Integration by Parts: Struggling with Homework

[the_dialogue]

Homework Statement

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.

Homework Equations

$$\int_{r_1}^{r_2} \overline{v}\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr})rdr$$

where $$\overline{v}$$ is the conjugate of $$v$$

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.

 

[Mark44]

I would simplilfy it first before trying to integrate.
$$\frac{d}{dr}(r\frac{du}{dr})~=~r\frac{d^2 u}{dr^2} + \frac{du}{dr}$$
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
$$\int_{r_1}^{r_2} \bar{v}(r\frac{d^2 u}{dr^2} + \frac{du}{dr})dr$$


Now split the above integral into two integrals. That should make the integration a little easier.

 

[the_dialogue]

I would simplilfy it first before trying to integrate.
$$\frac{d}{dr}(r\frac{du}{dr})~=~r\frac{d^2 u}{dr^2} + \frac{du}{dr}$$
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
$$\int_{r_1}^{r_2} \bar{v}(r\frac{d^2 u}{dr^2} + \frac{du}{dr})dr$$


Now split the above integral into two integrals. That should make the integration a little easier.

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?

 

[Mark44]

Is v_bar a constant in the integration or does it depend on r?

 

[the_dialogue]

Is v_bar a constant in the integration or does it depend on r?

V_bar has dependence on r.

 

[vela]

What's the expression you're trying to derive?

 

[the_dialogue]

Well the result should be:

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

Sorry the latex isn't updating, or at least it hasn't yet. Look at the code to see the rest of the equation...

 

[vela]

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

$$f=\overline{v} \mbox{ and } dg=\frac{d}{dr}(ru') dr$$

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

$$f=r\overline{v}' \mbox{ and } dg=u' dr$$

 

[the_dialogue]

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

$$f=\overline{v} \mbox{ and } dg=\frac{d}{dr}(ru') dr$$

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

$$f=r\overline{v}' \mbox{ and } dg=u' dr$$

For the second integration I get:

$$\left[\overline{v'}ru\right]_{r_1}^{r_2}-\int_{r_1}^{r_2}(u(\overline{v'}+r\frac{d\overline{v'}}{dr})dr$$Which is obviously different than what they got:

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

 

[vela]

Is this what you say they got?

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

If so, it looks the same to me. Why do you think they're different?

 

[the_dialogue]

Oh of course. Sorry about that. Many thanks.