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Simplify the follwoing Equation

  1. Jun 19, 2011 #1
    Sounds waves in a pipe of varying cross-section are described by the equation

    V2 d/dx (1/A dAu/dx) = d2u/dt2

    Where A = 0.2+0.3x

    So first I substituted A into the equation:

    V2 d/dx (1/(0.2+0.3x) d(0.2+0.3x)u/dx) = d2u/dt2

    V2 d/dx (0.3u/(0.2+0.3x) du/dx) = d2u/dt2


    This is as far as I can get, any help would be fantastic.
     
  2. jcsd
  3. Jun 19, 2011 #2
    I suppose this is V² d/dx (1/A d(Au)/dx) = d²u/dt², true? Then expand first before substituting, which gives
    d²u/dx² + d(ln A)/dx du/dx + d²(ln A)/dx² u = 1/V² d²u/dt².
    Now substitute to get
    d²u/dx² + 1/(x + 2/3) du/dx - 1/(x + 2/3)² u = 1/V² d²u/dt², which I think is about as simple as it gets.

    (PS: This has a solution u(x,t) = [ C1 x² (1 + x)/(2 + 3 x) + C2 (1 + (2 + 3 x)²)/(2 + 3 x) + C3 (
    (1 - (2 + 3 x)²))/(2 + 3 t) ] [ 1/2 C1 t² V² + C4 + t C5 ], where C1 to C5 are arbitrary constants. But that may not be the solution you are looking for.)
     
    Last edited: Jun 19, 2011
  4. Jun 19, 2011 #3

    hunt_mat

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    So the equation you have is:
    [tex]
    v^{2}\frac{\partial}{\partial x}\frac{1}{A(x)}\frac{\partial}{\partial x}(A(x)u)=\frac{\partial^{2}u}{\partial t^{2}}
    [/tex]
    Use:
    [tex]
    \frac{\partial}{\partial x}(Au)=A\frac{\partial u}{\partial x}+0.3u
    [/tex]
    Likewise for the 1/A term too.
     
  5. Jun 19, 2011 #4
    Thank you for the replies, hunt_mat could u please explain your post a little more please im confused :confused:
     
  6. Jun 19, 2011 #5

    hunt_mat

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    I will explain the second point:
    [tex]
    \frac{\partial}{\partial x}(A(x)u)=A\frac{\partial u}{\partial x}+u\frac{\partial A}{\partial x}=A\frac{\partial u}{\partial x}+0.3u
    [/tex]
     
  7. Jun 19, 2011 #6
    Ok using what you have said I have obtained:

    V2 d/dx (0.3u/0.2+0.3x + du/dx)

    V2 d/dx (0.15u+ u/x + du/dx)

    Is this correct??
     
  8. Jun 19, 2011 #7

    hunt_mat

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    You have to use the quotient rule which is:
    [tex]
    \frac{d}{dx}\left(\frac{X}{A}\right) =\frac{A\frac{dX}{dx}-X\frac{dA}{dx}}{A^{2}}
    [/tex]
    Where:
    [tex]
    X=\frac{\partial }{\partial x}(A(x)u))
    [/tex]
    So, no. Your answer isn't correct.
     
  9. Jun 19, 2011 #8
    Ok so I should use the quotient rule on:

    0.3u/0.2+0.3x

    (0.2+0.3x) (0.3) - 0.3u (0.3) / (0.2+0.3x)2

    0.6 +0.9x -0.9u / (0.2+0.3x)2

    am I on the right track??
     
  10. Jun 19, 2011 #9

    hunt_mat

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    Not quite, you should have:
    (0.3du/dx-(0.3)^2u)/(0.2+0.3x)2
     
  11. Jun 19, 2011 #10
    So the answer is:

    V2 d/dx ((0.3du/dx-(0.3)2u)/(0.2+0.3x)2 + du/dx)
     
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