulting trajectory as a layer on top of the contour plot and gradient field. Make the time bounds inintegrateODE()
large enough to get very close to the high-point in the contour plot.Modify the differential equations so that they correspond to gradient descent rather than ascent. Using the initial condition
, numerically integrate the differential equations and, as in (1), plot the trajectory as a layer on the contour-plot/gradient-field. Make the time bounds large enough to get very close to a local minimum of .From (2), change the initial condition to
and plot the trajectory. Whatβs different from the result in (2).
Solution containing functions x(t), y(t).
Solution containing functions x(t), y(t).
Solution containing functions x(t), y(t).