# Approximate solution for an ODE

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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## migrated from physics.stackexchange.comJun 17 '13 at 3:44

This question came from our site for active researchers, academics and students of physics.

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

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

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

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

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
@AlexNelson What are words? –  TheDoctor Sep 22 '14 at 0:13