Consider the system $$\frac{dx}{dt}=-2x+y\sin (xy), \\ \frac{dy}{dt}=-2y+x\sin (xy), \\ \frac{dz}{dt}=-2x$$
I have to show that the equilibrium point is asymptotically stable.
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The non-homogeneous term $g=(y\sin (xy), x\sin (xy))$ doesn't satisfy the condition $\frac{\|g\|}{\|\overline{x}\|}\rightarrow 0$ while $x\rightarrow 0$, does it?
So, we take the Taylor expansion at $(0,0)$: $$\sin (xy)=xy-\frac{x^3y^3}{3!}+\frac{x^5y^5}{5!}+\dots $$
So, the system becomes $$\frac{dx}{dt}=-2x+xy^2+O(x^3y^4), \\ \frac{dy}{dt}=-2y+x^2y+O(x^4y^3), \\ \frac{dz}{dt}=-2x$$
So, the system of first approximation is of the form $$\frac{dx}{dt}=-2x+xy^2, \ \ \frac{dy}{dt}=-2y+x^2y, \ \ \frac{dz}{dt}=-2x$$
How can we find the characteristic equation?
I got sruck because of the terms $xy^2$ and $x^2y$.
Could you give me a hint?
do we maybe look for a lyapunov function?