18
$\begingroup$

Motivation

I know that in a finite dimensional Euclidean space $\Bbb{E}^n$, for every basis $G=\{g_1,g_2,...,g_n\} \subset \Bbb{E}^n$ we can define a dual basis $G'=\{g^1,g^2,...,g^n\} \subset \Bbb{E}^n$ such that

$$g_i \cdot g^j = \delta_{i}^{j} \tag{1}$$

Also, it can be proved that such a basis exists and it is unique. The main advantage of dual bases is that when we write an arbitrary vector as a linear combination of the original basis $G$ then we can obtain the coefficients of the linear combination by just using the orthogonality property of $G$ and $G'$ like below

$$\begin{align} x &= \sum_{i=1}^{n}x^i g_i \\ x \cdot g^j &= \sum_{i=1}^{n} x^i g_i \cdot g^j = \sum_{i=1}^{n} x^i \delta_{i}^{j} = x^j \\ x &= \sum_{i=1}^{n}(x \cdot g^i) g_i \end{align} \tag{2}$$

Now, let us go into the infinite dimensional space of infinitely differentiable functions $f(x)$ over the interval $[-a,a]$. We all know that the eigen-functions of the Sturm-Liouville operator can form an orthonormal basis for such functions with proper boundary conditions at the end points. However, this is really a nice operator, a self-adjoint one which produces orthonormal eigen-functions. In some Partial Differential Equations (PDEs), we encounter non-self-adjoint operators and we should expand our solution and boundary data in terms of their eigen-functions which unfortunately are not orthogonal anymore!

Just to give an example, consider the following biharmonic boundary value problem (BVP)

$$\begin{array}{lll} \Delta^2 \Phi=0 & -a \le x \le a & -b \le y \le b \\ \Phi(a,y)=0 & \Phi_x(a,y)=0 & \\ \Phi(-a,y)=0 & \Phi_x(-a,y)=0 & \\ \Phi(x,b)=f(x) & \Phi_y(x,b)=0 & \\ \Phi(x,-b)=f(x) & \Phi_y(x,-b)=0 & \\ \end{array} \tag{3}$$

where we have the symmetry $f(-x)=f(x)$. Also, for the sake of continuity of boundary conditions at the corners we require that

$$f(a)=f(-a)=f'(a)=f'(-a)=0 \tag{4}$$

Then solving this BVP leads to the following eigen-value problem

$$ \left( \frac{d^4}{dx^4}+2\omega^2\frac{d^2}{dx^2}+\omega^4 \right)X(x)=0 \\ X(a)=X(-a)=X'(a)=X'(-a)=0 \tag{5}$$

which has a non-self adjoint operator. The eigen-functions are known as Papkovich-Fadle eigen-functions. They can form a basis for the infinite dimensional space of infinitely differentiable functions $f(x)$ satisfying $(4)$ over the interval $[-a,a]$. As I told before, these eigen-functions are not orthogonal and this makes finding the coefficients $c_i$ of the expansions

$$f(x)= \sum_{i=1}^{\infty} c_i X(\omega_i;x) \tag{6}$$

really difficult leading to solve an infinite system of linear algebraic equations for the unknown coefficients $c_i$!


Questions

$1.$ Is there a dual basis thing for the basis $X(\omega_i;x)$'s that can make the computation of $c_i$'s easier? To be more specific, is there some basis $Y(\omega_j;x)$ such that

$$\int_{-a}^{a} X(\omega_i;x) Y(\omega_j;x) dx =\delta_{ij} \tag{7}$$

which can be considered to have the similar role of $g^j$. If such a thing existed then we could compute the $c_i$'s by using $(6)$ and the orthogonality in $(7)$ easily

$$\int_{-a}^{a} f(x) Y(\omega_j;x) dx = \sum_{i=1}^{\infty} c_i \int_{-a}^{a} X(\omega_i;x) Y(\omega_j;x) dx = \sum_{i=1}^{\infty} c_i \delta_{ij} = c_j \tag{8}$$

$2.$ If the answer to question $1$ is YES, how it can be computed? If NO, Why?

$\endgroup$
  • $\begingroup$ Which scalar product are you referring to when you talk about (non-)selfadjointness and orthogonality? If it's the $L^2([-a,a]\times[-b,b]])$ scalar product, why not consider the $L^2$ as the underlying space? (Also: in which way is the series supposed to converge?) It's not a definte help for your question, but for this Hilbert space, we know at least what the dual looks like. $\endgroup$ – Roland Dec 6 '15 at 12:51
  • $\begingroup$ @Roland: Sorry, I am not much familiar with the language of functional analysis (FA) yet. :) However, by the scalar product of $f$ and $g$ over $[a,b]$ I mean $<f,g>=\int_{a}^{b} f \cdot g \, dx$. Also, I am interested in peice-wise continuous functions or maybe just continuous over $[a,b]$. I am not sure but the proper notation in FA should be $C^0$. About the convergence of the series, I think it is some conditional convergence! You can see this paper and other works of the author for details. :) $\endgroup$ – H. R. Dec 6 '15 at 13:24
  • $\begingroup$ Have you seen this? link $\endgroup$ – nik Dec 11 '15 at 12:45
  • $\begingroup$ @nik: Yes, It is mentioned in my question! :) $\endgroup$ – H. R. Dec 11 '15 at 12:47
  • 1
    $\begingroup$ I had already noticed this question. Unfortunately, it is my opinion that being able to compute that dual basis is as difficult as computing the coefficient of an arbitrary expansion. You still need to solve an infinite system of linear equations (the one you labelled (7) ). Maybe you are lucky and someone already computed that for you. But this is not the right place to ask. You might have more luck with engineers of physicists. $\endgroup$ – Giuseppe Negro Dec 16 '15 at 14:46
1
$\begingroup$

Not a concrete answer, but some generic pointers to your problem.

General Remarks on the dual space of $L^2$

It seems reasonable to consider your operator as an operator in $\mathcal H= L^2([-a,a]\times[-b,b])$, which is the Hilbert space of all measurable functions where $$\langle f,f\rangle=\int_{[-a,a]\times [-b,b]}f(x,y)\overline{f(x,y)}dx dy < \infty.$$

This Hilbert space has a nice relation to its dual, namely you can identify it with itself:

For $f \in \mathcal H$, we can define an element of the dual via

$$f'(g):= \langle g,f \rangle =\int_{[-a,a]\times [-b,b]}g(x,y)\overline{f(x,y)}dx dy,$$

defined on all $g\in \mathcal H.$

Vice versa, for every $f \in \mathcal H'$ there's an $f \in \mathcal H$ such that the equation above holds (cf., e.g. Conway, A Course in Functional analysis, 1990, Theorem 3.4 (The Riesz Representation Theorem)).

Note that neither $C^0$ nor the space of piecwise continuous functions equipped with this innner product is a Hilbert space. They are subsets of $L^2$, and thus, their duals are even larger than the dual of $L^2$. (Here, I'm assuming a bounded domain $[-a,a]\times [-b,b]$ and only finitely many jump points for the piecewise functions).

Remarks on your operator

Your differential operator (let's call it $A$) can be considered as an operator in $\mathcal H$. In this case, you need to specify the domain of $A$ (as a subset of $\mathcal H$) such that the functions $f$ are in $\mathcal H$ and $Af$ still lies in $\mathcal H$. The vector space of piece-wise continuous functions is a subspace of $\mathcal H$, but the first derivative doesn't need to exists. (I'm no PDE expert, but I'd expect some subset of $C^1$ with some additional constraints that the second derivative also exists.

In the case of Sturm-Liouville differential expressions (like $-(pf')'+qf$), one usually takes functions $f$ which are absolutely continuous (i.e. the first derivative is locally summable), where $pf'$ is absolutely continuous and $-(pf')'+qf$ lies in $L^2([a,b])$.

The boundary conditions you listed are an additional important restriction of your domain, but it's important that you state which kind of solutions you're allowing for your PDE.

Disclaimer

This is just presenting a general framework how to treat differential operators in Hilbert spaces. I don't know if this framework will help you to solve your PDE problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.