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I have to consider the following Cauchy problem $$\begin{cases}u''(t)+u(t)=a(t)u(t)\\ u(0)=1\\ u'(0)=0,\end{cases}$$ where $a\in C([0,+\infty))$ and $$\int_0^{+\infty}|a(t)|\mathrm dt<\infty.$$ I am asked to show that THE solution to the Cauchy problem above is bounded on $[0,+\infty)$.

In my attempt I tried to avoid any information about the RHS so I called it $f(t)=a(t)u(t)$ and I used the method of the variation of the parameters. What I got is that the solution to the Cauchy problem $$\begin{cases}u''(t)+u(t)=f(t)\\ u(0)=1\\ u'(0)=0,\end{cases}$$ is given by $$u(t)=-\cos(t)\int_0^t f(\xi)\sin(\xi)\mathrm d\xi+\sin(t)\int_0^tf(\xi)\cos(\xi)\mathrm d\xi+\cos(t).$$

Then I substitute back $a(\xi)u(\xi)=f(\xi)$ in the above and I tried to deduce some consequences, but unfortunately with no success. Am I on the right direction? Can you give any Hint please? even answer are accepted of course.

Regards,

-Guido-

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    Ciao Guido. Have you tried employing [Gronwall's lemma](http://en.wikipedia.org/wiki/Gronwall%27s_inequality#Integral_form_for_continuous_functions)?2012-06-02

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You are on the right track; however, your representation formula for $u$ is too complicated. I have obtained this:

$$u(t)=1+\int_0^t \sin(t-\tau)a(\tau)u(\tau)\, d\tau$$

So the absolute value of $u$ satisfies the integral inequality

$$\lvert u(t)\rvert \le 1+ \int_0^t \lvert a(\tau)\rvert\, \lvert u(\tau)\rvert \, d\tau,$$

hence, by Gronwall's lemma, it is dominated by the solution of the integral equation

$$v(t)=1+\int_0^t \lvert a(\tau)\rvert v(\tau)\, d\tau, $$

that is

$$\lvert u(t)\rvert\le \exp\left( \int_0^t\lvert a(\tau)\rvert\, d\tau\right).$$

In particular $\lvert u(t)\rvert \le \exp\left(\int_0^\infty \lvert a (\tau)\rvert\, d\tau\right)$ for all $t\ge0$.

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    @Giuseppe.... lol.. I have to remember just trigonometric identities it seems... :) to expiate it i will write down a thousand times $$\sin(t-\xi)=\sin(t)\cos(\xi)-\cos(t)\sin(\xi).$$ Grazie btw2012-06-03