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 Post subject: Laplacian Problem
PostPosted: Tue, 1 Nov 2011 00:36:00 UTC 
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Joined: Wed, 29 Sep 2010 08:29:49 UTC
Posts: 48
\begin{document}
Given

u(r,t)$=$Ct$^{\mathrm{-a\thinspace +\thinspace mb}}$ r$^{\mathrm{-m}}$

thus

u$^{\mathrm{g}} \quad =$C$^{\mathrm{g}}$ t$^{\mathrm{(-a\thinspace +\thinspace 
mb)g}}$ r$^{\mathrm{-mg}}$

I am getting:

Lapl u$^{\mathrm{g}} \quad =$C$^{\mathrm{g}}$ t$^{\mathrm{(-a+mb)\thinspace g}}$ 
[(-mg)(-mg-1)] r$^{\mathrm{-mg-2}}$ 

The correct answer is:

Lapl u$^{\mathrm{g}} \quad =$C$^{\mathrm{g}}$ t$^{\mathrm{(-a+mb)\thinspace g}}$ 
[(-mg)(-mg-1) $+$ (n-1)(-mg)] r$^{\mathrm{-mg-2}}$ 

Can you help make clear what I am not understanding about the Laplacian that is causing me to loose the (n-1)(-mg)?

Thanks!

\end{document}


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 Post subject: Re: Laplacian Problem
PostPosted: Tue, 1 Nov 2011 00:37:43 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12169
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Spartan95 wrote:
\begin{document}
Given

u(r,t)$=$Ct$^{\mathrm{-a\thinspace +\thinspace mb}}$ r$^{\mathrm{-m}}$

thus

u$^{\mathrm{g}} \quad =$C$^{\mathrm{g}}$ t$^{\mathrm{(-a\thinspace +\thinspace 
mb)g}}$ r$^{\mathrm{-mg}}$

I am getting:

Lapl u$^{\mathrm{g}} \quad =$C$^{\mathrm{g}}$ t$^{\mathrm{(-a+mb)\thinspace g}}$ 
[(-mg)(-mg-1)] r$^{\mathrm{-mg-2}}$ 

The correct answer is:

Lapl u$^{\mathrm{g}} \quad =$C$^{\mathrm{g}}$ t$^{\mathrm{(-a+mb)\thinspace g}}$ 
[(-mg)(-mg-1) $+$ (n-1)(-mg)] r$^{\mathrm{-mg-2}}$ 

Can you help make clear what I am not understanding about the Laplacian that is causing me to loose the (n-1)(-mg)?

Thanks!

\end{document}


It's impossible to tell unless you show us how you got there. You could have just forgotten it from the get-go, or lost it somewhere in the middle. Post your work and we should be able to find the problem

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