4
$\begingroup$

What analytical techniques are available for finding solutions to the nonlinear ODE $$y'' y' =A y' y + B (x-1) y,$$ with boundary conditions $$y(0) = 1, \quad y(1) = 0,$$ where $A$ and $B$ are positive, real constants? Unfortunately, neither $A$ nor $B$ are necessarily small.

Does assuming that $y'(x) \neq 0$ allow further progress?

$\endgroup$
5
  • $\begingroup$ Note that $$y(x)=0$$ is one solution. $\endgroup$ Jul 25, 2018 at 15:35
  • 1
    $\begingroup$ $y(x) = 0$ does not satisfy the boundary condition $y(0) = 1$ $\endgroup$
    – mjr
    Jul 25, 2018 at 15:41
  • 1
    $\begingroup$ Did you try the Bellman-Kalaba Quasi-linearization method? $\endgroup$
    – Cesareo
    Jul 25, 2018 at 16:09
  • $\begingroup$ Thanks for the idea - I haven't tried this yet. Is rand.org/pubs/reports/R438.html the standard reference for the technique? I cannot access this paper through my university without paying for a copy. $\endgroup$
    – mjr
    Jul 25, 2018 at 16:55
  • 2
    $\begingroup$ i can send you a copy if that helps $\endgroup$
    – user577488
    Jul 25, 2018 at 21:31

1 Answer 1

0
$\begingroup$

Hint:

Let $t=x-1$ ,

Then $y'y''=Ayy'+Bty$ with $y(-1)=1$ and $y(0)=0$

$y''=Ay+\dfrac{Bty}{y'}$ with $y(-1)=1$ and $y(0)=0$

$\dfrac{d^2y}{dt^2}=Ay+\dfrac{Bty}{\dfrac{dy}{dt}}$ with $y(-1)=1$ and $y(0)=0$

$\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)=Ay+\dfrac{Bty}{\dfrac{dy}{dt}}$ with $y(-1)=1$ and $y(0)=0$

$\dfrac{d}{dt}\left(\dfrac{1}{\dfrac{dt}{dy}}\right)=Ay+Bty\dfrac{dt}{dy}$ with $t(1)=-1$ and $t(0)=0$

$\dfrac{d}{dy}\left(\dfrac{1}{\dfrac{dt}{dy}}\right)\dfrac{dy}{dt}=Ay+Bty\dfrac{dt}{dy}$ with $t(1)=-1$ and $t(0)=0$

$-\dfrac{\dfrac{d^2t}{dy^2}}{\left(\dfrac{dt}{dy}\right)^3}=Ay+Bty\dfrac{dt}{dy}$ with $t(1)=-1$ and $t(0)=0$

$\dfrac{d^2t}{dy^2}=-Ay\left(\dfrac{dt}{dy}\right)^3-Bty\left(\dfrac{dt}{dy}\right)^4$ with $t(1)=-1$ and $t(0)=0$

You can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=429

$\endgroup$
2
  • 1
    $\begingroup$ Thanks. Do you have a further hint? This substitution for $x$ doesn't seem to have much effect on the form of the equation. I've been looking for substitutions for $y$ without much luck so far $\endgroup$
    – mjr
    Jul 26, 2018 at 13:44
  • $\begingroup$ It would be great to have some more details. I found the book you linked to, although somewhere else because the link seems broken, and unfortunately the section (2.6, and specifically 2.6.4) that discusses equations of the relevant form doesn't appear to cover the actual combination of terms in the question - unless I missed it! Do you know if it's covered in the book? Thanks $\endgroup$
    – mjr
    Jul 27, 2018 at 17:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .