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2nd derivative over 1st derivative squared & Dirichlet Problem

  1. Mar 21, 2013 #1
    1. The problem statement, all variables and given/known data
    First of all, for a different problem, I have the following written down:
    [tex]\frac{\frac{{\partial{}}^2\varphi{}}{\partial{}u^2}}{{\left(\frac{\partial{}\varphi{}}{\partial{}u}\right)}^2}[/tex]
    Where phi is a function of u, which is a function of x and y. Also, u is harmonic.
    Now, I want to rewrite this in some suggestive form, if there is one.

    Now, onto my other question:
    Naamloos.png
    2. Relevant equations
    Uhm, I can't really think of any at this point. Maybe the rotation matrix to rotate the strip?


    3. The attempt at a solution
    Concerning the rewriting: I already thought of
    [tex]\frac{\partial{}}{\partial{}u}\left(\frac{-1}{\frac{\partial{}\varphi{}}{\partial{}u}}\right)[/tex] but I somehow want it to be a function of phi, and I don't think this qualifies as such.

    Alright, and then the dirichlet problem. In my course, we haven't done any complicated PDE's yet whatsoever, so I checked with the teacher and he doesn't want a general formula, just a specific solution that works.
    For question a, I found a solution:
    (a+b)/2 - (a-b)/2 * X
    seems to satisfy the conditions given.

    Now for B, I somehow need to rotate and translate and scale the strip to the same one as the first. However, I don't understand how. If I understand correctly, I need to 'move' the plane bounded by y = x and y = x + 2 to the plane bounded by x = -1 and x = 1.
    In order to use the rotation matrix, I have to make y = x and y = x + 2 into a vector first though. I don't really understand how, so it would be great if someone could give me a hint!
     
  2. jcsd
  3. Mar 22, 2013 #2
    Alright, so for b this is what I have thought of so far. I need to apply the rotation-45-degree matrix, rescale, and then shift. As I have to rescale anyway, I'll just take the matrix to be
    1 -1
    1 1

    Instead of the square root two factors. However, I don't really know which points to use to find out exactly how it is done. I suppose (0,0) of the line y=x is still (0,0) after the rotation, and this just has to be shifted by 1 in the x direction. Does this mean that my transformation is just Sqrt(2)*(Rotation matrix)*original vector + a vector (1,0)?
    So then x' = x - y + 1 and y' = x + y

    But how do I use this to solve the differential equation?
     
    Last edited: Mar 22, 2013
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