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I've tried to solve a little problem that goes as follows:

Consider a system of ODEs: $$x'=y-x^3\text{ and } y'=-x^3-y^3$$ And the function $$L(x,y)=\frac{1}{2}y^2+\frac{1}{4}x^4.$$

Now I shall show that $(0,0)$ is not linearly stable, and that $L$ is a Lyapunov function.

I tried to investigate $\frac{dL}{dy}$ and $\frac{dL}{dx}$ and check if this is smaller or greater than $0$ at (0,0). However, these two terms are just $\frac{dL}{dy}=y,\frac{dL}{dx}=x^3$, so it seems we have to solve for $x,y$ explicitly? I'm confused because this looks like one of those exercise where brute-force can be avoided (And we did not deal with non-linear ODEs yet)?

yours, Marie

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You can investigate "the derivative of L with respect to your system", $\nabla L \cdot f(y)$, where $f(y)$ is your system. See the example here. The system is guaranteed to be stable if this derivative is negative definite. – tentaclenorm May 9 '12 at 22:37
up vote 2 down vote accepted

For $L(x,y)$ to be a Lyapunov function, what you want to do is consider $$\dfrac{dL}{dt} = y \dfrac{dy}{dt} + x^3 \dfrac{dx}{dt} = y (-x^3 - y^3) + x^3 (y - x^3)$$ Do you see why $\dfrac{dL}{dt} \le 0$, with equality only at $(0,0)$? And why $L(x,y) \ge 0$, with equality only at $(0,0)$?

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Awesome! So for the linear stability, we also have to differentiate w.r.t. $t$, not $x,y$ and hence the two exercises were connected! :) – Marie. P. May 9 '12 at 23:06
Wait - I see how $\frac{dL}{dt}<0$ and $L(x,y)>0$ except equalities at $0$, but if the derivative is negative, then how can the function start at $0$ and then be always greater than zero? – Marie. P. May 9 '12 at 23:16
If you start at $(0,0)$, then obviously you stay there. If you start somewhere else, then as time goes on $L(x,y)$ decreases so you must approach $(0,0)$. – Robert Israel May 10 '12 at 0:34
@Marie.P.: Also, that second equality there is the reason we say that the derivative ${dL\over dt}$ is computed along a solution since we substitute back in the dynamics from the ODE there. – JohnD Dec 14 '12 at 3:22

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