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Post by mondo on Aug 13, 2023 23:04:08 GMT
On page 532 in subsection 3 "The method of images" author gives a an example of a flow of a source with strength $2\pi$ located at $(2+i)$ for which the complex potential is $\Omega_u = log(z - 2 -i)$. I calculated this myself and got a slightly different result.
On page 454 author defines source as $\frac{1}{2\pi}(\frac{1}{\overline{z} - \overline{A}})$ hence for our example on page 532 the vector field of that source should be defined by $H(z) = \frac{1}{\overline{z} -2 -i)}$. Next we find $\Omega$ by $\Omega = \int{H(z)dz} = log(\overline{z} -2 -i)$. So the difference is that I got conjugate of $z$. Where did I make a mistake here?
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Post by mondo on Aug 17, 2023 5:45:58 GMT
I think this is because $\int \overline{z}dz = \frac{z^2}{2}$ and not $\frac{\overline{z}^2}{2}$ as I thought, but why?
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Post by Admin on Aug 18, 2023 14:27:30 GMT
Mondo
It's easy to make this mistake because Needham, on page 494 in section III paragraph 2 writes:
"..., we shall continue to write the vector field as $\overline{H}$"
So the vector field $V$ of the source on page 455, using the notation in chapter 11 should be written as
$\displaystyle\overline{H}=V=\frac{S}{2\pi}\bigg(\frac{1}{\overline{z}-\overline{A}}\bigg)$, so that we can then use all the results in chapter 11.
Vasco
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Post by Admin on Aug 18, 2023 14:57:49 GMT
I think this is because $\int \overline{z}dz = \frac{z^2}{2}$ and not $\frac{\overline{z}^2}{2}$ as I thought, but why? Mondo The function $\overline{z}$ has no antiderivative. Vasco |
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Post by mondo on Aug 18, 2023 19:27:32 GMT
Mondo It's easy to make this mistake because Needham, on page 494 in section III paragraph 2 writes: "..., we shall continue to write the vector field as $\overline{H}$" So the vector field $V$ of the source on page 455, using the notation in chapter 11 should be written as $\displaystyle\overline{H}=V=\frac{S}{2\pi}\bigg(\frac{1}{\overline{z}-\overline{A}}\bigg)$, so that we can then use all the results in chapter 11. Vasco It is very confusing. So you say when we need Polya vector of a source vector field we do $\overline{\overline{H}} = H$ and this is our polya vector field of the source? I think this is because $\int \overline{z}dz = \frac{z^2}{2}$ and not $\frac{\overline{z}^2}{2}$ as I thought, but why? Mondo The function $\overline{z}$ has no antiderivative. Vasco How to show that? Since $\overline{z}$ is just a reflection of $z$ in the real line, I would expect the derivative and antiderivative to exist. |
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Post by Admin on Aug 18, 2023 19:54:42 GMT
mondo Mondo It's easy to make this mistake because Needham, on page 494 in section III paragraph 2 writes: "..., we shall continue to write the vector field as $\overline{H}$" So the vector field $V$ of the source on page 455, using the notation in chapter 11 should be written as $\displaystyle\overline{H}=V=\frac{S}{2\pi}\bigg(\frac{1}{\overline{z}-\overline{A}}\bigg)$, so that we can then use all the results in chapter 11. Vasco It is very confusing. So you say when we need Polya vector of a source vector field we do $\overline{\overline{H}} = H$ and this is our polya vector field of the source? Yes. Each is the Polya vector of the other.Mondo The function $\overline{z}$ has no antiderivative. Vasco How to show that? Since $\overline{z}$ is just a reflection of $z$ in the real line, I would expect the derivative and antiderivative to exist. Vasco |
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Post by mondo on Aug 24, 2023 5:59:33 GMT
Vasco, I see the problem with $\overline{z}$ not being analytic. However, there is one thing that is questionable - on page 532 we have equation (13) that describes potential $\Omega_u = \frac{(1+i)}{z-2}$. This means in order to recover a vector field we need to take a derivative of potential function WRT to $z$ -> $\Omega_u = \frac{-(1+i)}{(z-2)^2}$ But this is supposed to be a dipole which equation is $\frac{d}{\overline{z} - \overline{A}}$ - a conjugate is lost. Should I expect to get a potential function if I integrate dipole field WRT to $z$? I think so right? But then how do we deal with the conjugate? Looks like it the book it is ignored and whenever we calculate a potential function there is $\overline{z} = z$
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Post by Admin on Aug 24, 2023 7:57:30 GMT
Vasco, I see the problem with $\overline{z}$ not being analytic. However, there is one thing that is questionable - on page 532 we have equation (13) that describes potential $\Omega_u = \frac{(1+i)}{z-2}$. This means in order to recover a vector field we need to take a derivative of potential function WRT to $z$ -> $\Omega_u = \frac{-(1+i)}{(z-2)^2}$ But this is supposed to be a dipole which equation is $\frac{d}{\overline{z} - \overline{A}}$ - a conjugate is lost. Should I expect to get a potential function if I integrate dipole field WRT to $z$? I think so right? But then how do we deal with the conjugate? Looks like it the book it is ignored and whenever we calculate a potential function there is $\overline{z} = z$ Mondo The vector field is $\overline{H}$ and so from page 503 $\displaystyle\Omega_u=\frac{(1+i)}{z-2}$ so $\displaystyle H=\Omega_u'=-\frac{(1+i)}{(z-2)^2}$, and finally The vector field is $\displaystyle\overline{H}=-\frac{(1-i)}{(\overline{z}-2)^2}$ Vasco |
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Post by mondo on Aug 24, 2023 21:45:59 GMT
Ok, I think as described in the first paragraph of page 502, we "recognize" that this derivative is $H$. It makes sense because the integral which gave us $\Omega$ was calculated using $H$ - equation (11) page 483 and Polya vector field just allowed us to simplify the calculations.
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