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In literature I am using, to solve the following differential equation:

$$\frac{d^2 u}{d \rho^2} = \bigg(1 -\frac{\rho_0}{\rho} + \frac{l(l+1)}{\rho^2} \bigg)u$$ where $l$ is a constant integer, we examine the asymptotic form of the solutions as $\rho \to \infty$ and $\rho \to 0$. As $\rho \to \infty$ the constant term in brackets dominates, so (approximately) $$\frac{d^2 u}{d \rho^2} = u~~~ \text{which has general solution}~~u(p) = Ae^{- \rho} + Be^{\rho}$$ for large $\rho$. We want a normalizable solution, so we take $B=0$. Evidently, $$u(\rho) \approx Ae^{- \rho} $$ for large $\rho$. On the other hand, as $\rho \to 0$ the third term dominates so then we have the approximation: $$\frac{d^2 u}{d \rho^2} = \frac{l(l+1)}{\rho^2}u.$$ The general solution is $$u(\rho) = C \rho^{l+1} + D \rho^{- l}$$ again we take $D =0$, thus $$u(\rho) \approx C \rho^{l+1}$$ for small $\rho$. The next step is said to "peel off the asymptotic behaviour" by introducing a new function $v(\rho)$: $$u(\rho) = \rho^{l+1}e^{- \rho}v(\rho).$$ Is this a standard method of solving differential equations and what is the advantage of writing $u$ as a new function $v$? Is the the last question valid because we can account for the limit behaviour of $u$ by requiring $v(\rho) \to A \rho^{-(l+1)}$ for large $\rho$ and $ v(\rho) \to Ce^{\rho}$ for small $\rho$?

Thanks for assistance.

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  • $\begingroup$ @SimpleArt Yes sorry, typo and $l$ is a constant integer value. Edited now. $\endgroup$ – user100431 Aug 5 '16 at 19:56
  • $\begingroup$ This ODE is the Coulomb wave equation which general solution is the linear combination of the Coulomb wave functions of first and second kind. This is a particular case of confluent hypergeometric function. The solution can also be expressed with the Whittaker functions. mathworld.wolfram.com/CoulombWaveFunction.html $\endgroup$ – JJacquelin Aug 6 '16 at 9:09
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One cannot say that this is "a standard method of solving differential equations". This is one way among several. When facing a difficult differential equation, it is of use to try several approches until finding a successful one. Of course, the search could be unsuccessful and numerical calculus is often required.

For example, it is a common use to change of function $y(x)=x^ae^{bx}g(x)$ and determine the parameters $a$ and $b$ which leads to an ODE simpler than the initial one. This is often valuable for ODE's of generalized Bessel kind or more generally of hypergeometric kind. Of course, this isn't successful in many other cases of ODEs.

In the present case, considering the limit solutions at $0$ and at $\infty$ is advantageous because it avoid boring calculus to determine the above parameters $a=-(l+1)$ and $b=-1$. Nevertheless the transformed ODE with unknown function $v(\rho)$ remains complicated. It is much simpler to directly refer to a known form of ODE which solutions are some known special functions : The Coulomb wave functions or the Whittaker functions in the present case.

http://mathworld.wolfram.com/CoulombWaveFunction.html

http://mathworld.wolfram.com/WhittakerDifferentialEquation.html

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  • $\begingroup$ Thanks for your answer and references. Just to confirm one point, you say that the change of function $y(x) = x^{a} e^{bx} g(x)$ is something that is often used. Is it motivated by the same kind of limit argument as in this specific case? $\endgroup$ – user100431 Aug 6 '16 at 14:18
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    $\begingroup$ No, it isn't motivated by any limit argument in general. But eventually, a limit argument can renforce the idea to try such a change of function. By the way, saying that it is "often used" is too much : "sometimes used" is better. Also, sometimes, other functions than $x^a$ and $e^{bx}$ can be tried as factors if they appear favorable, for example considering a limit or something else. $\endgroup$ – JJacquelin Aug 6 '16 at 15:01
  • $\begingroup$ If you follow these substitution methods for solving PDE's are the final solutions just approximations? $\endgroup$ – user100431 Aug 12 '16 at 13:05
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    $\begingroup$ These substitution methods are sometimes successful, sometimes not. If it is successful, you get the exact final solution. If it is not successful, you get no solution and you have to try other methods. $\endgroup$ – JJacquelin Aug 12 '16 at 16:56

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