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Smooth solution to PDE?

  1. Apr 20, 2007 #1
    What exactly is a smooth solution to PDEs. I couldn't find the definition in my books, googled that and came up empty handed. I suspect the solution must be continuous with all the deriviatives.
     
  2. jcsd
  3. Apr 20, 2007 #2
    You are right. A function is said to be smooth if all its derivatives are continous. The vector space of such functions is denoted [itex]C^\infty(\Omega)[/itex], where [itex]\Omega[/itex] is the domain where the function is defined.

    I dont believe there is a formal definition of smoothness, but in general, when a text is talking about a smooth function, is implying that the function has as many continous derivatives (not necesarily all) required for something to occur.
     
    Last edited: Apr 20, 2007
  4. Apr 21, 2007 #3
    Thanks alot.
     
  5. Apr 22, 2007 #4

    HallsofIvy

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    The "formal" definition of "smooth" that I have seen is "first derivative is continuous". I have also seen the phrase "sufficiently smooth" meaning as many continuous derivatives as you need.
     
  6. Apr 22, 2007 #5

    Chris Hillman

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    Hi all, the standard definition of "smooth" (without qualification) in most textbooks on manifold theory is indeed that derivatives of all orders exist and are continuous. Examples include bump functions (as in partitions of unity) and frequency components of wave packets. Contrast real analytic solutions, which are far more "rigid"!

    But Halls is also right, in the sense that many books/papers refer to "smooth of order such and such", or even "sufficiently smooth", meaning what AiRAVATA said: sufficiently smooth (usually short for "derivatives exist to order n and are continuous to order n-1") to ensure you can use whatever theorem you plan to quote in a proof.
     
    Last edited: Apr 22, 2007
  7. Apr 29, 2007 #6
    It depend if your are working on Sobolev sapces or others
     
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