# Advance Integration By Parts

1. Sep 1, 2010

### timman_24

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

Integrate the following by parts twice

$\int_{a}^{b}\frac{d}{dr}(r\frac{dT(r)}{dr})\psi(r)dr$

and show that it can be written as $-\lambda^2\bar{T}$ , where

$\bar{T}=\int_{a}^{b}r\psi(r)T(r)dr$

and the function $\psi$ satisfies the following equation

$\frac{1}{r}\frac{d}{dr}(r\frac{d\psi(r)}{dr})+\lambda^2\psi(r)=0$

Relevant equations and attempt

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

$\psi(r)r\frac{dT(r)}{dr}-\int_{a}^{b}\psi\prime(r)r\frac{dT(r)}{dr}dr$

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 $\psi\prime(r)r$ 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. Sep 1, 2010

### timman_24

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

3. Sep 2, 2010

### jackmell

I got thoughts but it's not exactly as you put it. For starters, I assume you mean the following problem:

Let $\psi(r)$ satisfy the following DE:

$$\frac{1}{r}\frac{d}{dr}(r\frac{d\psi(r)}{dr})+\lambda^2\psi(r)=0$$

then show:

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

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

$$r\psi \frac{dT}{dr}-rT\psi'+\int T(r\psi''+\psi')$$

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

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

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