Qualitative Analysis for Autonomous Second-Order ODE

Qualitative analysis is a first step in solving second-order ode. Qualitative
analysis can be used to verify numerical and analytic solutions. Even if a
explicit formula for a solution is known, qualitative analysis is useful, since
it can give a visual picture of the behavior of solutions to an ode.

In this section we consider qualitative analysis (or phase plane analysis) of
autonomous second-order ode. These are differential equations of the form:

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Note that the differential equation does not contain the independent variable t
explicitly. Our treatment here is introductory.

Example

It is convenient to illustrate the procedure with an example. Consider the
ode for a nonlinear pendulum: y''=-(g/L)sin(y), where g is the acceleration
due to gravity, L is the length of the pendulum, and y is the angle between
the pendulum and the vertical. For simplicity we set g/L=1. So the
differential equation is y''= -sin(y). To solve this problem completely an
initial angle y(0) and an initial angular velocity y'(0) must be specified.

Conversion to a System of ODE

The first step is to convert the original second-order ode into two first-order
ode. Define v=y'. Then v'=y''=-sin y. Hence, we get the following differential
equations:

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The variable v corresponds to the angular velocity of the pendulum. If initially
y(0) and y'(0) are specified, then v(0)=y'(0).

Fixed-Points

Fixed points for systems of differential equations are points in the y-v plane
where y'=0 and v'=0 simultaneously. Fixed points are analogous to constant
solutions for autonomous first-order ode. For our example the fixed points
are the solutions to the equations:

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Hence, the fixed points are

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If initially (y(0),v(0)) is a fixed point, then the solution is y(t)=y(0) and
v(t)=v(0). The system remains at the fixed point. This is straightforward
to verify. For example, if y(0)=0 and v(0)=0, then v(t)=0 and y(t)=0 is a
solution to the system of differential equations.

We can give a physical interpretation for the fixed points for the nonlinear
pendulum. If y is an even power of pi and v=0, the pendulum is in the resting
position as shown in the figure below. Obviously, in the absence of
additional forcing, the pendulum will remain in this position for all time.
If y is an odd power of pi and v=0, the pendulum is in the inverted position.
If the pendulum is perfectly balanced it will remain in the inverted position
for all time.

Vector Field

The system of differential equations for the nonlinear pendulum defines a
vector field in the y-v plane. At each point, the differential equation gives
y' and v'. For example, at y=0 and v=1, we have y'=1 and v'=0. These
derivatives indicate that v is not changing and y is increasing at a rate of
1 unit per unit time. This information is expressed by the vector (1,0).
The vector indicates the direction of motion of the system when y=0 and v=1.

We can assign a vector for each point in the y-v plane in this manner. This
is shown in the figure below. The small circle at (-pi,0), (0,0) and (pi,0) are
the fixed points.

Sketching Solutions

Solutions to the differential equation are tangent to vectors at each point in
the y-v plane. This is shown in the figure below. Notice that the solution
curves that are plotted do not give y or v as a function of time. The velocity
v is plotted as a function of y.

The solution curves in the figure are 2pi periodic in y. That means that the
structure in the interval -pi<=y<=pi repeats itself in the intervals
pi<=y<=3pi, -3pi<=y<=pi, and so on.

A physical explanation can be attached to the solution curves. The closed
curves centered about (0,0) correspond to oscillations about y=0. The
curves at the top and bottom of the figure correspond to whirling motion.
In this case the energy is sufficiently large for the pendulum to turn
continuously in the clockwise or counterclockwise direction. The top curve,
where y is increasing, corresponds to counter-clockwise whirling. The
bottom curve corresponds to clockwise whirling.


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