# Choosing boundary conditions for $(\frac{-d^2}{dx^2})^m$ on $H^m((0,1))$?

Consider the differential operator $D:$ $$Du:=\frac{-d^2}{dx^2}u$$ on the function space $$C=\{ u\in C^2([0,1]):u(0)=u(1)=0\}.$$

It's not hard to find the eigenvalues and eigenvectors(eigenfunctions) for $D$ by soloving the eigenvalue problem: $$-u''=\lambda u\qquad u(0)=u(1)=0.$$

Here are my questions:

• For each of the differential operators $D^m(m=2,3,4,\dots)$, what boundary conditions should one choose to ensure that $D^m$ and $D$ share the same eigenfunctions? I have no idea for even $m=2$.

• What if $$Du:=\frac{-d^2}{dx^2}u+u?$$

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This has been put on math.SE for several days without an answer. –  Jack May 3 '12 at 16:08

This is easy. If the boundary condition for D is, say, $Lu=0$, then for $D^m$ you should pose the boundary conditions $$Lu=LDu=LD^2u=...=LD^{m-1}u=0.$$
Then the corresponding function space for $D^m$ is $\{u\in C^m((0,1)):Lu=LDu=\cdots =LD^{m-1}=0\}$, isn't it? I'm afraid I'm not able to follow your "hint" to give a proof that $D^m$ and $D$ have same eigenfunctions. It seems that the results are the same for any linear differential operator $D$, right? –  Jack May 3 '12 at 17:51
Suppose $Du=\lambda u$ and $u$ satisfies the boundary condition $Lu=0$. Then $D^2u=\lambda Du=\lambda^2u$ and $Lu=LDu=\lambda Lu=0$. Now iterate this argument. –  Michael Renardy May 3 '12 at 18:07
Fair enough. Another problem is that whether the "reverse" is also true: Denote the boundary condition (in your answer) for $D^m$ as $\tilde{L}u=0$ and $D^mu=\lambda^m u$; can one have $Du=\lambda u$? –  Jack May 3 '12 at 18:54