Find an energy functional for the nonlinear viscous oscillator $x' = v$, $v' =-b(v)v-k(x)x$, $t>0$ Consider the nonlinear viscous oscillator $$\begin{cases} x' = v\\\ v' =-b(v)v-k(x)x,\quad t>0, \\\ \end{cases}$$ where $(x,v)$ is the position and velocity of the oscillator. Here $b : \mathbb R\to[0,\infty)$ is a continuous nonlinear friction, and $k : \mathbb R\to(0,\infty)$ is a continuous nonlinear spring restitution. Find an energy functional for the oscillator and use it to show that $$sup(\vert x(t)\vert+\vert v(t)+\vert v'(t)\vert)\le C\lt \infty ,$$ where $C$ is a constant depending on the initial energy of the oscillator, $b$ and $k$.

Multiply the first equation by $x \cdot k(x)$ and the second by $v$, then sum both of them. You obtain $$ x k(x) x'+vv'=-b(v)v^2. $$ Let $V(x)$ such that $V'(x)=x k(x)$, then $$ \frac{d}{dt}\left(\frac{v^2}{2}+V(x)\right)=-b(v)2^2\leq 0 $$ Consider the energy function $E(x,v)=\frac{v^2}{2}+V(x)$. Since it is not increasing in time, $$ \frac{v^2}{2}+V(x)\leq E(0). $$ Now, by physical grounds, $k(x)\geq 0$, then $V(x)$ is bounded from below; we can assume $V(x)\geq 0$ without loss of generality. Therefore $$ |v(t)|\leq \sqrt{2E(0)}. $$ If $k(x)\geq c>0$ (the spring does not become soft for $x\to\infty$), from $V(x)\leq E(0)$ you can also deduce $|x|\leq c_1$. Then the bound $|v'(t)|\leq c_2$ follows from the second differential equation.