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PDE U_x+U_y=1

  1. Feb 8, 2014 #1
    1. The problem statement, all variables and given/known data

    U_x+U_y=1 with boundary condition U(y,y/2)=y



    2. Relevant equations


    3. The attempt at a solution

    Well first I solved the homogeneous solution and got ax-ay where a is just a constant. Then for the non homogeneous solution I got ax+(1-a)y. After adding them both together and pluggin in the boundary condition I got U=x+(0)y? Is that right?
     
  2. jcsd
  3. Feb 8, 2014 #2

    LCKurtz

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    It's easy enough to check it yourself. Does your solution ##U(x,y)=x## satisfy ##U_x+U_y = 1## and ##U(y,\frac y 2) = y##?
     
  4. Feb 8, 2014 #3
    It does. But I'm suspicious whenever a solution to a PDE comes out that nice.
     
  5. Feb 8, 2014 #4

    LCKurtz

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    Hey, if you have the solution you have the solution. There's nothing left worry about. Good going.
     
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