I am being introduced to the Lyapunov functions in order to determine the stability of a given system. I know that finding a Lyapunov function is not easy, so I would like to ask for any trick or hint in order to find a Lyapunov function for $$ \left\{\begin{array}{l}x'=-4y+x^2,\\y'=4x+y^2\end{array}\right. $$ at $(0,0)$. I have tried combinations of $x^{2n}$ and $y^{2m} $ and also products of $x$ and $y$ but got nothing clear. Also, I've searched the phase plot for the system and it is clear that $(0,0)$ is a stable point (not asymptotically stable). Thanks in advance.

  • $\begingroup$ Using the symmetry of $-4y$ and $4x$, the gradient of the Lyapunov function $V(x,y)$ should be like $C(x,y) \cdot (x,y)$ in order to cancel both terms, but I don't know how to take advantage of this. $\endgroup$ – user55268 Dec 14 '15 at 22:19
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    $\begingroup$ I suggest to take a look at my answer: math.stackexchange.com/questions/1577274/… . There are some reasons why Lyapunov method is not the right name to call it here. You might find only the function for which time derivative with respect to a system will be zero in the neighbourhood of origin, which is really the first integral in this case. $\endgroup$ – Evgeny Dec 15 '15 at 22:41
  • $\begingroup$ @Evgeny Strange that Perko would add this as an exercise (5(d) in Differential Equations and Dynamical Systems, section 2.9), claiming that one should use appropriate Liapunov functions to determine stability... $\endgroup$ – Alex Provost Dec 15 '15 at 22:54
  • $\begingroup$ @A.P. I agree with you. I just feel a little bit nervous when derivative of Lyapunov function w.r.t. system of ODEs is identical zero :) technically it's still a Lyapunov function, but it is also a first integral... $\endgroup$ – Evgeny Dec 15 '15 at 23:00

It is not quite clear why you put a bounty on this question, since @Evgeny answered it in the best possible way. However, if you are looking for a Lyapunov function, here it is (up to an additive constant): $$ L(x,y)=\frac{2^{2/3} \left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right) \left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}+1\right) \left(\left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right) \log \left(2^{2/3} \left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}\right)\right)+\left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}-1\right) \log \left(2\ 2^{2/3} \left(\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}+1\right)\right)-3\right)}{9 \left(-\frac{\left(x^2-12 x+8 y+48\right)^3}{2 \left(x^2-4 y\right)^3}+\frac{3 x (x+4) \left(x^2-12 x+8 y+48\right)}{2 \sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\right)}-1\right)}-\frac{x (x+4) \left(4 \log \left(x^2-4 x+16\right)+x\right)}{18 \sqrt[3]{2} \sqrt[3]{x^3 (x+4)^3}} $$

  • $\begingroup$ I know, but how did you found out this function? $\endgroup$ – user55268 Jan 17 '16 at 22:50
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    $\begingroup$ I just asked Mathematica to solve the differential equation $dy/dx=(4x+y^2)/(-4y+x^2)$. The solution $L(x,y)=Const$ is a first integral and hence, as it was explained, the Lyapunov function. here is the nasty nature of your example: to find the Lyapunov function is tantamount to solving first order ODE explicitly, which cannot be done very often, but even if it could be done (see my answer), this is not of a great help. $\endgroup$ – Artem Jan 17 '16 at 22:59

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