Check sturm-liouville orthogonality problem

[ognik]

Functions u, v satisfy the S-L eqtn $ [py']'+\lambda wy=0 $. u,v satisfy boundary conditions that lead to orthogonality. Prove that for appropriate boundary conditions, u' and v' are orthogonal with p as weighting factor.

I'm sure I need to use the orthogonality integral $ \langle u'|v' \rangle_p $ and show $ \int_{a}^{b}u'^*v'p \,dx=0 $, then u',v' orthogonal and p is the weighting factor

By parts, and taking u,v real for now to save writing the *, leaving off the limits and dx for brevity, $ \int_{a}^{b}u'v'p \,dx = v'pu-\int u(v'p)' $
This is promising because from boundary conditions the 1st term = 0 for both limits.

$ \int u(v'p)' = uv'p - \int v'pu' $ and again the 1st term vanishes from boundary conditions.

This leaves me with $ \int_{a}^{b}u'v'p \,dx = - \int_{a}^{b}u'v'p \,dx $, so that $ \int_{a}^{b}u'v'p \,dx $ must = 0. (u,v complex makes no difference to the above.)

Does this all work OK please?

 

[Ackbach]

What are the BC's? If you look carefully, your two by-parts integrations are telling you exactly the same thing. That is, your second by-parts produced the same equation as the first one. I don't think it's warranted to conclude that $\displaystyle \int_{a}^{b}u'v'p \,dx = - \int_{a}^{b}u'v'p \,dx.$ You might need the original S-L equation. And you might need the orthogonality of the eigenfunctions themselves w.r.t. the weighting function $w$.

 

[ognik]

Thanks Ackbach, the BCs I am using are $ pu^*v'|_{x=a} = 0$ , ditto for x=b. Those make the 1st terms in my integrals vanish (0 - 0= 0).

I was consciously trying a repeated integral approach, could you give me a bit more detail on why you are not comfortable with it?

(repeated integral example: $ I = \int e^{x} sin(x) dx = -e^xcos(x) + e^xsin(x) - ∫ e^xsin(x) dx $
then $ 2I= -e^xcos(x) + e^xsin(x),$
then $ \int e^{x} sin(x) dx = \frac{1}{2} \left( -e^xcos(x) + e^xsin(x)\right) $ In my case I get $ 2I=0 $...

The S-L eqtn I am aware of is $ [pu']' - qy + \lambda w y = 0$ , if this is what you mean, I tried multiplying each by the opposite function and then subtracting - but couldn't make that work ...

I had used the orthogonality by applying the orthogonality integral, are there other orthogonality conditions I don't know of (or didn't think of)?

 

[Ackbach]

Yes, while the repeated by-parts trick can work in some situations, it doesn't work here. The problem is that you get an identity. So, just to make things clear, you have
\begin{align*}
\int_{a}^{b}u'v'p \,dx &= (v'pu)|_{a}^{b}-\int_a^b u(v'p)' \, dx \\
\int_a^b u(v'p)' \, dx &= (uv'p)|_a^b - \int_a^b v'pu' \, dx.
\end{align*}
Plugging the second by-parts into the first yields
$$\int_{a}^{b}u'v'p \,dx=(v'pu)|_{a}^{b}-\left[(uv'p)|_a^b - \int_a^b v'pu' \, dx\right]=\int_a^b v'pu' \, dx, $$
an identity. That is, you got back to where you started. So the repeated by-parts fails in such a case.

You do have an additional orthogonality condition: $u$ and $v$ are orthogonal w.r.t. the weight function $w$. That is,
$$\int_a^b uvw \, dx=0.$$
Now, using the fact that $[py']'+\lambda wy=0,$ or $[py']'=-\lambda wy$, I wonder if you can massage your first by-parts to get what you want?

 

[ognik]

Hows this?

$ [pv']=-\lambda vw$ times by u: $ uvw=-\frac{1}{\lambda}[pv']u $

$\therefore \int uvw=-\frac{1}{\lambda} \int [pv']u =0, \lambda \ne 0$ then $ \int[pv']'u = 0 $

By parts: $ upv'-\int pv'u' = 0 $ but by BC's, upv'= 0 for both limits, so $\int pv'u' =0 $ therefore u', v' orthog.

 

[ognik]

I just noticed I don't need to move the $\lambda$ across. Otherwise would appreciate just a yes if this is OK?

 

[Ackbach]

Hows this?

$ [pv']=-\lambda vw$ times by u: $ uvw=-\frac{1}{\lambda}[pv']u $

Sorry, I had forgotten about this thread! This needs to be $[pv']'=-\lambda vw$. You've got the right idea, though.

$\therefore \int uvw=-\frac{1}{\lambda} \int [pv']u =0, \lambda \ne 0$ then $ \int[pv']'u = 0 $

By parts: $ upv'-\int pv'u' = 0 $ but by BC's, upv'= 0 for both limits, so $\int pv'u' =0 $ therefore u', v' orthog.

I think if you fix the above issue, you'll have it.

 

[ognik]

Oops, sorry - just typos leaving out the '. Your help always appreciated, while I'm a bit more confident these days, I still wish I had answers to the books exercises so I didn't have to bother you guys as much ...