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Schlichting exact solution

  1. Mar 26, 2005 #1


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    After spending large time trying to extract the exact solution of this ODE, I haven't been able to demonstrate the final result I'm given.

    The equation is :

    [tex] \frac{f'f}{\eta^2}-\frac{f'^2}{\eta}-\frac{ff''}{\eta}=\Big(f''-\frac{f'}{\eta}\Big)'[/tex]

    where [tex] f=f(\eta)[/tex]

    Boundary conditions are:

    [tex] \eta=0[/tex] ; [tex] f=f'=f''=0[/tex]

    [tex]\eta\rightarrow\infty[/tex]; [tex]f'=0[/tex]

    I am supposed to obtain [tex] f=\frac{4c\eta^2}{1+c\eta^2}[/tex] with c= unknown constant.

    but I don't find the way to gather the derivatives and solve the equation.

    It corresponds to the exact similarity solution of the far field of a round laminar jet.
  2. jcsd
  3. Mar 27, 2005 #2


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    Why do u have 4 boundary conditions for a 3-rd order ODE...?

  4. Mar 27, 2005 #3


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    The left-hand-side reduces to:
    [tex] \frac{f'f}{\eta^2}-\frac{f'^2}{\eta}-\frac{ff''}{\eta}=-\frac{d}{d\eta}(\frac{1}{\eta}\frac{d}{d\eta}\frac{1}{2}f^{2})[/tex]
    since we have:
    Evidently the right-hand-side reduces to:
    Maybe that helps..
    Last edited: Mar 27, 2005
  5. Mar 27, 2005 #4


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    You're right. One of them is redundant. I think [tex]f '=0[/tex] is not needed at [tex]\eta=0[/tex]

    Arildno, it helps me a lot.

    Let's see, according what you've said:

    [tex] \Big[\eta\Big(\frac{f'}{\eta}\Big)'\Big]'=-\frac{1}{2}\Big[\frac{f^{2'}}{\eta}\Big][/tex]

    integrating once and imposing the boundary constraint at [tex]\eta\rightarrow\infty[/tex]


    that can be reshaped developing the right derivative to:

    [tex] \frac{1}{2}f^{2'}=2f'-(\eta f')'[/tex]

    integrating it another time and imposing [tex] f=0[/tex] at [tex]\eta=0[/tex]:

    [tex]\frac{f^2}{2}=2f-\eta f'[/tex]

    which leads to:

    [tex] \frac{df}{-f^2/2+2f}=\frac{d\eta}{\eta}[/tex]

    the right side can be calculated as:


    So finally I obtain:


    where [tex]A=ln(c)[/tex] is a constant which I haven't found the way to determinate with the boundary conditions because the the logarithms are not defined in [tex]\eta=0[/tex]. This constant is determined with the Integral Conservation Law of the Momentum Flux as Schlichting stated.

    Anyway it gives:


    Thank you very much for helping me. :wink:
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