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\centerline{Mathematics 304 -- Ordinary Differential Equations}
\centerline{Discussion/Lab 3 -- Selected Solutions}
\centerline{November 29, 2004}
\bigskip
\noindent
F.  It will save some work here to find the equilibrium
points and the Jacobian matrix of the systems:
$$\cases{{dx\over dt} = x - 4x^2 - xy - h & \cr
         {dy\over dt} = y - 2y^2 - 5xy & \cr}$$
in general (i.e. as functions of $h$), before looking
at the specific $h$-values given in the problem.
First, setting the right-hand side of the second
equation equal to zero and factoring, we get
$$y(1 - 2y - 5x) = 0$$
so at the equlibrium points, either $y = 0$, or 
$y = {-5\over 2}x + {1\over 2}$.  

Substituting $y = 0$ into the first equation we get
$4x^2 - x + h =0$, so solving by the quadratic formula
$$x = {1 \pm \sqrt{1 - 16h}\over 8},$$
and we have two equilibrium points
$$(x,y) = \left({1 \pm \sqrt{1 - 16h}\over 8},0\right)$$
(as long as $h < 1/16$).

Similarly, substituting $y = {-5\over 2}x + {1\over 2}$
into the first equation, we get another quadratic 
equation in $x$:
$${3\over 2}x^2 - {1\over 2}x + h = 0$$
so by the quadratic formula again,
$$x = {{1\over 2} \pm \sqrt{{1\over 4} - 6h} \over 3}
= {1 \pm \sqrt{1 - 24h} \over 6}.$$
We have possibly two equilibrium points from this as well
(as long as $h < 1/24$).

The Jacobian matrix of the system is independent of $h$:
$$J(x,y) = \pmatrix{1 - 8x - y & -x\cr
                     -5y & 1 - 4y - 5x}.$$

Substituting $h = 0$, we have four equilibrium points:
$$(0,0),\ (1/4,0),\ (0,1/2),\ (1/3,-1/3).$$
The last is not in the first quadrant of the $(x,y)$-plane,
however.  Hence it does not represent a realistic pair
of population values, and we will disregard it.  Checking
the other three critical points, we find:
\item{1)}  $J(0,0) = \pmatrix{1&0\cr 0 & 1\cr}$, so
$(0,0)$ is a {\it source}.  (This is always true for
these systems with no harvesting term ($h = 0$).) 
\item{2)}  $J(1/4,0) = \pmatrix{-1&-1/4\cr 0 & -1/4\cr}$, so
$(1/4,0)$ is a {\it sink} ($tr(J) < 0$ and 
$det(J) > 0$, so eigenvalues are both negative). 
\item{3)}  $J(0,1/2) = \pmatrix{1/2&0\cr -5/2 & -1\cr}$, so
$(0,1/2)$ is a {\it saddle point}  (eigenvalues have opposite
signs).

\noindent
The phase portrait plot for this system should show all
solutions starting from
points in the first quadrant
tending to $(1/4,0)$ as $t\to +\infty$, unless 
the initial condition has $x = 0$.  Note:  This is
{\it not a case of stable coexistence} because
$y \to 0$ as $t\to +\infty$.  The competition is
driving that species to extinction.

Substituting $h = 1/32$, we have four equilibrium points:
$$\eqalign{
({1+\sqrt{1/2}\over 8},0)\doteq (.213,0),&\ 
({1-\sqrt{1/2}\over 8},0)\doteq (.037,0)\cr
(1/4,-1/8),&\ (1/12,-7/24)\cr}.$$
The third one now is not in the first quadrant of the $(x,y)$-plane, 
and we will disregard it.  Checking
the other three critical points, we find:
\item{1)}  $J(.213,0) \doteq \pmatrix{-.707&-.213\cr 0 & -.067\cr}$, so
$(.213,0)$ is a {\it sink} (both eigenvalues negative). 
\item{2)}  $J(.037,0) \doteq \pmatrix{.707&-.037\cr 0 & .817\cr}$, so
$(.037,0)$ is a {\it source} (eigenvalues are both positive). 
\item{3)}  $J(1/12,7/24) = \pmatrix{1/24&-1/12\cr -35/24 & -7/12\cr}$, so
$(1/12,7/24)$ is a {\it saddle point}  (eigenvalues have opposite
signs since determinant is negative).

\noindent
The phase portrait plot for this system should show 
solutions starting from
points far enough to the right in the first quadrant
tending to $(.213,0)$ as $t\to +\infty$ and other
solutions starting from points with $x \le .037$
tending toward the $y$-axis.  Note:  This is
{\it not a case of stable coexistence} either because
either $x \to 0$ or $y \to 0$ as $t\to +\infty$.  
The competition is
driving one or other of the two species to extinction.
The one exception would be if the initial condition put
us exactly on one of the separatrix curves leading
into the saddle at $(1/12,7/24)$, but that is not
a stable situation because nearby initial conditions
would lead to very different outcomes.

With $h = 5/32$, all of the equilibria found above
will have complex coordinates, so we disregard all of
them.  All solutions for this system tend toward the 
$y$-axis as $t\to \infty$.  Thus, $x$ is going to 
zero.  The effect of the larger harvesting term is
to ``shift the terms of the competition'' in favor
of $y$.

\bigskip
\noindent
G.  Note: I was actually looking for a lot more here than
any of the groups presented in their solutions, so
this question was not counted for the total on the
assignment. 

The main question here was supposed to be:  {\it Under
what conditions on $a,b,c,d,M_1,M_2$ can stable
coexistence take place?}  From question A, this means:
when will the system have an equilibrium point 
$(x_0,y_0)$ in the
interior of the first quadrant (so $x_0,y_0 > 0$)
that is a {\it sink}?

\bigskip
\proclaim Claim.  This is true provided that
$ac - bd > 0, cM_2 - dM_1 > 0, aM_1 - bM_2 > 0$.
Equivalently, this is true if the fourth equilibrium
point of the system lies in the first quadrant, but
{\it outside} the triangle with vertices at the 
other three equilibria $(0,0),(M_1,0),(0,M_2)$.   
\par

\noindent
Before showing this, we recall the two systems
from question B:
\item{$\bullet$} In system 1, the fourth equilbrium
was at $(1,1)$,  which is {\it inside} the triangle
with vertices $(0,0), (2,0), (0,3)$.  This system
had a saddle point at $(1,1)$, and there 
was no stable coexistence.
\item{$\bullet$}  In system
2, the fourth equilibrium was at $(2,2)$,
which is {\it outside} the triangle with vertices
at $(0,0), (3,0), (0,3)$.  In this case, $(2,2)$
was a sink, which made for a stable coexistence
situation at those population levels

\bigskip
Now, we show why the claim is true.  From the factorizations
of the right-hand sides in the basic system, we know
that the fourth equilibrium is the intersection 
of the lines 
$$\cases{ax + by = aM_1 &\cr
         dx + cy = cM_2 &\cr}$$
or 
$$(x_0,y_0) = \left({c(aM_1 - bM_2)\over ac - bd}, 
{a(cM_2 - dM_1)\over ac - bd}\right).$$
By the hypotheses in the claim (and the general hypotheses
that $a,c > 0$), we see that $x_0 > 0$ and $y_0 > 0$.
So the fourth equilibrium is in the interior of the
first quadrant.  Moreover, computing the Jacobian 
matrix of the system at $(x_0,y_0)$, we find
$$J(x_0,y_0) = 
\pmatrix{{-ac(aM_1 - bM_2)\over ac - bd} &
{-bc(aM_1 - bM_2)\over ac - bd}\cr
{-ad(cM_2 - dM_1)\over ac - bd} &
{-ac(cM_2 - dM_1)\over ac - bd}\cr}$$
This gives, under the hypotheses of the Claim,
$$\det(J(x_0,y_0)) = {ac\over (ac - bd)}(aM_1-bM_2)(cM_2-dM_1) > 0$$
and 
$${\rm tr}(J(x_0,y_0)) = {-ac\over (ac-bd)}((aM_1-bM_2) + (cM_2-dM_1)) < 0$$
These inequalities show that $(x_0,y_0)$ is a sink.

(It is not difficult to see that the condition
$cM_2 - dM_1 > 0$ also implies that $(M_1,0)$
is a saddle point, and similiarly, $aM_1 - bM_2 > 0$
implies that $(0,M_2)$ is a saddle point.)

To see the equivalence with the second form in the Claim,
note that the conditions can be rewritten as 
$${a\over b} > {M_2\over M_1} >  {d\over c} > 0$$
or 
$${-a\over b} < {-M_2\over M_1} < {-d\over c} < 0.\leqno(*)$$
Geometrically, ${-a\over b}$ is the slope 
of the line $ax + by = aM_1$ from the first ODE in the system.
Also, ${-M_2\over M_1}$ is the slope of the line between
the equilibrium points $(M_1,0)$ and $(0,M_2)$.
Finally, ${-d\over c}$ is the slope 
of the line $dx + cy = cM_2$ from the second ODE.
The inequalities (*) are equivalent to saying that
the fourth equilbrium point $(x_0,y_0)$ (the intersection
of the two lines from the ODE's) lies outside the triangle with
vertices $(0,0), (M_1,0), (0,M_2)$.
\bye
    
