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I'm struggling here, trying to understand how to do this, and after 4 hours of reading, i still can't get around the concept and how to use it.

Basically, i have this problem:

A={(x,y) / x≥0, 0≤y≤pi

So U(x,0) = B; U(x,pi) = C; U'x(0,y) = 0;

I know that inside A, the laplace operator of U is 0. So i have to find U, and U must meet those requirements.

I don't have to use any form of differential equation. I'm supposed to find some sort of conformal transformation in order to make the domain a little more.. easy to understand. And then i should just get a result.

The problem is, i think i don't know how to do that. If any of you could help me understand how to solve this one, i might get the main idea and i could try to reproduce the resolution in similar cases.

Thank you very much, and i'm sorry for my english.

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Perhaps you could try to map your domain conformally to the unit disk and then use Poisson's formula? – Alex R. Sep 8 '12 at 19:22
I haven't seen Poisson's Formula in class, so i don't think it's necessary. Or, ar least, it can be solved without it. I thought about converting that into a disk, using the sin(z) transformation (as y is constant, it could give me an elipse). But i will have a condition in the point (0,0) and another in the border. And i don't think i know how to know the function i'm looking for with those two conditions (without using poisson, i think). Thank you for your answer! – Fede Sep 8 '12 at 20:06
up vote 0 down vote accepted

The domain is simple enough already. Observe that there is a function of the form $U=\alpha y+\beta$ which satisfies the given conditions.

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Oh, but.. what about the restriction of x≥0? That's my main problem. I solved one before, but without that restiction (it was a infinite region, limited by y=0 and y=pi . But.. i don't see how it could be applied here. Could you give me some more.. clues? Thank you! – Fede Sep 10 '12 at 16:40
@Fede I don't see what's the problem with having a smaller domain. If $U$ satisfies the equation in a larger domain, it also satisfies it in the smaller domain. The only extra thing to check is the boundary condition $U_x=0$ when $x=0$. But this is obviously true because $U_x$ is identically zero. – user31373 Sep 10 '12 at 16:54
i was writing a whole response to that, saying that I didn't get it. Then it hit me. U'x. Not U'y. Nobody cares if U varies in 'y'. This whole time I was thinking that as U'x was 0, U had to take a constant D.. but.. oh, I feel very stupid now. Thank you very, very much. You are awesome. (And sorry for my english). – Fede Sep 10 '12 at 17:39

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