For which values of the parameter α is the rest point…
For which values of the parameter $\alpha$ is the rest point $(0,0,0)$ of the system $$\dot{x}=\alpha x-y,\qquad\dot{y}=\alpha y-z,\qquad\dot{z}=\alpha z-x$$ stable?
Explanation of Method
The eigenvalues of the coefficient matrix are computed as explicit functions of the free parameter. The sign of their real part — via the discriminant of the characteristic equation, or the Routh–Hurwitz criterion applied directly to the characteristic polynomial — is then analysed case by case to pin down exactly which values of the parameter make the rest point stable, asymptotically stable, or unstable.
Step 1 · Coefficient matrix
$$A=\begin{pmatrix}\alpha&-1&0\\0&\alpha&-1\\-1&0&\alpha\end{pmatrix}.$$
Step 2 · Characteristic equation
Setting $\mu=\alpha-\lambda$, the determinant reduces to $\mu^3-1=0$, so $$(\alpha-\lambda)^3=1.$$
Step 3 · Solving the characteristic equation
The cube roots of unity are $1,\omega,\omega^2$ with $\omega=-\tfrac12+\tfrac{\sqrt3}{2}i$, giving the three eigenvalues $$\lambda_1=\alpha-1,\qquad \lambda_{2,3}=\alpha+\frac{1}{2}\mp\frac{\sqrt{3}}{2}\,i.$$
Step 4 · Stability conditions
Asymptotic stability requires $\operatorname{Re}(\lambda_k)<0$ for all $k$. From $\lambda_1$: $\alpha<1$. From $\lambda_{2,3}$: $\alpha<-\tfrac12$. The more restrictive condition is $$\alpha<-\frac{1}{2}.$$
Step 5 · Verification via the Routh--Hurwitz criterion
The characteristic polynomial is $\lambda^3-3\alpha\lambda^2+3\alpha^2\lambda+1-\alpha^3=0$, i.e.\ $a_1=-3\alpha$, $a_2=3\alpha^2$, $a_3=1-\alpha^3$. The conditions $a_1>0$, $a_3>0$, $a_1a_2-a_3>0$ give $\alpha<0$, $\alpha<1$, and $\alpha<-\tfrac12$ respectively; the binding condition is $\alpha<-\tfrac12$, confirming the eigenvalue analysis.
Step 6 · Boundary cases
At $\alpha=-\tfrac12$: $\lambda_{2,3}=\pm\tfrac{\sqrt3}{2}i$, $\lambda_1=-\tfrac32<0$ — a centre in the plane of the complex eigenvectors, stable but not asymptotically stable. At $\alpha=1$: $\lambda_1=0$, $\lambda_{2,3}=\tfrac32\mp\tfrac{\sqrt3}{2}i$ — unstable.
Step 7 · Summary table
| Range of $\alpha$ | Eigenvalue real parts | Stability of $(0,0,0)$ |
|---|---|---|
| $\alpha<-\tfrac{1}{2}$ | $\alpha-1<0,\;\alpha+\tfrac{1}{2}<0$ | Asymptotically stable |
| $\alpha=-\tfrac{1}{2}$ | $-\tfrac{3}{2}<0,\;\operatorname{Re}(\lambda_{2,3})=0$ | Stable (centre), not asympt.\ stable |
| $-\tfrac{1}{2}<\alpha<1$ | $\alpha-1<0,\;\alpha+\tfrac{1}{2}>0$ | Unstable |
| $\alpha=1$ | $\lambda_1=0,\;\operatorname{Re}(\lambda_{2,3})=\tfrac{3}{2}>0$ | Unstable |
| $\alpha>1$ | $\alpha-1>0,\;\alpha+\tfrac{1}{2}>0$ | Unstable |
Step 8 · Conclusion
$$\boxed{ \text{The rest point }(0,0,0)\text{ is \textbf{asymptotically stable} if and only if }\alpha<-\dfrac{1}{2}. }$$ The three eigenvalues are $\lambda_1=\alpha-1$ and $\lambda_{2,3}=\alpha+\tfrac12\mp\tfrac{\sqrt3}{2}i$, and the binding stability condition is $\operatorname{Re}(\lambda_{2,3})=\alpha+\tfrac12<0$, i.e.\ $\alpha<-\tfrac12$. At $\alpha=-\tfrac12$ the origin is a non-asymptotically-stable centre; for $\alpha>-\tfrac12$ it is unstable.
Summary Table
| Range of $\alpha$ | Eigenvalue real parts | Stability of $(0,0,0)$ |
|---|---|---|
| $\alpha<-\tfrac{1}{2}$ | $\alpha-1<0,\;\alpha+\tfrac{1}{2}<0$ | Asymptotically stable |
| $\alpha=-\tfrac{1}{2}$ | $-\tfrac{3}{2}<0,\;\operatorname{Re}(\lambda_{2,3})=0$ | Stable (centre), not asympt.\ stable |
| $-\tfrac{1}{2}<\alpha<1$ | $\alpha-1<0,\;\alpha+\tfrac{1}{2}>0$ | Unstable |
| $\alpha=1$ | $\lambda_1=0,\;\operatorname{Re}(\lambda_{2,3})=\tfrac{3}{2}>0$ | Unstable |
| $\alpha>1$ | $\alpha-1>0,\;\alpha+\tfrac{1}{2}>0$ | Unstable |
Additional Notes
- The boundary values of the parameter (where the discriminant vanishes, or a real part passes through zero) mark the transitions between qualitatively different phase portraits, and deserve special attention alongside the open ranges.
Result
The final classification is: