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How can I solve this question: Find the values of the positive constants $a, b$ and $c$, so that $v(x,y)=ax^2+bx^4+cy^2$ is a Lyapunov function for the stable equation $d^2x/dt^2 + 5dx/dt + x +2x^3=0$.

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What have you tried? And is this a homework question? –  Thomas E. Jan 13 '13 at 8:46
    
No it is not a homework ,but I am having an exam and this is one of the questions that I have found but I could not solve it –  JmD Jan 13 '13 at 8:53

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The idea is to make your $2$nd order differential equation to a $1$st order system, and then solve by differentiation.

I will denote $\dot{x}=\frac{dx}{dt}$ for short. Your differential equation is then of the form $\ddot{x}+5\dot{x}+x+2x^3=0$. By doing the standard substitution $y=\dot{x}$, we get the equaivalent $1$st order system $\dot{x}=y$ and $\dot{y}=-5y-x-2x^3$ (note that $\dot{y}=\ddot{x}$).

Then by differentiating the Lyapunov function $v$, substituting the differential equation system and partial derivatives of $v$ to it, it follows that \begin{align*} \dot{v}(x,y) &=\frac{\partial v}{\partial x}\dot{x}+\frac{\partial v}{\partial y}\dot{y}=(2ax+4bx^3)y+2cy(-5y-x-2x^3) \\ &=2axy+4bx^3 y-10cy^2-2cxy-4cx^3y \\ &=xy(2a-2c)+x^3y(4b-4c)-10cy^2. \end{align*} To obtain asymptotic stability, you want $\dot{v}(x,y)<0$ for all $(x,y)\in\mathbb{R}^{2}\setminus\{(0,0)\}$. This guarantees that the solutions converge to origin as $t\to\infty$, since the solutions pass smaller level surfaces of $v$ in $\mathbb{R}^{2}$ as the time elapses. This happens for example when $a=b=c>0$, since in this case $\dot{v}(x,y)=-10cy^2<0$. I don't know if this is the maximal set of values for $a,b$ and $c$, or if you're even looking for all of them (which would, in fact, be very unusual for these exercises), but at least all these values do the job.

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