I made the following system from the typical $x_1=x, x_2=x'$ transformation:
$x_1'=x_2\\x_2'=\sin^2(x_1)-x_2\cos^2(x_1)$
I then used the general energy function
$E(t,x_1,x_2)=\frac{1}{2}(x_2)^2+\int_{0}^{x_1}g(s)ds$ when $x''+f(x)x'+g(x)=0$.
Here, $E(t,x_1,x_2)=\frac{1}{2}(x_2)^2-\int_{0}^{x_1}\sin^2(s)ds$
And $\frac{dE}{dt}=-(x_2)^2\cos^2(x_1)≤0$ after applying the FTC, substituting, and cancelling terms.
So the system is always dissipating energy, that is, for any time t, $E(t)≤E(0)$.
I want to show the boundedness of $x_2$, that is, $|x_2(t)|≤K,$ for some constant $K$ and $t≥0$.
I get to the inequality: $\frac{1}{2}(x_2(t))^2-\int_{0}^{x_1(t)}\sin^2(s)ds≤\frac{1}{2}(x_2(0))^2-\frac{x_1(0)}{2}+\frac{1}{4}\sin(2x_1(0))$
Or if you prefer: $\frac{1}{2}(x_2(t))^2-\frac{x_1(t)}{2}+\frac{1}{4}\sin(2x_1(t))≤\frac{1}{2}(x_2(0))^2-\frac{x_1(0)}{2}+\frac{1}{4}\sin(2x_1(0))$.
I cannot figure out how to remove the terms with $x_1(t)$ to isolate $|x_2(t)|$ while also keeping the ≤ inequality...
Note: $-\frac{x_1(t)}{2}+\frac{1}{4}\sin(2x_1(t))<0$, for all $x_1(t)$
Any help?
Update: I've since shifted my efforts to first showing $x_1$ is bounded and then moving terms around in the inequality to show $x_2$ is bounded. I think this might be the only way to do it, but I'm not confident my method was correct...
If you use a more general idea of "energy function" then start with the standard $$ E=\frac12(x'^2+x^2) $$ with time derivative $$ E'=x'(x''+x)=x'(-\cos^2xx'+\sin^2x+x)\le x'(\sin^2x+x) $$ Using mean inequalities it follows that $$ E'\le x'(1+x)\le |x'|\sqrt{2(1+x^2)}\le\frac{1+x'^2+x^2}{\sqrt2}\\ E'\le \frac1{\sqrt2}+\sqrt2E\\ $$ which then integrates on $[0,\infty)$ to $$ 1+2E(t)\le e^{\sqrt2t}(1+2E_0) $$ Thus as long as the solution exists it remains exponentially bounded, and can thus be extended obeying the same bound. No divergence in finite time is possible.