Complicated exercise on ODE

I have this exercise extracted from a examination of qualitative theory of ODE (in which we study the existence and uniqueness of solutions, and stability using the function of Lyapunov)

I don’t know how to solve it: For 1) it’s ok but from 2) I don’t have any idea.

Given three parameters $L,a$ and $\alpha$, 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 $a>0$ and $\alpha \geq 0$.

a) Establish the existence of a positive constant $C$ such that :

$\displaystyle \frac{a}{4}x^2+\frac{y^2}{2}\leq C+1+\frac{L^2}{a}$
(You can use the functional $V(x,y)=\frac12 y^2+\frac{a}{2}x^2-L
x-\cos x)$

(b) Deduce that the solutions of $ (E) $ are bounded when $ t \rightarrow + \infty $ .

(3) consider the modified function $V_{\delta}(x,y)= V(x,y) +\delta xy (\delta >0)$.

(a) Write the equation satisfied by: $\dfrac{dV_{\delta}}{dt}$.

(b) Show that, for $\delta$ small enough, $$\frac{ax^2}{8}+\frac{y^2}{8}\leq V_{\delta}(x,y)+1+\frac{2L^2}{a} $$

(c) Deduce that if $a>0$ and $\alpha >0$, then there exists a constant $M=M(\alpha,a,L)$ (regardless of baseline values​​) such that
$$\forall (x_0,y_0)\in \mathbb{R}^2,\ \exists ~ T_0 ,\text{ such that }
> \forall t\geq T_0,\ x^2(t)+y^2(t)\leq M.$$

For 1) and 2) I don’t have a problem thank’s to @robjohn :Small question about ODE

For 3)

a) I have that : $\displaystyle \frac{\mathrm{d}V_{\delta}}{\mathrm{d}t}=(\delta-\alpha)y^2-\delta y’$

but I don’t know how to solve c)

Can someone help me?

Thank you.

Solutions Collecting From Web of "Complicated exercise on ODE"

For $a,\alpha>0$ we can prove that $x,\dot{x},\ddot{x}$ are bounded, $\dot{x}$ is square integrable, $\lim_{t\rightarrow +\infty}\dot{x}(t)=\lim_{t\rightarrow +\infty}\ddot{x}(t)=0$, $\lim_{t\rightarrow +\infty}\dot{x}(t)=x^*$ with $x^*$ a root of the equation $ax+\sin x=L$ and (a)-(c).

Proof: From the system equation (E) we have

$\frac{d}{dt}\big[\frac{1}{2}\dot{x}^2(t)+\alpha\int_0^t{\dot{x}^2(s)ds}+\frac{a}{2}x^2(t)-\cos( x(t))-Lx(t)\big]=0$.

Hence

$\frac{1}{2}\dot{x}^2(t)+\alpha\int_0^t{\dot{x}^2(s)ds}+\frac{a}{2}x^2(t)-\cos( x(t))-Lx(t)=c.$

Using the inequality $-Lx(t)\geq -\frac{a}{4}x^2(t)-\frac{L^2}{a}$ in the identity above we have

$\min\{\frac{1}{2},\frac{a}{4}\}[\dot{x}^2(t)+x^2(t)]\leq \frac{1}{2}\dot{x}^2(t)+\alpha\int_0^t{\dot{x}^2(s)ds}+\frac{a}{4}x^2(t)\leq c+1+\frac{L^2}{a}$

i.e. (c) holds true and $x\in L_{\infty}$, $\dot{x}\in L_{\infty}\cap L_{2}$. From (E) we also have $\ddot{x} \in L_{\infty}$. Now Barbalat lemma can be used to obtain
$\lim_{t\rightarrow +\infty}\dot{x}(t)=0$. Also $\int_0^{\infty}{\ddot{x}(s)ds}=-\dot{x}(0)$ and $\ddot{x}$ uniformly continuous and therefore $\lim_{t\rightarrow +\infty}\ddot{x}(t)=0$. As a result $x$ converges towards a root of the equation $ax+\sin(x)=L$.