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 $$\dot{x}=2x-y+2z,\qquad\dot{y}=5x-3y+3z,\qquad\dot{z}=-x-2z.$$

Step 1 · Coefficient matrix

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

Step 2 · Characteristic equation

Expanding $\det(A-\lambda I)=0$ gives $$\lambda^3+3\lambda^2+3\lambda+1=0,$$ which is recognised as a perfect cube: $(\lambda+1)^3=0$. Hence there is a single eigenvalue $\lambda=-1$ of algebraic multiplicity 3.

Step 3 · Geometric multiplicity

Row-reducing $A+I$ shows $\operatorname{rank}(A+I)=2$, so the nullity is 1: the geometric multiplicity of $\lambda=-1$ is only 1, less than its algebraic multiplicity 3. The matrix is defective, requiring a Jordan chain of length 3.

Step 4 · Jordan chain

Solving successively $(A+I)\mathbf v^{(1)}=0$, $(A+I)\mathbf v^{(2)}=\mathbf v^{(1)}$, $(A+I)\mathbf v^{(3)}=\mathbf v^{(2)}$ gives $$\mathbf v^{(1)}=(-1,-1,1)^\top,\qquad \mathbf v^{(2)}=(-1,-2,0)^\top,\qquad \mathbf v^{(3)}=(0,1,0)^\top.$$

Step 5 · General solution

For a triple eigenvalue $\lambda=-1$ with a single Jordan block of size 3: $$\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix} =e^{-t}\!\left[ C_1\mathbf{v}^{(1)} +C_2\!\left(t\,\mathbf{v}^{(1)}+\mathbf{v}^{(2)}\right) +C_3\!\left(\tfrac{t^2}{2}\,\mathbf{v}^{(1)}+t\,\mathbf{v}^{(2)}+\mathbf{v}^{(3)}\right) \right].$$ In components, each of $x(t),y(t),z(t)$ has the form $p(t)e^{-t}$ where $p(t)$ is a polynomial of degree at most 2 in $t$, e.g. $$z(t)=e^{-t}\!\left[C_1+C_2\,t+\tfrac{C_3}{2}t^2\right].$$

Step 6 · Behaviour of trajectories

Since $\lambda=-1<0$ and every component has the form (polynomial in $t$)$\cdot e^{-t}$, and $t^k e^{-t}\to0$ for any $k\geq0$ as $t\to+\infty$, every component tends to zero regardless of the constants $C_1,C_2,C_3$ — though the decay is not purely exponential, since for large $t$ the dominant term is $\tfrac{C_3}{2}t^2e^{-t}$. As $t\to-\infty$, $e^{-t}\to+\infty$ and all solutions diverge.

Step 7 · Stability analysis

Since the unique eigenvalue $\lambda=-1$ satisfies $\operatorname{Re}(\lambda)=-1<0$, the rest point is stable and asymptotically stable: every trajectory converges to the origin as $t\to+\infty$, since $t^k e^{-t}\to0$ for all $k\geq0$. This is a stable improper (degenerate) node in three-dimensional phase space, arising from the single Jordan block of size 3.

Step 8 · Conclusion

$$\boxed{ (0,0,0)\text{ is a \textbf{stable improper node} (asymptotically stable).}\quad \lambda=-1\;(\text{algebraic multiplicity }3,\;\text{geometric multiplicity }1). }$$ The single Jordan block of size 3 for $\lambda=-1<0$ produces solutions of the form $p(t)e^{-t}$ where $p(t)$ is a polynomial of degree at most 2. All trajectories converge to the origin as $t\to+\infty$, confirming asymptotic stability, but the approach is not purely exponential due to the polynomial pre-factors inherited from the Jordan structure.

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.