 Solution

Use the usual inner product $⟨u,v⟩={\int }_0^1\overline{u}vdx$. Moving four derivatives by integration by parts leaves $L^{\ast }=D^4+1$ and gives the fourth-order boundary identityThus the differential expression is formally self-adjoint. If both functions satisfy the clamped boundary conditions, every displayed endpoint product vanishes, proving symmetry on the specified domain.

To check full self-adjointness, also identify the adjoint domain. Allow $v$ to run through the clamped domain. At either endpoint, $v^{\prime \prime }$ and $v^{\prime \prime \prime }$ can be chosen independently, for instance using a polynomial near that endpoint multiplied by a smooth cutoff function. The remaining boundary form is $[\overline{u}{v}^{\prime \prime \prime }-\overline{u^{\prime }}{v}^{\prime \prime }{]}_0^1$. It vanishes for every such $v$ precisely when $u=u^{\prime }=0$ at both endpoints. These are exactly the original boundary conditions. On the regular finite interval, the adjoint domain consequently equals $\{u\in H^4(0,1):u(0)=u^{\prime }(0)=u(1)=u^{\prime }(1)=0\}$, rather than a larger domain. Here $H^4$ is the Sobolev space of functions whose weak derivatives through order four are square-integrable.

The operator together with these boundary conditions is fully self-adjoint. As a further consistency check, integration by parts gives $⟨y,Ly⟩={\int }_0^1(|y^{\prime \prime }{|}^2+|y{|}^2)dx>0$ for every nonzero admissible $y$. The reusable calculation is the Lagrange identity for a fourth-order differential operator.