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Homework Help: Advance Integration By Parts

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

    Integrate the following by parts twice

    [itex]\int_{a}^{b}\frac{d}{dr}(r\frac{dT(r)}{dr})\psi(r)dr[/itex]

    and show that it can be written as [itex]-\lambda^2\bar{T}[/itex] , where

    [itex]\bar{T}=\int_{a}^{b}r\psi(r)T(r)dr[/itex]

    and the function [itex]\psi[/itex] satisfies the following equation

    [itex]\frac{1}{r}\frac{d}{dr}(r\frac{d\psi(r)}{dr})+\lambda^2\psi(r)=0[/itex]

    Relevant equations and attempt

    Of course the integration by parts equation, but I used the tabular method to get the first integration of:

    [itex]\psi(r)r\frac{dT(r)}{dr}-\int_{a}^{b}\psi\prime(r)r\frac{dT(r)}{dr}dr[/itex]

    But as you can see in the next step it gets more complicated because the v choice now needs integration by parts as well and since the T(r) portion will end up being an integral without a definite solution, I don't know where to take it from there. I tried to get things to cancel but haven't found a way yet. Am I going about this the wrong way?

    On the second integration by parts I tried to group [itex]\psi\prime(r)r[/itex] together under u and it cleaned up v, but u became a mess without a way to cancel.

    Any help would be greatly appreciated. I am a little rusty on this stuff and this was provided as a refresher problem to me.
     
  2. jcsd
  3. Sep 1, 2010 #2
    I've given it my best shot and have tried just about every way possible. There must be some trick or technique that I haven't learned yet to deal with this. If anyone has any thoughts about how I might tackle it, it would be much appreciated.

    Thanks
     
  4. Sep 2, 2010 #3
    I got thoughts but it's not exactly as you put it. For starters, I assume you mean the following problem:

    Let [itex]\psi(r)[/itex] satisfy the following DE:

    [tex]\frac{1}{r}\frac{d}{dr}(r\frac{d\psi(r)}{dr})+\lambda^2\psi(r)=0[/tex]

    then show:

    [tex]\int_{a}^{b}\frac{d}{dr}(r\frac{dT(r)}{dr})\psi(r) dr=-\lambda^2\int_a^b r\psi T dr[/tex]

    But that's not what I get. You did parts once, how about do it again to obtain:

    [tex]r\psi \frac{dT}{dr}-rT\psi'+\int T(r\psi''+\psi')[/tex]

    now, using the differential equation, this looks to me to be:

    [tex]\int_{a}^{b}\frac{d}{dr}(r\frac{dT(r)}{dr})\psi(r) dr=\left(r\psi T'-rT\psi'\right)_a^b-\lambda^2\int_a^b r\psi T[/tex]

    and this is consistent with numerical calculations letting [itex]\psi(r)=\text{BesselJ}(0,\lambda r)[/itex] and [itex]T(r)=r^2+2r[/itex]
     
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