1
$\begingroup$

Let us consider the following differential equation $$ \dot{x}(t)=a - b\sin(x(t)), \quad a,b\in\mathbb{R}. $$

My question. Suppose $a>|b|$ and $x(0)=x_0\in\mathbb{R}$. Can the solution to the above equation be written in the form $$ x(t) = at + r(a,t), $$ where the term $r(a,t)$ is such that $r(a,t)\to 0$ for $a\to \infty$?

PS: Of course, the explicit solution can be computed via symbolic software tools, such as Wolfram Alpha. However, the symbolic expression (see here) looks quite messy and does not give much information on the behavior of the solution for $a\to\infty$...

Improve this question
$\endgroup$
9
  • $\begingroup$ I believe that my question is a research-level one, so I don’t understand why it received a downvote. Anyway, if the MO community think this is not actually the case, could someone please migrate it to MSE in order to avoid cross-postings? Thanks! $\endgroup$
    – Ludwig
    Commented Apr 3, 2018 at 22:20
  • $\begingroup$ The questions does not seem obvious to me. Unless you know it is a standard result in ODE's don't vote it down. $\endgroup$
    – Piotr Hajlasz
    Commented Apr 3, 2018 at 22:28
  • $\begingroup$ "The" solution? There are infinitely many, unless you set an initial condition. Do you mean "some solution", or "every solution" instead of "the solution"? Also, if $|a| \le |b|$ all solutions are bounded. Do you mean to restrict to the case $a > |b|$? $\endgroup$
    – Robert Israel
    Commented Apr 3, 2018 at 23:35
  • $\begingroup$ @RobertIsrael: You are right, thanks! I've just edited my question according to your comments. $\endgroup$
    – Ludwig
    Commented Apr 3, 2018 at 23:47
  • $\begingroup$ Did you try writing x in the desired form and plugging it into the equation? I don't see how you could then reach your conclusion of r going to zero. Gerhard "The Derivative Changes Too Much" Paseman, 2018.04.03. $\endgroup$
    – Gerhard Paseman
    Commented Apr 4, 2018 at 0:09

2 Answers 2

Reset to default
4
$\begingroup$

Consider the case $a > |b|$, so the solutions are unbounded. The equation is separable, and we get implicit solutions of the form

$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)} $$

which we can expand in a series in $1/a$ (uniformly convergent in $s$). Absolute convergence justifies interchanging sum and integral, so (for fixed $x$ and $b$)

$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (b \sin(s))^k a^{-1-k}\cr &= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}b^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$

Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Perhaps, shouldn’t be $b^k$ instead of $(-b)^k$? $\endgroup$
    – Ludwig
    Commented Apr 7, 2018 at 2:44
  • $\begingroup$ Oops, yes. Editing $\endgroup$
    – Robert Israel
    Commented Apr 8, 2018 at 9:14
2
$\begingroup$

Take the ODE $$ \tag{$*$} \dot{x}(t) = a - b \sin(x(t)) $$ on the one-dimensional torus $\mathbb{R}/2 \pi \mathbb{Z}$. The vector field has no zeros, so there is a unique periodic orbit, with period $$ \int\limits_{0}^{2 \pi} \frac{d\xi}{a - b \sin(\xi)} = \frac{2 \pi}{\sqrt{a^2 - b^2}}. $$ Return now to $(*)$ on the real line $\mathbb{R}$. We have $$ x\!\left(t + \frac{2 \pi}{\sqrt{a^2 - b^2}}\right) = x(t) + 2 \pi \quad \text{for all }t \in \mathbb{R}. $$ Put $$ h(t) := x(t) - \sqrt{a^2 - b^2} t. $$ $h$ is easily seen to be periodic, with period $2\pi/\sqrt{a^2 - b^2}$. Consequently, we have, for any solution $x(\cdot)$ to $(*)$, $$ x(t) = \sqrt{a^2 - b^2} t + h(t). $$ So, $$ r(a, t) = (\sqrt{a^2 - b^2} - a)t + h(t), $$ which, for a fixed $t$, converges to $h(t)$ as $a \to \infty$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .