After introducing and proving the stable manifold theorem in $\mathbb{R}^2$, my instructor gave an example of a system for which it was possible to explicitly compute the stable manifold. I am trying to understand the solution, step by step.
For the system vector field $F$ defined by $x'=-x$, and $y'=y+x^2$, we have a rest point at the origin and a linearized system around the origin of $DF_{0,0}=\begin{bmatrix} -1&0\\0&1\end{bmatrix}$, we have two eigenvalues, $\lambda= 1,-1$ which have nonzero real part and are saddle type, therefore insuring the presence of a stable curve at some point tangent to the eigenvector corresponding to $\lambda_{-1}$. This will be a flow (or union of two flows) in from some boundary of an open set around the rest point at the origin, and passing through the rest point. Is all of this correct?
To explicitly find the stable curve, we hypothesize that there is some parametrized curve $\gamma(t)=(t,v(t))$ for some unknown function $v(t)$. Why do we know that the x coordinate of this curve will vary linearily with t? Does this come from the fact that the vector field is clearly linear in the first coordinate?
Next, we use the condition that this flow or curve will differentiate to the vector field at any point and any $t$, i.e. $\gamma'(t)=(1,v'(t))=F(\gamma(t))=(-t, v(t)+t^2)$.
Yielding a new ode based on the slope of our gamma curve: $v'(t)=(-1/t)(v(t)+t^2)$.
Then we solve by separation of variables AND variation of parameters and I get a bit lost: Separation of variables: $v'=\frac{-v}{t}$ yielding $v(t)=-1/t$. Why was this done? Why was the $t^2/t=t$ term ignored?
Then, she writes $v=\frac{-g}{t}$ for i assume some hypothetical funciton of t, $g$ to solve for.
Yielding: $v'=-\frac{1}{t}*-\frac{g}{t}-\frac{g'}{t}$ and then setting $-\frac{g'}{t}=-t$, but why? why do we know that these two terms are equal?
Finally we get $g(t)=\frac{t^3}{3}+c$ yielding a general solution by plugging in $g$ into our expression for $v$, $v(t)=-\frac{t^2}{3}-\frac{c}{t}$.
Also, a more general question, when can we solve for a stable curve like this? Say we are in $R^{3}$, would we need 2 of the variables to vary linearly with t, and not be "intertwined" with the other two variables? Also, when is the stable manifold just the linearized systems stable eigenvector? Finally, the stability referred to here is asymptotic stability since we need to be center free correct?
Much simpler: write $y=ax^2+bx^3+cx^4+\cdots$; it follows from $y'=y+x^2$ that $$ y'=x'(2ax+3bx^2+4cx^3\cdots)=-x(2ax+3bx^2+4cx^3\cdots)=ax^2+bx^3+cx^4\cdots+x^2. $$ Hence, $-2a=a+1$, $-3b=b$, $-4c=c$, $\ldots$, that is, $a=-\frac13$, $b=c=\cdots=0$ and $y=-\frac13 x^2$.
In other words, the stable manifold is the curve $y=-\frac13 x^2$. More generally, you can do the same locally whenever you can apply the stable manifold theorem, and in any dimension. In this specific example you can in fact do it globally.
The origin is clearly not asymptotically stable in the example.