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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](http://mathworld.wolfram.com/LyapunovFunction.html). The system is guaranteed to be stable if this derivative is negative definite.2012-05-09

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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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    @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.2012-12-14