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:
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)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''= -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:
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:
Hence, the fixed points are
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.