- #1
HeisenbergsDog
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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.
##[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.