explanation for the next steps in other words (dipole)

I am trying to understand the lamb-chaplygin dipole solution.
can someone explain me the green lines in other words please?

this is the proof:

When introducing the stream function, the steps that you usually take are as follows:
1) replace u and v by the stream function.
2) Derive the horizontal momentum equation (for u) with respect to y and the other with respect to x.
3) Eliminate the pressure term, to end up with a single equation in ψ.

so:

v = -∂ψ/∂x, u = ∂ψ/∂y

(∂^2)ψ/∂x^2 + (∂^2)ψ/∂y^2 = f(ψ)

if we put f(ψ) = -(k^2)*ψ. where k is a constant, and transform to polar coordinates r, θ, we get:

(d^2)ψ/ dr^2 + (1/r) * (dψ/dr) + (1/r^2) * (d^2)ψ/ dθ^2 + (k^2)*ψ = 0

which is satisfied by:


where Js is a bassel's function.


any help appreciated! thanks!

 

I would have thought baeckmo to be all over this...

 

Hi Tom, what do you mean? I didn't understand.. sorry :/

 

Baeckmo is a member on this forum, he usually has a lot to say on matters such as these.

I'm surprised he hasn't spoken up.

You could try sending him a personal message.

 

ok, I will send him a message.. thank you very much!

 

What you are looking at is called a mathematical contrivance. I hope you have had a course in the solution of differential equations or this is going to get messy
What we need to do is find a finction (f) of ψ so that when we evaluate it, a solution of the stream function can be found. When we have a dipole, the stream function is a function of location of the point we evaluate, either x,y from the dipole, or the distance (r) and angle (theta) to the pair. What the proof did was to first convert the axis from a Cartesian coordinate system to a polar one. Notice that the (∂^2)ψ/∂x^2 + (∂^2)ψ/∂y^2 = f(ψ) term can be separated similar to the x^2+y^2 = r^2 of the rectangular to polar conversion. If we set r^2 = f(ψ) then all we need is a f such that it is easy to extract an r and theta...i.e. -(k^2)ψ (notice that when we take the square root of this we will end up with a complex conjugate necessary for a polar solution and also that v = -∂ψ/∂x, u = ∂ψ/∂ satisfies the Cauchy-Riemann conditions). Now we do the substitution and differential gradient (▼ couldn't find a font with the grad symbol) to get a form of the Bessel equation (x^2*y''+x*y'+(x^2-n^2)y=0) where the Bessel’s Function (a polar version of Jn(x)) is a solution.

Or at least that's how i remember it.

 

wow, thank you very much! you helped me a lot.. I think I start understanding it! thank you!! you explained very clear!

 

John,

You never cease to surprise me!

 

The solution methods of linear and non-linear differential equations is part and parcel of a modern engineering (Naval Arctitecture) degree. When I was in school we had to do this kind of stuff (dipole flow solution) by hand, even stuff like work out multiple dipole couplets to give a near net hull shape. It always helps to know the underlying math when evaluating computer progam output.

FWIW, it is always better to understand the math because mathematics is the language of engineering, science, architecture, art, and as Einstien asserted, God. Even da Vinci, 500 years ago, had the following statements that I keep up on my office wall:

and

So I won't give Heinlein's Lazarus Long quote I also keep up on the wall .

 

Amen.

 

hahah Amen!!! :]] and thank you again!