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Let $b(t) \in C^1([0,+\infty))$. I have to find a formula for the solution of this Cauchy's problem:

$$ \left\{\begin{aligned} x''(t)+x(t)&=b(t) \\ x(0)&=x_0\\ x'(0)&=x_1 \end{aligned}\right. $$ I have solved this part of the question and the formula is: $$x(t) = x_0\cos(t)+x_1\sin(t) -\cos(t)\int_0^tb(s)\sin(s)\,ds+\sin(t)\int_0^tb(s)\cos(s)\,ds.$$ Then the problem asks to demonstrate that if $b(t)$ is bounded and monotone also $x(t)$ is bounded. It also asks to find a counterexample if $b(t)$ is bounded but not monotone. I can solve the second part by picking for example $b(t) = \cos(t)$ but I'm not able to demonstrate the first fact. I have tried to estimate $|x(t)|$ but the only inequality I come up with is $|x(t)| \leq |x_0|+|x_1|+2Mt$ where $M:=\max_{[0,+\infty)}|b(t)|.$

Thank you in advance for your help.

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    $\begingroup$ $x$ will in general oscillate. How does one understand "monotone" in that context? $\endgroup$ – LutzL Jan 10 at 18:27
  • $\begingroup$ By "limitated", do you mean "bounded"? $\endgroup$ – Adrian Keister Jan 10 at 18:27
  • $\begingroup$ @Keister Yes i mean bounded $\endgroup$ – edo1998 Jan 10 at 18:34
  • $\begingroup$ @Lutzl Sorry i mistiped. I had to demonstrate that x is bounded! I'll edit the question $\endgroup$ – edo1998 Jan 10 at 18:35
  • $\begingroup$ If $(b_k)$ is monotonically increasing and bounded, then $b_*=\lim_kb_k$ exists and $\sum_{k=1}^{2n}(-1)^kb_k=\sum_{k=1}^{2n}(-1)^k(b_k-b_*)$ which converges by the Leibniz rule, making the sequence of partial sums bounded. You will need to use something similar. $\endgroup$ – LutzL Jan 10 at 18:43
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With $b\in C^1$ and monotonous, the sign of $b'$ is constant. Partial integration gives $$ \int_0^t\sin(t-s)b(s)ds=[\cos(t-s)b(s)]_0^t+\int_0^t\cos(t-s)b'(s)ds $$ Thus \begin{align} \left|\int_0^t\sin(t-s)b(s)ds\right|&\le |b(t)-b(0)\cos(t)|+\int_0^t|b'(s)|ds \\ &\le|b(0)|+|b(t)|+|b(t)-b(0)|\le 2b^*=2\sup_{s\ge0}|b(s)|. \end{align}

In some way this is an integral version of the Dirichlet test for series.

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