Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$

Question

Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!

Find expressions for the local stable and local unstable manifolds for the following system of equations in a neighborhood of the origin, $$x' = y+y^2 - 2xy + x^2$$ $$y'=x+y^2 - 2xy + x^2.$$

$\textbf{Solution:}$ Subtracting $x' = y+y^2 - 2xy + x^2$ from $y'=x+y^2 - 2xy + x^2$ gives us $y'-x' = x-y$ implies the following $$ x'+x=y'+y. \hspace{35pt} (1)$$ Integrating $x' = y+y^2 - 2xy + x^2$ with respect to $y$ and $y'=x+y^2 - 2xy + x^2$ with respect to $x$ gives us $$x = \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y \hspace{35pt} (2)$$ $$y=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3}. \hspace{35pt} (3)$$

Applying (2) and (3) to (1) gives us $$\frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y +y^2 -2xy+x^2$$ $$=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x + y^2 -2xy + x^2$$ $$\implies \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y = \frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x. \hspace{35pt}(4)$$

So, equation (4) denotes a function which if we replace $y$ with $x$ equation will be the same throughout. Thus, if a function of the form $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} - t^3 + t^2 + t \text{ where } t = x, y$$ implies $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} + t. \hspace{35pt} (5)$$

So equations (1), (4), and (5) define the stable and unstable points around the origin. So, $f'(x,y) >0$ as $$f'(x,y) = t + t^2 + 1 = (t+\frac{1}{2})^2 + \frac{3}{4}.$$ Therefore, it will be unstable and we are done.

Answer 1

$$x'=y+(x-y)^2 $$ $$y'=x+(x-y)^2$$ HINT : $$y'-x'=x-y\quad\implies\quad \frac{y'-x'}{y-x}=-1\quad\implies\quad y-x=c_1e^{-t}$$ $$y=x+c_1e^{-t}$$ $y'=x+(c_1e^{-t})^2=x'-c_1e^{-t}$ $$x'-x=c_1e^{-t}+c_1^2e^{-2t}$$ $$x(t)=c_2e^t-\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}e^{-2t}$$ $$y(t)=c_2e^t+\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}e^{-2t}$$

This is the explicit solution $x(t)$ and $y(t)$ .

I suppose that you can take it from here about stability.

Note : Eliminating $t$ from the above equations gives the trajectory equation $\quad 2(y-x)^3+3(y^2-x^2)=C \quad;\quad C=6c_1c_2$ .

Answer 2

As $\{x^2,y^2, xy\}$ go to zero more quickly by one order degree than $\{x,y\}$ we have the near zero the dynamical system behaves as

$$ \dot x = y\\ \dot y = x $$

The Jacobian at the origin is

$$ J = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) $$

characterizing a saddle point which is unstable.

Answer 3

The stable/unstable manifolds $W^s$/$W^u$ of the equilibrium point $0$ are the manifolds tangent to the stable/unstable eigenspace $E^s$/$E^u$ at $0$ (with the same dimension as the corresponding eigenspace) s.t. the trajectories starting in $W^s$/$W^u$ converge to $0$ when $t \to \pm \infty$.

After the change of variables $(x, y) = (\zeta - \xi, \zeta + \xi)$, the system becomes $$\begin {aligned} \dot \xi &= -\xi \\ \dot \zeta &= \zeta + 4 \xi^2. \end {aligned}$$ Find $E^s$ and $E^u$ first. Check that $W^u$ coincides with $E^u$. To find $W^s$, start with a series approximation. Substitute $\zeta = A \xi^\alpha$ into the equations and eliminate $\dot \xi$. The result is $$A (\alpha + 1) \xi^\alpha + 4 \xi^2 = 0.$$ Equating the powers of $\xi$ gives $\alpha$ and equating the coefficients gives $A$. Verify that this happens to give the exact equation for $W^s$.