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I've been working on a problem where I ended up with the following nasty first ODE: $$\frac{(-1+e^{l^2})(b^2+l^2)}{l^2}\phi'(l)+\frac{b^2 \left(1-e^{l^2}+l^2\right)+l^2 \left(-2+2 e^{l^2}+l^2\right)}{l^3}\phi(l)=k\frac{1-2b^2-2l^2}{b^2 + l^2},$$ where $k \in \mathbb{R}$ and $b \in \mathbb{R_{\geq 0}}$.

One can somehow simplify the equation and write it as: $$\phi'+\frac{b^2 \left(1-e^{l^2}+l^2\right)+l^2 \left(-2+2 e^{l^2}+l^2\right)}{\left(-1+e^{l^2}\right)\left(b^2+l^2\right)l}\phi= k \frac{l^2 \left(1-2 b^2-2 l^2\right)}{\left(-1+e^{l^2}\right) \left(b^2+l^2\right)^2}.$$

Writting the above as $\phi'+h(l)\phi=g(l)$ the solution is given by: $$\phi(l) = e^{-\int_1^l h(\xi) d\xi} \int_1^l g(\zeta)~e^{\int_1^\zeta h(\xi) d\xi} d\zeta+C_1 e^{ -\int_1^l h(\xi) d\xi},$$

where $C_1$ is an integration constant.

The problem is in solving that big integral in the previous expression. I can't solve it and Mathematica tells me that it can't be put in terms of standard math functions. I was then hoping that somebody could help me find an approximate solution maybe using the method of matched asymptotic expansions since I know that $ \phi(l)=O(1/l^2)$ as $l \to \infty$. I also know from symmetry that $\phi(0)=0$.

I'm still not very good with approximation methods, nor know many, so I would be really grateful for any help you could give me.

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1  
Have you tried the math stackexchange? This site is for conceptual physics questions. Yours seems to be pure math. If you can physically motivate your problem that might lead to a useful approximation method. For instance, how do you know the leading behaviour of $\phi(l)$? What does it mean? –  Michael Brown Jun 12 '13 at 14:40
    
I tried to put the problem within a physics background by editing the question but looking at the closing votes it wasn't enough. The behaviour of $\phi$, the gravitational field, for large $l$ - $l$ is some kind of distance - is because of the classical limit. The spacetime is asymptotically flat by construction so for large $l$: $\phi=O(1/l^2)$. –  PML Jun 12 '13 at 14:50
    
Why not just make the approximation that $\pm1+\exp(l^{2})\approx\exp(l^{2})$ and $\pm l^{2}+\exp(l^{2})\approx\exp(l^{2})$? –  Alex Nelson Jun 12 '13 at 21:07
    
@AlexNelson Hum, in the sense of the asymptotic expansion? I mean, take those approximations for $l \ll 1$ and get a result. Then for $l\gg 1$ those are still very good approximations plus $b^2+l^2\approx l^2$. Then get the result and then do the matching? If so it's actually simple. I thought I couldn't do that because of how it's explained in the wiki article... –  PML Jun 12 '13 at 23:29
    
@PML posting a bounty could be a good idea –  Dilaton Jun 13 '13 at 23:50

1 Answer 1

$\dfrac{(-1+e^{l^2})(b^2+l^2)}{l^2}\phi'(l)+\dfrac{b^2(1-e^{l^2}+l^2)+l^2(-2+2e^{l^2}+l^2)}{l^3}\phi(l)=k\dfrac{1-2b^2-2l^2}{b^2+l^2}$

$\dfrac{(e^{l^2}-1)(l^2+b^2)}{l^2}\phi'(l)+\dfrac{(e^{l^2}-1)(2l^2-b^2)+l^2(l^2+b^2)}{l^3}\phi(l)=\dfrac{k}{l^2+b^2}-2k$

$\phi'(l)+\biggl(\dfrac{2l^2-b^2}{l(l^2+b^2)}+\dfrac{l}{e^{l^2}-1}\biggr)\phi(l)=\dfrac{kl^2}{(l^2+b^2)^2(e^{l^2}-1)}-\dfrac{2kl^2}{(l^2+b^2)(e^{l^2}-1)}$

I.F. $=e^{\int\bigl(\frac{2l^2-b^2}{l(l^2+b^2)}+\frac{l}{e^{l^2}-1}\bigr)dl}=e^{\frac{3}{2}\ln(l^2+b^2)-\ln l+\frac{1}{2}\ln(1-e^{-l^2})}=\dfrac{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}{l}$

$\therefore\biggl(\dfrac{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}{l}\phi(l)\biggr)'=\dfrac{kle^{-l^2}}{\sqrt{l^2+b^2}\sqrt{1-e^{-l^2}}}-\dfrac{2kl\sqrt{l^2+b^2}e^{-l^2}}{\sqrt{1-e^{-l^2}}}$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\dfrac{kle^{-l^2}}{\sqrt{l^2+b^2}\sqrt{1-e^{-l^2}}}dl-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\dfrac{2kl\sqrt{l^2+b^2}e^{-l^2}}{\sqrt{1-e^{-l^2}}}dl$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\dfrac{kle^{-l^2}}{\sqrt{l^2+b^2}}\sum\limits_{n=0}^\infty\dfrac{(2n)!e^{-nl^2}}{4^n(n!)^2}dl-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int2kl\sqrt{l^2+b^2}e^{-l^2}\sum\limits_{n=0}^\infty\dfrac{(2n)!e^{-nl^2}}{4^n(n!)^2}dl$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{kl(2n)!e^{-(n+1)l^2}}{4^n(n!)^2\sqrt{l^2+b^2}}dl-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{kl(2n)!\sqrt{l^2+b^2}e^{-(n+1)l^2}}{2^{2n-1}(n!)^2}dl$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)u}}{2^{2n+1}(n!)^2\sqrt{u+b^2}}du-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!\sqrt{u+b^2}e^{-(n+1)u}}{4^n(n!)^2}du~(\text{Let}~u=l^2)$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!v^2e^{-(n+1)(v^2-b^2)}}{2^{2n-1}(n!)^2}dv~(\text{Let}~v=\sqrt{u+b^2})$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!v}{4^n(n!)^2(n+1)}d(e^{-(n+1)(v^2-b^2)})$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^n(n!)^2(n+1)}-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int e^{-(n+1)(v^2-b^2)}~d\left(\sum\limits_{n=0}^\infty\dfrac{k(2n)!v}{4^n(n!)^2(n+1)}\right)$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2}dv+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^n(n!)^2(n+1)}-\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{k(2n)!e^{-(n+1)(v^2-b^2)}}{4^n(n!)^2(n+1)}dv$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^nn!(n+1)!}+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\int\sum\limits_{n=0}^\infty\dfrac{kn(2n)!e^{(n+1)b^2}e^{-(n+1)v^2}}{4^n(n!)^2(n+1)}dv$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!ve^{-(n+1)(v^2-b^2)}}{4^nn!(n+1)!}+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=1}^\infty\dfrac{\sqrt\pi kn(2n)!e^{(n+1)b^2}\text{erf}(\sqrt{n+1}v)}{2^{2n+1}(n!)^2(n+1)^{\frac{3}{2}}}+\dfrac{Cl}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}$

$\phi(l)=\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=0}^\infty\dfrac{k(2n)!\sqrt{l^2+b^2}e^{-(n+1)l^2}}{4^nn!(n+1)!}+\dfrac{l}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}\sum\limits_{n=1}^\infty\dfrac{\sqrt\pi kn(2n)!e^{(n+1)b^2}\text{erf}(\sqrt{n+1}\sqrt{l^2+b^2})}{2^{2n+1}(n!)^2(n+1)^{\frac{3}{2}}}+\dfrac{Cl}{(l^2+b^2)^{\frac{3}{2}}\sqrt{1-e^{-l^2}}}$

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Perhaps you should use some words... –  Alex Nelson Jul 11 '13 at 14:03
    
I think you may need a constant of integration. –  Graham Hesketh Jul 11 '13 at 17:14
    
Sorry, I forgot to notice that $0<e^{-l^2}\leq1$ for $l\in\mathbb{R}$ , so in fact it should be relieved to expand $\dfrac{1}{\sqrt{1-e^{-l^2}}}$ . –  doraemonpaul Aug 27 '13 at 7:43

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