Evaluate integral of second order differential

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Homework Help Overview

The discussion revolves around evaluating the integral of a second-order differential equation, specifically the integral of the second derivative of a function over a specified interval. The context includes concepts related to calculus and differential equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the evaluation of the integral and discuss the implications of the function being defined on a closed contour. There are attempts to relate the problem to simpler integrals and considerations of the properties of the function involved.

Discussion Status

Participants are actively engaging with the problem, with some suggesting that the integral may be negligible due to the periodic nature of the function. There is a recognition of the need to understand harmonic functions and their constraints, indicating a productive exploration of the topic.

Contextual Notes

There are mentions of issues with LaTeX formatting in the thread, which may have affected the clarity of the mathematical expressions presented. Additionally, the integral is part of a larger problem, which adds complexity to the discussion.

billiards
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Homework Statement



Evaluate
[tex]\int^{2\pi}_{0}\frac{1}{r}\frac{d^{2}f}{d\varphi^{2}}d\varphi[/tex]

Homework Equations



n/a

The Attempt at a Solution



I get the integral as

[tex]\frac{1}{r}\frac{df}{d\varphi}[/tex]

Not sure how to evaluate this.
 
Last edited:
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Why don't you do a simple one first. Say:

[tex]f(x)=x^3+2x[/tex]

so that:

[tex]\frac{df}{dx}=3x^2+2[/tex]

and so if I want to integrate:

[tex]\int_0^{2\pi}\frac{df}{dx}dx=\int_0^{2\pi} (3x^2+2)dx=(x^3+2x)\biggr|_0^{2\pi}=f(2\pi)-f(0)[/tex]

edit: so I see you got that part. Then it's just the first derivative at the end points right? So just write it as such:

[tex]\frac{df}{d\phi}\biggr|_{0}^{2\pi}[/tex]
 
Thanks for quick response. (I had some terrible trouble with Latex, I don't know if other users were seeing what I was, but when I tried to preview my post it would not refresh the Latex even though I had changed the coding.)

Is there not any way that I can take this further. This is actually a small part of a larger problem in which I have to show that this integral is negligible.

Hey, I think I get it now! The function is defined on a disk, so it must be the same at 0 and 2 pi. Therefore this term is negligible.

Does that sound right?
 
I'd say if the derivative is analytic throughout the disk and you're integrating over a closed contour, then the integral is zero.
 
Thanks. This is part of a problem on harmonic functions (I still don't really know what they are) -- I guess that term puts some constraints on the function, hopefully the ones you mentioned above.

Cheers
 
billiards said:
Thanks for quick response. (I had some terrible trouble with Latex, I don't know if other users were seeing what I was, but when I tried to preview my post it would not refresh the Latex even though I had changed the coding.)
This is a known problem that has yet to be fixed. The problem seems to be that when you preview a post with LaTeX in it, the previewer grabs whatever is in a cache somewhere, so if you have made changes, they won't show up in a preview. The only way around this that I know is to refresh the page after you have clicked Preview Post.
 
[tex]\sqrt{Thanks Mark}[/tex]

Thant trick works for me! :smile:
 

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