Problem 21
Easy Linearization

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

Examine the character of the rest point $(0,0)$ of the system $$\dot{x}=-3x+\tfrac{7}{2}y,\qquad\dot{y}=-x+y.$$

Explanation of Method

Since the system is already linear and homogeneous, its coefficient matrix is written down directly. The eigenvalues are computed from the characteristic equation, and the rest point is classified according to the standard theory of planar linear systems: a node, saddle, focus, or centre, depending on the sign and nature of the eigenvalues.

Step 1 · Coefficient matrix

$$A=\begin{pmatrix}-3&\tfrac{7}{2}\\-1&1\end{pmatrix}.$$

Step 2 · Characteristic equation and eigenvalues

$$\det(A-\lambda I)=0\implies\lambda^2+2\lambda+\tfrac12=0 \implies2\lambda^2+4\lambda+1=0.$$ Discriminant $\Delta=8>0$, so $$\lambda_1=\frac{-2+\sqrt{2}}{2}\approx-0.293,\qquad \lambda_2=\frac{-2-\sqrt{2}}{2}\approx-1.707.$$

Step 3 · Classification of the rest point

Both eigenvalues are real, distinct, and negative. The rest point $(0,0)$ is a stable node: all trajectories approach the origin as $t\to+\infty$.

Step 4 · Eigenvectors

For $\lambda_1$: eigenvector $(4-\sqrt2,\,2)^\top$, giving the slower nodal axis $y=\tfrac{4+\sqrt2}{7}x$. For $\lambda_2$: eigenvector $(4+\sqrt2,\,2)^\top$, giving the faster nodal axis $y=\tfrac{4-\sqrt2}{7}x$.

Step 5 · General solution

$$x(t)=C_1(4-\sqrt2)e^{\lambda_1 t}+C_2(4+\sqrt2)e^{\lambda_2 t},\qquad y(t)=2C_1e^{\lambda_1 t}+2C_2e^{\lambda_2 t}.$$

Step 6 · Behaviour of trajectories near the origin

Since both eigenvalues are negative, all trajectories converge to the origin as $t\to+\infty$. For generic initial conditions ($C_1\neq0$) the approach is tangential to the slower eigendirection $y=\tfrac{4+\sqrt2}{7}x$ (since $e^{\lambda_1 t}$ decays more slowly than $e^{\lambda_2 t}$); only trajectories exactly on the faster eigendirection approach along $y=\tfrac{4-\sqrt2}{7}x$.

Step 7 · Stability analysis

Since both eigenvalues satisfy $\operatorname{Re}(\lambda_{1,2})<0$, the rest point is stable and asymptotically stable.

Step 8 · Conclusion

$$\boxed{ (0,0)\text{ is a \textbf{stable node} (asymptotically stable).} }$$ The eigenvalues $\lambda_1\approx-0.293$ and $\lambda_2\approx-1.707$ are both real, distinct, and negative. All trajectories approach the origin tangentially to the slower eigendirection $\mathbf v^{(1)}=(4-\sqrt2,2)^\top$, except those lying exactly on the faster eigendirection $\mathbf v^{(2)}=(4+\sqrt2,2)^\top$.


Summary Table

QuantityValue
Eigenvalues$\det(A-\lambda I)=0\implies\lambda^2+2\lambda+\tfrac12=0 \implies2\lambda^2+4\lambda+1=0.$
Classificationstable node

Additional Notes

Result

The final classification is:

$$\boxed{ (0,0)\text{ is a \textbf{stable node} (asymptotically stable).} }$$