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Uniqueness of Laplace's equation

  1. Jan 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove the uniqueness of Laplace's equation
    Note that V(x,y,z) = X(x) Y(y) Z(z))

    2. Relevant equations
    [tex] \frac{d^2 V}{dx^2} + \frac{d^2 V}{dy^2}+ \frac{d^2 V}{dz^2} = 0 [/tex]

    3. The attempt at a solution
    Suppose V is a solution of Lapalce's equation then let V1 also be a solution of Laplace's equation.

    then V - V1 is also a solution of laplace's equation
    [tex] \frac{d^2 (V-V_{1})}{dx^2} + \frac{d^2 (V-V_{1})}{dy^2}+ \frac{d^2 (V-V_{1})}{dz^2} = 0 [/tex]

    Are we alowed to say that
    [tex] \frac{d^2 (V-V_{1})}{dx^2} = \frac{d^2 V}{dx^2} - \frac{d^2 V_{1}}{dx^2} = 0 [/tex]
    [tex] \frac{d^2 V}{dx^2} = \frac{d^2 V_{1}}{dx^2}[/tex]
    because V is solved by separation of variables?
    Since their derivatives are equal thus V must be the same as V1.

    Is this a satisfactory solution??
     
  2. jcsd
  3. Jan 16, 2007 #2

    StatusX

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    No, that's no good, for a bunch of reasons. You need more information, specifically, boundary conditions. Otherwise the statement isn't true. For example, all linear equations are solutions to laplace's equation, but they aren't all equal.
     
  4. Jan 16, 2007 #3
    ah
    what if V and V1 both satisfied the same boundary conditions? Would that be acceptable??
     
  5. Jan 16, 2007 #4

    StatusX

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    It depends what the boundary conditions are. Where did you get this question?
     
  6. Jan 16, 2007 #5
    The original question is
    (a physics problem)
    Consider an empty 3D rectangular cube having all sides at a 0 potnetial. What is the potnetial inside the cube and how do you know this is the only possible answer.

    I solved the problem quite easily and got 0 for the Cn,m

    To prove taht this is the only possible solution i am invoking the uniqueness theorem of laplace's equation
     
  7. Jan 16, 2007 #6

    StatusX

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    You'll need to use the fact that a solution of Laplace's equation has no local maxima or minima. So if it's zero on the boundary of some closed region, then inside the region...
     
  8. Jan 19, 2007 #7
    so i dont need to solve this problem using separation of variables??
     
  9. Jan 19, 2007 #8

    HallsofIvy

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    No, in general, it is not necessary to solve an equation in order to prove that it has unique solutions (with appropriate boundary conditions).

    StatusX's point is that if [itex]\Phi(x)[/itex] and [itex]\Psi(x)[/itex] satisfy exactly the same boundary conditions, then [itex]\Phi(x)- \Psi(x)[/itex] is 0 at every point of the boundary. Now, what happens if you substitute [itex]\Phi(x)- \Psi(x)[/itex] into Laplace's equation?
     
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