System of ODE's

1. Nov 2, 2011

saul goodman

The question asks me to consider the system of differential equations:

$\frac{dx}{dt} = 1 - 2x + x^2y$
$\frac{dy}{dt} = x-x^2y$

It asks me to find the fixed point(s), and determine their stability, also to draw the phase plane.

So to find the fixed points, I set both equations equal to zero:
$1 - 2x + x^2y = 0 \Rightarrow x^2y = 2x-1$
$x-x^2y = 0 \Rightarrow x^2y = x$

Setting them equal to each other:

$2x-1 = x \Rightarrow x=1$ and from this we can deduce $y=1$. So the fixed point is (1,1).

Now this is where I'm not really sure on what to do and get confused by my lecture notes. I believe I have to use some Jacobian to work out the stability yet I have no clue why i'm doing this (I know eigenvalues/eigenvectors are involved somehow). Setting $$f(x,y)= 1 - 2x + x^2y$$ and $$g(x,y)= x-x^2y$$ I worked out the Jacobian at (1,1) to be:

\left|
\begin{array}{cc}
0 & 1 \\
-1 & -1 \end{array}
\right|

(Sorry about the lack of detail about this part but I'm new to latex and struggled with doing matrices, the jacobian is fx and fy on the first row and gx and gy on the second row.

Following lecture notes, I work out the 'trace' (T) of this matrix which is simply (a11+a22) which is -1 in this case, and also work out the determinant which is 1.

Since $T^2 - 4D = -3 < 0$, we have complex eigenvalues and I believe this means the phase plane will be a spiral, and since T < 0 this means the general trajactory will be moving towards the fixed point (stable). But do I need to work out the eigenvectors to see exactly what this spiral will look like? I'm not sure if I've worked out enough to complete the question. Also any help on actually explaining why I'm working out the Jacobian stuff would be great.

Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 2, 2011

jackmell

The Jacobian is just the 2D Taylor series for the right-hand sides truncated to the linear terms. That's how you "linearize" the non-linear system and for values close to the fixed-point, the linear version is usually a good approximation to the non-linear version. I think you're correct in the type of fixed-point you have: if the real part of the eigenvalues are negative, then the fixed-points are sinks. Then do a StreamPlot of the system in Mathematica to get the slope field or rather do one or two manually first to get the idea down, then StreamPlot.

Last edited: Nov 2, 2011