|
|
Post by mondo on Aug 28, 2023 0:12:54 GMT
I have several questions to the subsection 4 on pages 538 - 540:
1. In the first paragraph of page 539 "If we want the flow round $E$ to be uniform far from $E$ then we must further demand that $f(z)$ behaves like a multiple of the identity far from $C: f(z) \approx cz $ if $|z|$ is large". I don't quite get that. If the the flow is supposed to be uniform, which means it is constant speed then it should be $z$ not $cz$.
2. Next question comes from the sentence just below the above quote: "Assigning equal values of $\Psi$ to the original and image streamlines, the image flow far from $E$ will then be uniform with speed $1/|c|$ and direction $c$ [why?]" - does it follow from conformal invariance section described on p513?
3. Right above figure [23] "Thus $f$ should map the flow round $C$ to the flow that results from inserting $E$ into a uniform flow of velocity (2/3)." I believe this prediction is based on the fact that speed of $2/3$ should "offset" $3/2$ to make it unity. However, how do we know that $3/2$ signifies the speed of the flow? At the end of this page, we take the leading term of complex potential to mean the same, how is it calculated?
4. Second to last formula on page 539 if is for $z$, and author asks why there is a plus sign between the terms in the braces. Is that because the other solution, with a minus sign, would cause $z$ to be much smaller than $w$ and we know that $w$ is very close to $z$?
5. On page 540, in the top paragraph that starts with "As another illustration.." author says that the complex potential resulting from inserting $E$ into a uniform flow in the direction of $e^{i\phi}$ is $\Omega_u(z) = e^{-i\phi}z$ - I don't see it, how was it calculated? If the flow were just $1$ then the complex potential would be $\Omega = z$ so in the case of a uniform flow in the direction of $e^{i\phi}$ I would expect $\Omega = e^{i\phi}z$, so why the got a negative exponent there?
|
|
|
|
Post by mondo on Aug 28, 2023 19:19:05 GMT
I think I found an answer to my question #5 above. The uniform flow is in the direction $e^{i\phi}$ and hence there is an obstacle (unit disc $C$) inserted at the origin the undisturbed flow can be described by $\frac{1}{z} = e^{-i\phi}$ then $\Omega = e^{-i\phi}z$. However there are two things I am not sure here:
a. Why is the magnitude taken as $1$? The fact that the flow is uniform doesn't mean the flow can't actually be $|z|e^{i\phi}$
b. The undisturbed flow, described in the book by $\Omega_u$ described the flow inside the unit disc $C$ right?
|
|
|
|
Post by Admin on Aug 29, 2023 9:19:53 GMT
Mondo
Your first question:
I don't understand this question. Can you explain it again please? It all looks OK to me.
Thanks
Vasco
|
|
|
|
Post by Admin on Aug 29, 2023 9:46:40 GMT
Mondo
Your second question:
In the book Needham writes [why?], but this is not a suggested exercise. He provides the answer in the paragraphs which follow on pages 539-540.
Vasco
|
|
|
|
Post by Admin on Aug 29, 2023 10:09:01 GMT
Mondo
Your third question:
We apply the transformation to the case when the obstacle is $C$ the unit circle. This is described in subsection 2 on pages 527-528.
Vasco
|
|
|
|
Post by Admin on Aug 29, 2023 11:11:01 GMT
Mondo
Your fourth question:
Think of the situation far from $E$, when $|w|$ is very large.
Vasco
|
|
|
|
Post by Admin on Aug 29, 2023 12:48:20 GMT
Mondo Your fifth question: No, you have misunderstood what he is saying. He is talking initially about the undisturbed potential function for $C$ not $E$. A few years ago I produced my own copy of figure 24 on page 540 with Desmos. RHS: www.desmos.com/calculator/4g49lbnu4uVasco |
|
|
|
Post by Admin on Aug 29, 2023 19:30:44 GMT
|
|
|
|
Post by Admin on Sept 2, 2023 10:58:22 GMT
Mondo
I have updated my reply #6 above. In my opinion there is no error at the top of page 540.
Vasco
|
|
|
|
Post by mondo on Sept 2, 2023 22:41:15 GMT
Mondo Your first question: I don't understand this question. Can you explain it again please? It all looks OK to me. Thanks Vasco Sure, so what I say is at the beginning of page 539 author says we know a uniform flow round $C$, (a unit circle) and now base on that flow we want to find another one that would give us a uniform flow far from $E$ (an ellipse). Author concludes that the requirement to get that is $C: f(z) \approx cz $ - is that because the flow we try to find may not have a velocity $1$ and therefore we need to multiply by $c$? Mondo Your fourth question: Think of the situation far from $E$, when $|w|$ is very large. Vasco Good suggestion, so then, for large $w$ this whole expression would be close to $0$ if we have a minus sign there in the formula for $z$. Mondo Your fifth question: No, you have misunderstood what he is saying. He is talking initially about the undisturbed potential function for $C$ not $E$. A few years ago I produced my own copy of figure 24 on page 540 with Desmos. RHS: www.desmos.com/calculator/4g49lbnu4uVasco I understand the example he gives here. Basically the only difference with the previous one is the direction of the flow. And I don't understand why the undisturbed complex potential $\Omega_u = e^{-i\phi}z$. First, it would mean that the flow is unity $H(z) = 1$ but in the book author only says the flow is uniform. Second even if we assume the velocity of the flow is $1$ we should get $\Omega_u = e^{i\phi}z$ instead. So why the negative sign? Mondo I have updated my reply #6 above. In my opinion there is no error at the top of page 540. Vasco Thank you, what error do you see there? |
|
|
|
Post by Admin on Sept 6, 2023 16:11:23 GMT
Mondo Your first question: I don't understand this question. Can you explain it again please? It all looks OK to me. Thanks Vasco Sure, so what I say is at the beginning of page 539 author says we know a uniform flow round $C$, (a unit circle) and now base on that flow we want to find another one that would give us a uniform flow far from $E$ (an ellipse). Author concludes that the requirement to get that is $C: f(z) \approx cz $ - is that because the flow we try to find may not have a velocity $1$ and therefore we need to multiply by $c$? We are only demanding that the flow be uniform far from $c$.Mondo Your fourth question: Think of the situation far from $E$, when $|w|$ is very large. Vasco Good suggestion, so then, for large $w$ this whole expression would be close to $0$ if we have a minus sign there in the formula for $z$. Mondo Your fifth question: No, you have misunderstood what he is saying. He is talking initially about the undisturbed potential function for $C$ not $E$. A few years ago I produced my own copy of figure 24 on page 540 with Desmos. RHS: www.desmos.com/calculator/4g49lbnu4uVasco I understand the example he gives here. Basically the only difference with the previous one is the direction of the flow. And I don't understand why the undisturbed complex potential $\Omega_u = e^{-i\phi}z$. First, it would mean that the flow is unity $H(z) = 1$ but in the book author only says the flow is uniform. Second even if we assume the velocity of the flow is $1$ we should get $\Omega_u = e^{i\phi}z$ instead. So why the negative sign? Note that $e^{-i\phi}z=y\cos\phi-x\sin\phi+i(y\cos\phi-x\sin\phi)$ and so $\Psi=(y\cos\phi-x\sin\phi)$ and so the streamlines are straight lines at an angle of $\pi/4$ to the real axis as in LHS of figure 24.Mondo I have updated my reply #6 above. In my opinion there is no error at the top of page 540. Vasco Thank you, what error do you see there? I was referring to $e^{-i\phi}$Vasco |
|
|
|
Post by mondo on Sept 7, 2023 21:45:47 GMT
I checked the newer version of this book, where most of the mistakes where fixed and it is still written as $e^{-i\phi}$ Maybe this undisturbed complex potential $\Omega_u(z) = \frac{1}{z} = e^{-i\phi}$ is for the unit circle?
|
|
|
|
Post by Admin on Sept 7, 2023 21:51:30 GMT
Mondo
Maybe I didn't make it clear, but I do not consider this to be an error as I explain in my post.
Vasco
|
|
|
|
Post by mondo on Sept 7, 2023 22:53:03 GMT
Mondo Maybe I didn't make it clear, but I do not consider this to be an error as I explain in my post. Vasco Right, sorry I somehow thought you said "there IS an error". I agree but so would be $\Psi=(y\cos\phi+x\sin\phi)$ -a straight lines at an angle $\pi/4$ to the real axis. So it doesn't justify $e^{-i\phi}$ does it? |
|
|
|
Post by Admin on Sept 7, 2023 23:06:31 GMT
Mondo
Yes it does. But you have written it wrong. The plus sign should be minus.
Vasco
|
|
