Laplace equation polar where does the ln constant come from

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SUMMARY

The discussion focuses on the derivation of the natural logarithm term, ln(r), in the context of solving the Laplace equation in polar coordinates. The solutions to the ordinary differential equations (ODEs) are identified as cosine/sine functions and the term \(r^n\), stemming from the Cauchy-Euler type equation. The inquiry specifically addresses the appearance of ln(r) when the eigenvalue lambda is set to zero, particularly in relation to the steady-state solution which is constant and periodic with a period of 2π.

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Dustinsfl
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So we have the two ODE solutions are the cosine/sine and $r^n$ since it was a Cauchy Euler type.
For the steady state, the solution is just a constant since it has to have period 2pi. But with $r^n$, how with lambda equal to zero does $\ln r$ come into play?

If my question is hard to follow, go here http://web.mit.edu/6.013_book/www/chapter5/5.7.html

Look at (9) and you will see ln r. I don't see we get that.
 
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dwsmith said:
So we have the two ODE solutions are the cosine/sine and $r^n$ since it was a Cauchy Euler type.
For the steady state, the solution is just a constant since it has to have period 2pi. But with $r^n$, how with lambda equal to zero does $\ln r$ come into play?

If my question is hard to follow, go here http://web.mit.edu/6.013_book/www/chapter5/5.7.html

Look at (9) and you will see ln r. I don't see we get that.

Note that when \(m=0\) we have,

\[r\frac{d}{dr}\left(\frac{1}{r}\frac{dR}{dr}\right)=0\]

Now try to find \(R\).
 

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