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Predicting the form of solution of PDE

  1. May 7, 2013 #1
    Predicting the functional form of solution of PDE

    How do you conclude that the solution of the PDE

    [itex]u(x,y)\frac{∂u(x,y)}{∂x}+\upsilon(x,y)\frac{∂u(x,y)}{∂y}=-\frac{dp(x)}{dx}+\frac{1}{a}\frac{∂^{2}u(x,y)}{dy^{2}}[/itex]

    is of the functional form

    [itex]u=f(x,y,\frac{dp(x)}{dx},a)[/itex] ?


    I know this seems obvious, but I think it might not be necessary. How can you predict this by using reason, mathematical manipulation, etc?


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    Update:
    I get the fact that [itex]u[/itex] will be a function of [itex]x, y,[/itex] and [itex]a[/itex], but what about d/dx p(x)? Won't it change to something like [itex]p(x)+C[/itex] leaving [itex]u[/itex] to be no longer a function of d/dx p(x)?
     
    Last edited: May 7, 2013
  2. jcsd
  3. May 7, 2013 #2

    hunt_mat

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    It's sort of obvious really. It has derivative in both x and y, so they must be part of the solution. If a is constant then you will know that will paramatrise the solution and it will be dependent on the forcing term (the pressure gradient)
     
  4. May 7, 2013 #3
    But won't the term [itex]\frac{d}{dx}p(x)[/itex] change to something like [itex]p(x)+constant[/itex] ?
     
    Last edited: May 7, 2013
  5. May 7, 2013 #4

    hunt_mat

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    You're not integrating anything, dp/dx will be your input function.
     
  6. May 7, 2013 #5
    You will have to integrate it in order to obtain a solution :)
     
  7. May 7, 2013 #6
    Substitute the functional form into the PDE, effectively replacing u(x,y) and its gradients. If the functional form is depending on P(x,y) directly, then P(x,y) should also appear in the PDE.
     
  8. May 7, 2013 #7
    I'm quite sure that this is connected to the Buckingham Pi theorem as well, which would give you a mathematical way to get all the independent variables in a formal way.
     
  9. May 7, 2013 #8

    hunt_mat

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    The pi theorem is okay if you know the general functional form but not for general problems
     
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