Problem 27
Medium 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 $$\dot{x}=x-\tfrac{1}{2}y,\qquad\dot{y}=\tfrac{53}{2}x-y.$$

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

$$A=\begin{pmatrix}1&-\tfrac{1}{2}\\[4pt]\tfrac{53}{2}&-1\end{pmatrix}.$$

Step 2 · Characteristic equation and eigenvalues

$$\det(A-\lambda I)=0\implies\lambda^2+\tfrac{49}{4}=0 \implies\lambda_{1,2}=\pm\tfrac{7}{2}\,i.$$

Step 3 · Classification of the rest point

The eigenvalues are purely imaginary. Since the system is linear, the rest point $(0,0)$ is a centre exactly. All trajectories are closed curves encircling the origin.

Step 4 · Direction of rotation

Evaluating $\dot y$ at $(1,0)$ gives $\dot y=\tfrac{53}{2}>0$, so the trajectories rotate counter-clockwise.

Step 5 · General real solution

With eigenvector $\mathbf v=(1,2)^\top+i(0,7)^\top$ for $\lambda=-\tfrac72i$: $$x(t)=C_1\cos\tfrac72t+C_2\sin\tfrac72t,$$ $$y(t)=(2C_1-7C_2)\cos\tfrac72t+(2C_2+7C_1)\sin\tfrac72t.$$

Step 6 · First integral

Seeking a quadratic invariant gives $a=53$, $b=-4$, $c=1$, so the family of integral curves is $$53x^2-4xy+y^2=K,\quad K>0,$$ a family of ellipses (positive definite since $16-212=-196<0$ with positive leading coefficient).

Step 7 · Behaviour of trajectories

All trajectories are closed ellipses $53x^2-4xy+y^2=K$, traversed counter-clockwise with period $T=\dfrac{2\pi}{7/2}=\dfrac{4\pi}{7}$. No trajectory approaches or recedes from the origin.

Step 8 · Stability analysis

The rest point is stable in the sense of Lyapunov (for any $\varepsilon>0$, choose $K$ small enough that the ellipse lies within the $\varepsilon$-ball) but not asymptotically stable.

Step 9 · Conclusion

$$\boxed{ (0,0)\text{ is a \textbf{centre} (stable, not asymptotically stable).}\quad \lambda_{1,2}=\pm\tfrac{7}{2}\,i. }$$ All trajectories are closed ellipses $53x^2-4xy+y^2=K$ ($K>0$) traversed counter-clockwise with period $T=\tfrac{4\pi}{7}$. The origin is Lyapunov stable but not asymptotically stable.


Summary Table

QuantityValue
Eigenvalues$\det(A-\lambda I)=0\implies\lambda^2+\tfrac{49}{4}=0 \implies\lambda_{1,2}=\pm\tfrac{7}{2}\,i.$
Classificationcentre

Additional Notes

Result

The final classification is:

$$\boxed{ (0,0)\text{ is a \textbf{centre} (stable, not asymptotically stable).}\quad \lambda_{1,2}=\pm\tfrac{7}{2}\,i. }$$