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PDE problem

  1. Apr 12, 2008 #1
    1. We look at a Laplace equation ( [tex]\Delta u(x,y) =o[/tex]) on a square [0, 1]* [0, 1]
    If we know that [tex]u_{x=o}[/tex]= siny , [tex]u_{x=1}[/tex]= cosy
    [tex]u_y|_{y=0}[/tex]= 0 , [tex]u_y|_{y=1}[/tex]= 0 we differentiate here by y. proove that |u|<=1.






    3. The attempt at a solution

    We now know that the maximum of u has to be on the boundary. If it is greater then one, then it has to be on either y=0 or y =1.
     
  2. jcsd
  3. Apr 12, 2008 #2
    Yes, that is true and I would say good enough for an applied math course, with just a little bit more about the maximum value theorem (or whatever it is called) to justify.
     
  4. Apr 12, 2008 #3
    OK)) Could you tell me what that "little bit " is)?
     
  5. Apr 12, 2008 #4

    Dick

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    The solution to any old PDE doesn't satisfy the maximum principle. The solutions to the Laplace equation do, because they are a special kind of function with a special name. I think mindscrape just wants to you point to the theorem that says that the maximum is on the boundary.
     
  6. Apr 12, 2008 #5
    Guys, I have solved it!
    No more help needed on it!
     
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