Determine whether the solution for the ODE: $y'(x) =(1-y(x)^2)(x^2+1)^{-n}\sin(x)$ is generally increasing or decreasing.

Question

The ODE is for $n\in \mathbb{N}$, The given initial condition is $y(0)=0$ thus the solution is between two singular points $y=-1, +1$ .

We have the relation: $\int_{0}^{y(x)} \frac{1}{1-y^2} = \int_0 ^t \frac{sin(x)}{(x^2+1)^n} $, and from the fact the initial condition is between the singular points the the solution $y(x)$ is defined for $x\in\mathbb R$ .

The integral $\int_0 ^\infty \frac{sin(x)}{(x^2+1)^n} < \infty $ using absolutely convergence of $\int_0 ^\infty \frac{|sin(x)|}{(x^2+1)^n}$ and by the fact that $\int_0 ^\infty \frac{|sin(x)|}{(x^2+1)^n} \le \int_0 ^\infty \frac{1}{(x^2+1)} = \arctan(x)|_0^\infty <\infty$.

We also know that: $\int_0 ^{-\infty} \frac{sin(x)}{(x^2+1)^n} = \int_{-\infty} ^0 -\frac{sin(x)}{(x^2+1)^n} =_{sin(-x)=-sin(x)} \int_{-\infty} ^0 \frac{sin(-x)}{(x^2+1)^n} = \int_0 ^\infty \frac{sin(x)}{(x^2+1)^n} $.

Lastly we get $\lim_{x\rightarrow \infty} y(x) <1 , \lim_{x\rightarrow -\infty} y(x) >-1$ cause the integral $\int_{0}^{+1 / -1} \frac{1}{1-y^2} =\infty$.

$y(x)$ changes between monotonic increasing to monotonic decreasing due to the fact $1-y^2 > 0 \forall y\in (-1,1)$ and $\frac{sin(x)}{(x^2+1)^n}$ changes signs with respect to $\sin(x)$'s sign. I don't succeed to determine whether $y(t)$ is generally increasing or decreasing or in other words, what is the relation (in terms of who is greater) between $\lim_{x\rightarrow \infty} y(x)$ and $\lim_{x\rightarrow -\infty} y(x)$ ,is it even possible to determine it from the given conditions?

Answer 1

Let $z(x)=y(-x)$. Then \begin{align} z'(x)&=-y'(-x)\\ &=-(1-y(-x)^2)((-x)^2+1)^{-n}\sin(-x)\\ &=(1-z(x)^2)(x^2+1)^{-n}\sin x \end{align} and $z(0)$. By uniqueness of solutions, $z(x)=y(x)$ for all $x$, that is $y$ is even, and the behavior at $+\infty$ and $-\infty$ is the same.

Answer 2

Regarding the question in the title of your question: The solution of $$ \begin{cases} y'(x) =(1-y(x)^2) \frac{\sin{x}}{(x^2+1)^n} \\ y(0) = 0 \end{cases} $$ is increasing on any of the integrals $[2k\pi, (2k+1)\pi]$ and decreasing on any of the intervals $[(2k-1)\pi, 2k\pi]$ (you already observed that $y(x) \in (-1, 1)$ for all $x \in \mathbb{R}$).

From the relation $$ \int\limits_{0}^{y(x)} \frac{d\eta}{1 - \eta^2} = \int\limits_{0}^{x} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^n} $$ we obtain $$ y(x) = \tanh{G(x)}, $$ where $$ G(x) = \int\limits_{0}^{x} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^n}. $$ The limit $$ \lim\limits_{x \to \infty} G(x) = \int\limits_{0}^{\infty} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^n} $$ can be expressed in terms of some special functions. For example, for first positive integers Mathematica gives $$ \int\limits_{0}^{\infty} \frac{\sin{\xi} \, d\xi}{\xi^2 + 1} = \frac{\text{Ei}(1)-e^2 \text{Ei}(-1)}{2 e}, $$ $$ \int\limits_{0}^{\infty} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^2} = \frac{\text{Ei}(1)}{2 e} $$ $$ \int\limits_{0}^{\infty} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^3} = \frac{1}{16} \left(-e \text{Ei}(-1)+\frac{7 \text{Ei}(1)}{e}-2\right), $$ $$ \int\limits_{0}^{\infty} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^4} = \frac{1}{96} \left(-5 e \text{Ei}(-1)+\frac{37 \text{Ei}(1)}{e}-14\right), $$ where $\mathrm{Ei}$ denotes the exponential integral.

Finally, since the integrand is an odd function, one has $$ \int\limits_{0}^{\infty} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^n} = \int\limits_{0}^{-\infty} \frac{\sin{\xi} \, d\xi}{(\xi^2 + 1)^n}, $$ consequently $$ \lim\limits_{x\to\infty} y(x) = \lim\limits_{x\to-\infty} y(x). $$