Need help figuring out what this question wants me to do. No need to do entire problem for me, just need a push in the right direction on how to solve
Here's my system: \begin{gather*} \frac{du}{dt}=u(1-u^2)-w \\ \frac{dw}{dt}=u \end{gather*}
Consider region $\Omega$ in $uw$-plane, and show that it is a trapping region for sufficiently large R.
Update: So I've come to the conclusion that this is way to go,
AB: $u=-R, w=0$ $\implies \frac{du}{dt}|_{(-R,0)}=-R(1-R^2)$ and $\frac{dw}{dt}|_{(-R,0)}=-R<0$. Here if $|R|<1$ then $\frac{du}{dt}|_{(-R,0)}=-R(1-R^2)<0$ and $\frac{du}{dt}|_{(-R,0)}=-R(1-R^2)>0$ otherwise
BC: $u=R, w=R$ $\implies\frac{du}{dt}|_{(R,R)}=-R^3<0$ and $\frac{dw}{dt}|_{(R,R)}=R>0$
CD: $u=0, w=-R$ $\implies\frac{du}{dt}|_{(0,-R)}=R>0$ and $\frac{dw}{dt}|_{(0,-R)}=0$
so on and so forth for each grid section...
Is my intuition on how to do this make sense?
On AB: You have $w=R$ and $u\in[-R,0]$. An outward pointing vector is $(0,1)$. The component of the vector field in that direction is $u\le 0$, thus it points inwards, or at least not outwards.
On BC: $u+w=R$, $u\in[0,R]$. The vector $(1,1)$ points outwards. The scalar product with the vector field is $$ u(1-u^2)-w+u=3u-u^3-R. $$ $3u-u^3$ has a maximum for positive $u$ at $u=1$ with value $2$. Thus for $R>2$ the vector field is pointing inwards on this segment.
On CD: $u=R$, $w\in[-R,0]$. An outwards pointing vector is $(1,0)$. The component of the vector field in that direction is $R(1-R^2)-w$. This is maximal for $w=-R$ with value $2R-R^3=R(2-R^2)$. For $R>2$ this is negative.
The other segments are point-symmetric (at the origin) to previous segments.