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2D Laplacian

  1. Jul 17, 2005 #1
    Does anybody out there know what the Laplacian is for two dimensions?
  2. jcsd
  3. Jul 17, 2005 #2


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    The Laplacian is the trace of the Hessian operator and is undoubtedly given by

    [tex] \Delta =g^{ij}\nabla_{i}\nabla_{j} [/tex]

    ,where [itex] i,j=1,2 [/itex]. Incidentally, for a manifold on which Christoffel's symbols are 0, it reduces to the standard form

    [tex] \Delta=g^{ij}\partial_{i}\partial_{j} [/tex]

  4. Jul 17, 2005 #3
    That is probably one of the most complicated answers to a simple question I've ever seen.
  5. Jul 17, 2005 #4


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    It depends upon the coordinate system. Dextercioby gave the general formula for any coordinate system, thereby making it, as MalleusScientarum said, "one of the most complicated answers to a simple question I've ever seen"!

    In Cartesian coordinates it is [tex]\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}[/tex].

    By the way, the Laplacian is the simplest differential operator that is "invarient under rigid motions".
    Last edited: Jul 17, 2005
  6. Jul 17, 2005 #5


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    lets see now, isn't the laplacian the differential operator that vanishes on harmonic functions? if so, wouldn't that say it is the one that governs steady state heat flow? and characterizes the real and imaginary parts of holomorphic functions?

    harmonic forms on the other hand are very sueful in discussing cohomology. the de rham theorem says that every cohomology class on a compact oriented? diff manifold can be represented by a smooth differential form, but there is no uniqueness.

    by imposing a metric and hence defining a laplacian, one can define harmonic foirms and thewn there is a unique harminic representative for each cohomology class.

    e.g. on an elliptic curve, formed as a quotient of the complex numbers by a lattice, one has the natural harmonic basis dz and "dzbar".
    Last edited: Jul 17, 2005
  7. Jul 17, 2005 #6
    yes that stuff about harmonic & holomorphic functions sounds right. i don't know anything about heat flow though.
  8. Jul 17, 2005 #7
    yes, the heat-diffusion equation is:

    [tex]T_p = k\Delta T[/tex]

    where in cartesian coordinates [itex]T_p[/itex] is the change in temperature at the point [itex]p[/itex]

    [tex]T(t,x,y,z)[/tex] and [tex]p: (x,y,z)[/tex]

    [tex]T_p = k(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial ^2 T}{\partial z^2})[/tex]
  9. Jul 17, 2005 #8
    Apologies, I guess I should specify for polar coordinates.
  10. Jul 18, 2005 #9
  11. Jul 19, 2005 #10
    That's ultimately what I decided to do, but it always makes me nervous when I do something like that without knowing what mathematical reasoning I could use to justify it. As for the original answer, how would one find the metric tensor for an arbitrary coordinate system? Examples would be greatly appreciated.
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