We have a Sturm-Liouville operator $$ L=\frac{1}{w(x)}\left[\frac{d}{dx}\left(p(x)\frac{d}{dx}\right)+q(x)\right] $$ and consider $$ \frac{\partial c}{\partial t}=Lc, $$ with homogeneous boundary conditions.

If we are now searching for solutions, the technique is to start with considering the homogeneous equation, i.e. $q(x)=0$ and solves the Eigenvalue problem $$ L\Phi=\lambda\Phi. $$ There are functions $\Phi$ - called eigenfunctions - that solve this eigenvalue problem, they exist since $L$ is self-adjoint under homogeneous boundary conditions. The eigenfunctions are orthogonal.

Moreover, the eigenfunctions $\Phi$ form a basis for the function space consisting of functions that satisfy the boundary conditions, meaning that any such function can be expressed as a linear combination of the eigenfunctions. So we can find solutions of the inhomogeneous equation by making the approach $u(x,t)=\sum_n A_n\Phi_n$, put this into the equation and determine the constants.

My question is how one can show/ see that the eigenfunctions form a basis of the function space consisting of functions that satify the boundary conditions.

More precisely, I think, the function space for which the eigenfunctions form a basis is supposed to be the function space containing all functions that

(i) are quadrat-integrable with respect to the weight function $w$ and

(ii) satisfy the boundary conditions.

Do not know exactly if we really need (i).

Wikipedia says that the proper setting is the Hilbert space $L^2([a,b], w(x)dx)$ and that in this space, $L$ is defined on sufficiently smooth functions that satisfy the boundary conditions.

Anyhow: How to show/ see that the eigenfunctions form a basis?

  • $\begingroup$ Hint: Stone-Weierstrass theorem. Also, to even consider "the eigenfunctions are orthogonal" one needs an inner product on the function space, so this statement only makes sense in $L^2([a,b],dw)$ $\endgroup$ – charlestoncrabb Dec 26 '15 at 17:52

There are other conditions you need in order to guarantee a discrete basis $\{ \Phi_1,\Phi_2,\Phi_3,\cdots \}$. A typical case leading to discrete eigenvalues and Fourier expansions would be where (a) $p$ is continuously differentiable and strictly positive on $[a,b]$, (b) $w$ is continuous and strictly positive on $[a,b]$, and (c) $q$ is absolutely integrable on $[a,b]$. When you impose homogeneous endpoint conditions of the form $$ \cos \alpha f(a)+\sin\beta f'(a) = 0, \\ \cos \beta f(b) + \sin\beta f'(b) = 0, $$ for some real $\alpha,\beta$, then there is an infinite sequence of eigenvalues $$ \lambda_0 < \lambda_1 < \lambda_2 < \cdots < \lambda_n <\cdots, $$ that tends to $\infty$ for which non-trivial solutions of $Lf_j = -\lambda_j f_j$ exist which satisfy the homogeneous conditions; and the solution space is one-dimensional for each $j$. These eigenfunctions $\{ f_j \}$ are mutually orthogonally and, when properly normalized, the set $\{ f_j \}$ is a complete orthonormal basis of $L^2_w[a,b]$.

The convergence of the generalized Fourier series for $f \in L^2_w[a,b]$ converges in the norm of $L^2$ to $f$. You don't necessarily get such convergence if $f \in L^1_w[a,b]$ instead; the failure of converge in $L^1_w[a,b]$ occurs for the simplest classical Fourier case $q=0$, $w=1$, $p=1$.

A proof of these facts is not short. It can't be trivial, because the simplest case includes the classical Fourier series.

Reference: M. A. Al-Gwaiz, Sturm-Liouville Theory and its Applications, 2008. (Springer-Verlag Undergraduate Series.) Pay particular attention to Chapter 2.

  • $\begingroup$ Can you mention some Nice resource for a proof? :) $\endgroup$ – Hosein Rahnama Dec 27 '15 at 9:22
  • 1
    $\begingroup$ @H.R. : I added a reference at the end for you. $\endgroup$ – Disintegrating By Parts Dec 28 '15 at 2:40

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