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

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

Step 1 · Separation of linear and nonlinear parts

Write $\dot x=f(x,y)$, $\dot y=g(x,y)$ with $$f(x,y)=-x+2y+\underbrace{(-3x^2)}_{=:f_2},\qquad g(x,y)=3x-2y+\underbrace{2x^2+y^4}_{=:g_2}.$$ Both remainders satisfy $f_2,g_2=O(r^2)=o(r)$ as $r\to0$, so the hypotheses of the linearisation theorem hold.

Step 2 · Linearised system

The Jacobian at $(0,0)$ is $$A=\begin{pmatrix}-1&2\\3&-2\end{pmatrix}.$$

Step 3 · Characteristic equation

$$\det(A-\lambda I)=0\implies\lambda^2+3\lambda-4=0 \implies(\lambda+4)(\lambda-1)=0.$$ Thus $\lambda_1=-4$, $\lambda_2=1$.

Step 4 · Classification of eigenvalues

Real, distinct, opposite sign: $\lambda_1=-4<0<\lambda_2=1$.

Step 5 · Application of the linearisation theorem

Lyapunov's First Method (instability): if at least one eigenvalue has strictly positive real part and the nonlinear terms are $o(r)$, the rest point of the full nonlinear system is unstable. Here $\lambda_2=1>0$ and $f_2,g_2=O(r^2)=o(r)$.

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

For $r\leq1$: $|-3x^2|\leq3r^2$, $|2x^2+y^4|\leq3r^2$, so $\|F_2\|/r\leq3r\to0$.

Step 7 · Eigenvectors and saddle structure

For $\lambda_1=-4$: eigenvector $\mathbf v^{(1)}=(2,-3)^\top$ (stable separatrix tangent to $y=-\tfrac32x$). For $\lambda_2=1$: eigenvector $\mathbf v^{(2)}=(1,1)^\top$ (unstable separatrix tangent to $y=x$).

Step 8 · General solution of the linearised system

$$\begin{pmatrix}\xi(t)\\\eta(t)\end{pmatrix} =C_1 e^{-4t}\begin{pmatrix}2\\-3\end{pmatrix} +C_2 e^{t}\begin{pmatrix}1\\1\end{pmatrix}.$$ For $C_2\neq0$, $\|(\xi,\eta)\|\sim|C_2|e^t\to\infty$: instability already at the linear level.

Step 9 · Routh--Hurwitz check

For $\lambda^2+3\lambda-4$: $a_1=3>0$ but $a_2=-4<0$, confirming a root with positive real part.

Step 10 · Saddle structure of the nonlinear system

By the Hartman--Grobman theorem, since the linearisation is a saddle (opposite-sign real eigenvalues, none zero), the nonlinear system is topologically conjugate to it near the origin: a genuine saddle with one-dimensional stable manifold tangent to $y=-\tfrac32x$ and one-dimensional unstable manifold tangent to $y=x$.

Step 11 · Stability analysis

Since $\lambda_2=1>0$, the rest point is unstable in the sense of Lyapunov.

Step 12 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{unstable saddle point} of the nonlinear system.} }$$ The eigenvalues of the linearisation are $\lambda_1=-4<0$ and $\lambda_2=1>0$ (real, opposite signs). By Lyapunov's First Method, the positive eigenvalue guarantees instability; the nonlinear terms $-3x^2=O(r^2)$ and $2x^2+y^4=O(r^2)$ are $o(r)$ and do not affect the conclusion. The stable manifold is tangent to $y=-\tfrac32x$ and the unstable manifold to $y=x$ at the origin.

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.