Eigenfunctions of fourth-order differential operator This a question where I have thought quite long about:
The eigenfunctions (or also normal modes) of an dry Euler beam subject to free-free boundary conditions are given by
$$ \frac{\partial^4\psi}{\partial x^4}=\lambda_k^4\psi\qquad(0\le x\le1)\,,$$
$$\frac{\partial^2\psi}{\partial x^2}=\frac{\partial^3\psi}{\partial x^3}=0\qquad(x=0\text{ and }x=1)\,, $$
The solution of this problem is given by 
$$\psi_0(x)=1\qquad\text{and}\qquad\psi_1(x)=\sqrt3(2x-1),$$
$$ \psi_k(x)=\frac{\cosh((\frac12-x)\lambda_k)}{\cosh(\frac12\lambda_k)}+\frac{\cos((\frac12-x)\lambda_k)}{\cos(\frac12\lambda_k)}\qquad (k\ge 2 \text{ even}),$$
$$
 \psi_k(x)=\frac{\sinh((\frac12-x)\lambda_k)}{\sinh(\frac12\lambda_k)}+\frac{\sin((\frac12-x)\lambda_k)}{\sin(\frac12\lambda_k)}\qquad (k\ge 2\text{ odd}),$$
where 
$$ \cosh(\lambda_k)\cos(\lambda_k)=1\qquad (k\ge2)\,,$$
This is a complete set of eigenfunctions. These eigenfunctions are orthogonal (it can be shown that they are even orthonormal) in terms of the standard scalar product, since the fourth order derivative is a self-adjoint operator subject to the given boundary conditions and a standard smoothness condition for $\psi$. I would like to know if this set of eigenfunctions is also an orthonormal basis in $L_2([0,1])$.
There seems to be a Sturm-Liouville theory for fourth order differential operators. Is there any standard book which discusses such a problem?
 A: You need to  specify $\lambda_k$ more clearly.  More precisely, what is the spectrum of  this operator?   The equation
$$ \cos x\cosh x =1 $$
seems to have a unique solution $\mu_k$ on any interval of the form $(k\pi/2, k\pi/2+\pi)$, $k\in\mathbb{Z}$,  so I assume the spectrum   might be $\mu_k^4$, $k\in\mathbb{Z}$?!? (Please     edit you question to remove this ambiguity.)  In any case  if the boundary value problem is elliptic (please check that) then  the spectrum  is discrete.  In particular it can be determined    by finding the eigenfunctions which means solving  some ode's.  My guess is that you found  all the eigenfunctions, i.e., the system you found is complete.
Update.    You need to  check two things: 1) the boundary value problem is elliptic 2) it is symmetric.  I'll deal with the 2nd issue first because it is faster.     Denote by $A$ the operator
$$A=\frac{d^4}{dx^4}. $$
A simple integration by parts shows that for any $u,v\in C^4([0,1])$ we have
$$ \int_0^1 \bigl(\; v(Au) -u(Av)\;\bigr) dx=\sum_{j=0}^3(-1)^j\bigl( u^{(3-j)}(1)v^{(j)}(1)- u^{(3-j)}(0) v^{(j)}(0)\;\bigr). $$
If the function $u$ satisfies your boundary conditions $u^{(k)}(x)=0$ for $k=2,3$, $x=0,1$ the above equality  simplifies a bit
$$ \int_0^1 \bigl(\; v(Au) -u(Av)\;\bigr) dx= \sum_{j=2}^3(-1)^j\bigl( u^{(3-j)}(1)v^{(j)}(1)- u^{(3-j)}(0) v^{(j)}(0)\;\bigr). $$
If the function $v$ satisfies the same boundary conditions  as $u$,  then the last  equality takes the very simple form
$$ \int_0^1 \bigl(\; v(Au) -u(Av)\;\bigr) dx= 0. $$
This says that the boundary value problem is symmetric, or formally selfadjoint.
The ellipticity of this problem is another issue.   The most readable account I could find  is in  Chap. 20 vol.3 of the book The Analysis of Linear Partial Differential Operators by the late great Lars Hormander.
The ellipticity   of the boundary  value problem requires that  the symbol of your operator $A$ be elliptic (which it is) and that the boundary value conditions should satisfy the so called Lopatinskii-Schapiro conditions.
In your case they are trivially satisfied because  you work on a one-dimensional space $[0,1]$. The upshot   is that in your case the boundary conditions are elliptic.     We can form the unbounded operator  $\newcommand{\bD}{\boldsymbol{D}}$
$$ \hat{A}: \bD(\hat{A})\subset L^2(0,1)\to L^2(0,1), u\mapsto \frac{d^4 u}{dx^4} $$
Where the domain $\bD(\hat{A})$ of $\hat{A}$  consists of functions in the  Sobolev space $L^{4,2}(0,1)$ (four weak derivatives in $L^2$) such that  $u^{(j)}(x)=0$  for $x=0,1$, $j=2,3$.
The results  in the above  monograph show that $\hat{A}$ viewed as an unbounded operator on the Hilbert space $L^2(0,1)$, is closed, densely defined, selfadjoint and has compact resolvent.   This is all you need.  Arguably, the above argument  is a bit heavy, and it feels like  hunting a mosquito using a bazooka.
There is a direct, more elementary approach to proving that $\hat{A}$ has compact resolvent.       Observe first that the above integration by parts formulae show   that the operator $\hat{A}+1$ is positive,  i.e.,
$$ (\hat{A}u,u)_2+(u,u)_2>0,\;\;\forall u\in \bD(\hat{A})\setminus 0, $$
where $(-,-)_2$ denotes the $L^2$-inner product. Hence $\hat{A}+1$ is injective. Then   follow the strategy in the proof of Theorem 8.22  in  Brezis' book Functional Analysis, Sobolev Spaces and Partial Differential Equations to prove that $\hat{A}+1$ is invertible and it's inverse is  compact as an operator $L^2(0,1)\to L^2(0,1)$.
A: Choose some $\lambda \ne \lambda_k$ for all $k$.  Consider the resolvent $R_\lambda = (\lambda-\partial_x^4)^{-1}$.  This is a compact operator on the Hilbert space $H = \{ \psi \in L^2([0,1]) \; | \; \psi''(0) = \psi''(1) = \psi'''(0) = \psi'''(1) = 0\}$.  Thus $H$ has a basis of eigenvectors of $R_\lambda$.  But $R_\lambda$ and $\partial_x^4$ share eigenvectors, thus $H$ has a basis of eigenvectors of $\partial_x^4$ as desired.  The key bit is using compactness to show that a maximizing sequence for the Rayleigh quotient converges to an eigenvector.  The argument can be found in Lax's Functional Analysis Ch 28 Thm 3.
