Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Schlichting exact solution

  1. Mar 26, 2005 #1

    Clausius2

    User Avatar
    Science Advisor
    Gold Member

    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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

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

    Daniel.
     
  4. Mar 27, 2005 #3

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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

    Clausius2

    User Avatar
    Science Advisor
    Gold Member

    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]

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

    [tex]\frac{1}{2}\Big(\frac{df}{f}+\frac{df}{4-f}\Big)[/tex]

    So finally I obtain:

    [tex]\frac{1}{2}ln\Big(\frac{f}{4-f}\Big)=ln\eta+A[/tex]

    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:

    [tex]f=\frac{4c\eta^2}{1+c^2\eta^2}[/tex]

    Thank you very much for helping me. :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Schlichting exact solution
Loading...