# Demonstrate that solution of differential equation is bounded

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 $$M:=\max_{[0,+\infty)}|b(t)|.$$

• $x$ will in general oscillate. How does one understand "monotone" in that context? – LutzL Jan 10 at 18:27
• 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. – LutzL Jan 10 at 18:43
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}