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

Examine the stability of the rest point $(0,0)$ of the system $$\dot{x}=-x+y+2x^4-y^6,\qquad\dot{y}=x-3y+11y^4,$$ by the method of first approximation (linearisation).

Step 1 · Separation of linear and nonlinear parts

Write the system as $\dot{x}=f(x,y)$, $\dot{y}=g(x,y)$ with $$f(x,y)=-x+y+\underbrace{2x^4-y^6}_{=:\,f_2},\qquad g(x,y)=x-3y+\underbrace{11y^4}_{=:\,g_2}.$$ The nonlinear remainders satisfy $f_2,g_2=O(r^4)=o(r)$ as $r=\sqrt{x^2+y^2}\to0$, so the hypotheses of the Lyapunov--Poincar\'e linearisation theorem are satisfied.

Step 2 · Linearised system

The Jacobian at $(0,0)$ is $$A=\begin{pmatrix}-1&1\\1&-3\end{pmatrix}.$$

Step 3 · Characteristic equation

$$\det(A-\lambda I)=0\implies\lambda^2+4\lambda+2=0 \implies\lambda=\frac{-4\pm\sqrt{8}}{2}=-2\pm\sqrt{2}.$$ Thus $\lambda_1=-2+\sqrt2\approx-0.586$, $\lambda_2=-2-\sqrt2\approx-3.414$.

Step 4 · Classification of eigenvalues

Both eigenvalues are real, distinct, and negative: $\lambda_2<\lambda_1<0$.

Step 5 · Application of the linearisation theorem

Lyapunov's First Method: if all eigenvalues of the Jacobian have strictly negative real parts and the nonlinear terms are $o(r)$, the rest point of the full nonlinear system is asymptotically stable. Both conditions hold here.

Step 6 · Verification that nonlinear terms are $o(r)$

For $r\leq1$: $|2x^4-y^6|\leq3r^4$ and $|11y^4|\leq11r^4$, so $\|F_2\|/r\leq Cr^3\to0$.

Step 7 · Eigenvectors and behaviour of the linearised system

For $\lambda_1=-2+\sqrt2$: eigenvector $\mathbf v^{(1)}=(1+\sqrt2,1)^\top$. For $\lambda_2=-2-\sqrt2$: eigenvector $\mathbf v^{(2)}=(1-\sqrt2,1)^\top$. The general solution $$\begin{pmatrix}\xi(t)\\\eta(t)\end{pmatrix} =C_1 e^{(-2+\sqrt2)t}\begin{pmatrix}1+\sqrt2\\1\end{pmatrix} +C_2 e^{(-2-\sqrt2)t}\begin{pmatrix}1-\sqrt2\\1\end{pmatrix}$$ decays to zero as $t\to+\infty$: a stable node of the linearisation.

Step 8 · Routh--Hurwitz verification

For $\lambda^2+4\lambda+2$: $a_1=4>0$, $a_2=2>0$ — both satisfied.

Step 9 · Stability of the nonlinear system

By Lyapunov's First Method, $(0,0)$ is Lyapunov stable and asymptotically stable, of node type (inherited from the linearisation).

Step 10 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{asymptotically stable node} of the nonlinear system.} }$$ The eigenvalues of the linearisation are $\lambda_1=-2+\sqrt2\approx-0.586<0$ and $\lambda_2=-2-\sqrt2\approx-3.414<0$, both real and negative. By Lyapunov's First Method, the nonlinear terms $2x^4-y^6=O(r^4)$ and $11y^4=O(r^4)$ do not affect stability, and the rest point is asymptotically stable.

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.