Welcome to our community

Be a part of something great, join today!

Algebraic manipulations for system of differential equations

kalish

Member
Oct 7, 2013
99
I have a problem I would like some guidance on.

I need to find the values of $k$ for which $x^2+ky^2$ is a Liapunov function for the system $$\dot{x}=-x+y-x^2-y^2+xy^2, \dot{y}=-y+xy-y^2-x^2y$$

**My attempt:** $$\dot{V} = \frac{\partial V}{\partial x} \times \frac{dx}{dt} + \frac{\partial V}{\partial y} \times \frac{dy}{dt} = 2x(-x+y-x^2-y^2+xy^2)+2ky(-y+xy-y^2-x^2y)=-2x^2+2xy-2x^3-2xy^2+2x^2y^2-2ky^2+2kxy^2-2ky^3-2kx^2y^2 \leq 0$$

The thing that throws me off the most is the plural **values**.

Now this turns into a minimization problem.

I keep running into terms like $xy, y^3, x^3$, which are not "sign" friendly.

Should I look for terms that will "knockout" each other?

Should I use the AM-GM-HM inequality to help?
 

kalish

Member
Oct 7, 2013
99
I have a problem I would like some feedback on. I have spent 6 hours on it examining various techniques (numerically and analytically).

I need to find the values of $k$ for which $x^2+ky^2$ is a Liapunov function for the system $$\dot{x}=-x+y-x^2-y^2+xy^2, \dot{y}=-y+xy-y^2-x^2y$$

**My attempt:** $$\dot{V} = \frac{\partial V}{\partial x} \times \frac{dx}{dt} + \frac{\partial V}{\partial y} \times \frac{dy}{dt} = 2x(-x+y-x^2-y^2+xy^2)+2ky(-y+xy-y^2-x^2y)=-2x^2+2xy-2x^3-2xy^2+2x^2y^2-2ky^2+2kxy^2-2ky^3-2kx^2y^2 \leq 0$$

**Note:** Global stability is NEVER established because of the counterexample $(x,y)=(-2,0)$. Then, $$\dot{V}=2x(-x+y-x^2-y^2+xy^2)+2ky(-y+xy-y^2-x^2y)=2(-2)(2+0-4-0+0)+0=8 \nleq 0.$$ So we are concerned with establishing local stability only. It turns out that if $k=1$, then $$\dot{V}=-2x^2+2xy-2x^3-2xy^2+2x^2y^2-2y^2+2xy^2-2y^3-2x^2y^2=-2x^2-2y^2-2x^3-2y^3+2xy \leq 0 \implies x^2+y^2+x^3+y^3-xy \geq 0 \implies (x^2+y^2-xy)(x+y+1) \geq 0.$$
Upon solving the inequality, we see that for REAL $x,y$, the solution is $y\ \geq -x-1$. It turns out the domain of stability of the origin for $k=1$ is the set $x^2+y^2=\frac{1}{2}$.

**Method 1:** I tried $$x=r\cos\theta, y=\frac{r}{k}\sin\theta.$$ Then assuming my calculations are correct (I verified them several times) I end up with the following equation after multiplying all terms by $k^2$:
$$r^2\cos^2\theta-kr^2\sin\theta\cos\theta+k^2r^3\cos^3\theta+r^3\cos \theta-r^3\cos^3\theta-r^4\cos^2\theta+r^4\cos^4\theta+kr^2-kr^2\cos^2\theta-kr^3\cos\theta+kr^3\cos^3\theta+r^3\sin^3\theta+kr^4\cos^2\theta-kr^4\cos^4\theta \geq 0.$$

Upon attempting to bound terms, I ran into great difficulty (I won't elaborate here but it suffices to say I didn't get what I needed). This expression is monstrous, with lots of differing terms and exponents! **How can I proceed?**

**Method 2:** I linearized the system at the origin to get $$\dot{x}=-x+y, \dot{y}=-y,$$ which has eigenvalues $-1,-1$. Then $$\dot{V}=2x(-x+y)+2ky(-y)=-2x^2+2xy-2ky^2=x^2-xy+ky^2 \geq 0 \implies$$

$$(x-\frac{y}{2})^2 = x^2-xy+\frac{y^2}{4} or (x-\sqrt{k}y)^2=x^2-2\sqrt{k}xy+ky^2 \implies k=1/4$$.

I proceeded to verify this using the critical point test for multivariable functions and evaluated the determinant of the Hessian, obtaining $(-2/3,0)$ as a saddle point and $(2.239,-2.517)$ as a saddle point for $k=1/4$. I also graphed the surface on Google Plot.

**Summary:** I have tried different things and obtained $k>0$. My question is: **Which of my methods are valid, and how can they be improved?**

Thanks.

(I have crossposted this question on differential equations - Algebraic manipulation of Lyapunov function - Mathematics Stack Exchange, but have gotten little input.)