Sorry, I was away for a few days. This is not a continuation to Count Iblis's posts
Omri said:
About the single source - I understand you guessed ln(r) because you wanted to achieve (1/r).
That's right

And the 1/r came from Gauss' theorem in 2 D (or, if you want, conservation of flow: the flux through a circle with radius r1 equals the flux through a bigger circle with radius r2, so the flux density must go in 1/r).
Sorry for the stupid question, but why does that describe 2D flow? And even though you know that the real part is ln(r), you could define the imaginary part as you wish - I realize that the complex logarithm with Im(ln(z))=i*arg(z) is the simplest choice, but why would you pick it?
There's not much choice in the extension of a real function into a complex
analytic function. In fact, most of the time it is unique, up to a constant. So if you find (by guessing) ONE solution, it is THE solution. The reason is the Cauchy-Riemann equations. If you know the real part over the complex plane, that means that the first derivative of the imaginary part is fixed everywhere, in both x and y direction. Pick a point somewhere where you fix the value of the imaginary part (that's the free constant), and then you have fixed the values along the parallel line with the y-axis (because you know the y-derivative). From every point along this y-axis, with the x-derivative, you find the values everywhere along all lines parallel with the x-axis, that start at a point on the y-line, in other words, everywhere. The only thing that can happen is that at some points, you diverge, and then you introduce cuts. That's like the choice of the "0" for the argument in the logarithm. But strictly speaking, at those cuts, your function is not analytic anymore.
About the two sources - that was one nice trick. :) But when you take a point that is not on the x-axis, isn't it affected by the mirror source you added?
No. The contributions to the flows of different sources are independent. The reason is that if you take a small circle around any source, what will flow in from another source in that circle will also flow out of it. So what flows out net is the contribution of the source INSIDE, and that's not affected by what is put in by any other source.
The point is again that you know that the solution to the problem is going to be unique. So if you find A solution, it is THE solution. The solution to the problem is the one that will satisfy:
1) Poisson's equation everywhere outside of the sources
2) Gauss' law
3) The boundary conditions (like a wall or something, which is represented by a streamline).
4) eventually some limit conditions if the region is open, such as the flow at infinity
Those conditions can be shown to fix the solution uniquely. That means that if you find A way to satisfy them, it is the unique solution.
Now, finding an analytic function (outside of the sources) guarantees (1). Taking the solution F. ln(z-z0) satisfies (2). If you can fiddle the stuff such that (3) and (4) are ok, then you have found the solution.
Finally, could you please show me an example of how to use conformal mapping in order to solve a "complicated" problem?
Right. In a previous message, I talked about sources. That's because in fact, most of my uses of conformal mappings are in the electrostatic domain, where I have charges which act as sources for the electric field. But we can consider other kinds of "simple" solutions, such as uniform flow. Consider the "potential": phi(z) = z. This means that the real potential is x, and the streamlines are given by the equation y = constant. We have constant, horizontal flow. We can eventually consider that this is only valid in the upper half plane. So the potential phi(z) = z gives us a solution of constant flow along a wall.
Now, consider the map w = z^n. This maps a sector in the z-plane with opening angle pi/n onto the upper half plane in the w-plane. And now the thing comes:
assume that you want to look at a flow in a "corner" with opening angle pi/n. Let us fix n, set it to 3. That means, you want to look at the flow in a corner with opening angle 60 degrees. You consider z a point in the z-plane (in the sector). The corresponding w in the w-plane will be w = z^3. The *potential* in the w-plane, phi_w(w) = w because it is a uniform flow in an upper half plane. We now take over that potential, but at the point z in the z-plane: phi_z(z) = phi_w(w) = w = z^3. So phi_z(z) = z^3 is a potential flow in the z-plane. It is an analytic function within the sector (theta = 0 to theta = 60 degrees), so that's already ok. The "walls" of the sector phi_z(r,theta=0) = cos(3x0) r^3 + i sin(3x0) r^3 give us 0 for the imaginary part, that's constant, so that's a stream line, ok, and for z = cos(60) r + i sin(60) r we have phi_z(z) = cos(180) r + i sin(180) r has also the imaginary part equal to 0, hence constant, so the wall at 60 degrees is also a stream line.
What we have is that we have an "incoming" flow from infinity along the 60 degree wall, which turns in the corner, and flows out towards infinity along the 0 degree wall (the positive x-axis).