I was working on this question:
Here is what I have so far:
$ (a) \ \
\text{We have the algebraic expression to find equilibrium points:} \\
\frac{dy}{dt} = ry \big[1 - \left(\frac{y}{K}\right) \big] - h = 0 \\
\therefore \ ry \left(1- \frac{y}{K} \right) - h = 0 \\
\therefore \ - \frac{ry^{2}}{K} + ry - h = 0 \\
\therefore \ y^{2} - yK + \frac{hK}{r} = 0 \\
\therefore \ y^{2} - yK = - \frac{hK}{r} \\
\therefore \ y^{2} - yK + \frac{K^{2}}{4} = \frac{K^{2}}{4} - \frac{hK}{r} \\
\therefore \ \left (y - \frac{K}{2} \right)^{2} = \frac{K^{2}}{4} - \frac{hK}{r} \\
\therefore \ y - \frac{K}{2} = \sqrt{ \frac{K^{2}}{4} - \frac{hK}{r} } \\
\therefore \ y = \frac{K}{2} \pm \sqrt{ \frac{K^{2}}{4} - \frac{hK}{r} } \\ \\
y_{1} = \frac{K}{2} - \sqrt{ \frac{K^{2}}{4} - \frac{hK}{r} } \\
y_{2} = \frac{K}{2} + \sqrt{ \frac{K^{2}}{4} - \frac{hK}{r} }
$
I have two questions:
I got the above two points after seeing that the critical / equilibrium points of the differential equation with the general form $ \frac{dy}{dt} = f(y) $ comes from locating its roots.
When I tried working on this problem initially, I got the following results:
$ \frac{dy}{dt} = ry \big[1 - \left(\frac{y}{K}\right) \big] - h = 0 \\ h = ry \big[1 - \left(\frac{y}{K}\right) \big] \\ \frac{h}{ry} = \big[1 - \left(\frac{y}{K}\right) \big] \\ \frac{y}{K} = 1 - \frac{h}{ry} \\ y_{2} = K \big[ 1 - \frac{h}{ry} \big] \\ \\ \text{We also have:} \\ \frac{dy}{dt} = ry \big[1 - \left(\frac{y}{K}\right) - \frac{h}{ry} \big] = 0 \\ ry = 0 \\ y_{1} = 0 $
Why are these two points not valid contestants for being critical points?
Secondly, in the second part of the equation, my first hunch was to find the second derivative and then substitute the the two I have in my first workings. However, I realized that maybe the faster way would simply be to sketch the solutions with the two equilibrium points as the initial conditions on MATLAB and seeing where the solutions go (i.e. whether they are attracting or repelling).
Is there any computational way (not graphical) to determine the stability type?
I'm asking this because finding a general solution in terms of y is notoriously hard because of the h term, if the h term was not there it is easy to find a partial fraction decomposition and integrate the terms. Wolfram Alpha seems to agree with me:
Lets begin with the first point as outlined in my comment. Having $$ ry\left[1-\frac{y}{K}-\frac{h}{ry}\right] = 0 $$ creates an issue for $y=0$ being a solution - since we have $0\cdot\left[\cdots+\frac{1}{0}\right]=?$ problem.
Now on to the second issue how to determine stability. Usually one linearises around the fixed points and see how the solution behaves. This can be thought of as $$ y = y_e + \delta y $$ where for $$ \dot{y} = f(y) $$ we have $$ f(y_e) = 0 $$ and all variation is within the $\delta y$ so we have $$ \dot{\delta y} = f(y_e) + \frac{df}{dy}\vert_{y=y_e}\delta y + O\left((\delta y)^2\right) $$ so we have $$ \dot{\delta y} = 0 + \left(r-\frac{2ry_e}{K}\right)\delta y $$ or $$ \delta y = A\mathrm{e}^{\left(r-\frac{2ry_e}{K}\right)t} = A\mathrm{e}^{\pm r\sqrt{1-\frac{4h}{rK}}t} $$ so we have conditions of $$ 1-\frac{4h}{rK} > 0 \\ 1-\frac{4h}{rK} < 0 $$ what do they imply about the way $\delta y$ behaves and hence $y$ varies initially?