Let $\xi \in \mathbb{R}^2$, $\Phi \in C^0([0, +\infty[, \mathbb{R}^{2x2})$ a bounded function.
Let $y:[0, +\infty[ \rightarrow \mathbb{R}^2$ a solution of the following integral equation,
$$ y(x) = e^{-x} \Phi(x) \xi + \int_0^x e^{-(x-t)} \Phi(x-t) \begin{pmatrix} y_2^3(t) \\ y_1^7(t) \end{pmatrix} dt $$
I would like to show that for each $\epsilon > 0$, $\exists \delta > 0$ such that we have, $$ ||\xi|| > \delta \implies ||y(x)|| < \epsilon, \quad \forall x \geq 0 $$
I tried the following steps:
- Because $\Phi$ is bounded, $\forall x \in [0, +\infty[$, $\exists M \in \mathbb{R}^{2 \times 2}$ such that,
$$ \Phi(x) \leq M $$
- Then we have the inequality,
$$ y(x) \leq e^{-x} M \xi + \int_0^x e^{-(x-t)} M \begin{pmatrix} y_2^3(t) \\ y_1^7(t) \end{pmatrix} $$
- Because $y$ is solution of the integral equation, $y$ is differentiable. Let's take the derivative with respect to $x$ on both side of the inequality,
$$ y'(x) \leq - e^{-x} M \xi + M \begin{pmatrix} y_2^3(t) \\ y_1^7(t) \end{pmatrix} $$
- I thought to apply a variation of the Gronwall lemma stating that if we have,
$$ y'(x) \leq h(y(x)), \quad h > 0 $$
Then, $$ y(x) \leq H^{-1}(x - x_0) $$ where, $$ H(x) := \int_{y_0}^{y} \frac{1}{h(t)}dt $$
But doesn't seem to help. Any ideas ?