# Riccati D.E., vertical asymptotes

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 $x_0$:

$$y'=x^2+y^2$$ $$x\in \left [ a,b \right ]$$ $$b> a> 0$$ $$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$ $$a^2+y^2\leq y'\leq b^2+y^2$$ $$y'\geq a^2+y^2$$ $$\frac{y}{a^2+y^2}\geq 1$$ $$\int \frac{dy}{a^2+y^2}\geq \int dx=x+c$$ $$\frac{1}{a}\arctan \frac{y}{a}\geq x+c$$ $$\arctan \frac{y}{a}\geq a(x+c)$$ $$\frac{y}{a}\geq\tan a(x+c)$$ $$y\geq a\tan a(x+c)$$ $$a(x+c)\simeq \frac{\pi}{2}$$

But where to from here?

## 3 Answers

1. $x_0$ exists

First note that $y'''(x)$ is increasing$^{}$. It is also easy to see that $y'(0)=y''(0)=0$ but $y'''(0)=2$$^{}, so by Taylor's theorem^{},$$ 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, $$which, since y(\epsilon)>0, implies y(x)\to\infty as x\to x_0<\infty for some x_0>\epsilon. Edits: : Since y'(x)=x^2+y(x)^2\ge 0, y is increasing. Since y\ge 0 and x\ge 0, we have y''(x)=2x+2y(x)y'(x)\ge 0, so y' is also increasing. In a similar way, we deduce that y'''(x)\ge 0 and y^{(4)}(x)\ge 0. : Since y(0)=0, we have y'(0)=0. Therefore, y''(0)=2x+2y(x)y'(x)|_{x=0}=0. On the other hand, y'''(0)=2+2y'(x)^2+2y(x)y''(x)|_{x=0}=2. : First note that y is smooth. Indeed, since y is continuous and y'(x)=x^2+y(x)^2, we see that y'(x) is continuous. Since y''(x)=2x+2y(x)y'(x) and the right hand side is continuous, so is y''. In a similar way, all derivatives of y are continuous. Since y is smooth, Taylor's theorem can be applied:$$ y(x)=y(0)+xy'(0)+\frac{1}{2}x^2y''(0)+\frac{1}{6}x^3 y'''(c),\qquad x>0, $$where c\in(0,x). But the first three terms are zero by , so (*) holds. 2. Lower bound: Since a finite x_0>0 exists, we get$$ 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. $$Then y(\zeta) does not exist, so x_0<\zeta. Note that z\approx 1.25. • Can you flesh out your methods for arriving at (*)? – liana1000 Jul 7 '17 at 5:33 • @liana1000 I added more details before 2.. Let me know if I can clarify. – user254433 Jul 7 '17 at 6:47 The usual trick to get a better manageable equation out of this Riccati equation is to substitute y=-\frac{u'}{u} which results in the linear ODE of second order$$ 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] $$while the numerator -u'(x)=\frac{x^3}{3}(1-\frac{x^4}{4·7}\pm…) has its first positive root above \sqrt[4\,]{28}. • A symbolic solution would have to use parabolic cylinder functions of a complex argument, that's not very practical. The smallest positive zero x_0 of u(x) can be found numerically with Newton iteration. A nice detail: since u''(x_0)=0 due to the differential equation, we have cubic convergence in this case. The approximation x_0=2.003147359426885 can be obtained in three steps (starting from 1.8), so the upper bound is rather good. – Professor Vector Jul 6 '17 at 11:24 From numerical solution comes out that x=2 and x=-2 are vertical asymptotes. Trying to solve as a Bernoulli equation gives a mess and the substitution w=\frac{1}{y} gives problems as$$w=\frac{1}{y};\;w'=-\frac{y'}{y^2}$$Divide the original equation by y^2$$\frac{y'}{y^2}=\frac{x^2}{y^2}+1\rightarrow -w'=x^2w^2+1w'+x^2w^2=-1\rightarrow w(x)=c\;e^{-\frac{x^3}{3}}+\frac{e^{-\frac{x^3}{3}} x \Gamma \left(\frac{1}{3},-\frac{x^3}{3}\right)}{3^{2/3} \sqrt{-x^3}}$$The problem is now with the initial value, as w\to\infty as x\to 0 Anyway the general solution is$$y=\frac{1}{w(x)}=\frac{3 e^{\frac{x^3}{3}}}{3 c+x E_{\frac{2}{3}}\left(-\frac{x^3}{3}\right)}$$where E_k(x) is the integral exponential function defined by$$E_k(x)=\int_1^{\infty }\frac{e^{-tx}}{t^k}\,dt$$and has a vertical asymptote for any$c\in\mathbb{R}$Hope this helps • This seems to be wrong, it looks pretty much like the general solution of$-w'=x^2w+1,\$ instead. – Professor Vector Jul 6 '17 at 8:04