Riccati D.E., vertical asymptotes

Solution 1:

1. $x_0$ exists

First note that $y'''(x)$ is increasing$^{[1]}$. It is also easy to see that $y'(0)=y''(0)=0$ but $y'''(0)=2$$^{[2]}$, so by Taylor's theorem$^{[3]}$, $$ y(x)=\frac{x^3}{6}y'''(c)\ge \frac{x^3}{3},\qquad (*) $$ for all $x>0$ such that $y$ is defined. Choose one such $x=\epsilon>0$. Then if $x>\epsilon$, we get $$ y'(x)\ge \epsilon^2+y(x)^2, $$ which, since $y(\epsilon)>0$, implies $y(x)\to\infty$ as $x\to x_0<\infty$ for some $x_0>\epsilon$.

Edits:

$[1]$: Since $y'(x)=x^2+y(x)^2\ge 0$, $y$ is increasing. Since $y\ge 0$ and $x\ge 0$, we have $y''(x)=2x+2y(x)y'(x)\ge 0$, so $y'$ is also increasing. In a similar way, we deduce that $y'''(x)\ge 0$ and $y^{(4)}(x)\ge 0$.

$[2]$: Since $y(0)=0$, we have $y'(0)=0$. Therefore, $y''(0)=2x+2y(x)y'(x)|_{x=0}=0$. On the other hand, $y'''(0)=2+2y'(x)^2+2y(x)y''(x)|_{x=0}=2$.

$[3]$: First note that $y$ is smooth. Indeed, since $y$ is continuous and $y'(x)=x^2+y(x)^2$, we see that $y'(x)$ is continuous. Since $y''(x)=2x+2y(x)y'(x)$ and the right hand side is continuous, so is $y''$. In a similar way, all derivatives of $y$ are continuous. Since $y$ is smooth, Taylor's theorem can be applied: $$ y(x)=y(0)+xy'(0)+\frac{1}{2}x^2y''(0)+\frac{1}{6}x^3 y'''(c),\qquad x>0, $$ where $c\in(0,x)$. But the first three terms are zero by [2], so (*) holds.

2. Lower bound:

Since a finite $x_0>0$ exists, we get $$ y'(x)\le x_0^2+y(x)^2, $$ which, since $y(0)=0$, implies $$ y(x)\le x_0 \tan (x_0\,x). $$ If it were true that $x_0^2<\pi/2$, then $y(x_0)<\infty$, so $x_0\ge\sqrt{\pi/2}=:z$.

3. Upper bound

For $x>z$, where $z$ is the lower bound, we have $$ y'(x)\ge z^2+y(x)^2, $$ which implies $$ y(x)\ge z\,\tan z(x+c), $$ where $$ c=-z+\frac{1}{z}\arctan\frac{y(z)}{z}\ge-z+\frac{1}{z}\arctan\frac{z^2}{3} $$ by inequality (*). Let $$ \zeta=\frac{\pi}{2z}-c\le \frac{\pi}{2z}+z-\frac{1}{z}\arctan\frac{z^2}{3}\approx 2.12. $$ Then $y(\zeta)$ does not exist, so $x_0<\zeta$. Note that $z\approx 1.25$.

Solution 2:

The usual trick to get a better manageable equation out of this Riccati equation is to substitute $y=-\frac{u'}{u}$ which results in the linear ODE of second order

$$ u''+x^2u=0,\quad u(0)=1,\, u'(0)=0 $$

While this still does not lead to a symbolic solution without involving (very) special functions (Convert $\frac{d^2y}{dx^2}+x^2y=0$ to Bessel equivalent and show that its solution is $\sqrt x(AJ_{1/4}+BJ_{-1/4})$), one can easily find a power series solution $$ u(x)=1-\frac{x^4}{3·4}+\frac{x^8}{3·4·7·8}-\frac{x^{12}}{3·4·7·8·11·12}\pm… $$ This is an alternating series with eventually monotonically falling absolute values of the terms. For $x<\sqrt7$ one gets the bounds by partial sums $$ 1-\frac{x^4}{3·4}\le u(x)\le1-\frac{x^4}{3·4}+\frac{x^8}{3·4·7·8}. $$ The first positive root of $u(x_0)=0$ is the location of the first pole of $y$. From the bounds one gets the root bounds

$$ \sqrt[4\,]{12}\le x_0\le \sqrt[4\,]{16+4(3-\sqrt7)} $$

which numerically gives the interval $$ [1.8612097182041991,\; 2.042882110200651] $$ while the numerator $-u'(x)=\frac{x^3}{3}(1-\frac{x^4}{4·7}\pm…)$ has its first positive root above $\sqrt[4\,]{28}$.