# Sturm-Liouville operator, operating on what?

1. Sep 23, 2015

### HeisenbergsDog

Consider a general Sturm-Liouiville problem

$[p(x) y'(x) ]' + [q(x) + \lambda \omega (x) ] y(x) = 0$

$\quad \Leftrightarrow \quad \hat{L} y(x) = \lambda \omega(x) y(x) \quad \text{with} \quad \hat{L} y(x) = -[p(x) y'(x)]' - q(x) y(x)$

where $p(x), p'(x), q(x), \omega(x)$ are real-valued functions, continuous on the interval $[a,b]$, and $p(x), \omega(x) \geq 0 \, \forall \,x \in [a,b]$

with the boundary conditions
$\alpha_1 y(a) + \alpha_2 y'(a) = 0$
$\beta_1 y(b) + \beta_2 y'(b) = 0$

where $\alpha_1, \alpha_2 \in \mathbb{R}$, not both zero, and $\beta_1, \beta_2 \in \mathbb{R}$, not both zero.

In the various litterature it is claimed that $\hat{L}$ is a linear operator on a Hilbert space of functions defined by the boundary conditions. Can someone give me a rigorous definition of the function space in question?

For $\hat{L}$ to be a linear operator, surely it must be an endomorphism on some vector space, right? This is where I get confused.

Suppose the boundary conditions are $f(a)=f(b)=0$. This supposedly defines a Hilbert space (but for which functions? continuous? square-integrable?). Next, if $\hat{L}$is an endomorphism, then the function $\hat{L}f(x)$ should be a member of that space also. Then we should have $\hat{L}f(a) = \hat{L}f(b)$, but I don't see why this would be the case. For example, with $f(x)$ in the space, there's no reason for $f'(x)$ to be in the space.

I would like a clear definition of the Hilbert space in question, and in what sense $\hat{L}$ truly is an operator (=endomorphism) on that space.

(Bare in mind that I know very little about functional analysis and Hilbert spaces; I haven't had proper courses in these subjects yet.)

Thank you.

2. Sep 24, 2015

### Svein

$\hat{L}$ is a linear operator on the space of twice differentiable functions u(x) on the interval [a, b], subject to certain homogeneous boundary conditions (for example u(a) = 0, u(b) = 0).
The space is defined above: It is a Hilbert space with the inner product given by $(u, v)=\int_{a}^{b}u(x)v(x)dx$. Verifying that $\hat{L}$ is a linear operator is not complicated (an integral is a linear operator).

3. Sep 24, 2015

### HeisenbergsDog

This I can buy. Perfectly good vector space. Let's call it $V$.

Here I get into trouble. Let $u \in V$. Then $u$ is twice differentiable and $u(a)=u(b)=0$. We have no other restrictions on $u$. Now let $\hat{L}$ be defined as above. If $\hat{L}$ were to be an operator, then we must have $\hat{L} u \in V \,\, \forall \,u \in V$. But $\hat{L}u = -[pu']' - qu =: v$ and I fail to see why this function $v$ should be in $V$. The function $v$ need not satisfy $v(a)=v(b)=0$, or even be differentiable?

4. Sep 24, 2015

### Svein

Why? I did not state that $\hat{L}: V\rightarrow V$. In reality $\hat{L}: V\rightarrow L_{2}[a, b]$.

5. Sep 24, 2015

### HeisenbergsDog

Oh ok. So $\hat{L}$ is in general not an endomorphism.

I thought the term "linear operator" on $V$ was the same as endomorphism on $V$. That's the definition I've seen on wikipedia, and in Axler's linear algebra book, among others.

Is the same true for "operators" in quantum mechanics? They are not endomorphisms either?

For finite-dimensional vector spaces, eigenvalues and eigenvectors are defined only for endomorphisms. So I get confused about this in the infinite case. I really need to take a functional analysis course, but that's a few years ahead I'm afraid.

6. Sep 24, 2015

### Svein

Well, that depends on how you look at it. Since $V\subset L_{2}[a, b]$ and the inner product is defined on the whole of $L_{2}$, you have $\hat{L}: V\subset L_{2}[a, b]\rightarrow L_{2}[a, b]$.

7. Sep 25, 2015

### HeisenbergsDog

I see. But $V \subset L_2[a,b]$ only if $a,b$ are finite, right? That is, we get into trouble if the interval is all of $\mathbb{R}$.

8. Sep 25, 2015

### Svein

Well, $L^{2}<-\infty, \infty>$ is defined and exists, but the Sturm-Liouville problem is defined only on a finite interval (see https://en.wikipedia.org/wiki/Sturm–Liouville_theory).

9. Sep 25, 2015

### Orodruin

Staff Emeritus
There is still the possibility of having a singular SL problem on an infinite interval. This is also mentioned on the wiki page.

10. Sep 25, 2015

### Svein

Yes, but that is too complicated for a first-timer.

11. Sep 25, 2015

### HeisenbergsDog

Ok, so thank you Svein for making me understand this a little better. I'm not quite there yet, though.

In linear algebra (with finite dimensional vector spaces) we define an endomorphism to be self-adjoint if <u,Lv> = <Lu,v>. It doesn't make sense for an operator $T: U\to V$ to be self-adjoint.

So when we say $\hat{L}$ is self-adjoint, then surely we must regard it as an endomorphism, right?

12. Sep 26, 2015

### Svein

Why not? Especially if U⊂V...

13. Sep 26, 2015

### HeisenbergsDog

If $U,V$ are finite dimensional inner product spaces and $T:U\to V$ a linear map, then by definition, $T^\dagger: V \to U$. Thus if $T=T^\dagger$ then we must have $U=V$.

Suppose $U$ is a subspace of $V$, then we may have <v,Tu> = <Tv,u> for all $u,v \in U$ so that $T^\dagger v = Tv$ for all $v \in U$. But $T^\dagger v \in U$ for all $v \in V$ (and hence for all $v \in U$) so that $Tv \in U$ for all $v \in U$. Hence $T: U \to U$.

Please prove me wrong or point out my mistake.

14. Sep 29, 2015

### Xiuh

Just a quick note: The space $\{u\in C^2[a,b]\} : u(a)=u(b)=0\}$ with the inner product $(u,v)_{L^2} = \int_a^b uv$ is not a Hilbert space (it is not complete). In this case we should work on a so called Sobolev space, in this case the space $H_0^1(a,b) = \{u\in L^2(a,b) : u'\in L^2,u(a)=u(b)=0\}$. This space is a Hilbert space with the inner product given by
$$(u,v)_{H^1} = \int_a^b u'v'$$

You are right about the spaces being equal. I'll try to explain the definition in the infinite-dimensional case. Suppose you have an operator $T: dom(T)\subset V_1\rightarrow V_2$ such that $dom(T)$ is dense in $V_1$ ($V_1$ and $V_2$ are Hilbert spaces). Then $dom(T^*)$ is defined as the set of all $v\in V_2$ such that there exists $w\in H_1$ such that

$$(v,Tu)_{V_2} = (w,u)_{V_1},\qquad u\in dom(T)$$

Since $dom(T)$ is dense, $w$ is uniquely determined by $v\in dom(T^*)$. We can then define the operator $T^*:dom(T^*)\subset V_2 \rightarrow V_1$ by $T^*v := w$. The pair $(dom(T^*),T^*)$ is called the (Hilbert) adjoint of $(dom(T),T)$ (the domain is important!!).

Now, an operator $T: dom(T)\subset H \rightarrow H$ is called symmetric (notice we have only one space $H$) if
$$(Tu,v)=(u,Tv), \qquad u,v\in dom(T).$$

With this, we finally define an operator $(dom(T),T)$ to be self-adjoint if $(dom(T),T)=(dom(T^*),T^*)$, that is, if $T$ is symmetric and $dom(T)=dom(T^*)$. You can prove that if a linear operator $T: H \rightarrow H$ is continuous (as in the finite-dimensional case), then it is self-adjoint if and only if it is symmetric. In conclusion, you had the right idea, the domains must be equal.

As for the Sturm-Liouville problem, suppose we have the equation
$$-(pu')' + qu = f, \text{ on } (a,b),$$

with boundary conditions $u(a)=u(b)=0$. It is possible to prove that for $f\in L^2$ there exists a solution in $H_0^1$ (the space I defined earlier). We can then consider the operator $f\mapsto u$ defined from $L^2$ to $L^2$ (or in $H_0^1$). This is the operator we prove is self-adjoint (and with that obtain the solution to the eigenvalue problem, for instance). Let me know if you want more details, hope this helps.