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For a sufficiently regular solution, put $w=-\Delta u$, with $\Delta$ the Laplacian. Then $-\Delta u-w=0$ and $u-\Delta w=u+{\Delta }^{2}u=f$. The zero Dirichlet boundary conditions on $u$ and $\Delta u$ become zero Dirichlet boundary conditions on $u$ and $w$. Conversely, eliminating $w$ from these two equations recovers the fourth-order equation with the biharmonic operator. For the weak solution we will construct, these interior identities hold as distributional identities; zero Sobolev traces are imposed on $u$ and $w$, without assuming classical fourth derivatives of $u$.

Let $V=H_0^1(\Omega )\times H_0^1(\Omega )$, the Cartesian product of two zero-boundary Sobolev spaces, with normThis is a Hilbert space. Applying integration by parts separately to the two equations gives the weak formulationEquivalently, for $U=(u,w)$ and $T=(v,z)$, define the bilinear form and linear functionalWe seek $U\in V$ with $b(U,T)=F(T)$ for every $T\in V$.

The skew-coupled elliptic system has a useful bounded bilinear form estimate with constant one. Indeed, write the integrand as the dot product of $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$. The Cauchy-Schwarz inequality givesFor equal arguments, the two zeroth-order coupling terms cancel. If $K$ is the constant in the supplied Poincaré inequality, thenThus $b$ is a coercive bilinear form, although it is not a symmetric bilinear form. The Lax-Milgram theorem applies and yieldsIn particular, $\Vert (u,w){\Vert }_V\le (1+K)\Vert f{\Vert }_0$. Testing with compactly supported smooth test functions identifies $-\Delta u=w$ and $-\Delta w=f-u$ as distributional identities. This gives the asserted relation to the original boundary value problem at the available regularity.

Take $V_h={\mathcal{U}}_h\times {\mathcal{W}}_h\subset V$. The Galerkin method asks for $U_h=(u_h,w_h)\in V_h$ such thatSince a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. The bounded bilinear form and coercive bilinear form estimates in part (a) restrict to $V_h$ with the same constants. The Lax-Milgram theorem therefore gives exactly one $U_h$ for every $h$.

Put $E=U-U_h=(e_u,e_w)$. Subtracting the continuous and discrete weak formulations gives Galerkin orthogonality:For any $Y_h=(v_h,z_h)\in V_h$, the difference $Y_h-U_h$ also belongs to $V_h$, soThe last step uses the constant-one Cauchy-Schwarz inequality estimate from part (a). Bounding the four terms separately by a sum would obscure the precise constant requested here.

The two minimization problems over ${\mathcal{U}}_h$ and ${\mathcal{W}}_h$ are independent. Each has a best approximation because its vector subspace is finite-dimensional. Consequently, ifthen the preceding inequality gives exactlyBoth errors belong to the zero-boundary Sobolev space. Applying the Poincaré inequality componentwise now yieldsWhen $U=U_h$ the desired estimate is automatic; otherwise divide by the nonzero norm. ThusThe constant $C=1+K$ is independent of $h$. This is the Céa lemma estimate for this particular skew-coupled elliptic system.

A skew-coupled elliptic system has zeroth-order couplings with opposite signs. For the displayed example with homogeneous Dirichlet boundary conditions, its weak formulation on $V=H_0^1(\Omega )\times H_0^1(\Omega )$ has the bilinear formThe coupling cancels for equal arguments, so $b((p,q),(p,q))=\Vert \nabla p{\Vert }_0^2+\Vert \nabla q{\Vert }_0^2$. On a domain satisfying the Poincaré inequality, this is a coercive bilinear form. With the product first-order Sobolev norm, the Cauchy-Schwarz inequality applied to $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$ gives $|b(P,T)|\le \Vert P{\Vert }_V\Vert T{\Vert }_V$. The Lax-Milgram theorem therefore applies even though the bilinear form is not symmetric. Eliminating $w=-\Delta u$ recovers ${\Delta }^{2}u+u=f$, involving the biharmonic operator.