ChiralSuperfields said:
Thank you for your replies
@fresh_42 ! Do you please know whether when the solution is a stable node if rs is large and or when rs is small then origin is asymmetrically stable?
Thanks!
We have a trace ##-(g+d)=-3.68## and a determinant ##gd-rs=3.165## and two real eigenvalues
$$
\lambda =\dfrac{1}{2}\left(-3.68 \pm \sqrt{(2.12)^2-4\cdot 4.3\cdot 0.21}\,\right) \in \{-2.74\ ,\ -4.62\}.
$$
\begin{align*}
x'(t)&=0 \Longrightarrow y=-\dfrac{g}{r}\,x=-0.6744 \,x\\[6pt]
y'(t)&=0 \Longrightarrow y=\dfrac{s}{d}\,x=0.269\,x
\end{align*}
\begin{align*}
0&=x'' + 3.68x' +3.165=3.68\cdot \left(\dfrac{1}{3.68} x'' + x' + \dfrac{3.165}{3.68}\right)\\[6pt]
x(t)&=-\dfrac{1}{3.68} \cdot c_1 \cdot e^{-3.68\, t} +c_2 + \dfrac{3.165}{3.68} \cdot t\\[6pt]
x(t)&=-0.27174 \cdot c_1 \cdot e^{-3.68\, t} +c_2 + 0.86\cdot t\\[6pt]
x'(0)&=c_1-\dfrac{\det (A)}{\operatorname{trace}(A)}=c_1+\dfrac{gd-rs}{g+d}=c_1+0.86\\[6pt]
x(0)&=c_2 + \dfrac{1}{\operatorname{trace}(A)}\cdot c_1 =c_2-\dfrac{1}{g+d}\cdot c_1=c_2-\dfrac{c_1}{3.68}\\
&= c_2+\dfrac{1}{\operatorname{trace}(A)}\cdot \left(x'(0)+\dfrac{\det (A)}{\operatorname{trace}(A)}\right)
\end{align*}
This is in terms of the parameters
$$
x(t)=-\dfrac{1}{g+d} \cdot c_1 \cdot e^{-(g+d)\,t} +c_2 + \dfrac{gd-rs}{g+d} \cdot t
$$
If you solve the ODE for ##y(t)## in the same way, you should have all the information you can get. Now you can play around with different initial conditions, the variation of the parameters, and whatever you want to find. I would draw an ##(x,y)## coordinate system for various initial values and its vector field. The variation of parameters is probably a mess. We have ##4## parameters ##g,d,r,s## which can be either increased or decreased which are already ##16## cases, plus the fact that the eigenvalues (that also depend on the parameters) can become only one eigenvalue if the discriminant of the quadratic equation is ##0,## or become complex if the discriminant of the quadratic equation is negative.
