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Question about a boundary-value problem (electrostatics)

  1. Mar 25, 2016 #1
    Laplacian for polars:

    $$\frac{1}{r}\frac{\partial}{\partial r}\left( r\frac{\partial \phi}{\partial r}\right) + \frac{1}{r^{2}}\frac{\partial^{2} \phi}{\partial \theta^{2}} = 0$$

    This is in relation to a problem relating to a potential determined by the presence of a wedge shaped metallic conductor (see image).
    kajx4g.png
    Working through the problem:

    $$\phi (r,\theta) = R(r)\psi (\theta )$$ with $$R(r) \propto r^{v}$$ and Dirichlet boundary conditions $$\phi (r,0) = \phi (r,\alpha ) = V_{0}$$

    Substituting into the Laplacian:
    $$\frac{1}{r}\frac{\partial}{\partial r}\left( rvr^{v-1}\psi (\theta)\right) + \frac{1}{r^2}r^{v}\psi^{''}(\theta )$$
    $$ = r^{v-2}\left( \psi^{''}(\theta ) + v^{2}\psi (\theta )\right) = 0$$
    Therefore
    $$\psi^{''}(\theta ) + v^{2}\psi (\theta ) = 0$$
    Solution is of the form
    $$\psi (\theta) = Acos(v\theta ) + Bsin(v\theta )$$
    Applying BCs:
    $$\phi (r,\theta ) = r^{v}\left(Acos(v\theta ) + Bsin(v\theta )\right)$$
    $$\phi (r,0) = V_0 \rightarrow A = V_{0}r^{-v}$$
    $$\phi (r,\alpha) = V_0 \rightarrow B = \frac{V_{0}r^{-v}(1-cos(v\alpha ))}{sin(v\alpha )}$$ ...which leaves me with utter crap.

    However if we choose V0 such that it is zero then we can reapply the BCs:
    $$\phi (r,0) = V_0 \rightarrow A = 0$$
    $$\phi (r,\alpha) = V_0 \rightarrow sin(v\alpha ) = 0 \rightarrow v = \frac{n\pi }{\alpha}$$

    Resulting in a more pleasant $$\phi(r,\theta ) = Br^{\frac{n\pi }{\alpha}}sin(\frac{n\pi }{\alpha}\theta )$$

    So there a few things here that I'm struggling to understand:
    If I don't set V0 = 0 then how can I solve the problem?
    I should have a unique solution, yet B isn't determined?
    How might I correct my solution so that the potential was V0 on the boundaries?
     
  2. jcsd
  3. Mar 25, 2016 #2

    Twigg

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    Gold Member

    Notice you're short two boundary conditions to solve this problem. Add in ##\phi(r,\theta) < \infty## everywhere on the region of integration. First, test that condition as r goes to infinity: ##\phi(r,\theta) < \infty## as ##r \to \infty## implies ##\nu \leq 0##. Second, test the same condition as r goes to zero: ##\phi(r,\theta) < \infty## as ##r \to 0## implies ##\nu \geq 0##. Meaning, there's only one term in the solution: the 0th order term, simply ##\phi(r,\theta)=A=V_{0}##, ##\nu = 0##. The reason you didn't see it when you performed separation of variables is because ##\nu = 0## makes ##\frac{\partial}{\partial r}\phi(r,\theta) = 0##. If you had handled the zeroth-order term separately, you would've gotten ##\frac{\partial^{2}}{\partial\theta^{2}}\psi(\theta) = 0##, which implies ##\frac{\partial}{\partial\theta}\psi(\theta) = \frac{\psi(\alpha) - \psi(0)}{\alpha}=\frac{V_{0} - V_{0}}{\alpha} = 0##. That tells you that ##\psi(\theta) = \psi(0) = V_{0}##.
     
  4. Mar 26, 2016 #3
    Thank you for your reply.

    Wouldn't that lead to the conclusion that ##\phi (r,\theta ) = V_{0}## everywhere? The solution must have some ##r## dependence..? Shouldn't the solution be of the form that I provided when ##V_{0} = 0##?
     
  5. Mar 26, 2016 #4

    Twigg

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    Gold Member

    If having the unbounded powers of r weren't an issue, you'd have a superposition of solutions like you got for different values of n, plus the constant ##V_{0}##. There wouldn't be a unique solution because you're still short at least one boundary condition for r at infinity.
     
  6. Mar 26, 2016 #5

    TSny

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    Homework Helper
    Gold Member

    ##A## is required to be a constant in the process of solving by separation of variables. It cannot be a function of ##r##.

    When you write ##R(r) = r^{\nu}##, you are not allowing for all possible forms of ##R(r)##.

    I suggest you go back through the process of separation of variables to find the differential equation that ##R(r)## must satisfy and see if you can find a solution that is not of the form ##r^{\nu}##.
     
    Last edited: Mar 26, 2016
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