How to take the real part of this function

[makethings]

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.

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

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

appreciate any help. Thank you

 

[gabbagabbahey]

The real part of any function can be found using;

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

where the overline denotes complex conjugation...does that help?

 

[makethings]

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 $$e^{iz} = cosz + i sinz$$ and perhaps take the conjugate as that cos - i sin ?? But with a function with multiple terms containing i, what would I do?

 

[cup]

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 $$e^{iz} = cosz + i sinz$$ and perhaps take the conjugate as that cos - i sin ?? But with a function with multiple terms containing i, what would I do?

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.

 

[makethings]

To clarify, is it changing the sign of the terms containing i, or i itself?

 

[cup]

(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):

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

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

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

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

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

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

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

...I think. You should check it carefully.

 

[cup]

To clarify, is it changing the sign of the terms containing i, or i itself?

Example:

If

$$z = \frac{a-bi}{c+id}e^{-ig}$$

then

$$z^* = \frac{a+bi}{c-id}e^{+ig}$$

 

[makethings]

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.

 

[cup]

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.

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.