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I have the following dynamical system:

$$ \begin{align} \dot{x}&=-x-2y^2, \\ \dot{y}&=-x^2y-y^3. \end{align} $$ My task is to show that, for the dynamical system, the set $$S=\left\{ (x,y) \in \mathbb{R}^2:x \leq0 \right\}$$ is positive invariant.

My first thought is to use a Liapunov function defined by $$L:S \to \mathbb{R}, \: \: L(x,y)=-x,$$ which is positive definite. However, calculating $\dot{L}$ gives

$$\dot{L}=L_x\dot{x}+L_y\dot{y}=-\left(-x-2y^2\right)=x+2y^2,$$ from which I cannot seem to deduce anything.

Any help would be great!

Rodrigo de Azevedo
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George Wilson
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  • Might not simplified to something nice, but maybe try converting the system into polar coordinates? – Chee Han May 10 '17 at 21:01

1 Answers1

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Hint: Check how vector field points along the boundary of your domain of interest. If it points inside, then Bony-Brezis theorem can be applied. Or you just can say that vector field along the boundary doesn't let trajectories go out of domain $x \leqslant 0$.

Evgeny
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  • Oh right okay, so because on $x=0$, $\dot{x}=-2y^2<0$ for all $y$, the vector field will be pointing to the left, so trajectories can't 'escape' once they're in the domain $x \leqslant 0$? – George Wilson May 12 '17 at 17:03
  • Yep :) The only weak point in this kind of reasoning (not only yours, but mine too) is when vector field tangent to the boundary at some points. This is where Bony-Brezis theorem needed: it covers such cases. In all other cases "vector field along the boundary doesn't let trajectories to get out of domain" works nicely :) – Evgeny May 12 '17 at 17:26
  • You are welcome :) Should I add anything else to the answer? – Evgeny May 12 '17 at 17:41
  • I don't think there's any need; after all, I asked for a hint, not a full solution! – George Wilson May 12 '17 at 20:04