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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?


Added: Perhaps something to do with self-adjointness of the operator $D^4-\mu$?

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Since the ODE admits simple closed form solution you should find the equation for $\mu_n$, that is what have you done already? – Sasha Oct 26 '11 at 12:39
@Sasha: Thanks, the closed form is as such. But I still don't see how it works. Also, I suspect the nature of $\mu_n$ can be deduced from the nature of the operator (maybe analogous to the Sturm-Liouville operator or something of the sort?) – Rahul Oct 26 '11 at 12:57
You have your closed form, and your constants $C_1$, $C_2$, etc. should be adjusted to satisfy the boundary conditions. You get four equations. Did you do this? – abatkai Oct 26 '11 at 13:26
up vote 1 down vote accepted

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: enter image description here

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 $n$.

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