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A vanishing boundary condition sets a specified boundary quantity to zero. The quantity must be named: zero boundary value is a homogeneous Dirichlet boundary condition, while zero outward normal derivative is a homogeneous Neumann boundary condition. These are different restrictions.

Which boundary terms disappear under integration by parts depends on the operator and the permitted test functions. For a second-order weak problem, zero test-function trace eliminates ${\int }_{\partial U}({\partial }_{n}u)v$. A fourth-order problem can also involve derivative traces, so a zero value alone does not remove every boundary term. On $[0,1]$, $u(x)=x(1-x)$ vanishes at both endpoints but has nonzero endpoint first derivatives; the clamped boundary condition would require those slopes to vanish as well.