Biharmonic operator (${\Delta }^2$)

The biharmonic operator applies the Laplacian twice. Under clamped boundary conditions, two integrations by parts give the positive quadratic form

The clamped problem uses the weak solution identity $\int \Delta u\Delta v=\int fv$ for every $v$ in the clamped second-order Sobolev space. For $f\in L^2(U)$, the clamped Hessian identity and the Poincaré inequality make this a bounded coercive bilinear form. The Lax-Milgram theorem gives a unique weak solution without a mean-zero condition on $f$.

The biharmonic equation is ${\nabla }^{4}f=0$, where the biharmonic operator is the square of the Laplacian. In planar Stokes flow, taking the curl of the momentum equation and using a Cartesian streamfunction produces this equation. Its decaying spatial Fourier series modes have form $(A+By)e^{-|k|y}{e}^{ikx}$.

If $H,G$ are holomorphic functions, both real and imaginary parts of $\bar{z}H(z)+G(z)$ solve the biharmonic equation. In Wirtinger derivatives, $\Delta =4{\partial }_z{\partial }_{\bar{z}}$, so $\Delta (\bar{z}H+G)=4H^{\prime }$ is harmonic and its next Laplacian is zero. This construction gives local representations of planar solutions of the biharmonic equation; it is useful for the biharmonic stream function for planar Stokes flow.

For a smooth three-dimensional biharmonic function regular throughout a ball, its surface average has exactly this form. Taylor expansion and rotationally invariant spherical moments turn each even term into a multiple of an iterated Laplacian; terms from the fourth derivative onward vanish when ${\Delta }^{2}f=0$. This is the mean-value identity that converts the reciprocal surface average into Faxén's first law.