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I need to prove that the following equation

$y'=\sin (xy)\tag{1}$

Has a solution $y\not\equiv 0$ such that $\lim\limits_{x\to+\infty}y=0$. I was able to conclude that any solution of this equation (except for $y\equiv 0$) cannot cross the line $y=0$, because by the theorem of existence and uniqueness for any $x_0\in\mathbb R$ there may only exist one solution satisfying starting condition $y(x_0)=0$, which is $y\equiv 0$. This means that, for example, if a solution has a point with a value greater than zero, then the entire solution is greater than zero and it is bounded from below.

However, I am not sure where to go next. Even if I could prove that a solution is monotonically decreasing starting at some point, being bounded by zero doesn't guarantee that the limit equals zero. Any help would be appreciated

jabuj
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  • $y'=\sin(xy)$ implies $y''=\cos(xy)(xy'+y),$ which implies $y''/(xy'+y)=\cos(xy).$ $y''=\cos(xy)(xy'+y)$ also implies $y'''=-\sin(xy)(xy'+y)^2+\cos(xy)(xy''+2y')=-y'(xy'+y)^2+y''(xy''+2y')/(xy'+y).$ Therefore, $$y'''=-y'(xy'+y)^2+\frac{y''(xy''+2y')}{xy'+y}.$$ This is now a third-order equation, and is still non-linear, but we are no longer dealing with the obnoxious trigonometric function. This may be helpful. I cannot guarantee it yet, which is why it is not an answer, but hopefully someone can work something out from this. – Angel Dec 22 '21 at 13:50
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    This equation was discussed in this question. A paper by Mills, Wesfieler and Krall is cited where it is shown that (main theorem), for $y(0)=A$ with $A>0$, the solution oscillates until it intersects the line $y=x$. Passed this point, it approaches an hyperbola $y=B/x$ asymptotically. The values of $B$ and the way the hyperbola is approached are given in the theorem. Then all solutions converge to $0$ for $x\to\infty$. – Paul Enta Dec 22 '21 at 13:56

1 Answers1

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Whenever $x$ is sufficiently large and $xy$ is close to $(2n+1)\pi$, the equation becomes approximately $$ y'(x)=(-1)^{2n+1}\sin(xy-(2n+1)\pi)\approx(2n+1)\pi-xy, $$ using $\sin(u+k\pi)=(-1)^k\sin(u)$ and the small-angle approximation for $xy-(2n+1)\pi\approx 0$.

The approximate equation has a solution via integrating factor $e^{x^2/2}$ $$ y(x)=Ce^{-x^2/2}+(2n+1)\pi e^{-x^2/2}\int_0^xe^{s^2/2}ds=Ce^{-x^2/2}+(2n+1)\pi\sqrt{2}D(x/\sqrt2) $$ where $D$ is the Dawson function, a reduction of the imaginary error function. As can be found at the link, for large $x$ one has asymptotically $D(x)\sim\frac1{2x}$. As the first term vanishes quickly, this results in $y(x)\sim\frac{(2n+1)\pi}{x}$, so that the solution remains close to equilibrium of the right side, and thus also inside the assumptions of the initial approximation of the DE.

A suitable numerical visualization confirms this result, the plot below contains the plots of $xy(x)/\pi$ for several initial values $y(10)$. It is well visible that there are attracting lines close to the odd integers while the even integers are unstable.

enter image description here

Lutz Lehmann
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  • Could you please provide a little more detail on why for large $x$ and $xy\approx (2n+1)\pi$ the equation becomes approximately $y'=(2n+1)\pi-xy$? This may be simple, but I'm not quite getting it – jabuj Dec 22 '21 at 17:45
  • Because of the small-angle approximation of the sine, I'll add a step. – Lutz Lehmann Dec 22 '21 at 17:49
  • Ah, I see, thank you. Also another question: I don't really understand why considering $xy\approx(2n+1)\pi$ is sufficient to make a conclusion about the limit of $y$. Firstly, isn't it possible that $xy$ is never close enough for the approximation to give reasonable results? Like if $xy$ oscillates around $2\pi$ for example, this doesn't seem plausible, yet it is not obvious to me, why this could not be the case. Also even if there are such points, why talking only about them is enough? Maybe between those the function is not close to 0 and the limit doesn't exist at all – jabuj Dec 22 '21 at 18:05
  • But then the result would still hold. If $xy$ never comes close to an odd multiple of $\pi$, then this value is bounded between the next two such multiples. Oscillation means crossing, which is also not possible without coming close. As the plot shows, $x=10$ is already sufficiently large to invoke the approximation of the Dawson function, even $x=5$ is sufficient. $x=1$ is too small. – Lutz Lehmann Dec 22 '21 at 18:14