For few days I am working on Hopf bifurcation of a system like this:
$$ \frac{dx}{dt}=\alpha\frac{x^2 y + a x y}{x^2+bx+1}-1\\ \frac{dy}{dt}=\frac{1-y-cx^2 y }{1+x^2} $$
above $\alpha$ is the bifurcation parameter and $a,\;b,\;c$ are real parameters. I want to determine the parameter value that the Hopf bifurcation would occur. I have 2 sources (a thesis and an article focusing on another set of equations) for finding the parameters, I tried to recreate their results but I failed miserably. Found steady state values as $x=x(\alpha),\;y=y(x)$
$$ 0=\alpha\frac{x^2 y + a x y}{x^2+bx+1}-1\\ 0=\frac{1-y-cx^2 y }{1+x^2}\\ $$ then for combining two equations from y $$ y=\frac{1}{\alpha}\frac{x^2+bx+1}{x^2 + a x }\\ y=\frac{1}{1+cx^2} $$
and got an equation for $x$ and $\alpha$ only
$$ {\alpha}=\frac{(1+cx^2)(x^2+bx+1)}{x^2 + a x } $$
now stated bifurcation parameter in terms of the stated state. Then for
$$ F(x,y)=\alpha\frac{x^2 y + a x y}{x^2+bx+1}-1\\ G(x,y)=\frac{1-y-cx^2 y }{1+x^2} $$
wrote the characteristic equation and solve for lambda to find eigenvalues of the system. Meanwhile replaced
$$ y\to\frac{1}{1+cx^2}\\ {\alpha}\to\frac{(1+cx^2)(x^2+bx+1)}{x^2 + a x } $$ at the characteristic equation solution. For finding Hopf bifurcation I look for the the values of $x$ making the eigenvalue of the Jacobian $0$, such that by replacing those $x$ values can find the exact Hopf b. value from $$ {\alpha}=\frac{(1+cx^2)(x^2+bx+1)}{x^2 + a x } $$ however it does give wrong answers. Can you trace something wrong here? (Let's say everything looks good can you please suggest a document where each step in the computation is clear so that I can go through and realize where i do wrong)
Thank you for your time.
Note that the variable $x$ complicates the equilibrium calculation, so we set $x=x_0$ and $y=y_0$, and solve as follows for $\alpha$ and $c$:
$$ (\alpha^*,c^*)=\left(\frac{b x_0+x_0^2+1}{x_0 y_0 \left(a+x_0\right)},\frac{1-y_0}{x_0^2 y_0} \right) $$
In this way, by setting $\alpha=\alpha^*$ and $c=c^*$, your equilibrium point is given by $(x_0,y_0)$. Now, for example, if we consider that $x_0=\frac{1}{2}$ and $y_0=\frac{1}{2}$, and considering $a=1$ and $b=1$, we obtain two non-trivial equilibria, as shown in the following figure:
On the other hand, the linear approximation at $(1/2,1/2)$ is given by
$$ A=\left( \begin{array}{cc} \frac{4 (3 a+b+4)}{(2 a+1) (2 b+5)} & 2 \\ -\frac{8}{5} & -\frac{8}{5} \\ \end{array} \right) $$
Considering $b$ as bifurcation parameter, we have that the critical Hopf bifurcation value is given by
$$ b_0=\frac{5 (2-a)}{8 a-1}\hspace{0.5cm}\text{and}\hspace{0.5cm}\text{det}(A(a,b_0))=\frac{16}{25}>0. $$
The transversality condition:
$$ \frac{d}{db}(-\text{tr}(A(a,b_0))=\frac{4 (1-8 a)^2}{75 (2 a+1)^2}>0. $$
Therefore, the equilibrium point $(1/2,1/2)$ is locally asymptotically stable for $b<b_0$ and exhibits a Hopf bifurcation when $b\geq b_0$. You can verify that the eigenvalues at critical bifurcation value are purely imaginary and are the following:
$$ \lambda=\frac{4}{5}i\hspace{0.5cm}\text{and}\hspace{0.5cm}\bar{\lambda}=-\frac{4}{5}i $$
The limit cycle considering $a=1$ and $b=b_0+1/10000$:
Only the calculation of the first Lyapunov coefficient is needed to have a complete proof of the existence of the Hopf bifurcation.
If you want a more complete proof, don't assign numerical values to $x_0$ and $y_0$ and proceed in the same way. You just have to take care that the determinant of the linear approximation at the critical bifurcation value is strictly positive.