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I can't figure out if there is an analytic solution to the following integral

$x^2f^{''}[x]-xf^\prime[x]+f[x]\left(1-\frac{5/2x^2}{x+f^2[x]}\right)=0 $

with the boundary conditions $f[0]=1$ and $f[1]=0$. I can solve this using mathematica but I would like to see if there is an analytic method to solve this.

Thanks for any help in advance.

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  • $\begingroup$ Analytic solution seems impossible.CAS like Maple cannot do it, either. $\endgroup$ – Mariusz Iwaniuk Aug 3 at 9:34
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Hint:

Let $r=x^2$ ,

Then $\dfrac{df}{dx}=\dfrac{df}{dr}\dfrac{dr}{dx}=2x\dfrac{df}{dr}$

$\dfrac{d^2f}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{df}{dr}\right)=2x\dfrac{d}{dx}\left(\dfrac{df}{dr}\right)+2\dfrac{df}{dr}=2x\dfrac{d}{dr}\left(\dfrac{df}{dr}\right)\dfrac{dr}{dx}+2\dfrac{df}{dr}=2x\dfrac{d^2f}{dr^2}2x+2\dfrac{df}{dr}=4x^2\dfrac{d^2f}{dr^2}+2\dfrac{df}{dr}=4r\dfrac{d^2f}{dr^2}+2\dfrac{df}{dr}$

$\therefore4r^2\dfrac{d^2f}{dr^2}+2r\dfrac{df}{dr}-2r\dfrac{df}{dr}+f\left(1-\dfrac{5r}{2\sqrt r+2f^2}\right)=0$ with $f(0)=1$ and $f(1)=0$

$4r^2\dfrac{d^2f}{dr^2}=\dfrac{5rf}{2\sqrt r+2f^2}-f$ with $f(0)=1$ and $f(1)=0$

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  • $\begingroup$ wow thanks, that is really smart. Mathematica gives me back an analytic solution for that. $\endgroup$ – SAMCRO Aug 3 at 22:54

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