Self-adjoint fourth-order scalar differential operator ($D^4+p{D}^2+p^{\prime }D+r$)

For real coefficient functions, the fourth-order scalar operatoris formally self-adjoint exactly when $q=p^{\prime }$. Indeed, $(p{D}^2{)}^{\ast }=p{D}^2+2p^{\prime }D+p^{\prime \prime }$ and $(qD{)}^{\ast }=-qD-q^{\prime }$. Under clamped boundary conditions, its quadratic form is

For $L=D^4+q(x)$ where $q$ is a real continuous function on $[a,b]$, integration by parts givesThe potential terms cancel, and differentiating the boundary expression verifies the identity. Under clamped boundary conditions $u=u^{\prime }=v=v^{\prime }=0$ at both endpoints, this form vanishes. Conversely, requiring its vanishing against every clamped test function forces the adjoint function and its first derivative to vanish at both endpoints, because the second and third test-function derivatives are independent endpoint traces. Thus, on a regular finite interval with the usual fourth-order Sobolev domain, the clamped realization is self-adjoint. The equality of domains is part of this conclusion; formal self-adjointness of the expression alone does not supply it.