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

Examine the stability of the rest point $(0,0)$ of the system $$\dot{x}=-\sin x+3y+x^5,\qquad\dot{y}=\tfrac{1}{4}x-2y-\tfrac{1}{6}y^3,$$ by the method of first approximation (linearisation).

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

At $(0,0)$: $-\sin0+3\cdot0+0=0$ and $\tfrac14\cdot0-2\cdot0-0=0$.

Step 2 · Separation of linear and nonlinear parts

Using $-\sin x=-x+\tfrac{x^3}{6}+O(x^5)$: $$\dot x=-x+3y+\underbrace{\tfrac{x^3}{6}+\tfrac{119x^5}{120}+\cdots}_{=:f_2}, \qquad \dot y=\tfrac14x-2y+\underbrace{\left(-\tfrac16y^3\right)}_{=:g_2}.$$ Both remainders satisfy $f_2,g_2=O(r^3)=o(r)$ as $r\to0$.

Step 3 · Jacobian matrix at $(0,0)$

$$A=\begin{pmatrix}-1&3\\\tfrac14&-2\end{pmatrix}.$$

Step 4 · Characteristic equation

$$\det(A-\lambda I)=0\implies4\lambda^2+12\lambda+5=0 \implies\lambda=\frac{-12\pm8}{8}.$$ Thus $\lambda_1=-\tfrac12$, $\lambda_2=-\tfrac52$.

Step 5 · Classification of eigenvalues

Both real, distinct, negative: $\lambda_2=-\tfrac52<\lambda_1=-\tfrac12<0$.

Step 6 · Verification of the $o(r)$ condition

For $r\leq1$: $|f_2|\leq\tfrac{139}{120}r^3$, $|g_2|\leq\tfrac16r^3$, so $\|F_2\|/r\leq Cr^2\to0$.

Step 7 · Application of the linearisation theorem

Since both eigenvalues have strictly negative real part and the nonlinear terms are $o(r)$, Lyapunov's First Method gives asymptotic stability of the nonlinear rest point.

Step 8 · Routh--Hurwitz verification

For $\lambda^2+3\lambda+\tfrac54$: $a_1=3>0$, $a_2=\tfrac54>0$ — both satisfied.

Step 9 · Eigenvectors and nodal structure

For $\lambda_1=-\tfrac12$: eigenvector $\mathbf v^{(1)}=(6,1)^\top$ (slower axis, tangent to $y=\tfrac16x$). For $\lambda_2=-\tfrac52$: eigenvector $\mathbf v^{(2)}=(-2,1)^\top$ (faster axis, tangent to $y=-\tfrac12x$).

Step 10 · General solution of the linearised system

$$\xi(t)=6C_1e^{-t/2}-2C_2e^{-5t/2},\qquad \eta(t)=C_1e^{-t/2}+C_2e^{-5t/2}.$$ Both terms decay; the dominant (slower) rate $e^{-t/2}$ governs the approach to the origin along $y=\tfrac16x$ — a stable node.

Step 11 · Behaviour of the nonlinear system

By the Hartman--Grobman theorem (real, negative, distinct eigenvalues, none zero), the nonlinear system is topologically conjugate to its linearisation near the origin: a genuine stable node.

Step 12 · Stability analysis

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

Step 13 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{asymptotically stable node} of the nonlinear system.} }$$ The eigenvalues of the Jacobian at the origin are $\lambda_1=-\tfrac12<0$ and $\lambda_2=-\tfrac52<0$, both real and negative. By Lyapunov's First Method, the nonlinear terms $f_2=O(r^3)$ and $g_2=O(r^3)$ do not affect the conclusion, and the rest point is asymptotically stable, with trajectories approaching it tangentially to $y=\tfrac16x$.

Summary Table

Additional Notes

• The Hartman–Grobman theorem guarantees that, because no eigenvalue of the Jacobian is purely imaginary, the nonlinear phase portrait near the rest point is topologically equivalent to that of its linearisation — so the type found for the linear system (node, saddle, or focus) carries over unchanged to the full nonlinear system.