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Consider the first order nonlinear ODE problem: $$ y'(x)=\frac{1+ay(x)x}{1+by(x)x}, \quad x>0 $$ where $a, b>0$ are some constants. I would like to know if these kind of equations were studied somewhere else (i.e., existence of global solutions, uniqueness etc). Any references will be helpful.

Remarks:

Mathematica does not give me anything.

I have tried to use Maple and it shows me that $y$ is a solution of an equation involving some Whittaker functions but I am not sure about this.

If $a=b$ then $y=x+C$.

If $a=1, b=2$ then one can take $y(x) = \frac{1}{2}\left(x-\frac{1}{x} \right)$

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    $\begingroup$ If $y > 0$, the RHS is bounded by $\max(1, a/b)$ and is positive, this shows that for every $y_0 \geq 0$ there exists a unique global solution to your ODE with $\lim_{x\to 0}y(x) = y_0$. The argument also tells you that you can only change sign once. So you are down to analyzing the (backwards) initial value problem for $y(x_1) = 0$ and $x_1 > 0$ and solving in the interval $(0,x_1)$, and looking for purely negative solutions. $\endgroup$
    – Willie Wong
    Commented Jun 11, 2016 at 1:43
  • $\begingroup$ Is it possible to have a representation for the global solution? For example, I don't quite understand how maple arrived to Whitakker's functions. $\endgroup$
    – Paata Ivanishvili
    Commented Jun 11, 2016 at 2:04
  • $\begingroup$ If b>a then (going backwards to $x_{1}$) where y is negative and before it reaches the point $y'=\infty$ (assume it exists) we have $y' \leq 1/(1+bxy)$. On the other hand one can solve the equation f'=1/(1+bxf) explicitly and see what happens and then compare. I think it can be done. $\endgroup$
    – Paata Ivanishvili
    Commented Jun 11, 2016 at 2:59
  • $\begingroup$ What I wrote in my previous comment it does not quite work. Right now it is not quite clear for me what happens on the interval $(0,x_{1})$. I understand that y<0, and it is increasing, but when is it globally defined on (0,x_{1}) in terms of $a,b,x_{1}$? $\endgroup$
    – Paata Ivanishvili
    Commented Jun 11, 2016 at 14:00
  • $\begingroup$ Write the ODE in the form $$\underbrace{\left(1 + a \, x \, y\right)}_{=: A (x,y)} \, \mathrm{d}x + \underbrace{\left(-1 - b \, x \, y\right)}_{=: B (x,y)} \, \mathrm{d}y = 0$$ As $\partial_y A \neq \partial_x B$, we have an inexact equation. Thus, we look for an integration factor $\mu (x,y)$ such that $\partial_y (\mu A) = \partial_x (\mu B)$, which yields a PDE. $\endgroup$
    – Rodrigo de Azevedo
    Commented Jun 11, 2016 at 17:26

3 Answers 3

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You might find it useful to make a change of variables to reduce the equation to a more familiar form. For example, if we assume, as we may, that $a$ and $b$ are not equal, then we can substitute $y = (z+ax)/b$, where $z$ is a new unknown, and then the equation can be written in the form $$ \frac{dx}{dz} = \frac{(1 + xz + a x^2)}{b-a}, $$ which is a Riccati equation for $x$ as a function of $z$. Consequently, we understand all of the blow-up conditions since, by the usual linearization of Ricatti equations, we can express the solutions as the ratio of linear combinations of two solutions of a linear equation. Moreover, if we are given a particular, solution, we can now get the general solution by quadrature.

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    $\begingroup$ Blow-up for the original equation means you approach the curve $1+bxy=0$, which after your rewrite is the same as your numerator being zero. This is quite different from blow-up for the Riccati equation for $x(z)$, as usually interpreted. $\endgroup$
    – Christian Remling
    Commented Jun 11, 2016 at 15:27
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Mathematica does give solutions (in terms of the Lambert $W$ function) for specific values of $a, b.$ it's just that it finds different numbers of solutions for different $a, b$ so can't formulate a general answer. for example, for $a=3, b=7,$ it gives:

$$\left\{\left\{y(x)\to \frac{1}{21} \left(-4 W\left(-\frac{7}{4} \sqrt[4]{e^{-9 c_1-9 x}}\right)-7\right)\right\},\left\{y(x)\to \frac{1}{21} \left(-7-4 W\left(-\frac{7}{4} i \sqrt[4]{e^{-9 c_1-9 x}}\right)\right)\right\},\left\{y(x)\to \frac{1}{21} \left(-7-4 W\left(\frac{7}{4} i \sqrt[4]{e^{-9 c_1-9 x}}\right)\right)\right\},\left\{y(x)\to \frac{1}{21} \left(-4 W\left(\frac{7}{4} \sqrt[4]{e^{-9 c_1-9 x}}\right)-7\right)\right\}\right\}$$

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  • $\begingroup$ hm, it may be rewritten as $-(21y+7)/4=W(C e^{-3x/4})$, using definition of Lambert function we rewrite this as $W(C e^{-3x/4})=W(-(21y+7)/4 e^{-(21y+7)/4})$, $(3y+1)e^{(3x-21y)/4}=const$, but taking derivative I get another equation. $\endgroup$
    – Fedor Petrov
    Commented Jun 11, 2016 at 11:28
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    $\begingroup$ @Fedor: Check your algebra: You should be working with $e^{-9x/4}$, not $e^{-3x/4}$. $\endgroup$
    – Robert Bryant
    Commented Jun 11, 2016 at 13:20
  • $\begingroup$ @RobertBryant you are correct, thank you, but for all coefficients I get some different type of equation. $\endgroup$
    – Fedor Petrov
    Commented Jun 11, 2016 at 13:27
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Global existence (or not) can be clarified without any tools by just taking a good look at your equation. Clearly, the RHS $f(x,y)$ is bounded away from the curve $1+bxy=0$, where it is undefined. So global existence fails for a given solution precisely if this solution approaches $1+bxy=0$ in finite time (in particular, global existence for a solution $y(x)$ is guaranteed as soon as $y(x_0)\ge 0$ for some $x_0$).

This won't happen if $a\le b$, as we see by just checking what the ODE does close to our curve. For example, if $1+bx_0y(x_0)=\epsilon>0$, then $y'(x_0)>0$, so we're moving away from the curve.

If $a>b$, then, by the same argument, there are solutions that reach $1+bxy=0$ in finite time (those with sufficiently negative initial value $y(0)$).

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  • $\begingroup$ You are right, one should take a look to the phase portrait first :). Initially I was wondering (let $a<b$) if there is a global solution squeezed between the curves $x=0$ and $1+bxy=0$ such that $y^{*}(x)\sim -\frac{1}{bx}$ as $x \to 0+$, where $y^{*}(x)$ is a solution of the ODE. $\endgroup$
    – Paata Ivanishvili
    Commented Jun 11, 2016 at 16:30
  • $\begingroup$ @PaataIvanisvili: That doesn't happen because $y*'>0$ for such a solution (eventually), so it reaches $y>0$. $\endgroup$
    – Christian Remling
    Commented Jun 11, 2016 at 16:44
  • $\begingroup$ It happens in the example $a=1, b=2$ then $y^{*}(x)=\frac{1}{2}(x-\frac{1}{x})$ so when $x \to 0+$ then $y^{*}(x) \approx -\frac{1}{2x} = -\frac{1}{bx}$. So I was wondering if this may happen for different $a<b$ when the solution behaves as the isngular curve $-\frac{1}{bx}$ going backwards to $0$. Of course it will be always $y^{*}(x)>-\frac{1}{bx}$ but what in the asymptotic sense? $\endgroup$
    – Paata Ivanishvili
    Commented Jun 11, 2016 at 16:49
  • $\begingroup$ @PaataIvanisvili: Right, that's a separate issue, I was discussing what happens going forward in time. $\endgroup$
    – Christian Remling
    Commented Jun 11, 2016 at 17:03

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