How Do You Determine the Direction of Oscillation in Differential Equations?

Click For Summary
SUMMARY

The direction of oscillation in differential equations can be determined by analyzing the associated matrix and its eigenvalues. For the equation \(\frac{d^2 y}{dt^2} = -2y\), the eigenvalues are \(\pm \sqrt{2}i\), indicating oscillatory behavior. However, the eigenvalues alone do not reveal the direction of rotation; this can be established by evaluating the system at a specific point. For instance, using the point (1, 0) with the matrix \(\left(\begin{array}{cc}0&2\\-2&0\end{array}\right)\) shows that \(x\) increases while \(y\) decreases, confirming a counter-clockwise rotation.

PREREQUISITES
  • Understanding of differential equations and their solutions
  • Familiarity with eigenvalues and eigenvectors
  • Knowledge of matrix operations and characteristics
  • Basic concepts of rotation in the Cartesian plane
NEXT STEPS
  • Study the properties of eigenvalues and eigenvectors in dynamical systems
  • Learn about the trace and determinant of matrices and their implications
  • Explore the use of phase portraits to visualize oscillatory behavior
  • Investigate the impact of varying coefficients in differential equations on oscillation direction
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are studying differential equations and their applications in modeling oscillatory systems.

stunner5000pt
Messages
1,447
Reaction score
5
how would one go about finding the direction of oscillation for a differential equation?

for example \frac{d^2 y}{dt^2} = -2y
has eigenvalues \pm \sqrt{2}i
and the corresponding matrix is \left(\begin{array}{cc}0&2\\-2&0\end{array}\right) so the solution curves will form closed loops around the origin. But what about the direction -i.e. counter clockwise or clockwise, how would one go about figuring this out??

please help! I have an exam tomorrow on this stuff and this is where i am snagged!
 
Physics news on Phys.org
the two associated eigenvectors would be computed as follows

AX=\lambda X


\left(\begin{array}{cc}0&2\\-2&0\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = i \sqrt{2} \left(\begin{array}{cc}x\\y\end{array}\right)

which gives the equation
2y = i \sqrt{2} x and
-2x = i \sqrt{2} y
which gives me eigenvector which is 2y = i \sqrt{2} x which gives me V_{1} = \left(\begin{array}{cc}2\\i \sqrt{2}\end{array}\right) and V_{2} = \left(\begin{array}{cc}2\\-i \sqrt{2}\end{array}\right)
is this correct?? Am i supposed to simplify the eigenvectors??
Please help!
 
In this situation, the eigenvalues are not sufficient to tell you the direction of oscillation (I would say "rotation").
Consider the general example: \left(\begin{array}{cc}a&b\\c&d\end{array}\right).
The characteristic equation is \left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|= \lambda^2- (a+d))\lamba+ ad-bc= 0.
The solutions to that depend on two numbers, not 3. Specifically, a+ d, the "trace" of the matrix and ad-bc, the "determinant" of the matrix. Any two matrices having the same trace and matrix will have the same eigenvalues. In particular, \left(\begin{array}{cc}0&-2\\2&0\end{array}\right) has exactly the same eigenvalues but opposite rotation. In this case, since the eigenvectors are complex and you are working in the real plane, they don't help.

You can, easily, determine the direction by looking at anyone point. In particular, just because it is easy, suppose x= 1, y= 0. Then multiplication of (1, 0) by the matrix gives dx/dt= 2, dy/dt= -2. That is, x is increasing, y is decreasing- that "y is decreasing" tells us that the rotation is counter-clockwise.
 

Similar threads

Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
7
Views
3K
  • · Replies 26 ·
Replies
26
Views
4K