Timeline for Nonlinear ODE: $y'=(1+axy)/(1+bxy)$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2016 at 17:26 | comment | added | Rodrigo de Azevedo | 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. | |
| Jun 11, 2016 at 15:41 | answer | added | Christian Remling | timeline score: 2 | |
| Jun 11, 2016 at 15:26 | vote | accept | Paata Ivanishvili | ||
| Jun 11, 2016 at 14:10 | answer | added | Robert Bryant | timeline score: 6 | |
| Jun 11, 2016 at 14:00 | comment | added | Paata Ivanishvili | 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 $a,b,x_{1}$? | |
| Jun 11, 2016 at 8:39 | answer | added | Igor Rivin | timeline score: 2 | |
| Jun 11, 2016 at 2:59 | comment | added | Paata Ivanishvili | 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. | |
| Jun 11, 2016 at 2:04 | comment | added | Paata Ivanishvili | 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. | |
| Jun 11, 2016 at 1:43 | comment | added | Willie Wong | 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. | |
| Jun 11, 2016 at 1:31 | history | edited | Willie Wong |
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| Jun 10, 2016 at 22:50 | history | asked | Paata Ivanishvili | CC BY-SA 3.0 |