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.