Physics Help Forum Reynold's Theorem in One Dimensional Space

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 Jul 2nd 2018, 02:21 PM #1 Senior Member   Join Date: Jun 2010 Location: NC Posts: 406 Reynold's Theorem in One Dimensional Space I'de be grateful for a URL to a proof of the Reynold's Transport Theorem reduced to one dimensional space. Must be a one dimensional representation no hand-waving explanation. Where is this written, I wonder? Thank ye, M'boys... JP
 Jul 3rd 2018, 05:32 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 Hi, Jim Reynold's theorem, like Leibnitz integral theorem, which is the result of Leibnitz rule, stems from the differentiation of a product of two functions, u and v of the dependent variable(s). This is why it is said that Leibnitz is the one dimensional version. http://mathworld.wolfram.com/Reynold...rtTheorem.html So I assume by one dimensional you mean you have only one independent variable and one dependent one. ie you have some funtion y = f(x) so you can write the expression in the form $\displaystyle \frac{d}{{dx}}\left[ I \right] = \frac{d}{{dx}}\int_u^v {ydx}$ where u and v are functions of x alone. Can you confirm if this is what you mean or give an example if it is not? I can then supply the proof you seek which is based on the Pascal's triangle (or the binomial expansion) and induction. By the way, why did you post in High school Physics? This is hardly that although I would say that the above proof scheme would have been accessible of a UK student of the old fashioned A level. Last edited by studiot; Jul 3rd 2018 at 05:37 AM.
 Jul 3rd 2018, 10:04 AM #3 Senior Member   Join Date: Jun 2010 Location: NC Posts: 406 Studiot, Hi... Thanks. I'll look this over. I didn't mean to use a "HS Physics" thread? I got it wrong. So I'm still learning the ways of PHF. Thanks, JP
 Jul 5th 2018, 02:53 PM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 I'm still hoping for the example I asked for. But meanwhile thinking about it let us consider the transport theorem in multi dimensions and see if we can make any sense of a one dimensional version. Basically the theorem says that if we have a control volume then the time rate of change of the total of some quantity is given by In 3 dimensions The volume integral of the quantity taken over that volume plus or minus the rate of change of amount of that quantity entering or leaving through the boundary surface of that volume. In 2 dimensions we have an area integral over the area plus or minus the the rate of change of the amount passing across the boundary curve of the area. In one dimension we have the integral along one interval plus or minus the rate of change of amount entering or leaving via the two boundary points of the interval. So can we discover a suitable quantity that it makes sense to consider along a line (ie one axis or dimension)? Well if you consider say mass or momentum flow the answer is yes. Think about a fluid flowing along a very thin pipe. There is fluid in an interval of that pipe and fluid is entering at one end (boundary point) and leaving at the other.
 Jul 5th 2018, 03:44 PM #5 Senior Member   Join Date: Jun 2010 Location: NC Posts: 406 Sir Studiot... Hi. I've not forgotten nor lost interest. I expected no response about Reynolds. Knowing you will read what I write, encourages me. I'll be along shortly. I have to do some editing of what I have. It is a subtle topic. Thank you, jp
 Jul 6th 2018, 12:05 PM #6 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 I think in 1 D you are talking about the fundamental theorem of calculus, which all these theorems stem from (Gauss, Green, Stokes etc) Perhaps a real world example would be this example of unrolling a carpet. Ignore example 1 about trigonometry. Attached Thumbnails

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