Gary
GaryVasco
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Post by Gary on Jan 1, 2016 3:47:09 GMT
Vasco, This was a case where I did not fully understand your reasoning to the proof, so I pursued a somewhat different course. I am happy to be corrected if you see problems. Gary P. S. The attachment has been revised in response to Vasco's comments. Attachments:nh.ch5.ex7.pdf (135.52 KB)
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Post by Admin on Jan 2, 2016 9:34:58 GMT
Gary
I have read through your solution to exercise 7 and here are my comments:
1. After $\partial_xU(x)=\partial_yV(y)$, you say that "...both sides must be equal to a complex constant, say $A$ (after p231)....". Since $U$ and $V$ are both real valued functions then this constant must be real. Also I think there is a typo: p231 should be p221.
2. So when you arrive at $f=Az+B$ about 2/3 of the way down page 1, $A$ is real and $B$ is complex. As a consequence of this your reference to $a$ on page 2 as a complex number is also wrong since $A=a=$real. This is an important point (see below).
3. However, I think the rest of page 1 and the part of page 2 which comes before your proof by contradiction is unnecessary except as a verification of your original assumptions that f maps horizontal parallel lines to horizontal parallel lines. Just as an aside, the fact that $A$ is real means that it actually maps any set of parallel lines to the same set of parallel lines, not just horizontal parallel lines.
4. In your proof by contradiction you say, on the last line of page 2, that $\partial_xU(x,y)=\partial_yV(x,y)=$constant. Can you say why this is true as I cannot see it for myself?
Vasco
PS I would be interested to know which part of my argument you found difficult to follow.
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 2, 2016 19:25:04 GMT
Vasco,
Thank you for looking at ex. 7.
Regarding your points:
1. Yes, a typo.
2. I meant to say “a as the real part of a complex number a + ib”, but “a” would probably suffice. I did have some confusion over what was intended to be real and what was complex.
3. I’m a little puzzled here. Apparently the purpose of beginning with horizontal lines was to generate a general equation for parallel to parallel lines? But I think you are right that some of it is unnecessary. Also, I see now that Az + B takes parallel lines to parallel lines where A is real.
4. The lines of $e^{-i\phi}h$ are horizontal. Hence V(x,y) is a constant, and V(x,y) must also be a constant to satisfy CR, which leads to the contradiction. There really is no V(x,y) or U(x,y) because they are constants. I elaborated a bit in the revised attachment.
I lost the argument of your solution beginning at line -8 or the last paragraph of page 1. I think I get it now, but the first time through I could not see it. It seems to hinge on showing that if there is a function g(z) that takes parallel lines at any angle to parallel lines at an arbitrary angle, then g can be written in linear form. It is shown by sending the lines $g(e^{i\theta}z)$ to horizontal lines and using f = Az + B which permits g to be written in the linear form (a + bi)(x + iy) + B.
It’s true, I think, that it proves that parallel lines at any angle can be sent to parallel lines at any other angle with a linear mapping. But does this suffice to show that the linear mapping is “the only mapping that sends parallel lines to parallel lines”? Don’t we also have to show that non-linear mappings can’t do the job?
Gary
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Post by Admin on Jan 3, 2016 8:16:35 GMT
Gary
I have just had a very quick look at your amended solution and I notice that you are still saying in the middle of page 1 that
"This can only be true if both sides equal a (complex) constant, say $A$..."
But $A$ has to be a real constant since $U,V$ are by definition real functions and therefore their derivatives must be real functions.
Also on line -6 of page 1 you write
"in the case of horizontal lines, $z$ consists only of $x$."
But for the horizontal line through the point $(4,0)$ for example, $z=x+4i$. Maybe it would be better to say "for a given horizontal line, $z$ only depends on $x"$.
I will continue reading the new document and also answer your other points as soon as I can.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 3, 2016 16:45:36 GMT
Vasco,
Quite right. I’ll fix those items. When I wrote “z depends only on x”, I was thinking of moving the iy constant into B.
Gary
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Post by Admin on Jan 3, 2016 16:59:17 GMT
Gary I have written a document which I will attach to this post in the the next few minutes regarding some of your questions. Vasco GaryVCA5-07.pdf (25.43 KB) |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 3, 2016 21:30:40 GMT
Vasco,
I reread pp. 219-222, and I can see that the language foreshadows exercise 7, so it follows that your answer is correct and proof by contradiction becomes redundant. But this is based upon Needham’s assumption which you have nicely made explicit in your document:
If we can find a differential equation that represents a transformation we are interested in, and solve this differential equation to obtain its most general solution then this solution is the only solution for the transformation we are seeking. This is the key to the whole idea of this section and also to the solution of exercise 7.
There is a sense in which this seems obvious, given the validity of the CR equations, because all of the transformations can be demonstrated to be linear by integrating the partials and rewriting the result, but I also think a theorem or a proof by contradiction might lie behind it. In Gellert, et al. (1977) The VNR Concise Encyclopedia of Mathematics, I find the following which might be relevant:
The real part u and imaginary part v of a holomorphic function f = u + iv satisfy the Laplace differential equation [harmonic-GBP]…. Conversely, in a simply-connected domain D for a function u that is a solution in D of the Laplace differential equation there is a function v, uniquely determined apart from an additive constant, which together with u determines in D the holomorphic function f = u + iv. (p. 523)
No proof is given that v is uniquely determined.
Am I missing the obvious? I have done so many proofs in which I failed to complete the "only if" part, that I feel bound to seek the basis for the "only" in this problem.
Gary
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Post by Admin on Jan 3, 2016 23:29:30 GMT
Vasco, I reread pp. 219-222, and I can see that the language foreshadows exercise 7, so it follows that your answer is correct and proof by contradiction becomes redundant. But this is based upon Needham’s assumption which you have nicely made explicit in your document: If we can find a differential equation that represents a transformation we are interested in, and solve this differential equation to obtain its most general solution then this solution is the only solution for the transformation we are seeking. This is the key to the whole idea of this section and also to the solution of exercise 7.Gary I wouldn't call this an assumption by Needham - as you have written yourself above in your post, it is in fact clear that if we can solve the differential equation then this solution is the only solution. This is also the argument used in my solution to exercise 28 part (iv) of chapter 1. Notice that the solution of the differential equations is not linear for the example worked in the book, which involves the $log$ function, so this is not about linear and non-linear, but about whether the solution obtained is the only solution. I notice that you have resisted calling $A$ a real constant in your editing of your solution, just calling it a constant. I think it is important to point out that $A$ is real because $Az$ then involves no rotation, whereas if it is complex then $Az$ does involve rotation. Do you have a particular reason for not pointing out that it is real? Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 4, 2016 1:59:33 GMT
Vasco, some responses:
I agree that this is the only solution to be obtained by means of solving the differential equation. I just failed understand how we can be sure that this is the only possible solution in complex analysis. I suppose the answer lies in the fact that the mapping is analytic, which is defined by the CR equations, so we are working within a closed system where the most general solution provides the only answer.
A good point.
I think I was taking the phrasing from p. 221, para. 2, lines 2-3: The only way out of this is for both of these real quantities to equal a constant, say A. I should have taken the implication rather than the phrasing.
A good point that I had not considered. I had no particular reason for not pointing out that it is real.
Gary
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Post by Admin on Jan 4, 2016 7:48:38 GMT
Gary
You wrote
Yes, we are looking for a mapping with a certain property. In the example in the book this is the property of sending concentric circles to parallel lines. We also want the mapping to be analytic. This leads us via the C-R to differential equations with a unique solution. Done.
Vasco
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