For homework, I have to say something about the stability of the zero solution of the differential equation $v''+v+f(v')=0$, where $f$ is a differentiable function satisfying $f(0)=0$ and $f'\geq0$. I am asked to use the linearization method and if it leads nowhere, then try the Lyapunov method.
The second one seems easier, I think that a function of the type $\frac{1}{2}\left ( (v')^{2}+v^2 \right )$, or something like that including $f$ somehow, will offer a solution. But as far as the first method is concerned, I am stuck. How am I supposed to turn this system in the familiar form $\dot{y}=g(y)$ and linearize it? Any help would be very much appreciated.
If you linearize the above equation, we obtain $$ v''+v=-bv' $$ where $b=f'(0)\geq0$. This the equation for a damped harmonic oscillator. By assuming a solution $e^{\lambda t}$, you can show that $Re\lambda<0$. Alternatively, you can make the transform $v(t)=e^{-bt/2}y(t)$, and find the equation $$ y''=-(1-b^2/4)y, $$ whose solutions are easy to obtain. You will find 3 regimes:overdamping $(b>2)$, critical damping $(b=2)$, and underdamping $(b<2)$. Hence, $v=v'=0$ is a stable fixed-point.
If you use the Lyapunov method, you multiply on both sides by $v'$ and obtain $$ \frac12\frac{d}{dt}\left(v'^2+v^2\right)=-bv'^2\leq0 $$ So you have a positive-definite function $L(v,v')=(v'^2+v^2)/2$ (in the dynamical systems sense), here it is the mechanical energy or the Lyapunov function, that is always decreasing in time. The level sets of $L$ are concentric circles centered at the origin. The flow always points inwards. Hence, $v=v'=0$ is a stable fixed point. If you take the original system and do the same approach, you obtain $$ \frac12\frac{d}{dt}\left(v'^2+v^2\right)=-f(v')v' $$ One gets global stability if the function $f(v')$ and $v'$ have the same sign. The Lyapunov function remains the same on both systems.