The Question
The equation $$y''(t)+qy'(t)-v^2\sin(y(t))=-py(t-r)$$ describes a damped pendulum with a delayed negative feedback restoring force. Where $y(t)$ is the deviation from the "up" state, and the parameters $q,p,v$, and $r$ are all positive constants.
Determine the stability of $y=0$ and possible Hopf Bifurcation of this equation.
What I need help with
I am having trouble with finding the characteristic equation (See the end of the linearization section for the system I am having trouble finding the characteristic equation of).
Once I have the characteristic equation how do I find when the Hopf bifurcation occours?
Work Thus Far
First we find the linearization of the delayed differential equation (as $\sin(y(t))$ is nonlinear in $y(t)$), then the characteristic equation from that.
Linearize the equation
We introduce the function $x(t)$ which is defined from the following procedure. \begin{align*} -py(t-r)&=y''(t)+qy'(t)-v^2\sin(y(t)) \\ x'(t)=y''(t)&=-qy'(t)+v^2\sin(y(t))-py(t-r) \\ &=-qx(t)+v^2\sin(y(t))-py(t-r) \end{align*} So so get the equivalent two dimensional system \begin{cases} y'(t)=x(t) \\ x'(t)=-qx(t)+v^2\sin(y(t))-py(t-r) \end{cases}
Let now let $f(x(t),y(t),y(t-r))=-qx(t)+v^2\sin(y(t))-py(t-r)$. We linearize $f(x(t),y(t),y(t-r))$. Note that the equilibrium point is $(\bar{x},\bar{y},\bar{y}(t-r))=(0,0,0)$. \begin{align*} f(x(t),y(t),y(t-r)) &\approx \frac{\partial f(\bar{x},\bar{y},\bar{y}(t-r))}{\partial x}x+\frac{\partial f(\bar{x},\bar{y},\bar{y}(t-r))}{\partial y}y+\frac{\partial f(\bar{x},\bar{y},\bar{y}(t-r))}{\partial y(t-r)}y(t-r) \\ &=-q x(t)+v^2 \cos(\bar{y}) y(t)-py(t-r) \\ &=-q x(t)+v^2 y(t)-py(t-r) \end{align*} We now have the linearized system: \begin{cases} y'(t)=x(t) \\ x'(t)=-q x(t)+v^2 y(t)-py(t-r) \end{cases}
Characteristic Equation
I do not know the method/formula for finding the characteristic equation of a 2 dimensional system. I know in the 1D case once the system is linearized we let $x(t)=e^{\lambda t}$, then find the roots of characteristic equation in terms of $\lambda$.
I think that the characteristic equation that I get in the end should be of the form $$P(\lambda)+Q(\lambda)e^{-r \lambda}=0 $$ with $P(\lambda),Q(\lambda)$ being polynomial functions. Once it is in this form I have theorems(See F. Braur [JDE, 69(1987),185-191]) that will help me determine criteria for absolute stability.
Note that $$y''(t)=x'(t)=-qx(t)+v^2y(t)-py(t-r)\\ =-qy'(t)+v^2y(t)-py(t-r)$$ If you take now the Laplace transform of this equation you will obtain $$\lambda^2+q\lambda+v^2+p e^{-r\lambda}=0$$ which is the desired characteristic equation. This equation has an infinite number of roots. You need to find when the roots $\lambda=j\omega$ appear in order to identify the bifurcation point.