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[SOLVED] Laplace equation polar where does the ln constant come from

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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 \(R\).