Examine the character of the rest point (0,0,0) of the…

Examine the character of the rest point $(0,0,0)$ of the system $$\dot{x}=x-y-z,\qquad\dot{y}=x+y-3z,\qquad\dot{z}=x-5y-3z.$$

Step 1 · Coefficient matrix

$$A=\begin{pmatrix}1&-1&-1\\1&1&-3\\1&-5&-3\end{pmatrix}.$$

Step 2 · Characteristic equation

Expanding $\det(A-\lambda I)=0$ gives $$\lambda^3+\lambda^2-18\lambda+12=0.$$

Step 3 · Finding the roots

No rational root exists (testing $\lambda=1,2,3,-6$ all fail), so the depressed cubic is solved trigonometrically (casus irreducibilis, since the discriminant is positive, giving three distinct real roots): $$\lambda_1\approx3.332,\qquad\lambda_2\approx-5.041,\qquad\lambda_3\approx0.712.$$ These satisfy $\lambda^3+\lambda^2-18\lambda+12=0$ to within numerical accuracy.

Step 4 · Classification of the rest point

All three eigenvalues are real and distinct: $\lambda_2\approx-5.041<0<\lambda_3\approx0.712<\lambda_1\approx3.332$. Two eigenvalues are positive and one is negative, so the rest point is a saddle point of index 1 (one-dimensional stable manifold, two-dimensional unstable manifold). The origin is unstable.

Step 5 · Invariant subspaces

The unstable manifold $E^u$ (two-dimensional) is spanned by the eigenvectors of $\lambda_1$ and $\lambda_3$; trajectories in $E^u\setminus\{0\}$ diverge to infinity as $t\to+\infty$. The stable manifold $E^s$ (one-dimensional) is spanned by the eigenvector of $\lambda_2$; trajectories in $E^s\setminus\{0\}$ converge to the origin as $t\to+\infty$.

Step 6 · General solution

$$\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix} =C_1 e^{\lambda_1 t}\mathbf{v}^{(1)} +C_2 e^{\lambda_2 t}\mathbf{v}^{(2)} +C_3 e^{\lambda_3 t}\mathbf{v}^{(3)}.$$ For generic initial conditions the solution is dominated by $e^{\lambda_1 t}$ as $t\to+\infty$ and diverges.

Step 7 · Stability analysis

Since two eigenvalues have positive real parts, the rest point is unstable in the sense of Lyapunov; only trajectories starting exactly on the one-dimensional stable manifold $E^s$ converge to the origin.

Step 8 · Conclusion

$$\boxed{ (0,0,0)\text{ is a \textbf{saddle point} (unstable).} }$$ The three real eigenvalues of $A$ satisfy $\lambda^3+\lambda^2-18\lambda+12=0$ and are approximately $$\lambda_1\approx3.332>0,\qquad \lambda_2\approx-5.041<0,\qquad \lambda_3\approx0.712>0.$$ Two positive and one negative eigenvalue give a saddle of index~1: the two-dimensional unstable manifold and the one-dimensional stable manifold intersect only at the origin.

Summary Table

Additional Notes

• In three dimensions the equilibrium is classified by the mix of eigenvalue signs (or, for a repeated eigenvalue, by its Jordan structure), which determines the dimensions of the stable and unstable manifolds.