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enter image description here

The idea is through conformal transformations satisfying the conditions requested of the problem make this an easier problem to deal,but i don't know which be this transformation. I would appreciate any hint how to solve this. thank you very much

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Looking at my copy of Churchill, Complex Variables with Applications, the appendix on conformal transformations suggests that applying

$$ w = f(z) = \log \left(\frac{z-1}{z+1} \right) $$

will transform the region shown (for the case $a = 1$) into a strip in the $w$-plane, where the $V = 1$ rays are mapped to $\Im(w) = \pi$ and the $V = 0$ segment is mapped to $\Im(w) = 0$.

If you take $g(z) = \frac{1}{\pi} f(z)$, you map this region to one in which the $V = 1$ segment maps to $\Im(w) = 1$, and the $V = 0$ segment maps to $\Im(w) = 0$.

I seem to recall that this is what's needed in this kind of problem, but it's as far as my shoddy memory takes me. (The inverse conformal map for $f$, by the way, is $z = -\coth \frac{w}{2}$.)

OK. I went and read the relevant chapter of Churchill ... and found this problem in the exercises!

I think that having come up with the transformation, and the potential in the $w$-plane, I'm going to leave the conversion back to a potential in the $z$-plane up to you, although I confess I don't see offhand how to get from the $\log$-expression back to something involving $\arctan$.

In comments, I've been asked to add some pics to show off what the log-transform is doing. Let's look at this diagram:The point $z$ in the upper half plane, relative to $-1$ and $+1$

Now the "log" that I've used above is really $$ w = \log |\frac{z-1}{z+1}| + i \arg \frac{z-1}{z+1}. $$

Let's look at how that gets computed. First, the first ratio is just the ratio of distances to the two point $+1$ and $-1$; I'll come back to that. The second term -- the imaginary part, is the difference of the args of $z-1$ and $z - (-1)$, i.e., the angle labeled $\theta_1 - \theta_2$ in the diagram.

What's that look like on (or very near) the real axis? Well, for $z > 1$, $\theta_1$ and $\theta_2$ will both be nearly zero; for $z < -1$, they'll both be nearly $\pi$. For $-1 < z < 1$, we have $\theta_2 \sim 0$ and $\theta_1 \sim \pi$, so the difference is approximately $\pi$. In other words, the red segments of the real axis will be sent to $v = 0$ in the $w$-plane, but the blue segment will be sent to $v = \pi$. Note that the remote ends of the red segments (near $\pm \infty$) will be sent to $u = 0$, while the near ends (near $\pm 1$) are sent off to infinity.

What about the ratio of lengths? Well, along the $x = 0$ line, we'll have $r_1 = r_2$, so that will map to $u = 0$. The other $u = constant$ lines will become hyperbolic-like curves in the $xy$-plane.

Finally, the transform we really want is $$ w = \frac{1}{\pi}\log |\frac{z-1}{z+1}| + \frac{i}{\pi} \arg \frac{z-1}{z+1}, $$ which sends the blue segment to the $v = 1$ line in the $w$-plane.

Now since the imaginary part of $w$ is $v$, and we can substitute $x+iy$ for $z$, we get \begin{align} v &= \frac{1}{\pi} \arg \frac{z-1}{z+1} \\ &= \frac{1}{\pi} \arg \frac{(x-1) + iy}{(x+1) + iy} \\ &= \frac{1}{\pi} \arg \frac{x^2 + y^2 - 1 + i 2y}{(x+1)^2 + y^2} \\ &= \frac{1}{\pi} \arg {x^2 + y^2 - 1 + i 2y}\\ &= \frac{1}{\pi} \arctan \frac{2y}{x^2 + y^2 - 1}, \end{align} which is an expression for the potential at point $(x, y)$.

Finally, I'm going to show off the effect of the conformal transformation by showing an image in the $z$-plane, and the transformed image in the $w$-plane, courtesy of Matlab. The problem is that the points near $z = \pm 1$ get sent to infinity, so parts of the transformed image will be missing.

The starting image (which extends from $-2.5$ to $2.5$ in $x$, and from $0.05$ to $2.0$ in $y$ (with $y = 0$ near the top, because of Matlab's idiosyncracies!) shows a bunch of peppers: enter image description here

The transformed image, extending from -10 to 10 in $u$ and $0$ to $1$ in $v$, is this: enter image description here

If you can make sense of those, then bless you. I just followed the description given here: http://www.mathworks.com/help/images/examples/exploring-a-conformal-mapping.html to make the images. Here's my code:

function confImage

close all
A = imread('peppers.png');
A = A(31:330, 1:500, :);
figure, imshow(A)
title ('Original Image', 'FontSize', 14);

conformal = maketform('custom', 2, 2, [], @conformalInverse, []);

uData = [ -10   10];  % Bounds for REAL(w)
vData = [  0  1.0];  % Bounds for IMAG(w)
xData = [ -2.5    2.5 ];  % Bounds for REAL(z)
yData = [  0.05   2.0 ];  % Bounds for IMAG(z)


B = imtransform( A, conformal, 'cubic', ...
                'UData', uData,'VData', vData,...
                'XData', xData,'YData', yData,...
                'Size', [300 360], 'FillValues', 255 );
figure, imshow(B)
title('Transformed Image','FontSize',14)

function U = conformalInverse(X, t)

Z = complex(X(:,1),X(:,2));
W = (Z + 1./Z)/2;
U(:,2) = (1/pi) * angle((Z - 1) ./ (Z + 1))
U(:,1) = (1/pi)* log( abs((Z - 1) ./ (Z + 1)));  %real(W);
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    $\begingroup$ Should be $$\log \left(\frac{z-a}{z+a}\right),$$ I think. $\endgroup$ – Daniel Fischer Feb 10 '14 at 21:55
  • $\begingroup$ Yes. (I specified that I was handling the case $a = 1$, but this is the more general case.) $\endgroup$ – John Hughes Feb 10 '14 at 22:10
  • $\begingroup$ Ah, I overlooked that parenthesis. $\endgroup$ – Daniel Fischer Feb 10 '14 at 22:11
  • $\begingroup$ Hmm, a few pictures to see what's happening would be nice .. $\endgroup$ – Han de Bruijn Feb 11 '14 at 11:21
  • $\begingroup$ Done. By the way, the explanation is copied almost verbatim from section 82 of Churchill (the OP's textbook), "Steady Temperatures in a Half Plane", so if s/he wants to read it in more detail, it's there. $\endgroup$ – John Hughes Feb 11 '14 at 21:37
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For completeness of the record, I've decided to formulate my previously given comment as an answer. Yes, the mapping in John's answer is a Schwarz-Christoffel mapping :

enter image description here
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enter image description here
In general we have: $$\frac{dw}{dz}=K(z-a)^{\alpha/\pi-1}(z-b)^{\beta/\pi-1}$$ With $a=-1$ , $b=+1$ and $\alpha=\beta=0$ , giving: $$dw\;=\;K(z-1)^{-1}(z+1)^{-1}dz \;=\;\frac{K}{2}\left(\frac{dz}{z-1}-\frac{dz}{z+1}\right)$$ So indeed, integration gives apart from constants that can be determined (later): $$w=\log\left(\frac{z-1}{z+1}\right)$$

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