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

Examine the stability of the rest point $(0,0)$ of the system $$\dot{x}=-4y-x^3,\qquad\dot{y}=3x-y^3,$$ by the method of first approximation (linearisation).

Step 1 · Verification that $(0,0)$ is a rest point

At $(0,0)$ both right-hand sides vanish.

Step 2 · Separation of linear and nonlinear parts

$$\dot x=-4y+\underbrace{(-x^3)}_{=:f_2},\qquad\dot y=3x+\underbrace{(-y^3)}_{=:g_2},$$ with $f_2,g_2=O(r^3)=o(r)$.

Step 3 · Jacobian matrix at $(0,0)$

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

Step 4 · Characteristic equation

$$\det(A-\lambda I)=0\implies\lambda^2+12=0\implies\lambda_{1,2}=\pm2\sqrt3\,i.$$

Step 5 · Classification of eigenvalues — critical case

The eigenvalues are purely imaginary: $\operatorname{Re}(\lambda_{1,2})=0$. This is a centre for the linearisation, but the linearisation theorem is inconclusive here — the nonlinear terms may turn the centre into a focus (stable or unstable) or preserve it.

Step 6 · Lyapunov function approach

Consider $V(x,y)=3x^2+4y^2$, positive definite. Along trajectories: $$\dot V=6x\dot x+8y\dot y=6x(-4y-x^3)+8y(3x-y^3) =-24xy-6x^4+24xy-8y^4=-6x^4-8y^4.$$

Step 7 · Analysis of $\dot V$

$\dot V(x,y)=-6x^4-8y^4\leq0$, with equality only at $(0,0)$: $V$ is a strict Lyapunov function.

Step 8 · Lyapunov's direct method — stability theorem

Since $V$ is continuously differentiable, positive definite, and $\dot V<0$ for $(x,y)\neq(0,0)$, Lyapunov's stability theorem gives asymptotic stability of $(0,0)$.

Step 9 · Why the linearisation was inconclusive

The linear part of $\dot V$ is $6x(-4y)+8y(3x)=-24xy+24xy=0$: it cancels exactly (as expected for a linear centre), and the sign of $\dot V$ is determined entirely by the nonlinear terms $-6x^4-8y^4<0$. This is a paradigmatic example where linearisation gives a centre (purely imaginary eigenvalues) but the full nonlinear system is actually an asymptotically stable focus.

Step 10 · General solution of the linearised system

The linearised system $\dot\xi=-4\eta$, $\dot\eta=3\xi$ has trajectories $3\xi^2+4\eta^2=C$ (ellipses), traversed with angular frequency $\omega=2\sqrt3$ and period $T=\tfrac{\pi\sqrt3}{3}$.

Step 11 · Direction of rotation

At $(\varepsilon,0)$: $\dot y=3\varepsilon>0$, so trajectories (and the linearised ellipses) are traversed counter-clockwise.

Step 12 · Stability analysis

By Lyapunov's direct method: the level sets $V=c$ are positively invariant, and $\dot V<0$ everywhere except the origin forces $V(x(t),y(t))\to0$, hence $(x(t),y(t))\to(0,0)$: both Lyapunov stable and asymptotically stable.

Step 13 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{asymptotically stable focus} of the nonlinear system.} }$$ The eigenvalues of the linearisation are $\lambda_{1,2}=\pm2\sqrt3\,i$ (purely imaginary), so the first-approximation method is inconclusive. Stability is established instead by Lyapunov's direct method, using the strict Lyapunov function $V(x,y)=3x^2+4y^2$ with $\dot V=-6x^4-8y^4<0$ for $(x,y)\neq(0,0)$. The cancellation of all linear terms in $\dot V$, combined with the strict negativity of the remaining quartic terms, confirms asymptotic stability with counter-clockwise spiralling trajectories.

Summary Table

Additional Notes

• This is a case where the linearisation is inconclusive (purely imaginary eigenvalues) but the true nonlinear behaviour — here, an asymptotically stable focus rather than a centre — is revealed only by exhibiting a genuine Lyapunov function for the full nonlinear system.