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Finding solution of a PDE

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Assume we are in the open first quadrant in the (x,y) plane
    Say we have u(x,y) a C1 function in the closed first quadrant that satisfies the PDE:

    [tex]u_{y}=3u_{x}[/tex] in the open first quadrant

    Boundary Conditions:
    u(0,y)=0 for t greater than or equal to 0
    u(x,0)= g(x) for x greater than or equal to 0

    and g(x) = 0 for [tex]0\preceq x\preceq1[/tex]
    g(x) = (x-1)^5 for x greater than or equal to 1

    Is there a solution? And if there is, is the the only solution?


    2. Relevant equations

    For first order PDE: solution is f(Ay-Bx) where the PDE is AUx + BUy = 0

    3. The attempt at a solution

    [tex]u(x,y) = f(-x-3y)[/tex]

    so that means

    [tex]u(0,y) = f(-3y)=0[/tex]

    [tex]u(x,0) = f(-x)=g(x)[/tex]

    set -x = w

    f(w) = g(-x)

    therefore f(-x-3y) = g(x+3y)

    and u(x,y) = g(x+3y)

    This solution works for u(x,0) but I can't find it to work with u(0,y), if y > 1/3, then the solution would be (x-1)^5 with x greater than or equal to 1, which does not satisfy u(0,y) = 0

    so I concluded that there is no solution....this doesn't seem right for me. Any help? I feel that I dont actually have to try to find u in order to determine if a solution exists...
     
    Last edited: Mar 25, 2009
  2. jcsd
  3. Mar 25, 2009 #2
    Hello, maybe in a traditional pde view, it would be more helpful to think of y as t for time?
     
  4. Mar 25, 2009 #3
    bump, am I being too confusing here?
     
  5. Mar 25, 2009 #4
    bump....I have kind of a part two question depending on how correct my answer is.
     
    Last edited: Mar 25, 2009
  6. Mar 25, 2009 #5
    Oh, nevermind, I got it =]
     
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