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Neumann boundary conditions on a PDE

  1. Jan 4, 2012 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    From a previous exercise (https://www.physicsforums.com/showthread.php?t=564520), I obtained [itex]u(r,\phi) = \frac{1}{2}A_{0} + \sum_{k = 1}^{\infty} r^{k}(A_{k}cos(k\phi) + B_{k}sin(k\phi))[/itex] which is the general form of the solution to Laplace equation in a disk of radius a.
    I must find [itex]u(r, \phi )[/itex] for the Neumann's conditions ([itex]\frac{\partial u}{\partial n} \big | _c=f(\phi )[/itex]) where:
    1)[itex]f(\phi )=A[/itex]
    2)[itex]f(\phi )=Ax[/itex]
    3)[itex]f(\phi )=A(x^2-y^2)[/itex]
    4)[itex]f(\phi )=A \cos \phi +B[/itex]
    5)[itex]f(\phi )=A \sin \phi +B \sin ^3 \phi[/itex]

    2. Relevant equations
    Already given I think.


    3. The attempt at a solution
    I don't know how to start. :/
    What do they mean with the notation "_c"? Ah, in the contour I guess?
     
  2. jcsd
  3. Jan 4, 2012 #2
    "_c" means that you are evaluating the normal derivative along the boundary of the disk (ie. the circle of radius a). Hence that just means, compute the normal derivative of u and evaluate it at r=a. It might be more convenient to write [itex]\frac{\partial u}{\partial n} \big | _c=\nabla u\cdot\widehat{n}[/itex]. So now all you need to do is evaluate the gradient of u (in polar coordinates) and take the inner product with the normal (in polar coordinates), which is just the unit vector pointing radially outward (ie. a normal to a circle). Its then just a matter of using the given functions to work out the required coefficients.
     
    Last edited: Jan 4, 2012
  4. Jan 4, 2012 #3
    In fact you have pretty much already solved this problem in your last problem. Since it the boundary C is a circle the normal derivative is just [itex]\frac{\partial{u}}{\partial{r}}(a,\phi)[/itex]. Simply replace your [itex]f(\phi)[/itex] in each case with this expression and you'll see that the problem is nearly identical to what you have already done.

    Just a side note 1) and 4) might be problematic because of those two constant terms so keep an eye out for those two.
     
  5. Jan 5, 2012 #4

    fluidistic

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    Ok thanks a lot for the explanation. Indeed, [itex]\vec \nabla u \cdot \hat n = \left ( \frac{\partial u }{\partial r } \hat r + \frac{1}{r} \frac{\partial u }{\partial \phi } \hat \phi \right ) \cdot \hat n[/itex]. But [itex]\hat n = \hat r[/itex] so that, as you said: [itex]\vec \nabla u \cdot \hat n =\frac{\partial u }{\partial r}[/itex].
    I tried to calculate [itex]\frac{\partial u }{\partial r} (a, \phi)[/itex]. This gave me [itex]\sum _{k=1}^{\infty } a^{k-1} [A_k \cos (k \phi )+B_k \sin (\phi ) ]=A[/itex]. By inspection, [itex]A_1 \cos \phi + B_1 \sin \phi =A[/itex]. But I don't think it's possible because sin and cos are linearly independent, so a linear combination of them can't reach a constant. So I probably made an error somewhere...
     
  6. Jan 6, 2012 #5
    You are missing a factor of k in your sum (come down from derivative). You are absolutely right however that it is going to be impossible to get a constant term from a sum of cosines and sines. For the other questions, in which no constant (unlike in 1 and 4) is involved you should have no problem finding the coefficients using the general solution you found to Laplace's equation. However, there is something subtle going on with those constant terms. Think about what [itex]\frac{\partial u }{\partial r} (a, \phi)[/itex] means. After that look back at how you came to the general solution for Laplace's equation, and recall why you omitted certain solutions.
     
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