i have this question :
Given three parameters $L,a$ et $\alpha$, we consider the differential equation : $$(E)\qquad x''+\alpha x' +a x + \sin x =L, \ > t\geq0$$
1) Show that the maximal solutions of $(E)$ are defined on all $\mathbb {R}$.
2) Assume that $a>0$ et $\alpha \geq 0$.
a) Establish the existence of a positive constant $C$ such that :
$\displaystyle \frac{a}{4}x^2+\frac{y^2}{2}\leq C+1+\frac{L^2}{a^2}$
(You can use the functional $V(x,y)=\frac12 y^2+\frac{a}{2}x^2-L x-\cos x)$
I suppose that $y(t)=x'(t)$ , i have $\frac{dV}{dt}=-\alpha y^2$ so $V$ is decreasing but i don't know how to prove the existence of a positive $C$ such that :$\displaystyle \frac{a}{4}x^2+\frac{y^2}{2}\leq C+1+\frac{L^2}{a^2}$
I just find that : $\displaystyle\frac{y^2}{2}+\frac{a}{4}x^2 \leq V(x_0;y_0) + 1 + Lx-\frac{a}{4}x^2$
but how to find $\displaystyle\frac{L^2}{a^2}$ ??????????
Please help me
Thank you .
Multiply by $x'$ and integrate to get $$ \frac12x'^2+\alpha\int x'^2\,\mathrm{d}t+\frac12ax^2-\cos(x)=Lx+C\tag{1} $$ Since $\alpha\ge0$ and $\cos(x)\le1$, $(1)$ implies $$ \frac12x'^2+\frac12ax^2\le C+1+Lx\tag{2} $$ Subtracting $\frac14ax^2$ from both sides $$ \frac12x'^2+\frac14ax^2\le C+1+Lx-\frac14ax^2\tag{3} $$ maximum of $Lx-\frac14ax^2$ is $\frac{L^2}{a}$ at $x=\frac{2L}{a}$. Therefore, I get $$ \frac12x'^2+\frac14ax^2\le C+1+\frac{L^2}{a}\tag{4} $$