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Integration by Parts

  1. Mar 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Here is a question I'm struggling with. I encountered it in a paper, and although a solution is provided I'm not so sure I understand where they're coming from.

    2. Relevant equations

    [tex]\int_{r_1}^{r_2} \overline{v}\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr})rdr[/tex]

    where [tex]\overline{v}[/tex] is the conjugate of [tex]v[/tex]

    3. The attempt at a solution

    I tried using the traditional integration by parts learned in undergraduate calculus courses, but I'm not getting anything close to the solution.

    Any help getting started would be appreciated.
  2. jcsd
  3. Mar 7, 2010 #2


    Staff: Mentor

    I would simplilfy it first before trying to integrate.
    [tex]\frac{d}{dr}(r\frac{du}{dr})~=~r\frac{d^2 u}{dr^2} + \frac{du}{dr}[/tex]
    Here I'm using the product rule.

    Using the differentiated expression above, and doing some other simplification (cancelling the 1/r and r), your first integral becomes
    [tex]\int_{r_1}^{r_2} \bar{v}(r\frac{d^2 u}{dr^2} + \frac{du}{dr})dr[/tex]

    Now split the above integral into two integrals. That should make the integration a little easier.
  4. Mar 7, 2010 #3
    That's right -- I got that far. At this point I'm not sure how to go on. I have some trouble understanding the proper notation.

    Could someone assist in continuing?
  5. Mar 7, 2010 #4


    Staff: Mentor

    Is v_bar a constant in the integration or does it depend on r?
  6. Mar 7, 2010 #5
    V_bar has dependence on r.
  7. Mar 7, 2010 #6


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    What's the expression you're trying to derive?
  8. Mar 7, 2010 #7
    Well the result should be:

    [tex] \left[\overline{v}r\frac{du}{dr}\right]_{r_1}^{r_2}-\left[\overline{r}u\frac{\overline{dv}}{dr}\right]_{r_1}^{r_2}+\int_{r_1}^{r_2}(\overline{\frac{d^2 v}{dr^2}+\frac{1}{r}\frac{dv}{dr})rudr[/tex]

    Sorry the latex isn't updating, or at least it hasn't yet. Look at the code to see the rest of the equation...
  9. Mar 7, 2010 #8


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    OK, they integrated by parts twice. The first time, they used

    [tex]f=\overline{v} \mbox{ and } dg=\frac{d}{dr}(ru') dr[/tex]

    That gets you the first term and the integral of [itex]ru'\overline{v}'[/itex]. For the second integration by parts, they used

    [tex]f=r\overline{v}' \mbox{ and } dg=u' dr[/tex]
  10. Mar 7, 2010 #9
    For the second integration I get:


    Which is obviously different than what they got:

    [tex]-\left[\overline{r}u\frac{\overline{dv}}{dr}\right]_{r_1}^{r_2}+\int_{r_1}^{r_2}(\overline{\frac{d^2 v}{dr^2}+\frac{1}{r}\frac{dv}{dr})rudr[/tex]
    Last edited: Mar 7, 2010
  11. Mar 7, 2010 #10


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    Is this what you say they got?

    [tex]\int_{r_1}^{r_2} (\overline{\frac{d^2 v}{dr^2}+\frac{1}{r}\frac{dv}{dr}}) ru dr[/tex]

    If so, it looks the same to me. Why do you think they're different?
  12. Mar 8, 2010 #11
    Oh of course. Sorry about that.

    Many thanks.
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