Why do these functions form complete orthogonal systems in the Hilbert space?

[member 428835]

Hi PF!

A text states that the following two functions
$$
\psi^o_k = \sin(\pi(k-1/2)x)\cosh(\pi(k-1/2)(z+h)): k\in\mathbb{N},\\
\psi^e_k = \cos(\pi kx)\cosh(\pi k(z+h)): k\in\mathbb{N}
$$
each form complete orthogonal systems in two mutually orthogonal subspaces, which compose the Hilbert space.

Can someone explain this to me? Why are these orthogonal systems? Specifically, $\int_0^1 \psi^e_k \psi^o_k \, dx \neq 0$. And why is it that each by itself does not form a Hilbert space but together they do (is it because they are orthogonal systems, one cannot form a Hilbert space unless at least the other is present too)?

 

[eys_physics]

What is $z$? It seems like the basis functions depend on two variables. In that case I don't understand why you only integrate over $x$.

 

[member 428835]

What is $z$? It seems like the basis functions depend on two variables. In that case I don't understand why you only integrate over $x$.

Yes, $z$ is a variable with domain $z\in[-h,0]$. Evaluating the double integral in this domain for $z$ and $[0,1]$ for $x$ still doesn't equal zero. Any ideas?

 

[eys_physics]

Yes, $z$ is a variable with domain $z\in[-h,0]$. Evaluating the double integral in this domain for $z$ and $[0,1]$ for $x$ still doesn't equal zero. Any ideas?

It is not an requirement that your basis functions are orthogonal. You need to have a well-defined scalar product. Additionally, your set of basis functions has to complete. I guess that this second point explains why you need both $\psi^o_k$ and $\psi^e_k$ to have a Hilbert space. The functions $\psi^e_k$ are for example even with respect to $x$. So you cannot expand an odd function in this subspace.

 

[member 428835]

It is not an requirement that your basis functions are orthogonal. You need to have a well-defined scalar product. Additionally, your set of basis functions has to complete. I guess that this second point explains why you need both $\psi^o_k$ and $\psi^e_k$ to have a Hilbert space. The functions $\psi^e_k$ are for example even with respect to $x$. So you cannot expand an odd function in this subspace.

Cool, this is what I was thinking too, regarding the even/odd argument. So chalk it up to Fourier series theory that these two together are complete?

Why are these orthogonal though?

 

[eys_physics]

I think the issue here is that the range of $x$ is between 0 and 1. That is, not a full period. If your limits would be $x\in [-1,1]$ your basis functions would be orthogonal. This is also what you would have in Fourier series theory, where your range is a full period.

 

[member 428835]

I think the issue here is that the range of $x$ is between 0 and 1. That is, not a full period. If your limits would be $x\in [-1,1]$ your basis functions would be orthogonal. This is also what you would have in Fourier series theory, where your range is a full period.

In this case the domain of $z$ wouldn't matter, right, since sine times cosine is odd, integrated over a symmetric domain, always gives zero.

 

[eys_physics]

In this case the domain of $z$ wouldn't matter, right, since sine times cosine is odd, integrated over a symmetric domain, always gives zero.

Yes, you are correct.

 

[member 428835]

So to summarize: $\psi^o$ and $\psi^e$ are both orthogonal systems $\int_{-1}^1 \psi^e\psi^o\,dx = 0$ in two mutually orthogonal subspaces (even and odd). Since they are orthogonal, at least both are required to form a Hilbert space (possibly more). Since these two are systems essentially Fourier modes, it is well documented that both are complete and form a Hilbert space without extra systems.

Does that sound right?

 

[eys_physics]

So to summarize: $\psi^o$ and $\psi^e$ are both orthogonal systems $\int_{-1}^1 \psi^e\psi^o\,dx = 0$ in two mutually orthogonal subspaces (even and odd). Since they are orthogonal, at least both are required to form a Hilbert space (possibly more). Since these two are systems essentially Fourier modes, it is well documented that both are complete and form a Hilbert space without extra systems.

Does that sound right?

I agree with you except for one detail. Even if they would not be orthogonal you would need both odd and even basis functions. But, you can always use the Gramm-Schmidt procedure to construct a orthogonal basis. So, this is only a small detail.

 

[member 428835]

Thanks!

 

[WWGD]

I think the issue re the orthogonality is that a vector space (an inner-prod space, actually) can be expressed as the direct sum of a subspace and the Ortho complement of the subspace. Simplest example, consider the plane with the x-axis as a subspace and the y-axis as its Ortho complement so that $\mathbb R^2= x(+)y$. A subspace of a Hilbert space is not always itself a Hilbert space.

 

[WWGD]

Re your first question, were you given an explicit formula for the inner-product, or a description of the Hilbert space for which you were given these bases?A basis for a Hilbert space is a maximal orthohonal set.

 

[member 428835]

Re your first question, were you given an explicit formula for the inner-product, or a description of the Hilbert space for which you were given these bases?A basis for a Hilbert space is a maximal orthohonal set.

I was not. I can say that these basis functions are integrated at $z=0$ over $x\in[0,1]$ to approximate eigenvalues from a differential equation. Does that help?

 

[WWGD]

Somewhat. I can tell it is countably-infinite dimensional from the basis. Give me some time and I will go over the paper you linked.

 

[member 428835]

Somewhat. I can tell it is countably-infinite dimensional from the basis. Give me some time and I will go over the paper you linked.

Thanks!

 

[WWGD]

@joshmccraney :I was looking for the link you referred to, to look up some details. What happened to it?

 

[member 428835]

@joshmccraney :I was looking for the link you referred to, to look up some details. What happened to it?

Do you have access to it here?
https://link.springer.com/article/10.1007/BF01205747

 

[WWGD]

Got it, thanks.

 

[WWGD]

Yes, hi again, this is just the space $L_2(a,b)$ of square-integrable functions with the inner product you described: $<f,g>:= \int_a^b fg$. Remember that Hilbert spaces have the special property that the metric is generated by the inner-product. In Hilbert spaces, bases are maximal orthogonal systems. I am not clear if H is a subspace of a Hilbert space or a Hilbert space itself.