# Need help with nonlinear differential equations

1. May 16, 2005

### andrew410

The problem is on competing species. For the problem, I am supposed to find the critical points. For each critical point, I need to find the eigenvalues and eigenvectors and classify the type of critical point and its stability.

dx/dt = x(1.5 - x - 0.5y)
dy/dt = y(2 - y -0.75x)

I got all the critical points. They are (0, 0), (1.5, 0), (0, 2), (4/5, 7/5).
I understand how to get the eigenvalues, but I can't seem to get the eigenvectors.

So first you need to do that jacobian thing to the dx/dt and dy/dt.
After getting that for the corresponding critical point, I find A-rI for the Jacobian matrix and then do det(A-rI) to find the eigenvalues.
Now, I try to get the eigenvectors by subbing in the first eigenvalue into the A-rI matrix to get a new matrix. I use this new matrix to get the eigenvector, but the answer I get is incorrect.

For example, at critical point (0, 0), the eigenvalues are 1.5 and 2. The book tells me that the eigenvector is $$\left(\begin{array}{cc}1\\0\end{array}\right)$$ for eigenvalue 1.5 and $$\left(\begin{array}{cc}0\\1\end{array}\right)$$ for eigenvalue 2.

2. May 16, 2005

### Zurtex

Well the critical points are correct. The Jacobian matrix should look something like:

$$J= \left( \begin{array}{cc} \frac{1}{2}(3 - 4x - y) & -\frac{x}{2}\\ -\frac{3y}{4} & \frac{1}{4}(8 - 3x - 8y) \end{array} \right)$$

So at the critical point (0,0):

$$J_{(0,0)} = \left( \begin{array}{cc} \frac{3}{2} & 0\\ 0 & 2 \end{array} \right)$$

Making the characteristic polynomial:

$$\left( \frac{3}{2} - \lambda \right) (2 - \lambda) = 0$$

So as you say the eigenvalues are 2 and 3/2. So to work this out:

$$\left( \begin{array}{cc} \frac{3}{2} & 0\\ 0 & 2 \end{array} \right) \left( \begin{array}{ca} x \\ y \end{array} \right) = \lambda \left( \begin{array}{ca} x \\ y \end{array} \right)$$

Last edited: May 16, 2005
3. May 16, 2005

### HallsofIvy

Staff Emeritus
It's hard to tell what you are doing wrong when you don't tell us what you did!

You say "Now, I try to get the eigenvectors by subbing in the first eigenvalue into the A-rI matrix to get a new matrix. I use this new matrix to get the eigenvector, but the answer I get is incorrect." Exactly how are you using "this new matrix"? That will work if you do it correctly. Notice that Zurtex used the orginal matrix. That's all good.

4. May 16, 2005

### saltydog

Andrew, where you goin' with this? Who's eating who? Or do they just want the same nut in the tree? May I introduce the general expression as this may serve to "encapsulate" the overall theory.

$$x^{'}=P(x,y)$$

$$y^{'}=Q(x,y)$$

$$P(x,y)=1.5x-x^2-0.5xy$$

$$Q(x,y)=2y-y^2-0.75xy$$

Usually, the first step in analyzing general first-order systems like this is to find the equilibrium points as you did.

The next step is to "introduce" the slope field as this serves to provide a global picture of the dynamics. You know that already? Anyway, for others that don't know, the slope field is just a picture with little arrows showing the slope $\frac{dy}{dx}$ for this system throughout a specific portion of the x-y plane. The arrows show the general "flow" of x and y (parametrically). Thus looking at the slope field, one can immediately tell, for any starting point of x and y, what the "general" behavior of the system will be. That's very helpfull right? So I've included a plot of the slope field and dots to represent the equilibrium (fixed) points.

So Andrew, what's this to do with Jacobians, eigenvectors, and eigenvalues?

#### Attached Files:

• ###### competingspecies.JPG
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5. May 16, 2005

### Zurtex

saltydog I assume that course Andrew is covering is leading to drawing phase plane diagrams, this is best done by finding the nature and stability of the critical points, you need eigenvalues to classify these sometimes.

6. May 16, 2005

### saltydog

Suppose I shouldn't have introduced the slope field then. Cart before the, oh nevermind. Can we work this one through to completion though? How about you Andrew? You're not going to tell us, "dude, I just want the eigenvectors and get out of here" are you?

Edit: That is, I'd like to see it through to completion. It helps me understand it better and perhaps others to.

Last edited: May 16, 2005
7. May 16, 2005

### saltydog

Hello guys. I'm working on one with complex eigenvalues just as a prelude for the one above. It's:

$$x^{'}=-2x-3y$$

$$y^{'}=3x-2y$$

I realize it's linear but if a non-linear one can be "linearized" to this form via the Jacobian, then it will behave like this one near it's critical point. I mean, that's the point for using the Jacobian right? Anyway, it turns out for this one, the phase portrait is a spirial sink: all solutions (in the phase-plane) spiral around the origin. I've plotted two such solutions below in which the initial conditions differ slightly from one another. You'll note that they pretty much stick together. The system is stable for this reason.

Why are some differential equations and thus the real-world phenomenon they model NOT like this? That is, very small differences in the initial conditions result in very different long-term behaviors? This has immense significance to us. For example, ever wonder why although humans and chimps have . . .but I digress. Differential equations . . . they rock.

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8. May 18, 2005

### saltydog

So, using the Jacobian, we linearize the systems about each fixed point and thus for small values of x(t), and y(t), the non-linear system behaves like the corresponding linear system. The eigenvalues of the resulting linear system determine the overal global behavior of x(t) and y(t) since solutions to these sytems are in the form of exponential functions which have as their exponents, the eigenvalues of the sytem.

The various forms of the eigenvalues are:

Real, L1<0<L2: Saddle (orbits move towards the fixed point, then veer off away from it). This is an unstable fixed point.
Real, L1<L2<0: Sink (all orbits tend to the fixed point). This fixed point is stable.
Real, 0<L1<L2: Source (all orbits move away from the fixed point). The fixed point is unstable as nearby orbits move away from it.

Complex eigenvalues of the form (a+bi):
a>0: Spiral source. Orbits move away.
a<0: Spiral sink. Orbits tend to the fixed point.
a=0: Center: orbits circle center and are periodic;

Repeated eigenvalues:
If <0, orbits tend to the fixed point.
If >0, orbits fly of to infinity

One eigenvalue 0, the other non-zero:
The eigenvalue of 0 represents a line of equilibrium points.
If other is >0, then solutions move away from this line.
If other is<0, then solutions move towards this line.

The following table summarizes these results for this system:

$$\begin{array}{1|c|c|c|c|c|} \text{Fixed Point}&\text{Jacobian}&\text{Eigenvalues}&\text{Eigenvectors}&\text{type}\\ \hline (0,0)& \left(\begin{array}{cc}1.5 & 0 \\ 0 & 2 \end{array}\right)& 2,1.5 & \left(\begin{array}{ca}0 \\ 1 \end{array}\right) \left(\begin{array}{ca}1 \\ 0 \end{array}\right)& \text{Source}\\ \hline (0,2)& \left(\begin{array}{cc}0.5 & 0 \\ -1.5 & -2 \end{array}\right)& -2,0.5 & \left(\begin{array}{ca}0 \\ 1 \end{array}\right) \left(\begin{array}{ca}0.86 \\ -0.51 \end{array}\right)& \text{Saddle}\\ \hline (1.5,0)& \left(\begin{array}{cc}-1.5 & -0.75 \\ 0 & 0.88 \end{array}\right)& -1.5,0.88 & \left(\begin{array}{ca}1 \\ 0 \end{array}\right) \left(\begin{array}{ca}-0.3 \\ 0.95 \end{array}\right)& \text{Saddle}\\ \hline (4/5,7/5)& \left(\begin{array}{cc}-0.8 & -0.4 \\ -1.05 & -1.4 \end{array}\right)& -1.8,-0.38 & \left(\begin{array}{ca}0.37 \\ 0.93 \end{array}\right) \left(\begin{array}{ca}0.69 \\ -0.72 \end{array}\right)& \text{Sink}\\ \hline \end{array}$$

I've attached plots for the slope field of the source, saddle, and sink above along with an example solution. Andrew, and if I was your teacher, I'd want the arrow heads at the tip of those solutions too . . . thanks Zurtex.

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