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PDE's - Finding a certain solution to Laplace's equation on a circle

  1. Sep 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Find a solution of Laplace's equation [tex] u_{xx} + u_{yy} = 0 [/tex] of the form [tex] u(x,y) = Ax^2 + Bxy + Cy^2 \ (A^2 + B^2 + C^2 \not= 0 ) [/tex] which satisfies the boundary condition [tex] u(cos(\theta),sin(\theta)) = cos(2\theta) + sin(2\theta) [/tex] for all points [tex] (cos(\theta),sin(\theta)) [/tex] on the circle, [tex] x^2 + y^2 = 1 [/tex].


    2. Relevant equations
    Listed above.


    3. The attempt at a solution
    first, I found [tex] u_{xx} [/tex] and [tex] u_{yy} [/tex]

    [tex]u_{xx} = 2A [/tex]
    [tex]u_{yy} = 2C [/tex]

    From [tex] u_{xx} + u_{yy} = 0 [/tex] and the above results, I can get [tex] 2A + 2C = 0 [/tex].

    Now, I plugged in the boundary condition:
    [tex] cos(2\theta) + sin(2\theta) = Acos^2(\theta) + Bcos(\theta) sin(\theta) + Csin^2(\theta). [/tex]

    I tried various trig substitutions here and couldn't seem to get anywhere. However, with this equation and the one above, I have two equations (but there are three unknowns). I am pretty sure I have to use the [tex] x^2 + y^2 = 1[/tex] to write another equation so that I can solve for A, B, and C, but I do not know how to use the circle information.

    Any help would be greatly appreciated.

    Thanks in advance.
     
  2. jcsd
  3. Sep 2, 2008 #2

    Dick

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    Just use the double angle formulas for cos(2*theta) and sin(2*theta). I don't really see what the problem is.
     
  4. Sep 2, 2008 #3
    All right, let me see what I can do. I had tried this, but it wasn't simplifying, maybe there's something i'm not seeing.

    okay, so using those double angle identities:

    [tex]cos^2(\theta) - sin^2(\theta) + 2sin(\theta)cos(\theta) = Acos^2(\theta) + Bcos(\theta)sin(\theta) + Csin^2(\theta) [/tex]

    Oh, and I can solve that for
    [tex] A = 1,B = 2,C = -1 [/tex]

    So my solution is :

    [tex] u(x,y) = x^2 + 2xy - y^2 [/tex]

    This is the answer in the back of the book.

    I was just wondering why they put the [tex] x^2 + y^2 = 1 [/tex] in there? Is that something I didn't need to figure out the answer?

    Thanks so much for that little nudge too. I really appreciate that.
     
  5. Sep 2, 2008 #4

    Dick

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    No, you don't need it. If (x,y)=(cos(theta),sin(theta)) on the boundary, then of course x^2+y^2=1.
     
  6. Sep 2, 2008 #5
    Thanks so much!
     
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