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How to take the real part of this function

  1. Nov 9, 2008 #1
    Hello. I have been trying to solve Stokes Second Problem modified with a stationary upper plate (with oscillating bottom plate with velocity u(0,t) = U cos nt). I found my solution for the velocity profile but I can't simplify it since my math knowledge is limited.

    Now you don't need to know what Stokes Second Problem is, but I need some help trying to take the Real part of this solution.

    [tex]

    u(y,t) = Re\{U\frac{e^{a(H-y)} e^{i[a(H-y) +nt]} - e^{-a(H-y)} e^{-i[a(H-y) +nt]}}{ e^{aH} e^{iaH} - e^{-aH} e^{-iaH}}\}[/tex]

    where U, a, H, are constants
    the variable is y

    appreciate any help. Thank you
     
  2. jcsd
  3. Nov 9, 2008 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    The real part of any function can be found using;

    [tex]\text{Re} [f(y)]=\frac{f(y)+\overline{f(y)}}{2}[/tex]

    where the overline denotes complex conjugation....does that help?
     
  4. Nov 9, 2008 #3
    You might have to point me to a tutorial on taking complex conjugates of a function as complicated as mine. I know I can break down [tex] e^{iz} = cosz + i sinz [/tex] and perhaps take the conjugate as that cos - i sin ?? But with a function with multiple terms containing i, what would I do?
     
  5. Nov 9, 2008 #4

    cup

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    Assuming that U, a, H are all real, it is very simple to take the complex conjugate: Just change the signs of all the i's.
     
  6. Nov 9, 2008 #5
    To clarify, is it changing the sign of the terms containing i, or i itself?
     
  7. Nov 9, 2008 #6

    cup

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    (In the following, z* denotes the complex conjugate of z.)
    The first thing to do is to make the denominator real (you will see why this makes sense):

    [tex]
    U
    \frac{
    \left(
    e^{a(H-y)}
    e^{i[a(H-y) +nt]}
    -
    e^{-a(H-y)}
    e^{-i[a(H-y) +nt]}
    \right)
    \left(
    e^{aH}
    e^{iaH}
    -
    e^{-aH}
    e^{-iaH}
    \right)^{*}
    }
    {
    \left(
    e^{aH}
    e^{iaH}
    -
    e^{-aH}
    e^{-iaH}
    \right)
    \left(
    e^{aH}
    e^{iaH}
    -
    e^{-aH}
    e^{-iaH}
    \right)^{*}
    }
    =
    [/tex]

    [tex]
    U
    \frac{
    \left(
    e^{a(H-y)}
    e^{i[a(H-y) +nt]}
    -
    e^{-a(H-y)}
    e^{-i[a(H-y) +nt]}
    \right)
    \left(
    e^{aH}
    e^{-iaH}
    -
    e^{-aH}
    e^{iaH}
    \right)
    }
    {
    \left(
    e^{aH}
    e^{iaH}
    -
    e^{-aH}
    e^{-iaH}
    \right)
    \left(
    e^{aH}
    e^{-iaH}
    -
    e^{-aH}
    e^{iaH}
    \right)
    }
    =
    [/tex]

    [tex]
    U
    \frac{
    e^{a(2H-y)}
    e^{i[-ay+nt]}
    -
    e^{-ay}
    e^{i[a(2H-y)+nt]}
    -
    e^{ay}
    e^{-i[a(2H-y)+nt]}
    +
    e^{-a(2H-y)}
    e^{-i[-ay+nt]}
    }
    {
    e^{2aH}
    -
    e^{i2aH}
    -
    e^{-i2aH}
    +
    e^{-2aH}
    }
    =
    [/tex]

    [tex]
    U
    \frac{
    e^{a(2H-y)}
    e^{i[-ay+nt]}
    -
    e^{-ay}
    e^{i[a(2H-y)+nt]}
    -
    e^{ay}
    e^{-i[a(2H-y)+nt]}
    +
    e^{-a(2H-y)}
    e^{-i[-ay+nt]}
    }
    {
    2\cosh(2aH)
    -
    2\cos(2aH)
    }
    [/tex]

    The real part of this expression is now easily seen (as described above) to be

    [tex]
    U
    \frac{
    e^{a(2H-y)}
    \cos[-ay+nt]
    -
    e^{-ay}
    \cos[a(2H-y)+nt]
    -
    e^{ay}
    \cos[a(2H-y)+nt]
    +
    e^{-a(2H-y)}
    \cos[-ay+nt]
    }
    {
    \cosh(2aH)
    -
    \cos(2aH)
    }
    =
    [/tex]

    [tex]
    2U
    \frac{
    \cosh(a(2H-y))
    \cos[-ay+nt]
    -
    \cosh(ay)
    \cos[a(2H-y)+nt]
    }
    {
    \cosh(2aH)
    -
    \cos(2aH)
    }
    [/tex]

    ...I think. You should check it carefully. :smile:
     
    Last edited: Nov 9, 2008
  8. Nov 9, 2008 #7

    cup

    User Avatar

    Example:

    If

    [tex]
    z = \frac{a-bi}{c+id}e^{-ig}
    [/tex]

    then

    [tex]
    z^* = \frac{a+bi}{c-id}e^{+ig}
    [/tex]
     
  9. Nov 9, 2008 #8
    Thanks..this math is tedious. I was working on it at the same time you were and we have ever so slightly different answers ( I got some - where you have + and I have a sinh appear).. now I got to nit pick and see where you or I went wrong. ugh.. lol Thanks though for doing that grunt work. appreciate it very much.
     
  10. Nov 10, 2008 #9

    cup

    User Avatar

    I find that it's easier to do these kinds of calculations on a computer, using Tex.
    As long as you use a "vertical" coding style (one term per line only), it won't get messy.

    The ability to copy and paste really helps.
     
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