Examine the stability of the rest point of the system…

Examine the stability of the rest point of the system $$\dot{x}=e^{x+2y}-\cos3x,\qquad\dot{y}=\sqrt{4+8x}-2e^y,$$ by the method of first approximation (linearisation).

Step 1 · Finding the rest point

At $(0,0)$: $e^{0}-\cos0=1-1=0$ and $\sqrt4-2e^0=2-2=0$: $(0,0)$ is a rest point.

Step 2 · Jacobian matrix at $(0,0)$

With $f=e^{x+2y}-\cos3x$, $g=\sqrt{4+8x}-2e^y$: $$\frac{\partial f}{\partial x}\Big|_{(0,0)}=1,\quad \frac{\partial f}{\partial y}\Big|_{(0,0)}=2,\quad \frac{\partial g}{\partial x}\Big|_{(0,0)}=2,\quad \frac{\partial g}{\partial y}\Big|_{(0,0)}=-2,$$ so $$A=\begin{pmatrix}1&2\\2&-2\end{pmatrix}.$$

Step 3 · Separation of linear and nonlinear parts

Expanding $e^{x+2y}=1+x+2y+O(r^2)$, $\cos3x=1-\tfrac{9x^2}{2}+O(r^4)$, $\sqrt{4+8x}=2+2x+O(r^2)$, $2e^y=2+2y+O(r^2)$ gives $$f(x,y)=x+2y+\underbrace{\tfrac{9x^2}{2}+O(r^2)}_{=:f_2},\qquad g(x,y)=2x-2y+\underbrace{O(r^2)}_{=:g_2},$$ with $f_2,g_2=O(r^2)=o(r)$ as $r\to0$.

Step 4 · Characteristic equation

$$\det(A-\lambda I)=0\implies\lambda^2+\lambda-6=0 \implies(\lambda+3)(\lambda-2)=0.$$ Thus $\lambda_1=-3$, $\lambda_2=2$.

Step 5 · Classification of eigenvalues

Real, distinct, opposite sign: $\lambda_1=-3<0<\lambda_2=2$.

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

For $r\leq1$: $|f_2|\leq9r^2$, $|g_2|\leq Cr^2$, so $\|F_2\|/r\leq Cr\to0$.

Step 7 · Application of Lyapunov's First Method

Since $\lambda_2=2>0$ and $f_2,g_2=O(r^2)=o(r)$, the instability theorem applies: the nonlinear rest point is unstable.

Step 8 · Eigenvectors and saddle structure

For $\lambda_1=-3$: eigenvector $\mathbf v^{(1)}=(1,-2)^\top$ (stable separatrix tangent to $y=-2x$). For $\lambda_2=2$: eigenvector $\mathbf v^{(2)}=(2,1)^\top$ (unstable separatrix tangent to $y=\tfrac12x$).

Step 9 · General solution of the linearised system

$$\xi(t)=C_1e^{-3t}+2C_2e^{2t},\qquad\eta(t)=-2C_1e^{-3t}+C_2e^{2t}.$$ For $C_2\neq0$: $\|(\xi,\eta)\|\sim|C_2|\sqrt5\,e^{2t}\to\infty$, confirming instability already at the linear level.

Step 10 · Saddle structure of the nonlinear system

By the Hartman--Grobman theorem, the nonlinear system is topologically conjugate to its (saddle) linearisation near the origin: a one-dimensional stable manifold tangent to $y=-2x$, a one-dimensional unstable manifold tangent to $y=\tfrac12x$, and all other trajectories escaping along the unstable manifold.

Step 11 · Routh--Hurwitz check

For $\lambda^2+\lambda-6$: $a_1=1>0$ but $a_2=-6<0$, confirming a root with positive real part.

Step 12 · Stability analysis

Since $\lambda_2=2>0$, the rest point is unstable in the sense of Lyapunov.

Step 13 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{unstable saddle point} of the nonlinear system.} }$$ The eigenvalues of the Jacobian at the origin are $\lambda_1=-3<0$ and $\lambda_2=2>0$, real with opposite signs. By Lyapunov's First Method, the positive eigenvalue guarantees instability; the nonlinear remainders $f_2,g_2=O(r^2)=o(r)$ do not affect the conclusion. The stable manifold of the nonlinear saddle is tangent to $y=-2x$ and the unstable manifold to $y=\tfrac12x$ 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.