inequation

Hi!
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$<br /> <br /> I = (t_0, t_0 + T) \subset \mathbb{R}, T > 0<br /> <br />$

$u:I \to \mathbb{R}$

$u \in C^0(\overline{I})$

$f,g:I \to \mathbb{R}$

$f,g \in C^0(\overline{I}) , f(t) \ge 0 \ \forall t \in I$

Show that:

if (1)::: $u(t) - u_0 \le \int^t_{t_0} ((f(s)-u(s)+g(s))ds \forall t \in I$ and $u (t_0) = u_0$

=> $u(t) \le exp(\int^t_{t_0} f(x) dx) [u_0 + \int_{t_0}^t g(s)*exp(- \int^s_{t_0} f(x) dx)ds]$

Hint: $\epsilon > 0$

Define
$z_{\epsilon} (t) = exp(\int^t_{t_0} f(x) dx) [u_0 + \epsilon + \int^t_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx) ds]$

Show that
$u(t) < z_{\epsilon} (t) \forall t \in I$

by assuming that $\exists t_1 \in I : t_1 > t_0$ and $u(t_1) = z_{eps}(t_1)$
Therefor use (1) and
$\frac{d}{dt} z_{\epsilon}(t), z_{\epsilon} (t_0)$ and
$u(t) < z_{\epsilon}(t) \forall t \in [t_0,t_1)$

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I tried:

$z_{\epsilon}(t_0) = u_0 + \epsilon$

$\exists t_1 \in I : t_1 > t_0$ and $u(t_1) = z_{eps}(t_1)$

$u(t_1) \le \int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0$

multiplied by 1

$= [exp(\int^{t_1}_{t_0} f(x) dx)][-exp(\int^{t_1}_{t_0} f(x) dx)] * [\int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0]$

$u(t_1) = z_{\epsilon}(t_1)$

$\Rightarrow [exp(\int^{t_1}_{t_0} f(x) dx)][-exp(\int^{t_1}_{t_0} f(x) dx)] * [\int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0]$

$= exp(\int^{t_1}_{t_0} f(x) dx) [u_0 + \epsilon + \int^{t_1}_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx) ds]$

Divide by $[exp(\int^{t_1}_{t_0} f(x) dx)]$

$[-exp(\int^{t_1}_{t_0} f(x) dx)] * [\int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0]$

$= u_0 + \epsilon + \int^{t_1}_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx) ds]$

I dont have any clue what to do now.

Im not sure about this, too

$\frac{d}{dt} z_{\epsilon}(t)$

$= exp(f(t))*(u_0 + \epsilon + \int^t_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx)ds)$
$+ exp (\int^t_{t_0} f(x) dx) [g(t) *exp(-\int^s_{t_0}f(x) dx)]$

any help would be really apprectiated

thanks!