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\centerline
{
\bf Solutions to Attendance Quiz for  Lecture 17 of Dr. Z.'s Dynamical Models in Biology
class 
}


\bigskip

{\bf Name}: Dr. Z.

{\bf 1.}   Find all the {\bf stable equilibrium points} of the 2D {\it continuous} dynamical system given by the differential equations

$$
\frac{dx}{dt} =1 -\frac{2x}{1+y} \quad,
$$
$$
\frac{dy}{dt} =1 -\frac{2y}{1+x} \quad .
$$

{\bf Sol. of 1}: To find the equilibrium points we have to solve

$$
1 -\frac{2x}{1+y}=0 \quad,
$$
$$
1 -\frac{2y}{1+x}=0 \quad .
$$

Simplifying (you do it!) you get a system of two equations with unknowns $x$ and $y$,  whose solution is
$$
(x,y)=(1,1) \quad.
$$

So the set of equilibrium points  is $\{(1,1) \}$ .

Next we take the  Jacobian of the transformation from $R^2$ to $R^2$
$$
(x,y) \rightarrow (1 -\frac{2x}{1+y} , 1 -\frac{2y}{1+x}) \quad.
$$
$$
J(x,y)= \left[ \matrix{-frac{2}{1+y} & \frac{2x}{(1+y)^2} \cr
                 \frac{2y}{(1+x)^2}   & -frac{2}{1+x}} \right]
$$
Plugging-in $x=1$, $y=1$, we get

$$
J(1,1)= \left[ \matrix{-frac{2}{1+1} & \frac{2 \cdot 1 }{(1+1)^2} \cr
                 \frac{2\cdot 1}{(1+1)^2}   & -frac{2}{1+1}} \right]
$$
$$
\left[ \matrix{-1 & \frac{1}{2} \cr
                 \frac{1}{2}   & -1} \right] \quad .
$$

The characteristic equation is:
$$
(1+\lambda)^2- \frac{1}{4}=0
$$
$$
(1+\lambda)^2- (\frac{1}{2})^2=0
$$
$$
(\frac{1}{2} + \lambda)(\frac{3}{2}+ \lambda)=0
$$
The eigenvalues are $-\frac{1}{2}$ and $-\frac{3}{2}$ since they are both negative (in particular, they have a negative real parts)
this is {\bf stable}.

{\bf Ans.}:  So the set of stable equilibrium points  is $\{(1,1) \}$ .



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