## 3 Results
### The Case \(\tau>0\)  
We are interested in stability switches due to the presence of a delay \(\tau>0\) in (8). Consider the case \(n=m=2\) and \(x_{th}=y_{th}\), and let \((\tilde{x},\tilde{y})\) be one equilibrium point of (8). The linearized system about \((\tilde{x},\tilde{y})\) is given by  
\[\dot{Z}(t)=AZ(t)+BZ(t-\tau), \tag{15}\]  
with  
\[Z(t)= \begin{pmatrix}z_{1}(t)\\ z_{2}(t)\end{pmatrix} ,A= \begin{pmatrix}a&b\\ 0&d\end{pmatrix} ,B= \begin{pmatrix}0&0\\ c&0\end{pmatrix} ,\]  
and  
\[ \begin{split} a&=&-\gamma-\delta\tilde{y}+2\beta x_{th}^{2} \frac{\tilde{x}}{(x_{th}^{2}+\tilde{x}^{2})},\\ b&=&-\delta\tilde{x}\leq 0,\\ c&=& 2\rho x_{th}^{2}\frac{\tilde{x}}{(x_{th}^{2}+\tilde{x}^{2})} \geq 0\\ d&=&-\omega<0.\end{split}  \tag{16}\]  
The characteristic equation corresponding to (15) is given by  
\[det\left(\lambda\mathbb{I}-A-Be^{-\lambda\tau}\right)=0,\]  
or equivalently,  
\[\lambda^{2}-\lambda(a+d)+ad-bce^{-\lambda\tau}=0. \tag{17}\]  
Characteristic equations of this and more general type have been studied in [24]. In the following, we report results from [24], adapting them to our specific example. We apply standard methods for the analysis of characteristic equations and switches with respect to increasing delays, hence we consider purely imaginary roots, \(\lambda=i\varphi,\varphi>0\). Separating real and imaginary parts in (17) we obtain  
\[ \begin{split}\varphi^{2}-ad&=-bc\cos(\varphi t)\\ \varphi(a+d)&=bc\sin(\varphi t).\end{split} \]  
Now we square left- and right-hand sides and sum up the two equations, obtaining  
\[\varphi^{4}+\varphi^{2}(a^{2}+d^{2})+a^{2}d^{2}=b^{2}c^{2}. \tag{18}\]  
Its roots are  
\[\varphi_{\pm}^{2}=\frac{1}{2}\left(-(a^{2}+d^{2})\pm\sqrt{(a^{2}-d^{2})^{2}+4 b^{2}c^{2}}\right).\]  
It can be seen from (18) that the parabola in \(\varphi^{2}\) is open upwards and it has:  
* no positive intercept with the horizontal axis, if \(a^{2}d^{2}-b^{2}c^{2}>0\), _i.e._, if \(|ad|>-bc\);
* one positive intercept (\(\varphi_{+}\)) with the horizontal axis, if \(|ad|<-bc\). In the first case, there is no stability switch with respect to \(\tau\); that is, the stability of the equilibrium point \((\tilde{x},\tilde{y})\) remains the same for any \(\tau\geq 0\), and it is sufficient to study the ODE case (12). In the case \(|ad|<-bc\), there is one root (\(\varphi_{+}\)) with positive imaginary part, hence one stability switch. In order to find out in which direction the stability switch occurs, we study the sign of the real part \(\Re\lambda(\tau)\) in \(\lambda=i\varphi_{+}\), for \(\tau>0\). From (17) we have  
\[\left\{2\lambda-(a+d)+\tau bce^{-\lambda\tau}\right\}\frac{d\lambda(\tau)}{d\tau }=\lambda bce^{-\lambda\tau}.\]  
It follows  
\[\text{sign}\left\{\frac{d\Re\lambda(\tau)}{d\tau}\right\}_{ \lambda=i\varphi_{+}} =\text{sign}\left\{\Re\left(\frac{d\lambda}{d\tau}\right)^{-1} \right\}_{\lambda=i\varphi_{+}}\] \[=\text{sign}\left\{\Re\left(\frac{2\lambda-(a+d)}{\lambda( \lambda^{2}-(a+d)\lambda+ad)}\right)\right\}_{\lambda=i\varphi_{+}}\] \[=\text{sign}\left\{\frac{2(\varphi_{+}^{2}-ad)+(a+d)^{2}}{( \varphi_{+}^{2}(a+d)^{2}+(ad-\varphi_{+}^{2})^{2}}\right\}\] \[=\text{sign}\left\{2(\varphi_{+}^{2}-ad)+(a+d)^{2}\right\}\] \[=\text{sign}\left\{2\varphi_{+}^{2}+a^{2}+d^{2}\right\}\] \[=+1.\]  
Roots cross the imaginary axis from the left to the right, indicating stability loss. If the solution \((\tilde{x},\tilde{y})\) is asymptotically stable for \(\tau=0\), then it is uniformly asymptotically stable for all \(\tau<\tau_{c}\) and unstable for \(\tau>\tau_{c}\), where  
\[\tau_{c}=\frac{\theta_{c}}{\varphi_{+}}, \tag{19}\]  
with \(\theta_{c}\) implicitly defined by  
\[\varphi_{\pm}^{2}=\frac{1}{2}\left(-(a^{2}+d^{2})\pm\sqrt{(a^{2}-d^{2})^{2}+4 b^{2}c^{2}}\right).\]0  
All in all, we have shown the following result. **Theorem 2**. : _Let \((\tilde{x},\tilde{y})\) be one equilibrium point of (8), with \(\tau>0\), \(n=m=2\) and \(x_{th}=y_{th}\). Assume that \(|ad|<bc\), with \(a,b,c,d\) given in (16).