Region of attraction and stability via Liapunov's function

Question

EXERCISE:

Estimate the region of stability for the stationary point $O(0,0)$ given the differential system: $$x'=y$$ $$y'=x^7-2\cdot x-y$$ using the liapunov's funcion $V(x,y)=\dfrac{1}{2}\cdot x^2+\dfrac{1}{2}\cdot(x+y)^2$

Attempt: We have that $V'=V_x \cdot x'+V_y \cdot y'$

So,$V'=(2x+y) \cdot y +(x+y)\cdot (x^7-2x -y)=x^8-2x^2+x^7y-xy $

So,how can i find the sign of V'.I think that i have to find it $V'<0$ everywhere outside the origin and the stationary point $O(0,0)$ will be asymptotic stable!

After,that how can i find the region of attraction?

score 3 · Accepted Answer · answered Jan 23 '18 at 16:31

3

For very small $x$ you can, for a first picture, ignore the higher degree terms and just consider the quadratic terms. As $$ -2x^2-xy=-\frac18(4x+y)^2 +\frac18y^2 $$ you see that on the line $y=-4x$ the whole expression is $$ 2x^2-3x^8 $$ which is positive for $x^6<\frac23$. Thus your Lyapunov function does not generate level sets where the solutions always flow downwards.

In contrast $V=\frac12y^2+x^2-\frac18x^8$ leads to $\dot V=-y^2$.

answered Jan 23 '18 at 16:31

Lutz Lehmann

131,652

Τhe problem is that i have to use this ill-adapted liapunov's function! – Paris K. Patsogiannis Jan 23 '18 at 16:36
@ParisK.Patsogiannis: Well, this answer clearly shows that you can't, so either you've misread the problem, or whoever gave it to you messed up. – Hans Lundmark Jan 23 '18 at 18:24
@HansLundmark i re-read the problem again but it's not my fault.Maybe,the author didn't check his own exercise! – Paris K. Patsogiannis Jan 23 '18 at 19:33

Region of attraction and stability via Liapunov's function

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