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 $a\to\infty$...

  • $\begingroup$ 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 Apr 3 '18 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 Apr 3 '18 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 Apr 3 '18 at 23:35
  • $\begingroup$ @RobertIsrael: You are right, thanks! I've just edited my question according to your comments. $\endgroup$ – Ludwig Apr 3 '18 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 Apr 4 '18 at 0:09

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.

  • $\begingroup$ Perhaps, shouldn’t be $b^k$ instead of $(-b)^k$? $\endgroup$ – Ludwig Apr 7 '18 at 2:44
  • $\begingroup$ Oops, yes. Editing $\endgroup$ – Robert Israel Apr 8 '18 at 9:14

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 $a \to \infty$.


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