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Finding the streamlines of a complex potential for a Joukowsky transform from a unit circle to a flat plate.

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In page 36 of the paper ''On dynamic interactions between body motion and fluid motion by Frank T. Smith, Samire Balta, Kevin Liu and Edward R. Johnson'', there is the equation (65) giving the complex velocity potential. I am taking the imaginary part of this equation to reproduce Figure 18 on the next page.
Here is my Matlab code:
x = -2:0.02:2;
y = -1:0.01:1;
[X, Y] = meshgrid(x, y);
z = X + i.*Y;
zc = X - i.*Y;
a = sqrt(z.^2 - 1);
w = -i.*(z + a + log(1./(z + a)));
f = imag(w)
hold on
[C, h] = contour(X, Y, f, 'k');
y = 0;
line([-1,1],[y,y], 'linewidth', 2, 'color', 'k');
axis equal
hold off
For some reason starting off at x = 0.22 and carrying on the negative x values, I can’t get the same plot as in Figure 18.Can you see where have I gone wrong in my code?
Best regards,
Marialis Simoni

Accepted Answer

Alan Stevens
Alan Stevens on 22 Oct 2020
Edited: Alan Stevens on 22 Oct 2020
There are some typos in the paper by Smith et al, and some steps left out. I think the following generates figure 18:
x = [-2:0.01:-0.01 0.01:0.01:2];
y = -1:0.01:1;
[X, Y] = meshgrid(x, y);
z = X + 1i.*Y;
zeta = -z + sqrt(z.^2-1); % Equation (59)
idx = X<0;
zeta(idx) = 1./zeta(idx);
w = 1i.*(1./zeta - log(zeta)); % First line of equation (65)
f = imag(w);
hold on
[C, h] = contour(X, Y, f, 'k');
line([-1,1],[0,0], 'linewidth', 2, 'color', 'k');
axis equal
hold off
This results in


Show 5 older comments
Rik on 23 Oct 2020
I don't think you require my surname to have a productive conversation. You can find or deduce it if you look around a bit. It is not secret, but I do keep it slightly more private than I used to.
If you don't want your work public, you shouldn't be using a public forum. When you posted your question you agreed to the terms of use, which state that you publish everything on Matlab Answers under CC-by-sa 3.0. That means you already gave permission to everybody interested to use your work. If you need private consultation you should hire a private consultant.
I don't think plagiarism detectors will pose a problem, as you seem to be using your actual name. From what I have heard, most institutions will have a manual look if their software finds something. If they don't you can fight any accusation of plagiarism with great ease when it comes to this.
So, no, I don't think your reasons are enough to warrant deletion.
Also, the site admins are generally able to restore the original post, and we can also re-post it from the Google cache capture that Stephen posted.
Alan Stevens
Alan Stevens on 23 Oct 2020
@Rik In fact I had already answered. However, because I did so on my phone with its limited view, I inadvertently replied as a second answer rather than as a comment!
@Marialis Your question could not possiby be considered to be original research as it simply asked to reproduce a previously published result.
Marialis Simoni
Marialis Simoni on 23 Oct 2020
Hi everyone,
I reposted the question. Hope everything is fine now.
Also, Alan I meant what I wrote in a future context, when I produce my own original figures.
Thanks everyone for your constructive criticism.

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More Answers (1)

Alan Stevens
Alan Stevens on 22 Oct 2020
x=0:. There was a strange vertical line that O thought was related to this. However, you could put it back and see if it makes any difference.
It was clear from the plot without inverting zeta for negative X that the contours were inverted. I don't have a technical justification for it!


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