Asymptotic behavior: (d^2 u)/(dp^2) = [1 - (p_0)/p + {L(L+1)}/(p^2)]u

Viona

New member
Joined
Nov 22, 2018
Messages
21
Hello every one,

I was reading in a physics book where we want to solve this differential equation:
456.jpg
I guess it is difficult to solve directly so they examined asymptotic behavior:
4562.jpg
457-458.jpg
459.jpg

460.jpg
The thing what I did not get is equation (4.60). It will be nice if some one helped me to understand how they did this step.
Thanks.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,237
Hello every one,

I was reading in a physics book where we want to solve this differential equation:
View attachment 10618
I guess it is difficult to solve directly so they examined asymptotic behavior:
View attachment 10619
View attachment 10620
View attachment 10621

View attachment 10622
The thing what I did not get is equation (4.60). It will be nice if some one helped me to understand how they did this step.
Thanks.
I think they are implicitly defining something. Very obscure if that is what is going on. That is, I guess they are doing this:

\(\displaystyle \text {Define } \nu( \rho ) = \dfrac{\mu ( \rho )}{ \rho^{(L + 1)} e^{- \rho}} \implies \mu ( \rho ) = \rho^{(L+1)} e^{- \rho}\nu ( \rho ).\)

This is why I could not stand physics.
 

Viona

New member
Joined
Nov 22, 2018
Messages
21
I think they are implicitly defining something. Very obscure if that is what is going on. That is, I guess they are doing this:

\(\displaystyle \text {Define } \nu( \rho ) = \dfrac{\mu ( \rho )}{ \rho^{(L + 1)} e^{- \rho}} \implies \mu ( \rho ) = \rho^{(L+1)} e^{- \rho}\nu ( \rho ).\)

This is why I could not stand physics.
Interesting guess. It works for me. Thanks
 
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