fluid dynamics - Explanation for the next steps of chaplygin dipole

this post is the Chaplygin dipole, it's an interesting issue.

Can someone explain me these steps in other words please? any Explanation of any step will help me, I hope that together I will understand all! please explain me the steps in other words because I didn't understand what is written

In 1903 Chaplygin published another remarkable paper (Chaplygin 1903) devoted to the motion associated with a compact vorticity distribution in a two-dimensional unbounded inviscid flow. In the introduction of that paper he gave a precise formulation of the problem: Consider an unbounded mass of incompressible fluid in which the motion is parallel to the OXY plane; let the motion outside some circular cylinder be irrotational, the velocity being equal to zero at infinity. The question is to find a distribution of vortex lines inside the cylinder that gives rise to a uniformly translating vortex column with a continuous velocity distribution and with a positive pressure all around.

As a first example of the solution Chaplygin considered in detail a case of rectilinear motion of a circular vortex of radius $a$ with a constant translation velocity $v_0$. By superimposing on the whole fluid fluid a uniform velocity $-v_0$ he obtained a stationary problem of a steady vortex cylinder placed in a potential flow with uniform velocity at infinity. By choosing the polar coordinate system $(r,\theta)$ , with the origin at the centre of the cylinder, the stream function $\psi_1$ for the potential flow around the cylinder is written as:

Answer

Equation (2.5) expresses the velocity field in function of the stream function. It's not clear to me it really should be presented at this stage in the process, I guess it's useful to impose the conditions at infinity.

Equation (2.6) expresses that the two pieces of the total solution, the one inside the disk $\psi$ and the one outside $\psi_1$, have to match in some sense. The first part asks that the actual values of both solutions coincide on the circle, the second part asks that the derivatives be the same too. Nothing is said about higher order derivatives.

Obtaining (2.7) requires "some algebra", which often means it's quite a tricky process, but it's not really interesting how it's done (in the author's opinion at least). In the end, once you have a solution you can just plug it into the differential equation and see if it works.

There'll probably be some latitude in choosing various constants once a general solution is found. Equation (2.6) should help fix these. Note that $\psi_1$ vanishes at $r=a$, so that $\psi$ must too. It so happens that the Bessel function $J_1$ vanishes when its argument equals $b$, so that's probably the reason why it appears under the form given in (2.7). We have that $\partial_r \psi_1 = v_0 (1+\frac{a^2}{r^2})\sin{(\theta)}$, so that at $r=a$ we have $\partial_r \psi_1\vert_{r=a} = v_0 2\sin{(\theta)}$. Matching $\partial_r \psi$ to that value explains why there's the constant $J'_1(b)^{-1}$.