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I have the differential equation $y''+by=\cos(t)$. The general solution is

$y(t)=\begin{cases}c_1\cos(\sqrt{b}t)+c_2\sin(\sqrt{b}t)+\frac{1}{b-1}\cos(t), b>0, b\neq 1 \\ c_1\cos(t)+c_2\sin(t)+\frac{1}{2}t\sin(t), b =1 \\ c_1t+c_2+\frac{1}{b-1}\cos(t), b=0 \\ c_1 e^{\sqrt{|b|t}}+c_2e^{-\sqrt{|b|t}}+\frac{1}{b-1}\cos(t), b<0\end{cases}$

How do I determine if the solutions are bounded on the interval $[0,\infty)$?

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  • $\begingroup$ You have three options (depending on $b$), so you should check if each one is bounded or not. $\endgroup$
    – Cbjork
    May 4, 2018 at 23:38
  • $\begingroup$ How do I check if they are bounded? Do I simply use what I know about the boundedness of the constituent functions? $\endgroup$
    – blargen
    May 4, 2018 at 23:39
  • $\begingroup$ What does bounded mean? $\endgroup$
    – Cbjork
    May 4, 2018 at 23:41
  • $\begingroup$ $\exists M s.t. f(x) \leq M$ for all x $\endgroup$
    – blargen
    May 4, 2018 at 23:43
  • $\begingroup$ There is an absolute value you're missing, but yes. $\endgroup$
    – Cbjork
    May 4, 2018 at 23:44

1 Answer 1

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This is a very informal answer, but you can expand on it if you wish.


First, let's identify which functions are bounded and which are not.

  • A constant solution like $f(t)=c$ is obviously always bounded

  • A linear combination of trig functions is always bounded, since $|\sin t|, |\cos t| \le 1$.

  • A decaying exponential is bounded on $[0,\infty)$ since $0< e^{-t} \le 1 $

  • A growing exponential is unbounded on $[0,\infty)$ since $\lim_\limits{t\to\infty} e^t = \infty$

  • Polynomials of degree $1$ and higher are never bounded, since $\lim_\limits{t\to \infty} t^n = \infty$


Let's look at the cases

  • $b > 0 \ne 1$ is a linear combination of bounded functions, so is it always bounded. More formally $$ \big|y(t)\big| \le |c_1|\big|\cos(\sqrt{b}t)\big| + |c_2|\big|\cos(\sqrt{b}t)\big| + \frac{1}{|b-1|}|\cos t| \le |c_1| + |c_2| + \frac{1}{|b-1|} $$

  • $b=1$ grows without bounds, since it contains the term $ \frac{1}{2}t\sin t $ which has no bound over $[0,\infty)$

For the remaining cases, the particular solution is always bounded, while the general solutions both contain one unbounded term. Therefore, they will either be bounded or unbounded under certain conditions.

  • $b=0$ contains a linear term, so it will be bounded if $c_1 = 0$. For example, when $y'(0) = 0$

  • Similarly, $b < 0$ will be unbounded if $c_1 = 0$. For example, when $\lim_\limits{t\to\infty}y(t)=0$

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  • $\begingroup$ The last equality in your equation for $b>0$ should also be an inequality, as you replace oscillating functions by constant bounds. $x\sin x$ has no limit, usually $\lim_{x\to\infty} u(x)=\infty$ means that $u(x)>N$ for $x>x_N$ for all $N>0$, which is a formalization of "grows over all boundaries". Of course, maximizing over the oscillations works again, $\lim_{x\to\infty}\max_{t\in [0,x]}t\sin t=\infty$. $\endgroup$ May 6, 2018 at 11:55
  • $\begingroup$ Thanks for the corrections. That first equal sign was a typo, my bad. The limit was meant to be for $|y(t)|$ $\endgroup$
    – Dylan
    May 6, 2018 at 13:04
  • $\begingroup$ But even $|y(t)|$ oscillates between $0$ and $t$ (at $t=(k+0.5)\pi$), so there is no real convergence there, the symbol $\lim$ is not really applicable since its use implies the existence of a limit. You can say $\lim |y((k+0.5)\pi)|=\infty$ and that proves unboundedness. $\endgroup$ May 6, 2018 at 13:20
  • $\begingroup$ Hmm you're right. I'm just going to keep it vague $\endgroup$
    – Dylan
    May 6, 2018 at 15:12

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