Boundary value problem and eigenvalues This boundary value problem has (countably) infinitely many eigenvalues $\mu_n\in \mathbb R$:
$$y^{(4)}-\mu y=0$$ with boundary conditons: $y(0)=y(1)=y'(0)=y'(1)=0$
What happens to the values of $\mu_n$ as $n\to \infty$? Does it tend to infinity too? Why?
Thanks.
Added: Perhaps something to do with self-adjointness of the operator $D^4-\mu$?
 A: Let's consider $y^{(4)}(x) - 4 \mu y(x)=0$ for convenience. This equation is solved by
$$
   y(x) = c_1 \sin( \lambda x ) \sinh( \lambda x) + 
          c_2 \cos( \lambda x ) \sinh( \lambda x) + 
          c_3 \sin( \lambda x ) \cosh( \lambda x) +
          c_4 \cos( \lambda x ) \cosh( \lambda x)
$$
where $\lambda = \mu^{1/4}$. Initial condition $y(0) = c_4 = 0$ requires $c_4 = 0$, and $y^\prime(0) = \lambda ( c_2 + c_3)$ requires $c_3 = -c_2$. Further 
$$ 
 y(1) = c_1 \sin( \lambda) \sinh( \lambda) + c_2 \left( \cos(\lambda) \sinh(\lambda) - \sin(\lambda) \cosh(\lambda) \right)
$$
This gives
$$
    c_1 = c_2 \left( \cot(\lambda) - \coth(\lambda) \right)
$$
Using 
$$
 0 =  y^\prime(1) = 2 c_2 \lambda \sin(\lambda) \sinh(\lambda) + c_1 \lambda \left(
    \sin(\lambda) \cosh(\lambda) + \cos(\lambda) \sinh(\lambda)
   \right)
$$
gives the final equation for the eigenvalue:
$$
  \lambda \frac{\sinh^2(\lambda) - \sin^2(\lambda)}{\sin(\lambda) \sinh(\lambda)} = 0
$$
For every solution $\lambda$, the eigenvalue is $\mu = \lambda^{1/4}$. The only real solution of this equation is $\lambda = 0$.
Now let $\lambda = (1 \pm i) \kappa$, then the equation for eigen-$\lambda$ becomes:
$$
   0 = \lambda \frac{\sinh^2(\lambda) - \sin^2(\lambda)}{\sin(\lambda) \sinh(\lambda)} \stackrel{\lambda = (1\pm i) \kappa}{=} 2(\mp 1-i) \kappa \frac{ \cosh(2 \kappa) \cos(2 \kappa) - 1}{\cos(2\kappa) - \cosh(2 \kappa)} 
$$
Here is the plot, showing zeros:
 
Since those zeros correspond to solutions of $\cos(2 \kappa) = \frac{1}{\cosh(2 \kappa)}$, they are being located close to $\kappa = \frac{1}{2} \left( \frac{\pi}{2} + \pi n \right)$ for large $n$. 
