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 $$\frac{dx}{dt}=2y-z,\qquad \frac{dy}{dt}=3x-2z,\qquad \frac{dz}{dt}=5x-4y.$$

Step 1 · Coefficient matrix

$$A=\begin{pmatrix}0&2&-1\\3&0&-2\\5&-4&0\end{pmatrix}.$$

Step 2 · Characteristic equation

Expanding $\det(A-\lambda I)=0$ gives $$-\lambda^3+9\lambda-8=0\quad\Longrightarrow\quad\lambda^3-9\lambda+8=0.$$

Step 3 · Solving the characteristic equation

$\lambda=1$ is a root: $1-9+8=0$. Factoring, $\lambda^3-9\lambda+8=(\lambda-1)(\lambda^2+\lambda-8)$, and the quadratic gives $\lambda=\dfrac{-1\pm\sqrt{33}}{2}$. The three eigenvalues are $$\lambda_1=1,\qquad \lambda_2=\frac{-1+\sqrt{33}}{2}\approx2.372,\qquad \lambda_3=\frac{-1-\sqrt{33}}{2}\approx-3.372.$$

Step 4 · Classification of the rest point

All three eigenvalues are real and distinct, with mixed signs: $\lambda_3<0<\lambda_1<\lambda_2$. Two eigenvalues are positive and one is negative, so the origin is a saddle point of index 1: unstable, with a two-dimensional unstable manifold and a one-dimensional stable manifold.

Step 5 · Eigenvectors and invariant subspaces

For $\lambda_1=1$: eigenvector $(1,1,1)^\top$. The corresponding eigenvectors for $\lambda_2,\lambda_3$ can be computed similarly by solving $(A-\lambda_k I)\mathbf v=0$. The unstable manifold $E^u$ (two-dimensional) is spanned by the eigenvectors of $\lambda_1$ and $\lambda_2$; the stable manifold $E^s$ (one-dimensional) is spanned by the eigenvector of $\lambda_3$.

Step 6 · General solution

$$\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix} =C_1 e^{t}\mathbf{v}^{(1)} +C_2 e^{\lambda_2 t}\mathbf{v}^{(2)} +C_3 e^{\lambda_3 t}\mathbf{v}^{(3)}.$$ For generic initial conditions ($C_1\neq0$ or $C_2\neq0$), the solution is dominated by the fastest growing mode $e^{\lambda_2 t}$ and diverges to infinity as $t\to+\infty$.

Step 7 · Stability analysis

Since two eigenvalues have positive real part, the rest point is unstable in the sense of Lyapunov: there exist initial conditions arbitrarily close to the origin whose trajectories leave any fixed neighbourhood. 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 eigenvalues are $$\lambda_1=1>0,\quad \lambda_2=\frac{-1+\sqrt{33}}{2}\approx2.372>0,\quad \lambda_3=\frac{-1-\sqrt{33}}{2}\approx-3.372<0.$$ 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.