$ y' = x^2 + y^2 $ asymptote

Question

For the D.E. $$ y' = x^2 + y^2 $$ show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$.

I have also been given the hint that a crude estimate for the bounds is $$ \sqrt{\frac{\pi}{2}} < x_0 < 2\sqrt{\frac{\pi}{2}} $$ however I am unsure about how to obtain these bounds.

My approach so far has been to compare the differential equation above to the equation: $$y' = 1 + y^2 $$ which we know has the solution of $ y = \tan(x + c) $ and asymptotes at $x = \pm \frac{\pi}{2}$. But apart from that I am stuck.

Answer 1

We may check that the solution is given by:

$$ y(x) = x \cdot \frac{J_{3/4}(x^2/2)}{J_{-1/4}(x^2/2)}\tag{1}$$

where $J_n$ is a Bessel function. Now the least positive root of $J_{-1/4}$ is very close to $2$: numerically, the life span of $y(x)$ is turns out to be between $2$ and $2.01$.

It is interesting to recall that the ratio of two contiguous Bessel functions of the first kind has an extremely nice continued fraction representation:

$$\frac{J_\nu(z)}{J_{\nu-1}(z)}=\frac{z}{2\nu-\frac{z^2}{2(\nu+1)-\frac{z^2}{2(\nu+2)-\ldots}}}\tag{2}$$

so, for instance, our solution is approximated by $\frac{7x^3}{21-x^4}.$

Answer 2

As $f(x,y)=x^2+y^2$ is locally Lipschitz for the second variable, the equation has one and only one solution with the given initial condition defined for $x>0$ on an interval $[0,b)$.

We need to prove $b<+\infty$.

As $f$ is stricly positive for $(x,y) \neq (0,0)$, the solution $y(x)$ is strictly increasing. Hence for $0<a<x<b$ we have $$\frac{1}{x^2+y^2} \le \frac{1}{a^2+y^2}$$ and $$dx= \frac{dy}{x^2+y^2} \le \frac{dy}{a^2+y^2}$$ Therefore $$x - a= \int_{y(a)}^{y(x)} \frac{dy}{a^2+y^2} \le \frac{\pi}{2a} $$ and finally $$x \le \frac{\pi}{2a} +a$$ proving that $x$ is bounded by a value $x_0$ for which the solution has a vertical asymptote (the solution cannot have a limit for $x=x_0$, because in that case we would be able to extend the solution on the right).

To find upper and lower bounds for $x_0$ you need "to play" with the inequality $x \le \frac{\pi}{2a} +a$.

Answer 3

I will say nothing more, but since I've done the calculation I'm posting my answer.

Assume that y(x) is a solution: then, as its derivative is non negative, $\forall x$, $y(x) \geq 0$, so $y(1) \geq 0= \tan 0$. Note moreover that $\forall x \geq 1$, $y(x) \geq y'(x)^2 +1$. This forces $y(x)$, for $x \geq 1$ to be bigger than $\tan (x-1)$, which is the solution of the problem $$y(1)=0 \ \ \ y'(t)=y(t)^2+1,$$ and so the solution cannot exist beyond $1+\frac{\pi}2 $.