Jtdylktuy

I am considering the dynamical system:
$u'=v-0.25(v-u)^2$
$v'=u(1+v)+0.25(u+v)^2$

I have calculated the linear stable and unstable manifold as,
$E^s=sp(1,1)$ and $E^u=sp(-1,1)$ for eigenvalues $-1$ and $1$ respectively. I did this by calculating the linear eigenvalues and vectors and then using the defintions for the linear stable and unstable manifolds:

$E^s$
is the eigenspace corresponding to the eigenvalues with negative real parts. Let $v_1, . . . , v_k ⊂ R^n$
be the eigenvectors corresponding to the eigenvalues of $D(bar{x})f$ having negative real parts.
$E^s = sp{v_1, . . . , v_k}$.

$E^u$ is equivalent for a positive eigenvalue.

I am now trying to calculate an approximation for the unstable manifold but believe I should transform my system so that the linear unstable manifold is of the form ${(x,0):xinRe} but I am unsure on how to this.

My initial idea was to let $y=tilde{y}-1$ but that doesn't satisfy the above for all y, which is where I am confused.

begin{align}
u'&=v-frac14(v-u)^2\
v'&=u(1+v)+frac14(u+v)^2
end{align}
With the transformation $U=u+v$, $V=u-v$ (there is no real need to use rotations, the scaling here doesn't hurt), you get
begin{align}
U'&=U-frac12V^2+frac12U^2\
V'&=-V-frac12U^2\
end{align}
The unstable manifold $mathcal{U}$ is, to linear approximation, $V=0$.
So to obtain better approximations, we can impose an ansatz
$$
V=a_2U^2+a_3U^3+a_4U^4+dots
$$
and see what invariance gives us. Differentiating, we get $V'=(2a_2U+3a_3U^2+4a_4U^3+dots)U'$. Substituting for $U',V'$ from the differential equation (by invariance), we have
$$
-V-frac12U^2 =(2a_2U+3a_3U^2+4a_4U^3+dots)(U-frac12V^2+frac12U^2)
$$
and substituting our ansatz $V=a_2U^2+a_3U^3+a_4U^4+dots$,
begin{multline}
-(a_2U^2+a_3U^3+a_4U^4+dots)-frac12U^2 =(2a_2U+3a_3U^2+4a_4U^3+dots)times\left(U-frac12(a_2U^2+a_3U^3+a_4U^4+dots)^2+frac12U^2right)
end{multline}
Equating coefficients,
begin{align}
-a_2-frac12&=2a_2&a_2&=-frac16\
-a_3&=a_2+3a_3 & a_3&=frac1{24}\
-a_4&=frac32a_3+4a_4 & a_4&=-frac1{80}\&&vdots
end{align}
Finally, you might want to transform back to the original $u,v$ coordinates.