Using a linear approximation to deduce that phase-paths are closed

Question

Considering the plane autonomous system below:

$$\frac{dx}{dt} = ye^{xy}$$ $$ \frac{dy}{dt} = 1-x^{4} - y^{2}$$

I have been asked to show that the point $(1,0)$ is a critical point (it is a centre), and then to use $x = 1 + u$ and $y= v$ for $u,v$ sufficiently small to deduce that approximately $$ \frac{du}{dt} = v \hspace{1cm} \frac{dv}{dt} = -4u.$$

I am not sure how to deduce the approximate result above, my working is as follows:

$$ \frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt} = ve^{(1+u)v}$$ and $$ \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = 1 - (1+u)^{4} - v^{2}$$ How does the approximation result from this?

Answer 1

You get $$ ve^{u+uv}=v(1+u+u^2/2+...)(1+uv+u^2v^2/2+...) $$ and $$ 1-(1+u)^4-v^2=1-(1+4u+6u^2+4u^3+u^4)-v^2. $$ Removing terms of degree 2 and higher leaves you with the linearization.

Answer 2

The first-order approximation of $x$ is $x=x_0+\Delta x= x_0+x'\Delta t$

Here, $x_0 = 1$, so $x=1+x'\Delta t$, hence $u=x'\Delta t$. Similarly, $v = y'\Delta t$.

Then $\frac{du}{dt}=x'\Delta t=(ye^{xy})'\Delta t =[ y'e^{xy}+(x'y+xy')ye^{xy})]\Delta t$

Substituting in $x=1,y=0$, everything disappears except $y'\Delta t$, which is equal to $v$.

Similarly, $\frac{dv}{dt} = (1-x^4-y^2)'\Delta t = (-4x'x^3-2yy')\Delta t = -4x'\Delta t = -4u$