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Solutions to Laplace's equation

  1. Mar 15, 2009 #1
    [solved] solutions to Laplace's equation

    1. The problem statement, all variables and given/known data
    Find all solutions f(x,y) that satify Laplace's equation that are of the form:
    ax^3 + bx^2y + cxy^2 + dy^3

    2. Relevant equations
    Laplace states that fxx + fyy = 0

    3. The attempt at a solution
    fxx = 6ax + 2by
    fyy = 6dy + 2cx
    so 6ax + 2by + 6dy + 2cx = 0
    (3a+c)x + (3d+b)y = 0

    What do I do from here?
     
    Last edited: Mar 15, 2009
  2. jcsd
  3. Mar 15, 2009 #2

    gabbagabbahey

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    Well, if f(x,y) is a solution to Laplace's equation, then it satisfies it for all x and y....what does that tell you about (3a+c) and (3d+b)?
     
  4. Mar 15, 2009 #3
    Well then:
    (3a+c) = -(3d+b) = 0
    a = -d
    c = -b
    c = -3a
    b = 3a

    Plugging in:
    f(x,y) = ax^3 + 3ax^y - 3axy^2 - ay^3

    But if you take the double partials:
    fxx = 6x
    fyy = 6y

    and 6x-6y =/= 0 for all real numbers....

    EDIT: Wow, I'm stupid. I took the double partials incorrectly so I thought I had done it wrongly.. only I hadn't. Thanks!
     
    Last edited: Mar 15, 2009
  5. Mar 15, 2009 #4

    gabbagabbahey

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    c=-3a and b=-3d are necessary, but I'm not sure why you think a=-b and b=-d are.

    f(x,y)=ax^3-3dx^2y-3axy^2+dy^3 satisfies Laplace's equation for all a and d not just a=d
     
  6. Mar 15, 2009 #5
    If a=-b adn b=-d, then my above function satisfies for all a - that's good enough, yes?
     
  7. Mar 15, 2009 #6

    gabbagabbahey

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    It does, but its not the most general solution...which is what you are looking for.

    (3a+c) = -(3d+b) = 0 does not mean that a=-b and b=-d....the only requirement is that c=-3a and b=-3d.
     
  8. Mar 15, 2009 #7
    I see. Thanks a bunch!
     
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