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Finding a solution of this PDE

  1. Jul 28, 2011 #1
    Hi,

    i'm having trouble finding a solution to this PDE,

    [tex] \frac{d U(x,y,t)}{dt} = A(x) \frac{\partial U(x,y,t)}{\partial y} + B(y) \frac{\partial U(x,y,t)}{\partial x}[/tex]

    with only knowledge of the initial condition U(x,y,0)=F(x,y).

    I've tried to solve this using characteristics but the only examples i can find in books is for the case when the left hand side is zero. Tried following the method from some books but can only solve it for when the L.H.S is zero. I'm not sure where to go next

    Any help would be fantastic.

    Thanks.
     
  2. jcsd
  3. Jul 28, 2011 #2

    hunt_mat

    User Avatar
    Homework Helper

    For the LHS do you mean:
    [tex]
    \frac{\partial U}{\partial t}
    [/tex]
     
  4. Jul 28, 2011 #3
    The LHS should read ∂U(x,y,t)/∂t not dU(x,y,t)/dt.

    Then, for (parameter) s∈I⊂ℝ:

    d/ds[U(x(s),y(s),t(s))]= ∂U/∂x·dx/ds + ∂U/∂y·dy/ds + ∂U/∂t·dt/ds≡ B(y)∂U/∂x + A(x)∂U/∂y - ∂U/∂t= 0.

    You seek, U(x(s),y(s),t(s))= constant.

    ADDENDUM:
    Hint: dx/B = dy/A = dt/-1.
     
    Last edited: Jul 28, 2011
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