Derive the power series expansion for he local stable and unstable manifolds of the ODE $x'=(-x_1+x_2^2,2x_2+x_1x_2)$ for the singularity $(0,0)$

Question

We are asked to find the power series expansion of the local stable and unstable manifolds of $x'=f(x)$ for the singularity $(0,0)$ where $f(x_1,x_2)=(-x_1+x_2^2,2x_2+x_1x_2)$
I have tried to solve this-
Let's first compute $E^s$ and $E^u$ for the singularity $(0,0)$ of $f$ (Note a point $x_0\in\Bbb{R}^2$ is called singularity or the equilibrium point of $f$ if $f(x_0)=(0,0)$.
$E^s$ is the subspace generated by the generalized eigen vectors of the matrix $Df(0,0)$ corresponding to the eigen values with negetive real part and $E^u$ is the subspace generated by the generalized eigen vectors of the matrix $Df(0,0)$ corresponding to the eigen values with positive real part.
Now, $Df(0,0)=\begin{pmatrix} -1&0\\ 0&2 \end{pmatrix}$. So, it's easy to see the eigen vector corresponding to $-1$ (eigen value with negetive real part) is $e_1=(1,0)$ and the eigen vector corresponding to $2$ (eigen value with positive real part) is $e_2=(0,1)$. Hence, $E^s=L(\{e_1\})=\{(x_1,0)|x_1\in\Bbb{R}\}$ and $E^u=L(\{e_2\})=\{(0,x_2)|x_2\in\Bbb{R}\}$.
So, let $\psi_s:E^s\cap U\to E^u$ and $\psi_u:E^u\cap U\to E^s$ be the local stable and manifold manifolds where $U$ is a neighberhood of $(0,0)$.
Let, $x_2=\psi_s(x_1)$ and $x_1=\psi_u(x_2)$.
Now try to find $\psi_s$ first.
Let, the power series expansion of $\psi_s$ be $x_2=\psi_s(x_1)=a_2x_1^2+a_3x_1^4+\cdots$ (as $x_2(0)=0$, and $x_2'(0)=0$ the coefficients of $1,x_1$ have to be $0$).
Now, $x_2'=\psi_s'(x_1)x_1'$
$\implies 2x_2+x_1x_2=(2a_2x_1+3a_3x_1^2+\cdots)(-x_1+x_2^2)$
$\implies 2(a_2x_1^2+a_3x_1^3+\cdots)+x_1(a_2x_1^2+a_3x_1^3+\cdots)=(2a_2x_1+3a_3x_1^2+\cdots)(-x_1+(a_2x_1^2+a_2x_1^3+\cdots)^2)$
But it seems to be very difficult to compute the coefficients by comparing different powers of $x_1$ in the above equality. (Basically due to the $x_2^2$ part)
There is another way of computing $\psi_s$ and $\psi_u$ that I know is to check whenever $\lim_{t\to\infty} \phi_t(x_1,x_2)=0$ and $\lim_{t\to-\infty}\phi_t(x_1,x_2)=0$ respectively where $\phi_t$ is the flow generated by the vector field $f$. But in that case I can't compute the flow.
Can anybody solve my problem? Thanks for assistance in advance.

Answer 1

The equations are $$ x_1'=-x_1+x_2^2,\quad x_2'=2x_2+x_1x_2. $$ Note that for $x_2=0$ the second equation is satisfied and so this condition gives the stable manifold.

For the unstable manifold write $x_1=a_2x_2^2+a_3x_2^3+\cdots$, which gives $$ x_1'=-x_1+x_2^2=x_2'(2a_2x_2+3a_3x_2^2+\cdots) $$ and so $$ -a_2x_2^2-a_3x_2^3+\cdots+x_2^2=(2x_2+(a_2x_2^2+a_3x_2^3+\cdots)x_2)(2a_2x_2+3a_3x_2^2+\cdots). $$ For example, $-a_2+1=4a_2$ and so $a_2=\frac15$. Then $-a_3=6a_3$ and so $a_3=0$. Etc.

Of course, by symmetry we know that there are only even powers so you might as well as consider and try to see the pattern $$ -a_2x_2^2-a_4x_2^4+\cdots+x_2^2=(2x_2+(a_2x_2^2+a_4x_2^4+\cdots)x_2)(2a_2x_2+4a_4x_2^3+\cdots). $$