Sturm-Liouville operator, operating on what?

[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.

 

[Svein]

In the various litterature it is claimed that L^\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?

\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).

I would like a clear definition of the Hilbert space in question,

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).

 

[HeisenbergsDog]

... 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).

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

Verifying that \hat{L} is a linear operator is not complicated (an integral is a linear operator).

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?

 

[Svein]

If L^\hat{L} were to be an operator, then we must have L^u∈V∀u∈V\hat{L} u \in V \,\, \forall \,u \in V.

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

 

[HeisenbergsDog]

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

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.

 

[Svein]

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

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].

 

[HeisenbergsDog]

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].

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 $\mathbb{R}$.

 

[Svein]

I see. But V⊂L2[a,b]V \subset L_2[a,b] only if a,ba,b are finite, right? That is, we get into trouble if the interval is all of R\mathbb{R}.

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).

 

[Orodruin]

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

 

[Svein]

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

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

 

[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?

 

[Svein]

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→VT: U\to V to be self-adjoint.

Why not? Especially if U⊂V...

 

[HeisenbergsDog]

Why not? Especially if U⊂V...

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.

 

[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&#039;\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&#039;v&#039;

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&#039;)&#039; + 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.