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[SOLVED] Sturm-Liouville operator

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Poirot

Banned
Feb 15, 2012
250
What does it mean for Sturm-Liouville operator to be symmetric w.r.t an inner product?

I was reading in a book that it is symmetric but that was about a certain integral being zero and inner products had not even been mentioned.
 

Ackbach

Indicium Physicus
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Jan 26, 2012
4,191
You have to generalize the concept of "inner product". In normal 3D-space, it's called the dot product:
$$\langle \vec{x}|\vec{y}\rangle=\vec{x}\cdot\vec{y}=\sum_{j=1}^{3}x_{j}y_{j}.$$
But there are other "spaces" out there: metric spaces, normed spaces, inner product spaces, Banach spaces, Hilbert spaces, Sobolev spaces. They each have different axioms with which you start. The inner product space is fairly general, and the "vectors" can be the usual vectors in 3D space, or they could be functions in a function space. The usual inner product defined in a function space is
$$\langle f|g\rangle:=\int_{A} \overline{f} \,g\,d\mu.$$
Here $\overline{f}$ indicates the complex conjugate of $f$, and the $d\mu$ indicates that we've defined this integral to be a Lebesgue integral. $A$ is the set over which the function space is defined.

Now we have the background to answer your question. A Sturm-Liouville operator
$$L=\frac{1}{w(x)}\left(-\frac{d}{dx}\left[p(x)\,\frac{d}{dx}\right]+q(x)\right)$$
is symmetric (more properly, Hermitian) w.r.t. the inner product
$$\langle f|g\rangle:=\int_{A}\overline{f}\,g\,w(x)\,d\mu,$$
if and only if for every $f, g$ in the inner product space, it is true that
$$\langle Lf|g\rangle=\langle f|Lg\rangle.$$
With the Sturm-Liouville operator, you need to show that
$$\int_{A}\overline{\left\{\frac{1}{w(x)}\left(-\frac{d}{dx}\left[p(x)\,\frac{df}{dx}\right]+q(x)f\right)\right\}}\,g\,w(x)\,d\mu=\int_{A} \overline{f} \left\{\frac{1}{w(x)}\left(-\frac{d}{dx}\left[p(x)\,\frac{dg}{dx}\right]+q(x)g\right)\right\}\,w(x)\,d\mu.$$

You can do this using simple integration by parts twice. The boundary terms vanish because of the conditions on them. (Note that the boundary conditions are considered to be part of the operator.)

I note you marked this thread as solved. That is good. Perhaps this post will throw in a few helpful concepts.
 
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Poirot

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Feb 15, 2012
250
Sorry you took your time. I should perhaps do a bit more reasearch before asking here.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Sorry you took your time. I should perhaps do a bit more reasearch before asking here.
No, that's all right. Incidentally, there are some concepts here that might help you with your other problem. Take a look at how $w(x)$ appears in the Sturm-Liouville operator, as well as how it shows up in the inner product w.r.t. which the Sturm-Liouville operator is symmetric.