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Laplace's equation in two dimensions.

  1. May 21, 2013 #1
    Hi all,
    Had a doubt regarding Laplace's equation.
    In many textbooks, the general solution to the two dimensional Laplace's equation is mentioned as:

    [itex]\Phi(\rho,\phi) = A_{0} + B_{0}ln(\rho) + \sum_{n=1}^{\infty}\rho^n(A_ncos(n\phi) + B_n sin(n\phi)) + \sum_{n=1}^{\infty}\rho^{-n}(C_n cos(n\phi) + D_n sin(n\phi))[/itex]

    in polar coordinates.

    For convenience, I will name the two summation terms as:

    [itex]T_1 = \sum_{n=1}^{\infty}\rho^n(A_ncos(n\phi) + B_n sin(n\phi))[/itex]

    [itex]T_2 = \sum_{n=1}^{\infty}\rho^{-n}(C_n cos(n\phi) + D_n sin(n\phi))[/itex]

    Not much has really been mentioned on whether these two series, T1 and T2 converge or not. When the solution space does not include either zero or infinity it is generally implicitly assumed that they converge. When the solution space includes the origin, generally the procedure has been to equate the coefficients of the second series, T2, to zero, in other words, to exclude the second series from the solution.
    In many physical problems of interest however, the solution space includes both the origin of the coordinate system and is an unbounded region in all directions. My question is, how can we use the above expansions to handle such cases? Especially when there is a boundary condition of zero at infinity. The radius of convergence doesn't seem to be known a priori as the coefficients are unknown.
    For example, if you consider the case of an infinitely long conducting cable of arbitrary but uniform cross-section, (Arbitrary meaning that it need not be something geometrical or symmetric like a circle or a square, and could even be concave; and uniform meaning that the shape and area of the cross-section does not change in the 'z' direction so it can be reduced to a two dimensional problem.) maintained at a constant potential 'V', and where the potential is taken to be zero at infinity. It may often be convenient to choose the origin outside the cable. In this case, if I exclude the term T2 (for finiteness at the origin), again the boundary condition [itex]\Phi\rightarrow0[/itex] as [itex]\rho\rightarrow\infty[/itex] will imply that the coefficients in the term 'T1' become zero as well. So the solution appears to be trivially zero(obviously wrong!) even before I apply the boundary condition [itex]\Phi = V[/itex] at the surface of the cable.
    Can anyone let me know what I am doing wrong? Can anybody throw some light on this?
    Last edited: May 21, 2013
  2. jcsd
  3. May 21, 2013 #2

    Ben Niehoff

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    When you move the origin outside the cylinder, you are forgetting that ##\Phi## is still allowed to blow up inside the cylinder, since the region of interest is only outside.

    Also, in a 2-d problem, I don't think it is always consistent to set the potential to zero at infinity. For example, your cylinder, no matter what its shape, is going to look like a thin wire from very far away; therefore, at infinity the dominant term will be the logarithm.
  4. May 22, 2013 #3
    Thank you for your response. I agree that [itex]\Phi[/itex] is allowed to blow up inside the cylinder. But how is that helpful here? I agree with you that you can have the logarithm term in the expression for the potential. But that still does not appear to resolve the problem. We would not be able to have positive integral powers of [itex]\rho[/itex] in the solution. (As these could lead to the electric field being constant or infinite at infinity. While we expect the electric field to approach that of a system of line charges at infinity.) So if we assume that T1 = 0, we would get the solution as: [itex]\Phi = A_0 + B_0ln(\rho)[/itex] . Now we have only two unknown constants but applying the boundary conditions [itex]\Phi = V[/itex] at all points on the surface of the cable would result in an inconsistency(Infinite no. of equations but only two unknowns.). So what is going wrong here?
    Last edited: May 22, 2013
  5. May 27, 2013 #4

    Jano L.

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    What do you mean by inconsistency? The solution of the problem is

    \Phi = V \ln \frac{\rho}{R}.
  6. May 28, 2013 #5
    This solution is not consistent with [itex]\Phi = V [/itex] everywhere on the surface of the conductor. If it equals V at one point on the surface of the conductor, it would not equal V at another point on the surface of the conductor which has a different value of [itex]\rho[/itex].
  7. May 28, 2013 #6

    Jano L.

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    Perhaps I misunderstood the geometry you have in mind.

    The above solution is intended for the region outside a long metallic cylinder. Inside the cylinder, the potential is constant and equal to the potential on the surface.
  8. May 29, 2013 #7
    Hi, my original question was a general question for any conductor with arbitrary cross-section. This is related to the discussion we had on the other thread. I am trying to solve the 2 dimensional case first before attempting the 3 d one. The exact problem I had in mind was that of an incomplete(I will clarify the meaning of this) conducting cylinderical shell of infinite length and negligible thickness maintained at a potential V and of radius 'a'. This is not a complete cylinder but has an infinitely long slit from ##\phi = -\alpha## to ##\phi=+\alpha##. We would expect the potential to asymptotically approach that of a line charge at infinity, in other words ##\Phi \rightarrow k ln(\rho)## as ##\rho \rightarrow \infty## . Also, we would expect the field to be continuous at ##\rho=a## for ##\phi## = -##\alpha## to +##\alpha##.
  9. May 29, 2013 #8

    Jano L.

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    That seems like a difficult mathematical problem. I would try to solve it numerically first. Generally, there may be some problems due to sharp edges - if they are too sharp, the Laplace equation may not have a solution - but I am not sure what will happen in your example.
  10. May 29, 2013 #9
    Hi Jano,
    Thanks for your response and highlighting the point that this problem has sharp edges. It is a bit disappointing that even relatively simple problems of practical importance are difficult to solve analytically.
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