# Draw the trajectory in the x-xdot plane

1. Jul 23, 2008

### franky2727

got a question here saying first to find the equlib points of the system xdot=x(x+1) so there obviously 0 and -1

then part 2 is draw the trajectory in the x-xdot plane

the answers are increasing above 0 decreasing between -1 and 0 and increasing at <-1 again for obvious reasons. then i am asked to discuss the nature of the equlibrium points, don't remember doing this and i've lost all of my notes for this topic, it states that the x=0 point is unstable and the x=-1 point is asemptoticaly (i think thats the word) stable, could someone please explain this area, where it comes from and also the other possible answers apart from the 2 above for different situations, thanks

2. Jul 23, 2008

### franky2727

sorry to be a pain but i now have a similar problem with the next question of "for each of te following pairs of equations, classify the equlibrium point" writen next to this question i have wrote list the different types! meaning when is it a saddle point unstable spiral etc, but with only 3 practice questions to work from i have no where near enough information to get this from and again, no notes. :(

3. Jul 23, 2008

### HallsofIvy

An "equilibrium point" is a point where the derivative is 0 (so the function value is constant). Fior this problem that happens when x(x+1)= 0 so x= -1 or x= 0 as you say. For x< -1, both x and x+1 are negative so their product x(x+1) and so dx/dt is positive. That means that if x starts with a value less than -1, it increases. Since two solutions of such a differential equation cannot cross (uniqueness theorem), the solution must increase toward -1 and then "level off" becoming asymptotic to it.

If -1< x< 0, then x is negative but x+ 1 is positive. x(x+ 1) is negative so dx/dt is negative: x is decreasing. If x starts with a value between -1 and 0 it decreases but, again, cannot cross x= -1 so becomes asymptotic to it. Notice that if x is close to -1, on either side, it moves toward -1: x= -1 is a stable equilibrium. Think of this as the case of a ball at the bottom of a hole. At the bottom of the hole, there is no force toward either direction (as opposed to being on the side of the hole); that is what "equilibrium point" means. If the ball is moved slightly away from the bottom, it will quickly roll back to the bottom; that is what "stable" equilibrium point means.

If x> 0, then both x and x+1 are positive so x(x+1) is positive: dx/dt is positive and x is increasing. That means that if x is very close to 0, whether less than 0 or larger than 0, x moves away from 0: x= 0 is an unstable equilibrium. Think of this as a ball sitting at the top of hill. As long as the ball is at the top, it doesn't move-there is no resultant force on (as opposed to being on the side of the hill); again, that is what "equilibrium point" means. But if the ball is moved slightlly, it will roll quickly away from the top; that is what "unstable" equilibrium means.

That depends strongly on the particular system of equations. Do you know how to write the system of equations as a matrix equation? Do you know how to find the eigenvalues of the matrix?

4. Jul 25, 2008

### franky2727

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 isnt 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
(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 dont 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

5. Jul 27, 2008

### franky2727

anyone?

6. Jul 27, 2008

### HallsofIvy

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".

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.

7. Jul 29, 2008