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Homework Help: Laplace equation, cylindrical 2D

  1. Mar 28, 2008 #1
    [SOLVED] Laplace equation, cylindrical 2D

    1. The problem statement, all variables and given/known data
    I am given the Laplace eq. in cylindrical coord. (2D), and I am told that we can assume the solution u(rho, Phi) = rho^n * Phi(phi).

    Find the general solution.

    3. The attempt at a solution
    My teacher says that the general solution is u(rh,Phi) = SUM [A*cos(...) + B*sin(...)]*rho^n.

    It is tis version of the Laplace equation on the first equation on page 12 in here:


    Why do we not differentiate rho?
  2. jcsd
  3. Mar 28, 2008 #2


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    You do differentiate with respect to rho! Just plug in the general solution and you will see that it works, at the condition of choosing the constant inside the trig function correctly . You cannot put any old constant multiplying phi inside the trig function.
    write the argument of the trig function as [tex] c \phi [/tex] and plug the general solution in Laplace's equation and you will see that it does work at the condition of choosing "c" to be some particular value (in terms of "n").
  4. Mar 28, 2008 #3
    Ok, now I have substituted u(rho,Phi) in Laplaces equation, divided through by rho^n*Phi and multiplied by rho^2:

    \frac{\rho }{{\rho ^n }}\frac{\partial }{{\partial \rho }}\left( {\rho \frac{{\partial \rho ^n }}{{\partial \rho }}} \right) + \frac{1}{\Phi }\frac{{\partial ^2 \Phi }}{{\partial \phi ^2 }} = 0

    The rho-part does not equal rho^n. Or am I wrong here?
  5. Mar 28, 2008 #4


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    Look again at the laplacian in cylindrical coordinates. There are two mistakes in the expression you wrote above!
  6. Mar 28, 2008 #5
    Duh, I meant plane polar, not cylindrical - sorry!

    So the above is written in plane polar, but it still doesn't work.

    EDIT: The only thing that changes when going to plane polars from cylindrical is that the z-part is missing. So that counts for one of the errors. So one error is remaining, but there is none?
    Last edited: Mar 28, 2008
  7. Mar 28, 2008 #6


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    This is what you want.

    \rho \frac{\partial }{{\partial \rho }}\left( {\rho \frac{{\partial \Phi }}{{\partial \rho }}} \right) + \frac{{\partial ^2 \Phi }}{{\partial \phi ^2 }} = 0


    Just take the cylindrical form, drop z and multiply by rho^2. It does work.
  8. Mar 28, 2008 #7
    By doing that, I get [tex]

    {\frac{\rho}{\rho ^n} \frac{\partial }{{\partial \rho }}\left( {\rho \frac{{\partial \rho ^n}}{{\partial \rho }}} \right) + \frac{1}{\Phi}\frac{{\partial ^2 \Phi }}{{\partial \phi ^2 }} = 0.


    What I did was to insert u(rho, Phi) = rho^n*Phi in the expression Dick wrote to the right of the partial d's, right? Then I divide by rho^n and Phi, and the above expression is what I end up with.

    EDIT: Yeah, it really does work. Thanks to both of you!
    Last edited: Mar 28, 2008
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