franky2727 said:
thanks for the help in the first part and a really nice example, helped me get my head around it. as for the second part it isn't a question with matracies i knows this much as i have done this in a different topic with eivenvectors and values.
the 3 questions for this part are as follows
For each of the following pairs of equations classify the equlibrium point
(a) Xdot=-x-3y ydot=-x
It might be better to interpret this in terms of matrices:
\left[\begin{array}{c}x' \\ y'\end{array}\right]= \left[\begin{array}{cc}-1 & -3 \\ -1 & 0\end{array}\right]\left[\begin{array}{c} x \\ y\end{array}\right]
The equilibrium points are where the derivatives are equal to 0: -x- 3y= 0 and -x= 0. It is easy to see that the only solution to that is x= 0, y= 0.
Now, look for the eigenvalues of the coefficient matrix: The eigenvalue equation is
\left|\begin{array}{cc}-1-\lambda & -3 \\ -1 & -\lambda\end{array}\right|= \lamda^2+ \lambda- 3= 0
which has roots (from the quadratic equation)
\lambda= \frac{-1\pm\sqrt{13}}{2}
Since 13> 0, both roots are real. Further, since \sqrt{13} is larger than 1 it is easy to see that one root will be positive and one negative. Now that means there will be one line (an eigenvector corresponding to the negative root) where the "flow" will be toward the origin and another line (an eigenvector corresponding to the positive root) where the "flow" will be away from the origin. Since the "flow" will be both toward and away from the equilibrium point, that is a "saddle".
(b)Xdot=8x+2y ydot=-8x+8y
(c)Xdot=-2x-y ydot=2x
now i know how to get to the answers however i have kind of the same problem as i did in the first part, i don't understand how and why the points are saddle points, unstable spirals etc, could you or someone else who happens to be reading this please give me a list of the possible outcomes and how to choose the correct one and why. thanks
In general, a real positive eigenvalue means there is a line upon which the "flow" is away from the equilibrium point, a real negative eigenvalue means there is a line upon which the "flow" is toward the equilibrium point. Putting those together,
1) If both eigenvalues are positive, we have an unstable focus: all flow is away form the equilibrium point.
2) If both eigenvalues are negative, we have a stable focus: all flow is toward the equilibrium point.
3) If one eigenvalue is positive, the other negative, we have a saddle point.
A complex eigenvalue (and the eigenvalues must be complex conjugates in this case) means a circular "flow". Whether toward or away from the equilibrium point depends upon the real part.
4) Complex eigenvalues with 0 real part (imaginary eignvalues) means a center: circular "flow".
5) Complex eigenvalues with positive real part means an unstable spiral: roughly circular "flow" but tending away from the equilibrium point.
6) Comples eigenvalues with negative real part means a stable spiral: roughly circular "flow" but tending toward the equilibrium point.
If one or both of the eigenvalues is 0, then the coefficient matrix is not invertible and we will not have just one or separated equilibrium points- we may have an entire line or even an entire plane of equilibrium points.
