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Laplace Equation in Cartesian Coor.

  1. Dec 8, 2011 #1
    Solving the Laplace equation in Cartesian Coordinates leads to the 2nd order ODEs:
    \frac{X''}{X}=k_1, \qquad \frac{Y''}{Y}=k_2 \qquad \frac{Z''}{Z}=k_3
    In each case the sign of [itex]k_i[/itex] will determine if the solution (to the particular ODE) is harmonic or not.

    Hence, if two people solve the Laplace equation, one may get a solution that is harmonic in [itex]x[/itex], and the other harmonic in [itex]z[/itex].

    How could this be?
    Does this simply mean that both solutions satisfy the Laplace equation, and hence the solution is not unique?

    If we have non-uniqueness then when considering an actual physical situation (such as the potential of an electric field) how can we be sure that the solution we get is the true solution?

  2. jcsd
  3. Dec 8, 2011 #2


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    Because PDEs (or even ODEs) are always accompanied by boundary/limit conditions which restrict (even really heavily) the dimension of the space of solutions. Giving no boundary condition for the solution means a great deal of arbitrary in the shape of the solutions.
  4. Dec 17, 2011 #3
    So if I have the BC then how do I know which of the 6 (i think) possible solutions is the actual solution?
    Trial and error?
  5. Dec 29, 2011 #4
    Your boundary condition will specify which one you have to choose. Try solving the equation in a 3D domain where you specify the BCs ;)
  6. Jan 14, 2012 #5
    Thank you, both!
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