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Boundary Value Problem

  1. Oct 16, 2004 #1
    I need some help starting off on this question.

    Electrostatic potential [tex]V (x,y)[/tex] in the channel [tex]- \infty < x < \infty, 0 \leq y \leq a[/tex] satisfies the Laplace Equation

    [tex]\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2}= 0[/tex]

    the wall [tex]y = 0[/tex] is earthed so that

    [tex]V (x,0) = 0[/tex]

    while the potential on the wall [tex]y = a[/tex]

    [tex]V (x,a) = V_0 \cos{kx}[/tex] where [tex]V_0 , k[/tex] are positive constants.

    By seeking a soln of an appropriate form, find [tex]V (x,y)[/tex] in the channel.
     
  2. jcsd
  3. Oct 16, 2004 #2

    arildno

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    When in doubt, try separation of variables!
    V(x,y)=F(x)G(y).
    Inserting this into Laplace, yields, by rearrangement:
    [tex]\frac{G''(y)}{G(y)}=-\frac{F''(x)}{F(x)}[/tex]
     
  4. Oct 16, 2004 #3
    Thanks, I was trying to get it in the form [tex]V (x,y) = F (a) V_0 \cos{kx}[/tex] then sub the values into the Laplace eqn. Is that going about it the wrong way?
     
  5. Oct 16, 2004 #4

    arildno

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    You DO mean (in my notation!)
    [tex]V(x,y)=G(y)V_{0}\cos(kx)[/tex]???
    If so, then it will work.
    Note that the separation of variables method in this case implies:
    [tex]G''(y)=k^{2}G(y)[/tex]
     
  6. Oct 16, 2004 #5
    Hmm im not really keeping up with you here sorry. How can I find things out about F(x) and G(y) (in your notation) when we have just introduced them?

    --------edit----------

    oh yes sorry im with you, then differentiate V (x,y) = G(y) V_0 cos (kx) to get the different bits to go in the Laplace eqn? ahh!
     
    Last edited: Oct 16, 2004
  7. Oct 16, 2004 #6

    arildno

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    Your lightbulb switched on in your edit, I see..:wink:
    Hint:
    You should be able to see that G(y)=ASinh(ky), where A is some constant, and Sinh() the hyperbolic sine function.
     
  8. Oct 16, 2004 #7
    Now I've got G''(y) - k^2 G (y) = 0, still on the right track yeh?
     
  9. Oct 16, 2004 #8

    arildno

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    Yes, you are.
     
  10. Oct 17, 2004 #9
    From this how do I get to G(y)=ASinh(ky)?
     
  11. Oct 17, 2004 #10

    arildno

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    Note that:
    [tex]\frac{d}{dy}ASinh(ky)=kACosh(ky),\frac{d^{2}}{dy^{2}}ASinh(ky)=k^{2}ASinh(ky)[/tex]
    This shows that ASinh(ky) is a solution for G(y).
    Similarly, you may show that BCosh(ky) is another solution for G(y).
    Your general solution is therefore:
    G(y)=ASinh(ky)+BCosh(ky)
    Apply the boundary condition at y=0 to prove that B=0
     
  12. Oct 17, 2004 #11
    Thanks for being so patient. How can I show that ASinh(ky) is a soln for G(y)? Im used to looking at the differential eqn and substituting

    G(y) = e^{ry} so G'(y) = re^{ry} and G''(y) = r^2. e^{ry}

    Is there no such substitution to show that ASinh(ky) & BCosh(ky) are solns?
     
  13. Oct 17, 2004 #12

    arildno

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    Your approach is ABSOLUTELY correct!
    Do you agree with the following result, using your method:
    [tex]r=\pm{k}[/tex]??
     
  14. Oct 17, 2004 #13

    Yeh which gives gen soln G (y) = Ae^{ky} + Be^{-ky}

    but the boundary condition at y=0 doesnt allow for B so we are left with

    G(y) = Ae^{ky} ?
     
  15. Oct 17, 2004 #14

    arildno

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    A bit too fast there..
    Let's write the general solution for G(y) as follows:
    [tex]G(y)=K_{+}e^{ky}+K_{-}e^{-ky}[/tex]
    where the K's are constants to be determined by boundary conditions.
    Prior to that step, however, let's rewrite the general solution as:
    [tex]G(y)=(K_{+}+K_{-})\frac{e^{ky}+e^{-ky}}{2}+(K_{+}-K_{-})\frac{e^{ky}-e^{-ky}}{2}[/tex]
    1. We now set [tex]B=K_{+}+K_{-},A=K_{+}-K_{-}[/tex]
    (Clearly, A and B are as arbitrary as the K's!)
    2) We now recognize:
    [tex]Cosh(ky)=\frac{e^{ky}+e^{-ky}}{2}[/tex]
    [tex]Sinh(ky)=\frac{e^{ky}-e^{-ky}}{2}[/tex]
    Or, we may rewrite G(y) as:
    [tex]G(y)=ASinh(ky)+BCosh(ky)[/tex]
     
  16. Oct 17, 2004 #15
    right now I need to find

    B: iv done this and basically its because cosh can never be zero which implies B must be zero.

    A: at y=a V(x,a) = V_0 cos{kx}

    therefore for y=a, G(y)=1

    ASinh(ka) = 1 is that right?

    hmm im not sure about this.
     
  17. Oct 17, 2004 #16

    arildno

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    Yep, so your solution is:
    [tex]V(x,y)=V_{0}\frac{Sinh(ky)\cos(kx)}{Sinh(ka)}[/tex]
     
  18. Oct 17, 2004 #17
    Thank you so much, I understand much more now than I did even yesterday!
     
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