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This is the closure of the compactly supported test functions $C_c^{\infty }(U)$ in the $H^2(U)$ norm. On a smooth bounded domain it consists precisely of the $H^2$ functions with zero value and zero normal derivative traces. Tangential first derivatives then vanish as well. The clamped Hessian identity makes the Laplacian norm equivalent to the full $H^2$ norm on this space, enabling a Lax-Milgram theorem treatment of the clamped biharmonic problem.

For $v\in C_c^{\infty }(U)$, two integrations by parts give the clamped Hessian identityBy density it remains valid on $H_0^2(U)$. Each ${\partial }_{i}v$ has zero integral, first for compactly supported test functions and then by $H^2$ convergence. Applying the Neumann-Poincare inequality to each ${\partial }_{i}v$ givesThe zero-boundary Poincare inequality also gives $\Vert v{\Vert }_2^2\le C_D\Vert \nabla v{\Vert }_2^2$. Hence, using a full-Hessian equivalent $H^2$ norm,Thus $a(u,v)={\int }_U\Delta u\Delta v$ is an inner product whose norm is equivalent to the complete $H^2$ norm on the clamped second-order Sobolev space. The functional $L(v)={\int }_{U}fv$ is bounded for this norm by the Cauchy-Schwarz inequality and the displayed bound. Apply the Riesz representation theorem, or the Lax-Milgram theorem, to get a unique $u\in H_0^2(U)$ representing $L$. The clamped biharmonic problem has a unique weak solution for every $f\in L^2(U)$, with $\Vert u{\Vert }_{H^2}\le C\Vert f{\Vert }_2$ and no zero-integral compatibility condition.

The difference $w$ of two weak solutions lies in the clamped second-order Sobolev space. Testing with $w$ yields ${\int }_U(\Delta w{)}^2=0$, so $\Delta w=0$. Since $w\in H_0^1(U)$, integration by parts givesThe Poincare inequality for zero boundary values now implies $\Vert w{\Vert }_2=0$. The clamped biharmonic problem has at most one weak solution. Unlike the Neumann Poisson problem, no additive constant is allowed by these boundary traces.

The clamped second-order Sobolev space is $H_0^2(U)={\overline{C_c^{\infty }(U)}}^{H^2}$. On a smooth bounded domain the Sobolev trace theorem characterizes it by zero value and zero normal derivative on the boundary. In particular, its whole first-order boundary jet is zero, since tangential derivatives of the zero trace also vanish. As above, assume $f\in L^2(U)$ and classical regularity up to the boundary.

For a smooth weak solution, compactly supported test functions and two integrations by parts giveso ${\Delta }^{2}u=f$ pointwise. Membership in $H_0^2(U)$ supplies $u={\partial }_{n}u=0$ on $\partial U$. Thus it is a classical solution of the clamped biharmonic problem.

Conversely, a $C^4(\overline{U})$ classical solution with these traces belongs to $H_0^2(U)$. For every compactly supported test function, two integrations by parts give ${\int }_U\Delta u\Delta v={\int }_{U}fv$. Both sides are continuous for the $H^2$ norm, so the defining density of $C_c^{\infty }(U)$ in $H_0^2(U)$ extends this equality to every required test. The two notions agree under the stated smoothness.