Deriving the Exact Solution of the Schlichting ODE for Laminar Jet Flow

[Clausius2]

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 :

\frac{f'f}{\eta^2}-\frac{f'^2}{\eta}-\frac{ff''}{\eta}=\Big(f''-\frac{f'}{\eta}\Big)'

where f=f(\eta)

Boundary conditions are:

\eta=0 ; f=f'=f''=0

\eta\rightarrow\infty; f'=0

I am supposed to obtain f=\frac{4c\eta^2}{1+c\eta^2} 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.

 

[dextercioby]

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

Daniel.

 

[arildno]

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 :

\frac{f'f}{\eta^2}-\frac{f'^2}{\eta}-\frac{ff''}{\eta}=\Big(f''-\frac{f'}{\eta}\Big)'

where f=f(\eta)

Boundary conditions are:

\eta=0 ; f=f'=f''=0

\eta\rightarrow\infty; f'=0

I am supposed to obtain f=\frac{4c\eta^2}{1+c\eta^2} 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.

The left-hand-side reduces to:
\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})
since we have:
\frac{d}{d\eta}\frac{1}{2}f^{2}=ff'
and:
f'^{2}+ff''=\frac{d}{d\eta}ff'
Evidently the right-hand-side reduces to:
\frac{d}{d\eta}(f''-\frac{f'}{\eta})=\frac{d}{d\eta}(\eta\frac{d}{d\eta}\frac{f'}{\eta})
Maybe that helps..

 

[Clausius2]

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

Daniel.

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

Arildno, it helps me a lot.

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

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

integrating once and imposing the boundary constraint at \eta\rightarrow\infty

\Big(\frac{f'}{\eta}\Big)'=-\frac{f^{2'}}{2\eta^2}

that can be reshaped developing the right derivative to:

\frac{1}{2}f^{2'}=2f'-(\eta f')'

integrating it another time and imposing f=0 at \eta=0:

\frac{f^2}{2}=2f-\eta f'

which leads to:

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

the right side can be calculated as:

\frac{1}{2}\Big(\frac{df}{f}+\frac{df}{4-f}\Big)

So finally I obtain:

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

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

Anyway it gives:

f=\frac{4c\eta^2}{1+c^2\eta^2}

Thank you very much for helping me.