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Dirichlet problem

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Find function ##u(x,y)## that is harmonic on the upper half-plane ##0<Im(z)##. Note that
    ##u(x,0)=0##, ##x<0##
    ##u(x,0)=-1##, ##0<x<1## and
    ##u(x,0)=1##, ##x>1##.


    2. Relevant equations

    ##u(z)=\frac{1}{2\pi i}\int _{I}\gamma '(s)u(\gamma (s))\frac{\alpha '(\gamma (s))}{\alpha (\gamma (s))}Re(\frac{\alpha (\gamma (s))+\alpha (z)}{\alpha (\gamma (s))-\alpha (z)})ds##

    Where ##\alpha ## is holomorphic function that maps ##Im(z)>0## into open unit disk, and ##\gamma (s) ## parameterization of the edge.

    3. The attempt at a solution

    For this problem ##\alpha (z)=\frac{1-iz}{1+iz}## and it is also obvious that I will have to make at least 3 integrals.

    Let's firstly take a look at ##u(x,0)=0##, ##x<0##:

    This tells me that ##u(z)=\frac{1}{2\pi i}\int _{I}\gamma '(s)u(\gamma (s))\frac{\alpha '(\gamma (s))}{\alpha (\gamma (s))}Re(\frac{\alpha (\gamma (s))+\alpha (z)}{\alpha (\gamma (s))-\alpha (z)})ds## will be equal to ##0## for all ##x<0##.

    But the question here is: What about ##y## ? None of the conditions tell anything about ##y##. What do I do here?

    The same question is for ##u(x,0)=-1##, ##0<x<1##:

    Here the integral for ##x## goes from ##0## to ##1##. But again, ##y## can be anything from ##0## to ##\infty ##. ???

    I guess my question here is: What should I do with ##y## to get ##u(z)##?
     
  2. jcsd
  3. Apr 30, 2014 #2
    Ok, I found out that my relevant equation in original post was a very stupid equation to begin with. I think it would wiser to use the Poisson integral formula for the upper half plane: ##u(x,y)=\frac{y}{\pi }\int_{-\infty }^{\infty }\frac{u(t,0)}{y^2+(x-t)^2}dt## where I am guessing ##t## stands for some kind of parameterization of real axis? http://www.diss.fu-berlin.de/diss/s...ionid=76E8A9CC9C7EE802A569873649AAF2F7?hosts= page 59.

    Anyway, now with that in mind:

    ##u(x,y)=\int_{-\infty }^{0}...+\int_{0}^{1}...+\int_{1}^{\infty} ## where the first integral is obviously ##0##.

    ##u(x,y)=-\frac{y}{\pi }\int_{0}^{1}\frac{1}{y^2+(x-t)^2}dt+\frac{y}{\pi }\int_{1}^{\infty }\frac{1}{y^2+(x-t)^2}dt##

    ##u(x,y)=\frac{y}{\pi }[\frac{1}{y}arctan(\frac{u}{y})\mid _{u=x}^{u=x-1}-\frac{1}{y}arctan(\frac{u}{y})\mid _{u=x-1}^{u=-\infty }]##

    and finally

    ##u(x,y)=\frac{1}{\pi }[2arctan(\frac{x-1}{y})-arctan(\frac{x}{y})+\frac{\pi }{2}]##

    This should be a bit better than my first attempt.
     
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